The Math Thread

[Kharakov]

Ok, so it's a form of  synthetic division. That would have been an easy answer... although I like the way I do it better (it seems more intuitive, but then again, it's what I made up so of course it seems that way to me).

It produces consistent results, and a lot of the sequences I've generated with it in the last day have turned up as being discovered by Cauchy, Euler, Bernoulli, etc.

The "harmonic" zeros are all generated by multiplying the Harmonic series by a*(1-1)^n, which is the same thing as dividing it by (1-1)^-n, which happens to be what I was doing. So I'm multiplying and dividing by well defined zeros.

1/(1-1)^2= 1/(1- 2 +1) = 1 +2 +3 +4 +5 +6....

So 1 / (the binomial expansion of (1-1)^n) = the various simplex numbers, as long as you do the synthetic division method that I'm assuming everyone uses because of the sequences that keep popping up.

There are a LOT of zeros. 1/harmonic series gives you 1- |Gregory coefficient| series=0.

It's easy to do square roots using synthetic division. Probably how they taught people to do them in high school.

Maybe it's time to see a couple of rotational patterns 1/(1-i) = 1+ i -1 -i + 1 + i....

 

[beero1000]

Ok, so it's a form of  synthetic division. That would have been an easy answer... although I like the way I do it better (it seems more intuitive, but then again, it's what I made up so of course it seems that way to me).

It produces consistent results, and a lot of the sequences I've generated with it in the last day have turned up as being discovered by Cauchy, Euler, Bernoulli, etc.

The "harmonic" zeros are all generated by multiplying the Harmonic series by a*(1-1)^n, which is the same thing as dividing it by (1-1)^-n, which happens to be what I was doing. So I'm multiplying and dividing by well defined zeros.

1/(1-1)^2= 1/(1- 2 +1) = 1 +2 +3 +4 +5 +6....

So 1 / (the binomial expansion of (1-1)^n) = the various simplex numbers, as long as you do the synthetic division method that I'm assuming everyone uses because of the sequences that keep popping up.

There are a LOT of zeros. 1/harmonic series gives you 1- |Gregory coefficient| series=0.

It's easy to do square roots using synthetic division. Probably how they taught people to do them in high school.

Maybe it's time to see a couple of rotational patterns 1/(1-i) = 1+ i -1 -i + 1 + i....

I still don't see why what you're doing is better than just saying

\(\frac{\lim_{n \to \infty} \sum_{i=0}^n A_i }{\lim_{n \to \infty} \sum_{i=0}^n B_i} = \lim_{n \to \infty} \frac{\sum_{i=0}^n A_i}{\sum_{i=0}^n B_i} = 0\)

whenever

\( \sum_{i=0}^n A_i = o(\sum_{i=0}^n B_i)\)

 

[Kharakov]

I didn't think of using division to divide series. Standard long division works. I just never thought about keeping elements separate.

What's o?

The fact that 1/(1-4+7-4) = 1 +4 +9 +16 +25... or that you can calculate the square root of a number really easily using patterns in a division grid combined with expansions of the number like 1+999/1000 +999/1000^2... for square root of 2 or 1+1+999/1000 +999/1000^2... for sqrt(3)?? I didn't know that.

Show

I think they might have been talking about this in high school, but by the time they were (10th grade, advanced math) I had already given up on life. Rich, spoiled kids had all the good stuff, and I could see that I was going to end up working for them. Fuck- it's what I do now. And it sucks- every job I do, I feel like I am having my face pushed into a toilet full of shit by a bully. I dreaded supporting the people who had everything, and never being able to get them back for the way they made me feel as a kid, but the poor won't buck the system, and the rich are all sweet as can be, because they got away with what they did to people like me- they didn't have to do anything but be normal kids, and they'd break a few people, and it's all good, they're still off to the top, and the people they fuck up end up working for them anyway, so they both ruined their lives, and they get to use them to their hearts content. Not only that, they own the brainwashed fucking US Government (and Military), who either are in on it or too fucking stupid to know or care.

Fuck I hope I die soon. /endrant

Anyway...

What's o?

 

[Juma]

Here's an example of a harmonic 0: (1 + 1/2 + 1/3...) / (1+1+1...) = 1 -1/2 -1/6 -1/12.... = 0

Sorry but i dont get it. Can you explain exactly what you really are doing?

To me (1 + 1/2 + 1/3...) isnt less or bigger than (1+1+1...)

If you clearly specifies what you mean, for example:
That you talk about a specific serie and compare same number of terms:
As in:
If a = a(n-1) + 1/n, a(0)=0 and b = b(n-1) + 1, b(0)=0
Then a > b for all n.
Then I agree.
But you dont specify that..

 

[Kharakov]

Here's an example of a harmonic 0: (1 + 1/2 + 1/3...) / (1+1+1...) = 1 -1/2 -1/6 -1/12.... = 0

Sorry but i dont get it. Can you explain exactly what you really are doing?

To me (1 + 1/2 + 1/3...) isnt less or bigger than (1+1+1...)

To Cantor as well. Divide 1+1+1.../ (1+1/2+1/3....). Now divide 1+1/2+1/3.../ (1+1+1...). Think non- collapsed series when you divide. So 1/(1-1)= 1+1+1... is well defined, but 1/0 is not. 1+1+1.../ (1+2+3...) is well defined, infinity/infinity is not.

For which do you get a larger value? One is zero, the other looks pretty much divergent to me. Generally if you divide a larger infinity by a smaller infinity, you'll get that kind of result.

I need to get ready for bed, division examples are posted the page before, I will walk you through it tomorrow if you like.

I actually think the whole prohibition against dividing by zero is bullshit that corrupt assholes put out there to obfuscate math.

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It's by ivory tower types who like to bullshit the populace and make their work seems mysterious and out of reach, but necessary. Which is why anyone who speaks in an obfuscating manner should be tortured then killed, because they are deliberately fucking with the populace to hold them down so they can charge them to learn the obfuscated stuff.

Crap, I wonder if I pose a danger to the plan to keep the poor as wage slaves to the well off? Probably not. The poor are too damn stupid to stand up for themselves, band together, and take what they deserve and give the corrupt rich what they deserve (not every rich person is a good for nothing piece of shit, but a lot of the middle class are (a middle class which includes "poor on paper" who get "miraculous windfalls from God") and a lot of rich people are).

 

[Juma]

Here's an example of a harmonic 0: (1 + 1/2 + 1/3...) / (1+1+1...) = 1 -1/2 -1/6 -1/12.... = 0

Sorry but i dont get it. Can you explain exactly what you really are doing?

To me (1 + 1/2 + 1/3...) isnt less or bigger than (1+1+1...)

To Cantor as well. Divide 1+1+1.../ (1+1/2+1/3....). Now divide 1+1/2+1/3.../ (1+1+1...). Think non- collapsed series when you divide. So 1/(1-1)= 1+1+1... is well defined, but 1/0 is not. 1+1+1.../ (1+2+3...) is well defined, infinity/infinity is not.

For which do you get a larger value? One is zero, the other looks pretty much divergent to me. Generally if you divide a larger infinity by a smaller infinity, you'll get that kind of result.

I need to get ready for bed, division examples are posted the page before, I will walk you through it tomorrow if you like.

I actually think the whole prohibition against dividing by zero is bullshit that corrupt assholes put out there to obfuscate math.

Show

It's by ivory tower types who like to bullshit the populace and make their work seems mysterious and out of reach, but necessary. Which is why anyone who speaks in an obfuscating manner should be tortured then killed, because they are deliberately fucking with the populace to hold them down so they can charge them to learn the obfuscated stuff.

Crap, I wonder if I pose a danger to the plan to keep the poor as wage slaves to the well off? Probably not. The poor are too damn stupid to stand up for themselves, band together, and take what they deserve and give the corrupt rich what they deserve (not every rich person is a good for nothing piece of shit, but a lot of the middle class are (a middle class which includes "poor on paper" who get "miraculous windfalls from God") and a lot of rich people are).

You cannot divide 1+1/2+1/3... with (1+1+1...) since neither of these series converge.

I assume that what you really mean is that

a = (1+1/2+1/3 +...+ 1/n)/n -> 0 when n -> oo.

But that is not the same as to say ”the number 1+1/2+1/3 +... divided with the number 1+1+1..” which is obvious nonsens.

Thus you obviously mean something else then normal division of two numbers.
More something about how a behaves when n-> oo.

 

[lpetrich]

I actually think the whole prohibition against dividing by zero is bullshit that corrupt assholes put out there to obfuscate math.

Let's see what happens if we try. Let's see what division is -- it is the operation that undoes multiplication. x = a/b means that x is the solution of b*x = a. So let us look for solutions for b = 0.

For 0/0, we must solve 0*x = 0. Since 0*x is always 0, then we get 0 = 0 for all x. Thus, all x is a solution.

For 1/0, for example, we must solve 0*x = 1. Since 0*x is always 0, then we get 1 = 0 for all x. Thus, no x is a solution.

It is to avoid dividing by zero that mathematicians invented infinitesimals, and then limits. Infinitesimals are some ordinary number multiplied by an infinitesimal unit, e. That number has the property that e² = 0 and likewise for higher powers. It is evident that e is not a "normal" number. Limits were later invented to avoid introducing such "abnormal" numbers.

 

[beero1000]

What's o?

Landau notation:  Little-o notation. f = o(g) just means that g grows much faster than f asymptotically.

To Cantor as well. Divide 1+1+1.../ (1+1/2+1/3....). Now divide 1+1/2+1/3.../ (1+1+1...). Think non- collapsed series when you divide. So 1/(1-1)= 1+1+1... is well defined, but 1/0 is not. 1+1+1.../ (1+2+3...) is well defined, infinity/infinity is not.

Those are all not well-defined. The partial sum ratio limit can be, but not the infinite divergent sum ratio (without explicitly explaining what those symbols mean).

I actually think the whole prohibition against dividing by zero is bullshit that corrupt assholes put out there to obfuscate math.

Show

It's by ivory tower types who like to bullshit the populace and make their work seems mysterious and out of reach, but necessary. Which is why anyone who speaks in an obfuscating manner should be tortured then killed, because they are deliberately fucking with the populace to hold them down so they can charge them to learn the obfuscated stuff.

Crap, I wonder if I pose a danger to the plan to keep the poor as wage slaves to the well off? Probably not. The poor are too damn stupid to stand up for themselves, band together, and take what they deserve and give the corrupt rich what they deserve (not every rich person is a good for nothing piece of shit, but a lot of the middle class are (a middle class which includes "poor on paper" who get "miraculous windfalls from God") and a lot of rich people are).

Wut?

 

[Kharakov]

To Cantor as well. Divide 1+1+1.../ (1+1/2+1/3....). Now divide 1+1/2+1/3.../ (1+1+1...). Think non- collapsed series when you divide. So 1/(1-1)= 1+1+1... is well defined, but 1/0 is not. 1+1+1.../ (1+2+3...) is well defined, infinity/infinity is not.

For which do you get a larger value? One is zero, the other looks pretty much divergent to me. Generally if you divide a larger infinity by a smaller infinity, you'll get that kind of result.

You cannot divide 1+1/2+1/3... with (1+1+1...) since neither of these series converge.

Division method:

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Simplex convolution series division (Inverse Cauchy Product????), results are r_whatever:
\(\begin{tabular}{ l | c c c c c } & a & b & c & d & e ... \\ \div & A & B & C & D & E ... \\ \hline r_0 & r_0 A & r_0 B & r_0 C & r_0 D & r_0 E \\ r_1 & &r_1 A & r_1 B & r_1 C & r_1 D \\ r_2 & &&r_2 A & r_2 B & r_2 C \\ r_3 & &&&r_3 A & r_3 B \\ r_4 & &&&&r_4 A \\ \end{tabular} a= A r_0 b= B r_0 + A r_1 c= C r_0 + B r_1 + A r_2 ... r_0=\frac{a}{A} r_1=\frac{b-B r_0}{A} r_2=\frac{c- C r_0 - B r_1}{A} .... \)

\( \begin{tabular}{ r | r r r r r r} & 1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & ... \\ \div & 1 & 1 & 1 & 1 & 1& ... \\ \hline 1 & 1 & 1 & 1 & 1 & 1 \\ -\frac {1}{2} && -\frac {1}{2} & -\frac {1}{2} & -\frac {1}{2} & -\frac {1}{2} &... \\ -\frac {1}{6} &&& -\frac {1}{6} & -\frac {1}{6} &-\frac {1}{6} &... \\ -\frac {1}{12} &&&& -\frac {1}{12} & -\frac {1}{12}&... \\ -\frac {1}{20} &&&&& -\frac {1}{20}&... \\ \end{tabular} \) ................ \(\begin{tabular}{ r | r r r r r r } & 1 & 1 & 1 & 1 & 1& ... \\ \div & 1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5}& ... \\ \hline 1 & 1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5}& ... \\ \frac {1}{2} && \frac {1}{2} & \frac {1}{4} &\frac {1}{6} &\frac {1}{8} & ... \\ \frac {5}{12} &&& \frac {5}{12} & \frac {5}{24} & \frac {5}{36} & ... \\ \frac {3}{8} &&&& \frac {3}{8} & \frac {3}{16}&... \\ \frac {251}{720} &&&&& \frac {251}{720} &... \\ \end{tabular} \)
The first one collapses to 0 (1- (1/2+1/6+1/12+1/20) = 1-1), the second one is divergent (it's the series for -x/[(1-x)log(1-x)] at x=1...), Google +OEIS got me what the series was, although I

But that is not the same as to say ”the number 1+1/2+1/3 +... divided with the number 1+1+1..” which is obvious nonsens.

The series/the series.

Thus you obviously mean something else then normal division of two numbers.
More something about how a behaves when n-> oo.

1/(1+1) gives you the  Grandi series.
(1-1+1)/(1+1) gives you 1-2 -3 *(Grandi series), which equals 1/2 when collapsed.

1/(1+4+9+16....) = 1-4+7-8+8-8+8.... =1-4+7-8*Grandi= 1-4+7-4 (4 is the collapsed Grandi*8)=0 (collapsed whole thing)

So if you divide 1/(1-4+7-4) you get 1+4+9+16... which is what you get if you divide 1/(1-4+7-8+8-8+8-8+8...), so they are sort of the same, and work. But you lost some information when you collapsed -8+8-8+8... to 4, like you do when you collapse 1-1 to 0.

When you multiply 1+4+9+16... * 1-4+7-8+8-8+8.... you get 1. When you multiply 1+4+9+16.. * 1-4+7-4 you get 1+0+0+4+8+12... because you left out some information.
\( \begin{tabular}{ l | r r r r r r r r } &1 & 4 & 9 & 16 & 25 & ... \\ \times & 1 & -4 & 7 & -8 & 8 & ... \\ \hline & 1 & 4 & 9 & 16 & 25 &36 & 49& ... \\ + & & -4 & -16 & -36 & -64 & -100&-144&...\\ +& & & 7 & 28 & 63 & 112 & 175&...\\ +& & & & -8 & -32 &-72 &-128&... \\ +& & & & & 8 &32&72&...\\ +& & & & & &-8&-32&...\\ +& & & & & &&8&...\\ \hline & 1 & 0 & 0 & 0 & 0&0 &0 &... \\ \end{tabular} \)|||||||||| \( \begin{tabular}{ l | r r r r r r r r } &1 & 4 & 9 & 16 & 25 & 36 &49&...\\ \times & 1 & -4 & 7 & -4 & & \\ \hline & 1 & 4 & 9 & 16 & 25 &36 & 49& ... \\ + & & -4 & -16 & -36 & -64 & -100&-144&...\\ +& & & 7 & 28 & 63 & 112 & 175&...\\ +& & & & -4 & -16 &-36 &-64&... \\ \hline & 1 & 0 & 0 & 4 & 8&12&16&... \\ \end{tabular} \)


Basically, if you leave out some information because you assumed series collapse wouldn't impact the system (like quantum mechanics leaving out parts that they don't think are essential for the operation of the universe), you can't always reverse the process. This doesn't mean that you can't figure out what parts led from one to another later, but...

 

[Kharakov]

I actually think the whole prohibition against dividing by zero is bullshit that corrupt assholes put out there to obfuscate math.

Let's see what happens if we try. Let's see what division is -- it is the operation that undoes multiplication. x = a/b means that x is the solution of b*x = a. So let us look for solutions for b = 0.

For 0/0, we must solve 0*x = 0. Since 0*x is always 0, then we get 0 = 0 for all x. Thus, all x is a solution.

For 1/0, for example, we must solve 0*x = 1. Since 0*x is always 0, then we get 1 = 0 for all x. Thus, no x is a solution.

It is to avoid dividing by zero that mathematicians invented infinitesimals, and then limits. Infinitesimals are some ordinary number multiplied by an infinitesimal unit, e. That number has the property that e² = 0 and likewise for higher powers. It is evident that e is not a "normal" number. Limits were later invented to avoid introducing such "abnormal" numbers.

To make things consistent, right? Like the last time you said this to me. Wow... me dumb.

 

[Kharakov]

Those are all not well-defined. The partial sum ratio limit can be, but not the infinite divergent sum ratio (without explicitly explaining what those symbols mean).

So the following isn't multiplication of series with a valid convolution, with each series starting at the starting point of the lower series:

\( \begin{tabular}{ l | r r r r r r r r } &1 & 1 & 1 & 1& 1& ... \\ \times & a & -b &c & & &\\ \hline & a & a &a& a & a &... \\ & & -b& -b & -b & -b &... \\ & & & c & c & c &... \\ \hline & a & a-b & a-b+c & a-b+c & a-b+c&a-b+c &... \\ \end{tabular} \)

That's how the stuff meshes, but only for certain things, like the binomial expansion of a(1-1)^-n. So basically I'm doing a shift for every multiplication or division, to keep the series defined. It works for finite sums, for convergent series, etc.

So 2*1+1+1+1... is different than (1+1) * (1+1+1+1....) or (1+0+1) * (1+1+1+1...). I basically created a semi consistent system that works, but of course it introduces all kinds of new concepts that make things more complicated, like "series collapse", etc.

So I don't know if it could occur naturally- I was looking for some weird infinities that could occur naturally, without someone having to make up excuses for them to be well defined. The continuum is divided by nothing, after all (yeah, equivocation, untermensche style. I'm surprised he isn't arguing that I'm correct, but I am talking about math).


I find myself wondering if there is justification for them, but maybe I'm just trying to find some low hanging fruit... for someone of my intellectual stature.

 

[Kharakov]

Some rotation groups. You might notice something that looks like Petrie polys in the rotating, shifted (like infinite series), tetrahedron. Ohh, looks like a 2 simplex at first because of the angle.

 

[Kharakov]

Happy holidays. Here's a box for your cat.

 

[steve_bank]

Infinity by definition is not quantifiable, it is not a number.

 

[Jokodo]

There should be a dumb math questions thread.
I'm failing to work out the angle between the sides of a pyramid (constructed with equal triangles), given the angle of the top corner of its individual triangles.

Some time ago, I failed to work out how to derive the directional vector expressed in x-y-z format of a great circle (not in spherical coordinates!), given inclination relative to the sphere's equator (for simplicity, defined as the x-y-plane), angular distance from due east, and angular distance from the ascending node. I did work out the vertical and planar component of that vector, and got away with pretending that the planar component in turn was composed like the two-dimensional vector of an equatorial orbit, but only because I was able to restrict myself to low inclinations where the error thus introduced remained small.

 

[ruby sparks]

There should be a dumb math questions thread.
I'm failing to work out the angle between the sides of a pyramid (constructed with equal triangles), given the angle of the top corner of its individual triangles.

I am impressed by your thinking of that as a dumb question. I can think of dumber ones. You are talking about an intermediate thread at least, surely.

But I'll play. My dumb question is....how is the answer not always 90^o regardless of the top angle?

 

[beero1000]

There should be a dumb math questions thread.
I'm failing to work out the angle between the sides of a pyramid (constructed with equal triangles), given the angle of the top corner of its individual triangles.

Some time ago, I failed to work out how to derive the directional vector expressed in x-y-z format of a great circle (not in spherical coordinates!), given inclination relative to the sphere's equator (for simplicity, defined as the x-y-plane), angular distance from due east, and angular distance from the ascending node. I did work out the vertical and planar component of that vector, and got away with pretending that the planar component in turn was composed like the two-dimensional vector of an equatorial orbit, but only because I was able to restrict myself to low inclinations where the error thus introduced remained small.

You can make one!

The dihedral angle for the tetrahedron with top vertex face angles A, is arccos(tan(A/2)cot(A)).

There should be a dumb math questions thread.
I'm failing to work out the angle between the sides of a pyramid (constructed with equal triangles), given the angle of the top corner of its individual triangles.

I am impressed by your thinking of that as a dumb question. I can think of dumber ones. You are talking about an intermediate thread at least, surely.

But I'll play. My dumb question is....how is the answer not always 90^o regardless of the top angle?

The faces won't be perpendicular to each other. Imagine the top angle being very close to 120 degrees so that the pyramid is basically flat, and the faces are nearly parallel.

 

[steve_bank]

Well, if you bring relativity, in a planetary gravity field there id no such thing as a right angle. Planar trig is an approximation.

 

[lpetrich]

There should be a dumb math questions thread.
I'm failing to work out the angle between the sides of a pyramid (constructed with equal triangles), given the angle of the top corner of its individual triangles.

That's fairly easy for me, because I have Mathematica. It does algebra, but I can do analytic geometry with it, even if not Euclid-style constructions.

h = (height) / (half of base length)
cos(angle between edges at the top of the pyramid: a) = h² / (1 + h²)
cos(angle between sides: b) = 1 / (1 + 2*h²)

h = sqrt(cos(a)/(1 - cos(a))
cos(b) = (1 - cos(a))/(1 + cos(a))

Some time ago, I failed to work out how to derive the directional vector expressed in x-y-z format of a great circle (not in spherical coordinates!), given inclination relative to the sphere's equator (for simplicity, defined as the x-y-plane), angular distance from due east, and angular distance from the ascending node. I did work out the vertical and planar component of that vector, and got away with pretending that the planar component in turn was composed like the two-dimensional vector of an equatorial orbit, but only because I was able to restrict myself to low inclinations where the error thus introduced remained small.

 Orbital elements has the solution to that problem.

cos(asn)*cos(pla) - sin(asn)*sin(pla)*cos(inc)
sin(asn)*cos(pla) + cos(asn)*sin(pla)*cos(inc)
sin(pla)*sin(inc)

asn = angle from {1,0,0} in the equator to the ascending node
inc = inclination of orbit plane to equator
pla = angle from the ascending node to the object position

 

[ruby sparks]

I am impressed by your thinking of that as a dumb question. I can think of dumber ones. You are talking about an intermediate thread at least, surely.

But I'll play. My dumb question is....how is the answer not always 90^o regardless of the top angle?

The faces won't be perpendicular to each other. Imagine the top angle being very close to 120 degrees so that the pyramid is basically flat, and the faces are nearly parallel.

Ah. Gotcha. Thanks. Nowhere near as easy as I mistakenly thought.

Ok here is my guess. Don't mind if I'm wrong. The angle between the faces is the same as the angle at the top of each triangle.