Multiplication of Taylor series

Mar 2010
31
0
The question is:

Use multiplication of Taylor series to find the quartic Taylor polynomial about 0 for the function:

\(\displaystyle f(x)=\frac{sinx}{\sqrt{1+x}}\) evaluating the coefficients.

For sin x the standard Taylor series about 0 is:
\(\displaystyle x-\frac{1}{3!}x^3+...\)

For \(\displaystyle \sqrt{1+x}\) this can be rearranged to fit the standard Talor series
\(\displaystyle (1+x)^\alpha\), which is

\(\displaystyle 1+\alpha x+\frac{\alpha(\alpha-1)}{2!}x^2+\frac{\alpha(\alpha-1)(\alpha-2)}{3!}x^3\) ...

Multiplying these two together the answer I get is

\(\displaystyle \frac{1}{2}x+\frac{3}{8}x^2+\frac{5}{96}x^3+\frac{35}{128}x^4\)

Does this look along the right lines? (Worried)
 

Ackbeet

MHF Hall of Honor
Jun 2010
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No, that's incorrect, I'm afraid. You need to get the series for \(\displaystyle \frac{1}{\sqrt{1+x}},\) not \(\displaystyle \sqrt{1+x}.\) Multiply those two series together to get the result.
 
Mar 2010
31
0
No, that's incorrect, I'm afraid. You need to get the series for \(\displaystyle \frac{1}{\sqrt{1+x}},\) not \(\displaystyle \sqrt{1+x}.\) Multiply those two series together to get the result.
We have a list of 6 standard Taylor series about 0 that we have been told to use. There isn't one for \(\displaystyle \frac{1}{\sqrt{1+x}}\), which is why I used the one for \(\displaystyle (1+x)^\alpha\)
 

Ackbeet

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Well, then use \(\displaystyle \alpha=-\frac{1}{2}.\) Is that what you used?
 
Mar 2010
31
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Well, then use \(\displaystyle \alpha=-\frac{1}{2}.\) Is that what you used?
That's what I used, but haven't got the correct answer. In the next part of the question we have to use a software programme that we're given to check we have the correct answer, so I know what it should be, but can't seem to get it right
 

Ackbeet

MHF Hall of Honor
Jun 2010
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Hmm. We have

\(\displaystyle \displaystyle{\sin(x)=x-\frac{x^{3}}{3!}+\dots,\) as you had before. We don't need more terms because we're only looking for the first four terms. According to your formula there, we have

\(\displaystyle \displaystyle{(1+x)^{\alpha}=1+\alpha x+\frac{\alpha(\alpha-1)}{2!}x^2+\frac{\alpha(\alpha-1)(\alpha-2)}{3!}x^3},\)

so

\(\displaystyle \displaystyle{(1+x)^{-1/2}=1+(-1/2) x+\frac{(-1/2)(-1/2-1)}{2!}x^2+\frac{(-1/2)(-1/2-1)(-1/2-2)}{3!}x^3+\dots}\)
\(\displaystyle \displaystyle{=1-x/2+\frac{3x^{2}}{8}-\frac{5 x^{3}}{16}+\dots}\)

We don't need more terms than these, because the lowest power of x in the sin series is to the first power; hence, the cubic in this series will go to a fourth power.

Both of these are correct. Moving on, then:

\(\displaystyle \displaystyle{\frac{\sin(x)}{\sqrt{1+x}}=\left(x-\frac{x^{3}}{3!}+\dots\right)\left(1-x/2+\frac{3x^{2}}{8}-\frac{5 x^{3}}{16}+\dots\right)}\)

\(\displaystyle \displaystyle{=x-\frac{x^{2}}{2}+\frac{3x^{3}}{8}-\frac{5x^{4}}{16}-\frac{x^{3}}{6}+\frac{x^{4}}{12}+\dots}\)

\(\displaystyle \displaystyle{=x-\frac{x^{2}}{2}+\frac{5x^{3}}{24}-\frac{11x^{4}}{48}+\dots}\)

Can you see where your mistake is now?
 
Mar 2010
31
0
Hmm. We have

\(\displaystyle \displaystyle{\sin(x)=x-\frac{x^{3}}{3!}+\dots,\) as you had before. We don't need more terms because we're only looking for the first four terms. According to your formula there, we have

\(\displaystyle \displaystyle{(1+x)^{\alpha}=1+\alpha x+\frac{\alpha(\alpha-1)}{2!}x^2+\frac{\alpha(\alpha-1)(\alpha-2)}{3!}x^3},\)

so

\(\displaystyle \displaystyle{(1+x)^{-1/2}=1+(-1/2) x+\frac{(-1/2)(-1/2-1)}{2!}x^2+\frac{(-1/2)(-1/2-1)(-1/2-2)}{3!}x^3+\dots}\)
\(\displaystyle \displaystyle{=1-x/2+\frac{3x^{2}}{8}-\frac{5 x^{3}}{16}+\dots}\)

We don't need more terms than these, because the lowest power of x in the sin series is to the first power; hence, the cubic in this series will go to a fourth power.

Both of these are correct. Moving on, then:

\(\displaystyle \displaystyle{\frac{\sin(x)}{\sqrt{1+x}}=\left(x-\frac{x^{3}}{3!}+\dots\right)\left(1-x/2+\frac{3x^{2}}{8}-\frac{5 x^{3}}{16}+\dots\right)}\)

\(\displaystyle \displaystyle{=x-\frac{x^{2}}{2}+\frac{3x^{3}}{8}-\frac{5x^{4}}{16}-\frac{x^{3}}{6}+\frac{x^{4}}{12}+\dots}\)

\(\displaystyle \displaystyle{=x-\frac{x^{2}}{2}+\frac{5x^{3}}{24}-\frac{11x^{4}}{48}+\dots}\)

Can you see where your mistake is now?
Thank you so much for all your help. just one final question (I promise!). I worked out that I needed to take away \(\displaystyle \frac{1}{6}x^3\) from \(\displaystyle \frac{3}{8}x^3\) to get the correct coefficient and add \(\displaystyle \frac{1}{12}x^4\)to \(\displaystyle \frac{-5}{16}x^4\), but I'm not sure where the \(\displaystyle \frac{1}{12}x^4\) comes from? Thanks again (Happy)
 

Ackbeet

MHF Hall of Honor
Jun 2010
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2,433
CT, USA
The \(\displaystyle \frac{x^{4}}{12}\) comes from multiplying the \(\displaystyle -\frac{x^{3}}{3!}=-\frac{x^{3}}{6}\) in the sin expansion with the \(\displaystyle -\frac{x}{2}\) in the square root expansion.
 
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