Maclaurin Series of $\sqrt[a]{x+1}$ Without Derivatives

Question

I am trying to find the Maclaurin series of $\sqrt[a]{x+1}$ where $a > 1$.

My approach so far is:

$$\sqrt[a]{x+1} = c_0 + c_1x + c_2x^2 + c_3x^3 ...$$ $$Let \; x \rightarrow 0$$ $$c_0 = 1$$ $$\sqrt[a]{x+1} - 1 = c_1x + c_2x^2 + c_3x^3 + c_4x^4...$$ $$\frac {\sqrt[a]{x+1} - 1} x = c_1 + c_2x^1 + c_3x^2 + c_4x^3...$$ $$Let \; x \rightarrow 0$$ $$\lim_{x \rightarrow 0}{\frac {\sqrt[a]{x+1} - 1} x} = c_1$$

But then I get stuck at solving this limit, as my goal is avoid derivatives and L'Hopital's rule. If I can find the Maclaurin series of $\sqrt[a]{x+1}$, I'm hoping I can find Binomial Coefficients (or rows of Pascal's Triangle) for all real powers (I originally said fractional powers in the question, my mistake).

Is there a way to solve this limit without L'Hoptial (e.g. squeeze theorem) or is there an easier approach to finding Binomial Coefficients of fractional power without derivatives?

P.S. This is my first time writing in LaTeX, so I appologize if any formatting is off

Answer 1

Suppose that $\frac{1}{a} = \frac{p}{q} > 0$ is a rational number, i.e. $p,q>0$ are integers. Suppose that

$\sqrt[a]{1+x} = (1+x)^{p/q} = c_0 + c_1 x + c_2 x^2 + \ldots$

Raising both sides to the power $q$ yields

$(1+x)^p = (c_0 + c_1 x + \ldots)^q$

and by truncating to the relevant order and using the ordinary binomial theorem we find

$1 + \binom{p}{1} x + \binom{p}{2} x^2 + \ldots = c_0^q + c_0^{q-1} c_1 \binom{q}{1} x + c_0^{q-2} c_1^2 \binom{q}{2} + c_0^{q-1} c_2 \binom{q}{1} x^2 + \ldots$

where the terms indicated by '$\ldots$' are all cubic or higher. Hence, by comparing terms, we obtain

$c_0 = 1$,

$c_1 = p/q$, and

$c_2 = \frac{1}{q} \binom{p}{2} - \frac{1}{q} c_1^2 \binom{q}{2} = \frac{p}{2q} \left(\frac{p}{q} - 1\right)$.

One can in principle go to higher orders, but in practice it is increasingly tedious.

Finally, to say something about the case when $\frac{1}{a}$ is irrational, note that $g(a;x) = (1+x)^{1/a}$ is a continuous function of $a$. Hence, the expression derived above for rational $1/a>0$ must also apply for irrational $a$, just with $1/a$ substituted for the ratio $p/q$.

Answer 2

We'll do $n=2.$ $$\begin{align} \sqrt{1+x}-1&=\sqrt{1+x}-(1+x/2)+x/2 \\&=-\frac{-x^2}{\sqrt{1+x}+1+x/2} +\frac{x}2. \end{align}\tag1$$

The second step you are using $a-b=\frac{a^2-b^2}{a+b},$ with $a=\sqrt{1+x},b=1+x/2.$

Dividing the right side of $(1)$ by $x$ and letting $x\to 0,$ you get a limit of $\frac12.$

For general $n,$ we use $$a-b=\frac{a^n-b^n}{a^{n-1}+a^{n-2}b+\cdots+b^{n-1}}.$$

But each individual step after this is harder and harder, and this assumes you know the value $c_1$ already.

The next term you want $c_2$ such that $(1+x/n+c_2x^2)^n$ to start with $1+x+0x^2+\cdots.$ That, in this case, means $nc_2+ \frac1{n^2}\binom{n}2 =0.$

Answer 3

Let us show that $$\lim_{x \to 0} \frac{(x+1)^a - 1}{x} = a$$ without using L'Hospital.

If $a \geq 1$ use Bernoulli's inequality for real exponent. Let us concentrate on case where $x > 0$. (Case x < 0 is symmetric).

We have $\frac{(x+1)^a - 1}{x} \geq \frac{1 + ax - 1}{x} = a$ and, on the other hand, use famous inequality $(1+x)^a \leq 1 + ax + \frac{a(a-1)}{2}x^2$ (which holds for all $x > 0$ and $a\geq1$) to obtain that $\frac{(x+1)^a - 1}{x} \leq a + \frac{a(a-1)}{2}x \to a, x \to 0+$.

If $0 < a < 1$. Analogously, consider only $x > 0$.

In this case we use Bernoulli's inequality for real exponent $0<a \leq 1$ and obtain $\frac{(x+1)^a - 1}{x} \leq \frac{1 + ax - 1}{x} = a$

And use inequality $(1+x)^a \geq 1 + ax + \frac{a(a-1)}{2}x^2$ (which holds for all $x > 0$ and $0<a\leq1$) to get $\frac{(x+1)^a - 1}{x} \geq a + \frac{a(a-1)}{2}x \to a, x \to 0+$

Answer 4

Let $y:=x+1$ and $z:=y^{p/q}$ for conciseness.

$$\frac{z-1}{x}=\frac{z-1}{x}\frac{z^{q-1}+z^{q-2}+\cdots 1}{z^{q-1}+z^{q-2}+\cdots 1}=\frac{z^q-1}{x(z^{q-1}+z^{q-2}+\cdots 1)}=\frac{y^p-1}{x(z^{q-1}+z^{q-2}+\cdots 1)}=\frac{x(y^{p-1}+y^{p-2}+\cdots 1)}{x(z^{q-1}+z^{q-2}+\cdots 1)}.$$

Now if you let $x$ vanish, $y$ and $z$ tend to $1$ and the ratio to $\dfrac pq$.

Answer 5

By $y=\exp(\frac1a\ln(x+1))$ and well-known Maclaurin series, we have $$y=\sum_{m=0}^\infty\frac1{a^mm!}\left(\sum_{n=1}^\infty\frac{(-1)^{n+1}x^n}{n}\right)^m =1+\frac1a\sum_{n=1}^\infty\frac{(-1)^{n+1}x^n}n+\frac1{2a^2} \left(\sum_{n=1}^\infty\frac{(-1)^{n+1}x^n}n\right)^2+... $$