Taylor and Maclaurin Series

Question

Find the Taylor series for f(x)= e^8x expanded at about x= 4 and prove that the series converges to e^8x for all x

Answer 1

Given e^x = 1 + x + x^2/2! + x^3/3!...etc. all you need to do is to show substitute in 8x everywhere you see x and simplify.

For proving that the series converges for 8x, we need to show that it converges for x as any real number because multiplying x (any real number) by 8 still gives you a real number. So we take the easier route using the ratio test. However, when dealing with a power series, we are not asking IF a series is convergent (aka <1) but rather WHERE the series converge - is is the center only (all power series converge at the center, an interval around the center, or all real numbers. There are only three possibilities by theorem.

So take lim x goes to infinity abs(a(n+1)/a(n) <1 and describe the x's that make that statement true. To find a(n+1) you need to write the formula for e^x in terms of the position number. Aka, can you describe the position you see based on the position number (n) where it is located. Here a(n) = x^n/n! So a(n+1) = (x+1)^n+1 / (n+1)!. When you do the reduction, you'll find that x can be any real number which implies that 8x also works. Hope this helps. It's a bit complex for a single question in this forum.