Extraneous Solutions – Part 2 of 3

Solving an Equation as a Sequence of Equation Replacement Operations

Part 1 was so long because I wanted to be extremely thorough and to present things to an audience that perhaps hadn’t thought much about the logic of equation solving at all. Since we’re now all experts, perhaps it’s worth it to summarize everything very succinctly.

Given an equation in one free variable, we want to find the solution set. To do this, we replace that equation with an equivalent equation whose solution set is more obvious.

(1) 8x - 5 = 5x + 1

(2) 8x = 5x + 6

(3) 3x = 6

(4) x = 2

If in the transition from (1)-(2), from (2)-(3), and from (3)-(4) we are careful to replace each equation with an equivalent equation, then by the transitivity of equivalence, the original equation and terminal equation are guaranteed to be equivalent. Since the solution set of the terminal equation is obvious, we know the solution set of the original equation, as well. Thus solving an equation consists of establishing that certain equation replacement operations are indeed equivalence preserving and having the creativity and experience to know which ones to apply and in what order.

What are the Equivalence-Preserving Operations on Equations?

If a = b, then f(a) = f(b) for any well-defined function f. If a and b are expressions containing a free-variable, then any value of that variable which satisfies a = b will also satisfyf(a) = f(b). In other words, if you find it useful, feel free to replace any equation with a new equation which is the result of applying any function to both sides of the original equation. Any solution to the original equation will also be a solution to the new equation.

If the function f is also one-to-one, then by definition, f(a) = f(b) \Rightarrow a = b so any solution of f(a) = f(b) will also be a solution to a = b. Thus applying f to both sides of an equation is equivalence-preserving. If f is not one-to-one, then in general, the operation is not equivalence-preserving.

In solving equation (1), we applied f(n) = n + 5, g(n) = n - 5x and h(n) = n/2 in that order. Since all three of the functions are one-to-one, we are assured that (1) and (4) are equivalent. If we had cause to apply a non-one-to-one function, then we should be vigilant for extraneous solution.

A More Interesting Example


(5) \sqrt{6x-2} - \sqrt{x+1} = 2

As I mentioned in the other post, these square roots are begging to be squared, but since there are two of them, one squaring will not be enough. Even though it’s not necessary to do so, it’s helpful to move one radical expression to the other side.

(6) \sqrt{6x-2} = 2 + \sqrt{x+1}

(7) 6x - 2 = 4 + 4\sqrt{x + 1} + x + 1 We squared!

(8) 5x - 7 = 4\sqrt{x+1}

(9) 25x^2 - 70x + 49 = 16x + 16 We squared again!

(10) 25x^2 - 86x + 33 = 0

(11) (25x - 11)(x - 3) = 0

So x \in \{\frac{11}{25}, 3\}

Since in the transition from (6)-(7) and again in the transition from (8)-(9) we had reason to apply the non-one-to-one function f(n) = n^2, we should be vigilant for extraneous solutions. [Note: since both sides of (6) are necessarily positive, applying f(n) = n^2 is equivalence-preserving, so no extraneous roots will be created there.] By checking back in the original equation, we see that 3 is a solution, but \frac{11}{25} is not. I am more or less content to leave it at that. But some may ask for more clarity as to exactly what happened and when, so let’s indulge them.

I will now list each equation in reverse order along with its solution set:

(11) (25x - 11)(x - 3) = 0                            \{\frac{11}{25}, 3\}

(10) 25x^2 - 86x + 33 = 0                             \{\frac{11}{25}, 3\}

(9) 25x^2 - 70x + 49 = 16x + 16                 \{\frac{11}{25}, 3\}

(8) 5x - 7 = 4\sqrt{x+1}                                     \{3\}

Since 5\cdot\frac{11}{25} - 7 = \frac{11}{5} - \frac{35}{5} = -\frac{24}{5} \neq 4\sqrt{\frac{11}{25} + 1} = 4\sqrt{\frac{11}{25} + \frac{25}{25}} = 4\sqrt{\frac{36}{25}} = 4\cdot\frac{6}{5} = \frac{24}{5}

So we have isolated the precise moment when the extraneous solution x = \frac{11}{25} is created and it appears exactly when we would expect it, in the transition from (8) to (9) as we replaced (8) with the result of applying the non-one-to-one function f(n) = n^2 to both sides.

More specifically, if x = \frac{11}{25}, (8) reads - \frac{24}{5} = \frac{24}{5}, which is false, but (9) reads (- \frac{24}{5})^2 = ( \frac{24}{5})^2, which is true. For this particular value of x, we squared both sides and turned a false statement into a true statement. In retrospect, we can say that x =\frac{11}{25} is not a solution to (8) or to any previous equation in the solving sequence, but is a solution to (9) and thus to all subsequent equations in the solving sequence.

(7) 6x - 2 = 4 + 4\sqrt{x + 1} + x + 1                          \{3\}

Since both sides of (7) are positive when x = 3, it does not surprise us that,

(6) \sqrt{6x-2} = 2 + \sqrt{x+1}                               \{3\}

(5) \sqrt{6x-2} - \sqrt{x+1} = 2                                \{3\}

By fully analyzing the logic behind each step of our equation replacement sequence, we not only:

  • confirm that x = 3 is a solution and that x = \frac{11}{25} is not and
  • understand that squaring both sides may produce an extraneous solution

but also

  • isolate the precise step in the solving sequence in which this extraneous solution was created answering the why, how, and when for this problem
  • confirm that the non-solution status of x = \frac{11}{25} is not merely due to an error of algebra or arithmetic, but is a direct result of that fact that this value produces an equation (8) of the form a = -a

That last point is crucial in distinguishing the phenomenon of extraneous roots from the phenomenon of user error in algebra or arithmetic. If our equation solving sequence consists solely of equivalence-preserving operations, we do not even need to check to see if solutions to our terminal equation are also solutions to our original equation. If we do decide to check, perhaps out of an abundance of caution, and find a discrepancy, then user error must be to blame.

On the other hand, if a solver does employ solution-set-enlarging operations in the solving sequence and finds that a solution to the terminal equation is not a solution to the original equation, is this because the solution is extraneous or due to user error? One could perform an analysis like I did above and confirm that the non-solution is not due to user error, but instead to the logic of the process.

Extraneous Solutions – Part 1 of 3?


Within my small inner circle of math teachers, the mystery of extraneous solutions seems to be the issue of the year. I have so much to say on this topic (algebraic, logical, pedagogical, historical, linguistic) that I don’t really know where to begin. My only disclaimer is that I’m not really sure if this topic is all that important.

Solving an Equation with a Radical Expression

Consider the following equation:

(1) 2\sqrt{x+8} +5 = 11

One hardly needs algebra skills or prior knowledge to solve this, but prior experience suggests trying to isolate x.

(2) 2\sqrt{x+8} = 6 (we subtract 5 from both sides)

(3) \sqrt{x+8} = 3 (we divide both sides by 2)

Now, if the square root of something is 3, then that something must be 9, so it immediately follows that

(4) x+8 = 9

(5) x = 1 (we subtract 8 from both sides)

Squaring Both Sides

In my transition from (3) to (4), I used a bit of reasoning. Some conversational common sense told me that “if the square root of something is 3, then that something must be 9”. But that logic is usually just reduced to an algebraic procedure: “squaring both sides”. If we square both sides of equation (3), we get equation (4).

On the one hand, this seems like a natural move. Since the meaning of \sqrt{a} is “the (positive) quantity which when squared is a“, the expression \sqrt{a} is practically begging us to square it. Only then can we recover what lies inside. A quantity “which when squared is a” is like a genie “which when summoned will grant three wishes”. In both cases you know exactly what to do next.

Unfortunately, squaring both sides of an equation is problematic. If a = b is true, then a^2 = b^2 is also true. But the converse does not hold. If a^2 = b^2, we cannot conclude that a = b because opposites have the same square.

This leads to problems when solving an equation if one squares both sides indiscriminately.

A Silly Equation Leads to Extraneous Solutions

Consider the equation,

(6) x = 4

This is an equation with one free variable. It’s a statement, but it’s a statement whose truth is impossible to determine. So it’s not quite a proposition. Logicians would call it a predicate. Linguistically, it’s comparable to a sentence with an unresolved anaphor. If someone begins a conversation with the sentence “He is 4 years old”, then without context we can’t process it. Depending on who “he” refers to, the sentence may be true or false. The goal of solving an equation is to find the solution set, the set of all values for the free variable(s) which make the sentence true.

Equation (6) is only true if x has value 4. So the solution set is \left\{4 \right\} . But if we square both sides for some reason…

(7) x^2 = 16 has solution set \left\{4, -4\right\}

We began with x = 4, “did some algebra”, and ended up with x^2 = 16. -4 is a solution (by inspection) to x^2 = 16, but not to the original equation which we were solving, so we call -4 an “extraneous solution”. [Extraneous – irrelevant or unrelated to the subject being dealt with]

Note that he appearance of the extraneous solution in the algebra of (6)-(7) did not involve the square root operation at all. But this example was also a bit silly because no one would square both sides when presented with equation (6), so let’s look at a slightly less silly example.

Another Radical Equation

(8) 2\sqrt{x+8} + 5 = -1

(9) 2\sqrt{x+8} = -6

(10) \sqrt{x+8} = -3

People paying attention might stop here and conclude (correctly) that (10) has no solutions, since the square root of a number can not be negative. Closer inspection of the logic of the algebraic operations in (8)-(10) enables us to conclude that the original equation (8) has no solutions either. Since a = b \iff a - 5 = b -5, any solution to (8) will also be a solution to (9) and vice versa. Since a = b \iff a/2 = b/2, any solution to (9) will also be a solution to (10) and vice versa. So equations (8), (9), and (10) are all “equivalent” in the sense that they have the same solution set.

But what if the equation solver does not notice this fact about (10) and decides to square both sides to get at that information hidden inside the square root?

(11) x+8 = 9

(12) x = 1

Again we have an extraneous solution. x = 1 is a solution to (12), but not to the original equation (8). Where did everything go wrong? By the previous logic, (8), (9), and (10) are all equivalent. (11) and (12) are also equivalent. So the extraneous solution somehow arose in the transition from (10) to (11), by squaring both sides.

So unlike subtracting 5 from both sides or dividing both sides by 2, squaring both sides is not an equivalence-preserving operation. But we tolerate this operation because the implication goes in the direction that matters. If a = b, then a^2 = b^2, so if a and b are expressions containing a free variable x, any value of x that makes a = b true will also make a^2 = b^2 true.

In other words, squaring both sides can only enlarge the solution set. So if one is vigilant when squaring both sides to the possible creation of extraneous solutions, and is willing to test solutions to the terminal equation back into the original equation, the process of squaring both sides is innocent and unproblematic.

Those Who are Still Not Satisfied

Still there are some who are not satisfied with this explanation: “Why does this happen? What is really going on? Where do the extraneous solutions come from? What do they mean?”

One source of the problem is the square root operation itself. \sqrt{a} is, by definition, the positive quantity which when squared is a. The reason that we have to stress the positive quantity is that there are always two real numbers that when squared equal any given positive real number. There are a few slightly different ways of making this same point. The operation of squaring a number erases the evidence of whether that number was positive or negative, so information is lost and we are not able to reverse the squaring process.

We can also phrase the phenomenon in the language of functions. Since squaring is a common and useful mathematical practice, information will often come to us squared and we’ll need an un-squaring process to unpack that information. f(x) = x^2, for all the reasons just mentioned, is not a one-to-one function, so strictly speaking, it is not invertible. But un-squaring is too important, so we persevere. As with all non-one-to-one functions, we first restrict the domain of f(x) = x^2 to [0, \infty) to make it one-to-one. This inverse, f^{-1}(x) = \sqrt{x} thus has a positive range and so the convention that \sqrt{a} \geq 0 is born. So every use of the square root symbol comes with the proviso that we mean the positive root, not the negative root. We inevitably lose track of this information when squaring both sides.

[Note: Students can easily lose track of these conventions. After a lot of practice solving quadratic equations, moving from x^2 = 9 effortlessly to x = \pm 3, students will often start to report that \sqrt{9} = \pm 3.]

The convention that we choose the positive root is totally arbitrary. In a world in which we restricted the domain of  f(x) = x^2 to (-\infty, 0] before inverting, \sqrt{9} would be -3. In that world, x = 1 is a perfectly good solution to 2\sqrt{x+8} + 5 = -1, not extraneous at all.

A Trigonometric Equation which Yields an Extraneous Solution

For parallelism, consider the (somewhat artificial) equation:

(13) \arccos(2x-1) = \frac{4\pi}{3}

Like in (10), careful and observant solvers might notice that the range of the \arccos(x) function is [0, \pi] and correctly conclude that the equation has no solutions. But there seems to be a lot going on inside that \arccos expression, so many will rush ahead and try to unpack it by “cosineing”. Indeed, since a=b \Rightarrow \cos(a) = \cos(b), this seems innocent.

(14) 2x - 1 = -\frac{1}{2}

(15) 2x = \frac{1}{2}

(16) x = \frac{1}{4}

But x = \frac{1}{4} is an extraneous solution since \arccos(-\frac{1}{2}) = \frac{2\pi}{3} not \frac{4\pi}{3}.

The explanation for this extraneous solution will be similar to the logic we used above. If a = b, then \cos(a) = \cos(b), so if a and b are expressions containing a free variable x, any value of x that makes a = b true will also make \cos(a) = \cos(b) true. So we will not lose any solutions by “taking the cosine of both sides”. But as the cosine function is not one-to-one, \cos(a) = \cos(b) does not imply that a = b. So taking the cosine of both sides, just like squaring both sides, can enlarge the solution set.

The above paragraph explains why extraneous solutions could appear in the solution of (13), but maybe not why they do appear. For that, we again must look to the presence of the \arccos function. Since \cos is not one-to-one, we had to arbitrarily restrict its domain to [0, \pi] prior to inverting. So every use of the \arccos symbol comes with its own proviso that we are referring to a number in a particular interval of values. In a world in which we had restricted the domain of \cos to [\pi, 2\pi] prior to inverting, x = \frac{1}{4} would be a perfectly good solution to \arccos(2x-1) = \frac{4\pi}{3}, not extraneous at all.

The above examples seem to suggest that one can avoid dealing with extraneous solutions by carefully examining one’s equations at each step. But in practice, this really isn’t possible. I saved the fun examples for the end, but as this post is already way way too long, they will have to wait for a bit later.

-Will Rose


Thanks to John Chase for letting me guest post on his blog. Thanks to James Key for encouraging me again and again to think about extraneous solutions.

Proving identities – what’s your philosophy?

What happens in your classroom when you give students the following task?

Prove 1+\frac{1}{\cos{\theta}}=\frac{\tan^2{\theta}}{\sec{\theta}-1}.

Sometimes the command is Verify or Show instead of Prove, but the intent is the same.


Two non-examples

Here are two ways that a student might work the problem.

Method 1






Method 2





How do you feel about these methods? In my opinion, both methods represent a fundamental misunderstanding of the prompt. Method 1 is especially grotesque, but Method 2 also leaves a lot to be desired. Let me explain. And if you think the above methods are perfectly fine, please be patient and hear me out.

This is the crux of the issue:

The prompt was to prove the statement. But if the first line of our work is the very thing we’re out to prove, then we are already assuming the thing we want to prove. We’re Begging the Question.

It’s as if someone demands,

“Prove Statement X, please!”

and we reply,

“Well, let’s first start by assuming that Statement X is true.”

This is nonsense.

What went wrong?

So what is the proper way to engage this proof? Let’s roll back a bit.

The error in these approaches seems to stem from a desire to perform algebraic operations on both sides of an equation in the same way that you might if you were solving an equation.

When we “do algebra” and write Equation B below another Equation A without any words, we always mean that Equation A implies Equation B. That is, when we write

Equation A

Equation B

Equation C


we mean that Equation C follows from Equation B, which follows from Equation A.

Some might claim that each line should be equivalent to the last. But, again, when we “do algebra” by performing algebraic manipulations to both sides of an equation to transform it from equation A into equation B, we always mean A\Rightarrow B, we don’t mean A\iff B. Take, for example, the following algebra which results in an extraneous solution:






x=2 \text{ or } x=-1

In this example, each line follows from the previous, however reversing the logic doesn’t work. But we accept that this is the usual way we do algebra (A\Rightarrow B\Rightarrow C\Rightarrow \cdots). Here the last line doesn’t hold because only one solution satisfies the original equation (x=2). Remember that our logic is still flawless, though. Our logic just says that IF \sqrt{x+2}=x for a given xTHEN (\sqrt{x+2})^2=x^2.

As we move through the algebra line by line, we either preserve the solution set or increase its size. In the case above, the solution set for the original equation is {2}, and as we go to line 2 and beyond, the solution set is {2,-1}.

For more, James Tanton has a nice article about extraneous solutions and why they arise, which I highly recommend.

So if this is the universal way we interpret algebraic work, which is what I argue, then it is wrong to construct an argument of the form A\Rightarrow B\Rightarrow C in order to prove statement A is true from premise C. The argument begs the question.

Both Method 1 and Method 2 make this mistake.


How does a proof go again?

I want to actually make a more general statement. The argument I gave above regarding how we “do algebra” is actually how we present any sort of deductive argument. We always present such an argument in order, where later statements are supported by earlier statements.

ANY time we see a sequence of statements (not just equations) A, B, C that is being put forward as a proof, if logical connectives are missing, the mathematical community agrees that “\Rightarrow” is the missing logical connection.

That is, if we see the proof A,B,C as a proof of statement C from premise A, we assume that the argument really means A\Rightarrow B\Rightarrow C.

This is usually the interpretation in the typical two-column proof, as well. We just provide the next step with a supporting theorem/definition/axiom, but we don’t also go out of our way to say “oh, and line #7 follows from the previous lines.”

Example: Given a non-empty set E with lower bound a and upper bound b, show that a\leq b.

1. E is non-empty and a and b are lower and upper bounds for E. (given)
2. Set E contains at least one element x. (definition of non-empty)
3. a\leq x and x\leq b. (definitions of lower and upper bound)
4. a\leq b. (transitive property of inequality)

Notice I never say that one line follows from the next. And also notice that it would be a mistake to interpret the logical connectives as biconditional.

The path of righteousness

I encourage my students to work with only ONE side of the expression and manipulate it independently, in its own little dark box, and when it comes out into the light, if it looks the same as the other side, you’ve proved the equivalence of the expressions.

For example, to show that \log\left(\frac{1}{t-2}\right)-\log\left(\frac{10}{t}\right)=-1+\log\left(\frac{t}{t-2}\right) for t>2, I would expect this kind of work for “full credit”:

\text{LHS }=\log\left(\frac{1}{t-2}\right)-\log\left(\frac{10}{t}\right)



= -1 + \log\left(\frac{t}{t-2}\right)

=\text{ RHS}

Interestingly, I WOULD also accept an argument of the form A\iff B\iff C as justification for conclusion A from premise C, but I would want a student to say “A is true if and only if B is true, which is true if and only if C is true.” Even though it provides a valid proof, I discourage students from using this somewhat cumbersome construction.

So let’s return to the original problem and show a few ways a student could do it correctly.

Three examples

Method A – A direct proof by manipulating only one side






Method B – A proof starting with a known equality






Method C – Carefully specifying biconditional implications


\text{if and only if}


\text{if and only if}


\text{if and only if}


\text{if and only if}


While all of these are now technically correct, I think we all prefer Method A. The other methods are cool too. But please, please, promise me you won’t use Methods 1 or 2 which I presented in my introduction.

In conclusion

Some might argue that the heavy criticism I’ve leveled against Methods 1 and 2 is nitpicking. But I disagree. This kind of careful reasoning is exactly the business of mathematicians. It’s not good enough to just produce “answers,” our job is to produce good reasoning. Mathematics, remember, is a sense-making discipline.

Thanks for staying with me to the end of this long-winded post. Can you tell I’ve had this conversation with a lot of students over the last ten years?

Further reading

  1. Dave Richeson has a similar rant with a similar thesis here.
  2. This article was originally inspired by this recent post on Patrick Honner’s blog. A bunch of us fought about this topic in the comments, and in the end, Patrick encouraged me to write my own post on the subject. So here I am. Thanks for pushing me in the right direction, Mr. Honner!


Happy π day!

Today at our school we had to have the obligatory π day celebrations. Here are the ways we observed π at RM:

  • To raise money for the math honors society, students purchased π-grams for one another and a piece of pi with a note were delivered to the recipients during first period.
  • There was a pie eating contest on “main street”, also sponsored by the math honors society, and guess who won? Check out my crown!!

Get the joke?? On the back it says "We are the 3.14%."

Action shot

Mr. Chase wins!

  • In each of my classes kids bring in food and we have a π day party. I pass out a sign up sheet a few days before and kids agree to bring all sorts of round food. Here’s the sign up sheet. Feel free to use it, steal it, modify it. Also available in pdf format. Here’s one of my favorite food items from this year:

  • I showed these youtube videos.
  • I did a mini lecture on the history of π and gave some interesting facts about the number.
  • Students got to find out at what digit of π their birthday appears. You can too!

One more thing you can still do, if you haven’t yet observed π day:


Happy π day!!!


Also, on an unrelated note, today’s Google logo is great. If you’re interested in the mathematics of origami, you probably know who Robert Lang is. Today’s Google logo is an origami piece created by Lang in honor of the late Akira Yoshizawa, world famous origami artist.

Pi R Squared

[Another guest blog entry by Dr. Gene Chase.]

You’ve heard the old joke.

Teacher: Pi R Squared.
Student: No, teacher, pie are round. Cornbread are square.

The purpose of this Pi Day note two days early is to explain why \pi is indeed a square.

The customary definition of \pi is the ratio of a circle’s circumference to its diameter. But mathematicians are accustomed to defining things in two different ways, and then showing that the two ways are in fact equivalent. Here’s a first example appropriate for my story.

How do we define the function \exp(z) = e^z for complex numbers z? First we define a^b for integers a > 0 and b. Then we extend it to rationals, and finally, by requiring that the resulting function be continuous, to reals. As it happens, the resulting function is infinitely differentiable. In fact, if we choose a to be e, the \lim_{n\to\infty} (1 + \frac{1}{n})^n \, not only is e^x infinitely differentiable, but it is its own derivative. Can we extend the definition of \exp(z) \, to complex numbers z? Yes, in an infinite number of ways, but if we want the reasonable assumption that it too is infinitely differentiable, then there is only one way to extend \exp(z).

That’s amazing!

The resulting function \exp(z) obeys all the expected laws of exponents. And we can prove that the function when restricted to reals has an inverse for the entire real number line. So define a new function \ln(x) which is the inverse of \exp(x). Then we can prove that \ln(x) obeys all of the laws of logarithms.

Or we could proceed in the reverse order instead. Define \ln(x) = \int_1^x \frac{1}{t} dt . It has an inverse, which we can call \exp(x) , and then we can define a^b as \exp ( b \ln (a)). We can prove that \exp(1) is the above-mentioned limit, and when this new definition of a^b\, is restricted to the appropriate rationals or reals or integers, we have the same function of two variables a and b as above. \ln(x) can also be extended to the complex domain, except the result is no longer a function, or rather it is a function from complex numbers to sets of complex numbers. All the numbers in a given set differ by some integer multiple of

[1] 2 \pi i.

With either definition of \exp(z), Euler’s famous formula can be proven:

[2] \exp(\pi i) + 1 = 0.

But where’s the circle that gives rise to the \pi in [1] and [2]? The answer is easy to see if we establish another formula to which Euler’s name is also attached:

[3] \exp(i z) = \sin (z) + i \cos(z).

Thus complex numbers unify two of the most frequent natural phenomena: exponential growth and periodic motion. In the complex plane, the exponential is a circular function.

That’s amazing!

Here’s a second example appropriate for my story. Define the function on integers \text{factorial (n)} = n! in the usual way. Now ask whether there is a way to extend it to (some of) the complex plane, so that we can take the factorial of a complex number. There is, and as with \exp(z), there is only one way if we require that the resulting function be infinitely differentiable. The resulting function is (almost) called Gamma, written \Gamma. I say almost, because the function that we want has the following property:

[4] \Gamma (z - 1) = z!

Obviously, we’d like to stay away from negative values on the real line, where the meaning of (–5)! is not at all clear. In fact, if we stay in the half-plane where complex numbers have a positive real part, we can define \Gamma by an integral which agrees with the factorial function for positive integer values of z:

[5] \Gamma (z) = \int_0^\infty \exp(-t) t^{z - 1} dt .

If we evaluate \Gamma (\frac{1}{2}) we discover that the result is \sqrt{\pi} .

In other words,

[6] \pi = \Gamma(\frac{1}{2})^2 .

Pi are indeed square.

That’s amazing!

I suspect that the \pi arises because there is an exponential function in the definition of \Gamma, but in other problems involving \pi it’s harder to find where the \pi comes from. Euler’s Basel problem is a good case in point. There are many good proofs that

1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ... = \frac{\pi^2}{6}

One proof uses trigonometric series, so you shouldn’t be surprised that \pi shows up there too.

\pi comes up in probability in Buffon’s needle problem because the needle is free to land with any angle from north.

Can you think of a place where \pi occurs, but you cannot find the circle?

George Lakoff and Rafael Núñez have written a controversial book that bolsters the argument that you won’t find any such examples: Where Mathematics Comes From. But Platonist that I am, I maintain that there might be such places.

The Arc Cotangent Controversy

I love this discussion at squareCircleZ. All my readers should check it out. Which is the graph of arccot(x)?

from squarecircleZ

from squarecircleZ

I especially like this controversy because some big players have weighed in on each side. Mathcad and Maple prefer the first interpretation, Mathematica and Matlab prefer the second.

For a more thorough treatment, check out the original post here. Three cheers for great math blogging! 🙂