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March 25th, 2017, 03:58 PM   #1
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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.
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March 25th, 2017, 04:09 PM   #2
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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.
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Last edited by skipjack; March 26th, 2017 at 08:35 AM.
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March 25th, 2017, 05:03 PM   #3
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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)
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March 25th, 2017, 06:34 PM   #4
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$$\begin{align*}
&& f(x) + xf(1-x) &= x \\
&\text{set $x=0$ to get} & f(0) &= 0 \\
&\text{set $x=1$ to get} & f(1) + f(0) &= 1 \implies f(1) = 1 \\[8pt]
&\text{set $x=2$ to get} & f(2) + 2f(-1) &= 2 \quad (\text{eq }1) \\
&\text{set $x=-1$ to get} & f(-1) - f(2) &= -1 \quad (\text{eq }2) \\[8pt]
&\text{then $(1) - 2(2)$ gives} & 3f(2) &= 4
\end{align*}$$

Note that the first two lines are not necessary, they just illustrate the method: look for potentially useful values.
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Last edited by v8archie; March 25th, 2017 at 06:46 PM.
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March 25th, 2017, 06:37 PM   #5
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Quote:
Originally Posted by vlekje5 View Post
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.
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March 25th, 2017, 06:45 PM   #6
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Quote:
Originally Posted by vlekje5 View Post
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 .
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March 26th, 2017, 02:15 AM   #7
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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
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March 26th, 2017, 05:43 AM   #8
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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.

$f(2) + 2 * f(1 - 2) = 2 \implies f(2) = 2 - 2 * \dfrac{1}{3} = \dfrac{6 - 2}{3} = \dfrac{4}{3}.$
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March 26th, 2017, 07:10 AM   #9
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Quote:
Originally Posted by vlekje5 View Post
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 $(\text{eq }1)$.
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March 26th, 2017, 09:17 AM   #10
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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.
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