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J Pi Maths
United Kingdom
Приєднався 11 лип 2020
Hi,
I'm Jaymin
I like maths
I study maths at Oxford University
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Jaymin. Jaymout.
I'm Jaymin
I like maths
I study maths at Oxford University
Subscribe pls
Jaymin. Jaymout.
Why is 90 a 'sunny' number?
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Відео
The fundamental theorem of MAT
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A square? I think not...
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Two and Half Squares
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A twist to the classic four square theorem by Lagrange. 🎓 www.jpimathstutoring.com 📷 jpimaths Contact me: jpimaths@gmail.com
4 unknowns, 2 equations?!
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Surprisingly Straight Forward: Oxford Maths Problem
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The closest I'll come to solving the Riemann Hypothesis!
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An inequality involving the Riemann-Zeta function 🎓 www.jpimathstutoring.com 📷 jpimaths Contact me: jpimaths@gmail.com
The Deceiving Oxford MAT problem!
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These are BOTH powers of 5?!
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When does this have solutions?
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A past TMUA problem 🎓 www.jpimathstutoring.com 📷 jpimaths Contact me: jpimaths@gmail.com
Oxbridge Personal Statement Tips from an Oxford graduate!
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A trigonometric integral without using integration by parts
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Can you integrate tan^6(x) without using integration by parts or recurrence relations/equations. 🎓 www.jpimathstutoring.com 📷 jpimaths Contact me: jpimaths@gmail.com
The Korfballer's Dilemma (PUTNAM 2004)
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a DIFFERENT approach to this integral
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2 unknowns, 1 equation ... (BMO 2010)
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There are primes hidden in this equation...
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There are primes hidden in this equation...
An infinite sum to get you into CAMBRIDGE
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An infinite sum to get you into CAMBRIDGE
SOLVED: How many squares are there in a rectangle?
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SOLVED: How many squares are there in a rectangle?
n lines, k intersections... Is it possible?
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n lines, k intersections... Is it possible?
The product of ANY k consecutive integers is divisible by k!
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The product of ANY k consecutive integers is divisible by k!
THIS is how you solve a CIRCLE problem!
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THIS is how you solve a CIRCLE problem!
How do you explicitly define this function?
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How do you explicitly define this function?
The Difference of Two Squares featuring @blackpenredpen
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The Difference of Two Squares featuring @blackpenredpen
Trigonometric Identities used Beautifully!
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Trigonometric Identities used Beautifully!
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!
@@wesleydeng71 thank you!
Love your new flat. Wonderful views.
@@cycklist hahaha thank you 😂
If the upper limit is anything but the identity function, should you apply the chain rule when taking the derivative?
@@Hussain-px3fc yep, the chain rule does the job!
Oh now i understand why we don't use x to indicate multiplication
@@saaah707 haha yep, otherwise you end up trying to distinguish between x and x
This is type of question, I like how he just straight up know the answer and then goes on to prove it.
@@YunruiHu ah well, I didn't know it straight away, but looking at expression mod 4 or 8 is normally a good way to start when attempting to see if something could be square
why 4m and not another value?
In theory you could do 8k+r, where r is 0,1,...,7 but this requires a little more work
Can we use mod 4 to get the same result? Since every squared number is 0 or 1 mod 4
Good question, not quite as 3^n +2*17^n would be 1 or 3 mod 4, but 1 could be square!
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
No.. They are describing the solution to a set of two equations
2 equations and 1 extra info being abcd r integers
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!?!
There’s a theorem (from Lagrange I think) that claims that every integer can be written as the sum of 4 squares or less, so for some of them 2 is impossible
@@clementfradin5391 Thanks, awfully good of you. Yes, I remember reading something of that sort... on one of the popular maths books. As it happens so many times a bit late, I have just spotted that 257 is is sum of two squares: 1 and 16.
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.
@@quite_unknown_1 💪💪💪
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
y =( 2n+1)pi/2 + x would be the eqn
@@mahimaprasadbehera4723 that's incorrect
(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
I agree, there was more potential to make this problem more interesting by changing the options
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
Nice! Works just as well - good spot!
just assume f(x) to be a constant, you get f(x) = 1/sqrt(2)
The uniqueness theorem would be required in the event of a proof.
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 .
@@RSLT thank you I really appreciate it!!
Wait, seems like a well thought out proof. What more is there to the solution? Surely this cannot be the whole problem?
@@OCD.Reader I'm not sure what you mean by this - I proved the statement I set out to prove...
@@JPiMaths Sorry, my bad. I should have phrased it better. As someone not familiar with the mathematical statement of the riemann hypothesis, I was wondering what more is left to the hypothesis beyond what you proved in the video. I agree that you proved the statement you set out for. Maybe I missed something. I was just wondering what needs to happen after this for the hypothesis to be solved as well.
@@OCD.Reader His statement about the Riemann hypothesis was very tongue-in-cheek, this proof has almost nothing to do with actually proving the Riemann hypothesis. (The only thing both have in common is that they involve the Riemann zeta function.) The Riemann hypothesis says that all non-trivial zeroes of the Riemann zeta function have the real part 1/2.
@@OCD.Reader in short... a lot is yet to be proved. The Riemann Hypothesis is a theorem that involves the Riemann-Zeta function that I mention in this video, but there's a lot more to the theorem than this. The comment section is not large enough for me to go into detail about the proof of the Riemann Hypothesis
Great channel, well done Jaymin! 👍
@@dipshah1984 thank you very much!!!
Yes this is all correct. But one thing to remember is S+O+N+A = 🤘🏾
@@dipshah1984 I don't understand this but okay 😂
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?
@@srinivaschillara4023 poetic
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
@@ayushrudra8600 nice spot!
Didn't see the positive restriction on the domain, substituted x=0 and got the right answer :P
@@amazing-qq3wi interesting! I reckon you could probably somewhat allow that argument if you assume f is continuous and take the limit as x goes to 0.
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
@@HarmonicEpsilonDelta ooh yeah this is a nice little argument; I definitely didn't spot this, but it's very slick!
@[account_that_deleted_the_comment] the LHS is nondecreasing with respect to y but strictly increasing with respect to f(y), which is why “f(y)” can only attain one value
This is clever
K²+1/sqrt(2).k-1=0; k=sqrt(2)/2. How do u get 1/sqrt (2)
It is the same idi*t. @@Dav.02
I mean f(x) = 1/sqrt(2) is a working solution so just assume its the only one and done :p
@@quite_unknown_1 to be honest if you spot it, that's a good way to solve this in the exam under timed pressure!