My Math Forum guessing the base function of a real function that meets certain requirements

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 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(1-x)=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 $f(x) + x * f(1-x)$ 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. Thanks from vlekje5 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(1-x) 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(1-x) &= 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. Thanks from topsquark, JeffM1 and vlekje5 Last edited by v8archie; March 25th, 2017 at 06:46 PM.
March 25th, 2017, 06:37 PM   #5
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Quote:
 Originally Posted by vlekje5 was it a random example?
Yes. I wasn't clear. I was showing you how to work with the notation, but I did not give any useful information about solving the problem you posted, which is more tricky.

March 25th, 2017, 06:45 PM   #6
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Quote:
 Originally Posted by vlekje5 okay so the title is really long. i'm kind of ashamed to be asking 2 questions in one night.
If you have truly attempted a problem yourself, and truly attempted to understand it, then there is no reason whatsoever to be ashamed.

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(1-0)=0 since 0 times f(1-0) is still 0 you can write it as f(0)=0 set x= 1 to get f(1)+1*f(1-1)=1 shorter writing: f(1)+f(1-1)=1 f(1)+f(0)=1 now i'l try the tricky ones set x = 2 to get f(2)+2f(1-2)= 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(1--1)=-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? so in conclusion this is still unclear to me -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. $f(2) + 2 * f(1 - 2) = 2 \implies f(2) = 2 - 2 * \dfrac{1}{3} = \dfrac{6 - 2}{3} = \dfrac{4}{3}.$ Thanks from vlekje5
March 26th, 2017, 07:10 AM   #9
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Quote:
 Originally Posted by vlekje5 Code:  set x = 2 to get f(2)+2f(1-2)= 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(1--1)=-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? so in conclusion this is still unclear to me -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
$(\text{eq }1)$ and $(\text{eq }2)$ are just labels. We solve the system of simultaneous equations
$$f(2)+2f(-1)=2 \\ f(-1)-f(2) = -1$$ 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(1-x) ≡ 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. Thanks from vlekje5

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