trigonometry
inequalities
inequalities solved

Let x1, x2, .. xn be the numbers from interval (0, pi/2) such that:
tgx1 + tg2 + .. +tgxn <=n.
Prove that: sinx1.sinx2...sinxn <= 2n/2.

Regarding the one with sin and tg; are you sure the product of the sines isn't supposed to be <= 2^(-n/2), because you wrote 2^(n/2), and the product is obviously <= 1. I'm going to solve it for 2^(-n/2).

(sinx)^2=(tgx)^2/((tgx)^2+1). Let's denote tgxi=ai, so (sinxi)^2 = ai^2/ai^2+1. So we have to prove that product from i=1 to n of ai^2/(ai^2+1) <= 2^(-n) if  \sum ai <= n. But we use the fact that ai^2+1>=2ai, so our product is <= 2^(-n)*(product from i=1 to n of ai) <= 2^(-n) because  \sum ai <= n and we use AM-GM.

An "other" solution for N9 (the one with sin and tan) :

ai = tan(xi)
bi = sin(xi) = ai / sqrt (1+ai^2)

f(x) = ln_2 (x / sqrt(1+x^2))
f''(x) = -(3*x^2+1)/((1+x^2)^2*x^2*ln(2)) < 0

so f is strictly concave.

Jensen : f(sum xi / n) = -n/2 >= sum f(xi) / n

Taking 2^ gives the wanted ineg.


If $a, b, c, d$ are the solutions of the equation  $x^4-bx-3=0$, then an equation whose solutions are:

\[ \frac{a +b+ c}{d^2}, \frac{a +b+ d}{c^2}, \frac{a+c+d}{b^2}, \frac{b+c+d}{a^2} \]

$\text{(A)} 3x^{4}+bx+1 = 0 \qquad \text{(B)} 3x^{4}-bx+1 = 0 \qquad \text{(C)} 3x^{4}+bx^{3}-1 = 0$
$\text{(D)} 3x^{4}-bx^{3}-1 = 0 \qquad \text{(E) None of these}$

I think the problem actually gives the solutions of the new equation as:

$\frac{a+b+c}{d^2},~~\frac{a+b+d}{c^2},~~\frac{a+c+d}{b^2},~~\frac{b+c+d}{a^2}$

Can you take it from here?

The answer would be [b]D[/b]

[hide]Since the coefficient of $x^3$ in the polynomial function $f(x) = x^4-bx-3$ is zero, the sum of the roots of f(x) is zero, and therefore,

$\frac {a+b+c}{d^2}$ = $\frac {a+b+c+d-d}{d^2}$ = $\frac {-1}{d}$

Similarly,

$\frac {a+c+d}{b^2}$ = $\frac {-1}{b}, \frac {a+b+d}{c^2}$ = $\frac {-1}{c}, \frac {b+c+d}{a^2}$ = $\frac {-1}{a}$

Hence the equation f(-1/x) = 0 has the specified solutions:

$\frac {1}{x^4}$+$\frac {b}{x}-3$ = $0$

$1+bx^3-3x^4=0$

$3x^4-bx^3-1=0$[/hide]


Can you add a choice that says "Both"? :lol:

meh i really don't care... i don't like geek... sounds too geeky... i'll take nerd i guess cuz that's what everyone calls me although i really wish they didn't :) so i guess nerd or neither... borderline

nerd means lacking social skills or boringly studious. I'm about 1/3 there.

geek means someone who has an eccentric devotion to a particular interest. If you call a bookshelf full of math books eccentric, then yes.

My friend has a shirt that she wears around me that says "Geeks are HOT!" so I said geek!

geeks r geeky, and nerds r nerdy :D . but do both still apply the same 4 girls?

[quote]Whats the difference between a geek and a nerd?
[/quote]

See 1=2's post.  Take it as seriously as you want.

if thats what they r, then i'm neither. i'm mor of a teacher's pet than anything. but then again, most of my teachers dont know that i dont do my homework a lot... lol

Geek mainly. I am kinda social, but really geeky.

hmm.... ppl call me professor, assassin, or mercenary.... ppl who see me in  academics call me professor, people who see me in math call me assassin, and people who see me in sports call me mercenary....

i'm sorta the oracle
i really don't like that name
so where does that go?

um in my opinion, in a _really_ simple sense, nerds are geeks, minus having a life, and most probably any sense of humor.

myyellowducky:Read what 1=2 said.

I'm kind of both, but more of a nerd, cause I mostly don't have a life. But, w/e.

Nerd -- at least, that's what everyone calls me...

I think people call me a geek, or a nerd, I forgot......but thats only because I like math. I think I am a geek, I don't know. I think I have lots of characteristics that geeks don't have and regular people have.

[quote="undefined117"]any sense of humor[/quote]

If I tell a joke, everyone gets annoyed. But if I don't tell a joke that was invented, and invent one myself, the whole room could be laughing.

I consider myself a nerd, and I never thought that implies that I'm boringly studious or antisocial, it just means I'm more academicaly oriented than most...

I don't think other people really called me either of them, though, so maybe [b]1=2[/b]'s harsher definition is more widely used.

you should add another radio button with the "crazy" option.... i'm sure it will be the most voted one :P

i said geek b/c besides math, i really like computers.
geeks actually are ppl that have a big sense of computers.. at least thats how i hear it

Im everything
& I hang out with everyone: nerds, geeks, jocks, in-betweens, cute girls, wierd people

In other words I have friends in all categories and its fun :D

Most people call me a geek, I used to prefer nerd, but i guess i don't care.

Here's an interesting link:

[url]http://ask.yahoo.com/20060818.html[/url]

recently since its summer my self-opinion has changed. now i'm pretty much both: i have no life because i always obsess in front of the computer screen in my spare time and completely ignore my surroundings. buuuut i dont really care. 'cause even if i had a fight with one of my friends/not-really-friend, sometimes animehurl.com and some ice cream are the only friends you need. :D  :D  :D

Based on the link in post #25:

[b]Nerd:[/b] [color=green]Negative[/color]
[i]Not unattractive[/i]

[b]Geek:[/b] [color=green]Negative[/color]
[i]Not hot on technology[/i]

[b]Dork:[/b] [color=green]Negative[/color]
[i]Not stupid[/i]

 :wink:

um... you put the opposite there

from that I would go... learn some more technology stuff so I can vote geek


1) What product(s) will we obtain after heating a mixture of m-xylene, tert-butyl chloride, and aluminium chloride for several hours?

2) When p-methyl-N,N-dimethylaniline is treated with nitric acid and excess sulphuric acid what product(s) will be obtained?

I think i am missing something here. someone correct me.
I am guessing the obvious
(1) A mixture of 1-isobutyl 2,4 dimethyl benzene  and alpha,alpha dimethyl isobutyl-benzene


In triangle ABC, 3sin(A)+4cos(B)=6 and 4sin(B)+3cos(A)=1.  Find the measure of angle C.

I tried this:
Add both equations together and you get 3(sinA+cosA)+4(sinB+cosB)=7.
If both angles were pi/2, it'd work, but since they're in a triangle they can't be... anyone got a bright idea?

[quote="beta"]C=30 or 150.[/quote]

If C=150 then A <=30. So  4sin(B) + 3cos(A) >= 3cos(A)  > 1. So C=150 is not possible.

[quote="Wumbate"]Thanks.  Out of curiousity, where'd you get the idea to square them?[/quote]

I don't know, maybe I just felt like it. Actually, we wanted to get sin(A+B) or cos(A+B) somehow.

Well, when you have cos terms and sin terms, then squaring usually is helpful. Since we know (sin x)^2 + (cos x)^2 = 1.


I am having the problem described in the Asymptote: Help - Troubleshooting problem #2.
[i]"I have everything installed and the hotkey doesn't give me an error in TeXnicCenter, but when I hit the hotkey to compile my code, all that happens is that a window flashes briefly then disappears. Nothing else seems to happen."[/i]

I've done everything the solution suggested, but I am still having the same problem.

Any more ideas?

It may be that TexnicCenter can't find Asymptote. If you have installed it in a directory different from C:\Program Files\Asymptote then see http://www.artofproblemsolving.com/Forum/viewtopic.php?t=170406 for what to do.

If that's not the problem then you need to stop the window closing so that you can read the error message. Try the method suggested in the [b]first[/b] post at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=148557. If you see an error message which you can't deal with then post it here.

I did manage to get a screenshot of the window at the exact time when it came up.[/list]

That's exactly the same message as in http://www.artofproblemsolving.com/Forum/viewtopic.php?t=170406 which means that Asymptote can't work out where it is. So to tell it (assuming you are using Windows XP):
Go to  C:\Documents and Settings\your user name\.asy
1. if there's no config.asy file there make a new one with the following text
[code]import settings;
dir="C:\your path to Asymptote";[/code]
2. if there is already a config.asy file then edit or add the line
[code]dir="C:\your path to Asymptote";[/code]

I guess I should use LaTex to create config and save it as an .asy?

Yes, you can use any text editor you like including TexnicCenter if that's what you mean by 'using LaTeX'. Just make absolutely sure that it is really saved as config.asy and not config.asy.tex :)

Great! As at least 2 others have had the problem I'll add the solution to the Wiki.

Now I am getting:
cannot load module:'asymptote'

That's odd because, as far as I'm aware, there isn't an asymptote module. What asymptote code are you using? Perhaps you could tell us something about your setup?

Ok, I was trying to incorporate Asymptote into Latex by making it a package. I created a folder, Asymptote and put in it asymptote.sty and asycolors.sty. Then tried to compile:[url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Advanced_Configuration#Using_Asymptote_in_LaTeX]this[/url] but it didn't work so I put back the asymptote.sty and asycolors.sty  Giving me this error.

Here is the code I was trying to compile:
draw((0,0)--(100,100));
draw((0,100)--(100,0));
dot((50,50));
label("$ P$",(50,50),S);

[quote="hunter34"]Giving me this error.[/quote]You forgot to give 'this error'.
Creating the new asymptote folder has no effect on the way asymptote is run; asymptote itself uses the C:\Program Fies\Asymptote directory. The purpose of adding the sty files to a latex directory is to allow the asymptote package to create the asy file which happens the first time you compile latex => pdf. Then asymptote is run on the asy file and finally recompiling the tex file adds the picture to the output.

It's vital, after copying the sty files, that the file name database is refreshed. If it isn't there will be an error about asymptote.sty not being found and the asy file is not created.

So try again with the Using Asymptote in LaTeX instructions and report exactly what you've done and what errors you get - you can copy them from the tex file log.


I encountered this problem problem on a website (source: http://www.ping.be/~ping1339/Pvect.htm):

[quote]
[b]Question:[/b]
S = {(2,5,3)} and T = {(2,0,5)}
The intersection of span(S) and span(T) is a vector space.
Find this space.
[/quote]

[hide="Answer"]
span(S) = {r.(2,5,3) | r in R}
span(T) = {r'.(2,0,5) | r'in R}
The only common vector of span(S) and span(T) is (0,0,0).
The intersection of span(S) and span(T) is a vector space {(0,0,0)}.
This space is called the zero space or the null space.
[/hide]

The way I did this problem was that span(S) = {(1, 0, 0), (0, 1, 0), (0, 0, 1)} and span(T) = {(1, 0, 0), (0, 0, 1)}. Therefore the intersection R is R ={(1, 0, 0), (0, 0, 1)}.

I understand that span(S) can also be {(2, 5, 3)} since the one vector in S can be constructed a linear combination of itself, and same with the one vector in T, but there seem to be multiple answers to what the span(S) and span(T) are, i.e. they are not functions, and depending on what answer you provide you get a different vector space for the intersection of span(S) and span(T).

Can anyone clarify? Do you use the smallest basis for span so that it's a function? (By smallest I mean the one with the least amount of vectors.)

A span is not a function - a span is a set!

If we have vectors $v_{1},v_{2},\dots,v_{m}$ then

$\text{span}\{v_{1},v_{2},\dots,v_{m}\}=\{a_{1}v_{1}+a_{2}v_{2}+\cdots+a_{m}v_{m}: a_{1},a_{2},\dots,a_{m}\in\mathbb{R}\}$

- that is, the set of all possible linear combinations of these vectors.

So, now suppose $S=\{(2,5,3)\}.$ What is that? That's a set which consists of a single vector $u=(2,5,3)\in\mathbb{R}^{3}.$ What does that make the span of $S?$

$\text{\span}(S)=\{tu: t\in\mathbb{R}\}.$

Geometrically, that's a line through the origin. It is a one-dimensional subspace of $\mathbb{R}^{3}.$

Now, $T$ also consists of a single vector $v.$ Its span consists of everything of the form $sv.$ What is the intersection of these two spans? That would consist of all of the points that can be represented as both $tu$ and as $sv.$ So when does $tu=sv?$ If we could do that with $t\ne0,$ then we'd have $u=\left(\frac{s}{t}\right)v,$ which would make $u$ a scalar multiple of $v.$ But for $u=(2,5,3)$ and $v=(2,0,5),$ $u$ is not a scalar multiple of $v$ and $v$ is not a scalar multiple of $u.$ Hence the only way we can have $tu=sv$ is to have $t=s=0,$ which means that $\text{span}(S)\cap\text{span}(T)=\{0\}$ - that is the set which contains the zero vector and nothing else.

[I see now that you quoted this solution, but didn't understand it. What you said after that makes no sense.]

[quote]Do you use the smallest basis for span so that it's a function? (By smallest I mean the one with the least amount of vectors.)[/quote]

In that quote, the words "so that it's a function" are complete nonsense, and should never have appeared in the question. For a given subspace a spanning set with the fewest possible number of elements would be a basis for that subspace. A spanning set with more elements than that would be a dependent set and would not be a basis.

And for the record, the only possible basis for the subspace $\{0\}$ is the empty set - a basis with zero elements to span a zero-dimensional space.

Are you taking a linear algebra class? If so, I would suggest you consult with your instructor, as you would appear to be completely lost. If you're trying to learn this on your own, I'd suggest acquiring a standard textbook with a title like "Elementary Linear Algebra" or "Linear Algebra and Its Applications." There are a number of such books on the market, many with pretty good explanations. (The one I'm using this semester is by David C. Lay.)


[quote="mathcrazed"]Not too hard:

1) Show that $ \sin 1 < \log_{3}\sqrt {7}$

2) Show that $ 3 < 6\log 2 \plus{} \log 31 < 4$

3) Find the value of $ \log({\sqrt {3 \minus{} \sqrt {5}} \plus{} \sqrt {3 \plus{} \sqrt {5}}})$

4) If $ a \equal{} \log_{7}({2\sqrt {2} \plus{} 1}) \plus{} \log_{2}({\sqrt {2} \minus{} 1})$, express $ \log_{7}({2\sqrt {2} \minus{} 1}) \plus{} \log_{2}({\sqrt {2} \plus{} 1})$ in terms of $ a$.

5) Find the value of $ 6^{\log 40}*5^{\log 36}$

6) Show that in the decimal expansion of $ (\sqrt {65} \minus{} 8)^{100}$, no non-zero digit appears until after 121 zeros, including the zero in the ones place.[/quote]

[color=darkblue]I don't understand this problem:

2) Show that $ 3 < 6\log 2 \plus{} \log 31 < 4$

3) Find the value of $ \log({\sqrt {3 \minus{} \sqrt {5}} \plus{} \sqrt {3 \plus{} \sqrt {5}}})$

5) Find the value of $ 6^{\log 40}*5^{\log 36}$

[u]Note:[/u] Define about logarite is $ \log_{a}{x}$, but you give me $ \log{x}$[/color]

[hide="2"]$ 6\log2 \plus{} \log31 \equal{} \log2^6 \plus{} \log31 \equal{} \log(2^6\cdot31) \equal{} \log1984$
$ 3 \equal{} \log1000$ and $ 4 \equal{} \log10000$

and since
$ 1000 < 1984 < 10000$, $ 3 < 6\log2 \plus{} \log31 < 4$[/hide]

sidenote:
when you say $ \log x$ it is supposed to be comprehended as $ \log_{10}x$

[color=darkblue]3) Find the value of $ \log({\sqrt {3 \minus{} \sqrt {5}} \plus{} \sqrt {3 \plus{} \sqrt {5}}})$

it's easy:
$ {\sqrt {3 \minus{} \sqrt {5}} \plus{} \sqrt {3 \plus{} \sqrt {5}}}\equal{}\frac{{\sqrt {6 \minus{} 2\sqrt {5}} \plus{} \sqrt {6 \plus{} 2\sqrt {5}}}}{\sqrt{2}}$

$ \equal{}\frac{{\sqrt {(\sqrt {5}\minus{}1)^2} \plus{} \sqrt {(\sqrt {5}\plus{}1)^2}}}{\sqrt{2}} \equal{}\frac{2\sqrt{5}}{\sqrt{2}}\equal{}\sqrt{10}$

Then OK![/color]

[quote="thanhnam2902"][color=darkblue]3) Find the value of $ \log({\sqrt {3 \minus{} \sqrt {5}} \plus{} \sqrt {3 \plus{} \sqrt {5}}})$

it's easy:
$ {\sqrt {3 \minus{} \sqrt {5}} \plus{} \sqrt {3 \plus{} \sqrt {5}}} \equal{} \frac {{\sqrt {6 \minus{} 2\sqrt {5}} \plus{} \sqrt {6 \plus{} 2\sqrt {5}}}}{\sqrt {2}}$

$ \equal{} \frac {{\sqrt {(\sqrt {5} \minus{} 1)^2} \plus{} \sqrt {(\sqrt {5} \plus{} 1)^2}}}{\sqrt {2}} \equal{} \frac {2\sqrt {5}}{\sqrt {2}} \equal{} \sqrt {10}$

Then OK![/color][/quote]

but you hafta take it $ \log_{10}$, which is therefore $ \frac {1}{2}$

and you just beat me to it :(

[color=darkblue]4) If $ a \equal{} \log_{7}({2\sqrt {2} \plus{} 1}) \plus{} \log_{2}({\sqrt {2} \minus{} 1})$, express $ \log_{7}({2\sqrt {2} \minus{} 1}) \plus{} \log_{2}({\sqrt {2} \plus{} 1})$ in terms of $ a$.

Let $ b\equal{}\log_{7}({2\sqrt {2} \minus{} 1}) \plus{} \log_{2}({\sqrt {2} \plus{} 1})$.

We get $ a\plus{}b\equal{}\log_{7}({2\sqrt {2} \minus{} 1})({2\sqrt {2} \plus{}1}) \plus{} \log_{2}({\sqrt {2} \minus{} 1})({\sqrt {2} \plus{} 1})$

$ \Longrightarrow a\plus{}b\equal{}\log_{7}{7}\plus{}\log_{2}{2}\equal{}2$

$ \Longrightarrow b\equal{}2\minus{}a$

OK! :P [/color]

[quote="thanhnam2902"][color=darkblue]Oh! in my country note that $ \log_{10}{x}$ is $ \lg{x}$. So I don't understand.[/color][/quote]

ah perhaps you come from a different country. In America, $ \log x$ is almost always interpreted as $ \log_{10}x$

1. $ \sin 1 < \sin 30 \equal{} \dfrac{1}{2}$
Also,
$ \log_3 \sqrt{7} \equal{} \dfrac{\log 7}{2 \log 3}$
Therefore, it is true that $ \dfrac{1}{2} < \dfrac{\log 7}{2 \log 3}$, because $ \log 3< \log 7$
Hence, we are done.

[quote="thanhnam2902"][color=darkblue]4) If $ a \equal{} \log_{7}({2\sqrt {2} \plus{} 1}) \plus{} \log_{2}({\sqrt {2} \minus{} 1})$, express $ \log_{7}({2\sqrt {2} \minus{} 1}) \plus{} \log_{2}({\sqrt {2} \plus{} 1})$ in terms of $ a$.

Let $ b \equal{} \log_{7}({2\sqrt {2} \minus{} 1}) \plus{} \log_{2}({\sqrt {2} \plus{} 1})$.

We get $ a \plus{} b \equal{} \log_{7}({2\sqrt {2} \minus{} 1})({2\sqrt {2} \plus{} 1}) \plus{} \log_{2}({\sqrt {2} \minus{} 1})({\sqrt {2} \plus{} 1})$

$ \Longrightarrow a \plus{} b \equal{} \log_{7}{7} \plus{} \log_{2}{2} \equal{} 2$

$ \Longrightarrow b \equal{} 2 \minus{} a$

OK! :P [/color][/quote]

err. $ (\sqrt {2} \minus{} 1)(\sqrt {2} \plus{} 1) \equal{} 1$

For number 3, you could also square it to get 10. So it's obviously $ \sqrt {10}$

$ \sin 1$ has the $ 1$ in radians.

5. NOTE all are Log to the base 10

let k= 6^log40*5^log36
logK= log(6^log40 * 5^log36)
logK = log6*log40 + log5*2log6
logk= log6( log40 + log25)
logk = log6(log 1000)
logk = Log6(3)
k= 10^Log 216
K= 216.

$ \sin 1 < \sin 60^o \Rightarrow 3^{\sin 1} < 3^{\frac{\sqrt 3}{2}} < 3^{\frac {7}{4}}$

Now $ 3^{\frac{7}{4}} < 7$ iff $ 2187^\frac{1}{4} < 7$ iff $ 2187<7^4$ which is true

Hence $ 3^{\sin 1}< 7$ from which the result follows

[quote="mathcrazed"]
$ \sin 1$ has the $ 1$ in radians.[/quote]

[hide="solution"]Ok, so $ \sin 1 < \sin \dfrac{\pi}{3} \equal{} \dfrac{\sqrt {3}}{2}$ because $ \pi > 3$
Hence, we need to show that $ \dfrac{\sqrt {3}}{2} < \dfrac{\log7}{2\log 3}$
Thus, multiplying through by $ 2\log 3$ gives $ \sqrt {3}\log 3 < \log 7$
Then we must show that:
$ 3^{\sqrt {3}} < 7$
But, $ \sqrt {3} < \dfrac{7}{4}$ because if we square both sides and multiply through by 16, we obtain $ 48 < 49$, which is true.
Thus, we need to show that $ 3^{\sqrt {3}} < 3^{\dfrac{7}{4}} < 7$ is true. But, raising both sides to the fourth power gives  $ 3^7 \equal{} 2187 < 7^4 \equal{} 2801$. Hence, we are now done.  :D [/hide]
EDIT: hsbhatt... you beat me! :D

[quote="mathcrazed"]

6) Show that in the decimal expansion of $ (\sqrt {65} \minus{} 8)^{100}$, no non-zero digit appears until after 121 zeros, including the zero in the ones place.[/quote]

No one?

Can someone (mathcrazed included  :wink: ) please tell me if the just previous solution is right or not?


A glass cube, tied with fishing line at 4 of its vertices such that it hangs somewhere in the middle of the tank, is inserted inside a fish tank. In the middle of the cube there is a worm.
Three fish start at the same point inside the fish tank. The first fish swims so that its new position is 3 units to the right and 2 units into the depth of the tank from where it started. The second fish swims so that its new position is 5 units to the right and 4 units upwards from where it started. The third fish swims so that its new position is 1 unit to the right, 8 units upwards and 4 units into the depth of the tank from where it started. Now all three fish are on a vertex of the glass cube.
How far are the fish from the worm?

Assume that the fish started at the origin. Their new positions are:

(3, 0, 2); (5, 4, 0); (1, 8, 4).

The squares of the distances between each pair of points are: (distance formula)

24, 48, 72

In a cube with side length a, the only possible distances between any two vertices are a, a*sqrt(2), and a*sqrt(3). Therefore, the only posible squares of distances are: (distance formula)

a^2, 2a^2, and 3a^2.

a^2=24; 2a^2=48; 3a^2=72.

The distance from a vertix to the center of a cube with side length a is, by the distance formula:

d^2=3(a/2)^2 = (3a^2)/4

d^2=3*24/4=18

d=sqrt(18)=3sqrt(2)


geometry
perimeter
inequalities
ceiling function
triangle inequality

How many non-congruent triangles with integer sides and perimeter 1999 can be constructed?

Suppose the sides are x <= y <= z

By triangle inequality, x+y >= z.  so 1999 >= 2z.  It implies 999>= z.

Also, z >= 667.

Now y can be any value from z to ceil[(1999-z)/2].  The choices of y fix x.

So we have $\sum\limits_{z=667}^{999} = z - \lceil(1999-z)/2 \rceil$.  The first part is easily calculated.  The second part is $- \sum\limits_{z = 1000}^{1332} \lceil z/2 \rceil$ = $-[ 500 + 501 + 501 + 502 + 502 + \dots 666 + 666]$ which is also easily calculated.

Let $S_k = \sum\limits_{j = 1}^k = \dfrac{k(k+1)}{2}$.  Then it is $S_{999} - S_{666} - 500 -2( S_{666} - S_{500})$ which I am too lazy to calculate.


i heard it from my classmates,i wanna konw what karamate is.who can help me?

Actually it's Karamata (named after [url=http://en.wikipedia.org/wiki/Jovan_Karamata]Jovan Karamata[/url]).

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14975]Karamata's inequality[/url]


[quote="kabi"]HINT::
try to manipulate it in this manner::
use in place of sin^4(x) + cos^4(x) = (sin^2x+cos^2x)^2 - 2(sinxcosx)^2
                                                 =1-2(sin(x)cos(x))^2
 sin^6(x)+cos^6(x)=(sin^2x+cos^2x)(sin^4x+cos^4x-sin^2(x)cos^2(x))
u will get the answer[/quote]

are you sure that
 sin^6(x)+cos^6(x)=(sin^2x+cos^2x)(sin^4x+cos^4x-sin^2(x)cos^2(x))
?

i did the multiplication and i didn't get sin^6(x)+cos^6(x)

sin x=a
cos x=b
a^6+b^6= ( a^2)^3+(b^2)^3
            =  (a^2+b^2) ( a^4+b^4-(ab)^2)           {x^3 + y^3=(x+y)(x^2+y^2-xy)

however let me show it multiplication also
(sin^2x+cos^2x)(sin^4x+cos^4x-(sinxcos)^2)
=sin^6x + cos^2xsin^4x + sin^2xcos^cos^4x + cos^6x-sin^4xcos^2x-cos^4xsin^2x
=sin^6x+cos^6x

is it fine now?

hello, we have $ 3(\sin(x)^4\plus{}\cos(x)^4)\equal{}6\cos(x)^4\minus{}3\minus{}6\cos(x)^2$ (1) and $ 2(\sin(x)^6\plus{}\cos(x)^6)\equal{}2\minus{}6\cos(x)^2\plus{}6\cos(x)^4$ (2) and from her we get (1)-(2)=1.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, we have $ 3(\sin(x)^4 \plus{} \cos(x)^4) \equal{} 6\cos(x)^4 \minus{} 3 \minus{} 6\cos(x)^2$ (1) and $ 2(\sin(x)^6 \plus{} \cos(x)^6) \equal{} 2 \minus{} 6\cos(x)^2 \plus{} 6\cos(x)^4$ (2) and from her we get (1)-(2)=1.
Sonnhard.[/quote]

could you explain how you simplified these equations pls?
im new to all this trig stuff and am really confused.

thank you so much


inequalities
trigonometry
inequalities unsolved

Let ABC be an acute-angled triangle.  Prove that:

$\sqrt{\frac{3}{2}} \cdot \frac{a+b+c}{\sqrt{a^2+b^2+c^2}} \leq \frac{ \sin{A} }{\sqrt{ \sin^2A+ \sin^2B}}+ \frac{ \sin{B}}{\sqrt{\sin^2C+\sin^2A}}+ \frac{ \sin{C}}{\sqrt{ \sin^2A+ \sin^2B}}\leq   \sqrt {\frac{m_a^2+m_b^2+m_c^2}{3Rr}}$

At first I think you made a typo in the first denominator of the middle part of the inequality.

This means that the inequality should be:

$\displaystyle \sqrt{\frac{3}{2}} \cdot \frac{a+b+c}{\sqrt{a^2+b^2+c^2}} \leq \frac{ \sin{A} }{\sqrt{ \sin^2B+ \sin^2C}}+ \frac{ \sin{B}}{\sqrt{\sin^2C+\sin^2A}}+ \frac{ \sin{C}}{\sqrt{ \sin^2A+ \sin^2B}}\leq   \sqrt {\frac{m_a^2+m_b^2+m_c^2}{3Rr}}$.

Okay, after the Sine Law we have a / sin A = b / sin B = c / sin C, so this is equivalent to

$\displaystyle \sqrt{\frac{3}{2}}\cdot \frac{a+b+c}{\sqrt{a^{2}+b^{2}+c^{2}}}\leq \frac{a}{\sqrt{b^{2}+c^{2}}}+\frac{b}{\sqrt{c^{2}+a^{2}}}+\frac{c}{\sqrt{a^{2}+b^{2}}}\leq \sqrt{\frac{m_{a}^{2}+m_{b}^{2}+m_{c}^{2}}{3Rr}}$.

Since $m_{a}^{2}+m_{b}^{2}+m_{c}^{2}=\frac{3}{4}\left( a^{2}+b^{2}+c^{2}\right)$ and $\displaystyle Rr=\frac{abc}{2\left( a+b+c\right) }$, we can see $\displaystyle \frac{m_{a}^{2}+m_{b}^{2}+m_{c}^{2}}{3Rr}=\frac{\left( a+b+c\right) \left( a^{2}+b^{2}+c^{2}\right) }{2abc}$, and hence, we are left with

$\displaystyle \sqrt{\frac{3}{2}}\cdot \frac{a+b+c}{\sqrt{a^{2}+b^{2}+c^{2}}}\leq \frac{a}{\sqrt{b^{2}+c^{2}}}+\frac{b}{\sqrt{c^{2}+a^{2}}}+\frac{c}{\sqrt{a^{2}+b^{2}}}\leq \sqrt{\frac{\left( a+b+c\right) \left( a^{2}+b^{2}+c^{2}\right) }{2abc}}$.

Well, this looks much simpler...

  Darij

[quote="darij grinberg"]
$\displaystyle \sqrt{\frac{3}{2}}\cdot \frac{a+b+c}{\sqrt{a^{2}+b^{2}+c^{2}}}\leq \frac{a}{\sqrt{b^{2}+c^{2}}}+\frac{b}{\sqrt{c^{2}+a^{2}}}+\frac{c}{\sqrt{a^{2}+b^{2}}}\leq \sqrt{\frac{\left( a+b+c\right) \left( a^{2}+b^{2}+c^{2}\right) }{2abc}}$.

Well, this looks much simpler...[/quote]

Indeed, this is much simpler,

$\displaystyle \frac{a}{\sqrt{b^{2}+c^{2}}}+\frac{b}{\sqrt{c^{2}+a^{2}}}+\frac{c}{\sqrt{a^{2}+b^{2}}}
\leq \frac{a}{\sqrt{2bc}}+\frac{b}{\sqrt{2ca}}+\frac{c}{\sqrt{2ab}} = \frac{a\sqrt{a} + b\sqrt{b}+ c\sqrt{c}}{\sqrt{2abc}}$

Apply Cauchy-Schwarz to the numerator, we have $\displaystyle {a\sqrt{a} + b\sqrt{b}+ c\sqrt{c}} \leq  \left( a+b+c\right)^{1/2} \left( a^{2}+b^{2}+c^{2}\right)^{1/2} $

This proves the second part of the inequality.

For the first part, WLOG, We may assume $a+b+c=1$, Apply Jensen to the convex functon $\displaystyle f(x) = \frac{1}{\sqrt{x}}$, we have

$\displaystyle \frac{a}{\sqrt{b^{2}+c^{2}}}+\frac{b}{\sqrt{c^{2}+a^{2}}}+\frac{c}{\sqrt{a^{2}+b^{2}}} = a f(b^{2}+c^{2})+b f({c^{2}+a^{2})+c f({a^{2}+b^{2})}}$

$\displaystyle\geq  f(a(b^{2}+c^{2})+b({c^{2}+a^{2})+c({a^{2}+b^{2})) = f\left({\sum_{sym}a^2b\right) = \frac{1}{\sqrt{\sum_{sym}a^2b}}}}}$

So it suffices to prove $\displaystyle \frac{1}{\sqrt{\sum_{sym}a^2b}} \geq \sqrt{\frac{3}{2}}\cdot \frac{1}{\sqrt{a^{2}+b^{2}+c^{2}}}$.

It is equivalent to, $\displaystyle 3\sum_{sym}a^2b \leq 2(a^2+b^2+c^2)= 2(a^2+b^2+c^2)(a+b+c) = \sum_{sym}a^3 + 2\sum_{sym}a^2b$

It reduces to $\displaystyle \sum_{sym}a^2b \leq  \sum_{sym}a^3$ which is pretty obvious by AM-GM or Muirhead.

Q.E.D.

BTW, Fuzzylogic, your nice proofs show that the inequality

[tex]\sqrt{\frac{3}{2}}\cdot \frac{a+b+c}{\sqrt{a^{2}+b^{2}+c^{2}}}\leq \frac{a}{\sqrt{b^{2}+c^{2}}}+\frac{b}{\sqrt{c^{2}+a^{2}}}+\frac{c}{\sqrt{a^{2}+b^{2}}}\leq \sqrt{\frac{\left( a+b+c\right) \left( a^{2}+b^{2}+c^{2}\right) }{2abc}}[/tex]

holds for arbitrary positive reals a, b, c, not just for the sides of an acute triangle!

  Darij


solution in $N$
we have
$x^2 + y(x+1) = 17$
$y^2 + x(y+1) = 13$
then $(x-y)(x+y-1)=4$
then
$x-y=1$ and $x+y-1=4$  $(x=3 and y=2)$
or  $x-y=2$ and $x+y-1=2$(impossible in $N$)
the solution in $N$ are $(x=3;y=2)$

[quote="1234567890"]They don't have to be integers, do they??? :?[/quote]
I'll post what I have, if eminem is asking for real numbers solution...
[hide]Add both equations to get $x^2+2xy+y^2+x+y=30$

Subtract second from first to get $(x-y)(x+y-1)=4$

Recall $(x+y+1)^2=x^2+2xy+y^2+2x+2y+1$

$x^2+2xy+y^2+x+y=30$

$x^2+2xy+y^2+2x+2y+1=(x+y+1)+30$

$(x+y+1)^2-(x+y+1)-30=0$

$x+y+1=6$ or $x+y+1=-5$

If $x+y+1=6$, you get Fermat -Euler's answer $x=3,y=2$

If $x+y+1=-5$, you get $x=-\frac{23}7,y=-\frac{19}7$[/hide]
Edit: Fixed typo, missing 0 in 30

[hide]Yes there are two solutions  $(3,2)$ and $(-23/7,-19/7)$
Easiest  is add them and  get $(x+y)^2+(x+y)-30 = 0$ this  gives $(x+y) =- 6 \ or \ 5$
substituting the value of  $y$ ( $5-x$ or $-6-x$)   in first equation will give the required values.[/hide]

[quote="1234567890"]Nice, 10000th User. No way I could have thought of that factoring, but now I know how to be slick! I think you have a slight error, however. It's probably just a typo, though.
[quote="10000th User"]
$(x+y+1)^2-(x+y+1)-3=0$[/quote][/quote]
Thank you, I fixed typo


The diagonals of quadrilateral ABCD intersect perpendicularly at O. What can be said about quadrilateral ABCD?

Rhombus, because a rhombus's diagonals are perpendicular.

Well you can call ABCD an orthodiagonal quadrilateral. You can also say that $ (AB)^2 \plus{}(CD)^2 \equal{} (BC)^2 \plus{} (AD)^2$. That's all I know about that.

@gaussintraining: A square [i]is[/i] a rhombus.

[quote="gaussintraining"]But a rhombus is not necessarily a square.[/quote]

Thats why rhombus is an all inclusive statement and therefore a better answer than square.

All right, point conceded.

(Oh, btw, I never meant to say that a square was a better answer than a rhombus, just that it worked in addition to a rhombus).

modularmarc101 is correct - the conditions do not prove that the quadrilateral is a rhombus.  Easy contradiction: $ A\equal{}(0,0),B\equal{}(7,1),C\equal{}(8,0),D\equal{}(7,\minus{}1)$.

Yes, a rhombus requires all sides to be the same.

http://en.wikipedia.org/wiki/Rhombus

Can someone supply a proof that the sums of the squares of opposite sides are equal?

[quote="TachyonPulse"]The proof is straightforward using multiple applications of the Pythagorean Theorem on the 4 right triangles that are created by the diagonals.[/quote]

I see that proof.  I'm wondering if there's something more elegant.

Most of the other comments on this board are incorrect.  For example, a dart, similar in appearance to the Federation logo in Star Trek, satisfies the given constraints and isn't a rhombus.

[quote="AIME15"]modularmarc101 is correct - the conditions do not prove that the quadrilateral is a rhombus.  Easy contradiction: $ A \equal{} (0,0),B \equal{} (7,1),C \equal{} (8,0),D \equal{} (7, \minus{} 1)$.[/quote]

Also, the conditions do not prove that the quadrilateral is convex.

Contradiction:  $ A \equal{} (0,\minus{}1), B \equal{}  (2,0), C \equal{} (0,1), D \equal{} (1,0)$.

Polygons
As applied to a polygon, a diagonal is a line segment joining any two non-consecutive vertices.
Therefore, a quadrilateral has two diagonals, joining opposite pairs of vertices. For any convex
polygon, all the diagonals are inside the polygon, but for re-entrant polygons, some diagonals
are outside of the polygon.

*** Note:  Figure 1 mentioned in the following would not transfer over with this
cut-and-pasted text.

Perpendicular
Fig. 1: The line AB is perpendicular to the line CD, because the two angles it creates
(indicated in orange and blue, respectively) are each 90 degrees.  For other uses, see
 Perpendicular (disambiguation).
In geometry, two lines or planes (or a line and a plane), are considered perpendicular
(or orthogonal) to each other if they form congruent adjacent angles
(an L-shape). The term may be used as a noun or adjective. Thus, referring to Figure 1,
the line AB is the perpendicular to CD through the point B. Note that by definition,
a line is infinitely long, and strictly speaking AB and CD in this example represent line
segments of two infinitely long lines.
>>> Hence the line segment AB does not have to intersect line segment CD to be considered
perpendicular lines, because if the line segments are extended out to infinity, they would still
form congruent adjacent angles. <<<

Sources of both paragraphs:
Wikipedia

[quote="Peterhi"]A dart *is* concave.[/quote]

As Inspector Clouseau would say, "I know that.  I know that."

I was already working on sending my post before your post went through.

And I had more details that I wanted left on the screen beyond your four sentences.


geometry
rhombus
incenter
geometry proposed

The incircle of the triangle $\triangle ABC$ touches the sides BC, CA, AB at D, E, F respectively. Each pair from the incircles of the triangles $\triangle AEF, \triangle DBF, \triangle DEC$ has two common external tangents, one of which is a side of the triangle $\triangle ABC$. Show that the other three tangents are concurrent.

Nice problem, with solution similar to that of problem 18 (see [url=http://www.mathlinks.ro/Forum/topic-82064.html]Rhombus from tangents[/url]) from the same year. Neither problem has a solution on [url=http://www.kalva.demon.co.uk/russian/rus99.html]Kalva homepage[/url].

You can readily show that this is equivalent to http://www.mathlinks.ro/Forum/viewtopic.php?t=46240 .

  dg

they are concurrent at incenter of triangle.

just and angle chase .


calculus
derivative
trigonometry
function
calculus computations

Hello there,

I am able to calculate the derivative for the following parametric equations, but my calculations for the second derivative are wrong.

Could someone please set me on the right track?

Thanks.

---

1. Determine $ \frac {dy}{dx}$ and $ \frac {d^2y}{dx^2}$ for the parametric equations $ x \equal{} 2t \minus{} 3$ and $ y \equal{} \tan t$.

$ \frac {dy}{dt} \equal{} \sec^2t$

$ \frac {dx}{dt} \equal{} 2$

$ \frac {dy}{dx} \equal{} \frac {sec^2t}{2}$

To get the second derivative, I would need to find $ \frac {d}{dt}$ (Derivative of first derivative) and then divide by derivative of original x term:

$ h(x) \equal{} f(g(x))$

$ g(x) \equal{} \sec^2t$

$ f(x) \equal{} \frac {x}{2}$

Now, I am not sure how $ g'(x) \equal{} \sec t \tan t$.

Also, can $ \sec^2t$ be rewritten as $ 2(\sec t)$?

[quote="hash_include"]while calculating 2nd derivative, what you are basically doing is calculating
$ \frac {d(\frac {dy} {dx})}{dx}$

by applying Chain rule of differentiation, the above is nothing but:
$ \frac {d(\frac {dy} {dx})}{dt} . \frac {dt}{dx}$

$ \equal{} \frac {d (\frac {\sec^2t}{2})}{dt}. \frac 1 2$

$ \equal{} \frac {\sec^2t \tan t } {2}$
:)[/quote]

Thank you for your reply, hash_include. As I am studying this by myself, would you care to show how you got $ \frac {d(\frac {dy} {dx})}{dt} . \frac {dt}{dx}$ from the Chain Rule?

Also, how did you get $ \frac {\sec^2t \tan t } {2}$ from $ \frac {d (\frac {\sec^2t}{2})}{dt}. \frac 1 2$ ?

[quote="catalan"]
would you care to show how you got $ \frac {d(\frac {dy} {dx})}{dt} . \frac {dt}{dx}$ from the Chain Rule?
[/quote]
well, that's the basic idea of chain rule itself. you see, since $ \frac {dy}{dx}$ is a function of $ t$, i can take some $ g(t)\equal{}\frac {dy}{dx}$
so what i'm doing is calculating $ \frac {d(g(t))}{dx}$, which by chain rule, is
$ \frac {d(g(t))}{dt} . \frac {dt}{dx}$
Perhaps [url=http://en.wikipedia.org/wiki/Chain_rule]this[/url] will help you understand the idea behind chain rule. :)
[quote="catalan"]
Also, how did you get $ \frac {\sec^2t \tan t } {2}$ from $ \frac {d (\frac {\sec^2t}{2})}{dt}. \frac 1 2$ ?[/quote]
aah yes.  this too is an application of chain rule.
here,
$ \frac {d (\frac {\sec^2t}{2})}{dt} \equal{} \frac {d(\frac {\sec^2t}{2})}{d(\sec t)}. \frac {d(\sec t)}{dt}$
take $ \sec t$ to be some $ p$
we get $ \frac {d(\frac {\sec^2t}{2})}{d(\sec t)} \equal{} \frac {d(\frac {p^2}{2})}{dp}$

$ \equal{} p \equal{}  \sec t$
Also,
$ \frac {d(\sec t)}{dt} \equal{} \sec t \tan t$ (which i'm sure you must be knowing by now)

therefore,
$ \frac 1 2 \frac {d(\frac{\sec^2t)}{2}}{d(\sec t)}. \frac {d(\sec t)}{dt} \equal{} \frac 1 2 . \sec t \sec t \tan t$
which is nothing but $ \frac {\sec^2t \tan t } {2}$
:)

The beauty of first differntials is that you can actually manipulate them like fractions. Though sometimes it might not be conceptually right, it still works.

$ \frac {dy}{dx}$ = $ \frac {dy}{dt} . \frac {dt}{dx}$

Note: The same cannot be said for 2nd differntials.


[quote="Silverfalcon"]Two right circular cones have parallel bases (that is, their bases lie in parallel planes), and the apex of each is (at) the center of the base of the other.  The cones intersect in circle $ C$.  If the areas of the bases are 400 and 900, compute the area of circle $ C$.[/quote]
[hide="Solution"]
Let the centers of the bases be $ O$ and $ P$.  Now, also let $ A$ be on cone $ O$ and $ B$ be on cone $ P$ such that $ AO\perp OP$ and $ PB\perp OP$.  Let $ PA$ and $ OB$ intersect at $ Y$.  Now, the distance from $ Y$ to $ OP$ is the radius of circle $ C$.  Let $ OA = x$ and $ PB = y$ and pick $ Z$ on $ OP$ such that $ YZ\perp OP$.  Hence, $ \triangle ZYB\sim \triangle AOB\implies \frac {ZB}{AB} = \frac {YZ}{OA}$ and $ \triangle YAZ\sim \triangle PAB\implies \frac {AZ}{AB} = \frac {YZ}{PB}$.  Hence, $ 1 = \frac {AZ}{AB} + \frac {ZB}{ZB} = YZ(\frac {1}{OA} + \frac {1}{PB})\implies \frac {OA\cdot PB}{OA + PB} = YZ$.  We can find that $ OA = \frac {20}{\sqrt {\pi}}$ and $ PB = \frac {30}{\sqrt {\pi}}$ using the information about the areas of the circles.  Hence, $ YZ^2\pi = [(\frac {600}{\pi})\cdot ({\frac {\sqrt {\pi}}{50})]^2\pi = \boxed{144}}$ [/hide]

Good job, but perhaps, this solution is easier and faster.

[hide="Pole Problem"]
Before proceeding to the problem, I want to tell about the pole problem. Basically, if you have two poles as shown in the diagram, you have the following identity:

$ \frac{1}{x} \plus{} \frac{1}{y} \equal{} \frac{1}{z}$

This is one of frequently occurring ideas. The proof is very simple (just using similar triangles) and I suggest that you attempt to do it in spare time. [/hide]

[hide="Solution"]
Radius of the one circle is $ \frac{20}{\sqrt{\pi}}$ and $ \frac{30}{\sqrt{\pi}}$.  Consequently, the radius of the circle $ C$, or $ z$ in the diagram, becomes $ \frac{12}{\sqrt{\pi}}$. Substituting this as $ r$ for area, we know that the area of the circle $ C$ is $ \fbox{144}$. [/hide]


calculus
integration
calculus computations

How do I calculate the integral of:

$ \int_ { \minus{} \infty}^{1} xy \plus{} y^2\,dy.$

edit: wrong integral, see my next post down.

Thanks

[quote="Carcul"]That integral diverges.[/quote]

Sorry the full question was:

find c such that:

$ c\int_ {0}^{\infty}\ \int_ { \minus{} \infty}^{1} x \plus{} 2y\,dydx. \equal{} 1$

The solution says: $ c \equal{} \frac {2}{3}$.

How can I do it? :)

$ \iint_{D}F \equal{} \int_{0}^{2}\int_{2}^{4}c(6 \minus{} x \minus{} y)dydx \equal{} c8$ thus $ c \equal{} 1/8$


function
search
algebra unsolved
algebra

Prove that there exists a unique function $f:R^+ \rightarrow R^+$ such that
$f(f(x)) = 6x-f(x)$ and $f(x)>0$ for all $x>0$

Has been posted before, just search for it.

[hide]Well, the idea is to fix $x$ and consider the sequence $f(x)$, $f(f(x)$, $f(f(f(x)))$ etcetera (which is called the orbit of $x$, if I remember correctly? Anyone? Or the trace maybe? I can't remember.) Anyway, it's easy to see that that is a linear recursive sequence, so look at the recurrence relation - you will find a general formula. Now, keep in mind that all values involved are actually positive, so that gives you some extra conditions on the recurrence relation, if you get what I mean. From this you can easily derive the result.[/hide]

I think it's orbit . The only function that satisfies the equatin is f(x)=2x and it obviously satisfies the condition too.


[quote="joml88"]If $\{a_1,a_2,a_3,\ldots,a_n\}$ is a set of real numbers, indexed so that $a_1<a_2<a_3<\cdots<a_n,$ its complex power sum is defined to be $a_1i+a_2i^2+a_3i^3+\cdots+a_ni^n,$ where $i^2=-1.$  Let $S_n$ be the sum of the complex power sums of all nonempty subsets of $\{1,2,\ldots,n\}.$  Given that $S_8=-176-64i$ and $S_9=p+qi,$ were $p$ and $q$ are integers, find $|p|+|q|.$[/quote]

[hide]
On looking at the problem, the thing that jumps out is that there is a bijection between the number of subsets with n and without n in $\{1,2,3,\ldots,n\}$. Furthermore, we note that $a_1 < a_2< a_3 < \ldots$ and thus when $n$ is appended it will alwyas be the last term in the subsets.

At this point - we make a table:
[code]
size | terms added
1          9i
8C1      9i^2
8C2       9i^3
...
8C8       9i^9
[/code]

Now we simply add this to $2S_8$ to find $S_9$ because of the bijection.

we find that the 9s will give $9i(1+i)^8 = 9i * 2^4 * \cos{(45 * 8)} = 144$ so $S_9 = -352 + 144 - 128i = -352 + 16i$

Therefore the final answer is $368$[/hide]

[quote="agolsme"]On looking at the problem, the thing that jumps out is that there is a bijection between the number of subsets with n and without n in $\{1,2,3,\ldots,n\}$.[/quote]

What exactly do you mean by this? :?

Say you have the set ${\{1,2,3,\ldots,n}$ There are $2^{n-1}$ subsets of the elements $\{1,2,3\ldots,n-1\}$. There are also $2^{n-1}$ subsets which contain $n$. This bijection is evident when you consider all the subsets without $n$ and attaching $n$ to it to create a subset with n.

[quote="agolsme"][quote="joml88"]If $ \{a_1,a_2,a_3,\ldots,a_n\}$ is a set of real numbers, indexed so that $ a_1 < a_2 < a_3 < \cdots < a_n,$ its complex power sum is defined to be $ a_1i \plus{} a_2i^2 \plus{} a_3i^3 \plus{} \cdots \plus{} a_ni^n,$ where $ i^2 \equal{} \minus{} 1.$  Let $ S_n$ be the sum of the complex power sums of all nonempty subsets of $ \{1,2,\ldots,n\}.$  Given that $ S_8 \equal{} \minus{} 176 \minus{} 64i$ and $ S_9 \equal{} p \plus{} qi,$ were $ p$ and $ q$ are integers, find $ |p| \plus{} |q|.$[/quote]

[hide]
On looking at the problem, the thing that jumps out is that there is a bijection between the number of subsets with n and without n in $ \{1,2,3,\ldots,n\}$. Furthermore, we note that $ a_1 < a_2 < a_3 < \ldots$ and thus when $ n$ is appended it will alwyas be the last term in the subsets.

At this point - we make a table:
[code]
size | terms added
1          9i
8C1      9i^2
8C2       9i^3
...
8C8       9i^9
[/code]

Now we simply add this to $ 2S_8$ to find $ S_9$ because of the bijection.

we find that the 9s will give $ 9i(1 \plus{} i)^8 \equal{} 9i * 2^4 * \cos{(45 * 8)} \equal{} 144$ so $ S_9 \equal{} \minus{} 352 \plus{} 144 \minus{} 128i \equal{} \minus{} 352 \plus{} 16i$

Therefore the final answer is $ 368$[/hide][/quote]

$ 9i(1 \plus{} i)^8 \equal{} 9i * 2^4 * \cos{(45 * 8)} \equal{} 144$ ?
it should be 144i

I don't seem to understand the problem. If we add a 9th term, can't it be anything? Can someone example it without the complicated notation? Thanks

[hide]$S_8$ is the sum of many terms, that is $2^8-1$. $S_9$ has $2^9-1$ terms. Clearly we should try to relate the two.

If we write out $S_8$ as a sum of each $2^8-1$ terms, we have a choice of whether we want to add a 9 to each set or not.

Then each of the terms in $S_8$ we duplicate, one without a 9 in it and one with a 9 in it. But this only gets us $2^9-2$ terms. The last term in $S-9$ comes from the empty set with a 9 added to it.

We can write out the sum without having to find each term.

There are $2^8-1$ terms that are those of originally $S_8$. There are $2^8$ terms that are those of $S_8$, but with $9i^k$ added to each of them. (We put the empty set in with these terms). So our total sum is $2S_8+$ Some sum of $9i^k$

We clearly have to divide up the terms in $S_8$ into those that have $9i$, $-9$, $-9i$, or $9$ added to them. These have to have had  0 mod 4, 1 mod 4, 2 mod 4, and 3 mod 4 terms in it respectively before the terms are added.

There are:
$8C0+8C4+8C8=72$ terms 0 mod 4. Contributes $9i*72$

$8C1+8C5=64$ terms 1 mod 4. Contributes $-9*64$

$8C2+8C6=56$ terms 2 mod 4. Contributes $-9i*56$

$8C3+8C7=64$ terms 3 mod 4. Contrubutes $9*64$

Total sum is 144i for these terms. Now we add them to $2S_8+144i=-352+16i$ Our answer is $\boxed{368}$.[/hide]

p10: Note that the amount of subsets containing a $9$ is equal to the amount not containing a $9$, so we need to multiply $S_8$ by $2$ and add that to what we are about to find. We will think about going from an old set, a subset of the 8-element set, to a new set, a subset of the 9-element set. If our old set had $0$ elements. There are $\binom{8}{0}$ ways to pick the set, and we're just going to be adding $9i^1$. Similarly if the old set had $1$ element, there would be $\binom{8}{1}$ ways to pick it and we'd add $9i^2$. Our expression is hence $9i^1 \cdot \binom{8}{0} + 9i^2 \cdot \binom{8}{1} + \ldots + 9i^9 \cdot \binom{8}{8}$. We can factor $9i$ to get $i^0 \cdot \binom{8}{0} + i^1 \cdot \binom{8}{1} + \ldots + i^8 \cdot \binom{8}{8}$ multiplied by $9i$, which by the Binomial Theorem is $(1+i)^8 \cdot 9i$. We have $(1+i)^2 = 2i$, and $(2i)^4 = 16$, so $(1+i)^8 = 16$. Our expression is hence $144i$. But recall that we need to multiply $S_8$ by $2$, giving $-352i+16 \implies 352+16 = \boxed{368}$.


irrational number
number theory
relatively prime

prove that for $p_n$ as the nth irrational number, $0.p_1p_2p_3....p_n.......$ is irrational.

If anyone knows Dirichlt's theorem that says for an + b with relatively prime a and b, there are an infinite number of primes for n = 1, 2, 3...., you can use that.[/tex]

You mean $p_n$ is the $n^{th}$ prime number, don't you? Because there is no numeration of the irrationals with the integers...
Just use that there are an infinite number of primes of the form $10^{k+2}\cdot n + 10^{k+1}+1$ (which indeed follows from Dirichlet's theorem). Therefore, for each $k$, you will have an infinite number of strings of exactly $k$ consecutive digits $0$ in the decimal expansion of the given number. It follows that this decimal expansion is not periodic so that the given number is irrational.

Pierre.

I'm afraid I don't understand fully.  Doesn't k need to be some constant, and n the only variable in Dirichlet's?  Then, we have $10^(k+1)*n + 1$ but k + 1 is some number so it's like having $10^{8234582} * n + 1$ but we don't know which values of n will make this prime so the period might just be very very large.

Another thing in your answer, can't we say that it's just ${10^{k + 1} n + 1} $ since 1 is relatively prime to $10^{k + 1} $?  Or am I missing something here... [/tex]

Yes $k$ is a given number, and $n$ is the variable (Dirichlet's theorem deals with primes in arithmetic progressions).

No the period is not very large : There is no period at all.
Assume there is such a period, and let $d$ be its length. Note that the periodicity is not supposed to start from the beginning.
For each given $k$ there an infinite (!) number of primes of the form I said (even if I do not know where they are), thus the sequence of $k$ digits $0$ appears an infinite number of time, and so after any digit of given rank (and then after the period would start). Then, the period has to contain a string of $k$ consecutive $0$. Since this is true for each $k$, it follows that the periodic sequence of the digits is only made by $0$ digits (use $k > d$) which is absurd, since any prime number always contains at least some non-zero digit in its decimal expansion.

Yes, you are right about $10^{k+1} n + 1$. Using that, you don't need Dirichlet's theorem in its full form besause this special case can be proved using elementary means (even it is not so short). In another hand, the proof is the same only saying that the period contains string of at least (and not necessary exactly) $k$ consecutive digits $0$.

Pierre.


How do you do this problem:

While monitoring the distance from the space shuttle jupiter from the moon, the computer at NASA relates 2 stes of data, sets A and B. Set A contains all integers that can be written in the from n squared+n+1 where 1 is a positive integer between 1 and 100, inclusive. Set B contains all numbers whose absoluse value is less than or equal to 100. how many elements of set a coincide with those of set b?

please include an explanation

[hide]Define f(n) to equal $n^2$ + n + 1

So for integers in the sets to coincide,  f(n) has to be less than 100, and n has to be a pos integer less than 101.

n=1 works, and n= 9 does too. All teh ones between 1 and 9 work also.
n=10 doesn't work, because f(10) = 111
So 1 through 9 work: that's 9 integers.[/hide]

[hide]There's lots of things just thrown into this problem to make it seem harder.  We can rephrase it as:

How many positive integers n are there such that $n^2+n+1\le 100$?

 Looking at it, we see that if n fits the equation, then all integers less than n also fit.  Since 10 doesn't work ($100+10+1>100$) and 9 does work ($81+9+1<100$), there are 9 integers that fit.[/hide]

[color=blue]Moderator kyyuanmathcount's note: Again, please hide, thanks![/color]

[hide]since n has to be a positive integer between 1 and 100, wouldn't everything in set A be in set B?[/hide]

[hide="edit"]oh, i get it now. i was trying to find n instead of x. since everything that would fit in set A would fit in set B, you can take that part out. n has to be less than 10, because10^2 is already 100, and you have to add more things to it. n=9 works, because 81 + 9 + 1 = 91.[/hide]

no, because $n^2+n+1>100$ for $n\geq 10$
[hide="solution"]
We have to find all of the elements in A that are less than or equal to 100.
$n^2+n+1\leq 100$
$n^2+n\leq 99$
$n(n+1)\leq 99$
$n\leq 9$
We also know that $n\geq 1$
Thus n=1-9 works, and there are 9[/hide]


Let n be any number which has 2006 number of 1's in its binary representation.Prove that $2^{n-2006}$ divides $n!$.


geometry
rectangle
Pythagorean Theorem
power of a point

Let ABCD be a square of side length 10 with a circle inscribed in it. Let M be the midpoint of segment AB. Find the length of MC outside of the circle.

I tried using power of a point with pythagorean theorem, but i ended up with a complex number... :oops:

Power of a Point is actually not even needed for this. You only need the Pythagorean Theorem.

The diameter of the circle intersects with the midpoints of the two sides, so you'll have a rectangle with dimensions $ 5$ and $ 10$. Now you simply find the length of the diagonal.

no.. ur looking for the length of the diagonal OUTSIDE of the circle.

[geogebra]21e8f17b1b905bfec9265398877de291c39d768d[/geogebra]
Let $ N$ be the midpoint of $ CD$, and let $ Y$ be the intersection of the circle and $ MC$. (See geogebra figure.) By Power of a Point, $ CN^2\equal{}CP\cdot CM \Rightarrow 25\equal{}CP\cdot 5\sqrt{5} \Rightarrow CP\equal{}\sqrt{5}$.


Let $ f(x),g(x)$ be integrable on $ [a,b]$
show that $ \min\{f(x),g(x)\}$ and $ \max\{f(x),g(x)\}$be integrable on $ [a,b]$

[hide="Hint"]$ min\{f(x),g(x)\} \equal{} \frac {f(x) \plus{} g(x) \minus{} |f(x) \minus{} g(x)|}{2}$
$ max\{f(x),g(x)\} \equal{} \frac {f(x) \plus{} g(x) \plus{} |f(x) \minus{} g(x)|}{2}$[/hide]


AMC
AIME
USA(J)MO
USAMO
conics
geometry
geometric transformation

What was your score for the AMC 12 B? Did you find it harder or easier than the 12 A?

itd be neat to post A scores too and compare.

i thought it was near the difficulty level of A...i just screwed up more on A

129 B, wasnt much different than my A score, a bit lower...  Definitely harder than A.  23 questions were accessible (to me, of course) on the A, while on the B the number was more like 21, at best...maybe i just suck.

I got 105 on 12A, 108.5 on 12B

How many do I need on the AIME to have a shot at USAMO? (Yes, it will be an insanely high number.... but hopefully still 11?)

#21 was a bit scary... conics w/ rotations! I don't believe I've ever seen a conic w/ rotation problem from the AMC the past 20 years.

i got a 110, and i thought it was harder than the A

webhappy, i'm in the same boat as you.. i need 11+ on the AIME.. oh this will be fun

oh i forgot to post mine! 130 AMC 12A, 132.5 AMC 12B. B definitely harder than A.

120.5 amc 12A
One twenty something amc 12B

Yes B was definitely much harder than A

Not enough geometry problems... I purposefully studied up on geometry (did some problems off of some Alabama state contest) the night before, expecting that to be my weakness. Hm... somehow I got the 3-D truncated cone one wrong. I thought my similar triangles worked.

Why does #22 look deceptively easy? I was running around in circles, though.

Our coach didn't let us keep the tests...what was #22??

I had a 123, I thought I had a 129 but then I missed the one of the 19 problems that I did.  I'm still happy, I'm only a freshman.

126.5

19 right, 5 unanswered, missed the probability one (grr, overcounted)

22 was magic square one, where the products of rows and columns and diagonals is equal.

I did horrible, 93.5. No way I made the AIME, but this one is alot harder than any AMCs to me. Do they give special considerations to 8th graders?

hm, is it powers of 2? I thought it was for randomly distributed things, like the beginning digit of lengths of rivers in some previously specified measurement, for some reason. whatever...

[quote="ComplexZeta"]Yeah, it is for randomly distributed things. But why wouldn't powers of two be randomly distributed?[/quote]

That's a pretty shady assumption...

[quote="ComplexZeta"]Do you think it's shady? It's off by less than .8 from the actual answer. That's much closer than you would need to be able to be convinced that they're randomly distributed. In general, things are randomly distributed unless there's a good reason that they shouldn't be.[/quote]

That's not entirely true. A lot of things that seem "random" aren't really all that random. But Benford's Law is amazing because it applies to an amazingly diverse set of things - including the leading digits of powers of two. In fact they were one of the early pieces of evidence that he used to form his conjecture.

But there are a lot of different measures of so-called "randomness". Even things that behave by Benford's Law don't all have the same random features as one another when looked at through different statistical measures.

The study of randomness and it's various meanings is an area of deep interest in mathematics (as well as philosophy, computer science, and psychology).

What I mean is that statistically if you look at digits of powers of two (or any other positive integer except one that clearly doesn't work), you'll get the usual distribution that you would by picking numbers at random. Of course, if you look at the units digit, that isn't true. But if you look at the leading digit or the next one or the next one, they'll all be randomly distributed. I can't prove it, but I'm sure it's true.

But what does it mean to "pick numbers at random"? The very meaning of randomness is a hotly debated topic, even among professional mathematicians, computer scientists, and philosophers. It's not as obvious a thing as it seems.

Hm. You have a good point. I guess by random I mean that they obey Bedford's Law. That isn't a very good way of defining things though.

I haven't seen this AMC test, so I can't comment on Question 25, but there are some mathematical theorems that confirm Benford's law in certain cases.  For example, let n be a (large) positive integer.  Consider the harmonic probability distribution on 1..n, which assigns the integer k a probability proportional to 1/k.  Then the probability that a random number in this distribution has leading digit d approaches log10(1 + 1/d) as n approaches infinity.

i got 109

i did 1-15, got #13 wrong because i subtracted a and b and got 0 (stupid!)

i'm pretty happy with my score since i'm a freshman and i've never taken an AMC before

...and i didn't take the 12A (my school didn't offer it)

I realized that for powers, Benford's law is easy to prove.  For example, I will prove that the density of nonnegative integers k such that 2k has leading digit d is log10(1 + 1/d).  I will rely on the following well-known result.  Let x be an irrational number.  Then the multiples of x, modulo 1, are uniformly distributed between 0 and 1.

Note that 2k has leading digit d if and only if log10(2k) mod 1 is between log10(d) [inclusive] and log10(d + 1) [exclusive].  But log10(2k) = k log10 2, and log10 2 is irrational.  Thus log10(2k) mod 1 is uniformly distributed between 0 and 1.  In particular, the density of its being between log10(d) and log10(d + 1) is log10(d + 1) - log10(d), which is log10(1 + 1/d).

Yeah. Also, for powers of two specifically, I think it's especially easy to prove since the density of numbers beginning with 1 is clearly log102 (since it has to hit every number of digits exactly once).

Good point.  Here's a challenge problem.  Among powers of two, what is the density of those whose second digit (from the left) is 1?

It will still follow Benford's Law, so it will be log(12/11)+log(22/21)+...+log(92/91).

Yes! It's around 11.4%.  The second digit is a lot closer to being uniformly distributed than is the first digit.


function
limit
calculus
derivative
calculus computations

Suppose the function $ g$ is defined by:
$ g(x) \equal{} k\sqrt {x \plus{} 1}$ for $ 0 \leq x \leq 3$

and $ mx \plus{} 2$ for $ 3 < x \leq 5$ (the function has the bracket but I"m not sure how to put in the latex.)

where $ k$ and $ m$ are constants. If $ g$ is differentiable at $ x \equal{} 3$, what are the values of $ k$ and $ m$?

Here is the definition of diferentiability I found in my book:
We assume $ f$ is differentiable at a point $ x \equal{} a$. THen we know that $ f'(a)$ exists where:
$ f'(a)$ = $ \lim_{x\to a}\frac {f(x) \minus{} f(a)}{x \minus{} a}$.

I have no idea how to show that it exists...

A function is differentiable on an interval $ [a, b]$ if it is continous on that interval and if its derivative exists everywhere on that interval.

For the function you are talking about, if it is differentiable at $ x \equal{} 3$, then the left-handed value of the function equals the right-handed value of the function AND the left-handed value of the derivative of the function equals the right-handed value of the derivative of the function.

So, basically, find $ g'(x)$... and then set each expression in the piecewise definition of $ g(x)$ equal to each other with $ x \equal{} 3$; then, set each expression in the piecewise definition of $ g'(x)$ equal to each other with $ x \equal{} 3$.  Thus, you get a system of two equations in two unknowns ($ m$ and $ k$) that can be solved by the method of your choice.

ok, I get it. Thanks!

In my book it says that a function fails to be differentiable at a point if:
1) The function is not continuous at the point.
2) The graph has a sharp corner at that point.
3) The graph has a vertical tangent line.

so by setting the pieces in g(x) equal to each other with x=3 solves 1) and by setting the derivatives of the pieces in g(x) equal to each other with x=3 solves 2) and 3).

I'm still confused with the bounds. Since x=3 is only for the first function ($ 0\leq x \leq 3$), why can we set both of the functions $ x \equal{} 3$?


The factors of $ x^4\plus{}64$ are:

$ \textbf{(A)}\ (x^2\plus{}8)^2 \qquad
\textbf{(B)}\ (x^2\plus{}8)(x^2\minus{}8) \qquad
\textbf{(C)}\ (x^2\plus{}2x\plus{}4)(x^2\minus{}8x\plus{}16) \\
\textbf{(D)}\ (x^2\minus{}4x\plus{}8)(x^2\minus{}4x\minus{}8) \qquad
\textbf{(E)}\ (x^2\minus{}4x\plus{}8)(x^2\plus{}4x\plus{}8)$

[hide="Click for solution"]
$ x^4\plus{}64\equal{}(x^4\plus{}16x^2\plus{}64)\minus{}16x^2\equal{}(x^2\plus{}8)^2\minus{}(4x)^2$ or $ (x^2\plus{}4x\plus{}8)(x^2\minus{}4x\plus{}8)$ which is $ \boxed{\textbf{(E)}}$.
[/hide]


floor function
logarithms
number theory proposed
number theory

Let $p$ be a prime number and $m$ be a positive integer. Show that there exists a positive integer $n$ such that there exists $m$ consecutive zeros in the decimal representation of $p^{n}$.

i have the following solution for $p>5$ but i think something's definitely wrong. please point it out

[hide="my solution"]
note that the probelm is equivalent to proving the existance of $N<p^{n}$ such that $p^{n}-N\equiv 0(\bmod 10^{m})$
let $\phi (10^{m})=4\cdot 10^{m-1}=k$. since $p>5$, $p^{k}\equiv 1(\bmod 10^{m})$
as a result $p^{n}\equiv p^{n}-p^{k}+1(\bmod 10^{m})$
so just take $N=p^{n}-p^{k}+1$.[/hide]

[quote="t0rajir0u"]Don't you mean $p^{n}-N \equiv 0 \bmod 10^{m+\lfloor \log_{10}N \rfloor+1}$?[/quote]

Well, why is that ?  :maybe:
I think Nayel's solution for $p > 5$ is true..

Well, nayel's equivalent problem is trivial.  For any $p^{n}> 10^{m}$ we will have $N$ such that $p^{n}\equiv N \bmod 10^{m}$ and $N < p^{n}$.  This proves nothing since we don't know how long $N$ is.


analytic geometry
graphing lines
slope
function
calculus
derivative
integration

Hey,

[b]1. The problem statement, all variables and given/known data.[/b]
I was reading through the Differential Equation portion of my textbook and didn't quite understand the following paragraph.
[quote="Textbook"]
[b]Slope Fields[/b]
Solving a differential equation analytically can be difficult or even impossible.
However, there is a graphical approach you can use to learn a lot about the solution of a differential equation. Consider a differential equation of the form
$ {{y}^{\prime}} \equal{} {{F}{(x, y)}}$
At each point $ {(x, y)}$ in the $ {xy}$-plane where $ {F}$ is defined, the differential equation determines the slope $ {{y}^{\prime}} \equal{} {{F}{(x, y)}}$ of the solution at that point. If you draw a short line segment with slope $ {{F}{(x, y)}}$ at selected points $ {(x, y)}$ in the domain of $ {F}$, then these line segments form a [b]slope field[/b] or a [i]direction field[/i] for the differential equation
$ {{y}^{\prime}} \equal{} {{F}{(x, y)}}$
Each line segment has the same slope as the solution curve through that point. A slope field shows the general shape of all the solutions.
[/quote]
The above paragraph seemed a little confusing since, conventionally,
$ {y} \equal{} {f(x)}$
Where,
$ {{\frac{{d}^{}}{d{x}^{}}}{\Big[y\Big]}} \equal{} {{{f}^{\prime}}{(x)}}$
so that it is understood that $ {{{f}^{\prime}}{(x)}}$ like $ {f(x)}$ is a function of [u]only[/u] $ {x}$. However, in the above paragraph it is insisted that $ {{\frac{{d}^{}}{d{x}^{}}}{\Big[y\Big]}}$ must be a function of $ {x}$ and $ {y}$. Why is that?

[b]2. Relevant equations.[/b]
Knowledge Differential Equations and Slope Fields.

[b]3. The attempt at a solution.[/b]
[hide="My Attempt at a Solution."]
If we begin from the conventional notation that,
$ {z} \equal{} {f(x, y)}$
I run in to the problem that I don't know how to explicitly find $ {{z}^{\prime}}$. Specifically, I don't know how to differentiate $ {f(x, y)}$ with respect to $ {x}$ and $ {y}$ [u]simultaneously[/u]. So that the derivative is actually a function of $ {x}$ and $ {y}$, like $ {f(x, y)}$. How would I differentiate $ {z}$ with respect to $ {x}$ and $ {y}$ [u]simultaneously[/u]?

I note however, that in this particular case we're talking about $ {{y}^{\prime}}$ as opposed to $ {{z}^{\prime}}$.
Noting this I recall that we can rewrite,
$ {\frac{dy}{dx}} \equal{} {{{f}^{\prime}}{(x)}}$
as
$ {dy} \equal{} {{{{f}^{\prime}}{(x)}}{dx}}$
Where $ {y}$ is found by integrating both sides of the above equation.

However, when I try to do this with the equation given,
$ {{y}^{\prime}} \equal{} {F(x, y)}$
Which can be rewritten as,
$ {dy} \equal{} {F(x, y)dx}$
and when integrated is,
$ {y} \equal{} {{\int_{}^{}}{F(x, y)dx}}$
I find that I do not know how to evaluate the RHS. How would I evaluate it?
[/hide]
Thanks,

-[u][b][url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=33715]APSStudent[/url][/b][/u]

Hey,
[quote="akech"]
The keyword is the Jacobian, that is, the derivative of a function of more than 1 variables can be expressed as a certain matrix.
[/quote]
Well, we can also define the Jacobian as,
$ {{\frac {\partial}{{\partial}{(u, v)}}}{\bigg[(x, y)\bigg]}} \equal{} {{\frac {{{\partial}^{}}{x}}{{\partial}{{u}^{}}}}{\frac {{{\partial}^{}}{y}}{{\partial}{{v}^{}}}} \minus{} {\frac {{{\partial}^{}}{y}}{{\partial}{{u}^{}}}}{\frac {{{\partial}^{}}{x}}{{\partial}{{v}^{}}}}}$

Where we see that we're taking the partial derivative as opposed to a normal derivative. In addition, we're taking the partial derivative of the variables $ {x}$ and $ {y}$ as opposed to a [b]function[/b] of $ {x}$ and $ {y}$ (ie. $ {f(x, y)}$). The latter is which I wanted to know how to do, in other words how would one define,
$ {{\frac {{d}^{}}{{d}{{(x, y)}^{}}}}{\bigg[{f(x, y)}\bigg]}} \equal{} {\text{???}}$

Where it is understood that we want the derivative of $ {f(x, y)}$ with respect to $ {x}$ and $ {y}$ simultaneously.

Thanks,

-[u][b][url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=33715]APSStudent[/url][/b][/u]

$ y$ is a function of $ x$, so $ F(x,y) \equal{} F(x,y(x)) \equal{} G(x)$. Of course, this is totally useless for helping you determine what $ y$ is since you need to know what $ y$ is beforehand. (As stated in the first line of what you quoted, it's difficult and sometimes impossible to solve general differential equations analytically, ie, the way you're proceeding.)

Could anyone explain the question of APSStudent ?, I also want to understand that.


Let $(a_n)$ be a sequence of real numbers with the property that $\sum_{n\geq 1}{a_nb_n}$ converges for any sequence $(b_n)$ of positive real numbers which decreases to 0. Then $\sum_{n\geq 1}{a_n}$ converges;

First of all, we may assume $a_n\ge 0$, because we can change the sign of the terms of $(b_n)$ as we please.

Assume $\sum a_n$ is not convergent. Then there are sequences $(k_n),(p_n)$ s.t. $k_n\to\infty,p_n\in\mathbb N^*$ and $|a_{k_n+1}+\ldots+a_{k_n+p_n}|\ge t$ for some positive $t$. Since $k_n\to\infty$, we can assume $k_{n+1}>k_n+p_n$. Now take $b_{k_n+1}=b_{k_n+2}=\ldots=b_{k_n+p_n}=\frac 1n$ (and anything you want for the rest of the indices, as long as $b_n$ converges to $0$ and is $\ge 0$; remember, we assumed $a_n\ge 0$), and when we sum everything up we get something larger than $t\cdot$ the harmonic series, which diverges.

grobber, you're answering a different question than the one harazi asked. Note the restriction to positive decreasing $b_n.$ We can't change signs as we please.

I don't see anything in the original statement about $a_n$ being positive. And with that, I believe the orginal statement is false.

Let $a_n=(-1)^n$. Then it is indeed true (it's the Alternating Series Test) that $\sum a_nb_n$ converges for all positive $(b_n)$ decreasing to zero, but $\sum a_n$ diverges.

[quote="grobber"]First of all, we may assume $a_n\ge 0$, [b]because we can change the sign of the terms of $(b_n)$ as we please[/b].

[/quote]

I think this shows pretty clearly what it is that I did wrong: I did not notice that $b_n$ had to be positive. I just assumed $b_n$ could be any sequence which converges to $0$.


Given $ a, b, c \geq\ 0$. Prove that: $ \frac{3(a\plus{}b\plus{}c)}{2(ab\plus{}bc\plus{}ca)} \geq\ \frac{a}{a^2\plus{}b^2} \plus{} \frac{b}{b^2\plus{}c^2} \plus{} \frac{c}{c^2\plus{}a^2}$

It's stronger than the well-known:
$ \sum\ \frac{a^3}{a^2\plus{}b^2} \geq\ \frac{\sqrt{3(a^2\plus{}b^2\plus{}c^2)}}{2}$
But I haven't found any solution for this stronger  :(

See here: http://www.mathlinks.ro/viewtopic.php?t=193004 (Read carefully the proof in my attachment, nguoivn ;) )


Arthur walks six blocks east and then 12 blocks north. If each block is one-third of a mile, how many miles did he walk?

6+12=18 blocks
each block is a 1/3 of a mile.
1/3*18=6

ANSWER: 6 BLOCKS.

[quote="nikeballa96"]6+12=18 blocks
each block is a 1/3 of a mile.
1/3*18=6

ANSWER: [b]6 BLOCKS[/b].[/quote]

i think u mean 6 miles


The problem asks to prove that all entire functions which are also injective take the form of $ f(z) \equal{} az\plus{}b$. As a hint, the book says to use the Casorati-Weierstrass theorem to $ f(1/z)$.

(The Casorati-Weierstrass theorem says that if a function $ f(z)$ is holomorphic in a punctured disk with a singularity, then the image of the punctured disk is dense in $ \mathbb{C}$)

Any suggestions or ideas would be nice. Thanks in advance.


group theory
abstract algebra
algebra
polynomial
induction
Galois Theory
superior algebra

Prove that $ \left[\mathbb{Q}\left(\sqrt{p_{1}},\sqrt{p_{2}},\cdots,\sqrt{p_{n}}\right):\mathbb{Q}\right] \equal{} 2^{n}$, where $ p_{i}$ are different primes.

Or equivalently: $ \sqrt{p_{n}}\notin\mathbb{Q}\left(\sqrt{p_{1}},\sqrt{p_{2}},\cdots,\sqrt{p_{n\minus{}1}}\right)$.

There are other approaches (someone will call them elementary) for this, but here's the short way ;) :

As $\mathbb Q$ (like every other field) contains the $2$-nd roots of unity $\pm1$, we may apply Kummer theory for the operator $\mathfrak p : x \mapsto x^{2}$:
The multiplicative subgroups $H$ of $\mathbb Q^{*}$ such that $(\mathbb Q^{*})^{2}\subseteq H \subseteq \mathbb Q^{*}$ correspond bijectively to the abelian extensions $L | \mathbb Q$ of exponent $2$ by the mapping $H \mapsto \mathbb Q [\sqrt H]$, $\sqrt H$ being the set of square roots of elements of $H$.

Setting $H_{k}= \{ p_{1}, ... , p_{k}\}$ we see that the corresponding extensions $L_{k}$ of $H_{k}$ are all different, giving the result.

Another approach:
It suffices to show that $\sqrt{p_{n}}\not\in \mathbb Q_{p_{n}}[\sqrt{p_{1}},...,\sqrt{p_{n-1}}]$. This follows as $\mathbb Q_{p_{n}}[\sqrt{p_{1}},...,\sqrt{p_{n-1}}] | \mathbb Q_{p_{n}}$ is unramified but $\mathbb Q_{p_{n}}[\sqrt{p_{n}}] | \mathbb Q_{p_{n}}$ is ramified.

Here something more "elementary":

$\mathbb{Q}(\sqrt{p_{1}}, ..., \sqrt{p_{n}})\cong\mathbb{Q}[x_{1}, ..., x_{n}]/(x_{1}^{2}-p_{1}, ..., x_{n}^{2}-p_{n})$. Every map which maps $x_{i}$ to $\pm x_{i}$ can be extended to a ringautomorphism of $\mathbb{Q}[x_{1}, ..., x_{n}]/(x_{1}^{2}-p_{1}, ..., x_{n}^{2}-p_{n})$, so we have (at least) $2^{n}$ automorphisms of $\mathbb{Q}(\sqrt{p_{1}}, ..., \sqrt{p_{n}})$ and so the degree must be at least $2^{n}$. Since $2^{n}$ is also an upper bound for the degree, the degree is exactly $2^{n}$.

Where does the isomorphism at the beginning come from¿
In fact, you never used any property of the $p_{k}$, so this can't be right as setting $p_{1}=p_{2}$ shows for example.

[quote="zhaobin"]Is it right?
Prove:$[\mathbb{Q}(\sqrt{p_{1}},\sqrt{p_{2}},\cdots,\sqrt{p_{n}}): Q]=2^{n}$
Where $p_{i}$ are different primes.

Or equivalently:$\sqrt{p_{n}}\notin \mathbb{Q}(\sqrt{p_{1}},\sqrt{p_{2}},\cdots,\sqrt{p_{n-1}})$[/quote]
(First accecpt induction.)

This is a repeated application of the theorem:
$[K: F] =[K: E][E: F]$

So it sufficies to show,
$\sqrt{p_{n}}\notin \mathbb{Q}(\sqrt{p_{1}},\sqrt{p_{2}},\cdots,\sqrt{p_{n-1}})$
Because that would imply that each field extension is of degree two, there are $n$ of them and hence $2^{n}$ is the resulting field extension.

But the problem is how do we show that. The idea is as follows:
1)Assume not.
2)Find an irreducible polynomial which has $\sqrt{p_{1}}+\sqrt{p_{2}}+...+\sqrt{p_{n}}$ are a root over $\mathbb{Q}$
3)Show that the $\mbox{deg}(\sqrt{p_{1}}+...+\sqrt{p_{n}}, \mathbb{Q}) = m$ and where $m \not | 2^{n-1}$.
4)This would contradiction a well know theorem.
5)Hence $\sqrt{p_{1}}+...+\sqrt{p_{n}}\not \in \mathbb{Q}(\sqrt{p_{1}},....,\sqrt{p_{n-1}})$.
6)Thus, $\sqrt{p_{n}}\not \in \mathbb{Q}(\sqrt{p_{1}},...,\sqrt{p_{n-1}})$.

But the though thing is step 2 and 3.

I am sorry for being sloppy because I got to run.

@ZetaX:
The properties of the p are contained in this isomorphism.

@Gockel: that's clear to me (why else should I have pointed to this...).
But what about proving that isomorphism¿ It's like directly claiming that the Galois group is $ (\mathbb{Z}/2\mathbb Z)^{n}$ and concluding from this. I don't think that isomorphism is completely trivial, so prove it please...

[quote="sylow_theory"]
2)Find an irreducible polynomial which has $\sqrt{p_{1}}+\sqrt{p_{2}}+...+\sqrt{p_{n}}$ are a root over $\mathbb{Q}$
3)Show that the $\mbox{deg}(\sqrt{p_{1}}+...+\sqrt{p_{n}}, \mathbb{Q}) = m$ and where $m \not | 2^{n-1}$.
4)This would contradiction a well know theorem.
.[/quote]
Thank you,but can you show me more detail?What is the well-known theorem in (4) you want to say?

@Zetax
Thank you very much,but I haven't learn Kummer theory,but I will learn it these days.
(And I am also not learn something about ramified.) :(

@-oo-
Can you translate it now?

[quote="zhaobin"]Thank you,but can you show me more detail?What is the well-known theorem in (4) you want to say?
[/quote]
I still dont know why you gave me a perfect six cuz I just gave a suggestion. But thanks!

Okay, typing all these \mathbb gets really confusing and I got to run I hope it wrote it understandable.

[b]Theorem:[/b] Let $F$ be a field and $E$ its extension field. Let $\alpha$ be algebraic over $F$ and form the simple extension $F\leq F(\alpha) \leq E$. If $\beta \in F(\alpha)$ then $\mbox{deg}(\beta, F )$ divides $\mbox{\deg}(\alpha, F)$.

Usually we use the contrapositive of this theorem to show something [b]is not[/b] contained in a field.

For example, say we have $\mathbb{Q}(\sqrt{p})$ where $p$ is a prime. And we we want to show that $\sqrt{q}\not \in \mathbb{Q}(\sqrt{p})$ where $q$ is another prime and distinct from $p$.
We will form $\mathbb{Q}(\sqrt{p}+\sqrt{q})$ and find its degree over $\mathbb{Q}$. So let $x = \sqrt{p}+\sqrt{q}$ then $x^{4}-2x^{2}(p+q)+[(p+q)^{2}-4pq]$. This polynomial is irreducible over $\mathbb{Q}$ :maybe: . That means, $\sqrt{p}+\sqrt{q}\not \in \mathbb{Q}(\sqrt{p})$ because if it were by theorem $4$ would divide $2$ which is implossible. Thus, $\sqrt{p}+\sqrt{q}\not \in \mathbb{Q}(\sqrt{p})$ which means $\sqrt{q}\not \in \mathbb{Q}(\sqrt{p})$.

[quote="sylow_theory"]$x^{4}-2x^{2}(p+q)+[(p+q)^{2}-4pq]$. This polynomial is irreducible over $\mathbb{Q}$ :maybe: . .[/quote]
Hi,Thank you,I just think it is a nice approach that consider the sum of the square have the degree $2^{n}$ over $Q$.So I give you six cuz.

But I don't think that the polynomial above  is irreducible over $\mathbb{Q}$ is trivial when we do the $2^{n}$ case.

ok, here the "translation".  :-)

to be honest: I don't think that the other approaches mentioned here work, of course except the ones 'beating' by means of kummer or ramnification theory :-).

Thank you -oo- .
That is a technological induction :)

[quote="calc rulz"]Can't we just adjoin roots one at a time, show that the extension is degree $ 2$ each time, and then use the tower law?[/quote]
That is exactly what the entire discussion is about, yes.

brute force solution.
by induction: veiw an element of $ Gal(Q(\sqrt{p_1},\cdots,\sqrt{p_{n\minus{}1}})/Q)$ a correspondence to a subset of N={1,2,……,n-1}.
denote $ \alpha_J\equal{}\prod_{k\in J\subseteq N} \sqrt{p_k}$.
if $ \sqrt{p_n}\equal{}\sum_{J\subseteq N}C_J\alpha_J$, where at least one of $ C_J\in Q$ is non zero.

suppose $ p_1\in J'$ that $ C_J'\neq 0$. since $ \sigma (\sum_{J\subseteq N}C_J\alpha_J)\equal{}\pm \sum_{J\subseteq N}C_J\alpha_J$, if $ p_1\in J'$ it must also appear in other non-zero term. apply this for other $ p_k$, we know that there is only one term in the right.

now the proposition follows from the well-known fact that $ \prod \sqrt{p_i}\notin Q$.


What is the escape speed from an asteroid of diameter 370 m  with a density of 2130 kg/m^3 ?

Escape velocity is given by $ v^{2} \equal{} \frac{2Gm}{r}$.

the answer is different for different place!in fact we must find the number of masive  bodys That have gravitional field.but the  escape velocity for this asteroid in solar system is:
delta vee=(2^0.5-1)((6.67e-11*1.99e30)/R^2)^0.5
 :wink:

could u actually use latex!
it really quite confusing?

I actually thought the way airno wrote the equation was quite understandable. LaTeX would make it a little easier on the eyes, but it's not that important in this case. I think what airno meant to say was that escape velocity also had to take into account all of the other objects in the vicinity with considerable gravitational pull. Usually, though, the pull of the other objects (sun, moon, asteroids, etc.) is negligible because of their distance, at least in the case of escape velocity.

Also, just as a heads-up to rt_08: this topic was four months old. We don't generally contribute to topics as old as that unless we have new, important information to add. Though your desire and enthusiasm to help were commendable, your suggestion to use LaTeX didn't really contribute to the discussion.

ok now i would take this point into consideration


function
integration
calculus
real analysis
real analysis unsolved

Let a seismic disturbance travel with a speed which is a linear function of its depth $ \eta$ below the earth's surface. It travels along a subterranean path which minimizes its time of travel between fixed points $ ((a,a_1),(b,b_1))$.
Assume that a typical planar path joining the points is of a continuously differentiable function $ y(x)\equal{}\eta(x)\plus{}\sigma, a \leq x \leq b$. ($ \sigma$ is a constant and $ \eta (x)$ is the positive local depth of the path.

a)Show the time of travel, for some constant $ p$, is $ T(y)\equal{}p \int_a^b \left( \frac{\sqrt{1\plus{}y'(x)^2}}{y(x)}\right )dx$

I am stuck on deriving a formula for the speed. I am not sure if its $ v(\eta(x))\equal{}\alpha \eta(x) \plus{} \beta$    or    $ v(y(x))\equal{}\alpha \eta(x) \plus{} \beta$ and what to even do from there.

The arc length part (numerator) is straight forward.
Thanks!

Apparently they suggest that you assume that the speed at depth $ d$ is just $ c(d\plus{}\sigma)$. Otherwise I do not see how the formula you quoted can hold.

I take it you're using $ c(d \plus{} \sigma)$ as $ v(\eta(x) \plus{} \sigma)$ for $ v$ being the speed function? Or are each of those letters a constant.
Could you show me how it does not hold otherwise? Just to see whats going on.

Here is a url for the problem [url=http://books.google.ca/books?id=7RpC-h9pV5sC&pg=PA34&lpg=PA34&dq=Variational+calculus+and+optimal+control:+optimization+with+elementary+convexity+seismic&source=bl&ots=VLeQvqn7RO&sig=2se8l4hIWgltKu8wE1FL8dyT6-s&hl=en&ei=XXvBSriNBMOj8AbJ78WFBg&sa=X&oi=book_result&ct=result&resnum=1#v=onepage&q=&f=false]Seismic wave[/url]

[quote="Fermat's Little Turtle"]I take it you're using $ c(d \plus{} \sigma)$ as $ v(\eta(x) \plus{} \sigma)$ for $ v$ being the speed function? [/quote]
I mean, if you are given the depth, you get the speed at this depth by adding $ \sigma$ and multiplying by a constant factor. The expression $ \eta(x)$ is the depth of the trajectory at the horizontal position $ x$, so, [i]for this particular trajectory[/i], the speed will be a constant times the sum $ \eta(x)\plus{}\sigma$, i.e., a constant multiple of $ y(x)$.

Since the time is the distance divided by the speed, and the numerator has the meaning of the distance, the denominator $ y(x)$ must be the speed (or a constant multiple thereof) for the formula to hold.


geometry
circumcircle
incenter
projective geometry
cyclic quadrilateral
geometry theorems

[quote][color=darkred][b]Lemma I.[/b] The circle $w$ is internally tangent in the point $T$ to the circle $\mathcal C$. For a point $M\in w$ denote the chord $[AB]$ of the circle $\mathcal C$ which is tangent in the point $M$ to the circle $w$. Then the ray $(TM$ is the (internal) bisector of the angle $\widehat{ATB}$.
[hide="Proof of lemma I."]Denote the second intersection $N$ of the line $TM$ with the circle $\mathcal C$, the tangent $TX$ to the circle $w$ and the tangent $NY$ to the circle $\mathcal C$ so that the line $\overline{TMN}$ doesn't separate the points $X\ ,\ Y$. From the relations $\widehat{TNY}\equiv\widehat{NTX}\equiv\widehat{TMB}$ obtain $\widehat{TNY}\equiv\widehat{TMB}$, i.e. $AB\parallel NY$. Therefore, $NA=NB$, i.e. the ray $(TN$ is the (internal) bisector of the angle $\widehat{ATB}$.[/hide]
[b]Lemma II.[/b] The circle $\Omega$ is externally tangent in the point $T$ to the circle $\mathcal C$. For a point $M\in \Omega$ denote the chord $[AB]$ of the circle $\mathcal C$ which is tangent in the point $M$ to the circle $\Omega$. Then the ray $(TM$ is the external bisector of the angle $\widehat{ATB}$.[hide="Proof of lemma II."]Denote the second intersection $N$ of the line $TM$ with the circle $\mathcal C$, the tangent $TX$ to the circle $w$ and the tangent $NY$ to the circle $\mathcal C$ so that the line $\overline{TMN}$ doesn't separate the points $X\ ,\ Y$. From the relations $\widehat{TNY}\equiv\widehat{NTX}\equiv\widehat{TMB}$ obtain $\widehat{TNY}\equiv\widehat{TMB}$, i.e. $AB\parallel NY$. Therefore, $NA=NB$, i.e. the ray $(TN$ is the external bisector of the angle $\widehat{ATB}$.[/hide]

$1\blacktriangleright$ Let $ABC$ be a triangle and consider the circle $\omega$ which is tangent to the [b]segments[/b] $(AB) ,\ (AC)$ in the points $M\ ,\ N$ respectively and which is [b]internally[/b] tangent in the point $T$ to the circumcircle of the triangle $ABC$. Prove that the line $MN$ passes through the [b]incenter[/b] $I$ of the triangle $ABC$.

$2\blacktriangleright$ Let $ABC$ be a triangle and consider the circle $\Omega$ which is tangent to the [b]rays[/b] $(AB\ ,\ (AC$  in the points $M\ ,\ N$ respectively and which is [b]externally[/b] tangent in the point $T$ to the circumcircle of the triangle $ABC$. Prove that the line $MN$ passes through the[b] exincenter[/b] $I_{a}$ of the triangle $ABC$.[/color][/quote]

[color=darkblue][b][u]Proof.[/u][/b]

$1\blacktriangleright$ Denote the second intersections $U\ ,\ V$ of the circumcircle of $\triangle ABC$ with the lines $TM\ ,\ TN$ respectively.

[b]Is well-known (see Lemma I) [/b]that $I\in BV\cap CU$. Apply the [u]Pascal's theorem[/u] to the cyclical hexagon $CABVTU$ : $\left\{\begin{array}{c}N\in CA\cap VT\\\ M\in AB\cap TU\\\ I\in BV\cap UC\end{array}\right\|\implies I\in MN$.

$2\blacktriangleright$ Denote the second intersections $U\ ,\ V$ of the circumcircle of $\triangle ABC$ with the lines $TM\ ,\ TN$ respectively.

Denote the diameters $[UR]\ ,\ [VS]$ in the circumcircle of $\triangle ABC$. [b]Is well-known (see Lemma II) [/b]that $I\in BS\cap CR$ and $I_{a}\in BV\cap CU$.

Apply the [u]Pascal's theorem[/u] to the cyclical hexagon $CABVTU$ : $\left\{\begin{array}{c}N\in CA\cap VT\\\ M\in AB\cap TU\\\ I_{a}\in BV\cap UC\end{array}\right\|\implies I_{a}\in MN$.[/color]

[color=darkblue]I saw that the proof of the lemma II is unclearly with some mistakes. Lower I modified it.[/color]

[color=darkred][b]Lemma II.[/b] The circle $ \Omega$ is externally tangent in the point $ T$ to the circle $ \mathcal C$. For a point $ M\in \Omega$ denote the chord $ [AB]$ of the circle $ \mathcal C$ which is tangent in the point $ M$ to the circle $ \Omega$. Then the ray $ (TM$ is the external bisector of the angle $ \widehat{ATB}$.[/color]

[color=darkblue][b][u]Proof.[/u][/b] Denote the second intersection $ N$ of the line $ TM$ with the circle $ \mathcal C$, the common tangent $ t$ to these circles in the common point $ T$, the tangent $ n$ in the point $ N$ to the circle $ \mathcal C$ and the intersection $ X\in t\cap n$. Choose a point $ Y\in t$ so that $ T\in (XY)$. From the relations $ \widehat{TNX}\equiv\widehat{NTX}\equiv\widehat{MTY}\equiv\widehat {TMA}$ obtain $ \widehat{TNX}\equiv\widehat{TMA}$, i.e. $ AB\parallel NX$. Therefore, $ NA \equal{} NB$, i.e. the ray $ (TN$ is the external bisector of the angle $ \widehat{ATB}$.[/color]

[b][u]Alternate proof of Lemma 1:[/u][/b]
From an inversion with centre $T$ and radius $\lambda>1$, the larger circle maps to a line $n$ and the smaller circle maps to a line $m$ parallel to it. $A$ and $B$ maps to $A'$ and $B';$ and the line $AB$ maps to the circumcircle of $TA'B'$ which tangents $m$ at $P'$. Let $X$ be a point on $m$ such that $\angle XP'A'<\angle XP'B'$. From alternate segments theorem, $\angle XP'A'=\angle A'TP'$ and also since $m\parallel n,$ we have $\angle XP'A'=\angle P'A'B'$. From the above two equations, we have $\angle A'TP'=\angle P'A'B'=\angle P'TB'$ and when we invert the figure back, the lines $A'T, P'T$ and $B'T$ are mapped to itself and hence, by conformality, $\angle A'TP'=\angle P'TB'\Longrightarrow \angle ATP=\angle PTB\Longrightarrow PT$ bisects $\angle ATB$ as required.

[b][u]Alternate proof of Lemma 2:[/u][/b]
Let $TB$ extended meet $\Omega$ at $K$. From an inversion with centre $T$ and radius $\lambda>1$, let $X$ map to $X'$ for any arbitrary figure $X$. So, $\Omega'$ and $\mathcal{C}'$ are parallel lines on opposite sides of $T$ and $K'TB'$ is a transversal. $A'$ is some point on $\mathcal{C}'$ and the circumcircle of $A'TB'$ tangents $\Omega'$ at $M'$. By conformality, it suffices to prove that $\angle M'TK'=\angle M'TA'$. By alternate segments theorem, $\angle M'A'T=\angle TM'K'$. Since $\Omega'\parallel\mathcal{C}',$ we have $\angle A'B'K'=\pi-\angle M'K'T=\angle K'M'T+\angle M'TK'=\angle M'A'T+\angle M'TK'$. In cyclic quadrilateral $M'TB'A'$, we have $\angle TM'A'=\pi-\angle AB'K'=\pi-(\angle M'A'T+\angle M'TK')$. Now, consider $\angle M'TA'=\pi-(\angle M'A'T+\angle TM'A')=\angle M'TK'$ and hence, we have proved the result.

I found proofs without using the lemmas for your problems.

[quote="Virgil Nicula"][color=darkred]$1\blacktriangleright$ Let $ABC$ be a triangle and consider the circle $\omega$ which is tangent to the [b]segments[/b] $(AB) ,\ (AC)$ in the points $M\ ,\ N$ respectively and which is [b]internally[/b] tangent in the point $T$ to the circumcircle of the triangle $ABC$. Prove that the line $MN$ passes through the [b]incenter[/b] $I$ of the triangle $ABC$.[/color][/quote]

Consider an inversion with centre $A$ and radius $\lambda>1$. $B$ maps to $B'$; $C$ maps to $C'$ in the lines $AB$ and $AC$ respectively. $M, N$ map to $M', N'$ in the exterior of $AB', AC'$ respectively. The circumcircle of triangle $ABC$ maps to the line $B'C'$ and $\omega$ maps to the excircle of $AB'C'$ tangential to $AB', AC', B'C'$ at $M', N', T'$ respectively. Let $I_A$ be the excentre of triangle $AB'C'$ opposite $A$. So, $A, I, I_A$ are collinear. We have $\angle ABI=\frac{B}{2}$. Since $AB\cdot AB'=AC\cdot AC', B, C, C', B'$ are concyclic$\Longrightarrow \angle AB'C'=\angle ACB$ and hence $\angle AI_AB'=\frac{B}{2}$. Also, $\angle BAI=\angle B'AI_A\Longrightarrow \Delta BIA\sim I_AB'A$ and hence, $AB\cdot AB'=AI\cdot AI_A$ which means $I$ maps to $I_A$. In quadrilateral $AN'I_AM', \angle AN'I_A=\angle AM'I_A=90^{\circ}\Longrightarrow AN'I_AM'$ is cyclic which means when we invert the points back, the preimages of $M', I_A, N'$ are collinear and hence $M, N, I$ are collinear as required.

[quote="Virgil Nicula"][color=darkred]$2\blacktriangleright$ Let $ABC$ be a triangle and consider the circle $\Omega$ which is tangent to the [b]rays[/b] $(AB\ ,\ (AC$  in the points $M\ ,\ N$ respectively and which is [b]externally[/b] tangent in the point $T$ to the circumcircle of the triangle $ABC$. Prove that the line $MN$ passes through the[b] exincenter[/b] $I_{a}$ of the triangle $ABC$.[/color][/quote]

Consider an inversion with centre $A$ and radius $\lambda>1$. $B$ maps to $B'$; $C$ maps to $C'$ in the lines $AB$ and $AC$ respectively. $M, N$ map to $M', N'$ in the interior of $AB', AC'$ respectively. The circumcircle of triangle $ABC$ maps to the line $B'C'$ and $\Omega$ maps to the incircle of $AB'C'$ tangential to $AB', AC', B'C'$ at $M', N', T'$ respectively. Let $I$ be the incentre of triangle $AB'C'$. So, $A, I, I_A$ are collinear. We have $\angle AI_AB=\frac{C}{2}$. Since $AB\cdot AB'=AC\cdot AC', B, C, C', B'$ are concyclic$\Longrightarrow \angle AB'C'=\angle ACB$ and hence $\angle AIB'=\frac{C}{2}$. Also, $\angle BAI_A=\angle B'AI\Longrightarrow \Delta BI_AA\sim IB'A$ and hence, $AB\cdot AB'=AI_A\cdot AI$ which means $I_A$ maps to $I$. In quadrilateral $AN'IM', \angle AN'I=\angle AM'I=90^{\circ}\Longrightarrow AN'IM'$ is cyclic which means when we invert the points back, the preimages of $M', I, N'$ are collinear and hence $M, N, I_A$ are collinear as required.

Dear Virgil and Mathlinkers,
the first problem is a nice application of the lemme 1. Other applications?
Just for making a little recall, a generalization of this problem is the Sawayama's lemma.
For this one you can see in English
http://perso.orange.fr/jl.ayme  vol. 1 the Sawayama and Thebault theorem.
Sincerely
jean-Louis

See the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=31739&start=0]generalisation of problem 1[/url]. Comparing my proof for that, we can see that for the second problem, $I_A$ is the midpoint of $MN$


function
topology
real analysis
real analysis theorems

let fn be a sequence of functions
if fn are continuous and uniformly converge to f on EACH COMPACT SUBSET of a metric space X, does that mean that it's also true for all of X, since {x} is a compact subet of X for all x in X?

(i am ultimately trying to show that f itself is continuous, and I thought I would try to make use of the theorem which says that if fn are continuous and converge uniformly to f on X, then f is continuous on X.

any suggestions will be appreciated. thanks.

[quote="badatanalysis"]let fn be a sequence of functions
if fn are continuous and uniformly converge to f on EACH COMPACT SUBSET of a metric space X, does that mean that it's also true for all of X, since {x} is a compact subet of X for all x in X?[/quote]

Uniform convergence is something happening globally: for all epsilon, there exists delta such that for every $ x$, etc.  Just looking at a single value of $ x$ doesn't tell you anything about this.  For each epsilon and each $ x$, you'll have a different delta, and that doesn't necessarily give you a way to choose a global delta.  This is, in fact, why we need a notion of uniform convergence in the first place: because sometimes looking at each individual point isn't enough.

No. $ f_k(x)\equal{}x/k$ converges uniformly (to $ 0$) on every compact subset of $ \mathbb R$ but not on the entire $ \mathbb R$. By the way, unless you are somebody from a faraway place who specially came to Madison to post this question and not who I think you are, the exam including this problem is due tomorrow so you'll have to wait for the full answer until Tuesday ;) Here is a small hint though: you can check continuity by considering values on convergent sequences.

i had suspicions that you were magic, but now i know for sure!  thanks (and may no one ever climb up the beanstock and infiltrate your magical castle :ninja:)

This might be a little late but you need $ X$ to be compact and then your statement is true, I think.

That would be assuming away most of the problem, though.

I tried to find a counterexample in a topological space, but  :|