Hi guys, another problem. If anybody could guide me through the solution that would be great!

See screen shot on following link....

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- Aug 13th 2006, 08:46 AMc00kyMath problem, help needed.
Hi guys, another problem. If anybody could guide me through the solution that would be great!

See screen shot on following link....

CLICK HERE - Aug 13th 2006, 08:54 AMtopsquarkQuote:

Originally Posted by**c00ky**

$\displaystyle 1 - \frac{1}{2!}x^2 + \frac{1}{4!}x^4 - \frac{1}{6!}x^6$

for several values of x between 0 and $\displaystyle 2 \pi$ and see what happens when you compare that value to cos(x) when x is in radians.

What you will find is that this function only works well when x is "small." I don't know how well the approximation works for x near, say $\displaystyle \pi $, but for x less than, say, 1 it's probably pretty decent.

For the record, as a Taylor series about x = 0 we may approximate cos(x) as

$\displaystyle \sum_{r = 0}^{\infty}(-1)^r \frac{x^{2r}}{(2r)!} = 1 - \frac{1}{2!}x^2 + \frac{1}{4!}x^4 + ... $

so this approximation should be valid.

-Dan - Aug 13th 2006, 09:23 AMc00ky
You have lost me, i'm still a little confused.

- Aug 13th 2006, 09:46 AMQuick
Let me help.

you have expression: $\displaystyle (-1)^r\times\frac{x^{2r}}{(2r)!}$

expand to four terms:

$\displaystyle (-1)^r\times\frac{x^{2r}}{(2r)!}\quad+\quad(-1)^r\times$$\displaystyle \frac{x^{2r}}{(2r)!}\quad+\quad(-1)^r\times$$\displaystyle \frac{x^{2r}}{(2r)!}\quad+\quad(-1)^r$$\displaystyle \times\frac{x^{2r}}{(2r)!}$

now substitute 0,1,2 and 3 for "r"

$\displaystyle (-1)^0\times\frac{x^{2(0)}}{(2(0))!}\quad+\quad(-1)^1$$\displaystyle \times\frac{x^{2(1)}}{(2(1))!}\quad+\quad(-1)^2$$\displaystyle \times\frac{x^{2(2)}}{(2(2))!}\quad+\quad(-1)^3$$\displaystyle \times\frac{x^{2(3)}}{(2(3))!}$

simplify:

$\displaystyle 1\times\frac{x^{0}}{0!}\quad-\quad1$$\displaystyle \times\frac{x^{2}}{2!}\quad+\quad1\times$$\displaystyle \frac{x^{4}}{4!}\quad-\quad1\times$$\displaystyle \frac{x^{6}}{6!}$

simplify:

$\displaystyle \frac{1}{1}\quad-\quad\frac{x^{2}}{2!}\quad+\quad\frac{x^{4}}{4!}

\quad

-\quad\frac{x^{6}}{6!}$

simplify:

$\displaystyle 1$$\displaystyle -\frac{1}{2!}x^{2}$$\displaystyle +\frac{1}{4!}x^{4}$$\displaystyle -\frac{1}{6!}x^{6}$

now substitute a small value for x, then compare it to cos(x) and you will see they're similar - Aug 13th 2006, 09:56 AMThePerfectHackerQuote:

Originally Posted by**c00ky**

$\displaystyle f(x) = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...$ so, $\displaystyle f(0)=1$

Note that, $\displaystyle f(x)$ has radius of convergence of infinite. And,

$\displaystyle f'(x)=-\frac{x}{1!}+\frac{x^3}{3!}-\frac{x^5}{5!}+...$ so, $\displaystyle f'(0)=0$

Also,

$\displaystyle f''(x)=-1+\frac{x^2}{2!}-\frac{x^4}{4!}+...=-f(x)$

Thus, you have that,

$\displaystyle f''(x)+f(x)=0$ with, $\displaystyle f(0)=1,f'(0)=0$ by the existence theorem there is a unique function satisfing this equation. Thus, whatever is the solution must have that expansion stated above. Using various methods we find that the solution is,

$\displaystyle f(x)=\cos x$

----

Now, if you expand it to 4 terms you have the approximation for cosine,

$\displaystyle \cos x \approx 1- \frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}$

To find the accurary of the approximation note that, the remainder term is,

$\displaystyle R_4=\frac{-\sin (z) x^{5}}{5!}$ where $\displaystyle z$ is some number on the interval. If we take the interval $\displaystyle [-\pi/4,\pi/4]$ the maximum error is,

$\displaystyle \left| \frac{-\sin z (\pi/4)^5}{120}} \right| \leq \left| \frac{ |1|\cdot|1|^5}{120}\right| \leq .008$

Which is accurate to 2 decimal places. - Aug 13th 2006, 09:57 AMc00ky
I see, excellent. Thank you very much.