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February 1st, 2016, 07:14 PM   #31
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Quote:
Originally Posted by Denis
Geezzz...can someone pleeezzzze close this thread...
Okay, Denis. I'll close it with the final line:

1 × 2 = 2 (or in the order that you had it: 2 × 1 = 2)

Last edited by Timios; February 1st, 2016 at 07:18 PM.
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October 28th, 2016, 09:51 PM   #32
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The '=' is a relation operator which is used for assignment, and although we most commonly assign equality, it doesn't change what it truly is. That said ...

It turns out the ‘?’ can be whatever we want it to be!

Allow me to illustrate. Take for example, f(x)=(1/40)x^4-(13/20)x^3+(291/40)x^2-(553/20)x+42, and evaluate f(8), f(7), f(6), f(5), f(3). It turns out that:

f(8)=56,
f(7)=42,
f(6)=30,
f(5)=20,
f(3)=9.

Wait, what sorcery is this? Turns out that although the popular rule f(x)=x(x-1), which gives f(x)=6, satisfies the known values in the sequence, that f(x)=(1/40)x^4-(13/20)x^3+(291/40)x^2-(553/20)x+42 also satisfies them–except with a different value of f(3)!

Here’s another one that also works but gives f(3)=12:
f(x)=(1/20)x^4-(13/10)x^3+(271/20)x^2-(543/10)x+84

And here is one where f(3)=π
f(x)=(1/120)(π-6)x^4-(13/60)(π-6)x^3+(1/120)(251π-1386)x^2+(1/60)(3138-533π)x+14(π-6)

Finally, in general if you want the ‘?’=k, i.e., f(3)=k where k is the value of your choice, then
(1/120)(k-6)x^4-(13/60)(k-6)x^3+(1/120)(251k-1386)x^2+(1/60)(3138-533k)x+14(k-6)

More details here: https://www.scribd.com/doc/260182194...tary-Sequences
For a non-polynomial rule see here:

Last edited by skipjack; December 13th, 2016 at 08:22 PM.
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