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Taylor Polynomials

While linear approximations (Linearization) can give us great approximation, we can get even better approximations by using more derivatives!

The nth Degree Taylor Polynomial

The polynomial of degree n which approximates f(x)f\left(x\right) aroundx=x0x=x_0 is the nth degree Taylor Polynomial for f(x)f\left(x\right) at x0x_0, and is given by
Tn=f(x0)+f′(x0)1!(x−x0)+f′′(x0)2!(x−x0)2+...+f(n)(x0)n!(x−x0)n\boxed{T_n=f\left(x_0\right)+\frac{f'\left(x_0\right)}{1!}\left(x-x_0\right)+\frac{f''\left(x_0\right)}{2!}\left(x-x_0\right)^2+...+\frac{f^{\left(n\right)}\left(x_0\right)}{n!}\left(x-x_0\right)^n}
The Taylor Polynomial about x0=0x_0=0 is called the Maclaurin Polynomial.

Note: Sometimes the letter aa will be used instead of x0x_0.

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Example: Taylor Polynomials

Find the third degree Taylor polynomial of f(x)=ln⁡(1−2x) at x=0.f(x)=\ln(1−2x)\ \text{at}\ x=0.

f(0)=ln⁡(1−2(0))=0f′(x)=11−2x(−2)=−2(1−2x)−1  ⟹  f′(0)=−2(1−2(0))−1=−2f′′(x)=2(1−2x)−2(−2)  ⟹  f′′(0)=−4(1−2(0))−2=−4f′′′(x)=8(1−2x)−3(−2)  ⟹  f′′′(0)=−16((1−2(0)))−3=−16  ⟹  T3=0+−21!(x−0)1+−42!(x−0)2+−163!(x−0)3=−2x−2x2−83x3\begin{array}{l} \displaystyle f\left(0\right)=\ln\left(1-2\left(0\right)\right)=\boxed0 \\ \\ \displaystyle f'\left(x\right)=\frac{1}{1-2x}\left(-2\right)=-2\left(1-2x\right)^{-1} \implies f'\left(0\right)=-2\left(1-2\left(0\right)\right)^{-1}= \boxed{-2} \\ \\ f''\left(x\right)=2\left(1-2x\right)^{-2}\left(-2\right) \implies f''\left(0\right)=-4\left(1-2\left(0\right)\right)^{-2}=\boxed{-4} \\ \\ f'''\left(x\right)=8\left(1-2x\right)^{-3}\left(-2\right) \implies f'''\left(0\right)=-16\left(\left(1-2\left(0\right)\right)\right)^{-3}=\boxed{-16} \\ \\ \\ \implies \displaystyle T_3=\boxed0+\frac{\boxed{-2}}{1!}\left(x-0\right)^1+\frac{\boxed{-4}}{2!}\left(x-0\right)^2+\frac{\boxed{-16}}{3!}\left(x-0\right)^3 \\ \\ \displaystyle =-2x-2x^2-\frac{8}{3}x^3 \end{array}
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Example: Maclaurin Polynomials

Find the 4th degree Maclaurin Polynomial for f(x)=cos⁡3xf(x)=\cos{3x}.


Remember a Maclaurin Polynomial is a Taylor Polynomial with x0=0x_0=0.

f(0)=cos⁡(0)=1f′(x)=−3sin⁡(3x)  ⟹  f′(0)=−3sin⁡(0)=0f′′(x)=−9cos⁡(3x)  ⟹  f′′(0)=−9cos⁡(0)=−9f′′′(x)=27sin⁡(3x)  ⟹  f′′′(0)=27sin⁡(0)=0f(4)(x)=81cos⁡(3x)  ⟹  f(4)(0)=81cos⁡(0)=81  ⟹  M4(x)=1+0(x−0)11!+−9(x−0)22!+0(x−3)33!+81(x−0)44!=1−9x22!+81x44!\begin{array}{l} f(0)=\cos (0)=\boxed1 \\ \\ f'(x)=-3\sin(3x) \implies f'(0)=-3\sin(0)=\boxed0 \\ \\ f''(x)=-9 \cos(3x) \implies f''(0)= -9\cos(0)=\boxed{-9} \\ \\ f'''(x)=27\sin(3x) \implies f'''(0)=27\sin(0)=\boxed0 \\ \\ f^{(4)}(x)=81\cos(3x) \implies f^{(4)}(0)=81\cos(0)=\boxed{81} \\ \\ \\ \displaystyle \implies M_4(x)=\boxed1+\frac{\boxed0 (x-0)^1}{1!} +\frac{\boxed{-9}(x-0)^2}{2!}+\frac{\boxed0(x-3)^3}{3!}+\frac{\boxed{81}(x-0)^4}{4!} \\ \\ =1 - \dfrac{9x^2}{2!}+\dfrac{81x^4}{4!} \end{array}



Find the 2nd degree Taylor polynomial of f(x)=e2xf\left(x\right)=e^{2x}centered at a=ln⁡3a=\ln\sqrt{3}.

Extra Practice