# Using Taylor's Theorem and the Constancy Theorem, solve the following proof.

Using Taylor's Theorem and the Constancy Theorem prove that

$\sqrt{1+x}=1+\frac{x}{2}+\sum_{n=2}^{\infty}(-1)^{n-1} \frac{1}{2n} \frac{(1- \frac{1}{2})(2- \frac{1}{2}) ... ((n-1)- \frac{1}{2})}{(n-1)!}x^n$

Constancy Theorem Let $f: (a,b) \rightarrow \mathbb{R}$ be differentiable, and satisfy $f'(t)=0$, $\forall t\in(a,b)$. Then, $f$ is constant on $(a,b)$.

I am complete rubbish with Taylor Theorems. Could someone explain the general approach one takes with these sorts of questions. I have largely been given questions asking to only consider a Taylor Expression going out to the $n$-th derivative, where $n$ is small. In those cases, I have just brute forced it by taking derivatives out to the $n$-th derivative.

Clearly, there is a more elegant way to do these things. Help.

• Did you type the statement correctly? This can't possibly be right. The summation doesn't depend on $x$, so it's a constant. You're saying $\sqrt{1+x} = 1 + \frac{x}{2} + C$ holds true for all values of $x$, which is impossible. Commented Mar 9, 2015 at 23:04
• Very sorry. I have fixed the question Commented Mar 10, 2015 at 0:12
• Well, the right hand side is a power series. You should be able to just look at the general definition of a Taylor series, and match the coefficients. See if you can come up with a general formula for the $n$-th derivative of $f(x) = \sqrt{1+x}$. If you take a few derivatives of $f(x)$, you should see a pattern start to emerge. Then plug your result into the Taylor series definition and try to match it to the right hand side of the equation you're trying to prove. Commented Mar 10, 2015 at 12:51

For general real $α$ you can show that Newton's binomial series $$(1+x)^α=\sum_{k=0}^{\infty}\binom{α}{k}x^k$$ converges for $|x|<1$.
In this special case, $$\binom{1/2}{k}=\frac{\frac12(\frac12-1)...(\frac12-k+1)}{k!} =\frac{(-1)^{k-1}}{2^k}\frac{(2k-3)(2k-5)...3·1}{k!}$$ which has the same factors as your form of the coefficients, only differently arranged.