# Multiplication of Taylor series

#### cozza

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
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.

#### cozza

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

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

#### cozza

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
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?

#### cozza

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
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.

• cozza