How does a Taylor series relate to the Maclaurin series?

The Maclaurin series is the Taylor series with a=0.

Taylor series of \sin x about \frac{\pi}{6} to \left(x-\frac{\pi}{6}\right)^3?

\frac12+\frac{\sqrt3}{2}\left(x-\frac{\pi}{6}\right)-\frac14\left(x-\frac{\pi}{6}\right)^2-\frac{\sqrt3}{12}\left(x-\frac{\pi}{6}\right)^3.

First four terms of \ln x about x=2?

\ln2+\frac12(x-2)-\frac18(x-2)^2+\frac1{24}(x-2)^3.

When is a truncated Taylor series accurate?

When x is close to a, so (x-a)^r is small.

1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots

Series for \sin x and \cos x?

\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots, \cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\dots

Series for \arctan x?

x-\frac{x^3}{3}+\frac{x^5}{5}-\dots

Method for a limit of the form \frac00 using series?

Expand top and bottom, cancel the lowest power of x, then let x\to0.

How do you get the series for e^{2x^2}?

Replace x by 2x^2 in e^x: 1+2x^2+2x^4+\dots

Taylor series and limitsEdexcel A-Level Further Maths: Flashcards

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State the Taylor series of $f(x)$ about $x=a$.

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State the Taylor series of $f(x)$ about $x=a$ .: $f(a)+(x-a)f'(a)+\frac{(x-a)^2}{2!}f''(a)+\frac{(x-a)^3}{3!}f'''(a)+\dots$
What is the coefficient of $(x-a)^r$ in a Taylor series?: $\frac{f^{(r)}(a)}{r!}$ .
How does a Taylor series relate to the Maclaurin series?: The Maclaurin series is the Taylor series with $a=0$ .
Taylor series of $\sin x$ about $\frac{\pi}{6}$ to $\left(x-\frac{\pi}{6}\right)^3$ ?: $\frac12+\frac{\sqrt3}{2}\left(x-\frac{\pi}{6}\right)-\frac14\left(x-\frac{\pi}{6}\right)^2-\frac{\sqrt3}{12}\left(x-\frac{\pi}{6}\right)^3$ .
First four terms of $\ln x$ about $x=2$ ?: $\ln2+\frac12(x-2)-\frac18(x-2)^2+\frac1{24}(x-2)^3$ .
When is a truncated Taylor series accurate?: When $x$ is close to $a$ , so $(x-a)^r$ is small.
Series for $\mathrm{e}^x$ ?: $1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots$
Series for $\sin x$ and $\cos x$ ?: $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots$ , $\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\dots$
Series for $\arctan x$ ?: $x-\frac{x^3}{3}+\frac{x^5}{5}-\dots$
Method for a limit of the form $\frac00$ using series?: Expand top and bottom, cancel the lowest power of $x$ , then let $x\to0$ .
$\displaystyle\lim_{x\to0}\frac{x-\arctan x}{x^3}$ ?: $\frac13$ .
$\displaystyle\lim_{x\to0}\frac{\mathrm{e}^{2x^2}-1}{x^2}$ ?: $2$ .
$\displaystyle\lim_{x\to0}\frac{1-\cos x}{x^2}$ ?: $\frac12$ .
How do you get the series for $\mathrm{e}^{2x^2}$ ?: Replace $x$ by $2x^2$ in $\mathrm{e}^x$ : $1+2x^2+2x^4+\dots$

Exam questions on Taylor series and limits

The function $f(x)=\ln x$ is expanded as a Taylor series in ascending powers of $(x-2)$ .
Use the series up to and including the term in $(x-2)^3$ to estimate $\ln2.2$ , giving your answer to 4 decimal places.2 marks
The Maclaurin series $\mathrm{e}^{x}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots$ , $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots$ and $\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\dots$ may be used. All limits are as $x\to0$ .
Find $\displaystyle\lim_{x\to0}\frac{\mathrm{e}^{2x^2}-1}{x^2}$ .2 marks
Let $f(x)=\cos x$ .
Find the Taylor series of $f(x)$ in ascending powers of $\left(x-\frac{\pi}{3}\right)$ , up to and including the term in $\left(x-\frac{\pi}{3}\right)^3$ .3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).