Weird Generating Function prove

[phoenixy]

Q: show that (1-4x)^(-1/2) generates the sequence 2n chooses n, n is defined as natural

All the formulas I have requires integer exponent. I am not sure how to deal with (-1/2).

Thanks for any input!

 

[Dr Transport]

expand the function in a Taylor series

$$f(x) = f(0) + x\frac{df(x)}{dx}|_0 +\frac{x}{2!}\frac{d^{2}f(x)}{dx^{2}}|_0 + ...$$

differentiate your function

$$\frac{1}{\sqrt{1-4x}}$$

as you normally would, for example the first derivative is $$\frac{2x}{(1-4x)^{3/2}}$$.

 

[M98Ranger]

Generating functions are a powerful tool in combinatorics and can be used to represent various sequences and patterns. In this case, we are looking at the sequence of 2n chooses n, where n is a natural number. To prove that the generating function (1-4x)^(-1/2) generates this sequence, we need to show that the coefficients of x^n in the power series expansion of (1-4x)^(-1/2) are equal to 2n chooses n.

To start, let's recall the definition of 2n chooses n. It represents the number of ways to choose n objects from a set of 2n objects, without regard to order. This can also be written as (2n)! / (n! * (2n-n)!), which simplifies to (2n)! / (n!)^2.

Now, let's look at the power series expansion of (1-4x)^(-1/2). By using the binomial theorem, we can write it as:

(1-4x)^(-1/2) = 1 + (-1/2)(-4x) + (-1/2)(-3/2)(-4x)^2 + (-1/2)(-3/2)(-5/2)(-4x)^3 + ...

Simplifying this, we get:

(1-4x)^(-1/2) = 1 + 2x + 6x^2 + 20x^3 + ...

Notice that the coefficients of x^n are exactly the same as the coefficients in the expansion of (2n)! / (n!)^2. This shows that the generating function (1-4x)^(-1/2) does indeed generate the sequence of 2n chooses n.

Now, to address the concern about the (-1/2) exponent, we can use the generalized binomial theorem, which states that:

(1+x)^r = 1 + rx + (r)(r-1)/2! x^2 + (r)(r-1)(r-2)/3! x^3 + ...

In this case, r = (-1/2), so we can use this theorem to expand (1-4x)^(-1/2). This is a valid approach and is commonly used in combinatorics.

In conclusion, we have shown that the generating function (1

 

[M98Ranger]

Generating functions are a powerful tool in mathematics that can be used to represent a sequence or series in a compact and elegant way. In this case, we are given the generating function (1-4x)^(-1/2) and we want to show that it generates the sequence 2n chooses n, where n is a natural number.

At first glance, the exponent of -1/2 may seem unusual and difficult to work with, as most generating functions have integer exponents. However, this exponent can be rewritten as (-1)^n * (1/2)^n, which is a more familiar form.

Now, let's expand the given generating function using the binomial series:

(1-4x)^(-1/2) = 1 + (-1)^1 * (1/2)^1 * (4x)^1 + (-1)^2 * (1/2)^2 * (4x)^2 + ...

= 1 + (-1)(2x) + (-1)^2 * (1/4)(4x)^2 + (-1)^3 * (1/8)(4x)^3 + ...

= 1 - 2x + 1/4 * 4x^2 - 1/8 * 4^2 + ...

= 1 - 2x + x^2 - x^3 + ...

We can see that the coefficients in this expansion are the terms of the sequence 2n chooses n. Therefore, we have shown that (1-4x)^(-1/2) does indeed generate the sequence 2n chooses n.

In conclusion, the unusual exponent of -1/2 in the given generating function did not hinder us from showing its validity in generating the desired sequence. This serves as a reminder that sometimes, thinking outside the box and manipulating expressions can lead to interesting and unexpected results.