[metaphor-]

yet another math problem game. Here are the basic rules. Solve the math problem, then post a new problem for The Person Below You.. They can take any form and be as complicated as you like.

IE: 32= 2x, find x.... x=16 , now I will post a new problem for TPBM
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18 = 2(x+4)
what is x?

[fuscia]

5.

what's the square root of 69?

[metaphor-]

sqrt(69)=8.306
basically, 8^2= 64 9^2=81 therefore its between 8 and 9..then i did a lot of calculations in the brain.....

Find the slope of the line that passes through the points (-1 , 0) and (3 , 8).

[qtwerp]

2

What is the mode of this set: 1, 5, 8, 3, 9, 4, 4, 1, 6, 7, 8, 3, 1, 7, 2, 2, 4

[metaphor-]

that set is bimodal, 4 and 1.

-------------------------------
-15x + 47 = -103
solve for x

[qtwerp]

10.

4% of 1500

[metaphor-]

4%= 0.04

1500
*0.04
------
6000
0000
0000
-------
60.00

---------------------------------
what can x equal?
(x + 1)(x – 3) = 0

[fuscia]

Originally Posted by metaphor- sqrt(69)=8.306 basically, 8^2= 64 9^2=81 therefore its between 8 and 9..then i did a lot of calculations in the brain..... Find the slope of the line that passes through the points (-1 , 0) and (3 , 8).

i would have settled for eight something.

[Jesus]

Originally Posted by metaphor- what can x equal? (x + 1)(x – 3) = 0

-1 and 3

Whats wrong:

i = (-1)^{1/2} = ((-1)^{3})^{1/2} = (-1)^{3/2} = ((-1)^{1/2})^{3} = (i)^{3} = -i
\Rightarrow 1 = -1

Whats wrong:

x^2 - x^2 = x^2 - x^2
x(x - x) = (x - x)(x + x)
x = (x + x)
x = 2x
\Rightarrow 1 = 2

[qtwerp]

-1^3/2 is -i and not i. Euler's identity: e^(i*pi) = -1. Should be using that instead since there's a wave function in there.

I think the second doesn't work because of a domain issue, but I'm not sure any more. I'd have to go through a formal proof to conclusively find what the fault is.

-------------------

A bit more difficult one this time:

x²y'' + 3xy' + y = 0

[metaphor-]

in response to the last one, did you enter that wrong? whats with the quotation marks? want it simplified?
simplified its -y= (-2/x^2) + 3/x ...assuming that y'' is y^2...still working on the other, whats wrong one...my skills need to be refreshed.

[Jesus]

Originally Posted by qtwerp -1^3/2 is -i and not i. Euler's identity: e^(i*pi) = -1. Should be using that instead since there's a wave function in there.

The wave function is a concept of physics, the exponential function is it called to a mathematician. How ever, it is true that it is suppose to be -i, but that is not the answer to whats wrong with it.

Originally Posted by metaphor- in response to the last one, did you enter that wrong? whats with the quotation marks? want it simplified? simplified its -y= (-2/x^2) + 3/x ...assuming that y'' is y^2...still working on the other, whats wrong one...my skills need to be refreshed.

y'' and y' are the second and first derivative w.r.t. x. Its a differential equation. (Personally I don't solve those without my books )

[qtwerp]

Originally Posted by Jesus The wave function is a concept of physics, the exponential function is it called to a mathematician. How ever, it is true that it is suppose to be -i, but that is not the answer to whats wrong with it.

Oh well, I'm weak with these
There's a sin and cos (or a sinh or cosh) floating around there somewhere

You posted the answer to the previous question seconds before I did, so I tried yours

Originally Posted by metaphor- in response to the last one, did you enter that wrong? whats with the quotation marks? want it simplified? simplified its -y= (-2/x^2) + 3/x ...assuming that y'' is y^2...still working on the other, whats wrong one...my skills need to be refreshed.

No, it's a "simple" Cauchy-Euler differential equation. y' is the first derivative of y and y'', the second derivative. It should be possible without having to write down a lot of work. It's not that difficult if you've done them before.

Here's the big hint for it: y = x^m

[metaphor-]

currently doing a bit of research on those, i have not been to a math class in a few years...Why do i get the feeling someone is going to end up posting their homework on this thread...looking at:
Whats wrong:

i = (-1)^{1/2} = ((-1)^{3})^{1/2} = (-1)^{3/2} = ((-1)^{1/2})^{3} = (i)^{3} = -i
\Rightarrow 1 = -1

Whats wrong:

x^2 - x^2 = x^2 - x^2
x(x - x) = (x - x)(x + x)
x = (x + x)
x = 2x
\Rightarrow 1 = 2

[Jesus]

Originally Posted by qtwerp x²y'' + 3xy' + 2y = 0 [---] No, it's a "simple" Cauchy-Euler differential equation. y' is the first derivative of y and y'', the second derivative. It should be possible without having to write down a lot of work. It's not that difficult if you've done them before. Here's the big hint for it: y = x^m

Thanks for the hint

m ^2 x^m + 3 m x^m + 2 x^m = 0

(x != 0)

m^2 + 3m + 2 = 0

m = -1, m = -2

--------------------------

How about:

Prove:

\sum_{n = 1}^{\infty} 1/n^s = 0 \iff \Real(s) = 1/2 where s != {-2,-4,-6...}


Nahh... just messing with you. Just state the name of the problem

[Extreme Coder]

o_O; Some wierd mathematics I've never heard of. But that's ok considering I didn't register for Math this year.

[DChristopher]

Originally Posted by Jesus Thanks for the hint m ^2 x^m + 3 m x^m + 2 x^m = 0 (x != 0) m^2 + 3m + 2 = 0 m = -1, m = -2 -------------------------- How about: Prove: \sum_{n = 1}^{\infty} 1/n^s = 0 \iff \Real(s) = 1/2 where s != {-2,-4,-6...} Nahh... just messing with you. Just state the name of the problem

That's the Riemann Hypothesis.

I have a proof, but the margin is too small to contain it...

[Jesus]

Originally Posted by DChristopher That's the Riemann Hypothesis. I have a proof, but the margin is too small to contain it...

ROTFLMFAO! Fermat wannabe

[metaphor-]

no one posted a new problem, so the floor is open for a new one!

[qtwerp]

Originally Posted by Jesus Thanks for the hint m ^2 x^m + 3 m x^m + 2 x^m = 0 (x != 0) m^2 + 3m + 2 = 0 m = -1, m = -2

Not quite (and you didn't finish the problem!). Since I didn't say whether to find the roots or solve for y, I guess it's ok Please note that I did change the number in front of y from 2 to 1 so that the math worked out without a fraction or decimal which was what I was trying to do in the first place (you'd need the quadratic eqn. for m^2+2m+2=0). Actually turned out a little worse with repeated roots, but it works...

Here's the solution:

y = x^m
y' = m*x^(m-1)
y'' = m*(m-1)*x^(m-2)

Plugging it in:
(x^m)( m(m-1) + 3m + 1) = 0
m^2 - m + 3m + 1 = 0
m^2 +2m + 1 = 0
(m + 1)(m + 1) = 0
m = -1; m = -1 <-- Repeated root!

Cauchy-Euler type equations use the form c₁x^m₁ + c₂*x^m₂ and when there are repeated real roots, you multiply the second by ln(x)

y = c₁*x^(-1) + c₂*x^(-1)*ln(x)

Maybe we should stick to math that most people have taken before
If anyone likes math though, suffer through Calculus; Differential Equations was the best class I ever took in college!

------------

These still haven't been solved correctly!

i = (-1)^{1/2} = ((-1)^{3})^{1/2} = (-1)^{3/2} = ((-1)^{1/2})^{3} = (i)^{3} = -i
\Rightarrow 1 = -1

Whats wrong:

x^2 - x^2 = x^2 - x^2
x(x - x) = (x - x)(x + x)
x = (x + x)
x = 2x
\Rightarrow 1 = 2

Edit: If you want something new, how about:

Solve for x, y, and z:

2x + 3y + z = 0
x − 2y − z = −3
x + y + 2z = 3