Please help! Can you find f(x) from the following sequences?

Question

Problem No. 1:

x , y
0 , 1
1 , 3
2 , 15
3 , 37
4 , 69
5 , 111

Problem No. 2:

x , y
0 , -3
1 , 4
2 , 65
3 , 234
4 , 565
5 , 1112

How can I find f(x) from them? Is there a general rule or technique that I could use to derive f(x)?

I would much appreciate if you could teach me "how to" derive than give me instant answers.

Thank you very much in advance!!

Thanks guys for the effort in answering my problem. You sure are very helpful!

Answer 1

Yes, I think I can help you with your math problem and provide a general “technique” which I have used and proven to be useful. The technique that I am giving you is not the best technique but could solve problems like the one you’ve posted.

Problem No. 1:

Let’s start by observing a pattern – we subtract adjacent values from y and denote it as ∆y’. Then, again we subtract the adjacent values from ∆y’ and denote the values as ∆y’’.

Thus we have

x, y, ∆y', ∆y''
0, 1
1, 3, 2
2, 15, 12, 10
3, 37, 22, 10
4, 69, 32, 10
5, 111, 42, 10

From these data, I conclude that f(x) can be derive from second-degree simultaneous equations since we already have a constant in ∆y''.

That is f(x) = ax² + bx + c = d.

All we have to do is substitute values for x and solve for a, b and c from the linear equations.

Thus

c = 1, when x = 0
a + b + c = 3, when x = 1
4a + 2b + c = 15, when x = 2

From here, it is very easy to solve for a, b and c.

Without showing how we derive for a, b and c, the solution for Problem 1 should be
f(x) = 5x² - 3x + 1.

Problem No. 2:

The solution for Problem No. 2 is similar to what we did in solving for Problem No. 1.

We have

x, y, ∆y', ∆y'', ∆y'''
0, -3
1, 4, 7
2, 65, 61, 54
3, 234, 169, 108, 54
4, 565, 331, 162, 54
5, 1112, 547, 216, 54

Here we have f(x) = ax³ + bx² + cx + d = e --- this time it's third degree! Not quadratic!!

As usual, we simply plug the values for x and e and solve for a, b, c and d from the resulting simultanoues equations in order to get the solution!

I think you could do the rest by yourself Buddy!

If you get it right, the answer must be f(x) = 9x³ - 2x - 3.
HAVE A VERY NICE DAY!!

Answer 2

There are some general rules for certain types of functions, such as looking for "first differences" and "second differences" and the like.

These look like (hopefully) quadratic functions, so you can try to fit a function of y= Ax^2+Bx+C. To do this, pick the 1st, fourth and last points to form three equations (problem 1)
x=0, C=1
x=3: 9A+3B+1= 37
x=5: 25A+5B+1= 111
then 45A+15B+5=185 (5x the x=3 eqtn)
75A+15B+3=333 (3x the x=5 eqtn)
subtract: 30A -2 = 148 and A=5
B=-3. So y= 5x^2-3x+1

Answer 3

Last time I did a problem like this I used Excel (you can also use a graphing calculator). You plot the points and then the program can perform a best fit line with an approximate equation that will satisfy your needs.

That's my quick answer without really looking at the data (too much Calculus 3 homework to try it by hand!!!).