Approximation By Polynomial Functions

Sine Reduction Formula

Main Content

\[\int \sin^{n}x dx=-\frac{1}{n}\sin^{n-1}x\cos x+\left(\frac{n-1}{n}\right)\int\sin^{n-2}x dx\]

Proof.

\[\text{Let }n\in\mathbb{Z}^{+},\ \text{Given }\int \sin^{n}x dx\] \[\begin{align} u&=\sin^{n-1}x\\ du&=(n-1)\sin^{n-2}x\cos xdx\\ dv&=\sin xdx\\ v&=-\cos x\\ \int \sin^{n}x dx&=-\sin^{n-1}x\cos x-\int -\cos x(n-1)\sin^{n-2}x\cos xdx \\ &=-\sin^{n-1}x\cos x+\int (n-1)\sin^{n-2}x\cos^{2} xdx\\ &=-\sin^{n-1}x\cos x+\int (n-1)\sin^{n-2}x(1-\sin^{2}x) xdx\\ \int \sin^{n}x dx&=-\sin^{n-1}x\cos x+(n-1)\int \sin^{n-2}x - (n-1)\int \sin^{n}x dx\\ (1+(n-1))\int \sin^{n}x dx&=-\sin^{n-1}x\cos x+(n-1)\int \sin^{n-2}x \\ \int \sin^{n}x dx&=-\frac{1}{n}\sin^{n-1}x\cos x+\left(\frac{n-1}{n}\right)\int\sin^{n-2}x dx\ \ \ \ \blacksquare\end{align}\]

Wallis’s Theorem

Main Content

Theorem:

\[\prod_{n \mathop = 1}^\infty \frac {2 n} {2 n - 1} \cdot \frac {2 n} {2 n + 1} = \frac 2 1 \cdot \frac 2 3 \cdot \frac 4 3 \cdot \frac 4 5 \cdot \frac 6 5 \cdot \frac 6 7 \cdot \frac 8 7 \cdot \frac 8 9 \cdots = \frac \pi 2\]

Proof.

\[\text{From Sine Reduction Formula: }\] \[\begin{align} \int \sin^{n}x dx&=-\frac{1}{n}\sin^{n-1}x\cos x+\left(\frac{n-1}{n}\right)\int\sin^{n-2}x dx\\ \int_b^a \sin^{n}x dx&=\left(-\frac{1}{n}\sin^{n-1}x\cos x\right)\Big|_b^a+\left(\frac{n-1}{n}\right)\int_b^a\sin^{n-2}x dx\\ \int_0^{\frac{\pi}{2}} \sin^{n}x dx&=\left(-\frac{1}{n}\sin^{n-1}x\cos x\right)\Big|_0^{\frac{\pi}{2}}+\left(\frac{n-1}{n}\right)\int_0^{\frac{\pi}{2}}\sin^{n-2}x dx \\ \int_0^{\frac{\pi}{2}} \sin^{n}x dx&=0+\left(\frac{n-1}{n}\right)\int_0^{\frac{\pi}{2}}\sin^{n-2}x dx \\ \int_0^{\frac{\pi}{2}} \sin^{n}x dx&=\left(\frac{n-1}{n}\right)\int_0^{\frac{\pi}{2}}\sin^{n-2}x dx \\ \\\\ \int_0^{\frac{\pi}{2}} \sin^{2n+1}x dx&=\frac{2n}{2n+1}\times\int_0^{\frac{\pi}{2}}\sin^{2n-1}x dx \\ &=\frac{2n}{2n+1}\times\frac{2n-2}{2n-1}\times\int_0^{\frac{\pi}{2}}\sin^{2n-3}x dx \\ &=\frac{2n}{2n+1}\times\frac{2n-2}{2n-1}\times\ldots\times\int_0^{\frac{\pi}{2}}\sin^{3}x dx \\ &=\frac{2n}{2n+1}\times\frac{2n-2}{2n-1}\times\ldots\times\frac 2 3 \times\int_0^{\frac{\pi}{2}}\sin x dx \\ &=\frac{2n}{2n+1}\times\frac{2n-2}{2n-1}\times\ldots\times\frac 2 3 \times 1\\ \int_0^{\frac{\pi}{2}} \sin^{2n}x dx&=\frac{2n-1}{2n}\times\int_0^{\frac{\pi}{2}}\sin^{2n-2}x dx \\ &=\frac{2n-1}{2n}\times\frac{2n-3}{2n-2}\times\int_0^{\frac{\pi}{2}}\sin^{2n-4}x dx \\ &=\frac{2n-1}{2n}\times\frac{2n-3}{2n-2}\times\ldots\times\int_0^{\frac{\pi}{2}}\sin^{2}x dx \\ &=\frac{2n-1}{2n}\times\frac{2n-3}{2n-2}\times\ldots\times\frac 1 2 \times\int_0^{\frac{\pi}{2}}\sin^{0}x dx \\ &=\frac{2n-1}{2n}\times\frac{2n-3}{2n-2}\times\ldots\times\frac 1 2 \times\frac \pi 2\\ \int_0^{\frac{\pi}{2}} \sin^{2n+2}x dx&=\frac{2n+1}{2n+2}\times\int_0^{\frac{\pi}{2}}\sin^{2n}x dx \\ \\\\\end{align}\] \[\text{Let }\ I = \frac{\int_0^{\frac{\pi}{2}} \sin^{2n+1}x dx}{\int_0^{\frac{\pi}{2}} \sin^{2n}x dx} = \frac{2n}{2n+1}\times\frac{2n}{2n-1}\times\frac{2n-2}{2n-1}\times\frac{2n-2}{2n-3}\times\ldots\times\frac 2 3 \times\frac 2 1 \times\frac 2 \pi\] \[\begin{align}\forall x\in\left[0,\frac \pi 2\right],\ &\sin x \leq 1\\&\implies \text{higher powers of } \sin x \text{ yield smaller values}\\&\implies\sin^{2n}x\geq\sin^{2n+1}x\geq\sin^{2n+2}x\end{align}\] \[\begin{align}\int_0^{\frac{\pi}{2}}\sin^{2n}xdx&\geq\int_0^{\frac{\pi}{2}}\sin^{2n+1}xdx\geq\int_0^{\frac{\pi}{2}}\sin^{2n+2}xdx\\ \frac{\int_0^{\frac{\pi}{2}}\sin^{2n}xdx}{\int_0^{\frac{\pi}{2}}\sin^{2n}xdx}&\geq\frac{\int_0^{\frac{\pi}{2}}\sin^{2n+1}xdx}{\int_0^{\frac{\pi}{2}}\sin^{2n}xdx}\geq\frac{\int_0^{\frac{\pi}{2}}\sin^{2n+2}xdx}{\int_0^{\frac{\pi}{2}}\sin^{2n}xdx}\\ 1&\geq I \geq \frac{\frac{2n+1}{2n+2}\times\int_0^{\frac{\pi}{2}}\sin^{2n}x dx }{\int_0^{\frac{\pi}{2}}\sin^{2n}xdx}\tag{Plug in formulas}\\ 1&\geq I \geq \frac{2n+1}{2n+2}\\ 1&\geq I \geq 1-\frac{1}{2n+2}\\\\\end{align}\] \[\text{Plugging in for $I$:}\] \[1\geq \frac{2n}{2n+1}\times\frac{2n}{2n-1}\times \ldots\times\frac 2 3 \times\frac 2 1 \times\frac 2 \pi \geq 1-\frac{1}{2n+2}\] \[\frac \pi 2 (1)\geq \frac \pi 2 \left(\frac{2n}{2n+1}\times\frac{2n}{2n-1}\times \ldots\times\frac 2 3 \times\frac 2 1 \times\frac 2 \pi \right)\geq \frac \pi 2 \left(1-\frac{1}{2n+2}\right)\\ \frac \pi 2 \geq \frac{2n}{2n+1}\times\frac{2n}{2n-1}\times \ldots\times\frac 2 3 \times\frac 2 1 \geq \frac \pi 2 \left(1-\frac{1}{2n+2}\right)\\ \lim_{n\to\infty}\left(\frac \pi 2\right) \geq \lim_{n\to\infty}\left(\frac{2n}{2n+1}\times\frac{2n}{2n-1}\times \ldots\times\frac 2 3 \times\frac 2 1\right) \geq \lim_{n\to\infty}\left(\frac \pi 2 \left(1-\frac{1}{2n+2}\right)\right)\\ \frac \pi 2 \geq \lim_{n\to\infty}\left(\frac{2n}{2n+1}\times\frac{2n}{2n-1}\times \ldots\times\frac 2 3 \times\frac 2 1\right) \geq \frac \pi 2\\\\\] \[\text{By Squeeze Theorem: }\] \[\lim_{n\to\infty}\left(\frac{2n}{2n+1}\times\frac{2n}{2n-1}\times \ldots\times\frac 2 3 \times\frac 2 1\right) = \frac \pi 2\ \ \ \ \blacksquare\]

Factorial Function

Main Content

\[f(t)=\int_{0}^{\infty}x^t e^{-x}dx=tf(t-1)\]

Proof.

\[\begin{align} \int x^t& e^{-x}dx\\ u&=x^t\\du&=tx^{t-1}dx\\dv&=e^{-x}dx\\v&=-e^{-x}\\ \int x^t e^{-x}dx &=-x^t e^{-x}\Big|^{\infty}_0+t\int_0^{\infty}x^{t-1}e^{-x}dx\\ &=0+t\int_0^{\infty}x^{t-1}e^{-x}dx&&\left(\lim_{x\to\infty}e^{-x}=0\right)\\ &=tf(t-1)\\ f(t)&=tf(t-1)\\\\ \implies f(0)=1,f(1)=1,&f(2)=2,f(3)=6,f(4)=24\\ \implies \forall t\in\mathbb{Z}^+ &\cup\{0\},\ f(t)=t!\ \ \ \ \blacksquare \end{align}\]

Applications

• 1.) Find \(\left(\frac{1}{2}\right)!\)
  • \[\text{Use Factorial Function}\]
  • Answer: \(\int_0^\infty \sqrt{x}e^{-x}dx\)
• 2.) Can you compute fractional factorials?
  • Answer: You can compute any fractional factorial as long as you can compute the integral from plugging in the fraction into the Factorial Function.

Gamma Function

Main Content

\[\begin{align} \Gamma(t)&=\int_0^\infty x^{t-1}e^{-x}dx\\ &=f(t-1)\\ \implies \Gamma(1)=1,\ \Gamma(2)&=1,\ \Gamma(3)=2,\ \Gamma(4)=6\\ \implies t\Gamma(t)&=\Gamma(t+1)\end{align}\]

Calculating Area Under Bell Curve

Main Content

\[\begin{align} \text{Bell Curve: }& e^{-x^2}\\ \text{Area Under Half: }& \int_0^\infty e^{-x^2}dx=\frac{\sqrt{\pi}}{2} \end{align}\]

Proof.

\[\text{Calculate reduction formula:}\] \[\begin{align} \int&(1-x^2)^ndx\\ u&=(1-x^2)^n\\du&=n(1-x^2)^{n-1}(-2x)dx\\dv&=1dx\\v&=x\\ \int(1-x^2)^ndx &= x(1-x^2)^n+2n\int x^2(1-x^2)^{n-1}dx\\ &=x(1-x^2)^n+2n\int(1-(1-x^2))(1-x^2)^{n-1}dx\\ \int(1-x^2)^ndx &= x(1-x^2)^n+2n\int(1-x^2)^{n-1}dx-2n\int(1-x^2)^ndx\\ (2n+1)\int(1-x^2)^ndx&=x(1-x^2)^n+2n\int(1-x^2)^{n-1}dx\\ \int(1-x^2)^ndx&=\frac{x(1-x^2)^n}{2n+1}+\frac{2n}{2n+1}\int(1-x^2)^{n-1}dx\\\\ \end{align}\] \[\text{Calculate $f(n)$: }\] \[\begin{align} f(n)&=\int_0^1(1-x^2)^ndx\\ &=\left(\frac{x(1-x^2)^n}{2n+1}\right)\Big|_0^1+\frac{2n}{2n+1}\int_0^1(1-x^2)^{n-1}dx\\ &=0+\frac{2n}{2n+1}\int_0^1(1-x^2)^{n-1}dx\\ &=\frac{2n}{2n+1}\int_0^1(1-x^2)^{n-1}dx\\ f(n)&=\frac{2n}{2n+1}f(n-1)\\\\\end{align}\] \[\text{Calculate reduction formula:}\] \[\begin{align} \int\frac{1}{(1+x^2)^n}&=\int\frac{1+x^2}{(1+x^2)^n}-\frac{x^2}{(1+x^2)^n}dx\\ &=\int\frac{1}{(1+x^2)^{n-1}}dx-\int\frac{x^2}{(1+x^2)^n}dx\\ \text{Solving for: }& \int\frac{x^2}{(1+x^2)^n}dx\\ u&=x\\du&=1dx\\dv&=\frac{x}{(1+x^2)^n}\\v&=\int\frac{x}{(1+x^2)^n}dx =\frac{1}{2-2n}\left(\frac{1}{(1+x^2)^{n-1}}\right)&&\text{(U-Sub)}\\ \int\frac{1}{(1+x^2)^n}=\int\frac{1}{(1+x^2)^{n-1}}dx&-\left(\frac{1}{2-2n}\left(\frac{x}{(1+x^2)^{n-1}}\right)-\frac{1}{2-2n}\int\frac{1}{(1+x^2)^{n-1}}dx\right)\\ &=\frac{1}{2-2n}\left(\frac{x}{(1+x^2)^{n-1}}\right)+\left(1+\frac{1}{2-2n}\right)\int\frac{1}{(1+x^2)^{n-1}}dx\\ &=\frac{1}{2-2n}\left(\frac{x}{(1+x^2)^{n-1}}\right)+\frac{2n-3}{2n-2}\int\frac{1}{(1+x^2)^{n-1}}dx\\\\ \end{align}\] \[\text{Calculate $g(n)$: }\] \[\begin{align} g(n)&=\int_0^\infty\frac{1}{(1+x^2)^n}\\ &=\left(\frac{1}{2-2n}\left(\frac{x}{(1+x^2)^{n-1}}\right)\right)\Big|_0^\infty+\frac{2n-3}{2n-2}\int_0^\infty\frac{1}{(1+x^2)^{n-1}}dx\\ &=0 + \frac{2n-3}{2n-2}\int_0^\infty\frac{1}{(1+x^2)^{n-1}}dx\\ &=\frac{2n-3}{2n-2}\int_0^\infty\frac{1}{(1+x^2)^{n-1}}dx\\ g(n)&=\frac{2n-3}{2n-2}g(n-1) \end{align}\] \[\text{Using Wallis's Theorem:}\] \[\begin{align} \frac \pi 2 &\approx \frac 2 1 \times \frac 2 3 \times \frac 4 3 \times \frac 4 5 \times \times\ldots\times\frac{2n}{2n-1}\times\frac{2n}{2n+1}\\ \implies \pi &\approx\frac{2\times 2\times 4\times 4\times\ldots\times 2n\times 2n}{1\times 1\times 3\times 3\times \ldots\times (2n-1)\times(2n-1)}\times\frac{2n}{2n+1}\\ \implies \sqrt{\pi}&\approx\frac{2\times 4\times\ldots\times 2n}{1\times 3\times \ldots\times (2n-1)}\times\frac{1}{\sqrt{n}}\\ \text{Let}\ k(n) &= \frac{2\times 4\times\ldots\times 2n}{1\times 3\times \ldots\times (2n-1)}\times\frac{1}{\sqrt{n}}\\ &\implies\lim_{n\to\infty}k(n)=\sqrt{\pi} \end{align}\] \[\text{Finding $f(n)$ and $g(n)$ in terms of $k(n)$:}\] \[\begin{align} f(n)=\int_0^1(1-x^2)^ndx&=\frac{2n}{2n+1}f(n-1)\\ &=\frac{2n}{2n+1}\times\frac{2n-2}{2n-1}\times\ldots\times\frac 4 5 \times \frac 2 3\\ &=\frac{1}{2n+1}\times\sqrt{n}\times k(n)&&\text{(plug in $k(n)$ to verify)}\\ g(n)=\int_0^\infty\frac{1}{(1+x^2)^n}dx&=\frac{2n-3}{2n-2}g(n-1)\\ &=\frac{2n-3}{2n-2}\times\frac{2n-5}{2n-4}\times\ldots\times\frac 1 2 \times\int_0^\infty\frac{1}{(1+x^2)^1}dx\\ &=\frac{2n-3}{2n-2}\times\frac{2n-5}{2n-4}\times\ldots\times\frac 1 2 \times\frac{\pi}{2}&&\text{($\lim_{x\to\infty}\tan^{-1}x=\frac{\pi}{2}$)}\\ &=\frac{1}{\sqrt{n}\times k(n)}\times\frac{2n}{2n-1}\times\frac{\pi}{2}&&\text{(plug in $k(n)$ to verify)}\\ \end{align}\] \[\begin{align} &\text{Absolute min of $e^x-x-1$ is $(0,0)$}\\ &\implies e^x-x-1\geq 0\\&\implies \forall x\ e^x\geq1+x\\ &\implies e^{-x^2}\geq 1-x^2&&\text{(plug in $-x^2$ for $x$)}\\ &\implies e^{x^2}\geq 1+x^2 &&\text{(plug in $x^2$ for $x$)}\\ &\implies e^{-x^2}\leq\frac{1}{1+x^2}&&\text{(take inverse)}\\ &\implies (1-x^2)\leq e^{-x^2}\leq\frac{1}{1+x^2}&&\text{(combine previous)}\\ &\implies(1-x^2)^n\leq e^{-nx^2}\leq\frac{1}{(1+x^2)^n}\\ \end{align}\] \[\text{Proving some intermediate results:}\] \[\begin{align} &\int_0^1 e^{-nx^2}dx\\ y &= x\sqrt{n},\ dy=\sqrt{n}dx\\ \int_0^1 e^{-nx^2}dx&=\frac{1}{\sqrt{n}}\int_0^\sqrt{n}e^{-y^2}dy&&\text{($x=1\implies y=\sqrt{n}$)}\\ &=\frac{1}{\sqrt{n}}\int_0^\sqrt{n}e^{-x^2}dx&&\text{(change $y$ to $x$)}\\\\ &\int_0^\infty e^{-nx^2}dx\\ y &= x\sqrt{n},\ dy=\sqrt{n}dx\\ \int_0^\infty e^{-nx^2}dx&=\frac{1}{\sqrt{n}}\int_0^\infty e^{-y^2}dy&&\text{($x=\infty\implies y=\infty$)}\\ &=\frac{1}{\sqrt{n}}\int_0^\infty e^{-x^2}dx&&\text{(change $y$ to $x$)}\\\\ \frac{1}{\sqrt{n}}\int_0^\sqrt{n}e^{-x^2}dx&\leq\frac{1}{\sqrt{n}}\int_0^\infty e^{-x^2}dx&&\text{(area of $\int_0^\infty >$ area of $\int_0^\sqrt{n}$)}\\\\ \end{align}\] \[\text{Using all previous results:}\] \[\begin{align} (1-x^2)^n\leq &e^{-nx^2}\leq\frac{1}{(1+x^2)^n}\\ \implies\int_0^1(1-x^2)^ndx\leq&\int_0^1 e^{-nx^2}dx = \frac{1}{\sqrt{n}}\int_0^\sqrt{n}e^{-x^2}dx\\ &\leq\frac{1}{\sqrt{n}}\int_0^\infty e^{-x^2}dx=\int_0^\infty e^{-nx^2}dx\leq\int_0^\infty\frac{1}{(1+x^2)^n}dx\\ \implies\int_0^1(1-x^2)^ndx\leq&\frac{1}{\sqrt{n}}\int_0^\sqrt{n} e^{-x^2}dx\leq\int_0^\infty\frac{1}{(1+x^2)^n}dx\\ \implies\frac{1}{2n+1}\times\sqrt{n}\times k(n)\leq&\frac{1}{\sqrt{n}}\int_0^\sqrt{n} e^{-x^2}dx\leq\frac{1}{\sqrt{n}\times k(n)}\times\frac{2n}{2n-1}\times\frac{\pi}{2}\\ \implies\frac{n}{2n+1}\times k(n)\leq&\int_0^\sqrt{n} e^{-x^2}dx\leq\frac{1}{k(n)}\times\frac{2n}{2n-1}\times\frac{\pi}{2}\\ \implies\lim_{n\to\infty}\frac{n}{2n+1}\times k(n)\leq&\lim_{n\to\infty}\int_0^\sqrt{n} e^{-x^2}dx\leq\lim_{n\to\infty}\frac{1}{k(n)}\times\frac{2n}{2n-1}\times\frac{\pi}{2}\\ &\lim_{n\to\infty}\frac{n}{2n+1}\times k(n)=\frac 1 2 \times \sqrt{\pi}=\frac{\sqrt{\pi}}{2}\\ &\lim_{n\to\infty}\frac{1}{k(n)}\times\frac{2n}{2n-1}\times\frac{\pi}{2}=\frac{1}{\sqrt{\pi}}\times 1\times\frac{\pi}{2}=\frac{\sqrt{\pi}}{2}\\ \implies \frac{\sqrt{\pi}}{2}\leq &\lim_{n\to\infty}\int_0^\sqrt{n} e^{-x^2}dx\leq \frac{\sqrt{\pi}}{2}\\ \implies \lim_{n\to\infty}&\int_0^\sqrt{n} e^{-x^2}dx = \frac{\sqrt{\pi}}{2}\\ \implies &\int_0^\infty e^{-x^2}dx = \frac{\sqrt{\pi}}{2}\ \ \ \ \blacksquare \end{align}\]

Wallis’s Applied to Coin Flip

Main Content

Probability when flipping \(2n\) coins and getting \(n\) heads and \(n\) tails:

\[=\frac{ {2n}\choose{n}}{2^{2n}}\approx\frac{1}{\sqrt{\pi n}}\]

Proof.

\[\text{Using Wallis's Theorem:}\] \[\begin{align} \frac \pi 2 &\approx \frac 2 1 \times \frac 2 3 \times \frac 4 3 \times \frac 4 5 \times \times\ldots\times\frac{2n}{2n-1}\times\frac{2n}{2n+1}\\ \implies \pi &\approx\frac{2\times 2\times 4\times 4\times\ldots\times 2n\times 2n}{1\times 1\times 3\times 3\times \ldots\times (2n-1)\times(2n-1)}\times\frac{2n}{2n+1}\\ \implies \sqrt{\pi}&\approx\frac{2\times 4\times\ldots\times 2n}{1\times 3\times \ldots\times (2n-1)}\times\frac{1}{\sqrt{n}}\\ \implies &\lim_{n\to\infty}\frac{2\times 4\times\ldots\times 2n}{1\times 3\times \ldots\times (2n-1)}\times\frac{1}{\sqrt{n}}=\sqrt{\pi}\\\\ &\frac{2\times 4\times\ldots\times 2n}{2\times 4\times\ldots\times 2n}\times\left(\frac{2\times 4\times\ldots\times 2n}{1\times 3\times \ldots\times (2n-1)}\times\frac{1}{\sqrt{n}}\right)\\ &=\left(\frac{2^2\times 4^2\times\ldots\times 2n^2}{1\times 3\times \ldots\times (2n-1)\times2\times 4\times\ldots\times 2n}\times\frac{1}{\sqrt{n}}\right)\\ &=\frac{2^{2n}(n!)^2}{(2n)!}\approx \sqrt{\pi}\times\sqrt{n}=\sqrt{\pi n}\\ &\frac{(2n)!}{2^{2n}(n!)^2}\approx\frac{1}{\sqrt{\pi n}}\\ &\frac{ {2n}\choose{n}}{2^{2n}}\approx\frac{1}{\sqrt{\pi n}}\ \ \ \ \blacksquare \end{align}\]

Approximating \(\pi\) Using Wallis’s

Main Content

According to Wallis’s Theorem:

\[\begin{align} &\lim_{n\to\infty}\frac 2 1 \times \frac 2 3 \times \frac 4 3 \times \frac 4 5 \times \times\ldots\times\frac{2n}{2n-1}\times\frac{2n}{2n+1}=\frac{\pi}{2}\\ &\implies \pi = 2\times \lim_{n\to\infty}\frac 2 1 \times \frac 2 3 \times \frac 4 3 \times \frac 4 5 \times \times\ldots\times\frac{2n}{2n-1}\times\frac{2n}{2n+1} \end{align}\]

To get better and better approximations of \(\pi\), multiply more of these products together:

\[\begin{align} &\pi\approx2\times\frac 2 1 = 4\\ &\pi\approx2\times\frac 2 1\times \frac 2 3 = 2.\bar{6}\\ &\pi\approx2\times\frac 2 1\times \frac 2 3 \times \frac 4 3 = 3.\bar{5}\\ &\pi\approx2\times\frac 2 1\times \frac 2 3 \times \frac 4 3 \times \frac 4 5 = 2.8\bar{4}\\ &\pi\approx\ldots \end{align}\]

Approximating \(\tan^{-1}a\)

Main Content

We can approximate \(\tan^{-1}a\ \ (\forall\ |a|\leq 1)\) to the necessary accuracy, \(z\), by using the following steps:

• Find \(n\) such that

\[\frac{a^{2n+3}}{2n+3}\leq z\]

• Find approximation by plugging \(n\) and \(a\) into:

\[a-\frac{a^3}{3}+\frac{a^5}{5}-\frac{a^7}{7}+\ldots+(-1)^n\frac{a^{2n+1}}{2n+1}\]

• To check answer, approximation should be within \(z\) of actual value of \(\tan^{-1}a\)

Proof.

\[\begin{align} \tan^{-1}a&=\int_0^a\frac{1}{1+x^2}dx\\\\ \text{Using long division}&\text{ to find $\frac{1}{1+x^2}$:}\\ \end{align}\] \[\begin{align} \require{enclose} 1-x^2+x^4-\ldots \\[-3pt] 1+x^2\phantom{0} \enclose{longdiv}{1\phantom{00000000000000}}\kern-.2ex \\[-3pt] \underline{1+x^2}\phantom{0000000000} \\[-3pt] -x^2\phantom{0000000000} \\[-3pt] \underline{-x^2-x^4}\phantom{00000} \\[-3pt] x^4\phantom{00000} \\[-3pt] \underline{x^4+x^6} \\[-3pt] -x^6 \\[-3pt] \ldots \\[-3pt]\\ \end{align}\] \[\begin{align} \frac{1}{1+x^2}&=1-\frac{x^2}{1+x^2}\\ &=1-x^2+\frac{x^4}{1+x^2}\\ &=1-x^2+x^4-\frac{x^6}{1+x^2}\\ &=1-x^2+x^4-x^6+\ldots+(-1)^nx^{2n}+(-1)^{n+1}\frac{x^{2n+2}}{1+x^2}\\ &\implies \text{$\frac{1}{1+x^2}$ can be expressed as a geometric series}\\ &\phantom{000000}\text{with 1st term 1, ratio $-x^2$, converging to $\frac{1}{1+x^2}$}\\\\ \end{align}\] \[\text{Integrating previous result:}\] \[\begin{align} \int_0^a\frac{1}{1+x^2}dx&=\int_0^a\left(1-x^2+x^4-x^6+\ldots+(-1)^nx^{2n}+(-1)^{n+1}\frac{x^{2n+2}}{1+x^2}\right)dx\\ &=\left(x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\ldots+(-1)^n\frac{x^{2n+1}}{2n+1}\right)\Big|_0^a+\int_0^a(-1)^{n+1}\frac{x^{2n+2}}{1+x^2}dx\\ &=\left(a-\frac{a^3}{3}+\frac{a^5}{5}-\frac{a^7}{7}+\ldots+(-1)^n\frac{a^{2n+1}}{2n+1}\right)+(-1)^{n+1}\int_0^a\frac{x^{2n+2}}{1+x^2}dx\\ \implies \tan^{-1}a&=\left(a-\frac{a^3}{3}+\frac{a^5}{5}-\frac{a^7}{7}+\ldots+(-1)^n\frac{a^{2n+1}}{2n+1}\right)+(-1)^{n+1}\int_0^a\frac{x^{2n+2}}{1+x^2}dx\\\\ \end{align}\] \[\begin{align} \text{Let } E_n(a)=(-1)^{n+1}&\int_0^a\frac{x^{2n+2}}{1+x^2}dx\\ \frac{1}{1+x^2}&\leq 1\\ \implies\frac{x^{2n+2}}{1+x^2}&\leq x^{2n+2}\\ \implies \int_0^a\frac{x^{2n+2}}{1+x^2}dx&\leq\int_0^a x^{2n+2}dx\\ \implies\int_0^a\frac{x^{2n+2}}{1+x^2}dx&\leq\frac{a^{2n+3}}{2n+3}\\ \implies\left|(-1)^{n+1}\int_0^a\frac{x^{2n+2}}{1+x^2}dx\right|&\leq\frac{a^{2n+3}}{2n+3}\\ \implies|E_n(a)|&\leq\frac{a^{2n+3}}{2n+3}\\\\ \end{align}\] \[\text{Taking limit of previous result:}\] \[\begin{align} |a|\leq1&\implies\lim_{n\to\infty}|E_n(a)|\leq\lim_{n\to\infty}\frac{a^{2n+3}}{2n+3}\\ &\implies\lim_{n\to\infty}|E_n(a)|\leq0\\ &\implies\lim_{n\to\infty}|E_n(a)|=0\\ |a|>1&\implies\lim_{n\to\infty}|E_n(a)|\leq\lim_{n\to\infty}\frac{a^{2n+3}}{2n+3}\\ &\implies\lim_{n\to\infty}|E_n(a)|\leq\infty\\ &\implies\lim_{n\to\infty}|E_n(a)|\ \ \text{diverges}\\\\ \end{align}\] \[\text{Apply results to $\tan^{-1}a$:}\] \[\begin{align} E_n(a)&=(-1)^{n+1}\int_0^a\frac{x^{2n+2}}{1+x^2}dx\\ \implies \tan^{-1}a &= \left(a-\frac{a^3}{3}+\frac{a^5}{5}-\frac{a^7}{7}+\ldots+(-1)^n\frac{a^{2n+1}}{2n+1}\right) +E_n(a)\\ \implies \lim_{n\to\infty}\tan^{-1}a &=\lim_{n\to\infty}\left(a-\frac{a^3}{3}+\frac{a^5}{5}-\frac{a^7}{7}+\ldots+(-1)^n\frac{a^{2n+1}}{2n+1}\right) +\lim_{n\to\infty}E_n(a)\\ \implies \tan^{-1}a &=\lim_{n\to\infty}\left(a-\frac{a^3}{3}+\frac{a^5}{5}-\frac{a^7}{7}+\ldots+(-1)^n\frac{a^{2n+1}}{2n+1}\right) +\lim_{n\to\infty}|E_n(a)|\\ |a|\leq1&\implies\lim_{n\to\infty}|E_n(a)|=0\\ &\implies\tan^{-1}a =\lim_{n\to\infty}\left(a-\frac{a^3}{3}+\frac{a^5}{5}-\frac{a^7}{7}+\ldots+(-1)^n\frac{a^{2n+1}}{2n+1}\right) +0\\ &\implies\tan^{-1}a\ \ \text{can be approximated to }\left(a-\frac{a^3}{3}+\frac{a^5}{5}-\frac{a^7}{7}+\ldots+(-1)^n\frac{a^{2n+1}}{2n+1}\right)\\ |a|>1&\implies\lim_{n\to\infty}|E_n(a)|\ \ \text{diverges}\\ &\implies\tan^{-1}a \ \ \text{cannot be approximated}\ \ \ \ \blacksquare \end{align}\]

Applications

• 1.) \(\tan^{-1}\left(\frac 1 2\right)\) approximated to \(10^{-3}\)
  • \[\text{Find $n$ such that:}\]
  • \[\begin{align} \frac{\left(\frac 1 2\right)^{2n+3}}{2n+3}&\leq 10^{-3}\\ n&=2\\\\ \end{align}\]
  • \[\text{Find approximation by plugging $n=2$ and $a=\frac 1 2$ into:}\]
  • \[a-\frac{a^3}{3}+\frac{a^5}{5}-\frac{a^7}{7}+\ldots+(-1)^n\frac{a^{2n+1}}{2n+1}\\ =\frac 1 2 -\frac{\left(\frac 1 2 \right)^3}{3}+\frac{\left(\frac 1 2 \right)^5}{5}\\ =0.46458\bar{3} \\\\\]
  • \[\text{To check answer, approximation should be within $10^{-3}$ of actual value of $\tan^{-1}\left(\frac 1 2\right)$:}\]
  • \[\begin{align} \text{Approximated }\tan^{-1}\left(\frac 1 2\right) &= 0.46458\bar{3} \\ \text{Actual }\tan^{-1}\left(\frac 1 2\right) &= 0.463647609\ldots \\ \text{Difference between Actual and Approx.} &= 0.00093572433\ldots \\ 0.00093572433\ldots &< 10^{-3} \\\\ \end{align}\]
  • Answer: \(0.46458\bar{3}\)
• 2.) Approximate \(\pi\)
  • \[\text{Plugging $a=1$ into $\tan^{-1}a$ approximation formula:}\]
  • \[\tan^{-1}(a)=\left(a-\frac{a^3}{3}+\frac{a^5}{5}-\frac{a^7}{7}+\ldots+(-1)^n\frac{a^{2n+1}}{2n+1}\right)\\ \tan^{-1}(1)=\left(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\ldots+(-1)^n\frac{1}{2n+1}\right)\\ \frac \pi 4 = \left(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\ldots+(-1)^n\frac{1}{2n+1}\right)\\ \pi = 4\left(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\ldots+(-1)^n\frac{1}{2n+1}\right) \\\\\]
  • Answer: Approximate \(\pi\) by choosing an \(n\) and plugging it into \(4\left(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\ldots+(-1)^n\frac{1}{2n+1}\right)\)

“Polynomial Bases”

Introduction Problems

• 1.) Convert 2018 to base 9
  • \[\begin{align}2018\div9&=224\ R\ 2 \\224\div9&=24\ R\ 8 \\24\div9&=2\ R\ 6 \\\implies 2018 \text{ in base } 9 &= 2682\end{align}\]
  • Answer: 2682
• 2.) Convert \(x^3+x+1\) to a polynomial in \((x-2)\)
  • The process is identical to converting numerical bases
  • \[\begin{align}x^3+x+1&=11+(x-2)(x^2+2n+5) \\&=11+(x-2)(13+(x-2)(x+4)) \\&=11+(x-2)(13+(x-2)(6+1(x-2))) \\&=11+13(x-2)+6(x-2)^2+1(x-2)^3\end{align}\]
  • Answer: \(11+13(x-2)+6(x-2)^2+1(x-2)^3\)

Main Content

A polynomial \(P(x)\) of degree \(n\) can be written as a polynomial in \((x-a)\):

\[\forall \text{ unique constants } a_0,a_1,\ldots,a_n\\ P(x)=a_0+a_1(x-a)+a_2(x-a)^2+\ldots+a_n(x-a)^n\]

The constants can be calculated using:

\[a_k=\frac{P^{(k)}(a)}{k!}\]

where \(P^{(k)}(x)\) is the function \(P(x)\) after taking \(k\) derivatives.

And the polynomial can be expressed as:

\[P(x)=\sum_{k=0}^n\frac{P^{(k)}(a)}{k!}(x-a)^k\]

Proof.

\[\text{WWTP: }a_k=\frac{P^{(k)}(a)}{k!}\] \[\text{Plugging in $x=0$ to $P(x)$:}\] \[\begin{align} P(x) &= a_0+a_1(x-a)+a_2(x-a)^2+\ldots+a_n(x-a)^n \\ P(a) &= a_0 \end{align}\] \[\text{Then use derivatives:}\] \[\begin{align} P'(x) &= a_1+2a_2(x-a)+3a_3(x-a)^2+\ldots+na_n(x-a)^{n-1} \\ P'(a) &= a_1 \\ P''(x) &= 2a_2+6a_3(x-a)+12a_4(x-a)^2+\ldots+n(n-1)a_n(x-a)^{n-2} \\ P''(a) &= 2a_2 \\ &\ldots \\ \end{align}\] \[\text{After taking $k$ derivatives of $P(x)$,} \\ \text{non-zero values of $P^{(k)}(x)$ start from $a_k(x-a)^k$:} \\\] \[P^{(k)}(x)=\left(\frac{d}{dx}\right)^k(a_k(x-a)^k+\ldots)\] \[\text{and k derivatives of $a_k(x-a)^k$ is $k!a_k$:}\] \[\begin{align} P^{(k)}(x) &= k!a_k+\ldots \\ P^{(k)}(a) &= k!a_k \\ a_k &= \frac{P^{(k)}(a)}{k!}\ \ \ \ \blacksquare \end{align}\]

Proof.

\[\text{WWTP: }P(x)=\sum_{k=0}^n\frac{P^{(k)}(a)}{k!}(x-a)^k\] \[\text{Given by previous proof:}\] \[\begin{align} a_k &= \frac{P^{(k)}(a)}{k!} \\ P(x) &= a_0+a_1(x-a)+a_2(x-a)^2+\ldots+a_n(x-a)^n \\ &= \frac{P^{(0)}(a)}{0!}(x-a)^0+\frac{P^{(1)}(a)}{1!}(x-a)^1+\ldots+\frac{P^{(n)}(a)}{n!}(x-a)^n \\ &= \sum_{k=0}^n\frac{P^{(k)}(a)}{k!}(x-a)^k\ \ \ \ \blacksquare \end{align}\]

Applications

• 1.) Express \(P(x)=x^3+x+1\) in \((x-2)\)
  • \[\begin{align}& & P(2)&=11 \\P'(x)&=2x^2+1 & P'(2)&=13\\P''(x)&=6x & P''(2)&=12\\P'''(x)&=6 & P'''(2)&=6\\\\\end{align}\]
  • \[P(x) = \sum_{k=0}^n\frac{P^{(k)}(a)}{k!}(x-a)^k \\\\\text{Degree of $P(x)=n$ and $a=2$} \\\implies P(x) = \sum_{k=0}^3\frac{P^{(k)}(2)}{k!}(x-2)^k \\= \frac{P^{(0)}(2)}{0!}(x-2)^0+\frac{P^{(1)}(2)}{1!}(x-2)^1+\frac{P^{(2)}(2)}{2!}(x-2)^2+\frac{P^{(3)}(2)}{3!}(x-2)^3 \\= \frac{11}{1}(x-2)^0+\frac{13}{1}(x-2)^1+\frac{12}{2!}(x-2)^2+\frac{6}{3!}(x-2)^3 \\= 11+13(x-2)+6(x-2)^2+1(x-2)^3 \\\\\]
  • Answer: \(11+13(x-2)+6(x-2)^2+1(x-2)^3\)
• 2.) Express \(P(x)=x^4\) in \((x+3)\)
  • \[\begin{align}& & P(-3)&=81 \\P'(x)&=4x^3 & P'(-3)&=-108\\P''(x)&=12x^2 & P''(-3)&=108\\P'''(x)&=24x & P'''(-3)&=-72\\P''''(x)&=24 & P''''(-3)&=24\\\\\end{align}\]
  • \[P(x) = \frac{81}{0!}+\frac{-108}{1!}(x+3)+\frac{108}{2!}(x+3)^2+\frac{-72}{3!}(x+3)^3+\frac{24}{4!}(x+3)^4 \\= 81-108(x+3)+54(x+3)^2-12(x+3)^3+1(x+3)^4\\\\\]
  • Answer: \(81-108(x+3)+54(x+3)^2-12(x+3)^3+1(x+3)^4\)

Taylor Polynomials

Main Content

Definition:

A function \(f(x)\) is \(\underline{\textbf{infinitely differentiable}}\) if you can keep taking derivative after derivative of \(f(x)\) and obtain functions that are defined

Definition:

Let \(f(x)\) be infinitely differentiable at \(x=a\), the \(\underline{n^{\text{th}} \textbf{ order taylor polynomial of } f(x) \textbf{ at } x=a}\) is:

\[T_{(n,f,a)}(x)=\sum_{k=0}^n\frac{f^{(k)}(a)}{k!}(x-a)^k\]

Applications

• 1.) Find the \(n^{\text{th}}\) order taylor polynomial of \(f(x)=-\cos x\) at \(x=0\)
  • \[\begin{align}& & f(0)&=-1 \\f'(x)&=\sin x & f'(0)&=0\\f''(x)&=\cos x & f''(0)&=1\\f'''(x)&=-\sin x & f'''(0)&=0\\&\ldots & &\ldots \\\\\end{align}\]
  • \[\text{Since the pattern repeats for more derivatives:}\]
  • \[\begin{align}k\equiv 0\pmod 4 &\implies f^{(k)}(0)=-1 \\k\equiv 1\pmod 4 &\implies f^{(k)}(0)=0 \\k\equiv 2\pmod 4 &\implies f^{(k)}(0)=1 \\k\equiv 3\pmod 4 &\implies f^{(k)}(0)=0 \\\\\end{align}\]
  • \[\text{Calculating $T_{(n,f,0)}(x)$ for sample $n$:}\]
  • \[\begin{align} T_{(0,f,0)}(x) &= \frac{f(0)}{0!}(x-0)^0=-1 \\ T_{(1,f,0)}(x) &= \frac{f(0)}{0!}(x-0)^0+\frac{f'(0)}{1!}(x-0)^1=-1+0 \\ T_{(2,f,0)}(x) &= \frac{f(0)}{0!}(x-0)^0+\frac{f'(0)}{1!}(x-0)^1+\frac{f''(0)}{2!}(x-0)^2=-1+0+\frac{x^2}{2!} \\ T_{(3,f,0)}(x) &= -1+0+\frac{x^2}{2!}+\frac{f'''(0)}{3!}(x-0)^3=-1+0+\frac{x^2}{2!}+0 \\ T_{(4,f,0)}(x) &= -1+0+\frac{x^2}{2!}+0-\frac{x^4}{4!} \\ T_{(6,f,0)}(x) &= -1+0+\frac{x^2}{2!}+0-\frac{x^4}{4!}+\frac{x^6}{6!} \\ T_{(2n,f,0)}(x) &= \sum_{k=0}^n(-1)^{k+1}\frac{x^{2k}}{(2k)!} \\ T_{(n,f,0)}(x) &= -1+0+\frac{x^2}{2!}+0-\frac{x^4}{4!}+\frac{x^6}{6!}-\ldots \\\\ \end{align}\]
  • Answer: \(T_{(n,f,0)}(x) = -1+0+\frac{x^2}{2!}+0-\frac{x^4}{4!}+\frac{x^6}{6!}-\ldots\)
• 2.) Find the \(n^{\text{th}}\) order taylor polynomial of \(f(x)=e^x\) at \(x=1\)
  • \[f^{(k)}(x)=e^x\implies f^{(k)}(1)=e^1=e\]
  • \[T_{(n,f,1)}(x)=\sum_{k=0}^n\frac{f^{(k)}(1)}{k!}(x-1)^k=\sum_{k=0}^n\frac{e}{k!}(x-1)^k \\\\\]
  • Answer: \(T_{(n,f,1)}(x)=\sum_{k=0}^n\frac{e}{k!}(x-1)^k\)
• 3.) Find the \(n^{\text{th}}\) order taylor polynomial of \(f(x)=e^x\) at \(x=0\)
  • \[f^{(k)}(x)=e^x\implies f^{(k)}(0)=e^0=1\]
  • \[T_{(n,f,0)}(x)=\sum_{k=0}^n\frac{f^{(k)}(0)}{k!}(x-0)^k=\sum_{k=0}^n\frac{x^k}{k!} \\\\\]
  • Answer: \(T_{(n,f,0)}(x)=\sum_{k=0}^n\frac{x^k}{k!}\)
• 4.) Find the \(n^{\text{th}}\) order taylor polynomial of \(f(x)=\frac{1}{x}\) at \(x=1\)
  • \[\begin{align} f(x) &= \frac{1}{x} = x^{-1} \\ f'(x) &= -x^{-2} \\ f''(x) &= 2x^{-3} \\ f'''(x) &= -6x^{-4} \\ f^{(k)}(x) &= (-1)^kk!x^{-k-1} \\ f^{(k)}(1) &= (-1)^kk!1^{-k-1}= (-1)^kk! \\ a_k &= \frac{f^{(k)}(1)}{k!}=(-1)^k \\ T_{(n,f,1)}(x) &= 1-(x-1)+(x-1)^2-(x-1)^3+\ldots+(-1)^n(x-1)^n \\\\ \end{align}\]
  • Answer: \(T_{(n,f,1)}(x) = 1-(x-1)+(x-1)^2-(x-1)^3+\ldots+(-1)^n(x-1)^n\)
• 5.) Find the \(n^{\text{th}}\) order taylor polynomial of \(f(x)=\log(1+x)\) at \(x=0\)
  • \[\begin{align} & & f(0)&=0 \\ f'(x) &= \frac{1}{1+x}=(1+x)^{-1} & f'(0)&=1\\ f''(x) &= -1(1+x)^{-2} \cdot \frac{d}{dx}(1+x)=-(1+x)^{-2} & f''(0)&=-1\\ f'''(x) &= 2(1+x)^{-3} \cdot \frac{d}{dx}(1+x)=2(1+x)^{-3} & f'''(0)&=2\\ f''''(x) &= -6(1+x)^{-4} & f''''(0)&=-6\\ f^{(5)}(x) &= 24(1+x)^{-5} & f^{(5)}(0)&=24\\ &\ldots & &\ldots \\ f^{(k)}(x) &= (-1)^{k-1}(k-1)!(1+x)^{-k}\ \ (\forall k > 0) & f^{(k)}(0)&=(-1)^{k-1}(k-1)!\ \ (\forall k > 0)\\\\ \end{align}\]
  • \[T_{(n,f,0)}(x) = \sum_{k=0}^n\frac{f^{(k)}(0)}{k!}(x-0)^k \\\\ \text{Evaluating $k>0$ and $k=0$ separately:} \\ \text{by splitting the summation:}\]
  • \[\begin{align} T_{(n,f,0)}(x) &= \sum_{k=0}^0\frac{f^{(k)}(0)}{k!}(x-0)^k + \sum_{k=1}^n\frac{f^{(k)}(0)}{k!}(x-0)^k \\ &= \frac{f^{(0)}(0)}{0!}(x-0)^0 + \sum_{k=1}^n\frac{f^{(k)}(0)}{k!}(x-0)^k \\ &= 0 + \sum_{k=1}^n\frac{f^{(k)}(0)}{k!}(x-0)^k \\ &= \sum_{k=1}^n\frac{(-1)^{k-1}(k-1)!}{k!}x^k \\ &= \sum_{k=1}^n\frac{(-1)^{k-1}}{k}x^k \\\\ \end{align}\]
  • Answer: \(T_{(n,f,0)}(x)=\sum_{k=1}^n\frac{(-1)^{k-1}}{k}x^k\)
• 6.) Find the \(n^{\text{th}}\) order taylor polynomial of \(f(x)=xe^x\) at \(x=0\)
  • \[\begin{align} & & f(0)&=0 \\ f'(x) &= e^x+xe^x & f'(0)&=1\\ f''(x) &= e^x+e^x+xe^x & f''(0)&=2\\ &\ldots & &\ldots \\ f^{(k)}(x) &= ke^x+xe^x & f^{(k)}(0)&=k\\\\ \end{align}\]
  • \[\begin{align} T_{(n,f,0)}(x) &= \sum_{k=0}^n\frac{f^{(k)}(0)}{k!}(x-0)^k \\ &= \sum_{k=0}^n\frac{k}{k!}x^k \\\\ \end{align}\]
  • \[\text{Evaluating $k>0$ and $k=0$ separately,} \\ \text{by splitting the summation:}\]
  • \[\begin{align} T_{(n,f,0)}(x) &= \sum_{k=0}^0\frac{k}{k!}x^k + \sum_{k=1}^n\frac{k}{k!}x^k \\ &= \frac{0}{0!}x^0 + \sum_{k=1}^n\frac{k}{k!}x^k \\ &= 0 + \sum_{k=1}^n\frac{k}{k!}x^k \\ &= \sum_{k=1}^n\frac{x^k}{(k-1)!} \\\\ \end{align}\]
  • Answer: \(T_{(n,f,0)}(x)=\sum_{k=1}^n\frac{x^k}{(k-1)!}\)
• 7.) Find the \(n^{\text{th}}\) order taylor polynomial of \(f(x)=e^{2x}\) at \(x=0\)
  • \[\begin{align} f'(x) &= 2e^{2x} \\ f''(x) &= 4e^{2x} \\ f'''(x) &= 8e^{2x} \\ &\ldots \\ f^{(k)}(x) &= 2^ke^{2x} \\ f^{(k)}(0)&=2^k \\\\ \end{align}\]
  • \[\begin{align} T_{(n,f,0)}(x) &= \sum_{k=0}^n\frac{f^{(k)}(0)}{k!}(x-0)^k \\ &= \sum_{k=0}^n\frac{2^k}{k!}x^k \\\\ \end{align}\]
  • Answer: \(T_{(n,f,0)}(x)=\sum_{k=0}^n\frac{2^k}{k!}x^k\)
• 8.) Generalize results from Problem 7
  • \[\text{Finding the $n^{\text{th}}$ order taylor polynomial of $f(x)=e^{Cx}$,} \\ \text{at $x=0$ for some constant $C$:}\]
  • \[\begin{align} f'(x) &= Ce^{Cx} \\ f''(x) &= C^2e^{Cx} \\ f'''(x) &= C^3e^{Cx} \\ &\ldots \\ f^{(k)}(x) &= C^ke^{Cx} \\ f^{(k)}(0)&=C^k \\\\ \end{align}\]
  • \[\begin{align} T_{(n,f,0)}(x) &= \sum_{k=0}^n\frac{f^{(k)}(0)}{k!}(x-0)^k \\ &= \sum_{k=0}^n\frac{C^k}{k!}x^k \\\\ \end{align}\]
  • Answer: \(T_{(n,f,0)}(x)=\sum_{k=0}^n\frac{C^k}{k!}x^k\)
• 9.) Find the \(n^{\text{th}}\) order taylor polynomial of \(f(x)=e^x-\cos x\) at \(x=0\)
  • \[\begin{align} & & f(0)&=1-1=0 \\ f'(x)&=e^x+\sin x & f'(0)&=1+0=1\\ f''(x)&=e^x+\cos x & f''(0)&=1+1=2\\ f'''(x)&=e^x-\sin x & f'''(0)&=1-0=1\\ &\ldots & &\ldots \\\\ \end{align}\]
  • \[\text{Since the pattern repeats for more derivatives:}\]
  • \[\begin{align} k\equiv 0\pmod 4 &\implies f^{(k)}(0)=0 \\ k\equiv 1\pmod 4 &\implies f^{(k)}(0)=1 \\ k\equiv 2\pmod 4 &\implies f^{(k)}(0)=2 \\ k\equiv 3\pmod 4 &\implies f^{(k)}(0)=1 \\\\ \end{align}\]
  • \[\text{Calculating $T_{(n,f,0)}(x)$ for sample $n$:}\]
  • \[\begin{align} T_{(0,f,0)}(x) &= \frac{f(0)}{0!}(x-0)^0=0 \\ T_{(1,f,0)}(x) &= \frac{f(0)}{0!}(x-0)^0+\frac{f'(0)}{1!}(x-0)^1=0+x \\ T_{(2,f,0)}(x) &= \frac{f(0)}{0!}(x-0)^0+\frac{f'(0)}{1!}(x-0)^1+\frac{f''(0)}{2!}(x-0)^2=0+x+\frac{2}{2!}x^2 \\ T_{(3,f,0)}(x) &= 0+x+\frac{2}{2!}x^2+\frac{f'''(0)}{3!}(x-0)^3=0+x+\frac{2}{2!}x^2+\frac{1}{3!}x^3 \\ T_{(n,f,0)}(x) &= 0+x+\frac{2}{2!}x^2+\frac{1}{3!}x^3+\ldots \\\\ \end{align}\]
  • Answer: \(T_{(n,f,0)}(x) = 0+x+\frac{2}{2!}x^2+\frac{1}{3!}x^3+\ldots\)
• 10.) Generalize results from Problem 8
  • \[\text{Using results from Problem 8:}\]
  • \[\begin{align} f(x) &= e^x-\cos x \\ T_{(n,f,0)}(x) &= 0+x+\frac{2}{2!}x^2+\frac{1}{3!}x^3+\ldots \\\\ \end{align}\]
  • \[\text{Using results from Problem 1:}\]
  • \[\begin{align} g(x)&=-\cos x \\ T_{(n,g,0)}(x) &= -1+0+\frac{x^2}{2!}+0-\frac{x^4}{4!}+\frac{x^6}{6!}-\ldots \\\\ \end{align}\]
  • \[\text{Using results from Problem 3:}\]
  • \[\begin{align} h(x)&=e^x \\ T_{(n,h,0)}(x)&=\sum_{k=0}^n\frac{x^k}{k!} \\\\ \end{align}\]
  • \[\text{Combining results:}\]
  • \[\begin{align} f(x) &= e^x-\cos x \\ &= h(x)+g(x) \\\\ \end{align}\]
  • \[\text{Rewrite $T_{(n,f,0)}(x)$:}\]
  • \[\begin{align} T_{(n,f,0)}(x) &= 0+x+\frac{2}{2!}x^2+\frac{1}{3!}x^3+\ldots \\ &= \left(-1+0+\frac{x^2}{2!}+0-\frac{x^4}{4!}+\frac{x^6}{6!}-\ldots\right) + \left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots\right) \\ &= \left(-1+0+\frac{x^2}{2!}+0-\frac{x^4}{4!}+\frac{x^6}{6!}-\ldots\right) + \sum_{k=0}^n\frac{x^k}{k!} \\ &= T_{(n,g,0)}(x) + T_{(n,h,0)}(x) \\\\ \end{align}\]
  • \[\text{Therefore we can conclude:}\]
  • \[f(x)=g(x)+h(x)\implies T_{(n,f,0)}(x) = T_{(n,g,0)}(x) + T_{(n,h,0)}(x) \\\\\]
  • Answer: \(f(x)=g(x)+h(x)\implies T_{(n,f,0)}(x) = T_{(n,g,0)}(x) + T_{(n,h,0)}(x)\)
• 11.) Find the \(n^{\text{th}}\) order taylor polynomial of \(f(x)=79e^x\) at \(x=0\)
  • \[\begin{align} f'(x) &= 79e^x \\ f''(x) &= 79e^x \\ f^{(k)}(x) &= 79e^x \\ f^{(k)}(0) &= 79 \\\\ \end{align}\]
  • \[\begin{align} T_{(n,f,0)}(x) &= \sum_{k=0}^n\frac{f^{(k)}(0)}{k!}x^k \\ &= \sum_{k=0}^n\frac{79}{k!}x^k \\\\ \end{align}\]
  • Answer: \(T_{(n,f,0)}(x) = \sum_{k=0}^n\frac{79}{k!}x^k\)
• 12.) Generalize results from Problem 11
  • \[\text{Finding the $n^{\text{th}}$ order taylor polynomial of $f(x)=Ce^x$,} \\ \text{at $x=0$ for some constant $C$:}\]
  • \[\begin{align} f'(x) &= Ce^x \\ f''(x) &= Ce^x \\ f^{(k)}(x) &= Ce^x \\ f^{(k)}(0) &= C \\\\ \end{align}\]
  • \[\begin{align} T_{(n,f,0)}(x) &= \sum_{k=0}^n\frac{f^{(k)}(0)}{k!}x^k \\ &= \sum_{k=0}^n\frac{C}{k!}x^k \\\\ \end{align}\]
  • Answer: \(T_{(n,f,0)}(x) = \sum_{k=0}^n\frac{C}{k!}x^k\)

Theorem

Introduction

\[\text{Let $f(x)=e^x$:} \\ T_{(n,f,0)}(x)=\sum_{k=0}^n\frac{x^k}{k!} \\ \text{To find the following limits we apply L'Hôpital's rule (several times),} \\ \text{because both the numerator and denominator $\to 0$ as $x\to0$:}\] \[\begin{align} \lim_{x\to0}\frac{e^x-T_{(0,f,0)}(x)}{x^0}&=\lim_{x\to0}\frac{e^x-1}{x^0}=0 \\ \lim_{x\to0}\frac{e^x-T_{(1,f,0)}(x)}{x^1}&=\lim_{x\to0}\frac{e^x-(1+x)}{x}=0 \\ \lim_{x\to0}\frac{e^x-T_{(2,f,0)}(x)}{x^2}&=\lim_{x\to0}\frac{e^x-(1+x+\frac{x^2}{2!})}{x^2}=0 \\ \lim_{x\to0}\frac{e^x-T_{(3,f,0)}(x)}{x^3}&=\lim_{x\to0}\frac{e^x-(1+x+\frac{x^2}{2!}+\frac{x^3}{3!})}{x^3}=0 \\\\ \end{align}\] \[\text{$\implies e^x-(1+x+\frac{x^2}{2!}+\frac{x^3}{3!})$ is very small compared to $x^3$ as $x\to0$} \\\\ \text{This leads us to generalize to the following theorem...}\]

Main Content

Theorem:

\[\lim_{x\to a}\frac{f(x)-T_{(n,f,a)}(x)}{(x-a)^n}=0\]

Proof.

\[T_{(n,f,a)}(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!} (x-a)^k \\ \lim_{x\to a}\frac{f(x)-T_{(n,f,a)}(x)}{(x-a)^n} = \lim_{x\to a} \frac{f(x) - \sum_{k=0}^n \frac{f^{(k)}(a)}{k!} (x-a)^k}{(x-a)^n} \\\\\] \[\text{Take note of the following limits:}\] \[\begin{align} \lim_{x\to a} \sum_{k=0}^n \frac{f^{(k)}(a)}{k!} (x-a)^k &= \lim_{x\to a} \sum_{k=0}^0 \frac{f^{(k)}(a)}{k!} (x-a)^k + \lim_{x\to a} \sum_{k=1}^n \frac{f^{(k)}(a)}{k!} (x-a)^k \\ &= \lim_{x\to a} \frac{f^{(0)}(a)}{0!} (x-a)^0 + \lim_{x\to a} \sum_{k=1}^n \frac{f^{(k)}(a)}{k!} (x-a)^k \\ &= f(a) + \sum_{k=1}^n \frac{f^{(k)}(a)}{k!} \lim_{x\to a} (x-a)^k \\ &= f(a) + \sum_{k=1}^n \frac{f^{(k)}(a)}{k!} \cdot 0 &&\text{(since $k\geq 1$)}\\ &= f(a) \\\\ \lim_{x\to a} \left(f(x) - \sum_{k=0}^n \frac{f^{(k)}(a)}{k!} (x-a)^k \right) &= \lim_{x\to a} f(x) - \lim_{x\to a} \sum_{k=0}^n \frac{f^{(k)}(a)}{k!} (x-a)^k \\ &= f(a)-f(a) \\ &= 0 \\\\ \lim_{x\to a} (x-a)^n &= 0 \\\\ \end{align}\] \[\text{Since both numerator and denominator $\to 0$, apply L'Hôpital's rule:}\] \[\begin{align} \lim_{x\to a} \frac{f(x) - \sum_{k=0}^n \frac{f^{(k)}(a)}{k!} (x-a)^k}{(x-a)^n} &= \lim_{x\to a} \frac{\frac{d}{dx}\left( f(x) - \sum_{k=0}^n \frac{f^{(k)}(a)}{k!} (x-a)^k \right)}{\frac{d}{dx}\left( (x-a)^n \right)} \\ &= \lim_{x\to a} \frac{f'(x) - \sum_{k=1}^n \frac{f^{(k)}(a)}{(k-1)!} (x-a)^{k-1}}{n(x-a)^{n-1}} \\\\ \end{align}\] \[\text{Apply L'Hôpital's rule $n$ times, and take $n$ derivatives:}\] \[\begin{align} &= \lim_{x\to a} \frac{f^{(n)}(x)-f^{(n)}(a)}{n!} \\ &= 0 \ \ \ \ \blacksquare\\\\ \end{align}\]

Theorem

Main Content

Definition:

a function \(\underline{f(x)\textbf{ and } g(x) \textbf{ are equal up to order } n \textbf{ at } a}\) if \(\lim_{x\to a} \frac{f(x)-g(x)}{(x-a)^n} = 0\)

Theorem:

Let \(P\) and \(Q\) be two polynomials in \((x-a)\), of degree \(\leq n\), and suppose that \(P\) and \(Q\) are equal up to order \(n\) at \(a\). Then \(P=Q\).

Proof

\[\text{Let $R=P-Q$:}\] \[\begin{align} deg(P)\leq n\ \land\ deg(Q)\leq n &\implies deg(R)\leq n \\ &\implies R(x) = b_0 + \ldots + b_n(x-a)^n \end{align}\] \[\text{$P$ and $Q$ equal up to order $n$ at $a$:}\] \[\begin{align} &\implies \lim_{x\to a} \frac{P(x)-Q(x)}{(x-a)^n} = 0 \\ &\implies \lim_{x\to a} \frac{R(x)}{(x-a)^n} = 0 \\\\ \end{align}\] \[\text{WWTP: $R(x)=0\implies P-Q = 0 \implies P=Q$} \\\\ \lim_{x\to a} \frac{R(x)}{(x-a)^n} = 0 \implies \forall\ 0 \leq i \leq n,\ \lim_{x\to a} \frac{R(x)}{(x-a)^i} = 0 \\\\ \text{For $i=0$:} \\ \lim_{x\to a} \frac{R(x)}{(x-a)^0} = 0 \\ \lim_{x\to a} R(x) = 0 \\\\\] \[\text{Using $R(x)$ equation:}\] \[\begin{align} R(x) &= b_0 + \ldots + b_n(x-a)^n \\ \lim_{x\to a} R(x) &= \lim_{x\to a} \left( b_0 + \ldots + b_n(x-a)^n \right) \\ \lim_{x\to a} R(x) &= b_0 \\ 0 &= b_0 \\\\ \end{align}\] \[\text{Using the fact $b_0=0$:}\] \[\begin{align} R(x) &= b_1(x-a) + \ldots + b_n(x-a)^n \\ \frac{R(x)}{(x-a)} &= b_1 + \ldots + b_n(x-a)^{n-1} \\\\ \end{align}\] \[\text{For $i=1$:} \\ \lim_{x\to a} \frac{R(x)}{(x-a)^1} = 0 \\\\\] \[\text{Using $R(x)$ equation:}\] \[\begin{align} \frac{R(x)}{(x-a)} &= b_1 + \ldots + b_n(x-a)^{n-1} \\ \lim_{x\to a} \frac{R(x)}{(x-a)} &= \lim_{x\to a} \left( b_1 + \ldots + b_n(x-a)^{n-1} \right) \\ 0 &= b_1 \\\\ \end{align}\] \[\text{After continuiing in this way, $n$ times later:}\] \[\begin{align} b_0 &= b_1 = \ldots = b_n = 0 \\ \implies R(x) &= 0 + \ldots + 0(x-a)^n \\ &= 0 \ \ \ \ \blacksquare\\\\ \end{align}\]

Corollary

Main Content

Corollary:

Let \(f\) be \(n\)-times differentiable at \(a\), and suppose \(P\) is a polynomial in \((x-a)\) of degree \(\leq n\), which equals \(f\) up to order \(n\) at \(a\). Then \(P=T_{(n,f,a)}\).

Proof.

\[\text{By the previous theorem:} \\ \lim_{x\to a} \frac{f(x)-T_{(n,f,a)}(x)}{(x-a)^n} = 0 \\\\\] \[\text{By assumption,} \\ \text{$P$ is equal to $f$ up to order $n$ at $a$.}\] \[\begin{align} &\implies \lim_{x\to a} \frac{f(x)-P(x)}{(x-a)^n} = 0 \\ &\implies \lim_{x\to a} \frac{f(x)-P(x)}{(x-a)^n} = \lim_{x\to a} \frac{f(x)-T_{(n,f,a)}(x)}{(x-a)^n} \\ &\implies \lim_{x\to a} \frac{T_{(n,f,a)}(x)-P(x)}{(x-a)^n} = 0 \\ &\implies T_{(n,f,a)}(x) \text{ and } P(x) \text{ are equal up to order } n \text{ at } a \\\\ \end{align}\] \[\text{By previous theorem:} \\ \implies T_{(n,f,a)}(x) = P(x) \ \ \ \ \blacksquare\\\\\]

Applications

• 1.) Find the Taylor polynomial of \(f(x)=e^{(x^2)}\)
  • \[\text{From previously derived Talyor polynomial of $e^x$:} \\ \lim_{x\to 0} \frac{e^x - \left( 1+x+\frac{x^2}{2!}+\ldots+\frac{x^n}{n!} \right)}{x^n} = 0 \\\\\]
  • \[\text{Since $x^2\to 0$ as $x\to 0$, plug in for $x^2$:}\]
  • \[\begin{align} &\implies \lim_{x\to 0} \frac{e^{(x^2)} - \left( 1+x^2+\frac{x^4}{2!}+\ldots+\frac{x^{2n}}{n!} \right)}{x^{2n}} = 0 \\ &\implies e^{(x^2)} \text{ is equal to } \left( 1+x^2+\frac{x^4}{2!}+\ldots+\frac{x^{2n}}{n!} \right) \text{ up to order } 2n \text{ at } 0 \\ &\implies T_{(2n,f,0)}(x) = 1+x^2+\frac{x^4}{2!}+\ldots+\frac{x^{2n}}{n!} \\\\ \end{align}\]
  • Answer: \(T_{(2n,f,0)}(x) = 1+x^2+\frac{x^4}{2!}+\ldots+\frac{x^{2n}}{n!}\)
• 2.) Find the Taylor polynomial of \(f(x)=xe^{(x^2)}\)
  • \[\text{From previously derived Talyor polynomial of $e^{(x^2)}$:} \\ \lim_{x\to 0} \frac{e^{(x^2)} - \left( 1+x^2+\frac{x^4}{2!}+\ldots+\frac{x^{2n}}{n!} \right)}{x^{2n}} = 0 \\\\\]
  • \[\text{Multiplying top and bottom by $x$:}\]
  • \[\begin{align} &\implies \lim_{x\to 0} \frac{xe^{(x^2)} - x\left( 1+x^2+\frac{x^4}{2!}+\ldots+\frac{x^{2n}}{n!} \right)}{x^{2n+1}} = 0 \\ &\implies \lim_{x\to 0} \frac{xe^{(x^2)} - \left( x+x^3+\frac{x^5}{2!}+\ldots+\frac{x^{2n+1}}{n!} \right)}{x^{2n+1}} = 0 \\ &\implies xe^{(x^2)} \text{ is equal to } \left( x+x^3+\frac{x^5}{2!}+\ldots+\frac{x^{2n+1}}{n!} \right) \text{ up to order } 2n+1 \text{ at } 0 \\ &\implies T_{(2n+1,f,0)}(x) = x+x^3+\frac{x^5}{2!}+\ldots+\frac{x^{2n+1}}{n!} \\\\ \end{align}\]
  • Answer: \(T_{(2n+1,f,0)}(x) = x+x^3+\frac{x^5}{2!}+\ldots+\frac{x^{2n+1}}{n!}\)
• 3.) Generalize results from Problems 1 and 2
  • Answer: To find Taylor polynomials of functions that are similar to \(e^x\), utilize Taylor polynomial of \(e^x\) and either plug in for \(x\) values that approach 0, or multiply top and bottom of limit by chosen value.

Remainder Term

Main Content

Definition:

If \(f\) is a function for which \(T_{(n,f,a)}\) exists, we define the \(\underline{\textbf{remainder term}}\ R_{(n,f,a)}\) by

\[\begin{align} f(x) &= T_{(n,f,a)} + R_{(n,f,a)} \\ &= f(a) + f'(a)(x-a) + \ldots + \frac{f^{(n)}(a)}{n!}(x-a)^n + R_{(n,f,a)} \\ \end{align}\]

Theorem:

\[R_{(n,f,a)} = \int_a^x \frac{f^{(n+1)}(t)}{n!} (x-t)^n dt\]

Proof.

\[\text{Refer to rough notes page 3, Spivak textbook pages 422-423 for proof.}\]

Theorem:

Suppose that \(f',\ldots,f^{(n+1)}\) are defined on \([a,x]\), and that \(R_{(n,f,a)}\) is defined by

\[f(x) = f(a) + f'(a)(x-a) + \ldots + \frac{f^{(n)}(a)}{n!}(x-a)^n + R_{(n,f,a)}\]

Then the \(\underline{\textbf{Lagrange form of the remainder}}\) is

\[R_{(n,f,a)} = \frac{f^{(n+1)}(t)}{(n+1)!} (x-a)^{n+1}\ \text{for some $t$ in $(a,x)$}\]

Proof.

\[\text{Refer to rough notes page 3, Spivak textbook pages 422-424 for proof.}\]

Theorem:

\(e\) is irrational.

Proof.

\[\text{Refer to rough notes page 4, Spivak textbook page 429 for proof.}\]

Applications

• 1.) Apply Lagrange remainder to find formulas for \(\sin x, \cos x, e^x\)
  • \[\text{Refer to Spivak textbook page 424 for answer.}\]
• 2.) Approximate \(\sin(2)\) with an error of less than \(10^{-4}\)
  • \[\text{Refer to Spivak textbook page 427 for answer.}\]
• 3.) Approximate \(\int_0^{0.1}e^{(x^2)}dx\) with an error of less than \(10^{-4}\)
  • \[\text{Refer to rough notes page 4 for answer.}\]