1. ## differential equations

i wonder if this will sound like a completely ignorant question, i hope not.. i'm learning this from some sort of self study course which is based on doing exercises and studying solutions, and it's really over my actual school level.

i think i can kind of know what to do for the front half part, but look at
(ii) When f(x)=0, f(x)=0 satisfies (1)
how is that so? $f(x) \cdot f(y)$ will become $0$, but then how do i know that $f(x+y)=0$?

2. ## Re: differential equations

If you understood (i) and not (ii) then you're probably thinking about (ii) too hard. I would say it's simply because both sides of the equation have to be equal. If $f(x+y)=f(x)\cdot f(y)$ and $f(x)=0$ then you must have $f(x+y)=0$.

$f(x+y)=f(x)\cdot f(y)$

$f(x+y)=0 \cdot f(y)=0$

Therefore $f(x+y)=0$

It's the way they're wording it that's confusing you, they are saying it "satisfies" the equation, but we often refer to zeros as trivial solutions. If I have a diff eq. like $\frac{dy}{dx}=y$ for instance then one obvious "solution" is $y=0$ since you can plug it into the diff eq. and it's true. It's just uninteresting. We're more interested in the solns. obtained by separating variables and integrating.

3. ## Re: differential equations

If you understood (i) and not (ii) then you're probably thinking about (ii) too hard. I would say it's simply because both sides of the equation have to be equal. If $f(x+y)=f(x)\cdot f(y)$ and $f(x)=0$ then you must have $f(x+y)=0$.

$f(x+y)=f(x)\cdot f(y)$

$f(x+y)=0 \cdot f(y)=0$

Therefore $f(x+y)=0$

It's the way they're wording it that's confusing you, they are saying it "satisfies" the equation, but we often refer to zeros as trivial solutions. If I have a diff eq. like $\frac{dy}{dx}=y$ for instance then one obvious "solution" is $y=0$ since you can plug it into the diff eq. and it's true. It's just uninteresting. We're more interested in the solns. obtained by separating variables and integrating.
oh so... taking what u said about trivial solutions.
a function f(x)=0 means that it always gives 0 no matter what, so putting (x+y) in the function will be 0, just as using any other value x or y would.
whatever is it, f(x+y), or f(x) or f(y) all equals 0 so obviously $f(x+y)=f(x) \cdot f(y)$
is that correct? thanks!

4. ## Re: differential equations

Originally Posted by muddywaters
oh so... taking what u said about trivial solutions.
a function f(x)=0 means that it always gives 0 no matter what, so putting (x+y) in the function will be 0, just as using any other value x or y would.
whatever is it, f(x+y), or f(x) or f(y) all equals 0 so obviously $f(x+y)=f(x) \cdot f(y)$
is that correct? thanks!
Yes, you can think of it that way too.
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 21, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}