Hi, everyone!

Finally, I am done with my second article

This time, it is - Investigating Binomial Theorem

Now, I am confused because I have got multiple ideas with me to write an article on. So, I would like to know from you. Which topic you would recommend?

And don't forget to try out my advanced problems. Also, you will find more of my problems in the upcoming Proofathon Calculus Contest. I couldn't add them to the paper because the contest hasn't started yet.

Any kind of suggestions/feedback are appreciated.

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TopNewestadvanced inequality – Samuel Ayinde · 2 years, 5 months ago

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Really awesome article....Though it will take me a little while to completely understand it. – Shivam Mishra · 2 years, 2 months ago

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How can we submit our problems in contests like Proofathon,Jomo etc.? – Gautam Sharma · 2 years, 5 months ago

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here

For Proofathon, you can read all the rulesFor JOMO, go to the website and click the Rules option.

On the rules page, you can find the method of submission and all other necessary information you may require. – Kishlaya Jaiswal · 2 years, 5 months ago

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– Gautam Sharma · 2 years, 5 months ago

I didn't find any way to submit my problems?How did you submit problems in proofathon?Log in to reply

Once you have created an account, you'll find that a user id has been assigned to you. (You can find this user id on the ongoing contests page under the contest file button). So, when a contest is going on, you need to email your solutions (include your user id) to proofathon@gmail.com. Now, there are two ways of writing up your solutions - either you can create a pdf (via LaTeX) or you can even submit your handwritten solutions. (although we prefer that you create a pdf using LaTeX) (but the way of submission you choose, won't effect your grades) – Kishlaya Jaiswal · 2 years, 5 months ago

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– Gautam Sharma · 2 years, 5 months ago

Clarification:I am asking how can i submit problems created by me so that they appear in the question paper. I am not asking how to submit solutions.Log in to reply

Well, we aren't accepting user submitted problems for the time being. But anyways, I'll discuss it once again with the staff and notify you.I'm sorry once again.

Thanks. – Kishlaya Jaiswal · 2 years, 5 months ago

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– Gautam Sharma · 2 years, 5 months ago

How did you get into that committee?Log in to reply

@Cody Johnson (owner of Proofathon) posted on Brilliant about his idea and starting a new online competition. There, he asked all members, who wish to contribute or wanted to help to come forward. So, that's the story of how Proofathon started and how we all formed a team. – Kishlaya Jaiswal · 2 years, 5 months ago

More than a year ago,Log in to reply

I am rather confused , this I have'nt tried till now ( You wrote this under proving the Basel theorem)

How can you square it , its an inequality , sorry I have'nt done this before , it may be very simple , but I am confused right now.

Now this is always true - 1> -1

as you squared then my inequality will yield 1>1 , but 1 = 1

Squaring means multiplying terms by itself so for getting csc^2 on left we have to multiply csc throughout the inequality and take the domain of x as the expression remains positive which is only possible in first quadrant , we also have to see the angles when it becomes negative.

Please clear my this small doubt , I hav'nt tried squaring inequalities so I may be completely wrong. Thanks – Megh Choksi · 2 years, 5 months ago

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@megh choksi Well, I guess, you didn't notice that I am considering the inequality only in the domain \((0,\frac{\pi}{2})\) which means, all of the three terms (in the inequality) are positive and we can safely square out the terms in the inequality iff all the terms are positive.

The example, you describe, has a negative term viz. \(-1\), so that prohibits us from squaring out the inequality \(1 > -1\).

I hope, you would have understood now, why have I squared it out.

Thanks. – Kishlaya Jaiswal · 2 years, 5 months ago

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– Curtis Clement · 2 years, 5 months ago

To back up Kishlaya's point suppose \({x}\) < y such that \({x}\) + \({k}\) = y. This means that: \[(x+k)^2 = y ^ 2 \implies\ x^2 + (2xk + k^2) = y^2\] Now this means that: \[x\geq 0 \implies\ x^2 < y^2 \] but \({x}\) < 0 doesn't not imply that \(x^{2}\) < \(y^{2}\), as their difference may be negative.Log in to reply

How can we download the papers of previous contests? – Megh Choksi · 2 years, 5 months ago

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In that case, go to the following page - Past Contests

Click on "2013-14" link and there you can download the past contests paper. – Kishlaya Jaiswal · 2 years, 5 months ago

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I'm recommending:

(a) Non-trivial positive integers such that \( Fibonacci(x) = x^n\), for \( n \geq 3 \). For instance, \( n=1 \to x = 5 \) and \( n = 2 \to x = 12 \);

(b) Big values of \( n \) such that the base-10 logarithm of the first \( n \) primes is an integer;

(c) Non-trivial functions \( f(x) \) such that \( \underset{n \; \text{times}}{f(f(f\cdots(x))))} = f(x) \) for integers \( n \geq 2 \);

(d) Call me on Facebook for more! – Guilherme Dela Corte · 2 years, 5 months ago

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@Guilherme Dela Corte

Thanks,I'll take this topic as my next challenge. In meantime, if I don't get to something really more interesting, I'll try to com up with some non-trivial facts and numbers. But definitely, that will surely take lot of time to investigate. So, I guess, there will be a long pause on my articles till then. – Kishlaya Jaiswal · 2 years, 5 months ago

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