Taylor polynomials get better as their degree rises. Let the degree grow
without end, and the polynomial becomes a power series, the Taylor series
of f f f a a a
∑ n = 0 ∞ f ( n ) ( a ) n ! ( x − a ) n = f ( a ) + f ′ ( a ) ( x − a ) + f ′ ′ ( a ) 2 ! ( x − a ) 2 + ⋯ \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x - a)^n = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \cdots n = 0 ∑ ∞ n ! f ( n ) ( a ) ( x − a ) n = f ( a ) + f ′ ( a ) ( x − a ) + 2 ! f ′′ ( a ) ( x − a ) 2 + ⋯ Centered at a = 0 a = 0 a = 0 Maclaurin series . A handful of these come up
constantly:
Function Maclaurin series Converges for e x e^x e x 1 + x + x 2 2 ! + x 3 3 ! + ⋯ 1 + x + \tfrac{x^2}{2!} + \tfrac{x^3}{3!} + \cdots 1 + x + 2 ! x 2 + 3 ! x 3 + ⋯ all x x x sin x \sin x sin x x − x 3 3 ! + x 5 5 ! − ⋯ x - \tfrac{x^3}{3!} + \tfrac{x^5}{5!} - \cdots x − 3 ! x 3 + 5 ! x 5 − ⋯ all x x x cos x \cos x cos x 1 − x 2 2 ! + x 4 4 ! − ⋯ 1 - \tfrac{x^2}{2!} + \tfrac{x^4}{4!} - \cdots 1 − 2 ! x 2 + 4 ! x 4 − ⋯ all x x x 1 1 − x \tfrac{1}{1 - x} 1 − x 1 1 + x + x 2 + x 3 + ⋯ 1 + x + x^2 + x^3 + \cdots 1 + x + x 2 + x 3 + ⋯ − 1 < x < 1 -1 < x < 1 − 1 < x < 1 ln ( 1 + x ) \ln(1 + x) ln ( 1 + x ) x − x 2 2 + x 3 3 − ⋯ x - \tfrac{x^2}{2} + \tfrac{x^3}{3} - \cdots x − 2 x 2 + 3 x 3 − ⋯ − 1 < x ≤ 1 -1 < x \le 1 − 1 < x ≤ 1 arctan x \arctan x arctan x x − x 3 3 + x 5 5 − ⋯ x - \tfrac{x^3}{3} + \tfrac{x^5}{5} - \cdots x − 3 x 3 + 5 x 5 − ⋯ − 1 ≤ x ≤ 1 -1 \le x \le 1 − 1 ≤ x ≤ 1
Example 1 — The series for e^x
Find the Maclaurin series of e x e^x e x
Every derivative of e x e^x e x e x e^x e x e 0 = 1 e^0 = 1 e 0 = 1 1 n ! \tfrac{1}{n!} n ! 1
e x = ∑ n = 0 ∞ x n n ! = 1 + x + x 2 2 + x 3 6 + ⋯ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots e x = n = 0 ∑ ∞ n ! x n = 1 + x + 2 x 2 + 6 x 3 + ⋯ At x = 1 x = 1 x = 1 e = 1 + 1 + 1 2 + 1 6 + ⋯ e = 1 + 1 + \tfrac{1}{2} + \tfrac{1}{6} + \cdots e = 1 + 1 + 2 1 + 6 1 + ⋯
A Taylor series is built from derivatives at a single point, so it is fair to
ask whether it really adds up to f ( x ) f(x) f ( x ) P n ( x ) P_n(x) P n ( x ) 0 0 0 n n n e x e^x e x e ∣ x ∣ ∣ x ∣ n + 1 ( n + 1 ) ! e^{\lvert x \rvert}\tfrac{\lvert x \rvert^{n+1}}{(n + 1)!} e ∣ x ∣ ( n + 1 )! ∣ x ∣ n + 1 x x x 1 1 − x \tfrac{1}{1 - x} 1 − x 1
Polynomials from the series for 1/(1 − x)
The curve y = 1/(1 − x), rising toward an asymptote at x = 1, with the polynomials 1 + x + x² and 1 + x + ⋯ + x⁶. Between the dashed lines at x = −1 and x = 1 the polynomials close in on the curve as the degree rises; outside that interval they swing away from it.
-1 1 -1 1 2 3 4 5 6 x y
Polynomials from the series for 1/(1 − x)
A Taylor series equals its function exactly where the errors of its
polynomials shrink to zero. For all six series in the table, that is the
whole interval of convergence.
Derivatives can get messy, and there is rarely a need to compute them. A known
series can be changed into a new one:
Substitute: replace x x x − x 2 -x^2 − x 2 2 x 2x 2 x x 3 x^3 x 3 Multiply by a power of x x x Differentiate or integrate term by term, which keeps the radius of
convergence (the endpoints need checking again).
Example 2 — Substitution
Find the Maclaurin series of e − x 2 e^{-x^2} e − x 2
Replace x x x − x 2 -x^2 − x 2 e x e^x e x
e − x 2 = ∑ n = 0 ∞ ( − x 2 ) n n ! = 1 − x 2 + x 4 2 ! − x 6 3 ! + ⋯ e^{-x^2} = \sum_{n=0}^{\infty} \frac{(-x^2)^n}{n!} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \cdots e − x 2 = n = 0 ∑ ∞ n ! ( − x 2 ) n = 1 − x 2 + 2 ! x 4 − 3 ! x 6 + ⋯ It converges for all x x x
Example 3 — Integrating term by term
Find the Maclaurin series of arctan x \arctan x arctan x
Substitute − x 2 -x^2 − x 2 1 1 + x 2 = 1 − x 2 + x 4 − x 6 + ⋯ \tfrac{1}{1 + x^2} = 1 - x^2 + x^4 - x^6 + \cdots 1 + x 2 1 = 1 − x 2 + x 4 − x 6 + ⋯ ∣ x ∣ < 1 \lvert x \rvert < 1 ∣ x ∣ < 1 arctan x = ∫ 0 x d t 1 + t 2 \arctan x = \int_0^x \tfrac{dt}{1 + t^2} arctan x = ∫ 0 x 1 + t 2 d t
arctan x = x − x 3 3 + x 5 5 − x 7 7 + ⋯ \arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots arctan x = x − 3 x 3 + 5 x 5 − 7 x 7 + ⋯ At x = 1 x = 1 x = 1 π 4 = 1 − 1 3 + 1 5 − ⋯ \tfrac{\pi}{4} = 1 - \tfrac{1}{3} + \tfrac{1}{5} - \cdots 4 π = 1 − 3 1 + 5 1 − ⋯
Example 4 — Differentiating term by term
Find a series for 1 ( 1 − x ) 2 \tfrac{1}{(1 - x)^2} ( 1 − x ) 2 1
It is the derivative of 1 1 − x \tfrac{1}{1 - x} 1 − x 1 1 + x + x 2 + x 3 + ⋯ 1 + x + x^2 + x^3 + \cdots 1 + x + x 2 + x 3 + ⋯
1 ( 1 − x ) 2 = 1 + 2 x + 3 x 2 + 4 x 3 + ⋯ = ∑ n = 1 ∞ n x n − 1 ∣ x ∣ < 1 \frac{1}{(1 - x)^2} = 1 + 2x + 3x^2 + 4x^3 + \cdots = \sum_{n=1}^{\infty} n x^{n-1} \qquad \lvert x \rvert < 1 ( 1 − x ) 2 1 = 1 + 2 x + 3 x 2 + 4 x 3 + ⋯ = n = 1 ∑ ∞ n x n − 1 ∣ x ∣ < 1 Example 5 — An integral with no antiderivative
Estimate ∫ 0 1 e − x 2 d x \int_0^1 e^{-x^2}\,dx ∫ 0 1 e − x 2 d x
e − x 2 e^{-x^2} e − x 2
∫ 0 1 ( 1 − x 2 + x 4 2 − x 6 6 + ⋯ ) d x = 1 − 1 3 + 1 10 − 1 42 + ⋯ \int_0^1 \left(1 - x^2 + \frac{x^4}{2} - \frac{x^6}{6} + \cdots\right)dx = 1 - \frac{1}{3} + \frac{1}{10} - \frac{1}{42} + \cdots ∫ 0 1 ( 1 − x 2 + 2 x 4 − 6 x 6 + ⋯ ) d x = 1 − 3 1 + 10 1 − 42 1 + ⋯ Four terms give 0.7429 0.7429 0.7429 1 216 ≈ 0.005 \tfrac{1}{216} \approx 0.005 216 1 ≈ 0.005 0.7468 0.7468 0.7468
Example 6 — A limit
Find lim x → 0 sin x − x x 3 \displaystyle\lim_{x \to 0} \frac{\sin x - x}{x^3} x → 0 lim x 3 sin x − x
sin x − x = − x 3 6 + x 5 120 − ⋯ \sin x - x = -\tfrac{x^3}{6} + \tfrac{x^5}{120} - \cdots sin x − x = − 6 x 3 + 120 x 5 − ⋯
sin x − x x 3 = − 1 6 + x 2 120 − ⋯ ⟶ − 1 6 \frac{\sin x - x}{x^3} = -\frac{1}{6} + \frac{x^2}{120} - \cdots \longrightarrow -\frac{1}{6} x 3 sin x − x = − 6 1 + 120 x 2 − ⋯ ⟶ − 6 1 Common mistake
Substituting only part of the term. The series for sin 2 x \sin 2x sin 2 x ∑ ( − 1 ) n ( 2 x ) 2 n + 1 ( 2 n + 1 ) ! \sum \tfrac{(-1)^n (2x)^{2n+1}}{(2n + 1)!} ∑ ( 2 n + 1 )! ( − 1 ) n ( 2 x ) 2 n + 1 2 2 2 x x x 2 ∑ ( − 1 ) n x 2 n + 1 ( 2 n + 1 ) ! 2\sum \tfrac{(-1)^n x^{2n+1}}{(2n + 1)!} 2 ∑ ( 2 n + 1 )! ( − 1 ) n x 2 n + 1 2 sin x 2\sin x 2 sin x
Common mistake
Using a series outside its interval. 1 + x + x 2 + ⋯ 1 + x + x^2 + \cdots 1 + x + x 2 + ⋯ 1 1 − x \tfrac{1}{1 - x} 1 − x 1 ∣ x ∣ < 1 \lvert x \rvert < 1 ∣ x ∣ < 1 x = 2 x = 2 x = 2 1 1 − 2 = − 1 \tfrac{1}{1 - 2} = -1 1 − 2 1 = − 1
Common mistake
Forgetting the constant after integrating. Integrating a series term by
term produces a constant of integration. Fix it with a known value: for
arctan x \arctan x arctan x arctan 0 = 0 \arctan 0 = 0 arctan 0 = 0
Find the Maclaurin series of e 3 x e^{3x} e 3 x
Answer
∑ n = 0 ∞ 3 n x n n ! \sum_{n=0}^{\infty} \tfrac{3^n x^n}{n!} ∑ n = 0 ∞ n ! 3 n x n
Full solution
Substitute 3 x 3x 3 x x x x e x e^x e x ( 3 x ) n n ! = 3 n x n n ! \tfrac{(3x)^n}{n!} = \tfrac{3^n x^n}{n!} n ! ( 3 x ) n = n ! 3 n x n
Find the Maclaurin series of x sin x x\sin x x sin x
Answer
∑ n = 0 ∞ ( − 1 ) n x 2 n + 2 ( 2 n + 1 ) ! \sum_{n=0}^{\infty} \tfrac{(-1)^n x^{2n+2}}{(2n + 1)!} ∑ n = 0 ∞ ( 2 n + 1 )! ( − 1 ) n x 2 n + 2
Full solution
Multiply each term of the series for sin x \sin x sin x x x x x 2 − x 4 3 ! + x 6 5 ! − ⋯ x^2 - \tfrac{x^4}{3!} + \tfrac{x^6}{5!} - \cdots x 2 − 3 ! x 4 + 5 ! x 6 − ⋯
Find the Maclaurin series of cos ( x 2 ) \cos(x^2) cos ( x 2 )
Answer
∑ n = 0 ∞ ( − 1 ) n x 4 n ( 2 n ) ! \sum_{n=0}^{\infty} \tfrac{(-1)^n x^{4n}}{(2n)!} ∑ n = 0 ∞ ( 2 n )! ( − 1 ) n x 4 n
Full solution
Substitute x 2 x^2 x 2 cos x \cos x cos x 1 − x 4 2 ! + x 8 4 ! − ⋯ 1 - \tfrac{x^4}{2!} + \tfrac{x^8}{4!} - \cdots 1 − 2 ! x 4 + 4 ! x 8 − ⋯
Starting from 1 1 + x = 1 − x + x 2 − ⋯ \tfrac{1}{1 + x} = 1 - x + x^2 - \cdots 1 + x 1 = 1 − x + x 2 − ⋯ ln ( 1 + x ) \ln(1 + x) ln ( 1 + x )
Answer
x − x 2 2 + x 3 3 − ⋯ x - \tfrac{x^2}{2} + \tfrac{x^3}{3} - \cdots x − 2 x 2 + 3 x 3 − ⋯
Full solution
ln ( 1 + x ) = ∫ 0 x d t 1 + t \ln(1 + x) = \int_0^x \tfrac{dt}{1 + t} ln ( 1 + x ) = ∫ 0 x 1 + t d t 0 0 0 ln 1 = 0 \ln 1 = 0 ln 1 = 0
Find the Taylor series of e x e^x e x 1 1 1
Answer
∑ n = 0 ∞ e n ! ( x − 1 ) n \sum_{n=0}^{\infty} \tfrac{e}{n!}(x - 1)^n ∑ n = 0 ∞ n ! e ( x − 1 ) n
Full solution
Every derivative of e x e^x e x 1 1 1 e e e
Find the sums ∑ n = 0 ∞ 1 n ! \sum_{n=0}^{\infty} \tfrac{1}{n!} ∑ n = 0 ∞ n ! 1 ∑ n = 0 ∞ ( − 1 ) n 2 n + 1 \sum_{n=0}^{\infty} \tfrac{(-1)^n}{2n + 1} ∑ n = 0 ∞ 2 n + 1 ( − 1 ) n
Answer
e e e π 4 \tfrac{\pi}{4} 4 π
Full solution
The first is the series for e x e^x e x x = 1 x = 1 x = 1 arctan x \arctan x arctan x x = 1 x = 1 x = 1 arctan 1 = π 4 \arctan 1 = \tfrac{\pi}{4} arctan 1 = 4 π
Find lim x → 0 1 − cos x x 2 \displaystyle\lim_{x \to 0} \frac{1 - \cos x}{x^2} x → 0 lim x 2 1 − cos x
Answer
1 2 \tfrac{1}{2} 2 1
Full solution
1 − cos x = x 2 2 − x 4 24 + ⋯ 1 - \cos x = \tfrac{x^2}{2} - \tfrac{x^4}{24} + \cdots 1 − cos x = 2 x 2 − 24 x 4 + ⋯ 1 2 − x 2 24 + ⋯ → 1 2 \tfrac{1}{2} - \tfrac{x^2}{24} + \cdots \to \tfrac{1}{2} 2 1 − 24 x 2 + ⋯ → 2 1
For f ( x ) = cos ( x 2 ) f(x) = \cos(x^2) f ( x ) = cos ( x 2 ) f ( 4 ) ( 0 ) f^{(4)}(0) f ( 4 ) ( 0 )
Answer
− 12 -12 − 12
Full solution
The coefficient of x 4 x^4 x 4 − 1 2 -\tfrac{1}{2} − 2 1 f ( 4 ) ( 0 ) 4 ! \tfrac{f^{(4)}(0)}{4!} 4 ! f ( 4 ) ( 0 ) f ( 4 ) ( 0 ) = 24 ⋅ ( − 1 2 ) f^{(4)}(0) = 24 \cdot \left(-\tfrac{1}{2}\right) f ( 4 ) ( 0 ) = 24 ⋅ ( − 2 1 )
Estimate ∫ 0 0.5 sin ( x 2 ) d x \int_0^{0.5} \sin(x^2)\,dx ∫ 0 0.5 sin ( x 2 ) d x
Answer
About 0.04148 0.04148 0.04148
Full solution
sin ( x 2 ) = x 2 − x 6 6 + ⋯ \sin(x^2) = x^2 - \tfrac{x^6}{6} + \cdots sin ( x 2 ) = x 2 − 6 x 6 + ⋯ 0.5 3 3 − 0.5 7 42 + ⋯ ≈ 0.041667 − 0.000186 \tfrac{0.5^3}{3} - \tfrac{0.5^7}{42} + \cdots \approx 0.041667 - 0.000186 3 0. 5 3 − 42 0. 5 7 + ⋯ ≈ 0.041667 − 0.000186
A student writes the Maclaurin series of sin 2 x \sin 2x sin 2 x 2 x − 2 x 3 3 ! + 2 x 5 5 ! − ⋯ 2x - \tfrac{2x^3}{3!} + \tfrac{2x^5}{5!} - \cdots 2 x − 3 ! 2 x 3 + 5 ! 2 x 5 − ⋯
Hint
Substitute 2 x 2x 2 x x x x
Answer
The 2 2 2 2 x − ( 2 x ) 3 3 ! + ( 2 x ) 5 5 ! − ⋯ 2x - \tfrac{(2x)^3}{3!} + \tfrac{(2x)^5}{5!} - \cdots 2 x − 3 ! ( 2 x ) 3 + 5 ! ( 2 x ) 5 − ⋯
Full solution
The student’s series is 2 sin x 2\sin x 2 sin x 2 x 2x 2 x 2 x − 8 x 3 6 + 32 x 5 120 − ⋯ 2x - \tfrac{8x^3}{6} + \tfrac{32x^5}{120} - \cdots 2 x − 6 8 x 3 + 120 32 x 5 − ⋯
A check: the derivative of sin 2 x \sin 2x sin 2 x 0 0 0 2 cos 0 = 2 2\cos 0 = 2 2 cos 0 = 2 − 8 cos 0 = − 8 -8\cos 0 = -8 − 8 cos 0 = − 8 x 3 x^3 x 3 − 8 3 ! \tfrac{-8}{3!} 3 ! − 8