Derivate using taylor polynomial

[Anakin_99]

Hi guys I can't understand how use taylor polynomials to solve derivates, I think I understood the method but I can't understand how a n-derivate can be 0 and the n+1 derivate is != 0
Sorry if my english is so bad, let me give u an exemple of what I am saying.

I need to find the f(18), 18th derivate of x^(7)*sin(x^4).
Developing sin(x^4) using taylor I find sin(x^4) = x^4 -[ x^(12)]/6
so f(x) = x^11- [x^19]/6
so the f(18) of f(x) should be 0, but how can the f(19) be 19!/6 if the derivate of 0 is 0 ?
Thank u if u can explain this to me

 

[Dr.Peterson]

Hi guys I can't understand how use taylor polynomials to solve derivates, I think I understood the method but I can't understand how a n-derivate can be 0 and the n+1 derivate is != 0
Sorry if my english is so bad, let me give u an exemple of what I am saying.

I need to find the f(18), 18th derivate of x^(7)*sin(x^4).
Developing sin(x^4) using taylor I find sin(x^4) = x^4 -[ x^(12)]/6
so f(x) = x^11- [x^19]/6
so the f(18) of f(x) should be 0, but how can the f(19) be 19!/6 if the derivate of 0 is 0 ?
Thank u if u can explain this to me

I think you mean f^(18), not f(18), which would be the first derivative evaluated at 18.

Also, I think you are expanding about 0, so you are asking for f^(18)(0), not f^(18)(x). I understand your English, but you must be more careful about the notation!

Let's take a simple example, where the first derivative (at a given point) can be 0, but the second derivative is non-zero at the same point.

Consider f(x) = x^2. Then f'(x) = 2x, so f'(0) = 0; but f''(x) = 2, so f''(0) = 2.

Does that help?

 

[Anakin_99]

I think you mean f^(18), not f(18), which would be the first derivative evaluated at 18.

Also, I think you are expanding about 0, so you are asking for f^(18)(0), not f^(18)(x). I understand your English, but you must be more careful about the notation!

Let's take a simple example, where the first derivative (at a given point) can be 0, but the second derivative is non-zero at the same point.

Consider f(x) = x^2. Then f'(x) = 2x, so f'(0) = 0; but f''(x) = 2, so f''(0) = 2.

Does that help?

yes thank u very much

 

[Eugenio]

Hi guys I can't understand how use taylor polynomials to solve derivates, I think I understood the method but I can't understand how a n-derivate can be 0 and the n+1 derivate is != 0
Sorry if my english is so bad, let me give u an exemple of what I am saying.

I need to find the f(18), 18th derivate of x^(7)*sin(x^4).
Developing sin(x^4) using taylor I find sin(x^4) = x^4 -[ x^(12)]/6
so f(x) = x^11- [x^19]/6
so the f(18) of f(x) should be 0, but how can the f(19) be 19!/6 if the derivate of 0 is 0 ?
Thank u if u can explain this to m

You need to use the Taylor series of sin(x). This is

[MATH]sin(x)=\sum_{n=0}^{\infty }\frac{x^{2n}}{(2n)!}[/MATH]
So,

[MATH]sin(x^4)=\sum_{n=0}^{\infty }\frac{x^{4*2n}}{(2n)!}=\sum_{n=0}^{\infty }\frac{x^{8n}}{(2n)!}[/MATH]
And,

[MATH]x^7sin(x^4)=\sum_{n=0}^{\infty }\frac{x^{8n+7}}{(2n)!}=x^7+\frac{x^{15}}{2!}+\frac{x^{23}}{4!}+...[/MATH]
And the nth derivate evaluated in x = 0 for this function is the following:

[MATH]f^{(n)}(0)\neq 0 \Leftrightarrow n=7;15;23;...8k+7[/MATH]
Finally,

[MATH]f^{(18)}(0)=0 [/MATH]