Find the Taylor Series expansion of \(f(x)=\frac{1}{1-x}\)
If \(f(x)=\frac{1}{1-x}\), then…
\(f'(x)=(-1)(1-x)^{-2}(-1)=(-1)^2(1-x)^{-2}\\f''(x)=(-2)(-1)^2(1-x)^{-3}(-1)=(2)(-1)^4(1-x)^{-3}\\f'''(x)=(-3)(2)(-1)^4(1-x)^{-4}(-1)=(3)(2)(-1)^6(1-x)^{-4}\\f''''(x)=(-4)(3)(2)(-1)^6(1-x)^{-5}(-1)=(4)(3)(2)(-1)^8(1-x)^{-5}\)
We can see from this pattern that the \(n\)th derivative of \(f(x)\), \(f^{(n)}(x)\), will be \(f^{(n)}(x)=n!(-1)^{2n}(1-x)^{-n-1}=n!(1-x)^{-n-1}\)
When \(x=0\), \(f^{(n)}(x)=n!\) since \((1-0)^{-n-1}\) must be 1, and this results in the following expansion:
\(1+\frac{1!}{1!}x+\frac{2!}{2!}x^2+\frac{3!}{3!}x^3+...\\=1+x+x^2+x^3+...\)
Therefore, the Taylor Series expansion of \(f(x)=\frac{1}{1-x}\) is \(\sum\limits_{n=0}^\infty x^n\)
Find the Taylor Series expansion of \(f(x)=e^x\)
If \(f(x)=e^x\), then the \(n\)th derivative of \(f(x)\), \(f^{(n)}(x)\), will be \(f^{(n)}(x)=e^x\) for all \(n\).
When \(x=0\), \(f^{(n)}(x)=1\) since \(e^0=1\), and this results in the following expansion:
\(1+x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+\frac{1}{4!}x^4+...\)
Therefore, the Taylor Series expansion of \(f(x)=e^x\) is \(\sum\limits_{n=0}^\infty \frac{1}{n!}x^n=\sum\limits_{n=0}^\infty \frac{x^n}{n!}\)
Find the Taylor Series expansion of \(f(x)=ln(1+x)\)
If \(f(x)=ln(1+x)\), then…
\(f'(x)=(1+x)^{-1}\\f''(x)=(-1)(1+x)^{-2}\\f'''(x)=(2)(-1)^2(1+x)^{-3}\\f''''(x)=(3)(2)(-1)^3(1+x)^{-4}\)
We can see from this pattern that the \(n\)th derivative of \(f(x)\), \(f^{(n)}(x)\), will be \(f^{(n)}(x)=-1^{n+1}(n-1)!(1+x)^{-n}\)
When \(x=0\), \(f^{(n)}(x)=-1^{n+1}(n-1)!\) since \((1+0)^{-n-1}\) must be 1, and this results in the following expansion:
\(0+(-1)^2\frac{0!}{1!}x+(-1)^3\frac{1!}{2!}x^2+(-1)^4\frac{2!}{3!}x^3+(-1)^5\frac{3!}{4!}x^4+...\\=0+x-\frac{1}{2}x^2+\frac{1}{3}x^3-\frac{1}{4}x^4+...\)
Therefore, the Taylor Series expansion of \(f(x)=ln(1+x)\) is \(\sum\limits_{n=1}^\infty -1^{n+1}\cdot \frac{1}{n}x^n=\sum\limits_{n=1}^\infty -\frac{(-x)^n}{n}\)
Find the Taylor Series expansion of \(f(x)=x^{1/2}\)
If \(f(x)=x^{1/2}\), then…
\(f'(x)=\frac{1}{2}(x)^{-1/2}\\f''(x)=(-1)\frac{1}{4}(x)^{-3/2}\\f'''(x)=(-1)^2\frac{3}{8}(x)^{-5/2}\\f''''(x)=(-1)^3\frac{15}{16}(x)^{-7/2}\)
When \(x=1\), this results in the following expansion:
\(1+\frac{1}{2\cdot 1!}x-\frac{1}{4\cdot 2!}x^2+\frac{3}{8\cdot3!}x^3-\frac{15}{16\cdot4!}x^4+...\\=1+\frac{1}{2}x-\frac{1}{8}x^2+\frac{1}{16}x^3-\frac{5}{128}x^4+...\)