Please Help solve this taylor series problem

Question

im stuck at this question and dont know how to solve it. please help

Answer 1

Let e^x-a=∑a_n(x-a)ⁿ=a₀+a₁(x-a)+a₂(x-a)²+a₃(x-a)³+... for integer n≥0.

When we differentiate this as a series we get:

a₁+a₂(x-a)+a₃(x-a)²+...=∑na_n(x-a)^n-1 (n≥1) and when we differentiate e^x-a we get e^x-a. In fact, every derivative of e^x-a is going to be the same, but the derivative of the series augments the coefficient of x-a like this:

n(n-1)(n-2)...a_n etc., while the power of x-a decreases. The first term in the differentiated series will always be a constant (that is, won't contain x), so by setting x to a particular value on both sides of the equation so that all other terms become zero, it's easy (in this case) to find the coefficients a_n.

To find a_n we need an equation we can solve. 

Take the second derivative as an example: n(n-1)a_n(x-a)^n-2. (When n=2, this reduces to n(n-1)a₂=2!a₂.)

Go back to e^x-a=∑a_n(x-a)ⁿ=a₀+a₁(x-a)+a₂(x-a)²+a₃(x-a)³+... When x=a, e^x-a becomes e⁰=1 and the series becomes a₀ (because all the other terms become zero). Therefore a₀=1.

Similarly, the first derivative becomes 1 when x=a, so we have a₁=1. The second derivative is 2a₂=1. The third derivative is 6a₃=3!a₃=1. And so on. From this we get a_n=1/n!.

So e^x-a=∑(x-a)ⁿ/n!. (Note that 0!=1.) But e^x-a can be written e^x/e^a, therefore e^x/e^a=∑(x-a)ⁿ/n! and e^x=e^a∑(x-a)ⁿ/n!=e^a(1+(x-a)+(x-a)²/2!+(x-a)³/3!+..., so proving the given statement.

I've probably given you more than you asked for, because the question assumes you know the Taylor series for e^x=1+x+x²/2!+x³/3!+... and by replacing x with x-a on both sides you come up with same proof. However, I wanted to show you how straightforward it is to derive the series for e^x. It's not as frightening as you might have thought and it may help you to understand better the basic concepts behind deriving Taylor or Maclaurin series coefficients.