what-Maclaurin-series-of-function-tan-x-

Question Number 83864 by jagoll last updated on 07/Mar/20

what Maclaurin series of function

\tan (x) ?

Commented by niroj last updated on 07/Mar/20

Solution :

let, f (x) = \tan x, f_{0} (0) = 0

f_{1} (x) = \sec^{2} x, f_{1} (0) = 1

f_{2} (x) = 2 secx . secx . \tan x, f_{2} (0) = 0

= {2 \sec}^{2} xtanx

f_{3} (x) = {4 \sec}^{2} x . \tan^{2} x + {2 \sec}^{4} x, f_{3} (0) = 2

f_{4} (x) = {8 \sec}^{2} {xtan}^{3} x + {8 \sec}^{4} x . tanx ., f_{4} (0) = 0

f_{5} (x) = {16 \sec}^{2} {xtan}^{4} x . + {8 \sec}^{2} x . {3 \tan}^{2} x . \sec^{2} x +

{32 \sec}^{3} x . \sec x \tan^{2} x + {8 \sec}^{6} x, f_{5} (0) = 8

now, we know {maclaurin}^{'} s series of

expansion :

f (x) = f (0) + \frac{x}{1!} f_{1} (0) + \frac{x^{2}}{2!} f_{2} (0) + \frac{x^{3}}{3!} f_{3} (0) + \frac{x^{4}}{4!} f_{4} (0) + \frac{x^{5}}{5!} f_{5} (0)

\tan x = 0 + x . (1) + \frac{x^{2}}{2} (0) + \frac{x^{3}}{6} . (2) + \frac{x^{4}}{24} (0) + \frac{x^{5}}{120} . (8)

= x + \frac{1}{3} x^{3} + \frac{2}{15} x^{5} + \dots . .

hence proved {maclaurin}^{'} s function

of \tan x .

Commented by jagoll last updated on 07/Mar/20

thank you sir

Commented by niroj last updated on 07/Mar/20

y o u m u s t w e l c o m e s i r .

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