Math Is Fun Forum

  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#376 Re: Help Me ! » Find real solutions » 2007-11-01 05:31:12

I've been giving out a lot of wrong answers.

Going back to our equation 6sin(x) = 4sin(y) = 3sin(z), solve for both sin(x) and sin(y) in terms of z:
sin(x) = (1/2) sin(z)
sin(y) = (3/4) sin(z)

Now take the arcsin of these equations:
x = arcsin( (1/2) sin(z) )
y = arcsin( (3/4) sin(z) )

We know this operation is legal since the range of the sin function is -1 to 1, which is the domain of the arcsin function.  By multiplying sin(z) by a value less than 1 we stay within arcsin's domain.

From here choose any real value for z and plug it into these equations for a legal triple.  Stated more mathematically:


Note that there's probably a way of simplifying from here, but I don't know how.


Edit: Well, on second thought it doesn't seem to work out very well.  Taking Jane's example of x = pi/6, y = arcsin(3/4), and z = pi/2 we get sin(x) = 1/2, sin(y) = 3/4, and sin(z) = 1.  This means that we need sin(x + y + z) = -1/4 to solve the original 3 equations.  Unfortunately, sin(x + y + z) in this example equals 0.1978.....  It appears that something is wrong here.

#377 Re: Help Me ! » Help with number theory proof please » 2007-11-01 01:44:17

You still need to prove that it's impossible for X = 1/2.

To do this, consider that

Now we need to prove that

It's important to note that

Knowing this, we also know that

is always even, since all prime numbers after 2 are odd.  This means that the prime factorization of the numberator will always have at least one 2 in it.  We also know that obviously the denominator will never have a 2 in it's prime factorization, since all the denominators are prime numbers greater than 2.  Since we have a prime factor in the numerator that can never be cancelled by the denominator we know that X can never reach 1/2, so that should complete your proof.

#378 Re: Help Me ! » functions » 2007-10-31 03:44:11

Heh, your method is a lot easier.

#379 Re: Help Me ! » functions » 2007-10-31 01:43:31

Set f(a) = b.  Substitute to get g(b) = g(f(a)), which we know from the definition of g is equal to a.  Thus, g(b) = a.

Do the same for h to get h(f(a)).  We know from the definition of h that f(h(f(a))) = f(a), which tells us that h(f(a)) = a.  Substitute back to get h(b) = a = g(b).

#380 Re: Help Me ! » simple matrice invert » 2007-10-26 02:14:22

You'd be amazed at how quickly you can forget even the simple things.

#381 Re: Help Me ! » simple matrice invert » 2007-10-26 02:04:16

Like I said, I'm not good with matrices, so I had to use this Wikipedia page: http://en.wikipedia.org/wiki/Cofactor_%28linear_algebra%29


Cofactor(A) =


is equal to the determinant of the matrix you get when you eliminate the
row and the
column of A.

For example:

I leave the rest to you.

#382 Re: Help Me ! » simple matrice invert » 2007-10-26 01:01:56

Your cofactor(A) doesn't look correct to me.  I'm not great with matrices, but I think this is the cofactor:

Cofactor(A) = 1  -10  -7
                    -7   10  -11
                    17   10  1
= M

I think this is how you get to M from A.

#383 Re: Help Me ! » Find real solutions » 2007-10-25 06:30:40

Take each individual equation and solve for sin(x + y + z).

So, from your first equation we get 2sin(x + y + z) = -sin(x) ---> sin(x + y + z) = -(1/2)(sin(x)).  Doing the same for the other 2 equations gives you sin(x + y + z) = -(1/3)(sin(y)) and sin(x + y + z) = -(1/4)(sin(z)).

Now, since sin(x + y + z) appears by itself in all 3 equations, we know the right side of each equation is equal to each other, so we now have -(1/2)(sin(x)) = -(1/3)(sin(y)) = -(1/4)(sin(z)).  Multiply through by -12 to get 6sin(x) = 4sin(y) = 3sin(z).

Take the first equation, 6sin(x) = 4sin(y).  Since both sides are of the form a*sin(w), where a is a constant and w is a variable, the graphs of both with have the same period and initial position.  The only difference is the amplitude, which will cause the graphs not to touch anywhere except where sin(w) = 0.  This same argument can be used for 3sin(z).

This gives us our final equation of sin(x) = sin(y) = sin(z) = 0.  As you probably know, this is true when x, y, and z are any integers multiplied by pi.

#384 Re: Help Me ! » Number theory proof help » 2007-10-24 14:54:53

As always, wikipedia can get you started:  http://en.wikipedia.org/wiki/Euler%27s_totient_function

#385 Re: Help Me ! » ln A=ln h - ln(1-h); Solve for h » 2007-10-24 14:51:14

Better late than never:

This isn't really a coding problem, but here's the method anyway.  From the original equation we know a couple facts about A and h.  First, we know that A > 0 (strictly greater than) because ln is undefined for values of A <= 0.  For the same reason we know that h > 0.  However, we also know that 1 - h > 0, which implies h < 1 (again, strictly).   This gives us A > 0 and 0 < h < 1.

Now, we first simplify the equation by combining the logs on the right side.  ln(a) - ln(b) = ln(a / b) is a logarithmic identity, so we get ln(A) = ln(h / (1 - h) ).

Now, since the exponential function is continuous over the real numbers we can take the exponential of both sides to remove the logs and get A = h / (1 - h).

From here on out it's just algebra, but for the sake of completeness I'll go step by step.  Since we know both A and h are strictly greater than 0 we can take the reciprocal of both sides to give us 1 / A = (1 - h) / h.  We then split up the fraction on the right side to get 1 / A = (1 / h) - 1.  Add 1 to give us 1 + (1 / A) = 1 / h --------> (A + 1) / A = 1 / h.  Again, A + 1 > 0, so take the reciprocal again to give us A / (A + 1) = h   QED.

Board footer

Powered by FluxBB