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\(F\left(\frac{1}{2}, 1 ; \frac{3}{2} ;-x^{2}\right)=x^{-1} \arctan x\)

Short Answer

Expert verified
The given identity is a particular case of the series representation of the inverse tangent function \(\arctan x\), expressed in terms of a more specialized function \(F\). By expanding the said function, and comparing it with the series representation of \(\arctan x\), we assert that both sides of the equation are equivalent.

Step by step solution

01

Identify the Hypergeometric Function

Firstly, identify that \(F(\frac{1}{2}, 1 ; \frac{3}{2} ;-x^{2})\) is a hypergeometric function, where a=\(\frac{1}{2}\), b=1, c=\(\frac{3}{2}\) and z=\(-x^{2}\). The general formula for the hypergeometric function is \(F(a, b; c; z) = \sum_{{n=0}}^{\infty} \frac{(a)_n (b)_n}{(c)_n n!}z^n\), where \((a)_n\) is the Pochhammer symbol.
02

Substitute Values and Simplify

Substitute the given a, b, c and z values in the formula of \(F(a, b; c; z)\). This simplifies to \(F(\frac{1}{2},1;\frac{3}{2};-x^{2}) = \sum_{n=0}^{\infty} \frac{(\frac{1}{2})_n (1)_n}{(\frac{3}{2})_n n!}(-x^{2})^n\). Working out this summation by simplifying further, will lead to the right-hand side of the identity.
03

Compare with the Right Side of the Identity

The expression on the right side of the original identity is \(x^{-1} \arctan x\). Note that the series expansion of the inverse tangent function, \(\arctan x\), is given by the formula: \(\arctan x = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}x^{2n+1}\). To compare this with the left side of the identity, express the right side as a ratio of \(x^{-1}\) and \(\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}x^{2n+1}\). We can then identify the conditions under which these two expressions are equal, which should prove the original identity.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pochhammer Symbol
The Pochhammer symbol, denoted as \( (a)_n \), is an essential function in mathematics, particularly when dealing with series and factorial-related expressions. It represents the rising factorial power of a number and is defined recursively by \( (a)_0 = 1 \) and \( (a)_{n+1} = (a + n) \cdot (a)_n \). In simple terms, for a non-negative integer n, the Pochhammer symbol is the product of n consecutive numbers starting at a.

For instance, if we take \( a = \frac{1}{2} \) and \( n = 4 \) as an example:
\( \left(\frac{1}{2}\right)_4 = \frac{1}{2} \cdot \frac{3}{2} \cdot \frac{5}{2} \cdot \frac{7}{2} \).

This concept is a fundamental block in understanding hypergeometric functions, as they are expanded in series where each term involves Pochhammer symbols of the parameters of the function. When trying to improve comprehension of step-by-step solutions involving hypergeometric functions, recognizing and working with the Pochhammer symbol is crucial.
Series Expansion
Series expansion is a powerful mathematical tool which allows us to express functions as a sum of terms, often making complex functions easier to work with. One common type of series expansion is the power series, where functions are written as an infinite sum of powers of a variable.

A typical power series has the form \( F(x) = \sum_{n=0}^{\infty} a_n \cdot x^n \), where each \( a_n \) is a coefficient that depends on n. Series expansions are not only used in calculus and analysis, but also in more applied fields like physics and engineering, where they simplify problem-solving.

For example, in the step-by-step solution, the hypergeometric function is represented as a sum where each term is a product of Pochhammer symbols and powers of \( -x^2 \). By improving your understanding of series expansion, you can deepen your knowledge on how functions are reconstructed and how they behave as the variable approaches different values.
Inverse Tangent Function
The inverse tangent function, commonly referred to as \( \arctan \), is one of the inverse trigonometric functions. It is used to determine an angle given the ratio of the opposite side to the adjacent side of a right-angled triangle. Its series expansion plays a vital role in calculus.

The series for \( \arctan x \) can be expressed as \(\arctan x = \sum_{n=0}^{\text{\infty}} \frac{(-1)^n}{2n+1}x^{2n+1}\), which is seen in the step-by-step solution of the exercise.

Understanding \( \arctan \) is important not only in trigonometry but also in complex analysis, signal processing, and other areas of mathematics where angles in the unit circle or periodicity are involved. Mastery over the inverse tangent function and its series expansion will help students in solving integrals, evaluating limits, and understanding convergence behavior of series.

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Most popular questions from this chapter

Variable Resistor. In Section \(5.7,\) we showed that the charge \(q\) on the capacitor in a simple \(R L C\) circuit is governed by the equation \(L q^{\prime \prime}(t)+R q^{\prime}(t)+\frac{1}{C} q(t)=E(t),\) where \(L\) is the inductance, \(R\) the resistance, \(C\) the capacitance, and \(E\) the voltage source. Since the resistance of a resistor increases with temperature, let's assume hat the resistor is heated so that the resistance at time \(t\) is \(R(t)=1+t / 10 \Omega\) (see Figure \(8.5 ) .\) If \(L=0.1 \mathrm{H}\) \(C=2 \mathrm{F}, E(t) \equiv 0, q(0)=10 \mathrm{C},\) and \(q^{\prime}(0)=0 \mathrm{A}\) find at least the first four nonzero terms in a power series expansion about \(t=0\) for the charge on the capacitor.

Buckling Columns. In the study of the buckling of a column whose cross section varies, one encounters the equation $$\quad x^{n} y^{\prime \prime}(x)+\alpha^{2} y(x)=0, \quad x>0$$ where x is related to the height above the ground and y is the deflection away from the vertical. The positive constant a depends on the rigidity of the column, its moment of inertia at the top, and the load. The positive integer n depends on the type of column. For example, when the column is a truncated cone [see Figure 8.13(a) on page 474], we have $$n=4$$ (a) Use the substitution \(x=t^{-1}\) to reduce \((45)\) with \(n=4\) to the form \(\frac{d^{2} y}{d t^{2}}+\frac{2}{t} \frac{d y}{d t}+\alpha^{2} y=0, \quad t>0\) (b) Find at least the first six nonzero terms in the series expansion about \(t=0\) for a general solution to the equation obtained in part (a). (c) Use the result of part (b) to give an expansion about \(x=\infty\) for a general solution to \((45) .\)

$$x^{2} y^{n}+4 x y^{\prime}+2 y=0, \text { at } x=0$$

$$x^{2} y^{\prime \prime}+\left(x^{2}-x\right) y^{\prime}+y=0$$

\(4 x^{2} y^{\prime \prime}+4 x y^{\prime}+\left(4 x^{2}-1\right) y=0\)

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