A question about definite integrals and series limits

[KungPeng Zhou]

Homework Statement

$a_{n}=\int_{0}^{2-\sqrt{3}}\frac{1-x^{4n}}{1+x^{2}}dx$, evaluate $\underset{n\rightarrow \infty} {\lim}a_{n}$

Relevant Equations

FTC

In my opinion , if it can be shown that this is a monotonically bounded sequence, one can confirm that there is a limit.
First，we know $$ \frac{1-x^{4n}}{1+x^{2}}dx=(1-x^{2}) (1+x^{2}) ^{n-1}=(1-x^{4}) ^{n-1}(1+x^{2}).$$
According to the integral median theorem，we can get $$a_n=(2- \sqrt{3} ) (1-\alpha^{4})^{n-1}(1+x^{2}), \alpha\in[0,2-\sqrt{3}]$$
But I don't know how to continue with the question.

 

[PeroK]

Can you fix your Latex?. You need two dollar signs.

 

[fresh_42]

Can you fix your Latex?. You need two dollar signs.

Fixed it. I think.

 

[Mark44]

First，we know $$ \frac{1-x^{4n}}{1+x^{2}}dx=(1-x^{2}) (1+x^{2}) ^{n-1}=(1-x^{4}) ^{n-1}(1+x^{2}).$$

I don't think we know this at all.
If n = 2, we have this:
$$\frac{1-x^8}{1+x^2} = \frac{(1 - x^4)(1 + x^4)}{1+x^2}$$
$$=\frac{(1 + x^2)(1 - x^2)(1 + x^4)}{1 + x^2} =(1 - x^2)(1 + x^4)$$

I don't see how my final result can be manipulated to get either of your two results.

Minor nit: Your equation shouldn't include dx.

 

[pasmith]

Homework Statement: $a_{n}=\int_{0}^{2-\sqrt{3}}\frac{1-x^{4n}}{1+x^{2}}dx$, evaluate $\underset{n\rightarrow \infty} {\lim}a_{n}$
Relevant Equations: FTC

In my opinion , if it can be shown that this is a monotonically bounded sequence, one can confirm that there is a limit.
First，we know $$ \frac{1-x^{4n}}{1+x^{2}}dx=(1-x^{2}) (1+x^{2}) ^{n-1}=(1-x^{4}) ^{n-1}(1+x^{2}).$$

Are you sure about that? What you have written for the numerator is $$1 - x^{4n} = (1 - x^{2n})(1 + x^{2n}).$$ Did you mean $(1 - x^4)^n$?

If so, you end up with a polynomial in $x$, which you can integrate analytically: $$\begin{split} a_n &= \int_0^{2 - \sqrt{3}} (1 - x^4)^{n-1}(1 - x^2)\,dx \\ &= \sum_{k=0}^{n-1} \binom{n-1}{k}(-1)^k \int_0^{2-\sqrt{3}} x^{4k} - x^{4k + 2}\,dx. \end{split}$$

 

[Mark44]

Are you sure about that? What you have written for the numerator is $$1 - x^{4n} = (1 - x^{2n})(1 + x^{2n}).$$ Did you mean $(1 - x^4)^n$?

As far as I can tell, no, that's not what the OP meant.
The original text of the numerator, before it was formatted by another mentor, was "1 - x^{4n}". The braces around 4n suggest to me that the intended exponent on x was 4n. I.e., that the numerator wasn't intended to be $(1 - x^4)^n$.

 

[KungPeng Zhou]

Can you fix your Latex?. You need two dollar signs.

Are you sure about that? What you have written for the numerator is $$1 - x^{4n} = (1 - x^{2n})(1 + x^{2n}).$$ Did you mean $(1 - x^4)^n$?

If so, you end up with a polynomial in $x$, which you can integrate analytically: $$\begin{split} a_n &= \int_0^{2 - \sqrt{3}} (1 - x^4)^{n-1}(1 - x^2)\,dx \\ &= \sum_{k=0}^{n-1} \binom{n-1}{k}(-1)^k \int_0^{2-\sqrt{3}} x^{4k} - x^{4k + 2}\,dx. \end{split}$$

Maybe I have found a better way to solve the question. $a_n=\int_{0}^{2-\sqrt{3}}\frac{1-x^{4n}}{1+x^{2}}dx=\int_{0}^{2-\sqrt{3}}\frac{1}{1+x^{2}}dx-\int_{0}^{2-\sqrt{3}}\frac{x^{4n}}{1+x^{2}}dx=arctan(2-\sqrt{3}) -arctan0-\int_{0}^{2-\sqrt{3}}\frac{x^{4n}}{1+x^{2}}dx=\frac{π}{12}-\int_{0}^{2-\sqrt{3}}\frac{x^{4n}}{1+x^{2}}dx$
It is apparently that we know,
$0<\int_{0}^{2-\sqrt{3}}\frac{x^{4n}}{1+x^{2}}dx<\int_{0}^{2-\sqrt{3}}x^{4n}dx<\int_{0}^{1}x^{4n}dx<\frac{1}{1+4n}$，according to Squeeze Theorem, we can get the answer easily. So the answer is $\frac{π}{12}$

 

[Mark44]

It is apparently that we know, $0<\int_{0}^{2-\sqrt{3}}\frac{x^{4n}}{1+x^{2}}dx<\int_{0}^{2-\sqrt{3}}x^{4n}dx<\int_{0}^{1}x^{4n}dx<\frac{1}{1+4n}$

Can you justify the last inequality?
IOW, can you explain why $\int_{0}^{1}x^{4n}dx<\frac{1}{1+4n}$ must be true?

 

[KungPeng Zhou]

Can you justify the last inequality?
IOW, can you explain why $\int_{0}^{1}x^{4n}dx<\frac{1}{1+4n}$ must be true?

Sorry,maybe there is an error $\int_{0}^{1}x^{4n}dx=\frac{1}{1+4n}$
So, we know
$$0<\int_{0}^{2-\sqrt{3}}\frac{x^{4n}}{1+x^{2}}dx<\frac{1}{1+4n}$$

 

[haruspex]

Since |x|<1 throughout the range, why not take the limit first, then integrate?

 

[KungPeng Zhou]

Since |x|<1 throughout the range, why not take the limit first, then integrate?

It's very good. We can evaluate it easier with the way.

 

[pasmith]

Sorry,maybe there is an error $\int_{0}^{1}x^{4n}dx=\frac{1}{1+4n}$
So, we know
$$0<\int_{0}^{2-\sqrt{3}}\frac{x^{4n}}{1+x^{2}}dx<\frac{1}{1+4n}$$

$$\left| a_n - \frac{\pi}{12}\right| = \int_0^{2-\sqrt{3}} \frac{x^{4n}}{1 + x^2}\,dx < \int_0^{2-\sqrt{3}} x^{4n}\,dx = \frac{(2 - \sqrt{3})^{4n+1}}{4n+1}$$ is a tighter bound; depending on what you're doing convergence as $(4n+1)^{-1}$ may not be adequate.

 

[KungPeng Zhou]

$$\left| a_n - \frac{\pi}{12}\right| = \int_0^{2-\sqrt{3}} \frac{x^{4n}}{1 + x^2}\,dx < \int_0^{2-\sqrt{3}} x^{4n}\,dx = \frac{(2 - \sqrt{3})^{4n+1}}{4n+1}$$ is a tighter bound; depending on what you're doing convergence as $(4n+1)^{-1}$ may not be adequate.

Yes, it's tighter. It seems like the definition of limits.