It's been good to be back in math mode this week. Tonight I finished up the last two "hard problems" of my first post-midterm math assignment as well as the second half of assignment two. So I am now 2/9 finished with the second half of the homeworks. I know I will have to do it again on the exam, but I am looking forward to never having to determine the value of a definite integral by developing the Reimann sum. OK, I do feel sort of studly that I could do this, but now that I've proven it to myself (and soon, my instructor) I'll be quite ready to move on to other things.
Need I say that I took my midterm one month ago today and still have not received my grade or comments sheet back? Overall, I'm glad I'm not too dependent on this tour company to get me through my journey into the Land of Great Adventure. Because my instructor has (to my understanding) already told me my grade, I'm not freaking out about this, but (1) I would like to see this grade in official form and (2) I am interested in seeing the comments on my test.
I have to admit that my instructor's emphasis on the proofy bits has started to impact my thinking. This evening when I asked Robert a question on my homework that turned out to be about logarithms, I found myself asking him what was the proof. Fortunately, this proof was supplied by a later section of my textbook. It's amusing that they expect us to know the answer to something in section 5.3 that isn't brought up until section 6.1.
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2 comments:
I remember Riemann sums from doing them in vector calc. In that case, you are looking for the area under a surface, so you divide the bottom into squares and multiply the area of each square by the height of the surface at some (consistently chosen) point above the square. Same type of thing :-)
My response to Sally's demand for proof was, in its entirety, "uh, it has to do with how exponents work", which was quite useless, though trivially accurate. The proof in question was prove:
log(x)+log(y)=log(xy)
An equation near and dear to my heart. (When you take the log of the equation:
Y=A*X^B*e
you get
ln(y)=ln(A)+B*lnX+ln(e), where ln(A), B, and ln(e) can all be estimated through regression, and as a special bonus, B is the elasticity of Y with respect to X in economics.)
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