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Taylor series of āˆš(1 + x) up to degree 1

The Taylor polynomial of degree 1 for āˆš(1 + x) about 0 (the Maclaurin polynomial) is 1 + x/2. The full series converges for |x| < 1.

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āˆš(1 + x) and its degree 1 Taylor polynomial-1.5-1-0.50.511.50.20.40.60.811.21.41.61.8āˆš(1 + x)degree 1
āˆš(1 + x) and its degree 1 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)=āˆ‘k=0āˆžf(k)(0)k!ā€‰xkf(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!}\,x^k
  2. 2

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

    f(0)(0)=1,f(1)(0)=12f^{(0)}(0) = 1,\quad f^{(1)}(0) = \frac{1}{2}
  3. 3

    Divide each by k! and keep the terms up to degree 1:

    1+xā‰ˆ1+12x\sqrt{1 + x} \approx 1 + \frac{1}{2}x
  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
\[ \sqrt{1 + x} \approx 1 + \frac{1}{2}x \]
\[ f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!}\,x^k \]
\[ f^{(0)}(0) = 1,\quad f^{(1)}(0) = \frac{1}{2} \]

Degree 1 polynomial against the real āˆš(1 + x)

xPolynomialExactError
0.11.05000001.04880881.19e-3
0.51.25000001.22474492.53e-2
0.91.45000001.37840497.16e-2
22.00000001.73205082.68e-1

Questions

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

1 + x/2 - xĀ²/8 + xĀ³/16 - 5xā“/128 + 7xāµ/256 - 21xā¶/1024 + 33xā·/2048 - 429xāø/32768 + ā€¦

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

1 + x/2

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

1

What is the Maclaurin series?

A Taylor series centred at 0.

Other degrees of āˆš(1 + x)

Degree 1 of other functions

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