# Further Maths - Maclaurin Series

> Source: https://ollybritton.com/notes/a-level/further-maths/topics/maclaurin-series/ · Updated: 2021-02-09 · Tags: further-maths, differentiation, series

##### How could you imagine any function $f(x)$ could be written as a polynomial??
$$
f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + ... + a_r x^r
$$

##### $f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + ... + a_r x^r$ If you wanted to work out $a_0$, what could you set $x$ equal to??
$$
0
$$

##### $f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + ... + a_r x^r$ What happens if you substitute in $x = 0$??
$$
f(x) = a_0 + 0 + 0 + 0 + ...
$$

##### If you wanted to approximate the complicated function $f(x)$ around $0$, what value could you use as a best guess??
$$
f(x) = f(0)
$$

##### If you wanted to improve your current approximation of a function $f(x)$ around $0$, $f(x) = f(0)$, what could you add on so it incorporates the way the function is currently "moving"??
$$
f(x) = f(0) + f'(0)x
$$

##### Why is $f(x) = f(0) + f'(0)x$ a better approximation of the function $f(x)$ around $0$ than just $f(x) = f(0)$??
Because the rate of change of the approximation is the same as the actual function's.

##### How could you improve your guess for $f(x) = f(0) + f'(0)x$??
$$
f(x) = f(0) + f'(0)x + f''(0)x^2
$$

##### Why is $f(x) = f(0) + f'(0)x + f''(0)x^2$ a better approximation of the function $f(x)$ around $0$ than $f(x) = f(0) + f'(0)x$??
Because the rate of change of the rate of change is the same as the actual function.

##### What principle does the Maclaurin Series use for approximating a function??
If the $n$-th derivative of a function is the same as the the $n$-th derivate of the approximation of the function, then they should be roughly equal.

##### What are the three criteria for using a Maclaurin series for $f(x)$??
* $f(x)$ can be differentiated an infinite number of times
* $f^{(n)}(0)$ always converges
* The series formed converges

##### What are the first few terms of the Maclaurin series expansion of a function $f(x)$??
$$
f(x) = \frac{f(0)}{0!} + \frac{f'(0)}{1!}x^1 + \frac{f''(0)}{2!}x^2 + ...

$$

##### What is the general term $r$ of a Maclaurin series for $f(x)$??
$$
\frac{f^{(r)}(0)}{r!}x^r
$$

##### $$f(0) + f'(0)x + \frac{f''(0)}{2!}x^2$$ How could you describe this??
A Maclaurin polynomial of degree $2$.

### 2021-02-11
##### $$e^x = 1 + x + \frac{x^2}{2!} + ...$$ How could you work out the Maclaurin series for $e^{2x}$??
$$
1 + (2x) + \frac{(2x)^2}{2!} + ...
$$

##### $$\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} + ...$$ What substitution for $x$ could you make in order to work out the Maclaurin series for $\ln(1 - x)$??
$$
-x
$$

##### $$\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} + ...$$ How could you rewrite $$\frac{\ln(1 - x)}{\ln(1 + x)}$$ in order to work out the Maclaurin series??
$$
\ln(1 - x) - \ln(1 + x)
$$

##### $$\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} + ...$$ How could you rewrite $$\ln(\sqrt{1 + 2x})$$ in order to work out the Maclaurin series??
$$
\frac{1}{2}\ln(1 + 2x)
$$

##### $$\ln(1 + x) \to -1 < x \le 1$$ If the Maclaurin expansion of $\ln$ is valid for that interval, what interval is $\ln(1 + 2x)$ valid for??
$$
-\frac{1}{2} < x \le \frac{1}{2}
$$

#### $$\ln(x^2 + 6x + 9) \equiv 2\ln(x+3)$$ You managed to get this far in simplifying $\ln$ to use the Maclaurin expansion. What should the next step be??
$$
2(\ln(3) + \ln(1 + \frac{x}{3})
$$

#### Why is $$\ln(x^2 + 6x + 9) \equiv 2(\ln(3) + \ln(1 + \frac{x}{3})$$ better than $$\ln(x^2 + 6x + 9) \equiv 2(\ln(1 + 2x)) in order to apply the Maclaurin expansion$$??
In the later you have to expand a bracket up to the 4th degree, with the first you can just use a straight substitution.

### 2022-01-11
##### What's the trick you can use for finding the series expansion of something like $\sec(x)$ or $\frac{1}{\ln(1+x)}$??
Rewrite it in the form $(1 + g(x))^{-1}$ and then use binomial expansion.

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