Question

f^(1) (n) = g^(n) (0) holds good for some functions f(x) and g(x). Now let the coordinate axes containing graph g(x) be rotated by 30 degrees clockwise, then the corresponding Taylor series for the transformed g(x) is?

(a) g(0)+\(\frac{e^x – 1}{\sqrt{3}}+\frac{\sum_{n=1}^\infty f^{(1)}(n)x^n}{n!}\)

(b) \(g(0) + \frac{g^{(1)}.x}{1!} + \frac{g^{(2)}(1).x^2}{2!}+…\infty\)

(c) No unique answer exist

(d) Such function is not continuous

This question was posed to me during an interview.

My doubt is from Taylor Mclaurin Series topic in division Differential Calculus of Engineering Mathematics

Answer 1

Correct answer is (a) g(0)+\(\frac{e^x – 1}{\sqrt{3}}+\frac{\sum_{n=1}^\infty f^{(1)}(n)x^n}{n!}\)

Easiest explanation: We have f^(1) (n) = g^(n) (0)

As the coordinate axes containing f(x) is rotated the tan(30) term gets added to the derivative of f^(1)new(n)=g^(n)(0)-tan(30)

We have

g^(n)(0)=f^(1)(n)+tan(30)

The Taylor expansion centered at 0 for g(x) is given by

g(x)=\(g(0)+\frac{g^{(1)}(0).x}{1!}+\frac{g^{(2)}(0).x^2}{2!}+…..\infty \)

Now substituting g^(n)(0)=f^(1)(n)+tan(30) we have

g(x)=\(g(0)+\frac{(f^{(1)}(1)+tan(30)).x}{1!}+\frac{(f^{(1)}(2)+tan(30)).x^2}{2!}+….\infty\)

g(x)=\(g(0)+(\frac{f^{(1)}(1).x}{1!}+\frac{f^{(1)}(2).x}{2!}+…\infty)+tan(30)\times(\frac{x}{1!}+\frac{x^2}{2!}+….\infty)\)

=\(g(0)+\frac{\sum_{n=1}^{\infty}f^{(1)}(n).x^n}{n!}+\frac{1}{\sqrt{3}}\times(e^x-1)\)