The Binomial Series: Extending the Validity?

[PFuser1232]

The binomial series $(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!} x^2 + ...$ only converges for $|x| < 1$ right?
Is it true that writing $(1 + x)^n$ differently (i.e. $x^n (1 + \frac{1}{x})^n$) extends the validity of this series to include values of $x$ such that $|x| > 1$?

 

[Svein]

The binomial series (1+x)n=1+nx+n(n−1)2!x2+...(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!} x^2 + ... only converges for |x|<1|x| < 1 right?

No. A binomial series is only defined for a given n, Whatever the value of x, there exists an N such that nx&gt;1000 for n>N.

Is it true that writing (1+x)n(1 + x)^n differently (i.e. xn(1+1x)nx^n (1 + \frac{1}{x})^n) extends the validity of this series to include values of xx such that |x|>1|x| > 1?

No.

 

[micromass]

The binomial series $(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!} x^2 + ...$ only converges for $|x| < 1$ right?

Yes. Although the series might converge for more values than just $|x|<1$. Complete details are here: http://en.wikipedia.org/wiki/Binomial_series#Conditions_for_convergence

Is it true that writing $(1 + x)^n$ differently (i.e. $x^n (1 + \frac{1}{x})^n$) extends the validity of this series to include values of $x$ such that $|x| > 1$?

Yes, If $|x|>1$, then your trick can be used. If $|x|<1$, then your trick doesn't work, but the original series does of course. If $|x|=1$ then the situation is a bit annoying.

 

[PFuser1232]

No. A binomial series is only defined for a given n, Whatever the value of x, there exists an N such that nx&gt;1000 for n>N.

No.

Yes. Although the series might converge for more values than just $|x|<1$. Complete details are here: http://en.wikipedia.org/wiki/Binomial_series#Conditions_for_convergence
Yes, If $|x|>1$, then your trick can be used. If $|x|<1$, then your trick doesn't work, but the original series does of course. If $|x|=1$ then the situation is a bit annoying.

Interesting. I once watched a video in which someone proves that the sum of all natural numbers is $-\frac{1}{12}$, and in one of the steps, he substituted $x = 1$ into the binomial series. Apparently, anything is possible if you're Euler!

 

[micromass]

Interesting. I once watched a video in which someone proves that the sum of all natural numbers is $-\frac{1}{12}$, and in one of the steps, he substituted $x = 1$ into the binomial series. Apparently, anything is possible if you're Euler!

Yes, that is valid, but only under conditions. For the standard convergence of series, plugging in $x=1$ is forbidden. But there are other definitions where series do not converge how we they usually do. Under those definitions, you do get $-1/12$.

 

[PFuser1232]

Yes, that is valid, but only under conditions. For the standard convergence of series, plugging in $x=1$ is forbidden. But there are other definitions where series do not converge how we they usually do. Under those definitions, you do get $-1/12$.

Is it possible to approximate $\sqrt{2}$ using the binomial series?

 

[micromass]

Yes, but not directly. What you can do (for example) is to use $x=-1/2$ to approximate $1/\sqrt{2} = \frac{\sqrt{2}}{2}$. And then you can easily find $\sqrt{2}$.

Edit: I guess you can even do it directly, but convergence won't be very rapid.