Function max and min values

[Mark53]

Homework Statement

Let f [a, b] → R be a non-decreasing function; that is, f(x1) ≤ f(x2) for any x1, x2 ∈ [a, b] with x1 ≤ x2. So f attains a minimum value of m = f(a) and a maximum value of M = f(b) on [a, b]. Let Pn be a regular partition of [a, b] into n subintervals, each of length ∆x = (b − a)/n, and let mi and Mi be the minimum and maximum values of f on the i-th subinterval respectively for each i = 1, 2, . . . , n.

(a) Explain why Mi = mi+1 for each i = 1, 2, . . . , n − 1.
(b) Hence show that U(f,Pn) − L(f,Pn) = (Mn − m1) ∆x.
(c) Express (Mn−m1) ∆x in terms of f, a, b, n and use this to explain why f is integrable on [a, b].

The Attempt at a Solution

[/B]
A) this is because Mi must be less than mi as it is a non-decreasing function that is why Mi = mi+1

b)

=Mi∆x-mi∆x

=(Mi-mi)∆x

is this part correct?

c)

(Mn−m1) ∆x

(f(a)-f(b))((b-a)/n)

how do i show that it is integrable from here?

 

[Mark44]

Homework Statement

Let f [a, b] → R be a non-decreasing function; that is, f(x1) ≤ f(x2) for any x1, x2 ∈ [a, b] with x1 ≤ x2. So f attains a minimum value of m = f(a) and a maximum value of M = f(b) on [a, b]. Let Pn be a regular partition of [a, b] into n subintervals, each of length ∆x = (b − a)/n, and let mi and Mi be the minimum and maximum values of f on the i-th subinterval respectively for each i = 1, 2, . . . , n.

(a) Explain why Mi = mi+1 for each i = 1, 2, . . . , n − 1.
(b) Hence show that U(f,Pn) − L(f,Pn) = (Mn − m1) ∆x.
(c) Express (Mn−m1) ∆x in terms of f, a, b, n and use this to explain why f is integrable on [a, b].

The Attempt at a Solution

[/B]
A) this is because Mi must be less than mi as it is a non-decreasing function that is why Mi = mi+1

This is not an explanation. It's what we call "hand-waving."

b)

=Mi∆x-mi∆x

=(Mi-mi)∆x

is this part correct?

No. How does this show that U(f, P_n) - L(f, P_n) = (M_n - m₁)∆x?
Also, you started this off with "=Mi∆x-mi∆x". What is on the left side of this equation? Also, what does Mi∆x-mi∆x represent in relation to the function f and the partition?

c)

(Mn−m1) ∆x

(f(a)-f(b))((b-a)/n)

?

how do i show that it is integrable from here?

How is the term "integrable" defined?

One more thing. This is obviously a calculus problem. Please don't post such problems in the precalculus section.

 

[Mark53]

No. How does this show that U(f, P_n) - L(f, P_n) = (M_n - m₁)∆x?
Also, you started this off with "=Mi∆x-mi∆x". What is on the left side of this equation? Also, what does Mi∆x-mi∆x represent in relation to the function f and the partition?
One more thing. This is obviously a calculus problem. Please don't post such problems in the precalculus section.

Starting from LHS

U(f,Pn) − L(f,Pn)

=(f(b))*((b − a)/n)-(f(a))*((b − a)/n)

=Mi∆x-mi∆x

=(Mi-mi)∆x

=RHS

Does this show that they are equal to each other now?

 

[Mark44]

Starting from LHS

U(f,Pn) − L(f,Pn)

=(f(b))*((b − a)/n)-(f(a))*((b − a)/n)

You are essentially saying here that U(f, P_n) = f(b)Δx and that L(f, P_n) = f(a)Δx. This isn't true.
How are U(f, P_n) and L(f, P_n) defined? You should be using the definitions of these quantiities.

=Mi∆x-mi∆x

=(Mi-mi)∆x

This is not the right-hand side. The expression in parentheses is the max. value in the i-th subinterval minus the min. value in the same subinterval. Notice that in the OP, the expression on the right side is (M_n - m₁)∆x, not what you have above.

=RHS

Does this show that they are equal to each other now?

No.

 

[Mark53]

You are essentially saying here that U(f, P_n) = f(b)Δx and that L(f, P_n) = f(a)Δx. This isn't true.
How are U(f, P_n) and L(f, P_n) defined? You should be using the definitions of these quantiities.
This is not the right-hand side. The expression in parentheses is the max. value in the i-th subinterval minus the min. value in the same subinterval. Notice that in the OP, the expression on the right side is (M_n - m₁)∆x, not what you have above.

No.

U(f , P) = M1∆x + M2∆x + ··· + Mn∆x.

L(f , P) = m1∆x + m2∆x + ··· + mn∆x

is this correct?

 

[Mark44]

U(f , P) = M1∆x + M2∆x + ··· + Mn∆x.

L(f , P) = m1∆x + m2∆x + ··· + mn∆x

is this correct?

Yes. Each one is a sum of n terms. In summation notation, U(f, P_n) is $\sum_{i = 1}^n M_i\Delta x$, and similar for the lower sum L(f, P_n).

 

[Mark53]

Yes. Each one is a sum of n terms. In summation notation, U(f, P_n) is $\sum_{i = 1}^n M_i\Delta x$, and similar for the lower sum L(f, P_n).

would that mean if we take the max value of U as i=n and the minimum value of L which is i= 1 it will lead to the following:

=Mn∆x-M1∆x

=(Mn-M1)∆x

=RHS

 

[Mark44]

would that mean if we take the max value of U as i=n and the minimum value of L which is i= 1 it will lead to the following:

=Mn∆x-M1∆x

No. For one thing, don't start your work off with "=".
More importantly, you need to use the expressions for U and L that you have in post #5, and what you have above isn't what you're supposed to show.

It might be helpful to draw a picture, including the graph of a nondecreasing function, together with your partition.

=(Mn-M1)∆x

=RHS

There is still part a) to do...

 

[Mark53]

No. For one thing, don't start your work off with "=".
There is still part a) to do...

Does this have something to do with that the function is non decreasing and that the upper sums are greater than the lower sums but by adding one to the lower sum it would make them equal as they would be corresponding to the same value of the function

thus making the statement true: Mi = mi+1 for each i = 1, 2, . . . , n − 1

 

[Mark44]

Does this have something to do with that the function is non decreasing and that the upper sums are greater than the lower sums but by adding one to the lower sum it would make them equal as they would be corresponding to the same value of the function

No.

In fact, the statement in part a, as you wrote it, is incorrect, I believe.
Isn't this what it says in your textbook?
(a) Explain why M_i = m_i+1 for each i = 1, 2, . . . , n − 1

IOW, the right side isn't m_i + 1 -- it's m_{i + 1}, the minimum value in the i + 1st subinterval. You do need to know that the function is nondecreasing, however.

thus making the statement true: Mi = mi+1 for each i = 1, 2, . . . , n − 1

 

[Mark53]

No.

In fact, the statement in part a, as you wrote it, is incorrect, I believe.
Isn't this what it says in your textbook?
(a) Explain why M_i = m_i+1 for each i = 1, 2, . . . , n − 1

IOW, the right side isn't m_i + 1 -- it's m_{i + 1}, the minimum value in the i + 1st subinterval. You do need to know that the function is nondecreasing, however.

Sorry didn't know how to write the proper notation in my post you are right it should say m_{i + 1}

I am still confused on how i go about explaining it though

 

[Mark44]

Sorry didn't know how to write the proper notation in my post you are right it should say m_{i + 1}

I am still confused on how i go about explaining it though

Draw a picture, or better yet, two pictures. A non-decreasing function is one that is either constant (graph is a horizontal line) or strictly increasing (graph rises throughout the interval). Divide the interval into equal subintervals, and locate the minimum and maximum values in each subinterval.