Question

Let f^(1) (n) = g^(n) (0) for some functions f(x) and g(x). Now let the coordinate axes having graph f(x) be rotated by 45 degrees (clockwise). Then the corresponding Mclaurin series of transformed g(x) is?

(a) g(x)=g(0)+(e^x-1)+f(x)-f(0)

(b) τ(f(x+tan(45)))=τ45(g(x))

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

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

This question was addressed to me in unit test.

My question comes from Taylor Mclaurin Series topic in division Differential Calculus of Engineering Mathematics

Answer 1

Right option is (a) g(x)=g(0)+(e^x-1)+f(x)-f(0)

Explanation: The general expansion of Mclaurin is given by

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

We know that  f^(1) (n) = g^(n) (0)

After the graph f(x)  is rotated by 45  degrees we have an additional amount of tan45 gets added to slope of g(x)

f^(1) (0) + tan(45) = g^(n) (0)

After substituting this we have

 g(x) = g(0) + (f^(1) (1) + tan(45)) \(\times \frac{x}{1!}\) +  (f^(1) (2) + tan(45)) \(\times \frac{x^2}{2!}+…\infty\)

g(x) = g(0) + \((f^{(1)}(1).\frac{x}{1!} + f^{(1)}(2).\frac{x^2}{2!} + f^{(1)}(3).\frac{x^3}{3!}+…\infty)+(\frac{x}{1!}+\frac{x^2}{2!}+…\infty)\)

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

g(x)=g(0)+(e^x-1)+f(x)-f(0)