1. ## Taylor Polynomials

Now that CaptainBlack and Jhevon have helped me out with Latex notation I can ask my question.

I was working on taylor polynomials and I ran into a little snag which has me stumped.

For the function $f(x) = \cos (x)$ we get the taylor series:
$\cos (x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - ... + \frac{(-1)^n f^{(n)} (0) x^{2n}}{(2n)!}$
So for $\cos (x^6)$ we get
$\cos (x^6) = 1 - \frac{x^{12}}{2!} + \frac{x^{24}}{4!} - ... + \frac{(-1)^n f^{(n)} (0) x^{12n}}{(2n)!}$

Yet if we construct the taylor polynomial for $\cos (x^6)$
from scratch:
$f (x) = \cos (x^6)$
$f^{'} (x) = -6x^5\sin (x^6)$
$f^{''} (x) = -30x^4\sin (x^6) - 36x^{10}\cos (x^6)$
$f^{'''} (x) = -120x^3\sin (x^6) - 180x^9\cos (x^6) - 360x^9\cos (x^6) + 216x^{15}\sin (x^6)$

for x= 0
$f (0) = \cos (0^6) = 1$
$f^{'} (0) = -6(0^5)\sin (0^6) = 0$
$f^{''} (0) = -30(0^4)\sin (0^6) - 36(0^{10})\cos (0^6) = 0$
$f^{'''} (0) = -120(0^3)\sin (0^6) - 180(0^9)\cos (0^6) - 360(0^9)\cos (0^6) + 216(0^{15})\sin (0^6)= 0$
in fact we dont see a non zero constant term until the 12th order differential:
Which looks quite crazy I've just realised. Some of the terms are $-36.10! - 18.10! - 8.10! - 3.10!$.
So if we get this, then the numerator on the second term of the taylor polynomial should be massive. So why does letting $x = x^6$ work? I understand the logic behind it I suppose, as the taylor polynomial work for any value of $x$ and $x^6$ is just a finite number (for any finite $x$ and for this $cos$ function, I know it doesn't hold for all functions). But why does the method of constructing the taylor polynomial from the formula:
$\sum_{a=0}^{\infty} \frac{f^{(a)} (0) x^{(a)}}{a!}$ fail?

2. Originally Posted by behemoth100
Now that CaptainBlack and Jhevon have helped me out with Latex notation I can ask my question.

I was working on taylor polynomials and I ran into a little snag which has me stumped.

For the function $f(x) = \cos (x)$ we get the taylor series:
$\cos (x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - ... + \frac{(-1)^n f^{(n)} (0) x^{2n}}{(2n)!}$
So for $\cos (x^6)$ we get
$\cos (x^6) = 1 - \frac{x^{12}}{2!} + \frac{x^{24}}{4!} - ... + \frac{(-1)^n f^{(n)} (0) x^{12n}}{(2n)!}$

Yet if we construct the taylor polynomial for $\cos (x^6)$
from scratch:
$f (x) = \cos (x^6)$
$f^{'} (x) = -6x^5\sin (x^6)$
$f^{''} (x) = -30x^4\sin (x^6) - 36x^{10}\cos (x^6)$
$f^{'''} (x) = -120x^3\sin (x^6) - 180x^9\cos (x^6) - 360x^9\cos (x^6) + 216x^{15}\sin (x^6)$

for x= 0
$f (0) = \cos (0^6) = 1$
$f^{'} (0) = -6(0^5)\sin (0^6) = 0$
$f^{''} (0) = -30(0^4)\sin (0^6) - 36(0^{10})\cos (0^6) = 0$
$f^{'''} (0) = -120(0^3)\sin (0^6) - 180(0^9)\cos (0^6) - 360(0^9)\cos (0^6) + 216(0^{15})\sin (0^6)= 0$
in fact we dont see a non zero constant term until the 12th order differential:
Which looks quite crazy I've just realised. Some of the terms are $-36.10! - 18.10! - 8.10! - 3.10!$.
So if we get this, then the numerator on the second term of the taylor polynomial should be massive. So why does letting $x = x^6$ work? I understand the logic behind it I suppose, as the taylor polynomial work for any value of $x$ and $x^6$ is just a finite number (for any finite $x$ and for this $cos$ function, I know it doesn't hold for all functions). But why does the method of constructing the taylor polynomial from the formula:
$\sum_{a=0}^{\infty} \frac{f^{(a)} (0) x^{(a)}}{a!}$ fail?
It does not fail, it gives exactly the same results as substituting $x^6$ into the expansion of $\cos(x)$.

Now the direct computation is long winded and it is error prone, and depends
on the cancellation of large factors, but it does give the same reslt.

RonL

3. Hmmm... but from my working you have... oh wait are you saying that the numerator will equal $\frac{12!}{2!}$?

4. Originally Posted by behemoth100
Hmmm... but from my working you have... oh wait are you saying that the numerator will equal $\frac{12!}{2!}$?
When I run this on a Computer Algebra System, the coefficients of the term
in $x^{12}$ does indeed cancels down to $-1/2$

RonL