Mathematics Compilations

Let us take a look at a problem that came in JEE Mains 2018 CBT. If $|z-3+2i|\le 4$ then the difference between the greatest value and the least value of $|z|$ is: 1. $4+\sqrt{13}$ 2. $2\sqrt{13}$ 3. $8$ 4. $\sqrt{13}$ This can be approached using the conventional formula to get the answer. But what if you are not sure about the formula? Even if you do remember the formula, what if you want to arrive at the solution faster? There is an excellent graphical method to solve this problem. Consider $z=x+3+(y-2)i$. Note that I am considering this way so that the inequality is neat and clean: $$\sqrt{x^2+y^2}\le 4\\ \therefore x^2+y^2\le 4^2$$ The red region here represents the inequality shown. The region thus contains all and the only values that $x$ and $y$ can take. Let's find out $|z|$ now. $$|z|=\sqrt{(x+3)^2+(y-2)^2}$$ In a graphical point of view, $|z|$ represents the distance between any point in the red area and $(-3,2)$. It is obvious that the point $(-3,2)$ it...

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...

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