Wherein Mr. Engel loses
Jan. 19th, 2005 08:42 pm![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
sniffnoyMr. Holbrook got these new problem books (I forget the title, but it's by Arthur Engel, IIRC) that have not been edited at all.
Find all f:R→R st ∀x,y∈R, xf(y)+yf(x)=(x+y)f(x)f(y)
Its answer: Pick any set A⊆R st ∀x∈A, -x∈A. Then let f(x)=1 iff x∈A, 0 elsewise.
Mr. Engel[0], did you ever try plugging that back in? I can find considerably more restrictions on f than that... and that's for when f *is* of that form. I can give you a whole family of functions that fit the equation but are not 0 or 1 everywhere... in fact, of all the solutions to that functional equation, only 3 are of that form.
Similarly with the next question, find all f:R→R st ∀x,y∈R, f(x-y)=f(x)f(y). It does the same thing for its answer, though this time with 1 and -1. It's really not hard to show that f has to be identically 1. These are answers that would be given by someone who doesn't understand the idea of proof. Not only that, Mr. Engel[0], but in your solution, right next to each other, are the statements that f(0) has to be 1, and that f is identically -1 is one solution. I guess this is what "not edited at all" means.
ADDENDUM FROM 2 YEARS IN THE FUTURE: Doing this second problem again, I see f could be identically 0 as well (but that's the only other solution). Engel is still horribly wrong.
-Sniffnoy
[0]Of course, really, he didn't write the problems or the solutions; he got them from various Olympiads. But still...
--
"What I'm saying is Sluggy's going to be like a box o chocolates. You
never know what yer gunna get! I'm personally hoping for "chocolates,"
only because that's what it says on the box."
-Pete Abrams
Find all f:R→R st ∀x,y∈R, xf(y)+yf(x)=(x+y)f(x)f(y)
Its answer: Pick any set A⊆R st ∀x∈A, -x∈A. Then let f(x)=1 iff x∈A, 0 elsewise.
Mr. Engel[0], did you ever try plugging that back in? I can find considerably more restrictions on f than that... and that's for when f *is* of that form. I can give you a whole family of functions that fit the equation but are not 0 or 1 everywhere... in fact, of all the solutions to that functional equation, only 3 are of that form.
Similarly with the next question, find all f:R→R st ∀x,y∈R, f(x-y)=f(x)f(y). It does the same thing for its answer, though this time with 1 and -1. It's really not hard to show that f has to be identically 1. These are answers that would be given by someone who doesn't understand the idea of proof. Not only that, Mr. Engel[0], but in your solution, right next to each other, are the statements that f(0) has to be 1, and that f is identically -1 is one solution. I guess this is what "not edited at all" means.
ADDENDUM FROM 2 YEARS IN THE FUTURE: Doing this second problem again, I see f could be identically 0 as well (but that's the only other solution). Engel is still horribly wrong.
-Sniffnoy
[0]Of course, really, he didn't write the problems or the solutions; he got them from various Olympiads. But still...
--
"What I'm saying is Sluggy's going to be like a box o chocolates. You
never know what yer gunna get! I'm personally hoping for "chocolates,"
only because that's what it says on the box."
-Pete Abrams
