Recent content by Wiz14

If f(y) = f(x) then substituting f(y) into the original equation gives f(f(y) = f(x) + y = f(x) + x, then subtracting the f(x) from the last equation gives x = y, is this correct? Thanks for the help.

If f(f(x)) = 0, independently of the argument f(x), then doesn't substituting f(y) for f(x) give f(f(y)) = 0?

sorry I am new to this stuff, but do you mean f(f(x)) = 0 implies f(f(y)) = 0 which implies f(f(x)) = f(f(y)) ? If yes then how does this prove f(x) = f(y)?

Homework Statement The function from R to R satisfies x + f(x) = f(f(x)) Find all Solutions of the equation f(f(x)) = 0. Part of the problem solution says that if f(x) = f(y), then "obviously" x = y. I understand the rest of the solution, but why does f(x) = f(y) imply that x = y?

Homework Statement dY/dt = y(c - yb) C and B are constants. Im supposed to find and explicit solution for y, but I am having trouble. Homework Equations The Attempt at a Solution dY/y(c - yb) = dt ∫(1/c)dy/y + ∫(b/c)dY/c - yb = ∫dt (i used partial fraction decompositions)...

Thank you for your response, it makes sense to me now. I also have another question. After you find the general solution, how do you find these singular solutions? How can you know that you are missing solutions or if you have all of them?

My textbook defines a solution to a differential equation to be a function f(x) such that when substituted into the equation gives a true statement. What I'm confused about are singular solutions. For example the logistic equation: dP/dt = rP(1-P/K) where r and K are constants. My textbook says...

Thank you for your response. The main reason I posted this question, and the reason I doubt my solution is correct, is because as you said we haven't used the fact that p is prime. But I cannot find any mistakes. I will try to better explain my answer. By definition, a divides p+1 means that...

Let p be a prime number and 1 <= a < p be an integer. Prove that a divides p + 1 if and only if there exist integers m and n such that a/p = 1/m + 1/n My solution: a|p+1 then there exists an integer m such that am = p+1 Dividing by mp a/p = 1/m + 1/mp So if I choose n = mp(which is...

I am reading that proof now but where is the flaw in my reasoning? A ≤B and B ≤ A is like saying A = B or A is strictly less than B and B is strictly less than A, which is a contradiction, so A must = B.

1.There exists an injection from A to B ⇔ A ≤ B 2.There exists an injection from B to A ⇔ B ≤ A 3.If A ≤ B and B ≤ A, then A = B Does this prove the Cantor Bernstein theorem? Which says that if 1 and 2 then there exists a Bijection between A and B (A = B) And if it does, why is there a...

but how to exclude the rest by considerations mod 3 and 8? thank you.

Thank you for your answer, but can you be more specific? How can we prove that your solutions and my solutions are the only solutions?

For each integer n > 1, let p(n) denote the largest prime factor of n. Determine all triples (x; y; z) of distinct positive integers satisfying  x; y; z are in arithmetic progression,  p(xyz) <= 3. So far I have come up with 22k + 1, 22k + 1 + 22k, and 22k + 2 other than the solutions...

The Following is defined, log2004(log2003(log2002(log2001x))) where x > c, what is c? Answer is 2001^2002, but how to obtain it? I do not know much about logs as I am only in precalculus.