Math problem, help needed.

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Aug 13th 2006, 08:46 AM
c00ky

Math 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 AM
topsquark

Quote:

Originally Posted by c00ky

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

See screen shot on following link....

CLICK HERE

Well, the best thing to do would be to work out
$1 - \frac{1}{2!}x^2 + \frac{1}{4!}x^4 - \frac{1}{6!}x^6$
for several values of x between 0 and $2 \pi$ 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 $\pi$ , 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
$\sum_{r = 0}^{\infty}(-1)^r \frac{x^{2r}}{(2r)!} = 1 - \frac{1}{2!}x^2 + \frac{1}{4!}x^4 + ...$
so this approximation should be valid.

-Dan
Aug 13th 2006, 09:23 AM
c00ky

You have lost me, i'm still a little confused.
Aug 13th 2006, 09:46 AM
Quick

Let me help.

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

expand to four terms:

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

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

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

simplify:

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

simplify:

$\frac{1}{1}\quad-\quad\frac{x^{2}}{2!}\quad+\quad\frac{x^{4}}{4!}<br /> \quad<br /> -\quad\frac{x^{6}}{6!}$

simplify:

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

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 AM
ThePerfectHacker

Quote:

Originally Posted by c00ky

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

See screen shot on following link....

CLICK HERE

The Taylor Expansion for $\cos x$ is given by,
$f(x) = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...$ so, $f(0)=1$
Note that, $f(x)$ has radius of convergence of infinite. And,
$f'(x)=-\frac{x}{1!}+\frac{x^3}{3!}-\frac{x^5}{5!}+...$ so, $f'(0)=0$
Also,
$f''(x)=-1+\frac{x^2}{2!}-\frac{x^4}{4!}+...=-f(x)$
Thus, you have that,
$f''(x)+f(x)=0$ with, $f(0)=1,f'(0)=0$ 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,
$f(x)=\cos x$
----
Now, if you expand it to 4 terms you have the approximation for cosine,
$\cos x \approx 1- \frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}$
To find the accurary of the approximation note that, the remainder term is,
$R_4=\frac{-\sin (z) x^{5}}{5!}$ where $z$ is some number on the interval. If we take the interval $[-\pi/4,\pi/4]$ the maximum error is,
$\left| \frac{-\sin z (\pi/4)^5}{120}} \right| \leq \left| \frac{ |1|\cdot|1|^5}{120}\right| \leq .008$
Which is accurate to 2 decimal places.
Aug 13th 2006, 09:57 AM
c00ky

I see, excellent. Thank you very much.