geometry incenter trigonometry circumcircle inradius

Can any show me how to figure out the raduis of an inscribed circle inside a triangle, or a circumscriced circle around a triangle (I think that's what it's called)?

I know that in a right triangle, the hypotenuse is the diameter of the circumscribed circle, but that's all I know.

Let there be triangle ABC with an inscribed circle with center P. Now draw radii PD, PE, and PF to sides AB, BC, and AC. Let the radius be r. Now draw PA, PB, and PC. Since the circle is tangent to all sides PD, PE, and PF are all the altitudes of triangles APB, BPC, and CPA, respectively. So, we can say that

Q (Area of ABC) = Area of triangles APB, BPC, and CPA
Q = PD(AB)/2 + PE(BC)/2 + PF(AC)/2
Q = AB(r)/2 + BC(r)/2 + AC(r)/2
2Q = r(AB+BC+AC)
2Q/(AB+BC+AC) = r

So, one could think of it this way...

The radius of an inscribed circle in a triangle is equal to two times the area divided by the perimeter.

Hope this helps!

The incentre of a triangle is the point at which the three angle bisectors of the triangle meet.

The circumcentre of a triangle is the point at which the three perpendicular bisectors of the triangle meet.

[quote="Rep123max"]For the circumscribed circle, you can get the area in terms of the three sides and R for the radius. Basically, you have to re-work $A=\frac{ab(\sin c)}{2}$ by using the Law of Sines which states that $\frac{a}{\sin a}=\frac{b}{\sin b}=\frac{c}{\sin c}=2r$.[/quote]

Is there an explanation that doesn't use sine, cosines, and tangents?  I know what they are, but I have no idea on how to use them on un-right-angle triangles.

[tex]\displaystyle{{abc}\over{4R}}=sr=K[/tex], where a, b, c are the three sides of the triangle, [tex]s = {{1}\over{2}}(a+b+c)[/tex], R is the circumradius, r is the inradius, and K is the area of the triangle.

A little more info on circumscribed circles:

http://mcraefamily.com/MathHelp/GeometryTriangleCircumscribedCircle.htm


group theory
abstract algebra
superior algebra
superior algebra unsolved

a group G is called T-groups if every subnormal subgroup of G is normal.
Why the groups abelian are T-groups?
Why the groups simple are T-groups?
many thanks
sastra :D

Thats kind of trivial...
Every subgroup of an abelian group ist normal and therefore subnormal.
Since there are just the trivial normal subgroups in simple groups, there are also just two subnormal subgroups, namely the trivial ones {1} and G itself.


limit
inequalities
real analysis
real analysis unsolved

Let $ (x_n)_n\ge{1}$ so that $ \lim_{n\to\infty} \frac{(x_1)^2\plus{}(x_2)^2\plus{}(x_3)^2\plus{}\cdot \cdot \cdot \plus{} (x_n)^2}{n}\equal{}0$

Calculate $ \lim_{n\to\infty} \frac{n\cdot{x_1}\plus{}(n\minus{}1)\cdot{x_2}\plus{}(n\minus{}2)\cdot{x_3}\plus{}\cdot \cdot \cdot 1 \cdot{x_n}}{n^2\plus{}n}$

Sorry for the mistake
Let $ (x_n)_{n\ge{1}}$ so that $ \lim_{n\to\infty} \frac{(x_1)^2\plus{}(x_2)^2\plus{}(x_3)^2\plus{}\cdot \cdot \cdot \plus{} (x_n)^2}{n}\equal{}0$

Calculate $ \lim_{n\to\infty} \frac{n\cdot{x1}\plus{}(n\minus{}1)\cdot{x_2}\plus{}(n\minus{}2)\cdot{x_3}\plus{} \cdot \cdot \cdot \plus{}1 \cdot{x_1}}{n^2\plus{}n}$

After you've made a post, an "edit" button appears (on the AoPS skin, it's right next to the button the "quote" button).  This allows you to edit your post after submitting it.  (This ability goes away after a short period of time (36 hours?).)  Before you submit, there is a "Preview" button that allows you to check that your post is correct and complete.

From Cauchy Schwartz inequality we have :

$ \frac{x_{1}^2\plus{}...\plus{}x_{n}^2}{n} \geq \left( \frac{x_{1}\plus{}...x_{n}}{n} \right)^2$

Therefore $ \lim _{n \rightarrow \infty }a_n \equal{} 0$ where $ a_n \equal{} \frac{1}{n} \sum _{k\equal{}1}^n x_{k}$
Since

$ \frac{nx_{1}\plus{}...\plus{}x_{n}}{n^2\plus{}n}<\frac{n}{n\plus{}1}\cdot a_n$ the limit we seek is 0

Why    $ \frac{ n\cdot{x_1}\plus{}\cdot \cdot \cdot \plus{} x_n}{n^2\plus{}n}<\frac{n}{n\plus{}1} \cdot a_n$ because if $ (x_n)_{n\ge{1}}$ has negative elements the inequality isn't true

[quote="tarta"]Why    $ \frac { n\cdot{x_1} \plus{} \cdot \cdot \cdot \plus{} x_n}{n^2 \plus{} n} < \frac {n}{n \plus{} 1} \cdot a_n$ because if $ (x_n)_{n\ge{1}}$ has negative elements the inequality isn't true[/quote]

Yes you are right.I m sorry.I  solved it with x_k >0 in my mind.


Let $ a\equal{}b\equal{}c \rightarrow 0^{\plus{}}$ then the inequality is wrong.

[quote="lipako"]$ \sum\frac {b^5 \plus{} 3a^3 \plus{} c^5 \plus{} abc}{b^3 \plus{} c^3}\geq\frac {19}{3}$

with

$ a,b,c > 0$[/quote]

Dear Lipako the inequality is correct if $ a\plus{}b\plus{}c\equal{}1$.


how do u convert polar equations to rectangular?

r=4/1-cosX

r^2=cosX

$ r\equal{}\sqrt{ x^{2} \plus{} y^{2} }$ and $ \theta \equal{} \arctan \left( \frac{y}{x} \right)$?

i no like for the first one i crossed multiply and got r-rcosX=4 which turns into r-x=4 then can you make the r (square root x^2 +y^2) -x=4 and leave it like that??


how do you become a mod for a forum topic

The admins choose. If you ask you will probably get rejected. Instead, if they have open spots, they look for people who are active and helpful on the forum, and also look at things like rating, post content, etc. to choose.

Of course, levans said here they're no longer really looking for mods :(
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=219311[/url]

Which is sad cause I wanted to be one maybe in 5 years or so..............

That post is for For the Win mods. This is not related to message board moderators. Although I don't think we are currently seeking any message board moderators either.

So, well, even though I'm not a mod... I made something I can live by on AoPS.
[b]Be responsible, be helpful, be smart.[/b]
Take that to mind, I joined a couple months ago and started out as a semi-spammer but now I've become more helpful to other people.


If $ x,y,z$ are nonnegative numbers such that $ x \plus{} y \plus{} z \equal{} 3,$ then

$ \left(x \plus{} y^2\right)\left(y \plus{} z^2\right)\left(z \plus{} x^2\right)\leq13 \minus{} 7xyz \plus{} 2x^2y^2z^2,$

with equality if and only if $ x \equal{} y \equal{} z \equal{} 1.$
[hide="Proof."]$ 13\left(\frac {x \plus{} y \plus{} z}{3}\right)^6 \minus{} 7xyz\left(\frac {x \plus{} y \plus{} z}{3}\right)^3 \plus{} 2x^2y^2z^2$

$ \minus{} \left[x\left(\frac {x \plus{} y \plus{} z}{3}\right) \plus{} y^2\right]\left[y\left(\frac {x \plus{} y \plus{} z}{3}\right) \plus{} z^2\right]\left[z\left(\frac {x \plus{} y \plus{} z}{3}\right) \plus{} x^2\right]$

$ \equiv\frac {F(x,y,z)}{729}.$

$ F(x,y,z) \equal{} F(x,x \plus{} s,x \plus{} t)$

$ \equal{} 243x^4\left(s^2 \minus{} st \plus{} t^2\right) \plus{} 162x^3(s \plus{} t)\left(4s^2 \minus{} 7st \plus{} 4t^2\right)$

$ \plus{} 81x^2\left[(s \minus{} t)^2\left(7s^2 \plus{} 12st \plus{} 7t^2\right) \plus{} s^2t^2\right]$

$ \plus{} \frac {9x}{800000}\left[(19s \minus{} 20t)^2\left(37513s^3 \plus{} 93255s^2t \plus{} 132600st^2 \plus{} 34000t^3\right) \plus{} 57807s^5 \plus{} 444825s^4t\right]$

$ \plus{} \frac {s \plus{} t}{3125}\left[(4s \minus{} 5t)^2\left(2453s^3 \plus{} 1995s^2t \plus{} 10725st^2 \plus{} 1625t^3\right) \plus{} 1377s^5 \plus{} 16200s^4t\right]$

$ \geq 0,$

which is clearly true for $ x \equal{} \min\{x,y,z\}.$[/hide]
With the same conditions, the stronger inequality

$ 5\left(x \plus{} y^2\right)\left(y \plus{} z^2\right)\left(z \plus{} x^2\right)\leq64 \minus{} 33xyz \plus{} 9x^2y^2z^2$

holds too.

See also here : http://www.mathlinks.ro/viewtopic.php?t=208388


geometry
3D geometry
tetrahedron
sphere
trigonometry
trig identities
Law of Cosines

A tetrahedron $ABCD$ is inscribed in the sphere $S$. Find out the locus of the points $P$, situated in $S$, for which
\[ \frac{AP}{PA_{1}} + \frac{BP}{PB_{1}} + \frac{CP}{PC_{1}} + \frac{DP}{PD_{1}} = 4 \]  where$\ A_{1}, B_{1}, C_{1}, D_{1}$ are the other intersection points of $AP$,$BP$,$CP$,$DP$ with $S.$

Let the $S$ have center $O=(0,0,0)$ and radius $r$.

We multiply both sides of the condition by the power of $P$ with respect to the sphere $S$ to obtain \[AP^{2}+BP^{2}+CP^{2}+DP^{2}= 4(r^{2}-OP^{2}).\] We now use the law of cosines to obtain \begin{eqnarray*}\sum_{\mbox{cyc}}[r^{2}+OP^{2}-2 OA \cdot OP \cos{(AOP)}] &=& 4(r^{2}-OP^{2})\\ 4 OP^{2}&=& \sum_{\mbox{cyc}}[OA \cdot OP \cos{(AOP)}] \end{eqnarray*} Now if we let $P = ( x,y,z ), A = (a_{1}, a_{2}, a_{3}), B = ( b_{1}, b_{2}, b_{3})$, etc., we can write this as \[4 ( x^{2}+y^{2}+z^{2}) = ( a_{1}+b_{1}+c_{1}+d_{1})x+( a_{2}+b_{2}+c_{2}+d_{2})y+( a_{3}+b_{3}+c_{3}+d_{3})z,\] whereupon it becomes apparent that if we complete the square, we will obtain an equation for a sphere with center halfway between $O$ and the centroid of $ABCD$, and with radius half of the distance from $O$ to the centroid of $ABCD$.  Therefore this is the desired locus, Q.E.D.


I recall that a while back there were a few religion-related polls, and I was thinking that perhaps we could have a more in-depth discussion here.  I'd like to start this off by asking those interested to respond to a couple of questions:
1.  What are your religious beliefs?
2.  Why?  (This can perhaps start off with little being said and develop as the discussion goes on.)

I hope that eventually this will develop into a discussion/debate, so I'd like to try to lay a few ground rules so that we can all get along.
1.  Please make coherent posts.  Grammar, spelling, etc. is helpful.
2.  No name-calling or general insults.
3.  Try to repeat yourself as little as possible.  We want this to be an active discussion, but we don't want to have so much going on that no one can keep up with it.

I'll start this off.  I consider myself to be a moderate to conservative Protestant Christian (although I'm more liberal politically - not that that is particularly relevant to the topic).  I belong to a Baptist church, but I don't like what the SBC has been doing recently, esp. as regards its position on women.   I may elaborate some more later.
Why (a quick overview):  Logical consistency of the doctrines, and the fact that for the most part they "ring true" with the world and philosophical expectations (The purely theistic part especially with philosohphical expectations, but to an extent many specifically Christian doctrines (the Trinity, etc.) as well.  I may elaborate on this later too.)  The evidence for the truth of the historical claims made (ministry, crucifiction, empty tomb, etc.   The resurrection itself has not so much historical evidence as testimonial: Paul refers to more than 500 people who had seen the risen Christ (1 Cor. 15:7), and the dedication of the early church to that doctrine - see the following:), esp. regarding the fact that the early church was situated immediately thereafter when such claims could be easily refuted - and were not.  The success and dedication of the early church to the unpopular (wow, that is an understatement) belief in Jesus as Savior and as divine, in such a potentially hostile (evidentially) and certainly hostile (socially) environment, especially when the claims made by the Apostles were made based on evidence and prior theology (Acts 17:11, Acts 18:28, for instance), rather than as is common today a "just feel it and believe" attitude, is remarkable evidence for the truth of their beliefs.  (There is more evidence than that which I have not mentioned, of course.  Just thought I'd mention that. ;)  Yes, I did just use the Bible as a history of the early church.  That's what Acts is, and what Paul's letters can serve as.  This is different than believing [i]doctrine[/i] simply "because the Bible says so."  The basics, at least, of doctrine need more support than the "just believe it" argument.)  Finally, there is personal experience which, though certainly not conclusive in and of itself, together with the philosophical/logical and historical evidence forms the primary basis for my faith.

Wow, that was one long paragraph.

Let's hear from everyone else now!

[i]Grammar edit.[/i] :oops:

i have no religious ties to anything and don't really believe in a supreme being. my reasoning- it comes from a simpsons book.
I think that god doesn't have the ability to do everything.
If god can finish eating anything, not matter how big, then god can't make a hot dog so big, that even he can't eat.

Denominationally, I am Unitarian Universalist.  This at once says a lot about my religious beliefs and doesn't say very much, so I'll have to elaborate.  For those who aren't familiar with UUism, we're known as the "religion without a creed" because we encompass atheists, agnostics, theists, deists, pantheists, and people who also identify as christian, jewish, buddhist, pagan, humanist, etc... but we share a common love and respect for the this world, all its creatures and each person's quest for spiritual fulfillment.  We aren't much into the concept of religion as received from an infallible authority on high, be it a religious leader or the literal word of the Bible (hence the "religion without a creed" deal).

Okay, that's enough about my denomination...  my personal religious beliefs are harder to pin down.  My one-word answer is "agnostic", but that hardly comes close to covering the range of my beliefs.  Our church had a wonderful minister who told me that the way to find what I really believed was not to shop among the religious traditions and find which philosophies fit me best, but to look at what in this world matters to me, what moves me spiritually.  I haven't paid enough attention to his advice, which is one reason why I can't come up with a coherent theology here.  The "agnostic" bit is that I find it hard to believe in a personal God... I am open to many visions of a Supreme Being transcending space and time, but have no reason to hold to one in specific, and don't believe I will ever be able to presume upon His nature (to use the standard sexist pronoun :-P).  There is much in this world I find worthy of reverence, the miracle of the cosmos, and of life, and love, and humanity's ability to reach out to each other.  (Oh, and mathematics :-)  Too often I find myself habituated to it all... dragged down by the daily details of life.  Even though I do not hold strictly to traditional religion, there is much in the sacred writings of other religions that I find inspiring (e.g. the Sermon on the Mount).

Well, that's my statement of confused faith, and I am sure my beliefs will change over time.

I am a Christian. More specifically, a Mennonite. An no, that does NOT mean that I am Amish. Too many people confuse those. Amish are more conservative Mennonite. I am Mennonite, because I was born into a Mennonite family. I agree with the beliefs for the most part. I was baptised in September of this past year of my own accord. I don't believe in baptism at birth. God is my rock, and is always there. Whenever I need Him.

I guess I call myself a "sparesly practicing conservative Jew".  That may seem a bit weird, since the "next step up" from non-practicing would be reform. However, I really don't agree with reform.

For example, they often play music in a reform synagogue on Saturday, which you're not supposed to do. I don't think that it's terrible for Jews to not follow all the rules of Shabat, (I certainly don't follow them), but I think that in a synagogue, which in a way represents the Jewish faith, they should follow all the reasonable customs. I say reasonable because there's a good argument against some Orthodox customs, such as seperating men and women at service. But there is no good argument why music should be played against the custom.

It seems that conservative Jews believe that the Orthodox customs are right, but that it is not feasible for most people to follow them. It seems that the reform Jews believe that the Orthodox customs are wrong, old-fashioned, and unnecessary. This is why I side with being conservative.

I'd also like to point out that being an Orthodox Jew means only that you follow the laws and customs of Judaism, (which actually don't amount to that much except on Saturday). It does not necessarily mean that your whole life revolves around the study of Judaism. Thus a person could be an Orthodox Jew and you wouldn't really be able to tell except on Saturday.

I like the philosophical aspects of Judaism more than the actual customs. I like the idea that God is completely incomprehensible and undescribable. (Many would actually argue that my previous statement was a description and therefore invalid). I also like the idea that every person is equally far and equally close to God as any other person. Unlike common belief, a rabbi is not the Jewish equivalent of a priest. A priest's status is religious and comes with religious obligations. A rabbi, however, is no different from any other person. A rabbi is of course well educated in the Torah and Talmud, but he, unlike a priest, need to be ordained in order to become a rabbi. All a person needs to become a rabbi is a congregationwilling to hire him to lead the service.

The fact that I like the philosophical aspect of Judaism is one reason why I am non-practicing. At a service, there is a long religious procedure, after which the Torah is read and the rabbi discusses it. What the rabbi says is of course very interesting and gives you more insight into the Torah. Then there is a conluding religious procedure, and then you eat. When I lived in New Haven, however, I attended a slightly different service, which I liked and came to every Saturday. The congregation was small there, maybe about 20 people, and on most days the rabbi was not there, so instead of the rabbi discussing the Torah, all of us discussed it. This made the discussion much more dynamic and interesting. There don't seem to be any such congregations here, and I don't go to any others because the nondynamic discussion isn't intriguing enough.

I made a lot of statements in this post that I think are true, but may not be. So if you see anything false here, please correct me.

I am Jewish, but in a more spiritual sense than a purely religious sense.  I do not believe in a god so strongly that I almost just wrote I know there is no god.  But I must say, I hate these discussions, because they rely on blind faith than actual logic.  If anything insightful is said on this thread, the people who most need to read it will be too dense to care.

Regarding yif man's argument:  This is actually a very old line of reasoning (its "traditional" form is "Can God make a rock so big He can't lift it?"), and it is based on a misconception.  The problem here is the definition of omnipotence.  Although many people think that it means simply "ability to do everything," a more proper conception would be "ability to do anything logically possible."  Placing "God can" in front of a meaningless statement doesn't make it meaningful: a rock so big that God can't lift it cannot possibly exist, thus nothing, not even omnipotence, can make it so.  This is really not a limit on God, the fact is simply that a logically inconsistent world is meaningless.

I disagree with Magnara in his opinion that religion relies only on "blind faith."  I can't speak for other religions, but those who believe Christianity based on "just because I do" are not following the Apostolic model.  Christianity is believed because the weight of evidence points toward its truth, and because it makes sense of the world.  Incidentally, "faith" in Christianity is (supposed to be) [i]loyalty and dedication[/i] rather than simply blind belief.  Thomas, for instance, is rebuked not for not believing blindly in Jesus's resurrection but for not believing even given sufficient evidence (Jesus's miracles, the reliable testimony of the other Apostles, etc.).  The Christian assertion is that the evidence - philosophical, cosmological, historical, personal - is there if you are willing to look; that one should examine the evidence and believe (and be loyal) based on it.

I am an Orthodox Christian (not to be confused with Orthodox Judaism.) My family is not particularly religious - that is; my parents do believe in God, but they do not practice what is generally asked of Orthodox Christians. Personally, I "discovered" my religion several years earlier when I started reading the Bible. At first I read out of curiosity, but I gained a lot from it eventually. I'm not going to defend my religion against those who disagree with it, but Setofallsets has essentially described my point of view. His defense of his faith is eloquently stated.

I am Roman Catholic Christian.

I am a very stong believer in the Lord and this might be because I was born into a rather unreligious family but I attended a Catholic private school for 3 years.  That made me realize the improtance of religion and that the Catholic faith is acctually what I believe in unlike some members of my family who claim that they are "Catholic" but they do not believe in the statements of faith or the Gospel.

[quote="Setofallsets"]Regarding yif man's argument:  This is actually a very old line of reasoning (its "traditional" form is "Can God make a rock so big He can't lift it?"), and it is based on a misconception.  The problem here is the definition of omnipotence.  Although many people think that it means simply "ability to do everything," a more proper conception would be "ability to do anything logically possible."  Placing "God can" in front of a meaningless statement doesn't make it meaningful: a rock so big that God can't lift it cannot possibly exist, thus nothing, not even omnipotence, can make it so.  This is really not a limit on God, the fact is simply that a logically inconsistent world is meaningless.[/quote]
Yes, i agree. Note that I got my argument from a The Simpsons book, so it shouldn't be taken as anything more than a joke. I totally agree with you on this, except I really don't believe in a single supreme diety. I guess I'm still searching for a faith.

[quote="yif man12"]Note that I got my argument from a The Simpsons book, so it shouldn't be taken as anything more than a joke.[/quote]

I had a feeling that that was the case, but some people see this (or a variant) as a significant argument, so I thought it was worth responding to.

I am sympathetic to Southern Baptists, but have had a hard time figuring out what I believe.  Particularly, this one question has bugged me, about omnipotence.  I have been told that God can do anything all my life.  I ask why does sin and corruption even exist.  They tell me because God gave us free choice and we screwed it up.  But, God knew we were going to screw up because he is omniscient.  So, he created us knowing we were going to blow it.  Now, why didn't he just create the ones that would be faithful?  My youth minister told me that 'lost' people were created for the glory of God.  But, they cause him so much pain.  I suppose Christians believe God put on blinders when he created man.

Omnipotence, Omniscience, and free will do not seem to mix with me.  God decided our choices before we made them because he created everything.

Magnara, if you have opinions, you have just as much 'blind faith' as anybody else.  To develop opinions, you must start somewhere.  Most people probably trust their own reasoning, which I believe to be utterly insecure.  Others can trust virtually nothing, and thus develop few opinions because they are always doubting.

[quote="juggle7"]Omnipotence, Omniscience, and free will do not seem to mix with me.  God decided our choices before we made them because he created everything.[/quote]

Philosophy has a long tradition of wondering about this (and other closely related) questions.

I am a secular humanist.  My extended family includes Jews, UUs, and assorted other Protestant denominations.  No Southern Baptists, though.  (In fact, no southern anythings.)

I agree with God's omnipotence, but not omniscience. Since God is omnipotent, he could choose to be omniscient, but I believe he would not choose to do this since he wouldn't want the future to be predetermined, because if the future was predetermined, people would not have free choice.

Another point: Imagine if God gave everyone a billion dollars. Then money would be worthless. Similarly, if God made it so that everyone was righteous, then rightousness would be worthless. So that is why God lets sin and corruption exist. It's not because he wants to give us free choice between rightousness and sin. God limits our choices every day. We can't fly, for example. We can't even control our own heartbeat. We can't telekinetically move objects. God could easily limit people to righteous acts and still allow us some degree of free choice. However, if he did this, rightousness would be worthless.

[quote="juggle7"]Now, why didn't he just create the ones that would be faithful?[/quote]

I assume that by "faithful," you mean those who eventually turn back to God and are saved, since you contrast this designation with "lost."  (If you mean instead "those who would never fall," I point you to the fact that many who fall can still be saved, and so were worth creating.)  If that is the case, you have asked an extremely difficult question (I bet you knew that already, though.).  There is no given explanation in traditional Christian dotrine, so I can't give any definite answers to this question, but I will attempt to show why this isn't a fatal conundrum.  I will posit a couple scenarios which would, if true, answer this question and which are [i]consistent[/i] with Christian doctrine and theology.  Keep in mind that what follows is [i]not[/i] doctrine, it is merely [i]possible[/i] truth (I don't want to lead anyone astray here.).

Solution 1:  Universal Salvation.  This is the one we hope is true -- that everyone will be saved, and the question is moot because there will be no one who is finally lost.

Solution 2:  The existence of certain lost people is a necessary condition for certain people's salvation.  This sounds a bit funny, but it is not inconsistent with Christian theology and doctrine.  It is a "best of all possible worlds" scenario;  the premise being that the only way to guarentee the salvation of all of those who will be saved involves the existence of many who will not.  God's purpose, after all (or so we believe), is to save as many as is possible.  This of course brings up the secondary question: 'So why do those who end up "lost" need to actually exist?  God could just "fool" us into thinking they do in order to save those in need of their existence.'  The answer to this is, first: maybe that is the case, although it is unlikely, since we do not believe that God deceives - and second: perhaps (and we have reason to believe this is so) God thinks it better for a person to exist and reject Him than to not exist at all (although it is of course infinitely better to turn back to God).  Thus He still creates those whom He knows will not choose to serve Him, for both their good and the salvation of others.  If you don't agree with the way He has set things up and think you could do it better, remember that people's lives exhibit sensitive dependence and that God has omniscience to work with, while we don't.

Most other scenarios that would answer your question are variants on (2).  If it seems difficult to understand, well, it is.  Creation was a job that only Omniscience could do perfectly.

This in addition brings up the question as to why we haven't been told what the answer to your question is.  Aside from the simple fact that we don't need to know, there are many ways in which it could be detrimental (or self-contradictory) for us to know the answer.  As an example, our not knowing it might be a necessary condition to bring about the salvation of certain people: those we then preach to, not knowing whether they will be saved but hoping so; those who, not knowing, thus dedicate themselves in order to ensure their salvation, the list goes on.  (If this latter seems shallow, remember, God will have us return to Him any way he can.  He is not proud.  He is, however, perfect and Just, and so we must actually submit to Him.  Salvation must be accepted.)

[quote="juggle7"]Omnipotence, Omniscience, and free will do not seem to mix with me. God decided our choices before we made them because he created everything.[/quote]

Since I disbelieve in Predestination (the common interperetation, in that we do not have free will), I feel I have to respond to this.  Simply because God created us and knows what our choices will be does not mean that we do not make those choices.  God's knowing what you are going to do does not mean that you [i]can not[/i] choose to do something else, it means that you [i]will not[/i].  God created you, a free agent.  He is omniscient, so He knows what you will choose to do (and can base other things on knowing what you will do), but you still will [i]choose[/i] to do what you do, and are not forced to do it by God.

Again, as I have mentioned, these are difficult questions, and, there may be (make that "are," since I believe that God's answer will be the best, as it will be certainly true) better answers than those that I posted that I do not know of.  I do think that those that I have given should suffice to show that the questions posed can be answered consistently with Christian doctrine.

Appendum (almost cross-posted with Yasha and wanted to respond to some of his comments.):

[quote="Yasha"]Since God is omnipotent, he could choose to be omniscient, but I believe he would not choose to do this since he wouldn't want the future to be predetermined, because if the future was predetermined, people would not have free choice. [/quote]

See my above paragraph on why Omniscience does not preclude free will.  Foreknowledge by God of a choice does not make it any less of a choice.

[quote="Yasha"]Similarly, if God made it so that everyone was righteous, then rightousness would be worthless.[/quote]

I agree with you, but not for the reason you posit.  If God made it so that everyone was "righteous," it would not be merely a limitation on our acts (as are the other things you mention), but a limitation on our will.  If "God made it so that everyone was righteous," the righteousness would not be worthless because everyone was righteous, but because people would not be righteous [i]by choice[/i].  (And the only [i]true[/i] righteousness is that which is not forced from outside; ie, by choice.  We cannot love God meaningfully if we do not have the choice to reject Him instead.)  If I am misinterpreting your position here I apologize.

[quote="Setofallsets"][quote="Yasha"]Since God is omnipotent, he could choose to be omniscient, but I believe he would not choose to do this since he wouldn't want the future to be predetermined, because if the future was predetermined, people would not have free choice. [/quote]

See my above paragraph on why Omniscience does not preclude free will.  Foreknowledge by God of a choice does not make it any less of a choice.[/quote]
If I understand correctly, you are saying that my choice to make this post was a free choice, but God has known about it ever since the beginning of the world. I will have to agree with your argument that people can have free choice even if God is omniscient. However, though your argument disproves my argument that God is not omniscient, it still does not prove that God is omniscient. So now I don't even know what I want to believe. I want God to be able to know everything when he makes his decisions, but I don't want the future to be unchangeable.

[quote="Setofallsets"][quote="Yasha"]Similarly, if God made it so that everyone was righteous, then rightousness would be worthless.[/quote]

I agree with you, but not for the reason you posit.  If God made it so that everyone was "righteous," it would not be merely a limitation on our acts (as are the other things you mention), but a limitation on our will.  If "God made it so that everyone was righteous," the righteousness would not be worthless because everyone was righteous, but because people would not be righteous [i]by choice[/i].  (And the only [i]true[/i] righteousness is that which is not forced from outside; ie, by choice.  We cannot love God meaningfully if we do not have the choice to reject Him instead.)  If I am misinterpreting your position here I apologize.[/quote]
I agree with your point, but what if God only created the people who would be righteous? In that case, God would not be limiting anyone's will; he would just be limiting who existed. I think that in this scenario righteousness would not have much value. (btw, I don't think people are divided into righteous and not righteous. I think righteousness is a multidimensional concept, and that people can change how righteous they are through their actions. I'm just generalizing for the sake of argument).

[quote="Yasha"]If I understand correctly, you are saying that my choice to make this post was a free choice, but God has known about it ever since the beginning of the world.[/quote]

That is correct.

[quote="Yasha"]However, though your argument disproves my argument that God is not omniscient, it still does not prove that God is omniscient.[/quote]

This is true, but I didn't intend it to prove God's omniscience.  I do think that the weight of argument is for the omniscience of God, but I acknowledge that this is no proof.
Some reasons, in brief, to believe (absent valid disproofs) that God is omniscient:  Your point that an omnipotent God could choose to have omniscience;  God is perfect in all ways, knowledge (and thus omniscience) is a good and a perfection, ergo God is omniscient *; the fact that omniscience would be necessary for perfectly good behavior, etc.   Again, these are not proofs, but are good arguments and evidence for God's omniscience.

* I'm sure someone will accuse me of playing semantic games with this one, as it begins to sound a lot like the ontological "proof" of God's existence, which is no proof but one of the weaker arguments, actually.  The reasoning is valid here (provided that omniscience is not self-contradictory, which I think I have shown it is not), but subject to agreement with the premises (that God is perfect, and that knowledge is good).   I repeat, I am not purporting to give a rock-solid [i]proof[/i] but strong evidence.

[quote="Yasha"]I don't want the future to be unchangeable. [/quote]

Depends on how you define change.  The future can't "have been" one thing, and then, as a result of someone's actions "change" to be something else.  It only happens once, and whatever it is, it is (this is even independent of the omniscience of God).  I suppose what you really mean is that you want to be able to affect the future with your actions ([i]not[/i] "change" it, since it never "was" anything to begin with); I am holding that indeed you can, but God already knows how you will do so.

[quote="Yasha"]but what if God only created the people who would be righteous?[/quote]

Assuming, for the sake of argument, that one can group people thus, I would claim that their righteousness would not in fact be worthless.  The value of some things is not comparative.  Good is valuable; evil, its lack or negation, lacks value.  This is, however, a moot point, as God clearly thinks it good to create and save those who are not completely righteous by themselves.

I'm Orthodox Jewish, but like someone said you wouldn't know except for Saturdays and that I keep kosher.  I went to a conservative Jewish middle-school, though.  Sometimes the religious observance of saturdays is difficult-I miss out on some events and it is a real pain to take the SATs on a sunday, but otherwise it's ok.

I'm an agnostic, meaning I believe that there is no proof for or against God's existence. To be more precise, I am a 'weak agnostic', because a 'strong agnostic' believes that finding proof for or against God's existence is impossible, and I just believe there is no proof available yet. Why? Well, because that's just what I believe. I can't think of any other way to explain it.

I am:

Agnostic (fairly strong)
Atheistic
Somewhat existentialist

Now before you ask me, like the kids on my little sister's middle school bus, whether I worship the devil or have little wooden idols on my mantle, allow me to explain myself.  I prefer science to religion, but that is my personal choice.  In history, the human perception of a deity has led to various misconceptions about the universe around us, such as the geocentric theory, the existence of dinosaurs, etc.  In addition, I have seen numerous people who claim to be Christian but evidently have little to no interest in the actual religious aspects of it.  I've also seen some people stretching religion beyond reasonable limits, to the point of being totally illogical.  As a result, I stepped away from that and became an atheist, because the presence of a deity no longer made sense to me, and the pseudo-religious were so insincere that I longer desired to be associated with them.  Please don't mistake "atheistic" to mean "immoral"; The Ten Commandments, with the exception of those religiously affiliated, still hold sway with me, and I do my best to keep virtue in mind.  At this point, I do not believe in a deity, but feel I am prepared to be proven wrong when I die.  I don't participate in organized religion, but encourage religion for those who draw support from it.  I celebrate Christmas in that I celebrate the joy of others, and it seems improbably that I would be able to avoid Christmas even if I tried.


algebra
polynomial
Gauss
superior algebra
superior algebra unsolved

(a)What are the irreducible polynomials in:

(i) [b]C[/b][x]

(ii) [b]R[/b][x]

(iii) [b]Z[/b][x]

in terms of the irreducibe of [b]Z[/b] and [b]Q[/b][x]

(b) Factorize x^5 + 2x^4- 3x^2 - 4x -2 into irreducible in each of (i) [b]Z[/b][x],(ii) [b]Q[/b][x], (iii)[b]R [/b][x],  (iv) [b]C[/b][x]

Can you give me some hints please?

Thank you very much

HINT (i) use the factor theorem and that $ \mathbb{C}$ is algebraically closed
(ii) an odd degree poly over $ \mathbb{R}$ has a root in $ \mathbb{R}$ by IVT
(iii) Gauss' Lemma


For a training ride, a bicyclist first warms up by biking at 12 mph for 10 minutes and at the end cools down by biking at 8 mph for 15 minutes.  In between the warm-up and cool down, he bikes at an average of 24 mph.  In a two-hour training ride, how many total miles does he bike?

A.42	B.43	C.44	D.45	E.46

[hide]He bikes 24 mph for $ \frac{19}{12}$ hours.
$ 24\cdot\frac{19}{12}\plus{}12\cdot\frac16\plus{}8\cdot\frac14\equal{}\boxed{42}$
A[/hide]


[hide="The winner was..."]SplashD[/hide]

[hide="The top 6"]SplashD:150
Xantos C Guin:144
JSteinhardt:127.5
pianoforte:124.5
(tie) Nrz and Fireninja0:123[/hide]

[hide="The answers"]
1) E
2) A
3) D
4) E
5) B
6) D
7) D
8) D
9) C
10) free question
11) D
12) D
13) A
14) C
15) B
16) D
17) A
18) B
19) C
20) D
21) B
22) C
23) free question
24) E
25) C[/hide]

Nice contest.

How did I manage to miss #3?

Oh well,  :winner_second: isn't bad at all.

yay for forth :clap2: !

I like the contest (except for questions 9 and 10, those were annoying...).

@mustafa- How many participants were there?

How do you think this mock compared in difficulty to the real AMC 10?

Also, another question to ponder, did you all think the questions resembled the level for 10th graders and below?

Were they well written?

organized?

I don't know about this test because I didn't participate.

What would you guys give this test on a scale of 1 to 10?

[quote="pianoforte"]@mustafa- How many participants were there?[/quote]

10 people sent in answers


If there is a fish tank in the shape of a perfect cube with a side of 36cm and it takes 2 minuets to fill it with one liter of water, how long will it take to fill the whole tank. to the nearest minuet.

did anybody get 93 minuets???

[hide]Simply find the volume of the cube and convert the volume to liters instead of cubic centimeters. Then, you end up with 93.312[/hide]


algebra
polynomial
trigonometry
function
complex analysis
real analysis
real analysis unsolved

Maybe the following lemma is usefull:

\[ P(z_0)=\frac 1 n \sum_{k=0}^{n-1} P(z_0+\omega \varepsilon_k),\forall \omega \in \mathbb{C} \]

Where $\varepsilon_k=\cos \frac{2k\pi}{n}+i \sin\frac{2k\pi}{n}$ the n-th roots of unity.

[quote="Moubinool"]P polynomial of $C[X]$
U an open of $C$

prove that $sup_{x\in U}|P(x)|=sup_{x\in Fr(U)}|P(x)|$

Fr(U) frontière de U[/quote]
Isn't this a direct implication of Maximum modulus principle: if function $f$ is holomorphic in $D$ and continious in $\bar D$, then $|f|$ attains it's maximum on the $FrD$.

How to prove it with only 'math-spé' knowledge ?

\barD seems to have an surprisingly large effect. Please fix this (I think just adding a space would do it).


Bow to natural solitions of the equations system
x^y+y=2059,y^x+x=123

Partial solution(after the first view):
From $ y^x\plus{}x\equal{}123$ we take that if $ y\ge12$ then $ x\equal{}1$-contradiction.
So $ 0<y\le11$
For $ y\equal{}11,x\equal{}2$ works
Analogously after taking some cases we can solve the problem,but i'm pretty sure that there must be a better way.

Dimitris has almost done this.....
Let $ x$ be even.
Easy to see for $ y\ge 12$ no naturals shall satisfy the given eqn.....
$ y^x \plus{} x \equal{} 123 \\
\ y^2\equiv 1(mod3) \\
\implies y^{2a}\equiv 1(mod3)$
So $ y^{2a} \plus{} 2a\equiv (2a \plus{} 1)(mod3)$
So let's see the smallest natural satisfying this - which is $ a \equal{} 1 \implies \boxed {x \equal{} 2}$
So for even 'x' no other solutions, lets see for odd 'x'!  :roll:


There are several (finite) cities in a country. Some of the cities are connected by one-way airlines. It is known that there exists a city from which not every city can be reached. Prove that one can select a group of cities such that none of the cities in the group can be reached from any city not in the group.

Assume that there is no path from $ A$ to $ B$.
Then, consider the set $ S$ of towns formed by $ B$ and the towns $ M$ for which there is a path from $ M$ to $ B$. Let $ S'$ be the set formed by the remaining towns. Clearly both $ S$ and $ S'$ are not empty.
Now, let $ T$ be any town belonging to $ S'$. Then, from $ T$ we cannot reach any town in $ S$ otherwise we would be able to continue this path to reach $ B$, contradicting that $ T$ is not in $ S$.
Therefore, none of the towns in $ S$ can be reach from any town not in $ S$.

Pierre.


Να προσδιορίσετε το πολυώνυμο $P(x)$ ώστε $P(1960)=1992$ και $P(x)=\sqrt{P(x^2+1)-33}+32,$ για κάθε $x\geq0.$

Παρατηρούμε ότι P(1960) = 1992 = 1960 + 32
Θα δείξω ότι P(x)= x+ 32 για άπειρα x και άρα θα είναι η μόνη λύση το P(x)=x+32.
Έχω ότι ισχύει για x=1960. Θέτοντας όπου x τo 1960 παρατηρώ ότι
1960^2 + 33 = 1960^2 + 1 + 32 = P(1960^2 + 1) . Άρα αν έχω μια λύση
xi και βρω μια x(i+1)>xi τότε επαγωγικά θα υπάρχουν άπειρες.
όμως αν xi μία τιμή για την οποία P(xi) = xi +32 τότε κάνοντας πράξεις
προκύπτει ότι και για x(i+1)=xi^2 +1 > xi ισχύει P(x(i+1)) = x(i+1) +32.
Θυμίζει λίγο σε σκέψη προηγούμενη πολυωνυμική που είχα βάλει.

Ξέχασα να πω ότι πρέπει να ελένξω αν ικανοποιεί τις συνθήκες μου το πολυώνυμο...


Let {a[size=75]n[/size]}[size=75]n>=1[/size] be a sequence with a[size=75]n[/size] >= 0 for all n >= 1, and let S[size=75]n[/size] = sum of a [size=75]k[/size] from k=1 to n.

a. show that if {a[size=75]n[/size]}[size=75]n>=1[/size] is decreasing and sum of a[size=75]n[/size] converges, then lim(n->infinity) n*a[size=75]n[/size]=0.
b. show that if a[size=75]1[/size] > 0 and sum of a[size=75]n[/size] for n>=1 diverges, then sum of a[size=75]n[/size]/S[size=75]n[/size] for n>=1 also diverges.

a. $ a_{N} \frac {N}{2} \leq \sum_{N/2 \leq n \leq N}a_{n} \equal{} o(1)$
b. I just post my pdf file
I  solved other  problem for my friend (colleague) , but there is also  your problem (see lema 1)


integration
linear algebra
matrix
calculus

1)Let $ a_1,a_2,..,a_n$ are $ n$ positive real numbers and $ A\equal{}[a_{ij}]_{1 \leq i,j \leq n}$ such that :
  $ a_{ij}\equal{}\frac{|a_i\plus{}a_j|\minus{}|a_i\minus{}a_j|}{a_ia_j(a_i\plus{}a_j)}$ for all $ 1 \leq i,j \leq n$
 Prove that : $ \det(A) \geq 0$
2) Let $ b_1,b_2,..,b_n$ are $ n$ positive integers numbers and $ B\equal{}[b_{ij}]_{1 \leq i,j \leq n}$ such that :
     $ b_{ij}\equal{}\frac{(\minus{}1)^{i\plus{}j}(i\plus{}j)!}{(b_i\plus{}b_j\plus{}1)^{i\plus{}j\plus{}1}}$ for all $ 1 \leq i,j \leq n$
Prove that : $ \det(B) \geq 0$

[hide="Hint"]
I can only prove for the case when $ a_1 > a_2 > a_3 > \cdots> a_n$
Note that we can write $ {a}_{ij}\equal{}{b}_{ij}.{c}_{ij}$ where $ {b}_{ij}\equal{}\frac {1}{a_i\plus{}a_j}$ and $ {c}_{ij}\equal{}\frac {2}{min(a_i,a_j)}$
Note that if $ B\equal{}[{b}_{ij}]$ and $ C\equal{}[{c}_{ij}]$ then $ B$ and $ C$ are semi positive definite matrix. Proof : Note that $ {b}_{ij}\equal{}\int\limits_{0}^{ \plus{} \infty}{e}^{(a_i\plus{}a_j)u)}\ du\equal{}\int\limits_{0}^{ \plus{} \infty}{e}^{(a_i)u)}.{e}^{(a_j).u}\ du\equal{}\int\limits_{0}^{ \plus{} \infty}f_i(u).f_j(u)\ du\equal{}<f_i(u).f_j(u)>$ as we write $ f_i(u)\equal{}{e}^{a_i.u}$ and by noting that $ \int\limits_{0}^{ \plus{} \infty}f_i(u).f_j(u)\ du\equal{}<f_i(u).f_j(u)>$ is an inner product and if $ A\equal{}[{a}_{ij}]$ where $ {a}_{ij}\equal{}<v_i,v_j>$ then $ A$ is semidefinite positive.
Note also that $ {c}_{ij}\equal{}\frac {2}{a_m}$ where $ m\equal{}min(i,j)$ by the assumption above. We will prove that $ B$ is a semidefinite positive matrix by using Mathematical Induction. For $ n\equal{}1$ it is obviously true. Then, if the statement is true for $ n\equal{}k$, for some natural number $ k$, then for $ n\equal{}k\plus{}1$, $ det({C}_{n\plus{}1})\equal{}\frac {1}{a_1}\frac {a_1}{a_2}det(C_n)\minus{}\frac {1}{a_1}.det(C_n)$ that is also greater than $ 0$, so $ C$ is also (semi)definite positive. I don't know how to continue and I am sorry if I am wrong...  :(
[/hide]

Carefull in calculate intergral , poof of your is false ?  :(
Solution of me large .I try to find a short solution


calculus
integration
analytic geometry
conics
ellipse
rotation
calculus computations

Calculate
\[ \oint\limits_C {x^2 ds}
\]
where
\[ C: \equal{} \left\{ {\left( {x,y,z} \right): x^2  \plus{} y^2  \plus{} z^2  \equal{} 1} \right\} \cap \left\{ {\left( {x,y,z} \right): x \plus{} y \plus{} z \equal{} 0} \right\}.
\]

Looks like a plane passing through the origin and intersecting the unit sphere. Eliminating $ z^2$ from both sets I get $ x^2+y^2+xy-1/2=0$.  Rotating coordinates, this becomes an ellipse rotated $ \pi/4$.  In the new u-v coordinate system (rotated $ \pi/4$ from the x-y coordinate system), the equation for $ C$ is:

$ u^2+1/3 v^2=1/3$

Using $ u=rCos(\theta)$ and $ v=rSin(\theta)$ I can express u and v parametrically as:

$ u(\theta)=\frac{Cos(\theta)}{\sqrt{3Cos^2(\theta)+Sin^2(\theta)}}$

$ v(\theta)=\frac{Sin(\theta)}{\sqrt{3Cos^2(\theta)+Sin^2(\theta)}}$

I can now express the original variables x and y in terms of $ \theta$ by the transformation

rules:

$ x(t)=\frac{u(t)}{\sqrt{2}}-\frac{v(t)}{\sqrt{2}}$

$ y(t)=\frac{u(t)}{\sqrt{2}}+\frac{v(t)}{\sqrt{2}}$

and substitute these into the formula:

$ \int_{C} x^2 ds=\int_{0}^{2\pi}
x(t)^2\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt$

giving me:

$ \int_0^{2\pi}\frac{1-2Cos(t)Sin(t)}{2(2+
Cos(2t)}\sqrt{\frac{5+4Cos(2t)}{(2+Cos(2t))^3}}dt$

which is approx 1.57193.  Not sure it's correct though and probably an easier way to do this yielding an exact answer.


function
floor function
real analysis
real analysis unsolved

Construct a monotone function on [0,1] which is discontinous at every rational point.

$ Q\cap [0,1]\equal{}\{q_1,q_2,q_3,...\}$
let $ A(x)\equal{}\{n: x\ge q_n\}$

then $ f(x): \equal{}\sum_{A(x)} \frac{1}{n^2} \le \sum^{\infty}_{n\equal{}1}\frac{1}{n^2} \equal{} \frac{\pi^2}{6},  \forall x\in[0,1]$

since $ x\le y \to A(x) \subset A(y) \to f(x)\le f(y)$ we get monotone increasing.

also, $ x\equal{}q_k\in Q\cap [0,1] \to k\in A(x)\backslash A(x\minus{}\epsilon)  \to$
$ \mid f(x)\minus{}f(x\minus{}\epsilon) \mid \equal{} \sum_{A(x)\backslash A(x\minus{}\epsilon)} \frac {1}{n^2} \ge \frac{1}{k^2} \to$  f discontinuous at x. similar argument for x=0 on the right side gives us what we want?
maybe?

A more explicit version of the same idea:

$ f(x)\equal{}\sum_{n\equal{}1}^{\infty} \frac{\lfloor n x \rfloor}{n^3}$.

\[ f(x)\equal{}\sum_{n\equal{}1}^\infty{\frac{\lfloor nx \rfloor}{2^n}}\]

It can be easily verified that $ f$ is continuous at $ x_0\in [0,1]$  iff $ x_0\notin\mathbb{Q}$.  Actually this function is well known, but its name escapes me at the moment.

thanx all of u

after posting this question i was able to think up a solution similar to kenn's only I used [Sum (1/k^2)+ 1/n^2] with k<n


A $3000\times 3000$ square is tiled by dominoes (i. e. $1\times 2$ rectangles) in an arbitrary way. Show that one can color the dominoes in three colors such that the number of the dominoes of each color is the same, and each dominoe $d$ has at most two neighbours of the same color as $d$. (Two dominoes are said to be [i]neighbours[/i] if a cell of one domino has a common edge with a cell of the other one.)

This is a double-post.  Moderator, please delete.


Let $ X$ be a metric space, and $ A$ a nonempty subset of $ X$.  Define
\[ U(A,\epsilon)\equal{}\{x\in X : d(x,A)<\epsilon\}.\]
Let $ A$ be compact.  Show that for any open $ U$ containing $ A$, there exists an $ \epsilon$ such that $ U(A,\epsilon)\subset U$.

$ U^c$ is closed.  Let $ \delta \equal{} \inf\{d(x,y): x\in A, y\in U^c\}$.  Since $ A$ is compact, we actually have that  $ \delta \equal{} \min\{d(x,y): x\in A, y\in U^c\}$ (think about continuous functions on compact sets), and the minimum is actually attained.  On the other hand, we clearly have that $ \delta > 0$, since $ A\subset U$ and $ U\cap U^c \equal{} \varnothing$.  Hence it suffices to take $ \varepsilon \equal{} \delta/2$.


limit
function
logarithms
calculus
integration
geometric series
calculus computations

Find the next limit:
$ \lim_{n\to\infty}{(1\minus{}\frac{1}{2}\plus{}\frac{1}{3}\minus{}.........\plus{}\frac{1}{2n\minus{}1}\minus{}\frac{1}{2n})}$

[hide="Solution with Auxiliary Function"]$ 1 \minus{} x \plus{} x^2 \minus{} x^3 \ldots \equal{} \frac {1}{1 \plus{} x}$ by sum of geometric series

Integrating, $ x \minus{} \frac {x^2}{2} \plus{} \frac {x^3}{3} \minus{} \frac {x^4}{4} \ldots \equal{} \ln |1 \plus{} x| \plus{} C$

Plug in $ x \equal{} 0$ in the equation above to get $ C \equal{} 0$

Then plug in $ x \equal{} 1$ to get $ 1 \minus{} \frac {1}{2} \plus{} \frac {1}{3} \minus{} \frac {1}{4} \ldots \equal{} \ln |1 \plus{} 1| \plus{} 0 \equal{} \ln 2$[/hide]

[hide="Solution with Technique Above"]
$ \gamma \equal{} \lim_{m \rightarrow \infty } \left[ \left( \sum_{k\equal{}1}^m \frac{1}{k} \right)  \minus{} \ln m\right]$

Thus for large $ m$, we can write $ \sum_{k\equal{}1}^m \frac{1}{k} \equal{} \ln m \plus{} \gamma$

$ 1 \minus{} \frac {1}{2} \plus{} \frac {1}{3} \minus{} ......... \plus{} \frac {1}{2n \minus{} 1} \minus{} \frac {1}{2n}\equal{}\frac{1}{n\plus{}1}\plus{}\frac{1}{n\plus{}2}\plus{}\cdots \plus{}\frac{1}{2n}$

$ \equal{} \sum_{k\equal{}1}^{2n} \frac{1}{k} \minus{} \sum_{k\equal{}1}^n \frac{1}{k} \equal{} ( \ln 2n \plus{} \gamma) \minus{} (\ln n \plus{} \gamma) \equal{} \ln 2$
[/hide]

[quote="The Zuton Force"]Then plug in $ x \equal{} 1$ to get ...[/quote]
This isn't as innocent as a "plug in." It's a quote of a major and important theorem, Abel's Theorem.

I tink this method is easier....

as kunny said this is nothing but the sum that he says( i am sorry i don know to use latex)...

THen this limit of sum can be converted into an integral......

and any limit of sum of the form can be converted to an integral...

so the integral comes as $ \int_0^1\frac {1}{1 \plus{} x} dx$ which is nothing but $ {ln2}$

is that not just from studying the power series of log (1+x)?


algebra
polynomial
modular arithmetic
Primitive Roots

Suppose that $p>3$ is prime. Prove that the products of the primitive roots of $p$ between $1$ and $p-1$ is congruent to $1$ modulo $p$.

Just note that if $g$ is primitive root then $\frac{1}{g}$ is a primitive root as well. Because $-1$ can't be a primitive root (since the order of $-1$ is 2) the conclusion follows.

Alternatively, if $g$ is a primitive root, then all the other primitive roots are $g^k$, with $\gcd(p-1,k) = 1$ (there are $\varphi(p-1)$ of them), so their product is
\[\prod_{\gcd(p-1,k) = 1} g^k = g^{\sum_{\gcd(p-1,k) = 1} k}.\]
Since $\gcd(p-1,k) = 1$ if and only if $\gcd(p-1,p-1-k) = 1$, it follows that $\sum_{\gcd(p-1,k) = 1} k = \dfrac {1} {2} (p-1) \varphi(p-1)$, a multiple of $p-1$ (since $p>3$, it follows $\varphi(p-1)$ is even), hence the product is equal to $1$.

To be punctilious, the above post states that $-1$ cannot be a primitive root (without in so many words referencing the given condition $p>3$); othervise $g$ and $\dfrac {1} {g}$will be the same element (to wit, $-1$).

[hide="Solution"]
It is well known that $(\mathbb{Z}/p\mathbb{Z})^{\times}\equiv C_{p-1}$
Represent $C_{p-1}=\{1,a,\cdots,a^{p-2}\}$ for some arbitrary generator $a$ (we treat $a$ as an element in $\mathbb{F}_p$ when we refer to multiplication at the end of the proof).
Clearly, the set of primitive roots can be expressed as:
\[ S=\{a^n| (n,p-1)=1\} \]
(in particular, $|S|=\phi(p-1)$)
Then, the original question is to compute \[\prod_{s\in S}s\]
Indeed, we use the existence of a handy involution to do so.
Define an involution $f:S\to S$ to take the element $a^n$ to $a^{p-1-n}$. This is well-defined since $(n,p-1)=1\iff (n,p-1-n)=1$. Furthermore, this involution has no fixed points.
Indeed, the only possible fixed point is $\frac{p-1}{2}$. However, we can see that \[a^{\frac{p-1}{2}}\not\in S\]
Because $p>3$, $\frac{p-1}{2}>1$. Thus, $\left(\frac{p-1}{2},p-1\right)=\frac{p-1}{2}>1$, so by the definition of $S$, $a^{\frac{p-1}{2}}\not\in S$.
Thus, $f$ induces a partition of $S$ into pairs of the form $\{s,f(s)\}$. We compute as follows:
 \[\prod_{s\in S}s=\prod_{\text{pairs}}s\cdot f(s)=\prod a^n\cdot a^{p-1-n}=\prod a^{p-1}=\prod 1=1\]
This yields the requested result.
[/hide]

Another fancy way (which is much more justified at the question asking for their sum) is to see that the number in question is the constant coefficient of the cyclotomic polynomial $\Phi_{p-1}$.

Uhm... this seems pretty trivial if I'm not mistaken. It's not hard to show that if $g$ is a primitive root modulo $p,$ then so must be $g^{-1}$ ([hide=" just in case..."])

Note that $g$ is a primitive root $\pmod{p}$ iff $g^m \equiv g^n \pmod{p} \iff m \equiv n \pmod{p-1}$. Now,
\[(g^{-1})^m \equiv (g^{-1})^n \pmod{p} \iff g^{-m} \equiv g^{-n} \pmod{p} \\ \iff -m \equiv -n \pmod{p-1} \iff m \equiv n \pmod{p-1},\]
so $g^{-1}$ is a primitive root modulo $p$ as well. [/hide]), so we just pair up the primitive roots with their inverses. The only exceptions are 1 and -1, which are their own inverses, but fortunately they aren't primitive roots $\pmod{p}$ for $p>3.$

let $g$ be a primitive root of $p$ then we shall assert that $g^{-1}$ is also a primitive root,

assume that $(g^{-1})^x \equiv 1 \mod p$ that $x<p-1$

$(g)^x(g^{-1})^x \equiv g^x \equiv 1 \mod p$ which is a contradiction so $x=p-1$ and $g^{-1}$

is also a primitive root. thereupon because

$(g^{-1})^{-1} =g$ and $g \not\equiv \pm 1$, it follows the the conclusion straightly!


geometry
rectangle
combinatorics proposed
combinatorics

For any positive integer $ n$, let $ [n] = \{1, 2, \ldots, n\}$.  In how many ways is it possible to partition the set $ [2n]\times[2n]$ into $ n$ sets of 4 elements each such that each set is of the form $ \{(a, b), (a+k, b), (a, b+k), (a+k, b+k)\}$?  That is, how many ways can we decompose a finite lattice into squares?

Possible variations to consider are to replace squares with rectangles, replace the square grid with a rectangular grid, or both.

Smaller cases are tractable:
[hide="2 by 2n with squares"]Clearly, there is only one such decomposition.[/hide]
[hide="2 by 2n with rectangles"]It's not very hard to see that the answer is $ (2n-1)!! = (2n-1)\cdot(2n-3)\cdot\cdots\cdot3\cdot1$.[/hide]
[hide="4 by 2n with squares"]Let $ a_{n}$ be the number of ways to accomplish this task.  Consider the square which covers $ (1, 1)$.  It may also cover $ (2, 1)$, in which case the points $ (3, 1)$ and $ (4, 1)$ must be covered by the same square.  Then we have left over a $ 4 \times (2n-2)$ strip.  If the square that covers $ (1, 1)$ also covers $ (3, 1)$ then $ (2, 1)$ and $ (4, 1)$ must be covered by the same square, as must the pairs $ \{(1, 2), (3, 2)\}$ and $ \{(2, 2), (4, 2)\}$ which leaves us with a $ 4 \times (2n-4)$ strip.  Finally, if the square that covers $ (1, 1)$ also covers $ (4, 1)$, then $ \{(2, 1), (3, 1), (2, 2), (3, 2)\}$ must be a square, so $ \{(1, 2), (4, 2), (1, 5), (4, 5)\}$, $ \{(1, 3), (4, 3), (1, 6), (4, 6)\}$, $ \{(2, 3), (3, 3), (2, 4), (3, 4)\}$ and $ \{(2, 5), (3, 5), (2, 6), (3, 6)\}$ must be as well, leaving us a $ 4 \times (2n-6)$ strip.  Thus $ a_{n}= a_{n-1}+a_{n-2}+a_{n-3}$ with initial conditions $ a_{0}= a_{1}= 1$ and $ a_{2}= 2$; in other words, the tribonacci numbers.[/hide]
[hide="4 by 2n with rectangles"]The recurrence that results is $ b_{n}= 3\sum_{k = 1}^{n}\binom{2n-1}{2k-1}\cdot (2k-1)! \cdot b_{n-k}$ with $ b_{0}= 1$ which gives rise to $ b_{n}= (2n-1)!! \cdot (2n+1)!!$ (courtesy of the [url=http://www.research.att.com/~njas/sequences/A079484]OEIS[/url])[/hide]


darkquantum wrote:Circle O_1 of radius 105 is tangent to both sides of a 60 degree angle. For all n greater than 1, Circle O_(n+1) is tangent to circle O_n and both sides of the angle, and is smaller than circle O_n. Find the sum of the radii of all of the circles O_i.

[hide]Ok...infinite geometric sequence... 60 degree angle is ABC

O_1 is tangent at A and B... so, you draw O_1A and O_1C which is perpendicular to BA BC respectively... so then also because O_1A and O_1C are equal (105 units) then BO_1 is the bisector of angle ABC... so it forms 30-60-90 triangle BAO_1. and then you draw O_2 with radius r... and draw the same perpendicular lines to the tangent point from the center. And then they are similar, so then

r:105 = BO_2: BO_1

BO_2 x sin 30 = r

so BO_2 = 2r

r:105 = 2r:(2r+r+105)  (r+105 = O_2O_1)

r(3r+105) = 210r

3r^2 + 105r = 210r

3r^2 - 105r = 0

3r(r - 35) = 0

so r = 35... and the ratio between the radii is 1/3

So then the sum of all of them is... (105 - 0)/(2/3) = 157.5???

[/hide]

The problem statement should say "for all n greater than or equal to 1."

It's a geometric series equal to 315/2.

[quote="darkquantum"]Circle O_1 of radius 105 is tangent to both sides of a 60 degree angle. For all n greater than 1, Circle O_(n+1) is tangent to circle O_n and both sides of the angle, and is smaller than circle O_n. Find the sum of the radii of all of the circles O_i.[/quote]
I saw a problem similar to this in a Mathcounts warm up...

Just an interesting fact

[quote="teratomato"]The problem statement should say "for all n greater than or equal to 1."[/quote]

[color=cyan]Not particularly . . . since there were no circles O_n defined except for n positive integers, there isn't really any need to exclude those not defined.  I mean, maybe, if you were being fabulously technical, you might do it -- but it isn't really necessary.[/color]

It should say that because currently, he's defining O_1, then O_3 in terms of O_2, then O_4 in terms of O_3, etc. There's no O_2. We can hardly breathe.

[color=cyan]Oh, I see.  Sorry, I misunderstood you.  Yes, or the recusive definition should be O_n and O_(n-1).  Anyhow, you know what I meant :).[/color]


How many base three numbers with anywhere from 1 to 5 digits use only 1s and 0s.  Explain in detail.

[hide="Casework!"]
In base 3 numbers, there are only 3 digits, 0, 1, 2. We are told that we can't use 2.
If there is 1 digit, you have but one option, 1.
2 digits-10, 11
3- 111, 110, 101, 100
4- 1111, 1110, 1101, 1100, 1011, 1010, 1001, 1000

I see a pattern here. For the first option we have 1, 2nd is 2, 4, 8, and they keep increasing in powers of 2, so the 5th option would be 16. Adding all of these up we have 16*2=32-1=[b]31[/b][/hide]

[hide] There may a quicker way than casework:
Let's say that we are allowed to have a 0 as the first digit of our number.  Thus, there are two choices per digit.  This gives us 2*2*2*2*2=32.  However, we can't include 00000, so there are $31$. [/hide]

[hide="answer"]this question is equivalent to how many base 2 #s are there with 5 or less diits. This is $2^{5}-1$, b/c of $00000$, or $\boxed{31}$.[/hide]


can someone solve this and show all the work, please and thank you!!!

Greg began his 500KM road trip to Montreal at 2am. His averaged speed for the first hour of the trip was 110km/hr. What average speed does Greg need to maintain if he is to reach Montreal by 6am?

Solve what? I don't see anything... :lol:

Seems the question is the average speed in the second (last) part of the trip.

It's easy to find the remainder length and time . Divide to find

[hide=click to see the result] 130 km/hr

You could try doing it now. If you'l still need help, please ask...


Go to http://www.mapleglobal.com and download the 175 MB-ish file. It's a really good, addicting, RPG. It's completely free.

My name in there's Spasmodical.

[url=http://invisionfree.com/directory.php?go=491280]
Maplestory forums[/url]

People here who play:

Spasmodical - Level 33 - Cleric
beny1991 - Level 27 - Rogue/Thief
Typhius - Level ? - Magician

thx for posting
I gotta download it
Maple, is the name of an amazing geometry software.
a game called Maple should be funny.

[quote="MithsApprentice"][quote="Treething"]Yeah, that's one of the first things I realized when dling it :P .

It's a 170 MB file, so it'll take about 20 minutes on cable.[/quote]

hmm...is it like a MUD? b/c 170 MB is definitely not a lot of room to fit loads of game and story then add good graphics and audio to it...[/quote]

Well,... if you put all the graphics and audio on a server, and only put the software needed to connect in the file then it might work.

[quote="yif man12"]But then why would you have to download anything? Why not just have it accessible on the internet?[/quote]

basically what  I mean by a MUD...

I'm wondering what that 170 MB is. If it's the whole game, then why don't they just put all the game data on the internet? If it is, it probably isn't such a great game.

And if they have everything on the internet, it would take rather...uh...good connection to download as you're playing...especially if you expect good graphics/audio/etc and no lag...besides, get a couple thousand players downloading gigs at a time and ur talking about major major server fees as well, especially for a free game.

Nonono. 170 MB is the size of the file it takes to install it.

http://invisionfree.com/directory.php?go=491280

right...we're discussing what's in that file exactly...

[quote="MithsApprentice"]hmm...is it like a MUD? b/c 170 MB is definitely not a lot of room to fit loads of game and story then add good graphics and audio to it...[/quote]

170 MB is pretty big, IMO (especially on my slow internet connection)...I think you could fit a decent game on there.

BTW, nice job on individual, Miths.  Ford knows we needed it.

I can always IM the file to you for a much quicker download :P ...

u know what else? MAPLE STORY SUX! its a complete copy of ragnarok and priston tale.. everything..

o_o

I've never played either of those, so I can't say that it's not, but...

When were those two released? Maplestory came out in Korea (or Japan, I don't remember) around 2000-2001, so...

In maple story....what is the whole point?like...what do you do the whole time, other than killling monsters? you're alone the whole time,(like the opposite of runescpae)?Just wondering, because i never played it before....

You eat monsters =D !

Erm, you go on quests, party with other people...

It's just version 0.21 (in Korea they're in like 1.41), so the game's not close to being finsihed.

Dude, hacks are illegal. *glomps him*

Level 41 =D .

buddy ol pal o treething... wanna help me lvl up? pls cause i either lvl to slow or the monster is too strong... eh i wish my ice was stronger.  I'll fame you... eh... what class are you again?

elvenchamp777

lol, I'm a cleric.

Train in either Land of the Wild Boar or Ant Tunnel IV :P .

yeah! i'll look for you XD

elvenchamp777

A friend of mine who is in the underground of quite a few sites pointed me to the following website where the private bypass is located for 0.32

http://rapidshare.de/files/810917/MapleBypassPrivate032.exe.html

He said it utilises the correct VB API in order to let u use your hacks again.
However this one is labeled private so dont start selling them on Maplestory!

...

Not tempted, and I could contact an admin...

hackers should burn in h,e, double hockey sticks XD  WE WOULD HAVE OSSYRIA BUT NOOOOOOOOOOOOO!!! You hackers have to screw it up by hacking and botting so they spend like no time on Ossyria so STOP HACKING!

on a dif. note who has been able to play without lag cause it always lags like heck everytime i try to play...

elvenchamp777

another reason not to download hacks is that you'll probably get keylogged and your ms account will get hacked

wats keylogged?  Anyway not many people hack now, everyone bots at pig beach i keep changing channels and there's always someone who bots or an iron hog so it's hard to train there.  yeah and DON'T HACK OR BOT!

elvenchamp777

a keylog keeps track of what you type in (a.k.a. your password)

[quote="moogra"]lvl 64 but i quit -_-[/quote]
david, you're not supposed to revive really old topics like this(I think).

p.s. Maple Story is the worst online game ever.

YAY!!!
I get a warning BEFORE I download it!!! :D


1. Mr.Haddick selects a baseball squad from among nine men. Each player except the pitcher and catcher can play any position. How many different fielding arrangements can he make?
I did like:9!, but the solution was 1*1*7!, I don't understand why the first two was 1,1?? Is there supposed to be 9 choices for each?
2. For a problem using permutation, it means that the order matters and it repeats,like ABC, BAC? but for combination, it never repeats? only ABC? That leads me to my next question, In how many ways can we select a collection of three red or three white marbles from a bag of six white marbles and four red marbles?
The solution was 4C3=4, 6C3=20, then add them together, but i am confused because why we use combination for each color here? Can we use permutation as well? because drawing out red from red marbles or white from white marbles, they are the same color, even if we use permutation, order wouldn't be detected?

1. There is 1 way to place the pitcher, 1 for the catcher, and 7! for everyone else.
2. Order doesn't matter because it is marbles....


[quote="sunken rock"]Three parallel lines drawn through the vertices of a triangle $ ABC$ intersect a straight line $ d$ at $ a, b$ and $ c$ respectively.
The parallel through $ b$ to $ AB$ intersects the parallel through $ c$ to $ AC$ at $ A'$, similarly finding the points $ B'$ and $ C'$.
Prove that the points $ A', B'$ and $ C'$ lie on a straight line and $ A'B' \parallel Aa$.

Best regards,
sunken rock[/quote]

u mean parallel line through $ A,B,C$ intersect $ d$ at $ a,b,c$ respectively?

Yes, you are right; see the attached drawing, sorry for not posting it before!
I hope it's better now!

Best regards,
sunken rock

I have a boring proof, with long calculations, so I am interested in other method(s).

Thanks & Best regards,
sunken rock

Let the drawn parallel line through $ A$ intersects $ BC$ at $ F$.
we draw a parallel line $ l$ through $ A'$ to $ AF$ and denote $ c_{BC} \cap l \equal{} B''$.we define $ c_{BC}$ the line thru $ c$ and parallel to $ BC$. similarly for the other parallel lines which go thru $ b,a$.
we will prove that $ B'' \equal{} B'$.
for this we must prove that $ B''a \parallel{} AB$ and so we'll get $ B''a \equal{} a_{AB}$
so it sufficies to prove that $ aB'' \parallel{} A'b \equal{} b_{BA}$
denote  $ K \equal{} c_{BC} \cap b_{AB}$
we must prove that $ \frac {ca}{cb} \equal{} \frac {cB''}{B''K}$.
we have $ \frac {ca}{cb} \equal{} \frac {CF}{FB}$
and also since $ b_{AB} \parallel{} AB$ and $ c_{AC} \parallel{} AC$ and $ c_{BC} \parallel{} BC$ we get $ \Delta A'Kc \sim \Delta ABC$.
and since $ A'B'' \parallel{} A'F$ so $ F,B''$ are two corresponding points.so the result is $ \frac {CF}{FB} \equal{} \frac {cB''}{B''K}$
so finallly we have what we want : $ \frac { cB''}{B''K} \equal{} \frac {ca}{cb}$.
and so $ A'B' \parallel{} AF$.the rest is done similarly.

Let $ K,\ L.\ M$ be ( instead of $ a,\ b,\ c$ ), the points of intersecttions of the line $ (d)$, as the problem states.

The lines through $ K,\ M$ and parallel to $ AB,\ AC$ respectively, intersect each other at point so be it $ A'$ and we denote as $ D,\ E,$ the points of intersections of $ A'M,\ A'K,$ from the lines through $ K,\ M$ respectively and parallel to $ BC.$

We draw the line through $ A'$ and parallel to $ AL,$ which intersects $ ME,\ KD,$ at points $ B',\ C',$ respectively and it is enough to prove that $ LB'\parallel AB$ and $ LC'\parallel AC.$

$ \bullet$ It is easy to show that $ \frac {B'E}{B'M} \equal{} \frac {C'K}{C'D} \equal{} \frac {ZB}{ZC}$ $ ,(1)$ where $ Z\equiv BC\cap AL.$

But, based on the [b]Thales theorem[/b], from $ BK\parallel ZL\parallel CM$ $ \Longrightarrow$ $ \frac {ZB}{ZC} \equal{} \frac {LK}{LM}$ $ ,(2)$

From $ (1),\ (2)$ $ \Longrightarrow$ $ \frac {B'E}{B'M} \equal{} \frac {LK}{LM} \equal{} \frac {C'K}{C'D}$ $ ,(3)$

From $ (3)$ $ \Longrightarrow$ $ LB'\parallel KE\parallel AB$ and $ LC'\parallel MD\parallel AC$ and the proof is completed.

Kostas Vittas.

Very nice proofs, indeed!
Thank you!

Best regards,
Stan Fulger


greatest common divisor
number theory solved
number theory

Prove that for any positive integer greater than 2 there are
infinitely many positive integers, such that the number
a^n - 1  is divisible by n. Is the same true for a = 2?

This was posted before here.

Here's the idea: for $a>2$ take a prime $p$ which divides $a-1$. Put $x_1=p,\ x_{n+1}=a^{x_n}-1$. You can easily see that the terms of the sequence $\{x_n\}$ satisfy $x_n|a^{x_n}-1$.

For the other part, assume we have $n|2^n-1$. Take $p$ to be the smallest prime divisor of $n$. Then $p|2^n-1,\ p|2^{p-1}-1$ (Fermat's thm).Then $p|(2^n-1,2^{p-1}-1)=2^{(n,p-1)}-1=2-1=1$, which is false (we used the fact that $(n,p-1)=1$ because $p$ was the [b]smallest[/b] prime divisor of $n$).

$(x,y)$ is the greatest common divisor of $n$.


LaTeX
logarithms
calculus
calculus computations

In this post the notation a(n) is used for the sequence a sub n , which means the usual meaning (because i don't know LATEX)

Let a(n) be real or complex, with a(n) different than -1 for all n. Suppose that the sum (from n=1 to infinity) of |a(n)| is convergent, say to S. Let p(n)=product (from r=1 to n) of [1+a(n)]. Prove that p(n)-p(n-1)=a(n)p(n-1) for n>=2 and |p(n)| <= exp(S) for all n.
Everything ok until now
Then it asks to deduce that the sum (from n=2 to infinity) of [p(n)-p(n-1)] is convergent, and hence that the sequence p(n) tends to a limit, say p, as n tends to infinity.
I have some trouble with those last bits, so your help would be more than valuable. Thanks

$ 0 \leq \sum_{n\equal{}2}^\infty |p_n \minus{} p_{n\minus{}1}| \equal{} \sum_{n\equal{}2}^\infty |a_n p_{n\minus{}1}| \leq e^S \sum_{n\equal{}2}^\infty |a_n|  \equal{} e^S \cdot (S \minus{} |a_1|) < \infty$.
So the sum is convergent.

Concerning LaTeX: the above is generated by typing

[code]
$0 \leq \sum_{n=2}^\infty |p_n - p_{n-1}| = \sum_{n=2}^\infty |a_n p_{n-1}| \leq e^S \sum_{n=2}^\infty |a_n|  = e^S \cdot (S - |a_1|) < \infty$
[/code]

Note the $ \$$ at the start and end to enter and exit Math mode.

Thanks Tyl,
any ideas about the proof that $ p_n$ tends to a limit?

Maybe this would help:

$ p_n \equal{} \prod ^\infty_{r\equal{}1}\left(1\plus{}a_r\right)$

$ \ln |p_n| \equal{} \sum ^\infty_{r\equal{}1}\ln \left(|1\plus{}a_r|\right) \le \sum ^\infty_{r\equal{}1}|a_r| < S \Rightarrow |p_n| < e^S$

[quote="milin"]Maybe this would help:

$ p_n \equal{} \prod ^\infty_{r \equal{} 1}\left(1 \plus{} a_r\right)$

$ \ln |p_n| \equal{} \sum ^\infty_{r \equal{} 1}\ln \left(|1 \plus{} a_r|\right) \le \sum ^\infty_{r \equal{} 1}|a_r| < S \Rightarrow |p_n| < e^S$[/quote]

Hmm, this has been used in the above, and the OP said that that wasn't a problem. However, we haven't yet proven that $ (p_n)$ tends to some limit, just that all of the terms in the sequence are finite.

I don't have full rigour by my side here, but I think it's along the lines of: "all terms are finite, the sum of the difference of consecutive terms of the sequence converges, hence the differences of consecutive terms tend to zero, hence $ (p_n)$ goes to some limit."

Edited to add brackets around $ p_n$. Not sure if you commented on my response or the previous one.

This is a question for the sequence $ (p_n)$, not the term $ p_n$, so we should prove that the sequence tends to a limit, and not just give a bound for the n-th term..

[quote="milin"]Maybe this would help:

$ p_n \equal{} \prod ^\infty_{r \equal{} 1}\left(1 \plus{} a_r\right)$

$ \ln |p_n| \equal{} \sum ^\infty_{r \equal{} 1}\ln \left(|1 \plus{} a_r|\right) \le \sum ^\infty_{r \equal{} 1}|a_r| < S \Rightarrow |p_n| < e^S$[/quote]

Oh, as a matter of fact, this should be:
$ p_n \equal{} \prod ^n_{r \equal{} 1}\left(1 \plus{} a_r\right)$

$ \ln |p_n| \equal{} \sum ^n_{r \equal{} 1}\ln \left(|1 \plus{} a_r|\right) \le \sum ^n_{r \equal{} 1}|a_r| \leq \sum ^\infty_{r \equal{} 1}|a_r| \equal{} S \Rightarrow |p_n| \leq e^S$


function search quadratics geometry rectangle algebra polynomial

"The number of cents pre kilometer it cost to drive a car depends on how gast you drive it. At low speeds the cost is high because the engine operates inefficiently, while at high speeds the cost is high because the engine must overcome wind resistence. At moderate sppes the cost reaches a minimum. Assume, therefore, that the number of cents per kilometer varies quadratically with the number of kilometer per hour (kph). "

then it asked: "Supppse that it costs 28, 21, and 16 cents pre kilometer to drive at 10, 20, 30 kph, respectively. Write the function"
I tried to plugged it into my graphing calculator by goin to STAT and edit and put 28, 21, 16 (in that order) in the L1 collumn. then i put 10, 20, 30 kph (in that order) in L2. then i went to STAT again and click on "LinReg(ax+b)" and i got a function.
my problem is that later on, it asked me "how much would you spend to drive at 150 kph?" I use the function and replace the y with 150 and look for X. i got a negative answer. It is not possible to pay a negative cents.

is my function wrong? if it is how would i get the right function. and what does varies quadratically means?

1) This is a homework question. We are not here to do your homework for you.
2) It seems like you simply don't know the concept that you are working with. I recommend you do a search for the term "quadratic" on the internet or on Mathworld. I'll give you a small hint, though: you assumed linear dependence. This was a bad assumption for three big reasons. The first reason is that the question didn't say anything about linear dependence of functions. The second is that those three points do not in fact lie on a straight line. The third is that the question went to great pains to give you some idea of what quadratic dependence is: "At low speeds the cost is high because the engine operates inefficiently, while at high speeds the cost is high because the engine must overcome wind resistence. At moderate sppes the cost reaches a minimum." That whole lead-in was not just for the amusement of the question-writer.

"Varies quadratically" means the function you are looking for is in the form cost per kilometer = C(s) = as^2 + bs + c where a, b, and c are constants and s is the speed in kilometers per hour.  You should not need a calculator from here, as three points can determine a unique quadratic function.  Indeed, I'm not sure if a calculator is capable of specifically quadratic regression.

Your calculator does have a quadratic regression function.
However, in addition to JBL's list of grievances, you should also be careful when you're extrapolating, especially finding something at 150, when you're given values for 10, 20, 30. How do you know that the pattern will follow so far from what you're given? I know this is probably a homework problem, but in the real world, you should be wary when extrapolating.

I am bad and so I will tell you how to get the correct equation without calculating it, but you need access to a computer with Microsoft Excel.

Start Excel and enter the speed values 10, 20, 30 mph into the 1st column A (cells A1, A2, A3). Then enter the cost values 28, 21, 16 cents per mile into the 2nd column B (cells B1, B2, B3). Highlight the rectangle A1-B3 filled with numbers (click the left mouse button on the upper left cell and dragg it to the lower right cell). Open the <Insert> menu and select <Chart>. In the dialog box, select the chart type <XY(Scatter)> and click the <Finish> button. A chart with 3 data point appears.

Select the data points by clicking any one data point with the left mouse button.  Open the <Chart> menu and select <Add Trendline>. In the dialog box, select the the trend/regression type <Polynomial> and make sure that the order (of the polynomial) is 2 - that is, quadratic. Click the <Options> tab, check the option "Display equation on chart" and click the <OK> button. You should see the quadratic equation - record it on a piece of paper.

Select the trendline by clicking it with the left mouse button and from the <Format> menu, select <Selected Trendline>. Alternately, click the trendline with the right mouse button and from the floating menu, choose <Format Trendline>. The previous dialog box shows up again. Click the <Options> tab and enter <Forcast>: Forward 120, Backward -10. This wil extend the trendline 10 mph below the lowest speed data point ( to 10 - 10 = 0 mph) and 120 mph above the highest speed data point ( to 30 + 120 = 150 mph). What do you see ? A minimum ?

Open the <Format Trendline> dialog box again (as in the previous paragraph), change the Trend/Regression Type to <Linear> and observe, what happened to the trendline and the equation. Compare the equation with the equation you got using the graphic calculator. Do you see a minimum now ?

You should have an equation

$y = ax^2 + bx + c$

where x is the speed in mph and y the cost in cents per mile. Hopefully, you remember that a quadratic equation has a single solution, when its discriminant $D = b^2 - 4ac = 0$ and the single solution is then $x = \frac{-b}{2a}$. This means that the parabola just touches the x-axis at its minimum. And the minimum is always at $x = \frac{-b}{2a}$, even if the parabola intersects the x- axis at 2 points or does not intersect it at all. What is the speed x, at which the cost y is minimum ?

Now, you can calculate the coefficients a, b, c and they should agree with what you got from Excel. You have 3 pairs of points $[x, y] = [10, 28], [20, 21], [30, 16]$. They all lie on the parabola and therefore, all 3 have to obey the equation of the parabola:

$28 = a \cdot 10^2 + b \cdot 10 + c$
$21 = a \cdot 20^2 + b \cdot 20 + c$
$16 = a \cdot 30^2 + b \cdot 30 + c$

These are 3 equations for 3 unknowns a, b, c and you should be able to solve them, to calculate a, b, c.

I hope that you will come back and report in details on what you did and what results you got. Otherwise, I will never ever answer a question from you again.

yetti


Hello,

Given that obviously there are many music players here in the forum, I was wondering if you think that some age is really too late to learn an instrument.

No body denies that if you start playing very young, say ten years old, probably you will be better than someone who started playing at twenty. I have heard, for example, people saying that after the 15 years old is almost imposible to learn how to play violin (I have never tried).

The truth is that I learned to play floute very young, I was never really good at it, and it has been about 12 years since the last time I practiced. I tried to learn how to play the guitar and piano, but I did not succeed on any of those, and in general I consider I am not good at plastic or musical arts, but may be there can be someone about my age or older (I am 24) who has never play any instrument, and suddenly learns and turn to be very good.

Moderator: I posted it here instead of The Round Table, because I thought it was more oriented to Music than to General Discussion.

I don't think it's too late anytime.  I know that my piano teacher used to teach, or maybe still does, teaches seniors, so I think 24 is an okay. age.  It just may be harder to learn.

I don't think its ever too late to learn, as long as you can really enjoy playing.

Yeah i think that it is never too late to learn... playing an instrument can really help when you're stressed out

you've probably got to be really dedicated to play an instrunment cause you don't have a parent to command you to play an instrunment anymore.

My piano teacher told me about a man who started to play when he was 26.  I can't remember the name though..(this was like 2 years ago)  I do remember him saying how he went on to play concert piano and was a very very respected performer.

I think the best neurological evidence is that anyone at any age who practices enough and listens to enough good music can learn to play a musical instrument. The advantage young learners have is mostly TIME--they are supported by someone else, usually, and thus have more opportunity to practice.

Studies have shown that people who begin their musical training early are more likely to hold onto their absolute pitch (perfect pitch) recognition abilities, but for the rest of the population, being successful in music requires dedication and talent. But mainly dedication.

As for age...there have been many stories of musical savants (ok, so most of us here are not musical savants) who can sit down at a piano for the first time at age 40 and are able to spontaneously play something they heard when they were 7 (i.e. Beethoven's 5th) without any prior musical training. So I don't think age has much to do with it.

Your brain develops a lot before you're 20...ESPECIALLY between the ages of 13 and 18 about.  So, by being exposed to an instrument and/or playing it at an early age, you'll get a better appreciation for it and that kind of music, and will have a better ear.  It is not impossible at all to start above the age of 20...it's just extremely hard.  Mostly the best people started at an age of about 3 years and were made to practice a lot.

Although I wanted to learn piano when I was young, I didn't have the opportunity until I was 28 years old.  The only other instrument I was familiar with was the recorder, which I picked up in 4th grade.  My first piano teacher was an enthusiastic Chinese woman from Taiwan who allowed me to move quickly through the beginning pieces.  I practiced every day and enjoyed the thrill of making music.

All went well until the first recital.  A stream of tiny kids preceded me onto the stage, all playing perfectly.  Then it was my turn.  I started my piece, which I knew inside out.  Or so I thought.  Then a wave of stage fright hit me for the first time in my life.  I stumbled through the rest of the piece, played the final chords, and flew off the stage, catching a glimpse of my teacher's shocked expression as I exited the building.

The end of my piano career?  Actually no.  I played just fine at home and a performance career had never been my goal so I went back to the piano.  I loved making music -- as long as I didn't have to play in front of a crowd.  We moved to another city and I found another excellent teacher, this time a Russian woman.  Soon I was playing Bach two-part and three-part inventions.

Then the subject of recitals came up again.  My Russian teacher refused to accept my self-diagnosis of recital-phobia so -- masochist that I was -- I performed again.  This time, I was prepared for the possibility of total meltdown so before the performance I focused on calming myself down and visualizing myself playing perfectly.  I performed a couple of short Tchaikovsky pieces, missed only a couple of notes, and sat down with an immense feeling of relief.  I had one more final recital, playing the lovely Chopin Nocturne in C-sharp minor, before retiring permanently.  When my toddler son insisted on climbing onto my lap every time I sat down at the piano, I hung up the keys, so to speak.

I envy my kids, who can play by ear, who are totally comfortable on stage.  I think the stage fright I experienced was partly because of my inability to play by ear.  Once I lost my place during that first performance, all those white keys looked remarkably the same.  My piano training lasted about four years.  I learned some difficult pieces during that time, especially the Bach.  These days I don't play, mostly because I never learned to sightread well.  I don't know if this is because I learned music as an adult, but even with lots of practice, I found sightreading difficult.

The educator John Holt wrote an excellent account of his efforts to learn cello as an adult, called [i]Never Too Late: My Musical Life Story[/i].  NPR host Noah Adams took up piano at age 51 and wrote about it in [i]Piano Lessons: Music, Love, and True Adventures[/i].

So, to end this very long message, any adult interested in picking up a new instrument should go for it!  I might go back to the piano myself.

I've been playing the violin since i was five. I think that it is never too late to learn, but as you grow up, it gets harder to learn new things...

Look, I think that it is VERY frustrating to learn music for the first time. I have had the painful experience of forcing myself through practicing guitar.

I have learned various things from this experience that I think will help you a lot  :lol: .

First, choose one instrument and stick to it. Second, getting a teacher may not seem like a good idea with the high costs but it is DEFINETLY worth it in the end. Third, posture is one of the most important thing for begginers, believe it or not. Finally, whenever you practice you should have a kind of agenda to follow through instead of randomly playing.

I don't know if that's all commen sense but I could have been much better if I knew this before I began playing.

I really sound... arrogant.. or something to that.. please let it pass.

There's no doubt that learning an instrument at a young age is easier than as an adult. I myself have been playing the piano for over ten years - since I was four. But, I will repeat what others have said before me: Give it a try! It's never too late to learn.

It's tooooooo late .....when you're dead. :D
Anyways , you can learn an instrument at any age , provided you have the time for it.....{ beyond a certain age , most people get involved with something or the other , and invariably end up not being able to practise properly.}

[quote=math92]it's never too late[/quote]


Holder, Cauchy, Cauchy...

${{(\sum_{\text{cyc}} \frac{a^3}{3a+2b+c})
(\sum_{\text{cyc}} 3a+2b+c})
(\sum_{\text{cyc}} 1}) \geq (a+b+c)^3$

${(\sum_{\text{cyc}} \frac{b^2}{3a+2b+c})
(\sum_{\text{cyc}} 3a+2b+c}) \geq (a+b+c)^2$

${(\sum_{\text{cyc}} \frac{c}{3+2a+b})
(\sum_{\text{cyc}} c(3a+2b+c)}) \geq (a+b+c)^2$

Using the fact that $a^2+b^2+c^2 \geq ab+bc+ca$ in the last inequality, divide and add 'em up!

Alternatively, you could prove the first part by Cauchy too...

${\displaystyle (\sum_{\text{cyc}} \frac{a^3}{3a+2b+c})
(\sum_{\text{cyc}} a(3a+2b+c)})
\geq (a^2+b^2+c^2)^2$

Now note that

$2\cdot 3(a^2+b^2+c^2)^2 \geq (a+b+c)^2 (a^2+b^2+c^2+ab+bc+ca)$

follows from

$3(a^2+b^2+c^2) \geq (a+b+c)^2$
$2(a^2+b^2+c^2) \geq a^2+b^2+c^2+ab+bc+ca$


complex analysis
real analysis
real analysis theorems

Hi..Who can guids me to a site or a n eBook that contains
solutions of excersices'Rudin book (real and complex analysis)
thanks.

I'm pretty sure that a complete set of solutions don't exist.  Do them yourself.

But I need it .. :o
What dose I do then?
Thanks.