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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Metadata
date: 2021-02-09 10:42
tags:
- '@?further-maths'
- '@?differentiation'
- '@?series'
- '@?public'
title: Further Maths - Maclaurin Series
