The function from hell.

[Q_Me]

f(xy) = f(x) + f(y)

f(2) = 0.3010

Find f(3).

[beros87]

I don't think that the function is correct in the first place because no way that f(xy)= f(x) + f(y). Or maybe the question is missing.

[alkatran]

Hint: log()

[Jacob Roman]

Let me get a whack at it:

f(xy) = f(x) + f(y)

f(2) = 0.3010

Find f(3). Right?

f(3) = 0.4771

So that would mean that

     f(xy) = f(x) + f(y)

     f(2*3) = f(2) + f(3)

     f(6) = f(2) + f(3)
 
     f(6) = 0.3010 + 0.4771

     0.7781 =  0.3010 + 0.4771

Let's double check: Log(6) = 0.7781

That was too easy!

Now try working out the math's behind cubic patches (aka Functions from Hell!!!):

X(a,c) = Axa³c³ + Bx3a³c²(1-c) + Cx3a³c(1-c)² + Dxa³(1-c)³ + Ex3a²(1-a)c³ +
            Fx9a²(1-a)c²(1-c) + Gx9a²(1-a)c(1-c)² + Hx3a²(1-a)(1-c)³ + Ix3a(1-a)²c³ + 
            Jx9a(1-a)²c²(1-c) + Kx9a(1-a)²c(1-c)² + Lx3a(1-a)²(1-c)³ +
            Mx(1-a)³c³ + Nx3(1-a)³c²(1-c) + Ox3(1-a)³c(1-c)² + Px(1-a)³(1-c)³       

Y(a,c) = Aya³c³ + By3a³c²(1-c) + Cy3a³c(1-c)² + Dya³(1-c)³ + Ey3a²(1-a)c³ +
            Fy9a²(1-a)c²(1-c) + Gy9a²(1-a)c(1-c)² + Hy3a²(1-a)(1-c)³ + Iy3a(1-a)²c³ + 
            Jy9a(1-a)²c²(1-c) + Ky9a(1-a)²c(1-c)² + Ly3a(1-a)²(1-c)³ +
            My(1-a)³c³ + Ny3(1-a)³c²(1-c) + Oy3(1-a)³c(1-c)² + Py(1-a)³(1-c)³  

Z(a,c) = Aza³c³ + Bz3a³c²(1-c) + Cz3a³c(1-c)² + Dza³(1-c)³ + Ez3a²(1-a)c³ +
            Fz9a²(1-a)c²(1-c) + Gz9a²(1-a)c(1-c)² + Hz3a²(1-a)(1-c)³ + Iz3a(1-a)²c³ + 
            Jz9a(1-a)²c²(1-c) + Kz9a(1-a)²c(1-c)² + Lz3a(1-a)²(1-c)³ +
            Mz(1-a)³c³ + Nz3(1-a)³c²(1-c) + Oz3(1-a)³c(1-c)² + Pz(1-a)³(1-c)³

[someone_259@hotmail]

Let me get a whack at it:

f(xy) = f(x) + f(y)

f(2) = 0.3010

Find f(3). Right?

f(3) = 0.4771

So that would mean that

     f(xy) = f(x) + f(y)

     f(2*3) = f(2) + f(3)

     f(6) = f(2) + f(3)
 
     f(6) = 0.3010 + 0.4771

     0.7781 =  0.3010 + 0.4771

Let's double check: Log(6) = 0.7781

That was too easy!

Now try working out the math's behind cubic patches (aka Functions from Hell!!!):

X(a,c) = Axa³c³ + Bx3a³c²(1-c) + Cx3a³c(1-c)² + Dxa³(1-c)³ + Ex3a²(1-a)c³ +
            Fx9a²(1-a)c²(1-c) + Gx9a²(1-a)c(1-c)² + Hx3a²(1-a)(1-c)³ + Ix3a(1-a)²c³ + 
            Jx9a(1-a)²c²(1-c) + Kx9a(1-a)²c(1-c)² + Lx3a(1-a)²(1-c)³ +
            Mx(1-a)³c³ + Nx3(1-a)³c²(1-c) + Ox3(1-a)³c(1-c)² + Px(1-a)³(1-c)³       

Y(a,c) = Aya³c³ + By3a³c²(1-c) + Cy3a³c(1-c)² + Dya³(1-c)³ + Ey3a²(1-a)c³ +
            Fy9a²(1-a)c²(1-c) + Gy9a²(1-a)c(1-c)² + Hy3a²(1-a)(1-c)³ + Iy3a(1-a)²c³ + 
            Jy9a(1-a)²c²(1-c) + Ky9a(1-a)²c(1-c)² + Ly3a(1-a)²(1-c)³ +
            My(1-a)³c³ + Ny3(1-a)³c²(1-c) + Oy3(1-a)³c(1-c)² + Py(1-a)³(1-c)³  

Z(a,c) = Aza³c³ + Bz3a³c²(1-c) + Cz3a³c(1-c)² + Dza³(1-c)³ + Ez3a²(1-a)c³ +
            Fz9a²(1-a)c²(1-c) + Gz9a²(1-a)c(1-c)² + Hz3a²(1-a)(1-c)³ + Iz3a(1-a)²c³ + 
            Jz9a(1-a)²c²(1-c) + Kz9a(1-a)²c(1-c)² + Lz3a(1-a)²(1-c)³ +
            Mz(1-a)³c³ + Nz3(1-a)³c²(1-c) + Oz3(1-a)³c(1-c)² + Pz(1-a)³(1-c)³

Wow, just wondering how do you get f(3) = 0.4771 ?

Thank you!

[Rassis]

Hint: log()

...said Alkatran.

log(3) = 0.4771

[Sharps4570]

From the original post, how did you recognize that f(2) = 0.3010 was log(2)?

[Brant]

Because if you know the properties of logarithms you know that Log2(xy) is the same as log2(x) + log2(y)

[Andrew Roberts]

f(xy) = f(x) + f(y) applies to any function. It's not normally used in trigonometric problems and is usually introduced with logarithms also he could have know log2 but if your not sure how to do it just try out f(2) using all the functions on your calculator.

[Rassis]

This thread is a good example of how giving it a strange title – regardless of the content – increases exponentially the number of viewers. Funny, the way we can be trapped by “words”, isn’t it? And marketeers know it very well, may we like it or not!