Help i need help on learning integer input and output tables

[IanMike]

What is the rule for this table?

In​	Out​
-11​	1​
-10​	2​
-7​	5​
-3​	9​
2​	14​
6​	18​

Rule: Add ?

 

[MarkFL]

What is the constant difference between the Out and In values?

 

[Dr.Peterson]

What are your thoughts? What do you have to add to -11 to get 1? Does that work for other inputs?

 

[Deleted member 4993]

@IanMike

You sent out an e-mail on Tuesday 12:01 AM - complaining that people are not responding to your e-mail. E-mail above you wrote at 11:39 PM. You expected an answer within 22 minutes.

https://www.freemathhelp.com/forum/threads/why-are-people-not-responding-to-me.118046/#post-466115

However, a question was put to you above (#3) and you did not respond yet (several days past). How come??!!

 

[HallsofIvy]

Given any finite number of data points there exist an infinite number of functions that give those data points. It is true that given n data points there exist a unique polynomial of degree n- 1 that gives those data points. You can find that polynomial using the 'Lagrange interpolating formula':
$\displaystyle P(x)= \sum_{i= 1}^n y_n\frac{x- x_1)(x- x_2)\cdot\cdot\cdot(x- x_{i-1})(x- s_{i+1}\cdot\cdot\cdot(x- x_n)}{(x_i- x_1)(x_i- x_2)\cdot\cdot\cdot(x_i- x_{i-1})(x_i-x_{i+1}\cdot\cdot\cdot(x_i-x_n)}$.

Here, with this data 6 data points, (-11, 1), (-10, 2), (-7, 5), (-3, 9), (2, 14), and (6, 18). The Lagrange interpolating polynomial is the fifth degree polynomial $\displaystyle 1\frac{(x+ 10)(x+ 7)(x+ 3)(x- 2)(x- 6)}{(-11+ 10)(-11+ 7)(-11+ 3)(-11- 2)(-11- 6)}+ 2\frac{x+ 11)(x+ 7)(x+ 3)(x- 2)(x- 6)}{(-10+ 11)(-10+ 7)(-10+ 3)(-10- 2)(-10- 6)}+ 5\frac{x+ 11)(x+ 10)(x+ 3)(x- 2)(x- 6)}{(-7+ 11)(-7+ 10)(-7+ 3)(-7- 2)(-7- 6)}+ 9\frac{x+ 11)(x+ 10)(x+ 7)(x- 2)(x- 6)}{(-3+ 11)(-3+ 10)(-3+ 7)(-3- 2)(-3- 6)}+ 14\frac{x+ 11)(x+ 10)(x+ 7)(x- 2)(x- 6)}{(2+ 11)(2+ 10)(2+ 7)(2+ 3)(2- 6)}+ 18\frac{(x+ 11)(x+ 10)(x+ 7)(x+ 3)(x- 2)}{(6+ 11)(6+ 10)(6+ 7)(6+ 3)(6- 2)}$. Of course that might not be the simplest function interpolating these points. As I said there are an infinite number of functions that will give these points.

 

[Deleted member 4993]

Given any finite number of data points there exist an infinite number of functions that give those data points. It is true that given n data points there exist a unique polynomial of degree n- 1 that gives those data points. You can find that polynomial using the 'Lagrange interpolating formula':
$\displaystyle P(x)= \sum_{i= 1}^n y_n\frac{x- x_1)(x- x_2)\cdot\cdot\cdot(x- x_{i-1})(x- s_{i+1}\cdot\cdot\cdot(x- x_n)}{(x_i- x_1)(x_i- x_2)\cdot\cdot\cdot(x_i- x_{i-1})(x_i-x_{i+1}\cdot\cdot\cdot(x_i-x_n)}$.

Here, with this data 6 data points, (-11, 1), (-10, 2), (-7, 5), (-3, 9), (2, 14), and (6, 18). The Lagrange interpolating polynomial is the fifth degree polynomial $\displaystyle 1\frac{(x+ 10)(x+ 7)(x+ 3)(x- 2)(x- 6)}{(-11+ 10)(-11+ 7)(-11+ 3)(-11- 2)(-11- 6)}+ 2\frac{x+ 11)(x+ 7)(x+ 3)(x- 2)(x- 6)}{(-10+ 11)(-10+ 7)(-10+ 3)(-10- 2)(-10- 6)}+ 5\frac{x+ 11)(x+ 10)(x+ 3)(x- 2)(x- 6)}{(-7+ 11)(-7+ 10)(-7+ 3)(-7- 2)(-7- 6)}+ 9\frac{x+ 11)(x+ 10)(x+ 7)(x- 2)(x- 6)}{(-3+ 11)(-3+ 10)(-3+ 7)(-3- 2)(-3- 6)}+ 14\frac{x+ 11)(x+ 10)(x+ 7)(x- 2)(x- 6)}{(2+ 11)(2+ 10)(2+ 7)(2+ 3)(2- 6)}+ 18\frac{(x+ 11)(x+ 10)(x+ 7)(x+ 3)(x- 2)}{(6+ 11)(6+ 10)(6+ 7)(6+ 3)(6- 2)}$. Of course that might not be the simplest function interpolating these points. As I said there are an infinite number of functions that will give these points.

Also, the out-put from the above function may not be integer (not explicitly demanded).