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.
 

FAQ: A question about definite integrals and series limits

What is a definite integral?

A definite integral is a mathematical expression that calculates the area under a curve within a specified interval. It is denoted as ∫[a,b] f(x) dx, where [a,b] represents the interval and f(x) is the function being integrated.

How do you evaluate a definite integral?

To evaluate a definite integral, you find the antiderivative (or indefinite integral) of the function, and then apply the Fundamental Theorem of Calculus. This involves computing the difference between the values of the antiderivative at the upper and lower limits of the interval.

What is a series limit?

A series limit refers to the value that the partial sums of an infinite series approach as the number of terms goes to infinity. If this value exists and is finite, the series is said to converge to that limit.

How do definite integrals relate to series limits?

Definite integrals and series limits are related in that they both can represent the sum of infinitely many infinitesimal quantities. For example, the definite integral can be seen as the limit of a Riemann sum, which is an infinite series of areas of rectangles under a curve.

Can you provide an example of evaluating a definite integral using a series limit?

Yes, consider the definite integral ∫[0,1] x^n dx. The antiderivative of x^n is (x^(n+1))/(n+1). Evaluating this from 0 to 1 gives (1^(n+1))/(n+1) - (0^(n+1))/(n+1), which simplifies to 1/(n+1). As n approaches infinity, the series limit of these integrals approaches 0.

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