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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.
The Taylor series for arctan is an infinite series representation of the arctangent function, which is defined as the inverse of the tangent function. It is given by the formula: arctan(x) = x - (x^3)/3 + (x^5)/5 - (x^7)/7 + ...
The Taylor series for arctan is derived by using the Taylor series expansion formula, which is based on the derivatives of a function evaluated at a specific point. In this case, the derivatives of the arctangent function are evaluated at x = 0, and the resulting series is centered at x = 0.
No, the Taylor series for arctan only converges for values of x within the interval [-1, 1]. This is because the arctangent function is undefined for values outside of this interval.
Yes, the Taylor series for arctan does converge at x = 1. This can be shown by substituting x = 1 into the series and using the alternating series test, which states that if the terms of an alternating series decrease in absolute value and approach 0, then the series converges.
The accuracy of the Taylor series for arctan at x = 1 depends on the number of terms used in the series. The more terms that are included, the closer the approximation will be to the actual value of arctan(1), which is equal to π/4. However, even with a large number of terms, the series will never be exact due to the nature of infinite series.