테일러·매클로린 급수

${p}_{4}(x)={(x+3)}^{3}-11{(x+3)}^{2}+39(x+3)-41;$ exact

$\sum_{n=0}^{\infty }\frac{{(-1)}^{n}{(3x)}^{2n}}{2n!}$

$\sum_{n=0}^{\infty }\frac{{(-1)}^{n}}{(2n)!}{\left(x-\frac{\pi }{2}\right)}^{2n}$

$\sum_{n=1}^{\infty }\frac{{(-1)}^{n}}{n!}{x}^{2n}$

$F(x)=\sum_{n=0}^{\infty }\frac{{(-1)}^{n}}{(2n+1)(2n+1)!}{x}^{2n+1}$

$\displaystyle 1+\frac x2-\frac{x^2}{8}+\frac{x^3}{16}-\frac{5x^4}{128}$

$\displaystyle 1-\frac x2+\frac{3x^2}{8}-\frac{5x^3}{16}+\frac{35x^4}{128}$

$\displaystyle 1+\frac x3-\frac{x^2}{9}+\frac{5x^3}{81}-\frac{10x^4}{243}$

$\displaystyle 1+4x+6x^2+4x^3+x^4$ (note the series is finite, and the formula still applies)

$\displaystyle \int_0^{\sqrt{\pi}} \sin \big(x^2\big)\ dx \approx \int_0^{\sqrt{\pi}} \left(x^2-\frac{x^6}6+\frac{x^{10}}{120}-\frac{x^{14}}{5040}\right) dx = 0.8877$

$\displaystyle \int_0^{\pi^2/4} \cos \big(\sqrt{x}\big)\ dx \approx \int_0^{\pi^2/4} \left(1-\frac x2+\frac{x^2}{24}-\frac{x^3}{720}\right)\ dx = 1.1412$. (Actual answer: $\pi-2$)

Following how a Taylor series is constructed, the Taylor series and the function agree at the point chosen as the centre. So, $T(5)=\arctan^3\left(e^5+7\right)$.

If we were evaluating a Taylor series at a point other than its centre, we would generally need to check that (a) the series converges, and (b) it converges to the same value as the function we used to create it.

Our limit has the indeterminate form $1^{\infty}$; as with l'Hopital's rule, we can change it to a friendlier form using the natural logarithm.

$$\begin{aligned} f(x)&=\left(1+x+x^2\right)^{2/x}\\ \log(f(x))&=\log\left[\left(1+x+x^2\right)^{2/x}\right]=\frac{2}{x}\log\left(1+x+x^2\right) \end{aligned}$$

Recall $\log(1+y)=\displaystyle\sum_{n=1}^\infty \frac{(-1)^{n+1}y^n}{n}$, and set $y=x+x^2$. The series converges when $|y|<1$, and since we only consider values of $x$ that are very close to 0, we can assume $|x+x^2|<1$.

$$\begin{aligned} \log(f(x))&=\frac{2}{x}\log\left(1+(x+x^2)\right)=\frac{2}{x}\sum_{n=1}^\infty\frac{(-1)^{n+1}(x+x^2)^n}{n}\\ &=\frac{2}{x}\left[(x+x^2)-\frac{(x+x^2)^2}{2}+\frac{(x+x^2)^3}{3}-\cdots\right]\\ &=2+2x-\frac{(x+x^2)^2}{2x}+\frac{(x+x^2)^3}{3x}-\cdots\\ &=2+2x-\frac{(x^2+x)(1+x)}{2}+\frac{(x^2+x)^2(1+x)}{3}-\cdots\\ \lim_{x \to 0}\log(f(x))&=\lim_{x \to 0}\left[2+2x-\frac{(x^2+x)(1+x)}{2}+\frac{(x^2+x)^2(1+x)}{3}-\cdots\right] \\ &=2+0+0\cdots=2\\ \lim_{x \to 0} f(x)&=\lim_{x \to 0}e^{\log f(x)}=e^2 \end{aligned}$$

Recall the Taylor series for arctangent is: $$\arctan x = \sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{2n+1}$$ There are similarities between this and our given series: skipping powers of $x$, and a denominator that's not factorial. We'll try to manipulate it to look like our series. First, we antidifferentiate, to get a factor of $(2n+2)$ on the bottom.

$$\begin{aligned} \int \arctan x\,\mathrm{d}x &= \sum_{n=0}^\infty (-1)^n\frac{x^{2n+2}}{(2n+1)(2n+2)}+C \\ \text{We can find the antiderivative of arctangent using integration by parts. Let $u=\arctan x$ and $\mathrm{d}v=\mathrm{d}x$; then $\mathrm{d}u=\frac{1}{1+x^2}\mathrm{d}x$ and $v=x$.} \\ \int \arctan x \,\mathrm{d}x&=x\arctan x - \int \frac{x}{1+x^2}\,\mathrm{d}x+C \\ \text{Now, we use the substitution $w=1+x^2$, $\mathrm{d}w=2x\mathrm{d}x$.} \\ &=x\arctan x - \frac{1}{2}\log(1+x^2)+C\\ \text{So, }\qquad \sum_{n=0}^\infty (-1)^n\frac{x^{2n+2}}{(2n+1)(2n+2)}&=x\arctan x - \frac{1}{2}\log(1+x^2)+C \\ \text{To find $C$, we evaluate both sides of the equation at $x=0$.} \\ 0&=0\arctan 0 -\frac{1}{2}\log(1)+C=C\\ \text{Therefore, }\qquad \sum_{n=0}^\infty (-1)^n\frac{x^{2n+2}}{(2n+1)(2n+2)}&=x\arctan x - \frac{1}{2}\log(1+x^2) \\ \text{Multiplying both sides by $x^2$,} \\ \sum_{n=0}^\infty (-1)^n\frac{x^{2n+4}}{(2n+1)(2n+2)}&=x^3\arctan x - \frac{x^2}{2}\log(1+x^2) \end{aligned}$$