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Taylor series of 1/(1 + x²) up to degree 8

The Taylor polynomial of degree 8 for 1/(1 + x²) about 0 (the Maclaurin polynomial) is 1 - x² + x⁴ - x⁶ + x⁸. The full series converges for |x| < 1.

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1/(1 + x²) and its degree 8 Taylor polynomial-1.5-1-0.50.511.50.20.40.60.811/(1 + x²)degree 8
1/(1 + x²) and its degree 8 Taylor polynomial. The polynomial hugs the curve inside the radius of convergence (1) and leaves it outside.

How to work it out

  1. 1

    The Taylor series of f about 0 uses the derivatives of f at 0:

    f(x)=∑k=0∞f(k)(0)k! xkf(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!}\,x^k
  2. 2

    For f(x) = 1/(1 + x²) the first derivatives at 0 are:

    f(0)(0)=1,f(1)(0)=0,f(2)(0)=−2,f(3)(0)=0,f(4)(0)=24,f(5)(0)=0,f(6)(0)=−720f^{(0)}(0) = 1,\quad f^{(1)}(0) = 0,\quad f^{(2)}(0) = -2,\quad f^{(3)}(0) = 0,\quad f^{(4)}(0) = 24,\quad f^{(5)}(0) = 0,\quad f^{(6)}(0) = -720
  3. 3

    Divide each by k! and keep the terms up to degree 8:

    11+x2≈1−x2+x4−x6+x8\frac{1}{1 + x^{2}} \approx 1 - x^{2} + x^{4} - x^{6} + x^{8}
  4. 4

    Radius of convergence: 1. Inside it, more terms give a better approximation; outside, the series does not settle.

LaTeX source of the formulas on this page
\[ \frac{1}{1 + x^{2}} \approx 1 - x^{2} + x^{4} - x^{6} + x^{8} \]
\[ f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!}\,x^k \]
\[ f^{(0)}(0) = 1,\quad f^{(1)}(0) = 0,\quad f^{(2)}(0) = -2,\quad f^{(3)}(0) = 0,\quad f^{(4)}(0) = 24,\quad f^{(5)}(0) = 0,\quad f^{(6)}(0) = -720 \]

Degree 8 polynomial against the real 1/(1 + x²)

xPolynomialExactError
0.10.990099010.990099019.90e-11
0.50.800781250.800000007.81e-4
0.90.745126210.552486191.93e-1
2205.000000.200000002.05e+2

Questions

What is the Taylor series of 1/(1 + x²)?

1 - x² + x⁴ - x⁶ + x⁸ + …

What is the degree 8 Taylor polynomial of 1/(1 + x²)?

1 - x² + x⁴ - x⁶ + x⁸

What is the radius of convergence of the series of 1/(1 + x²)?

1

What is the Maclaurin series?

A Taylor series centred at 0.

Other degrees of 1/(1 + x²)

Degree 8 of other functions

    Taylor series of 1/(1 + x²): degree 8 polynomial and error | WordGPT