I need Help please!!

Linear, quadratic, module, parametric equations

I need Help please!!

Postby Guest » Mon Jan 14, 2019 8:32 pm

I need an equation for a website I'm programming

Inputs = 4, 6, 8, 10, 12, 14, 16
b = 3, 4, 4, 5, 5, 5, 5
c = 7, 12, 15, 20, 24, 28, 31*

When 'a' is inputs then I need 'b' and 'c' to equal the number in the same order as the 'a' inputs.
Ex. If a = 4, then b = 3 and c = 7 If a = 6, then b = 4 and c = 12

*not a typo
Guest
 

Re: I need Help please!!

Postby Guest » Mon Jan 14, 2019 8:39 pm

Also, to help solve this. if the equation causes decimal places, I can choose to either round them or delete them.
So if the equation makes c's numbers look like 3.1, 4.1, 4.3, 4.8, 5.1, 5.3, 5.4 then that's fine. Thanks!!
Guest
 

Re: I need Help please!!

Postby Guest » Tue May 14, 2019 5:26 pm

Do you need a programming in c++
Guest
 

Re: I need Help please!!

Postby Guest » Sun Jun 02, 2019 5:45 am

Given any "n" data points there exist a polynomial of degree n-1 that fits those points. Here you have two sets of data points, (a, b) and (a, c), each with 7 data points so a 6 degree polynomial will fit each.

One method of getting that polynomial is "Newton's divided difference" scheme. If, for example, the data points are [tex](x_0, y_0)[/tex], [tex](x_1, y_1)[/tex], [tex](x_2, y_2)[/tex] then "Newton's divided difference" would give [tex]y= y_0 \frac{(x- x_1)(x- x_2)}{(x_0- x_1)(x_0- x_2)}+ y_1\frac{(x- x_0)(x- x_2)}{(x_1- x_0)(x_1- x_2)}+ y_2\frac{(x- x_0)(x- x_1)}{(x_2- x_0)(x_2- x_1)}[/tex]

For (a, b), we have data points (4, 3), (6, 4), (8, 4), (10, 5), (12, 5), (14, 5), (16, 5) so the 6th degree polynomial is
[tex]b= 3\frac{(x- 6)(x- 8 )(x- 10)(x- 12)(x- 14)(x- 16)}{(4- 6)(4- 8 )(4- 10)(4- 12)(4- 14)(4- 16)}+ 4\frac{(x- 4)(x- 8 )(x- 10)(x- 12)(x- 14)(x- 16)}{(6- 4)(6- 8 )(6- 10)(6- 12)(6- 14)(6- 16)}+ 4\frac{(x- 4)(x- 6)(x- 10)(x- 12)(x- 14)(x- 16)}{(8- 4)(8- 6)(8- 10)(8- 12)(8- 14)(8- 16)}+ 5\frac{(x- 4)(x- 6)(x- 8 )(x- 12)(x- 14)(x- 16)}{(10- 4)(10- 6)(10- 8 )(10- 12)(10- 14)(10- 16)}+ 5\frac{(x- 4)(x- 6)(x- 8 )(x- 10)(x- 14)(x- 16)}{(12- 4)(12- 6)(12- 8 )(12- 10)(12- 14)(12- 16)}+ 5\frac{(x- 4)(x- 6)(x- 8 )(x- 10)(x- 12)(x- 16)}{(14- 4)(14- 6)(14- 8 )(14- 10)(14- 12)(14- 16)}+ 5\frac{(x- 4)(x- 6)(x- 8 )(x- 10)(x- 12)(x- 14)}{(16- 4)(16- 6)(16- 8 )(16- 10)(16- 12)(16- 14)}[/tex]

And similarly for c.

Of course, there are infinitely many non-polynomial functions that will fit any finite number of data points.
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