The Standard Series Toolkit
Five Maclaurin series — for \(e^x\), \(\sin x\), \(\cos x\), \(\frac{1}{1-x}\), and \(\ln(1+x)\) — are worth memorizing outright: substitute, multiply, differentiate, or integrate them and you can generate a huge family of other series without ever computing a derivative.
By the end you'll be able to recall the five standard series with their intervals of convergence, and build new Maclaurin series from them by substitution and other term-by-term operations instead of recomputing derivatives.
Predict: substituting −x² into the eˣ series — do all powers of x survive in the new series, or only the even ones? Apply the default substitution below to check, then hover any generated term to see which term of the base series it came from.
Pick a base series card, choose a substitution, and watch the new series get built term by term. The dashed partial sum hugs the true function only inside the shaded interval of convergence — drag the terms slider up and watch the fit improve there (and only there).
New series for e−x² = 1 − x² + x⁴/2! − x⁶/3! + ⋯ · valid for all x
In the last lesson you built Taylor series the slow way — computing \(f^{(n)}(0)\) for every \(n\). This lesson is the shortcut: memorize five reference series once, then manufacture everything else from them by algebra.
The five building blocks, with where they converge: \[e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \cdots \quad (\text{all } x)\] \[\sin x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \quad (\text{all } x)\] \[\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \quad (\text{all } x)\] \[\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + \cdots \quad (|x| < 1)\] \[\ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots \quad (-1 < x \le 1)\] To find a series for a new function \(g\), first ask whether \(g\) is one of these five in disguise — via substitution, multiplication by a power of \(x\), addition or subtraction, differentiation, or term-by-term integration. Recognizing the right building block turns a tedious repeated-differentiation problem into simple algebra.
Why memorize these? Because most series you'll ever need are one move away from them. Need \(e^{-x^2}\)? Substitute into \(e^x\) — one line, no derivatives. Need \(\arctan x\)? Substitute \(-x^2\) into the geometric series and integrate. The toolkit approach also carries the convergence information with it: \(e^x\), \(\sin x\), and \(\cos x\) converge everywhere, so anything built from them by substitution does too, while anything built from \(\frac{1}{1-x}\) or \(\ln(1+x)\) inherits a restricted interval that must be re-checked after the substitution — exactly the shaded band you saw move around in the demo above.
Find the Maclaurin series for \(e^{-x^2}\). Replace \(x\) with \(-x^2\) in the \(e^x\) series: \[e^{-x^2} = \sum_{n=0}^{\infty} \frac{(-x^2)^n}{n!} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \cdots,\] valid for all \(x\), since substituting into a series that converges everywhere doesn't change "converges everywhere." Only even powers survive — each \((-x^2)^n\) is an even power of \(x\), which is what the default demo state above shows.
Find the Maclaurin series for \(x \sin x\). Multiply the \(\sin x\) series through by \(x\): \[x \sin x = x\left(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots\right) = x^2 - \frac{x^4}{3!} + \underline{\;\;?\;\;} - \cdots\] Fill in the third nonzero term, and write the general term.
Reveal the answer
Multiplying \(\frac{x^5}{5!}\) by \(x\) gives \(\dfrac{x^6}{5!}\) — every power just goes up by one. In general, \[x \sin x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+2}}{(2n+1)!},\] valid for all \(x\). Note the denominator stays \((2n+1)!\) — multiplying by \(x\) touches the powers, never the coefficients.
More info — where these five series come from, and why the shortcut is legal
These reference series aren't new objects: each one was originally built with the derivative-based coefficient formula \(c_n = f^{(n)}(0)/n!\) from the previous lesson — \(e^x\) is the easiest, since every derivative is \(e^x\) and every \(f^{(n)}(0) = 1\). The toolkit works because a function has only one Maclaurin series: however you produce a power series that equals \(g(x)\) near 0 — substitution, integration, multiplication — it must be the Maclaurin series of \(g\), the same one the derivative formula would have given you. So reuse is not an approximation shortcut; it's the exact same answer, obtained faster. The OpenStax section linked under Dive deeper works several of these constructions in full detail.
Check your understanding
Substituting into a standard series, the first three nonzero terms of the Maclaurin series for \(\sin(x^2)\) are:
You build a series for \(\dfrac{1}{1-2x}\) by substituting \(2x\) into the geometric series \(\sum x^n\). On what interval does the new series converge?
Which toolkit route produces the Maclaurin series \(\arctan x = \sum_{n=0}^{\infty} \dfrac{(-1)^n x^{2n+1}}{2n+1}\)?
Two ways to get the coefficient of \(x^2\) in the Maclaurin series of \(e^{5x}\): compute \(f''(0)/2!\) from the derivative-based Taylor coefficient formula, or substitute \(5x\) into the \(e^x\) series. Both give:
Recap
- Memorize five series: \(e^x = \sum \frac{x^n}{n!}\), \(\sin x = \sum \frac{(-1)^n x^{2n+1}}{(2n+1)!}\), \(\cos x = \sum \frac{(-1)^n x^{2n}}{(2n)!}\) (all \(x\)); \(\frac{1}{1-x} = \sum x^n\) (\(|x| < 1\)); \(\ln(1+x) = \sum \frac{(-1)^{n+1} x^n}{n}\) (\(-1 < x \le 1\)).
- Generate new series by substitution, multiplication by a power of \(x\), addition/subtraction, differentiation, or term-by-term integration — don't recompute derivatives from scratch.
- Re-check the interval of convergence after every substitution: series built from \(\frac{1}{1-x}\) or \(\ln(1+x)\) inherit a restricted interval (e.g. \(u = 2x\) shrinks \(|u| < 1\) to \(|x| < \frac{1}{2}\)).
- Spot-check the first two or three terms numerically — misindexed signs and factorials are the most common slips.
- Signature move: \(\arctan x = \sum \frac{(-1)^n x^{2n+1}}{2n+1}\) for \(|x| \le 1\), from substituting \(-x^2\) into the geometric series and integrating.
Dive deeper
- OpenStax Calculus Volume 2, §6.4 Working with Taylor Series Demonstrate building new Maclaurin series from standard ones.
- Paul's Online Math Notes — Taylor Series List the reference Maclaurin series for e^x, sin x, and cos x.
Sources
- Standard Maclaurin Series as Building Blocks