Skip to main content

The $\operatorname{erf}$ function

Can a function be 'defined' as the anti-derivative of another function?

$\operatorname{erf}$ is one such function. Even though it has various proper infinite series expansions (as in Wolfram MathWorld - Erf), it is defined by mathematicians as such:
$$\operatorname{erf}(z) \equiv \frac{2}{\sqrt{\pi}}\int_0^ze^{-t^2}dt$$
This function has extensive implications in statistics, and can be used to express the integral of $e^{-x^2}$ and $e^{x^2}$. See the post that tries to find integral of $e^{x^2}$ - we need not use integration by parts there, but rather can use the $\operatorname{erf}$ function.

Source: Wolfram MathWorld

Labels:

Comments

Post a Comment

Popular posts from this blog

Factorials, and the Gamma Function

What is the Gamma function? What does it have to do with the factorial of a number? And why does such a formula even work? The Factorial In our high school combinatorics course, we were introduced to the factorial of a natural number. We defined the factorial of a number $n$ to be the product of all the natural numbers till $n$. Or symbolically: $$n! = 1 . 2 . 3 ... (n-2)  (n-1)  n$$ This is a very useful function in Counting problems. This function also pops up in the exponential function: $$e^x = 1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+...$$ Setting this aside, we think: can we 'extend' this function to all real number? Surely, even if we were to come up with such a function, what would it mean to take this new 'factorial' of some real number? Well, let's take a look at one important property of the factorial function: $$n! = n(n-1)!$$ and this is true for all natural numbers $n$. So, even if we came up with this homologous function, we...
Post a Comment
Read more

Problems that deceive you #1

Let us look at one of the most easiest problems in calculus $$\int_0^{\frac{\pi}{2}}\sqrt{4\sin^2\frac{x}{2}-4\sin\frac{x}{2}+1}\; dx$$ Seems pretty simple and straightforward, doesn't it? Inside the square root, the expression evaluates to $(2\sin\frac{x}{2}-1)^2$, and thus the integral will be $$\int_0^{\frac{\pi}{2}}\Bigl(2\sin\frac{x}{2}-1\Bigr)dx$$ This then can be evaluated by the following: $$\biggl[-4\cos\frac{x}{2} -x\biggr]_0^{\frac{\pi}{2}}$$ Hold on for the surprise... Its wrong! What seemed like a really straightforward and easy problem, is designed in such a way to appear easy so that we approach the wrong way. Why is the approach wrong? We did not consider the fact that a simple square root symbol represents its principal square root i.e., the positive square root. That means each and every value the expression inside the original integral evaluates to a positive quantity. But that is not the case when we simplified the integral to  $$\int...
Post a Comment
Read more