Відео

WHAT THE SIGMA?!?!?!!?!??!!??!?!?!?

2^100*2^100*…2^100=2^10000

is it true (1/n)th root of x is equal to x^n ? given that : n is not 0 or complex. x is positive real number?

Just did 2^3! Vs (2^3)!

A simple thing can be the following to prove that (q+1) divides q! if (q+1) is composite. Write q!=(q+1)!/(q+1)... Now we have to prove that (q+1)! is divisible by (q+1)² if (q+1) is composite. Decompose (q+1) to m×n where m and n are two integers (which are less than (q+1)). Factors of (q+1) are of course less than (q+1) and since (q+1)! contains all integers 1≤k≤(q+1) then m,n will lie too. Hence m,n and (q+1) being present in (q+1)! proves that (q+1)! is divisible by (q+1)² and thus q! is divisible by (q+1) if it is composite.

Nice!

Love your new flat. Wonderful views.

If the upper limit is anything but the identity function, should you apply the chain rule when taking the derivative?

Oh now i understand why we don't use x to indicate multiplication

This is type of question, I like how he just straight up know the answer and then goes on to prove it.

why 4m and not another value?

Can we use mod 4 to get the same result? Since every squared number is 0 or 1 mod 4

I don’t care

here is how I did it. first, consider the graph sqrt x, so sqrt (x +p) is just the graph of sqrtx shifted over to the left by p units. Obviously, p >= 0, since sqrts sum to nonnegative numbers. Also, p = 0 is trivial. Finially, sqrt(x+p)+ sqrt(x) is defined only for x>=0, given that p>=0, and is continuous and has a range of sqrt p <= y < inf. So, we only have solutions if p > sqrt p, so p >= 1 and p = 0 from the earlier trivial solution.

I was able to do this because I recently solved this type of problem in quadratic where RHS is prime and product of LHS has to be equal to the number or 1

isn't this technically 3 equations not 2

out of interest, is there an integer solution to a sum of squares equalling 257 ? Is there some theorem saying something like any number canbe respresented as a sum of squares of two numbers!?!

Too many instances of “this is bigger than this, and that is bigger than this”. It’s not as descriptive as, for example, The left is bigger than the right, and the bound is larger than the right

What does Brahmagupta-Fibonnaci mean

POWER of exponential !

Go straight to the dad to lesson his boy.

I know just because of factorial GROW ALOT faster than exponential

ATTEMPT: f(1) + f(1) = 1/f(1). 2f(1)^2 = 1 f(1) = 1/sqrt(2) f(x) + 1/sqrt(2) = 1/f(x) f(x)^2 + f(x)/sqrt(2) - 1 = 0 (f(x) + sqrt(2))(f(x) - 1/sqrt(2)) = 0 because f(x) > 0, f(x) = 1/sqrt(2) for all x. f(2021) = 1/sqrt(2)

Brahmagupta-Fibonacci coming is clutch as always.

This is biggest than this and this is higher than this and just that is higher than this so just this just bredź to be lower than that so that can be lower than this - very clear

I looked at the thumbnail, set y=0, therefore f is constant since f(x) = 1/f(0) -f(0). Since f(x)>0 this is allowable. Then just compute any value and that’s the answer. For f(0),f(1) I found 1/root(2). Then I used f(1) to check and got 1/root(2).

I tried another way that might work. I expanded the cos^2(x) to 1-sin^2(x) and 1-sin^2(y), then rearranged both sides to see that sin^2(x)-sin(x)+1 = sin^2(y) - sin(y) + 1. If we let sin(x) = m, and sin(y) = q, then both quadratics on either side give the same expression, which means that there must be infinite solutions of their intersection!

the solution set you gave is not complete, the graph contains the family of lines x+y=(2n+1)pi and x=y +2npi where n is any integer

(2^3)! = 2^3 x 2^2 = 2^(3+2) = 2^5 2^(3!) = 2^(3x2) = 2^6 2^6 > 2^5 ; 2^(3!) > (2^3)! 2^(100!) > (2^100)!

Nice analysis, but must admit I was confounded. Maybe not my best day. Thanks for these sort of streching problems...

the hint makes it trivial

It would be a good problem, but as a multi-choice problem it's too easy and kills the fun. Straight forward to see that x = y + 2npi is in the graph

Ah yes, let's compare the size of a city with Sagitarius A*

Or just simply put all functions of y to one side and all functions of X to the other side. Than you have: sinx + cos²x = siny + cos²y And this is true whenever x = y, and just don't forget the periodicity. So e is correct answer

just assume f(x) to be a constant, you get f(x) = 1/sqrt(2)

I did some quick maths: (2^100)! would be 2^(100+99+98+….) = 2^5050. So I quickly reached the conclusion the left one is larger.

My dudes... just imagine 2^(100) as x....

Nice question. But I think using logarithms at start is a much cleaner way to reach the solution and avoid excessive writing. Log both sides: LHS: log(2^(100!)) = 100! * log(2) RHS: log(2^100!) Using the logarithm addition rule: log(2^100!) = Summation of log(n) where n goes from 1 to 2^100 Since the highest n is 2^100, we can use it as an upper bound for each term: RHS_upper bound = 2^100 * log(2^100) = 2^100 * 100 *log 2 So, we want to prove LHS > RHS_upper or LHS = k * RHS_upper where k > 1, which would mean it will be true for the actual RHS. Cancel log(2) and 100 on each side, we have: 99 * 98 * 97 * ... * 1 = k * 2^100 = k * 8 * 2^97 Since 99 > 8 and each of the terms like 98, 97,...2 is term wise greater or equal to 2 in 2^97. We need k>1 Hence LHS > RHS

Liked and subscribed and love to see more .

Wait, seems like a well thought out proof. What more is there to the solution? Surely this cannot be the whole problem?

Great channel, well done Jaymin! 👍

Yes this is all correct. But one thing to remember is S+O+N+A = 🤘🏾

2^100! - 100! factors (2^100)! - only 2^100 factors, 2^100! > (2^100)! More precisely, (2^100)!<(2^100)^(2^100)<2^(2^7•2^100)=2^2^107<2^2^108=2^4^54<2^(100•999•…•47)<2^(100!)

Kid named 2(100!)!:

Deceiving, nay, it is devious.... such are the kind monsters who set the MAT; Why kind, they gave a hint! Did they really?

i didn’t see the domain restriction since i only looked at the thumbnail but substituted y=0 and found that the function was constant

Didn't see the positive restriction on the domain, substituted x=0 and got the right answer :P

This and that not a variables 😂

Actually, once you have proven f(1)=… then you have f(1)+f(y)=1/f(y). The LHS is increasing with respect to f(y) but the RHS is decreasing, so it follows that f(y) must be constant in the connected region of the domain containing 1. To sum up, the quadratic argument of the video is better xddd but I found this observation interesting

I mean f(x) = 1/sqrt(2) is a working solution so just assume its the only one and done :p