Question

Find the value of e^^π⁄4√2

(a) 1.74

(b) 1.84

(c) 1.94

(d) 1.64

I have been asked this question in unit test.

This interesting question is from Taylor Mclaurin Series topic in portion Differential Calculus of Engineering Mathematics

Answer 1

The correct option is (a) 1.74

To explain: Let, f(x) = e^xSin(x), f(0) = 1

Now, the expansion of xSin(x) is \(x^2-\frac{x^3}{3!}+\frac{x^6}{5!}+….\)

Hence, \(e^xSin(x)=e^y=1+y+\frac{y^2}{2!}+\frac{y^3}{3!}+….\)

Hence,

\(e^{xSin(x)}=1+(x^2-\frac{x^4}{3!}+\frac{x^6}{6!}+…..)+\frac{(x^2-\frac{x^4}{3!}+\frac{x^6}{6!}+…..)^2}{2!}\)

\(+\frac{(x^2-\frac{x^4}{3!}+\frac{x^6}{6!}+…..)^3}{6}+….\)

\(e^{xSin(x)}=1+x^2-\frac{x^4}{3!}+\frac{x^6}{5!}+\frac{x^4}{2}-\frac{x^6}{6}+\frac{x^6}{6}+….\) (we neglect all other other terms by considering the options given)

Hence, \(e^{xSin(x)}=1+x^2+\frac{x^4}{3}+\frac{x^6}{120}+……\)

Putting, x = π/4,

We get,

f(π/4)=e^π/4 Sin(π/4)=e^π/(4√2)=1+(π/4)^2+1/3 (π/4)^4+….=1+.6168+.1268=1.74