Question

if f(x)=summation of x^4n/ (4n)!
prove that the fourth derivative of f(x)= f(x)

Answer 1

summation also known as sigma

Answer 2

*

Answer 3

$$
\Large{
f(x)=\Sigma {x^{4n}\over{(4n)!}}\\
=\cfrac{1}{1}+\cfrac{x^8}{8!}+\cfrac{x^{12}}{12!}+\cfrac{x^{16}{}}{16!}+\cdots
}
$$

Is this the problem?

Now find \(\large{\cfrac{\partial^4 f(x)}{\partial x^4}}\) and compare to f(x)

Note that
$$
\large{
f(x)=\Sigma {x^{4n}\over{(4n)!}}\\
= \cfrac{1}{2} (\cos(x)+\cosh(x)-2)
}
$$
which might be easier to differential 4 times.

Answer 4

Correction on expansion:
$$
\Large{
f(x)=\Sigma {x^{4n}\over{(4n)!}}\\
=\cfrac{1}{1}+\cfrac{x^4}{4!}+\cfrac{x^{8}}{8!}+\cfrac{x^{12}{}}{12!}+\cdots
}
$$

Answer 5

yes that is the correct problem.
so i do the expansion?

Answer 6

and then how do you get the 1/2 (cosx +coshx -2?

Answer 7

I used a table of series expansions, which is usually a legitimate method if you stumble upon it.

Answer 8

oh so from there you find that its 1/2(cosx+coshx -2) ? what does the h stand for though?

Answer 9

Another way, done by differentiating the power series:
\[\large\begin{align*}f(x)&=\sum_{n=0}^\infty\frac{x^{4n}}{(4n)!}\\\\
f'(x)&=\sum_{n=0}^\infty \frac{4nx^{4n-1}}{(4n)!}\\
&=\sum_{n=1}^\infty \frac{x^{4n-1}}{(4n-1)!}\\\\
f''(x)&=\sum_{n=1}^\infty \frac{(4n-1)x^{4n-2}}{(4n-1)!}\\
&=\sum_{n=1}^\infty \frac{x^{4n-2}}{(4n-2)!}\\\\
f'''(x)&=\sum_{n=1}^\infty\frac{(4n-2)x^{4n-3}}{(4n-2)!}\\
&=\sum_{n=1}^\infty\frac{x^{4n-3}}{(4n-3)!}\\\\
f^{(4)}(x)&=\sum_{n=1}^\infty\frac{(4n-3)x^{4n-4}}{(4n-3)!}\\
&=\sum_{n=1}^\infty\frac{x^{4n-4}}{(4n-4)!}
\end{align*}\]
Then check each sum term by term.

Answer 10

Wolfram insists that the starting index is \(n=0\) though...

Answer 11

cosh(x) is cosine hyperbolic its derivative is sinh(x) and derivative of sinh(x) is cosh(x)

Answer 12

@ssoleman,
\[\cosh x=\frac{e^x+e^{-x}}{2}~~\text{and}~~\sinh x=\frac{e^x-e^{-x}}{2}\]
are the definitions for the hyperbolic trig functions.

Answer 13

thank you sithsandgiggles.. ill look into that way too

but so the derivative of cosh(x) is not -sinh(x)?

Answer 14

no, d/dx cosh(x) = sinh(x)

Answer 15

\[\frac{d}{dx}\cosh x=\frac{d}{dx}\left[\frac{e^x+e^{-x}}{2}\right]=\frac{e^{x}-e^{-x}}{2}=\sinh x\]

Answer 16

okay thank you.. and now that i have found all the derivative until the fourth, where do I go from there with this ?

Answer 17

now compare to f(x)

Answer 18

so does it make sense that the fourth d of f(x) = 1/2(cos(x)+cosh(x)

Answer 19

which means that they equal eachother!