Can a Coarser Partition Yield a More Accurate Riemann Sum?

[sandra1]

Homework Statement

Give an example of a function, interval, and partition P for which a left-handed sum using P is closer to the actual value of the Riemann Integral than the left-handed sum using partition Q which is a refinement of P

Homework Equations

The Attempt at a Solution

f(x) = 2x, x is in [0,1]
partition P = {0,1/3,2/3,1}
partition Q = {0,1/3,2/3,3/4,1}

since f(x) is increasing on [0,1] we have f(x_k) is max value for f and f(x_k-1) is min value for f on each subinterval [x_k-1, x_k]

so L(P,f) = 0.0 + (1/3)(2/3) + (2/3)(4/3) = 10/9
and L(Q,f) = 0.0 + (1/3)(2/3) + (2/3)(4/3) + (3/4)(6/4) = 161/72

actual Riemann Integral Value = 1

so |L(P,f) - 1| = (10/9) - 1 = 1/9 *
|L(Q,f) - 1| = (161/72) - 1 = 89/72 **
* < ** --> left-handed sum using P is closer to the actual value of the Riemann Integral than the left-handed sum using partition Q.

I am not sure if this is right. So any comment would be much appreciated. Thanks all.

 

[Dick]

You aren't doing the Riemann sum right. The left handed Riemann sum is the sum of (interval width)*(value at left hand endpoint) isn't it? Your (interval widths) aren't right. You aren't going to have much luck finding an example with a strictly increasing function.

 

[sandra1]

you're right. i was wrong. so it must be that

L(P,f) = 2/3 and,
L(Q,f) = 17/24

so now L(Q,f) is closer so an increasing function wouldn't work.

So a decreasing function must work right? so how about f(x) = -2x using the same two partitions P,Q

then L(P,f) = 0 + (1/3)(-2/3) + (1/3)(-4/3) = -6/9 = -2/3
and L(Q,f) = 0 + (1/3)(-2/3) + (1/12)(-4/3) + (1/4)(-6/4) = -17/24

so |L(P,f) - 1| = |(-2/3) - 1| = 5/3 *
|L(Q,f) - 1| = |(-17/24) - 1| = 41/24 **

* < ** --> left-handed sum using P is closer to the actual value of the Riemann Integral than the left-handed sum using partition Q.

is it ok now? *.*

 

[lanedance]

note the value of the Riemann integral is now -1...

i think the mian point was strictly increasing... how about considering a periodic function and relate the period to the refined set?

 

[Dick]

I think you need a function that has both increasing and decreasing parts. Here's a hint. Look at x-x^3. It's zero at x=1 and x=(-1). What's the integral between those two points?

 

[sandra1]

I think you need a function that has both increasing and decreasing parts. Here's a hint. Look at x-x^3. It's zero at x=1 and x=(-1). What's the integral between those two points?

thanks very much both of you for the hint. you're right i need a function with both increasing and decreasing parts. So ok from your hint f(x) = x-x^3. x is in [-1,1]
So the real value of riemann integral is 0.

pick P = {-1,0,1}
Q = {-1,0,1/2,1}

left-handed sum using P = 1.(-1-(-1)^3) + 1.0 = 0
left-handed sum using Q = 1.(-1-(-1)^3) + (1/2).0 + (1/2)((1/2)-(1/2)^3)
= 0 + 0 + 3/16 = 3/16
0< 3/16 --> done.

Is it what you meant?

 

[Dick]

thanks very much both of you for the hint. you're right i need a function with both increasing and decreasing parts. So ok from your hint f(x) = x-x^3. x is in [-1,1]
So the real value of riemann integral is 0.

pick P = {-1,0,1}
Q = {-1,0,1/2,1}

left-handed sum using P = 1.(-1-(-1)^3) + 1.0 = 0
left-handed sum using Q = 1.(-1-(-1)^3) + (1/2).0 + (1/2)((1/2)-(1/2)^3)
= 0 + 0 + 3/16 = 3/16
0< 3/16 --> done.

Is it what you meant?

Yep, that's pretty much what I meant. Finding a case where the integral equals a Riemann sum makes it pretty easy to find another partition that approximates less well.

 

[sandra1]

Yep, that's pretty much what I meant. Finding a case where the integral equals a Riemann sum makes it pretty easy to find another partition that approximates less well.

thanks very much for your help.