Minor mistake at the end: the sum of r_n should be for n between 1 and 2000, not 2001. We've shown before that it's a 2000th degree polynomial, not 2001th degree, so it has 2000 roots instead of 2001. Doesn't change the answer either way. Cheers!
I notice the same detail. As soon as the polynomial is no more as 2000-th degree, there is one root lees. :) This is just a very small detail. Still a really nice video on a complicated and tricky subject. Thanks a lot for all your fascinating videos. Keep going... We keep living and learning... :)
I tried to do it on my own first before watching the video, and I went through a different path using the properties of complex numbers, First I proved that x can't be real, because for the equation's root to be real, it needs to fulfill that x=x-1/2, which leads to a contradiction. Therefore, the roots must all be complex numbers. Let x = a + bi, then necessarily the number needs to fulfill that |x| = |x-1/2| for it to be a root of the equation. This leads to a=1/4. For the b, I switched the numbers in the equation into their exponential form, cancelling the modules and multiplying the 2001 exponent into the exponent in the e. Keeping the cyclic nature of the exponents in complex numbers, that leads to 2001*arctan(4b)=2001*arctan(-4b)+2k*pi, for k in Z. Keeping in mind the inverse symmetry of the arctangent function, you can solve the equation and get that b=(tan(k*pi/2001))/4, for any integer k. As such, the roots are of the form x = 1/4 + [tan(k*pi/2001)/4]i, for k in the integer. However, we need to keep in mind two things. First, since x is not real, then b cannot be 0. Second, since the tangent is cyclic for every pi, going beyond k=1000 is only going to yield the same roots over and over, so in reality, discarding the values k=0 and k=2001 and for the reasons above, we have a total of 2000 roots. For convenience, we take the values of k between -1000 and 1000 to calculate the sum of the roots. The inverse symmetry of the tangent function makes it so half of the additions cancel out the other half. for the imaginary side, leaving only the real 1/4 to add ti itself 2000 times, thus resulting in a total addition of 500. Phew! Ok, that was a lot of work, but I hope it makes sense. Cheers!
@@ayaansiddiqui745 Complex numbers are intimately related to trigonometry, as a complex number can be represented by its xy coordinates in the form of a + bi, or they can be represented as a module and an angle, similar to the polar coordinates. Sometimes switching from one representation to another can open new ideas and paths to solve problems. :) I'm sure once you get the hang of trigonometry and complex numbers, the explanation will make more sense to you.
Dear Newton (sorry if your name is not Newton) I’m afraid there is a little problem in your solution. As we have x^2000, we should finish with 2000 roots instead of 2001. By the way, I appreciate your work very much! Cheers!
This is so useful! I have a premimition of sorts when it comes to olympiads and big numbers, since they always go hand in hand. Your videos feel like they branch out especially when they need to, thank you so much!
The answer is correct = 500. However, the equation is 2000 degree (not 2001 degree) because when we use Newton's binomial formula for (1/2 - x)^2001 it will contain the term (-x^2001) which will cancel analogical term with plus sign x^2001 and the polynomial as well as the equation will become 2000 degree and will have 2000 roots (not 2001)😊.
For any given term (f)(a^p)(b^q) in a binomial expansion, where f is the coefficient, the next term is [(f)(p)/(q+1)][a^(p-1)][b^(q+1)]. It’s always f x p divided by q+1 for that next term. At the outset, the first term of (a+b)^n = a^n = [1][a^n][b^0] = [1/0!]a^n The second term is then [1xn/(0+1] [a^(n-1)][b^(0+1)] = n[a^(n-1)][b^1] = [n/1!][a^(n-1)][b^1]. In the same way, the third term is [n(n-1)/(1+1)][a^(n-1)][b^(1+1) = [n(n-1)/2][a^(n-2)][b^2] = = [n(n-1)/2!][a^(n-2)][b^2], and the fourth term is [{n(n-1)(n-2)/(2)}/{3}][a^(n-3)][b^3] = {[n(n-1)(n-2)]/[(2)(3)]}{ [a^(n-3)][b^3]} = {[n(n-1)(n-2)]/[(3!]}{ [a^(n-3)][b^3]}, and on and on. If the first term is considered the zeroth term, then each i-th term is {[n(n-1)(n-2)…(n-i)]/[i!]}{ [a^(n-i)][b^i]}.
Hello again sir! Ive been studying summation and stumbled upon the Reimann Zeta Function. I would love a video on this topic to learn more about it and also i would be interested in hearing your thoughts on whether or not its possible to give a mathematical proof. Love this channel. Have a great day!
It is easy to see that if r is a root, then 1/2-r is a root, too. There are 2000 roots in total, because it is a polynomial of order 2000. That is 1000 pairs each of which sums to 1/2. So the total is 500.
Probably it’s a typo, but once the x^2001 term gets cancelled, the number of solutions should be 2000, right? So the end sum should be from n=1 to 2000, not 2001… or have I missed anything? Anyway, thanks a lot for your videos, I really enjoy them!
Multiplying (and even dividing by 8) is not the best idea as soon as it's not the final answer and we're going to divide by 2001/2 later( well don't exactly know, but it's quite predictable as soon as you know why you're doing what you're doing 😉), so it would be easier to calculate without calculator(with such big numbers there's a bigger chance to make a mistake while typing them and writing them out IMHO) But as for reducing the fraction with factorials that was also my thought why it wasn't reduced asap instead of keeping it till the very end.😊
@@lukaskamin755 So what, if we divide by 2001/2 ? You can still multiply (250*2001) by 2/2001, cancel out the 2001 and you are left with 250 * 2 = 500. I do agree though, that you should keep the (250 * 2001) and not the 500250, as we do have the 2001 factor elsewhere!
Shorter solution: In any math competition comes a level of intuition. Why in the world would they put a x^2001 as the first term? Surely it must be there for a reason. Well, it certainly is. Focus ONLY ON THE LAST TERM of (1/2-x)^2001. Notice how (-x)^2001 = -x^2001. This cancels with the first x^2001. So now you have (by the binomial theorem) 1/2^2001 - (nCr(2001, 1))*1/2^1999 x + .... -(nCr(2001, 1999))1/2^2 x^1999 + (nCr(2001, 2000))1/2 x^2000 Now, Vieta's formula tells us the sum of the roots is -b/a, where b is the coefficient of x^(n-1) and a is the coefficient of x^n. Here, our n = 2001. So looking at the above, our a and b are: a = (nCr(2001, 2000))1/2 = 2001/2 b = -(nCr(2001, 1999))1/2^2 = (2001*2000)/8 Reminder that nCr(a, b) = nCr(a, a-b), so something nasty like nCr(2001, 1999) = nCr(2001, 2), which is much easier to calculate. Now, -b/a = (2001*2000)/8 ÷ 2001/2 = 4000/8 = 500.
It is and easy solutions if we take (1/2-x) to the other side and divide whole hy x^2001 using exponents it would become (1/2x - 1) 2001 = 1 Now we know that 1 raised to the power of any number is 1 then 1/2x - 1 should be 1 hence the solution
This is a non-standard equation, it's impossible to solve it using elementary functions. The closest we can get to find the answer is by approximation :)
sinx = lnx will have 1 solution . This can be easily checked as sinx has a range of -1 to 1 while lnx has a range of Real Numbers . lnx must intersect the region between -1 to 1 once . Where ? That I can't tell . For this you need a graphing software
@@gileadedetogni9054 here is how I thought Aim : sinx = log x base 10 Try : Write log x base 10 as lnx/ln10 sinx = lnx/lnx10 ln10sinx = lnx Set both equal to y Y = ln10sinx Y = lnx ln10 is roughly 2.303 So we have Y = 2.303sinx So range of Y is around [-2.3 , 2.3] So lnx must also be in this range for solution to exist Turns out x must be from [0.1,10] for lnx to match the range of ln10sinx 0.1 is roughly like π/30 and 10 is roughly like 3π So it's like checking solution of 2.303sinx and lnx in [π/30 , 3π] Till π/2 sinx is increasing and at π/2 sinx is 2.3 while lnx is around 0.45 And at sin(π) the value is 0 while ln(π) = 1.4 so one solution must be there. Similarly it can be proved that sinx is greater than lnx Somewhere between π and 2π and it's symmetric around 3π/2 for π and 2π so 2 solutions there Aliter : I thought of finding zeroes of y = lnx - 2.303sinx But it was getting way difficult for me to prove that 1/x = 2.303cosx has 2 solutions One solution can be found of y by commenting on its monotonicity but then other will get stuck None the less this question according to me can not be solved without having any log table handy
