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Topic 1 of 6

The Taylor Formula

A Taylor series lets you write any smooth function as an infinite polynomial. If you know all the derivatives of f at a single point a, you know the function everywhere near a. This seemingly simple idea underlies nearly every approximation in physics.

Key Concepts

Taylor Series
The expansion f(x)=∑n=0∞f(n)(a)n!(x−a)nf(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n writes any smooth function as a power series around the expansion point aa.
Maclaurin Series
The special case of a Taylor series expanded around a=0a = 0. Most physics approximations are Maclaurin series — we expand near equilibrium, small angles, or zero coupling.
Taylor Coefficient
The coefficient of (x−a)n(x-a)^n is cn=f(n)(a)/n!c_n = f^{(n)}(a)/n!, determined entirely by the nn-th derivative of ff at the expansion point.
N-th Order Polynomial
Keeping the first N+1N+1 terms gives TN(x)T_N(x), an approximation valid when (x−a)(x-a) is small. Adding more terms improves accuracy and extends the valid range.
Analytic Function
A function is analytic at aa if its Taylor series converges to it in some open neighborhood of aa. All elementary functions — sin, cos, exe^x, polynomials — are analytic on their domains.

Key Equations

General Taylor Series
f(x)=∑n=0∞f(n)(a)n! (x−a)nf(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}\,(x-a)^n
Expands ff around the point aa using all its derivatives there.
Maclaurin Series
f(x)=f(0)+f′(0) x+f′′(0)2! x2+f′′′(0)3! x3+⋯f(x) = f(0) + f'(0)\,x + \frac{f''(0)}{2!}\,x^2 + \frac{f'''(0)}{3!}\,x^3 + \cdots
Taylor series centered at a=0a=0, the most common form in physics.
N-th Order Approximation
TN(x)=∑n=0Nf(n)(a)n! (x−a)nT_N(x) = \sum_{n=0}^{N} \frac{f^{(n)}(a)}{n!}\,(x-a)^n
The finite truncation, valid when (x−a)(x-a) is small.
Worked Example

Finding the Maclaurin Series of exe^x

Problem

Derive the Maclaurin series for f(x)=exf(x) = e^x using the formula cn=f(n)(0)/n!c_n = f^{(n)}(0)/n!.

Solution

All derivatives of exe^x are exe^x itself, so f(n)(0)=e0=1f^{(n)}(0) = e^0 = 1 for all n≥0n \geq 0.

cn=f(n)(0)n!=1n!c_n = \frac{f^{(n)}(0)}{n!} = \frac{1}{n!}

Substituting into the Maclaurin formula:

ex=∑n=0∞xnn!=1+x+x22!+x33!+x44!+⋯e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots

This converges for all real xx. At x=1x=1 we get e=1+1+1/2+1/6+1/24+…≈2.71828e = 1 + 1 + 1/2 + 1/6 + 1/24 + \ldots \approx 2.71828.

Answer ex=∑n=0∞xnn!e^x = \displaystyle\sum_{n=0}^{\infty} \frac{x^n}{n!}, with Taylor coefficients cn=1/n!c_n = 1/n!
Practice

Exercises

6 problems
1 of 6

What is the coefficient c3c_3 (the coefficient of x3x^3) in the Maclaurin series of exe^x? Give your answer as a decimal.

2 of 6

Evaluate the 2-term Maclaurin approximation ex≈1+xe^x \approx 1 + x at x=0.5x = 0.5.

3 of 6

The Maclaurin series of cos⁡(x)\cos(x) contains only even powers. What is the coefficient of x4x^4? (Give as a decimal.)

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4 of 6

Use the 1st-order Taylor expansion f(x)≈f(1)+f′(1)(x−1)f(x) \approx f(1) + f'(1)(x-1) with f(x)=xf(x) = \sqrt{x} to approximate 1.04\sqrt{1.04}.

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5 of 6

What is the coefficient of x5x^5 in the Maclaurin series of sin⁡(x)\sin(x)?

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6 of 6

Evaluate the 4-term series sin⁡(x)≈x−x3/6+x5/120−x7/5040\sin(x) \approx x - x^3/6 + x^5/120 - x^7/5040 at x=0.1x = 0.1. Round to 6 decimal places.

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Key Takeaways

  • The Taylor series f(x)=∑n=0∞f(n)(a)n!(x−a)nf(x) = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n encodes all derivative information at the expansion point aa.
  • The Maclaurin series is the special case a=0a=0, the most common form in physics.
  • The nn-th Taylor coefficient is cn=f(n)(a)/n!c_n = f^{(n)}(a)/n!.
  • Truncating to NN terms gives TN(x)T_N(x), a polynomial approximation that improves as ∣x−a∣|x-a| decreases.
  • Every smooth function is locally a polynomial — this is the fundamental insight behind nearly every physics approximation.