Solve Strange Function: Find the Equation

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The discussion revolves around finding the equation for a function defined by specific coordinates: (3,1), (4,1), (5,2), (6,2), (7,3), (8,3), (9,4), and (10,4). Participants suggest various approaches, including using a combination of a diagonal function and an oscillating function, as well as the Lagrange Interpolating Polynomial. A proposed solution is the function f(x) = floor((1/2)(x-1)), which fits the given points. The original poster expresses gratitude for the assistance received, indicating that the problem has been resolved. The discussion highlights collaborative problem-solving in mathematics.
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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.
 
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mubashirmansoor said:
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
 
You may want to investigate this: http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html" .
 
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radou said:
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
 
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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
 
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Thanks from everybody who posted.
The problem is solved & you were all a great help
Once again Thankyou...
 
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