newbie needing analytical description of a periodic function

**rossy54** · August 14th 2010, 05:31 AM

I have this function here.

newbie needing analytical description of a periodic function-untitled-1.jpg

Can someone please explain why from 4 < x < 10 it is 7 - x?

**chiph588@** · August 14th 2010, 06:41 AM

Well, two points make a line. Take the two endpoints and see what you get.

**rossy54** · August 14th 2010, 06:58 AM

not really sure. From 0 < x < 4 its 3x/4 but cant see why its 7 - x.

**chisigma** · August 14th 2010, 07:39 AM

You have a function...

$f(x)= a_{0} + a_{1}\ x$ (1)

... and is $f(4)=3$ and $f(10)=-3$ ... then You can find $a_{0}$ and $a_{1}$ in 'standard way'...

Kind regards

$\chi$ $\sigma$

**Vlasev** · August 14th 2010, 02:53 PM

rossy54, the function changes from place to place. From 0 to 4 it is indeed 3x/4. However, it immediately changes to 7-x when x is between 4 and 10. After that it is a constant.

**yeKciM** · August 14th 2010, 03:04 PM

to make equation of line through two points :

$(y-y_1)=\displasyle \frac{y_2-y_1}{x_2-x_1} \cdot (x-x_1)$

now your two points $M_1 (4,3) \; M_2(10,-3)$

so you will have

$\displaystyle (y-3)=\frac{-3-3}{10-4} \cdot ( x-10)$

$\displaystyle (y-3)=\frac{-6}{6} \cdot ( x-10)$

$\displaystyle y-3= 10-x$

$\displaystyle y=7- x$

if you don't trust me

just put any x from 4 to 10 and you will get exact value of y...
for example x=7 -> y=0 (as you see on graphic)

if u need to represent your function over step functions it would be like this

$x(t) = \frac{3t}{4}[u(t)-u(t-4)] + (7-t)[u(t-4) - u(t-10)]$

P.S. you can take any two points from that line and put in formula and you will get same result

(doesn't have to be "start" and "end")

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