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.
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.
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?
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
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.
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.
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.
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.
This channel is the home of smart maths.
Agreeeed 😊
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)
Thanks Dr.Peyam. I really enjoyed it.
You’re so welcome :)
Didn't expect Dr Peyam to say "Oh yeah, she ate." Great video!
Merci infiniment.. Bonne continuation.. tu es génial !!!
De rien :)
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?
Very cool solution
wow those clever transformations at the end🥳nice
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
nice one ! thank you (also, great use of the magical clicking fingers !)
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.
He does say it at the very start, but it’s understandably easy to miss
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.
I did it using base 2 logarithms
Thanks sir.
Completely understand 😊😊😊
Most welcome!!
Nice
,❤
These videos make me feel so stupid