can you solve this Berkeley Math Tournament quadratic problem?

Рет қаралды 6,293

Күн бұрын

Пікірлер: 24

@randomjin9392 Ай бұрын

It's not that xˣ is strictly increasing (because on (0, 1/e) it's decreasing), it's just that it's increasing for where we care about, i.e. positive integer N.

@joseluishablutzelaceijas928 Ай бұрын

That observation with the 2^160 = 32^32 was really unexpected and nice... I argued by taking the base 2 logarithm of both sides of 2^(20/n) >= 2*sqrt(n) to obtain 20/n >= 1+(log_{2}(n))/2. As the LHS is decreasing and the RHS increasing, I simply needed to find the smallest n s.t. the RHS is not smaller than the LHS anymore, which was easy to find as, given the many 2's, it is natural to try powers of 2. Thanks for the problem and its solution.

@henrymarkson3758 Ай бұрын

This channel is the home of smart maths.

@drpeyam Ай бұрын

Agreeeed 😊

@jackmeyergarvey Ай бұрын

Didn't expect Dr Peyam to say "Oh yeah, she ate." Great video!

@WahranRai Ай бұрын

0:53 Why did you consider +- . Just raise both sides to power 1/n ---> (((x^2 + n)/x)^n)^(1/n) = (2^20)^(1/n) ---> (x^2 + n)/x = 2^(20/n)

@slavinojunepri7648 Ай бұрын

Very cool solution

@Malekbih5918 Ай бұрын

Merci infiniment.. Bonne continuation.. tu es génial !!!

@drpeyam Ай бұрын

De rien :)

@harikishan5690 Ай бұрын

wow those clever transformations at the end🥳nice

@jeanfredericferte1128 Ай бұрын

nice one ! thank you (also, great use of the magical clicking fingers !)

@DanGRV Ай бұрын

I used a different approach: (x+n/x)^n has a minimum at x=sqrt(n) this means that 2^20 >= (2sqrt(n))^n after squaring and some other algebraic steps 2^(40-2n) >= n^n setting n=2^k, comparing exponents and after some other steps 40 >= (k+2) 2^k the right hand side grows really with k, only had to test k=1, 2 and 3 to find that n=2^3=8 is the last n which allows solutions

@Tletna Ай бұрын

Maybe I just missed it earlier in the video but why did you stop at N = 1 and not do N = 0 or negative values when summing all N's < or = to 8?? Maybe I'll just watch it again later. Also, you knew to manipulate the numbers so that it could more easily be solved because it is a known Berkeley math tournament problem. How would we know to look for this in other similar problems?

@szewing9038 Ай бұрын

Thanks Dr.Peyam. I really enjoyed it.

@drpeyam Ай бұрын

You’re so welcome :)

@ronbannon Ай бұрын

You should have stated before doing the problem that n is a natural number. Although n is often used to represent natural numbers, it could also mean integers. However, n could also represent any class of numbers, and there's no reason to assume such a restriction unless stated.

@redpepper74 Ай бұрын

He does say it at the very start, but it’s understandably easy to miss

@utilizator1701 Ай бұрын

Even though my final answer is the same, my calculation says that the grade 2 equation of X has real solutions if n is less or equal to a number n0 between 8 and 9 such that 2^(40/n0)-4*n0 = 0.