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This is one of our many quizzes on Taylor & Maclaurin series. Click on this to access more such competitive quizzes on Differential Calculus.

Taylor & Maclaurin series Quiz 2

mathematics

 Which of the following statements are correct based on the taylor series?

  It is the basic definition of Taylor’s theorem.

 Taylor series of a function is an infinite sum of terms expressed in terms of function derivatives at a single point.

 Expand the given function $f(x)=5{{x}^{2}}+13x+2$ at 

 This problem can be solved using the Taylor series. Based on Taylor’s theorem,  $\\ \text{f(x)=f(a)+{f}'(a)(x-a)+}\dfrac{{f}''(a)}{2!}{{(x-a)}^{2}}+.....+\dfrac{{{f}^{(n)}}(a)}{n!}{{(x-a)}^{n}}+...$ at $x=a$.

 $ \text{We know that for expanding a function in Taylor series we find derivatives of  f(x) upto the derivative of f(x) becomes zero.}\\ \text{Given }f(x)=5{{x}^{2}}+13x+2 \\ \text{Therefore }\\ {{f}^{1}}\left( x \right)=10x+3 \\ {{f}^{2}}(x)=10 \\ {{f}^{3}}(x)=0 \\ \text{Required series is }5{{x}^{2}}+10x+2\\ =\sum\limits_{n=0}^{\infty }{\dfrac{{{f}^{(n)}}(2)}{n !}}{{(x-2)}^{n}} \\ =f(2)+{{f}^{1}}(2)(x-2)+\dfrac{{{f}^{2}}(2){{(x-2)}^{2}}}{2!}+0 \\ =48+33(x-2)+5{{(x-2)}^{2}} $

 Calculate $\sqrt{1.001}$ in 6 decimal places

 A function f(x) can be written in a Maclaurin series, as shown below$f(x)=f(0)+{f}'(0)x+\dfrac{{f}''(0){{x}^{2}}}{2!}+........+\dfrac{{{f}^{(n)}}(0){{x}^{n}}}{n!}+.....$ This problem is solved using the Maclaurin series.

 $ \text{Consider the expansion of function using }{{\left( 1+x \right)}^{1/2}} \text{ and substitute }\\ x=0.001{{\left( 1+x \right)}^{1/2}}=1+\dfrac{x}{2}-\dfrac{{{x}^{2}}}{8}+\dfrac{{{x}^{3}}}{16}+......(1)\\ \text{Place }x=0.001\text{in equation (1), then we get}\\ \sqrt{1.001}=1+\dfrac{1}{2}\times 0.001-\dfrac{1}{8}\times {{(0.001)}^{2}}+\dfrac{1}{16}\times {{(0.001)}^{3}}-... \\  =1+0.005-0.0000125+0.0000000625-.... \\  =1.005012 $

 Use the maclaurin series for cosx to evaluate $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{x+\cos x}{x(x+2)}$

 Place the value of cos x based on the Maclaurin series in the given question. A function f(x) can be written in a Maclaurin series, as shown below$f(x)=f(0)+{f}'(0)x+\dfrac{{f}''(0){{x}^{2}}}{2!}+........+\dfrac{{{f}^{(n)}}(0){{x}^{n}}}{n!}+.....$.Then find the limit by placing zero. 

 $ \text{Expansion of cosx in maclaurin's theorem is }\\ \cos x=1-\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{4}}}{4!}-\dfrac{{{x}^{6}}}{6!}+...\\ \underset{x\to 0}{\mathop{\lim }}\,\dfrac{x+\cos x}{x(x+2)}$$=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{x+1-\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{4}}}{4!}-\dfrac{{{x}^{6}}}{6!}+...}{x(x+2)}=\dfrac{1}{0}\\$<br>So, the value of limit does not exist.

 Find the Maclaurin series of $f(x)=\sqrt{1+x+89{{x}^{2}}}$

 A function f(x) can be written in a Maclaurin series, as shown below$\\ f(x)=f(0)+{f}'(0)x+\dfrac{{f}''(0){{x}^{2}}}{2!}+........+\dfrac{{{f}^{(n)}}(0){{x}^{n}}}{n!}+.....$

 $ \text{We know that to write the function f(x) in a Maclaurin series, the below equation is followed.}\\ f(x)=f(0)+{f}'(0)x+\dfrac{{f}''(0){{x}^{2}}}{2!}+........+\dfrac{{{f}^{(n)}}(0){{x}^{n}}}{n!}+.....\\ \text{Given}\\ f(x)=\sqrt{1+x+89{{x}^{2}}}....(1) \\ \text{Differentiate on both sides, we get}\\ {f}'(x)=\dfrac{1+188x}{2\sqrt{1+x+89{{x}^{2}}}}....(2) \\ \text{Differentiate on both sides, we get}\\ {f}''(x)=\dfrac{2\sqrt{1+x+89{{x}^{2}}}(188x)-(1+188x)\left( \dfrac{1+188x}{2\sqrt{1+x+89{{x}^{2}}}} \right)}{4\left( 1+x+89{{x}^{2}} \right)}....(3) \\ \text{Place x=0 in equation (1), equation (2) and equation (3), then we get }\\ f(0)=\sqrt{1+0+89(0)}=1 \\ {f}'(0)=\dfrac{1+188(0)}{2\sqrt{1+(0)+89{{(0)}^{2}}}}=\dfrac{1}{2} \\ {f}''(0)=\dfrac{2\sqrt{1+0+89{{(0)}^{2}}}(188(0))-(1+188(0))\left( \dfrac{1+188(0)}{2\sqrt{1+0+89{{(0)}^{2}}}} \right)}{4\left( 1+0+89{{(0)}^{2}} \right)}=\dfrac{-1}{8} \\  \text{So,}\\ f(x)=1+\dfrac{x}{2}-\left( \dfrac{1}{8} \right)\dfrac{{{x}^{2}}}{2!}+............$