2017 partial derivatives for a function of 2017 variables

Calculus Level 4

If f ( x 1 , x 2 , x 3 , , x 2017 ) = ( x 1 x 2 ) ( x 2 x 3 ) ( x 3 x 4 ) ( x 2016 x 2017 ) ( x 2017 x 1 ) f(x_1, x_2, x_3, \ldots , x_{2017}) = (x_1 - x_2)(x_2 - x_3)(x_3 - x_4)\cdots(x_{2016} - x_{2017})(x_{2017} - x_1) find the value of n = 1 2017 f x n ( x 1 , x 2 , x 3 , , x 2017 ) = ( 1 , 2 , 3 , . . . , 2017 ) . \displaystyle \sum_{n=1}^{2017} \left. \dfrac{\partial f}{\partial x_{n}} \right|_{(x_1, x_2, x_3, \ldots ,x_{2017})=(1,2, 3, ..., 2017)}.


The answer is 0.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Ivan Koswara
Mar 21, 2017

The partial derivative with respect to x n x_n is the product of all ( x i x i + 1 ) (x_i - x_{i+1}) that don't involve x n x_n , multiplied by ( x n 1 2 x n + x n + 1 ) (x_{n-1} - 2x_n + x_{n+1}) , taking indices modulo 2017. (This can be easily shown; the terms in the form ( x i x i + 1 ) (x_i - x_{i+1}) are constant, leaving ( x n 1 x n ) ( x n x n 1 ) (x_{n-1} - x_n)(x_n - x_{n-1}) , and we use the product rule.)

Now, if n 1 , 2017 n \neq 1, 2017 , we have x n 1 2 x n + x n + 1 = ( n 1 ) 2 n + ( n + 1 ) = 0 x_{n-1} - 2x_n + x_{n+1} = (n-1) - 2n + (n+1) = 0 , so the latter factor is zero and so the partial derivative is zero. If n = 1 , 2017 n = 1, 2017 , all the terms in the form ( x i x i + 1 ) (x_i - x_{i+1}) are equal to 1 -1 , because the only term that isn't is ( x 2017 x 1 ) (x_{2017} - x_1) and it's not included in the product (because n = 1 , 2017 n = 1, 2017 ). So the partial derivative is simply ( 1 ) 2015 ( x n 1 2 x n + x n + 1 ) (-1)^{2015} (x_{n-1} - 2x_n + x_{n+1}) . When n = 1 n = 1 , this is 2017 -2017 , and when n = 2017 n = 2017 , this is 2017 2017 , so their sum is 0.

Thus in total, the sum of the partial derivatives is 0.

Exact same method! (+1)

Tapas Mazumdar - 4 years, 2 months ago

Well, u may encounter some thought experiment. There is a sweet chain that has been built here . It starts from a point and gets back to it , a circular chain . And for this kind of things partial differentiation is like eliminating each lattice of a crystal. let call these ' lattice' of the chain unit . So starting from the first one i.e x1 , we will found (-1)^(n-2)×(n-1)+(-1)^n . Let's call this result p And the next units(except the last one) will give us 0. Why ? Because we will found there a strange symmetry of distribution of signs that's because of the cyclic nature of this chain . And the last one will be just the opposite of the first one in sign . It's like seeing p in a "sign" mirror . And why so ? That's because the anchor xn is also lives in symmetry . The chain is (x1-x2).......(xn-1 - xn)(xn-1) . And that's because it generates the same result of first unit(p) but project it through a sign mirror? So it becomes -p . Why it uses the sign mirror . Because of it's ascending superiority xn>xn-1>xn-2>.......x3>x2>x1.

That's it! Pretty simple!

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...