Thread: Equations

x=0, x=sqrte.

Probably went wrong. :l

Differentiating ln(-x) is a bit weird.

I think the answers are
x=0 and x=e
See solution below

Spoiler for solution:

(x^2*ln(-x))' = x*ln(x)
2*x*ln(-x) + x^2*(1/-x)*(-1) = x*ln(x)
2*x*ln(-x) - x - x*ln(x) = 0 | :x (x/=0)
2*ln(-x) - 1 - ln(x) = 0
2*ln(-x) - ln(x) = 1
ln((-x)^2) - ln(x) = 1
ln((x^2)/x) = 1
ln(x) = 1
e^ln(x) = e^1
x = e

Test x=0
2*0*ln(0) + 0*(1/0)*(-1) = 0*ln(0)
0 = 0
Works (I think)

That wasn't actually the equation I intended to make, but so what. It worked

Oh yeah, x=e is right, I left the 2 in front of the ln when I simplified...

x=0 isn't actually a solution because ln0 is undefined.

Originally Posted by Xei

x=0 isn't actually a solution because ln0 is undefined.

Yeah, I was wondering about that too. But it's timed with 0, and since anything timed 0 equals 0, wouldn't also undefined*0 equal 0?
But at any rate that's why I added the '(I think)' in the solution :p

That's a topic for analysis I think... but basically lnx tends to -infinity as x tends to 0, so you've got 0 * - infinity; this is ambiguous because 0 times anything is 0 but infinity times anything is infinity.

cos(x)^2 + cos(x)^2 = 1

somebody that's not named after a greek letter solve for x....

cos^2(x) + cos^2(x) = 1
2cos^2(x) = 1
cos^2(x) = (1/2)
cos(x) = sqrt(1/2)
x = +/-45˚ + n360˚

yup.

I meant to say.

cos(x)^2 + sin(x)^2 = 1

It is a trick question.

Originally Posted by aaronasterling

yup.
cos(x)^2 + sin(x)^2 = 1

Hehe, that one is pretty easy.
x € <<--, -->>
Isn't that how one writes it?

New equation:
(y')^2 = 8x
Solve the equation when y = 0 and c = -16

Heres a related one:

Give a polynomial equation which vanishes on the image of the function

f: R -> R x R
x -> (cos(x), sin(x))

Spoiler for hint:

This is just a partial restatement of the theorem that cos(x)^2 + sin(x)^2 = 1. The challenge is to complete that restatement by giving the polynomial equation

i cant see clearly and am not used to the notations we have available to us. thats...

(dy/dx)^2 = 8x ?

if so then x = 4. been too long since I've done this stuff.

Originally Posted by aaronasterling

i cant see clearly and am not used to the notations we have available to us. thats...

(dy/dx)^2 = 8x ?

Yeah, you could write it like that. I'm pretty sure you could also write
y' (dy/dx) = 8x
Not sure though.
Edit:
If so, then it's easy. If not, then I have no idea how to solve it. Or if it can be solved.

Originally Posted by aaronasterling

Heres a related one:

Give a polynomial equation which vanishes on the image of the function
f: R -> R x R
x -> (cos(x), sin(x))

I don't think I've seen something like that before. Is that an inequality?

The email notification didn't let me see that you'd responded. wierd.

I wrote it a different way out of habit....

R is the real numbers. R x R also written R^2 can be thought of as the cartesian plane. It's usual to be explicit about what the domain and codomain of a function is. so all I'm saying is f(x) = (cos(x), sin(x)) where the right hand side is a vector.

does that help?

Originally Posted by khh

Yeah, you could write it like that. I'm pretty sure you could also write
y' (dy/dx) = 8x
Not sure though.
Edit:
If so, then it's easy. If not, then I have no idea how to solve it. Or if it can be solved.

I got the answer 4: see the post where I asked. it is trivial. I am terrible at differential equations though so I don't mind

Originally Posted by aaronasterling

R is the real numbers. R x R also written R^2 can be thought of as the cartesian plane. It's usual to be explicit about what the domain and codomain of a function is. so all I'm saying is f(x) = (cos(x), sin(x)) where the right hand side is a vector.

Well, all the words make sense but my mind can't seem to put them together. Guess someone else will have to solve it.
I think I got 2 when I solved the differential.

(y')^2 = 8x
y' (dy/dx) = 8x
y'dy = 8xdx
y = 4x^2 + c
16 = 4x^2
x = sqrt(4) = 2

I don't know if step 1 --> 2 is legal, though.

Originally Posted by khh

Well, all the words make sense but my mind can't seem to put them together. Guess someone else will have to solve it.
I think I got 2 when I solved the differential.

(y')^2 = 8x
y' (dy/dx) = 8x
y'dy = 8xdx
y = 4x^2 + c
16 = 4x^2
x = sqrt(4) = 2

I don't know if step 1 --> 2 is legal, though.

funny, I messed up then. I got:

(y')^2 = 8x
y' = (8x)^1/2
y = 2x^3/2 - 16 and then solve that for 4. i'll look at it more. I'd at least like to know if i messed up my calculus or algebra like I said, it's been years.

EDIT: damn it, now I'm gonna have to hit my calculus book and solve all the problems. thanks alot khh the problem is that I just glossed over the part where you magically move the dx to the other side because I couldn't justify it and figured i'd wait till i could to study that technique. I got into algebra instead.

as far as the one that I posed goes, the answer is easy if you understand the question: it's

Spoiler for answer:

x^2 + y^2 - 1 = 0

say we let g(x, y) = x^2 + y^2 - 1. Then gf(x) = cos(x)^2 + sin(x)^2 - 1 = 1 - 1 = 0.

the function f(x) = (cos(x), sin(x)) draws the unit circle in the plane!
the function g(x,y) = x^2 + y^2 - 1 vanishes on the unit circle in the plane.

IMO, cos and sin are most naturaly thought of as being the components of the function f that I gave: they relate linear measure to anglular measure.

Originally Posted by aaronasterling

funny, I messed up then. I got:

(y')^2 = 8x
y' = (8x)^1/2
y = 2x^3/2 - 16 and then solve that for 4. i'll look at it more. I'd at least like to know if i messed up my calculus or algebra like I said, it's been years.

I think you should get
y' = sqrt(8x)
y = (1/(3/2))(8x)^(3/2) + c
16 = 2/3*(8x)^(3/2)
(24)^2 = (8x)^(3)
(576/(9*8^3) = x^3
(576/4608) = x^3
x = (1/2)
...
Making equations is harder than I thought.

Originally Posted by khh

Making equations is harder than I thought.

especially differential equations.....

did the solution to the one I posed make any sense?

Originally Posted by aaronasterling

especially differential equations.....

did the solution to the one I posed make any sense?

It made some sense, but not enough so that I really understood it. I think the language barrier and the fact that I possibly haven't learned all of it yet was a bit hard to overcome :p

There's two different solutions I believe; you've got y'^2 = f(x) but you have to bear in mind that this means that y' = +/- f(x)^1/2