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[math H] Convergence Test (Calculus level)

Blogs > evanthebouncy!
Post a Reply
evanthebouncy!
Profile Blog Joined June 2006
United States12796 Posts
Last Edited: 2009-02-20 11:41:18
February 20 2009 09:07 GMT
#1
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?



Life is run, it is dance, it is fast, passionate and BAM!, you dance and sing and booze while you can for now is the time and time is mine. Smile and laugh when still can for now is the time and soon you die!
Nytefish
Profile Blog Joined December 2007
United Kingdom4282 Posts
February 20 2009 09:42 GMT
#2
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

No I'm never serious.
Cpt Obvious
Profile Blog Joined November 2006
Germany3073 Posts
February 20 2009 09:44 GMT
#3
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.
Nobody ever reads signatures of people like me, do they?
datscilly
Profile Blog Joined November 2007
United States528 Posts
Last Edited: 2009-02-20 09:52:08
February 20 2009 09:46 GMT
#4
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.
evanthebouncy!
Profile Blog Joined June 2006
United States12796 Posts
Last Edited: 2009-02-20 11:42:16
February 20 2009 09:50 GMT
#5
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.
Life is run, it is dance, it is fast, passionate and BAM!, you dance and sing and booze while you can for now is the time and time is mine. Smile and laugh when still can for now is the time and soon you die!
evanthebouncy!
Profile Blog Joined June 2006
United States12796 Posts
February 20 2009 09:52 GMT
#6
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? :/
Life is run, it is dance, it is fast, passionate and BAM!, you dance and sing and booze while you can for now is the time and time is mine. Smile and laugh when still can for now is the time and soon you die!
datscilly
Profile Blog Joined November 2007
United States528 Posts
February 20 2009 10:20 GMT
#7
On February 20 2009 18:50 evanthebouncy! wrote:
Show nested quote +
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


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?
fight_or_flight
Profile Blog Joined June 2007
United States3988 Posts
Last Edited: 2009-02-20 10:44:13
February 20 2009 10:31 GMT
#8
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]
Do you really want chat rooms?
Wonders
Profile Blog Joined September 2006
Australia753 Posts
February 20 2009 10:33 GMT
#9
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].
Kentor *
Profile Blog Joined December 2007
United States5784 Posts
February 20 2009 10:45 GMT
#10
it converges. but how?????
silynxer
Profile Joined April 2006
Germany439 Posts
February 20 2009 11:22 GMT
#11
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).
evanthebouncy!
Profile Blog Joined June 2006
United States12796 Posts
February 20 2009 11:41 GMT
#12
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.
Life is run, it is dance, it is fast, passionate and BAM!, you dance and sing and booze while you can for now is the time and time is mine. Smile and laugh when still can for now is the time and soon you die!
mikeymoo
Profile Blog Joined October 2006
Canada7170 Posts
February 20 2009 17:39 GMT
#13
You got your problem solved, so I'm gonna say gratz on new job!
o_x | Ow. | 1003 ESPORTS dollars | If you have any questions about bans please PM Kennigit
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