Read A Blog - Calculus: Problems and Solutions - Hip calculus discussion, with problems and solutions.

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Properties of Log

There are some properties of logarithms that you absolutely must know to get through calculus. Most students shriek or faint when they see a logarithm, but if you learn these simple rules they become much easier to work with. Get a little practice, and you’ll be on your way to mastering logarithms!

All you really need to know are two rules: first, logs turn multiplication into addition:

$displaystyle log ab = log a + log b$ .

Second, taking the log of a power turns into multiplication by the exponent:

$displaystyle log a^p = plog a$ .

From these, we can deduce a rule for quotients:

$displaystyle log frac{a}{b} = log a - log b$

Here’s an exercise: how could I have shown the third rule if I only knew the first two?

To get more examples of problems with logs, take a look at the 99 calculus solutions manual.

Limits That Don’t Exist

Some functions don’t have a limit as the value of their parameter approaches $pminfty$ . Here’s one example:

$displaystylelim_{x ightarrowinfty} sin x$

just doesn’t exist. Why is that? Because as $x$ increases, $sin x$ keeps oscillating between 1 and -1. It never approaches any one value, and so we can’t assign a limit to it.

Notice that this is different from a function having a limit of $infty$ or $-infty$ . In that case, the limit exists, and is equal to plus or minus infinity. In the case above, the limit doesn’t even exist.

Here’s another example that sometimes trips students up: what is the following limit?

$displaystyle lim_{t ightarrowinfty} xcos(pi x)$

Integration by Parts

When we looked at integrating the natural logarithm, we used a little trick: we took the product rule

$displaystyle (uv)' = u'v + uv'$ ,

integrated it to get

$displaystyle uv = int v,du + int u,dv$ ,

and made a clever choice of $u$ and $v$ to find the integral we wanted. This is called integration by parts, and it’s usually written like this:

$displaystyle int u,dv = uv - int v,du$ .

There are plenty of examples of integration by parts in my calculus I solutions guide. Have a look.

Textbook Rental

There’s a new service for students to rent textbooks for the semester: chegg.com. If you’re not planning on using your books later, this might be a good option for you. Check it out!

I should mention that if you’re in a science or engineering field, you will need your calculus book again at some point, so you should think about just buying one.

Integrating the Natural Logarithm

You’ll probably see this integral someday:

$displaystyle int ln x,dx$ .

It looks so simple, and you think, “gee, I probably was supposed to memorize that” or “oh I can do that, it looks so easy.” But then you don’t remember the antiderivative and get stuck. Cause, heck, what can you do with just $ln x$ anyway? It doesn’t decompose into anything nicer.

The trick is to use integration by parts. Let’s look at it backwards:

If we were really smart or really lucky, we might approach this problem by saying

$displaystyle frac{d}{dx}(xln x) = ln x + 1$ .

(Is this really true? Use the product rule to show it.) Then we can integrate both sides:

$displaystyle intfrac{d}{dx}(xln x),dx = intln x,dx + int 1,dx$ .

Now using the Fundamental theorem of calculus, the integral on the left is just equal to $xln x$ . Solving for $intln x,dx$ , we get

$displaystyle intln x = xln x - x + C$ .

Yay!

Remembering One Trig Identity

We’ve been talking about a way to figure out trig identities without having to just memorize them. Here is how I showed the identity

$displaystyle cos^2 x = frac{cos 2x + 1}{2}$ ,

by just knowing Euler’s formula and some facts on complex numbers. I started off by using Euler’s formula to write

$displaystyle cos^2 x = left(Re e^{ix} ight)^2$ .

Then I used the fact from complex numbers that $left(Re z ight)^2 = Re(z^2)+left(Im z ight)^2$ to say that

$displaystyle left(Re e^{ix} ight)^2 = Releft( (e^{ix})^2 ight) + left(Im e^{ix} ight)^2$
$displaystyle = cos 2x + sin^2 x$ .

Since $sin^2 x = 1-cos^2 x$ , this is in turn equal to

$displaystyle cos 2x + 1-cos^2 x$ .

Now we can add $cos^2 x$ to get

$2cos^2 x = cos 2x + 1$ ,

which is pretty much what we wanted. Cool, huh?

It turns out that you can figure out pretty much any of the identities you’re gonna see in calculus or differential equations by just using this technique.

Can you figure out how to show these identities?

$sin^2 x = frac{1-cos 2x}{2}$
$cos 2x = cos^2 x - sin^2 x$
For practice, try to figure this one out without looking up what the right answer is. $cos (x+y) = ??$

Some Facts About Complex Numbers

Let’s look at some properties of complex numbers, numbers of the form $a+ib$ where $a$ and $b$ are real numbers and $i^2=-1$ . We’ll use these together with Euler’s formula to deduce some trig identities.

First, if $z = a+ib$ , I’m going to call $a$ the real part of $z$ and $b$ the imaginary part. We write $a=Re z$ and $b=Im z$ ; the big goofy symbols are for historical reasons.

Now what if we start squaring $z$ ? What is $Re (z^2)$ ? Computing, we get $z^2 = a^2+2iab + i^2b = a^2-b^2+2iab$ , so
$Re (z^2) = a^2-b^2 = (Re z)^2-(Im z)^2$ .

Similarly, we can compute that $Im (z^2) = 2(Re z)(Im z)$ . Check these facts to make sure you understand.

Next time, we’ll use them for something!

Euler’s Formula

In the last post we talked about two important trig identities. I said you should memorize them (and you probably should), but what if you forget? I want to talk about a powerful way to derive trig identities by remembering just a couple of facts. This is great if you’re forgetful like me! To get started, though, we need to look at a wonderful formula discovered by Leonhard Euler. It’s called, appropriately enough, Euler’s Formula:

$displaystyle e^{ix} = cos x + isin x$

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