Solve Strange Function: Find the Equation

[mubashirmansoor]

A strange function...

Last night I was faced with a strange function & I couldn't figure out the equation of the function hence I thought I can get some help from the math experts over here...

The function passes through these cordinates:

(3,1) (4,1) (5,2) (6,2) (7,3) (8,3) (9,4) (10,4) .....
it's logicallt very simple to predict it's next terms but I still couldn't find the equation...

I'd be really really glad if someone would help me figure this out...
Tahnks.

 

[kesh]

Last night I was faced with a strange function & I couldn't figure out the equation of the function hence I thought I can get some help from the math experts over here...

The function passes through these cordinates:

(3,1) (4,1) (5,2) (6,2) (7,3) (8,3) (9,4) (10,4) .....
it's logicallt very simple to predict it's next terms but I still couldn't find the equation...

I'd be really really glad if someone would help me figure this out...
Tahnks.

add two functions. a diagonal ascending one like y = x/2, and an oscilating one like sin. fiddle with coefficients to get it right

 

[radou]

You may want to investigate this: http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html" .

 

[kesh]

You may want to investigate this: http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html" .

trouble with that is it covers a finite number of points. the sequence in the post is infinite



i reckon (2x - 3 - Cos[Pi x])/4 will do it. Cos taking radian arguments. silly errors aside

 

[Edgardo]

Try using the floor function:

\left(x, \lfloor \frac{1}{2}\left( x-1 \right) \rfloor \right)

That is the function might look like this:

f: [3,\infty) \rightarrow \Re; x \mapsto \lfloor \frac{1}{2}\left(x-1\right) \rfloor

 

[mubashirmansoor]

Thanks from everybody who posted.
The problem is solved & you were all a great help
Once again Thankyou...