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Show that $$\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots$$
Because ##\dfrac{1}{1+z^2}## only converges for ##|z| < 1##?wrobel said:The expansion ##\arctan z=z-\frac{z^3}{3}+\ldots## follows from general theory for ##|z|<1.## The assertion ##\arctan(1)=1-\frac{1}{3}+\ldots## needs some additional work. Namely, one must show that the series in the right side converges to ##\arctan(1)## indeed.
Because it is insufficient to prove that 1-1/3+... converges, one must prove that it converges to ##\arctan 1##julian said:Because ##\dfrac{1}{1+z^2}## only converges for ##|z| < 1##?
Well, since arctanz is Complex-analytic; entire actually, that should be enough.wrobel said:Because it is insufficient to prove that 1-1/3+... converges, one must prove that it converges to ##\arctan 1##
By which theorem is it enough? Consider me as a wet blanket but I think that if we are solving a math problem, we must refer to mathematical theorems and apply them. For example, the result follows fromWWGD said:Well, since arctanz is Complex-analytic; entire actually, that should be enough.
I thought we were working over the Complexes, where analiticity is guaranteed by differentiability.wrobel said:By which theorem is it enough? Consider me as a wet blanket but I think that if we are solving a math problem, we must refer to mathematical theorems and apply them. For example, the result follows from
Theorem (Abel). Assume we have an analytic function
$$f(z)=\sum_{k=0}^\infty a_kz^k,\quad a_k\in\mathbb{R}$$ and
the Taylor series is convergent in ##\{|z|<1\}##. If a series
##a=\sum_{k=0}^\infty a_k## converges then
$$\lim_{\mathbb{R}\ni x\to 1-}f(x)=a.$$
We are working over complexes. Please present your argument as a formal proof.WWGD said:I thought we were working over the Complexes, where analiticity is guaranteed by differentiability.
You mean that a differentiable function is infinitely-differentiable and analytic?wrobel said:We are working over complexes. Please present your argument as a formal proof.
I mean posts #5, 6WWGD said:You mean that a differentiable function is infinitely-differentiable and analytic?
The issue is whether the Taylor series for ##\arctan## converges to the function at the radius of convergence (that is to say at ##x = 1##). We can take that the Taylor series converges to the function value within the radius of convergence, but not at the radius itself. That's a standard result. In this case we have:WWGD said:Isn't this one of the mainstream results in Complex Analysis? Differentiability implies ##C^{\infty}##, implies Analytic?
Thanks for clarifying. I missed that obvious point( not being sarcastic) . Embarrassed.PeroK said:The issue is whether the Taylor series for ##\arctan## converges to the function at the radius of convergence (that is to say at ##x = 1##). We can take that the Taylor series converges to the function value within the radius of convergence, but not at the radius itself. That's a standard result. In this case we have:
$$\arctan(x) = \sum_{n = 0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1} \ \ \ (|x| < 1)$$The question is what happens at ##x = 1##? The series for ##x = 1## converges, by the alternating series test, but does it converge to ##\arctan(1)##?
Note that I posted a proof from Wiki above (post #9), that uses integration over the interval ##(0, 1)## to prove the equality at the endpoint.
I'd also be interested to see an alternative proof using complex numbers, if you could supply one.