Taylor Series Math Example 1

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Example 1

easy
Find the Maclaurin series for exe^x up to the x4x^4 term.

Solution

  1. 1
    The Taylor series formula is f(x)=∑n=0∞f(n)(0)n!xnf(x) = \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n. Compute the derivatives of exe^x at x=0x=0.
  2. 2
    Since every derivative of exe^x is exe^x, we have f(n)(0)=e0=1f^{(n)}(0) = e^0 = 1 for all nn.
  3. 3
    Substitute into the formula: ex=∑n=0∞xnn!=1+x+x22!+x33!+⋯e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!} = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots

Answer

ex=1+x+x22+x36+x424+⋯e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \cdots
Every derivative of exe^x is exe^x, so all coefficients are 1/n!1/n!. Converges for all xx.

About Taylor Series

A representation of a function as an infinite sum of terms calculated from the function's derivatives at a single point: f(x)=∑n=0∞f(n)(a)n!(x−a)nf(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n
When a=0a = 0, it's called a Maclaurin series.

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More Taylor Series Examples

Example 2 hard

Find the Maclaurin series for [formula] and state the interval of convergence.

Example 3 easy

Write the first four non-zero Maclaurin terms for [formula].

Example 4 medium

Use three terms of the Maclaurin series for [formula] to approximate [formula].

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