Question

Find the first 5 terms of this series which centers at x=0

Answer 1

Summation

Answer 2

I'm thinking of starting with sin(x) = the series of (-1)^n * x^(2n+1) / (2n+1)!

Answer 3

thats a good start yes

Answer 4

the 1/5 is a constant so we can set it aside

Answer 5

Okay and then sin(x) / 5x would be 1/2 * the summation of (-1)^n * x^(2n) / (2n+1)!

Answer 6

1/5*

Answer 7

yep

Answer 8

Okay, so then you just plug in n=1,2,3,4,5...?

Answer 9

n=0,1,2,3,4 is what id do, the summation of sin starts at n=0 in your presentation there

Answer 10

Wait but isn't an integral? So we have to find the integral of that summation?

Answer 11

thats what you posted .... no take backs lol

Answer 12

Ah no, I was just catching my mistake xD

Answer 13

Okay how would you find the integral of this:

Answer 14

\[\sum_{}^{} (-1)^n * x^{2n} / (2n+1)!\]

Answer 15

you have a polynomial, how do you integrate powers of x?

Answer 16

all the other stuff is just constants

Answer 17

Ah you just add +1 to the powers of x...

Answer 18

and divide, yeah

Answer 19

\[\sum_{}^{} (-1)^n * x^{2n+1} / (2n+1)! + c\]

Answer 20

Okay but wolfram alpha states that there is a (2n+1) on the bottom.... how?

Answer 21

for starter, you didnt divide your exponent

x^n integrates to x^(n+1)/(n+1)

Answer 22

and there is no +C in this case, unless you want to define it as the constant first term.

Answer 23

Oh I get it! thanks :)

Answer 24

\[\int\frac{(-1)^n}{(2n+1)!}x^{2n}dx\]
the only thing that is affected by x, is x itself

\[\frac{(-1)^n}{(2n+1)(2n+1)!}x^{2n+1}\]
\[\frac{(-1)^n}{(2n+1)(2n+1)(2n)!}x^{2n+1}\]
\[\frac{(-1)^n}{(2n+1)^2(2n)!}x^{2n+1}\]
but thats just playing with algebra

Answer 25

if we simply list the first 5 terms

\[\frac{1}{1!},\frac{-1}{3!}x^2,\frac{1}{5!}x^4,\frac{-1}{7!}x^6,\frac{-1}{9!}x^8\]
integrate them
\[\frac{1}{1!}x,\frac{-1}{3.3!}x^3,\frac{1}{5.5!}x^5,\frac{-1}{7.7!}x^7,\frac{-1}{9.9!}x^9\]

Answer 26

lol, got a stray negative in there

Answer 27

dont forget the 1/5 laying outside