Skip to content
Exercise 7.4 · Q1

Q.Write the Maclaurin series expansion of the following functions:

(i) exe^x
(ii) sin⁡x\sin x
(iii) cos⁡x\cos x
(iv) log⁡(1−x)\log(1-x); −1≤x<1-1\le x<1
(v) tan⁡−1(x)\tan^{-1}(x); −1≤x≤1-1\le x\le1
(vi) cos⁡2x\cos^2x
Puducherry TnboardTextbookSubjectiveImportance★★★★★
21% · 31/148 Questions
✓ Free question

Each series is obtained by differentiating repeatedly, evaluating at x=0x=0, and substituting into f(x)=∑f(n)(0)n!xnf(x)=\sum\frac{f^{(n)}(0)}{n!}x^n; part (vi) is easiest via the double-angle identity rather than raw differentiation.

Step 1 (i). f(x)=exf(x)=e^x. Every derivative of exe^x is exe^x itself, so f(n)(0)=1f^{(n)}(0)=1 for all nn.

ex=1+x+x22!+x33!+⋯=∑n=0∞xnn!.e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots=\sum_{n=0}^{\infty}\frac{x^n}{n!}.

Step 2 (ii). f(x)=sin⁡xf(x)=\sin x. f(0)=0, f′(0)=cos⁡0=1, f′′(0)=−sin⁡0=0, f′′′(0)=−cos⁡0=−1, f(4)(0)=0,…f(0)=0,\,f'(0)=\cos0=1,\,f''(0)=-\sin0=0,\,f'''(0)=-\cos0=-1,\,f^{(4)}(0)=0,\ldots (period-4 cycle 0,1,0,−10,1,0,-1).

sin⁡x=x−x33!+x55!−⋯=∑n=0∞(−1)nx2n+1(2n+1)!.\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}.

Step 3 (iii). f(x)=cos⁡xf(x)=\cos x. f(0)=1, f′(0)=−sin⁡0=0, f′′(0)=−cos⁡0=−1, f′′′(0)=0,…f(0)=1,\,f'(0)=-\sin0=0,\,f''(0)=-\cos0=-1,\,f'''(0)=0,\ldots (cycle 1,0,−1,01,0,-1,0).

cos⁡x=1−x22!+x44!−⋯=∑n=0∞(−1)nx2n(2n)!.\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}.

Step 4 (iv). f(x)=log⁡(1−x)f(x)=\log(1-x). f′(x)=−11−x⇒f′(0)=−1f'(x)=\dfrac{-1}{1-x}\Rightarrow f'(0)=-1; f′′(x)=−1(1−x)2⇒f′′(0)=−1f''(x)=\dfrac{-1}{(1-x)^2}\Rightarrow f''(0)=-1; in general f(n)(0)=−(n−1)!f^{(n)}(0)=-(n-1)! for n≥1n\ge1, giving f(n)(0)n!=−1n\dfrac{f^{(n)}(0)}{n!}=-\dfrac1n.

log⁡(1−x)=−x−x22−x33−⋯ ,−1≤x<1.\log(1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\cdots,\qquad -1\le x<1.

Step 5 (v). f(x)=tan⁡−1xf(x)=\tan^{-1}x. f′(x)=11+x2f'(x)=\dfrac{1}{1+x^2}; expanding 11+x2=1−x2+x4−⋯\dfrac1{1+x^2}=1-x^2+x^4-\cdots (a geometric series in −x2-x^2) and integrating term by term (equivalently, tabulating f(n)(0)f^{(n)}(0)) gives

tan⁡−1x=x−x33+x55−⋯ ,−1≤x≤1.\tan^{-1}x=x-\frac{x^3}{3}+\frac{x^5}{5}-\cdots,\qquad -1\le x\le1.

Step 6 (vi). f(x)=cos⁡2xf(x)=\cos^2x. Use cos⁡2x=1+cos⁡2x2\cos^2x=\dfrac{1+\cos2x}{2} and substitute the Step 3 series with x→2xx\to2x: cos⁡2x=1−(2x)22!+(2x)44!−⋯=1−2x2+2x43−⋯\cos2x=1-\dfrac{(2x)^2}{2!}+\dfrac{(2x)^4}{4!}-\cdots=1-2x^2+\dfrac{2x^4}{3}-\cdots.

cos⁡2x=1+(1−2x2+2x43−⋯ )2=1−x2+x43−⋯ .\cos^2x=\frac{1+\left(1-2x^2+\tfrac{2x^4}{3}-\cdots\right)}{2}=1-x^2+\frac{x^4}{3}-\cdots.

✓Final answer

(i) ex=∑n=0∞xnn!e^x=\displaystyle\sum_{n=0}^\infty\frac{x^n}{n!}. (ii) sin⁡x=x−x33!+x55!−⋯\sin x=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\cdots. (iii) cos⁡x=1−x22!+x44!−⋯\cos x=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\cdots. (iv) log⁡(1−x)=−x−x22−x33−⋯ , −1≤x<1\log(1-x)=-x-\dfrac{x^2}2-\dfrac{x^3}3-\cdots,\ {-}1\le x<1. (v) tan⁡−1x=x−x33+x55−⋯ , −1≤x≤1\tan^{-1}x=x-\dfrac{x^3}3+\dfrac{x^5}5-\cdots,\ {-}1\le x\le1. (vi) cos⁡2x=1−x2+x43−⋯\cos^2x=1-x^2+\dfrac{x^4}{3}-\cdots.

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.