# The Problem of Approximations I Exercises

Hey everyone! I thought I would start a discussion for this week’s exercises. Feel free to post your questions and discoveries or tricks that helped you learn how to do the exercises!

Looking at exercise 1 and using the Matrix Formulation of the Galerkin Approximation, can we assume that N_1 and N_2 are (1;0) and (0;1)? I’m having trouble wrapping my head around the fact that f is in R^3 while we’re trying to create something in R^2 (3D to 2D) if I’m understanding correctly.

Hey, Kennen,

So a couple of suggestions here: you’re in R^3 and R^2 here, so note that the symmetric bilinear form will be the dot product, not the integrals that are in most of the examples in the notes. Also, if you’re doing a dot product between something in R^2 and R^3 you should fill in the third coordinate in the R^2 vectors with a zero so that you can perform the dot product as if they were both in R^3.

So yes, N_1 and N_2 are as you wrote them, but add a zero on for the third coordinate, or (equivalently) only use the first two entries of f in the dot product.

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By removing the 3rd coordinate in the R^3 vectors, is that the change of basis asked for in number 2?

No, this is not the change of basis referred to in problem 2. In problem 2 it is asking you to represent
f:=e_1+e_2+e_3

where

e_1=(1,0,0)^T,
e_2=(0,1,0)^T,
e_3=(0,0,1)^T,

in terms of the new basis

\tilde e_1=(1,0,0)^T,
\tilde e_2=(1,1,0)^T,
\tilde e_3=(1,1,1)^T.

Alright so I feel like I’m doing it wrong but I’m not sure. So in problem 1, f=e_1+e_2+e_3 = (1,1,1)^T right? So for problem 2, f= \tilde e_3? I guess I’m not seeing how this changes the problem really other than a different basis where N_1=(1,0)^T and N_2 = (1,1)^T. Am I missing something?

Part of the point of this exercise is be able to recognize what does and does not change. Specifically, we want to look at the matrix and the solution.

You are on the right track when you said that you’re not really seeing how this changes the problem. That is an important insight. As you move through the problem things might look different, but it is the same problem just in a different basis. In a way everything that you see is just a different Representation of the same thing that you saw previously

Ok thanks Kyle! That clears up a lot of my thoughts on these problems. Thanks for helping me understand it’s a little more meta than I thought.