Finding Critical Points of P: Sum Notation Explained

[hoffmann]

i have the following problem:

find the critical points of:

P = (x_{1} - 1)^{2} + (x_{n})^{2} + \sum(x_{k+1} - x_{k})

the bounds of the sum are from i = 1 to n-1.

so i differentiate P with respect to x and i set it equal to zero, and i eventually get the expression:

\sum(x_{k+1} - x_{k}) = 1 - x_{1} - x_{n}

what do i do from here / how do i differentiate the summation notation?

thanks!

 

[Dick]

How can you differentiate with respect to x when there is no x in the problem? I think you want to differentiate with respect to x1, x2, x3... xn and set them all equal to zero to get a critical point.

 

[hoffmann]

alright, so how would i do that within the summation notation?

 

[Dick]

You can write the summation in a form that's a lot simpler. Take the case n=4. Then the sum is (x4-x3)+(x3-x2)+(x2-x1). Do you see what I'm saying?

 

[hoffmann]

i'll write P as:

P = x^{2}_{1} - 2x_{1} + 1 + x^{2}_{n} + (x_{n} - x_{n-1}) + ... + (x_{2} - x_{1})

and now differentiating P wrt x1, ..., xn:

P^{'}_{x_{1}} = 2x_{1} -3 = 0, x_{1}= 3/2

P^{'}_{x_{2}} = -1 + 1= 0, x_{2}= 0

P^{'}_{x_{n}} = 2x_{n}+ 1 = 0, x_{n}= -1/2

so would the critical points be (3/2, 0, ..., 0, -1/2)? or am i doing something wrong...

 

[Dick]

Only a bit wrong. Your P'_x2 doesn't have any x2 in it. It's identically zero. No matter what x2 is. Doesn't that mean x2 can be anything? Same for x3...xn-1?

 

[hoffmann]

yep, that makes sense. why isn't x2, ..., xn-1 = 0 then? are you saying it should just be 1?

 

[Dick]

No, I'm saying that if x2...xn-1 don't even occur in your equations, then there is nothing to solve for and they could be ANYTHING.