# Series expansion question.

• Apr 21st 2008, 04:08 PM
bobak
Series expansion question.
The first part i had no trouble with.

Its just the explanation for part 2 for why it is not possible to express $g(x)$ as a series of non-negative powers of x. I was thinking it could be related to $g(x)$ being undefined at $x=0$ but then it can still be expressed as a series of non-negative power of (x-a) though, and why does it work for $xg(x)$? i am struggling to give a decent explanation for this part

Thanks

Bobak
• Apr 21st 2008, 04:19 PM
Mathstud28
Quote:

Originally Posted by bobak
The first part i had no trouble with.

Its just the explanation for part 2 for why it is not possible to express $g(x)$ as a series of non-negative powers of x. I was thinking it could be related to $g(x)$ being undefined at $x=0$ but then it can still be expressed as a series of non-negative power of (x-a) though, and why does it work for $xg(x)$? i am struggling to give a decent explanation for this part

Thanks

Bobak

Well...I know the reason that the the powers are even is when you add the two powers of x on the denominator then subtract one you get an even number
• Apr 21st 2008, 04:24 PM
bobak
Quote:

Originally Posted by Mathstud28
Well...I know the reason that the the powers are even is when you add the two powers of x on the denominator then subtract one you get an even number

I also know the reason the powers are even, thanks for answering a question I didn't ask. the function is even $-xg(-x) = xg(x)$
• Apr 21st 2008, 04:32 PM
Mathstud28
Quote:

Originally Posted by bobak
I also know the reason the powers are even, thanks for answering a question I didn't ask. the function is even $-xg(-x) = xg(x)$

I don't know if you are being sarcastic but if $-xg(-x)-xg(x)$ the function is odd...because if g(x) was even $g(-x)=g(x)$...so wouldnt $-xg(-x)=-xg(x)$ if it was even...sorry if I am being stupid
• Apr 21st 2008, 04:54 PM
bobak
Quote:

Originally Posted by Mathstud28
I don't know if you are being sarcastic but if $-xg(-x)-xg(x)$ the function is odd...because if g(x) was even $g(-x)=g(x)$...so wouldnt $-xg(-x)=-xg(x)$ if it was even...sorry if I am being stupid

I was being sarcastic, but $g(x)$ is odd and $xg(x)$ is even. why ? because $g(x)$ is the product of an odd and even function and $xg(x)$ is the product of an odd and odd function.

Bobak
• Apr 21st 2008, 04:57 PM
Mathstud28
Quote:

Originally Posted by bobak
I was being sarcastic, but $g(x)$ is odd and $xg(x)$ is even. why ? because $g(x)$ is the product of an odd and even function and $xg(x)$ is the product of an odd and odd function.

Bobak

I am sorry...I misunderstood your question...and you need not be so hostile
• Apr 21st 2008, 05:43 PM
PaulRS
$

\sinh \left( x \right) \cdot \cosh \left( {2x} \right) = \frac{{e^{3x} + e^{ - x} - e^x - e^{ - 3x} }}
{4} = \sum\limits_{n = 0}^\infty {\frac{{3^n + \left( { - 1} \right)^n - 1 - \left( { - 3} \right)^n }}
{{4 \cdot n!}}x^n }

$

Suppose $
\frac{1}
{{\sinh \left( x \right) \cdot \cosh \left( {2x} \right)}} = \sum\limits_{n = 0}^\infty {a_n \cdot x^n }
$

So multiplying we get: $
\left( {\sum\limits_{n = 0}^\infty {a_n \cdot x^n } } \right) \cdot \left[ {\sum\limits_{n = 0}^\infty {\left( {\tfrac{{3^n + \left( { - 1} \right)^n - 1 - \left( { - 3} \right)^n }}
{{4 \cdot n!}}} \right) \cdot x^n } } \right] = 1

$

Collecting terms: $
\sum\limits_{n = 0}^\infty {\left[ {\sum\limits_{k = 0}^n {a_k \cdot \left( {\tfrac{{3^{n - k} + \left( { - 1} \right)^{n - k} - 1 - \left( { - 3} \right)^{n - k} }}
{{4 \cdot \left( {n - k} \right)!}}} \right)} } \right] \cdot x^n } = 1

$

Thus we should have $

a_0 \cdot \left( {\tfrac{{3^0 + \left( { - 1} \right)^0 - 1 - \left( { - 3} \right)^0 }}
{{4 \cdot \left( 0 \right)!}}} \right) = 1

$
which is clearly absurd. (Wink)