Limits, yes another limits thread.

MathematicalPhysicist

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i need to compute lim (a^n-b^n)^(1/n) when a>b>0.
lim ((3n)!/((2^3n)n!(2n!)))^(1/n) where i need to use the lemma that:
if an>0 for every n, and lim(x_n+1/x_n)=L then lim x_n^(1/n)=L, how to use it here?

for the first i used practically everything i know, the formual for a^n-b^n, and the fact that 0<a^n-b^n<a^n and lots more algebraic techniques, apparently not everything.

your help is appreciated.
 

MathematicalPhysicist

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another thing, seemingly unrealted.
i need to show that for 0<x<2pi x>=sinx.
i tried to dissect it into parts, i.e:
x>1 and for x<1, for x>1 it's obvious, my problem is with x<1, then how to solve it?
 

arildno

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It is relatively simple to show that there IS a limit, and that it lies between a and b, however any more precise than that, I don't know.

Use the following identity in a clever manner:
[tex]a^{n}-b^{n}=(a-b)\sum_{i=0}^{n-1}a^{n-1-i}b^{i}[/tex]
 
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HallsofIvy

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The last one, to show that [itex]x\ge sin x[/itex] for all [itex]0\le x\le 2\pi[/itex] is easy. What is the derivative of f(x)= x- sin(x)? For what values of x is that positive? Since sin(0)= 0, what does that tell you?
 

MathematicalPhysicist

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oh, come on halls, you don't know a proof which doesnt employ derivatives, sure it's easy with derivative but i want to show it without.
 

arildno

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Well, what's wrong with a geometrical argument on the unit circle, then?
 

MathematicalPhysicist

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It is relatively simple to show that there IS a limit, and that it lies between a and b, however any more precise than that, I don't know.

Use the following identity in a clever manner:
[tex]a^{n}-b^{n}=(a-b)\sum_{i=0}^{n-1}a^{n-1-i}b^{i}[/tex]
as i said i tried this equality, i got something like this: (a-b)a^n>=(a-b)(a^(n-1)+a^(n-2)b+....+b^n-2a+b^(n-1))>=(a-b)(b^(n-1)+...+b^(n-1))=(a-b)b^n but as you said that's easy to show, the problem is to what does it converge?
any other tips?

btw, i also need some help in the other limit.

as always your help is appreciated.
(my custom mantra (-: ).
 

MathematicalPhysicist

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Well, what's wrong with a geometrical argument on the unit circle, then?
i dont follow you, how to use the unit circle here?
i mean the identity which seems to be good here is sin^2x+cos^2x=1
sin^2x=1-cos^2x
sinx/x=sqrt(1-cos^2x)/x<=1 when 0<sinx<=1
cause 1-x^2<=cos^2x<=1
is this correct?
 

arildno

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Remember that in general, we have NO infallible technique that enables us to "calculate" a limit.
For practical purposes (say when solving a non-linear system numerically), we use the Cauchy criterion to say we have "essentially" reached the limit.
But formally, of course, this is bogus.


What we DO have, is first and foremost a PROPERTY that the limit must have. It is by no means guaranteed by this that we actually manage to FIND that number.
 

MathematicalPhysicist

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so to sum up, you don't know how to calculate the limit, correct?
if it's any good, i got a hint to use the sandwich lemma.
 

arildno

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arildno

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i dont follow you, how to use the unit circle here?
i mean the identity which seems to be good here is sin^2x+cos^2x=1
sin^2x=1-cos^2x
sinx/x=sqrt(1-cos^2x)/x<=1 when 0<sinx<=1
cause 1-x^2<=cos^2x<=1
is this correct?
No. What length does "sin(x)" represent on the unit circle, and where can you find the length "x" on the same circle?
Compare those lengths!
 

MathematicalPhysicist

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ah, i see your point. x is the length of the arc and sinx is the length of the y coordinate on the coordiantion system.
 

arildno

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Of course, the geometrical argument rests also upon the assumption that the straight line is the shortest distance between two points..
 

MathematicalPhysicist

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about the limit with a and b, i think it converges to a, the problem is how to prove it, obviously [tex](a^n-b^n)^{\frac{1}{n}}<a[/tex]
one way to show that it converges to a is by the sandwich lemma, the problem is to find the suitable lower bound, i thought perhaps a^n/n can be a good candidate, but im not sure this is correct, can someone help?
 

matt grime

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It's always easier to get rid of two things that depend on n, and replace it with one.

What can you say about the convergence, or otherwise, of

[tex] (1-r^n)^{1/n}[/tex] where 0<r<1?

It is quite straight forward just using the most naive bounds there are for (1-r^n) (assuming we don't think 0 is a naive lower bound, I mean).
 
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MathematicalPhysicist

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it converges to 1, is it not?
anyway, i need to find a suitable lower bound for the limit.
 

MathematicalPhysicist

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It's always easier to get rid of two things that depend on n, and replace it with one.

What can you say about the convergence, or otherwise, of

[tex] (1-r^n)^{1/n}[/tex] where 0<r<1?

It is quite straight forward just using the most naive bounds there are for (1-r^n) (assuming we don't think 0 is a naive lower bound, I mean).
obviously it's smaller than 1, but from what it's bigger?
it's bigger than 1-r^n cause (1-r^n)^n<=1-r^n and 1-r^n<1.
will this also work fro number besides 1 and r<1, i mean:
0<b<a (a^n-b^n)^(1/n)<a, but is it: a^n-b^n<(a^n-b^n)^(1/n)
in order for to be true, a^n-b^n should be smaller than 1.
 

matt grime

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Think about it a little bit longer, please.... For instance, have you figured out what relation I might intend r to have with a and b? How did I get from 0<b<a to 0<r<1?


Why have you not used the fact that r^n<r , for instance? You don't appear to have thought the suggestion through at all.
 

MathematicalPhysicist

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well you replaced r with b and 1 with, and i did you use the fact that r^n<r when i said that (1-r^n)^n<1-r^n obviously r^n<r<1 and thus also what i wrote is correct.
anyway, i think i understand.
r^n<r
then 1-r^n>1-r so we have a lower bound and an upper.
but my question still stands, how can i use it here?
i mean a>b>0 so a^n>a^n-b^n but i cannot say anything from lower besides 0, cause im not given that b<1.

perhaps you want to dissect this problem into situations where 0<b<a<1
0<b<1<a 1<b<a, i thought about this and obviously this is the way to go, but i thought perhaps there's a starightforward approahc which doesnt employ splitting into three situations.
 

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