Mathematics Compilations

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...

The definite integral between two points $a$ and $b$ gives the area under the curve for the function between $a$ and $b$, textbooks say. A question arises - how? I had this doubt when I started to learn calculus in my school. And I decided to set out to prove it. Consider a function $\operatorname{f}$, and its derivative $\operatorname{f'}$. We have to prove that $\int_a^b \operatorname{f'}(x) = $ Area under the curve for $\operatorname{f'}(x)$ between $a$ and $b$ Let Area under the curve for $\operatorname{f'}(x)$ between $a$ and $b$ = $S$ Divide the shaded region into parts of very very small width $\delta \to 0$. Each part can be considered as a rectangle of width $\delta$ and height $\operatorname{f'}(c)$ where $c$ is the $x$ coordinate of the graph, $a\le c\le b$. $$\therefore S = \lim_{\delta \to 0}\Bigl(\delta \times \operatorname{f'}(a) + \delta \times \operatorname{f'}(a+\delta) + \delta \times \operatorname{f'}(a+2\delta) +...\\+ \...

Prime numbers are one of the most speculated topics in mathematics. Mathematicians have pondered about them enough to arrive at a proper conclusion to the above question. Let us look at the first few prime numbers: $$2\quad3\quad5\quad7\quad11\quad13\quad17\quad19\quad23\quad29\quad31\quad37\quad41$$ Let us now look at the first few prime numbers after $100000$: $$100003\quad100019\quad100043\quad100049\quad100057\quad100069\quad100103\\100109\quad100129\quad100151\quad100153\quad100169\quad100183$$ If you look closely, you may notice that the average separation between consecutive prime numbers between $100003$ and $100183$ is larger than that of $2$ and $41$. Naturally, our brain thinks that this separation increases on and on as we go to larger numbers, and at one point the set of prime numbers terminate. But that is not the case. It is proven that there are infinitely many prime numbers. The Proof: Euclid had the same question we have today, and he gave a beautiful proof t...

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

Imaginary numbers are not 'imaginary'. Real numbers are not 'real'. In fact, they're both in our heads... We are quite familiar with the logarithmic function. We define it as the inverse of the exponential function. That is, we say: $$\log{e^x} = e^{\log{x}} = x$$ (In fact, this is how the inverse of a function is even defined.) Now, we're all comfortable with the natural log. We use it for various results, but we always keep in mind that the number we're evaluating is POSITIVE. Why though? Let's consider some negative number $a$. What does it mean to take it's natural logarithm? It means to find a number such that $e$ raised to that number gives $a$. And surely since $e$ raised to any power always gives a positive value, it makes no sense to take the log of a negative number, right? Right? Complex Numbers: Function Domains Remember how we  extended  the existence of real numbers to the complex number system? What did we do? ...

Using the Taylor series is a straight forward method, but let us try to approach the problem in a different manner. Integration by Parts NOTE: This method requires working knowledge of  integration by parts . When I tried to solve this problem, I first started to expand it using the integration by parts method. The pattern I observed was the thing which helped me put myself on the right track to solve this problem. Here is the pattern: $$\int e^{x^2}dx = xe^{x^2} - 2\int x^2e^{x^2}dx$$ $$3\int x^2e^{x^2}dx = x^3e^{x^2} - 2\int x^4e^{x^2}dx$$ $$5\int x^4e^{x^2}dx = x^5e^{x^2} - 2\int x^6e^{x^2}dx$$ To make sure this pattern sustains, I decided to find the integral of $x^{2n}e^{x^2}$. $$Y = \int x^{2n}e^{x^2}dx$$ $$Y = x^{2n+1}e^{x^2} - \int x[2nx^{2n-1}e^{x^2}+2x^{2n+1}e^{x^2}]dx$$ $$(2n+1)Y = x^{2n+1}e^{x^2}-2\int x^{2n+2}e^{x^2}dx$$ $$\therefore Y = \frac{x^{2n+1}e^{x^2}}{2n+1}-\frac{2}{2n+1}\int x^{2n+2}e^{x^2}dx$$ Note that the $\int x^{2n+2}e^{x^2}dx$ is sim...

We know from the product rule that $d(xy) = xdy+ydx$. Integrating it, $$\int d(xy) = \int xdy + \int ydx$$ $$\therefore \int xdy = xy- \int ydx$$ This has applications in many different problems, and is known as integration by parts.

Consider the given equation... $$y^2-7=x^3=(x^{\frac{3}{2}})^2$$ $$y^2-(x^{\frac{3}{2}})^2=7$$ If $x^{\frac{3}{2}}$ is a decimal, then $y$ should also be some decimal to make the equation an integer, i.e., $7$. So $y$ will be a decimal in this case. But consider when $x^{\frac{3}{2}}$ is an integer. The only integers whose difference of squares is $7$, is $4$ and $3$ (why?). $$4^2 - 3^2 =7$$ But compare it to the equation $y^2-(x^{\frac{3}{2}})^2=7$. Then $$x^{\frac{3}{2}} = 3$$ $$x = 3^{\frac{2}{3}}$$ But here $x$ is a decimal! Therefore, at least one of the variables $x$ and $y$ is a decimal in the given equation, thus proving the fact that it has no integral solutions.

Assume there  is  a complex number whose square root is purely imaginary. $$\sqrt{a+ib} = ic$$ Squaring both sides, $$a+ib = (ic)^2 = -c^2$$ The RHS is a real number, whereas the LHS is a complex number. The derived equation is false. Therefore, our original assumption that  there can be a complex number whose square root is purely imaginary  is false.