i-want-the-formolla-of-taylor-and-maclorien-series-of-this-function-1-f-x-ln-x-2-f-x-sec-x-3-f-x-csc-x-4-f-x-cot-x-5-f-x-tan-x-6-f-x-sech-x

Question Number 145874 by Mrsof last updated on 09/Jul/21

$${i}\:{want}\:{the}\:{formolla}\:{of}\:{taylor}\:{and}\:{maclorien}\: \\ $$$${series}\:{of}\:{this}\:{function} \\ $$$$\left(\mathrm{1}\right){f}\left({x}\right)={ln}\left({x}\:\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{2}\right){f}\left({x}\right)={sec}\left({x}\right) \\ $$$$ \\ $$$$\left(\mathrm{3}\right){f}\left({x}\right)={csc}\left({x}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{4}\right){f}\left({x}\right)={cot}\left({x}\right) \\ $$$$ \\ $$$$\left(\mathrm{5}\right){f}\left({x}\right)={tan}\left({x}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{6}\right){f}\left({x}\right)={sech}\left({x}\right) \\ $$$$ \\ $$$$\left(\mathrm{7}\right){f}\left({x}\right)={tanh}\left({x}\right)\:\:\:\:\:\:\:\:\left(\mathrm{8}\right){f}\left({x}\right)={csch}\left({x}\right) \\ $$$$ \\ $$$$\left(\mathrm{9}\right){f}\left({x}\right)={coth}\left({x}\right)\:\:\:\:\:\:\:\:\:\left(\mathrm{10}\right){f}\left({x}\right)={cosh}\left({x}\right) \\ $$$$ \\ $$$${how}\:{can}\:{help}\:{me}\:{please} \\ $$

Commented by Mrsof last updated on 09/Jul/21

$$??????? \\ $$

Answered by Olaf_Thorendsen last updated on 09/Jul/21

(2) f(x) = sec(x) = (1/(cosx)) f(x) = Σ_(n=0) ^∞ (E_(2n) /((2n)!))x^(2n) (3) f(x) = csc(x) = (1/(sinx)) f(x) = (1/x)+Σ_(n=1) ^∞ ∣B_(2n) ∣((2(2^(2n−1) −1))/((2n)!))x^(2n) , 0<∣x∣<π (4) f(x) = cot(x) f(x) = (1/x)−Σ_(n=1) ^∞ ∣B_(2n) ∣(2^(2n) /((2n)!))x^(2n−1) , 0<∣x∣<π (5) f(x) = tan(x) f(x) = Σ_(n=1) ^∞ ∣B_(2n) ∣((2^(2n) (2^(2n) −1))/((2n)!))x^(2n−1) , ∣x∣<(π/2) (6) f(x) = sech(x) = (1/(ch(x))) f(x) = Σ_(n=0) ^∞ (−1)^n (E_(2n) /((2n)!))x^(2n) (7) f(x) = tanh(x) f(x) = Σ_(n=1) ^∞ B_(2n) ((2^(2n) (2^(2n) −1))/((2n)!))x^(2n−1) , ∣x∣<(π/2) (8) f(x) = csch(x) = (1/(sh(x))) = ((2e^x )/(e^(2x) −1)) f(x) = (1/x)−Σ_(n=1) ^∞ B_(2n) ((2(2^(2n) −1))/((2n)!))x^(2n−1) , 0<∣x∣<π (9) f(x) = coth(x) f(x) = (1/x)+Σ_(n=1) ^∞ B_(2n) (2^(2n) /((2n)!))x^(2n−1) , 0<∣x∣<π (10) f(x) = cosh(x) f(x) = Σ_(n=0) ^∞ (1/((2n)!))x^(2n)

$$\left(\mathrm{2}\right) \\ $$$${f}\left({x}\right)\:=\:\mathrm{sec}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{cos}{x}} \\ $$$${f}\left({x}\right)\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{E}_{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!}{x}^{\mathrm{2}{n}} \\ $$$$\left(\mathrm{3}\right) \\ $$$${f}\left({x}\right)\:=\:\mathrm{csc}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{sin}{x}} \\ $$$${f}\left({x}\right)\:=\:\frac{\mathrm{1}}{{x}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mid{B}_{\mathrm{2}{n}} \mid\frac{\mathrm{2}\left(\mathrm{2}^{\mathrm{2}{n}−\mathrm{1}} −\mathrm{1}\right)}{\left(\mathrm{2}{n}\right)!}{x}^{\mathrm{2}{n}} ,\:\mathrm{0}<\mid{x}\mid<\pi \\ $$$$\left(\mathrm{4}\right) \\ $$$${f}\left({x}\right)\:=\:\mathrm{cot}\left({x}\right) \\ $$$${f}\left({x}\right)\:=\:\frac{\mathrm{1}}{{x}}−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mid{B}_{\mathrm{2}{n}} \mid\frac{\mathrm{2}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!}{x}^{\mathrm{2}{n}−\mathrm{1}} ,\:\mathrm{0}<\mid{x}\mid<\pi \\ $$$$\left(\mathrm{5}\right) \\ $$$${f}\left({x}\right)\:=\:\mathrm{tan}\left({x}\right) \\ $$$${f}\left({x}\right)\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mid{B}_{\mathrm{2}{n}} \mid\frac{\mathrm{2}^{\mathrm{2}{n}} \left(\mathrm{2}^{\mathrm{2}{n}} −\mathrm{1}\right)}{\left(\mathrm{2}{n}\right)!}{x}^{\mathrm{2}{n}−\mathrm{1}} ,\:\mid{x}\mid<\frac{\pi}{\mathrm{2}} \\ $$$$\left(\mathrm{6}\right) \\ $$$${f}\left({x}\right)\:=\:\mathrm{sech}\left(\mathrm{x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{ch}\left({x}\right)} \\ $$$${f}\left({x}\right)\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \frac{{E}_{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!}{x}^{\mathrm{2}{n}} \\ $$$$\left(\mathrm{7}\right) \\ $$$${f}\left({x}\right)\:=\:\mathrm{tanh}\left({x}\right) \\ $$$${f}\left({x}\right)\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{B}_{\mathrm{2}{n}} \frac{\mathrm{2}^{\mathrm{2}{n}} \left(\mathrm{2}^{\mathrm{2}{n}} −\mathrm{1}\right)}{\left(\mathrm{2}{n}\right)!}{x}^{\mathrm{2}{n}−\mathrm{1}} ,\:\mid{x}\mid<\frac{\pi}{\mathrm{2}} \\ $$$$\left(\mathrm{8}\right) \\ $$$${f}\left({x}\right)\:=\:\mathrm{csch}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{sh}\left({x}\right)}\:=\:\frac{\mathrm{2}{e}^{{x}} }{{e}^{\mathrm{2}{x}} −\mathrm{1}} \\ $$$${f}\left({x}\right)\:=\:\frac{\mathrm{1}}{{x}}−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{B}_{\mathrm{2}{n}} \frac{\mathrm{2}\left(\mathrm{2}^{\mathrm{2}{n}} −\mathrm{1}\right)}{\left(\mathrm{2}{n}\right)!}{x}^{\mathrm{2}{n}−\mathrm{1}} ,\:\mathrm{0}<\mid{x}\mid<\pi \\ $$$$\left(\mathrm{9}\right) \\ $$$${f}\left({x}\right)\:=\:\mathrm{coth}\left({x}\right) \\ $$$${f}\left({x}\right)\:=\:\frac{\mathrm{1}}{{x}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{B}_{\mathrm{2}{n}} \frac{\mathrm{2}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!}{x}^{\mathrm{2}{n}−\mathrm{1}} ,\:\mathrm{0}<\mid{x}\mid<\pi \\ $$$$\left(\mathrm{10}\right) \\ $$$${f}\left({x}\right)\:=\:\mathrm{cosh}\left({x}\right) \\ $$$${f}\left({x}\right)\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}\right)!}{x}^{\mathrm{2}{n}} \\ $$

Commented by tabata last updated on 09/Jul/21

thank you sir but whats the mean B_(2n) and E_(2n)

$${thank}\:{you}\:{sir}\:{but}\:{whats}\:{the}\:{mean}\:{B}_{\mathrm{2}{n}} \:{and}\:{E}_{\mathrm{2}{n}} \\ $$

Commented by Olaf_Thorendsen last updated on 09/Jul/21

E_k = Euler numbers. B_k = Bernoulli numbers.

$$\mathrm{E}_{{k}} \:=\:\mathrm{Euler}\:\mathrm{numbers}. \\ $$$$\mathrm{B}_{{k}} \:=\:\mathrm{Bernoulli}\:\mathrm{numbers}. \\ $$

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