Calculus II - Taylor (Maclaurin) Series

12 flashcards · Mathematics · by luckydog789 (harrison holbrook)

Taylor series
Suppose fff is a function that has derivatives of all orders on the open interval (c−R,c+R)(c-R,c+R)(c−R,c+R) , 0"> R>0R>0R>0 . If fff can be represented by the power series ∑k=0∞ak(x−c)k\sum_{k=0}^\infty a_k(x-c)^k∑k=0∞ak(x−c)k , whose radius of convergence is RRR , then its Taylor series is f(x)=∑k=0∞f(k)(c)k!(x−c)k=f(c)+f′(c)(x−c)+f′′(c)2!(x−c)2...f(x)=\sum_{k=0}^\infty \frac{f^{(k)}(c)}{k!}(x-c)^k=f(c)+f'(c)(x-c)+\frac{f''(c)}{2!}(x-c)^2...f(x)=∑k=0∞k!f(k)(c)(x−c)k=f(c)+f′(c)(x−c)+2!f′′…
Maclaurin series
A Taylor series centered around 000 ( c=0c=0c=0 ), f(x)=∑k=0∞f(k)(0)k!xk=f(0)+f′(0)x+f′′(0)2!x2...f(x)=\sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!}x^k=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2...f(x)=∑k=0∞k!f(k)(0)xk=f(0)+f′(0)x+2!f′′(0)x2...
Taylor polynomial
The first nnn -terms of a Taylor series, Pn(x)=∑k=0nf(k)(c)k!(x−c)kP_n(x)=\sum_{k=0}^n \frac{f^{(k)}(c)}{k!}(x-c)^kPn(x)=∑k=0nk!f(k)(c)(x−c)k
Taylor's formula with the Lagrange form of the remainder
Let fff be a function whose first n+1n+1n+1 derivatives are continuous on an open interval III containing the number ccc . Then for every xxx in III , there is a number uuu between ccc and xxx for which f(x)=∑k=0nf(k)(c)k!(x−c)k+Rn(x)=Pn(x)+Rn(x)f(x)=\sum_{k=0}^n \frac{f^{(k)}(c)}{k!}(x-c)^k+R_n(x)=P_n(x)+R_n(x)f(x)=∑k=0nk!f(k)(c)(x−c)k+Rn(x)=Pn(x)+Rn(x) Where Rn(x)=f(n+1)(u)(n+1)!(x−c)n+1R_n(x)=\frac{f^{(n+1)}(u)}{(n+1)!}(x-c)^{n+1}Rn(x)=(n+1)!f(n+1)(u)(x−c)n+1 is the Lagrange form of t…
Convergence of a Taylor series
If a function fff has derivatives of all orders in an open interval I=(c−R,c+R)I=(c-R,c+R)I=(c−R,c+R) , 0"> R>0R>0R>0 , centered at ccc , and if lim⁡n→∞Rn(x)=lim⁡n→∞f(n+1)(u)(n+1)!(x−c)n+1=0\lim_{n\to\infty}R_n(x)=\lim_{n\to\infty}\frac{f^{(n+1)}(u)}{(n+1)!}(x-c)^{n+1}=0limn→∞Rn(x)=limn→∞(n+1)!f(n+1)(u)(x−c)n+1=0 for all numbers xxx in III , where uuu is a number between ccc and xxx , then f(x)=∑k=0∞f(k)(c)k!(x−c)kf(x)=\sum_{k=0}^\infty \frac{f^{(k)}(c)}{k!}(x-c)^kf(x)=∑k=0∞k!f(k)(c)(x−c)…
Expansion of 11−x\frac{1}{1-x}1−x1
11−x=∑k=0∞xk=1+x+x2...\frac{1}{1-x}=\sum_{k=0}^\infty x^k=1+x+x^2...1−x1=∑k=0∞xk=1+x+x2... for ∣X∣<1|X| < 1∣X∣<1
Expansion of ln⁡11−x\ln\frac{1}{1-x}ln1−x1
ln⁡11−x=∑k=1∞xkk=x+x22+x33...\ln\frac{1}{1-x}=\sum_{k=1}^\infty \frac{x^k}{k}=x+\frac{x^2}{2}+\frac{x^3}{3}...ln1−x1=∑k=1∞kxk=x+2x2+3x3... for [−1,1)[-1,1)[−1,1)
Expansion of ln⁡(1+x)\ln(1+x)ln(1+x)
ln⁡(1+x)=∑k=0∞(−1)kxk+1k+1\ln(1+x)=\sum_{k=0}^\infty \frac{(-1)^kx^{k+1}}{k+1}ln(1+x)=∑k=0∞k+1(−1)kxk+1 ......... for (−1,1](-1,1](−1,1]
Expansion for tan⁡−1x\tan^{-1}xtan−1x
tan⁡−1x=∑k=0∞(−1)kx2k+12k+1=x−x33+x55...\tan^{-1}x=\sum_{k=0}^\infty (-1)^k\frac{x^{2k+1}}{2k+1}=x-\frac{x^3}{3}+\frac{x^5}{5}...tan−1x=∑k=0∞(−1)k2k+1x2k+1=x−3x3+5x5... for [−1,1][-1,1][−1,1]
Expansion for exe^xex
ex=∑k=0∞xkk!=1+x+x22!+x33!...e^x=\sum_{k=0}^\infty \frac{x^k}{k!}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}...ex=∑k=0∞k!xk=1+x+2!x2+3!x3... for all R\mathbb{R}R
Expansion of sin⁡x\sin{x}sinx
sin⁡x=∑k=0∞(−1)kx2k+1(2k+1)!=x−x33!+x55!...\sin{x}=\sum_{k=0}^\infty\frac{(-1)^kx^{2k+1}}{(2k+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}...sinx=∑k=0∞(2k+1)!(−1)kx2k+1=x−3!x3+5!x5... for all R\mathbb{R}R
Expansion of cos⁡x\cos{x}cosx
cos⁡x=∑k=0∞(−1)kx2k(2k)!=1−x22!+x44!...\cos{x}=\sum_{k=0}^\infty\frac{(-1)^kx^{2k}}{(2k)!}=1-\frac{x^2}{2!}+\frac{x^4}{4!}...cosx=∑k=0∞(2k)!(−1)kx2k=1−2!x2+4!x4... for all R\mathbb{R}R