Question

@jtvatsim This should be a quick one

Answer 1

Basically im just unsure if its that simple... im doubting everything ive ever learned ever

Answer 2

The second to last line should have 1/x, but I think that looks good... let me look closer to confirm.

Answer 3

Wait which part? i dont see where a i/x would go because i already have an x*2n! in there

Answer 4

We know that cos(x) = 1 - x^2/2! + x^4/4! ...
Therefore, 1 - cos(x) = x^2/2! - x^4/4! + x^6/6! ....
Thus, [1- cos(x) ]/x = x/2! - x^3/4! + x^5/6! ...

Check and see if that's right.

Answer 5

i think maybe the x would go under the factorial part?

Answer 6

instead of how ive written it

Answer 7

The thing I'm wondering is why do you have so many 1- x^2, 1 - x^4 and so on?

Answer 8

Im not sure what you mean

Answer 9

OK, let's start over from the top... I think I might have missed something.

Answer 10

So we want to find the expansion of
\[f(x) = \frac{1-\cos(x)}{x}\]
correct?

Answer 11

yes

Answer 12

so what you were saying was good where we basically do 1- the cosx expansion

Answer 13

OK, so, we start with a known series
\[\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} \cdots\]

Answer 14

and then subtract 1 from it and then we would divide that thing by x for each term correct?

Answer 15

Well, we are subtracting IT from one... so that
\[1 - \cos(x) = 1 - (1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} \cdots)\]
and yes, then divide the result by x

Answer 16

hmmm ok

Answer 17

Giving us \[1 - cos(x) = \frac{x^2}{2!} - \frac{x^4}{4!} + \frac{x^6}{6!} \cdots\]

Answer 18

notice how the signs have changed since we took - cos(x)

Answer 19

and also the leading one vanished.

Answer 20

yeah gimme a sec on that part

Answer 21

kk

Answer 22

ok i so the signs flip because the one cancels out the first term and then we essentially multiply it all by -cosx

Answer 23

well not exactly -cosx but we multiple cosx expansion by -1

Answer 24

exactly, we are basically multiplying by -1

Answer 25

ok yeah i got that

Answer 26

and then just take the flipped signs series and divide by x?

Answer 27

Now, division by x is easy because there are only x powers in our expansion.

Answer 28

alright i got it now. Thanks again.

Answer 29

No worries! anytime

Answer 30

so how would the sum part of it look?

Answer 31

Yeah, that's fun. :)

Answer 32

\[\frac{x}{2!} - \frac{x^3}{4!} + \frac{x^5}{6!} \cdots = \sum_{n=1}^\infty \frac{(-1)^{n-1} x^{2n-1}}{(2n)!}\] I think

Answer 33

just replace the n with n-1 right?

Answer 34

ok yeah what you have

Answer 35

The annoying 2n-1 is required because of the odd exponents.

Answer 36

could the (-1)^(n-1) be the same as (-1)^(n+1)?

Answer 37

Yes. It's personal preference at that point.

Answer 38

ok just making sure. Thank you!

Answer 39

It could even by (-1)^(n+99)

Answer 40

No problem!