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Consider the function: $$f(x) = \frac{1}{x^2+10}$$ Find the first three non-zero terms of the Taylor series of

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    What is the relationship between a function and any given coefficient of its Taylor series? Surely you have heard this. – 2017-01-15
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    Unfortunately, I haven't – 2017-01-15
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    Then I don't think you should be working on this problem. I think you should _read_ about Taylor series. They are of little value if you don't know what they signify. – 2017-01-15
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    Does "A global Taylor’s theorem with integral remainder" have anything to do with this? – 2017-01-15
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    @user406921You just need to read the basic definition of [Taylor series](https://en.wikipedia.org/wiki/Taylor_series), which will tell you how to calculate the terms of the series. – 2017-01-15
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    I know how to find the Taylor series. However in the notes of this module we were given only a small section on taylor series which only included "A global Taylor’s theorem with integral remainder". This is why I was wondering if there was more to the question. – 2017-01-15
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    What prevented you to type the words `Taylor series` into the search engine of your choice? – 2017-01-15
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    @user406921 Does this have to do with the [Lagrange Remainder](http://mathworld.wolfram.com/LagrangeRemainder.html) perhaps? – 2017-01-15

2 Answers 2

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A Taylor series centered at $x=0$ (Also called a Maclaurin Series) is given by:

$$f(x)=f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+...=\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n \tag{1}$$

A Taylor series centered at $x=a$ is given by:

$$f(x)=f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^2+...=\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n \tag{2}$$

NB: I will assume that you want the Taylor Series centered at $x=0$ as on equation $(1)$.

Therefore, you can find the first 3 non-zero terms of the Taylor series by differentiating your function $f(x)$, and then substituting $x=0$ into those terms. Then, you can substitute this into equation $(1)$.

However, because the derivatives will not look nice (they will become large), we can make this simpler for ourselves by substituting $u=x^2$.

Therefore, you have your function:

$$f(u)=\frac{1}{u+10}$$

We find the derivatives:

$$f'(u)=-\frac{1}{(u+10)^2}$$

$$f''(u)=\frac{2}{(u+10)^3}$$

We substitute $u=0$ (since we intended to center it at $x=0$):

$$f(0)=\frac{1}{10}$$ $$f'(0)=-\frac{1}{100}$$ $$f''(0)=\frac{2}{1000}=\frac{1}{500}$$

Therefore, we can substitute this into:

$$f(u) \approx f(0)+\frac{f'(0)}{1!}u+\frac{f''(0)}{2!}u^2$$

$$f(u) \approx \frac{1}{10}-\frac{1}{100}u+\frac{1}{1000}u^2$$

We can substitute back for $u=x^2$ to obtain the first three terms of the Taylor Series:

$$\boxed{f(u) \approx \frac{1}{10}+\frac{1}{100}x^2+\frac{1}{1000}x^4}$$

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    Is there anyway to use "A global Taylor’s theorem with integral remainder" with this? – 2017-01-15
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    Can you show us where exactly you found this phrase? – 2017-01-15
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As commented before: $$f(x) = \frac{1}{x^2+10}=\dfrac{1}{10}\cdot \frac{1}{1+\dfrac{x^2}{10}}\underbrace{=}_{\text{geometric series}}\dfrac{1}{10}\left(1-\frac{x^2}{10}+\frac{x^4}{100}-\cdots\right).$$

Edit. Alternatively $$f(x)=f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^2+\cdots$$

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    Is there anyway to use "A global Taylor’s theorem with integral remainder" with this? – 2017-01-15