# Finding Critical Points of P: Sum Notation Explained

• hoffmann
In summary, the conversation discusses finding the critical points of a given function using differentiation and summation notation. Differentiating the function with respect to each variable and setting them equal to zero, the critical points are found to be (3/2, 0, ..., 0, -1/2). However, it is noted that some variables do not occur in the equations and therefore can be any value.
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!

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.

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

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?

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...

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?

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

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.

## What is a critical point in mathematics?

A critical point in mathematics is a point on a graph or function where the derivative is equal to zero. This means that the slope of the function at that point is flat, and the function is neither increasing nor decreasing.

## How do you find critical points using sum notation?

To find critical points using sum notation, you must first take the derivative of the function. Then, set the derivative equal to zero and solve for the variable. This will give you the critical points of the function.

## What is the purpose of finding critical points?

The purpose of finding critical points is to identify the local maximum and minimum points on a graph or function. These points can provide valuable information about the behavior and shape of the function.

## What is the difference between a local and global critical point?

A local critical point is a point on the function where the derivative is equal to zero, but it may not be the highest or lowest point on the entire function. A global critical point, on the other hand, is the highest or lowest point on the entire function.

## Can a function have multiple critical points?

Yes, a function can have multiple critical points. These points can be either local or global critical points, and they may occur at different locations on the function.

• Calculus and Beyond Homework Help
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
856
• Calculus and Beyond Homework Help
Replies
5
Views
1K
• Calculus and Beyond Homework Help
Replies
9
Views
3K
• Calculus and Beyond Homework Help
Replies
3
Views
433
• Calculus and Beyond Homework Help
Replies
6
Views
1K
• Calculus and Beyond Homework Help
Replies
27
Views
648
• Calculus and Beyond Homework Help
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
727