Expand the function f(x) = 1/x + 3 into a Taylor series around a = -2 and determine the interval of convergence."

Question

Hi, I dont really understand this questioon can you give me the answer step by step so I can understand how to solve this problem

Answer 1

Recall Taylor Series Formula

f(x) ≈ f(a) + f'(a)(x-a) + f''(a)(x-a)²/2! + f'''(a)(x-a)³/3! + ... + fⁿ(a)(x-a)ⁿ/n! = ∑[n=0,∞] fⁿ(a)(x-a)ⁿ/n!

f(a) = 1/(a+3) = 1/(-2+3) = 1

f'(a) = -1/(a+3)² = -1/(-2+3)² = -1

f''(a) = 2/(a+3)³ = 2/(-2+3)³ = 2

f'''(a) = -6/(a+3)⁴ = -6/(-2+3)⁴ = -6

...

f(x) ≈ 1 + -1(x-(-2)) + 2(x-(-2))²/2! + -6(x-(-2))³/3! + ...

f(x) ≈ 1 - (x+2) + (x+2)² - (x+2)³ + ...

f(x) = ∑[n=0,∞] (-1)ⁿ(x+2)ⁿ

Using the ratio test:

L = lim_n->∞ |a_n+1/a_n|

= lim_n->∞ |[(-1)ⁿ⁺¹(x+2)ⁿ⁺¹]/[(-1)ⁿ(x+2)ⁿ]|

= lim_n->∞ |(-1)(x+2)|

= |x+2|

Set L< 1:

|x+2| < 1

-1 < x+2 < 1

-3 < x < -1

Since x=-3 and x=-1 don't allow the series to converge, the interval of convergence would be (-3,-1) when centered at a=-2.

You can also use the fact that 1/(1+x) ≈ 1 - x + x² - x³ + ... and substitute "x" with "x+2" to get 1/(x+3) ≈ 1 - (x+2) + (x+2)² - (x+2)³ + ...

Either approach works, but if you're not confident with your known power series, you should follow the Taylor Series Formula.

Answer 2

I really hope that the x+3 was in a denominator :D

Let's begin. The first thing you want to do is always just write out the first three/four derivatives of your function. So: f'(x)= -1/(x+3)^2, f''(x)= 2/(x+3)^3, f''(x)= -6/(x+3)^4, and f''''(x)= 24/(x+3)^5. Now, use the standard taylor series equation: f(x) = f(a)+f′(a)!(x−a)+ f”(a)(x−a)^2/2!...+...f^n(a)(x−a)^n/n! and plug in the known values.

This leaves us with: f(x) = f(-2)+f′(-2)!(x+2)/2!+ f”(-2)2!(x+2)^2...+...f^n(a)(x−a)^n/n!

Now, plug in our actual numbers:

f(x) = 1-(x+2)+(x+2)^2-(x+2)^3+(x+2)^4...+...(x+2)^n(-1)^n

Now to find the interval of convergence we need to do the ratio test:

lim −›∞∑absval(a n+1/a n) (sorry for the formatting)

lim −›∞∑absval((x+2)^(n+1))/(x+2)^(n)

which simplifies to

lim −›∞∑absval(x+2)

Now, since the ratio test has to have the absolute value of the term be less than one to converge, we can make the inequality: 1<x+2<1 which turns into: -3<x<-1 which is the final answer.

Sorry for any formatting issues, if you have any questions please let me know.