Maclaurin Series of Sqrt(1+x)
In this tutorial we shall derive the series expansion of $$\sqrt {1 + x} $$ by using Maclaurin’s series expansion function.
Consider the function of the form
\[f\left( x \right) = \sqrt {1 + x} \]
Using $$x = 0$$, the given equation function becomes
\[f\left( 0 \right) = \sqrt {1 + 0} = \sqrt 1 = 1\]
Now taking the derivatives of the given function and using $$x = 0$$, we have
\[\begin{gathered} f’\left( x \right) = \frac{1}{{2\sqrt {1 + x} }} = \frac{1}{2}{\left( {1 + x} \right)^{ – \frac{1}{2}}},\,\,\,\,\,\,\,\,\,\,f’\left( 0 \right) = \frac{1}{{2\sqrt {1 + 0} }} = \frac{1}{{2\sqrt 1 }} = \frac{1}{2} \\ f”\left( x \right) = – \frac{1}{4}{\left( {1 + x} \right)^{ – \frac{3}{2}}},\,\,\,\,\,\,\,\,\,\,f”\left( 0 \right) = – \frac{1}{4}{\left( {1 + 0} \right)^{ – \frac{3}{2}}} = – \frac{1}{4} \\ f”’\left( x \right) = \frac{3}{8}{\left( {1 + x} \right)^{ – \frac{5}{2}}},\,\,\,\,\,\,\,\,\,\,f”’\left( 0 \right) = \frac{3}{8}{\left( {1 + 0} \right)^{ – \frac{5}{2}}} = \frac{3}{8} \\ \cdots \cdots \cdots \; \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\ \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\ \end{gathered} \]
Now using Maclaurin’s series expansion function, we have
\[f\left( x \right) = f\left( 0 \right) + xf’\left( 0 \right) + \frac{{{x^2}}}{{2!}}f”\left( 0 \right) + \frac{{{x^3}}}{{3!}}f”’\left( 0 \right) + \cdots \]
Putting the values in the above series, we have
\[\begin{gathered} \sqrt {1 + x} = 1 + x\left( {\frac{1}{2}} \right) + \frac{{{x^2}}}{{2!}}\left( { – \frac{1}{4}} \right) + \frac{{{x^3}}}{{3!}}\left( { – \frac{3}{8}} \right) + \cdots \\ \sqrt {1 + x} = 1 + \frac{x}{2} – \frac{{{x^2}}}{8} + \frac{{{x^3}}}{{16}} – \cdots \\ \end{gathered} \]
Benjamin
January 16 @ 1:21 am
There is an error in the (-3/8) after ‘, we have’ , the – sign should not be there.
And two more terms show it is not just powers of two in the expansion:
sqrt(1+x) = 1 .. – 5/128 x^4 + 7/32 x^5 ..
Esfandiar Bandari
February 14 @ 11:54 am
I did the math and I think turns into calculating Gamma divided by factorial. Specifically, the formula is something like:
1+ sigma_{n=1 to infinity} \fract{\Gamma \left( n -1.5 \right) }{n\ !} * (-1)^n * x^n
Sorry my latex is a bit rusty. As for if the fraction can be simplified? There is this for calculating \Gamma \left( n +0.5 \right)
https://math.stackexchange.com/questions/1444967/proving-that-gamma-leftn-frac12-right-frac2n-sqrt-pi22n
Cheers, E