Further Maths - Maclaurin Series
How could you imagine any function $f(x)$ could be written as a polynomial?
$f(x) = a _ 0 + a _ 1 x + a _ 2 x^2 + a _ 3 x^3 + … + a _ r x^r$ If you wanted to work out $a _ 0$, what could you set $x$ equal to?
$f(x) = a _ 0 + a _ 1 x + a _ 2 x^2 + a _ 3 x^3 + … + a _ r x^r$ What happens if you substitute in $x = 0$?
If you wanted to approximate the complicated function $f(x)$ around $0$, what value could you use as a best guess?
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”?
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 $f(x) = f(0) + f’(0)x$?
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 $f(x)$?
What is the general term $r$ of a Maclaurin series for $f(x)$?
\[f(0) + f'(0)x + \frac{f''(0)}{2!}x^2\]
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 $e^{2x}$?
\[\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} + ...\]
What substitution for $x$ could you make in order to work out the Maclaurin series for $\ln(1 - x)$?
\[\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} + ...\]
How could you rewrite
\[\frac{\ln(1 - x)}{\ln(1 + x)}\]
in order to work out the Maclaurin series?
\[\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} + ...\]
How could you rewrite
\[\ln(\sqrt{1 + 2x})\]
in order to work out the Maclaurin series?
\[\ln(1 + x) \to -1 < x \le 1\]
If the Maclaurin expansion of $\ln$ is valid for that interval, what interval is $\ln(1 + 2x)$ valid for?
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\[\ln(x^2 + 6x + 9) \equiv 2\ln(x+3)\]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.