Find the first three non zero terms of the Maclaurin expansion and the radius of convergence of the following function: $$f(x)=(4-x)^{1/2}$$
First I found the following to be: $$f(0)=2$$
$$f^{'}(0)=\frac{-1}{4}$$
$$f^{''}(0)=-1/32$$
$$f^{'''}(0)=-3/256$$
The series is
$$2-\frac{x}{4}-\frac{1}{32}\frac{1}{2!}x^{2}-\frac{3}{256}\frac{1}{3!}x^{3}+…$$
Which I am pretty sure is right, the problem comes with finding the radius.
I tried and found a general solution which is: $$\frac{1.-1.-1.-3.-5…(-(2n-1))}{{2^{n}2^{2n-1}}n!}x^n$$ I would really appreciate if someone check the numerator of the formula, and also guide me on the easiest way to find the nth term equation of a series.
After appling the ratio test, I found $$R=∞$$
[Math] Find the three non zero terms of the Maclaurin expansion and the radius of convergence of the following function: $f(x)=(4-x)^{1/2}$
taylor expansion
Best Answer
There is a better way of finding the radius of convergence with a function such as this. Essentially, the radius of convergence for this function (and others that are made up of radicals, rational functions, etc.) is the distance between the center of the expansion (0, in this case, since it is a Maclaurian series) and the nearest point at which the function is ill behaved; that could be a singularity in the case of rational functions, or a place where the argument of the radical is zero.
In this case, the function is $\sqrt{4-x}$. The nearest point where this function is no longer (arbitrarily) differentiable is when $x = 4$. So the radius of convergence is $|4 - 0| = 4$.
Also, in terms of determining the coefficients, we can once again use other series that we know to make life easier. We can rewrite the function in question to be $$ \sqrt{4 - x} = \sqrt{4(1 - x/4)} = 2\sqrt{1 - x/4} $$ If you then know the Maclaurian series to $\sqrt{1 - u}$, then the substitution $u =x/4$ will give you the desired answer.