Question

How do you find the MacLaurin’s Formula for $ f\left( x \right) = \sinh x $ and use it to approximate $ f\left( {\dfrac{1}{2}} \right) $ within $ 0.01 $ ?

Answer 1

Hint: We have been given a hyperbolic function and we have to find the MacLaurin’s Formula or series for the given function. MacLaurin’s series is a Taylor series expansion of a function about the point $ 0 $ . The MacLaurin’s series is given as,
 $ f\left( x \right) = f\left( 0 \right) + \dfrac{{f'\left( 0 \right)}}{{1!}}x + \dfrac{{f''\left( 0 \right)}}{{2!}}{x^2} + \dfrac{{f'''\left( 0 \right)}}{{3!}}{x^3} + ...\;\left( {upto\;\infty \;terms} \right) $
We have to find the derivatives of the given function at the point $ 0 $ to find the series. And then we have to use the series to find the value of $ f\left( {\dfrac{1}{2}} \right) $ .

Complete step by step solution:
We have been given a function $ f\left( x \right) = \sinh x $ . We have to convert this function into MacLaurin’s formula. MacLaurin’s formula is the Taylor series expansion of the function about the point $ 0 $ , given as,
 $ f\left( x \right) = f\left( 0 \right) + \dfrac{{f'\left( 0 \right)}}{{1!}}x + \dfrac{{f''\left( 0 \right)}}{{2!}}{x^2} + \dfrac{{f'''\left( 0 \right)}}{{3!}}{x^3} + ...\;\left( {upto\;\infty \;terms} \right) $
First we find the value of the function at the point $ 0 $ .
 $ f\left( 0 \right) = \sinh 0 = 0 $
Now we have to find the derivatives of the function at the point $ 0 $ .
 $
  f'\left( x \right) = \dfrac{{d\left( {\sinh x} \right)}}{{dx}} = \cosh x \\
   \Rightarrow f'\left( 0 \right) = \cosh 0 = 1 \;
  $
Similarly,
 $
  f''\left( x \right) = \dfrac{{d\left( {\cosh x} \right)}}{{dx}} = \sinh x \\
   \Rightarrow f''\left( 0 \right) = \sinh 0 = 0 \;
  $
Further,
 $
  f'''\left( x \right) = \dfrac{{d\left( {\sinh x} \right)}}{{dx}} = \cosh x \\
   \Rightarrow f'''\left( 0 \right) = \cosh 0 = 1 \\
  f''''\left( x \right) = \dfrac{{d\left( {\cosh x} \right)}}{{dx}} = \sinh x \\
   \Rightarrow f''''\left( 0 \right) = \sinh 0 = 0 \;
  $
We can observe the pattern that the value of even derivatives at point $ 0 $ is $ 0 $ . And the value of odd derivatives at point $ 0 $ is $ 1 $ .
Thus, using the values of the function and derivatives at point the MacLaurin series becomes,
 $
  f\left( x \right) = f\left( 0 \right) + \dfrac{{f'\left( 0 \right)}}{{1!}}x + \dfrac{{f''\left( 0 \right)}}{{2!}}{x^2} + \dfrac{{f'''\left( 0 \right)}}{{3!}}{x^3} + \dfrac{{f''''\left( 0 \right)}}{{4!}}{x^4} + ... \\
   \Rightarrow f\left( x \right) = 0 + \dfrac{1}{{1!}}x + \dfrac{0}{{2!}}{x^2} + \dfrac{1}{{3!}}{x^3} + \dfrac{0}{{4!}}{x^4} + ... \\
   \Rightarrow f\left( x \right) = x + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^5}}}{{5!}} + ... \;
  $
Thus, we get the MacLaurin’s formula for the given function as,
 $ f\left( x \right) = x + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^5}}}{{5!}} + ... $
Now we will use this series to evaluate $ f\left( {\dfrac{1}{2}} \right) $ .
We can write,
 $ f\left( {\dfrac{1}{2}} \right) = \left( {\dfrac{1}{2}} \right) + \dfrac{1}{{3!}}.{\left( {\dfrac{1}{2}} \right)^3} + \dfrac{1}{{5!}}.{\left( {\dfrac{1}{2}} \right)^5} + ... $
We will simplify this further to get the value,
 $
  f\left( {\dfrac{1}{2}} \right) = \left( {\dfrac{1}{2}} \right) + \left( {\dfrac{1}{6} \times \dfrac{1}{8}} \right) + \left( {\dfrac{1}{{120}} \times \dfrac{1}{{32}}} \right) + ... \\
   \Rightarrow f\left( {\dfrac{1}{2}} \right) = \dfrac{1}{2} + \dfrac{1}{{48}} + \dfrac{1}{{3840}} + ... \;
  $
Since we have to find the value within $ 0.01 $ , we can neglect the terms after the second term.
 $
  f\left( {\dfrac{1}{2}} \right) \approx \dfrac{1}{2} + \dfrac{1}{{48}} \\
   \Rightarrow f\left( {\dfrac{1}{2}} \right) \approx \dfrac{{24 + 1}}{{48}} = \dfrac{{25}}{{48}} \approx 0.52 \;
  $
Thus, we get the value,
 $ f\left( {\dfrac{1}{2}} \right) \approx 0.52 $
So, the correct answer is “0.52”.

Note: MacLaurin’s series is a special case of Taylor series about point $ 0 $ . We use the value of the derivatives of the given function at a point to find the series. The function has to be infinitely differentiable at the given point as the MacLaurin’s series contains an infinite number of terms. To find the value of the function using MacLaurin’s series we use only 3-4 terms as required.