I need to remember to do this in 2016.

# Month: October 2014

# Inverse Functions: Logarithms and Complex Numbers (Part 30)

Ordinarily, there are no great difficulties with logarithms as we’ve seen with the inverse trigonometric functions. That’s because the graph of satisfies the horizontal line test for any or . For example,

,

and we don’t have to worry about “other” solutions.

However, this goes out the window if we consider logarithms with complex numbers. Recall that the trigonometric form of a complex number is

where and , with in the appropriate quadrant. This is analogous to converting from rectangular coordinates to polar coordinates.

Over the past few posts, we developed the following theorem for computing in the case that is a complex number.

**Definition**. Let be a complex number so that . Then we define

.

Of course, this looks like what the definition ought to be if one formally applies the Laws of Logarithms to . However, this complex logarithm doesn’t always work the way you’d think it work. For example,

.

This is analogous to another situation when an inverse function is defined using a restricted domain, like

or

.

The Laws of Logarithms also may not work when nonpositive numbers are used. For example,

,

but

.

This material appeared in my previous series concerning calculators and complex numbers: https://meangreenmath.com/2014/07/09/calculators-and-complex-numbers-part-21/

# Inverse Functions: Arcsecant (Part 29)

We now turn to a little-taught and perhaps controversial inverse function: arcsecant. As we’ve seen throughout this series, the domain of this inverse function must be chosen so that the graph of satisfies the horizontal line test. It turns out that the choice of domain has surprising consequences that are almost unforeseeable using only the tools of Precalculus.

The standard definition of uses the interval — or, more precisely, to avoid the vertical asymptote at — in order to approximately match the range of . However, when I was a student, I distinctly remember that my textbook chose as the range for .

I believe that this definition has fallen out of favor today. However, for the purpose of today’s post, let’s just run with this definition and see what happens. This portion of the graph of is perhaps unorthodox, but it satisfies the horizontal line test so that the inverse function can be defined.

Let’s fast-forward a couple of semesters and use implicit differentiation (see also https://meangreenmath.com/2014/08/08/different-definitions-of-logarithm-part-8/ for how this same logic is used for other inverse functions) to find the derivative of :

At this point, the object is to convert the left-hand side to something involving only . Clearly, we can replace with . As it turns out, the replacement of is a lot simpler than we saw in yesterday’s post. Once again, we begin with one of the Pythagorean identities:

So which is it, the positive answer or the negative answer? In yesterday’s post, the answer depended on whether was positive or negative. However, with the current definition of , we know **for certain** that the answer is the **positive** one! How can we be certain? The angle must lie in either the interval or else the interval . In either interval, is positive. So, using this definition of , we can simply say that

,

and we don’t have to worry about that appeared in yesterday’s post.

Turning to integration, we now have the simple formula

that works whether is positive or negative. For example, the orange area can now be calculated correctly:

So, unlike yesterday’s post, this definition of produces a simple integration formula.

So why isn’t this the standard definition for ? I’m afraid the answer is simple: with this definition, the equation

is no longer correct if . Indeed, I distinctly remember thinking, back when I was a student taking trigonometry, that the definition of seemed really odd, and it seemed to me that it would be better if it matched that of . Of course, at that time in my mathematical development, it would have been almost hopeless to explain that the range had been chosen to simplify certain integrals from calculus.

So I suppose that The Powers That Be have decided that it’s more important for this identity to hold than to have a simple integration formula for

# Inverse Functions: Arcsecant (Part 28)

We now turn to a little-taught and perhaps controversial inverse function: arcsecant. As we’ve seen throughout this series, the domain of this inverse function must be chosen so that the graph of satisfies the horizontal line test. It turns out that the choice of domain has surprising consequences that are almost unforeseeable using only the tools of Precalculus.

The standard definition of uses the interval , so that

Why would this be controversial? Yesterday, we saw that is both positive and negative on the interval , and so great care has to be used to calculate the integral:

Here’s another example: let’s use trigonometric substitution to calculate

The standard trick is to use the substitution . With this substitution:

- , and

Therefore,

At this point, the common mistake would be to replace with . This is a mistake because

Furthermore, for this particular problem, is negative on the interval . Therefore, for this problem, we should replace with .

Continuing the calculation,

So if great care wasn’t used to correctly simplify , one would instead obtain the incorrect answer .

# Inverse Functions: Arcsecant (Part 27)

We now turn to a little-taught and perhaps controversial inverse function: arcsecant. As we’ve seen throughout this series, the domain of this inverse function must be chosen so that the graph of satisfies the horizontal line test. It turns out that the choice of domain has surprising consequences that are almost unforeseeable using only the tools of Precalculus.

The standard definition of uses the interval , so that

Why would this be controversial? Let’s fast-forward a couple of semesters and use implicit differentiation (see also https://meangreenmath.com/2014/08/08/different-definitions-of-logarithm-part-8/ for how this same logic is used for other inverse functions) to find the derivative of :

At this point, the object is to convert the left-hand side to something involving only . Clearly, we can replace with . However, doing the same with is trickier. We begin with one of the Pythagorean identities:

So which is it, the positive answer or the negative answer? The answer is, without additional information, **we don’t know**!

- If (so that ), then is positive, and so .
- If (so that ), then is negative, and so .

We therefore have two formulas for the derivative of :

These may then be combined into the single formula

It gets better. Let’s now find the integral

Several calculus textbooks that I’ve seen will lazily give the answer

This answer works as long as , so that reduces to simply . For example, the red signed area in the above picture on the interval may be correctly computed as

However, the orange signed area on the interval is **incorrectly** computed using this formula!

This is patently false, as the picture clearly indicates that the above integral has to be negative. For this reason, careful calculus textbooks will often ask students to solve a problem like

and the caveat is needed to ensure that the correct antiderivative is used. Indeed, a calculus textbook that doesn’t include such caveats are worthy of any scorn that an instructor cares to heap upon it.

# Inverse Functions: Arcsecant (Part 26)

The standard definition of uses the interval — or, more precisely, to avoid the vertical asymptote at . This portion of the graph of satisfies the horizontal line test and, conveniently, matches almost perfectly the domain of . This is perhaps not surprising since, when both are defined, and are reciprocals.

Since this range of matches that of , we have the convenient identity

To see why this works, let’s examine the right triangle below. Notice that

.

Also,

.

This argument provides the justification for — that is, for — but it still works for and .

So this seems like the most natural definition in the world for . Unfortunately, there are consequences for this choice in calculus, as we’ll see in tomorrow’s post.

# Inverse Functions: Arctangent and Angle Between Two Lines (Part 25)

The smallest angle between the non-perpendicular lines and can be found using the formula

.

A generation ago, this formula used to be taught in a typical Precalculus class (or, as it was called back then, analytical geometry). However, I find that analytic geometry has fallen out of favor in modern Precalculus courses.

Why does this formula work? Consider the graphs of and , and let’s measure the angle that the line makes with the positive axis.

The lines and are parallel, and the axis is a transversal intersecting these two parallel lines. Therefore, the angles that both lines make with the positive axis are congruent. In other words, the is entirely superfluous to finding the angle . The important thing that matters is the slope of the line, not where the line intersects the axis.

The point lies on the line , which also passes through the origin. By definition of tangent, can be found by dividing the and coordinates:

.

We now turn to the problem of finding the angle between two lines. As noted above, the intercepts do not matter, and so we only need to find the smallest angle between the lines and .

The angle will either be equal to or , depending on the values of and . Let’s now compute both and using the formula for the difference of two angles:

Since the smallest angle must lie between and , the value of must be positive (or undefined if … for now, we’ll ignore this special case). Therefore, whichever of the above two lines holds, it must be that

We now use the fact that and :

The above formula only applies to non-perpendicular lines. However, the perpendicular case may be remembered as almost a special case of the above formula. After all, is undefined at , and the right hand side is also undefined if . This matches the theorem that the two lines are perpendicular if and only if , or that the slopes of the two lines are negative reciprocals.

# Inverse Functions: Arctangent and Angle Between Two Lines (Part 24)

Here’s a straightforward application of arctangent that, a generation ago, used to be taught in a typical Precalculus class (or, as it was called back then, analytical geometry).

Find the smallest angle between the lines and .

This problem is almost equivalent to finding the angle between the vectors and . I use the caveat *almost* because the angle between two vectors could be between and , while the smallest angle between two lines must lie between and .

This smallest angle can be found using the formula

,

where and are the slopes of the two lines. In the present case,

.

Not surprisingly, we obtain the same answer that we obtained a couple of posts ago using arccosine. The following picture makes clear why .

In tomorrow’s post, I’ll explain why the above formula actually works.

# Inverse Functions: Arccosine and Dot Products (Part 23)

The Law of Cosines can be applied to find the angle between two vectors and . To begin, we draw the vectors and , as well as the vector (to be determined momentarily) that connects the tips of the vectors and .

Using the usual rules for adding vectors, we see that , so that

We now apply the Law of Cosines to find :

We now apply the rule , convert the square of the norms into dot products. We then use the distributive and commutative properties of dot products to simplify.

We can now cancel from the left and right sides and solve for :

Finally, we are guaranteed that the angle between two vectors must lie between and (or, in degrees, between and ). Since this is the range of arccosine, we are permitted to use this inverse function to solve for :

The good news is that there’s nothing special about two dimensions in the above proof, and so this formula may used for vectors in for any dimension .

In the next post, we’ll consider how this same problem can be solved — but only in two dimensions — using arctangent.

# Inverse Functions: Arccosine and Dot Products (Part 22)

Here’s a straightforward application of arccosine, that, as far as I can tell, isn’t taught too often in Precalculus and is not part of the Common Core standards for vectors and matrices.

Find the angle between the vectors and .

This problem is equivalent to finding the angle between the lines and . The angle is not drawn in standard position, which makes measurement of the angle initial daunting.

Fortunately, there is the straightforward formula for the angle between two vectors and :

We recall that is the dot product (or inner product) of the two vectors and , while is the norm (or length) of the vector .

For this particular example,

In the next post, we’ll discuss why this actually works. And then we’ll consider how the same problem can be solved more directly using arctangent.