Find F(x) given X and Y values

[Drakkith]

Homework Statement

Find the function F(x) given the X and Y values in the table.

Relevant Equations

F(0) = 5
F(1) = 20
F(2) = 80
F(3) = 320
F(4) = 1280

Hey folks. I'm trying to help my daughter (9th grade) with a homework problem and I'm not sure how to go about finding the answer.
Given the values in the table, find the function F(x).

It's not a linear equation, so it can't be of the form $F(x)=ax+b$.
I thought it might be $F(x)=15x^z+5$, where $z$ is some exponent, but that doesn't seem to work either.
I noticed that the increase of the Y values increases by a factor of 4 each time (by 15 then 60 then 240 then 960), but I don't know how to write an equation that does that.
Any ideas?

 

[member 587159]

$$F(x)=5.4^x$$

 

[Drakkith]

$$F(x)=5.4^x$$

Sorry, that doesn't seem to work with our given values. And I don't think my daughter has done exponential functions yet in math.

 

[member 587159]

Sorry, that doesn't seem to work with our given values. And I don't think my daughter has done exponential functions yet in math.

It does work... You can always fit a fifth degree polynomial but I doubt that's what is intended.

 

[PeroK]

Homework Statement:: Find the function F(x) given the X and Y values in the table.
Relevant Equations:: F(0) = 5
F(1) = 20
F(2) = 80
F(3) = 320
F(4) = 1280

Hey folks. I'm trying to help my daughter (9th grade) with a homework problem and I'm not sure how to go about finding the answer.
Given the values in the table, find the function F(x).

It's not a linear equation, so it can't be of the form $F(x)=ax+b$.
I thought it might be $F(x)=15x^z+5$, where $z$ is some exponent, but that doesn't seem to work either.
I noticed that the increase of the Y values increases by a factor of 4 each time (by 15 then 60 then 240 then 960), but I don't know how to write an equation that does that.
Any ideas?

Every number is four times the previous. Noticing that should help.

 

[SammyS]

$$F(x)=5.4^x$$

@Drakkith

Rewriting what QED wrote:

F(x) = 5 ⋅ (4^x ) .

 

[Drakkith]

@Drakkith

Rewriting what QED wrote:

F(x) = 5 ⋅ (4^x ) .

Oh, is that $5*4^x$? I thought it was 5.4 to the $X$.

Every number is four times the previous. Noticing that should help.

Hmmm. If this was recursion it would, but I'm afraid I can't see how to make the equation do a factor of 4 increase each time. Not with a simple polynomial. It appears that $F(x)=5*4^x$ works, but I don't think they've done exponential functions yet.

 

[WWGD]

There is a recursion, the pattern F(x+1)=4F(x). Is that related to her class material?

 

[PeroK]

Hmmm. If this was recursion it would, but I'm afraid I can't see how to make the equation do a factor of 4 increase each time. Not with a simple polynomial. It appears that $F(x)=5*4^x$ works, but I don't think they've done exponential functions yet.

If the sequence was $2, 4, 8, 16$, what would you do?

 

[etotheipi]

In the discrete case it's a geometric series (just with the indices shifted one place to the left i.e. $F(0) = u_1, F(1) = u_2, \dots$ ). So you might say $$u_n = ar^{n-1} \implies F(k) = u_{k+1} = ar^k = 5 \times 4^k$$but there's also nothing to say it doesn't work if $k$ now comes from a continuous interval.

 

[Drakkith]

If the sequence was $2, 4, 8, 16$, what would you do?

Ok, I just realized an exponential function $F(x)=z^x$ increases by a factor of $z$.
So the answer would be $5*4^x$.

And after talking to my daughter again, there appears to have been some confusion as to which section the question came from. Turns out they HAVE done exponential functions after all.

Thanks everyone. You learn something new everyday!

 

[berkeman]

Rewriting what QED wrote:

F(x) = 5 ⋅ (4^x ) .

Oh, is that 5∗4x5*4^x? I thought it was 5.4 to the XX.

Yeah, me too!

 

[SammyS]

It does work... You can always fit a fifth degree polynomial but I doubt that's what is intended.

Actually a degree 4 polynomial will do.

$F(x) = \dfrac{135} 8 x^4 - \dfrac{315} 4 x^3 + \dfrac{1125} 8 x^2 - \dfrac{255} 4 x + 5$

Added in Edit:
My reply here was intended as an example of a polynomial which passes through the given values, and this polynomial does exactly that.

@kuruman's point, made in the post following this, is well taken.

 

[kuruman]

Actually a degree 4 polynomial will do.
$F(x) = \dfrac{135} 8 x^4 - \dfrac{315} 4 x^3 + \dfrac{1125} 8 x^2 - \dfrac{255} 4 x + 5$

I got $$F(x)=
891.445 \cos(x) + 2752.35 \cos(2 x) + 544.52 \cos(3 x) - 2570.62 \cos(4 x) - 1612.7 \cos(5 x).$$One can always fit $N$ data points to any model with $N$ adjustable parameters. See plot below showing the degree 4 polynomial (blue), the sinusoidal superposition (brown) and $F(x)=5\times 4^x$ (green). All three lines pass through the data points exactly. The simpler model is the exponential and we must choose it by applying Occam's razor.

I extended the plot to include $x=5$ and show that the three lines diverge. If we had $F(5)$, we would be able to distinguish which one is correct. If $F(5)$ is right on the exponential, @SammyS could add another term to the polynomial and I could another sine making both our models more complex while the exponential would retain its simplicity.