MTGjumper
Veteran
Do you know what the Euclidean algorithm is?
Mathematics Thread
- Thread starter IanTheCuber
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vcuber13
Veteran
i would do something like this (i dont know if this will always work or if it will yield all solutions)
simplify to \( y=\frac{90}{x}-2x \)
separate into the individual terms and analyze
\( \frac{90}{x} \) will give an integer solution when x is a divisor of 90: 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90, and their negatives
\( 2x \) will give an integer solution for all integers of x
combining the two, the 24 integer solutions are \( x = \pm1, \pm2, \pm3, \pm5, \pm6, \pm9, \pm10, \pm15, \pm18, \pm30, \pm45, \pm90, \)
5BLD
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Calculator/Math Thread
Thats what i did at first, trying to get a fraction of some sort etc.
Nice method.
However. Are you sure you haven't LOST any solutions by getting rid of the square root?
Thats what i did at first, trying to get a fraction of some sort etc.
Nice method.
However. Are you sure you haven't LOST any solutions by getting rid of the square root?
vcuber13
Veteran
what do you mean by get rid of the square root?
wolfram gets 24 solutions
http://www.wolframalpha.com/input/?i=+6x^2%2B3xy%3D270
wolfram gets 24 solutions
http://www.wolframalpha.com/input/?i=+6x^2%2B3xy%3D270
5BLD
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Calculator/Math Thread
Ah i was being silly. Nice work. I didnt actually realise i could use negative factors. Idk why.
Ah i was being silly. Nice work. I didnt actually realise i could use negative factors. Idk why.
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5BLD
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Calculator/Math Thread
A similar question i came up with is:
6x^2+3xy+1441=y^2
I did reduce it to stuff but it ended up with trying to find perfect squares that satisfy weird stuff and root(33)s being juggled around everywhere. The answer wolfram gives is interesting. And here they CAN be expressed as F(n) where n is any integer.
I will look at it once more tomorrow but id just like to see if there are any blindingly obvious ways i havent tried yet.
Maybe dividing sides by stuff rather than rooting may work. Idk ill try tomorrow.
Yeah diophantine equations fascinate me. I may try solving stuff like the 6x^2+3xy=270 with continued fractions. I heard its possible for x^2-ky^2=c....
Edit: quadratic formula may work, but it's kinda silly...
A similar question i came up with is:
6x^2+3xy+1441=y^2
I did reduce it to stuff but it ended up with trying to find perfect squares that satisfy weird stuff and root(33)s being juggled around everywhere. The answer wolfram gives is interesting. And here they CAN be expressed as F(n) where n is any integer.
I will look at it once more tomorrow but id just like to see if there are any blindingly obvious ways i havent tried yet.
Maybe dividing sides by stuff rather than rooting may work. Idk ill try tomorrow.
Yeah diophantine equations fascinate me. I may try solving stuff like the 6x^2+3xy=270 with continued fractions. I heard its possible for x^2-ky^2=c....
Edit: quadratic formula may work, but it's kinda silly...
ben1996123
Banned
ill try some stuff tomorrow if i feel like it.
CarlBrannen
Veteran
6x^2 + 3xy = 270 so
2x^2 +xy = 90 so
x(2x+y) = 90 = (2)(3)(3)(5)
Solutions are read off by noting that x can be any divisor of 90, i.e. the products {1,2}{1,3,9}{1,5} = { 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90}.
From that, you get y by putting y = (90/x) - 2x.
So I get 12 solutions, not 24. Did I do something wrong?
2x^2 +xy = 90 so
x(2x+y) = 90 = (2)(3)(3)(5)
Solutions are read off by noting that x can be any divisor of 90, i.e. the products {1,2}{1,3,9}{1,5} = { 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90}.
From that, you get y by putting y = (90/x) - 2x.
So I get 12 solutions, not 24. Did I do something wrong?
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vcuber13
Veteran
as i said you can use the negatives
ben1996123
Banned
5BLD
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Calculator/Math Thread
Mk. I was being silly again. People usually get N and Z mixed up but i get Z and Q mixed up.
Allsoux, i mean't elemen't not subse't
Mk. I was being silly again. People usually get N and Z mixed up but i get Z and Q mixed up.
Allsoux, i mean't elemen't not subse't
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vcuber13
Veteran
stannic
Regular
It shows three roots of form a + bi where |b| is very close to 0. Probably rounding error?
Edit: with last coeff -18 instead of -19, works nice. With last coeff -20, shows one real root and two complex.
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ben1996123
Banned
ok so, something really cool I found today (well, I think its cool because stuff like this really interests me).
\( 0.4^{0.4}\approx ln(2) \)
\( 0.4^{0.4}\approx 0.69314484\cdots \)
\( ln(2)\approx 0.69314718\cdots \)
\( ln(2)-0.4^{0.4}\approx 0.000002337\cdots \)
\( 0.4^{0.4}\approx ln(2) \)
\( 0.4^{0.4}\approx 0.69314484\cdots \)
\( ln(2)\approx 0.69314718\cdots \)
\( ln(2)-0.4^{0.4}\approx 0.000002337\cdots \)
gaussgenius
Member
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- Oct 20, 2012
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Might be helpful tool for you guys.... This is a project my friends and I are working on, and is academic/educational oriented (this is like TI-83 or TI-89 on steroids, and usable on your mobile devices):
http://rapidcalculator.com is a free mobile computing platform. You have the ability to perform heavy calculations on the web or on their mobile devices like the Iphone, Ipad, Android phones, Samsung Galaxy Tablet, etc. Perform intensive math/science/engineering calculations, matrix operations like multiplication/inverses, plot graphs of functions, operations on polynomials like multiplication/differentiation/integration, evaluate Fast Fourier Transforms, manipulate complex numbers and much more! Does not require any installation or plugin.
This tool is currently still under development, but we appreciate your feedback! Let us know what math functions that we should add, and what features would generally be useful for students or teachers. We hope this will be useful for students/teachers/researchers of math/science/engineering. And especially the Speedcubing community, where most people are interested in math and computation.
To start you guys off, you could perhaps expand the polynomial (x^2+2x+1)^999. This is hard by hand, but in RapidCalculator you will just type the command:
poly_x(1,2,1)^999
and see the results!
Your graphing calculator will probably crash on this, but not our platform!
Cube on!
http://rapidcalculator.com is a free mobile computing platform. You have the ability to perform heavy calculations on the web or on their mobile devices like the Iphone, Ipad, Android phones, Samsung Galaxy Tablet, etc. Perform intensive math/science/engineering calculations, matrix operations like multiplication/inverses, plot graphs of functions, operations on polynomials like multiplication/differentiation/integration, evaluate Fast Fourier Transforms, manipulate complex numbers and much more! Does not require any installation or plugin.
This tool is currently still under development, but we appreciate your feedback! Let us know what math functions that we should add, and what features would generally be useful for students or teachers. We hope this will be useful for students/teachers/researchers of math/science/engineering. And especially the Speedcubing community, where most people are interested in math and computation.
To start you guys off, you could perhaps expand the polynomial (x^2+2x+1)^999. This is hard by hand, but in RapidCalculator you will just type the command:
poly_x(1,2,1)^999
and see the results!
Your graphing calculator will probably crash on this, but not our platform!Cube on!
ben1996123
Banned
It's not correct though.
gaussgenius
Member
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- Oct 20, 2012
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Thanks for the feedback! Which one is not correct? Is it the poly_x(1,2,1)^999 ? Btw, the output is correct, just that the format is probably a little confusing, so to read the format
1 x^1998 + 1998 x^1997 + 1.995e+06 x^1996 + .....
The 1.995e+06 should read as 1.995*10^6 (this is the standard notation http://www.aaastudy.com/g6_71gx1.htm). There is some rounding involved in some of the coefficients due to some of them being extremely large in magnitude. Also, we need to allocate computational resources fairly, so in the case of extremely large computations some results might be rounded slightly.
For example, trying a smaller example will give the correct non-rounded value:
(x^2+2x+1)^2 can be evaluated using the command poly_x(1,2,1)^2 which gives 1 x^4 + 4 x^3 + 6 x^2 + 4x + 1
http://rapidcalculator.proalgorithms.com/syntax.html gives some examples on how to use the calculator for many other functions. For example, to plot the graph of x^2, you will do plot[x^2].
And for your post above, regarding ln(2) and 0.4^0.4, typing the command log(2)-0.4^0.4 gives the value
2.33740479894e-06
which is what you found above.
Do you have any suggestions for features that you would use in school? We could try to add it in!
http://rapidcalculator.com or http://rapidcalculator.proalgorithms.com --- try it out!
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ben1996123
Banned
\( (x^2+2x+1)^{999} \) gives stuff with negative signs in.
ThomasJE
Premium Member
How would you guys sole this without using trial and improvement?
\( x^3+2x=4 \)
I must have spent half an hour trying to figure this out, but I didn't get anywhere.
\( x^3+2x=4 \)
I must have spent half an hour trying to figure this out, but I didn't get anywhere.
