function
complex analysis
complex analysis unsolved

So this was on a problem set due today. It is quite unbelievable. I don't think it has ever been posted, but it is definitely mind boggling enough that it is worth a look at.

There exists an entire function $ f$ with the property that for every entire function $ g$, and for every $ \epsilon >0$, and for every compact set $ K$ in the plane, there is a $ c>0$ such that $ \sup_{z\in K}|g(z)\minus{}f(z\plus{}c)|<\epsilon$.

Basically there is a single entire function that on arbitrarily large sets looks like every other entire function just by shifting horizontally.

Not that I'm going to say that it isn't also amazing and fascinating, but the Riemann zeta function seems to have a weaker form of universality than this function. First, it does have a pole, so isn't entire. Also, our compact sets have to be contained in the 1/2 to 1 strip, and our function we are approximating can't have zeros in that region. I'm pretty sure that we cannot write the function I described in any nice explicit way like $ \zeta(s)$.

But I didn't know about this. Thanks! That is really cool.


I suspect that someone specific has made me a foe. Is there any way to see if this person really has or not?

I doubt it; Mr. Vornicu last I heard was going to have both friends and foes private, unlike on slashdot where they're public. I like /.'s way more.

Well it's not like I can pm this person and ask, because your foes can't see your pms.

Currently, foes option doesn't do anything.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ k$ $ \subset$ $ K$ be a finite Galois extension with $ G=(K/k)$, let $ L$ be a subfield of $ K$ containing $ k$ with $ H=G(K/L)$, and let $ L'$ be the compositum in $ K$ of the fields $ \sigma (L)$, $ \sigma$ $ \in$ $ G$.  Show that:
(a) $ L'$ is the unique smallest subfield of $ K$ that contains $ L$ and is Galois over $ k$
(b) $ G(K/L')=\cap_\sigma_\in_G$ $ \sigma$ $ H$ $ \sigma{}^-{}^1$

$ \tau \in G$ fixes $ L'$ if and only it fixes every $ \sigma(L)$ with $ \sigma \in G$, i.e. if and only if $ \sigma^{\minus{}1} \tau \sigma$ fixes $ L$. this shows (b).  since $ \cap_{\sigma \in G} \sigma H \sigma^{\minus{}1}$ is the largest normal subgroup if $ G$ which is contained in $ H$, (a) is a consequence of (b) using the galois correspondence.

[quote="verde"]Let $ k$ $ \subset$ $ K$ be a finite Galois extension with $ G = (K/k)$, let $ L$ be a subfield of $ K$ containing $ k$ with $ H = G(K/L)$, and let $ L'$ be the compositum in $ K$ of the fields $ \sigma (L)$, $ \sigma$ $ \in$ $ G$.  Show that:
(a) $ L'$ is the unique smallest subfield of $ K$ that contains $ L$ and is Galois over $ k$
(b) $ G(K/L') = \cap_\sigma_\in_G$ $ \sigma$ $ H$ $ \sigma^ - ^1$[/quote]

Why do you post your homework here ?


http://www.mathlinks.ro/Forum/viewtopic.php?t=430
Lagrangia gives a proof there.

May I know if the following proof is ok?
WLOG, let x>=y>=z
we have
(x,y,z,x+y+z/3,x+y+z/3,x+y+z/3) majorizes (x+y/2,y+z/2,z+x/2,y+z/2,x+z/2,x+y/2)
If f(x) is convex,
directly by Karamata inequality we have popoviciu inequality.
Is it ok?

I think it's not, ie because x+y+z/3 is greater than z, so you should change order of the elements which you use for majorization, but anyway it might still not work

Sorry i don't understand this?
[quote="Megus"]because x+y+z/3 is greater than z, so you should change order of the elements which you use for majorization[/quote]

[quote="Megus"]In majorization elements have to be placed in order which depends on their values. For example, you can't write {3,4,2,5} majorizes something, but {5,4,3,2}.[/quote]

oh i see thanks!

See the last post of this thread:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11894&highlight=


algebra
polynomial
real analysis
real analysis unsolved

Prove that all numbers of the form n^(p/q), where n, p and q are positive integers, are algebraic numbers.

The root of the polynomial $ x^{q}\minus{}n^{p}\equal{} 0$ is algebraic.  More exercises about the algebraic numbers:

- Prove that the sum of two algebraic numbers is algebraic.
- Prove that the product of two algebraic numbers is algebraic.


geometry
parallelogram
cyclic quadrilateral
perpendicular bisector
geometry proposed

Let $RSTU$ be a cyclic quadrilateral with $ST > RU$, $TU > RS$.  Take points $X$ and $Y$ on the sides $ST$ and $TU$, respectively, so that $SX = RU$ and $UY = RS$.  Let $M$ be the mid-point of $XY$.  Prove that $\angle SMU$ is a right angle.

Nice problem!

Let Q be a point such that the quadrilateral URSQ is a parallelogram. Then UQ = RS, so UY = RS yields UQ = UY. Thus, the triangle QUY is isosceles, and we can label its angles with $\measuredangle UQY=\measuredangle UYQ=\varphi $ and $\measuredangle QUY=180^{\circ }-2\varphi $. Similarly, the triangle QSX is isosceles, and we call its angles $\measuredangle SQX=\measuredangle SXQ=\psi $ and $\measuredangle QSX=180^{\circ }-2\psi $. Finally, we label the angles of the parallelogram URSQ with $\measuredangle RUQ=\measuredangle RSQ=\alpha $ and $\measuredangle URS=\measuredangle UQS=180^{\circ }-\alpha $. Since the quadrilateral URST is cyclic, we have

$180^{\circ }=\measuredangle RUT+\measuredangle RST=\left( \measuredangle RUQ+\measuredangle QUY\right) +\left( \measuredangle RSQ+\measuredangle QSX\right) $
$=\left( \alpha +\left( 180^{\circ }-2\varphi \right) \right) +\left( \alpha +\left( 180^{\circ }-2\psi \right) \right) =2\alpha +360^{\circ }-2\varphi -2\psi $,

what quickly leads to $2\left( \varphi +\psi -\alpha \right) =180^{\circ }$ and $\varphi +\psi -\alpha =90^{\circ }$. Now,

$\measuredangle XQY=360^{\circ }-\measuredangle UQY-\measuredangle UQS-\measuredangle SQX$
$=360^{\circ }-\varphi -\left( 180^{\circ }-\alpha \right) -\psi =180^{\circ }-\left( \varphi +\psi -\alpha \right) $
$=180^{\circ }-90^{\circ }=90^{\circ }$.

Thus, the point Q lies on the circle with diameter XY. The center of this circle is the midpoint M of the segment XY; hence, MQ = MX = MY. From MQ = MY, it follows that the point M lies on the perpendicular bisector of the segment QY. From UQ = UY, it follows that the point U lies on this perpendicular bisector, too. Thus, the line MU is the perpendicular bisector of the segment QY. We conclude $MU\perp QY$. Similarly, $MS\perp QX$. Since < XQY = 90, we have $QX\perp QY$. Thus, $MU\perp MS$, so that < SMU = 90 as claimed.

  Darij

I could go ahead and post my proof, but it's basicly the same thing, except that I constructed a poit $R$ s.t. $UXRY$ is a parallelogram :).

As soon as you constructed the point $Q$, I saw what was happening.  Nice.

Produced $TS$ and $TU$ at $P$ and $Q$ st $PS=SX=RU=a$, $QU=UY=RS=b$.
Also: $XY=d, PQ=c$.
Now the pairwise of triangles $(RSP,URY)$;$(RSX,QUR)$ are congruent, where:
$PR=RY=y,RQ=RX=x$, also: $PY//SM$,$QX//UM$.
In the $PQYX$ quadrilateral we shall prove that its diagonals are orthogonals.
We know:
$\measuredangle SRX=\measuredangle RQU,\measuredangle SXR=\measuredangle QRU,\measuredangle URY=\measuredangle RPS,\measuredangle RYU=\measuredangle PRS$,
$\measuredangle XRY+\measuredangle PRQ=180$.
In the triangles:
                 $PRQ:c^2=x^2+y^2-2cos\measuredangle PRQ$ ..(1)
                 $XRY:d^2=x^2+y^2-2cos\measuredangle XRY$ ..(2)
Median's theorem to triangles:
                 $PRX:x^2+y^2=2(a^2+b^2)$ ...(3)
                 $QRY:x^2+y^2=2(a^2+b^2)$ ...(4)
From (1),(2),(3),(4):
$PQ^2+XY^2=PX^2+QY^2$
Hence $PY\perp QX$ i.e. $\measuredangle SMU=90$. QED.


floor function
function
conics
hyperbola
analytic geometry
graphing lines
slope

I can believe such problem appeared in Arthur Engle's problem solving strategies.
Apparently, no one sees reasons to solve it...

Just coincidence maybe, but did you get your avater off a hacking website?

We have that the general term is 2(n-2) for all n, except when n is 1,2,3,4,5,6: then it is 1,2,4,5,7,9, as opposed to -2, 0 2, 4 ,6 ,8.  In these
cases, its +3, +2, +2, +1, +1, +1.

So we want 2(n-2) + 1 for n=1..6 , + 1 for n=1..3, +1 for n=1.

We try f(n) =  2(n-2) + floor(6/n).  It works for all n except we need -1 for n=3, -2 for n=2, and -3 for n=1.

Now try f(n) =  2(n-2) + floor(6/n) - floor(3/n).  It works for all n except n=1, in which case we need +2.

Now try f(n) = 2(n-2) + floor(6/n) - floor(3/n) + 2*floor(1/n), and we are done.

ps:
If its wrong, its because I rushed.  But the general idea is this: floor(k/n) = 0 for n > k.

[quote="seamusoboyle"]...Or, if you feel like it, test my one out, i didnt check it, but fairly sure it works....[/quote]

You should have tested it, because it does not work. This is what I get from your formula when plugging it into Excel:

$a_n = 1, 2, 4, 6, 8, 10, 12, 12, 14, 16, ...$

$a_n = 2(n - 2) + \lfloor{\frac{14}{2n + 1}}\rfloor - \lfloor{\frac 1 n}\rfloor$

This gives $a_n = 1, 2, 4, 5, 7, 9, 10, 12, 14, 16, ...$

Nice one, how did you derive your sequence, or was it hit or miss?

In a similar way as Singular. First, $2(n-2)$ is the main term. Then the initial 6 members of the sequence have to be adjusted by +3, +2, +2, +1, +1, +1. The importance of adjusting the terms is highest for $a_6$ and then decreases for $a_5, a_4, ..., a_1$. A linear function will not do - some remaining deviations will be always positive and some negative. Next thing to try is a hyperbola, which in combination with the floor function corrects most terms starting with $a_6$. The slope is adjusted by its numerator coefficient and/or by shifting the vertical asymptote left and right. I few attempts give the second term of the formula, which corrects everything but the least important member $a_1$. The remaining correction of the single least important term is trivial to deal with.

What about $a_n=2(n-2)+\lfloor \frac{6}{n}\rfloor -\lfloor \frac{2}{n}\rfloor-\lfloor \frac{1}{n}\rfloor$?

[quote="seamusoboyle"]Hi, i came across a promblem in Arthur Engle's problem solving strategies. However, the solution given is not correct. I have found a solution, but not the nicest one in the world. The problem states to find the general term for the sequence:

$1,2,4,5,7,9,10,12,14,16,18$ and goes up in twos from thereon. Floor values are allowed. The solution i found is:

$a_n=2n+ \frac{257 ( \lfloor \frac{7}{n} \rfloor )^3}{420}+\frac{1657 ( \lfloor \frac{7}{n} \rfloor )}{420}-(\frac{13 ( \lfloor \frac{7}{n} \rfloor )^4}{280}+\frac{703 ( \lfloor \frac{7}{n} \rfloor )^2}{280}+4)$

Anyone care to beat me on elegance? Or, if you feel like it, test my one out, i didnt check it, but fairly sure it works....[/quote]

Whats the symbol "L _l"  mean? (it looks like i have to learn LaTeX)

[quote="henryyangrui"]
Whats the symbol "L _l"  mean? (it looks like i have to learn LaTeX)[/quote]
if i'm not mistaken, it means: the integer equal to or closest to (but smaller than) the number within those symbols... crappy explanation  :?
e.g.:
$\lfloor {3} \rfloor=3$
$\lfloor {-14.2} \rfloor=-15$
$\lfloor {9.9999999...} \rfloor=9$
$\lfloor {-12.5431} \rfloor=-13$

Yes, $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$, also called the floor function/value. It's not really a $\LaTeX$ thing, but more of a math thing in general.


The mean of four distinct positive integers is 5. If the largest
integer is 13, what is the smallest integer?

If the mean of 4 integers is 5, that means the sum is $ 4 \times 5 \equal{} 20$. Since the largest is 13, the sum of the other 3 is $ 20 \minus{} 13 \equal{} 7$. That means we want to find 3 distinct integers such that the smallest one is as low as possible (and still positive.) Note that $ 1 \plus{} 2 \plus{} 4 \equal{} 7$, so $ \boxed{1}$ is the correct answer.


Provise a syntesis of the following compound starting from cyclohexane:

$ (c\minus{}Hex)\minus{}C \equiv C\minus{}(c\minus{}Hex)$

(where c-Hex is the cyclohexyl group).

Dunno if this is right :maybe:  :maybe:

1. Catalytic dehydrogenation to get benzene :maybe:
2. use excess of benzene from above reaction and react with Cl-CH2-CH2-Cl to get Ph-CH2-CH2-Ph
3. Hydrogenate with Ni as catalyst to get the cyclohexyl gps
4. Bromination; elimination using alc. KOH to get (c-hex)-CH=CH-(c-hex)
5. X2; heat with alc. KOH to get (c-hex)-CX=CH-(c-hex)
6. React with NaNH2

EDITED

[quote="Carcul"]Dehydrogenation of cyclohexane to get benzene doesn't look like a reasonable laboratory procedure.[/quote]

So, is there no easy method to convert cyclo-hexane to benzene???

You can do that by multiple halogenations and dehydrohalogenations, but that should not be the path here.

But elimination would be a favorable path since your halide is secondary.

Arent elimination and nucleophilic substitution roughly equal for a secondary halide though?

No. That depends on nucleophile, halide, and solvent.

[quote="jdlkem"]1. React 2 equiv c-Hex with NBS/h$ \nu$
This will produce 2 equivs of cyclohexyl bromide, bromide is a good leaving group.

2. React 1 equiv acetylene with 2 equiv NaNH2 in mixture.
This should give an acetylene dianion, where the two dianions can react with the two equivalents of cyclohexyl bromide to give the product.[/quote]
then carry out an ethynylation reaction. i.e react two moles of cyclohexanone with acetylene and nanh2 foll by red. hopefully the ketones do not react sluggishly.

I am new to organic chem and have been looking at some old problems posted to try out. this synthesis looked farely simple to me to start trying out. So heres my first attempt at a synthesis. Sorry if I offend anyone here by posting some unnatural chemistry...

1. Brominate Cyclohexane with nbs--> bromocyclohexane
2. Add two equivalents of bromocyclohexane to acetlyne in presence of NaNH2 --> product

Thoughts?

rawr different approach:

1. NBS the c-Hex.
2. Mg + ether to make grignard.
3. react w/ ethanedial.
4. acidify.
5. dehydration probably wouldnt work well. so, thionyl chloride! and then KOH @ 200 deg C (favors internal alkyne)

@ jdlkem

Arent less steps better?
btw congrats at IChO

[quote="Feynman_wannabe"]2. Add two equivalents of bromocyclohexane to acetlyne in presence of NaNH2 --> product[/quote]

Elimination would be the preferred pathway.

[quote="Jdlkem"]4. acidify.[/quote]

That would be tricky, since there is a strong tendency for the Pinacol Rearrangement.


I looked up the official AMC site, but I'm still confused since it's my first time taking this test.  :)
I found out that my school does not offer AMC. However, there's a university near where I live that offers the test. How do I register for AMC12 if I'm going to take the test at the university not at my high school? What's the deadline and cost for the registration?

There's not much concrete information here to provide any help.

1.  Have you read Frequently Asked Question #4?  Have you acted on the suggestions there?  What response did you get?  Have you found at least two other students in your school who are interested in taking an AMC contest?  Have you offered to help the school pay for the contests?  What was the reaction?  Have you taken your request to the Chair of the Math Department at your school?  How about getting all the schools in your district together for a joint session that benefits all of them?  For that, you may want to contact the Math Supervisor for your school district.  Every district has one.

2.  Have you contacted the person at the University near you offering the contest?  The contact information is on the listing?  What did that person say?  Did they give you information about taking the contest at the University site?

3.   Look again at the AMC site, the registration deadlines and cost for registration are at:
http://www.unl.edu/amc/b-registration/b1-archive/2008-2009/2009-H-Brochure.pdf
and
http://www.unl.edu/amc/b-registration/b1-archive/2008-2009/2009-H-Registrations.pdf
and
http://www.unl.edu/amc/b-registration/b1-archive/2008-2009/2009-H-Registrations.pdf
and
http://www.unl.edu/amc/whatswhen.shtml

Steve Dunbar
MAA Director, American Mathematics Competitions

One ongoing source of confusion with taking the B test at a local university is registration.  We have contacted both AMC and the local university and gotten conflicting info.

The university contact insists that the student must register with AMC in advance, and all he does is make sure that he's ordered enough tests.

AMC tells us that the university site does the registration, and the personal information gets filled out on test day before the test.

Please clarify.  Who gets money and when?  How does registration happen?  How will the student find out scores?

Sorry to be so obtuse, but I'm really confused.

The person at the University is mistaken.  Please refer them to the AMC Website instructions for Colleges and Universities registering as administration sites.  You can review the process yourself to make sure you are familiar with it before speaking to them again.

An overview of the process is at the 2-page flyer:
http://www.unl.edu/amc/b-registration/b1-archive/2008-2009/CU2009/2009-C+U,Messenger.pdf
Please pay particular attention to Section 4 on page 2, and even more so, the second paragraph of that section.

Then refer them to the Instructions booklet:
http://www.unl.edu/amc/b-registration/b1-archive/2008-2009/CU2009/2009-C+U,Booklet.pdf

It's too much to repeat here so have them read
Section II, Preliminary Instructions for Administering the AMC 10 B/AMC 12 B, on page 3.
Pay particular attention to items 1,2,4,5, and 6.  It's less than 2 booklet pages.

A person, generally an interested or involved faculty member has to take responsibility for the administration.  If you are talking to an administrative or reception person at the institution, they are likely to have never heard of the MAA.  A math faculty member should at least be aware of the MAA, if not already a member of the MAA.   They should be at least aware of the AMC.  You can talk to the faculty member and encourage them to sponsor an administration site.  Usually, the college or university will pay (AMC) for the registration, and do the ordering, administration and set-up.  Most will pay for it because it is good recruiting and outreach.  A few may "charge back" the students to defray their costs.  If the college does charge back, it should only be a few dollars, which is what it costs per student.  Of course, if the college includes a lunch, a T-shirt, and offers a grand prize, they may charge extra for those.  The process for finding out scores is detailed in the Instruction booklet.

Last year, over 30 colleges and universities in addition to about 4000 high schools participated without a single hitch.  It's easy, but if you can suggest alternative ways to make the process easier, less confusing, more productive and enjoyble for everyone, we welcome your suggestions.

Steve Dunbar
MAA Director, American Mathematics Competitions


This is a question in aops book that's part of Mandelbrot question. I think my answer key is wrong or I either didn't pick out the word.

Anyway,

Two bicyclists are seven-eighth of the way through a mile long tunnel when a train approaches the closer end at 40 mph.  The riders take off at the same speed in opposite directions, and each escapes the tunnel as the train passes them. How fast did they ride?

In my answer key, it's [hide]30[/hide] but I believe that's a mistake.

Here is my work:

Since we don't know the time, we have to figure out the time. Knowing that train have to go through all the tunnel, it's distance is 1. Speed is 40 so the time must be 1/40.

Now, we know that bicyclists were 7/8 when the closer end of the tunnel they were going came a train. So, they turned and speed up to go back 7/8 of the tunnel to get out.

Then we will get this:

7/8 = x(speed) * 1/40

And I believe that should be [hide]35[/hide].

I was really happy to see myself getting all the questions in the book right (most of them went pretty easy, and I can't believe I'm doing it myself!  ) And well, the book gave an explanation about one rider getting out of tunnel earlier than other one but nope, in my question, from beginnning to end, it said, "they" meaning dos, (in spanish) or two people.

Please check this out please.

Thank you.

Silverfalcon

Silverfalcon wrote:Now, we know that bicyclists were 7/8 when the closer end of the tunnel they were going came a train. So, they turned and speed up to go back 7/8 of the tunnel to get out.

Then we will get this:

7/8 = x(speed) * 1/40

And I believe that should be [hide]35[/hide].

First, how do you know that the bicyclist starts when the train is at the end of the tunnel?  What if they start before the train even enters the tunnel?  Think about it... (Btw, I agree with the answer from the key)

First of all, the train is still a certain distance away from the tunnel, so while it will indeed travel through the entire tunnel in 1/40 hour, it also needs to travel the distance between its current position and the beginning of the tunnel.

As for the confusion about two different times, the bicyclists are heading in opposite directions, one toward the closer (but also straight into the path of the train) end and one toward the farther end.  Thus, the one heading toward the closer end will reach his end of the tunnel faster than the other person - but they will both reach them at the same time as the train reaches their respective ends.

I don't know where this problem was in the AoPS books (I remember it, but not the location), so I think the easiest way to solve it is to say

[hide]Let r be the rate, in miles per hour, of the bicyclists.  Thus the first bicyclist will travel his 1/8 mile in (1/8)/r = 1/(8r) hours, and the second bicyclist will travel his 7/8 mile in (7/8)/r = 7/(8r) hours.  The train thus must be at the entrance of the tunnel at time t = 1/(8r) and at the exit at time t=7/(8r), so it took 7/(8r) - 1/(8r) = 3/(4r) hours to travel the length of the tunnel.  But that length was 1 mile and it was travelling at 40 mph, so we have

d = rt

1 = 40 * 3/(4r)

1 = 30/r

r = 30

Thus the bicyclists' rate is 30 mph.[/hide]


Find the maximum and minimum value of the following functions.

(1) $ f_1(x)\equal{}\cos 2x\plus{}2\sqrt{2}\sin x\ (0\leq x\leq \pi)$.

(2) $ f_2(x)\equal{}\sqrt{3}\sin x\plus{}\cos x\ \left(\frac{\pi}{6}\leq x\leq \frac{2}{3}\pi\right)$.

(3) $ f_3(x)\equal{}\cos ^ 2 x\minus{}4\cos x\sin x\minus{}3\sin ^ 2 x\ \left(0\leq x\leq \frac{\pi}{2}\right)$.

(4) $ f_4(x)\equal{}\sin x\plus{}\sin 2x\plus{}\cos x\ (0\leq x<2\pi)$.

hello, to (1) the maximum is $ 2$ and will be obtained by $ x\equal{}\frac{\pi}{4}$ and $ x\equal{}\frac{3\pi}{4}$
The minimum is $ 1$ and will be obtained by $ x\equal{}0$ and $ x\equal{}\pi$.
Sonnhard.

Dr Sonnhard could you post the whole solution? Here's the solution for #2 (all angles in degrees):
[hide]$ 2(\sqrt3/2\sin x \plus{} 1/2\cos x)\equal{}2(\sin(x\plus{}30)), 60\le x\plus{}30\le150$

So the max of $ 2$ occurs when $ x\plus{}30\equal{}90$ and the min of $ 1$ occurs when $ x\plus{}30\equal{}150$[/hide]

[hide="#1"]$ f_1(x)\equal{}1\minus{}(2\sin^2x\minus{}2\sqrt2\sin x)\equal{}2\minus{}(\sqrt2\sin x\minus{}1)^2$[/hide]
[hide="#3"]$ f_3(x)\equal{}2\cos 2x\minus{}2\sin 2x\minus{}1\equal{}2\sqrt2\cos(2x\plus{}\frac{\pi}{4})\minus{}1$[/hide]
[hide="#4"]$ f_4(x)\equal{}(\sin x\plus{}\cos x)^2\plus{}(\sin x\plus{}\cos x)\minus{}1\equal{}2\sin(x\plus{}\frac{\pi}{4})^2\plus{}\sqrt2\sin(x\plus{}\frac{\pi}{4})\minus{}1$[/hide]


I have a problem but now i cann't solve it:
[b]problem[/b]: let a,b,c, d are reals that satified $a^2+b^2+c^2+d^2\leq1$. Find maximal the expression: $T=(a+b)^4+(a+c)^4+(a+d)^4+(b+c)^4+(b+d)^4+(c+d)^4$

[quote="kunny"]arqady, I solved again, but I got another answer. Where is something wrong with this answer. :?

Use Jensen's inequality we have $\left(\frac{a+b}{2}\right)^4\leq \frac{a^4+b^4}{2}\Longleftrightarrow (a+b)^4\leq 8(a^4+b^4).$ etc.

Thus $T\leq 24(a^4+b^4+c^4+d^4)=24\left\{(a^2+b^2+c^2+d^2)^2-2(a^2b^2+a^2c^2+a^2d^2+b^2c^2+b^2d^2+c^2d^2)\right\}$
$\leq 24(a^2+b^2+c^2+d^2).$ Since $a^2+b^2+c^2+d^2\leq 1,$ yielding $T\leq 24.$[/quote]

I think you use $(a^2+b^2+c^2+d^2)^2\geq (\sum a^2)^2 -2\sum a^2b^2$ but the equality sign holds when $a=b=c=d=0$ which isnt actually a maximum

[quote="shyong"]
I think you use $(a^2+b^2+c^2+d^2)^2\geq (\sum a^2)^2 -2\sum a^2b^2$ but the equality sign holds when $a=b=c=d=0$ which isnt actually a maximum[/quote]

I don't think so. The equality holds for example when $a\neq 0$ and $b=c=d=0$ and $a^2+b^2+c^2+d^2=1.$
that is to say $a=\pm 1$ and $b=c=d=0.$

[quote="kunny"][quote="shyong"]
I think you use $(a^2+b^2+c^2+d^2)^2\geq (\sum a^2)^2 -2\sum a^2b^2$ but the equality sign holds when $a=b=c=d=0$ which isnt actually a maximum[/quote]

I don't think so. The equality holds for example when $a\neq 0$ and $b=c=d=0$ and $a^2+b^2+c^2+d^2\leq1.$
that is to say $-1\leq a\leq 1$ and $b=c=d=0.$[/quote]

but remember also you apply Jensen  :)  And the equality case for Jensen is $a=b$ .... etc

That's true. I can understand.Thank you, shyong.  :)


Tasaode hesabi namotenahi ba ghadre nesbate $d$ dar nazar begirid ,farz konid dar in donbaleh hatman ye adade morabae kamel va yek adade mookaabe kamel vojood dashte bashad sabet konid hatman yek adade tavan sheshome kamel niz vojood darad.

Soal e sakhtiye!
Romanian TST 1998.

see here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=658


1-1. $\displaystyle \text { there are integers a and b, } a,b>1,  \text { such that }  a^{9a} = b^{2b}.$

$\text { Find the minumum of } 2a + b$

1-6. $\displaystyle \text { In a function f(x), if } f(3) = 1, (x^2-x-1)f(x^2) = f(x).$

$\text { Then find 10m+n, where } f(-9) = \frac{n}{m}$

For the first one:
[hide]We see $a > b$, for otherwise the RHS is a larger number raised to a larger power.  Since $a > b$, we must have $2a < 9b$ or the LHS would be a larger number raised to a larger power.  Furthermore, we know that since a, b are integers and $a^x = b^y$ for integral x, y, there exists t, u, v such that $a = t^u, b=t^y$ and so b divides a.  Thus, it follows that $a = 2b, 3b$ or $4b$ and $t = 2$ or $3$.  In the first case, we have:
$2^{2(v+1)2^{v+1}}=2^{9v2^v}$ so $4(v+1)=9v$, an equation with no solutions in integers.
In the second case, we have
$3^{2(v+1)3^{v+1}}=3^{9v3^v}$ so $2(v+1)=3v$, $v = 2$, $a = 27, b = 9, 2a + b = 63$.
In the third case, we have
$2^{2(v+2)2^{v+2}} = 2^{9v2^v}$ so $8(v+2) = 9v$, $v = 16$, $a = 2^18, b = 2^16, 2a + b = reallybig$

and so the minimum is 63.[/hide]


Find the number of postive roots of the equation : $ x^2\plus{}\frac{1}{x^2}\minus{}6x\minus{}\frac{6}{x}\plus{}12\equal{}a$ for real number $ a$.

Note that
$ x^2 \plus{} \frac {1}{x^2} \minus{} 6x \minus{} \frac {6}{x} \plus{} 12 \equal{} t^2 \minus{} 6t \plus{} 10,\qquad t: \equal{} x \plus{} \frac {1}{x}$.
Now, you have transformed the problem to the solution of the equation
$ t^2 \minus{} 6t \plus{} 10 \minus{} a \equal{} 0,\, a\in\mathbb{R}$
which is solvable in $ \mathbb{R}$ only for $ a\ge 1$. For $ a < 1$ you obtain nonreal roots $ t_{1,2}$ and thus also $ x\not\in\mathbb{R}$ - a constradiction with the fact that $ x > 0$.

The rest is not so complicated.

Also, you need $ t \equal{} x \plus{} \frac 1x \ge 2\sqrt {x \cdot \frac {1}{x}} \equal{} 2$ (since $ x > 0$). The roots are $ t \equal{} 3 \pm 2\sqrt {a \minus{} 1}$, and for $ a > 2$ we have $ t_{2} \equal{} 3 \minus{} 2\sqrt {a \minus{} 1} < 2$. Thus for $ a \equal{} 1$ and $ a > 2$, we have two solutions, and for $ 1 < a \le 2$ we have four solutions for $ x$.


function
calculus
derivative
algebra
polynomial
quadratics
complex analysis

Recalling a brief flurry of activity among the grad students standing around in a department lounge, a long time ago ...

The theory of differentiable complex-valued functions of a single complex variable is a resounding success.

Could there be a satisfactory theory of differentiable quaternion-valued functions of a single quaternionic variable? Can we reconstruct the analogues of some of our favorite theorems? Or is there a fatal flaw that dooms the whole enterprise?

Hmm... I would say that the main problem lies in non-commutativity of multiplication. Of course, $f(q)=q$ is differentiable with the derivative $1$ in any reasonable sense of the word. But what would be the derivative of $f(q)=q^2$? (or, generally, what is the derivative of the product of 2 "differentiable" functions?).

All right, a sub-task. Can we make a coherent definition of the "formal derivative" of a polynomial? If we could do that, we'd have a chance of extending to power series.

If we take the simplest form of "linear operator", I think the answer is that the only differentiable polynomials in $q$ are of the form $aq+b$. That is, $qa$ and $q^2$ aren't differentiable.

The differential of $q^2$ is $q\cdot dq+dq\cdot q$, which is not properly linear. The differential of $qa$ is $dq\cdot a$.

We probably have that the only differentiable functions are of the form $aq+b$.

If you want any more, you have to open things up to linear operators of the form $df=u\cdot dq\cdot v$, and then how do you take a derivative of one of those?

Just as a postscript, I'll report what ended that long-ago grad student discussion. We tried to make sense of the formal derivative of a polynomial. For instance, the formal derivative of the arbitrary quadratic monomial $axbxc$ should be $abxc+axbc.$

But there's a problem even with this, as one of the professors in the room eventually pointed out to us. The representation of a polynomial (that is what terms with what coefficients it is to have) is not uniquely determined by its values as a function. Although I have forgotten the specific example, we did come up with two different expressions that were identical as functions but whose formal derivatives were different as functions.

[quote="fleeting_guest"][quote]Could there be a satisfactory theory of differentiable quaternion-valued functions of a single quaternionic variable? [/quote]

There is a version of complex analysis specific to the quaternions that was developed at the beginning of the last century (and, probably, redeveloped periodically by various people since then).  I think the name first associated with this is Feuter.

There is also a version for Clifford algebras of any dimension, called "Clifford analysis".  There is a cult of people who try to formulate physics in this language (Maxwell equations, quantum mechanics, etc), but otherwise it is not a very active subject.

You have to choose carefully the definitions to get any nonlinear functions in such a theory.  Over $\mathbb{C}$ there are various threads that magically intertwine, but are different or nonexistent in the Clifford/quaternionic situation:

power series
Cauchy-Riemann equations
$d/{d{\bar{z}}} = 0$
conformal maps
harmonic functions

e.g. in $\dim > 2$ there are no nonlinear conformal mappings except for inversions.[/quote]

A quaternion equivalent for the Cauchy-Riemann (C-R) Condition is possible with a quaternion differential operator.  An appropriate operator  is $X= d/cdt + \nabla$. Here "c" is the speed of light and "t" is time, thus giving a scalar derivative.  The scalar derivative can be simplified to $d/dw$  where $c=dw/dt$ and "w" is the scalar wavelength.

Using this operator and Hamilton's Rules for quaternions , the C-R Conditions are

$0 = XF = (d/dw + \nabla)(F_s + F_v) = ( dF_s/dw - \nabla \cdot F_v) + (dF_v/dw + \nabla \times F_v + \nabla F_s)$

As "Reals and Complex numbers" are a subsets of quaternions, this operator works for all those domains.


My buddy and I stopped in a sports restaurant the other day that had Keno going.  We got into a small debate about the probability of a single number coming up after n number of games.  If you don't know anything about Keno basically there are 80 possible numbers that can come out of a lottery-like ball machine and 20 of these numbers are returned each game.  My buddy stated that the chances of picking a number that is drawn each game is 25% since they draw 20 balls out of 80 possible.  He then stated that the probabilty of a paricular ball being drawn after n games provided that it has not already been drawn is .75^n.  On the surface this sounded reasonable, but I tended to disagree with him and actually witnesses a number that did not come up for 28 straight games.  By his calculations the chances of this happening are about .03%.  Something is fishy about this method.  Will someone give me the correct formula so I can show him.

Thanks,
Chad

I believe your friend is right - the chance of any given ball not coming up for 28 games running is $0.75^{28} \approx 0.0003$. However, when you take into account that there are 80 balls, the probability that this happens to any one of the balls is $(1-(1-(0.75^{28}))^{80}) \approx 0.025$ - still small, but feasible.

Thanks for the info.  I am still very confused and a complete amateur with probabilities.  Can you give me the underlying formulas that would apply to this situation?  What if there have been 10 games played without a certain number being drawn once.  What is the probability of the number being drawn in the next 10 games?

thanks,
Chad

Well, the probability that a single ball (say ball $x$) doesn't come up on the first round is 0.75. Subsequent rounds are independent of the first, so the probability of $x$ not coming up for the first 28 rounds is $0.75^{28}$ - let's call this number $p$ for convenience.
Now, the probability that $x$ does show up some time in those 28 rounds is clearly $1-p$. Moreover, you have 80 balls, all of them independent* - therefore the probability that every one of them shows up is $(1-p)^{80}$, and the probability that at least one doesn't show up is one minus the probability above, i.e. $1-((1-p)^{80})$.

[quote]
What if there have been 10 games played without a certain number being drawn once. What is the probability of the number being drawn in the next 10 games? [/quote]
The probability of the number (again, call it $x$) being drawn in the next 10 games is actually the same regardless of whether $x$ has appeared 0, 5 or 10 times in the first 10 - the first 10 games and the next 10 are entirely independent events. So we want the probability of $x$ being drawn at least once in 10 games. As above, the probability that it never gets drawn is $0.75^{10}\approx0.06$, so the probability that it does get drawn at least once is one minus that, about 94%.

* actually this isn't a perfectly accurate model, since exactly 20 balls show up at a time - but the difference shouldn't be too great, since that's the expected number anyway, and it simplifies things enormously.


geometry
circumcircle
trigonometry
trig identities
Law of Sines

A circle passes through the three vertices of an isosceles triangle that has two sides of length $ 3$ and a base of length $ 2$. What is the area of this circle?

$ \textbf{(A)}\ 2\pi\qquad \textbf{(B)}\ \frac {5}{2}\pi\qquad \textbf{(C)}\ \frac {81}{32}\pi\qquad \textbf{(D)}\ 3\pi\qquad \textbf{(E)}\ \frac {7}{2}\pi$

hello, by the theorem of Pythagorean we get the hight of the triangle as $ h\equal{}\sqrt{8}$. By a well-known formula we have $ R\equal{}\frac{2\cdot\cdot3\cdot3}{4\cdot\sqrt{8}}$, so we get the searched area as $ A\equal{}\frac{\pi\cdot 81}{32}$.
The answer is $ \boxed{C}$.
Sonnhard.

sonnard, what is this "well-known" formula

$ [ABC] \equal{} \frac {abc}{4R}$, where $ a$,$ b$, and $ c$ are opposite sides of $ A$,$ B$, and $ C$, respectively, and $ R$ is the circumradius. I personally haven't used this formula much, nor can I can remember it.

[hide="Different"]
Let $ \overline{BD}$ be an altitude of $ \triangle ABC$ and let $ O$ denote the center of the circle with radius $ r$ that passes through $ A$, $ B$, and $ C$.

Then $ BD\equal{}\sqrt{3^2\minus{}1^2}\equal{}2\sqrt{2}$ and $ OD\equal{}2\sqrt{2}\minus{}r$.  Because $ \triangle ADO$ is right, we have:

$ r^2\equal{}1^2\plus{}\left (2\sqrt{2}\minus{}4 \right )^2\equal{}1\plus{}8\minus{}4\sqrt{2}r\plus{}r^2$ and $ r\equal{}\frac{9}{4\sqrt{2}}\equal{}\frac{9}{8}\sqrt{2}$.

Thus, the area of the circle is:

$ \left (\frac{9}{8}\sqrt{2} \right )^2\pi\equal{}\frac{81}{32}\pi$, or $ \boxed{\textbf{(C)}}$.
[/hide]

[hide="Trigonometry"]
Drop an altitude that divides the isosceles triangle into two congruent right triangles. Call the angle opposite the altitude $ \theta$. The altitude is $ \sqrt{8}\equal{}2\sqrt{2}$ long, by the Pythagorean theorem.

We have that $ \sin\theta\equal{}\frac{2\sqrt{2}}{3}$.

By the extended Law of Sines, we have $ R\equal{}\frac{1}{2}\cdot\frac{3}{\frac{2\sqrt{2}}{3}}\equal{}\frac{1}{2}\cdot3\cdot\frac{3}{2\sqrt{2}}$
$ R\equal{}\frac{9}{4\sqrt{2}}$.

Therefore, $ \pi R^2\equal{}\pi\cdot\frac{81}{16\cdot 2}\equal{}\frac{81}{32}\pi$.
[/hide]


Hello,

Lynbrook Math is thinking of hosting a math tournament at Lynbrook

It will probably be from the AMC 12 to USAMO range, seeing as we already have a lot of easier-level math contests in this area.

Anyone interested?

Details to come, if I get this off the ground.

Thanks,
Timothy Chu

It would be structured similarly to the Bay Area Math Meet (which died in 2005), though all planning is still in its way preliminary stages and this event might not happen, depending on how things work out...

It would two-person event with 20 questions (AMC 10-12 level) at one-minute-per-question.
And then an individual test (1.5 hoursish, 20 questions, fun AMC 12-AIME level problems)
And then a countdown style thing (4 person per team)

Winners overall would then take part in a proof contest sort of thingie, where each team gets 1 hour to solve a bunch of questions, and then 1 hour to present. Ideally, there would be a panel of judges (although I've only gotten one willing to help so far =/) to judge the proofs. The winner of everything would be the winner of the last contest, and there would (hopefully) be some good prizes!

Of course, this might not actually happen, depending on how things work out...

[quote="Scrambled"]Winners overall would then take part in a proof contest sort of thingie, where each team gets 1 hour to solve a bunch of questions, and then 1 hour to present. Ideally, there would be a panel of judges (although I've only gotten one willing to help so far =/) to judge the proofs. The winner of everything would be the winner of the last contest, and there would (hopefully) be some good prizes![/quote]

Sounds like Math Battle. Interesting, but I did pretty badly last year.

About what time of the year will this be held?

Hopefully between February and March (before AIME), although we've got all the problems (original!) basically taken care of already!!!

Sounds like fun.  i'm always happy to hear of new contests starting!  Be sure to get in touch with me during your planning stages, Tim; i'm happy to share my insights regarding contest organization and administration (unless you've done this before, there are always unexpected logistical traps you have to watch out for).  And by all means, be sure to check with me about timing before you publicize anything because you'll want to schedule around other events that are already planned.  Sounds like you've got the academic side of it covered, so that's a good start.  Looking forward to see what you come up with..

-Tim

You'll also want to check the date with SMT.  As I recall, near the end of February/early March, Tim has two contests scheduled three weeks apart.  SMT has one of the dates in between, and the other would then probably be open.  Looks like SMT is currently scheduled for February 28th, but you should check that with Wendy to make sure it's final.

Thanks for the input.

The website said SMT is on 28th, so I assume thats final. In any case, I'd probably find a date early march, where no contests seem to collide.

[quote="not_trig"][quote="CatalystOfNostalgia"]whoa this is a good idea, tim chu
perhaps we should try it in mass, seeing that the greater boston math league has managed to get 6 problems wrong in their answer key so far this year[/quote]

hmm dude is that a student-run thing or not?

unfortunately student-run things that don't have 999 people who can do all the problems (in reasonable time) means there will inevitably be some wrong answers :([/quote]

it is not a student run event, rather an event run by (public) high school math teachers who have proven themselves to not be so competent

however the point is now moot as we are now going to do a middle school version of HMNT (or something similar). we will credit backho chu for inspiration.

Dude. This is a good idea. I would definitely participate.

Tim, we should also do MAO.

Yay so how does March 13 sound.

Three local professors have given tentative willingness to judge the final round of this, so that is a go! (hopefully)
Problems are just about done yay (although haven't gone through any rigorous review yet)

The Lynbrook MAth Olympiad! (LMAO)

Although again, this may or may not happen depending on the willingness of facility managers at Lynbrook to allow us space...

(Also Wenkai, if you could find a regional director who'd be willing to assume the MAO duties, then I'd be up for it, but as it happens I think Korsunsky has her hands full already)

[quote="Scrambled"]
The Lynbrook MAth Olympiad! (LMAO)
[/quote]

I love you mr. chu

that is all


algebra
polynomial
induction
combinatorics proposed
combinatorics

The sequence of polynomials $ (a_n)$ is defined as, $ a_0 \equal{} 0$, $ a_1 \equal{} x \plus{} 2$ and for all $ n > 1$ integers, $ a_n \equal{} a_{n \minus{} 1} \plus{} 3a_{n \minus{} 1}a_{n \minus{} 2} \plus{} a_{n \minus{} 2}$.
a.Show that for $ k,m$ positive integers, if $ k$ divides $ m$; $ a_k$ divides $ a_m$.
b.Find all $ n$ positive integers such that sum of the roots of polynomial $ a_n$ is an integer.

Edit: $ a_1$ is $ x\plus{}2$ not $ x\plus{}1$, but nothing changes.

the recursion relation is equivalent to $ 3a_{n} \plus{} 1 \equal{} (3a_{n \minus{} 1} \plus{} 1)(3a_{n \minus{} 2} \plus{} 1)$. therefore (by direct induction):
$ 3a_{n} \plus{} 1 \equal{} (3a_{n \minus{} k} \plus{} 1)^{F_{k \plus{} 1}}(3a_{n \minus{} (k \plus{} 1)} \plus{} 1)^{F_{k}}$, $ F_{k}$ is the Fibonacci sequence with $ F_{1} \equal{} F_{2} \equal{} 1$, so: $ 3a_{n}\plus{}1\equal{}(3a_{1}\plus{}1)^{F_{n}}(3a_{0}\plus{}1)^{F_{n\minus{}1}}\equal{}(3x\plus{}4)^{F_{n}}, a_{n}\equal{}\frac{(3x\plus{}4)^{F_{n}}\minus{}1}{3}$
I need to leave now... But I think the first part is now almost trivial.

bambaman has shown that $ a_{n} \equal{} \frac13 ((3x \plus{} 4)^{F_n} \minus{} 1)$

Now (a) follows from standard (Fibonacci) divisibility facts:
If $ k$ divides $ m$, then $ F_k$ divides $ F_m$.
If $ a$ divides $ b$, then $ y^a\minus{}1$ divides $ y^b\minus{}1$.
Set $ y: \equal{}3x\plus{}4$, and $ a: \equal{}F_k$, and $ b: \equal{}F_m$, and you are done.

For (b), we consider the polynomial $ b_n\equal{}(3x \plus{} 4)^{F_n} \minus{} 1$.
For $ n\equal{}1$, the sum of all roots is an integer.
For $ n\ge2$, the coefficient of the highest term is $ 3^{F_n}$, and
the coefficient of the second-highest term is $ 4F_n\cdot 3^{F_n\minus{}1}$.
Hence, the sum of all roots equals $ 4F_n/3$.
The answer to (b) is: The sum of the roots of polynomial $ a_n(x)$ is
an integer, if and only if $ n\equal{}1$ or $ n$ is a multiple of $ 3$.

You have a little mistake in the last part. $ 3$ divides $ F_m$ if and only if $ 4$ divides $ m$. Also $ n \equal{} 2$ satisfies because $ F_2 \equal{} 1$.
So the answer is $ 1$, $ 2$ and $ 4k$.


calculus
derivative
function
real analysis
real analysis unsolved

We recently started doing word problems with Calculus in class and I am having trouble using derivatives to solve them. I don't have a true example but they are mostly dealing with cost optimizing. Can any body provide hints to solve these problems especially because they are an augment of derivatives.

You create a function that models the situation you are dealing with and then you just take the derivative with respect to the correct variable, set it to zero, and solve.  There really aren't any special tricks, it's just looking at the problem, and seeing what you need to set up.  If you have an example, I could try to help you understand that.


Let ABC be an equilateral triangle, and let O be a point in its interior. Perpendiculars from O are dropped to AB, BC, and CA, at P, Q, and R, resp. with OP:OQ:OR = 1:2:3. Compute the ratio of the area of OPRA to the area of ABC.

I'm sure there is a very nice way to solve the problem. That being said:

[hide="Easy way to brute-force"]Set O as the origin, and construct P, Q, and R. Then, construct the perpendicular lines, and let their intersections be A, B, and C. Now, you know the coordinates of everything[/hide]


Prove or disprove: if $ a_i>0$ i=1,2,...,n where $ n>2$ and $ k,l\ge 1$ and $ a_1\equal{}a_{n\plus{}1}$, then

$ \sum_{i\equal{}1}^{n}\frac{a_ia_{i\plus{}1}}{ka_i^2\plus{}la_{i\plus{}1}^2}\le\frac{n}{k\plus{}l}$

[quote="MTristan"]Prove or disprove: if $ a_i > 0$ i=1,2,...,n where $ n > 2$ and $ k,l\ge 1$ and $ a_1 \equal{} a_{n \plus{} 1}$, then

$ \sum_{i \equal{} 1}^{n}\frac {a_ia_{i \plus{} 1}}{ka_i^2 \plus{} la_{i \plus{} 1}^2}\le\frac {n}{k \plus{} l}$[/quote]

The answer is disprove. Exam, when $ n\equal{}3, k\equal{}1, l\equal{}6$ and $ a\equal{}4,b\equal{}3,c\equal{}2$  :wink:

However, it's true for concrete values of $ n, k$ and $ l$. For example, with $ n\equal{}3$, I think the following inequality is true for all $ l\in\left[\frac1{5},5\right]$
\[ \frac {ab}{a^2\plus{}lb^2}\plus{}\frac {bc}{b^2\plus{}lc^2}\plus{}\frac {ca}{c^2\plus{}la^2}\le\frac3{1\plus{}l}\]


What is the remainder when $ 5^{30}$ is divided by 7?

hello, we have $ 5^3\equiv \minus{}1 \mod 7$ and from here we get $ (5^3)^{10}\equiv(\minus{}1)^{10}\equiv 1\mod 7$.
Sonnhard.


geometry
parallelogram
geometry unsolved

ABC is a triangle with s(A)=30, s(C)=90.  ABD, ACE and BCF are equilateral triangles such that D, E and F are points not contained in ABC.  AB=1 is given...

line DE intersects AB at point G. Find the area of DGF =?

1) D, B and F are collinear, hence $ DF \equal{} \frac {3}{2}  \ (\  1 \ )$
2) It's easy to see that $ \triangle BEF$ is isosceles, hence, if M is the midpoint of BF, then $ EM\equal{}EC\plus{}CM\equal{}\frac{\sqrt 3}{2}\plus{}\frac{\sqrt 3}{4}\equal{}\frac {3\sqrt 3}{4} \ ( \ 2 \ )$; also $ EM\perp DF$ ( 3 )
3) If N is the midpoint of AB, then $ DN \parallel{}\equal{}AE \perp AB$, hence ADNE is a parallelogram and G is the midpoint of DE.
From (1) and (2), with observation (3), $ S_{DEF}\equal{}\frac{9\sqrt 3}{16}$, while $ S_{DFG}\equal{}\frac{S_{DEF}}{2}\equal{}\frac{9 \sqrt 3}{32}$.

Best regards,
sunken rock


[img]http://i3.photobucket.com/albums/y57/lilshantypanty/prob3-1.jpg[/img]

Any help is appreciated! Thanks!

Since $ f(z)$ is analytic inside and on $ \partial D(0,1)$ with no zeros on $ \partial D(0,1)$, then the Argument Principle requires that the first integral give the number of zeros inside $ D(0,1)$ so let these zeros be:

$ z_0, z_1$.

Extending the Argument Principle to the case of:

$ \int_{\Gamma}g(z)\frac {f'(z)}{f(z)}dz$

where $ g(z)$ is analytic on an inside $ \Gamma$, the second integral gives their sum or:

$ z_0 \plus{} z_1 \equal{} 0 \Rightarrow z_0 \equal{} \minus{} z_1$

and likewise the third integral gives the sum of the squares of these zeros or:

$ z_0^2 \plus{} z_1^2 \equal{} 1/2$

or:

$ 2z_0^2 \equal{} 1/2$

This gives:

$ z_0 \equal{} 1/2$ and $ z_1 \equal{} \minus{} 1/2$ or:

$ f(z) \equal{} (z \minus{} 1/2)(z \plus{} 1/2)$


logarithms function calculus algebra domain

I tried that but still and I got near the answer but it comes a bit wrong :?

By definition $y = \log_ax \Leftrightarrow x = a^y \Leftrightarrow \log_{10}x = y\log_{10}a \Leftrightarrow y = \log_ax=\dfrac{\log_{10}x}{\log_{10}a}$

To be really careful you actually first have to do this with $\ln$ and thus show that all properties of $\ln$ hold true for $\log_a$ so that you can be sure that $\log_{10}(x^p) = p\log_{10}x$.

[quote="Kalle"]By definition $y = \log_ax \Leftrightarrow x = a^y \Leftrightarrow \log_{10}x = y\log_{10}a \Leftrightarrow y = \log_ax=\dfrac{\log_{10}x}{\log_{10}a}$

To be really careful you actually first have to do this with $\ln$ and thus show that all properties of $\ln$ hold true for $\log_a$ so that you can be sure that $\log_a(x^p) = p\log_ax$.[/quote]
Thanks :D

Is there anything incomplete or incorrect with this proof? Pardon the lack of code.

log_a(x) = 1 / log_x(a)
              = 1 / (log_10(a) / log_10(x) )
              = log_10(x) / log_10(a)

If I recall correctly, this was taught to me in my Grade 12 Introduction to Advanced Functions and Calculus class.

well u use the assumption with in the proof. i can prove 1=0 assuming 1=0 as well:
$1=1 \stackrel{0=1}{\implies }1-0=1-1\implies 1=0$ qed :P

I think this is in fact a special case of $\log_ab\cdot \log_bc = \log_ac$. This thing can be proven by calculating $a^\text{lefthand}$ and $a^\text{righthand}$:  both equal $c$. Use that and the fact that f : x -> a^x is a bijection between its domain and range.


Find the maximum value of

            [tex]    3sin(x)  + 4cos(x) [/tex]

I didn't know that typing [tex] at the start and [\tex] at the end is all you need to do this.

Call the quantity we want to maximize $Q$:

$(\sin^2 x+\cos^2x)(3^2+4^2)\geq Q^2$

therefore $Q_{max}=5$.  Equality is attained for $\sin x=\frac {3}{5}$, $\cos x=\frac {4}{5}$.

can ya explain how ya derived the two expressions and solved for it. what about the max value of

       [tex]    2\sin (x)  + 7\cos (x) [/tex]

In general, for positive integers $m,n$, let $Q(x)=m\sin x+n\cos x$. Then $Q_{\mathtt{max}}\Leftrightarrow Q_{\mathtt{max}}^2$.
Now, $Q^2=(m\sin x+n\cos x)^2$
Apply Cauchy-Schwarz and you get $Q^2\leq (m^2+n^2)(\sin^2x+\cos^2x)=m^2+n^2$ and so $Q\leq\sqrt{m^2+n^2}$


Prove that

$ \prod\frac{r_a\plus{}r_b}{r_c}\sum\frac{r_b\plus{}r_c}{r_a}\geq48$

.

[quote="Ligouras"]Prove that

$ \prod\frac {r_a \plus{} r_b}{r_c}\sum\frac {r_b \plus{} r_c}{r_a}\geq48$

.[/quote]

$ a \equal{} x \plus{} y,b \equal{} y \plus{} z,c \equal{} z \plus{} x$

$ \prod\frac {r_a \plus{} r_b}{r_c}\sum\frac {r_b \plus{} r_c}{r_a}\geq48$

$ \Longleftrightarrow (y^2z \plus{} y^2x \plus{} yz^2 \plus{} 8xyz \plus{} xz^2 \plus{} x^2y \plus{} x^2z)(y^2z \plus{} y^2x \plus{} yz^2 \minus{} 6xyz \plus{} xz^2 \plus{} x^2y \plus{} x^2z)\geq 0$

obvious true.


I know LaTex is built in to this forum, such that if I wanted to write 3x+2 I would just enclose it with dollar signs to get $ 3x \plus{} 2$.
But how do I get Latex available for use in acrobat to submit solutions for mathematics competitions? Is it particularly hard, because I need to be able to use it quickly.
Thank you in advance.

Oh yeah, I looked at the LaTeX tutorials and I'm still just a bit iffy. I know you still enclose things in dollar signs, but I want to know what program to download, etc.

Start at [[LaTeX]].  Then click "About" and read it if you haven't already.  Once you're done with that, click on "Downloads" to read about what you need to download and where you need to download it from.

Perhaps have a look at [url=http://texblog.net/latex-link-archive/distribution-editor-viewer/]Distributions, Editors, Viewers[/url], choose your operating system there. If you tell us your system we could recommend some software to you.

Stefan
--
[url=http://texblog.net]TeXblog.net[/url]


The shoelace theorem works because we can predictably decompose any plane polygon into triangles. There is a similar determinant formula for the volume of a tetrahedron, but decomposing a polyhedron into tetrahedra isn't so obvious.

isn't it that it only works if you have verticies, and verticies are 3D?

Keep thinking about this; you're on the right track.  The answer is yes and jmerrry is the closest.

Here are a couple hints:
In the expression for the shoelace problem there's a number:
$ \frac12(a_1b_2\minus{}a_2b_1)$.
What does this number do?  (The origin is somehow important.)

Now the 3D generalization of this number is
$ \frac16(a_1b_2c_3\minus{}a_1b_3c_2\plus{}a_2b_3c_1\minus{}a_2b_1c_3\plus{}a_3b_1c_2\minus{}a_3b_2c_1)$.
(ugly, huh?)

The second hint is that for every $ \textbf{edge}$ of the polygon we got a summand like above. . . .

The shoelace theorem is here, btw:
http://www.artofproblemsolving.com/Wiki/index.php/Shoelace_Theorem

[quote="AlTcApS"]is there such a thing like the shoelace theorem except for finding the area of polyhedra in the 3D plane?[/quote]
no it doesnt work

Read post before. There is a way, though it is quite ugly(in my perspective.)


$(3,2,4), \; (3,3,3), \; (1,3,5) \; \text{produce}\; 13=F_{7}$ instead of $F_{9}$. You sure you didn't mean $F_{a+b+c-2}$ or something like that?

It depends how you define the Fibonacci numbers.  Some people take $F_{0}= F_{1}= 1$; others take $F_{1}= F_{2}= 1$.  This proposition is true only in the former case; some sort of adjustment is necessary in the latter case.

:Note: edited in light of the fact that I said exactly the opposite of what was true.  (Thanks, Ubemaya!)

A hint for a much, much, much cleaner solution:

Consider a 1*n grid of squares, to be tiled with squares, which are 1*1, and dominoes, which are 1*2. How many ways are there to do this? Biject this to the Fibonacci numbers, then use a combinatorial argument to prove the assertion.

I'm pretty sure (a.k.a. [i]really[/i] sure) that this works in the $f_{0}= 1, f_{1}= 1$ scheme, not when $F_{1}= 1, F_{2}= 1$, because that's the way the dominos and squares tiling is defined.

Actually, I just checked $(3, 2, 4)$, and it DOES work.

For the proof (if you know about square and domino tilings):
[hide]
Consider an $a+b+c$ length board. Draw imaginary lines dividing it into lengths of $a$, $b$, and $c$, in that order (from left to right). There are 4 cases: No dominoes lie on the two imaginary lines, a domino lies on one or the other, or dominoes lie on both.

Case 1: No dominoes
There are $f_{a}, f_{b}, f_{c}$ ways to tile the board segment of length $a, b, c$ respectively. There are $f_{a}f_{b}f_{c}$ ways total.

Case 2: Domino on first line
The domino takes one square away from the board segment of length $a$ and $b$, so now we have $f_{a-1}f{b-1}f_{c}$ ways.

Case 3: Domino on second line
See Case 2. $f_{a}f_{b-1}f_{c-1}$

Case 4: Dominoes on both lines.
The first domino takes a square away from the board segments of length $a$ and $b$, and the second domino takes a square away from the board segments of length $b$ and $c$. $f_{a-1}f_{b-2}f_{c-1}$.

Total: There are $f_{a+b+c}$ ways to tile the whole board, so adding up the separate cases we get \[f_{a+b+c}= f_{a}f_{b}f_{c}+f_{a-1}f_{b-1}f_{c}+f_{a}f_{b-1}f_{c-1}+f_{a-1}f_{b-2}f_{c-1}\]
[/hide]

If you're unfamiliar with squares and dominoes, read this first.
[hide]
Take a board of length $n$ and unit width. Divide it into $n$ squares. Define $f_{n}$ to be the number of ways to tile this board with squares, of length $1$, and dominoes, of length $2$. Notice that the first tile has to be either a square or a domino, and we can break up $f_{n}$ into these two cases. If the first tile is a square, there is a board of length $n-1$ left, which by definition has $f_{n-1}$ tilings. If the first tile is a domino, there is a board of length $n-2$ left, which has $f_{n-2}$ tilings. So we have the fibonacci relationship: $f_{n}= f_{n-1}+f_{n-2}$. Since there is one way to tile a non-existent board and one way to tile a board of length $1$, we have $f_{0}= 1, f_{1}= 1$, and so $f_{n}$ is the $nth$ Fibonacci number.
[/hide]

[quote="Karth"][quote="diophantient"][quote="13375P34K43V312"]Show that $F_{a+b+c}=F_{a}F_{b}F_{c}+F_{a-1}F_{b-1}F_{c}+F_{a}F_{b-1}F_{c-1}+F_{a-1}F_{b-2}F_{c-1}$, where $F_{n}$ refers to the $n$-th Fibonacci number.[/quote]
[hide="well"]
Binet's Formula :D[/hide][/quote]

Are you serious  :maybe:[/quote]
Notice the :D icon :P
It's probably doable that way though, anyone want to give it a try :P


Prove that in an arithmetic progression of natural numbers can't have all term be primes, unless they are all equal.

In the sequence $ a, a\plus{}n, a\plus{}2n, ...$ consider the term $ a\plus{}a\cdot n\equal{}a(1\plus{}n)$.

Why of course...

Thanks for the help Vworldv.

That only works if $ a$ is not equal to $ 1$. If $ a\equal{}1$:

Consider the lowest prime $ p$ which is not a factor of $ n$. Now consider
$ 1,1\plus{}n,1\plus{}2n...,1\plus{}pn$. All of these have different remainders mod $ p$, hence one most be divisible by $ p$.

@malcolm:
the A.P. consists only of primes, so you need not worry about 1 being in it.


[hide]1-5/6*3/4=3/8=prob that a 3 shows up

prob that 3 shows up and 5 is the sum is 1/6*1/4+1/6*1/4=1/12

(1/12)/(3/8)=2/9,  I'm not sure thats right though[/hide]

[hide]I got that same answer, with the same method. The probability of a 3 is 3/8, and the probability of a sum of 5 (not considering the 3/8 yet) is 1/12. You divide the probabilities because the probability is 1/12 out of a probability of 3/8, so (1/12)/(3/8) = 2/9 is our answer. You can also list each possibility. There are 9 possibilities with a 3: (3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(1,3),(2,3), and (4,3) since we already counted (3,3). Out of these, only 2 of them, (3,2), and (2,3), sum to 5, so the answer is 2/9.[/hide]

[quote="236factorial"]Why did you divide the probabilities?[/quote]
You do this because it was given that at least one is a 3.  The probability that we are trying to find is that it sums to five and there is a 3.  But that probability is out of the total probability, which includes parts that contradict what was given, so we divide by the probability of getting at least one 3.

It kinda looks like this:

$\frac{\frac{\text{prob of sum=5 and one is 3}}{\text{total prob}}}{\frac{\text{prob of having at least one 3}}{\text{total prob}}}=\frac{\text{prob of sum=5 and one is 3}}{\text{prob of having at least one 3}}$  Since the part about one 3 was given it becomes the denominator or new total probability.

I hope that made sense

[quote="236factorial"]Zan tosses one fair 6-sided die with faces labeled 1 through 6 and one fair 4-sided die with faces labeled 1-4. If at least one die shows a "3", what is the probability that the sum is 5? Express your answer as a common fraction.[/quote]

  [hide="Alternate Solution"]

  We can reason in the following manner:

  Suppose the 6-sided die shows a 3.  Then the 4-sided die can be any of 4 different numbers but only 1 number (2) will add up to 5.  Now suppose the 4-sided die shows a 3.  Now the 6-sided die can show any of 5 numbers ( remember we counted when both were 3:   3,3  so we dont count it again)  but only 1 number will work (2).  Thus, the probability if at least one die shows a 3, there sum is 5. is:   $\frac{1+1}{4+5}=\frac{2}{9}$ [/hide]

[hide]2/9 because if the four sided one rolls three the six sided one has to roll two to make five for a total.Two ways to make five out of nine ways to roll three=2/9
 :idea:
 :) [/hide]

Thank you everyone. The solutions made it more clear to me (I wonder why I didn't think of that last night).


Would this be the best way to solve this problem?
Is there another way?

3x- 2yi = 2 + (x+y)i

1)

{3x = 2

{-2yi = (x+y)i

2)

{x = $\frac{2}{3}$

{-2yi = ($\frac{2}{3}$+y)i

3)

-2yi = $\frac{2}{3}$i + yi

4)

-3yi = $\frac{2}{3}$i

times 3

5)

-9yi = 2i

6)

divide by -9i

y = $\frac{-2}{9}$

x = $\frac{2}{3}$

I think the way you did is the best way. There might be another way though.


search
number theory unsolved
number theory

Show that there is an infinite sequence of positive integers $ a_1,a_2,a_3,...$ such that

(i) each positive integer occurs exactly once in the sequence, and
(ii) each positive integer occurs exactly once in the sequence
$ |a_1\minus{}a_2|,|a_2\minus{}a_3|,...$

The brute-force approach (start with 1, at each stage add the smallest positive integer that didn't already appear and whose difference from the last element hasn't appeared as a consecutive difference) yields:

[code]
1   2   4   7   3   8   14   5   12  20 ...
  1   2   3   4   5   6    9   7    8 ...
[/code]
(the top row is the sequence, the bottom row is the difference sequence). This sequence is on [url=http://www.research.att.com/~njas/sequences/?q=1%2C2%2C4%2C7%2C3%2C8%2C14%2C5&language=english&go=Search]OEIS[/url]where it's claimed that it's still unknown whether the difference sequence really ends up including all integers.

Anyway, [url=http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.pjm/1102811644]this paper[/url] presents a construction (see Theorem 2). There may be an easier construction, I don't know.


given the sequence with
  $a_{0}=0,a_{1}=1,a_{n+2}=2a_{n+1}+a_{n}$
 prove that $2^{k}|a_{n} \Leftrightarrow 2^{k}|n$

It give $a_n=\frac{1}{2\sqrt 2}[(1+\sqrt 2)^n-(1-\sqrt 2)^n]=\sum_{m=0}^{[(n-1)/2]} C_n^{2m+1}2^m$
Therefore if n is odd then $a_n$ is odd, if $2^k|n$ then $2^k|a_n$, exactly $ord_2(a_n)=ord_2(n)$. If $2^k|a_n$ then $2^k|n$.


Solve the equation $2z^{3}-(5+6i)z^{2}+9iz+1-3i = 0$, knowing that one of the solutions is real.

[hide]
If one root is real, we can separate the real and imaginary parts and set them equal to zero. So

$-6z^{2}+9z-3 = 0 \Rightarrow-3(2z-1)(z-1) = 0$

and

$2z^{3}-5z^{2}+1 = 0 \Rightarrow (2z-1)(z^{2}-2z-1) = 0$.

So clearly the real root is $z = \frac{1}{2}$. Then we factor the original equation into

$(2z-1)(z^{2}-(2+3i)z+(3i-1))$,

which factors (by the quadratic formula or otherwise) now into

$(2z-1)(z-(1+2i))(z-(1+i))$

so the three roots are $z = \frac{1}{2}, 1+2i, 1+i$.[/hide]


search
modular arithmetic
number theory
relatively prime

Let $ m$ and $ n$ be positive integers with $ 1 \le m < n$. In their decimal representations, the last three digits of $ 1978^m$ are equal, respectively to the last three digits of $ 1978^n$. Find $ m$ and $ n$ such that $ m \plus{} n$ has its least value.

Many of thee post-test problems are on the forum -- try a search: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=24809&search_id=1520143648

The other thread didn't have the correct answer, so...
[hide]
Basically, we want the smallest value of $ m\plus{}n$ such that $ 1978^n \equiv 1978^m \pmod{1000}$. This reduces to $ 978^n \equiv 978^m \pmod{1000}$. Divide both sides by $ 8$ (this requires $ n > m \ge 3$) to obtain $ 2^{n\minus{}3}489^n \equiv 2^{m\minus{}3}489^m \pmod{125}$. Since $ 2$ and $ 489$ are relatively prime to $ 125$, we can divide both sides by $ 2^{m\minus{}3}489^m$, making the equation $ 2^{n\minus{}m}489^{n\minus{}m} \equiv 978^{n\minus{}m} \equiv 103^{n\minus{}m} \equiv 1 \pmod{125}$. Since we want $ m\plus{}n$ to be minimal, it must equal $ \phi{(125)} \equal{} 100$. We also have the additional restriction that $ n > m \ge 3$, so the minimal value is obtained when $ m\equal{}3$ and $ n\equal{}103$. Thus, the answer is $ 106$.

I assumed that the smallest value $ n\minus{}m$ such that $ 103^{n\minus{}m} \equiv 1 \pmod{125}$ was $ 100$, since it produced the right answer, but how do I know if there's a smaller value $ n\minus{}m$?
[/hide]


Hey everybody.

I'm currently studying in the United States however, I'm only a Canadian citizen with an F-1 Visa.  Can I still participate in the USAMTS?

US Residents who are not US Citizens are allowed to participate - make sure you use your US residential address when registering.

Hello Mr. Rusczyk,

What is exactly a U.S. Resident?  I don't actually live in the U.S.; I just go to boarding school here.  The rest of my family is in Canada.

Could I still participate?  A No answer would be fine however; I just wanted to make sure I'm not missing out on anything Mathematics.

Going to school in the US is close enough for us :)  Make sure you use your school address.


Given the angle XOY, variable points M and N are considered on the arms OX and OY, respectively, so that OM+ON is constant. Determine the geometric locus of the mid-point of MN.

This is *very* hard to explain, so bear with me or just read the last 4 sentences and look at the picture :)

WLOG, let O be the origin of a coordinate plane and let OX be going along the x-axis.  Let angle XOY = $\theta$ and the fixed length OM + ON = a.  Let M be t units away from O and N be a-t units away from O.  By using the usual trig ratios, we can easily find that the x-coordinate of the midpoint is $\frac{(a-t)\cos \theta - x}{2} + x$ and the y-coordinate is $\frac{(a-t) \sin \theta}{2}$.  Both are linear since sin theta, cos theta, and a are constant.  Let R be the point on OX that is a units away from O and S be the point on OY that is a units away from O.  Letting M=R and N=O, OM + ON = a, and the midpoint of MN is the midpoint on OR.  Letting M=S and N=O, OM + ON = a, and the midpoint of MN is the midpoint of OS.  Since our locus is a straight line (already proven a few sentences before), and two points determine a line, our locus is the line connecting the midpoint of OR and the midpoint of OS.


[hide="possible answer.."]
consider $\log_{5}{x}= 0$. x must equal $1$ since $5^{0}= 1$

This means that $\log_{2}({\log_{3}{x}})= 1$

$\log_{2}{x}= 1 \Rightarrow x = 2$ since $2^{1}= 2$

Thus

$\log_{3}{x}= 2 \Rightarrow x = 9$ since $3^{2}= 9$

So our answer for x is $\framebox{9}$
[/hide]

The people who make these tests are evil. Evil.

[quote="Ignite168"]Someone didn't read the problem.  :whistling: I think the problem writers intentionally put it there to catch people who do that  :wink:[/quote]

lol darnit! I even have the real answer on my sheet of paper it was just when I was writing it out on here that I forgot.. i get so caught up in the latex I forgot the real answer >_<

[hide="answer"]

if $x = 9$ then $x^{-1}= \frac{1}{9}$
[/hide]


modular arithmetic
algebra unsolved
algebra

Let $ a_{1}\equal{} 19, a_{2}\equal{} 98$. For $ n \ge 1$, define $ a_{n\plus{}2}$ to be the remainder of $ a_{n}\plus{}a_{n\plus{}1}$ when it is divided by $ 100$. What is the remainder when

$ a_{1}^{2} \plus{} a_{2}^{2} \plus{} ... \plus{} a_{1998}^{2}$

is divided by $ 8$?

Yes, James, it's here: http://www.bmoc.maths.org/home/bmo1-1998.pdf

I hope this works as I'm having problems with my computer.

How did you get $ 0$?

I got a similar solution with Zuming Feng...Here it goes

Consider $ a_n \pmod{4}$ which is not changed by taking the remainder divided by $ 100$, there's the cycle $ 3,2,1,3,0,3$ which repeats $ 333$ times. Then  $ a_{1}^{2}\plus{}a_{2}^{2}\plus{}...\plus{}a_{1998}^{2} \equiv333(1\plus{}4\plus{}1\plus{}1\plus{}0\plus{}1) \equiv0 \pmod{8}$, as claimed.

you're welcome Andrew...BTW, zuming feng is an american-based chinese coach for usa team..

Thanks James. Where is Zuming Feng's solution?


i noticed that a general test is now an option...how will this affect how indivdual scores count toward the total score? will it still be the top four scores on each test?

also, it seems that each question will be worth one point this year. does anyone have the points awarded for each question in previous years?

In previous years it was 3, 3, 3, 4, 4, 4, 4, 5, 5, 5 for a total of 40.

ok, that still leaves the question of how individual scores will count towards a team's total scores.

also, will the general test be weighted the same as the other tests, or will it be scaled down (like at hmmt)?


Does anyone know how to find the inverse laplace transform of $ \frac{(s\plus{}1)^2}{(s\plus{}2)^4}$? I can't manipulate the expression to yield a suitable from of the laplace.

Write it as a linear combination of terms $ \frac{1}{(s\plus{}2)^k}$- you should know how to handle each of those terms.

jmerry is talking about partial fraction decomposition.  Equivalently, expand everything in terms of $ r \equal{} s \plus{} 2$ and divide.

Here's a formula that could help you:

IF:

$ \prod^{n}_{j\equal{}1}\frac{1}{(x\minus{}a_{j})^{p_j}}\equal{}\sum^{n}_{i\equal{}1}\sum^{p_{i}}_{k\equal{}1}\frac{c_{i,k}}{(x\minus{}a_i)^k}$

THEN:

$ c_{i,k}\equal{}\frac{1}{(p_i\minus{}k)!}\lim_{x\rightarrow a_i}\frac{d^{p_i\minus{}k}}{dx}\prod^{n}_{j\equal{}1}\frac{(x\minus{}a_i)^{p_i}}{(x\minus{}a_j)^{p_j}}$

I do realize that it looks awful, but this is a strictly-general-all-cases-in-one formula. It gets quite simple when you start getting it. Ad quite useful!


algebra
polynomial
function
topology
Functional Analysis
advanced fields
advanced fields unsolved

Given a transformation $ T: V\to V$ such that
$ T\left(f\left(x\right)\right)=xf^{\prime}\left(x\right)$,
where $ V=C^{\infty}\left(\mathbb{R},\mathbb{R}\right)=\left\{f: \mathbb{R}\to\mathbb{R}\mid f\text{ is differentiable infinitely many times}\right\}$.

Find $ \text{Spec}\left(T\right)$.

Hereby, $ \text{Spec}\left(T\right)=\left\{m\in K\mid P\left(m\right)=0\right\}$, where $ K=\left(K,+,\cdot\right)$ and where $ P\left(m\right)$ is the characteristic polynomial of the given trasformation.

I'm afraid you are thoroughly confused about what the spectrum is in the case when you have a linear mapping in an infinite-dimensional space :roll:. Since $C^\infty(\mathbb R)$ is not even a Banach space, the thing you are looking for is not something covered by standard definitions from functional analysis. Lacking a better idea, we can try to restate the problem in the following way. Consider the space of complex-valued infinitely smooth functions defined on the entire real line with the standard topology given by the seminorms $p_{a,n}(f)=\max_{[-a,a]}|f^{(n)}|$. For which $\lambda$ is does the operator $f\mapsto xf'-\lambda f$ have a bounded inverse in this space? It is clear that it is sometimes true (say, for $\lambda=-1$) and sometimes false (say, for $\lambda=0$), so, at least, the question is now meaningful but, I'm afraid, in this form it is not very interesting either: the answer to this seems to be "For negative integer $\lambda$ and only for them", so, in this sense $\text{spec\,}(T)=\mathbb C\setminus \mathbb Z_{-}$. Another reasonable interpretation would be to consider $T$ as an unbounded operator in the Hilbert space $L^{2}(\mathbb R)$. Then it becomes a good exercise in functional analysis and the topic goes straight to Advanced Fields. :lol:

It's not from so much time that I am studying linear algebra.. with infinit dimensions... thats why i asked help.. thnx for your answer... [b]even I did understood it at well how you found the spec(T)...[/b] . thnx for your answer, if you can give me some other information about that I will be gratefull..

At the begining the question that I had was just on polynomials.
Is given a trasformation T:R[t]-R[t]  such that :
T(P(t)) = t Pf'(t)
Find the spec(T) = ?

And the answer is ' spec(T) = Z non negative', I could resolve it proceding also with standard functional analysis.


topology
real analysis
real analysis unsolved

Hi, I'm stuck on a problem in Topology and I appreciate all the help I can get.
Let $\left\{ U_{\alpha}\right\}_{\alpha \in A}$ be a finite open cover of a compact metric space $X$.
Then I have already shown that there exists $\epsilon > 0$ such that for each $x \in X$, the open ball $B(x;\epsilon)$ is contained in one of the $U_{\alpha}$'s.

Now they call such a $\epsilon$ a Lebesgue number of the cover. In the next part of the problem they ask you to show
that if [u]at least one[/u] of the $U_{\alpha}$'s is a proper subset of $X$, then there is a largest Lebesgue number for the cover.

But if for some $\alpha \in A$, $U_{\alpha}= X$, then because $X$ is bounded you get $B(x;\epsilon) = X$ for sufficiently large $\epsilon$ so that the Lebesgue numbers are not bounded.
So they must mean that [u]all[/u] $U_{\alpha}$'s are proper subsets of $X$, right?

In that case I managed to prove that the Lebesgue numbers are bounded but I'm stuck on proving that the supremum of the Lebesgue numbers is again a Lebesgue number.

Hint : Let $\epsilon$ be the supremum of the Lebesgue numbers for the covering. Fix a point $x$. The open ball $B(x;\epsilon)$ is union of smaller open balls $B(x;\epsilon-1/n)$, $n=1,2,3,\ldots$. And each $B(x;\epsilon-1/n)$ is contained in some $U_{n}$.


since you're restricting the other major to "soft" sciences, then maybe economics

Yea, I'm thinking economics or business; however, I was hoping for some explanation as to why that would be the case.

Both of them give you a good basis for starting your own business, which is always a useful skill; and even if you don't end up starting your own business, understanding how they work is invaluable when working in or with businesses.

Why soft sciences? Tacking on a physics or math degree to that would make an uber resume.

[quote="nigma"]Which soft science degrees would, when paired with a B.S. in biochemistry, offer a large variety of career paths?[/quote]

not exactly science, but i would major in political science, and go on the law school, just because intellectual property in molecular biology is so messed up. IP lawyers who knows biochem will definitely be in demand.

[I'm replying to hcheng501 here, not nigma.]

Political science may be a popular major among pre-law students, but there's no reason you'd have to major in that. Law schools take students from all majors - what they want to see is evidence that you can reason and evidence that you can write. Any major, minor, or combination of courses that makes you develop your writing skills would be good. And (provided they see that evidence of verbal and writing skills), law school admissions committees tend to look favorably on science majors - they just don't see all that many of them.


geometry
3D geometry
tetrahedron
circumcircle
geometry proposed

Let a tetrahedron $ ABCD$ and $ A',B',C',D'$ be the circumcenters of triangles $ BCD,CDA,DAB,ABC$ respectively. Denote planes $ (P_A),(P_B),(P_C),(P_D)$ be the planes which pass through $ A,B,C,D$ and perpendicular to $ C'D',D'A',A'B',B'C'$ respectively. Prove that these planes have a common point called $ I.$ If $ P$ is the center of the circumsphere of the tetrahedron, must this tetrahedron be regular?

$
\mbox{The point }I (\{I\}=(P_A)\cap(P_B)\cap(P_C)\cap(P_D))\mbox{-is the isogonal of point }P\\
 \mbox{with to the tetrahedron }ABCD.$
[[b] View:[/b]
[i]Nathan Altshiller-Court[/i]: MODERN PURE SOLID GEOMETRY(secod edition); theorem 752, p.282.]
$
\mbox{If } I=P \mbox{, then tetrahedron } ABCD \mbox{ is isosceles, ie: }|AB|=|CD|, |AC|=|BD|\\
\mbox{and }|AD|=|BC|.
$


Let be $ a,b,c\in \mathbb{R}^{*}$ such that $ (a\plus{}b\plus{}c)\left(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\right)\equal{}1$ . Prove that :

$ (a\plus{}b)(b\plus{}c)(c\plus{}a)\equal{}0$

[hide="Solution"]Expand and multiply by $ abc$ to get

$ a^2c\plus{}b^2c\plus{}a^2b\plus{}bc^2\plus{}ab^2\plus{}ac^2\plus{}2abc\equal{}0$

$ c(a\plus{}b)^2\plus{}c^2(a\plus{}b)\plus{}ab(a\plus{}b)\equal{}0$

$ (a\plus{}b)(ac\plus{}bc\plus{}c^2\plus{}ab)\equal{}0$

$ (a\plus{}b)(a\plus{}c)(b\plus{}c)\equal{}0$[/hide]


articles
quadratics
integration
Pythagorean Theorem
geometry

I have no idea what board this would go in, because it may be well known or something of the sort.  A friend and I discussed this today, and I gave up after I couldn't solve it in 10 minutes  :).  Given a finite set of points in the plane, how would one find the point (probably not in the set) for which the sum of the distances from each of the other points is minimized?  I know that for a triangle it is the Fermat point.  Anyone know how to find beyond that?  And if you're up for a real challenge, try to generalize to either 3, 4, or n dimensions.

If any moderator wants to move this, you more than have my blessing.

A while ago, I saw an article about this on [url]http://mathworld.wolfram.com[/url], though I can't remember the exact name of the piece. I recall thinking that the triangle and its Fermat point were the only palatable linear distance-sum case. That said, the case for which an elementary approach is sufficient is the quadratic case; i.e., for some points $A_1, \dots, A_k$ and arbitrary point $X$, trying to minimize

$\sum_{i=1}^k A_iX^2$

or even the more general case for any finite union $\Gamma$ of continuous sets of points ,

$\int_{Y \in \Gamma} XY^2 dY$

[hide]Both have solutions $X$ given by the $n$-space center of mass. This can be shown by employing the Pythagorean Theorem in conjunction with the fact that $ax^2 + bx + c$ is minimized at $x = \frac{-b}{2a}$ if $a \ge 0$.[/hide]


function
ratio
LaTeX
limit
geometry
geometric transformation
calculus

I am to determine whether the given series diverges or converges

$\sum_{n=1}^{\infty} \frac{n!}{n^n}$

Just by looking at it (and using what Kent Merryfield suggested: intuition), I feel that it should converge, since n^n is a function that approaches infinity much faster than n factorial.

To prove this, the only tests I can think to use are the ratio test or the limit comparison test.

The ratio test gets really messy because of n^n and by intuition, I do not know off the top of my head what to compare it with as n approaches infinite like the comparison test says to...

Any type of help is appreciated, thanks

[hide]
using the ratio test (D'Alembert) we get

a_n+1 / a_n = 1/(1+1/n)^^n -->e^^-1 so it is convergent[/hide]

I`m just going to $Latex$ what Delta said: if $\lim_{n\to\infty} \sup \frac{a_{n+1}}{a_n} <1$ the the series $\sum_{n=1}^{\infty} a_n$ converges

[hide] on second thought n!/n^^n <1.2/n.n (n>3) and we are done![/hide]

An actual translation of delta's terrible notation:

$\lim\frac{a_{n+1}}{a_n}=\lim\frac1{(1+\frac1n)^n}=\frac1e$, so the series converges.

Second post:

On second thought, $\frac{n!}{n^n}\le \frac{1\cdot 2}{n\cdot n}=\frac2{n^2}$ for $n\ge 2$ and we are done by comparison to the convergent series $\sum\frac2{n^2}$.

For ankur87's intuition:

Both $n!$ and $n^n$ tend to infinity very rapidly, where by "very rapidly" I mean faster than any geometric sequence. You sense that $n^n$ is somehow "much" bigger - and that's on a scale in which things are already fast compared to the geometric. The issue is putting something concrete behind that sense.

There are five comparison tests commonly presented in a mainstream calculus book like Stewart: simple (one-sided) comparison, limit comparison, the integral test, the root test and the ratio test. (The final line in jmerry's post is a one-sided comparison.)

All of these comparison tests deal only with the size of the terms and are thus tests for absolute convergence. If you hang around these forums long enough, you'll see a sixth test: the condensation test. That's also a comparison test, and its range of applicability is similar to the integral test.

Limit comparison is the "de-clutterer," the test for stripping things down to their essence. That's not really our problem here - we're already at the essence, we just need to figure out what that essence is.

Notice that I listed the root test and the ratio test as comparison tests. Comparison to what? The answer is that they compare a given series to a geometric series. Once you realize this, you know when you can apply them: in cases where something in the term grows or shrinks geometrically fast or faster.

If we have $\sum\frac{\ln n}{n^2},$ then nothing in those terms grows geometrically fast. This cannot be usefully compared to a geometric series. Any attempt to use either the root test or the ratio test would be a pure waste of time, as those tests are doomed to be inconclusive in this case.

But $\sum{\frac{n!}{n^n}?}$ With both numerator and denominator growing at a faster than geometric rate, and with us seeking to balance them, we should definitely turn to either the root or the ratio test.

The choice between the root and the ratio test at this level is a tactical one - which one would lead to a more tractable computation? (Theortically, the root test - actually, the careful limsup version of the root test - has a slight advantage over the ratio test. That makes no difference here.)

At that tactical level the factorial strongly suggests using the ratio test. Factorials are well-suited to taking ratios. But the $n^n$ is awkward - more suited to the root test. The tactical problem with using the root test is that you would have to deal with $\sqrt[n]{n!}$ - a snap if you already know Stirling's asymptotic approximation for the factorial but not easy if you don't know that. So we slog through the algebra that results from the ratio test and eventually get the limit in jmerry's post.

Closely related question: what is the radius of convergence of the power series

$\sum_{n=1}^{\infty}\frac{n!x^n}{n^n}?$

(I just asked for the radius of convergence. The interval of convergence - that is, the question of convergence at the endpoints - does require knowledge of Stirling's approximation.)

[quote="ankur87"]I don't understand what you showed me, delta.

And for spix post, that is the ratio test defined, so it did not really help me much since I didn't get what delta was getting at...[/quote]

Suppose $\lim_{n\to\infty} \sup \frac{a_{n+1}}{a_n} <1$ then there exists $q<1$ such that $\frac{a_{n+1}}{a_n}<q,\forall n\geq N_0$ which gives us that $\sum_{n=1}^{\infty} a_n \leq a_1+a_2+...+a_{N_0-1}+ a_{N_0}(1+q+q^2+q^3+....)$ which solves the problem.

Kent, thank you so much for the time that you put into that post to help me understand. I really appreciate you taking the time to break things down for me, it helps me more than anything.

Those are the five tests we have learned in the book. We have also gone over alternating series test, I don't know if that falls under a category you've already listed.

I am just getting into power series now, and I do not know Stirling's approximation (but I did look at it at mathworld), so it might be a bit before I can answer that question....

Thanks again for the help.

_Ankur

[quote]We have also gone over alternating series test, I don't know if that falls under a category you've already listed. [/quote]
The alternating series test is different. It is the one test in a book like Stewart that [i]isn't[/i] a comparison test. In order for the test to be valid, the terms of the series have to go to zero, but the terms can go to zero at an arbitrarily slow rate, and if the other conditions of the alternating series test are met, then the series converges.

Since it's not a comparison, the alternating series test is the only one of the tests you've learned that is capable of proving convergence in the absence of absolute convergence. That means that the alternating series test can be used to prove that a series is conditionally convergent.

If you prove convergence by any comparison - comparison, limit comparison, integral, condensation, root, ratio - then you have proved absolute convergence. To prove conditional convergenc, you would first have to deny absolute convergence (by some comparison) and then prove convergence, presumably by the alternating series test or something like it.

There is room for "something like it," room to improve the alternating series test, room to find a generalization. Wouldn't you like to know whether

$\sum_{n=1}^{\infty}\frac{\sin n}n$ converges?

That one lies just outside the reach of all six of the tests you name, but can be approached by some theorems that generalize the alternating series test.


vector
quadratics
trigonometry
calculus
calculus computations

Hello,

Before I ask my question, I'd just like to say that this is from a current homework assignment in my Calc III class.  I don't expect any kind of exact answer, or really anything other than a basic tip to point me in the right direction.

The question:

Find all planes that pass through the points A(1,1,2) and B(2,0,1) and are a distance of (1/2) from the point (1,2,3).

I've been doing pretty well throughout this class, but for some reason can't seem to grasp how to begin this question.  I know the distance formula for a point to a plane, and how to solve it when I'm given the equation of the plane and the point.  Since here I'm only given two points, I'm really thrown off.

I know that I would sub in the (1/2) for distance, and the next logical step to me seems to be to sub in given points and solve the system of equations.  I'm only given two points though, and still have three unknowns.  Or maybe that's not even how to go about doing it.

This is a pretty basic question considering what we've been doing in class and with the homework/test problems, but I'm just having a complete brain fart (and know I'm just missing the obvious here) when it comes to this question.

Thanks all!

Two points determine a line, hence a family of planes.  Parameterize this family, and then use point-to-plane to figure out what member of the family you want.

This is a tricky problem, and you should not feel ashamed for not getting it. Suppose that $ P$ is a plane at distance $ 1/2$ from the point $ C(1,2,3)$. Let $ D$ be the point of $ P$ that is closest to $ C$. What do you know about the vector $ \overrightarrow{CD}$?

Where I'm really having trouble is viewing the family of possible planes.  I can visualize in my head exactly what's going on, but putting it on paper (in a coherant manner) has me really thrown.  We havn't really done anything except for deriving the equation of a plane through the vector equation (with a point and a vector normal to the plane).

[quote="mlok"]What do you know about the vector $ \overrightarrow{CD}$?[/quote]

Immediately, what comes to mind is that [b]CD[/b] has a magnitude of 1/2 and is normal to the plane who's equation I am supposed to be finding.

I hate brain farts.

[quote="Chunk3rs"]Where I'm really having trouble is viewing the family of possible planes.  I can visualize in my head exactly what's going on, but putting it on paper (in a coherant manner) has me really thrown.  We havn't really done anything except for deriving the equation of a plane through the vector equation (with a point and a vector normal to the plane).
[quote="mlok"]What do you know about the vector $ \overrightarrow{CD}$?[/quote][/quote]

The point $ D$ on the line $ AB: \ \frac {x \minus{} 1}{1} \equal{} \frac {y \minus{} 1}{ \minus{} 1} \equal{} \frac {z \minus{} 2}{ \minus{} 1} \equal{} t$ can be written as $ D(t \plus{} 1,\ \minus{} t \plus{} 1,\ \minus{} t \plus{} 2)$ for some $ t\in\mathbb{R}$.
$ \overrightarrow{CD}\bot \overrightarrow{AB}\Longleftrightarrow (t, \minus{} t \minus{} 1, \minus{} t \minus{} 1)\cdot(1, \minus{} 1, \minus{} 1) \equal{} 0\Longleftrightarrow t \equal{} \minus{} \frac {2}{3}.$

$ \therefore D\left(\frac {1}{3},\ \frac {5}{3},\ \frac {8}{3}\right)\Longrightarrow \overrightarrow{CD} \equal{} \left( \minus{} \frac {2}{3}, \minus{} \frac {1}{3}, \minus{} \frac {1}{3}\right)\parallel{}(2,\ 1,\ 1),$ which is a normal vector of the desired plane. Thus we can describe the plane as $ 2x \plus{} y \plus{} z \equal{} d$.

Then $ |\overrightarrow {CD}| \equal{} \frac {1}{2}\Longleftrightarrow \frac {|2\cdot 1 \plus{} 2 \plus{} 3 \minus{} d|}{\sqrt {2^2 \plus{} 1^2 \plus{} 1^2}} \equal{} \frac {1}{2}\Longleftrightarrow d \equal{} 7\pm \frac {\sqrt {6}}{2}.$

Ans. $ 2x \plus{} y \plus{} z \equal{} 7\pm \frac {\sqrt {6}}{2}.$

[quote="Chunk3rs"]Immediately, what comes to mind is that [b]CD[/b] has a magnitude of 1/2 and is normal to the plane who's equation I am supposed to be finding.[/quote]
Which gives three equations: $ |\overrightarrow{CD}|\equal{}1/2$, $ \overrightarrow{CD}\cdot \overrightarrow{AD}\equal{}0$, and $ \overrightarrow{CD}\cdot \overrightarrow{AD}\equal{}0$. Unfortunately, the equations are quadratic, and as far as I remember, there is no general method for solving a system of three quadratic equations in three unknowns.

Let's try to modify this approach: we know that $ D$ lies on the plane $ P_0$ that passes through $ C$ and is perpendicular to $ AB$. Let $ E$ be the intersection of $ P_0$ with the line $ AB$. Now $ CDE$ is a right triangle in the plane $ P_0$ with known hypotenuse $ CE$ and known length of $ CD$... this should be doable.

-----
The issue I have with [b]kunny[/b]'s proposed solution is that $ D$ need not lie on the line $ AB$.

[quote="Chunk3rs"]
Find all planes that pass through the points A(1,1,2) and B(2,0,1) and are a distance of (1/2) from the point (1,2,3).
[/quote]
that's an easy question.

the generic equation of a plane through $ (1,1,2)$ is given by
$ a(x \minus{} 1) \plus{} b(y \minus{} 1) \plus{} c(z \minus{} 2) \equal{} 0$,
where $ (a,b,c)$ are numbers to be determined (you should know they are proportional to the [i]direction cosines[/i] of the normal to the plane)

the plane also goes through $ (2,0,1)$, which means
$ a(2 \minus{} 1) \plus{} b(0 \minus{} 1) \plus{} c(1 \minus{} 2) \equal{} 0$
$ \Rightarrow\;\;\;a \minus{} b \minus{} c \equal{} 0$
$ \Rightarrow\;\;\;a \equal{} b \plus{} c\;\;\;\quad\;\;\;\quad(1)$

at this point, we need just $ 1$ more equation involving $ (a,b,c)$ (and [u]not[/u] $ 2$, mind you) to get our plane, and we shall get that using the information about the distance of the plane from the [b]3rd[/b] point which is $ (1,2,3)$.

using that information, we have,
$ \bigg|\;\dfrac{a(1 \minus{} 1) \plus{} b(2 \minus{} 1) \plus{} c(3 \minus{} 2)}{\sqrt {\;a^2 \plus{} b^2 \plus{} c^2\;}}\;\bigg| \equal{} \frac {1}{2}$

$ \Rightarrow\;\;\;4(b \plus{} c)^2 \equal{} (a^2 \plus{} b^2 \plus{} c^2)$

using the value of $ a$ from $ (1)$, we get
$ 4(b \plus{} c)^2 \equal{} \left[\;(b \plus{} c)^2 \plus{} b^2 \plus{} c^2\;\right]$
$ \Rightarrow\;\;b^2 \plus{} 3bc \plus{} c^2 \equal{} 0$

solving this quadratic in $ b$ will give
$ b \equal{} \frac { \minus{} 3\pm\sqrt 5}{2}\;c$

you can pick [i]either [/i]value of $ b$ as your answer here. let us pick $ b \equal{} \frac { \minus{} 3 \plus{} \sqrt 5}{2}\;c$, which then gives us $ a \equal{} \frac { \minus{} 1 \plus{} \sqrt 5}{2}\;c$

and so, the equation of the plane is,
$ \left(\frac { \minus{} 1 \plus{} \sqrt 5}{2}\right)\;c\;(x \minus{} 1) \plus{} \left(\frac { \minus{} 3 \plus{} \sqrt 5}{2}\right)\;c\;(y \minus{} 1) \plus{} c\;(z \minus{} 2) \equal{} 0$

[i]eliminating [/i]$ c$, we finally get
$ \boxed{\left(\frac { \minus{} 1 \plus{} \sqrt 5}{2}\right)(x \minus{} 1) \plus{} \left(\frac { \minus{} 3 \plus{} \sqrt 5}{2}\right)(y \minus{} 1) \plus{} (z \minus{} 2) \equal{} 0}$

[ PS: you can always check your answer by substituting back the information given in the question and seeing if they match the equation of the plane! ]

[quote="Chunk3rs"]Find all planes ...[/quote]
[quote="misan"]you can pick [i]either[/i] value of $ b$ as your answer here.[/quote]

The two values of $ b$ give you two different planes, both of which work.


It seems as when I pass a topic one day, I log on the next day and my percentage of mastery has gone down.  How come?

The same thing happens to me. It gets really annoying.

I had the same problem. The best way to fix it is always let alcumus choose your topic, if you decide to continue in your current topic sometimes you lose your progress. Untill this gets fixed just avoid using the continue in current topic button.

Our algorithm designer is looking into this.

We have attempted to fix this issue but we're not completely sure if that's the end of it. Please keep us posted on any irregularities.
 :)

same with me
it said i mastered it and i didn't dfo anything
then after 3minutes it said i only passed it and it was weird


$n$ people are standing in a circle.  Each one is labeled from $1$ to $n$.  On the first round, every other person is killed.  Without anybody changing position, every other person is killed again, and so on until there is one survivor.  How can you compute the label of the survivor?

Solve the problem and post solution in [hide="hidden"][/hide] text, then view my solution.

[hide="Solution"]
1) Translate $n$ into binary.
2) Shift the sequence of 0's and 1's to the left once, putting the leftmost number on the right.
3) Convert back to base 10
example: 22 :arrow: 10110 :arrow: 01101 :arrow: 1101 :arrow: 13
[/hide]

[hide="Slight misunderstanding"]
When you say multiply by two and add one, are you referring to the binary number or the base ten number?  Could you use my example of 22 :arrow: 13?
[/hide]

Operations like multiplication and addition have the same effect regardless of what base he's working in.  ;)

[quote="jrshoch"]These kind of problems are hard..... the solution I found has no explanation and is sort of messy  :(

[hide]Write $n$ as a binary number.  Then remove the first digit.  Multiply the remaining number by 2 and add 1.  That's the label of the survivor.[/hide]

Edit:  Our solution algorithms are different... but always arrive at the same number!  For anyone who already solved the problem or who has read both solutions, want to try to prove that they always produce the same number?  Remember to keep everything hidden![/quote]

Proving that they always arrive at the same number is fairly simple.

[hide]
The leftmost digit in a binary number is [u]always[/u] a 1, right? So with miyomiyo's algorithm, you essentially shift that 1 to the right, leaving everything else in the same order.

With jrshoch's method, you remove the leftmost 1, then add a 0 to the right, then change that 0 into a 1.

It's clearly the same thing.
[/hide]


geometry
abstract algebra
inequalities
advanced fields
advanced fields unsolved

I stuck at a proof in Jost book "Riemannian geometry and geometric analysis",

in the proof,

Corollary 2.2.1
There exists a constant $ c$, depending only on the Riemannian metric of $ M$, with the property that for all closed forms $ \beta$ that are
orthogonal to the kernel of  $ \delta$ ( codifferential ), we have $ (\beta, \beta) \leq c(\delta \beta , \delta \beta)$ .

I have no problem with this Corollary. But then the book said:

For $ \phi \in \Omega ^p (M)$ (the space of p-forms), let $ \pi (\phi)$ be the orthogonal projection onto the kernel of $ \delta$,
and let $ \psi \equal{} \phi \minus{} \pi (\phi)$ ; so in particular $ \delta \psi \equal{} \delta \phi$.
Then since  $ \psi$ is orthogonal to the kernel of  $ \delta$ , by Corollary 2.2.1
$ \| \psi \| _{L2} \leq c \| \delta \psi \| _{L2} \equal{} c \| \delta \phi \| _{L2}$...

I can't see why Corollary 2.2.1 can by applied here: how do we know $ \psi$ is closed, that is, do we have $ d \psi \equal{}0$ ?

If I'm understanding you correctly, we can apply the corollary because they're co-differential.  We're not saying that psi is closed, we're just saying that it's orthogonal to the kernel of delta (by construction, since we've performed an orthogonal projection).

Also, I don't believe there's any way psi can be closed, because of the way it's set up.  It's either zero or orthogonal to the kernel of delta.

kalazealot, I believe you are correct, and that there is a missing step in Jost's argument.  The missing step is that $ \varphi$ can be orthogonally decomposed into a closed form and a form orthogonal to all closed forms.  For the closed piece, you get the desired inequality between 2.2.15 and 2.2.16, meanwhile the other piece makes no contribution to the inequality because it is orthogonal to $ \eta$.


Given $ f(x)\equal{}x^{\frac{1}{3}}$ and $ g(x)\equal{}3\minus{}x$, find $ (f^{\minus{}1} \circ  g^{\minus{}1})(\minus{}1)$.

[hide="My Solution"]
We are looking for $ f^{\minus{}1}(g^{\minus{}1}(\minus{}1))$. Let's start with $ g^{\minus{}1}(\minus{}1)$. We know that its value is

$ 3 \minus{} x \equal{} \minus{}1 \implies x \equal{} 4$.

So we are now looking for $ f^{\minus{}1}(4)$. Doing the same thing as before, we have

$ x^{\frac{1}{3}} \equal{} 4 \implies x \equal{} \boxed{64}$
[/hide]


calculus
derivative
function
absolute value
calculus computations

Decide whether each conjecture is true or false. Provide an argument for those that are true and a counterexample for each one that is false.
( a ) If a derivative function is not constant, then the derivative must take on some irrational valus.
( b ) If $ f^{'}$ exists on an open interval, and there is some point c where $ f^{'}(c)>0$, then there exists a e-neighborhood $ V_{e}(c)$ in which $ f^{'}(x)>0$ for all $ x \in V_{e}(c)$.

As a counterexample to number one, how about the derivative of the absolute value function, given by $ f'(x) =  \frac{x}{|x|} $? Or is that not permissible because it's not continuous, or because it could be called a function with different parts (The term escapes me, when a function is given by different statements depending on the input value)

The problem with that example is that $ |x|$ isn't differentiable everywhere.

We have the intermediate value property for derivatives: if $ f'$ is defined everywhere on $ [a,b]$, the set $ \{f'(x): x\in [a,b]\}$ includes all of $ [f'(a),f'(b)]$. If this is ever nontrivial, there are irrational values.


modular arithmetic
number theory proposed
number theory

a) Prove that there exists a sequence of digits $\{c_n\}_{n\geq 1}$ such that or each $n\geq 1$ no matter how we interlace $k_n$ digits, $1\leq k_n\leq 9$, between $c_n$ and $c_{n+1}$, the infinite sequence thus obtained does not represent the fractional part of a rational number.

b) Prove that for $1\leq k_n\leq 10$ there is no such sequence $\{c_n\}_{n\geq 1}$.

[i]Dan Schwartz[/i]

At a) you have to find a sequence of digits $c_1,c_2,\ldots$ such that no matter how we put 1 or 2 or 3 or ... or 9 digits between $c_1$ and $c_2$, then between $c_2$ and $c_3$, etc. the resulting number ${0.c_1 \ a_1a_2\ldots a_{k_1}} \ c_2 \ b_1b_2\ldots b_{k_2} \ c_3 \ \ldots$ is not rational.

At b) you must prove that if we allow to include 10 digits between each two consecutive terms of $\{c_n\}_n$ we can create a rational number.

For a. It is quite clear.
Take, for example, $c_n=s_k\pmod{10}$ for all $10^{k!}<n\leq 10^{(k+1)!}$, where $s_k$ is a sequence s.t. for each ordered pair of digits $(a,b)$ there are infinitely many $i$ s.t. $s_i=a$ and $s_i=b$.
Suppose we can construct a rational number $r$ from it. The choice of $c_n$ guaratees that part of each digit in representation of $r$ is at least $\frac{1}{10}$. It follows that each digit 0,..., 9 meets exactly 10% of times in the period of $r$. Moreover it is clear we should have $r_n=r_{n+10}$, i.e. period is $d_1d_2...d_10$, where $\{d_1,...,d_10\}=\{0,1,2,...,9\}$.
On  the other hand it is an impossible. Indeed, suppose $c_n=d_1$ and $c_{n+1}=d_2$ (remember, there are infinitely many such $n$), then we should insert at least 10 numbers between them! Contradiction.

[quote="Myth"]For a. It is quite clear. [/quote]This appears to be the most difficult problem on the test, as far as the results go :)

For b. Indeed, I am dumb. We can always provide a period $0123456789$.

Also for (a) (terribly sorry if this is the same as Myth's solution).

Given the first few digits $c_1,c_2,\ldots,c_k$ and a rational, partition the decimal representation of the rational in groups of $9$ digits. From each such group, at least one digit is missing, and in each group we must have at least one of the $c_i$'s. Pick a digit $a$ which is missing from a group of $9$ which appears infinitely many times (which is part of the period, for example). If $t$ is large enough and $c_{k+1},c_{k+2},\ldots,c_{k+t}$ are all equal to $a$, then this rational cannot be formed by the procedure described in the problem. We now enumerate the rationals, and, at each step, we construct more of our sequence in the manner described above.

I hope it's Ok.