1.6 Series expansion
Nonlinear functions can be nasty, but polynomials are the “nicest” version. A
series expansion is a systematic approach to turn a general nonlinear
equation into a polynomial. Suppose we have some function f ( x ) f(x) f ( x )
f ( x ) = ∑ n = 0 ∞ c n ( x − x 0 ) n f(x) = \sum_{n=0}^{\infty} c_n(x-x_0)^n f ( x ) = n = 0 ∑ ∞ c n ( x − x 0 ) n To determine the coefficients c n c_n c n f and its
derivatives at x 0 x_0 x 0
f ( x ) = c 0 + c 1 ( x − x 0 ) + c 2 ( x − x 0 ) 2 + c 3 ( x − x 0 ) 3 + ⋯ f ( x 0 ) = c 0 f ′ ( x ) = c 1 + 2 c 2 ( x − x 0 ) + 3 c 3 ( x − x 0 ) 2 + ⋯ f ′ ( x 0 ) = c 1 f ′ ′ ( x ) = 2 c 2 + 6 c 3 ( x − x 0 ) + ⋯ f ′ ′ ( x 0 ) = 2 c 2 f ′ ′ ′ ( x ) = 6 c 3 ( x − x 0 ) + ⋯ f ′ ′ ( x 0 ) = 6 c 3 \begin{align}
f(x) &= c_0 + c_1(x-x_0) + c_2(x-x_0)^2 + c_3(x-x_0)^3 +
\cdots & f(x_0) &= c_0 \\
f'(x) &= c_1 + 2 c_2(x-x_0) + 3 c_3(x-x_0)^2 + \cdots & f'(x_0) &= c_1 \\
f''(x) &= 2 c_2 + 6 c_3(x-x_0) + \cdots & f''(x_0) &= 2 c_2 \\
f'''(x) &= 6 c_3(x-x_0) + \cdots & f''(x_0) &= 6 c_3
\end{align} f ( x ) f ′ ( x ) f ′′ ( x ) f ′′′ ( x ) = c 0 + c 1 ( x − x 0 ) + c 2 ( x − x 0 ) 2 + c 3 ( x − x 0 ) 3 + ⋯ = c 1 + 2 c 2 ( x − x 0 ) + 3 c 3 ( x − x 0 ) 2 + ⋯ = 2 c 2 + 6 c 3 ( x − x 0 ) + ⋯ = 6 c 3 ( x − x 0 ) + ⋯ f ( x 0 ) f ′ ( x 0 ) f ′′ ( x 0 ) f ′′ ( x 0 ) = c 0 = c 1 = 2 c 2 = 6 c 3 The pattern of matching derivatives gives rise to the Taylor series .
For example, let’s compute the Taylor series for f ( x ) = ln x f(x) = \ln x f ( x ) = ln x x 0 = 2 x_0 = 2 x 0 = 2
f ( x ) = ln x f ( 2 ) = ln 2 f ′ ( x ) = 1 x f ′ ( 2 ) = 1 2 f ′ ′ ( x ) = − 1 x 2 f ′ ′ ( 2 ) = − 1 4 \begin{align}
f(x) &= \ln x & f(2) &= \ln 2 \\
f'(x) &= \frac{1}{x} & f'(2) &= \frac{1}{2} \\
f''(x) &= -\frac{1}{x^2} & f''(2) &= -\frac{1}{4}
\end{align} f ( x ) f ′ ( x ) f ′′ ( x ) = ln x = x 1 = − x 2 1 f ( 2 ) f ′ ( 2 ) f ′′ ( 2 ) = ln 2 = 2 1 = − 4 1 Hence,
f ( x ) ≈ ln ( 2 ) + 1 2 ( x − 2 ) − 1 2 ⋅ 1 4 ( x − 2 ) 2 + ⋯ f(x) \approx \ln(2) + \frac{1}{2}(x-2) - \frac{1}{2}\cdot\frac{1}{4} (x-2)^2 + \cdots f ( x ) ≈ ln ( 2 ) + 2 1 ( x − 2 ) − 2 1 ⋅ 4 1 ( x − 2 ) 2 + ⋯ In general, we could determine the infinite series and general expressions for
coefficients, but for practical purposes, we will usually stop at either first
or second order polynomials since that is easier to work with.
For example, let’s find the two-term Taylor series for
f ( x ) = x e 2 x 2 f(x) = xe^{2 x^2} f ( x ) = x e 2 x 2 about x = 0 x = 0 x = 0 product rule calculations.
The value of the function is f ( 0 ) = 0 f(0) = 0 f ( 0 ) = 0
f ′ ( x ) = x ( e 2 x 2 4 x ) + e 2 x 2 = ( 4 x 2 + 1 ) e 2 x 2 \begin{align}
f'(x) &= x(e^{2x^2}4x) + e^{2x^2} \\
&= (4x^2 + 1)e^{2x^2}
\end{align} f ′ ( x ) = x ( e 2 x 2 4 x ) + e 2 x 2 = ( 4 x 2 + 1 ) e 2 x 2 so f ′ ( 0 ) = 1 f'(0) = 1 f ′ ( 0 ) = 1
f ′ ′ ( x ) = ( 4 x 2 + 1 ) ( e 2 x 2 4 x ) + 8 x e 2 x 2 = ( 16 x 3 + 12 x ) e 2 x 2 \begin{align}
f''(x) &= (4x^2+1)(e^{2x^2}4x)+8xe^{2x^2} \\
&= (16x^3+12x)e^{2x^2}
\end{align} f ′′ ( x ) = ( 4 x 2 + 1 ) ( e 2 x 2 4 x ) + 8 x e 2 x 2 = ( 16 x 3 + 12 x ) e 2 x 2 so f ′ ′ ( 0 ) = 0 f''(0) = 0 f ′′ ( 0 ) = 0
f ′ ′ ′ ( x ) = ( 16 x 3 + 12 x ) ( e 2 x 2 4 x ) + ( 48 x 2 + 12 ) e 2 x 2 = ( 64 x 4 + 96 x 2 + 12 ) e 2 x 2 \begin{align}
f'''(x) &= (16x^3+12x)(e^{2x^2}4x)+(48x^2+12)e^{2x^2} \\
&= (64x^4+96x^2+12)e^{2x^2}
\end{align} f ′′′ ( x ) = ( 16 x 3 + 12 x ) ( e 2 x 2 4 x ) + ( 48 x 2 + 12 ) e 2 x 2 = ( 64 x 4 + 96 x 2 + 12 ) e 2 x 2 so f ′ ′ ′ ( 0 ) = 12 f'''(0) = 12 f ′′′ ( 0 ) = 12
Putting it all together,
f ( x ) ≈ 0 + 1 ⋅ x + 1 2 ⋅ 0 ⋅ x 2 + 1 6 ⋅ 12 ⋅ x 3 + ⋯ = x + 2 x 3 + ⋯ f(x) \approx 0 + 1 \cdot x + \frac{1}{2} \cdot 0 \cdot x^2 +
\frac{1}{6} \cdot 12 \cdot x^3 + \cdots = x + 2x^3 + \cdots f ( x ) ≈ 0 + 1 ⋅ x + 2 1 ⋅ 0 ⋅ x 2 + 6 1 ⋅ 12 ⋅ x 3 + ⋯ = x + 2 x 3 + ⋯ Combining series ¶ Some of these calculations can sometimes be avoided by combining series
together. For example, the Taylor series for e x e^x e x
e x = ∑ n = 0 ∞ x n n ! = 1 + x + x 2 2 + ⋯ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \cdots e x = n = 0 ∑ ∞ n ! x n = 1 + x + 2 x 2 + ⋯ The argument of e in f is 2 x 2 2x^2 2 x 2 x is zero, so we
can substitute it in the known series to obtain
e 2 x 2 ≈ 1 + 2 x 2 + ( 2 x 2 ) 2 2 + ⋯ e^{2x^2} \approx 1 + 2x^2 + \frac{(2x^2)^2}{2} + \cdots e 2 x 2 ≈ 1 + 2 x 2 + 2 ( 2 x 2 ) 2 + ⋯ Finally, inserting this series into f and expanding
f ( x ) = x e 2 x 2 ≈ x ( 1 + 2 x 2 + 2 x 4 + ⋯ ) = x + 2 x 3 + ⋯ \begin{align}
f(x) &= x e^{2x^2} \approx x (1 + 2x^2 + 2x^4 + \cdots) \\
&= x + 2x^3 + \cdots
\end{align} f ( x ) = x e 2 x 2 ≈ x ( 1 + 2 x 2 + 2 x 4 + ⋯ ) = x + 2 x 3 + ⋯ When doing this process, it can be helpful to keep track of dropped terms using
O O O
e x = ∑ n = 0 ∞ x n n ! = 1 + x + x 2 2 + O ( x 3 ) e 2 x 2 = 1 + 2 x 2 + 2 x 4 + O ( x 6 ) x e 2 x 2 = x + 2 x 3 + O ( x 5 ) \begin{align}
e^x &= \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + O(x^3) \\
e^{2x^2} &= 1 + 2x^2 + 2x^4 + O(x^6) \\
xe^{2x^2} &= x + 2x^3 + O(x^5)
\end{align} e x e 2 x 2 x e 2 x 2 = n = 0 ∑ ∞ n ! x n = 1 + x + 2 x 2 + O ( x 3 ) = 1 + 2 x 2 + 2 x 4 + O ( x 6 ) = x + 2 x 3 + O ( x 5 ) Skill builder problems ¶ Expand the following to two terms:
The function and its derivatives at x = 0 x=0 x = 0
f ( x ) = cos ( x ) f ( 0 ) = 1 f ′ ( x ) = − sin ( x ) f ′ ( 0 ) = 0 f ′ ′ ( x ) = − cos ( x ) f ′ ′ ( 0 ) = − 1 \begin{align}
f(x) &= \cos(x) & f(0) &= 1 \\
f'(x) &= -\sin(x) & f'(0) &= 0 \\
f''(x) &= -\cos(x) & f''(0) &= -1
\end{align} f ( x ) f ′ ( x ) f ′′ ( x ) = cos ( x ) = − sin ( x ) = − cos ( x ) f ( 0 ) f ′ ( 0 ) f ′′ ( 0 ) = 1 = 0 = − 1 So:
cos x ≈ 1 − x 2 2 \cos x \approx 1-\frac{x^2}{2} cos x ≈ 1 − 2 x 2 The function and its derivatives at x = 0 x=0 x = 0
f ( x ) = sin ( x ) f ( 0 ) = 0 f ′ ( x ) = cos ( x ) f ′ ( 0 ) = 1 f ′ ′ ( x ) = − sin ( x ) f ′ ′ ( 0 ) = 0 f ( 3 ) ( x ) = − cos ( x ) f ( 3 ) ( 0 ) = − 1 \begin{align}
f(x) &= \sin(x) & f(0) &= 0\\
f'(x) &= \cos(x) & f'(0) &= 1 \\
f''(x) &= -\sin(x) & f''(0) &= 0 \\
f^{(3)}(x) &= -\cos(x) & f^{(3)}(0) &= -1
\end{align} f ( x ) f ′ ( x ) f ′′ ( x ) f ( 3 ) ( x ) = sin ( x ) = cos ( x ) = − sin ( x ) = − cos ( x ) f ( 0 ) f ′ ( 0 ) f ′′ ( 0 ) f ( 3 ) ( 0 ) = 0 = 1 = 0 = − 1 So:
sin x ≈ x − x 3 6 \sin x \approx x-\frac{x^3}{6} sin x ≈ x − 6 x 3 The function and its derivatives at x = 0 x=0 x = 0
f ( x ) = e x f ( 0 ) = 1 f ′ ( x ) = e x f ′ ( 0 ) = 1 \begin{align}
f(x)&=e^x & f(0) &= 1\\
f'(x)&=e^x & f'(0) &= 1
\end{align} f ( x ) f ′ ( x ) = e x = e x f ( 0 ) f ′ ( 0 ) = 1 = 1 So:
e x ≈ 1 + x e^x \approx 1+x e x ≈ 1 + x The function and its derivatives at x = 0 x=0 x = 0
f ( x ) = ln ( 1 + x ) f ( 0 ) = 0 f ′ ( x ) = 1 1 + x f ′ ( 0 ) = 1 f ′ ′ ( x ) = − 1 ( 1 + x ) 2 f ′ ′ ( 0 ) = − 1 \begin{align}
f(x)&=\ln(1+x) & f(0) &= 0\\
f'(x)&=\frac{1}{1+x} & f'(0) &= 1\\
f''(x)&=-\frac{1}{(1+x)^2} & f''(0) &= -1
\end{align} f ( x ) f ′ ( x ) f ′′ ( x ) = ln ( 1 + x ) = 1 + x 1 = − ( 1 + x ) 2 1 f ( 0 ) f ′ ( 0 ) f ′′ ( 0 ) = 0 = 1 = − 1 So:
ln ( 1 + x ) ≈ x − x 2 2 \ln(1+x) \approx x-\frac{x^2}{2} ln ( 1 + x ) ≈ x − 2 x 2 The function and its derivatives at x = 0 x=0 x = 0
f ( x ) = 1 1 + x f ( 0 ) = 1 f ′ ( x ) = − 1 ( 1 + x ) 2 f ′ ( 0 ) = − 1 \begin{align}
f(x)&=\frac{1}{1+x} & f(0) &= 1\\
f'(x)&=-\frac{1}{(1+x)^2} & f'(0) &= -1
\end{align} f ( x ) f ′ ( x ) = 1 + x 1 = − ( 1 + x ) 2 1 f ( 0 ) f ′ ( 0 ) = 1 = − 1 So:
1 1 + x ≈ 1 − x \frac{1}{1+x} \approx 1-x 1 + x 1 ≈ 1 − x Recall:
cos x ≈ 1 − x 2 2 \cos x \approx 1-\frac{x^2}{2} cos x ≈ 1 − 2 x 2 So:
cos ( 4 x ) ≈ 1 − ( 4 x ) 2 2 = 1 − 8 x 2 \cos(4x) \approx 1-\frac{(4x)^2}{2} = 1-8x^2 cos ( 4 x ) ≈ 1 − 2 ( 4 x ) 2 = 1 − 8 x 2 Recall:
cos ( x ) ≈ 1 − x 2 2 \cos(x) \approx 1-\frac{x^2}{2} cos ( x ) ≈ 1 − 2 x 2 So:
cos ( x − π ) ≈ 1 − ( x − π ) 2 2 \cos(x-\pi) \approx 1-\frac{(x-\pi)^2}{2} cos ( x − π ) ≈ 1 − 2 ( x − π ) 2 about x 0 = π x_0 = \pi x 0 = π
Recall:
e x ≈ 1 + x sin ( x ) ≈ x − x 3 6 \begin{align}
e^x &\approx 1+x \\
\sin(x) &\approx x-\frac{x^3}{6}
\end{align} e x sin ( x ) ≈ 1 + x ≈ x − 6 x 3 So:
e x sin x ≈ ( 1 + x + ⋯ ) ( x − x 3 6 + ⋯ ) = x + x 2 e^x \sin x \approx (1+x+\cdots)(x-\frac{x^3}{6}+\cdots) = x+x^2 e x sin x ≈ ( 1 + x + ⋯ ) ( x − 6 x 3 + ⋯ ) = x + x 2 The function and its derivatives at x = π x=\pi x = π
f ( x ) = cos ( x ) f ( π ) = − 1 f ′ ( x ) = − sin ( x ) f ′ ( π ) = 0 f ′ ′ ( x ) = − cos ( x ) f ′ ′ ( π ) = 1 \begin{align}
f(x) &= \cos(x) & f(\pi) &= -1 \\
f'(x) &= -\sin(x) & f'(\pi) &= 0 \\
f''(x) &= -\cos(x) & f''(\pi) &= 1
\end{align} f ( x ) f ′ ( x ) f ′′ ( x ) = cos ( x ) = − sin ( x ) = − cos ( x ) f ( π ) f ′ ( π ) f ′′ ( π ) = − 1 = 0 = 1 So:
cos x ≈ − 1 + 1 2 ( x − π ) 2 \cos x \approx -1+\frac{1}{2}(x-\pi)^2 cos x ≈ − 1 + 2 1 ( x − π ) 2 at x = π x = \pi x = π
The function and its derivatives at x = 2 x=2 x = 2
f ( x ) = 7 x 2 − 6 x + 1 f ( 2 ) = 17 f ′ ( x ) = 14 x − 6 f ′ ( 2 ) = 22 \begin{align}
f(x)&=7x^2-6x+1 & f(2) &= 17\\
f'(x)&=14x-6 & f'(2) &= 22
\end{align} f ( x ) f ′ ( x ) = 7 x 2 − 6 x + 1 = 14 x − 6 f ( 2 ) f ′ ( 2 ) = 17 = 22 So:
f ( x ) ≈ 17 + 22 ( x − 2 ) f(x) \approx 17+22(x-2) f ( x ) ≈ 17 + 22 ( x − 2 )