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Ah so it converges? I see. Thanks.
So I got a new tutor job(yay) and a student brought to me a question I cannot solve. I'd like to see how someone could do it though.
We want to test for convergence/divergence for
Integrate from 0 to 1 of function f(x)
where
f(x) = 1/[(1-x^4)^(1/3)]
So f(x) approaches infinity at x = 1, what we might want to do is to bound f(x) below with some other g(x), show that g(x) diverges, therefore f(x) also diverges. However I am unable to bound it with something easy enough to tell that it actually diverges...
The intuition is that the integration from 0 to 1 of f(x) diverges, because x^(4/3) < x if x is between 0 and 1, and online definite integrator tells me it diverges as well. But how do we bound it?
Any ideas?
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Can you compare with 1/[(1-x^4)^(1/4)] and look at a substitution x^2=sin(u)?
I don't think it'll work, but I gotta go so no time to try it out 
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Well, the integral over f(x)=x doesn't diverge though, not even if you integrate from minus infinity to infinity.
What you might want to do is find a function like g(x) = 1/1-x, show that g(x) < f(x) for x between 0 and 1, and then be finished because g(x) obviously diverges. The "<" is important here, you want to find a minorant that diverges, not a majorant that diverges, because that wouldn't prove anything (What matters here is the limes for x->1, which is +infinity. If it were -infinity, you'd still have to find a diverging majorant).
Basically look through all basic fractal functions that diverge at x=1, find one that is always < f(x) in the interval and diverges, be finished.
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I think that the integral of (1-x^4)^(-1/3) from 0 to 1 does in fact converge. Using this integrator, http://integrals.wolfram.com/index.jsp , we have
x*Hypergeometric2F1[1/4, 1/3, 5/4, x^4] as the indefinite integral. Then looking up , http://en.wikipedia.org/wiki/Hypergeometric_series , we find that 2F1(a, b; c; 1) converges when c-a-b>0.
I also used the first definite integrator to come up on googling "definite integrator", but it gives the result "NaN", and I'm guessing it's wrong- it's not powerful enough for this integral.
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On February 20 2009 18:46 datscilly wrote:I think that the integral of (1-x^4)^(-1/3) from 0 to 1 does in fact converge. Using this integrator, http://integrals.wolfram.com/index.jsp , and then looking up , http://en.wikipedia.org/wiki/Hypergeometric_series , we get that 2F1(a, b; c; 1) converges when c-a-b>0. I also used the first definite integrator to come up on googling "definite integrator", but it gives the result "NaN", and I'm guessing it's wrong- it's not powerful enough for this integral.
The result NaN means it diverges, and I doubt the student needs to know about hypergeometric_series considering she's having problems doing u_substitutions
Take it back. If it actually converges, we must bound it.
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On February 20 2009 18:44 Cpt Obvious wrote:
Well, the integral over f(x)=x doesn't diverge though, not even if you integrate from minus infinity to infinity.
What you might want to do is find a function like g(x) = 1/1-x, show that g(x) < f(x) for x between 0 and 1, and then be finished because g(x) obviously diverges. The "<" is important here, you want to find a minorant that diverges, not a majorant that diverges, because that wouldn't prove anything (What matters here is the limes for x->1, which is +infinity. If it were -infinity, you'd still have to find a diverging majorant).
Basically look through all basic fractal functions that diverge at x=1, find one that is always < f(x) in the interval and diverges, be finished.
Haha you are the obvious 
Give me some actual basic factual functions to try out, don't just say find some because no crap I already know I need to find some. I already stated in my question, I need some basic factual functions, so you got any suggestions? :/
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On February 20 2009 18:50 evanthebouncy! wrote:The result NaN means it diverges, and I doubt the student needs to know about hypergeometric_series considering she's having problems doing u_substitutions
I know that "NaN" means that it diverges. What I meant was that even though I acknowledge that the online definite integrator says it diverges, it actually converges. If the most powerful math software gives such a result, then it should be seriously considered, and it does say that it converges, with the caveat that the hypergeometric series has been understood correctly.
Both the TI-83 and Mathematica give the same answer: 1.16... for the TI and Gamma(5/4)*Gamma(2/3)/Gamma(11/12) = 1.16... for Mathematica. Is it possible that your student gave you was a challenge question of some sort?
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On February 20 2009 18:52 evanthebouncy! wrote:Show nested quote +On February 20 2009 18:44 Cpt Obvious wrote:
Well, the integral over f(x)=x doesn't diverge though, not even if you integrate from minus infinity to infinity.
What you might want to do is find a function like g(x) = 1/1-x, show that g(x) < f(x) for x between 0 and 1, and then be finished because g(x) obviously diverges. The "<" is important here, you want to find a minorant that diverges, not a majorant that diverges, because that wouldn't prove anything (What matters here is the limes for x->1, which is +infinity. If it were -infinity, you'd still have to find a diverging majorant).
Basically look through all basic fractal functions that diverge at x=1, find one that is always < f(x) in the interval and diverges, be finished. Haha you are the obvious Give me some actual basic factual functions to try out, don't just say find some because no crap I already know I need to find some. I already stated in my question, I need some basic factual functions, so you got any suggestions? :/
you mean fractal i think
yea same as poster below me
![[image loading]](http://img8.imageshack.us/img8/8908/fb6ae0ea607a05d6711dfb9yp3.png)
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Are you sure it diverges? Because I typed this into Mathematica:
Integrate[1/((1 - x^4)^(1/3)), {x, 0, 1}] (this just evaluates the integral),
and got
(Gamma[2/3] Gamma[5/4])/Gamma[11/12].
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Kentor
United States5784 Posts
it converges. but how?????
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Kind of lame to start posting again after hiding in a hole for years (out of shame for being banned for extreme stupidity) in a thread like this but if I'm not mistaken I solved the problem and it would have been a waste of time not to post the solution:
Consider the Integral over (1/(1-x^2)^(1/2))=arcsin(x)=PI/2 for x=1 and obviously the convergent majorant you are searching for (due to limitations of notation a not entirely formal correct description but I'm sure you can do the rest yourself).
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On February 20 2009 20:22 silynxer wrote:
Kind of lame to start posting again after hiding in a hole for years (out of shame for being banned for extreme stupidity) in a thread like this but if I'm not mistaken I solved the problem and it would have been a waste of time not to post the solution:
Consider the Integral over (1/(1-x^2)^(1/2))=arcsin(x)=PI/2 for x=1 and obviously the convergent majorant you are searching for (due to limitations of notation a not entirely formal correct description but I'm sure you can do the rest yourself).
Thanks.
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Canada7170 Posts
You got your problem solved, so I'm gonna say gratz on new job!
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EPT
