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Maclaurin Series (Edexcel A Level Further Maths: Core Pure): Flashcards

Exam code: 9FM0

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Define a Maclaurin series.

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Define a Maclaurin series.
It is a representation of a function as an infinite sum of ascending integer powers of $x$ , built from the value of the function and of all its derivatives at $x = 0$ .
With every one of its infinitely many terms included, it is exactly equal to the original function.
To build a Maclaurin series from first principles, what do you need to calculate?
The value of the function and of each of its derivatives at $x = 0$ , that is $f (0)$ , $f' (0)$ , $f'' (0)$ and so on.
You need them as far as the power of $x$ the question asks for, since the $r$ th derivative is what supplies the coefficient of $x^{r}$ .
What happens to a Maclaurin series when you stop it at a particular power of $x$ ?
It becomes an approximation to the function instead of being equal to it.
The more terms are kept the better that approximation is, so a series truncated at the $x^{7}$ term is more accurate than the same series truncated at $x^{3}$ .
True or False?
A truncated Maclaurin series is only an approximation, so it is never exactly equal to the function.
False.
At $x = 0$ every term containing $x$ vanishes, so a truncated series agrees with the function exactly there, however few terms it has.
Away from zero it is an approximation, and one that generally worsens as $x$ moves further from the origin.
Do all the standard Maclaurin expansions hold for every value of $x$ ?
No. The expansions of $e^{x}$ , $\sin x$ and $\cos x$ hold for all $x$ , but $\ln (1 + x)$ holds only for $- 1 < x \leq 1$ and $\arctan x$ only for $- 1 \leq x \leq 1$ .
Outside its interval a series does not converge to the function, so the expansion says nothing about it there.
How is the Maclaurin series of ${(1 + x)}^{n}$ related to its binomial expansion?
They are exactly the same series.
So unless a question insists on the general Maclaurin formula you may quote the binomial expansion instead, and the formula will just as happily rebuild the binomial series if you have forgotten it.
How do you find the Maclaurin series of a composite function such as $\sin 2 x$ or $e^{x^{2}}$ ?
Take the standard expansion of the outside function and substitute the inside function everywhere an $x$ appears in it.
Then expand the brackets and collect the powers of $x$ , keeping only those up to the power you were asked for.
The expansion of $\ln (1 + x)$ begins $x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \dots$ Fill in the first three terms of the expansion of $\ln (1 + 3 x)$ :
$\ln (1 + 3 x) =______-______+______- \dots$
The completed expansion is:
$\ln (1 + 3 x) = 3 x - \frac{9}{2} x^{2} + 9 x^{3} - \dots$
The brackets have to be expanded before any coefficient can be read off, so the second term is $\frac{{(3 x)}^{2}}{2} = \frac{9}{2} x^{2}$ and not $\frac{3}{2} x^{2}$ .
The expansion of $\ln (1 + x)$ is valid for $- 1 < x \leq 1$ . What is the valid interval for $\ln (1 + 3 x)$ ?
The condition applies to whatever fills the bracket, so it becomes $- 1 < 3 x \leq 1$ , which gives $- \frac{1}{3} < x \leq \frac{1}{3}$ .
Substituting into a standard expansion moves its interval of validity, so the new interval has to be worked out rather than carried across unchanged.
How do you find the Maclaurin series of a product of two functions, such as $e^{x} \sin x$ ?
Write down the series for each function separately, put each in brackets, then multiply them out and collect like powers of $x$ .
Anything reaching a power higher than the one asked for is discarded, and it is quite possible for a coefficient to come out as zero.
When multiplying two Maclaurin series together, how do you decide how many terms of each to write down?
Keep enough terms of each that every product which can still reach the required power is available, and no more than that.
A series whose lowest term is $x$ lifts everything it multiplies one power higher, so fewer of its terms are needed than of a series beginning with a constant.

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