I think the solution for the factorial problem was way too compilicate. We can rewrite the equation as: 2^b = c! - a! = a! * ((a+1) * (a+2) * ... * c - 1) Note that the contents of the outer brackets on the right side can only be even if the product (a+1)*... inside only has a single term a+1, otherwise it's impossible. And also that a! can't have any prime factors > 2. So there are 2 cases to consider: 1) 2^b is odd => 2^b = 1 => b = 0 => c! - a! = 1 => c = 2, a = 0 or a = 1 2) 2^b is even => 2^b = a! * ((a + 1) - 1) = a! * a => a = 2 => 2^b = 2! * 2 = 4 => b = 2 => c! = 2 + 4 => c = 3
It's my extremely subjective personal opinion, but calculus to me is smooth and connected, while number theory is pure chaos and disconnected ideas. Increasing a number by 1 is usually one of the most drastic changes possible in that context. It always feels like the problems have to be custom tailored to fit whatever random method only works for that problem. But again, this is really subjective and probably unfair.
@@DrR0BERT I'm not a researcher anymore and the subject is a good bit hairier than I can take on as a filthy casual. I saw someone present a proof of the prime number theorem once, but I was unable to understand beyond the first half in just the one sitting.
I did a math minor with analysis topics along with my calculus based engineering degree so I feel like I can give the calc problems more of a go on my own I usually don't know how to approach the number theory problems, but I learn a lot from the videos
Not done. You have to show that the indefinite integral converges. All you've shown is that if it converges then it's value is 0. Take the function f(x) = 1/x if x between 0 and 1 and -1/x if x is bigger than 1.
For the first problem, a simpler solution is to break into integral 0 -> 1 plus integral 1 -> inf, do a change of variables to x=1/u in the second integral, and then see the two are negatives of one another by the defining function property
I always prefered calculus. I tried once to get into the math team of my country, once they had questions of drawing 10 traingles with 7 lines, I realized it's not my cup of tea.
I studied mathematics many years ago, your videos are very good and rekindle my love for mathematics. if my voice has any weight i lean towards number theory. thank you Michael
I would say, pound for pound, "discrete" problems are harder than "continuous" probs. The present probs are a nicely contrasted pair. Once you've twigged the "trick" in the first prob (the hard step), the solution is easy to write. For the second problem, it's the other way round. Here, it's easy to come up with solutions just by playing around with numbers, but actually writing out a concise proof that you have obtained _all_ the solutions is significantly taxing Like Penn, I got c! - a! = 2^b, which I then wrote as [ c(c-1) ... (a+1) - 1 ] .a! = 2^b Clearly a! must be a power of 2 => a = 0, 1, 2. Also c(c-1) ... (a+1) can only be even by being 2, and since a+1 =< 3, it can only be odd by being 3 Thus: a = 0 => c = 2 => b = 0 a = 1 => c = a + 1 = 2 => b = 0 a = 2 => c = a + 1 = 3 => b = 2 and we have solutions a, b, c = 0, 0, 2; 1, 2, 0; 2, 2, 3
ab = 1 ==> b = 1/a was screaming to me to split the integral into two (one with a domain from 0 to 1 and one with a domain from 1 to infinity). When you do the same u-sub you showed on the latter integral, the bounds become 0 to 1 just like the former integral. You can then use identity xf(x) + (1/x)f(1/x) = 0 to finish. Interesting that you solved the problem by cloning the integral and multiplying by half.
Problem 2 is vastly simplified by noting that a < c, a=0,1 forces b to be 0, since c! has to be even. a=2, has a solution with b=2 and c=3, but if c > 3 then the equation breaks mod 4.
I'd be happy never to see another number theory problem. For my tastes, it's calculus and geometry forever! I guess you're blessed to have a range of viewers, Michael!
Well, look at his substitution of x→1/t. He then replaces t with x. This is because it's a definite integral. The numerical answer will be the same if we do it with t or u or θ or y or x. So he does it with x, to be able to combine it with the first integral, which is also in x and he wants to make it clear what he's doing.
Personally, I prefer the Calculus problem, but I have warmed up more to Number Theory over the years. (It helps that there are some truly excellent Undergraduate Texts available, such as Silverman's "A Friendly Introduction to Number Theory", or "An Introduction to the Theory of Numbers", 5th Edition, by Niven, Zuckerman and Montgomery. (BTW: Hugh Montgomery taught the Undergraduate Putnam Problem seminar that I audited at Michigan back in the 70's.) The question that I have, however, is whether or not there is a non-trivial example of the type of real-valued function in question, i.e., one that is not identically zero. Restricting ourselves to differentiable functions, I get that f(1) = 0, and all higher-order derivatives of f(x) at x = 1 can be expressed in terms of f'(1), e.g., f''(1) = -3f'(1). This suggests that a power series development around x = 1 is possible in the differentiable case. (Maybe I'll get back to this later.)
The condition on the function is actually fairly weak, so much so that any integrable function g defined on ]-1;1[ can be uniquely extended to a function f defined on R that satisfies the condition. If g is continuous and lim x-> +/-1 g(x)=0, then f is also continuous on R. For instance, if g is an even function with g(x)=x-1 for x in [0,1], then f(x) = 1/x^2 - 1/x^3 for x in [1, +inf[ is a continuous solution. You probably can play around a find a Cinf solution by choosing an adequate formula for g.
Small world. I had Dr. Montgomery for analytic number theory back in the day too. And I would agree with you on your two texts you suggest. I was taught out of Niven and Zuckerman (before Hugh got added on as co-author).
Gonna do a tier list for the best problems of the branched i encountered a little more: S tier: Ring theory A tier: Real analysis, Combinatorics, Number theory B tier: Group theory, Linear Algebra, point set topology C tier: Dif Equations D tier: Geometry
Actually the second problem doesn't state that b is integer. For a an c one may infer it from the fact that they are the argument of a factorial, but you can certainly raise 2 to a non-integer power.
@@__christopher__Duh! Ok, point taken. However, this is a "number theory" problem, which are pretty much always about integers. I'd be very surprised if the original problem didn't specify that all three were integers. Otherwise, there would be infinitely many solutions, b = ln(c!-a!) / ln 2 for all c>a
Never really understood why math guys with clean writing skills drop the units, ...from the length of a curve to it's confined area, where pi can be a relative scalar on one black board, and possess square units on another, and then later even have volume attributes somewhere else, and still be transcendental without disrepute, just so long as one can take the log and still get something that exists? The ultimate numerical politician beyond the trinity, PI, ...!
13:30 case2: 3>c>a
THIS ! 😂
@@CairosNaobum It's just a typo 👀
Glad not the only person that noticed the error.
I think the solution for the factorial problem was way too compilicate. We can rewrite the equation as:
2^b = c! - a! = a! * ((a+1) * (a+2) * ... * c - 1)
Note that the contents of the outer brackets on the right side can only be even if the product (a+1)*... inside only has a single term a+1, otherwise it's impossible. And also that a! can't have any prime factors > 2.
So there are 2 cases to consider:
1) 2^b is odd => 2^b = 1 => b = 0 => c! - a! = 1 => c = 2, a = 0 or a = 1
2) 2^b is even => 2^b = a! * ((a + 1) - 1) = a! * a => a = 2 => 2^b = 2! * 2 = 4 => b = 2 => c! = 2 + 4 => c = 3
It's my extremely subjective personal opinion, but calculus to me is smooth and connected, while number theory is pure chaos and disconnected ideas. Increasing a number by 1 is usually one of the most drastic changes possible in that context. It always feels like the problems have to be custom tailored to fit whatever random method only works for that problem. But again, this is really subjective and probably unfair.
Have you looked into analytic number theory? It's a beautiful subject.
@@DrR0BERT I'm not a researcher anymore and the subject is a good bit hairier than I can take on as a filthy casual. I saw someone present a proof of the prime number theorem once, but I was unable to understand beyond the first half in just the one sitting.
I am never turning down Number theory problems. They are always the best!!!!
Hello from Bangladesh. I work with youngsters for MO, and seeing something from our community here is joyful.
i like the calculus problems as well as the geometry problems
I am more of a fan of the combinatorics problems. Number theory and Geometry comes as a close second.
@@primenumberbuster404
I don't like combinatorics
I like number theory, geometry and calculus
Case 2: a
I did a math minor with analysis topics along with my calculus based engineering degree so I feel like I can give the calc problems more of a go on my own
I usually don't know how to approach the number theory problems, but I learn a lot from the videos
From 1/x^2 f(1/x) = - f(x), it immediately follows (with integral f(x) = integral f(1/x)/x^2) that integral of f(x) dx = - integral f(x) dx. Done.
Not done. You have to show that the indefinite integral converges. All you've shown is that if it converges then it's value is 0.
Take the function f(x) = 1/x if x between 0 and 1 and -1/x if x is bigger than 1.
For the first problem, a simpler solution is to break into integral 0 -> 1 plus integral 1 -> inf, do a change of variables to x=1/u in the second integral, and then see the two are negatives of one another by the defining function property
I always prefered calculus. I tried once to get into the math team of my country, once they had questions of drawing 10 traingles with 7 lines, I realized it's not my cup of tea.
I studied mathematics many years ago, your videos are very good and rekindle my love for mathematics. if my voice has any weight i lean towards number theory. thank you Michael
I would say, pound for pound, "discrete" problems are harder than "continuous" probs.
The present probs are a nicely contrasted pair. Once you've twigged the "trick" in the first prob (the hard step), the solution is easy to write. For the second problem, it's the other way round. Here, it's easy to come up with solutions just by playing around with numbers, but actually writing out a concise proof that you have obtained _all_ the solutions is significantly taxing
Like Penn, I got c! - a! = 2^b, which I then wrote as [ c(c-1) ... (a+1) - 1 ] .a! = 2^b
Clearly a! must be a power of 2 => a = 0, 1, 2. Also c(c-1) ... (a+1) can only be even by being 2, and since a+1 =< 3, it can only be odd by being 3
Thus:
a = 0 => c = 2 => b = 0
a = 1 => c = a + 1 = 2 => b = 0
a = 2 => c = a + 1 = 3 => b = 2
and we have solutions a, b, c = 0, 0, 2; 1, 2, 0; 2, 2, 3
I get most of your proof, but I don't understand this part: "Also c(c-1) ... (a+1) can only be even by being 2". Why?
If it was even and 4 or more, c(c-1) ... (a+1) - 1 would be odd, and 2^b doesn't contain an odd factor@@bjornfeuerbacher5514
Professor Michael Penn is a Master of Number Theory yet he lives in the heart and soul of contemporary Mathematical Science. Stay Blessed Michael.
ab = 1 ==> b = 1/a was screaming to me to split the integral into two (one with a domain from 0 to 1 and one with a domain from 1 to infinity). When you do the same u-sub you showed on the latter integral, the bounds become 0 to 1 just like the former integral. You can then use identity xf(x) + (1/x)f(1/x) = 0 to finish. Interesting that you solved the problem by cloning the integral and multiplying by half.
Problem 2 is vastly simplified by noting that a < c, a=0,1 forces b to be 0, since c! has to be even. a=2, has a solution with b=2 and c=3, but if c > 3 then the equation breaks mod 4.
I like both of them and functions equation as well!)
I'd be happy never to see another number theory problem. For my tastes, it's calculus and geometry forever! I guess you're blessed to have a range of viewers, Michael!
Now I realize that my country's Math Olympiad is not that much bad.
Never was, guys are putting a lot of effort here.
*realise *maths
@@cycklist 哈喽英国人。
Probably anyway.
Anyway, that's how the Americans spell those words.
😐😐😐@@cycklist
@@cycklist I still am dumbfounded the brits retained the s when shortening a word that isn't plural.
I LOVED Professor Michael Penn's pregnant pause at the 12:26 Mark !! I just ABSOLUTELY LOVED IT !!
Both of those are good, but the integral with complicated integrands I don't like, unless they come from real life.
That second question got me wondering. Are there any instances where a! - b! = c^2 for c > 2 ? Or a! - b! = c^n for c & n > 2 ?
Calc
Number theory, though I'm studying it only from your channel.
Thanks for the new content! Also: I'm confused where that 1/x^2 comes from/in, but I'm also usually mathematically clueless ... so ...
Other than that, the rest follows clearly enough.
Also (and I only point this out not to nitpick but because I make math errors as easily as eating too many Oreos),
why isn't it 3 > c > a?
Well, look at his substitution of x→1/t. He then replaces t with x. This is because it's a definite integral. The numerical answer will be the same if we do it with t or u or θ or y or x. So he does it with x, to be able to combine it with the first integral, which is also in x and he wants to make it clear what he's doing.
Thanks, but that's not my confusion. Why is it squared? @@xinpingdonohoe3978
Personally, I prefer the Calculus problem, but I have warmed up more to Number Theory over the years. (It helps that there are some truly excellent Undergraduate Texts available, such as Silverman's "A Friendly Introduction to Number Theory", or "An Introduction to the Theory of Numbers", 5th Edition, by Niven, Zuckerman and Montgomery. (BTW: Hugh Montgomery taught the Undergraduate Putnam Problem seminar that I audited at Michigan back in the 70's.) The question that I have, however, is whether or not there is a non-trivial example of the type of real-valued function in question, i.e., one that is not identically zero. Restricting ourselves to differentiable functions, I get that f(1) = 0, and all higher-order derivatives of f(x) at x = 1 can be expressed in terms of f'(1), e.g., f''(1) = -3f'(1). This suggests that a power series development around x = 1 is possible in the differentiable case. (Maybe I'll get back to this later.)
The condition on the function is actually fairly weak, so much so that any integrable function g defined on ]-1;1[ can be uniquely extended to a function f defined on R that satisfies the condition. If g is continuous and lim x-> +/-1 g(x)=0, then f is also continuous on R. For instance, if g is an even function with g(x)=x-1 for x in [0,1], then f(x) = 1/x^2 - 1/x^3 for x in [1, +inf[ is a continuous solution. You probably can play around a find a Cinf solution by choosing an adequate formula for g.
Small world. I had Dr. Montgomery for analytic number theory back in the day too. And I would agree with you on your two texts you suggest. I was taught out of Niven and Zuckerman (before Hugh got added on as co-author).
Gonna do a tier list for the best problems of the branched i encountered a little more:
S tier: Ring theory
A tier: Real analysis, Combinatorics, Number theory
B tier: Group theory, Linear Algebra, point set topology
C tier: Dif Equations
D tier: Geometry
I am curious, why do you like Ring Theory more tha. Group Theory. I find them quite similar
Those ideals sure be looking ideal
no. pythagorean theorem literally holds the world together.
Calculus for life!
Actually the second problem doesn't state that b is integer. For a an c one may infer it from the fact that they are the argument of a factorial, but you can certainly raise 2 to a non-integer power.
Will raising 2 to a non-integer power ever be equal to an integer?
@@gerryiles3925 of couse. 2^(ln 3/ln 2) = 3.
@@__christopher__Duh! Ok, point taken. However, this is a "number theory" problem, which are pretty much always about integers. I'd be very surprised if the original problem didn't specify that all three were integers. Otherwise, there would be infinitely many solutions, b = ln(c!-a!) / ln 2 for all c>a
14:22 the solution doesn't obey the inequality of case 2....
How is the step in 3:33 possible?
I prefer number theory problems for sure
very clever !
bro's gaslighting his class with that [ case2: 3
I liked both problems.
I'm definitely a calculus guy.
Sorry for unrelated question, but do we have non-real constants in mathematic? Complex, quaternion, hypercomplex etc.
Well, i is a complex constant, isn't it? ;)
@@bjornfeuerbacher5514 Of cource that true, but also if i is all, that seems strange.
@@DL7014 What about the zeroes of the Riemann zeta functions? These are infinitely many complex constants. ;)
Hi,
I prefer the first type of problems.
Hello
Love love love number theory problems
You're asking which of my children I prefer...
In case 2 your inequalities are the wrong way around
I enjoy the number theory problems more because I know less about it.
I am number problem guy because that question of calculas is very much predictable
number theory and geometry.
Definitely prefer number theory to calculus.
I’m a physicist, so I prefer calculus problems.
Calculus here. Is one of the foundations of the mathematical physics.
Never really understood why math guys with clean writing skills drop the units, ...from the length of a curve to it's confined area, where pi can be a relative scalar on one black board, and possess square units on another, and then later even have volume attributes somewhere else, and still be transcendental without disrepute, just so long as one can take the log and still get something that exists? The ultimate numerical politician beyond the trinity, PI, ...!
For NT problem it's simpler to write c!-a! as
a! * ( c!/a! - 1) = 2^b
Neither term can have three so a
Definitely a good strategy to factor to get multiplication then utilize fundamental theorem of arithmetic
Number theory.
Algebraic problems
Numerical calculations
Calculus
Calculus all the way
Love from India
To Integrate or not to integrate? To Diophantinate or not? That is the question?
Another vote for number theory.
Also, I suspect number theory lovers are less likely to vote here. Voting means contact with messy reality, uggh... :)
I’m a physics weenie, so I much prefer the calculus videos.
🔥
I think you overcomplicated your way to the solution of the first problem
Its not bengladesh its Bangladesh
man! it isnt bang laDesh, its bangladesh.
Why there is a flag on Thumbnail?
What's the point
It's the flag of Bangladesh. Why that flag is there should be rather obvious.
Michael used appropriate flags in his thumbnails already quite often.
Well... just watch 6 secs of this vid...
What is your problem for using Flag?
To Integrate or not to integrate? To Diophantinate or not? That is the question?