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Question: How do you write the Taylor series for \(f\left( x \right)=\cosh x\)?...

How do you write the Taylor series for f(x)=cosh⁡xf\left( x \right)=\cosh x?

Explanation

Solution

The Taylor series expansion of a function is given by ∑n=0∞fn(x0)n!(x−x0)n\sum\limits_{n=0}^{\infty }{\dfrac{{{f}^{n}}\left( {{x}_{0}} \right)}{n!}{{\left( x-{{x}_{0}} \right)}^{n}}}. For obtaining the Taylor series of the given function, we can choose x0=0{{x}_{0}}=0 and calculate f(x),f′(x),f′′(x),.......f\left( x \right),f'\left( x \right),f''\left( x \right),....... at x=0x=0 from the given function f(x)=cosh⁡xf\left( x \right)=\cosh x. On putting these into the expanded form of ∑n=0∞fn(x0)n!(x−x0)n\sum\limits_{n=0}^{\infty }{\dfrac{{{f}^{n}}\left( {{x}_{0}} \right)}{n!}{{\left( x-{{x}_{0}} \right)}^{n}}}, we will obtain the required Taylor series.

Complete step by step solution:
We know that the Taylor series expansion for a function f(x)f\left( x \right) is given by ∑n=0∞fn(x0)n!(x−x0)n\sum\limits_{n=0}^{\infty }{\dfrac{{{f}^{n}}\left( {{x}_{0}} \right)}{n!}{{\left( x-{{x}_{0}} \right)}^{n}}}. On expanding it we can write it as f(x0)0!(x−x0)0+f′(x0)1!(x−x0)1+f′′(x0)2!(x−x0)2+f′′′(x0)3!(x−x0)3+........\dfrac{f\left( {{x}_{0}} \right)}{0!}{{\left( x-{{x}_{0}} \right)}^{0}}+\dfrac{f'\left( {{x}_{0}} \right)}{1!}{{\left( x-{{x}_{0}} \right)}^{1}}+\dfrac{f''\left( {{x}_{0}} \right)}{2!}{{\left( x-{{x}_{0}} \right)}^{2}}+\dfrac{f'''\left( {{x}_{0}} \right)}{3!}{{\left( x-{{x}_{0}} \right)}^{3}}+......... According to the above question, the function is given as
⇒f(x)=cosh⁡x........(i)\Rightarrow f\left( x \right)=\cosh x........\left( i \right)
Differentiating the above equation with respect to x, we get
⇒f′(x)=sinh⁡x.......(ii)\Rightarrow f'\left( x \right)=\sinh x.......\left( ii \right)
Again on differentiating, we get
⇒f′′(x)=cosh⁡x.......(iii)\Rightarrow f''\left( x \right)=\cosh x.......\left( iii \right)
Let us choose x0=0{{x}_{0}}=0. Then from (i) we can write
⇒f(x0)=f(0) ⇒f(x0)=cosh⁡0 ⇒f(x0)=1 \begin{aligned} & \Rightarrow f\left( {{x}_{0}} \right)=f\left( 0 \right) \\\ & \Rightarrow f\left( {{x}_{0}} \right)=\cosh 0 \\\ & \Rightarrow f\left( {{x}_{0}} \right)=1 \\\ \end{aligned}
From (ii) we can write
⇒f′(x0)=f′(0) ⇒f′(x0)=sinh⁡0 ⇒f′(x0)=0 \begin{aligned} & \Rightarrow f'\left( {{x}_{0}} \right)=f'\left( 0 \right) \\\ & \Rightarrow f'\left( {{x}_{0}} \right)=\sinh 0 \\\ & \Rightarrow f'\left( {{x}_{0}} \right)=0 \\\ \end{aligned}
From (iii) we can write
⇒f′′(x0)=f′′′(0) ⇒f′′(x0)=cosh⁡0 ⇒f′′(x0)=1 \begin{aligned} & \Rightarrow f''\left( {{x}_{0}} \right)=f'''\left( 0 \right) \\\ & \Rightarrow f''\left( {{x}_{0}} \right)=\cosh 0 \\\ & \Rightarrow f''\left( {{x}_{0}} \right)=1 \\\ \end{aligned}
Similarly, we will get
⇒f′′′(x0)=0 ⇒f′′′′(x0)=1 \begin{aligned} & \Rightarrow f'''\left( {{x}_{0}} \right)=0 \\\ & \Rightarrow f''''\left( {{x}_{0}} \right)=1 \\\ \end{aligned}
Therefore, the Taylor series expansion for the given function about x0=0{{x}_{0}}=0 becomes

& \Rightarrow f\left( x \right)=\dfrac{f\left( 0 \right)}{0!}{{\left( x-0 \right)}^{0}}+\dfrac{f'\left( 0 \right)}{1!}{{\left( x-0 \right)}^{1}}+\dfrac{f''\left( 0 \right)}{2!}{{\left( x-0 \right)}^{2}}+\dfrac{f'''\left( 0 \right)}{3!}{{\left( x-0 \right)}^{3}}+\dfrac{f''''\left( 0 \right)}{4!}{{\left( x-0 \right)}^{4}}....... \\\ & \Rightarrow f\left( x \right)=\dfrac{1}{0!}+\dfrac{0}{1!}x+\dfrac{1}{2!}{{x}^{2}}++\dfrac{0}{3!}{{x}^{3}}+\dfrac{1}{4!}{{x}^{4}}....... \\\ & \Rightarrow f\left( x \right)=1+\dfrac{1}{2}{{x}^{2}}+\dfrac{1}{24}{{x}^{4}}....... \\\ \end{aligned}$$ Hence, the Taylor series expansion for the given function is obtained. **Note:** We must note that since the point of expansion was not mentioned in the question, we assumed it to be zero. But we can expand the given function about any point. It is necessary to write minimum three non zero terms of a Taylor series, as in the above solution. We can also use the definition of the cosine hyperbolic function given by $\cosh x=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}$, for finding out the value of the function and its derivatives at different points. Also, the sine hyperbolic function is defined as $\sinh x=\dfrac{{{e}^{x}}-{{e}^{-x}}}{2}$.