Series expansion question.

**bobak** · April 21st 2008, 03:08 PM

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

**Mathstud28** · April 21st 2008, 03:19 PM

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

**bobak** · April 21st 2008, 03:24 PM

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 $-xg(-x) = xg(x)$

**Mathstud28** · April 21st 2008, 03:32 PM

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

**bobak** · April 21st 2008, 03:54 PM

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

**Mathstud28** · April 21st 2008, 03:57 PM

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

**PaulRS** · April 21st 2008, 04:43 PM

$ \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.

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