# Are the sum and the hypergeometric equal?

• uart
In summary, the conversation discusses finding the series solution to a 2nd order differential equation and Wolfram's listing of the Airy functions in terms of the Hypergeometric function. The conversation then delves into exploring the equality of two different series solutions and the steps taken to prove their equivalence.
uart
I was looking at finding a series solution to a 2nd order DE the other day and came up with the following (for one of the solutions, and there was a somewhat similar series for the other solution).

$$\sum_{k=0}^{\infty} \frac{x^{3k}}{(3k)!} \prod_{m=1}^{k-1} (3m+1)$$

Wolfram said the solutions were the Airy functions $Ai$ and $Bi$, and since these can be defined in terms of the Hypergeometric function, $~_0F_1$, I suspected that the series I had might be equivalent to a hypergeometric. http://mathworld.wolfram.com/AiryFunctions.html

Wolfram lists one of the hypergeometrics involved in $Ai$ and $Bi$ as, $~_0F_1(\frac{2}{3}, \frac{x^3}{9})$. And when I expand this as per the definition here http://mathworld.wolfram.com/ConfluentHypergeometricLimitFunction.html I get the following.

$$\sum_{k=0}^{\infty} \frac{x^{3k}}{9^k \, k!} \left[ \prod_{m=0}^{k-1} (\frac{2}{3}+m) \right]^{-1}$$

When I sum these numerically in MATLAB I seem to get the same answer for both. I can see a lot of similarity but can't quite make out the equality of the two. Can anyone see any easy way to show they're equal?

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Ok I figured it out, and yes they are identical :)

I'm busy now but will post it later in anyone is interested.

OK, since I asked it I guess I should show a solution. It turned out to not be that difficult. Start in the obvious way by stripping away the bits that are already the same and writing just the bits that we require equality on.

$$(3k)!=9^k \, k! \prod_{m=0}^{k-1} (\frac{2}{3}+m) \prod_{m=1}^{k-1} (3m+1)$$

Splitting the $9^k$ into $3^k \, 3^k$ we can take one of these terms inside the fractional product to get.

$$(3k)! = 3^k \, k! \prod_{m=0}^{k-1} (3m + 2) \prod_{m=1}^{k-1} (3m+1)$$

Then with some simple rearrangements of the products we get.

$$(3k)! = 3^k \, k! \prod_{m=1}^{k} (3m - 1) \prod_{m=1}^{k-1} (3m+1)$$

$$(3k)! = 3^k \, k! \, \, (3k-1) \prod_{m=1}^{k-1} \left\{ (3m - 1) (3m+1) \right\}$$

From here it's pretty easy to see the equality. The product, $(3k-1) \prod_{m=1}^{k-1} \left\{ (3m - 1) (3m+1) \right\} = 2 \cdot 4 \cdot 5 \cdot 7 \cdot 8 \cdot 10 \cdot 11 \ldots (3k-1)$, provides all the non factor 3 terms of $(3k)!$, while the "$3^k \, k!$" fills in all the factor three terms as $3^k \, k! = 3 \cdot 6 \cdot 9 \cdot 12 \ldots 3k$.

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## 1. What is the difference between the sum and the hypergeometric?

The sum and the hypergeometric are two different mathematical concepts. The sum refers to the total of a set of numbers, while the hypergeometric refers to a specific type of probability distribution.

## 2. Can the sum and the hypergeometric be equal?

No, the sum and the hypergeometric cannot be equal because they are two different types of mathematical concepts. However, in certain cases, the sum of a set of numbers may be equal to the expected value of a hypergeometric distribution for a given set of parameters.

## 3. How is the hypergeometric distribution related to the sum?

The hypergeometric distribution is a probability distribution that calculates the probability of obtaining a certain number of successes in a specific number of draws from a finite population without replacement. The sum, on the other hand, is a basic mathematical operation that calculates the total of a set of numbers. These two concepts are related in that the expected value of a hypergeometric distribution can sometimes be equal to the sum of a set of numbers.

## 4. What are some real-world applications of the hypergeometric distribution?

The hypergeometric distribution has various applications in real-world scenarios, such as quality control in manufacturing, sampling in market research, and analyzing gene frequencies in population genetics. It is also used in statistical analysis to determine the likelihood of obtaining a specific sample from a population without replacement.

## 5. How can I calculate the hypergeometric distribution?

The hypergeometric distribution can be calculated using a specific formula: P(X = k) = (C(k,n) * C(N-k, n-m)) / C(N, n), where P(X = k) is the probability of obtaining exactly k successes, n is the number of draws, m is the number of successes in the population, and N is the total population size. Alternatively, you can use statistical software or online calculators to calculate the hypergeometric distribution.

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