A question about definite integrals and series limits

  • #1
KungPeng Zhou
22
7
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.
 
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  • #2
Can you fix your Latex?. You need two dollar signs.
 
  • #3
PeroK said:
Can you fix your Latex?. You need two dollar signs.
Fixed it. I think.
 
  • #4
KungPeng Zhou said:
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.
 
  • #5
KungPeng Zhou said:
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 [tex]
1 - x^{4n} = (1 - x^{2n})(1 + x^{2n}).[/tex] Did you mean [itex](1 - x^4)^n[/itex]?

If so, you end up with a polynomial in [itex]x[/itex], which you can integrate analytically: [tex]\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}[/tex]
 
  • #6
pasmith said:
Are you sure about that? What you have written for the numerator is [tex]
1 - x^{4n} = (1 - x^{2n})(1 + x^{2n}).[/tex] Did you mean [itex](1 - x^4)^n[/itex]?
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##.
 
  • #7
PeroK said:
Can you fix your Latex?. You need two dollar signs.

pasmith said:
Are you sure about that? What you have written for the numerator is [tex]
1 - x^{4n} = (1 - x^{2n})(1 + x^{2n}).[/tex] Did you mean [itex](1 - x^4)^n[/itex]?

If so, you end up with a polynomial in [itex]x[/itex], which you can integrate analytically: [tex]\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}[/tex]
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}##
 
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  • #8
KungPeng Zhou said:
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?
 
  • #9
Mark44 said:
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
$$0<\int_{0}^{2-\sqrt{3}}\frac{x^{4n}}{1+x^{2}}dx<\frac{1}{1+4n}$$
 
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  • #10
Since |x|<1 throughout the range, why not take the limit first, then integrate?
 
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  • #11
haruspex said:
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.
 
  • #12
KungPeng Zhou said:
Sorry,maybe there is an error ##\int_{0}^{1}x^{4n}dx=\frac{1}{1+4n}##
So, we know
$$0<\int_{0}^{2-\sqrt{3}}\frac{x^{4n}}{1+x^{2}}dx<\frac{1}{1+4n}$$

[tex]\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}[/tex] is a tighter bound; depending on what you're doing convergence as [itex](4n+1)^{-1}[/itex] may not be adequate.
 
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  • #13
pasmith said:
[tex]\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}[/tex] is a tighter bound; depending on what you're doing convergence as [itex](4n+1)^{-1}[/itex] may not be adequate.
Yes, it's tighter. It seems like the definition of limits.
 

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