Question

Find the Taylor Series expansion of sinh (x) centered around 5?

(a) \(\frac{x}{1!}+\frac{x^3}{3!}+\frac{x^5}{5!}+…..\infty\)

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

(c) \(1+\frac{x^2}{2!}+\frac{x^4}{4!}+…\infty\)

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

The question was posed to me by my college professor while I was bunking the class.

I'd like to ask this question from Taylor Mclaurin Series in portion Differential Calculus of Engineering Mathematics

Answer 1

Right option is (a) \(\frac{x}{1!}+\frac{x^3}{3!}+\frac{x^5}{5!}+…..\infty\)

The best I can explain: We know the Taylor series does not change the polynomial. Hence, be the polynomial centered at 5 or anywhere else would yield the same polynomial. In this case the polynomial centered at 0 has to be equal to the polynomial centered at 5.