Maclaurin series

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Prerequisite – Taylor theorem and Taylor series

We know that formula for expansion of Taylor series is written as:
$f(x)=f(a)+\sum_{n=1}^{\infty}\frac{f^n(a)}{n!}(x-a)^n$

Now if we put a=0 in this formula we will get the formula for expansion of Maclaurin series. T
hus Maclaurin series expansion can be given by the formula –
$f(x)=f(0)+\sum_{n=1}^{\infty}\frac{f^n(0)}{n!}(x)^n$

Maclaurin series expansion of some elementary functions :

Exponential function :
$f(x)=e^x$
Differentiating n times, $f^n(x)=e^x.$
So we get $f^n(0)=1$
Thus $e^x = 1+\frac{x}{1!}+ \frac{x^2}{2!}+ \frac{x^3}{3!}+....+ \frac{x^{n-1}}{(n-1)!}+\frac{x^n}{n!}$
f(x) = cos x
$\cosx= 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+$ …..
f(x) = sin x
$\sinx = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+....$
f(x) = (ax + b)^m
$(ax+b)^m=b^m[1+m(a/b)\frac{x}{1!}+m(m-1)(a/b)^2\frac{x^2}{2!}+m(m-1)(m-2)(a/b)^3\frac{x^3}{3!}+.....$
f(x) = ln(1+x)
$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}+.....+(-1)^{n-1}\frac{x^n}{n}+.....$
f(x) = ln(1-x)
$\ln(1-x)=-(x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}+.....+\frac{x^n}{n}+....)$

Example-1:
Find the first seven terms of f(x) = ln(sec x).

Explanation :
$f(x) = ln(\secx)$
$f(0) = ln(\sec0)=0$
Differentiating w.r.t. x,
$f'(x)= (1/secx).\secx.\tanx = \tan x$
$f'(0)= \tan 0 = 0$
$f''(x)= \sec^2x\scriptstyle\implies f''(0) = \sec^20=1$
$f'''(x)= 2\secx.\secx.\tanx=2sec^2x.tanx\scriptstyle\implies f'''(0) = 0$
$f''''(x)= 4\sec^2x.\tan^2x+2\sec^4x\scriptstyle\implies f''''(0) = 0+2 = 2$
$f'''''(x)= 8\sec^2x.\tan^3x+16\sec4x.\tanx\scriptstyle\implies f'''''(0) = 0$
$f''''''(x)= 16\sec^2x.\tan^4x+88\sec4x.\tan^2x+16\sec^6x\scriptstyle\implies f''''''(0) = 16$
Thus we get the Maclaurin series as –
$f(x) = f(0)+f'(0).x/1!+f''(0).x^2/2!+f'''(0).x^3/3!+.... \text{upto 7 terms}$
$f(x)=ln(\secx)=0+1.x^2/2!+0+2.x^4/4!+0+16x^6/6!+....$
$f(x)=\frac{x^2}{2}+\frac{x^4}{12}+\frac{x^6}{45}+....$

Example-2:
Evaluate Maclaurin series for tan x.

Explanation :
$f(x) = \tan x, f(0)=0$
$f'(x) = \sec^2x \scriptstyle\implies f'(0)=1$
$f''(x) = 2\sec^x.\secx.\tanx=2\sec^2x.\tanx=2(\tanx+\tan^3x) \scriptstyle\implies f''(0)=0$
$f'''(x) = 2+8\tan^2x+6\tan^4x \scriptstyle\implies f'''(0)=2$
$f''''(x) = 16\tanx+40\tan^3x+24\tan^5x \scriptstyle\implies f''''(0)=0$
$f'''''(x) = 16\sec^2x+120\tan^2x.sec^2x+120\tan^4x\sec^2x \scriptstyle\implies f'''''(0)=16$
Thus we get Maclaurin series as –
$\tanx=x+\frac{1}{3}x^3+\frac{2}{15}x^5+.......$

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Last Updated : 16 Jun, 2020

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