MTGjumper
Veteran
Rearrange to \( x^3 + 2x - 4 = 0 \) and then use the substitution \( x = y - \frac{2}{3y} \). Good things should happen.
Mathematics Thread
- Thread starter IanTheCuber
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conn9
Regular
We learned factor theorem and factorising cubics in maths recently so I'll show you the method we learned.
This only works for cubic equations with integer answers, which yours isn't.
This only works for cubic equations with integer answers, which yours isn't.
x^3 + 2x = 4
x^3 + 2x - 4 = 0
Factors of -4 (c value) are 1, -4, 2, -2, -1, 4.
Plug all of these into f(x) = x^3 + 2x - 4.
If one gives you an answer of 0, turn it into (x - a) where a is the value you plugged in to get 0. This is a factor.
You can either keep substituting into f(x) to find the three 0 values, or Do long division: (x^3 + 2 -4) / (x - a) to get a quadratic equation, and factorise it to get the remaining two factors.
You'll get f(x) = (x-a)(x-b)(x-c) where a, b and c are three values where f(x) = 0.
Then solve (x-a)=0, (x-b)=0, (x-c)=0, to get the three (it could be one or two) x values.
I guess cubic formula is more helpful.
x^3 + 2x - 4 = 0
Factors of -4 (c value) are 1, -4, 2, -2, -1, 4.
Plug all of these into f(x) = x^3 + 2x - 4.
If one gives you an answer of 0, turn it into (x - a) where a is the value you plugged in to get 0. This is a factor.
You can either keep substituting into f(x) to find the three 0 values, or Do long division: (x^3 + 2 -4) / (x - a) to get a quadratic equation, and factorise it to get the remaining two factors.
You'll get f(x) = (x-a)(x-b)(x-c) where a, b and c are three values where f(x) = 0.
Then solve (x-a)=0, (x-b)=0, (x-c)=0, to get the three (it could be one or two) x values.
I guess cubic formula is more helpful.
5BLD
Legend
Calculator/Math Thread
Quick question:
When is the real part of i^x equal to its imaginary part, and define its value. Don't gimme lots of decimal places just gimme exact valüę. Lets see who does it first.
As for something I have been thinking about... Is x^x really a function? I mean, it doesn't have an inverse function. Although y=>x is a one to many function the same goes for y=x^2, where x=±sqrt(x), a one to many function. But how about arcsinx. Hmmmmmm.
Also my new favourite function is (-1)^x because of its interesting graph.
@ben:
Hast youdunit? Sorry am not on skiep because well you know why. Show me what you've got when i am back on skiep.
Quick question:
When is the real part of i^x equal to its imaginary part, and define its value. Don't gimme lots of decimal places just gimme exact valüę. Lets see who does it first.
As for something I have been thinking about... Is x^x really a function? I mean, it doesn't have an inverse function. Although y=>x is a one to many function the same goes for y=x^2, where x=±sqrt(x), a one to many function. But how about arcsinx. Hmmmmmm.
Also my new favourite function is (-1)^x because of its interesting graph.
@ben:
Hast youdunit? Sorry am not on skiep because well you know why. Show me what you've got when i am back on skiep.
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stannic
Regular
i ^ (1/2) = sqrt(2)/2 + i * sqrt(2)/2
I think x[SUP]x[/SUP] is a function but it is not always invertible.
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5BLD
Legend
Calculator/Math Thread
Yeah thats right, nice. Prove it fully though.
As for x^x, well it acts weirdly. You can't integrate it... And some other funny stuff...
Yeah thats right, nice. Prove it fully though.
As for x^x, well it acts weirdly. You can't integrate it... And some other funny stuff...
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stannic
Regular
i = 0 + 1i = cos (pi/2) + i sin (pi/2)
i^(1/2) = cos ((pi/2 + 2*pi*k)/2) + i sin ((pi/2 + 2*pi*k)/2), k=0..1
i^(1/2) = cos (pi/4 + pi*k) + i sin (pi/4 + pi*k), k=0..1
k=0: pi/4 + pi*k = pi/4
i^(1/2) = cos (pi/4) + i sin (pi/4) = sqrt(2)/2 + i sqrt(2)/2
k=1: pi/4 + pi*k = 5*pi/4
i^(1/2) = cos (5*pi/4) + i sin (5*pi/4) = -sqrt(2)/2 - i sqrt(2)/2
i^(1/2) = cos ((pi/2 + 2*pi*k)/2) + i sin ((pi/2 + 2*pi*k)/2), k=0..1
i^(1/2) = cos (pi/4 + pi*k) + i sin (pi/4 + pi*k), k=0..1
k=0: pi/4 + pi*k = pi/4
i^(1/2) = cos (pi/4) + i sin (pi/4) = sqrt(2)/2 + i sqrt(2)/2
k=1: pi/4 + pi*k = 5*pi/4
i^(1/2) = cos (5*pi/4) + i sin (5*pi/4) = -sqrt(2)/2 - i sqrt(2)/2
MTGjumper
Veteran
5BLD: Argand diagrams make your question easy. Also, a function doesn't have to have an inverse to make it a function. In that case, it's a bijection.
ben1996123
Banned
another way of doing it:
i^x = e^ixpi/2 = cos(xpi/2) + isin(xpi/2)
Re(i^x)=Im(i^x) when cos(xpi/2)=sin(xpi/2)
cos(pi/4)=sin(pi/4), so xpi/2 = pi/4
x=1/2
Re(i^x)=Im(i^x) when cos(xpi/2)=sin(xpi/2)
cos(pi/4)=sin(pi/4), so xpi/2 = pi/4
x=1/2
well it dose but W(stuff) and tungstenizing stuff is stupid.
the Im(stuff) = 0 question? noux, not done anything on that since thou died from skiep. also i dost not know why thou died from skiep .
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gaussgenius
Member
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Thanks for the feedback! We limited the memory to the computational platform that is why some parts got approximated. We fixed that by increasing the memory limit way higher. In addition, we improved the command so that it is even more convenient to type polynomials. For x^2000-1 you used to have to type poly_x(1,0,0,...,-1). Now you can just type poly_x[x^2000-1]
And for the problem, poly_x[(x^2+2*x+1)^999] now gives the exact correct solution (you will have to wait a while though due to the intensive computation going on...)
By the way, thanks for letting us know!
The RapidCalculator Team
Try RapidCalculator at:
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http://rapidcalculator.proalgorithms.com
instructions on how to use at
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5BLD
Legend
Calculator/Math Thread
-that is how i did it, too, stannic your ways interesting tho.
-lol tungsten function. And Li(x) is the lithium function.
-cuz i am nothear and entonces just fohen. Meh i havent done much on it at all, cuz no skiep. Will do moar tomorrow when I am here
Yeah they do indeed, i did it algebraically but you can do it with the complex plane too as you say.
And yeah the function thing... I didn't say a function needs an inverse but i just thought it was strange that it did not have nor has an integral function...
-that is how i did it, too, stannic your ways interesting tho.
-lol tungsten function. And Li(x) is the lithium function.
-cuz i am nothear and entonces just fohen. Meh i havent done much on it at all, cuz no skiep. Will do moar tomorrow when I am here
Yeah they do indeed, i did it algebraically but you can do it with the complex plane too as you say.
And yeah the function thing... I didn't say a function needs an inverse but i just thought it was strange that it did not have nor has an integral function...
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IanTheCuber
Regular
I never knew this thread would become so popular: try and think of some math puzzles. Here's mine:
Pick a 3-digit number. However, the first and last digits must have a difference of more than 1. So, 493 Would not work sonce 4-3=1. Also, for you smart-alecs, you can't do 036 or 092 XD. So, your number is 427.
Flip the number and subtract so that the result will be a positive number. 724-427=297.
Now, flip the number and ADD the numbers. 792+297=1089. The answer is ALWAYS 1089. Why? Honestly it is VERY hard to put it in mathematical terms. So, if anyone else knows how, do it.
PS: To admins, I was joking around about the smart-alec thing.
Pick a 3-digit number. However, the first and last digits must have a difference of more than 1. So, 493 Would not work sonce 4-3=1. Also, for you smart-alecs, you can't do 036 or 092 XD. So, your number is 427.
Flip the number and subtract so that the result will be a positive number. 724-427=297.
Now, flip the number and ADD the numbers. 792+297=1089. The answer is ALWAYS 1089. Why? Honestly it is VERY hard to put it in mathematical terms. So, if anyone else knows how, do it.
PS: To admins, I was joking around about the smart-alec thing.
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5BLD
Legend
Calculator/Math Thread
let number be 100a+10b+c
by subtracting its reverse (100c+10b+a) we get
99(a-c)
[Or more like abs(the stuff)]
Which is always a three digit multiple of 99 assuming a and c differ by more than two (aha).
All multiples of 99 of 3 digits have the middle digit as 9, and the first and last digits always add up to 9. Hence you will always have 909+2*(90) when adding its reverse which is 1089
let number be 100a+10b+c
by subtracting its reverse (100c+10b+a) we get
99(a-c)
[Or more like abs(the stuff)]
Which is always a three digit multiple of 99 assuming a and c differ by more than two (aha).
All multiples of 99 of 3 digits have the middle digit as 9, and the first and last digits always add up to 9. Hence you will always have 909+2*(90) when adding its reverse which is 1089
vcuber13
Veteran
i also solved it, a bit differently though
but i dont see why you cant have a 0 or have a difference of 2:
number: 940
940-049=891
891+198=1089
number: 554
554-445=990
990+099=1089
but i dont see why you cant have a 0 or have a difference of 2:
number: 940
940-049=891
891+198=1089
number: 554
554-445=990
990+099=1089
ben1996123
Banned
ok so, today me and 5bld just torked about fractals and the mandelbrot set and julia sets. lol julia was a man. 5bld made a program to generate an image of the mandelbrot set so I decided to do it with opengl but then I didn't feel like doing any programming today. heres a cool image from 5blds program:
5BLD
Legend
Calculator/Math Thread
Ohai ben. Tomorrow (EDIT: today-- jeez it's late, or early, do you ever sleep?) lets both make julia set prodders. Also heres some stuff:
If |z|>2, |z^2+c|>=|z^2|+|c|>2|z|-|c|
If |z|>|c|, 2|z|-|c|>|z|
So |z^2+c|>|z|
Therefore its increasing
also, about 86% but imma hybërñæt it.
Ohai ben. Tomorrow (EDIT: today-- jeez it's late, or early, do you ever sleep?) lets both make julia set prodders. Also heres some stuff:
If |z|>2, |z^2+c|>=|z^2|+|c|>2|z|-|c|
If |z|>|c|, 2|z|-|c|>|z|
So |z^2+c|>|z|
Therefore its increasing
also, about 86% but imma hybërñæt it.
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ben1996123
Banned
hi can somepony help me with this question pree
find the telegraphic fourier set potential when the riemann surface begins to integrate through the pair of simultaneous multi-variable exponentially polynomial 5-parameter complex functions, and then from this deduce why there's no euler surface for this sort of plane. give your answer in terms of ∫.
find the telegraphic fourier set potential when the riemann surface begins to integrate through the pair of simultaneous multi-variable exponentially polynomial 5-parameter complex functions, and then from this deduce why there's no euler surface for this sort of plane. give your answer in terms of ∫.
5BLD
Legend
extension: invent a new theorem by pythagoras based on your findings
Ickathu
Veteran
So these are two problems in one of the Mandlebrot Competition tests that I can't solve. I'm in a mandelbrot club, and we were assigned these sets of problems, and I'm stuck on 2 of them.
The problems get progressively harder (8 in each test) and I think this is the best I've gotten so far on any of the practice tests.
"6. The quantity
(tan(pi/5)+i) / (tan(pi/5)-i)
is a tenth root of unity. In other words, it is equal to cos(2npi/10) + isin(2npi/10) for some integer n between 0 and 9 inclusive. Which value of n?"
No idea where to even start on that one.
"8. There is a unique triangle ABC such that AC = 14, cos(A) = 4/5, and the radius of the inscribed circle is r = 4. Find the area of this triangle."
I drew the picture, but I'm not really sure where to go from there.
The problems get progressively harder (8 in each test) and I think this is the best I've gotten so far on any of the practice tests.
"6. The quantity
(tan(pi/5)+i) / (tan(pi/5)-i)
is a tenth root of unity. In other words, it is equal to cos(2npi/10) + isin(2npi/10) for some integer n between 0 and 9 inclusive. Which value of n?"
No idea where to even start on that one.
"8. There is a unique triangle ABC such that AC = 14, cos(A) = 4/5, and the radius of the inscribed circle is r = 4. Find the area of this triangle."
I drew the picture, but I'm not really sure where to go from there.
CarlBrannen
Veteran
On (6) it's easy to see that (tan(pi/5) + i)^10 = (tan(pi/5) - i)^10 but the two sides are complex conjugates. For them to be equal they must both be real.
Therefore tan(pi/5) +i is a 10th root of a real number. But its squared magnitude is tan^2(pi/5) + 1 = sec^2(pi/5), so therefore:
(tan(pi/5) + i)/sec(pi/5) = cos(2n pi/10) + i sin(2n pi/10)
and so cos(pi/5) = sin(2n pi/10) which is easy to solve, one finds n=2:
sin (2x2 pi/10) = sin( 2 pi/5) = cos( pi/2 - 2 pi/5) = cos(pi/5), as desired.
Therefore tan(pi/5) +i is a 10th root of a real number. But its squared magnitude is tan^2(pi/5) + 1 = sec^2(pi/5), so therefore:
(tan(pi/5) + i)/sec(pi/5) = cos(2n pi/10) + i sin(2n pi/10)
and so cos(pi/5) = sin(2n pi/10) which is easy to solve, one finds n=2:
sin (2x2 pi/10) = sin( 2 pi/5) = cos( pi/2 - 2 pi/5) = cos(pi/5), as desired.
CarlBrannen
Veteran
(8) It's easy to make mistakes on this sort of problem. Here's a quick attempt. I'm sure it's the right way of doing it, but I'll leave it to you to figure out where any errors are.
It's very useful to know that if you choose an arc BC of a circle, the measure of the arc (which will be between 0 and 359.999... degrees) will always be twice the angle you would get by choosing any point on the arc A, and measuring the angle BAC (which will be between 0 and 179.999... degrees).
Thus the angular measure of side A is twice the arc cosine of 4/5. Half this angle gives half the measure of the arc. So this defines a "little right triangle" with ratios of sides of 3/5 to 4/5 to 1. The right angle of this "little right triangle" is on the midpoint of side A, so we can use this "little right triangle" to find the length of A. The hypotenuse of the "little right triangle" is a radius of the circle so has length 4. Thus the other sides have lengths 4(3/5)=12/5 and 4(4/5)=16/5. To figure out the length of A we want the 3/5 side (i.e. the sine side rather than the cosine side) so we have A = 2 (12/5) = 24/5. Since AC = 14, this gives C = 14/(24/5) = 35/12.
I don't recall all the trig formulas, but I do recall that sin(A)/A = sin(B)/B = sin (C)/C. And sin(A) = 3/5 so this gives me sin(C) = (3/5)(35/12)/(24/5) = 35/96.
Now to measure the area, break it into two parts by drawing a perpendicular from point B down to side B. By trig, this perpendicular has length C cos(A) = (35/12)(4/5) = 7/3. The area of the triangle is therefore half of (7/3) B = (7/6)B.
But B = C cos(A) +- A cos(C). I put +- because you don't know from the equations whether angle B is acute or obtuse. To figure that out, simply look at sin(A)=3/5 and sin(C) = 35/96. Does angle A and C add up to less than or more than 90 degrees? If less than, use the obtuse case, if more than, use the acute case. In the event, both sin(A) and sin(C) are smaller than sqrt(1/2) and so A and C are less than 90 degrees so they add up to less than 90 degrees. Therefore angle B is obtuse and we must use the + sign.
From here it's a matter of substitution.
It's very useful to know that if you choose an arc BC of a circle, the measure of the arc (which will be between 0 and 359.999... degrees) will always be twice the angle you would get by choosing any point on the arc A, and measuring the angle BAC (which will be between 0 and 179.999... degrees).
Thus the angular measure of side A is twice the arc cosine of 4/5. Half this angle gives half the measure of the arc. So this defines a "little right triangle" with ratios of sides of 3/5 to 4/5 to 1. The right angle of this "little right triangle" is on the midpoint of side A, so we can use this "little right triangle" to find the length of A. The hypotenuse of the "little right triangle" is a radius of the circle so has length 4. Thus the other sides have lengths 4(3/5)=12/5 and 4(4/5)=16/5. To figure out the length of A we want the 3/5 side (i.e. the sine side rather than the cosine side) so we have A = 2 (12/5) = 24/5. Since AC = 14, this gives C = 14/(24/5) = 35/12.
I don't recall all the trig formulas, but I do recall that sin(A)/A = sin(B)/B = sin (C)/C. And sin(A) = 3/5 so this gives me sin(C) = (3/5)(35/12)/(24/5) = 35/96.
Now to measure the area, break it into two parts by drawing a perpendicular from point B down to side B. By trig, this perpendicular has length C cos(A) = (35/12)(4/5) = 7/3. The area of the triangle is therefore half of (7/3) B = (7/6)B.
But B = C cos(A) +- A cos(C). I put +- because you don't know from the equations whether angle B is acute or obtuse. To figure that out, simply look at sin(A)=3/5 and sin(C) = 35/96. Does angle A and C add up to less than or more than 90 degrees? If less than, use the obtuse case, if more than, use the acute case. In the event, both sin(A) and sin(C) are smaller than sqrt(1/2) and so A and C are less than 90 degrees so they add up to less than 90 degrees. Therefore angle B is obtuse and we must use the + sign.
From here it's a matter of substitution.
