Find the power series representation for $ f(x) = arctan (e^x) $ and its interval of convergence
$begingroup$
friends.
As stated on the title, my question is: find the power series representation for $ f(x) = arctan (e^x) $ and its interval of convergence.
This question got me a bit confused due to the presence of the $ e^x $.
The first path I tried was taking the derivative of the function but I realized that it wouldn't take me anywhere. Then, I decided to do the following:
Let $ g(x) = arctan (x) $. Then:
$ g'(x) = frac{1}{1+x^2} $ and so $ g'(x) = sum_{n=0}^{infty} (-1)^n cdot x^{2n} $ for $ -1 < x < 1 $.
Integrating both sides we get:
$ arctan(x) = sum_{n=0}^{infty} (-1)^n cdot frac{x^{2n+1}}{2n+1} + C $ and because $ arctan(0) = 0 $, $ C $ also equals 0.
With that in mind, replacing $ x $ with $ e^x $ we have:
$ arctan(e^x) = sum_{n=0}^{infty} (-1)^n cdot frac{ (e^x)^{2n+1}}{2n+1} $
and now for the interval of convergence, using the Ratio Test:
$ lim_{n to infty} left| frac{(e^x)^{2n+3} }{2n+3} cdot frac{2n+1}{(e^x)^{2n+1}} right| = |e^{2x}| $
and so for the series to converge we must have:
$ -1 < e^{2x} < 1 $
Now, I have two questions:
Can I switch $ x $ with $ e^x $ without any problems?
And if so and assuming my solution and my ratio test are correct, is there any to simplify my interval of convergence?
Really appreciate any help.
Thanks in advance.
Pedro.
calculus sequences-and-series convergence power-series exponential-function
asked Oct 29 '15 at 3:47
Pedro CunhaPedro Cunha
6618
$endgroup$
|
show 5 more comments
$begingroup$
friends.
As stated on the title, my question is: find the power series representation for $ f(x) = arctan (e^x) $ and its interval of convergence.
This question got me a bit confused due to the presence of the $ e^x $.
The first path I tried was taking the derivative of the function but I realized that it wouldn't take me anywhere. Then, I decided to do the following:
Let $ g(x) = arctan (x) $. Then:
$ g'(x) = frac{1}{1+x^2} $ and so $ g'(x) = sum_{n=0}^{infty} (-1)^n cdot x^{2n} $ for $ -1 < x < 1 $.
Integrating both sides we get:
$ arctan(x) = sum_{n=0}^{infty} (-1)^n cdot frac{x^{2n+1}}{2n+1} + C $ and because $ arctan(0) = 0 $, $ C $ also equals 0.
With that in mind, replacing $ x $ with $ e^x $ we have:
$ arctan(e^x) = sum_{n=0}^{infty} (-1)^n cdot frac{ (e^x)^{2n+1}}{2n+1} $
and now for the interval of convergence, using the Ratio Test:
$ lim_{n to infty} left| frac{(e^x)^{2n+3} }{2n+3} cdot frac{2n+1}{(e^x)^{2n+1}} right| = |e^{2x}| $
and so for the series to converge we must have:
$ -1 < e^{2x} < 1 $
Now, I have two questions:
Can I switch $ x $ with $ e^x $ without any problems?
And if so and assuming my solution and my ratio test are correct, is there any to simplify my interval of convergence?
Really appreciate any help.
Thanks in advance.
Pedro.
calculus sequences-and-series convergence power-series exponential-function
asked Oct 29 '15 at 3:47
Pedro CunhaPedro Cunha
6618
$endgroup$
$begingroup$
You can't just replace $x$ with $e^x$, since the resulting expression isn't a power series. You would have to replace $x$ with the power series for $e^x$, but that gets complicated quickly..
$endgroup$
– Alex Zorn
Oct 29 '15 at 3:50
$begingroup$
Thanks for the quick reply, @Alex Zorn. You mean that the series would be $ sum_{n=0}^{infty} (-1)^n frac{sum_{k=0}^{infty}left( frac{x^k}{k!} right)^{2n+1}}{2n+1} $ or that replacement would have to have been made sooner? Also, if I can't replace, then what does this "true" statement from Wolfram|Alpha mean? solution
$endgroup$
– Pedro Cunha
Oct 29 '15 at 3:54
$begingroup$
I mean, if you replace x with e^x you get a valid equality. But it's just not a power series.
$endgroup$
– Alex Zorn
Oct 29 '15 at 3:54
$begingroup$
Also, since $e^0 = 1$ what you should really do is find the taylor series of arctan(x) about x = 1...
$endgroup$
– Alex Zorn
Oct 29 '15 at 3:56
1
$begingroup$
@PedroCunha you would want $$sum_{n=0}^{infty} (-1)^n frac{ left(sum_{k=0}^infty frac{x^k}{k!} right)^{2n+1}}{2n+1}$$
$endgroup$
– graydad
Oct 29 '15 at 3:56
|
show 5 more comments
$begingroup$
friends.
As stated on the title, my question is: find the power series representation for $ f(x) = arctan (e^x) $ and its interval of convergence.
This question got me a bit confused due to the presence of the $ e^x $.
The first path I tried was taking the derivative of the function but I realized that it wouldn't take me anywhere. Then, I decided to do the following:
Let $ g(x) = arctan (x) $. Then:
$ g'(x) = frac{1}{1+x^2} $ and so $ g'(x) = sum_{n=0}^{infty} (-1)^n cdot x^{2n} $ for $ -1 < x < 1 $.
Integrating both sides we get:
$ arctan(x) = sum_{n=0}^{infty} (-1)^n cdot frac{x^{2n+1}}{2n+1} + C $ and because $ arctan(0) = 0 $, $ C $ also equals 0.
With that in mind, replacing $ x $ with $ e^x $ we have:
$ arctan(e^x) = sum_{n=0}^{infty} (-1)^n cdot frac{ (e^x)^{2n+1}}{2n+1} $
and now for the interval of convergence, using the Ratio Test:
$ lim_{n to infty} left| frac{(e^x)^{2n+3} }{2n+3} cdot frac{2n+1}{(e^x)^{2n+1}} right| = |e^{2x}| $
and so for the series to converge we must have:
$ -1 < e^{2x} < 1 $
Now, I have two questions:
Can I switch $ x $ with $ e^x $ without any problems?
And if so and assuming my solution and my ratio test are correct, is there any to simplify my interval of convergence?
Really appreciate any help.
Thanks in advance.
Pedro.
calculus sequences-and-series convergence power-series exponential-function
asked Oct 29 '15 at 3:47
Pedro CunhaPedro Cunha
6618
$endgroup$
friends.
As stated on the title, my question is: find the power series representation for $ f(x) = arctan (e^x) $ and its interval of convergence.
This question got me a bit confused due to the presence of the $ e^x $.
The first path I tried was taking the derivative of the function but I realized that it wouldn't take me anywhere. Then, I decided to do the following:
Let $ g(x) = arctan (x) $. Then:
$ g'(x) = frac{1}{1+x^2} $ and so $ g'(x) = sum_{n=0}^{infty} (-1)^n cdot x^{2n} $ for $ -1 < x < 1 $.
Integrating both sides we get:
$ arctan(x) = sum_{n=0}^{infty} (-1)^n cdot frac{x^{2n+1}}{2n+1} + C $ and because $ arctan(0) = 0 $, $ C $ also equals 0.
With that in mind, replacing $ x $ with $ e^x $ we have:
$ arctan(e^x) = sum_{n=0}^{infty} (-1)^n cdot frac{ (e^x)^{2n+1}}{2n+1} $
and now for the interval of convergence, using the Ratio Test:
$ lim_{n to infty} left| frac{(e^x)^{2n+3} }{2n+3} cdot frac{2n+1}{(e^x)^{2n+1}} right| = |e^{2x}| $
and so for the series to converge we must have:
$ -1 < e^{2x} < 1 $
Now, I have two questions:
Can I switch $ x $ with $ e^x $ without any problems?
And if so and assuming my solution and my ratio test are correct, is there any to simplify my interval of convergence?
Really appreciate any help.
Thanks in advance.
Pedro.
calculus sequences-and-series convergence power-series exponential-function
calculus sequences-and-series convergence power-series exponential-function
asked Oct 29 '15 at 3:47
Pedro CunhaPedro Cunha
6618
asked Oct 29 '15 at 3:47
Pedro CunhaPedro Cunha
6618
asked Oct 29 '15 at 3:47
Pedro CunhaPedro Cunha
6618
asked Oct 29 '15 at 3:47
Pedro CunhaPedro Cunha
6618
asked Oct 29 '15 at 3:47
Pedro CunhaPedro Cunha
6618
6618
$begingroup$
You can't just replace $x$ with $e^x$, since the resulting expression isn't a power series. You would have to replace $x$ with the power series for $e^x$, but that gets complicated quickly..
$endgroup$
– Alex Zorn
Oct 29 '15 at 3:50
$begingroup$
Thanks for the quick reply, @Alex Zorn. You mean that the series would be $ sum_{n=0}^{infty} (-1)^n frac{sum_{k=0}^{infty}left( frac{x^k}{k!} right)^{2n+1}}{2n+1} $ or that replacement would have to have been made sooner? Also, if I can't replace, then what does this "true" statement from Wolfram|Alpha mean? solution
$endgroup$
– Pedro Cunha
Oct 29 '15 at 3:54
$begingroup$
I mean, if you replace x with e^x you get a valid equality. But it's just not a power series.
$endgroup$
– Alex Zorn
Oct 29 '15 at 3:54
$begingroup$
Also, since $e^0 = 1$ what you should really do is find the taylor series of arctan(x) about x = 1...
$endgroup$
– Alex Zorn
Oct 29 '15 at 3:56
1
$begingroup$
@PedroCunha you would want $$sum_{n=0}^{infty} (-1)^n frac{ left(sum_{k=0}^infty frac{x^k}{k!} right)^{2n+1}}{2n+1}$$
$endgroup$
– graydad
Oct 29 '15 at 3:56
|
show 5 more comments
$begingroup$
You can't just replace $x$ with $e^x$, since the resulting expression isn't a power series. You would have to replace $x$ with the power series for $e^x$, but that gets complicated quickly..
$endgroup$
– Alex Zorn
Oct 29 '15 at 3:50
$begingroup$
Thanks for the quick reply, @Alex Zorn. You mean that the series would be $ sum_{n=0}^{infty} (-1)^n frac{sum_{k=0}^{infty}left( frac{x^k}{k!} right)^{2n+1}}{2n+1} $ or that replacement would have to have been made sooner? Also, if I can't replace, then what does this "true" statement from Wolfram|Alpha mean? solution
$endgroup$
– Pedro Cunha
Oct 29 '15 at 3:54
$begingroup$
I mean, if you replace x with e^x you get a valid equality. But it's just not a power series.
$endgroup$
– Alex Zorn
Oct 29 '15 at 3:54
$begingroup$
Also, since $e^0 = 1$ what you should really do is find the taylor series of arctan(x) about x = 1...
$endgroup$
– Alex Zorn
Oct 29 '15 at 3:56
1
$begingroup$
@PedroCunha you would want $$sum_{n=0}^{infty} (-1)^n frac{ left(sum_{k=0}^infty frac{x^k}{k!} right)^{2n+1}}{2n+1}$$
$endgroup$
– graydad
Oct 29 '15 at 3:56
$begingroup$
You can't just replace $x$ with $e^x$, since the resulting expression isn't a power series. You would have to replace $x$ with the power series for $e^x$, but that gets complicated quickly..
$endgroup$
– Alex Zorn
Oct 29 '15 at 3:50
$begingroup$
You can't just replace $x$ with $e^x$, since the resulting expression isn't a power series. You would have to replace $x$ with the power series for $e^x$, but that gets complicated quickly..
$endgroup$
– Alex Zorn
Oct 29 '15 at 3:50
$begingroup$
Thanks for the quick reply, @Alex Zorn. You mean that the series would be $ sum_{n=0}^{infty} (-1)^n frac{sum_{k=0}^{infty}left( frac{x^k}{k!} right)^{2n+1}}{2n+1} $ or that replacement would have to have been made sooner? Also, if I can't replace, then what does this "true" statement from Wolfram|Alpha mean? solution
$endgroup$
– Pedro Cunha
Oct 29 '15 at 3:54
$begingroup$
Thanks for the quick reply, @Alex Zorn. You mean that the series would be $ sum_{n=0}^{infty} (-1)^n frac{sum_{k=0}^{infty}left( frac{x^k}{k!} right)^{2n+1}}{2n+1} $ or that replacement would have to have been made sooner? Also, if I can't replace, then what does this "true" statement from Wolfram|Alpha mean? solution
$endgroup$
– Pedro Cunha
Oct 29 '15 at 3:54
$begingroup$
I mean, if you replace x with e^x you get a valid equality. But it's just not a power series.
$endgroup$
– Alex Zorn
Oct 29 '15 at 3:54
$begingroup$
I mean, if you replace x with e^x you get a valid equality. But it's just not a power series.
$endgroup$
– Alex Zorn
Oct 29 '15 at 3:54
$begingroup$
Also, since $e^0 = 1$ what you should really do is find the taylor series of arctan(x) about x = 1...
$endgroup$
– Alex Zorn
Oct 29 '15 at 3:56
$begingroup$
Also, since $e^0 = 1$ what you should really do is find the taylor series of arctan(x) about x = 1...
$endgroup$
– Alex Zorn
Oct 29 '15 at 3:56
1
1
$begingroup$
@PedroCunha you would want $$sum_{n=0}^{infty} (-1)^n frac{ left(sum_{k=0}^infty frac{x^k}{k!} right)^{2n+1}}{2n+1}$$
$endgroup$
– graydad
Oct 29 '15 at 3:56
$begingroup$
@PedroCunha you would want $$sum_{n=0}^{infty} (-1)^n frac{ left(sum_{k=0}^infty frac{x^k}{k!} right)^{2n+1}}{2n+1}$$
$endgroup$
– graydad
Oct 29 '15 at 3:56
|
show 5 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Note that $f(0)=frac{pi}{4}$ and that
$$f'(x)=frac{e^x}{1+e^{2x}}=frac{1}{2cosh x}$$
The nearest singularities to zero are $pm ipi/2$, so the radius of convergence of the power series expansion of $f$ around $0$ is $pi/2$. Now, since $f'$ is even it has a power series expansion of the form $f'(x)=sum_{n=0}^infty a_nx^{2n}$. Using the fact that $cosh(x)f'(x)=frac{1}{2}$ we see that the coefficients $(a_n)$ can be inductively calculated by the formula:
$$a_0=frac{1}{2},qquad a_n=-sum_{k=0}^{n-1}frac{a_k}{(2n-2k)!}$$
In particular,
$$
a_0=frac{1}{2},a_1=-frac{1}{4},a_2=frac{5}{48},a_3=-frac{61}{1440}
$$
and then
$$f(x)=frac{pi}{4}+sum_{n=0}^inftyfrac{a_n}{2n+1}x^{2n+1}.$$
That is
$$f(x)=frac{pi }{4}+frac{x}{2}-frac{x^3}{12}+frac{x^5}{48}-frac{61 x^7}{10080}+cdots$$
edited Dec 20 '18 at 15:27
Felix Klein
1,6441416
answered Oct 29 '15 at 4:48
Omran KoubaOmran Kouba
24.8k13881
$endgroup$
$begingroup$
Perfect answer. Once the series for $cosh x$ is available it is an easy matter to find its reciprocal and integrate. But the best part was to use complex analysis to find the radius of convergence. +1
$endgroup$
– Paramanand Singh
Oct 29 '15 at 5:18
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
Note that $f(0)=frac{pi}{4}$ and that
$$f'(x)=frac{e^x}{1+e^{2x}}=frac{1}{2cosh x}$$
The nearest singularities to zero are $pm ipi/2$, so the radius of convergence of the power series expansion of $f$ around $0$ is $pi/2$. Now, since $f'$ is even it has a power series expansion of the form $f'(x)=sum_{n=0}^infty a_nx^{2n}$. Using the fact that $cosh(x)f'(x)=frac{1}{2}$ we see that the coefficients $(a_n)$ can be inductively calculated by the formula:
$$a_0=frac{1}{2},qquad a_n=-sum_{k=0}^{n-1}frac{a_k}{(2n-2k)!}$$
In particular,
$$
a_0=frac{1}{2},a_1=-frac{1}{4},a_2=frac{5}{48},a_3=-frac{61}{1440}
$$
and then
$$f(x)=frac{pi}{4}+sum_{n=0}^inftyfrac{a_n}{2n+1}x^{2n+1}.$$
That is
$$f(x)=frac{pi }{4}+frac{x}{2}-frac{x^3}{12}+frac{x^5}{48}-frac{61 x^7}{10080}+cdots$$
edited Dec 20 '18 at 15:27
Felix Klein
1,6441416
answered Oct 29 '15 at 4:48
Omran KoubaOmran Kouba
24.8k13881
$endgroup$
$begingroup$
Perfect answer. Once the series for $cosh x$ is available it is an easy matter to find its reciprocal and integrate. But the best part was to use complex analysis to find the radius of convergence. +1
$endgroup$
– Paramanand Singh
Oct 29 '15 at 5:18
add a comment |
$begingroup$
Note that $f(0)=frac{pi}{4}$ and that
$$f'(x)=frac{e^x}{1+e^{2x}}=frac{1}{2cosh x}$$
The nearest singularities to zero are $pm ipi/2$, so the radius of convergence of the power series expansion of $f$ around $0$ is $pi/2$. Now, since $f'$ is even it has a power series expansion of the form $f'(x)=sum_{n=0}^infty a_nx^{2n}$. Using the fact that $cosh(x)f'(x)=frac{1}{2}$ we see that the coefficients $(a_n)$ can be inductively calculated by the formula:
$$a_0=frac{1}{2},qquad a_n=-sum_{k=0}^{n-1}frac{a_k}{(2n-2k)!}$$
In particular,
$$
a_0=frac{1}{2},a_1=-frac{1}{4},a_2=frac{5}{48},a_3=-frac{61}{1440}
$$
and then
$$f(x)=frac{pi}{4}+sum_{n=0}^inftyfrac{a_n}{2n+1}x^{2n+1}.$$
That is
$$f(x)=frac{pi }{4}+frac{x}{2}-frac{x^3}{12}+frac{x^5}{48}-frac{61 x^7}{10080}+cdots$$
edited Dec 20 '18 at 15:27
Felix Klein
1,6441416
answered Oct 29 '15 at 4:48
Omran KoubaOmran Kouba
24.8k13881
$endgroup$
$begingroup$
Perfect answer. Once the series for $cosh x$ is available it is an easy matter to find its reciprocal and integrate. But the best part was to use complex analysis to find the radius of convergence. +1
$endgroup$
– Paramanand Singh
Oct 29 '15 at 5:18
add a comment |
$begingroup$
Note that $f(0)=frac{pi}{4}$ and that
$$f'(x)=frac{e^x}{1+e^{2x}}=frac{1}{2cosh x}$$
The nearest singularities to zero are $pm ipi/2$, so the radius of convergence of the power series expansion of $f$ around $0$ is $pi/2$. Now, since $f'$ is even it has a power series expansion of the form $f'(x)=sum_{n=0}^infty a_nx^{2n}$. Using the fact that $cosh(x)f'(x)=frac{1}{2}$ we see that the coefficients $(a_n)$ can be inductively calculated by the formula:
$$a_0=frac{1}{2},qquad a_n=-sum_{k=0}^{n-1}frac{a_k}{(2n-2k)!}$$
In particular,
$$
a_0=frac{1}{2},a_1=-frac{1}{4},a_2=frac{5}{48},a_3=-frac{61}{1440}
$$
and then
$$f(x)=frac{pi}{4}+sum_{n=0}^inftyfrac{a_n}{2n+1}x^{2n+1}.$$
That is
$$f(x)=frac{pi }{4}+frac{x}{2}-frac{x^3}{12}+frac{x^5}{48}-frac{61 x^7}{10080}+cdots$$
edited Dec 20 '18 at 15:27
Felix Klein
1,6441416
answered Oct 29 '15 at 4:48
Omran KoubaOmran Kouba
24.8k13881
$endgroup$
Note that $f(0)=frac{pi}{4}$ and that
$$f'(x)=frac{e^x}{1+e^{2x}}=frac{1}{2cosh x}$$
The nearest singularities to zero are $pm ipi/2$, so the radius of convergence of the power series expansion of $f$ around $0$ is $pi/2$. Now, since $f'$ is even it has a power series expansion of the form $f'(x)=sum_{n=0}^infty a_nx^{2n}$. Using the fact that $cosh(x)f'(x)=frac{1}{2}$ we see that the coefficients $(a_n)$ can be inductively calculated by the formula:
$$a_0=frac{1}{2},qquad a_n=-sum_{k=0}^{n-1}frac{a_k}{(2n-2k)!}$$
In particular,
$$
a_0=frac{1}{2},a_1=-frac{1}{4},a_2=frac{5}{48},a_3=-frac{61}{1440}
$$
and then
$$f(x)=frac{pi}{4}+sum_{n=0}^inftyfrac{a_n}{2n+1}x^{2n+1}.$$
That is
$$f(x)=frac{pi }{4}+frac{x}{2}-frac{x^3}{12}+frac{x^5}{48}-frac{61 x^7}{10080}+cdots$$
edited Dec 20 '18 at 15:27
Felix Klein
1,6441416
answered Oct 29 '15 at 4:48
Omran KoubaOmran Kouba
24.8k13881
edited Dec 20 '18 at 15:27
Felix Klein
1,6441416
edited Dec 20 '18 at 15:27
Felix Klein
1,6441416
edited Dec 20 '18 at 15:27
Felix Klein
1,6441416
1,6441416
answered Oct 29 '15 at 4:48
Omran KoubaOmran Kouba
24.8k13881
answered Oct 29 '15 at 4:48
Omran KoubaOmran Kouba
24.8k13881
answered Oct 29 '15 at 4:48
Omran KoubaOmran Kouba
24.8k13881
24.8k13881
$begingroup$
Perfect answer. Once the series for $cosh x$ is available it is an easy matter to find its reciprocal and integrate. But the best part was to use complex analysis to find the radius of convergence. +1
$endgroup$
– Paramanand Singh
Oct 29 '15 at 5:18
add a comment |
$begingroup$
Perfect answer. Once the series for $cosh x$ is available it is an easy matter to find its reciprocal and integrate. But the best part was to use complex analysis to find the radius of convergence. +1
$endgroup$
– Paramanand Singh
Oct 29 '15 at 5:18
$begingroup$
Perfect answer. Once the series for $cosh x$ is available it is an easy matter to find its reciprocal and integrate. But the best part was to use complex analysis to find the radius of convergence. +1
$endgroup$
– Paramanand Singh
Oct 29 '15 at 5:18
$begingroup$
Perfect answer. Once the series for $cosh x$ is available it is an easy matter to find its reciprocal and integrate. But the best part was to use complex analysis to find the radius of convergence. +1
$endgroup$
– Paramanand Singh
Oct 29 '15 at 5:18
add a comment |
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You can't just replace $x$ with $e^x$, since the resulting expression isn't a power series. You would have to replace $x$ with the power series for $e^x$, but that gets complicated quickly..
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– Alex Zorn
Oct 29 '15 at 3:50
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Thanks for the quick reply, @Alex Zorn. You mean that the series would be $ sum_{n=0}^{infty} (-1)^n frac{sum_{k=0}^{infty}left( frac{x^k}{k!} right)^{2n+1}}{2n+1} $ or that replacement would have to have been made sooner? Also, if I can't replace, then what does this "true" statement from Wolfram|Alpha mean? solution
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– Pedro Cunha
Oct 29 '15 at 3:54
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I mean, if you replace x with e^x you get a valid equality. But it's just not a power series.
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– Alex Zorn
Oct 29 '15 at 3:54
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Also, since $e^0 = 1$ what you should really do is find the taylor series of arctan(x) about x = 1...
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– Alex Zorn
Oct 29 '15 at 3:56
1
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@PedroCunha you would want $$sum_{n=0}^{infty} (-1)^n frac{ left(sum_{k=0}^infty frac{x^k}{k!} right)^{2n+1}}{2n+1}$$
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– graydad
Oct 29 '15 at 3:56