Numbers sequence problem

[OctangularAddict]

Hello, I can't figure out the answer to this exercise. It says:
190 -> 42
180 -> 37
160 -> 24
140 -> 16
122 -> 10
154 -> ?

Thank you for your help.

 

[Subhotosh Khan]

Hello, I can't figure out the answer to this exercise. It says:
190 -> 42
180 -> 37
160 -> 24
140 -> 16
122 -> 10
154 -> ?

Thank you for your help.

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This problem can be attempted in many ways. What methods have you been taught in your class.

 

[OctangularAddict]

We were given this problem as part of the summer homework and it doesn't seem to be related to anything we have done during the school year, I suspect we have to try and solve it in different ways but I haven't managed to do it. In our school we do math competitions and no way of solving is indicated.

 

[Subhotosh Khan]

Okay .... then .... what are the topics in math you had covered last academic session (semester)?

 

[Dr.Peterson]

Hello, I can't figure out the answer to this exercise. It says:
190 -> 42
180 -> 37
160 -> 24
140 -> 16
122 -> 10
154 -> ?

Is this the entire problem, word for word? If not, please quote it exactly.

But if it is, then it can't be answered!

It isn't really about a sequence at all, but a function: determining an "output" for a particular "input". And there is no one function that produces these outputs; in particular, if you graph these pairs, you'll see that they don't lie on a straight line or any other "nice" curve. The best you could do would be to "fit" the data to a particular kind of function (e.g. linear) to estimate the requested value. Or you could use what is called linear interpolation. Neither of those can claim to give "the correct answer"; but you could give reasons why your answer makes sense as a good guess.

 

[HallsofIvy]

Given any finite set of data, there exist an infinite number of possible function that satisfy that data so there is no one "correct" answer to this problem. It is true that given n data points, there exist a unique polynomial of degree n-1 that passes through those data points. Here you are given 5 data points so there exist a unique polynomial of degree 4 giving this data.

Any such polynomial can be written \(\displaystyle y= ax^4+ bx^3+ cx^2+ dx+ e\). Putting (x, y) from your data into that equation gives you five equations to solve for a, b, c, d, and e.

A variation of this same idea is to use "Newton's divided difference formula". A polynomial satisfying this data is \(\displaystyle 42\frac{(x- 180)(x- 160)(x- 140)(x- 122)}{(190- 180)(190- 160)(190- 140)(190- 122)}\)\(\displaystyle + 37\frac{(x- 190)(x- 160)(x- 140)(x- 122)}{(180- 190)(180- 160)(180- 140)(190- 122)}\)\(\displaystyle + 24\frac{(x- 180)(x- 190)(x- 140)(x- 122)}{(160- 190)(160- 180)(160- 140)(160- 122)}\)\(\displaystyle + 16\frac{(x- 190)(x- 180)(x- 160)(x- 122)}{(140- 190)(140- 180)(140- 160)(140- 122)}\)\(\displaystyle + 10\frac{(x- 190)(x- 180)(x- 160)(x- 140)}{(122- 190)(122- 180)(122- 160)(122- 140)}\).