Question

.

Answer 1

@EulerGroupie

Answer 2

@TuringTest

Answer 3

\[\sum_{n=0}^{3}\frac{(-1)^{n}n!}{2^{n}(n+2)}=\frac{(1)(1)}{1(2)}(x-5)^{0}+\frac{(-1)(1)}{2(3)}(x-5)+\frac{(1)(2)}{4(4)}(x-5)^{2}+\frac{(-1)(6)}{8(5)}(x-5)^{3}\]
\[\frac{1}{2}-\frac{1}{6}(x-5)+\frac{1}{8}(x-5)^{2}-\frac{3}{20}(x-5)^{3}\]

Answer 4

Your answer got cut off for me

Answer 5

But I think I got the pattern. Let me work it out

Answer 6

That summation isn't quite right either... it needs (x-5)^n in it.

Answer 7

The one the question gave us?

Answer 8

It gave f^n, but the Taylor expansion multiplies (x-c)^n

Answer 9

Oh true. How about part b?

Answer 10

It looks like ratio test... I'm working it.

Answer 11

Okay. Ill try the same thing

Answer 12

When you do the ratio test do you put (x+5) on the top and bottom?

Answer 13

(x-5)^(n+1) on top... (x-5)^(n) on the bottom reduces to (x-5)

Answer 14

oh yeah thanks

Answer 15

is it 0?

Answer 16

After a few steps:\[\lim_{x \rightarrow \infty}\left| \frac{n ^{2}+3n+2}{2n+6}(x-5) \right|<1\]then\[\frac{1}{2}\left| x-5 \right|<1\]\[\left| x-5 \right|<2\]
Radius of Convergence is 2 about a center of 5.

Answer 17

wait... oops

Answer 18

How are you getting 1/2

Answer 19

it will go to infiniti because the top has a higher power

Answer 20

That's the oops....

Answer 21

Okay so does that mean there is no radius of convergence?

Answer 22

I think so... its getting late and I'm fuzzy. I think that's the case ... it doesn't converge except at x=5, which is useless. ROC=0

Answer 23

@UnkleRhaukus do you agree with us?

Answer 24

How about part c?

Answer 25

Its an alternating series, so use alternating series error approximation.

Answer 26

Alright let me try that out

Answer 27

I'm fried... its 1am here. Good luck and good night.

Answer 28

Its 3 here.. haha

Answer 29

3am*

Answer 30

lol... die-hard!