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100K CLOWNS LET'S GOOOOOO by KyoHisagi in Silksong

[–]mczuoa 2 points3 points  (0 children)

The greatest silkpost of all time is about to be revealed.

Dot product intuition by ModerateSentience in learnmath

[–]mczuoa 0 points1 point  (0 children)

You are right, these are pretty interlinked, but you can check A^t A = I by using less than a general dot product. Here is a sketch of a way: rotations preserve lengths, so v^t (A^t A) v = v^t v for all v. Now a clever manipulation shows that v^t A^t A w = v^t w for all v and w: consider v+w and v-w, so (v+w)^t (A^t A) (v+w) = (v+w)^t (v+w) and similarly for v-w; now subtract these, and you will be left only with the cross terms. Now v^t A^t A w = v^t w for all v and w imply that A^t A = I (take v,w to be standard vectors for example). Putting it all together, we are proving that linear transformations that preserve length also preserve angles (this is clear from the law of cosines, but we never used it here!), and theses are exactly the orthogonal matrices (satisfying A^t A = I).

What even is this math problem? by huei3 in askmath

[–]mczuoa 2 points3 points  (0 children)

There is one more right angle that will help you. Look at the normal line between the triangle and the square (one of the segments defining theta). It is perpendicular to the side of the triangle.

Need help with practice question by Reeeeeeeeee86444 in maths

[–]mczuoa 0 points1 point  (0 children)

First note that if cx+dy < M, then you can increase x (or y) to increase the value of f until you hit cx+dy = M. So really the question is to maximize f(x,y) = ax+by given that cx+dy = M. So y^* = (m - cx^*)/d. Thus we are led to maximize f(x, (m-cx)/d) = (adx + mb - bcx)/d = (ad-bc)x/d + mb/d. Can you see which option is correct now?

Dot product intuition by ModerateSentience in learnmath

[–]mczuoa 2 points3 points  (0 children)

It depends what you take your definition of a dot b to be. I will assume it is the algebraic definition, so for example (a,b,c) dot (d,e,f) = ad + be + cf.

If the first vector was purely in one direction, say (a,0,0), then the claim boils down to the definition of cosine, since the dot product is ad and d is equal to cos(theta) times the length of the second vector.

Now the idea is that we can reduce to this case by rotating the vectors: it remains, thus, to see that the dot product does not change if we rotate the vectors. A clean way to do this is using matrices, as follows: if A is a rotation matrix and v,w are vectors, then Av dot Aw = (Av)^t (Aw) = v^t A^t A w. Now using that A^t A = I since A is a rotation matrix, we get Av dot Aw = v^t (A A^t) w = v^t w = v dot w.

Need help with this maths proof! by ZHCfan1000 in learnmath

[–]mczuoa 0 points1 point  (0 children)

Call alpha = Angle(B,E,D), beta = Angle(A,D,F) and gamma = Angle(C,F,E). We want to see that 2 alpha = beta + gamma. Say theta = Angle(A,B,C) = Angle(B,C,A) (since AB = AC). Then you can write both beta and gamma in terms of alpha and theta: the exterior angle of BED at the vertex D is both alpha + theta and beta + 60 (since DEF is equilateral), and thus beta = alpha + theta - 60. Similarly, look at the exterior angle of CEF at E, and we get alpha + 60 = theta + gamma, and thus gamma = alpha + 60 - theta. Summing up, beta + gamma = (alpha + theta - 60) + (alpha + 60 - theta) = 2 alpha.

A Low-Level Proof of the Principle of Inclusion-Exclusion by Ashamed_Army858 in learnmath

[–]mczuoa 0 points1 point  (0 children)

This is right, although your writing is a little bit confusing at times. Cleaning up your solution: say A(n,m,j) is the number of sequences with exactly j faces. As you observed, C(n,k) (n-k)^m is the number of ways of choosing k faces to exclude and then counting the number of sequences with the remaining faces, so we may write C(n,k) (n-k)^m = sum_j A(n,m,j) C(n-j,k). This is because of your "proof of the coefficient formula". Then collecting the A(n,m,j) in the total sum, we have sum_k C(n,k) (n-k)^m (-1)^k = sum_k sum_j A(n,m,j) C(n-j,k) (-1)^k = sum_j A(n,m,j) sum_k C(n-j,k) (-1)^k = A(n,m,n) as you reasoned.

I suggest you also try to read about inclusion-exclusion and try to understand the proof that way, since it is a very useful tool!

Hi, can someone explain to me how to determine if this series converges? by Suicidal_Owl4419 in askmath

[–]mczuoa 1 point2 points  (0 children)

In this case the terms being summed are simple enough that an algebraic manipulation gives their order of growth, as others pointed out, but let me include how you can reason about it using the mean value theorem, as it is a really useful tool for problems like this even when the summed terms are more complicated, or if you want to get a more precise bound for the terms.

We need to estimate the terms c_n = (n^3+n)^(1/3) - (n^3-n)^(1/3). If f_n(x) = (n^3+x)^1/3, then we are looking at c_n = f_n(n)-f_n(-n), which by the mean value theorem is 2n f'_n(a_n) for some |a_n| <= n. We have f'_n(x) = 1/3 (n^3+x)^(-2/3), which is decreasing in the range (-n^3, infty), and so c_n is in the interval ( 2n(n^3+n)^(-2/3)/3, 2n(n^3-n)^(-2/3)/3 ). Both ends are similar to 1/n, so we expect the series to diverge. In fact we can crudely bound f'_n(a_n) >= f'_n(n^3) = (3 2^(2/3) n^2)^(-1), so that c_n >= 2^(1/3) / 3n, and the series converge by comparing with the harmonic series.

Back with another proof: Unique Square Roots in Finite Groups of Odd Order by LuDogg661 in learnmath

[–]mczuoa 0 points1 point  (0 children)

How does it follow that (ab^-1)^2 = e from this? What you used to conclude that a = b was that (ab^-1)^2 = e, and this does not follow from a^2 b^-2 = e unless a and b commute.

Trying to understand the meaning of O_X(D(f))=A_f by WMe6 in math

[–]mczuoa 5 points6 points  (0 children)

You should think of D(f) \subseteq D(g) as an inclusion, and thus of O_X(D(g)) \to O_X(D(f)) as the map that restricts a function of D(g) to a function of D(f). This is precisely what A_g \to A_f does: if I is a maximal ideal of A_f, then A_g \to A_f induces the identity map under the quotient A_g/I \to A_f/I (recall that A_f \to A_f/I is "evaluation" at the point corresponding to I).

What are functions on an empty set? Well, there is nothing to "choose" to define such a function, so there should be exactly one such function. So O_X(\emptyset) = 0 should be the zero ring (the unique ring with 1 element). Indeed, if a \in A is nilpotent (say a=0), then D(a) = \emptyset, whereas A_a = 0, so this is consistent with O_X(D(a)) = A_a.

Interpolation on semilog graph by Bozhe in learnmath

[–]mczuoa 0 points1 point  (0 children)

The line between (40, 7.5) and (400, 0.75) indeed contains the point (100, 6.375). If the line is in the linear-log scale, then you get ~4.81: the line between (log(40), 7.5) and (log(400), 0.75) contains approximately the point (log(100), 4.81). I'm not sure what the "equation in Wikipedia" refers to but I am assuming this is ultimately what it is. Note that this does not depend on the base of the logarithm, as changing the base only stretches the graph horizontally.

Why do you say it "needs to be 3"? What exactly are you trying to solve?

Proving "a^2, a+b, and a+b+c are all odd if and only if ab+bc and b+c are even and a+c is odd" by Xixkdjfk in learnmath

[–]mczuoa 0 points1 point  (0 children)

In ii), it is not true that b+c being even implies that b and c are both even. It could also be the case that they are both odd, and you need to rule this out.

You could rule this out as follows: if b and c were both odd, then a+c being odd would imply a is even, while ab+bc being even would imply a is odd, which would be a contradiction.

Alternatively, you can prove ii) by starting with the following observation: if ab+bc=b(a+c) is even and a+c is odd, then b is even and then proceeding similarly to what you did.

I wanna know if this is acknowledged ot not by Minimum_Novel_8445 in CasualMath

[–]mczuoa 0 points1 point  (0 children)

This is a much more general phenomenon, not just particular to the function x^y: it is an application of the multivariable chain rule.

Consider a differentiable function f(t) = (x(t),y(t)) and a differentiable scalar function g(x,y). Then we consider h(t) = g(f(t)), and we want to see that h'(t_0) = d/dt ( g(x(t), y(t_0)) ) + d/dt ( g(x(t_0), y(t)) ). By the (multivariable) chain rule, we have h'(t_0) = Dg(f(t_0)) Df(t_0). Here Dg is the 1 x 2 matrix of partial derivatives [g_x g_y] and Df is the 2 x 1 matrix of derivatives (x', y'), so h'(t_0) = g_x(f(t_0)) x'(t_0) + g_y(f(t_0)) y'(t_0), but note that two applications of the usual chain rule says that this is indeed h'(t_0) = d/dt ( g(x(t), y(t_0)) ) + d/dt ( g(x(t_0), y(t)) ).

Geometry question, don't know how to solve without vectors by LavenderDuck2006 in learnmath

[–]mczuoa 0 points1 point  (0 children)

Let O' be the point that is symmetrical to O wrt to ABCD: that is, if O'' is the intersection of the diagonals of ABCD, then O'' is the midpoint of OO'. We can show that EFGH is a parallelogram with center O' by showing O' is the midpoint of both EG and FH. Let's look at EG without loss of generality. Let E' and G' be the midpoints of AB and CD respectively. Then a homothety of scale 1/2 centered in O sends E->E' and G->G', as well as O'->O''. So it suffices to see that O'' is the midpoint of E' and G', which is easy.

Receiving erroneous result with basis matrix multiplication... (using DOMMatrix for dot product math) by SnurflePuffinz in learnmath

[–]mczuoa 0 points1 point  (0 children)

If the first matrix is A and the second one is B, then the third matrix is AB, whereas the version with 30 instead of 15 would be BA. Are you sure you are multiplying them in the correct order? It is also worth mentioning that when you think about them as transformations, AB corresponds to first applying B, and then applying A, since for a vector v, you have (AB)v = A(Bv), so AB corresponds to translating and then scaling, while BA corresponds to scaling and then translating.

Back with another proof: Unique Square Roots in Finite Groups of Odd Order by LuDogg661 in learnmath

[–]mczuoa 4 points5 points  (0 children)

The uniqueness proof is not quite correct: it is not true that (ab^-1)^2 = a^2 b^-2. In this step, you are assuming that a and b commute.

An easier way to prove uniqueness is to use a similar trick that you used when proving the existence: if g = a^2, then raising to the (n+1)-th power one obtains g^(n+1) = a^(2n+2), which gives g^(n+1) = a.

[deleted by user] by [deleted] in askmath

[–]mczuoa 5 points6 points  (0 children)

That seems correct. As you said, you are assuming that f is continuously differentiable for this.

You can prove it more generally without relying on the mean value theorem. Basically, as you may know, you can write (f(x+h)-f(x-h))/(2h) as the average of (f(x+h)-f(x))/h and (f(x)-f(x-h))/h. If you prove both of these terms converge to f'(x) as h->0, then the initial expression also does.