I do not understand this.

What is the degree of the function that generates the data shown?

X | Y |

-3 | 159 |

-2 | 29 |

-1 | -1 |

0 | -3 |

1 | -1 |

2 | 29 |

3 | 159 |

I need someone to tell me how to understand this...

OfficialBubbleTanks Nov 30, 2017

#1**+3 **

Some of the other mathematicians have got much better techniques for doing these but I resorted to trial and error and just looking for patterns.

I started by adding 3 to the y value because 0 to the power of anything is 0

x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | |

y | 159 | 29 | -1 | -3 | -1 | 29 | 159 | |

y+3 | 162 | 32 | 2 | 0 | 2 | 32 | 162 | 32=2^5 but 162 is not a power of 3 .. but 81 is |

(y+3)/2 | 81 | 16 | 1 | 0 | 1 | 16 | 81 | |

[(y+3)/2]^(1/4) | 3 | 2 | 1 | 0 | 1 | 2 | 3 | |

so .. working backwards

\(y=2x^4-3\)

I worked this out quite quickly because I have had a lot of practice but you should get a better answer off someone else

Melody Nov 30, 2017

#2**+3 **

Very nice detective work, Melody!!!

I haven't seen this done before, but it appears to work!!!

Since the polynomial is symmetric about y and appears to intercept the x axis twice, I would have guessed that it was a quadratic......but yours perfectly models the data....

CPhill Nov 30, 2017

#3**+2 **

Thanks Chris,

Since the polynomial is symmetrical about the x axis, I guessed that the highest power of x (degreee) was even.

I could see a power of 2 would be too small but it wasn't likely to be much bigger.

I could see that if would look kind of like a parabola. With the vertex at (0,-3)

If I had been thinking clearly I could have easily surmised that the formula would be

\((y+3)=a(x+0)^n\\ y=ax^n-3 \qquad \text{Where n is even} \)

The values of a and n could be worked out easily enough the way I did it before.

-------------------------------------

I did think there was another routine way to work out just the degree.

I thought you knew how to do tit?

Geno is the first one I think I have seen do it.. but I thought I'd seen you and Heureka do it since then....

Melody
Nov 30, 2017

#5**+2 **

Geno's method is known as the sum of differences method......this would have probably worked here, but it's a bit of work.....I like your approach better......

CPhill Dec 1, 2017

#9**+2 **

I did try Gino's method as well but it didn't work. IPerhaps I have problems with primary school arithmetic

Heureka has it working below. I will check what he has done. Thanks Heureka.

Thanks also to Hectictar and Gino, yes that was what I was one of the posts I was referring too.

It was good to see Dragonlance's comment too. Another of our old favourites who has moved on.

Melody
Dec 1, 2017

#6**+3 **

\(\displaystyle \begin{array}{rrrrrr} x & y\\ -3 & 159\\ & & -130 \\ -2 & 29 & & 100\\ & & -30 & & -72\\ -1 & -1 & & 28 & & 48\\ & & -2 && -24 & \\ 0 & -3 & & 4 & & 48\\ & & 2 && 24 \\ 1 & -1 & & 28 & & 48 \\ & & 30 & &72 \\ 2 & 29 & & 100\\ & & 130\\ 3 & 159\\ \end{array} \)

If the function is a polynomial tabulated exactly (no decimal approximations), its degree can be found using a difference table, (see above).

If the function is linear, the entries in the first difference column (the third column in the table), will be constant

If it's a quadratic the entries in the second difference column will be constant, and so on.

In the table above, entries in the fourth difference column are constant, so it's a quartic.

The table with its differences can then be used to calculate the coefficients of the polynomial.

Guest Dec 1, 2017

edited by
Guest
Dec 1, 2017

#8**+2 **

Oh my god this is so helpful! Thanks so much guest! You should join Web2.0, the people here are very nice and very helpful!

OfficialBubbleTanks
Dec 1, 2017

#7**+3 **

**What is the degree of the function that generates the data shown?**

\(\displaystyle \begin{array}{|r|rrrrr|} \hline x & y\\ \hline -3 & 159 \\ & & -130 \\ -2 & 29 & & 100 \\ & & -30 & & -72 \\ -1 & -1 & & 28 & & 48 \\ & & -2 & & -24 & \\ 0 & -3 & & 4 & & 48 \\ & & 2 & & 24 & \\ 1 & \color{red}d_0=-1 & & 28 & & 48 \\ & & \color{red}d_1=30 & & 72 & \\ 2 & 29 & & \color{red}d_2=100 & & 48 \\ & & 130 & & \color{red}d_3=120 & \\ 3 & 159 & & 220 & & \color{red}d_4=48 \\ & & 350 & & 168 & \\ 4 & 509 & & 388 & & \\ & & 738 & & \\ 5 & 1247& & & & \\ \hline \end{array}\)

\(\displaystyle \begin{array}{|rcll|} \hline y(x) &=& \binom{x-1}{0}\cdot {\color{red}d_0 } + \binom{x-1}{1}\cdot {\color{red}d_1 } + \binom{x-1}{2}\cdot {\color{red}d_2 } + \binom{x-1}{3}\cdot {\color{red}d_3 } + \binom{x-1}{4}\cdot {\color{red}d_4 } \\\\ &=& \binom{x-1}{0}\cdot ( {\color{red}-1 } ) + \binom{x-1}{1}\cdot {\color{red}30 } + \binom{x-1}{2}\cdot {\color{red}100 } + \binom{x-1}{3}\cdot {\color{red}120 } + \binom{x-1}{4}\cdot {\color{red}48 } \\\\ &=& -1 \\ &&+~ \underbrace{30 \cdot(x-1)}_{\text{max degree } 1} \\ &&+~ \underbrace{100 \cdot\left(\frac{x-1}{2}\right)\cdot \left(\frac{x-2}{1}\right)}_{\text{max degree } 2} \\ &&+~ \underbrace{120 \cdot \left(\frac{x-1}{3}\right)\cdot \left(\frac{x-2}{2}\right)\cdot \left(\frac{x-3}{1}\right)}_{\text{max degree } 3} \\ &&+~ \underbrace{48 \cdot\left(\frac{x-1}{4}\right)\cdot \left(\frac{x-2}{3}\right)\cdot \left(\frac{x-3}{2}\right)\cdot \left(\frac{x-4}{1}\right)}_{\text{max degree } \color{red}4} \\ &=& \mathbf{2x^4-3} \\ \hline \end{array}\)

expand see **WolframAlpha:**

If the difference \(d_i\) is constant, so the degree of the polynomial to calculate the coefficients is \(i\).

\({\color{red}d_4}=48\) is constant, so the degree is \({\color{red}4}\)

If the difference \(d_i \) is constant, so the degree of the polynomial to calculate the sum of the coefficients is \(i+1\).

heureka Dec 1, 2017

#10**+2 **

Thanks to Geno, Melody, Heureka and Guest !!!

Here's an algebraic solution.......

As the Guest pointed out.....this will be a 4th degree polynomial in the form

ax^4 + bx^3 + cx^2 + dx + e

Since (0, -3) is on the graph, then e = - 3

We can first solve for a and c

So we have that

a(-2)^4 + b(-2)^3 + c(-2)^2 + d(-2) - 3 = 29

a(2)^4 + b(2)^3 + c(2)^2 + d(2) - 3 = 29

Adding these, we have that

32a + 8c - 6 = 58

32a + 8c = 64

4a + c = 8 (1)

And we have that

a(1)^4 + b(1)^3 + c(1)^2 + d(1) - 3 = -1

a(-1)^4 + b(-1)^3 + c(-1)^2 + d(-1) - 3 = -1

Adding these we get that

2a + 2c = 4

a + c = 2 (2)

Subtract (2) from (1)

3a = 6 ⇒ a = 2

And 2 + c = 2 ⇒ c = 0

And we have that

2 (1)^4 + b(1)^3 + d(1) - 3 = -1

2(-1)^4 + b(-1)^3 + d(-1) - 3 = - 1

Subtract these

2b + 2d = 0

b + d = 0 (3)

And

2(2)^4 + b(2)^3 + d(2) - 3 = 29

2(-2)^4 + b(-2)^3 + d(-2) - 3 = 29

Subtract these

16b + 4d = 0

4b + d = 0 (4)

Subtract (3) from (4)

3b = 0

b = 0

And b + d = 0 ⇒ 0 + d = 0 ⇒ d = 0

So......the function is

2x^4 - 3

CPhill Dec 1, 2017