I think this can be done a bit more simply: note (eg, by induction) that f(n) = n * sum (a_j / p_j). f(n) = n => sum (a_j / p_n) = 1, and the only way that can happen is if we have only one prime and a_1 = p_1, so n =p^p for some p.
Yep. That's exactly how I did. We should meet and start making math videos. Recently I watched a teacher using 4 videos to prove a Pithagorean relation associated with 4 consecutive Fibonacci numbers. I demonstrated it in one comment. Man, his choice of variables was very bad. Not clever at all.
@@erikhansbo5834 let P = prod p_j, then sum a_j/p_j = 1 => sum a_j P/p_j = P Each p_j must divide all terms. So it must divide a_j P/p_j, but doesn't divide P/p_j. So a_j = p_j k_j, with k_j nonnegative integer. Using that in the expression of the sum, we get sum k_j = 1 The only solutions is all k's equal 0 and k_j=1, for some j.
@@erikhansbo5834 If a/b and c/d are in their lowest form and b and d are coprime, then the denominator of a/b + c/d is bd [*]. Because the p_i are all coprime, the sum is only an integer if all the terms are integers -- otherwise the denominator must be greater than one. [*]: If not, then p | bd and p | ad + bc. WLOG p | b. Then p | ad, and because b and d are coprime, p | a, so a/b was not in lowest form.
@@samueldeandrade8535 So if you actually add every step to the proof (namely, that if the sum over a_i/p_i's is 1, then k=1 and a_i = p_i), you end up with about the same content as in the video. I do find your approach much more elegant, but the main difference is just giving mid results for granted.
Note that if you extend that power rule to p^p you get p*p^(p-1) or just p^p again - immediately when you wrote n’=n I started to look for the equivalent of exp(x) and it took me only a couple guesses to come to that result.
Mike! You might have probably heard this before: I suppose these videos are made such that you've already solved the problem and you present it to us. But for me it'd be really interesting to see you actually solving the problem from scratch. To see your way of thinking that leads to the solution and how long does it actually take to solve them. Best regards! (And of course: awesome video once again :) )
@@Minskeeeee actually, it is pretty simple: let P = prod p_j, then sum a_j/p_j = 1 => sum a_j P/p_j = P Each p_j must divide all terms. So it must divide a_j P/p_j, but doesn't divide P/p_j. So a_j = p_j k_j, with k_j nonnegative integer. Using that in the expression of the sum, we get sum k_j = 1 The only solutions is all k's equal 0 and k_j=1, for some j.
@@samueldeandrade8535 is that not just Michael's method? by multiplying by prod(p_j), it seems like you're just reversing that cancelation that the comment refers to why cancel that term just to bring it back in the proof? seems like it's not actually easier
To make it look more beautiful, we could also write the product of all the primes with their powers reduced by 1 in the general formula with a product series notation. I later got why he didn't do like that.
Is it possible to just have \sum_{j=1}^{k} a_j * rad(n) / p_j = rad(n) * \sum{a_j / p_j} which leads to \sum{a_j / p_j} = 1? I think I am missing something but don't want to think about it too much 🙂
1. Just wondering if I'm correct on this: you can extend the domain of f to N0 by setting f(1)=0... or did I miss something? 2. The divisibility arguments seem unnecessarily complicated. We can simply note that rad(n) is a constant with respect to the summation index j and take it out of the sum. Then we cancel all of the products and get 1=sum(a_j/p_j), which leads to k=1, a=p as the only solution. Or am I making an assumption I'm not supposed to?
I'm not exactly sure how you are avoiding the divisibility argument there? By what logic are you justifying that the only way the sum can equal 1 is k=1 a=p, without an essentially identical divisibility argument?
The question as you wrote it is incredibly open - yes, you can literally just define f(p) for all primes p, that gets you f(p^q), and then you can choose completely arbitrary values f(n) for any n that is not a power of a prime. It'd be more interesting if you wanted the function to have further required properties.
Am i missing something? The question as set only gives the result for primes, not for the general case of integers... 😮 Later: aha! he's deriving the result for composites 😊
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I think this can be done a bit more simply: note (eg, by induction) that f(n) = n * sum (a_j / p_j). f(n) = n => sum (a_j / p_n) = 1, and the only way that can happen is if we have only one prime and a_1 = p_1, so n =p^p for some p.
Yep. That's exactly how I did. We should meet and start making math videos. Recently I watched a teacher using 4 videos to prove a Pithagorean relation associated with 4 consecutive Fibonacci numbers. I demonstrated it in one comment. Man, his choice of variables was very bad. Not clever at all.
Could there not be several terms all adding up to 1 together? a_j / p_j could be a fraction, no?
@@erikhansbo5834 let P = prod p_j, then
sum a_j/p_j = 1
=> sum a_j P/p_j = P
Each p_j must divide all terms. So it must divide a_j P/p_j, but doesn't divide P/p_j. So a_j = p_j k_j, with k_j nonnegative integer.
Using that in the expression of the sum, we get
sum k_j = 1
The only solutions is all k's equal 0 and k_j=1, for some j.
@@erikhansbo5834 If a/b and c/d are in their lowest form and b and d are coprime, then the denominator of a/b + c/d is bd [*]. Because the p_i are all coprime, the sum is only an integer if all the terms are integers -- otherwise the denominator must be greater than one.
[*]: If not, then p | bd and p | ad + bc. WLOG p | b. Then p | ad, and because b and d are coprime, p | a, so a/b was not in lowest form.
@@samueldeandrade8535 So if you actually add every step to the proof (namely, that if the sum over a_i/p_i's is 1, then k=1 and a_i = p_i), you end up with about the same content as in the video. I do find your approach much more elegant, but the main difference is just giving mid results for granted.
Note that if you extend that power rule to p^p you get p*p^(p-1) or just p^p again - immediately when you wrote n’=n I started to look for the equivalent of exp(x) and it took me only a couple guesses to come to that result.
Yeah, in a test environment I suspect guessing that answer and verifying is the dominant strategy. Although this proof is pretty.
Mike! You might have probably heard this before: I suppose these videos are made such that you've already solved the problem and you present it to us. But for me it'd be really interesting to see you actually solving the problem from scratch. To see your way of thinking that leads to the solution and how long does it actually take to solve them. Best regards! (And of course: awesome video once again :) )
Isn’t it simpler to just cancel rad(n) at 16:55 and get sum{a_j/p_j} = 1 or am I missing something
Yes that's much faster
how does sum{a_j/p_j} = 1 imply that there must be a single a=p pair? Would the logic of proving that be easier than Michael's solution?
Dr. Penn makes that argument at th-cam.com/video/4oAda4DtEo4/w-d-xo.html
@@Minskeeeee actually, it is pretty simple:
let P = prod p_j, then
sum a_j/p_j = 1
=> sum a_j P/p_j = P
Each p_j must divide all terms. So it must divide a_j P/p_j, but doesn't divide P/p_j. So a_j = p_j k_j, with k_j nonnegative integer.
Using that in the expression of the sum, we get
sum k_j = 1
The only solutions is all k's equal 0 and k_j=1, for some j.
@@samueldeandrade8535 is that not just Michael's method? by multiplying by prod(p_j), it seems like you're just reversing that cancelation that the comment refers to
why cancel that term just to bring it back in the proof? seems like it's not actually easier
To make it look more beautiful, we could also write the product of all the primes with their powers reduced by 1 in the general formula with a product series notation. I later got why he didn't do like that.
I get that if f(n)=n there can be only one prime p dividing n and we must have n=p^p so the answer is 5^5=3125.
Ahhh ... the beauty of math.
16:20
wow... still here! amazing!
@@pragyanpranay3681 Yep
Why use cap notation, you could just use power 0 for this term
Is this an old video? I remember seeing this
Nah, u are remembering an older video he did on the arithmetic derivative.
Really loved this problem!
How would one show that this function is well defined?
Is it possible to just have \sum_{j=1}^{k} a_j * rad(n) / p_j = rad(n) * \sum{a_j / p_j} which leads to \sum{a_j / p_j} = 1? I think I am missing something but don't want to think about it too much 🙂
I got that also. From this we can deduce that a_j >= p_j so this implies that there is only one prime factor and a_j=p_j.
1. Just wondering if I'm correct on this: you can extend the domain of f to N0 by setting f(1)=0... or did I miss something?
2. The divisibility arguments seem unnecessarily complicated. We can simply note that rad(n) is a constant with respect to the summation index j and take it out of the sum. Then we cancel all of the products and get 1=sum(a_j/p_j), which leads to k=1, a=p as the only solution. Or am I making an assumption I'm not supposed to?
This function can be extended to all the rationals, to some irrationals, and even to some complex (non real) numbers :)
I'm not exactly sure how you are avoiding the divisibility argument there? By what logic are you justifying that the only way the sum can equal 1 is k=1 a=p, without an essentially identical divisibility argument?
Exists a function such that f(p^q)=p^f(q) for all p, q primes?
the identity works
That’s all?
The question as you wrote it is incredibly open - yes, you can literally just define f(p) for all primes p, that gets you f(p^q), and then you can choose completely arbitrary values f(n) for any n that is not a power of a prime.
It'd be more interesting if you wanted the function to have further required properties.
In this video:
How to solve an arithmetic differential equation ❌
The product rule is called the Leibniz Rule✔
by the definition of f, f(1 x 1)=1 x f(1) + 1 x f(1) => f(1)= 2 x f(1), which seems to be a problem. Is it?
No, because f(1)=0 satisfies that equation.
f(p) = f(1p) = 1f(p) + pf(1) = 1 implies f(1) = 0 right?
I think it was assumed that the unit is excluded from the set of primes.
it is. that makes it necessary to ask if one's value exists and if the function remains consistent.
@@jasonroberts2010 Yes. When I first looked at the problem I even was not sure if the function is well defined :)
Can't we just divide by P_j instead of using the "hat"?
Same suggestion
Cest magnifique
Interesting :)
How can I get more information about this function? f(x*y)=x*f(y)+y*f(x)
"Bolivia TST 2021" isn't enough for search more about this
Am i missing something?
The question as set only gives the result for primes, not for the general case of integers... 😮
Later: aha! he's deriving the result for composites 😊
arith-EMETIC? Was it intended?
Is this first?
nobody cares
@@opensocietyenjoyeri do
First what?