Your number theory videos are my favourites! Cool techniques used here!

Once you understand how multiplication of complex numbers is a way to say "similar", the conformal property of holomorphic functions is obvious.

Using the techniques used to bound a, and c<1, we can bound c to [(sqrt(5)-1)/2,1). Thus we can choose the lower bound, which is the golden ratio. So then we need 2sqrt(2)+sqrt(5)/2-3/2 <=5/2, or 4sqrt(2)+sqrt(5) <= 8.

Is this similar to Part (3) of Problem 723 of Ramanujan’s 3rd notebook: floor(sqrt(n)+sqrt(n+1))=floor(sqrt(4n+2))?

Beautiful problem and solution!

Using the well-known inequalities between geometric, arithmetic and squared means for not equal a, b, c: cbrt(abc) < (a+b+c)/3 < sqrt((a^2+b^2+c^2)/3) /cbrt stands for cube root/ for sqrt(n-1), sqrt(n), sqrt(n+1), we get 3*(cbrt(sqrt((n^3-n)) < sqrt(n-1)+sqrt(n)+sqrt(n+1) < 3*sqrt((n-1+n+n+1)/3) = sqrt(9n). For n>=4, it's easy to see that sqrt(9n-1) < 3*(cbrt(sqrt((n^3-n)), or equivalently (9n-1)^3 < 3^6(n^3-n): it leads to the quadratic 0 < (9n)^2 - 28*9n + 1/3 = (9n-14)^2 - 14^2 + 1/3, clearly true for n>=4. So for n>=4, sqrt(9n-1) < sqrt(n-1)+sqrt(n)+sqrt(n+1), and the same is easy to check "manually" for n=2, 3. Replacing n by n+1, we get the two desired inequalities: sqrt(9n+8) < sqrt(n)+sqrt(n+1)+sqrt(n+2) < sqrt(9n+9) implying the identity in question. Q.E.D.

Where were you 15 years ago when I was struggling with this lol

Your videos are wonderful <3 Do you have a big whiteboard in your home for recording? I'm a little envious if that's the case...

Beautiful solution, I always love inequality work.

In the second case you came up with a joint distribution, which is not uniform anymore. Does this make sense, when assuming, picking a point uniformly? For me the joint distribution has to have a density function which is constant not depending on any variable to have a equal weighting of each point.

I know that this isn't totally relevant because we are accepting all positive real values for n, but I thought it was interesting to note that 9n+8 can never be a aquare number for any integer valuenof n. 0,1,4,7 are the only square numbers mod 9

Great problem! Heer's a question: 3 sqrt(n) < sqrt(n) + sqrt(n+1) + sqrt(n+2) is true for all natural numbers; and sqrt(9n+8) < 3 sqrt(n+1) for all natural numbers. So why can't you just compare floor(3 sqrt(n)) < = floor(3 sqrt(n+1)), which is true for all natural numbers; hence, the stated problem is true for all natural numbers.

How was this problem come up with? The '9' coefficient makes since (1+1+1)^2 = 9, but the constant at the end seems tricky to pinpoint, and I assume there are other non-integer constants which still satisfy the inequality. Is it posssible to generalise?

I used desmos to graph the three functions anx actually the equation holds for x>=0.6782 whitch is the lower bound value that has been found. Great video as always.

It’s Friday already ???

always love seeing your video, especially ones about floor function.

Let _x = n(n + 1)(n + 2)(n + 3)_ Let _m = n + ³/₂_ ∴ _x = (m - ³/₂)(m + ³/₂)(m - ½)(m + ½)_ = _(m² - ⁹/₄)(m² - ¼)_ Let _t = m² - ⁵/₄_ ∴ _x = (t - 1)(t + 1)_ ⇒ _x = t² - 1_ Since _x ∈ ℕ_ then _t ∊ ℤ_ Suppose that for some integer _a:_ _x = a²_ ∴ _t² - a² = 1_ ⇒ _(t - a)(t + a) = 1_ ⇒ _t - a = t + a = ±1_ ⇒ _a = 0_ ⇒ _x ∉ ℕ_ ⇒ *_x ≠ a²_*

Nice. Very elegant elementary proof and one that is easy to remember.

I came up wtih this during my first semester at USM, despite not having done dual enrollment. I was supposed to go to the University of Michigan PhD program in mathematics.

If you're just looking for a way to derive the angle sum formulas for sin and cos, Euler's e^iθ = cosθ + i*sinθ provides a much more compact way of doing so, with fairly straightforward algebra.

Beautiful ❤

Goodness. All you need to know is that if x between 1 to 10, then x^5 and x have the same unit's digit: ie x^5 = x mod 10. Of course, it's clearly true for all x. now modulo 10: m^17 * n =( m^5)^3 * m^2 * n = m^3 * m^2 * n = m^5 * n = m * n exchanging m, n and taking the difference you get the result.

That's good, but I didn't know the starting formula for the area: half the sine of the angle times the product of the two adjacent sides. That's news to me. Where does that come from?

👍

You changed the problem statement in mid-discussion. At 4:30, "we've run into a problem" because the smallest sequence WITH NO REPEATS gives too large and answer. However, at 1:02, you state that repeats are allowed (2x2 x 3). Which rule do we follow? At 4:30, shouldn't we add 2x2 x 3 x 4 x 6 x ... 31 (value about 80 billion) to the list? How about 2x2x2x2 x 3 x 5 x 7 x . . . x 30 (value about 45 billion)? By your original definition of the problem, there ARE infinitely many such numbers, since we can always take out an appropriate number of the the largest factors in the sequential list and replace them with enough 2's or 3' or 5's or 7's to get us to the right order of magnitude.

This is much messier than stacking 2 right triangles, having hypotenuse of the first be also the long leg of the second. The stacking scheme relies only upon the definition of sine is opposite over hypotenuse, cosine is adjacent over hypotenuse. Other than that only simple geometry (similar triangles) is required. Using the sine triangle area rule is itself more complicated than the sine of angle sum. It also can't operate in other than the 1st quadrant.

Well ... No. This is not the best diagram. Also, the notation used is not good.

N e a t !

Nice I didn't want to learn all these (if you have other techniques for the other formulas I'm interested)

Your proofs are (of course) valid, but the wrong way round. We prove the area of a triangle 📐 theorem using the sin(a+b) rule. Likewise, the cosine rule comes way later in the study of trigonometry. Though far more difficult, the standard way of proving them using the "double triangle" and distance methods from first principles is more satisfying.

Nice

The best proof for angle sum identities is just a complex multiplication… note that i said proof not diagram

Lovely proof. Good approach to those who find it hard to remember the traditional methods.

I could derive the first result from your figure without resorting to the formula 1/2*a*b*sin(\theta+\phi) by using the law of sines instead: sin(\theta+\phi)/(c+d)=sin(90°-\theta)/a=h/(a*b) Multiplying both sides with (c+d) yields: sin(\theta+\phi)/(c+d)=h*c/(a*b) + h*d/(a*b)=sin(\theta)*cos(\phi)+cos(\theta)*sin(\phi).

You did not derive that Area of triangle =0.5×abSin(Theta+Phi)

This is a nice proof for small angles as you said

I love how simple this proof is. I also like the other proof with the constructed perpendicular lines and adding the lengths but the elegance of this proof is charming

I've never seen this argument before. It's incredible!

I love your videos, I'd love to see you do a video where you do something like recreating trig functions while trying to avoid standard methods, or, making analogous functions out of the remainder mod function (just distinguishing it from the absolute value "mod") I also think it may be fun to do something like this: I noticed the other day that for n-dimensional-squares you can count in binary up to the number of places for dimensions for vertices locations. I'd love to see some way of mathing stuff so that works with other shapes, and then trying to find the underlying pattern in how that changes based on the shape

Why is area for triangle 1/2*ab*sin(theta+phi) ?

Great pedagogical tool

Hey, I like your content a lot. But you should work on how you write 'n' and 'm'. They are hardly distinguishable.

Cool

Nice

Very interesting conclusions. Gladly it's only a semi-paradox

Very elegant.

Divisibility by 2 can be easily checked but for 5 theres a really clean method where you show 1^4=1 2^4=1 3^4=1 4^2=1 So a^16=1 (mod 5) Meaning a^17=a and mn^17-nm^17 reduces to mn-nm=0

Pretty sure the word for change in acceleration over time is jerk

math is good.....

So if b^2 = 4ac+1 or 4ac then and only then is the same answer is given by both formulations?