Question

TAYLOR EXPANSION: I have to find the first three terms of the Taylor expansion about x = 0 for f(x) = sin(x)/x. To get the first three terms I know I need to calculate up to the 4th derivative, which I did and verified by Mathematica. Here's what I know so far:

Answer 1

??

Answer 2

\[f(x) = \frac{ \sin(x) }{ x }\]
\[f'(x) = \frac{\cos\left(x\right)}{x}-\frac{\sin\left(x\right)}{x^2}\]
\[f''(x)=\frac{2\sin\left(x\right)}{x^3}-\frac{\sin\left(x\right)}{x}-\frac{2\cos\left(x\right)}{x^2}\]
\[f'''(x)=\frac{3\sin\left(x\right)}{x^2}+\frac{6\cos\left(x\right)}{x^3}-\frac{6\sin\left(x\right)}{x^4}-\frac{\cos\left(x\right)}{x}\]
\[f''''(x)=-\frac{12\sin\left(x\right)}{x^3}+\frac{4\cos\left(x\right)}{x^2}-\frac{24\cos\left(x\right)}{x^4}+\frac{24\sin\left(x\right)}{x^5}+\frac{\sin\left(x\right)}{x}\]

Answer 3

I was told to evaluate the limits, but for some of them I get a limit of infinity, which I can't really use...

Answer 4

\[f(x_0+\epsilon)=f(x_0)+\epsilon f'(x_0)+\frac{ 1 }{ 2! }\epsilon^2 f''(x_0)+...\]

This is what I'm using for the Taylor Expansion, where \[x_0=0\]

Answer 5

DX

Answer 6

Which one

Answer 7

They seem very correct to me...

Answer 8

But now what? I was told to do something with limits. The limit of f(x) is 1, which is the correct first term of the Taylor Expansion. However, when I do the limit for f'(x), I get infinity.... what do I do with that??

Answer 9

I don't think I'm doing my limits right

Answer 10

Calc 3

Answer 11

I should probably know this, but how did you come up with the first part

Answer 12

Nvm, makes sense

Answer 13

Go on

Answer 14

So regarding those first four terms of 1, 0, 0, 0, what do I do with those? Those aren't the first four terms of the taylor expansion

Answer 15

So you're saying the taylor expansion is simply 1?

Answer 16

I think so. However, give me few minutes, I review my notes.

Answer 17

Well, a quick search on Wolfram Alpha gives me an answer of \[1-\:\frac{x^2}{6}+\frac{x^4}{120}\] for the first three non-zero terms. Doesn't this mean the third term you found can't be zero?

Answer 18

oh, yes, it is, since \(lim_x\rightarrow 0 \dfrac{sinx}{x}=1\)

Answer 19

it is, if x =0

Answer 20

the all other terms but the first one =0 when x =0

Answer 21

What?

Answer 22

What what? you have it, right? it just confirms our work.

Answer 23

see it, at the very first line of the theorem, it says lim (sinx/x) as x-->0 =1
http://planetmath.org/sites/default/files/texpdf/39255.pdf

Answer 24

I know the lim of sin(x)/x is 1, but according to the taylor expansion calculated from Wolfram Alpha, the Taylor expansion is more than just 1 for the first couple orders. I understand what you did, but it doesn't come out to the "correct" answer. However, I don't understand how to get to the "correct" answer

Answer 25

you get \(sinx/x= 1-x^2/2!+x^4/5!-x^7/7!\), right?

so, just plug x=0 in
you have the first term is 1, second term is 0 and..... so on.

the first three terms of it are 1, 0, 0
Dat sit

Answer 26

Then I'm obviously missing something. How does that translate to the answer of \[1-\frac{ x^2 }{ 6 }+\frac{ x^4 }{ 120 }+...\]

Answer 27

hey!! 3! =6
and 5! =120

Answer 28

Thank you I gotta go to class now

Answer 29

ok