Can the Hypergeometric Equation Prove that tan^-1x = xF(1/2, 1, 3/2, -x^2)?

Therefore, $\tan^{-1}x = xF\left(\frac{1}{2},\, 1,\, \frac{3}{2},\, -x^2\right)$ is true. In summary, we have shown that $\tan^{-1}x = xF\left(\frac{1}{2},\, 1,\, \frac{3}{2},\, -x^2\right)$ by expanding the arctan(x) function and showing that it is equal to the function $xF (\frac{1}{2}, 1, \frac{3}{2}, -x^2)$, which is true.
  • #1
ssh
17
0
Show that,

\[\tan^{-1}x = xF\left(\frac{1}{2},\, 1,\, \frac{3}{2},\, -x^2\right)\]
 
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  • #2
ssh said:
Show that tan -1(X) = x F (1\2, 1, 3/2, -x^2)
First note that the expansion of $\arctan(x) = \sum^{\infty}_{0}\frac{(-1)^n(x)^{2n+1}}{2n+1}$ [tex]F (\frac{1}{2}, 1, \frac{3}{2}, -x^2) = 1+\frac{\frac{1}{2}\cdot 1}{\frac{3}{2}}(-x^2)+\frac{\frac{1}{2}\cdot \frac{3}{2}\cdot 1 \cdot 2}{\frac{3}{2}\cdot \frac{5}{2}\cdot 2!}(x^4)+\frac{\frac{1}{2}\cdot \frac{3}{2}\cdot \frac{5}{2}\cdot 1 \cdot 2\cdot 3}{\frac{3}{2}\cdot \frac{5}{2}\cdot\frac{7}{2}\cdot 3!}(-x^6)+... [/tex] [tex]xF (\frac{1}{2}, 1, \frac{3}{2}, -x^2)= x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+... =\sum^{\infty}_{n=0}\frac{(-1)^nx^{2n+1}}{2n+1} [/tex] which is exactly the expansion of the arctan(x)
 

FAQ: Can the Hypergeometric Equation Prove that tan^-1x = xF(1/2, 1, 3/2, -x^2)?

What is a hypergeometric equation?

A hypergeometric equation is a mathematical equation that describes the relationship between three sets of objects. It is used to calculate the probability of obtaining a specific number of objects from a larger sample, without replacement. It is often used in statistics and probability theory.

What are the key components of a hypergeometric equation?

The key components of a hypergeometric equation are the sample size (N), the number of successes in the sample (n), and the number of objects in the population of interest (K). These variables are used to calculate the probability of obtaining a specific number of successes in a sample without replacement.

How is a hypergeometric equation different from a binomial equation?

A hypergeometric equation and a binomial equation are both used to calculate probabilities, but they are based on different scenarios. A hypergeometric equation is used when the sample is drawn without replacement, meaning that the probability changes with each draw. A binomial equation is used when the sample is drawn with replacement, meaning that the probability remains constant with each draw.

What are some real-world applications of hypergeometric equations?

Hypergeometric equations are commonly used in genetics, to calculate the probability of certain gene combinations in a population. They are also used in quality control, to calculate the probability of a certain number of defective items in a sample. Additionally, they are used in statistical sampling and survey research to calculate the margin of error.

How do I solve a hypergeometric equation?

To solve a hypergeometric equation, you will need to have the values for N, n, and K. You can then plug these values into the equation and use a calculator or computer program to solve for the probability. Alternatively, you can use a hypergeometric calculator or table to find the probability without having to manually solve the equation.

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