Evaluating f(xy)=f(x)+f(y)-1

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f(0*y)=f(0)+f(y)-1

So f(y)=1 for every y in R

bartoche

1
because f(x) = 1, k = 1

try y=0
f(0) = f(x) + f(0) - 1
so f(x) = 1
k=1 so 4k-3 = 1

armacham

Let 2 = 16, and let f : {2} —> {k}. Then f(2) = f(16) = k. Q. E. D.

I was told the domain could be anything I wanted it to be.

angelmendez-rivera

One can actually find the explicit function f, by noting the rule is similar to that of logarithms, but not identical. Try substituting f(x)=ln(g(x)). Good luck in completing the steps to find f. See if you can reproduce the 4k-3 result.

meir.zeilig-hess

Here is my solution, together with some other observations (written before I saw the video).

EDIT: I have noted from the comments that if k≠1 then f cannot be defined at 0, otherwise f(2×0)=f(2)+f(0)-1, so f(0)=f(2)+f(0)-1, and f(2)=1, which contradicts f(2)=k.

As f(xy)=f(x)+f(y)-1
we have f(xy)-1=f(x)-1+f(y)-1
Let g(x)=f(x)-1 (so g(2)=k-1)
Then g(xy)=g(x)+g(y)
By induction, g(xⁿ)=ng(x) for n a positive integer (just as for log).
Hence f(xⁿ)=nf(x)-(n-1) for n a positive integer.
Hence f(16)=f(2⁴)=4f(2)-3=4k-3.

Having answered the question, let's investigate the functions f and g a little further.

By setting x=y=1, we get g(1)=g(1)+g(1), so g(1)=0. So the rule g(xⁿ)=ng(x) holds for n=0.

If n is a negative integer and x≠0 then setting "x"=xⁿ and y=x⁻ⁿ, we get g(xⁿx⁻ⁿ)=g(xⁿ)+g(x⁻ⁿ), so g(1)=g(xⁿ)+g(x⁻ⁿ), g(xⁿ)+(-n)g(x)=0, g(xⁿ)=ng(x).

Hence the rule g(xⁿ)=ng(x) (where x≠0 if n<0) holds for all integers n, just as for logs.

In particular, g(2ⁿ)=ng(2)=n(k-1) for any positive integer n.

Now let q be a positive integer and x a real number (where x≥0 if q is even) and let a=ᑫ√x, so x=aᑫ.
Then g(aᑫ)=qg(a), so g(x)=qg(a), g(a)=1/q g(x), and g(x^(1/q))=1/q g(x).

If p is a integer, q a positive integer and x a real number (where x≠0 if p<0 and x≥0 if q is even), then g(x).

In particular, If p is any integer and q any positive integer, then g(2^(p/q))=p/q g(2)=p/q (k-1).

Hence, just as for logs, we have g(xʳ)=rg(x), so f(xʳ)=rf(x)-(r-1), for any rational r.

But don't think that just because g satisfies the key law of logs that it is not defined for x=0 or x<0.
EDIT: as noted above, if k≠1, f cannot be defined at x=0.

For example, setting x=y=0, we get g(0)=g(0)+g(0), so g(0)=0.

Setting x=y=-1, we get g(1)=g(-1)+g(-1), so g(-1)=g(1)/2=0.

In general, if a<0, setting x=|a|, y=-1, we get g(a)=g(x)+g(-1)=g(x)+0=g(x).

So we have g(0)=0, so f(0)=1, and for x<0, g(x)=g(|x|), so f(x)=f(|x|).
EDIT: as noted above, if k≠1, f cannot be defined at x=0.

If it were given that f is continuous, the situation would change somewhat (assuming k≠1). We would then have g(xʳ)=rg(x), so f(xʳ)=rf(x)-(r-1), for any real r.

For any a>0, setting x=2 and r=log₂a (so xʳ=a), we get g(a)=g(xʳ)=rg(x)=log₂a g(2)=(k-1)log₂a, so f(a)=(k-1)log₂a +1.

We have g(1/2ⁿ)=-ng(2)=-n(k-1), f(1/2ⁿ)=-n(k-1)+1.
So if k≠1, we would have f(1/2ⁿ)→∞ if k<1 and f(1/2ⁿ)→-∞ if k>1.
Hence, if f is continuous, f cannot be defined at x=0.
EDIT: as noted above, if k≠1, f cannot be defined at x=0, even without the continuity condition.

With the continuity condition, f is defined elsewhere and completely specified:
For x≠0, f(x)=(k-1)log₂|x| +1.

Finally we note that this function does indeed satisfy the functional equation. This is simpler to check for g: g(x)=(k-1)log₂|x|:
+ log₂|y|)=(k-1)log₂|x| + (k-1)log₂|y|=g(x)+g(y).

MichaelRothwell

If 0 is in the domain, then f(0) = f (0*0) = 2f(0) - 1, which implies f(0)=1

Furthermore, for any x: f(0) = f(0 * x) = f(0) + f(x) - 1 = f(x). Thus f(x) = f(0) = 1.

Thus, if 0 is in the domain, the only function f satisfying the condition would be f(x) = 1

As a result f(2) = k is impossible for any k not equal to 1. And if k=1, then f(16)=1.

This looked way too easy. So, where did I go wrong??

arthur_p_dent

Chosing y=0 results in f(x)=1 for all x...

harrymattah

To the frustration of math profs everywhere, an application of this is complexity analysis in algorithm design from computer science. Big O notation is aptly named.
(:

ardiris

wanted to get the general solution, so set t=2^x and solved f(x) = (k-1)*log2(x) + 1.

I had originally been in agreement with the comments below that f(x) = 1 for all values of x. however, if the true general solution contains the log term, then the domain is limited.

MichaelJamesActually

Given:
f(xy) = f(x) + f(y) – 1
f(2) = k
To find:
f(16) in terms of k

As the given relation holds for any x & y, setting y = x = t:
f(t²) = f(t) + f(t) – 1
f(t²) = 2f(t) – 1 ― ①

Substituting t = 2:
f(2²) = 2f(2) – 1
f(4) = 2f(2) – 1.

Substituting f(2) = k:
f(4) = 2k – 1 — ②

Now substituting t = 4:
f(4²) = 2f(4) – 1
f(16) = 2f(4) – 1.

Substituting in ②:
f(16) = 2(2k – 1) – 1
f(16) = 4k – 3.

GirishManjunathMusic