The Standard Maclaurin Series

Start from the geometric series

$1+x+x^2+x^3+\dots$1 + x + x² + x³ + … is a geometric series with first term 1 and common ratio x. When |x| < 1 its sum is $\frac{1}{1-x}$1/(1 − x), so $\frac{1}{1-x}=1+x+x^2+x^3+\dots$1/(1 − x) = 1 + x + x² + x³ + … for −1 < x < 1. At x = 0.5 the running totals are 1, 1.5, 1.75, 1.875, 1.9375, closing in on $\frac{1}{1-0.5}=2$1/(1 − 0.5) = 2.

Outside that range the two sides part company. At x = 2 the left side is $\frac{1}{1-2}=-1$1/(1 − 2) = −1, but the series 1 + 2 + 4 + 8 + … diverges. An equation between a function and its series holds only where the series converges.

This is the Maclaurin series of $\frac{1}{1-x}$1/(1 − x), found without differentiating anything. Differentiating does give the same answer: the derivatives of $\frac{1}{1-x}$1/(1 − x) at 0 are 1, 1, 2, 6, 24, …, which are 0!, 1!, 2!, 3!, 4!, so every coefficient is n! divided by n!, which is 1.

The gold curve is $y=\frac{1}{1-x}$y = 1/(1 − x), and the plain one is the series stopped at the $x^8$x⁸ term. Near 0 the two run together: at x = 0.5 the polynomial gives 1.9961 against 2. Closer to x = 1 more terms are needed, and outside −1 < x < 1 no number of terms will do: at x = −1.2 the polynomial gives 2.7999 against 0.4545.

Substitute

Anything can be put in place of x, as long as it stays between −1 and 1. Replace x by −x: $\frac{1}{1+x}=1-x+x^2-x^3+\dots$1/(1 + x) = 1 − x + x² − x³ + …. The odd powers of −x are negative, so the signs now alternate. This holds when |−x| < 1, which is again −1 < x < 1.

Replace x by $-x^2$−x² instead, and $\frac{1}{1+x^2}=1-x^2+x^4-x^6+\dots$1/(1 + x²) = 1 − x² + x⁴ − x⁶ + …, also for −1 < x < 1. Replacing x by $x^2$x² gives $\frac{1}{1-x^2}=1+x^2+x^4+\dots$1/(1 − x²) = 1 + x² + x⁴ + …. Each one is a new series with no derivatives worked out.

Integrate

The derivative of ln(1 + x) is $\frac{1}{1+x}$1/(1 + x), so ln(1 + x) can be found by integrating the series for $\frac{1}{1+x}$1/(1 + x), one term at a time. 1 integrates to x, −x to $-\frac{x^2}{2}$−x²/2, $x^2$x² to $\frac{x^3}{3}$x³/3, and $-x^3$−x³ to $-\frac{x^4}{4}$−x⁴/4. So $\ln \left(1+x\right)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots +c$ln(1 + x) = x − x²/2 + x³/3 − x⁴/4 + … + c.

To find the constant, put x = 0. The left side is ln 1 = 0 and every term on the right is 0, so c = 0, and $\ln \left(1+x\right)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots$ln(1 + x) = x − x²/2 + x³/3 − x⁴/4 + …. The coefficients are $\frac{1}{n}$1/n, not $\frac{1}{n!}$1/n!.

Check at x = 0.2. The running totals are 0.2, 0.18, 0.182667, 0.182267 and 0.182331, and ln 1.2 = 0.182322.

The series holds for $-1<x\le 1$−1 < x ≤ 1. Inside, it converges as the series for $\frac{1}{1+x}$1/(1 + x) does. At x = 1 it is $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots$1 − 1/2 + 1/3 − 1/4 + …, which converges by the alternating series test, and its sum is ln 2. At x = −1 it is $-\left(1+\frac{1}{2}+\frac{1}{3}+\dots \right)$−(1 + 1/2 + 1/3 + …), the harmonic series made negative, which diverges, and ln 0 has no value. Beyond x = 1 the terms grow: at x = 2 the running totals are 2, 0, 2.667, −1.333, and after ten terms −64.825, nowhere near ln 3 = 1.098612.

The same move turns $\frac{1}{1+x^2}$1/(1 + x²) into $\arctan x=x-\frac{x^3}{3}+\frac{x^5}{5}-\dots$arctan x = x − x³/3 + x⁵/5 − …, because the derivative of arctan x is $\frac{1}{1+x^2}$1/(1 + x²) and arctan 0 = 0. At x = 0.5 the running totals are 0.5, 0.458333, 0.464583 and 0.463467, and arctan 0.5 = 0.463648.

The gold curve is y = ln(1 + x), and the plain one is its series stopped at the $x^{10}$x¹⁰ term. Over most of the stretch from −1 to 1 the two are hard to tell apart: at x = 0.5 the polynomial gives 0.4054 against ln 1.5 = 0.4055. Past x = 1 the polynomial turns away: at x = 1.5 it gives −2.4031 against ln 2.5 = 0.9163.

Differentiate

Start from $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots$sin x = x − x³/3! + x⁵/5! − x⁷/7! + … and differentiate each term. x gives 1. $\frac{x^3}{3!}$x³/3! gives $\frac{3x^2}{3!}$3x²/3!, and $\frac{3}{3!}=\frac{1}{2!}$3/3! = 1/2!, so it gives $\frac{x^2}{2!}$x²/2!. In the same way $\frac{x^5}{5!}$x⁵/5! gives $\frac{x^4}{4!}$x⁴/4!. So the derivative is $1-\frac{x^2}{2!}+\frac{x^4}{4!}-\dots$1 − x²/2! + x⁴/4! − …, and since the derivative of sin x is cos x, $\cos x=1-\frac{x^2}{2}+\frac{x^4}{24}-\dots$cos x = 1 − x²/2 + x⁴/24 − ….

Check at x = 0.5: 1 − 0.125 + 0.0026042 = 0.8776042, and cos 0.5 = 0.8775826.

Differentiating again gives $-x+\frac{x^3}{3!}-\frac{x^5}{5!}+\dots$−x + x³/3! − x⁵/5! + …, which is −sin x, as it should be. Differentiating the series for $e^x$eˣ term by term gives the same series back, which matches the fact that $e^x$eˣ is its own derivative.

The five to know

$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots$eˣ = 1 + x + x²/2! + x³/3! + …, for every x.

$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots$sin x = x − x³/3! + x⁵/5! − …, for every x, with x in radians.

$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\dots$cos x = 1 − x²/2! + x⁴/4! − …, for every x, with x in radians.

$\frac{1}{1-x}=1+x+x^2+x^3+\dots$1/(1 − x) = 1 + x + x² + x³ + …, for −1 < x < 1.

$\ln \left(1+x\right)=x-\frac{x^2}{2}+\frac{x^3}{3}-\dots$ln(1 + x) = x − x²/2 + x³/3 − …, for $-1<x\le 1$−1 < x ≤ 1.

The ranges come from the ratio test. For the first three, the factorial underneath drives the ratio to 0 whatever x is, so the radius is infinite. For the last two the radius is 1, and the endpoints were tested above.

New series from the five

Substitute into $e^x$eˣ: replacing x by $-x^2$−x² gives $e^{-x^2}=1-x^2+\frac{x^4}{2}-\frac{x^6}{6}+\dots$e^(−x²) = 1 − x² + x⁴/2 − x⁶/6 + …, for every x. At x = 0.5 these four terms give 0.7786458, and $e^{-0.25}=0.7788008$e^(−0.25) = 0.7788008.

Substitute into sin x: replacing x by 2x gives $\sin 2x=2x-\frac{{\left(2x\right)}^3}{3!}+\cdots =2x-\frac{4x^3}{3}+\dots$sin 2x = 2x − (2x)³/3! + … = 2x − 4x³/3 + …. Each power of 2x brings its own power of 2.

Differentiate $\frac{1}{1-x}=1+x+x^2+x^3+\dots$1/(1 − x) = 1 + x + x² + x³ + …: $\frac{1}{{\left(1-x\right)}^2}=1+2x+3x^2+4x^3+\dots$1/(1 − x)² = 1 + 2x + 3x² + 4x³ + …, for −1 < x < 1. At x = 0.5 the left side is $\frac{1}{0.25}=4$1/0.25 = 4, and the series 1 + 1 + 0.75 + 0.5 + 0.3125 + … adds up to 4.

The binomial series

One more series covers every power of 1 + x: ${\left(1+x\right)}^n=1+nx+\frac{n\left(n-1\right)x^2}{2!}+\frac{n\left(n-1\right)\left(n-2\right)x^3}{3!}+\dots$(1 + x)ⁿ = 1 + nx + n(n − 1)x²/2! + n(n − 1)(n − 2)x³/3! + …. When n is a whole number the coefficients reach 0 and the series stops, giving the binomial theorem, true for every x. For any other n it goes on forever, and it holds for −1 < x < 1.

With n = −1 the coefficients are −1, then $\frac{\left(-1\right)\left(-2\right)}{2!}=1$(−1)(−2)/2! = 1, then $\frac{\left(-1\right)\left(-2\right)\left(-3\right)}{3!}=-1$(−1)(−2)(−3)/3! = −1, so ${\left(1+x\right)}^{-1}=1-x+x^2-x^3+\dots$(1 + x)⁻¹ = 1 − x + x² − x³ + …, the series for $\frac{1}{1+x}$1/(1 + x) again.

With $n=\frac{1}{2}$n = 1/2 the coefficients are $\frac{1}{2}$1/2, then $\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2!}=-\frac{1}{8}$(1/2)(−1/2)/2! = −1/8, then $\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{3!}=\frac{1}{16}$(1/2)(−1/2)(−3/2)/3! = 1/16, so $\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}-\dots$√(1 + x) = 1 + x/2 − x²/8 + x³/16 − …. At x = 0.1 that gives 1 + 0.05 − 0.00125 + 0.0000625 = 1.0488125, and $\sqrt{1.1}=1.0488088$√1.1 = 1.0488088.

The usual mistakes

Using a series outside its range. $\ln \left(1+x\right)=x-\frac{x^2}{2}+\dots$ln(1 + x) = x − x²/2 + … is no use for ln 3, because x = 2 is outside $-1<x\le 1$−1 < x ≤ 1 and the terms grow.

Putting factorials into the logarithm series. Integrating $x^n$xⁿ gives $x^{n+1}$xⁿ⁺¹ divided by n + 1, so the coefficients of ln(1 + x) are 1, $\frac{1}{2}$1/2, $\frac{1}{3}$1/3, …, not $\frac{1}{n!}$1/n!.

Leaving out the constant of integration without checking it. For ln(1 + x) it is 0 because ln 1 = 0; for a function that is not 0 at x = 0 it would not be.

Confusing the sine and cosine series. Cosine starts at 1, because cos 0 = 1, and has the even powers; sine starts at x and has the odd powers.