# Power Series

1. Jul 1, 2005

### R_Y_A_N

Given:
/ (sinx)/x for x =/= 0​
g(x)=
\ 1 for x=0​

(26)
g(0), the 26th derivative of g at 0 (Hint: can you find a power series for g(x)?)

At zero the function is 1 right?...so the first or second or 26th derivative is zero...right? but that seems way to easy..please help me understand why i need to use a power series...thanks

2. Jul 1, 2005

### OlderDan

The first derivative at zero is zero, but not the second. The power series might help you identify the higher order derivatives. By symmetry you can argue that the odd derivatives are zero, but not the evens.

3. Jul 1, 2005

### Hurkyl

Staff Emeritus
This reasoning is wrong -- remember that the derivative of a function at a point depends on the values near that point... and the function isn't 1 everywhere near that point.

4. Jul 1, 2005

### lurflurf

The power series of sin (about 0) is sin(x)~x-x^3/3!+x^5/5!-x^7/7!+...+(-1)^k x^(2k+1)/(2k+1)!+...(k the number of the nonzero term)
x g(x)=sin(x)
So you should be able to find a power series for g. You will also see that all the derivatives are not zero.
I think what confused you is
f'(1) is not the same as (f(1))'
like if f(x)=x^2
f'(x)=2x f'(1)=2 (f(1))'=(1)'=0

5. Jul 1, 2005

### R_Y_A_N

but for a power series...you need to take all the derivatives of the function up to the 26th..and that get's messy..do i really have to take the 26th derivative of sinx/x?

6. Jul 1, 2005

### Hurkyl

Staff Emeritus
The goal of the problem is to use the power series to find the derivatives, not the other way around. You need to find the power series through some other method...

7. Jul 1, 2005

### R_Y_A_N

y' = (-x^-2)(sinx) + (x^-1)(cosx)
y'' = (2x^-3)(sinx) + (-2x^-2)(cosx) - (x^-1)(sinx)
y''' = (-6x^-4)(sinx) + (6x^-3)(cosx) + (3x^-2)(sinx) - (x^-1)(cosx)
i don't see a pattern..maybe i'm doing something wrong...or maybe i'm just blind

8. Jul 1, 2005

### R_Y_A_N

i'm getting very confused now...you use the derivatives to make a power series to approximate the value at that point..so to find a power series to find the 26th derivative would require me to find the 26th derivative to make the power series by definition right?

9. Jul 1, 2005

### HallsofIvy

Perhaps you should try paying attention to the responses you already have:
sin x= x- (1/3!)x^3+ (1/5!)x^5- 1/(7!)x^7+...- (1/27!)x^27+ ...
so (sin x)/x= 1- (1/3!)x^2+ (1/5!)x^4- 1/(7!)x^6+ ... - (1/27!)x^26+ ...

Of course, the nth coefficient of the MacLaurin series for any function g(n) is
(1/n!)g(n)(0). So what is g(26)(0)?

10. Jul 1, 2005

### lurflurf

The higher derivatives often get messy. For this reason it is often easier to deduce a power series from another known one. In this case you know the power series for sin(x) and you want to find the power series of g(x) where x*g(x)=sin(x).
for example (1-x)^-1~1+x+x^2+x^3+x^4+x^5+...
x/(1-x)~x+x^2+x^3+x^4+x^5+x^6+...
See a pattern?
Have you covered formal operations on power series such as multiplication, division, and addition?

11. Jul 1, 2005

### R_Y_A_N

so it's 0^26 or 1?

12. Jul 1, 2005

### R_Y_A_N

no i haven't unfortunately

13. Jul 1, 2005

### lurflurf

So the idea is is f and g have power series you can often find the power series of f/g by dividing their power series. This can some time be tricky but in your problem we have x which has a simple power series (just x). So the general term in sin(x) is (-1)^k*x^(2k+1)/(2k+1)! The general term in sin(x)/x is then (-1)^k*x^(2k)/(2k+1)! The 2k-th term is also $$g}^{(2k)}{(0)}*x^{2k}/(2k)!$$
So you should be able to figure out what the 26th derivative of g at 0 is.

Last edited: Jul 1, 2005
14. Jul 2, 2005

### R_Y_A_N

so, if i'm understanding this correctly, all the derivatives will be either -1 or 1, depending on whether (-1) is raised to an even or odd power, and in this case we raise (-1) to the 26/2 or 13 power and get -1 as the 26th derivative at zero

15. Jul 2, 2005

### HallsofIvy

No, lurflurf just told you that "the general term in sin(x) is (-1)^k*x^(2k+1)/(2k+1)!" so the general term in sin(x)/x is (-1)^k*x^(2k)/(2k+1)! and that "The 2k-th term is also $$g}^{(2k)}{(0)}*x^{2k}/(2k)!$$". Equate the two:
(-1)^k*x2k/(2k+1)!= g(2k)(0)x2k/(2k)!. Solving for g(2k), g2k(0)= (-1)k(2k+1).

16. Jul 2, 2005

### lurflurf

Here is another way to find the derivatives without power series.
x g(x)=sin(x)
take k+1 derivatives
$$x g}^{(k+1)}{(x)}+(k+1) g}^{(k)}{(x)}=sin}^{(k+1)}{(x)}$$
let x=0
$$(k+1) g}^{(k)}{(0)}=sin}^{(k+1)}{(0)}=cos{(k {\pi}/2)}$$
$$g}^{(k)}{(0)}=\frac{{sin}^{(k+1)}{(0)}}{k+1}=\frac{cos{(k {\pi}/2)}}{k+1}$$
Now try to see how it works with power series.
see that
g(0)=1
g'(0)=0
g''(0)=-1/3
g'''(0)=0
g''''(0)=1/5

Last edited: Jul 2, 2005