# [Calvin] Which solution would you feature (5)?

Previous Discussion.

Below, we present a problem from the 1/14 Algebra and Number Theory set, along with 3 student submitted solution. You may vote up for the solutions that you think should be featured, and should vote down for those solutions that you think are wrong (voting is anonymous!). Also, feel free to make remarks about these solutions, especially since threading of comments has been introduced :).

Solutions to a recursive function $$f$$ is a function from the reals to the reals, satisfying $$x + f(x) = f( f(x) )$$. In the interval $$[-10, 10]$$, how many solutions are there to $$f(x) = 0$$?

You may try the problem by clicking on the above link.

All solutions may have LaTeX edits to make the math appear properly. The exposition is presented as is, and has not been edited.

$\mbox{Remarks from Calvin}$

Solution A - This was the approach that was most often used. By 'clever' substitution, we seek to understand the value of $$f(0)$$ and $$f( f(0) )$$. We realize that $$f(0)$$ is a solution to this problem, and turns out to be the only solution (regardless of the domain, as long as it includes 0). This solution is presented by Zk.

Solution B - I couldn't understand this solution when I read it, and I see that others agree with me. There could possibly be typos around, as Omid pointed out that $$x=f(x)$$ didn't make sense. Oldrin gave an attempt to justify Solution B, but he added a lot more information than the 1 equation in the solution. Remember to double check that what you wrote is the same as what you are thinking. I am not a mind reader, certainly not across the vast distances.

Solution C - As the Key Technique, this solution lists 'injectivity', which is used to show that $$f(x) = 0$$ has at most one solution. In fact, we can show more than that. We can show the the function must be an injection. Proof: Suppose $$f(x) = f(y)$$, then $$f(f(x) = f( f(y) )$$ and hence $$x = f( f(x) ) - f(x) = f( f(y) ) - f(y ) = y$$. Thus, the function is injective. This solution is presented by Kyriakos.

Note by Calvin Lin
5 years, 7 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

Solution A - By substituting $$x=0$$, we get $$f(0)=f(f(0))$$. By substituting $$x= f(0)$$, we get $$f(0)+f(f(0))=f(f(f(0)))$$. But since $$f(0)= f(f(0))$$, the equation above is actually equivalent to $$f(0)+f(f(0))=f(f(0))$$. This implies that $$f(0)=0$$. Hence, whe know there exists at least one $$x$$ such that $$f(x)=0$$. By substituting $$f(x)=0$$ into our original equation, we get $$f(0)=x$$. Therefore, since $$f(0)$$ is a constant, there is only one possible $$x$$ (which is the value 0), hence the answer is 1. note: the interval $$[-10,10]$$ is not necessary.

Staff - 5 years, 7 months ago

Solution C - First, we will prove that $$f(x)$$ is an injective function.What we have to prove is that for any $$x_1,x_2\in\mathbb{R}$$ it holds that $${f(x_1)=f(x_2)\iff{x_1=x_2}}$$.We proceed as follows:

$f(x_1)=f(x_2)\Rightarrow f(f(x_1))=f(f(x_2))\Rightarrow f(x_1)+x_1=f(x_2)+x_2\Rightarrow x_1=x_2$ Since $$f$$ is not a multivalued function it is also true that $$x_1=x_2\Rightarrow{f(x_1)=f(x_2)}$$ Now that f is injective there can be no more than one solution to the equation f(x)=0 in the function's domain.

Setting x=0 in the recursive relationship we obtain: $$f(f(0))=f(0)\Leftrightarrow$$ $$f(0)=0$$ The former holds because f has been proven to be injective. So x=0 is a solution to f(x)=0 and obviously the only one.

Staff - 5 years, 7 months ago

Staff - 5 years, 6 months ago

Solution A and C are correct, in that they both prove existence and uniqueness of the solution x=1. On the other hand, it is not clear how Solution B found that $$x=f(x)$$ (it looks like this should be $$x=f(0)$$).

- 5 years, 7 months ago

If $$f(x)=0$$ and $$x+f(x)=f(f(x))$$, then it follows $$x=f(0)$$. As $$f(f(0))=f(0)+0$$ we substitute $$x=f(0)$$ to yield $$f(x)=f(0)$$. Thus $$x+f(0)=f(0)$$ and ultimately $$x=0$$.

- 5 years, 7 months ago

We've proven its existence and its uniqueness is apparent from $$x=f(0)$$ i.e. the definition of the recursive function.

- 5 years, 7 months ago

@O B I'm uncertain what your comments mean. Are you trying to justify Solution B?

Note that we don't accept other solutions, since the point of this discussion is to discuss the selected solutions, and not solicit for more submissions.

Staff - 5 years, 7 months ago

Calvin, I did not add any additional information. While perhaps solution B may have been phrased in a way that lacked clarity, and it most certainly not the most pleasing of the 3, it was correct.

- 5 years, 6 months ago

Yes, I am very clearly justifying Solution B...

- 5 years, 7 months ago

Solution A and Solution C are correct.

- 5 years, 7 months ago

Solution B - If $$f\left(x\right)=0$$, $$x+f\left(x\right)=x+0=x=f\left(x\right)$$. Again, we know that $$f\left(x\right)=0$$ so this can only occur at $$x=f\left(x\right)=0$$. $$0$$.

Staff - 5 years, 7 months ago