# Swiggety swog, what’s in the blog?

Today we learned how to use complex analysis to solve real-values integrals that would otherwise be very difficult to solve.

Example: No complex variables in sight in that integral, right (assuming x is real-valued, haha)? Well you can CONVERT THIS TO A COMPLEX-VALUED INTEGRAL AND HAVE AN EASIER TO SOLVE PROBLEM!

That freaking blew my mind this morning in class. I’d go through the details of how to do this, but I’m a lazyass and don’t want to use Word’s equation editor to make like 30 different equations showing the steps to solve. Instead, I’ll link to Dr. Datta’s notes from class. Go to page 10 in the PDF (the page labeled “161”) for this example.

FREAKING. AWESOME.

Side note: if any of you ever end up going back to UI or know anyone who will be taking some upper-division math classes there, I highly recommend Dr. Datta. She’s very clear at explaining things, good at giving examples, gives reasonable homework, and is always willing to help.

# “When Will I Use That?” – Calculus Edition

Alternate title: Claudia Makes Things Way More Complicated than They Need to Be Because She Sucks

We had this bonus question on our homework for Probability today:

Suppose X has a density defined by Let FX(x) be the cumulative distribution of X. Find the area of the region bounded by the x-axis, the y-axis, the line y = 1, and the curve y = FX(x).

And I was like, “Aw, sweet! Areas of regions! CALCULUS!”

So first, I had to find the cumulative distribution function (cdf) of X. Easy. It’s just the integral of the density fX(x) from negative infinity to a constant b. In this case: With 2 ≤ b ≤ 3. So that’s my curve y. The area I’m looking for, therefore, is this (the red part, not the purplish part): Now anyone with half a brain would look at this and go, “oh yeah, that’s easy. I can find the area of the rectangle formed by the two axes, the line y = 1, and the line x = 3, then find the area of the region below the curve from 2 to 3, and subtract the latter from the former to get the correct area.”

Which works. Area of rectangle = 3, area of region below FX(x) = .25, area of region of
interest = 2.75.

Or they could remember the freaking formula that was explicitly taught last week. Such areas can be calculated using: But did I see either of those? Nooooooope.

I looked at the graph and was like, “how the hell do you find that?” I tried a few things that didn’t work, then realized that it would be a lot easier to figure out if I changed the integral from being in terms of x (or b, rather) to being in terms of y. So then I just had to integrate. This gave me the right answer: 2.75!

Moral of the story: don’t complicate things. But if you do complicate things, you might actually end up in a scenario where you’ll use something that you were taught back in calculus I but didn’t ever suspect you’d actually use. I had appreciated learning the handy-dandy technique of changing variables, but I didn’t think I’d be in a situation where I’d apply it. Shows what I know, eh?

It was a nice refresher, at least. I’ve missed calculus.

# Who is Fubini?

I dig my calc III teacher. He’s awesome. But I wish he’d do what I wish all math teachers would do when they introduce a theorem or lemma or rule: tell us a little bit about the person responsible for it, especially if the theorem/lemma/rule is named after the dude.

Like today we talked a lot about Fubini’s Theorem. We used it in like three examples. I used it on the homework I did right after class.

All the while without knowing who the heck this Fubini guy was.

So I checked him out this afternoon. Guido Fubini was an Italian mathematician who lived from 1879 to 1943. He was pretty into geometry and calculus for most of his life and moved around in different professorships in Europe before accepting an invitation to teach at Princeton in 1939 (partially to get away from the Nazis; he was Jewish).

So what the heck is this theorem, anyway?

Well. Let’s just look at rectangular domains first (because that’s all we’ve learned so far, haha…we’re doing non-rectangular domains tomorrow). So let’s look at a pretty double integral to start.

(P.S. I’m loving this chapter on double integrals already, simply because it means I have to write more integral signs. I FREAKING LOVE THAT SYMBOL.)

Say some rectangular region R is defined by the intervals [a,b] x [c,d]. If a function of two variables z= f(x,y) is continuous over R, then we can write the volume of the solid that lies below the surface z = f(x,y) and above the rectangle R as:

Or: Iterated integrals!

Cool? Cool. So what does Fubini’s Theorem state? Again, assuming z = f(x,y) is continuous over R and R is a rectangular region, Fubini’s Theorem allows us to switch the order of integration while still getting the same correct result at the end: Which is pretty snazzy (there’s a few other statements in the theorem; I just chose this conclusion as the example to show here).

But what I found most interesting about this theorem is that while double integration has been around for quite a long time, this theorem was proved sometime during Fubini’s lifetime–sometime in the late 1800s or early 1900s.  (I can’t find an exact date for it, but that’s mainly because my internet’s deciding to be a bitch right now). Which makes sense, I guess, considering there exist cases where this doesn’t hold and so it may not have been an “obvious” thing or may not have been easily provable…but still. Interesting passage of time before we got to this theorem.

HUZZAH CALCULUS!

# Somebody needs to do this if it hasn’t been done yet

Imagine a creation story where the Cosmos gives us two brother gods: Integration and Differentiation. They are responsible for two components of the Universe.

Integration—”The Great Summer”—is in charge of unity and space (well, area, but let’s just go with space). He wields integral symbols as weapons and lives in the sky.

Differentiation—”The Great Changer”—is in charge of division and, of course, change. He’s able to take the smallest components of the universe (hence the “division” aspect) and create a degree of change in it*. He has armor made out of barbs tangent to his skin and lives in the earth.

Something to draw, maybe…?

*Yes, I know taking the derivative of a function does not cause the change measured. Just work with me here.

# Math-inclined friends, I need your help!

So we’re doing trigonometric integrals in calculus and one of our homework problems is this little dude: We rushed through this section of the chapter this morning ’cause we’re behind schedule and I’m a little shaky on them (also I’m dumb), so I went to the calc room in Polya to get some help. I showed one of the tutors in there this integral.

I told him how I thought we should start: since (1-cos2x) is the numerator of the half-angle formula for sin2x, we could multiply both sides of the half-angle formula to change (1-cos2x) to 2sin2x and then go from there.

He said he’d never even thought about solving it like that, but when I asked him what the “normal” method would be for this integral, he didn’t know.

So is there another way of solving this?