A problem on inequality

Using simple formula to prove a strong inequality.

Pigeonhole Principle

The numbers 1 to 20 are placed in any order around a circle. Prove that the sum of some 3 consecutive numbers must be at least 32!

This problem uses the alternate form of pigeon hole principle which is as follows:
 If the average of n positive numbers is t, then at least one of the numbers is greater than or equal to t. Further, at least one of the numbers is less than or equal to t.

The proof is very simple, assume the contradiction and proceed!

#Solution https://youtu.be/GLQg6cSAbms

Sum of first 'n, natural numbers

A simple way to calculate the sum of first 'n' natural numbers witout the use of calculator. Infact the same procedure is used to calculate the sum of n terms of any A.P series. See it and try to obtain the formula yourself

Inequality

A challenging problem based on the inequality that square of a real number is always greater than equal to zero. Learn the trick and prepare yourself for more challenging problems based on the same ideology.

Divisibility of 2222^5555+5555^2222 by 7

For any prime p, square root of p is irrational.

Complex Numbers in Solving Trigonometrical problems

Show that $ \cos \frac{\pi}{11}+\cos \frac{3\pi}{11}+\cos \frac{5\pi}{11}+\cos 7\frac{\pi}{11}+\cos \frac{9\pi}{11}= \frac{1}{2}$ $$$$ I have used complex numbers to solve the problems. Standard trigonometric methods will be very much difficult to apply and will involve lots of calculation!!! Also it is important to link the different topics and see the beauty of mathematics! Important problem for students in 10+2 level $$$$ Also the idea used to solve the problem can be used to calculate the value of $ \sum_{i=1}^{\frac{n-1}{2}} \cos \frac{(2i-1) \pi}{n}$ where $n$ is an odd positive integer. $$$$ See the video below for the explanation. $$$$

Equation Solving

Find all real number(s) $x$ such that $2^x+3^x+6^x-4^x-9^x=1 $ $$$$ This problem can be solved by transforming the equation where it can be written as sum of squares of number equals zero. From there the result will follow. See the video below for the explanation. $$$$

Sophie Germain Identity

Here is a problem of primality testing of a number. As you can see a direct computation will not yield an effective result and will take a much longer time! So, the question arises how to solve the problem in the right way. Here comes the roll of simple and elegant identity known as Sophie Germain Identity. Watch the video below for the solution

For solution of the problem above watch the video below.

Indian Statistical Institute ( ISI ) B.Math & B.Stat : Algebra

Let $a, b$ and $c$ be such that $a+b+c=0$ and \( P =\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}\) is defined. What is the value of $P$? $$$$ $Solution: $ \( 2a^2+bc = a^2 + a^2 -(a+c)c\), since $b=-(a+c)$. $$$$ \( \implies 2a^2+bc = a^2 + a^2 -ac-c^2 = (a-c)(a+c)+a(a-c) = (a-c)(2a+c)=(a-c)(2a-a-b)=(a-c)(a-b)\) $$$$ Therefore \( \frac{a^2}{2a^2+bc} = \frac{a^2}{(a-c)(a-b)}\) $$$$ Similarly, \( \frac{b^2}{2b^2+ac} = \frac{b^2}{(b-a)(b-c)} \) and \( \frac{c^2}{2c^2+ab} = \frac{c^2}{(c-a)(c-b)} \) $$$$ Therefore \( P = \frac{a^2}{(a-c)(a-b)} +\frac{b^2}{(b-a)(b-c)}+\frac{c^2}{(c-a)(c-b)} = \frac{1}{a-b} \big( \frac{a^2}{a-c} - \frac{b^2}{b-c} \big)+\frac{c^2}{(c-a)(c-b)}\) $$$$ \( \implies P = \frac{ab-ca-cb}{(c-a)(c-b)}+ \frac{c^2}{(c-a)(c-b)} = \frac{ab-ca-cb+c^2}{(c-a)(c-b)} = \frac{(c-a)(c-b)}{(c-a)(c-b)} = 1\)$$$$

Indian Statistical Institute ( ISI ) B.Math & B.Stat : Calculus

If \( A = \int_{0}^{\pi} \frac{cos x}{(x+2)^2}dx\), then show that \( \int_{0}^{\frac{\pi}{2}} \frac{\sin x \cos x}{(x+1)}dx = \frac{1}{2} \big( \frac{1}{2} +\frac{1}{\pi+2} - A\big)\). $$$$ Solution: \( A = \cos x \int_{0}^{\pi} \frac{dx}{(x+2)^2} - \int_{0}^{\pi} \bigg( \frac{d}{dx} \cos x \int \frac{dx}{(x+2)^2} \bigg) dx \) $$$$ \( \implies A = \frac{-\cos x}{x+2}\mid_{0}^{\pi} - \int_{0}^{\pi} \frac{\sin x}{(x+2)} dx \) $$$$ \( \implies A = \frac{1}{\pi +2} + \frac{1}{2} - \int_{0}^{\pi} \frac{2 \sin \frac{x}{2} \cos \frac{x}{2} }{2(1+\frac{x}{2})} dx \) $$$$ Putting $\frac{x}{2}=t$ in the last integral, we have $dx=2dt$ and $t$ varies from $0$ to $\frac{\pi}{2}$ $$$$ Therefore, \( A = \frac{1}{\pi +2} + \frac{1}{2} - 2 \int_{0}^{\frac{\pi}{2}} \frac{\sin t \cos t}{(1+t)}dt \)$$$$ \( \implies \int_{0}^{\frac{\pi}{2}} \frac{\sin t \cos t}{(1+t)}dt = \frac{1}{2} \bigg( \frac{1}{\pi +2} + \frac{1}{2} -A \bigg) \) $$$$ \( i.e., \) \( \int_{0}^{\frac{\pi}{2}} \frac{\sin x \cos x}{(1+x)}dx = \frac{1}{2} \bigg( \frac{1}{\pi +2} + \frac{1}{2} -A \bigg) \) $$$$

Indian Statistical Institute ( ISI ) B.Math & B.Stat : Calculus

Evaluate \( lim_{n \rightarrow \infty} \bigg(\prod_{r=1}^{r=n} (1+\frac{2r-1}{2n}) \bigg)^{\frac{1}{2n}} \) $$$$ Let \( A = \bigg(\prod_{r=1}^{r=n} (1+\frac{2r-1}{2n}) \bigg)^{\frac{1}{2n}} \) $$$$ \( \implies A = \frac{\bigg(\prod_{r=1}^{r=2n} (1+\frac{r}{2n}) \bigg)^{\frac{1}{2n}}}{\bigg(\prod_{r=1}^{r=n} (1+\frac{2r}{2n}) \bigg)^{\frac{1}{2n}}} \) $$$$ \( \implies ln A = \frac{1}{2n} \sum_{r=1}^{r=2n} ln ( 1+\frac{r}{2n})- \frac{1}{2n} \sum_{r=1}^{r=n} ln ( 1+\frac{r}{n})\) $$$$ \( \implies lim_{n \rightarrow \infty} ln A = \frac{1}{2} lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{r=2n} ln ( 1+\frac{r}{2n}) - \frac{1}{2} lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{r=n} ln ( 1+\frac{r}{n}) \) $$$$ \( \implies lim_{n \rightarrow \infty} ln A = \frac{1}{2} \int_{0}^{2} ln (1+ \frac{x}{2}) dx - \frac{1}{2} \int_{0}^{1} ln (1+ x ) dx \) $$$$ \( \implies lim_{n \rightarrow \infty} ln A = ln 4 - 1 - ( ln 2 - \frac{1}{2}) = ln 2 - \frac{1}{2}\)$$$$ \( \implies lim_{n \rightarrow \infty} A = e^{ln 2 - \frac{1}{2}} = \frac{2}{\sqrt{e}} \) $$$$ \( \implies lim_{n \rightarrow \infty} \bigg(\prod_{r=1}^{r=n} (1+\frac{2r-1}{2n}) \bigg)^{\frac{1}{2n}} = \frac{2}{\sqrt{e}} \) $$$$