If f ( x 1 , x 2 , x 3 , … , x 2 0 1 7 ) = ( x 1 − x 2 ) ( x 2 − x 3 ) ( x 3 − x 4 ) ⋯ ( x 2 0 1 6 − x 2 0 1 7 ) ( x 2 0 1 7 − x 1 ) find the value of n = 1 ∑ 2 0 1 7 ∂ x n ∂ f ∣ ∣ ∣ ∣ ( x 1 , x 2 , x 3 , … , x 2 0 1 7 ) = ( 1 , 2 , 3 , . . . , 2 0 1 7 ) .
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.
Exact same method! (+1)
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!
Problem Loading...
Note Loading...
Set Loading...
The partial derivative with respect to x n is the product of all ( x i − x i + 1 ) that don't involve x n , multiplied by ( x n − 1 − 2 x n + x n + 1 ) , taking indices modulo 2017. (This can be easily shown; the terms in the form ( x i − x i + 1 ) are constant, leaving ( x n − 1 − x n ) ( x n − x n − 1 ) , and we use the product rule.)
Now, if n = 1 , 2 0 1 7 , we have x n − 1 − 2 x n + x n + 1 = ( n − 1 ) − 2 n + ( n + 1 ) = 0 , so the latter factor is zero and so the partial derivative is zero. If n = 1 , 2 0 1 7 , all the terms in the form ( x i − x i + 1 ) are equal to − 1 , because the only term that isn't is ( x 2 0 1 7 − x 1 ) and it's not included in the product (because n = 1 , 2 0 1 7 ). So the partial derivative is simply ( − 1 ) 2 0 1 5 ( x n − 1 − 2 x n + x n + 1 ) . When n = 1 , this is − 2 0 1 7 , and when n = 2 0 1 7 , this is 2 0 1 7 , so their sum is 0.
Thus in total, the sum of the partial derivatives is 0.