§24.8 Series Expansions

Permalink:: http://dlmf.nist.gov/24.8

§24.8(i) Fourier Series
§24.8(ii) Other Series

§24.8(i) Fourier Series

Notes:: For (24.8.1)–(24.8.3) see Apostol (1976, p. 267). For (24.8.4) and (24.8.5) see Erdélyi et al. (1953a, p. 42).
Keywords:: Bernoulli polynomials, Euler polynomials
Permalink:: http://dlmf.nist.gov/24.8.i

If $n=1,2,\dots$ and $0\leq x\leq 1$ , then

24.8.1 $\mathop{B_{{2n}}\/}\nolimits\!\left(x\right)=(-1)^{{n+1}}\frac{2(2n)!}{(2\pi)^{{2n}}}\sum _{{k=1}}^{\infty}\frac{\mathop{\cos\/}\nolimits\!\left(2\pi kx\right)}{k^{{2n}}},$

Symbols:: $\mathop{B_{{n}}\/}\nolimits\!\left(x\right)$ : Bernoulli polynomials, $\mathop{\cos\/}\nolimits z$ : cosine function, $!$ : $n!$ : factorial, $k$ : integer, $n$ : integer and $x$ : real or complex
A&S Ref:: 23.1.18
Referenced by:: §24.8(i)
Permalink:: http://dlmf.nist.gov/24.8.E1
Encodings:: TeX, pMML, png

24.8.2 $\mathop{B_{{2n+1}}\/}\nolimits\!\left(x\right)=(-1)^{{n+1}}\frac{2(2n+1)!}{(2\pi)^{{2n+1}}}\sum _{{k=1}}^{\infty}\frac{\mathop{\sin\/}\nolimits\!\left(2\pi kx\right)}{k^{{2n+1}}}.$

Symbols:: $\mathop{B_{{n}}\/}\nolimits\!\left(x\right)$ : Bernoulli polynomials, $!$ : $n!$ : factorial, $\mathop{\sin\/}\nolimits z$ : sine function, $k$ : integer, $n$ : integer and $x$ : real or complex
A&S Ref:: 23.1.17
Permalink:: http://dlmf.nist.gov/24.8.E2
Encodings:: TeX, pMML, png

The second expansion holds also for $n=0$ and $0<x<1$ .

If $n=1$ with $0<x<1$ , or $n=2,3,\dots$ with $0\leq x\leq 1$ , then

24.8.3 $\mathop{B_{{n}}\/}\nolimits\!\left(x\right)=-\frac{n!}{(2\pi i)^{n}}\sum _{{\substack{k=-\infty\\ k\neq 0}}}^{\infty}\frac{e^{{2\pi ikx}}}{k^{n}}.$

Symbols:: $\mathop{B_{{n}}\/}\nolimits\!\left(x\right)$ : Bernoulli polynomials, $e$ : base of exponential function, $!$ : $n!$ : factorial, $k$ : integer, $n$ : integer and $x$ : real or complex
Referenced by:: §24.8(i)
Permalink:: http://dlmf.nist.gov/24.8.E3
Encodings:: TeX, pMML, png

If $n=1,2,\dots$ and $0\leq x\leq 1$ , then

24.8.4 $\mathop{E_{{2n}}\/}\nolimits\!\left(x\right)=(-1)^{n}\frac{4(2n)!}{\pi^{{2n+1}}}\sum _{{k=0}}^{\infty}\frac{\mathop{\sin\/}\nolimits\!\left((2k+1)\pi x\right)}{(2k+1)^{{2n+1}}},$

Symbols:: $\mathop{E_{{n}}\/}\nolimits\!\left(x\right)$ : Euler polynomials, $!$ : $n!$ : factorial, $\mathop{\sin\/}\nolimits z$ : sine function, $k$ : integer, $n$ : integer and $x$ : real or complex
A&S Ref:: 23.1.18
Referenced by:: §24.8(i), §24.9
Permalink:: http://dlmf.nist.gov/24.8.E4
Encodings:: TeX, pMML, png

24.8.5 $\mathop{E_{{2n-1}}\/}\nolimits\!\left(x\right)=(-1)^{n}\frac{4(2n-1)!}{\pi^{{2n}}}\sum _{{k=0}}^{\infty}\frac{\mathop{\cos\/}\nolimits\!\left((2k+1)\pi x\right)}{(2k+1)^{{2n}}}.$

Symbols:: $\mathop{E_{{n}}\/}\nolimits\!\left(x\right)$ : Euler polynomials, $\mathop{\cos\/}\nolimits z$ : cosine function, $!$ : $n!$ : factorial, $k$ : integer, $n$ : integer and $x$ : real or complex
A&S Ref:: 23.1.17
Referenced by:: §24.8(i)
Permalink:: http://dlmf.nist.gov/24.8.E5
Encodings:: TeX, pMML, png

§24.8(ii) Other Series

Notes:: See Berndt (1975b, pp. 176–178). (24.8.8) reduces to (24.8.6) when $\alpha=\beta=\pi$ and $n$ is odd.
Keywords:: Bernoulli polynomials
Permalink:: http://dlmf.nist.gov/24.8.ii

24.8.6 $\mathop{B_{{4n+2}}\/}\nolimits=(8n+4)\sum _{{k=1}}^{\infty}\frac{k^{{4n+1}}}{e^{{2\pi k}}-1},$ $n=1,2,\dots$ ,

Symbols:: $\mathop{B_{{n}}\/}\nolimits\!\left(x\right)$ : Bernoulli polynomials, $e$ : base of exponential function, $k$ : integer and $n$ : integer
Referenced by:: §24.8(ii)
Permalink:: http://dlmf.nist.gov/24.8.E6
Encodings:: TeX, pMML, png

24.8.7 $\mathop{B_{{2n}}\/}\nolimits=\frac{(-1)^{{n+1}}4n}{2^{{2n}}-1}\sum _{{k=1}}^{\infty}\frac{k^{{2n-1}}}{e^{{\pi k}}+(-1)^{{k+n}}},$ $n=2,3,\dots$ .

Symbols:: $\mathop{B_{{n}}\/}\nolimits\!\left(x\right)$ : Bernoulli polynomials, $e$ : base of exponential function, $k$ : integer and $n$ : integer
Permalink:: http://dlmf.nist.gov/24.8.E7
Encodings:: TeX, pMML, png

Let $\alpha\beta=\pi^{2}$ . Then

24.8.8 $\frac{\mathop{B_{{2n}}\/}\nolimits}{4n}\left(\alpha^{n}-(-\beta)^{n}\right)=\alpha^{n}\sum _{{k=1}}^{\infty}\frac{k^{{2n-1}}}{e^{{2\alpha k}}-1}-(-\beta)^{n}\sum _{{k=1}}^{\infty}\frac{k^{{2n-1}}}{e^{{2\beta k}}-1},$ $n=2,3,\dots$ .

Symbols:: $\mathop{B_{{n}}\/}\nolimits\!\left(x\right)$ : Bernoulli polynomials, $e$ : base of exponential function, $k$ : integer, $n$ : integer, $\alpha$ : parameter and $\beta$ : parameter
Referenced by:: §24.8(ii)
Permalink:: http://dlmf.nist.gov/24.8.E8
Encodings:: TeX, pMML, png

24.8.9 $\mathop{E_{{2n}}\/}\nolimits=(-1)^{n}\sum _{{k=1}}^{\infty}\frac{k^{{2n}}}{\mathop{\cosh\/}\nolimits\!\left(\tfrac{1}{2}\pi k\right)}-4\sum _{{k=0}}^{\infty}\frac{(-1)^{k}(2k+1)^{{2n}}}{e^{{2\pi(2k+1)}}-1},$ $n=1,2,\dots$ .

Symbols:: $\mathop{E_{{n}}\/}\nolimits\!\left(x\right)$ : Euler polynomials, $e$ : base of exponential function, $\mathop{\cosh\/}\nolimits z$ : hyperbolic cosine function, $k$ : integer and $n$ : integer
Keywords:: Euler polynomials
Permalink:: http://dlmf.nist.gov/24.8.E9
Encodings:: TeX, pMML, png

§24.8 Series Expansions

Contents

§24.8(i) Fourier Series

§24.8(ii) Other Series