Taylor series of √(1 + x) up to degree 7
The Taylor polynomial of degree 7 for √(1 + x) about 0 (the Maclaurin polynomial) is 1 + x/2 - x²/8 + x³/16 - 5x⁴/128 + 7x⁵/256 - 21x⁶/1024 + 33x⁷/2048. The full series converges for |x| < 1.
How to work it out
- 1
The Taylor series of f about 0 uses the derivatives of f at 0:
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For f(x) = √(1 + x) the first derivatives at 0 are:
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Divide each by k! and keep the terms up to degree 7:
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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
\[ \sqrt{1 + x} \approx 1 + \frac{1}{2}x - \frac{1}{8}x^{2} + \frac{1}{16}x^{3} - \frac{5}{128}x^{4} + \frac{7}{256}x^{5} - \frac{21}{1024}x^{6} + \frac{33}{2048}x^{7} \]
\[ f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!}\,x^k \]
\[ f^{(0)}(0) = 1,\quad f^{(1)}(0) = \frac{1}{2},\quad f^{(2)}(0) = -\frac{1}{4},\quad f^{(3)}(0) = \frac{3}{8},\quad f^{(4)}(0) = -\frac{15}{16},\quad f^{(5)}(0) = \frac{105}{32},\quad f^{(6)}(0) = -\frac{945}{64} \]Degree 7 polynomial against the real √(1 + x)
| x | Polynomial | Exact | Error |
|---|---|---|---|
| 0.1 | 1.0488088 | 1.0488088 | 1.21e-10 |
| 0.5 | 1.2247810 | 1.2247449 | 3.62e-5 |
| 0.9 | 1.3816380 | 1.3784049 | 3.23e-3 |
| 2 | 3.0000000 | 1.7320508 | 1.27e+0 |
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 7 Taylor polynomial of √(1 + x)?
1 + x/2 - x²/8 + x³/16 - 5x⁴/128 + 7x⁵/256 - 21x⁶/1024 + 33x⁷/2048
What is the radius of convergence of the series of √(1 + x)?
1
What is the Maclaurin series?
A Taylor series centred at 0.