March 25th, 2017, 03:58 PM  #1 
Newbie Joined: Mar 2017 From: leuven Posts: 18 Thanks: 0  guessing the base function of a real function that meets certain requirements
Okay, so the title is really long. I'm kind of ashamed to be asking 2 questions in one night. I've just started a course and am really stuck on some of the exercises that aren't explained in the book. So basically this is asked: Given is a real function f that meets these requirements: f(x)+x*f(1x)=x determine on basis of this f(2). So basically they are using f(x) in an actual equation. I have never ever seen anything like that before. Is there any rule behind solving things like these? I will need it further on in the book most likely and googling it in my language gave me irrelevant results. Greets, v Last edited by skipjack; March 26th, 2017 at 08:09 AM. 
March 25th, 2017, 04:09 PM  #2 
Senior Member Joined: Aug 2012 Posts: 1,528 Thanks: 364 
Start with a simple example. Say $f(x) = x^2$. Then $f(4) = 16$, etc. What is $f(x) + x$? Just plug in some values. $f(2) + 2 = 4 + 2 = 6$. What is $f(x) + x * f(1x)$ as in your problem? Plug in a number. $f(3) + 3 * f(2) = 9 + 3 * 4 = 21$. You just plug in some value for $x$ and that's how you evaluate it. That's what the notation means. Last edited by skipjack; March 26th, 2017 at 08:35 AM. 
March 25th, 2017, 05:03 PM  #3 
Newbie Joined: Mar 2017 From: leuven Posts: 18 Thanks: 0 
okay that looks clear enough, however in this excercise we don't actually get the first f(x)=x² so did you find this by purely guessing? is that maybe a universal rule? was it a random example? so it only shows us f(x)+x*f(1x) and then they say to fill in f(2) 
March 25th, 2017, 06:34 PM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,941 Thanks: 2267 Math Focus: Mainly analysis and algebra 
$$\begin{align*} && f(x) + xf(1x) &= x \\ &\text{set $x=0$ to get} & f(0) &= 0 \\ &\text{set $x=1$ to get} & f(1) + f(0) &= 1 \implies f(1) = 1 \\[8pt] &\text{set $x=2$ to get} & f(2) + 2f(1) &= 2 \quad (\text{eq }1) \\ &\text{set $x=1$ to get} & f(1)  f(2) &= 1 \quad (\text{eq }2) \\[8pt] &\text{then $(1)  2(2)$ gives} & 3f(2) &= 4 \end{align*}$$ Note that the first two lines are not necessary, they just illustrate the method: look for potentially useful values. Last edited by v8archie; March 25th, 2017 at 06:46 PM. 
March 25th, 2017, 06:37 PM  #5 
Senior Member Joined: Aug 2012 Posts: 1,528 Thanks: 364  
March 25th, 2017, 06:45 PM  #6  
Senior Member Joined: Feb 2016 From: Australia Posts: 1,322 Thanks: 453 Math Focus: Yet to find out.  Quote:
While shame or lack of confidence may lead to more rigorous attempts at a problem, it is not necessarily the most efficient way to learn .  
March 26th, 2017, 02:15 AM  #7 
Newbie Joined: Mar 2017 From: leuven Posts: 18 Thanks: 0  Code: okay so i think i understand the first two lines set x= 0 to get so you fill in the 0 on the x'es f(0)+0*f(10)=0 since 0 times f(10) is still 0 you can write it as f(0)=0 set x= 1 to get f(1)+1*f(11)=1 shorter writing: f(1)+f(11)=1 f(1)+f(0)=1 now i'l try the tricky ones set x = 2 to get f(2)+2f(12)= f(2)+2f(1) if i solve this normally i would get f(2)+f(2)=0 but i think that it is required to be equal to 2. so to achieve that you put (eq 1 ) in the brackets behind it f(2)+2f(1) =2 (eq 1) So what have we actually done here? did we somehow take out the 2f(1)? assuming this is the case the formula would like this in 2 to get f(2)+2f(0)=2 (eq 1) in the next set x we set it to x=1 to get f(1)+1f(11)=1 f(1)+1f(2)=1 eq(2) so now we take out the f(2) to still get it to be equal to 1 with the taken out it would look like this f(1)f(0)=1 eq(2): 2 is taken out now for the final part i have some troubles (1)2(2) gives 3f(2)=4 first for the to get part (1)  2(2) are these the numbers form (eq1) and (eq2)? where did you get these? and the 2(2) has a second 2 between the brackets, where does that come from? i also don't get how you get 3f(2)=4 from the final then. am i looking at this completely wrong? am i right about what the (eq)'s do? (taking out the second f(x) to get it to be correct) how do you form the final condition? the condition just above the last one (set x = 1 to get) is just filling in the 1 we just took out the last line right? and how do you get the final line of code from the condition (1)2(2) gives 3f(2)=4 
March 26th, 2017, 05:43 AM  #8 
Senior Member Joined: May 2016 From: USA Posts: 785 Thanks: 312 
Archie gave you the answer to what we think the question is, but his explanation was very concise. $f(x) + x * f(1  x) = x \implies f(x) = x\{1  f(1  x)\}.$ $(1  2) = \ 1\ and\ (1  \{\ 1\}) = 2.$ $y = f(2)\ and\ z = f(\ 1).$ $y = 2\{1  f(1  2)\} = 2\{1  f(\ 1)\} = 2(1  z) = 2  2z.$ $z = f(\ 1) = (\ 1)\{1  f(1  \{\ 1\})\} = f(1 + 1)  1 = f(2)  1 = y  1.$ $\therefore y = 2  2z = 2  2(y  1) = 2  2y + 2 = 4  2y \implies$ $3y = 4 \implies y = \dfrac{4}{3} \implies f(2) = \dfrac{4}{3}.$. Of course that means that $f(\ 1) = \dfrac{4}{3}  1 = \dfrac{1}{3}.$ Let's check. $f(2) + 2 * f(1  2) = 2 \implies f(2) = 2  2 * \dfrac{1}{3} = \dfrac{6  2}{3} = \dfrac{4}{3}.$ 
March 26th, 2017, 07:10 AM  #9  
Math Team Joined: Dec 2013 From: Colombia Posts: 6,941 Thanks: 2267 Math Focus: Mainly analysis and algebra  Quote:
$$f(2)+2f(1)=2 \\ f(1)f(2) = 1$$ for $f(2)$. One way to do this is by eliminating $f(1)$ by subtracting twice $(\text{eq }2)$ from $(\text{eq }1)$.  
March 26th, 2017, 09:17 AM  #10 
Global Moderator Joined: Dec 2006 Posts: 17,919 Thanks: 1386 
Consider the functional equation f(x) + x*f(1x) ≡ x, where x is any real. If x is real, 1  x is also real, so x can be replaced with 1  x to give f(1  x) + (1  x)f(x) ≡ 1  x, which implies x*f(1  x) + x(1  x)f(x) ≡ x(1  x). As x*f(1  x) ≡ x  f(x), x  f(x) + x(1  x)f(x) ≡ x(1  x), which implies f(x) ≡ (x(1  x)  x)/(x(1  x)  1) ≡ x²/(x²  x + 1) (because x²  x + 1 ≡ (x  1/2)² + 3/4, which cannot be zero). Substituting x = 2 in the above explicit formula gives f(2) = 4/3. 

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