1
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$
    – nukeguy
    Commented Mar 9, 2015 at 23:04
  • 1
    $\begingroup$ Very sorry. I have fixed the question $\endgroup$
    – Bob the Builds
    Commented Mar 10, 2015 at 0:12
  • $\begingroup$ 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. $\endgroup$
    – nukeguy
    Commented Mar 10, 2015 at 12:51

1 Answer 1

Reset to default
0
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .