Challenge

Who Needs Trig Sub? Part 2

I recently wrote about an ingenious integration performed by two of my students. But that wasn’t the whole story. Here’s the rest.

It begins with this integral.

Usually when challenged to evaluate this integral students will try a few substitutions, fail to find an antiderivative, and realize something new is needed. This sets the stage for trigonometric substitution, which we then use to untangle the algebraic obstacles in the integrand.

But this was an unusual year. Not only did two students surprise me with a purely geometric approach, a third student found another completely different solution I had never seen before. She used integration by parts!

This application of integration by parts leads to the following equation.

Which she simplified to this.

This is the point where I would probably abandon integration by parts, because it doesn’t look the situation has improved. But my student did something incredibly inspired. She rewrote the integral

like this

And this is definitely an improvement. The integral on the right is known: it’s just $arcsin(x)$ . And the other integral is the one we were trying to find in the first place. When this is substituted back into the integration by parts equation you get this

The integral we want to evaluate is now on both sides of the equation, so just collect like terms and solve

Before this year I had never seen any alternate derivation of this formula. This year students produced two completely new ones!

I can understand why students had never attempted this technique before. Integration by parts typically comes after trig substitution in the course, so it wouldn’t usually be an option for them. But this year, because of the way I rearranged the curriculum, integration by parts came first. And I’m thrilled that my student thought to try it. And persevered!

It also made me feel good about my integration by parts lesson. Usually the last problem I present in that lesson is the integral

But this year I almost didn’t. This problem requires repeated integration by parts and quite a bit of perseverance to solve. At the end of a long lesson I considered passing on it, but we pushed through. And it was worth it! Seeing this technique set up my student to find her own ingenious integral. And to teach me something new.

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By MrHonner, 2 years2 years ago

Challenge Teaching

sin(x) + cos(x)

Here is a fun little exploration involving a simple sum of trigonometric functions.

Consider f(x) = sin(x) + cos(x), graphed below.

Surprisingly, it appears as though sin(x) + cos(x) is itself a sine function. And while its period is the same as sin(x), its amplitude has changed and it’s been phase-shifted. Figuring out the exact amplitude and phase shift is fun, and it’s also part of a deeper phenomenon to explore.

Consider the function g(x) = a sin(x) + b cos(x). Playing around with the values of a and b is a great way to explore the situation.

On the way to a complete solution, a nice challenge is to find (and characterize) the values of a and b that make the amplitude of g(x) equal to one. It’s also fun to look for values of a and b that yield integer amplitudes: for example, 5sin(x) + 12cos(x) has amplitude 13, and 4sin(x) + 3cos(x) has amplitude 5.

Ultimately, this exploration leads to a really lovely application of angle sum formulas. Recall that

$sin(A + B) = sin(A) cos (B) + sin(B) cos(A)$

If we let A = x, we get

$sin(x + B) = sin(x) cos(B) + sin(B) cos(x)$

With a little rewriting, we have

$sin(x + B) = cos(B) sin(x) + sin(B) cos(x)$

which looks similar to our original function f(x) = sin(x) + cos(x), except for what’s in front of sin(x) and cos(x). We handle that with a clever choice of B.

Let $B = \frac{\pi}{4}$ . Now we have

$sin(x + \frac{\pi}{4}) = cos(\frac{\pi}{4})sin(x) + sin(\frac{\pi}{4})cos(x)$

$sin(x + \frac{\pi}{4}) = \frac{\sqrt{2}}{2} sin(x) + \frac{\sqrt{2}}{2}cos(x)$

And a little algebra gets us

$sin(x) + cos(x) = \sqrt{2}sin(x + \frac{\pi}{4})$

And so sin(x) + cos(x) really is a sine function! Not only does this transformation explain the amplitude and phase shift of sin(x) + cos(x), it generalizes beautifully.

For example, consider 5sin(x) + 12cos(x). We can rewrite this in the following way.

$5sin(x) + 12cos(x) = 13 ( \frac{5}{13} sin(x) + \frac{12}{13} cos(x))$

$5sin(x) + 12cos(x)= 13 ( cos(\beta) sin(x) + sin(\beta) cos(x))$

$5sinx + 12cosx = 13 sin (x + \beta)$

where $\beta = arcsin(\frac{12}{13}) = arccos(\frac{5}{13})$ .

There’s quite a lot of trigonometric fun packed into this little sum. And there’s still more to do, like exploring different phase shifts and trying the cosine angle sum formula instead. Enjoy!

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By MrHonner, 6 years6 years ago

Challenge Numbers

The 2017-18 Conjecture

Like many mathematicians and teachers, I often enjoy thinking about the mathematical properties of dates, not because dates themselves are inherently meaningful numerically, but just because I enjoy thinking about numbers.

A new year means a new number to think about. And one interesting fact about our new year, 2018, is that it is semiprime.

A number is semiprime if it is the product of exactly two prime factors: for example, 15 = 3 * 5 is semiprime, as is 49 = 7 * 7, but neither 13 nor 30 are. Semiprime numbers are also referred to as biprime, 2-almost prime, or pq–numbers.

Semiprimes are very interesting in and of themselves, particularly in cryptography, but what caught my attention is that the previous year, 2017, is a prime number. That means we have a semiprime number, 2018, adjacent to a prime number, 2017. How unusual is this?

I played around a bit and ended up writing some simple programs to find and analyze semiprimes. Among the first 500,000 integers, there are roughly 108,000 semiprimes and 41,500 primes. Of the 108,000 semiprimes, only about 2,500 (or 2.3%) are adjacent to a prime number. This seems low to me: there are 83,000 prime-adjacent spots among the first 500,000 integers, representing 18% of the spots semiprimes could occupy. But only about 2.3% of the 108,000 semiprimes end up in those spots. That seems unusual. * [See Update]

In thinking about what happens further out along the number line, I couldn’t help but wonder if there are infinitely many prime-semiprime pairs like 2017 and 2018. I certainly don’t know the answer, but I thought I would start the new year boldly, with a conjecture:

The 2017-18 Conjecture

There are infinitely many pairs of consecutive integers one of which is prime and one of which is semiprime.

I think this problem’s resemblance to the Twin Prime Conjecture led me to both imagine this conjecture and also suspect it’s true. As with virtually everything in mathematics, I’m sure someone has thought of this before, and I would love a reference if anyone can provide it.

Thinking ahead, I was excited to notice that next year will also be a semiprime!

But it appears that the Twin Semiprime Conjecture is already an existing open question, which means I have less than a year to come up with a new conjecture for 2019.

Happy New Year! 2018 has already inspired to me to do some number theory, tackle some computing challenges, and think about some new ideas for the classroom. It’s a good mathematical start to the new year, and here’s hoping 2018 only gets better.

UPDATE, 1/18/2018

In a comment, Brent pointed out that I undercounted the number of semiprimes adjacent to a prime. A recalculation is consistent with Brent’s numbers: among the 108,000 semiprimes up to 500,000, around 4,900 of them are adjacent to prime number. Thanks, Brent!

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By MrHonner, 6 years6 years ago

Challenge Geometry

Jason Merrill’s Lawnmower Puzzle

Jason Merrill recently posted a fun geometry puzzle inspired by his work on the Lawnmower Math activity for Desmos. Here’s my paraphrase of the puzzle:

Suppose a lawnmower is tethered to a circular peg in the middle of the lawn. As the lawnmower moves along its spiral path, the rope shortens as its winds around the peg. At the moment the lawnmower contacts the peg, how much rope remains uncoiled?

When I first considered this problem it seemed hard. After some thought, it seemed obvious. Then, after some more thought, it seemed hard again. That’s the sign of a compelling problem!

I enjoyed working out a solution, the heart of which I’ve included below. Jason graciously included my solution in his post sharing his own, and he also does a wonderful job describing the journey of making simplifying assumptions, both mathematical and physical, that allow us to start moving toward a solution. It’s the kind of work that often goes unmentioned in problem solving, especially in school mathematics, and this puzzle provides a nice opportunity to make that thinking transparent.

I highly recommend reading the puzzle and his solution at his blog. Thanks for the fun problem, Jason!

By MrHonner, 7 years7 years ago

Application Challenge

How Much Would You Pay for a 20% Discount?

A local Office Max is going out of business and is having a very interesting sale.

I’m not sure I’ve ever seen a sale where you earn a discount by purchasing a certain number of items. Of course, I immediately began exploring the mathematical consequences of the policy.

The first thing that occurred to me was that you can essentially purchase a 20% discount. Say you need to buy n items. Simply buying another 20 – n items earns you a 20% discount. The natural question is thus, “Under what circumstances would buying an additional 20 – n items be worth a 20% discount?”

There are a variety of factors to consider. For example, if you can just find an additional 20 – n items that you are happy to buy, it’s definitely worth it: you get the 20% discount, and you get items of value to you. Also, the answer likely depends on n: if you are only 1 item short of the discount, it’s easier to justify an unnecessary purchase than if you are, say, 19 items short.

As an extreme case thinker, I considered the following scenario. Suppose I wanted to buy one item; under what circumstances would I buy 19 items I didn’t want in order to get a 20% discount?

Obviously, the key to this strategy is finding a cheap item to purchase 19 times. I thought I had found the cheapest possible item here:

Nineteen composition books would cost me $14.06. If the 20% discount saved me more than $14.06, this strategy would be worth it. This sets the bar for my one item at $70.30.

However, I later realized I could do better here:

These paper folders cost more per item, but unlike the composition books above, the folders are themselves eligible for the 20% discount! Nineteen folders would cost $16.91, but they’ll be discounted 20% to $13.53. This means if my single item cost more than $67.65, this strategy would save me money.

I could have done a lot better if these Slim Jims were sold here, or these 10-cent envelopes! But this is the best I could find in the store.

Another interesting question to consider is “For what range of prices would buying nine additional items, to receive a 10% discount, be a better strategy than buying 19 additional items, to get the 20% discount?”

In any event, I appreciate Office Max giving me something interesting to think about as I waited in line. And as usual, I waited a very long time. Let’s just say it’s no surprise they are going out of business.

By MrHonner, 8 years8 years ago

The 2017-18 Conjecture

There are infinitely many pairs of consecutive integers one of which is prime and one of which is semiprime.

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