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Kharakov
Quantum Hot Dog
What's x(1), x(2)... x(x)? Assuming it's a functional equation like g(1), g(2)....
f(x)= x^2
if f(x) = g(1) + g(2) +g(3)+ ...+ g(x-2) + g(x-1)+ g(x) you're going to have a bunch of g': g'(x) + g'(x-1) + g'(x-2)+...
otherwise:
if f(x) = g(1) + g(2) +g(3) +... +g(n-1) + g
+ g(x) for the functions to be equal, since g(1) through g are independent of g(x), they all have to remain the same no matter what, sog(x) if a special function: g(x) = 0, unless g(x)=f(x); or g(1) + g(2) + g(3).... +g
= 0, yet g(x) = x^2;beero1000
Veteran Member
I believe he intended an analogous version of this "proof" that 1 = 2.
\(x = \underbrace{1 + 1 + 1 + \ldots + 1}_{x \textrm{ times}} = \underbrace{\frac{\mathrm{d}}{\mathrm{d}x}\left(x\right) + \frac{\mathrm{d}}{\mathrm{d}x}\left(x\right) + \frac{\mathrm{d}}{\mathrm{d}x}\left(x\right) + \ldots + \frac{\mathrm{d}}{\mathrm{d}x}\left(x\right)}_{x \textrm{ times}} = \frac{\mathrm{d}}{\mathrm{d}x}\underbrace{\left(x + x + x + \ldots + x\right)}_{x \textrm{ times}} = \frac{\mathrm{d}}{\mathrm{d}x}\left(x^2\right) = 2x\)
\(x = \underbrace{1 + 1 + 1 + \ldots + 1}_{x \textrm{ times}} = \underbrace{\frac{\mathrm{d}}{\mathrm{d}x}\left(x\right) + \frac{\mathrm{d}}{\mathrm{d}x}\left(x\right) + \frac{\mathrm{d}}{\mathrm{d}x}\left(x\right) + \ldots + \frac{\mathrm{d}}{\mathrm{d}x}\left(x\right)}_{x \textrm{ times}} = \frac{\mathrm{d}}{\mathrm{d}x}\underbrace{\left(x + x + x + \ldots + x\right)}_{x \textrm{ times}} = \frac{\mathrm{d}}{\mathrm{d}x}\left(x^2\right) = 2x\)
Alcoholic Actuary
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I like this one. I think the 'problem' lies in defining the function (x+x+...) for x number of times over a continuous interval. (What does it mean to add something pi number of times?). If it can't be defined over a continuous interval, then it isn't differentiable, whereas x^2 is differentiable.
aa
beero1000
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That's close, but I don't think that's the real problem. After all, we can think of multiplication as an (differentiable) extension of repeated addition, even when x and y are arbitrary.
On the other hand, thinking of it that way definitely emphasizes the real problem.
If \(F(x,y) = \underbrace{\left(x + x + x + \ldots + x\right)}_{y \textrm{ times}} = xy\) then the "proof" translates to \( x = F(1,x) = F(\frac{\mathrm{d}}{\mathrm{d}x}(x),x) = \frac{\mathrm{d}}{\mathrm{d}x} F(x,x) = \frac{\mathrm{d}}{\mathrm{d}x}(x^2) = 2x\). Then, you just need to think about what it means to differentiate a function of two variables...
Kharakov
Quantum Hot Dog
I'm leery of the part after the third equals sign....
ryan
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I actually don't know the answer to this. My professor gives us the answer tomorrow. I think it might have something to do with the fact that differentiating functions do not follow the associative law; dx^2 does not equal dx*dx which then wouldn't allow dx(1) + dx(2) ... dx(x)
beero1000
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The third equals sign itself is the problem. In particular, we cannot forget to differentiate the "x times" operation with respect to x. In other words, the chain rule is important!
\(\frac{\mathrm{d}}{\mathrm{d}x} F(u,v) = \frac{\partial}{\partial u} F(u,v) \frac{\mathrm{d}}{\mathrm{d}x} u + \frac{\partial}{\partial v} F(u,v) \frac{\mathrm{d}}{\mathrm{d}x} v\)
That means that the left part is not the same as the right part because \(\frac{\mathrm{d}}{\mathrm{d}x}\underbrace{\left(x + x + x + \ldots + x\right)}_{x \textrm{ times}} = \underbrace{\frac{\mathrm{d}}{\mathrm{d}x}\left(x\right) + \frac{\mathrm{d}}{\mathrm{d}x}\left(x\right) + \frac{\mathrm{d}}{\mathrm{d}x}\left(x\right) + \ldots + \frac{\mathrm{d}}{\mathrm{d}x}\left(x\right)}_{x \textrm{ times}} + \underbrace{\left(x + x + x + \ldots + x\right)}_{\frac{\mathrm{d}}{\mathrm{d}x}\left(x\right) \textrm{ times}} = \underbrace{1 + 1 + 1 + \ldots + 1}_{x \textrm{ times}} + \underbrace{\left(x + x + x + \ldots + x\right)}_{1 \textrm{ time}} = x + x = 2x\)
Kharakov
Quantum Hot Dog
There is another example here: 
Total_derivative#Differentiation_with_indirect_dependencies.
Still haven't found an inverse function set for the compound inverse nested functions I mentioned earlier. Maybe look for a good explanation for generating Laurent series, so I can come up with a generalized inverse function.
The idea would be to solve for x and n (x would vary with n) knowing y, of y= generalized series.
Alpha won't generate a generalized series, although it will generate the series for specific n.
Total_derivative#Differentiation_with_indirect_dependencies. Still haven't found an inverse function set for the compound inverse nested functions I mentioned earlier. Maybe look for a good explanation for generating Laurent series, so I can come up with a generalized inverse function.
The idea would be to solve for x and n (x would vary with n) knowing y, of y= generalized series.
Alpha won't generate a generalized series, although it will generate the series for specific n.
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ryan
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Nobody knew the answer, so he said he won't tell us until Monday :/.
Kharakov
Quantum Hot Dog
So, I'm not familiar with Mathematica or Alpha syntax. Wonder how to do something along the lines of:
solve([y=sqrt(x^2-x)], [x]); // Maxima solve command
Returns:
\($$x=-\frac{\sqrt{4\,{y}^{2}+1}-1}{2} \ \ \ x=\frac{\sqrt{4\,{y}^{2}+1}+1}{2}$$\)
I'd like to do the same with Alpha... something like:
solve ([Series[(2*x)^(7 + 1)*(x - Nest[Power[(# + x^2 - x), 1/2] &, 0, 7]), {x, Infinity, 5}]], [x]);
Might work if the Maxima build I'm using had the nesting commands...
I do have access to an ancient computer with 13 year old Mathematica 4.2 (or maybe 5!!!) on it, but I don't know if it has the various commands.
ETA: Ok, Alpha solves, but you can't use the series command or nesting. Mehh. I wonder about transferring the ancient Mathematica to this computer to test it out.
solve([y=sqrt(x^2-x)], [x]); // Maxima solve command
Returns:
\($$x=-\frac{\sqrt{4\,{y}^{2}+1}-1}{2} \ \ \ x=\frac{\sqrt{4\,{y}^{2}+1}+1}{2}$$\)
I'd like to do the same with Alpha... something like:
solve ([Series[(2*x)^(7 + 1)*(x - Nest[Power[(# + x^2 - x), 1/2] &, 0, 7]), {x, Infinity, 5}]], [x]);
Might work if the Maxima build I'm using had the nesting commands...
I do have access to an ancient computer with 13 year old Mathematica 4.2 (or maybe 5!!!) on it, but I don't know if it has the various commands.
ETA: Ok, Alpha solves, but you can't use the series command or nesting. Mehh. I wonder about transferring the ancient Mathematica to this computer to test it out.
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fast
Contributor
Is there a significance to an increasing or decreasing disparity between an average and a median? For instance, suppose the average household income and median household income kept getting farther and farther apart overtime. Does this say something informative?
beero1000
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That generally indicates a skewed distribution.
i.e. if the median household income is $50,000, while the mean household income is $70,000, that would indicate that half of all households make less than $50k, yet the half that makes more than $50k makes so much more that they drag the mean all the way up to $70k, etc.
Income is a good example of a right-skewed distribution. Most people make a middling amount, but there are some that make huge amounts. This doesn't affect the median, but strongly affects the mean.
An informative example: Suppose there is a room with 10 people of average income, say $50k each. Then Bill Gates walks into the room. The room's median income is still $50k, but the mean income changes from $50k to billions of dollars per year.
Alcoholic Actuary
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bilby
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If we define the distance from a vertex of the tetrahedron to the sphere as 1, then it varies from 0 to 1, depending on the point at which you measure.
You're welcome.
beero1000
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The distance to which part of the tetrahedron? Faces, edges, vertices?
Kharakov
Quantum Hot Dog
Well... everywhere.
How about just the distance from the origin, to an equilateral triangle that is orthogonal to the x-axis, pointed up (y is up for me...). Then a rotation matrix to reach one of the other faces. I'll figure it out from there.
It's for art man. Art! Unless you think easy transforms between Platonic solids are something that can be used for nefarious purposes.
Bomb#20
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Well, since it's for art...Well... everywhere.
How about just the distance from the origin, to an equilateral triangle that is orthogonal to the x-axis, pointed up (y is up for me...). Then a rotation matrix to reach one of the other faces. I'll figure it out from there.
It's for art man. Art! Unless you think easy transforms between Platonic solids are something that can be used for nefarious purposes.
I make the distance from the surface of the inscribed sphere to a vertex of the tetrahedron to be equal to the diameter of the sphere. For the distance from the sphere to an edge I'm getting (sqrt(3) - 1) times the radius of the sphere.
Go forth and be artistic.
(And yes, you'll be expected to post a photo...)