Multiplication of Taylor series

**cozza** · July 27th 2010, 09:54 AM

The question is:

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

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

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

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

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

Multiplying these two together the answer I get is

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

Does this look along the right lines?

**Ackbeet** · July 27th 2010, 09:57 AM

No, that's incorrect, I'm afraid. You need to get the series for $\frac{1}{\sqrt{1+x}},$ not $\sqrt{1+x}.$ Multiply those two series together to get the result.

**cozza** · July 27th 2010, 11:13 AM

Originally Posted by Ackbeet

No, that's incorrect, I'm afraid. You need to get the series for $\frac{1}{\sqrt{1+x}},$ not $\sqrt{1+x}.$ 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 $\frac{1}{\sqrt{1+x}}$ , which is why I used the one for $(1+x)^\alpha$

**Ackbeet** · July 27th 2010, 11:15 AM

Well, then use $\alpha=-\frac{1}{2}.$ Is that what you used?

**cozza** · July 27th 2010, 11:47 AM

Originally Posted by Ackbeet

Well, then use $\alpha=-\frac{1}{2}.$ 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** · July 27th 2010, 12:07 PM

Hmm. We have

$\displaystyle{\sin(x)=x-\frac{x^{3}}{3!}+\dots,$ 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

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

so

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

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:

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

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

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

Can you see where your mistake is now?

**cozza** · July 27th 2010, 12:27 PM

Originally Posted by Ackbeet

Hmm. We have

$\displaystyle{\sin(x)=x-\frac{x^{3}}{3!}+\dots,$ 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

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

so

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

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:

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

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

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

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 $\frac{1}{6}x^3$ from $\frac{3}{8}x^3$ to get the correct coefficient and add $\frac{1}{12}x^4$ to $\frac{-5}{16}x^4$ , but I'm not sure where the $\frac{1}{12}x^4$ comes from? Thanks again

**Ackbeet** · July 27th 2010, 12:30 PM

The $\frac{x^{4}}{12}$ comes from multiplying the $-\frac{x^{3}}{3!}=-\frac{x^{3}}{6}$ in the sin expansion with the $-\frac{x}{2}$ in the square root expansion.

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