Question

Find the value of S=\(\sum_{n=1}^\infty \frac{(-1)^{n+1} \times (2n-1)^3}{(2n-1)!}\) using n^th derivatives.

(a) – 2 * sin(1)

(b) 3 * sin(1)

(c) 3 * cos(1)

(d) – 3 * cos(1)

I got this question in semester exam.

This intriguing question comes from The nth Derivative of Some Elementary Functions topic in division Differential Calculus of Engineering Mathematics

Answer 1

Correct choice is (a) – 2 * sin(1)

To explain I would say: We have to consider the function f(x) = sin(e^x) in order to get the series in some way.

Expanding the given function into a Taylor series we have

f(x)=\(\frac{e^x}{1!}-\frac{e^{3x}}{3!}+\frac{e^{5x}}{5!}…\infty \)

Now observe that our series in question doesn’t have the exponential function, this gives us the hint that some derivative of this function has to be taken at x = 0

Observe that the term (2n – 1)^3 has exponent equal to 3

Hence we have to take the third derivative of the function to get the required series

Now taking the third derivative yields

f(x)=\(\frac{e^x}{1!}-\frac{3^3e^{3x}}{3!}+\frac{5^5e^{5x}}{5!}…\infty \)

Now substituting x=0 we get

\(\frac{1^3}{1!}-\frac{3^3}{3!}+\frac{5^3}{5!}…\infty=\sum_{n=1}^\infty \frac{(-1)^{n+1} \times (2n-1)^3}{(2n-1)!} \)

To find the value of this series we need to take the third derivative of original function at the required point, this is as follows

f^(3)(x) = -2e^xsin(e^x)

Substituting x = 0 we get

f^(3)(0) = -2sin(1).