Past, present, and future again

Calculus Level 5

Given the function f ( x 1 , x 2 , x 3 , … , x 2017 ) = ∑ 1 ⩽ n < m ⩽ 2017 x n x m f(x_1, x_2, x_3, \ldots, x_{2017} ) = \displaystyle \sum_{1 \leqslant n < m \leqslant 2017} x_n x_m

Find ∑ k = 1 2017 ∂ f ∂ x k ∣ ( x 1 , x 2 , x 3 , … , x 2017 ) = ( 1 , 2 , 3 , … , 2017 ) \displaystyle \sum_{k=1}^{2017} \left. \dfrac{\partial f}{\partial x_k} \right|_{(x_1, x_2, x_3, \ldots, x_{2017} )=(1, 2, 3, \ldots , 2017)}


The answer is 4102868448.

Hobart Pao by Hobart Pao , 23, USA

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.

1 solution

Tapas Mazumdar
Mar 27, 2017

We can write the function as another sum as

f ( x 1 , x 2 , ⋯ , x 2017 ) = ∑ 1 ≤ n < m ≤ 2017 x n x m = 1 2 ( ∑ n = 1 2017 ∑ m = 1 2017 x n x m − ∑ k = 1 2017 x k 2 ) \displaystyle f(x_1,x_2, \cdots , x_{2017}) = \sum_{1 \le n < m \le 2017} x_n x_m = \dfrac{1}{2} \bigg( \displaystyle \sum_{n=1}^{2017} \sum_{m=1}^{2017} x_n x_m - \sum_{k=1}^{2017} {x_k}^2 \bigg)

Now

∂ f ∂ x k = 1 2 ( ∑ n = 1 2017 ∑ m = 1 2017 ∂ ∂ x k ( x n x m ) − ∑ k = 1 2017 ∂ ∂ x k x k 2 ) = 1 2 ( 2 { 2017 ∑ k = 1 2017 x k } − 2 ∑ k = 1 2017 x k ) = 1 2 ( 4032 ∑ k = 1 2017 x k ) \begin{aligned} \dfrac{\partial f}{\partial x_k} &= \dfrac{1}{2} \bigg( \displaystyle \sum_{n=1}^{2017} \sum_{m=1}^{2017} \dfrac{\partial}{\partial x_k} \left( x_n x_m \right) - \sum_{k=1}^{2017} \dfrac{\partial}{\partial x_k} {x_k}^2 \bigg) \\ &= \dfrac{1}{2} \bigg( \displaystyle 2 \left\{ 2017 \sum_{k=1}^{2017} x_k \right\} - 2 \sum_{k=1}^{2017} x_k \bigg) \\ &= \dfrac{1}{2} \bigg( \displaystyle 4032 \sum_{k=1}^{2017} x_k \bigg) \end{aligned}

This is where I missed the factor of 1 2 \dfrac 12 and ended up with the answer as 8205736896 8205736896 .

Thus, we get

∑ k = 1 2017 ∂ f ∂ x k ∣ ( x 1 , x 2 , x 3 , … , x 2017 ) = ( 1 , 2 , 3 , … , 2017 ) = 1 2 ( 4032 ∑ j = 1 2017 j ) = 1 2 × 4032 × 2017 × 2018 2 = 4102868448 \displaystyle \sum_{k=1}^{2017} \left. \dfrac{\partial f}{\partial x_k} \right|_{(x_1, x_2, x_3, \ldots, x_{2017} )=(1, 2, 3, \ldots , 2017)} = \dfrac{1}{2} \bigg( \displaystyle 4032 \sum_{j=1}^{2017} j \bigg) = \dfrac{1}{2} \times 4032 \times \dfrac{2017 \times 2018}{2} = 4102868448

6 Helpful 0 Interesting 0 Brilliant 0 Confused

WOAHHH this is an extremely neat solution and very easy to understand! You deserved way more upvotes man! Keep it up!

Pi Han Goh - 4 years, 2 months ago

Log in to reply

Thank you sir. Unfortunately, I wasn't able to score points for this. Nevertheless, I believe it is what the approach of the very problem that I correctly got so I have no problems.

Tapas Mazumdar - 4 years, 2 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...