polymathematicCalcea Johnson and Ne’Kiya Jackson are two high school students at St. Mary's Academy in New Orleans who recently presented a new proof of the Pythagorean theorem at the Spring Southeastern Sectional Meeting of the American Mathematical Society. There are, of course, many proofs of the Pythagorean theorem, but what sets this one apart is that it's (mostly) trigonometric, but does not circularly rely on the Pythagorean identity. You can read the AMS abstract here (meetings.ams.org/math/spring2023se/meetingapp.cgi/Paper/23621), and you can read more about these remarkable young women here (theguardian.com/us-news/2023/mar/24/new-orleans-pythagoras-theorem-trigonometry-prove).
There's not a lot of information online yet about how their proof proceeds, but from a few of the slides you can pull out of news coverage (you can see de-skewed versions of three of them here: dropbox.com/s/h99ezl8l360xk4r/JOHNSON%20and%20JACKSON%20New%20Pyth%20Proof.pdf?dl=1), it seems the proof relies roughly on the definition of the sine ratio itself and the Law of Sines, along with a very clever "waffle cone" triangular shape.
If you'd like to play around with the waffle cone shape, I built a little illustration in Desmos that you van view here: desmos.com/calculator/tyw7ktgjpq. You can see the interplay of the infinitely many right triangles going down to the right from the isosceles triangle, and it also includes a folder with the proof.
Finally, some news coverage has gone a little overboard on stating the accomplishment here. It is true that a collection of proofs of the Pythagorean theorem from roughly a hundred years ago stated that a trigonometric proof was impossible (Elisha Loomis wrote "no trigonometric proof is possible" here: files.eric.ed.gov/fulltext/ED037335.pdf). But in 2009, Jason Zimba published just such a result in Forum Geometricorum: https://forumgeom.fau.edu/FG2009volume9/FG200925.pdf. You can see that proof and other related proofs at Cut the Knot:
Pythagoras Would Be Proud: High School Students New Proof of the Pythagorean Theorem [TRIGONOMETRY]polymathematic2023-04-04 | Calcea Johnson and Ne’Kiya Jackson are two high school students at St. Mary's Academy in New Orleans who recently presented a new proof of the Pythagorean theorem at the Spring Southeastern Sectional Meeting of the American Mathematical Society. There are, of course, many proofs of the Pythagorean theorem, but what sets this one apart is that it's (mostly) trigonometric, but does not circularly rely on the Pythagorean identity. You can read the AMS abstract here (meetings.ams.org/math/spring2023se/meetingapp.cgi/Paper/23621), and you can read more about these remarkable young women here (theguardian.com/us-news/2023/mar/24/new-orleans-pythagoras-theorem-trigonometry-prove).
There's not a lot of information online yet about how their proof proceeds, but from a few of the slides you can pull out of news coverage (you can see de-skewed versions of three of them here: dropbox.com/s/h99ezl8l360xk4r/JOHNSON%20and%20JACKSON%20New%20Pyth%20Proof.pdf?dl=1), it seems the proof relies roughly on the definition of the sine ratio itself and the Law of Sines, along with a very clever "waffle cone" triangular shape.
If you'd like to play around with the waffle cone shape, I built a little illustration in Desmos that you van view here: desmos.com/calculator/tyw7ktgjpq. You can see the interplay of the infinitely many right triangles going down to the right from the isosceles triangle, and it also includes a folder with the proof.
Finally, some news coverage has gone a little overboard on stating the accomplishment here. It is true that a collection of proofs of the Pythagorean theorem from roughly a hundred years ago stated that a trigonometric proof was impossible (Elisha Loomis wrote "no trigonometric proof is possible" here: files.eric.ed.gov/fulltext/ED037335.pdf). But in 2009, Jason Zimba published just such a result in Forum Geometricorum: https://forumgeom.fau.edu/FG2009volume9/FG200925.pdf. You can see that proof and other related proofs at Cut the Knot:
The empty set is weird because while it *contains* nothing, it is not itself *nothing*. So we have to be careful when working through whether the empty set (that is, not nothing but the set that contains nothing) is an element or subset of some other set. Today's video is about some basic true false questions we can answer using the empty set and the sets that contain it.
First of all, there are a few ways to symbolize the empty set, including a symbol that looks like a zero with a slash through it (∅) or just a pair of braces with nothing between them: {}. What can be a little confusing about the first symbol, which is definitely more common, is that it's easy to forget that it represents a set, not merely nothing itself. The empty set definitely isn't *nothing*, it's the set that *contains* nothing.
One way to think through that distinction is to compare the empty set to several other sets it may or may not be an element of. And so that's what we have with this little true/false quiz. Enjoy!
Follow Tim Ricchuiti: TikTok: tiktok.com/@polymathematic Mathstodon: https://mathstodon.xyz/@polymathematic Instagram: instagram.com/polymathematicnet Reddit: reddit.com/user/polymath-matic Facebook: facebook.com/polymathematicAre You Smarter Than OpenAI’s ChatGPT o1? The Prince and Princess Puzzlepolymathematic2024-09-26 | In today's video, we dive into a classic age-related puzzle involving a prince and princess with some tricky clues about their ages. This was a riddle that executives at OpenAI used to announce their new model for ChatGPT: o1 (currently in preview mode).
The initial confusion for the puzzle is understandable, but to try to solve it as a human, I begin with the tried-and-true strategy of students everywhere: "guess and check". Once we have a particular solution that works, however, we can turn to algebra to help us figure out the general case solution.
Interestingly enough, when I fed this back into ChatGPT using the o1-preview model, it got it wrong at first. But after some coaxing, It was back on track. Moral of the story: humans still have a slight edge (for now) in reasoning.
I first came across the announcement at John Gruber's Daring Fireball. I refrained from reading his solution until I came up with my own, but you can see we took pretty similar paths once algebra got involved. daringfireball.net/linked/2024/09/12/openai-o1
Follow Tim Ricchuiti: TikTok: tiktok.com/@polymathematic Mathstodon: https://mathstodon.xyz/@polymathematic Instagram: instagram.com/polymathematicnet Reddit: reddit.com/user/polymath-matic Facebook: facebook.com/polymathematicThe Two Hardest Two-Digit Numbers to Multiplypolymathematic2024-07-15 | Picking up on a series I started a couple summers ago, if we're multiplying in our head, no calculator, what are the two hardest two-digit numbers to multiply together? That is, what pair of numbers would be best at preventing us from using some clever trick to speed up our calculation?
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On one level, this depends a lot on whether we happen to know certain number tricks. But that doesn't mean this is entirely subjective. For example, we'll definitely want to avoid certain numbers that factor in several different ways, because that increases the number of options we have for manipulation. In today's example, 37×72, the fact that 72 has so many different factors allows us to manipulate the problem and end up with the considerably easier computation of 111×24. From there we use something called the 111 trick, though we could probably also turn it into some basic distribution and achieve the same result.
The answer is no, because if anything, zero is almost the opposite of a prime number. Whereas prime numbers have the *least* number of factors possible (other than 1 itself, which is not prime as we covered yesterday), 0 has the *most* number of factors possible. In a sense, zero has infinitely many factors.
Because of the special fact that anything times zero is zero, we can imagine that 0 is 0×1 or 0×2 or 0×3 or 0 times any other natural number. In that way, zero has as its factors every single natural number, and for sure is not prime.
The answer is no, but it's not necessarily a satisfying no: 1 is not a prime number because mathematicians say it's not.
There's no fundamental reality that demands we describe prime numbers in a particular way. Instead, mathematicians are simply defining prime numbers in the most useful way possible. In this case, the most useful way possible is to preserve what are called "unique factorizations". The idea is that we want to be able to uniquely describe the prime factorization of any particular number. With something like 12, we want to say that 12 is 2×2×3. But if we could 1 as a prime number, we would have to say that 12 is *also* 1×2×2×3. And for that matter that 12 is 1×1×2×2×3. And so on. If 1 is a prime number, it would break our ability to uniquely factor 12. And so we say that it's not!
Another way around this problem is to be a bit more careful when we define prime numbers in the first place. If we say prime numbers must have exactly two *distinct* factors, we would not include 1.
To illustrate just how vast this difference is, let's take a moment to break it down with some more familiar numbers. Imagine comparing something like a million to a billion. A million seconds is about 11.5 days, while a billion seconds is nearly 32 years. Now, multiply that sense of scale by an unimaginable factor, and you're starting to get the idea.
We covered an intuitive way to understand powers of two yesterday (youtube.com/shorts/OvZglaG9pVA), but what about a slightly less intuitive way? We typically think of powers as a multiplicative process, and as long as the exponent is a positive integer, that will definitely work. In that case, repeated multiplication will give us back the value we need.
But what if instead of letting our exponents get greater, we let them get lesser? Rather than repeatedly multiply by our base (in this video, 2), we end up repeatedly dividing. And most important, when we get all the way back to the zero exponent, we divide our base by itself. As any number divided by itself is 1, the zero power for any base yields one.
For any base, that is, except 0 itself. Zero is the one number that we are *not* allowed to divide by itself, at least not meaningfully, and so this understanding doesn't help us evaluate 0^0.
That doesn't mean that 0^0 can't be assigned a value. Depending on the expression that generated it, it often makes sense to assign it the value 1 anyway. But on other occasions, it makes more sense to assign it 0, or perhaps even "infinity".
When we first learn exponentiation, we typically learn it in terms of repeated multiplication. So something like 2³ is 8 because we take 2 and multiply it by itself three times: 2 × 2 × 2. This works fine for any positive integer powers, but what about raising something to the zero power?
It seems like if we tried multiplying 2 by itself "zero times", that would probably mean that the result should be zero. But in fact, it makes more sense to let the result be 1. There are several reasons for this, but this activity with folding paper gives us one intuitive reason to expect that something raised to the zero power is 1.
The rules of the game as simple. Take four numbers and combine them with addition, subtraction, multiplication, and division in whatever order you'd like. But you must use all four, and your result from those operations should be 24.
Twenty-four tends to be a good number to aim for because there are several numbers that are factors of 24 that you can target (2, 3, 4, 6, 8, and 12) and there are some other numbers close to 24 that also have several factors you can aim for (like 30), then follow that up with some addition or subtraction.
How about the one I left over from the video? How would you combine 2, 15, 11, and 7 to come up with 24?
It turns out, however, that that's not the case, and we've known for several hundred years now. Johann Lambert first proved that pi is irrational in the mid-18th century. And although the details are a little tough to work out in a 60-second short, the gist of it is that Lambert showed a particular infinite continuing fraction *had* to produce an irrational output if its input was rational (and non-zero). Contrapositively, if the output of that fraction was rational, that had to mean the input was irrational.
Conveniently, that infinite continuing fraction is equivalent to the tangent of the input. And since we know that various fractions of pi have tangent values of 1, we know it would produce a rational output in the infinite continuing fraction form. Therefore, that fraction of pi (π/4) must be irrational. And since we know it's not the 4 making it irrational, it must mean that pi is irrational.
This is, admittedly, a tough one to wrap your head around. Because of the nature of infinity, it can be just as hard to imagine pi containing every possible string of digits, and in that way encoding any information you want somewhere within its decimal expansion, as it is to imagine that it does NOT contain every possible string of digits. And, to be sure, it is totally possible that pi DOES contain every possible string of digits—not just once, but infinitely many times! Most mathematicians believe that's true, and they've even given that a name: normality. But no one has been able to prove that pi is normal, and just because it's an irrational number, that's not enough to guarantee it.
Consider the weird number I construct in this video: the digits 1 through 9 repeating over and over again in that order, but with a different number of zeros between them every time. This is an irrational number. Because the zeros are changing, it would be impossible to represent this number with a ratio between two integers. Its decimal expansion also contains every possible digit an infinite number of times. And yet, there are tons and tons of sequences that will never be in this number. For example, 10 will never be in this number.
Now, you might object that pi is just a typical number occuring all on its own out there in the world, and so it won't obey the weird rules this constructed number will. Maybe! I think that's probably true. But so far, no one can prove that it's true, and even more importantly, there's nothing about the nature of irrationality or any particular use of digits that *requires* it to be true.
I don't think so. I saw this post from @kylehill and while I get the logic, it doesn't strike me as quite right. It's totally true that when light leaves from the sun it takes about 500 seconds (or a little over 8 minute) to get to Earth. This is a simple enough calculation, taking the 150 million kilometers the sun is away from the earth and dividing by the speed of light (about 300,000 kilometers per second) to get roughly 500 seconds.
But the "eclipse" isn't merely some special light from the sun. The eclipse is that light's occlusion by the moon. At best, I think we can say the "eclipse" happens beginning from the moon, which is only about 384,400 kilometers away. When you do the same calculation for that distance, you get roughly 1.28 seconds. So I think it would make more sense to say that the eclipse "happens" 1.28 seconds before we see it.
Well, all except some irrational commenters who presumed it *must* be anything other than a really cool story about these two students' ingenuity.
Some commenters, like the one in this video, preferred to believe this was really a story about a couple high schoolers getting credit for an AI's work. First of all, this is exceedingly unlikely, as the only publicly available AI at the time couldn't even multiply properly, much less come up with novel proofs of something like the Pythagorean Theorem.
But even beyond that, there's something really sad about a person choosing to develop an ignorant cynicism over embracing a really cool story. I hope I can do my little part to disabuse people of that notion. Well done, students!
Love the rest of this video from @modernday_eratosthenes (which you can find below), but this bit at the end isn't quite right. Even though we don't know if pi is normal, and therefore we don't know if it will contain any arbitrary *finite* string of digits, we definitely *do* know that pi is irrational, and therefore it cannot possibly end in a string of infinitely many 8s.
One funny objection people sometimes have to this is that they say "well, of course pi is rational, it can be written as the ratio between a circle's circumference and its diameter." Which is totally true! That is the fundamental definition of pi. The trick is that's not what we mean by "rational". Any number *can be* written as a ratio, because every number can be written as itself divided by 1. What we mean when we say a number is rational is that it can be written as the ratio *between two integers*. Accordingly, an irrational number is a number that cannot be written as the ratio between two integers. And although we also couldn't prove that for a very long time, as of the mid-18th century, Johann Lambert was able to prove that pi is irrational by a continued fraction representation of the tangent function. But that will have to be a video for another day!
One thing that's always bothered me, and this is such a weird math teacher thing to be bothered by, is the idea that you can find any string of digits you want somewhere in pi. Not only that, but you can find that string of digits infinitely many times. And that could be true, but we actually don't *know* that it's true.
If we could prove that pi were a normal number, meaning a number that, in its decimal representation, can be expected to have every digit or grouping of digits occur with equal frequency, then we could definitely say this. But no one has yet been able to prove that pi is a normal number.
I did a video recently figuring out how much money Scrooge McDuck has using a net present value calculation from a silly comic panel about how he can lose a billion dollars a minute for 600 years before going broke (youtu.be/7rng56ZfCHQ). One commenter objected to the total I came up with because Scrooge McDuck's money is mostly gold rather than cash. Is that a good reason to come up with a different value?
No! This is akin to saying a pound of lead weighs more than a pound of feathers. Whatever value we come up with for Scrooge McDuck's money IS the amount of money he has. He might have that in gold, he might have it in cash, he might have it in stock, but it's worth the same amount no matter how he holds it.
Now, you can argue that his future value will differ depending on how he holds the money. Maybe you think gold is a good investment and his money will grow a lot if he keeps it in gold. Maybe you think gold is a terrible investment and his money will grow more slowly. But either way, once we calculate the amount of money he has, that's the amount of money he has, regardless of how he invests it.
Solving absolute value equations typically requires that we consider several different "cases", in which we represent the quantity we're taking the absolute value of as either a positive or a negative. For a single variable, this is fine, as it usually only requires two cases, but for multiple variables, this method can quickly become unwieldy. Can we do any better?
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Of course! Well, of course, depending on the particular equations we're solving. In today's video, we're solving two equations in which we can be a little clever about what we can discover for the variables before we fully consider the different cases. Doing so lets us considerably reduce the amount of we need to do in order to solve the system of equations.
There's a nifty little formula for the sum of the measures of the interior angles of a polygon, but what about the exterior angles. Is there a simple way to compute the sum of the measures of the exterior angles for a polygon of n sides or corners?
We can actually do even better than a formula. We have the answer! It turns out the sum of the measures of the exterior angles of a simple polygon is 360º no matter how many sides it has. To understand why, let's turn to a very determined, very slow turtle.
The slow turtle helps us think through what's happening at each exterior angle, regardless of the specifics of the polygon. The slow turtle will walk along the outside of the polygon, hugging the edge until he gets to each next corner. Then he'll turn just enough to get back to hugging the edge and begin walking forward again.
By the time the slow turtle gets back to where he started, regardless of how many times he had to stop and turn, he will have made one full revolution. Therefore, the sum of the measures of the exterior angles of a polygon will always be 360º.
If you need to square a two-digit number, there is a general trick, but it's not very good. i prefer to have the first 40 memorized and then use case-by-case tricks for the remainder. there are good ones for the squares 85 to 100 and for any square of a number ending in 5. but this trick for the numbers between 40 and 60 is a favorite of mine.
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I prefer simplifying here with exponent rules, which means that we need to use the exponential equivalent to the square root radical: the one-half power. Re-writing the nested radicals as repeated one-half powers allows us to combine the terms by either adding exponents together, applying the product of powers rule, or by multiplying exponents, applying the power of a power rule.
Finally, at the end, we have x to the 7/8 power equals 7, so we can raise each side of the equation to the 8/7 power in order to isolate x and get our solution at x equals 7 to the 8/7 power. Alternatively, if we prefer, we can convert that back into radical form and have x equals the 7th root of 7 to the eighth power.
Then you bought it back for $1100. You just spent $100 more than you sold it for, so you're $100 down on this transaction. Finally, you sold it again for $1300.
Have you made money or lost money?
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I think people make two main mistakes when they consider this problem. First of all, it's weird to think of making money buying and selling cows, as it's not a typical investment vehicle (for non-farmers anyway). Second of all, people tend to get caught up on where we're getting the money from to make these purchases.
So I took care of those two problems by first of all assuming we have a bank account with $2000, and then tracking the credits and debits in that account as we made our purchases and sales; and second of all, by changing the cow to shares of stock in Cow Incorporated. By the end, we've bought and sold two shares of Cow Inc., and our bank account has $2400 compared to the original $2000, showing us that we really have made $400.
Now, in terms of convincing people this is true, there are definitely pitfalls. People might object that we're introducing information that wasn't in the original problem. But no matter what amount you begin with in the bank account, or even if you think of it as buying the cow on credit, you'll always end up at the end $400 ahead of where you started.
Today we have a textbook from 1937 with a problem set on ratios. These are harder than your average ratio problems, so we can be confident this was testing more advanced students. But we can also see a stark contrast to the context-ified problems of modern textbooks. There is a tiny bit of context about a man's salary, but outside of that, this is basically purely symbolic manipulation (with a bit of geometry thrown in as well).
You can see Catriona's original problem on her Twitter: twitter.com/Cshearer41/status/1281873084836970496/. As always, I love that it initially appears there can't possible be enough information, but as you start to chase some of the lengths, everything comes together.
What about you? I'm pretty happy with the method I came up with, but are there any shortcuts I missed?
Follow Tim Ricchuiti: TikTok: tiktok.com/@polymathematic Mathstodon: https://mathstodon.xyz/@polymathematic Instagram: instagram.com/polymathematicnet Reddit: reddit.com/user/polymath-matic Facebook: facebook.com/polymathematicWhy is it difficult to undo exponentiation? #RootsVsLogspolymathematic2024-02-16 | A viewer asked me recently to explain the difference between logs and roots. We can think of either logs or roots as, in some sense, "inverses" of exponentiation, but right off the bat, that's weird, because why would we need two different inverses? If we need to undo addition, we just subtract. If we need to undo multiplication, we just divide. But all of a sudden for exponential equations, we need two separate inverse operations?
This video is an excerpt from my longer answer, which is the related video above, and can be found on my channel @polymathematic . I hope you'll watch!
#RootsVsLogs #ExponentiationBasics #LogarithmsExplainedInverting Exponentiation? Why Logarithms and Roots Are a Bit Weird.polymathematic2024-02-16 | What's the difference between logarithms and roots, and what's more, why does there have to be a difference at all?
It's natural to presume that exponentiation should be "undo-able" in the same way that operations like addition and multiplication are, namely, with a single inverse that returns either of the original addends or factors. However, when we try to apply the inverses of exponentiation in this way, we run into problems. For example, if we don't know the base of an exponential equation like x^3=125, we can recover that base by taking the cube root of each side of the equation. But if instead it were the exponent that was missing, a root would not help us. In an equation like 5^y=125, we can't simply take the "y-th" root of each side, because it's precisely y that we do not know. Instead we need a different sort of inverse called a logarithm. Ultimately, the reason that exponentiation requires two different sorts of inverses, when operations like addition and multiplication do not, is that while addition and multiplication are commutative, exponentiation is not.
And my apologies to @3blue1brown for messing up his channel's name. I don't know what I was thinking!
In today's video, I'm tackling the fascinating equation "π^4 + π^5 ≈ e^6" and bringing it to life through geometry. I'll show you how I turn this abstract mathematical statement into a tangible geometric figure - almost a right triangle with sides of π^2, π^(2.5), and a hypotenuse eerily close to e^3. Using Pythagoras' theorem as my guide, I dive into the relationship between π and e, dissecting how these transcendental numbers interact in a geometric setting. We'll walk through the algebra needed to morph our initial equation into this geometric wonder, all the while understanding the precision and boundaries of such an approximation. It's a journey from the realm of algebra into the visual world of geometry, shedding light on the intriguing dance between π and e. With great thanks to John Cook who first introduced me to this approximation. You can find more of his work here: johndcook.com/blog/.
If you want to see more ridiculous math calculations, check out my channel @polymathematic !
Absolutely love the How Ridiculous guys, my boys and I watch every video (all the way to the end—hey, 44 club!), so when I saw Brett asking about the circumference of this absolutely ridiculous circular area of t-shirts, I knew I had to put this video together.
I went through several options on calculating the radius of the circle. I thought about trying to use a pendulum equation and time the swing of the paint around the circle, but I couldn't find a clean enough clip in the original video to be confident in the timing. There's another spot where Brett mentions there are 6000 items of merch making up the circle. I kind of got an okay answer out of assuming roughly two square feet (no metric conversion just yet) per item, but it seemed like an overestimate to me. So I went with a tried and true estimate the radius directly from a known object. If I underestimated your height, Scott, I apologize.
In today's video video, let's take a look at a definite integral that leads to an absolutely delightful evaluation: the natural log of the golden ratio, phi. It's a journey that starts with the integral of 1/sqrt(x^2+1) from 0 to 1/2.
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I begin by substituting x = tan u and dx = sec^2 u du. This allows us to simplify the integrand to sec u du using a pythagorean identity, and to shift the bounds of integration to 0 and the inverse tangent of 1/2. From there, we're simply integrating sec u du, which you can do by multiplying it by sec u + tan u over itself.
With another substitution, this allows us to integrate and evaluate ln | sec u + tan u | from the previous bounds. Our lower bound evaluates to ln 1, which is 0, and therefore which goes away. Our upper bound turns into ln (sec (tan^-1 (1/2) + tan (tan^-1 (1/2))). The tan and inverse tan cancel out leaving 1/2, and we can set up a right triangle to figure out that sec (tan^-1 (1/2)) is the same as √5/2.
The grand finale? This gives us back ln (1/2 + √5/2), which is of course ln φ, the natural log of the golden ratio.
I did not discover this integral myself, instead I saw David Meyer share his work along with some extensions at: davidmeyer.github.io/qc/golden_ratio.pdf. It turns out you can derive essentially any integral leading to the natural log of an algebraic number by manipulating the bounds and the fact that our original integral is another expression for hyperbolic sine, which has an alternate expression as ln(x + √(x^2 + 1)).
Today, we're diving into the unexpected intersection of logarithms and candy-making. Ever heard of dragon's beard candy? It's a mesmerizing process where the candy-maker starts with a single loop of candy and then doubles it over and over. By the time she's done, she's created over a million loops of the candy.
This got me thinking – what if we knew the number of loops and we wanted to work immediately backwards to the number of folds that it look to make them. To determine that, we need to be able to count, but not count like 1, 2, 3, 4, 5. Instead, we need to be able to count by powers of 2.
Fortunately, there is such a tool that will help us count that way, and that tool is the logarithm. The logarithm will take a number like 1,048,576 as an input and return which power that number is of a certain base. The base of the logarithm, in this case 2, is the same as the exponential base that would generate the count in the first place. So, for example, log base 2 of 4 is 2 because 4 is the second power of 2. Log base 2 of 32 is 5 because 32 is the fifth power of 2. And finally, log base 2 of 1,048,576 gives our 20 folds precisely because 1,048,576 is the 20th power of 2.
The essence of understanding logarithms lies in recognizing them as exponents in disguise. Here, we're essentially asking, "What power of 4 gives us a certain number?" The answer, in our case, is 4², which simplifies to 16.
Transforming this logarithmic expression into its exponential form, we get 16 = 3x + 1. It's now a linear equation, a straightforward path leading us to find the value of x.
Initially, the series resisted simplification, a common occurrence in mathematical problems. However, through persistence and analytical thinking, we discovered that this series could be transformed into a more approachable form. This breakthrough came by rewriting it as a telescoping series: the sum from 1 to infinity of 1/((3/2)^n - 1) - 1/((3/2)^n+1 - 1).
Telescoping series are interesting because they have a natural way of simplifying themselves. In this series, each term consists of two parts that effectively cancel out the adjoining terms. This characteristic simplifies the calculation significantly, leaving us with just the first term and the limit of the final term as n approaches infinity.
Upon computing these values, the sum of the series is revealed to be 2. This outcome is a testament to the elegance and simplicity that often lies beneath the surface of complex mathematical problems.
This exploration underscores the value of perseverance and analytical thinking in mathematics. Problems that seem impenetrable at first can often be simplified and solved with the right approach.
Thank you for joining me in this mathematical journey. If you find such explorations insightful, please consider subscribing for more content. Here, we delve into the fascinating world of mathematics, uncovering the beauty and simplicity hidden within complex concepts.
Until next time, keep exploring and appreciating the wonders of mathematics. Your support and curiosity are what make these explorations worthwhile.
Ever been confidently strolling through a math test, only to be ambushed by a question that seems to mock the very notion of 'easy'? Well, strap in, because today we're demystifying one of those classic moments with a touch of algebraic elegance. In our latest video, we decipher the cryptic message hidden within the lines and variables of what appears to be a teacher's sly challenge: Finding GF (no, not that GF, we're talking algebra here). With a straight line, two segments, and a sprinkle of x's, we'll unwrap the mystery of the lengths, revealing the subtle dance between numbers and letters. It's a journey from the simplicity of '7' to the enigma of '3x' and '9x+1', where the 'x' is not just a letter, but the key to unlocking the test's secret. So, forget about the heartache of high school relationships for a second, and let's find GF together in a way that would make Pythagoras swipe right.
#gfequation #mathmemes #oneminutemathSolving Japanese Puzzle Master Nob Yoshigahara’s Masterpiece #oneminutepuzzlespolymathematic2023-12-05 | Check out the main channel! @polymathematic
Ever wondered what it takes to solve a puzzle designed by a master? In today's video, we step into the numerical dojo of Nob Yoshigahara, a legend in the puzzle community, as we tackle his ingeniously crafted masterpiece. This isn't just a test of arithmetic; it's a dance of digits where only the most astute can lead. Join us for a brisk brain workout that will leave you pondering the elegance of numbers long after the minute is up.
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We start off by setting the stage with our compound interest formula: A=P(1+r)^t. Here, 'A' represents the final amount, 'P' is the principal amount, 'r' is the annual growth rate, and 't' is time in years. With A set at $10,000,000, P as $1,000, and t spanning a 60-year investment horizon, the mission is to unravel the mysterious rate 'r' that made this financial magic possible. Spoiler alert: it turns out to be an impressive 16.6% annual growth rate!
But that's not all. Let's also use this formula to figure out when this investment first hit the million-dollar mark. Keeping A and P consistent, and armed with our calculated growth rate of 0.166, it's time to play around with 't'. This part involves a bit of logarithmic fun, showing how math is not just numbers and equations, but a powerful tool to decode real-world financial phenomena.
So, whether you're a math geek, a finance aficionado, or just someone curious about how money grows over time, this video is tailor-made for you. Join me in unraveling these calculations and gain a deeper understanding of compound interest - the secret sauce behind many investment success stories!
In this video, we dive straight into an algebraic conundrum: finding x³ + y³ given x + y = 4 and xy = 5. We begin by directly addressing the sum, raising both sides of the equation x + y = 4 to the third power. This expansion gives us x³ + 3x²y + 3xy² + y³ = 64, a mix of cubic and quadratic terms. Our next step is crucial: we rearrange this equation to isolate x³ + y³. By factoring, we find ourselves with x³ + y³ + 3xy(x + y) = 64. Here's where our given values come into play. We substitute 5 for xy and 4 for x + y, turning the term 3xy(x + y) into 60. With this substitution, our equation simplifies to x³ + y³ + 60 = 64. Subtracting 60 from both sides, we find the elegant solution: x³ + y³ = 4. This video showcases how, by cleverly manipulating and substituting in algebra, we can find specific values without needing to know every variable. It's a straightforward yet insightful journey into the world of algebraic equations.
Hey everyone! In today's video, let's dive into a fun shortcut: finding remainders without actually calculating a full quotient. We'll look at the remainder left over when 6^3 + 4^2 * 2^1 is divided by 8.
The trick is that we can tell each of these terms is actually a multiple of 8. 6^3 must have 2^3 as a factor, since 2 is a factor of 6. Since 2^3 is 8, we have our first multiple of 8. 4^2 * 2^1 is a multiple of 8 because 4 * 2 is a multiple of 8. This is helpful because any multiples of 8 must sum to another multiple of 8. Therefore, our remainder must be zero.
In this video, I dive into a fascinating mathematical challenge: find the lone factorial that, when removed, transforms the product of the first 100 factorials into a perfect square. It's a puzzle that might sound daunting at first, but stick with me, and you'll see the beauty in its complexity.
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The approach I take begins not by wrestling with gigantic numbers, but by playing a clever game with exponents when we write the product out differently. We can write that giant factorial product as 1^100 * 2^99 * 3^98 and so on, all the way to 100^1. Here's where the fun begins: I notice that all the even powers (the perfect squares) can be set aside. They're already contributing nicely to our quest for a perfect square.
The real action lies in the odd powers, all of which are conveniently on even bases. For example, 2^99 has just one '2' too many to be a perfect square. Similarly, 4^97 has just one '4' too many. This pattern repeats all the way through, turning our problem into a quest to deal with these pesky even bases.
Since it's all the even numbers up to 100 that are causing those problems, we want to investigate different ways to represent the product of those even numbers. There is actually a related function called the double factorial that represents this: 100!! And that double factorial can then transformed into 2^50 times 50!. Since 2^50 is already a perfect square, our culprit in disturbing the perfect square peace is none other than 50!.
'What's the difference between 82.1% of 954 and 92.1% of 954?' is a great percentage problem that reveals percentages are often really multiplication, and multiplication is distributive. That is, it's not just a simple percentage calculation, but a lesson in distribution.
We begin by converting these percentages into their decimal counterparts and set up the equation: 0.921 * 954 - 0.821 * 954. The twist comes when we apply the distributive property, turning the equation into 954 * (0.921 - 0.821). That makes our final calculation 954 * (0.10), which means this difference really comes down to finding 10% of 954.
Problems like this are an excellent reminder that seemingly complex calculations can often have elegant simplifications.
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To start, I use a Desmos graph to illustrate that the integral's value remains consistent, even when we employ u-substitution, as long as we correctly adjust the bounds. You can play around with that graph yourself here: desmos.com/calculator/9daw49oohf.
The journey begins by breaking down cos³(x) dx into cos(x) * cos²(x) dx. This step simplifies the expression, paving the way for the next crucial step: applying the Pythagorean identity. Here, I rewrite cos²(x) as (1 - sin²(x)), a common trick in trigonometric integrations that further simplifies our integral.
Then comes the pivotal part: setting u = sin(x) for u-substitution. This transforms our integral into the form of ∫(1 - u²) du. However, the most critical aspect here is updating the integration bounds. Then we can substitute the original bounds (0 to π/2) into the equation u = sin(x), which gives us new bounds for the integral in terms of u.
One of the keys to approaching these problems is to build a good intuitive sense of what kinds of situations require constant *ratios*. These can be represented as direct variation questions. Both of the quantities exhibit the same relationship, either both increasing or both decreasing. Then you'll want to note scenarios that require constant *products*. These can be represented as inverse variation questions. One quantity grows while the other lessens correspondingly.
Joint variation occurs in scenarios where we have more than just two quantities we're comparing, and we can tell that different pairs of quantities exhibit different relationships when we hold the other constant.
Today, we're diving into a tricky little integral that popped up in the qualifying rounds of the Berkeley Integration Bee (HT @blackpenredpen for the problem). This is a classic example of how the hard part of calculus is almost never the calculus, but instead the algebra (and in this case a couple trig identities).
We start by dissecting the integrand using the difference of squares, a neat algebra trick. This transforms our expression into ((cos(x))^2 - (sin(x))^2)((cos(x))^2 + (sin(x))^2). The beauty lies in the simplification: the second part equals 1, thanks to the trusty Pythagorean identity. And the first part? It elegantly turns into cos(2x) with the double angle identity.
The crux of the solution lies in a simple u-substitution, steering us towards a surprisingly neat answer: 1/2. It's a reminder that sometimes, the most elegant solutions in mathematics come from stepping back and looking at the problem from a different angle.
Picture this: a traffic cone soaring through the air, only to encounter the swift, precise cut of a lightsaber. Every way the light saber can cut the traffic cone is a different species of conic section. But we don't necessarily have a great intuition for light sabers and traffic cones, so let's visualize this using the Desmos 3D calculator instead.
When we let a plane intersect a cone, each slice unveils one of the four conic sections – the circle, ellipse, parabola, and hyperbola.
But why should we care about these shapes? Conic sections are not mere abstract concepts; they are the very language in which the universe writes its laws. From the orbits of celestial bodies to the design of telescopes, these shapes are everywhere. Our exploration is more than just a mathematical exercise; it's a journey into the heart of the patterns that govern our world.
This video is an invitation to look beyond the mundane and find magic in mathematics. It's for the curious souls who find beauty in equations and for those who never thought a traffic cone and a lightsaber could reveal the secrets of the cosmos. So, join us in this unique adventure, where humor and science converge to uncover the elegance of conic sections.
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I think there is, and to explore that, let's start with a question: How many ways can you arrange nothing?
Normally, a factorial tells us how many ways we can arrange a set of numbers. Take 5! for example; it's a product of all positive integers up to 5 and symbolizes the permutations of a five-element set. Now, if we add any number from 2 to 5 to 5!, we get a sequence of composites, as each addition creates a number divisible by that added integer.
When we reach 0!, we hit a unique scenario—how do you arrange a set with nothing in it? Here's the twist: there is one way to do nothing. The empty set, though bare, has one representation of itself. Hence, by the very definition we've settled upon, we get 0! = 1. This isn't a random decision; it's a deliberate definition to keep the algebra consistent, especially when factorials play a crucial role in equations and expressions throughout mathematics.
This might raise an eyebrow or two. How can doing nothing result in something? It's a bit like the sound of one hand clapping, a contemplation that leads us down the philosophical paths of mathematics. In defining 0! as 1, we're not just playing by the rules; we're acknowledging that even in the realm of zeros and emptiness, there's a place for structure and order.
Don't forget to like, share, and discuss in the comments: What's your take on the concept of zero factorial? Let's talk about the empty set and the unity that starts the count of all things in math!
While multiplying numbers slightly less than 100 might seem daunting, there's a surprisingly straightforward trick that makes it a cinch. This video demystifies this math magic, making it accessible to everyone.
The secret lies in the representation of the numbers. Consider two integers just under 100, represented as (100 - a) and (100 - b). By multiplying these binomials, we tap into a simple yet powerful algebraic principle. This approach not only simplifies the multiplication process but also offers a deeper insight into the relationship between these numbers and their proximity to 100.
In this video, we break down each step of this method, demonstrating how easily you can apply it to everyday calculations. From (100 - a)(100 - b) to the final product, you'll see the magic unfold, turning a seemingly complex multiplication into a straightforward task.
But we don't just stop at how; we delve into the why. Understanding the underlying principles enhances your mathematical intuition and turns this trick into a valuable tool for quick and efficient calculations.
Whether you're a student, a teacher, or just a curious mind, this video will change the way you look at multiplication near 100. Embrace this trick and watch as numbers just below 100 multiply with surprising ease!
👉 Stay tuned for more insightful math tricks and tips. Don't forget to hit like, share, and subscribe for a regular dose of math magic. Happy calculating! 💡🎓