I want a 10 hour video of him just doing every example up to n=100 or something stupid. It'll be the background sound of my life

@bugg4938

Жыл бұрын

Could skip straight to 100 with binomial theorem

@water_is_wet

Жыл бұрын

xd

@elmo6123

Жыл бұрын

XD

@bigfgreatsword

Жыл бұрын

Ah yes, binomial thereom as x approaches infinity, a infinitely long video

@bugg4938

Жыл бұрын

@@bigfgreatsword 🤤

5:26 when you click the wrong key when coding

As soon as I saw the title I knew it was gonna be this problem...

@Jeremiah_wtv

Жыл бұрын

Is this high school level questions?

@DELETEDUNUSED

Жыл бұрын

It was in the thumbnail

@utkarshkulshrestha8652

Жыл бұрын

Binomial theorem class 11 this concept has been used here

I found it way easier to start with a change of variable: x=k*y, which can be done without loss of generality. This causes the x to cancel out, leaving a single variable equation in k, not counting n.

@TheOoorrrr

Жыл бұрын

You are left with the same equation just for y=1 (the 'canonical form' if you will)

@AdamVB

Жыл бұрын

What does this accomplish? You are just going from two variables (x and y) to two variables (k and y). In fact, that just slightly increases the complexity

@dbliss314

Жыл бұрын

@@AdamVB I may have slightly misstated it. The x's get replaced by ky, and then the y's cancel, leaving only k. (And n of course)

@dbliss314

Жыл бұрын

@@TheOoorrrr Correct. The reason I took the more complicated route of making a new variable is just to prove that it can be done without loss of generality. You do end up with the exact same equation by setting y to 1.

@AdamVB

Жыл бұрын

@@dbliss314 I see what you are saying, Letting x = ky, Lhs: [Ky + y ]^n [Y(k+ 1) ]^n Y^n * (k + 1)^n Rhs: (Ky)^n + y^n K^n * y^n + y^n Y^n * (k^n + 1) So, (k + 1)^n = k^n + 1 But how is this useful?

This man thought me how to solve for (x+y)³, (x+y)⁴ and (x+y)⁵ in mere minutes while my teachers couldn't teach this to anyone in a class of 30.

@JLemast

Жыл бұрын

taught* ^

@Mrpallekuling

11 ай бұрын

There are unfortunately many poor teachers, from primary school up to, and including, university

doing the complex solutions to this sounds like it would be really interesting as well!

@AFastidiousCuber

Жыл бұрын

It would be surprising if they had a nice closed form.

@nikolakosanovic9931

Жыл бұрын

I thought you will find n such that (x+y)^n =x^n + y^n is always true

@muenstercheese

Жыл бұрын

absolutely!

@manstuckinabox3679

Жыл бұрын

You gave me a great Idea for an introduction presentation when I first meet my complex analysis prooffesor (pun there).

@epalegmail

Жыл бұрын

@@nikolakosanovic9931 n=0 or n=1

It’s pretty intuitive that this would be the solutions for reals, but I’d really like to see how it plays out over complex numbers.

@farfa2937

Жыл бұрын

@@emilyliedtke7059 Cool, I might also try myself when I have a while.

Thanks, I was trying to work this out yesterday.

Wow that was very thorough. Very precise. We did this in a proof class, but we just said if x=0 or if y=0, guess the author of the book missed the memo.

I suppose the much more interesting/useful case of this equation is in a ring of prime characteristic p. Then we get the Frobenius endomorphism.

@FreeGroup22

Жыл бұрын

yes, thats the point of the frobenius endomorphism lol

Thank you, professor

great video ! one quick question ...at 12:30 in the note, shouldnt it be f(0)=f(-y)=0 ? ^^

@alrightfolks7459

Жыл бұрын

I think you're right, I was confused at that line for a while

@schweinmachtbree1013

Жыл бұрын

yup, it gets sneakily fixed at 15:10

@adamlea6339

Жыл бұрын

Yes, I was lost there until it got stealthily edited.

Great terminology fix for this famous "dream." Have you done the proof that the binomial coefficients of (x+y)^p are divisible by p (i.e., "an advanced student's adventures in dreamland").

@advaykumar9726

Жыл бұрын

Fermats theorem is derived from that

@user-en5vj6vr2u

Жыл бұрын

Interesting but not a hard proof after you write down the n choose k formula

@bentoomey15

Жыл бұрын

@@user-en5vj6vr2u more than just interesting, it's important. I was trying to make an invitation to talk about characteristic p and the action of frobenius.

@marcushendriksen8415

Жыл бұрын

@@bentoomey15 should've just said that lol

@Lucaazade

Жыл бұрын

p!/k!(p-k)! is a multiple of p because it is the definition of prime that non-multiples of p don’t cancel multiples of p. Talking about more context sounds fun but obviously “The Proof” can’t be stretched into a video 😊

I really like the approach taken here. My own mathematical exploration normally manifests in five steps, of varying complexity: 1. Collect data / compute a few cases. 2. Try to determine whether there is a pattern to the results. 3. Represent the pattern with a formula or a proposition, if possible. 4. Try to prove it based on insight gained or techniques developed during step 1 or 2. 5. Manipulate the formula to give a useful result. This is exactly what happened in the video! Thank you for the clear presentation.

What about n=1? In this case it is true for all x and y, but where does the analysis for odd powers that excludes solutions where x is neither 0 nor -y go wrong? It is the step around 13:17, where m is zero, and the condition is x^0=(x+y)^0 which is true for all non-zero x, in which case f'(x) is zero for all x.

@fevesvfr

Жыл бұрын

I was coming here to comment about the special care needed for n=1 and n=2. Glad I am not the first one

@khoozu7802

Жыл бұрын

U forget x not equal to - y for the eqn x^0=(x+y)^0

@khoozu7802

Жыл бұрын

I think it is better to put m>=1

@KatePhiz

Жыл бұрын

Actually it's a bigger problem than that. Nowhere was n restricted to positive integers, nor even to just integers.

@easypezy2155

Жыл бұрын

@@KatePhiz (x+y)^n = x^n +y^n will always be true for values of n greater than 2. However, if n is less than 2 there is special behavior. You will notice that for 2 this will be false. This is due to the fact that x^2 + y^2 does not equal to (x + y)^2 for the values x, y, being negative or complex. Now if n = 1 then this is true. This is called the distributive law, because it distributes the coefficient from a polynomial over the sum of it's terms. So that a(x + y) = ax + ay. Now for the "Nowhere was n restricted to positive integers, nor even to just integers" The reason why I assumed n to be a positive integer is because this is what's required for the result to be correct. That being said, the distributive law still holds for complex values of n, and if n were a negative real value then the result would be false for values of x and y greater than 1. Furthermore, if n is an imaginary number or a complex number with negative imaginary part (i.e. n = ai with a > 0) then the result is true.

I have nothing to do with Maths right now and I still watch almost all videos.

Spivak’s Calculus is my favourite too!

@radadadadee

9 күн бұрын

the only way to learn calculus! ;)

I came here full of hope.

How about x and y are k by k matrix, quaternion, or such things? xy + yx = 0 cases are existent even if x and y are both non-zeros.

0:35 Why x can be 0, but not x ? (i think y will be the divisor of something in a calcul) (edit : ah nop, it's for something else ^^ )

In other words: For even exponent: x=0, so y^n = y^n For odd exponent: x = 0, as before, or x = -y, so 0^n = 0

n = 1 is one solution too

beginning is mispelled on the blackboard

In my office wall I have a meme which reads "Every time a student does (a + b)^2 = a^2 + b^2...... ......a kitten dies"

it is true when n is prime and the arithmetic is defined in Z/nZ.

What if x and y would be matrices?

Both cases need n > 0, or am I mistaken? The case where n=1 is trivial but still in that case (x+y)^n=x^n+y^n is always true.

I just landed up to your channel and loved this video 😊 Thank you for sharing it :)

Isn’t it just n=1 by inspection

what about non-integer cases of n?

What about if n=1

When n is 1.

I was hoping he would point out the obvious solutions right off the bat: n=1 and x=0. Rediscovering x=0 as a solution in all cases is painful when it's quite evident from the original equation.

it's more interesting to think about finite fields. Not R

Immediate thought without doing much work assuming n is a positive real integer. First trivial cases: x=0, n=1 For odd n, it should be for x = -y For even n, no obvious non-trivial case I have not yet proven to myself that those are the only solutions, or if there exist others. Thought process: Note pascals triangle is symetrical, ignore the boundary 1's as they are accounted for in x^n +y^n. Clearly due to signs there exists no value where x and y share the same sign. Note the coefficients for pascals triangle once we include the sign of x and y for each term. The right half is the left half flipped and multiplied by -1. So for |x| = |y|, all those terms would cancel out. However, for even n, the row of pascal's triangle will have an odd number of coefficients, and the coefficient along the line of symmetry will remain unless x=y=0, which is not allowed. My thought on what to do next might be to look at the coefficients ignoring the boundary 1's. Divide by xy, which removes the trivial case. And try to get the roots of the remaining polynomial, which should also be values that make the equation true.

He didn't say that _n_ is an integer greater than 1, though he solved it only for those cases. So I spent a few more minutes pondering over _n_ = 0, _n_ = 1, and negative integers. I'm getting myself a coffee now. Later, I'll ponder over the cases where _n_ is any real number. Some other folks here have mentioned extending the problem to complex numbers, matrices etc. But that's too much for me!

@filippopotame3579

Жыл бұрын

Same here, I started on n = 0 and n = 1, only to loose hope when tackling the real numbers. I was a bit disappointed when resuming the vid to see it was only solved for integers above 1, but the proof is fun.

@archon8255

Жыл бұрын

Bro he said Real Numbers. That means positive integers above 0, and excludes negative numbers. Well, still good if you tried with other things as well

@filippopotame3579

Жыл бұрын

@@archon8255 Our issue is that he specified x and y to be real and said nothing about n. And real numbers include positive integers, negative ones, rational numbers, irraltionals etc. it does not mean integers above 0.

@archon8255

Жыл бұрын

@@filippopotame3579 Ohh sorry. I might have misunderstood the comment that it referred to n. My bad!

@filippopotame3579

Жыл бұрын

@@archon8255 No problem bro! I was confused too. Have a good day.

Finally it doesn't answer the initial question. The only time wieh (x+y)^n = x^n + y^n is with imaginary numbers.

I never understood why anglo-saxons mathematicians don't use variation tables (and sign tables) as we do in France. They would make the last argument much more clear and readable in my opinion (of course I'm used to them so I'm a bit biased). Nice video anyway !

I saw the thumbnail, instantly reminded me of the calculus 2 test that I just failed

At 15:12, at first I thought it was odd that x = -y/2 and x = -y would appear in the positive x- axis but then I remember that y 0 (very similar) and drawn the three points on the negative x-axis. Also, the fact that f'(x) is increasing at x in (-y/2, + \infty) means that there is at most only one other zero.

@paull2937

Жыл бұрын

It’s annoying when you get a strikethrough in your comment when you type dashes representing negative symbols.

@JaybeePenaflor

Жыл бұрын

@@paull2937 Correct! I tried revising this over and over again but the strikethrough wouldn't go away.

To simplify further we can notice that the problem is equivalent to (1+u)^n=1+u^n via u=x/y after dividing by y^n. Also, binomial theorem shows that if u>0 this can't be true, but the case u

@TheEternalVortex42

Жыл бұрын

If u

@Hyakurin_

Жыл бұрын

@@TheEternalVortex42 the sign of y/x and x/y is the same

As a tutor this identity drives me nuts!

Can someone explain the bit just after 3:58? Why is the stuff under the radical always negative?

@bartimaus7912

Жыл бұрын

Because it is 9y^2-16y^2=-7y^2. By assumption y!=0, so y^2>0. Therefore -7y^2

I thought of some motivation, but that's alright very interesting one though

That's when n=1

By inspection, x == y == 1/2, n == 1.

16:05

Can you show way Kaprekar’s constant works?

@skullersky1210

Жыл бұрын

Very briefly, Kaprekar's constant is what it is because it's a fixed point of the algorithm. If you arrange the digits of 6174 greatest to least, and subtract the digits least to greatest you'll get 6174. Since this isn't true for any other 4-digit integer, as you iterate through the process, you'll eventually land on a number that gets you to 6174, and wont change. You can do something similar if you take to cosine of a random number, and keep taking the cosine of the result. As you iterate through, you'll approach the number whose cosine is itself.

I always use Pascal's Triangle to figure out the coefficients.

@marcushendriksen8415

Жыл бұрын

Lol I've just beaten the formula into my brain over the years: "n choose k is equal to n factorial divided by k factorial times n minus k factorial" 😅

Thought this would be about Frobenius endomorphism.

It is true sometimes

is it okay to consider only n = 2 case by the Fermat's Last Theorem?

@easypezy2155

Жыл бұрын

Fermat's Last Theorem only holds in the sense of integer values. So n = 2 is the only case that can be held under the context of Fermat's Last Theorem. The theorem states that every integer power greater than 2 cannot equal a sum of other positive integer powers. But if you want to apply this idea to primes there is the Euler theorem which applies this to integers that are congruent to 1 mod 4.

This is Binomial Theorem.

Genial, gràcies per compartir

Cool video

I honestly dont know how you can discuss this subject without mentioning the characteristic of a ring

Thanks for the video. But wondering if graph shown at 11:00 is correct?

@Duduvianna3

Жыл бұрын

It is and it isn't. It is correct as a general graph, it isn't because f(x) techinically doesn't have a graph since m is not defined. With an m defined you could graph it, for example with m=1 it would be a straight line since f(x)=-2xy, with m=2, x's exponent would be 3, so more like what he drew.

You have stated that x and y are real, but not n. Also if n=0 you have no solutions, per 0^0 being an impossible number.

(x+y)^p = x^p + y^p is true in an integral domain of characteristic p

Why can't you just write that negative numbers to odd powers are always negative, therefore, x^n + (-x)^n = 0 when n is odd

Yeah everytime I see (x+y)² I still wish it were so simple to deal with, and I'm 30

If we are dealing with a field where n is prime and px=0 for every x in the field then this is true

Genial!! ETS un crack

x and y equal any numbers. n = 1.

if x and y =2, and n=1, then it will be true as both sides are 4

Michael Penn, if you will edit your misspelling of "beginning," then that will be a good "start."

It's also an identity in tropical arithmetic, where addition is min and multiplication is ordinary addition.

Why don't we exclude x=0 case as well as we did with y=0 solution. Anyway It's a trivial solution for any n.

@marcushendriksen8415

Жыл бұрын

WLOG

@IoT_

Жыл бұрын

@@marcushendriksen8415 WLOG has nothing to do with what I was saying. I didn't ask why y was excluded instead of x. I asked why x wasn't excluded AS WELL AS y.

@marcushendriksen8415

Жыл бұрын

@@IoT_ lol all right man, my bad, I misread you

Just n = 1 for any x,y

How about 1 and -1…

My first response is that we are working over a field of characteristic p where p|n....

Just FYI, from 10:51 onwards there are mistakes that your sleepy editor didn't remove.

"But two is not equal to zero..." I love abstract algebra 😄

Aww it's real numbers. I thought this was going to be something about Fermat's last theorem. Which is that there are of course no integer solutions for n at least 3. I wonder about prime numbers to DIFFERENT integer powers though. Like p1^3 and p2^5, is there a solution for p3^7. Like 2^5 is 32, and 3^4 is 81, and they differ by 49 which is 7^2. Well except the 2nd power isn't enough, it should be more than that.

@easypezy2155

Жыл бұрын

As it turns out, this has actually been solved by Euler. The solutions exist if and only if (p - 1)/(p + 1) is equal an integer. This means the possible solutions are the numbers of the form p^n - 1 where p is any prime number. So that's pretty neat.

At first glance, when X, Y, and n all equal 1.

Well, it’s true if char(R) | n

Cái này ở Việt Nam chúng tôi đã học hồi lớp 7 lúc 13 tuổi rồi, tôi không biết bài toán này được dạy ở cấp bậc mấy ở nước ngoài

*On a side note, the topic of this video reminds me of a 2015 R/learnmath thread that I discovered about a year ago...* *Context:* OP was trying to differentiate f(x)= 2/x *OP's work:* f(x)= 2/x, f'(x)=? f(x) = 2 × (1/x) f(x) = 2x⁻¹ f'(x) = (f(x + h) - f(x)) / h , lim h-->0 f'(x) = (2(x + h)⁻¹ - 2x⁻¹) / h f'(x) = (2x⁻¹ + 2h⁻¹ - 2x⁻¹ ) / h f'(x) = 2h⁻¹ / h f'(x) = (2/h) / h f'(x) = 2/h × 1/h f'(x) = 2/h² *Top Commenter:* "From line 5 to line 6: (x+h)⁻¹ does not equal x⁻¹ + h⁻¹"

12:27 it should be f evaluated at -y

Freshman's Dream*

Why is John McEnroe teaching math?

Well, but hang on: if x = 0 then y^n = y^n (wow!) Or if x = -y, then (x+y)^n = (x-x)^n = 0, right? and then x^n + y^n = x^n - x^n = 0. So 0=0, wow!

When n=1

Soy de Argentina, la verdad siempre pensé que la notación matemática y las pruebas eran una perdida de tiempo. Pero realmente pude entender todo el proceso sin saber totalmente lo que decía el chico

@angelcaru

Жыл бұрын

Increíble

@radadadadee

9 күн бұрын

yo soy de boca y se contar hasta 10 de memoria

when x equal y may be

My solution was n=1

X=0 or y=0

Whenever you're working in characteristic n.

Let x be an integer. x!=(x+1)! Which integer is x such that the equation is true? (Press read more for answer) x=0, because both 0! and 1! equals to 1 for some reason

@filippopotame3579

Жыл бұрын

I almost lost my sanity on your riddle because you said positive, which is my book means

@supernt7852

Жыл бұрын

@@filippopotame3579 Whoops. I edited it to ‘an integer’ to avoid any confusion.

When x=y=n=1

I don't usually expect an explanation why 2 d.n.e. 0 lol.

So the student would be wrong almost everywhere. Ie. The set of values where the student is right has 0 Lebesgue measure.

n = 1

Fields of characteristic p are a beginner‘s dream… :-)

Похож на Марка Цукерберга.

Он не знает бином Ньютона.

At 12:30: f(0) = f(-x) = 0 doesn't make any sense

@ronniedahlgren2733

Жыл бұрын

Ah just noticed the change on the chalkboard at end of the video. Too bad the mistake is not captioned as I got completely lost at following the reasoning at first.

n=1,

orthogonal vectors, if you want to allow x and y to be stuff other than numberz

I have not seen the video 🖐🏻🖐🏻 My answer is : If one of them is 0, then it's true.