Maths

I'm not looking for help by any means
I just want to clarify that there is a numerical way to prove the form of the function given the inital values

>>8533827
Can you think of a polynomial with a different degree but that still fulfills all those conditions? If you can, then there's at least two polynomials of different degrees with the same information, and hence you know that that information is not enough to determine a polynomial's degree.

As a starting point, try examining the behaviour of simple polynomials of degree 2 and 4.

>>8533832
I did with a degree of 4th and found a solution.
I can't be done with a degree of 2 since the curve drawn out with my points is not symmetrical.

>>8533832
>>8533833
I'll try with a degree of 5 and 6 and see if I can find a solution. My teacher claims that there is one solution however.

>>8533835
I'd focus on 6 if I were you. You might want to read up a bit on overfitting to get some inspiration/insight into whether your question is possible or not.

>>8533838
I will try 6 soon. I want to know if there's a certain way without overfitting. I've attempted and there I found my solution. However, it had to be done guessing the degree of the polynomial.

>>8533827
I think it should be of degree >2 since it has at least 2 extremium. Also did you mean f''(0) is less, more or equal zero?

>>8533845
Yeah it has to be of a degree higher than 2 since f'' needs to be dependant of x in order to have 2 solutions.
f''(0) = 0

Is there a standard procedure or do I simply need to brute force it?

>>8533827
The minimum degree is 5, for which the polynomial is:
f(x) = -x^5/96 + 7*x^4/48 - x^3/2 + 2

For any degree above 5, there are infinitely many solutions; you can choose almost[1] any coefficients for the x^6 and above terms, with the coefficients for the x^3, x^4 and x^5 terms dictated by them. The constant term is always 2 (because f(0)=2), the x and x^2 terms are always zero (because f'(0)=0 and f''(0)=0).

[1] The only constraint being f''(4)>0, which is easier to meet if the coefficients of the high-degree terms are negative. E.g. for a degree-8 polynomial you need
1280 c6 + 22016 c7 + 240640 c8 - 8 < 0
(where c6,c7,c8 are the coefficients of the x^6,x^7,x^8 terms respectively).

>>8533871
Use a computer algebra system (CAS), e.g. Mathematica, Maple or Maxima (the last one is a bit clunky, but free, and entirely adequate for this). Or write a program to do it (evaluating and differentiating a polynomial represented as an array of coefficients is trivial, and solving systems of linear equations isn't exactly rocket science).

Define f(x) as a high-degree polynomial with variables as coefficients. Differentiate twice, Specifying that f(x)=y (or f' or f'') gives you a linear equation in the coefficients. Several such constraints give you a system of linear equations which can be solved for the coefficients.

An under-specified system (i.e. more terms than constraints) will have free variables, meaning that you have a parametric family of solutions rather than a single solution.

>>8533914
Yes thank you. I also realised it on my own.I think my teacher will be satisfied with prooving that if there's only 1 solution; the solution must be of the 5th grade. Thanks anon!

>>8533924
That's the thing. I can get many solutions using a computer system wich evaluates my inputs. By using equation system I can only find a solution for a polynomial of the 5th degree. I'll be satisfied with just showing the solution and describing that there's infinitely many other solution if the polynomial is of a higher degree.

File: SS_021.png (10KB, 794x590px) Image search: [Google]

10KB, 794x590px

So i started learning linear algebra and im doing some example of linear combinations
I need to find out if (3,2,1) can be presented as linear combination of a,b,c vectors.

So is the way i think correct way:
1. Ill need to set 1st number to be one by swaping R1 with R2
2. Ill need to make 1 in 2nd position in diagonal
3. I would swap R2 and R3 and then R2=2*R3+R2

Is this correct way or im doing shit?
Haven't done any math in like 2 years

>>8533952
you're trying to put things in reduced row echelon form, check out that phrase

>>8534430
I know, but i was just wondering if my way of thinking is ok

>>8534432
you can multiply a row by a scalar, swap rows, and add a scalar multiple of one row to another while maintaining row equivalence. it looks like you're doing these operations, so you're okay

>>8533952
That matrix has a non-zero determinant, so A.x=b has a solution x=A^-1.b for any b.

In this case, it's [1,-1/2,5/2].

[2 -2 0 | 3]
[1 -2 0 | 2]
[1 5 1 | 1]

Subtract row 2 from row 1

[1 0 0 | 1]
[1 -2 0 | 2]
[1 5 1 | 1]

Subtract row 1 from row 2

[1 0 0 | 1]
[0 -2 0 | 1]
[1 5 1 | 1]

Subtract row 1 from row 5

[1 0 0 | 1]
[0 -2 0 | 1]
[0 5 1 | 0]

Multiply row 2 by -1/2

[1 0 0 | 1 ]
[0 1 0 | -1/2]
[0 5 1 | 0 ]

Subtract 5 times row 2 from row 3

[1 0 0 | 1 ]
[0 1 0 | -1/2]
[0 0 1 | 5/2]

>>8533952
det(A)=2*(-2)+2*1+0=-2
so the matrix is invertible and any vector can be written as a linear combination of the three.