Question

PLEASE HELP!!!!!!
Show that (1/2)(Tn+Mn)=T2n
Trapezoidal and Midpoint rules

Answer 1

So, we're saying \(\dfrac{1}{2}\left(T_{n} + M_{n}\right) = T_{2n}\), where \(T_{n}\) is the Trapezoidal Rule with 'n' subdivisions and \(M_{n}\) is the Midpoint Rule with 'n' subdivisions. Is that wehat is required?

Do you have a working definition of the Sums required? Let's write out the Midpoint and Trapezoidal rules. Go!

Answer 2

Okay so the Midpoint rule is
\[Mn=\Delta x(f(x1)+f(x2).....+f(xn))\]
And the trapezoidal is
\[Tn=\Delta x/3[f(x0)+2f(x1)+2f(x2).....2f(xn-1)+f(xn)]\]

Answer 3

sorry trapezoidal is deltax/2 at the beginning

Answer 4

Fair enough, with the correction, but really a little unsatisfactory, anyway. What do you think? Do we NEED to know that this \(x_{i}\) are equally spaced? Obviously, with the \(\Delta x\) factored out, this is the intent. My real question, here is this, is \(x_{1}\) in the Midpoint Rule defintion EXACTLY the same as \(x_{1}\) in the Trepezoidal Rule definition?

Answer 5

No, they would be different because midpoint is going in-between the values while in the Trapezoidal method you are using the values. Like if we had from 1 to 4 then we would use f(1.5),f(2.5),f(3.5) for midpoint and for trapezoidal f(0),f(2),f(4)

Answer 6

Very good. It would help, then, if we had a notation so that they were on the same basis. No?

Answer 7

do you mean like this?
[f(1.5)+f(2.5)+f(3.5).......f(xn)]
(1/2)[f(1)+2f(2)+2f(3)......+2f(xn-1)+f(xn)]

Answer 8

Sort of. I was thinking more down the lines of this:

1) Unconfuse ourselves by using \(y_{i}\), rather than \(x_{i}\) for the Midpoint Rule.
2) Write a general expression for each \(y_{i}\) from the Midpoint Rule in terms of its surrounding \(x_{i}\) from the Trapezoidal Rule.

Can you give that a go?

Answer 9

So [f(y1)+f(y2)......f(yn)]
(1/2)[f(x0)+2f(x1)+2f(x2)......2f(xn-1)+f(xn)]

Answer 10

With some \(\Delta y\) and \(\Delta x\), yes.

Answer 11

how do we manipulate these then to make it work?

Answer 12

We know this: \(y_{i} = \dfrac{x_{i-1} + x_{i}}{2}\). Do you see that this is so?

Answer 13

yes

Answer 14

We're almost there. Good work hanging in there this far!

\(\dfrac{1}{2}(T_{n} + M_{n}) = \dfrac{1}{2}\left[\dfrac{\Delta x}{2}\left(f(x_{0}) + 2f(x_{1})+etc\right) + \Delta y\left(f(y_{1})+f(y_{2})+etc\right)\right]\)

Are you seeing it, yet?

Answer 15

I think I almost understand

Answer 16

We just need to notice, AND PROVE, that the Midpoint Rule provides EXACTLY what is needed to move from \(T_{n}\) to \(T_{2n}\). The only real question is the coefficients.

Answer 17

so then we need to turn the y's into terms of x?

Answer 18

First \(\Delta x = \Delta y\)

Sorting the terms by increasing value on the horizontal axis...

Then \(\dfrac{1}{2}\left(T_{n} + M_{n}\right) = \Delta x\left[¼f(x_{0}) + ½f(y_{1}) + ½f(x_{1}) + ½f(y_{2}) + ¼f(x_{2}) + etc\right]\)

Answer 19

so is the mn supposed to match the tn to make it n=2

Answer 20

Write out a few terms of \(T_{2n}\). To keep from confusing ourselves, use \(z_{i}\).

Answer 21

Whoops! I missed some coefficients. I was alternating 1, 2, 1, 2, 1 instead of the proper 1,2,2,2,1

Let's see if it works at all. Two (2) intervals of length 1 defined on [0,2]

T2 = ½(f(0) + 2f(1) + f(2))
M2 = (f(0.5) + f(1.5))

T2 + M2 = ½f(0) + f(0.5) + f(1) + f(1.5) + ½f(2)

½(T2 + M2) = ¼f(0) + ½f(0.5) + ½f(1) + ½f(1.5) + ¼f(2)
= ¼(f(0) + 2f(0.5) + 2f(1) + 2f(1.5) + f(2))
= [(1/2)/2](f(0) + 2f(0.5) + 2f(1) + 2f(1.5) + f(2))
= T4 on the same interval with half the step size.

Anyway, that's the idea. It's a little tricky doing all the LaTeX coding by hand,

Answer 22

Okay so then dividing it all my 2 gives us that (1/2)(Tn+Mn)=T2n?

Answer 23

This is not a difficult demonstration . The difficulty is in the indices - just keeping it all straight.