# Integral Convergence and Taylor Series

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For this integral:

If I was asked to work out the values of n for which the integral converges, then would the answer be the same if I was asked to work out the values of n for which the integral

Thank you!

For Taylor series, is this correct in the attached image.

In the previous questions, we had n = 1/2 and hence the square root.

If I was asked to work out the values of n for which the integral converges, then would the answer be the same if I was asked to work out the values of n for which the integral

**absolutely**converges, since the integrand is non-negative over the range of integration?Thank you!

For Taylor series, is this correct in the attached image.

In the previous questions, we had n = 1/2 and hence the square root.

Last edited by Takeover Season; 5 months ago

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(Original post by

For this integral:

If I was asked to work out the values of n for which the integral converges, then would the answer be the same if I was asked to work out the values of n for which the integral

**Takeover Season**)For this integral:

If I was asked to work out the values of n for which the integral converges, then would the answer be the same if I was asked to work out the values of n for which the integral

**absolutely**converges, since the integrand is non-negative over the range of integration?For that integral, the only bad point is just .

As we get that behaves like by simply expanding the cosine function about this point.

So .

Hence the integral behaves like

For which n does it converge? The substitution would put it in a more familiar form that you may be used to.

Last edited by RDKGames; 5 months ago

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#3

I am also unsure why you're integrating over since the function doesn't even exist for the upper half of the region.

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(Original post by

Yes. If over some interval then .

For that integral, the only bad point is just .

As we get that behaves like by simply expanding the cosine function about this point.

So .

Hence the integral behaves like

For which n does it converge? The substitution would put it in a more familiar form that you may be used to.

**RDKGames**)Yes. If over some interval then .

For that integral, the only bad point is just .

As we get that behaves like by simply expanding the cosine function about this point.

So .

Hence the integral behaves like

For which n does it converge? The substitution would put it in a more familiar form that you may be used to.

Yes, that is one of our standard test functions and so it converges for n - 1 < 1, so n < 2!

Last edited by Takeover Season; 5 months ago

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(Original post by

I am also unsure why you're integrating over since the function doesn't even exist for the upper half of the region.

**RDKGames**)I am also unsure why you're integrating over since the function doesn't even exist for the upper half of the region.

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(Original post by

hi

**RDKGames**)hi

as , then and so as converges for all n < 1, so does our original integral?

Last edited by Takeover Season; 5 months ago

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#7

(Original post by

Also, why can't we say that:

as , then and so as converges for all n < 1, so does our original integral?

**Takeover Season**)Also, why can't we say that:

as , then and so as converges for all n < 1, so does our original integral?

We have already shown that the integrand converges for n<2 so doing this method instead and showing that it diverges for n<1 misses out the values 1<n<2 for which the integral also converges for.

Approximating in the limit is more precise than just slapping on the global maximum.

Last edited by RDKGames; 5 months ago

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(Original post by

You can but it's not as precise.

We have already shown that the integrand converges for n<2 so doing this method instead and showing that it diverges for n<1 misses out the values 1<n<2 for which the integral also diverges for.

Approximating in the limit is more precise than just slapping on the global maximum.

**RDKGames**)You can but it's not as precise.

We have already shown that the integrand converges for n<2 so doing this method instead and showing that it diverges for n<1 misses out the values 1<n<2 for which the integral also diverges for.

Approximating in the limit is more precise than just slapping on the global maximum.

Also, do you mean "misses out the values 1<n<2 for which the integral also

**converges**for".

Also, if we use the Taylor's theorem, we can always ensure we get a

**correct**test whether using the limit comparison test or the direct comparison test right as it gives you the perfect bound? Like here we used it y = cos x is bounded above by y = 1.

But, Taylor's expansion would tell us that for , and then we can use the direct comparison test to show that and then use the direct comparison test for and show this converges for n < 2.

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(Original post by

Oh, so we can show it using the direct comparison test, but it may miss out some of the values for which the integral converges?

Also, do you mean "misses out the values 1<n<2 for which the integral also

Also, if we use the Taylor's theorem, we can always ensure we get a

But, Taylor's expansion would tell us that for , and then we can use the direct comparison test to show that and then use the direct comparison test for and show this converges for n < 2.

**Takeover Season**)Oh, so we can show it using the direct comparison test, but it may miss out some of the values for which the integral converges?

Also, do you mean "misses out the values 1<n<2 for which the integral also

**converges**for".Also, if we use the Taylor's theorem, we can always ensure we get a

**correct**test whether using the limit comparison test or the direct comparison test right as it gives you the perfect bound? Like here we used it y = cos x is bounded above by y = 1.But, Taylor's expansion would tell us that for , and then we can use the direct comparison test to show that and then use the direct comparison test for and show this converges for n < 2.

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(Original post by

Yes.

**RDKGames**)Yes.

If I wanted to test for the absolute convergence of an integral because it oscillates and it had an improper integral of the second kind, or it had a problem point either within the interval or at its endpoints. Then, I need to see the Taylor series of the absolute value of the function near the problem point. If I had f(x) was negative over [a,b], how do I find the taylor series of |f(x)| at x = a? Please look at my original post if you wish, I said my explanation there and I was wondering if the approach was correct.

I'd just find the taylor series of -f(x) at x = a and that'd give me the taylor series of |f(x)| at x = a?

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