Mixing
Trying to find Taylor expansion of $arcsin(x)$
Find $sum_{n=1}^{infty} frac {1^2times3^2times...times(2n-1)^2}{2^{2n}(2n+1)!}=?$
So my guess is this is some kind of $arcsin(x)$.
My attempt:
Let $f(x) = arcsin(x)$ then
$f'(x) = frac 1{sqrt{1-x^2}}$
We know that : $(1+x)^{alpha}=1+sum_{n=1}^{infty}frac{alpha(alpha-1)...(alpha-n+1)}{n!}x^n$.So then with $alpha=-frac 12$ and $xleftarrow-x^2$ we get that:
$$frac 1{sqrt{1-x^2}}=1+sum_{n=1}^{infty} frac {(-frac12)(-frac 32)(-frac 52)...(-frac {2n-1}2)}{n!}(-1)^nx^{2n}=1+sum_{n = 1}^{infty}frac{1times3times5times...times(2n-1)}{2^nn!}x^{2n}.$$
if we integrate both sides we get that:
$$f(x)=x+sum_{n=1}^{infty}frac {1times3times...times(2n-1)}{2^nn!(2n+1)}x^{2n+1}$$
Is there something wrong with my expansion? I think I might've made a mistake somewhere
real-analysis sequences-and-series power-series taylor-expansion
edited Nov 20 '18 at 23:13
Bernard
118k639112
asked Nov 20 '18 at 23:10
C. Cristi
1,476218
add a comment |
Find $sum_{n=1}^{infty} frac {1^2times3^2times...times(2n-1)^2}{2^{2n}(2n+1)!}=?$
So my guess is this is some kind of $arcsin(x)$.
My attempt:
Let $f(x) = arcsin(x)$ then
$f'(x) = frac 1{sqrt{1-x^2}}$
We know that : $(1+x)^{alpha}=1+sum_{n=1}^{infty}frac{alpha(alpha-1)...(alpha-n+1)}{n!}x^n$.So then with $alpha=-frac 12$ and $xleftarrow-x^2$ we get that:
$$frac 1{sqrt{1-x^2}}=1+sum_{n=1}^{infty} frac {(-frac12)(-frac 32)(-frac 52)...(-frac {2n-1}2)}{n!}(-1)^nx^{2n}=1+sum_{n = 1}^{infty}frac{1times3times5times...times(2n-1)}{2^nn!}x^{2n}.$$
if we integrate both sides we get that:
$$f(x)=x+sum_{n=1}^{infty}frac {1times3times...times(2n-1)}{2^nn!(2n+1)}x^{2n+1}$$
Is there something wrong with my expansion? I think I might've made a mistake somewhere
real-analysis sequences-and-series power-series taylor-expansion
edited Nov 20 '18 at 23:13
Bernard
118k639112
asked Nov 20 '18 at 23:10
C. Cristi
1,476218
1
The title of the question is a bit misleading.
– Berci
Nov 20 '18 at 23:33
1
It's correct. The first term, conventionally, corresponds to $n=0$ in the general formula. There is a more compact form: $;displaystylesum_{n=0}^inftyfrac{(2n-1)!!}{(2n)!!}frac{x^{2n+1}}{2n+1}$.
– Bernard
Nov 20 '18 at 23:34
add a comment |
Find $sum_{n=1}^{infty} frac {1^2times3^2times...times(2n-1)^2}{2^{2n}(2n+1)!}=?$
So my guess is this is some kind of $arcsin(x)$.
My attempt:
Let $f(x) = arcsin(x)$ then
$f'(x) = frac 1{sqrt{1-x^2}}$
We know that : $(1+x)^{alpha}=1+sum_{n=1}^{infty}frac{alpha(alpha-1)...(alpha-n+1)}{n!}x^n$.So then with $alpha=-frac 12$ and $xleftarrow-x^2$ we get that:
$$frac 1{sqrt{1-x^2}}=1+sum_{n=1}^{infty} frac {(-frac12)(-frac 32)(-frac 52)...(-frac {2n-1}2)}{n!}(-1)^nx^{2n}=1+sum_{n = 1}^{infty}frac{1times3times5times...times(2n-1)}{2^nn!}x^{2n}.$$
if we integrate both sides we get that:
$$f(x)=x+sum_{n=1}^{infty}frac {1times3times...times(2n-1)}{2^nn!(2n+1)}x^{2n+1}$$
Is there something wrong with my expansion? I think I might've made a mistake somewhere
real-analysis sequences-and-series power-series taylor-expansion
edited Nov 20 '18 at 23:13
Bernard
118k639112
asked Nov 20 '18 at 23:10
C. Cristi
1,476218
Find $sum_{n=1}^{infty} frac {1^2times3^2times...times(2n-1)^2}{2^{2n}(2n+1)!}=?$
So my guess is this is some kind of $arcsin(x)$.
My attempt:
Let $f(x) = arcsin(x)$ then
$f'(x) = frac 1{sqrt{1-x^2}}$
We know that : $(1+x)^{alpha}=1+sum_{n=1}^{infty}frac{alpha(alpha-1)...(alpha-n+1)}{n!}x^n$.So then with $alpha=-frac 12$ and $xleftarrow-x^2$ we get that:
$$frac 1{sqrt{1-x^2}}=1+sum_{n=1}^{infty} frac {(-frac12)(-frac 32)(-frac 52)...(-frac {2n-1}2)}{n!}(-1)^nx^{2n}=1+sum_{n = 1}^{infty}frac{1times3times5times...times(2n-1)}{2^nn!}x^{2n}.$$
if we integrate both sides we get that:
$$f(x)=x+sum_{n=1}^{infty}frac {1times3times...times(2n-1)}{2^nn!(2n+1)}x^{2n+1}$$
Is there something wrong with my expansion? I think I might've made a mistake somewhere
real-analysis sequences-and-series power-series taylor-expansion
real-analysis sequences-and-series power-series taylor-expansion
edited Nov 20 '18 at 23:13
Bernard
118k639112
asked Nov 20 '18 at 23:10
C. Cristi
1,476218
edited Nov 20 '18 at 23:13
Bernard
118k639112
asked Nov 20 '18 at 23:10
C. Cristi
1,476218
edited Nov 20 '18 at 23:13
Bernard
118k639112
edited Nov 20 '18 at 23:13
Bernard
118k639112
edited Nov 20 '18 at 23:13
Bernard
118k639112
118k639112
asked Nov 20 '18 at 23:10
C. Cristi
1,476218
asked Nov 20 '18 at 23:10
C. Cristi
1,476218
asked Nov 20 '18 at 23:10
C. Cristi
1,476218
1,476218
1
The title of the question is a bit misleading.
– Berci
Nov 20 '18 at 23:33
1
It's correct. The first term, conventionally, corresponds to $n=0$ in the general formula. There is a more compact form: $;displaystylesum_{n=0}^inftyfrac{(2n-1)!!}{(2n)!!}frac{x^{2n+1}}{2n+1}$.
– Bernard
Nov 20 '18 at 23:34
add a comment |
1
The title of the question is a bit misleading.
– Berci
Nov 20 '18 at 23:33
1
It's correct. The first term, conventionally, corresponds to $n=0$ in the general formula. There is a more compact form: $;displaystylesum_{n=0}^inftyfrac{(2n-1)!!}{(2n)!!}frac{x^{2n+1}}{2n+1}$.
– Bernard
Nov 20 '18 at 23:34
1
1
The title of the question is a bit misleading.
– Berci
Nov 20 '18 at 23:33
The title of the question is a bit misleading.
– Berci
Nov 20 '18 at 23:33
1
1
It's correct. The first term, conventionally, corresponds to $n=0$ in the general formula. There is a more compact form: $;displaystylesum_{n=0}^inftyfrac{(2n-1)!!}{(2n)!!}frac{x^{2n+1}}{2n+1}$.
– Bernard
Nov 20 '18 at 23:34
It's correct. The first term, conventionally, corresponds to $n=0$ in the general formula. There is a more compact form: $;displaystylesum_{n=0}^inftyfrac{(2n-1)!!}{(2n)!!}frac{x^{2n+1}}{2n+1}$.
– Bernard
Nov 20 '18 at 23:34
add a comment |
1 Answer
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$$sum_{ngeq 1}frac{(2n-1)!!^2}{4^n(2n+1)!}=sum_{ngeq 1}frac{binom{2n}{n}}{16^n (2n+1)}=int_{0}^{1}left(frac{2}{sqrt{4-x^2}}-1right),dx=2arcsinfrac{1}{2}-1=frac{pi}{3}-1. $$
answered Nov 21 '18 at 0:42
Jack D'Aurizio
287k33280657
Can you be more explicit at firs step?
– C. Cristi
Nov 21 '18 at 6:39
@C.Cristi: $$frac{(2n-1)!!^2}{(2n+1)!}=frac{left[(2n)!/(2^n n!)right]^2}{4^n(2n+1)!}=frac{(2n)!binom{2n}{n}}{4^n (2n+1)!}=frac{binom{2n}{n}}{4^n(2n+1)}.$$
– Jack D'Aurizio
Nov 21 '18 at 16:44
add a comment |
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1 Answer
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$$sum_{ngeq 1}frac{(2n-1)!!^2}{4^n(2n+1)!}=sum_{ngeq 1}frac{binom{2n}{n}}{16^n (2n+1)}=int_{0}^{1}left(frac{2}{sqrt{4-x^2}}-1right),dx=2arcsinfrac{1}{2}-1=frac{pi}{3}-1. $$
answered Nov 21 '18 at 0:42
Jack D'Aurizio
287k33280657
Can you be more explicit at firs step?
– C. Cristi
Nov 21 '18 at 6:39
@C.Cristi: $$frac{(2n-1)!!^2}{(2n+1)!}=frac{left[(2n)!/(2^n n!)right]^2}{4^n(2n+1)!}=frac{(2n)!binom{2n}{n}}{4^n (2n+1)!}=frac{binom{2n}{n}}{4^n(2n+1)}.$$
– Jack D'Aurizio
Nov 21 '18 at 16:44
add a comment |
$$sum_{ngeq 1}frac{(2n-1)!!^2}{4^n(2n+1)!}=sum_{ngeq 1}frac{binom{2n}{n}}{16^n (2n+1)}=int_{0}^{1}left(frac{2}{sqrt{4-x^2}}-1right),dx=2arcsinfrac{1}{2}-1=frac{pi}{3}-1. $$
answered Nov 21 '18 at 0:42
Jack D'Aurizio
287k33280657
Can you be more explicit at firs step?
– C. Cristi
Nov 21 '18 at 6:39
@C.Cristi: $$frac{(2n-1)!!^2}{(2n+1)!}=frac{left[(2n)!/(2^n n!)right]^2}{4^n(2n+1)!}=frac{(2n)!binom{2n}{n}}{4^n (2n+1)!}=frac{binom{2n}{n}}{4^n(2n+1)}.$$
– Jack D'Aurizio
Nov 21 '18 at 16:44
add a comment |
$$sum_{ngeq 1}frac{(2n-1)!!^2}{4^n(2n+1)!}=sum_{ngeq 1}frac{binom{2n}{n}}{16^n (2n+1)}=int_{0}^{1}left(frac{2}{sqrt{4-x^2}}-1right),dx=2arcsinfrac{1}{2}-1=frac{pi}{3}-1. $$
answered Nov 21 '18 at 0:42
Jack D'Aurizio
287k33280657
$$sum_{ngeq 1}frac{(2n-1)!!^2}{4^n(2n+1)!}=sum_{ngeq 1}frac{binom{2n}{n}}{16^n (2n+1)}=int_{0}^{1}left(frac{2}{sqrt{4-x^2}}-1right),dx=2arcsinfrac{1}{2}-1=frac{pi}{3}-1. $$
answered Nov 21 '18 at 0:42
Jack D'Aurizio
287k33280657
answered Nov 21 '18 at 0:42
Jack D'Aurizio
287k33280657
answered Nov 21 '18 at 0:42
Jack D'Aurizio
287k33280657
answered Nov 21 '18 at 0:42
Jack D'Aurizio
287k33280657
287k33280657
Can you be more explicit at firs step?
– C. Cristi
Nov 21 '18 at 6:39
@C.Cristi: $$frac{(2n-1)!!^2}{(2n+1)!}=frac{left[(2n)!/(2^n n!)right]^2}{4^n(2n+1)!}=frac{(2n)!binom{2n}{n}}{4^n (2n+1)!}=frac{binom{2n}{n}}{4^n(2n+1)}.$$
– Jack D'Aurizio
Nov 21 '18 at 16:44
add a comment |
Can you be more explicit at firs step?
– C. Cristi
Nov 21 '18 at 6:39
@C.Cristi: $$frac{(2n-1)!!^2}{(2n+1)!}=frac{left[(2n)!/(2^n n!)right]^2}{4^n(2n+1)!}=frac{(2n)!binom{2n}{n}}{4^n (2n+1)!}=frac{binom{2n}{n}}{4^n(2n+1)}.$$
– Jack D'Aurizio
Nov 21 '18 at 16:44
Can you be more explicit at firs step?
– C. Cristi
Nov 21 '18 at 6:39
Can you be more explicit at firs step?
– C. Cristi
Nov 21 '18 at 6:39
@C.Cristi: $$frac{(2n-1)!!^2}{(2n+1)!}=frac{left[(2n)!/(2^n n!)right]^2}{4^n(2n+1)!}=frac{(2n)!binom{2n}{n}}{4^n (2n+1)!}=frac{binom{2n}{n}}{4^n(2n+1)}.$$
– Jack D'Aurizio
Nov 21 '18 at 16:44
@C.Cristi: $$frac{(2n-1)!!^2}{(2n+1)!}=frac{left[(2n)!/(2^n n!)right]^2}{4^n(2n+1)!}=frac{(2n)!binom{2n}{n}}{4^n (2n+1)!}=frac{binom{2n}{n}}{4^n(2n+1)}.$$
– Jack D'Aurizio
Nov 21 '18 at 16:44
add a comment |
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1
The title of the question is a bit misleading.
– Berci
Nov 20 '18 at 23:33
1
It's correct. The first term, conventionally, corresponds to $n=0$ in the general formula. There is a more compact form: $;displaystylesum_{n=0}^inftyfrac{(2n-1)!!}{(2n)!!}frac{x^{2n+1}}{2n+1}$.
– Bernard
Nov 20 '18 at 23:34