Question

Picture will give explanation. I got the answer wrong and im not sure how..

Answer 1

gimme a sec i gotta post the pic

Answer 2

what grade?

Answer 3

calculus 2

Answer 4

taylor series expansion

Answer 5

cant help sorry

Answer 6

@jtvatsim

Answer 7

Looks good, your line when you write "substitute x^2 for x" contains an error - you shouldn't have x^5 and x^6 where they are right now. But your final answer looks correct.

Answer 8

Ok i think i started to understand some parts of it but im still sort of lost

Answer 9

which parts have you confused?

Answer 10

no i checked the book and the answer was wrong. I think i started my mistake by doing the integration of (1-x)^-1

Answer 11

because by chain rule it would have made the lnx negative and i didnt include that

Answer 12

id say the part i really dont get is the maclauren series of ln(1+x)

Answer 13

OK, as far as why it equals the expansion?

Answer 14

as far as why the expansion is how it is i think is what i mean to say. because shouldnt it just be the integral of (1+x)^-1?

Answer 15

and thats like x^n

Answer 16

which would make the integral of x^n equal to (1/n+1)(x^n+1)

Answer 17

but they just have n on the bottom not n+1

Answer 18

this is the answer

Answer 19

OK, I see what you mean. let me work something out real quick.

Answer 20

Still there?

Answer 21

yeah, sorry, had to step away from the computer for a minute

Answer 22

I'm starting to work out the derivation, but now I'm getting confused to! ;)

Answer 23

OK, so here is how we derive the Maclaurin series for ln|1 + x|

Answer 24

Yeah its frikken tough and I have my final next wednesday

Answer 25

f(x) = ln|1+x|
f'(x) = 1/(1+x) = 1/(1-[-x])

But we know that 1/(1-x) = 1 + x + x^2 + ...
So, 1/(1-[-x]) = 1 + (-x) + (-x)^2 + (-x)^3 + ... = 1 - x + x^2 - x^3 ...

Answer 26

yeah thats what i was thinking

Answer 27

Good, so then
f'(x) = 1 - x + x^2 - x^3 ...
implies that
f(x) = x - x^2/2 + x^3/3 - x^4/4 ...

OR

ln|1 + x| = x - x^2/2 + x^3/3 - x^4/4 ...

Answer 28

yeah i think im following you

Answer 29

In other words
\[\ln|1+x| = \sum_{n=1}^\infty \frac{(-1)^{n+1} x^n}{n}\]
although you could write this using n+1 if you prefer... like this

Answer 30

oh so it starts at 1?

Answer 31

wait yeah do the n+1 one

Answer 32

\[\ln|1+x| = \sum_{n=0}^\infty \frac{(-1)^n x^{n+1}}{n+1}\]

Answer 33

flutter YES

Answer 34

lol, these are equivalent series.

Answer 35

Ok yeah thats what i had from the beginning and then plugged in the x^2 to get (1/n+1)x^(2n+2)

Answer 36

ok then i guess thats where i was getting screwed because i didnt know they were equivalent and because the book had it the other way i just figured i was wrong

Answer 37

yep, you knew what you were doing, but you didn't know that you knew it... something like that. :)

Answer 38

i like cant thank you enough honestly

Answer 39

i was panicking like crazy thinking that everything i knew was wrong and that i was screwed

Answer 40

no worries, glad to save your view of the world... just don't divide by 0. :)

Answer 41

haha yeah wont be doing that. Thanks again

Answer 42

No probs Good luck!