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Taylor series of √(1 + x) up to degree 6

The Taylor polynomial of degree 6 for √(1 + x) about 0 (the Maclaurin polynomial) is 1 + x/2 - x²/8 + x³/16 - 5x⁴/128 + 7x⁵/256 - 21x⁶/1024. The full series converges for |x| < 1.

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√(1 + x) and its degree 6 Taylor polynomial-1.5-1-0.50.511.50.20.40.60.811.21.41.61.8√(1 + x)degree 6
√(1 + x) and its degree 6 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 + x) the first derivatives at 0 are:

    f(0)(0)=1,f(1)(0)=12,f(2)(0)=−14,f(3)(0)=38,f(4)(0)=−1516,f(5)(0)=10532,f(6)(0)=−94564f^{(0)}(0) = 1,\quad f^{(1)}(0) = \frac{1}{2},\quad f^{(2)}(0) = -\frac{1}{4},\quad f^{(3)}(0) = \frac{3}{8},\quad f^{(4)}(0) = -\frac{15}{16},\quad f^{(5)}(0) = \frac{105}{32},\quad f^{(6)}(0) = -\frac{945}{64}
  3. 3

    Divide each by k! and keep the terms up to degree 6:

    1+x≈1+12x−18x2+116x3−5128x4+7256x5−211024x6\sqrt{1 + x} \approx 1 + \frac{1}{2}x - \frac{1}{8}x^{2} + \frac{1}{16}x^{3} - \frac{5}{128}x^{4} + \frac{7}{256}x^{5} - \frac{21}{1024}x^{6}
  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
\[ \sqrt{1 + x} \approx 1 + \frac{1}{2}x - \frac{1}{8}x^{2} + \frac{1}{16}x^{3} - \frac{5}{128}x^{4} + \frac{7}{256}x^{5} - \frac{21}{1024}x^{6} \]
\[ f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!}\,x^k \]
\[ f^{(0)}(0) = 1,\quad f^{(1)}(0) = \frac{1}{2},\quad f^{(2)}(0) = -\frac{1}{4},\quad f^{(3)}(0) = \frac{3}{8},\quad f^{(4)}(0) = -\frac{15}{16},\quad f^{(5)}(0) = \frac{105}{32},\quad f^{(6)}(0) = -\frac{945}{64} \]

Degree 6 polynomial against the real √(1 + x)

xPolynomialExactError
0.11.04880881.04880881.49e-9
0.51.22465521.22474498.97e-5
0.91.37393111.37840494.47e-3
20.937500001.73205087.95e-1

Questions

What is the Taylor series of √(1 + x)?

1 + x/2 - x²/8 + x³/16 - 5x⁴/128 + 7x⁵/256 - 21x⁶/1024 + 33x⁷/2048 - 429x⁸/32768 + …

What is the degree 6 Taylor polynomial of √(1 + x)?

1 + x/2 - x²/8 + x³/16 - 5x⁴/128 + 7x⁵/256 - 21x⁶/1024

What is the radius of convergence of the series of √(1 + x)?

1

What is the Maclaurin series?

A Taylor series centred at 0.

Other degrees of √(1 + x)

Degree 6 of other functions

    Taylor series of √(1 + x): degree 6 polynomial and error | WordGPT