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Daftar deret matematika

Berikut adalah daftar deret matematika yang berisi tentang rumus untuk penjumlahan terhingga dan tak terhingga. Ini dapat digunakan bersama-sama dengan alat-alat lain untuk menghitung penjumlahan.Di sini, dianggap memiliki nilai adalah polinomial Bernoulli. adalah bilangan Bernoulli, dan di sini, , adalah bilangan Euler. adalah fungsi zeta Riemann. adalah fungsi gamma. adalah fungsi poligamma adalah polilogaritma. adalah koefisien binomial melambangkan eksponensial dari

Wikipedia article
Diperbarui 13 November 2025

Sumber: Lihat artikel asli di Wikipedia

Berikut adalah daftar deret matematika yang berisi tentang rumus untuk penjumlahan terhingga dan tak terhingga. Ini dapat digunakan bersama-sama dengan alat-alat lain untuk menghitung penjumlahan.

  • Di sini, 0 0 {\displaystyle 0^{0}} {\displaystyle 0^{0}} dianggap memiliki nilai 1 {\displaystyle 1} {\displaystyle 1}
  • B n ( x ) {\displaystyle B_{n}(x)} {\displaystyle B_{n}(x)} adalah polinomial Bernoulli.
  • B n {\displaystyle B_{n}} {\displaystyle B_{n}} adalah bilangan Bernoulli, dan di sini, B 1 = − 1 2 {\displaystyle B_{1}=-{\frac {1}{2}}} {\displaystyle B_{1}=-{\frac {1}{2}}},
  • E n {\displaystyle E_{n}} {\displaystyle E_{n}} adalah bilangan Euler.
  • ζ ( s ) {\displaystyle \zeta (s)} {\displaystyle \zeta (s)} adalah fungsi zeta Riemann.
  • Γ ( z ) {\displaystyle \Gamma (z)} {\displaystyle \Gamma (z)} adalah fungsi gamma.
  • ψ n ( z ) {\displaystyle \psi _{n}(z)} {\displaystyle \psi _{n}(z)} adalah fungsi poligamma
  • Li s ⁡ ( z ) {\displaystyle \operatorname {Li} _{s}(z)} {\displaystyle \operatorname {Li} _{s}(z)} adalah polilogaritma.
  • ( n k ) {\displaystyle {\binom {n}{k}}} {\displaystyle {\binom {n}{k}}} adalah koefisien binomial
  • exp ⁡ ( x ) {\displaystyle \exp(x)} {\displaystyle \exp(x)} melambangkan eksponensial dari x {\displaystyle x} {\displaystyle x}

Penjumlahan pangkat

Lihat rumus Faulhaber

  • ∑ k = 0 m k n − 1 = B n ( m + 1 ) − B n n {\displaystyle \sum _{k=0}^{m}k^{n-1}={\frac {B_{n}(m+1)-B_{n}}{n}}} {\displaystyle \sum _{k=0}^{m}k^{n-1}={\frac {B_{n}(m+1)-B_{n}}{n}}}

Beberapa nilai pertamanya adalahː

  • ∑ k = 1 m k = m ( m + 1 ) 2 {\displaystyle \sum _{k=1}^{m}k={\frac {m(m+1)}{2}}} {\displaystyle \sum _{k=1}^{m}k={\frac {m(m+1)}{2}}}
  • ∑ k = 1 m k 2 = m ( m + 1 ) ( 2 m + 1 ) 6 = m 3 3 + m 2 2 + m 6 {\displaystyle \sum _{k=1}^{m}k^{2}={\frac {m(m+1)(2m+1)}{6}}={\frac {m^{3}}{3}}+{\frac {m^{2}}{2}}+{\frac {m}{6}}} {\displaystyle \sum _{k=1}^{m}k^{2}={\frac {m(m+1)(2m+1)}{6}}={\frac {m^{3}}{3}}+{\frac {m^{2}}{2}}+{\frac {m}{6}}}
  • ∑ k = 1 m k 3 = [ m ( m + 1 ) 2 ] 2 = m 4 4 + m 3 2 + m 2 4 {\displaystyle \sum _{k=1}^{m}k^{3}=\left[{\frac {m(m+1)}{2}}\right]^{2}={\frac {m^{4}}{4}}+{\frac {m^{3}}{2}}+{\frac {m^{2}}{4}}} {\displaystyle \sum _{k=1}^{m}k^{3}=\left[{\frac {m(m+1)}{2}}\right]^{2}={\frac {m^{4}}{4}}+{\frac {m^{3}}{2}}+{\frac {m^{2}}{4}}}

Lihat konstanta zeta.

  • ζ ( 2 n ) = ∑ k = 1 ∞ 1 k 2 n = ( − 1 ) n + 1 B 2 n ( 2 π ) 2 n 2 ( 2 n ) ! {\displaystyle \zeta (2n)=\sum _{k=1}^{\infty }{\frac {1}{k^{2n}}}=(-1)^{n+1}{\frac {B_{2n}(2\pi )^{2n}}{2(2n)!}}} {\displaystyle \zeta (2n)=\sum _{k=1}^{\infty }{\frac {1}{k^{2n}}}=(-1)^{n+1}{\frac {B_{2n}(2\pi )^{2n}}{2(2n)!}}}

Beberapa nilai pertamanya adalahː

  • ζ ( 2 ) = ∑ k = 1 ∞ 1 k 2 = π 2 6 {\displaystyle \zeta (2)=\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}={\frac {\pi ^{2}}{6}}} {\displaystyle \zeta (2)=\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}={\frac {\pi ^{2}}{6}}} (Masalah Basel)
  • ζ ( 4 ) = ∑ k = 1 ∞ 1 k 4 = π 4 90 {\displaystyle \zeta (4)=\sum _{k=1}^{\infty }{\frac {1}{k^{4}}}={\frac {\pi ^{4}}{90}}} {\displaystyle \zeta (4)=\sum _{k=1}^{\infty }{\frac {1}{k^{4}}}={\frac {\pi ^{4}}{90}}}
  • ζ ( 6 ) = ∑ k = 1 ∞ 1 k 6 = π 6 945 {\displaystyle \zeta (6)=\sum _{k=1}^{\infty }{\frac {1}{k^{6}}}={\frac {\pi ^{6}}{945}}} {\displaystyle \zeta (6)=\sum _{k=1}^{\infty }{\frac {1}{k^{6}}}={\frac {\pi ^{6}}{945}}}

Deret pangkat

Polilogaritma orde rendah

Penjumlahan terhingga

  • ∑ k = i n z k = z i − z n + 1 1 − z {\displaystyle \sum _{k=i}^{n}z^{k}={\frac {z^{i}-z^{n+1}}{1-z}}} {\displaystyle \sum _{k=i}^{n}z^{k}={\frac {z^{i}-z^{n+1}}{1-z}}}, (deret geometrik)
  • ∑ k = 1 n k z k = z 1 − ( n + 1 ) z n + n z n + 1 ( 1 − z ) 2 {\displaystyle \sum _{k=1}^{n}kz^{k}=z{\frac {1-(n+1)z^{n}+nz^{n+1}}{(1-z)^{2}}}} {\displaystyle \sum _{k=1}^{n}kz^{k}=z{\frac {1-(n+1)z^{n}+nz^{n+1}}{(1-z)^{2}}}}
  • ∑ k = 1 n k 2 z k = z 1 + z − ( n + 1 ) 2 z n + ( 2 n 2 + 2 n − 1 ) z n + 1 − n 2 z n + 2 ( 1 − z ) 3 {\displaystyle \sum _{k=1}^{n}k^{2}z^{k}=z{\frac {1+z-(n+1)^{2}z^{n}+(2n^{2}+2n-1)z^{n+1}-n^{2}z^{n+2}}{(1-z)^{3}}}} {\displaystyle \sum _{k=1}^{n}k^{2}z^{k}=z{\frac {1+z-(n+1)^{2}z^{n}+(2n^{2}+2n-1)z^{n+1}-n^{2}z^{n+2}}{(1-z)^{3}}}}
  • ∑ k = 1 n k m z k = ( z d d z ) m 1 − z n + 1 1 − z {\displaystyle \sum _{k=1}^{n}k^{m}z^{k}=\left(z{\frac {d}{dz}}\right)^{m}{\frac {1-z^{n+1}}{1-z}}} {\displaystyle \sum _{k=1}^{n}k^{m}z^{k}=\left(z{\frac {d}{dz}}\right)^{m}{\frac {1-z^{n+1}}{1-z}}}

Penjumlahan tak terhingga, sah untuk | z | < 1 {\displaystyle \left|z\right|<1} {\displaystyle \left|z\right|<1} (lihat polilogaritma)

  • Li n ⁡ ( z ) = ∑ k = 1 ∞ z k k n {\displaystyle \operatorname {Li} _{n}(z)=\sum _{k=1}^{\infty }{\frac {z^{k}}{k^{n}}}} {\displaystyle \operatorname {Li} _{n}(z)=\sum _{k=1}^{\infty }{\frac {z^{k}}{k^{n}}}}

Berikut ini adalah sebuah sifat yang berguna untuk menghitung polilogaritma urutan bilangan bulat rendah secara rekursif dalam bentuk tertutup:

  • d d z Li n ⁡ ( z ) = Li n − 1 ⁡ ( z ) z {\displaystyle {\frac {d}{dz}}\operatorname {Li} _{n}(z)={\frac {\operatorname {Li} _{n-1}(z)}{z}}} {\displaystyle {\frac {d}{dz}}\operatorname {Li} _{n}(z)={\frac {\operatorname {Li} _{n-1}(z)}{z}}}
  • Li 1 ⁡ ( z ) = ∑ k = 1 ∞ z k k = − ln ⁡ ( 1 − z ) {\displaystyle \operatorname {Li} _{1}(z)=\sum _{k=1}^{\infty }{\frac {z^{k}}{k}}=-\ln(1-z)} {\displaystyle \operatorname {Li} _{1}(z)=\sum _{k=1}^{\infty }{\frac {z^{k}}{k}}=-\ln(1-z)}
  • Li 0 ⁡ ( z ) = ∑ k = 1 ∞ z k = z 1 − z {\displaystyle \operatorname {Li} _{0}(z)=\sum _{k=1}^{\infty }z^{k}={\frac {z}{1-z}}} {\displaystyle \operatorname {Li} _{0}(z)=\sum _{k=1}^{\infty }z^{k}={\frac {z}{1-z}}}
  • Li − 1 ⁡ ( z ) = ∑ k = 1 ∞ k z k = z ( 1 − z ) 2 {\displaystyle \operatorname {Li} _{-1}(z)=\sum _{k=1}^{\infty }kz^{k}={\frac {z}{(1-z)^{2}}}} {\displaystyle \operatorname {Li} _{-1}(z)=\sum _{k=1}^{\infty }kz^{k}={\frac {z}{(1-z)^{2}}}}
  • Li − 2 ⁡ ( z ) = ∑ k = 1 ∞ k 2 z k = z ( 1 + z ) ( 1 − z ) 3 {\displaystyle \operatorname {Li} _{-2}(z)=\sum _{k=1}^{\infty }k^{2}z^{k}={\frac {z(1+z)}{(1-z)^{3}}}} {\displaystyle \operatorname {Li} _{-2}(z)=\sum _{k=1}^{\infty }k^{2}z^{k}={\frac {z(1+z)}{(1-z)^{3}}}}
  • Li − 3 ⁡ ( z ) = ∑ k = 1 ∞ k 3 z k = z ( 1 + 4 z + z 2 ) ( 1 − z ) 4 {\displaystyle \operatorname {Li} _{-3}(z)=\sum _{k=1}^{\infty }k^{3}z^{k}={\frac {z(1+4z+z^{2})}{(1-z)^{4}}}} {\displaystyle \operatorname {Li} _{-3}(z)=\sum _{k=1}^{\infty }k^{3}z^{k}={\frac {z(1+4z+z^{2})}{(1-z)^{4}}}}
  • Li − 4 ⁡ ( z ) = ∑ k = 1 ∞ k 4 z k = z ( 1 + z ) ( 1 + 10 z + z 2 ) ( 1 − z ) 5 {\displaystyle \operatorname {Li} _{-4}(z)=\sum _{k=1}^{\infty }k^{4}z^{k}={\frac {z(1+z)(1+10z+z^{2})}{(1-z)^{5}}}} {\displaystyle \operatorname {Li} _{-4}(z)=\sum _{k=1}^{\infty }k^{4}z^{k}={\frac {z(1+z)(1+10z+z^{2})}{(1-z)^{5}}}}

Fungsi eksponensial

  • ∑ k = 0 ∞ z k k ! = e z {\displaystyle \sum _{k=0}^{\infty }{\frac {z^{k}}{k!}}=e^{z}} {\displaystyle \sum _{k=0}^{\infty }{\frac {z^{k}}{k!}}=e^{z}}
  • ∑ k = 0 ∞ k z k k ! = z e z {\displaystyle \sum _{k=0}^{\infty }k{\frac {z^{k}}{k!}}=ze^{z}} {\displaystyle \sum _{k=0}^{\infty }k{\frac {z^{k}}{k!}}=ze^{z}} (bandingkan rata-rata distribusi Poisson)
  • ∑ k = 0 ∞ k 2 z k k ! = ( z + z 2 ) e z {\displaystyle \sum _{k=0}^{\infty }k^{2}{\frac {z^{k}}{k!}}=(z+z^{2})e^{z}} {\displaystyle \sum _{k=0}^{\infty }k^{2}{\frac {z^{k}}{k!}}=(z+z^{2})e^{z}} (bandingkan momen kedua distribusi Poisson)
  • ∑ k = 0 ∞ k 3 z k k ! = ( z + 3 z 2 + z 3 ) e z {\displaystyle \sum _{k=0}^{\infty }k^{3}{\frac {z^{k}}{k!}}=(z+3z^{2}+z^{3})e^{z}} {\displaystyle \sum _{k=0}^{\infty }k^{3}{\frac {z^{k}}{k!}}=(z+3z^{2}+z^{3})e^{z}}
  • ∑ k = 0 ∞ k 4 z k k ! = ( z + 7 z 2 + 6 z 3 + z 4 ) e z {\displaystyle \sum _{k=0}^{\infty }k^{4}{\frac {z^{k}}{k!}}=(z+7z^{2}+6z^{3}+z^{4})e^{z}} {\displaystyle \sum _{k=0}^{\infty }k^{4}{\frac {z^{k}}{k!}}=(z+7z^{2}+6z^{3}+z^{4})e^{z}}
  • ∑ k = 0 ∞ k n z k k ! = z d d z ∑ k = 0 ∞ k n − 1 z k k ! = e z T n ( z ) {\displaystyle \sum _{k=0}^{\infty }k^{n}{\frac {z^{k}}{k!}}=z{\frac {d}{dz}}\sum _{k=0}^{\infty }k^{n-1}{\frac {z^{k}}{k!}}\,\!=e^{z}T_{n}(z)} {\displaystyle \sum _{k=0}^{\infty }k^{n}{\frac {z^{k}}{k!}}=z{\frac {d}{dz}}\sum _{k=0}^{\infty }k^{n-1}{\frac {z^{k}}{k!}}\,\!=e^{z}T_{n}(z)} dengan T n ( z ) {\displaystyle T_{n}(z)} {\displaystyle T_{n}(z)} adalah polinomial Touchard.

Fungsi trigonometrik, trigonometrik invers, hiperbolik, dan hiperbolik invers

  • ∑ k = 0 ∞ ( − 1 ) k z 2 k + 1 ( 2 k + 1 ) ! = sin ⁡ z {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}z^{2k+1}}{(2k+1)!}}=\sin z} {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}z^{2k+1}}{(2k+1)!}}=\sin z}
  • ∑ k = 0 ∞ z 2 k + 1 ( 2 k + 1 ) ! = sinh ⁡ z {\displaystyle \sum _{k=0}^{\infty }{\frac {z^{2k+1}}{(2k+1)!}}=\sinh z} {\displaystyle \sum _{k=0}^{\infty }{\frac {z^{2k+1}}{(2k+1)!}}=\sinh z}
  • ∑ k = 0 ∞ ( − 1 ) k z 2 k ( 2 k ) ! = cos ⁡ z {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}z^{2k}}{(2k)!}}=\cos z} {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}z^{2k}}{(2k)!}}=\cos z}
  • ∑ k = 0 ∞ z 2 k ( 2 k ) ! = cosh ⁡ z {\displaystyle \sum _{k=0}^{\infty }{\frac {z^{2k}}{(2k)!}}=\cosh z} {\displaystyle \sum _{k=0}^{\infty }{\frac {z^{2k}}{(2k)!}}=\cosh z}
  • ∑ k = 1 ∞ ( − 1 ) k − 1 ( 2 2 k − 1 ) 2 2 k B 2 k z 2 k − 1 ( 2 k ) ! = tan ⁡ z , | z | < π 2 {\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k-1}(2^{2k}-1)2^{2k}B_{2k}z^{2k-1}}{(2k)!}}=\tan z,|z|<{\frac {\pi }{2}}} {\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k-1}(2^{2k}-1)2^{2k}B_{2k}z^{2k-1}}{(2k)!}}=\tan z,|z|<{\frac {\pi }{2}}}
  • ∑ k = 1 ∞ ( 2 2 k − 1 ) 2 2 k B 2 k z 2 k − 1 ( 2 k ) ! = tanh ⁡ z , | z | < π 2 {\displaystyle \sum _{k=1}^{\infty }{\frac {(2^{2k}-1)2^{2k}B_{2k}z^{2k-1}}{(2k)!}}=\tanh z,|z|<{\frac {\pi }{2}}} {\displaystyle \sum _{k=1}^{\infty }{\frac {(2^{2k}-1)2^{2k}B_{2k}z^{2k-1}}{(2k)!}}=\tanh z,|z|<{\frac {\pi }{2}}}
  • ∑ k = 0 ∞ ( − 1 ) k 2 2 k B 2 k z 2 k − 1 ( 2 k ) ! = cot ⁡ z , | z | < π {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}2^{2k}B_{2k}z^{2k-1}}{(2k)!}}=\cot z,|z|<\pi } {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}2^{2k}B_{2k}z^{2k-1}}{(2k)!}}=\cot z,|z|<\pi }
  • ∑ k = 0 ∞ 2 2 k B 2 k z 2 k − 1 ( 2 k ) ! = coth ⁡ z , | z | < π {\displaystyle \sum _{k=0}^{\infty }{\frac {2^{2k}B_{2k}z^{2k-1}}{(2k)!}}=\coth z,|z|<\pi } {\displaystyle \sum _{k=0}^{\infty }{\frac {2^{2k}B_{2k}z^{2k-1}}{(2k)!}}=\coth z,|z|<\pi }
  • ∑ k = 0 ∞ ( − 1 ) k − 1 ( 2 2 k − 2 ) B 2 k z 2 k − 1 ( 2 k ) ! = csc ⁡ z , | z | < π {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k-1}(2^{2k}-2)B_{2k}z^{2k-1}}{(2k)!}}=\csc z,|z|<\pi } {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k-1}(2^{2k}-2)B_{2k}z^{2k-1}}{(2k)!}}=\csc z,|z|<\pi }
  • ∑ k = 0 ∞ − ( 2 2 k − 2 ) B 2 k z 2 k − 1 ( 2 k ) ! = csch ⁡ z , | z | < π {\displaystyle \sum _{k=0}^{\infty }{\frac {-(2^{2k}-2)B_{2k}z^{2k-1}}{(2k)!}}=\operatorname {csch} z,|z|<\pi } {\displaystyle \sum _{k=0}^{\infty }{\frac {-(2^{2k}-2)B_{2k}z^{2k-1}}{(2k)!}}=\operatorname {csch} z,|z|<\pi }
  • ∑ k = 0 ∞ ( − 1 ) k E 2 k z 2 k ( 2 k ) ! = sech ⁡ z , | z | < π 2 {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}E_{2k}z^{2k}}{(2k)!}}=\operatorname {sech} z,|z|<{\frac {\pi }{2}}} {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}E_{2k}z^{2k}}{(2k)!}}=\operatorname {sech} z,|z|<{\frac {\pi }{2}}}
  • ∑ k = 0 ∞ E 2 k z 2 k ( 2 k ) ! = sec ⁡ z , | z | < π 2 {\displaystyle \sum _{k=0}^{\infty }{\frac {E_{2k}z^{2k}}{(2k)!}}=\sec z,|z|<{\frac {\pi }{2}}} {\displaystyle \sum _{k=0}^{\infty }{\frac {E_{2k}z^{2k}}{(2k)!}}=\sec z,|z|<{\frac {\pi }{2}}}
  • ∑ k = 1 ∞ ( − 1 ) k − 1 z 2 k ( 2 k ) ! = ver ⁡ z {\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k-1}z^{2k}}{(2k)!}}=\operatorname {ver} z} {\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k-1}z^{2k}}{(2k)!}}=\operatorname {ver} z} (versinus)
  • ∑ k = 1 ∞ ( − 1 ) k − 1 z 2 k 2 ( 2 k ) ! = hav ⁡ z {\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k-1}z^{2k}}{2(2k)!}}=\operatorname {hav} z} {\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k-1}z^{2k}}{2(2k)!}}=\operatorname {hav} z}[1] (haversinus)
  • ∑ k = 0 ∞ ( 2 k ) ! z 2 k + 1 2 2 k ( k ! ) 2 ( 2 k + 1 ) = arcsin ⁡ z , | z | ≤ 1 {\displaystyle \sum _{k=0}^{\infty }{\frac {(2k)!z^{2k+1}}{2^{2k}(k!)^{2}(2k+1)}}=\arcsin z,|z|\leq 1} {\displaystyle \sum _{k=0}^{\infty }{\frac {(2k)!z^{2k+1}}{2^{2k}(k!)^{2}(2k+1)}}=\arcsin z,|z|\leq 1}
  • ∑ k = 0 ∞ ( − 1 ) k ( 2 k ) ! z 2 k + 1 2 2 k ( k ! ) 2 ( 2 k + 1 ) = arcsinh ⁡ z , | z | ≤ 1 {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}(2k)!z^{2k+1}}{2^{2k}(k!)^{2}(2k+1)}}=\operatorname {arcsinh} {z},|z|\leq 1} {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}(2k)!z^{2k+1}}{2^{2k}(k!)^{2}(2k+1)}}=\operatorname {arcsinh} {z},|z|\leq 1}
  • ∑ k = 0 ∞ ( − 1 ) k z 2 k + 1 2 k + 1 = arctan ⁡ z , | z | < 1 {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}z^{2k+1}}{2k+1}}=\arctan z,|z|<1} {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}z^{2k+1}}{2k+1}}=\arctan z,|z|<1}
  • ∑ k = 0 ∞ z 2 k + 1 2 k + 1 = arctanh ⁡ z , | z | < 1 {\displaystyle \sum _{k=0}^{\infty }{\frac {z^{2k+1}}{2k+1}}=\operatorname {arctanh} z,|z|<1} {\displaystyle \sum _{k=0}^{\infty }{\frac {z^{2k+1}}{2k+1}}=\operatorname {arctanh} z,|z|<1}
  • ln ⁡ 2 + ∑ k = 1 ∞ ( − 1 ) k − 1 ( 2 k ) ! z 2 k 2 2 k + 1 k ( k ! ) 2 = ln ⁡ ( 1 + 1 + z 2 ) , | z | ≤ 1 {\displaystyle \ln 2+\sum _{k=1}^{\infty }{\frac {(-1)^{k-1}(2k)!z^{2k}}{2^{2k+1}k(k!)^{2}}}=\ln \left(1+{\sqrt {1+z^{2}}}\right),|z|\leq 1} {\displaystyle \ln 2+\sum _{k=1}^{\infty }{\frac {(-1)^{k-1}(2k)!z^{2k}}{2^{2k+1}k(k!)^{2}}}=\ln \left(1+{\sqrt {1+z^{2}}}\right),|z|\leq 1}

Penyebut faktorial yang dimodifikasi

  • ∑ k = 0 ∞ ( 4 k ) ! 2 4 k 2 ( 2 k ) ! ( 2 k + 1 ) ! z k = 1 − 1 − z z , | z | < 1 {\displaystyle \sum _{k=0}^{\infty }{\frac {(4k)!}{2^{4k}{\sqrt {2}}(2k)!(2k+1)!}}z^{k}={\sqrt {\frac {1-{\sqrt {1-z}}}{z}}},|z|<1} {\displaystyle \sum _{k=0}^{\infty }{\frac {(4k)!}{2^{4k}{\sqrt {2}}(2k)!(2k+1)!}}z^{k}={\sqrt {\frac {1-{\sqrt {1-z}}}{z}}},|z|<1}[2]
  • ∑ k = 0 ∞ 2 2 k ( k ! ) 2 ( k + 1 ) ( 2 k + 1 ) ! z 2 k + 2 = ( arcsin ⁡ z ) 2 , | z | ≤ 1 {\displaystyle \sum _{k=0}^{\infty }{\frac {2^{2k}(k!)^{2}}{(k+1)(2k+1)!}}z^{2k+2}=\left(\arcsin {z}\right)^{2},|z|\leq 1} {\displaystyle \sum _{k=0}^{\infty }{\frac {2^{2k}(k!)^{2}}{(k+1)(2k+1)!}}z^{2k+2}=\left(\arcsin {z}\right)^{2},|z|\leq 1}[2]
  • ∑ n = 0 ∞ ∏ k = 0 n − 1 ( 4 k 2 + α 2 ) ( 2 n ) ! z 2 n + ∑ n = 0 ∞ α ∏ k = 0 n − 1 [ ( 2 k + 1 ) 2 + α 2 ] ( 2 n + 1 ) ! z 2 n + 1 = e α arcsin ⁡ z , | z | ≤ 1 {\displaystyle \sum _{n=0}^{\infty }{\frac {\prod _{k=0}^{n-1}(4k^{2}+\alpha ^{2})}{(2n)!}}z^{2n}+\sum _{n=0}^{\infty }{\frac {\alpha \prod _{k=0}^{n-1}[(2k+1)^{2}+\alpha ^{2}]}{(2n+1)!}}z^{2n+1}=e^{\alpha \arcsin {z}},|z|\leq 1} {\displaystyle \sum _{n=0}^{\infty }{\frac {\prod _{k=0}^{n-1}(4k^{2}+\alpha ^{2})}{(2n)!}}z^{2n}+\sum _{n=0}^{\infty }{\frac {\alpha \prod _{k=0}^{n-1}[(2k+1)^{2}+\alpha ^{2}]}{(2n+1)!}}z^{2n+1}=e^{\alpha \arcsin {z}},|z|\leq 1}

Koefisien binomial

  • ( 1 + z ) α = ∑ k = 0 ∞ ( α k ) z k , | z | < 1 {\displaystyle (1+z)^{\alpha }=\sum _{k=0}^{\infty }{\alpha \choose k}z^{k},|z|<1} {\displaystyle (1+z)^{\alpha }=\sum _{k=0}^{\infty }{\alpha \choose k}z^{k},|z|<1} (lihat teorema binomial)
  • ∑ k = 0 ∞ ( α + k − 1 k ) z k = 1 ( 1 − z ) α , | z | < 1 {\displaystyle \sum _{k=0}^{\infty }{{\alpha +k-1} \choose k}z^{k}={\frac {1}{(1-z)^{\alpha }}},|z|<1} {\displaystyle \sum _{k=0}^{\infty }{{\alpha +k-1} \choose k}z^{k}={\frac {1}{(1-z)^{\alpha }}},|z|<1}[3]
  • ∑ k = 0 ∞ 1 k + 1 ( 2 k k ) z k = 1 − 1 − 4 z 2 z , | z | ≤ 1 4 {\displaystyle \sum _{k=0}^{\infty }{\frac {1}{k+1}}{2k \choose k}z^{k}={\frac {1-{\sqrt {1-4z}}}{2z}},|z|\leq {\frac {1}{4}}} {\displaystyle \sum _{k=0}^{\infty }{\frac {1}{k+1}}{2k \choose k}z^{k}={\frac {1-{\sqrt {1-4z}}}{2z}},|z|\leq {\frac {1}{4}}}, menghasilkan fungsi bilangan Catalan[3]
  • ∑ k = 0 ∞ ( 2 k k ) z k = 1 1 − 4 z , | z | < 1 4 {\displaystyle \sum _{k=0}^{\infty }{2k \choose k}z^{k}={\frac {1}{\sqrt {1-4z}}},|z|<{\frac {1}{4}}} {\displaystyle \sum _{k=0}^{\infty }{2k \choose k}z^{k}={\frac {1}{\sqrt {1-4z}}},|z|<{\frac {1}{4}}}, menghasilkan fungsi koefisien binomial pusat[3]
  • ∑ k = 0 ∞ ( 2 k + α k ) z k = 1 1 − 4 z ( 1 − 1 − 4 z 2 z ) α , | z | < 1 4 {\displaystyle \sum _{k=0}^{\infty }{2k+\alpha \choose k}z^{k}={\frac {1}{\sqrt {1-4z}}}\left({\frac {1-{\sqrt {1-4z}}}{2z}}\right)^{\alpha },|z|<{\frac {1}{4}}} {\displaystyle \sum _{k=0}^{\infty }{2k+\alpha \choose k}z^{k}={\frac {1}{\sqrt {1-4z}}}\left({\frac {1-{\sqrt {1-4z}}}{2z}}\right)^{\alpha },|z|<{\frac {1}{4}}}[3]

Bilangan harmonik

(Lihat bilangan harmonik yang didefinisikan H n = ∑ j = 1 n 1 j {\textstyle H_{n}=\sum _{j=1}^{n}{\frac {1}{j}}} {\textstyle H_{n}=\sum _{j=1}^{n}{\frac {1}{j}}})

  • ∑ k = 1 ∞ H k z k = − ln ⁡ ( 1 − z ) 1 − z , | z | < 1 {\displaystyle \sum _{k=1}^{\infty }H_{k}z^{k}={\frac {-\ln(1-z)}{1-z}},|z|<1} {\displaystyle \sum _{k=1}^{\infty }H_{k}z^{k}={\frac {-\ln(1-z)}{1-z}},|z|<1}
  • ∑ k = 1 ∞ H k k + 1 z k + 1 = 1 2 [ ln ⁡ ( 1 − z ) ] 2 , | z | < 1 {\displaystyle \sum _{k=1}^{\infty }{\frac {H_{k}}{k+1}}z^{k+1}={\frac {1}{2}}\left[\ln(1-z)\right]^{2},\qquad |z|<1} {\displaystyle \sum _{k=1}^{\infty }{\frac {H_{k}}{k+1}}z^{k+1}={\frac {1}{2}}\left[\ln(1-z)\right]^{2},\qquad |z|<1}
  • ∑ k = 1 ∞ ( − 1 ) k − 1 H 2 k 2 k + 1 z 2 k + 1 = 1 2 arctan ⁡ z log ⁡ ( 1 + z 2 ) , | z | < 1 {\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k-1}H_{2k}}{2k+1}}z^{2k+1}={\frac {1}{2}}\arctan {z}\log {(1+z^{2})},\qquad |z|<1} {\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k-1}H_{2k}}{2k+1}}z^{2k+1}={\frac {1}{2}}\arctan {z}\log {(1+z^{2})},\qquad |z|<1}[2]
  • ∑ n = 0 ∞ ∑ k = 0 2 n ( − 1 ) k 2 k + 1 z 4 n + 2 4 n + 2 = 1 4 arctan ⁡ z log ⁡ 1 + z 1 − z , | z | < 1 {\displaystyle \sum _{n=0}^{\infty }\sum _{k=0}^{2n}{\frac {(-1)^{k}}{2k+1}}{\frac {z^{4n+2}}{4n+2}}={\frac {1}{4}}\arctan {z}\log {\frac {1+z}{1-z}},\qquad |z|<1} {\displaystyle \sum _{n=0}^{\infty }\sum _{k=0}^{2n}{\frac {(-1)^{k}}{2k+1}}{\frac {z^{4n+2}}{4n+2}}={\frac {1}{4}}\arctan {z}\log {\frac {1+z}{1-z}},\qquad |z|<1}[2]

Koefisien binomial

  • ∑ k = 0 n ( n k ) = 2 n {\displaystyle \sum _{k=0}^{n}{n \choose k}=2^{n}} {\displaystyle \sum _{k=0}^{n}{n \choose k}=2^{n}}
  • ∑ k = 0 n ( − 1 ) k ( n k ) = 0 , {\displaystyle \sum _{k=0}^{n}(-1)^{k}{n \choose k}=0,} {\displaystyle \sum _{k=0}^{n}(-1)^{k}{n \choose k}=0,} dengan n ≥ 1 {\displaystyle n\geq 1} {\displaystyle n\geq 1}
  • ∑ k = 0 n ( k m ) = ( n + 1 m + 1 ) {\displaystyle \sum _{k=0}^{n}{k \choose m}={n+1 \choose m+1}} {\displaystyle \sum _{k=0}^{n}{k \choose m}={n+1 \choose m+1}}
  • ∑ k = 0 n ( m + k − 1 k ) = ( n + m n ) {\displaystyle \sum _{k=0}^{n}{m+k-1 \choose k}={n+m \choose n}} {\displaystyle \sum _{k=0}^{n}{m+k-1 \choose k}={n+m \choose n}} (lihat multihimpunan)
  • ∑ k = 0 n ( α k ) ( β n − k ) = ( α + β n ) {\displaystyle \sum _{k=0}^{n}{\alpha \choose k}{\beta \choose n-k}={\alpha +\beta \choose n}} {\displaystyle \sum _{k=0}^{n}{\alpha \choose k}{\beta \choose n-k}={\alpha +\beta \choose n}} (lihat identitas Vandermonde)

Fungsi trigonometrik

Penjumlahan fungsi sinus dan kosinus muncul dalam deret Fourier.

  • ∑ k = 1 ∞ sin ⁡ ( k θ ) k = π − θ 2 , 0 < θ < 2 π {\displaystyle \sum _{k=1}^{\infty }{\frac {\sin(k\theta )}{k}}={\frac {\pi -\theta }{2}},0<\theta <2\pi } {\displaystyle \sum _{k=1}^{\infty }{\frac {\sin(k\theta )}{k}}={\frac {\pi -\theta }{2}},0<\theta <2\pi }
  • ∑ k = 1 ∞ cos ⁡ ( k θ ) k = − 1 2 ln ⁡ ( 2 − 2 cos ⁡ θ ) , θ ∈ R {\displaystyle \sum _{k=1}^{\infty }{\frac {\cos(k\theta )}{k}}=-{\frac {1}{2}}\ln(2-2\cos \theta ),\theta \in \mathbb {R} } {\displaystyle \sum _{k=1}^{\infty }{\frac {\cos(k\theta )}{k}}=-{\frac {1}{2}}\ln(2-2\cos \theta ),\theta \in \mathbb {R} }
  • ∑ k = 0 ∞ sin ⁡ [ ( 2 k + 1 ) θ ] 2 k + 1 = π 4 , 0 < θ < π {\displaystyle \sum _{k=0}^{\infty }{\frac {\sin[(2k+1)\theta ]}{2k+1}}={\frac {\pi }{4}},0<\theta <\pi } {\displaystyle \sum _{k=0}^{\infty }{\frac {\sin[(2k+1)\theta ]}{2k+1}}={\frac {\pi }{4}},0<\theta <\pi },
  • B n ( x ) = − n ! 2 n − 1 π n ∑ k = 1 ∞ 1 k n cos ⁡ ( 2 π k x − π n 2 ) , 0 < x < 1 {\displaystyle B_{n}(x)=-{\frac {n!}{2^{n-1}\pi ^{n}}}\sum _{k=1}^{\infty }{\frac {1}{k^{n}}}\cos \left(2\pi kx-{\frac {\pi n}{2}}\right),0<x<1} {\displaystyle B_{n}(x)=-{\frac {n!}{2^{n-1}\pi ^{n}}}\sum _{k=1}^{\infty }{\frac {1}{k^{n}}}\cos \left(2\pi kx-{\frac {\pi n}{2}}\right),0<x<1}[4]
  • ∑ k = 0 n sin ⁡ ( θ + k α ) = sin ⁡ ( n + 1 ) α 2 sin ⁡ ( θ + n α 2 ) sin ⁡ α 2 {\displaystyle \sum _{k=0}^{n}\sin(\theta +k\alpha )={\frac {\sin {\frac {(n+1)\alpha }{2}}\sin(\theta +{\frac {n\alpha }{2}})}{\sin {\frac {\alpha }{2}}}}} {\displaystyle \sum _{k=0}^{n}\sin(\theta +k\alpha )={\frac {\sin {\frac {(n+1)\alpha }{2}}\sin(\theta +{\frac {n\alpha }{2}})}{\sin {\frac {\alpha }{2}}}}}
  • ∑ k = 0 n cos ⁡ ( θ + k α ) = sin ⁡ ( n + 1 ) α 2 cos ⁡ ( θ + n α 2 ) sin ⁡ α 2 {\displaystyle \sum _{k=0}^{n}\cos(\theta +k\alpha )={\frac {\sin {\frac {(n+1)\alpha }{2}}\cos(\theta +{\frac {n\alpha }{2}})}{\sin {\frac {\alpha }{2}}}}} {\displaystyle \sum _{k=0}^{n}\cos(\theta +k\alpha )={\frac {\sin {\frac {(n+1)\alpha }{2}}\cos(\theta +{\frac {n\alpha }{2}})}{\sin {\frac {\alpha }{2}}}}}
  • ∑ k = 1 n − 1 sin ⁡ π k n = cot ⁡ π 2 n {\displaystyle \sum _{k=1}^{n-1}\sin {\frac {\pi k}{n}}=\cot {\frac {\pi }{2n}}} {\displaystyle \sum _{k=1}^{n-1}\sin {\frac {\pi k}{n}}=\cot {\frac {\pi }{2n}}}
  • ∑ k = 1 n − 1 sin ⁡ 2 π k n = 0 {\displaystyle \sum _{k=1}^{n-1}\sin {\frac {2\pi k}{n}}=0} {\displaystyle \sum _{k=1}^{n-1}\sin {\frac {2\pi k}{n}}=0}
  • ∑ k = 0 n − 1 csc 2 ⁡ ( θ + π k n ) = n 2 csc 2 ⁡ ( n θ ) {\displaystyle \sum _{k=0}^{n-1}\csc ^{2}\left(\theta +{\frac {\pi k}{n}}\right)=n^{2}\csc ^{2}(n\theta )} {\displaystyle \sum _{k=0}^{n-1}\csc ^{2}\left(\theta +{\frac {\pi k}{n}}\right)=n^{2}\csc ^{2}(n\theta )}[5]
  • ∑ k = 1 n − 1 csc 2 ⁡ π k n = n 2 − 1 3 {\displaystyle \sum _{k=1}^{n-1}\csc ^{2}{\frac {\pi k}{n}}={\frac {n^{2}-1}{3}}} {\displaystyle \sum _{k=1}^{n-1}\csc ^{2}{\frac {\pi k}{n}}={\frac {n^{2}-1}{3}}}
  • ∑ k = 1 n − 1 csc 4 ⁡ π k n = n 4 + 10 n 2 − 11 45 {\displaystyle \sum _{k=1}^{n-1}\csc ^{4}{\frac {\pi k}{n}}={\frac {n^{4}+10n^{2}-11}{45}}} {\displaystyle \sum _{k=1}^{n-1}\csc ^{4}{\frac {\pi k}{n}}={\frac {n^{4}+10n^{2}-11}{45}}}

Fungsi rasional

  • ∑ n = a + 1 ∞ a n 2 − a 2 = 1 2 H 2 a {\displaystyle \sum _{n=a+1}^{\infty }{\frac {a}{n^{2}-a^{2}}}={\frac {1}{2}}H_{2a}} {\displaystyle \sum _{n=a+1}^{\infty }{\frac {a}{n^{2}-a^{2}}}={\frac {1}{2}}H_{2a}}[6]
  • ∑ n = 0 ∞ 1 n 2 + a 2 = 1 + a π coth ⁡ ( a π ) 2 a 2 {\displaystyle \sum _{n=0}^{\infty }{\frac {1}{n^{2}+a^{2}}}={\frac {1+a\pi \coth(a\pi )}{2a^{2}}}} {\displaystyle \sum _{n=0}^{\infty }{\frac {1}{n^{2}+a^{2}}}={\frac {1+a\pi \coth(a\pi )}{2a^{2}}}}
  • ∑ n = 0 ∞ 1 n 4 + 4 a 4 = 1 8 a 4 + π ( sinh ⁡ ( 2 π a ) + sin ⁡ ( 2 π a ) ) 8 a 3 ( cosh ⁡ ( 2 π a ) − cos ⁡ ( 2 π a ) ) {\displaystyle \displaystyle \sum _{n=0}^{\infty }{\frac {1}{n^{4}+4a^{4}}}={\dfrac {1}{8a^{4}}}+{\dfrac {\pi (\sinh(2\pi a)+\sin(2\pi a))}{8a^{3}(\cosh(2\pi a)-\cos(2\pi a))}}} {\displaystyle \displaystyle \sum _{n=0}^{\infty }{\frac {1}{n^{4}+4a^{4}}}={\dfrac {1}{8a^{4}}}+{\dfrac {\pi (\sinh(2\pi a)+\sin(2\pi a))}{8a^{3}(\cosh(2\pi a)-\cos(2\pi a))}}}
  • Suatu deret tak terhingga dari setiap fungsi rasional n {\displaystyle n} {\displaystyle n} dapat direduksi menjadi suatu deret terhingga dari fungsi poligamma, dengan menggunakan dekomposisi pecahan parsial.[7] Fakta ini juga berlaku pada deret terhingga dari fungsi rasional, yang memungkinkan hasilnya dihitung dalam waktu konstanta bahkan jika deret tersebut memiliki banyak suku.

Fungsi eksponensial

  • 1 p ∑ n = 0 p − 1 exp ⁡ ( 2 π i n 2 q p ) = e π i / 4 2 q ∑ n = 0 2 q − 1 exp ⁡ ( − π i n 2 p 2 q ) {\displaystyle \displaystyle {\dfrac {1}{\sqrt {p}}}\sum _{n=0}^{p-1}\exp \left({\frac {2\pi in^{2}q}{p}}\right)={\dfrac {e^{\pi i/4}}{\sqrt {2q}}}\sum _{n=0}^{2q-1}\exp \left(-{\frac {\pi in^{2}p}{2q}}\right)} {\displaystyle \displaystyle {\dfrac {1}{\sqrt {p}}}\sum _{n=0}^{p-1}\exp \left({\frac {2\pi in^{2}q}{p}}\right)={\dfrac {e^{\pi i/4}}{\sqrt {2q}}}\sum _{n=0}^{2q-1}\exp \left(-{\frac {\pi in^{2}p}{2q}}\right)}(lihat relasi Landsberg–Schaar)
  • ∑ n = − ∞ ∞ e − π n 2 = π 4 Γ ( 3 4 ) {\displaystyle \displaystyle \sum _{n=-\infty }^{\infty }e^{-\pi n^{2}}={\frac {\sqrt[{4}]{\pi }}{\Gamma \left({\frac {3}{4}}\right)}}} {\displaystyle \displaystyle \sum _{n=-\infty }^{\infty }e^{-\pi n^{2}}={\frac {\sqrt[{4}]{\pi }}{\Gamma \left({\frac {3}{4}}\right)}}}

Lihat pula

  • Deret (matematika)
  • Daftar integral
  • Notasi Sigma § Identitas
  • Deret Taylor
  • Teorema binomial
  • Deret Gregory
  • On-Line Encyclopedia of Integer Sequences

Catatan

  1. ↑ Weisstein, Eric W. "Haversine". MathWorld. Wolfram Research, Inc. Diarsipkan dari versi aslinya tanggal 2005-03-10. Diakses tanggal 2015-11-06.
  2. 1 2 3 4 Wilf, Herbert R. (1994). generatingfunctionology (PDF). Academic Press, Inc.
  3. 1 2 3 4 "Theoretical computer science cheat sheet" (PDF).
  4. ↑ "Bernoulli polynomials: Series representations (subsection 06/02)". Wolfram Research. Diakses tanggal 2 June 2011.
  5. ↑ 1 + \\frac{1}{2^2} + \\frac{1}{3^2} + \\dots = \\frac{\\pi^2}{6} </math> and related identities"},"url":{"wt":"http://homepage.univie.ac.at/josef.hofbauer/02amm.pdf"},"access-date":{"wt":"2 June 2011"}},"i":0}}]}' id="mwAXg"/>Hofbauer, Josef. "A simple proof of 1 + 1 2 2 + 1 3 2 + ⋯ = π 2 6 {\textstyle 1+{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}+\dots ={\frac {\pi ^{2}}{6}}} {\textstyle 1+{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}+\dots ={\frac {\pi ^{2}}{6}}} and related identities" (PDF). Diakses tanggal 2 June 2011.
  6. ↑ Sondow, Jonathan; Weisstein, Eric W. "Riemann Zeta Function (eq. 52)". MathWorld—A Wolfram Web Resource.
  7. ↑ Abramowitz, Milton; Stegun, Irene (1964). "6.4 Polygamma functions". Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. hlm. 260. ISBN 0-486-61272-4.

Referensi

  • Banyak buku-buku dengan sebuah daftar integral juga memiliki sebuah daftar deret.

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Daftar Isi

  1. Penjumlahan pangkat
  2. Deret pangkat
  3. Polilogaritma orde rendah
  4. Fungsi eksponensial
  5. Fungsi trigonometrik, trigonometrik invers, hiperbolik, dan hiperbolik invers
  6. Penyebut faktorial yang dimodifikasi
  7. Koefisien binomial
  8. Bilangan harmonik
  9. Koefisien binomial
  10. Fungsi trigonometrik
  11. Fungsi rasional
  12. Fungsi eksponensial
  13. Lihat pula
  14. Catatan
  15. Referensi

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