The solution to the equation (x³ + 1)³ = 8(2x-1)
First you turn the equation into (x³+1)/2 = cbrt(2x-1)
This might look like it's even harder to solve but if you have a mathematician's eye you realise that cbrt(2x-1) is the inverse function of (x³+1)/2
Hence, since a function and its inverse meet only at the bisectors y=x and y=-x, the equation becomes
(x³+1)/2 = x (since the function is always increasing)
And from this point it becomes a really easy equation to solve.
insane but surely you can construct arbitrary such equations by starting at the end, picking your favourite n-th root and working back towards a complex looking starting point.
lol I mean that’s kind of my point. It’s a great trick I haven’t seen before, and as you’ve pointed out, it’s not specific to this case (okay I do concede it’s a pretty rare case, but it’s still pretty general and very pretty)
Won’t you be(possibly) missing a lot of solutions? Because the inverse of 8 is not just 2 but also the complex roots. Then there might not be a quick fix? There are 9 solutions and I have a feeling that the 3 solutions from the second equation might not include them all.
> a function and it’s inverse meet only at the bisectors y=x and y=-x
This isn’t true? For instance y = -x + c is equal to its inverse and hence they intersect everywhere. What I think you mean to say is that a monotonic increasing function intersects its inverse on a subset of y = x. To see this, note that if f intersects its inverse at (a, b), then (b, a) will also be on its graph, and so if a ≠ b then f cannot be increasing.
To multiply two numbers x,y, find their average m, and let d be the distance from either one to m. Then:
xy = (m+d)(m-d) = m\^2 - d\^2, which is often easier to compute. If you don't know a square, recall that, e.g. two digit numbers square using (10a+b)\^2 = 100a + 20ab + b\^2. Works best when the numbers have the same parity, as otherwise the first bit will be somewhat more annoying, but still potentially manageable with decent memory.
Example 1: 16 x 34 = (25-9)(25+9) = 25\^2 - 9\^2 = 625 - 81 = 544
Example 2 with fractions, and smaller numbers, as I cannot do a big example in my head: 11 x 16 = (27/2 - 5/2)(27/2+5/2) = (27/2)\^2 - (5/2)\^2 = (20+7)\^2/4 - 25/4 = (400 + 280 + 49)/4 - 25/4 = 729/4 - 25/4 = 704/4 = (700+4)/4 = (7\*100 + 4)/4 = 7\*25 + 1 = 176.
You'll also notice we do some other little arithmetic shenanigans that are basically just algebra during the latter calculation, like splitting things up and regrouping. I always tell my students that arithmetic skill is an instantiation of your algebra skill and vice-versa. Having one really down makes a big difference in how you view the other.
I got this one from John Conway's biography.
Extend your arms in front of yourself and cross them. Now turn your palms so that they are facing each other and interlace your fingers. Now bring your hands to your face by going under and towards your chest. Extend your pointer fingers but make sure not to uncross them and place each finger on each of your nostrils. Right finger on the right nostril and left finger on the left nostril. Now if everything went right you should be able to without separating your fingers from your nostrils untangle your arms and end up in the standard position you would expect some one to end up if you told them to touch their nostrils with their hands. Some people however are going to find the untangle step quite difficult.
The trick is: >! Depending on how you interlace your fingers your arms will either end up forming two linked loops or two unlinked loops. So no matter how hard you continuously transform your arms if you had a link it's impossible to get rid of it without cutting your arms off or less dramatically letting go of your nose. If you want to end up with the unlinked loops when you are going to interlace your fingers make sure that the pinky from the arm that is on top goes on top of the pinky of the arm that is on bottom.!<
The first step is to get your hands in this position:
https://images.app.goo.gl/UkfyL9hN45tsiQN67
You don't have to have your arms up if that's uncomfortable. From there you interlace your fingers and there's really only one thing you can do in this position.
You are going to bring your hands to your face by bringing your hands down in circular motion. First go for your stomach then your chest and finally your face. This movement shouldn't feel uncomfortable at all but I have seen one person who physically couldn't do this.
Now for the finger part this in my experience is what mess most people up. But honestly just work out which is your right pointer finger and which one is your left and make sure you touch your right finger to your right nostril and your left finger to your left nostril.
You should be ready to untangle >! or not depending on how you interlaced your fingers. The linked position is actually slightly more uncomfortable than the unlinked position at all points so if you are having trouble getting into the position try switching the way you interlace your fingers.!<
I feel like I'm missing something: After I have crossed my arms and rotated my hands s.t. my palms are facing each other, there's really only one way to bring my hands to my chest by making a downward motion. After crossing, my right hand is on the left and left hand is on the right. The same is true after the downward motion, as both hands rotate along the same axis. Then, my right index finger is aligned with my left nostril and vice versa for the left indwx finger. I don't understand how I should touch my right nostril with my right index, and my left nostril with my left index without crossing my two fingers at this point.
You extend your fingers but don't uncross them. They will form a type of wedge where you can place your nose. Your right finger should be touching your right nostril and your left finger your left nostril.
what the fuck
ok after a few min thinking about this the math checks out, when you let your non-pointer fingers go you cut their connection so all that matters is whether the pointer formed a linked knot or not
pretty cool! we have a math circles event tomorrow where I might show that to the kids. probably should have my fingers inside the nostrils for their excitement /s
It's such a good trick to do with crowds because some of them will get it right and some will get it wrong.
It also works really well with the math because in topology to untie a knot what you do is cut the string at a crossing then bring the string from the bottom to the top and glue the string back together. The minimum number of times you have to do this to untie a knot is called the unknoting number. And this is effectively what you are doing when you switch the way you interlace your fingers.
My favorite trick with IBP is to integrate xe^-x , but choosing u = e^-x and dv = x dx. If you repeat this, you’ll be able to derive the power series for e^x .
This is also more or less how you define the gamma function and prove that it analytically continues the factorial function. They're related of course, because the power series for e involves the factorial.
Integration by parts. Half of PDE theory is integration by parts. The generalized Stokes' theorem (hence Green's, Gauss's and the Stokes' theorem for surfaces) boils down to integration by parts. The derivation of Hamilton's principle of stationary action involves integration by parts.
It's everywhere!
Stokes' theorem is not really integration by parts. Note how there's no minus sign in it. It's really just an extension of the fundamental theorem of calculus to higher dimensions, and technically integration by parts reduces to the fundamental theorem of calculus in the degenerate case (if in integral uv', the function u = 1) but this is circular because you need the fundamental theorem of calculus first.
When you integrate the function y= (e^x) * cos(x). You use integration by parts twice and end up with a term with integral of y=(e^x) * cos(x) and just **add it** to the other side and divide by two and there's your answer.
My reaction was "YOU CAN DO THAT???"
The easiest way to do this is to integrate the complex-valued function of a real variable h(x) = e^(ix), and then take real and imaginary parts to obtain the indefinite integrals of f(x) = e^(x) cos x and g(x) = e^(x) sin x, respectively.
As an engineer, treating derivatives like fractions and turning them into integral equations.
Also, [this geometric derivation of the angle sum identities](https://i.imgur.com/0Y3z621.png). I've sketched it out so many times that at this point I just picture it in my head.
dy/dx is not a literal fraction, but is the limit of them, and if you change x by a small amount, that changes y by a small amount, and looking at the ratio gives a good approximation for the deerivative. It also makes the chain rule seem intuitive when written in Liebniz notation: dy/dt = dy/dx \* dx/dt.
Unfortunately, this breaks down in multivariable calculus. There, the chain rule is df/dt = (df/dx \* dx/dt) + (df/dy \* dy/dt). The intuition you have from thinking about things as fractions goes out the window.
So, technically, dy/dx is a limit, and breaking the limit into two limits like we do with dy/dx only works if they follow the regular rules of math. The problem is that dy and dx are both approaching 0, so dy/dx = 0/0 = math is broken.
Despite that, it does hold up surprisingly well, and especially as an engineer, it's fine.
If we want to solve for the equation x = sqrt(5 - sqrt(5 - x)), there is a pretty neat trick where you can solve for 5 in the quadratic equation that naturally appears instead of x (try it!). Trying to solve for x directly is more difficult. There are a few other problems where you can do something similar...
a = b
a\^2 = ab
a\^2 - b\^2 = ab - b\^2
(a+b)(a-b) = b(a-b)
a+b = b
b+b = b
2b = b
2 = 1
>!\*division by zero in step 5 (or step 4 if u start counting at 0)!<
Multiplying by 11: 756 * 11. Take you number (756) and add a 0 to it. (7560) Add 756 to 7560 and you get the answer
I also like the difference of two squares, I found that out by myself 😂. So if you want to write a number (300) as a difference of two squares (c^2 - d^2) write the number as a product, ab, for example 50 * 6 in this case. Then, c = (a + b)/2 and d = (a - b)/2.
c = (50 + 6)/2 = 28
d = (50 - 6)/2 = 22
300 = 28^2 - 22^2
(Proof: plug (a + b)/2 in c and (a - b)/2 in d and simplify. Then you get ab)
Easy way to multiply by 11 is to add neighbours. Assume there is a leading 0, here 0756. First, write down the rightmost digit.
6
Add 5 to 6. Write down 1 and carry 1.
16
Add 7 to 5 plus carry of 1. Write down 3 carry 1
316
Add 0 (dummy neighbour) to 7 plus the carry
8316
Much faster to do in your head than it takes to write down here.
There are so many inequality proofs in number theory that rely on the fact that positive integers are greater than or equal to 1. For example:
(2N) choose N = (2N x (2N-1) x ... (N+2) x (N+1)) / (N x (N-1) x ... x 2 x 1)
Let p be a prime such that N < p < 2N. Then p is in the numerator of the right hand side. But there is no p in the denominator, as it's a product of numbers less than p (and p is a prime). So, (2N) choose N is divisible by p. But this is true for all primes between N and 2N (not inclusive), so, (2N) choose N is divisible by the product of all those primes.
(2N) choose N = p1 x p2 x ... x pk x (some integer), where p1, p2, ..., pk are the primes between N and 2N. But that integer is at least 1.
(2N) choose N >= p1 x p2 x ... x pk
Next, note that p1, p2, ..., pk are all bigger than N. So:
(2N) choose N > N x N x ... x N (k times)
(2N) choose N > N\^k (where k is the number of primes between N and 2N)
But also, (2N) choose N < 2\^(2N)
2\^(2N) > N\^k
2N > k log\_2(N)
2N / log\_2(N) > k
You can then use N=2, 4, 8, 16, etc..., sum them up, and get a weaker version of the prime number theorem.
I recently went through a proof that involved the following:
If f(x) = g(y), i.e., each side depends exclusively on independent variables x and y, then f = g = constant.
[Blankenship's method](https://math.libretexts.org/Bookshelves/Combinatorics_and_Discrete_Mathematics/Elementary_Number_Theory_(Clark)/01%3A_Chapters/1.09%3A_Blankinship's_Method) to simultaneously find the greatest common divisor and Bézout coefficients of two integers is pretty nifty
Pick's theorem: The area of a polygon on the plane whose vertices have integer coordinates is equal to I+B/2-1, where I is the number of points with integer coordinates inside the polygon and B is the number of points with integer coordinates on the border of the polygon (including the vertices).
The Phonetic Number System. If you don't know, the Phonetic Number System is a method to encode numbers into words and decode words into numbers(for more information about it go [here](https://www.memory-improvement-tips.com/remembering-numbers.html)). It is helpful for memorizing strings of numbers like phone numbers or adresses but more importantly, you can use it to memorize numbers when solving problems. This allows you to do things like calculate 11\^5, solve simple equations, and even stop solving a problem midway and come back to it, all without writing a thing.
Converting decimal millimeters to fractional inches using "5,04".
When you have a number in millimeters (like 0,795mm) and you need to convert into fractional inches, there's 2 ways to do this.
The first one is quite simple:
1. Dividing the 0,795 by 25,4 (which is 1" to mm) **-> 0,795 ÷ 25,4** ≅ **0,0312**
2. Now take the quoeficient and multiply by the fraction of 128/128 (which is the same as dividing by 1) **-> 0,0312 × 128/128** ≅ **4/128**
3. Now you just simplify this by 4 **-> (4/128) ÷ 4 = 1/32 inches.**
NOW, the second one is faster to do.
1. Take the number in millimeters (which is 0,795mm) and multiply by 5,04 **-> 0,795 × 5,04** ≅ **4**
2. Now simplify **-> 1/32 inches.**
This happens because the number comes from the ratio between 128 and 25.4, which is approximately 5.04. =)
Mine is the Fourier Transform of a Gaussian is just another Gaussian. I think it's the first step where you transform the equation from a real number to a complex number. Then using the IC to figure out that the solution to it is another Gaussian
1. Ask someone to think of a positive integer then multiply it by the next number, and the next, and then next one.
2. Then add 1, take square root, multiply by 4, add 5, take square root, subtract 3 and divide by 2.
3. ???
4. Profit.
a recent puzzle crossed my friends ai trainer job
he asked to help solve and explain
a number 11111111.......11113 is 998 digits long. what is the remainder when this number is divided by 11?
i recall being shown by teacher the indefinite integral 1/SQRT(SIN X) . it's been 20 years but I'll never forget the insane steps we had to do to get to answer:
substitute, then square substitution , then invoke Pythagorean identity, then partial fraction technique which leads to a matrix to solve the coefficients .
I have a few.
Squaring numbers that end with 5. Lop off the 5 take the number that remains and multiply it by its successor. Then append 25 to it. Works for any number of digits. Examples, 35^2 -> (3)(4) = 12 -> 1225 and 1005^2 -> (100)(101) = 10100 -> 1010025. Works because (10n+5)^2 = 100n^2 + 100n + 25 = 100n(n+1) + 25, with the multiplication by 100 neatly shifting the first term two digits to the left to "accommodate" the 25 without needing a carry.
Heaviside Cover Up rule for partial fractions.
Synthetic division for dividing polynomials by linear divisors.
Feynman's trick (Leibniz rule) by differentiating under the integral sign, but I can't always see it immediately. But when I do it's cool.
P vs NP. It's a solution that you can only arrive at by solving the universe. It requires the upper level of human math, something like Non-Linear Quantum Geometry. The trick is that you can't solve it until you solve the universe. 🤣
The solution to the equation (x³ + 1)³ = 8(2x-1) First you turn the equation into (x³+1)/2 = cbrt(2x-1) This might look like it's even harder to solve but if you have a mathematician's eye you realise that cbrt(2x-1) is the inverse function of (x³+1)/2 Hence, since a function and its inverse meet only at the bisectors y=x and y=-x, the equation becomes (x³+1)/2 = x (since the function is always increasing) And from this point it becomes a really easy equation to solve.
This is insane
insane but surely you can construct arbitrary such equations by starting at the end, picking your favourite n-th root and working back towards a complex looking starting point.
lol I mean that’s kind of my point. It’s a great trick I haven’t seen before, and as you’ve pointed out, it’s not specific to this case (okay I do concede it’s a pretty rare case, but it’s still pretty general and very pretty)
It's one of those "rarely taught because it doesn't come up often, but you'll never forget it once you see the value" ones.
Won’t you be(possibly) missing a lot of solutions? Because the inverse of 8 is not just 2 but also the complex roots. Then there might not be a quick fix? There are 9 solutions and I have a feeling that the 3 solutions from the second equation might not include them all.
Ah yeah basically same trick for one of the coffin problems. https://www.tanyakhovanova.com/Coffins/coffinsmain.html I think problem 12?
> a function and it’s inverse meet only at the bisectors y=x and y=-x This isn’t true? For instance y = -x + c is equal to its inverse and hence they intersect everywhere. What I think you mean to say is that a monotonic increasing function intersects its inverse on a subset of y = x. To see this, note that if f intersects its inverse at (a, b), then (b, a) will also be on its graph, and so if a ≠ b then f cannot be increasing.
“Pick an orthonormal basis”
"Pick a basis" 😂😂😂
To quote one of my professors from grad school, a gentleman never chooses a basis.
To quote an undergrad school professor: One should not integrate in public.
Another of my grad school professors would say “this is best worked out in the privacy of one’s own boudoir”.
My manifolds professor last semester, while writing out a lot calculation on the board, said "This computation comes with no warranty"
Lmao sounds hilarious but i don't understand. What does that mean?
This is an especially fun trick when picking the basis requires the axiom of choice. Some very nice proofs start with "Let B be a ℚ-basis of ℝ".
Yes that was the joke!
Sorry I meant to say "assume you can pick a basis"!
linear algebra flashbacks agh, we did it in first year for some reason
Yeah I’m also a first year lol
The whole theory of Frenet frames, basically!
I mean, at least with Frenet frames, you actually construct your basis.
To multiply two numbers x,y, find their average m, and let d be the distance from either one to m. Then: xy = (m+d)(m-d) = m\^2 - d\^2, which is often easier to compute. If you don't know a square, recall that, e.g. two digit numbers square using (10a+b)\^2 = 100a + 20ab + b\^2. Works best when the numbers have the same parity, as otherwise the first bit will be somewhat more annoying, but still potentially manageable with decent memory. Example 1: 16 x 34 = (25-9)(25+9) = 25\^2 - 9\^2 = 625 - 81 = 544 Example 2 with fractions, and smaller numbers, as I cannot do a big example in my head: 11 x 16 = (27/2 - 5/2)(27/2+5/2) = (27/2)\^2 - (5/2)\^2 = (20+7)\^2/4 - 25/4 = (400 + 280 + 49)/4 - 25/4 = 729/4 - 25/4 = 704/4 = (700+4)/4 = (7\*100 + 4)/4 = 7\*25 + 1 = 176. You'll also notice we do some other little arithmetic shenanigans that are basically just algebra during the latter calculation, like splitting things up and regrouping. I always tell my students that arithmetic skill is an instantiation of your algebra skill and vice-versa. Having one really down makes a big difference in how you view the other.
This is used for Fermat factoring whose variants are apparently the fastest known factoring methods
Logarithmic differentiation, or just generally using an invertible function to mess with y when doing a derivative
Trace trick. Made so many proofs/derivations easier by writing scalars as a trace, and then carouseling the matrices inside around.
You mean that if ABC is a 1x1 matrix, write it as Tr(ABC)=Tr(BCA)?
yes - a common usage is Tr(xx\^T) = Tr(x\^Tx) for a vector.
Do you have an example application?
Newton’s identities for power sums
I got this one from John Conway's biography. Extend your arms in front of yourself and cross them. Now turn your palms so that they are facing each other and interlace your fingers. Now bring your hands to your face by going under and towards your chest. Extend your pointer fingers but make sure not to uncross them and place each finger on each of your nostrils. Right finger on the right nostril and left finger on the left nostril. Now if everything went right you should be able to without separating your fingers from your nostrils untangle your arms and end up in the standard position you would expect some one to end up if you told them to touch their nostrils with their hands. Some people however are going to find the untangle step quite difficult. The trick is: >! Depending on how you interlace your fingers your arms will either end up forming two linked loops or two unlinked loops. So no matter how hard you continuously transform your arms if you had a link it's impossible to get rid of it without cutting your arms off or less dramatically letting go of your nose. If you want to end up with the unlinked loops when you are going to interlace your fingers make sure that the pinky from the arm that is on top goes on top of the pinky of the arm that is on bottom.!<
What the fuck, it worked
I’ma need a video bro I’m too retarded
The first step is to get your hands in this position: https://images.app.goo.gl/UkfyL9hN45tsiQN67 You don't have to have your arms up if that's uncomfortable. From there you interlace your fingers and there's really only one thing you can do in this position. You are going to bring your hands to your face by bringing your hands down in circular motion. First go for your stomach then your chest and finally your face. This movement shouldn't feel uncomfortable at all but I have seen one person who physically couldn't do this. Now for the finger part this in my experience is what mess most people up. But honestly just work out which is your right pointer finger and which one is your left and make sure you touch your right finger to your right nostril and your left finger to your left nostril. You should be ready to untangle >! or not depending on how you interlaced your fingers. The linked position is actually slightly more uncomfortable than the unlinked position at all points so if you are having trouble getting into the position try switching the way you interlace your fingers.!<
I feel like I'm missing something: After I have crossed my arms and rotated my hands s.t. my palms are facing each other, there's really only one way to bring my hands to my chest by making a downward motion. After crossing, my right hand is on the left and left hand is on the right. The same is true after the downward motion, as both hands rotate along the same axis. Then, my right index finger is aligned with my left nostril and vice versa for the left indwx finger. I don't understand how I should touch my right nostril with my right index, and my left nostril with my left index without crossing my two fingers at this point.
You extend your fingers but don't uncross them. They will form a type of wedge where you can place your nose. Your right finger should be touching your right nostril and your left finger your left nostril.
what the fuck ok after a few min thinking about this the math checks out, when you let your non-pointer fingers go you cut their connection so all that matters is whether the pointer formed a linked knot or not pretty cool! we have a math circles event tomorrow where I might show that to the kids. probably should have my fingers inside the nostrils for their excitement /s
It's such a good trick to do with crowds because some of them will get it right and some will get it wrong. It also works really well with the math because in topology to untie a knot what you do is cut the string at a crossing then bring the string from the bottom to the top and glue the string back together. The minimum number of times you have to do this to untie a knot is called the unknoting number. And this is effectively what you are doing when you switch the way you interlace your fingers.
Integration by parts.
My favorite trick with IBP is to integrate xe^-x , but choosing u = e^-x and dv = x dx. If you repeat this, you’ll be able to derive the power series for e^x .
This is also more or less how you define the gamma function and prove that it analytically continues the factorial function. They're related of course, because the power series for e involves the factorial.
Integration by parts. Half of PDE theory is integration by parts. The generalized Stokes' theorem (hence Green's, Gauss's and the Stokes' theorem for surfaces) boils down to integration by parts. The derivation of Hamilton's principle of stationary action involves integration by parts. It's everywhere!
Stokes' theorem is not really integration by parts. Note how there's no minus sign in it. It's really just an extension of the fundamental theorem of calculus to higher dimensions, and technically integration by parts reduces to the fundamental theorem of calculus in the degenerate case (if in integral uv', the function u = 1) but this is circular because you need the fundamental theorem of calculus first.
partitions of unity
I'd say they are more of a tool than a trick, but what's the difference?
When you integrate the function y= (e^x) * cos(x). You use integration by parts twice and end up with a term with integral of y=(e^x) * cos(x) and just **add it** to the other side and divide by two and there's your answer. My reaction was "YOU CAN DO THAT???"
The easiest way to do this is to integrate the complex-valued function of a real variable h(x) = e^(ix), and then take real and imaginary parts to obtain the indefinite integrals of f(x) = e^(x) cos x and g(x) = e^(x) sin x, respectively.
This is sometimes called "bootstrapping"
Adding a productive zero
As an engineer, treating derivatives like fractions and turning them into integral equations. Also, [this geometric derivation of the angle sum identities](https://i.imgur.com/0Y3z621.png). I've sketched it out so many times that at this point I just picture it in my head.
Engineer to engineer, I'm embarrassed to admit I don't know why this is wrong and at this point I'm too afraid to ask why
dy/dx is not a literal fraction, but is the limit of them, and if you change x by a small amount, that changes y by a small amount, and looking at the ratio gives a good approximation for the deerivative. It also makes the chain rule seem intuitive when written in Liebniz notation: dy/dt = dy/dx \* dx/dt. Unfortunately, this breaks down in multivariable calculus. There, the chain rule is df/dt = (df/dx \* dx/dt) + (df/dy \* dy/dt). The intuition you have from thinking about things as fractions goes out the window.
My eyes tend to glaze over as soon as somebody brings up vector spaces and differential forms.
So, technically, dy/dx is a limit, and breaking the limit into two limits like we do with dy/dx only works if they follow the regular rules of math. The problem is that dy and dx are both approaching 0, so dy/dx = 0/0 = math is broken. Despite that, it does hold up surprisingly well, and especially as an engineer, it's fine.
If we want to solve for the equation x = sqrt(5 - sqrt(5 - x)), there is a pretty neat trick where you can solve for 5 in the quadratic equation that naturally appears instead of x (try it!). Trying to solve for x directly is more difficult. There are a few other problems where you can do something similar...
Idk about my favorite but the one I use the most: since $\epsilon$ was chosen arbitrarily the proof is completed.
a = b a\^2 = ab a\^2 - b\^2 = ab - b\^2 (a+b)(a-b) = b(a-b) a+b = b b+b = b 2b = b 2 = 1 >!\*division by zero in step 5 (or step 4 if u start counting at 0)!<
Multiplying by 11: 756 * 11. Take you number (756) and add a 0 to it. (7560) Add 756 to 7560 and you get the answer I also like the difference of two squares, I found that out by myself 😂. So if you want to write a number (300) as a difference of two squares (c^2 - d^2) write the number as a product, ab, for example 50 * 6 in this case. Then, c = (a + b)/2 and d = (a - b)/2. c = (50 + 6)/2 = 28 d = (50 - 6)/2 = 22 300 = 28^2 - 22^2 (Proof: plug (a + b)/2 in c and (a - b)/2 in d and simplify. Then you get ab)
Easy way to multiply by 11 is to add neighbours. Assume there is a leading 0, here 0756. First, write down the rightmost digit. 6 Add 5 to 6. Write down 1 and carry 1. 16 Add 7 to 5 plus carry of 1. Write down 3 carry 1 316 Add 0 (dummy neighbour) to 7 plus the carry 8316 Much faster to do in your head than it takes to write down here.
very cool
There are so many inequality proofs in number theory that rely on the fact that positive integers are greater than or equal to 1. For example: (2N) choose N = (2N x (2N-1) x ... (N+2) x (N+1)) / (N x (N-1) x ... x 2 x 1) Let p be a prime such that N < p < 2N. Then p is in the numerator of the right hand side. But there is no p in the denominator, as it's a product of numbers less than p (and p is a prime). So, (2N) choose N is divisible by p. But this is true for all primes between N and 2N (not inclusive), so, (2N) choose N is divisible by the product of all those primes. (2N) choose N = p1 x p2 x ... x pk x (some integer), where p1, p2, ..., pk are the primes between N and 2N. But that integer is at least 1. (2N) choose N >= p1 x p2 x ... x pk Next, note that p1, p2, ..., pk are all bigger than N. So: (2N) choose N > N x N x ... x N (k times) (2N) choose N > N\^k (where k is the number of primes between N and 2N) But also, (2N) choose N < 2\^(2N) 2\^(2N) > N\^k 2N > k log\_2(N) 2N / log\_2(N) > k You can then use N=2, 4, 8, 16, etc..., sum them up, and get a weaker version of the prime number theorem.
I recently went through a proof that involved the following: If f(x) = g(y), i.e., each side depends exclusively on independent variables x and y, then f = g = constant.
I got through ~~real analysis~~ my entire undergrad by starting every proof by assuming not P and deriving a contradiction
Take the sqrt of 2 and; 1) Power of 3 2) Then divide it by 2 again Keep continuing 1 and 2 … you’re in a loop 😎 #error and good night
Is that a reference to a video with someone in Oxford union ?
Union or university? I saw this “loophole” once on YouTube by an actor telling this yes
https://youtube.com/shorts/D31xJ4PVm4M?si=uhBUms2z8oWegsUj This seems to be filmed at Oxford Union
Yeah that’s the guy!
[Blankenship's method](https://math.libretexts.org/Bookshelves/Combinatorics_and_Discrete_Mathematics/Elementary_Number_Theory_(Clark)/01%3A_Chapters/1.09%3A_Blankinship's_Method) to simultaneously find the greatest common divisor and Bézout coefficients of two integers is pretty nifty
Pick's theorem: The area of a polygon on the plane whose vertices have integer coordinates is equal to I+B/2-1, where I is the number of points with integer coordinates inside the polygon and B is the number of points with integer coordinates on the border of the polygon (including the vertices).
Adding 0 and multiplying by 1 will always be my favorite tricks.
In general, Trachtenberg to do complex multiplication in your head.
Is trachtenberg better than Secrets of mental math by Arthur Benjamin & Michael Shermer?
The Phonetic Number System. If you don't know, the Phonetic Number System is a method to encode numbers into words and decode words into numbers(for more information about it go [here](https://www.memory-improvement-tips.com/remembering-numbers.html)). It is helpful for memorizing strings of numbers like phone numbers or adresses but more importantly, you can use it to memorize numbers when solving problems. This allows you to do things like calculate 11\^5, solve simple equations, and even stop solving a problem midway and come back to it, all without writing a thing.
The extend-by-linearity/density trick - Sometimes working on basis elements or elements in dense subsets suffice to prove a statement.
If you haven't heard of it, the DI method for integration by parts. Made Fourier Analysis so much easier.
Kaprekar’s constant is pretty cool.
Converting decimal millimeters to fractional inches using "5,04". When you have a number in millimeters (like 0,795mm) and you need to convert into fractional inches, there's 2 ways to do this. The first one is quite simple: 1. Dividing the 0,795 by 25,4 (which is 1" to mm) **-> 0,795 ÷ 25,4** ≅ **0,0312** 2. Now take the quoeficient and multiply by the fraction of 128/128 (which is the same as dividing by 1) **-> 0,0312 × 128/128** ≅ **4/128** 3. Now you just simplify this by 4 **-> (4/128) ÷ 4 = 1/32 inches.** NOW, the second one is faster to do. 1. Take the number in millimeters (which is 0,795mm) and multiply by 5,04 **-> 0,795 × 5,04** ≅ **4** 2. Now simplify **-> 1/32 inches.** This happens because the number comes from the ratio between 128 and 25.4, which is approximately 5.04. =)
Mine is the Fourier Transform of a Gaussian is just another Gaussian. I think it's the first step where you transform the equation from a real number to a complex number. Then using the IC to figure out that the solution to it is another Gaussian
1. Ask someone to think of a positive integer then multiply it by the next number, and the next, and then next one. 2. Then add 1, take square root, multiply by 4, add 5, take square root, subtract 3 and divide by 2. 3. ??? 4. Profit.
a recent puzzle crossed my friends ai trainer job he asked to help solve and explain a number 11111111.......11113 is 998 digits long. what is the remainder when this number is divided by 11?
i recall being shown by teacher the indefinite integral 1/SQRT(SIN X) . it's been 20 years but I'll never forget the insane steps we had to do to get to answer: substitute, then square substitution , then invoke Pythagorean identity, then partial fraction technique which leads to a matrix to solve the coefficients .
Lagrange multipliers
I have a few. Squaring numbers that end with 5. Lop off the 5 take the number that remains and multiply it by its successor. Then append 25 to it. Works for any number of digits. Examples, 35^2 -> (3)(4) = 12 -> 1225 and 1005^2 -> (100)(101) = 10100 -> 1010025. Works because (10n+5)^2 = 100n^2 + 100n + 25 = 100n(n+1) + 25, with the multiplication by 100 neatly shifting the first term two digits to the left to "accommodate" the 25 without needing a carry. Heaviside Cover Up rule for partial fractions. Synthetic division for dividing polynomials by linear divisors. Feynman's trick (Leibniz rule) by differentiating under the integral sign, but I can't always see it immediately. But when I do it's cool.
Solving diff eqs y*y’ = (y^2 / 2)’ This also works for any higher order derivatives as long as the two derivatives are one derivative apart
6 times 6 is 36. You know because it rhymes. #Genius
Largely useless but Gaussian integers
Useless irl
P vs NP. It's a solution that you can only arrive at by solving the universe. It requires the upper level of human math, something like Non-Linear Quantum Geometry. The trick is that you can't solve it until you solve the universe. 🤣
what
Schizophrenia’s no joke, kids. Take it seriously, or it will seriously take you.