What do you think about this conjecture?

Question

Beforehand, please know that I'm a little bit of an amateur mathemitician, so this could be very wrong. But, I have tested it over and over and over again and it seems to be plausible, but there's still other things that I haven't tried that might possibly prove it wrong. That's why it's a 'conjecture' and not a 'theorem' or definitely not a 'law'.

This conjencture involves functions. Any function. Let's say we have the function $f(x)$. Obviously, this doesn't mean I just multiply f by x, it is a function of x. But, what if there was a way to separate the two? Well, I came up with this simple equation as the base of this entire conjecture:

$$f(x)=f*Ƨx$$ where $f$ is the function modifier and $Ƨx$ is the terminal, meaning what the function modifier is in terms of. The first problem with this is $Ƨx$. This is not a number or anything like that. It is its own concept, but it can be utilized in an equation. The second problem with this is $f$. How do he know that f is used in a function, and that it's not variable? Well, we don't know, unless it is blatantly defined as a function modifier. Usually, it is assumed that it's a function modifier, since it is in front of $Ƨx$ and it's 'f', probably the most commonly used function modifier. Simply put, $f(x)$ is the same as a function modifier $f$ multiplied by a value that has $Ƨ$ in front of it. There is no operation that can add $Ƨ$ in front of a value, $Ƨ$ can only be taken from an actually function.

Since: $$f(x) = f*Ƨx$$ Therefore: $$f = f*Ƨx/Ƨx$$ and Also: $$Ƨx = f*Ƨx/f$$

With this conjecture, I can find the solution to some problems:

1. find $g(x)$ in terms of $a$ and $b$, assuming $f$ and $g$ are both function modifiers $$f(a)+f(b)=g(a+b)$$ Solve: $$f(a)+f(b)=g*Ƨ(a+b)$$ $$(f(a)+f(b))/Ƨ(a+b)=g$$ $$(f(a)+f(b))/Ƨ(a+b)*Ƨx = g*Ƨx$$ $$(f(a)+f(b))Ƨx/Ƨ(a+b)=g(x)$$ Assuming $f/f=1$ (which it should) $$f/f *(f(a)+f(b))Ƨx/Ƨ(a+b)=g(x)$$ $$((f(a)+f(b))f*Ƨx)/f*Ƨ(a+b)=g(x)$$ $$((f(a)+f(b))f(x))/f(a+b)=g(x)$$ So my solution is: $$((f(a)+f(b))f(x))/f(a+b)=g(x)$$

Now after some tests, this proves to be right. I'll try to prove it here by saying $f(x)=2x-1$. This next problem will be similar to the last problem, excpet that $f(x)$ will now be defined.

2. find $g(x)$, assuming $f$ and $g$ are both function modifiers. $f(x)=2x-1$ $$f(a)+f(b)=g(a+b)$$ Solve: $$((f(a)+f(b))f(x))/f(a+b)=g(x)$$ $$((2a-1)+(2b-1))(2x-1)/(2(a+b)-1)=g(x)$$ $$(2a-1+2b-1)(2x-1)/(2a+2b-1)=g(x)$$ $$(2a+2b-2)(2x-1)/(2a+2b-1)=g(x)$$ $$(4ax-2a+4bx-2b-2x+2)/(2a+2b-1)=g(x)$$ Now let's see if $g(x)$ really is equal to that... $$f(a)+f(b) = 2a-1+2b-1 = 2a+2b-2$$ Let's plug in $a+b$ into $g(x)$ to see if it's equal to $f(a)+f(b)$ $$((2a-1)+(2b-1))(2(a+b)-1)/(2(a+b)-1)=g(a+b)$$ The $2(a+b)-1$ in the nominator cancels out with the $2(a+b)-1$ in the denominator: $$(2a-1)+(2b-1)=g(a+b)$$ $$2a+2b-1=g(a+b)$$ And, it works! If you're thinking this, I also thought that it was a little suspicious that the nominator and the denominator cancel out, but considering I used $Ƨx$ as the basis, that suspicion decreased.

I've also used this conjecture to find another way to define the Gamma Function($\Gamma(x)$) using this definition: $$\Gamma(1-x)\Gamma(x)=\pi/sin(\pi*x)$$ $$\Gamma*Ƨ(1-x)\Gamma*Ƨ(x)=\pi/sin(\pi*x)$$ $$\Gamma^2*Ƨ(1-x)(Ƨx)= \pi/sin(\pi*x)$$ $$\Gamma^2*Ƨx=\pi/sin(\pi*x)Ƨ(x-1)$$ $$\Gamma^2*Ƨx*Ƨx=\pi/sin(\pi*x)Ƨ(x-1) *Ƨx$$ $$(\Gamma*Ƨx)^2=\pi/sin(\pi*x)Ƨ(x-1) *Ƨx$$ $$(\Gamma(x))^2=\pi/sin(\pi*x)Ƨ(x-1) *Ƨx$$ $$\Gamma(x)=sqrt((\pi*Ƨx)/(sin(\pi*x)Ƨ(x-1)))$$ However, this solution can not be used to directly utilize the function, since we don't know what $Ƨx$ and $Ƨ(x-1)$ are, so it stays like that. Since it doesn't involve any other function, I can't multiply the numerator and denominator by a function operator.

I've also used this conjecture to attempt to a find a way to define antiderivation. Now I COULD be very wrong on this because I don't really take into account the limit, which I have a feeling I should be, and I'm only assuming that $h$ is still approaching $0$ after all of this.

$$f'(x)= lim(h>0) (f(x+h)-f(x))/h$$ $$f'(x)= lim(h>0) (f*Ƨ(x+h)-f*Ƨ(x))/h$$ $$f'(x)= lim(h>0) (f*(Ƨ(x+h)-Ƨ(x)))/h$$ $$f'(x)*h= lim(h>0) f*(Ƨ(x+h)-Ƨ(x))$$ $$(f'(x)*h)/(Ƨ(x+h)-Ƨx) = lim(h>0) f$$ $$(f'(x)*h)/(Ƨ(x+h)-Ƨx)*Ƨx = lim(h>0) f*Ƨx$$ $$(hf'(x)*Ƨx)/(Ƨ(x+h)-Ƨx) = lim(h>0) f(x)$$ $$ f(x) = lim(h>0)(hf'(x)*Ƨx)/(Ƨ(x+h)-Ƨx)$$

This is proven to be wrong, since h is not defined outside of the limit, but still, it shows the use of my conjecture.

And so, this has been my conjecture. I don't know if it's solid, and I hope it doesn't get disproved (most likely will), but even if I'm redefining something that I didn't know existed, I hope it shows another way of thinking it.

Answer 1

Your 'solution' is just

$$f(x)\frac{f(a)+f(b)}{f(a+b)}=g(x)$$ So of course plugging in $x=a+b$ yields: $$f(a+b)\frac{f(a)+f(b)}{f(a+b)}=f(a)+f(b)$$ In the same way as $$b\frac{a}{b}=a$$

However there is no reason to expect $f(c)+f(d)=g(c+d)$ for any two $c,d$. In your own example, set $x=c+d$: $$\frac{4a(c+d)−2a+4b(c+d)−2b−2(c+d)+2}{2a+2b−1}=g(c+d)$$ And now we see that in this case of course there is no cancellation, and so in general $$f(c)+f(d)\not=g(c+d)$$ this is only true if $c=a, d=b$, and in that case it's almost a tautology that it's true.

Answer 2

I think I see what you're trying to do here, however, it doesn't work. From what I understand now, you are 'seperating' the 'input to the function': $Ƨx$, from the 'body of the function': $f$. However this leads to some strange and wrong conclusions, for example:

$$\frac{f(x)}{g(x)}=\frac{fƧx}{gƧx}=\frac{f}{g}$$ Which is a meaningless result (if I understand your idea correctly), because it's just a 'function modifier' on it's own, without the input. I'm sure there are other more enlighening example of this concept not working out, but right I cannot really think of any of the top of my head.

Ok here is another example:

$$1=\frac{fgƧ^2x}{fgƧ^2x}=\frac{f(g(x))}{g(f(x))}\Leftrightarrow f(g(x))=g(f(x))$$ However this is clearly not true for arbitrary $f,g$.