# Math Help - Series expansion question.

1. ## 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

2. 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

3. 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 $-xg(-x) = xg(x)$

4. 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

5. 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

6. 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

7. $

\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.