I have noticed that if there is an easy way to prove something, then Michael will point it out and then prove it a harder, but more entertaining and often a more general way :-).
That's true. In mathematics important theorems are often proved in different ways using different techniques. Examples are e.g. theorem of pythagoras or the law of quadratic reciprocity. By this you can understand relations between different areas of mathematics.
@@gabriel7233 Additionally, if you know both techniques, the work for induction in this case is easier. 2^(n-7) > 128n > 128 implies 2^((n+1)-7)=2*2^(n-7)>128n+128=128(n+1).
I think that this is one of the beauty of maths. There's so much ways to interpret something that you can always found a new one. For example, one of the most relevant theorems of math is The Fundamental Theorem of Algebra and one of the most important consequences is the fact that links algebra with the study of functions (calculus).
This is a high school math problem, so derivatives are an overkill. There is a simpler way, and for a kid trying these problems that solution is more useful.
@@vedhase2370 cuz let's say 1=n^(1/n-7) Then taking log on bs we get 0=(1/n-7)xlog(n) Now (1/n-7) is non zero as n>=8 => logn =0 => n=1 Which is a contradiction. QED
I don't quite get the fact he is doing too much to prove that sentence. There is no number greater than 8 multiplied by itself n-7 times going to give a number lesser or equal than 1, enough proof.
@@AnindyaMahajan I don't see why. a^y means a is a real constant and y is the input. Here a = x (so a is not a constant) and y = 1/(x-7). After x>=8, the function (x^(1/(x-7))) is strictly decreasing. Apart from looking at the graph what other proof is there? Edit: Actually after giving it some thought I see it is obvious since x^(1/(x-7)) = 1 when 1/(x-7) = 0 which leads to: 1 = 0 So x^(1/(x-7)) is never 1 and since the function is strictly decreasing it will never be less than 1 either.
Without induction, in my opinion the easiest way is to use binomial coefficients. n < 2^(n-7) for n >=11 is equivalent to n+7 < 2^n for n>=4. We check n=4 and n=5 by hand and for n>5 we have 2^n = (1+1)^n > (n choose 0) + (n choose 1) + (n choose n-1) + (n choose n) = 1 + n + n + 1 = 2 + 2n > n + 7.
@@shurik3nz346 I wasn't advocating strapping people to chairs and prying their eyes open Clock of Orange style; I meant that people would benefit but most people are ignorant of this channel's existence.
Why the equivalence is false? I had seen other comments claiming the same thing, one of them said that f(x) is negative for some real numbers, but how can this be true for x>= 11?
@@Nicolas-zf3pv It's not clear that if f(n)>0 for n>=11, n integer then f(x)>0 for x>11, x real. In this case the equivalency holds because f is continuous and monotonous. But it doesn't hold for any f. He proves later that it's monotonous, so with that bit from the proof (and the fact that the function is clearly continuous) he could prove that the equivalency holds as well. He doesn't need to, because the other direction is sufficient. But the way he writes it, it's not sufficient.
when i first looked at the problem, i found it pretty clear that either the exponents needs to be 1 (which gives 8 as a solution) or n needs to be a perfect power (because otherwise we would never get an integer by taking an (n-7)th root). the rest would be quite similar to what you did, just showing that f(x)=x^(1/(x-7)) is decreasing for x>=8.
Finding the solutions is easy; the hard part is proving that there are no other solutions. It may be easy enough to grasp intuitively, but math requires more proof than just intuition.
set f(x) =x^(1/x-7) arguing the monotony of the function on [8,+∞) let a,b≥8 be real numbers s.t a1/b-7 ⇒ f(a)>f(b) ⇒the function is strictly decreasing on [8,+∞). therefore max(f(x))=f(8)=8. We have 0
8:20 Another nice way to see 2^11 - 128(11) > 0 is to factor a 2^7 out of each term and observe it's true iff (2^4 - 11) > 0. (Even if it's not memorized, 128=2^7 was used earlier in the problem.)
First impression: Holy moly, that's gonna be hard Second impression: But wait, that's gonna be impossible very soon Solution: It's actually just the first 2 allowed numbers I'd say the perceived difficulty of this exercise is a strictly decreasing function of time.
I had the opposite experience. My immediate impression (even before I noticed the ≥ 8 condition) was 8 works, 9 works, and then that's it. The more he explained why higher numbers don't work, the more I started to think that there's going to be some other number, probably a multiple of 7, where it equals 1, but that it would be really hard to prove.
6:55 I don't think it is equivalent because you could take such a function that it would be positive on integers but not always positive for real numbers but anyway that doesn't change the solution
He can't do backflips in videos any more, due to discontinuities outside the domain of the video in the neighborhood of his head. (That is, hanging microphones.)
I tried to solve this from the thumbnail, so I missed that we were solving over the natural numbers. I got n = - W(-1, -ln(z) / z^7 ) / ln(z) where z is an integer and W is the Lambert W function aka the ProductLog. Apparently, we need the -1 branch. n. n^(1/n-7) 10.375 2 9 3 8.54777 4 8.31611 5 8.17246 6 8.07331 7 8 8 7.94315 9 7.89749 10 7.85982 11 7.82809 12 7.80089 13 7.77725 14 7.75646 15 7.73799 16...
In order for it to be an element of N then n should be written as n=m^(n-7), with m>=3>e cuz n>8>e (the case of n=8) is trivial) also m=n^m when we play around with the inequality and by replacing values we get n
@@vedhase2370 well, any number to the zero ith power is equal to one, ex 3^0 = 1 and 7^0 = 1. The problem states that n is greater than or equal to eight. 1/(n-7) is always going to be greater than 0 if n is greater than eight. That means that the equation basically boils down to n^(a number greater than 0). In order for the whole equation to be less then or equal to one, n must be raised to the power of 0, or less, but since 1/(n-7) is always greater than zero due to the parameters of the question, we know that n^(1/(n-7)) is greater than one. Hope this clears things up
An arguably quicker way of proving 2^n>128n could have been by plugging n=11 and checking that 2^n is indeed larger and then seeing that the slope of 2^n at n=11 is 1024*11 (chain rule) and that the slope of 128n is 128. Plus the slope of 2^n only keeps on growing whereas 128n has a constant (and lower) slope. Combining f(11)>g(11) and f'(x)>g'(x) for x in [11,∞), you know that f(x)>g(x) ,if f(x)=2^x and g(x)=128x
I just directly differentiated the original function extended to the reals, showed that was negative for n>=8 and then by exploration found that 8&9 were the only solutions.
My solution substitute x = n - 7 and n = a^x for some integer a to obtain a^x = x + 7. Then obviously for x >= 4 we need a < 2 and we only need to evaluate x = 1,2,3. The only satisfactory pairs (a,x) are (8,1) and (3,2)
Love the channel, interesting problems and crystal clear explanations! for proving ln 2 > 1/2, maybe it is a bit more straightforward to do: ln 2 = 1/2 * 2 * ln 2 = 1/2 * ln (2^2) = 1/2 * (lg 4 / lg e) > 1/2 ?
@@ddognine I think that's quite straightforward as e=2.7*** < 4. of course, for formal proof one should probably claim on monotonicity of lg function etc.
Observe that for n^(1÷m) to be a natural number all of n's prime factors must have multiplicity that's divisible by m. The smallest such number (other than 1) has prime factorization 2^m (because 2 is the smallest prime) which obviously grows faster than m+7 once you get past about m=4
... and obviously 1 isn't the mth root of any number m+7 because it's not the mth root of any number other than 1. Proving this is a waste of time. Unless they score your answer based on how fundamental your proofs go?
I believe you could also use binomial expansion for 2^m to show the inequality as well. Use 2^m = SUM( m choose k) from k=0 to m and use however many terms you want to show that 2^m > 128m
The sequence decreases on its domain and is asymptotic to 1. We know a_8 = 8, a_9 = 3, so you need to consider whether there's an n such that n^(1/(n-7)) =2; it is relatively easy to show that there are no natural number solutions to this equation.
@@angelmendez-rivera351 ohk, I didn't know the meme was that old, relatively new here; but just saying upload time and posting time does not have significant time lag and creating a new account doesn't take more than, say, 10 minutes max.
Excellent! You did your number eight like me until I were 8 years old. I remember my teacher scolded me for it. So far, I do my eights "normally". It was just to annoy you but you're doing a great job!
11:19 : After reading Richard Feynman's brilliant "Surely You're Joking, Mr. Feynman!" I can't forget that ln 2 = 0.69... which is clearly bigger than 0.5
Please correctly state the question by adding: n ∈ N (natural number). Otherwise, there are more real number solutions for n, since the real function f(x) = x ^ (1/x-7) is a continuous function, and we have f(8)=8 and f(11)
Wonderful video, always a pleasure to watch. I was wondering if you Could make a video on the “points in the space” type of problem (I believe it’s called misc but I’m not sure). An example to this type of problem is “There are a 1000 points in the space, prove you could always pass a line such that half the points would be on one side and the other half on the other side”.
you need not use derivative of f(x). it's enough that just prove f(n+1) > f(n) for all n > 7，which is much easier: 2^(n+1)-128(n+1)-(2^n-128n)=2^n-128 > 0
The function f(x)=x^{1/(x-7)}=e^{ln x/(x-7)} is greater than 1 for x>7 and monotone decreasing, since f'(x)=f(x)/[x(x-7)^2]*[-7-x(ln x -1)]10 we have 1
Unless n is in natural, there has to be a solution for f(n) = n^(1/(n-7)) = 2. Note that f is monotonic, continous in (7,+inf> and has an asymptote at f = 1.
Let n=b^x. Then we need to observe when b^[x/(b^x-7)] is an integer. That means b^x-7 divides x. For that to be possible, b^x-7 =< x. If x=1, b^x=8. If x>1, then 2^x-7 =< x so x
I used a slightly different approach. I took the derivative of x^(1/(x-7)), and showed it was decreasing for all x greater than or equal to 8. Since we know 1 is our lower bound, and the n=11 case is less than 2, then we know we only need to check 8, 9, and 10
Set k = n-7 (k+7)^(1/k) in N, k > 0 (k+7) = m^k, m,k in N k = 1 and 2 are trivial solutions, with m = 8 and 3. Notice that (k+7)^(1/k) = m is strictly decreasing with k, so any other solution would have m = 2. 2^k = k+7 has a solution between 3 and 4, so not in integers. Hence, n in {8,9} are the only solutions.
If n^(1/(n-7)) is an integer, then n = m^(n-7) for some integer m greater than or equal to 2. Take a prime divisor p of n and let e be its exponent in the factorization of n. Since n = m^(n-7), the exponent of p in the factorization of m^(n-7) is greater than or equal to n-7 and hence so is a. It thus follows that 2^a - 7 ≦ n - 7 ≦ a. Therefore 2^a - a ≦ 7．Because 2^x - x is increasing on x ≧ 1 and 2^4 - 4 > 7, we have the inequality a ≦ 3; Hence n ≦ 10.
I think it's easier how I solved it. So if n ^ (n-7) is natural, it means that n = k ^ (n-7), with k a natural number. It is immediately verified that n = 8 and n = 9 are solutions and k = 8 and k = 9. For n = 10, the only close k could be 2 becouse 2 ^ 3 = 8 and 3 ^ 3 is 27, much greater than 10. Therefore we observe that for n> = 11, n = 11, n = 2. By induction, P (m) => P (m + 1) So we have m
after watching Numberphile: *I NOW KNOW AND UNDERSTAND MATH! I CAN SOLVE ANY MATH PROBLEMS NOW, I CAN DO IT!* after watching serious Math like this channel: 😭😭😭😭😭😭😭
isn't it easier to take both sides to the power n-7 so we're looking for x^(n-7)=n . x positive integer. since x=1^(n-7) is always 1 and the smallest x is then 2 and 2^(n-7)>n for all n>11 is pretty obvious with no heavy duty math. The induction is just that for n+1, 2*(previous>n)>n+1 .
I really like the ending you make in each video. That's a good place to stop. It feels like maths has an infinite domain one can explore. It's like the opposite of claustrophobia.
I don't want to split hairs, but it seems to me that the last equivalence at 6:49 is not true, or at least not that evident.. And in any case, a simple
A suggestion for a quick and easy proof (it feels almost too easy - are there any flaws in my reasoning?) : f(n)=n^(1/(n-7) >= 8^(1/(n-7)) = 2^(3/(n-7) > 1. When does f(n) get >=2? 2^(3/(n-7))>=2 3/(n-7) >=1 n
Well 14:14 may be a good place to stop. It is also an excellent place to start. Why? There are interesting ways forward that include asymptotic approximations concepts where for given epsilons and deltas there are Ns where n>N provide acceptable values. Admitted not pointwise answers other than N but mathematically acceptable (?) answers to tolerances appreciated by mathematicians, physicists and engineers alike. It seems good in the round to bring these things together and Michael's video is an excellent venue for that purpose alone. It also seem healthy introducing real analysts to ordinary world of numerical readings and associated tolerance values or error bounds? Debatable? I used Qualculate v4.9.0 a free software calculator with graphic facilities that make it easy to graph and change input range and so forth. So for answers accurate to 1 within error bound 10^(-4) at n=20000 and n=20001 yielded a interesting descending staircase graph. Obtained by plotting the graph where n is replaced by x and real x is approximated by setting sampling rate at 1001 with minimum value on x = 20000 and maximum value of x = 20001. Of course these equate to n=20000 and n=20001 respectively. An interesting change also happens between x=20000 and x=20002 I chose to present that range as it is the first time I noticed graph switch from smooth line to digital steps. In this case descending staircase. Important concepts? numerical methods, graphical methods, practical application of epsilons and deltas within acceptable error bounds. Thus! (I claim) fuller answer is 8, 9, infinity and N for infinitely many n>N where N defines finite range of n directly according to unacceptable error bound and infinitely many n>N within acceptable error bound.
0:52 A wonderful example of why you have to be careful making generalizations in mathematics, as this is a false statement. n=1 could be a potential solution (if it were not excluded by the n greater than or equal to 7 condition), as 1^(1/(1-7))=1^(-1/6)=1/(1^6)=1 is in fact an positive integer.
Just from think about it: n must be greater than 7 if it is less than that we have 1/(n^(7-n)) which isn't an integer, 8 and 9 both work 8^(1/(8-7))=8^(1/1)=8^1=8 and 9^(1/(9-7))=9^(1/2)=3 but are there more?
0:45 : I can't see why the obvious solution n=1 is ignored, as 1^(1-7) = 1^(-6) = 1 is definitely a natural number. I don't really appreciate fiddling with derivatives and natural logarithms instead of the induction proof (which is trivial even for me, and 90% of the things on this channel are way too complicated for me). Otherwise, a great video, thank you!
Wouldn't it be enough to see that the real function f=x^(1/(x-7)) is continuous for x>7 and approaches 1 from above as x goes to infinity? This proves that when x is large enough the inequality 1
Why doesn't 1 work? Edit : I know the question says "n more than 8", but he explains at 0:48 that all integers between 1 and 7 are not solutions anyway
You can also take the derivative of the function itself. It takes a bit of implicit differentiation but it's not too complex and you get y((x-7)/x-ln(x))/(x-7)^2=y'. Then since you know y is always positive and so is (x-7)^2 (at least whenever x>8, which is the relevant range anyways) you can show (x-7)/x-ln(x) is always negative since (x-7)/x approaches 1 and ln(x) approaches infinity (a bit trickier than just that but that's the idea). Positive * positive * negative = negative. Since the derivative is always negative, and you know for x=10, y8 or whatever the function is always greater than 1, you know the value will always be between 1 and 2 and therefore can't be an integer.
Can't we just say for the last part: ln(2) > 1/2 e ^ (ln(2)) > e^(1/2) [e to the power of both sides] 2 > e^(1/2) [by definition/properties of logarithm] 4 > e [squaring both sides of the logarithm] Or should x^(log_x(y)) = y (where log_x is log base x) be proven in olympiad setting as well?
Not clear whether the n >= 8 requirement was added or part of the original problem. It sounded like Michael said all n < 8 didn't work. Actually n = 1 also works.
I really don't know why he did the ln2 > 1/2 thing. "It's well known that e is 2.718..)" but wouldn't that also require a proof? I could just say "well my calculator says ln2 is 0.693..". Other than that, showing the gradient is always positive was good. edit: now I see it, he wanted to make it easy to understand
"similar things happen if n is equal to 5, 4, 3, 2, or 1. True for 5, 4, 3, and 2, but 1? 1^(1/1-7) = 1^(-1/6) which is 1. So I don't see the issue here. Did I miss something?
check small values, you gen n = -+1,8,9. prove by induction, if for some n >9 n^(n-7) > n, look at n+1: (n+1)^(n-6)= (n+1)^(n-7) * (n+1) us n+1 what is compare (n+1)^(n-7) v 1, what is obviously bigger then one.
Also the fact that the minimum prime factorization can be log2(n) indicates that the forth root is only available at the integer 16 at minimum is sort of a proof.
8:35
underlining inequality signs is usually not the best way to emphasize them ;)
I think that's why he's using a different colour to do this:)
I will draw a line through this equals sign to emphasize it ≠
The best way to emphasize a capital "I" is to draw a horizontal line on top of it. At least, T think so.
@@RadicalCaveman / think extreme italics work well.
😂😂@@RadicalCaveman
Can't believe I'm watching this for entertainment
lool me too
I'm watching this to fall asleep
ikr, it's super fun
Lol me too
That's because he is (somewhat) entertaining
I have noticed that if there is an easy way to prove something, then Michael will point it out and then prove it a harder, but more entertaining and often a more general way :-).
That's true. In mathematics important theorems are often proved in different ways using different techniques. Examples are e.g. theorem of pythagoras or the law of quadratic reciprocity. By this you can understand relations between different areas of mathematics.
All he did in this proof was use Calculus instead of induction. You could even say he did it the easy way.
@@wospy1091
Induction is easier.
@@davidbrisbane7206 not for everyone but it's clearly more elementary
@@gabriel7233 Additionally, if you know both techniques, the work for induction in this case is easier. 2^(n-7) > 128n > 128 implies 2^((n+1)-7)=2*2^(n-7)>128n+128=128(n+1).
He explained so many stuffs in a single video. This shows how different math problems are related to each other!!
I think that this is one of the beauty of maths. There's so much ways to interpret something that you can always found a new one.
For example, one of the most relevant theorems of math is The Fundamental Theorem of Algebra and one of the most important consequences is the fact that links algebra with the study of functions (calculus).
@@ckeimel True to say
May I know why it isn't equal to 1?? I am not really good at math. Someone, please help me out.
@@vedhase2370 wich part of the solution troubles you?
I like the method that was chosen to prove the inequality - it proves it not only for natural numbers, but for all real numbers as well.
Same!
This is a high school math problem, so derivatives are an overkill. There is a simpler way, and for a kid trying these problems that solution is more useful.
May I know why it isn't equal to 1?? I am not really good at math. Someone, please help me out.
@@vedhase2370 The first thing you need to know are the statement gaving, by the problem.
@@vedhase2370 cuz let's say 1=n^(1/n-7)
Then taking log on bs we get
0=(1/n-7)xlog(n)
Now (1/n-7) is non zero as n>=8
=> logn =0
=> n=1
Which is a contradiction. QED
What I've learnt from this channel is to start every math problem by considering case n=1.
N^(1/n-7)>1
The proof of this is obvious and left as an exercise to the reader
It is obvious. a^x is a strictly increasing function for a>1 and equals 1 at only x=0.
*n^(1/(n-7))>1
I don't quite get the fact he is doing too much to prove that sentence. There is no number greater than 8 multiplied by itself n-7 times going to give a number lesser or equal than 1, enough proof.
@@josem138 Not multiplied by itself (n-7) times, but 1/(n-7) times. But yes the claim is still trivial.
@@AnindyaMahajan I don't see why. a^y means a is a real constant and y is the input. Here a = x (so a is not a constant) and y = 1/(x-7). After x>=8, the function (x^(1/(x-7))) is strictly decreasing. Apart from looking at the graph what other proof is there?
Edit: Actually after giving it some thought I see it is obvious since
x^(1/(x-7)) = 1
when
1/(x-7) = 0
which leads to:
1 = 0
So
x^(1/(x-7)) is never 1 and since the function is strictly decreasing it will never be less than 1 either.
Fun fact, we know that 0
Ha ha! How?
@@35571113 You take 5th root of pi and 17 by hand, everybody is so impressed they ignore the fact that you didn't prove anything yet.
@@volodyanarchist lmao
May I know why it isn't equal to 1?? I am not really good at math. Someone, please help me out.
@@vedhase2370 I was wondering that too.
7:10 is not equivalent, but you only need the reverse implication here. f(x) >0 everywhere => f(n) > 0 for natural numbers
Without induction, in my opinion the easiest way is to use binomial coefficients. n < 2^(n-7) for n >=11 is equivalent to n+7 < 2^n for n>=4. We check n=4 and n=5 by hand and for n>5 we have
2^n = (1+1)^n > (n choose 0) + (n choose 1) + (n choose n-1) + (n choose n) = 1 + n + n + 1 = 2 + 2n > n + 7.
beautiful idea
great as always. love when you just underline something and mark it as "clear." that's how you know you're watching a real mathematician
Showing the "obvious" parts and "the long way" is so so good. It's how we all learn. Very good job!
Although it was excluded right at the beginning, n=1 would be a solution as 1^x=1\in N for any x.
This channel is such a gem, more people should be watching this stuff.
No, a person.can watch what they want on youtube.
Abacus, should and must are different
Abacus , ahh I see. Thanks for enlightening me.
@@shurik3nz346 I wasn't advocating strapping people to chairs and prying their eyes open Clock of Orange style; I meant that people would benefit but most people are ignorant of this channel's existence.
@@stevenwilson5556 haha alright
I love this channel, awesome the way you spent a several seconds and a couple of sentences in proving that 10 is not a perfect cube.
the equivalence at 7:30 is false.
But the implication b=>a is enough to prove the result (b result with x in R).
Why the equivalence is false? I had seen other comments claiming the same thing, one of them said that f(x) is negative for some real numbers, but how can this be true for x>= 11?
@@Nicolas-zf3pv It's not clear that if f(n)>0 for n>=11, n integer then f(x)>0 for x>11, x real. In this case the equivalency holds because f is continuous and monotonous. But it doesn't hold for any f. He proves later that it's monotonous, so with that bit from the proof (and the fact that the function is clearly continuous) he could prove that the equivalency holds as well. He doesn't need to, because the other direction is sufficient. But the way he writes it, it's not sufficient.
Man who I wish I would have had such videos in high-school almost 20 years ago.
when i first looked at the problem, i found it pretty clear that either the exponents needs to be 1 (which gives 8 as a solution) or n needs to be a perfect power (because otherwise we would never get an integer by taking an (n-7)th root). the rest would be quite similar to what you did, just showing that f(x)=x^(1/(x-7)) is decreasing for x>=8.
Love your videos, participating in my first olympiad in a month or two
Good luck!
What olympiad are you attending?
@@James_Moton Norwegian one
All the best✌️👍👍
Good luck!
I felt this much easier than any other question
For example question 6 of IMO 2020
Yes
Bcz it is
@@roboto12345 I can only solve Q1 of imo2020
Finding the solutions is easy; the hard part is proving that there are no other solutions. It may be easy enough to grasp intuitively, but math requires more proof than just intuition.
set f(x) =x^(1/x-7)
arguing the monotony of the function on [8,+∞) let a,b≥8 be real numbers s.t a1/b-7 ⇒ f(a)>f(b)
⇒the function is strictly decreasing on [8,+∞). therefore max(f(x))=f(8)=8.
We have 0
8:20 Another nice way to see 2^11 - 128(11) > 0 is to factor a 2^7 out of each term and observe it's true iff (2^4 - 11) > 0. (Even if it's not memorized, 128=2^7 was used earlier in the problem.)
Me: Sees 1/n-7
*Ezee*
Me: Sees n just below it
*crap*
Same
1/n-7 is integer only for n=6,8
Dorulo ili kvo
May I know why it isn't equal to 1?? I am not really good at math. Someone, please help me out.
@@vedhase2370 very late,but since 1/(n-7) is positive, n^(1/(n-7)) > n^(0) = 1
First impression: Holy moly, that's gonna be hard
Second impression: But wait, that's gonna be impossible very soon
Solution: It's actually just the first 2 allowed numbers
I'd say the perceived difficulty of this exercise is a strictly decreasing function of time.
It has a spike like x^-2, a discontinuity that occurs when he mentions that we're not using induction
Well if feel like doing a challenging prob
You can try this
ua-cam.com/video/igdy05LZj90/v-deo.html
I had the opposite experience. My immediate impression (even before I noticed the ≥ 8 condition) was 8 works, 9 works, and then that's it. The more he explained why higher numbers don't work, the more I started to think that there's going to be some other number, probably a multiple of 7, where it equals 1, but that it would be really hard to prove.
6:55 I don't think it is equivalent because you could take such a function that it would be positive on integers but not always positive for real numbers but anyway that doesn't change the solution
This was a really lovely problem. Well presented Michael, loving your stuff
The work is pleasingly clear and labeled
A backflip would have been great.
I love these videos. It makes me kinda sad I didn’t go for pure instead of applied math.
He can't do backflips in videos any more, due to discontinuities outside the domain of the video in the neighborhood of his head. (That is, hanging microphones.)
@@iabervon Excellent !
I probably would have turned back after taking the derivative. Thanks for an unusually clear and direct work-through!
I tried to solve this from the thumbnail, so I missed that we were solving over the natural numbers. I got n = - W(-1, -ln(z) / z^7 ) / ln(z) where z is an integer and W is the Lambert W function aka the ProductLog. Apparently, we need the -1 branch.
n. n^(1/n-7)
10.375 2
9 3
8.54777 4
8.31611 5
8.17246 6
8.07331 7
8 8
7.94315 9
7.89749 10
7.85982 11
7.82809 12
7.80089 13
7.77725 14
7.75646 15
7.73799 16...
In order for it to be an element of N then n should be written as n=m^(n-7), with m>=3>e cuz n>8>e (the case of n=8) is trivial) also m=n^m when we play around with the inequality and by replacing values we get n
IMO is happening now, it would be fun to feature some of the questions from this year
Love number theory
The problem with underlining an inequality to emphasize its strictness is then you make it nonstrict 🤣
May I know why it isn't equal to 1?? I am not really good at math. Someone, please help me out.
@@vedhase2370 Maybe he considered the fact that 1^n = 1 for all real values of n not worth commenting on.
@@vedhase2370 well, any number to the zero ith power is equal to one, ex 3^0 = 1 and 7^0 = 1. The problem states that n is greater than or equal to eight. 1/(n-7) is always going to be greater than 0 if n is greater than eight. That means that the equation basically boils down to n^(a number greater than 0).
In order for the whole equation to be less then or equal to one, n must be raised to the power of 0, or less, but since 1/(n-7) is always greater than zero due to the parameters of the question, we know that n^(1/(n-7)) is greater than one.
Hope this clears things up
I thought of that too, he almost managed to confuse me!
An arguably quicker way of proving 2^n>128n could have been by plugging n=11 and checking that 2^n is indeed larger and then seeing that the slope of 2^n at n=11 is 1024*11 (chain rule) and that the slope of 128n is 128. Plus the slope of 2^n only keeps on growing whereas 128n has a constant (and lower) slope. Combining f(11)>g(11) and f'(x)>g'(x) for x in [11,∞), you know that f(x)>g(x) ,if f(x)=2^x and g(x)=128x
I did another induction:
n < 2^(n-7) is true for n=11, then we can say:
2^(n+1-7) = 2*2^(n-7) > 2n = n+n > n+1 (for n>=11)
So therefore: n+1
I just directly differentiated the original function extended to the reals, showed that was negative for n>=8 and then by exploration found that 8&9 were the only solutions.
My solution substitute x = n - 7 and n = a^x for some integer a to obtain a^x = x + 7. Then obviously for x >= 4 we need a < 2 and we only need to evaluate x = 1,2,3. The only satisfactory pairs (a,x) are (8,1) and (3,2)
That thinking of searching for n>11 is really brilliant.
Love the channel, interesting problems and crystal clear explanations!
for proving ln 2 > 1/2, maybe it is a bit more straightforward to do:
ln 2 = 1/2 * 2 * ln 2 = 1/2 * ln (2^2) = 1/2 * (lg 4 / lg e) > 1/2
?
But, now you have to prove that ln 4 is greater than 1.
@@ddognine I think that's quite straightforward as e=2.7*** < 4. of course, for formal proof one should probably claim on monotonicity of lg function etc.
Hey, Michael. Can you please make a video on problem #6 from the 2009 IMO.
Observe that for n^(1÷m) to be a natural number all of n's prime factors must have multiplicity that's divisible by m. The smallest such number (other than 1) has prime factorization 2^m (because 2 is the smallest prime) which obviously grows faster than m+7 once you get past about m=4
... and obviously 1 isn't the mth root of any number m+7 because it's not the mth root of any number other than 1. Proving this is a waste of time. Unless they score your answer based on how fundamental your proofs go?
This was my approach as well, and I'm surprised I haven't seen more comments about it. It seems way easier than any other solution I have seen
I believe you could also use binomial expansion for 2^m to show the inequality as well. Use 2^m = SUM( m choose k) from k=0 to m and use however many terms you want to show that 2^m > 128m
The sequence decreases on its domain and is asymptotic to 1. We know a_8 = 8, a_9 = 3, so you need to consider whether there's an n such that n^(1/(n-7)) =2; it is relatively easy to show that there are no natural number solutions to this equation.
14:14
I'm starting to think that this is an alt account made by Michael Penn himself
Gareth Ma Use induction, this is really trivial
@@BerfOfficial he's not asking in reference to the problem
ANIMExFAN It’s even easier using Gougu theorem
@@angelmendez-rivera351 ohk, I didn't know the meme was that old, relatively new here; but just saying upload time and posting time does not have significant time lag and creating a new account doesn't take more than, say, 10 minutes max.
Excellent! You did your number eight like me until I were 8 years old. I remember my teacher scolded me for it. So far, I do my eights "normally". It was just to annoy you but you're doing a great job!
11:19 : After reading Richard Feynman's brilliant "Surely You're Joking, Mr. Feynman!" I can't forget that ln 2 = 0.69... which is clearly bigger than 0.5
n = 1 is a soln
trivial
good point, missed that.
The problem states that n>=8.
@@AnkhArcRod But he says the reason the problem says n>=8 is that it doesn't work for n
AnkhArcRod yea, but he said that n=1 wasn't a solution when he was talking about why the bounds of the problem were n>=8
0:50 "Similar things happen with n=5, 4, 3, 2, and 1..." A very minor correction, but actually 1⁻¹/⁶ = 1 so n=1 works. 🙂
0:47 Actually, n = 1 will work (if not for the restriction on n)
Me after seeing thumbnail: "Oh that's easy, it's 8"
Me after clicking on video: "Ah, the old bait and switch"
From e < 4, we get 1 < 2 ln 2 by taking the natural log on both sides, and then dividing both sides by 2 yields the inequality.
Please correctly state the question by adding: n ∈ N (natural number). Otherwise, there are more real number solutions for n, since the real function f(x) = x ^ (1/x-7) is a continuous function, and we have f(8)=8 and f(11)
Wonderful video, always a pleasure to watch. I was wondering if you Could make a video on the “points in the space” type of problem (I believe it’s called misc but I’m not sure). An example to this type of problem is “There are a 1000 points in the space, prove you could always pass a line such that half the points would be on one side and the other half on the other side”.
0:24 0:48 For n=1, n^(1/(n-7))=1^(-1/6)=1
you need not use derivative of f(x). it's enough that just prove f(n+1) > f(n) for all n > 7，which is much easier: 2^(n+1)-128(n+1)-(2^n-128n)=2^n-128 > 0
Yes, that’s the proof by induction.
Actually he mentioned the method in the video (proof by induction)
He mentioned PMI and said he would prove it using a "different" method.
What about n=1? It's also a solution since 1^(-1/6)=1...
The function f(x)=x^{1/(x-7)}=e^{ln x/(x-7)} is greater than 1 for x>7 and monotone decreasing, since
f'(x)=f(x)/[x(x-7)^2]*[-7-x(ln x -1)]10 we have 1
0:51 Wouldn't n=1 work as 1^(-1/6)=1 ?
n has to be higher or equal to 8 though
Unless n is in natural, there has to be a solution for f(n) = n^(1/(n-7)) = 2. Note that f is monotonic, continous in (7,+inf> and has an asymptote at f = 1.
it's viable also for n=1 as 1^any number=1 which is an integer.
What if we enter n=1? We would get 1, or am I wrong? 0:52 you said it's clearly not natural
Let n=b^x. Then we need to observe when b^[x/(b^x-7)] is an integer. That means b^x-7 divides x. For that to be possible, b^x-7 =< x. If x=1, b^x=8. If x>1, then 2^x-7 =< x so x
Finally, something I managed to solve before watching the video. Beautiful solution.
Could you please do IMO 2003 P2 its a cool question
Nice suggestion
There’s a lot to learn from that problem
Would be grateful if Michael could solve it for us
Yeah great , I hope sir does it
Fun problem
I used a slightly different approach. I took the derivative of x^(1/(x-7)), and showed it was decreasing for all x greater than or equal to 8. Since we know 1 is our lower bound, and the n=11 case is less than 2, then we know we only need to check 8, 9, and 10
Set k = n-7
(k+7)^(1/k) in N, k > 0
(k+7) = m^k, m,k in N
k = 1 and 2 are trivial solutions, with m = 8 and 3.
Notice that (k+7)^(1/k) = m is strictly decreasing with k, so any other solution would have m = 2.
2^k = k+7 has a solution between 3 and 4, so not in integers.
Hence, n in {8,9} are the only solutions.
One of the best: creative and not too hard!
If n^(1/(n-7)) is an integer, then n = m^(n-7) for some integer m greater than or equal to 2. Take a prime divisor p of n and let e be its exponent in the factorization of n. Since n = m^(n-7), the exponent of p in the factorization of m^(n-7) is greater than or equal to n-7 and hence so is a. It thus follows that 2^a - 7 ≦ n - 7 ≦ a. Therefore 2^a - a ≦ 7．Because 2^x - x is increasing on x ≧ 1 and 2^4 - 4 > 7, we have the inequality a ≦ 3; Hence n ≦ 10.
I think it's easier how I solved it. So if n ^ (n-7) is natural, it means that n = k ^ (n-7), with k a natural number. It is immediately verified that n = 8 and n = 9 are solutions and k = 8 and k = 9. For n = 10, the only close k could be 2 becouse 2 ^ 3 = 8 and 3 ^ 3 is 27, much greater than 10. Therefore we observe that for n> = 11, n = 11, n = 2. By induction, P (m) => P (m + 1) So we have m
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I feel like the claim at 13:25 needs more justification since we cant put brackets or anything as the series is not absolutely convergent
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Right? This is nuts
isn't it easier to take both sides to the power n-7 so we're looking for x^(n-7)=n . x positive integer. since x=1^(n-7) is always 1 and the smallest x is then 2 and 2^(n-7)>n for all n>11 is pretty obvious with no heavy duty math. The induction is just that for n+1, 2*(previous>n)>n+1 .
I really like the ending you make in each video. That's a good place to stop. It feels like maths has an infinite domain one can explore. It's like the opposite of claustrophobia.
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I don't want to split hairs, but it seems to me that the last equivalence at 6:49 is not true, or at least not that evident.. And in any case, a simple
A suggestion for a quick and easy proof (it feels almost too easy - are there any flaws in my reasoning?) : f(n)=n^(1/(n-7) >= 8^(1/(n-7)) = 2^(3/(n-7) > 1. When does f(n) get >=2? 2^(3/(n-7))>=2 3/(n-7) >=1 n
By thinking of it as an n-7th root you can also show that n>=2^(n-7) as n!=1 and hence 8
Well 14:14 may be a good place to stop. It is also an excellent place to start. Why? There are interesting ways forward that include asymptotic approximations concepts where for given epsilons and deltas there are Ns where n>N provide acceptable values. Admitted not pointwise answers other than N but mathematically acceptable (?) answers to tolerances appreciated by mathematicians, physicists and engineers alike. It seems good in the round to bring these things together and Michael's video is an excellent venue for that purpose alone.
It also seem healthy introducing real analysts to ordinary world of numerical readings and associated tolerance values or error bounds? Debatable?
I used Qualculate v4.9.0 a free software calculator with graphic facilities that make it easy to graph and change input range and so forth.
So for answers accurate to 1 within error bound 10^(-4) at n=20000 and n=20001 yielded a interesting descending staircase graph.
Obtained by plotting the graph where n is replaced by x and real x is approximated by setting sampling rate at 1001 with minimum value on x = 20000 and maximum value of x = 20001. Of course these equate to n=20000 and n=20001 respectively. An interesting change also happens between x=20000 and x=20002
I chose to present that range as it is the first time I noticed graph switch from smooth line to digital steps. In this case descending staircase.
Important concepts? numerical methods, graphical methods, practical application of epsilons and deltas within acceptable error bounds.
Thus! (I claim) fuller answer is 8, 9, infinity and N for infinitely many n>N where N defines finite range of n directly according to unacceptable error bound and infinitely many n>N within acceptable error bound.
One of the few number theory problems I silved on my own
0:52 A wonderful example of why you have to be careful making generalizations in mathematics, as this is a false statement. n=1 could be a potential solution (if it were not excluded by the n greater than or equal to 7 condition), as 1^(1/(1-7))=1^(-1/6)=1/(1^6)=1 is in fact an positive integer.
Yeah Sir please do IMO questions also like IMO 2003 P2 number theory and algebra questions
Fun problem
Yeah that would be great
11:10 - like... with a... calculator?? Or can't we use one?
For ln(2)>1/2 can't we just operate on both sides with e. Get 2>√e then square both for 4>e. That's true, so the starting inequality was true.
Just from think about it: n must be greater than 7 if it is less than that we have 1/(n^(7-n)) which isn't an integer, 8 and 9 both work 8^(1/(8-7))=8^(1/1)=8^1=8 and 9^(1/(9-7))=9^(1/2)=3 but are there more?
Thanks for the great videos!
Simpler final proof (edit: oops, it's the same):
ln(2)>1/2
2>e^(1/2)
4>e
Lol, it's the same proof he used, nothing different.
@@MarcoMate87 Oh, you're right. I just saw that he started differently and wrote this down quickly.
not sure it's a double arrow there at the bottom at 7:31.
0:45 : I can't see why the obvious solution n=1 is ignored, as 1^(1-7) = 1^(-6) = 1 is definitely a natural number.
I don't really appreciate fiddling with derivatives and natural logarithms instead of the induction proof (which is trivial even for me, and 90% of the things on this channel are way too complicated for me). Otherwise, a great video, thank you!
Wouldn't it be enough to see that the real function f=x^(1/(x-7)) is continuous for x>7 and approaches 1 from above as x goes to infinity?
This proves that when x is large enough the inequality 1
Amazing explanation
Why doesn't 1 work?
Edit : I know the question says "n more than 8", but he explains at 0:48 that all integers between 1 and 7 are not solutions anyway
You are correct it would be a solution, but he probably didn't think it was necessary to mention it
He was simply wrong to claim that 1 would not be a solution.
You can also take the derivative of the function itself. It takes a bit of implicit differentiation but it's not too complex and you get y((x-7)/x-ln(x))/(x-7)^2=y'. Then since you know y is always positive and so is (x-7)^2 (at least whenever x>8, which is the relevant range anyways) you can show (x-7)/x-ln(x) is always negative since (x-7)/x approaches 1 and ln(x) approaches infinity (a bit trickier than just that but that's the idea). Positive * positive * negative = negative. Since the derivative is always negative, and you know for x=10, y8 or whatever the function is always greater than 1, you know the value will always be between 1 and 2 and therefore can't be an integer.
Can't we just say for the last part:
ln(2) > 1/2 e ^ (ln(2)) > e^(1/2) [e to the power of both sides] 2 > e^(1/2) [by definition/properties of logarithm] 4 > e [squaring both sides of the logarithm]
Or should x^(log_x(y)) = y (where log_x is log base x) be proven in olympiad setting as well?
Not clear whether the n >= 8 requirement was added or part of the original problem. It sounded like Michael said all n < 8 didn't work. Actually n = 1 also works.
I really don't know why he did the ln2 > 1/2 thing. "It's well known that e is 2.718..)" but wouldn't that also require a proof? I could just say "well my calculator says ln2 is 0.693..".
Other than that, showing the gradient is always positive was good.
edit: now I see it, he wanted to make it easy to understand
Do those olympiads even allow calculators?
ln2=0.3010
"similar things happen if n is equal to 5, 4, 3, 2, or 1. True for 5, 4, 3, and 2, but 1? 1^(1/1-7) = 1^(-1/6) which is 1. So I don't see the issue here. Did I miss something?
check small values, you gen n = -+1,8,9.
prove by induction, if for some n >9 n^(n-7) > n, look at n+1:
(n+1)^(n-6)= (n+1)^(n-7) * (n+1) us n+1 what is compare
(n+1)^(n-7) v 1, what is obviously bigger then one.
2:19 Why would anyone write 8 like that way?
I've been writing my 8s like this for a long time, and always have felt like a child for doing so, and I feel better now :)
You forgot about solution n=1 while discussing n
Also the fact that the minimum prime factorization can be log2(n) indicates that the forth root is only available at the integer 16 at minimum is sort of a proof.
well if you include limit of n as it approaches infinity, then it would equal 1