Mixing
Series expansion of $logfrac{sin z}{z}$
$begingroup$
What is the series expansion of $f(z)= logfrac{sin z}{z}$ upto $z^6$ terms?
Since $f(z)$ is an even function,so the coefficients of all the odd powers of z must be $0$ and I calculated the coefficient of $z^2 $ to be $-1/4$ (am I correct) but what will be the expansion till $z^6$?
complex-analysis power-series
edited Jan 12 at 9:42
Kemono Chen
3,0721743
asked Jan 12 at 9:03
Nilesh KhatriNilesh Khatri
195
$endgroup$
add a comment |
$begingroup$
What is the series expansion of $f(z)= logfrac{sin z}{z}$ upto $z^6$ terms?
Since $f(z)$ is an even function,so the coefficients of all the odd powers of z must be $0$ and I calculated the coefficient of $z^2 $ to be $-1/4$ (am I correct) but what will be the expansion till $z^6$?
complex-analysis power-series
edited Jan 12 at 9:42
Kemono Chen
3,0721743
asked Jan 12 at 9:03
Nilesh KhatriNilesh Khatri
195
$endgroup$
add a comment |
$begingroup$
What is the series expansion of $f(z)= logfrac{sin z}{z}$ upto $z^6$ terms?
Since $f(z)$ is an even function,so the coefficients of all the odd powers of z must be $0$ and I calculated the coefficient of $z^2 $ to be $-1/4$ (am I correct) but what will be the expansion till $z^6$?
complex-analysis power-series
edited Jan 12 at 9:42
Kemono Chen
3,0721743
asked Jan 12 at 9:03
Nilesh KhatriNilesh Khatri
195
$endgroup$
What is the series expansion of $f(z)= logfrac{sin z}{z}$ upto $z^6$ terms?
Since $f(z)$ is an even function,so the coefficients of all the odd powers of z must be $0$ and I calculated the coefficient of $z^2 $ to be $-1/4$ (am I correct) but what will be the expansion till $z^6$?
complex-analysis power-series
complex-analysis power-series
edited Jan 12 at 9:42
Kemono Chen
3,0721743
asked Jan 12 at 9:03
Nilesh KhatriNilesh Khatri
195
edited Jan 12 at 9:42
Kemono Chen
3,0721743
asked Jan 12 at 9:03
Nilesh KhatriNilesh Khatri
195
edited Jan 12 at 9:42
Kemono Chen
3,0721743
edited Jan 12 at 9:42
Kemono Chen
3,0721743
edited Jan 12 at 9:42
Kemono Chen
3,0721743
3,0721743
asked Jan 12 at 9:03
Nilesh KhatriNilesh Khatri
195
asked Jan 12 at 9:03
Nilesh KhatriNilesh Khatri
195
asked Jan 12 at 9:03
Nilesh KhatriNilesh Khatri
195
195
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Using steps, start with
$$log left(frac{sin (x)}{x}right)=log left(1+left(frac{sin (x)}{x}-1right)right)$$ Let $t=frac{sin (x)}{x}-1$ and use
$$log(1+t)=t-frac{t^2}{2}+frac{t^3}{3}-frac{t^4}{4}+frac{t^5}{5}-frac{t^6}{6}+frac{t^7}{
7}+Oleft(t^8right)$$
Now
$$t=frac{sin (x)}{x}-1=-frac{x^2}{6}+frac{x^4}{120}-frac{x^6}{5040}+Oleft(x^8right)$$ I suppose that the binomila expansion will be simple since you are not asked for too many terms.
answered Jan 12 at 9:14
Claude LeiboviciClaude Leibovici
121k1157133
$endgroup$
add a comment |
$begingroup$
I will post another approach with Bernoulli numbers. This answer provides the general terms of the coefficients.
$$lnfrac{sin x}x=int_0^xleft(cot t-frac1tright)dt\
=int_0^xsum_{nge 1}frac{(-4)^nB_{2n}}{(2n)!}t^{2n-1}dt\
=sum_{nge 1}frac{(-4)^nB_{2n}}{(2n)!(2n)}x^{2n}\
=-frac16x^2-frac1{180}x^4-frac1{2835}x^6-cdots$$
answered Jan 12 at 9:48
Kemono ChenKemono Chen
3,0721743
$endgroup$
$begingroup$
Nice way to do it for sure ! $to +1$
$endgroup$
– Claude Leibovici
Jan 12 at 10:23
$begingroup$
This is a really good method.
$endgroup$
– Szeto
Jan 13 at 2:59
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Using steps, start with
$$log left(frac{sin (x)}{x}right)=log left(1+left(frac{sin (x)}{x}-1right)right)$$ Let $t=frac{sin (x)}{x}-1$ and use
$$log(1+t)=t-frac{t^2}{2}+frac{t^3}{3}-frac{t^4}{4}+frac{t^5}{5}-frac{t^6}{6}+frac{t^7}{
7}+Oleft(t^8right)$$
Now
$$t=frac{sin (x)}{x}-1=-frac{x^2}{6}+frac{x^4}{120}-frac{x^6}{5040}+Oleft(x^8right)$$ I suppose that the binomila expansion will be simple since you are not asked for too many terms.
answered Jan 12 at 9:14
Claude LeiboviciClaude Leibovici
121k1157133
$endgroup$
add a comment |
$begingroup$
Using steps, start with
$$log left(frac{sin (x)}{x}right)=log left(1+left(frac{sin (x)}{x}-1right)right)$$ Let $t=frac{sin (x)}{x}-1$ and use
$$log(1+t)=t-frac{t^2}{2}+frac{t^3}{3}-frac{t^4}{4}+frac{t^5}{5}-frac{t^6}{6}+frac{t^7}{
7}+Oleft(t^8right)$$
Now
$$t=frac{sin (x)}{x}-1=-frac{x^2}{6}+frac{x^4}{120}-frac{x^6}{5040}+Oleft(x^8right)$$ I suppose that the binomila expansion will be simple since you are not asked for too many terms.
answered Jan 12 at 9:14
Claude LeiboviciClaude Leibovici
121k1157133
$endgroup$
add a comment |
$begingroup$
Using steps, start with
$$log left(frac{sin (x)}{x}right)=log left(1+left(frac{sin (x)}{x}-1right)right)$$ Let $t=frac{sin (x)}{x}-1$ and use
$$log(1+t)=t-frac{t^2}{2}+frac{t^3}{3}-frac{t^4}{4}+frac{t^5}{5}-frac{t^6}{6}+frac{t^7}{
7}+Oleft(t^8right)$$
Now
$$t=frac{sin (x)}{x}-1=-frac{x^2}{6}+frac{x^4}{120}-frac{x^6}{5040}+Oleft(x^8right)$$ I suppose that the binomila expansion will be simple since you are not asked for too many terms.
answered Jan 12 at 9:14
Claude LeiboviciClaude Leibovici
121k1157133
$endgroup$
Using steps, start with
$$log left(frac{sin (x)}{x}right)=log left(1+left(frac{sin (x)}{x}-1right)right)$$ Let $t=frac{sin (x)}{x}-1$ and use
$$log(1+t)=t-frac{t^2}{2}+frac{t^3}{3}-frac{t^4}{4}+frac{t^5}{5}-frac{t^6}{6}+frac{t^7}{
7}+Oleft(t^8right)$$
Now
$$t=frac{sin (x)}{x}-1=-frac{x^2}{6}+frac{x^4}{120}-frac{x^6}{5040}+Oleft(x^8right)$$ I suppose that the binomila expansion will be simple since you are not asked for too many terms.
answered Jan 12 at 9:14
Claude LeiboviciClaude Leibovici
121k1157133
answered Jan 12 at 9:14
Claude LeiboviciClaude Leibovici
121k1157133
answered Jan 12 at 9:14
Claude LeiboviciClaude Leibovici
121k1157133
answered Jan 12 at 9:14
Claude LeiboviciClaude Leibovici
121k1157133
121k1157133
add a comment |
add a comment |
$begingroup$
I will post another approach with Bernoulli numbers. This answer provides the general terms of the coefficients.
$$lnfrac{sin x}x=int_0^xleft(cot t-frac1tright)dt\
=int_0^xsum_{nge 1}frac{(-4)^nB_{2n}}{(2n)!}t^{2n-1}dt\
=sum_{nge 1}frac{(-4)^nB_{2n}}{(2n)!(2n)}x^{2n}\
=-frac16x^2-frac1{180}x^4-frac1{2835}x^6-cdots$$
answered Jan 12 at 9:48
Kemono ChenKemono Chen
3,0721743
$endgroup$
$begingroup$
Nice way to do it for sure ! $to +1$
$endgroup$
– Claude Leibovici
Jan 12 at 10:23
$begingroup$
This is a really good method.
$endgroup$
– Szeto
Jan 13 at 2:59
add a comment |
$begingroup$
I will post another approach with Bernoulli numbers. This answer provides the general terms of the coefficients.
$$lnfrac{sin x}x=int_0^xleft(cot t-frac1tright)dt\
=int_0^xsum_{nge 1}frac{(-4)^nB_{2n}}{(2n)!}t^{2n-1}dt\
=sum_{nge 1}frac{(-4)^nB_{2n}}{(2n)!(2n)}x^{2n}\
=-frac16x^2-frac1{180}x^4-frac1{2835}x^6-cdots$$
answered Jan 12 at 9:48
Kemono ChenKemono Chen
3,0721743
$endgroup$
$begingroup$
Nice way to do it for sure ! $to +1$
$endgroup$
– Claude Leibovici
Jan 12 at 10:23
$begingroup$
This is a really good method.
$endgroup$
– Szeto
Jan 13 at 2:59
add a comment |
$begingroup$
I will post another approach with Bernoulli numbers. This answer provides the general terms of the coefficients.
$$lnfrac{sin x}x=int_0^xleft(cot t-frac1tright)dt\
=int_0^xsum_{nge 1}frac{(-4)^nB_{2n}}{(2n)!}t^{2n-1}dt\
=sum_{nge 1}frac{(-4)^nB_{2n}}{(2n)!(2n)}x^{2n}\
=-frac16x^2-frac1{180}x^4-frac1{2835}x^6-cdots$$
answered Jan 12 at 9:48
Kemono ChenKemono Chen
3,0721743
$endgroup$
I will post another approach with Bernoulli numbers. This answer provides the general terms of the coefficients.
$$lnfrac{sin x}x=int_0^xleft(cot t-frac1tright)dt\
=int_0^xsum_{nge 1}frac{(-4)^nB_{2n}}{(2n)!}t^{2n-1}dt\
=sum_{nge 1}frac{(-4)^nB_{2n}}{(2n)!(2n)}x^{2n}\
=-frac16x^2-frac1{180}x^4-frac1{2835}x^6-cdots$$
answered Jan 12 at 9:48
Kemono ChenKemono Chen
3,0721743
answered Jan 12 at 9:48
Kemono ChenKemono Chen
3,0721743
answered Jan 12 at 9:48
Kemono ChenKemono Chen
3,0721743
answered Jan 12 at 9:48
Kemono ChenKemono Chen
3,0721743
3,0721743
$begingroup$
Nice way to do it for sure ! $to +1$
$endgroup$
– Claude Leibovici
Jan 12 at 10:23
$begingroup$
This is a really good method.
$endgroup$
– Szeto
Jan 13 at 2:59
add a comment |
$begingroup$
Nice way to do it for sure ! $to +1$
$endgroup$
– Claude Leibovici
Jan 12 at 10:23
$begingroup$
This is a really good method.
$endgroup$
– Szeto
Jan 13 at 2:59
$begingroup$
Nice way to do it for sure ! $to +1$
$endgroup$
– Claude Leibovici
Jan 12 at 10:23
$begingroup$
Nice way to do it for sure ! $to +1$
$endgroup$
– Claude Leibovici
Jan 12 at 10:23
$begingroup$
This is a really good method.
$endgroup$
– Szeto
Jan 13 at 2:59
$begingroup$
This is a really good method.
$endgroup$
– Szeto
Jan 13 at 2:59
add a comment |
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