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Maclaurin series

  • Last Updated : 16 Jun, 2020
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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 :

  1. 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!}
  2. f(x) = cos x
    \cosx= 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+…..
  3. f(x) = sin x
    \sinx = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+....
  4. 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!}+.....
  5. 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}+.....
  6. 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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