# Tailor series question

1. Jan 5, 2008

2. Jan 5, 2008

### Rainbow Child

I don't any other way by forming the derivatives!

But for your function I would tried to write $\sin x$ by expotentials

$$\sin x=\frac{e^{i\,x}-e^{-i\,x}}{2\,i}\Rightarrow f(x)=\frac{e^{(i+1)\,x}-e^{(-i+1)\,x}}{2\,i}-x^2-x$$

which would yelds exp with derivation, but the calculations of $$f^{(n)}(0)$$ will involve complex numbers

By the way the first non-zero term is

$$\frac{f^{(3)}(0)}{3!}\,x^3=\frac{1}{3}\,x^3$$

For the (b) part, just plug the series in the place where $f(x)$ was.

3. Jan 5, 2008

### morphism

You can also plug in the expansions for e^x and sin(x). It shouldn't be too hard to get the first nonzero term then.

4. Jan 6, 2008

### transgalactic

how did you transformed from my expresion to what you showed

5. Jan 6, 2008

### Gib Z

If your aim is to find that limit, the easiest way is NOT finding the Taylor series of that function, but express $$\lim_{x\to 0} \frac{f(x)}{x^2}$$ as $$\lim_{x\to 0} \left( \frac{e^x-1}{x} \frac{\sin x}{x} - 1 \right)$$

That limit is easy to evaluate; The only bit I would imagine you would have any trouble with is $$\lim_{x\to 0} \frac{e^x-1}{x}$$, which you can use exp(x)'s Taylor series/L'Hopitals Rule (their all the same) to work out.

Last edited: Jan 6, 2008
6. Jan 6, 2008

### transgalactic

i didnt understand how am i soppose to comeup with the first object that is not
equals to 0
without making a million derivatives

7. Jan 6, 2008

### Rainbow Child

Why make million derivatives?

$$\sin x=\frac{e^{i\,x}-e^{-i\,x}}{2\,i}\Rightarrow f(x)=\frac{e^{(i+1)\,x}-e^{(-i+1)\,x}}{2\,i}-x^2-x$$

$$f'(x)=\frac{1}{2\,i}\left((i+1)\,e^{(i+1)\,x}-(-i+1)\,e^{(-i+1)\,x}\right)-2\,x-1\Rightarrow f'(0)=0$$

$$f''(x)=\frac{1}{2\,i}\left((i+1)^2\,e^{(i+1)\,x}-(-i+1)^2\,e^{(-i+1)\,x}\right)-2\Rightarrow f''(0)=0$$

$$f'''(x)=\frac{1}{2\,i}\left((i+1)^3\,e^{(i+1)\,x}-(-i+1)^3\,e^{(-i+1)\,x}\right)\Rightarrow f'''(0)=2$$

8. Jan 6, 2008

### transgalactic

those derivatives are with complex numbers

i didnt study it yet

so from your answer i get that the only way is to make derivatives

9. Jan 6, 2008

### morphism

Read my post. You don't need to take derivatives if you know the expansions of sin(x) and e^x.

10. Jan 7, 2008

### transgalactic

what is the
expansions for e^x and sin(x) ??

you ment their tailor series

11. Jan 7, 2008

### Rainbow Child

$$e^x=1+x+\frac{x^2}{2!}+\dots+\frac{x^n}{n!}+\dots$$

$$\sin x=x-\frac{x^3}{3!}+\dots+(-1)^{n}\frac{x^{2\,n+1}}{(2\,n+1)!}+\dots$$

12. Jan 7, 2008

### transgalactic

how do i substitute the expantion in the limit

i dont know where to stop
there is no limit to the number of members in the expantion
??

13. Jan 7, 2008

### Gib Z

Is that a problem :( ? You don't really have to stop anywhere, write it as $$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} , \sin x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}$$. Even though theres an infinite number of terms, think about what happens to them when you take the limit.

EDIT: If you look at my previous post, you will see the form that makes it the easiest to evaluate. The problem is even easier if you do the limit in little chunks, ie Find $$\lim_{x\to 0} \frac{ e^x -1}{x}, \lim_{x\to 0} \frac{\sin x}{x}$$ first. You can do this!

Last edited: Jan 7, 2008
14. Jan 7, 2008

### HallsofIvy

Staff Emeritus
Your problem was to find the first member (i.e. coefficient of xn). Take as many terms of the Taylor series of ex and sin(x) as you need! It probably won't be more than 4 or 5.

15. Jan 7, 2008

### transgalactic

the first one that you showed solvse by lhopital answer :e^x
the second one is a formula answer :1
can you give me an example of solving limit problems with a substitution?

16. Jan 8, 2008

### Gib Z

What idea do you have in mind, when you use the word 'substitution'? For the two limits you just did, instead of L'hopitals rule or remembering the limit, you could have 'substituted' e^x and sin x with the series given above. But that is not a real substitution, because the series and and e^x or sin (x) are actually identical.

If your thinking of 'substitution' in the way I usually think of it, try showing;

$$\lim_{n\to \infty} \left( 1 + \frac{x}{n}\right)^n = \lim_{u \to 0} \left( 1 + ux\right)^{(\frac{1}{u})}$$.

17. Jan 8, 2008

### transgalactic

i ment solving a limit proble, using tailor series
can you show me an example of that
??

18. Jan 8, 2008

### Gib Z

Well lets show that as x --> 0, (sin x)/x is 1.

$$\sin x = x - x^3/3!....$$

$$\frac{\sin x}{x} = 1 - x^2/3!...$$

As x goes to zero, the limit is 1.

19. Jan 8, 2008

### transgalactic

why you desided to stop on the third power
??

20. Jan 8, 2008

### Gib Z

I didn't actually stop at the third power, the dots were meant to indicate that the general pattern of the terms continued until infinity. Even if I took a million terms, don't you see how only the first one is a constant, and all the others involve x and will become 0 when you take the limit?