Taylor/Maclaurin Series
Calculus
Intro: We display symbolic terms and a numeric check for a given x.
Worked example
- Expand $e^x$ at $x=0$ to $N=5$
- Goal: Find the Maclaurin (Taylor at 0) series for $f(x)=e^x$ and list the first $N=5$ nonzero terms.
- Definition (Maclaurin): $\displaystyle f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n$.
- Derivatives of $e^x$: $f^{(n)}(x)=e^x$ for all $n\ge0$, hence $f^{(n)}(0)=e^0=1$ for all $n$.
- Plug into the formula: $\displaystyle e^x=\sum_{n=0}^{\infty}\frac{1}{n!}x^n=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots$.
- List the first 5 terms (through $n=5$): $\displaystyle 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!}$.
- Compact result up to $N=5$: $\displaystyle \boxed{e^x \approx 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120}}$.
- Remainder (Lagrange form, optional): $\displaystyle R_5(x)=\frac{e^{\xi}}{6!}x^6$ for some $\xi$ between $0$ and $x$. This bounds the truncation error after 5 terms.
- Quick numeric check at $x=1$ (optional): partial sum $S_5=1+1+\tfrac{1}{2}+\tfrac{1}{6}+\tfrac{1}{24}+\tfrac{1}{120}=2.716666\ldots$ while $e\approx2.7182818\ldots$, error $\approx1.615\times10^{-3}$, consistent with a small positive remainder.
FAQs
Remainder term?
We can estimate truncation error if requested.
Why choose Mathix?
- Get clear, step-by-step solutions that explain the “why,” not just the answer.
- See the rules used at each step (power, product, quotient, chain, and more).
- Optional animated walk-throughs to make tricky ideas click faster.
- Clean LaTeX rendering for notes, homework, and study guides.
Taylor expansion
- f(x)=\sum_{n=0}^\infty f^{(n)}(0)x^n/n!
- For e^x, every derivative at 0 equals 1
- 1 + x + x^2/2! + x^3/3! + x^4/4! + x^5/5!