User:Oh Isaac/Hypergeomtric Function2

This report is a summary of the discussion between Professor Loc Vu-Quoc and Isaac Lin about hypergeometric functions.

Given
Hypergeomtric function could be write as a series:

in which:

$$(q)_n = \left\{ \begin{array}{ll} 1                    & \mbox{if } n = 0 \\ q(q+1) \cdots (q+n-1) & \mbox{if } n > 0 \end{array} \right .$$

provided that c does not equal 0, −1, −2, ... . Here (q)n is the Pochhammer symbol and notice that the series terminates if either a or b is a nonpositive integer. For complex arguments z with |z| ≥ 1 it can be analytically continued along any path in the complex plane that avoids the branch points 0 and 1.

Find
In question R5.11 of in the course of EGM 6321 Principles of Engineering Analysis 1, programs made by Matlab code shows that the formula $$\,_2F_1(\frac{1}{3},1;\frac{4}{3},x)$$ has imaginary parts, which is also shown in the website of WolframAlpha as below.

So here comes the question: How could hypergeometric function, which can be expressed as the sum of a real number series, has complex part?

Background Knowledge
The Hypergeometric function$$F(a,b;c;x)$$satisfies the hypergeometric equation:

In the equation the argument, x, and three paramenters, a,b, and c, are not restricted to be real only. This equation has three regular singular points at $$z=0,1$$and $$\infty$$, unless certain relations are applied to three parameters.

Near the point $$z=0$$ the hupergeometric equation has two independent solutions, one is the series expansision of $$, but a more formal form of the items in it is :

where $$\Gamma$$is the Gamma function. Gamma function defined as$$\Gamma(n) = (n-1)!\,$$, when n is real.

Another solution of hypergeomtric equation near 0 is

The series $$converges for |x|<1, and could be expressed as a uniformly converging integral,

Where$$Re[c]>Re[b]>0, |x-1|>0, \mbox{arg}(1-x)<\pi.$$

Accoring to YUPAI P. HSU, the integral in $$is analytic in x for every t. Accordingly, the hypergeometric function, $$F(a,b;c;x)$$is analytic in the whole x-plane cut along $$[0,\infty]$$, not just within the original |x|<1. And So can it derives to the whole x-plane. Regarded as a function of the three complex paramenters, the hypergeometric funciton is an entire function of both a and b, and a meromorphic function of c, with only simple poles at $$c=0,-1,-2,\cdots$$, which is shown in $$caused by the singularities of parameters in gamma function $$\Gamma(c)$$.

The hypergeometric function can be analytically extended to other region from within the unit disk |x|<1 by substitute the original argument to other forms. For example, by substitute t to s-1, $$can be transformed

In order to keep convergence, we need $$|\frac{z}{z-1}|<1$$, for complex number z

So the available z extends to any $$Re[z]<\frac{1}{2}$$, there is $$|\frac{z}{z-1}|<1$$. On the right hand side a possible convergent series is obtained which can be used to calculate. There are twenty four valid transformations like $$obtained by Kummer, called Kummer's Solution.

To do this, the real axis was divided in to six intervals and in each interval the argument w is strictly limited in the convergence range so that the series in powers of w is kept converged rapidly, which are shown in the table below. Details of the whole transformations could refer to Rorbert C. Forrey.

Explanation of the Problem
In the problem referred before, the complex value appear in the domain that $$x>2$$in $$\,_2F_1(\frac{1}{3},1;\frac{4}{3},x)$$. In this case, when calculating the value of it, the substitution of argument is from x to $$\frac{1}{x}$$, and the transformation formula is

with constraints that

so we get a convergent series to calculate the value when |x|>2.According to $$i=\sqrt{-1}$$, once a or b is proper fraction and x is positive,$$will generate imaginary part from $$(-x)^{a}$$and $$(-x)^{b}$$. This can explain why complex parts appear in the problem.

Conclusion
So we know from above that the hypergeometric series $$is a expression to represent hypergeometric function when argument x is limited in |x|<1, but is not the only one to describe it. $$and its transformations are also used to describe hypergeometric function in different domains and the transformations may generate complex value. All of them above are the solutions of hypergeometric equation.

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Trial Explanation 1
by Loc Vu-Quoc

i thought that there could be a series of real numbers that converges to an imaginary number since $$i=\sqrt{-1}$$. so if you consider the power series of, say, $$(1+x)^{-1/2}$$, and let $$x\rightarrow-2$$, then you would get the imaginary unit i. you may want to think along this line.

of course, you had an example of an infinite series of real numbers converging to complex numbers: Review the exponential of a matrix lecture in which i gave an example of a real matrix whose exponent contained complex numbers (oscillations).

re the converged result of a series as a complex number, think of the exponential of a matrix for the case of oscillations that i mentioned to you and that i had in my lecture notes. if you truncate the series (no matter how high the number of terms you took, 400, 500, 1 million, 1 trillion, etc.), you would always get a real matrix ! the converged series, i.e., with infinitely many terms, is a matrix with complex coefficients.

Problem
by Lin

I understand the the exponential of a matrix oscillates because it multiplies by i by itself, but in this case it seems to involved more complex matrix computations and we should have a way to find the imaginary part in it like we can use Euler formula to separate sin and cos into the form contains i. But in the $$I don't find any component of i. I don't know whether I understand your meaning.

$$expm\left[\begin{vmatrix} 0&-1 \\ 1&0 \end{vmatrix}t\right]=\begin{vmatrix} \cos{x} & -\sin{x}\\ \sin{x} & \cos{x} \end{vmatrix}=\begin{vmatrix} \frac{e^{-ti}+e^{ti}}{2}& \frac{-e^{-ti}+e^{ti}}{2}i\\ \frac{e^{-ti}-e^{ti}}{2}i&\frac{e^{-ti}+e^{ti}}{2} \end{vmatrix}$$

the value of this matrix also contains imaginary part when t change:

Trial Explanation 2
by Lin

According to Rorbert C. Forrey, the hypergeometric fuction$$\,_2F_1$$ is defined by series $$, which converges for$$|x|<1 $$as can be seen by the ratio test. For z outside the circle of convergence, it is necessary to transform z into a new form of hypergeomtric function with argument w that $$|x|<1 $$.

To do this, the real axis was divided in to six intervals and in each interval the argument w is strictly limited in the convergence range so that the series in powers of w is kept converged rapidly.

For example:

When $$-\infty<x<-1$$, the transformation of argument is $$w=\frac{1}{1-x}$$, and the transformation of function is:

In this way, the range is limited and divergence is almost avoid. (They also use finite difference and similar function to avoid infinite and divergence caused by Gamma function. The Gamma function and other transformation function may generate imaginary part). So the value of hypergeomtric function is not calculated only by the sum of a series of real numbers but get involved with complex functions, which made the result complex when software uses this method to calculate the value.

Trial Explanation 3
by Lin

Maybe it is just a extension tips automatically caused by the computer or website. Consider a series:

And we know:

The WolframAlpha shows that this series has a imaginary part. But as we know, $$diverge in the range $$|x|>1$$,so it has no value. But WolframAlpha show a complex part of it which in fact should be the complex graph of $$, not the original series.

Problem
by Lin I think the value of andare equal just in a particular range, they themselves are not the same thing. But when I chose "real-value plot" in WolframAlpha the imaginary part disappeared.I don't know what is the difference.

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