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Chapter 11: Q3P (page 548)

Prove that erf(x) is an odd function of x. Hint: Put t = -s in (9.1) .

Short Answer

Expert verified

It has been proved that the erf(x) is an off function of x .

Step by step solution

01

Definition of error of function

The error of the function is defined as $(x) = \frac{2}{\sqrt{π}} \int_{0}^{x} e^{- t^{2}} d t$ .

02

Prove that error is odd

The value of integration is $e r f (x) = \frac{2}{\sqrt{π}} \int_{0}^{x} e^{- t^{2}} d t$ .

Substitute -x in place of x .

$e r f (- x) = \frac{2}{\sqrt{π}} \int_{0}^{- x} e^{- t^{2}} d t$

Let u = -s , then du = -ds .

The equation becomes as follows:

$e r f (- x) = \frac{2}{\sqrt{π}} \int_{0}^{x} e^{- s^{2}} (- d s) e r f (- x) = - e r f (x)$

Hence, the statement has been proved.

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Most popular questions from this chapter

Show that the four answers given in Section 1 for $\int_{0}^{\frac{π}{2}} \frac{dθ}{\sqrt{cosθ}}$ are all correct. Hints: For the beta function result, use(6.4). Then get the gamma function results by using (7.1) and the various Γ function formulas. For the elliptic integral, use the hint of Problem 17 with $α = \frac{π}{2}$ .

In Problem 4 to 13, identify each of the integral as an elliptic (see Example 1 and 2). Learn the notation of your computer program (see Problem 3) and then evaluate the integral by computer.

12. $\int_{0}^{\frac{1}{2}} \frac{\sqrt{100 - t^{2}}}{\sqrt{1 - t^{2}}} d t$

In A uniform solid sphere of density $\frac{1}{2}$ is floating in water. (Compare Chapter 8, Problem (5.37).) It is pushed down just under water and released. Write the differential equation of motion (neglecting friction) and solve it to obtain the period in terms of $K (5^{\frac{- 1}{2}})$ . Show that this period is approximately 1.16 times the period for small oscillations.

Using (5.3) with (3.4) and (4.1), find $Γ (3 / 2)$ , $Γ (- 1 / 2)$ , and $Γ (- 3 / 2)$ in terms of $\sqrt{π}$ .

In the Table of Laplace Transforms (end of Chapter 8, page 469), verify $Γ$ the function results for L5 and L6. Also show that $L (1 / \sqrt{t}) = \sqrt{π / p}$ .

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