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) (7)f(x)=tanh(x) (8)f(x)=csch(x) (9)f(x)=coth(x) (10)f(x)=cosh(x) how can help me please


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

$i w a n t t h e f o r m o l l a o f t a y l o r a n d m a c l o r i e n$ $s e r i e s o f t h i s f u n c t i o n$ $(1) f (x) = l n (x) (2) f (x) = s e c (x)$ $(3) f (x) = c s c (x) (4) f (x) = c o t (x)$ $(5) f (x) = t a n (x) (6) f (x) = s e c h (x)$ $(7) f (x) = t a n h (x) (8) f (x) = c s c h (x)$ $(9) f (x) = c o t h (x) (10) f (x) = c o s h (x)$ $h o w c a n h e l p m e p l e a s e$
Commented by Mrsof last updated on 09/Jul/21

$? ? ? ? ? ? ?$
	Answered by Olaf_Thorendsen last updated on 09/Jul/21

	$(2)$ $f (x) = \sec (x) = \frac{1}{\cos x}$ $f (x) = \underset{n = 0}{\sum^{\infty}} \frac{E_{2 n}}{(2 n)!} x^{2 n}$ $(3)$ $f (x) = \csc (x) = \frac{1}{\sin x}$ $f (x) = \frac{1}{x} + \underset{n = 1}{\sum^{\infty}} ∣ B_{2 n} ∣ \frac{2 (2^{2 n - 1} - 1)}{(2 n)!} x^{2 n}, 0 <∣ x ∣< π$ $(4)$ $f (x) = \cot (x)$ $f (x) = \frac{1}{x} - \underset{n = 1}{\sum^{\infty}} ∣ B_{2 n} ∣ \frac{2^{2 n}}{(2 n)!} x^{2 n - 1}, 0 <∣ x ∣< π$ $(5)$ $f (x) = \tan (x)$ $f (x) = \underset{n = 1}{\sum^{\infty}} ∣ B_{2 n} ∣ \frac{2^{2 n} (2^{2 n} - 1)}{(2 n)!} x^{2 n - 1}, ∣ x ∣< \frac{π}{2}$ $(6)$ $f (x) = sech (x) = \frac{1}{ch (x)}$ $f (x) = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \frac{E_{2 n}}{(2 n)!} x^{2 n}$ $(7)$ $f (x) = \tanh (x)$ $f (x) = \underset{n = 1}{\sum^{\infty}} B_{2 n} \frac{2^{2 n} (2^{2 n} - 1)}{(2 n)!} x^{2 n - 1}, ∣ x ∣< \frac{π}{2}$ $(8)$ $f (x) = csch (x) = \frac{1}{sh (x)} = \frac{2 e^{x}}{e^{2 x} - 1}$ $f (x) = \frac{1}{x} - \underset{n = 1}{\sum^{\infty}} B_{2 n} \frac{2 (2^{2 n} - 1)}{(2 n)!} x^{2 n - 1}, 0 <∣ x ∣< π$ $(9)$ $f (x) = \coth (x)$ $f (x) = \frac{1}{x} + \underset{n = 1}{\sum^{\infty}} B_{2 n} \frac{2^{2 n}}{(2 n)!} x^{2 n - 1}, 0 <∣ x ∣< π$ $(10)$ $f (x) = \cosh (x)$ $f (x) = \underset{n = 0}{\sum^{\infty}} \frac{1}{(2 n)!} x^{2 n}$
	Commented by tabata last updated on 09/Jul/21

	$t h a n k y o u s i r b u t w h a t s t h e m e a n B_{2 n} a n d E_{2 n}$
	Commented by Olaf_Thorendsen last updated on 09/Jul/21

	$E_{k} = Euler numbers .$ $B_{k} = Bernoulli numbers .$
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