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If I have this pattern:

1 goes to 40

1000 goes to 1360

10000 goes to 3480

How do I ask Wolfram Alpha what 20,000,000 goes to?

This is my attempt: http://www.wolframalpha.com/input/?i=1-%3E+40%2C+1000-%3E+1360+%2C+10000+-%3E3480%2C+20000000+-%3E+y+

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Is that an arithmetic or geometric progression? Could you elaborate on what you're trying to do? – R.K. Jun 2 '11 at 11:16
I don't have know the terminology to articulate what I am trying to do. Given that pattern I have shown, what notation would I use to ask wolfram alpha what 20million goes to. – Dan Jun 2 '11 at 13:06
Do you know the formula used to get those results? – Stimy Jun 3 '11 at 17:05
Nope. I was hoping I could insert the pattern and wolfram could plot the graph? – Dan Jun 5 '11 at 8:46

3 Answers

up vote 0 down vote accepted

So I am pretty sure that Wolfram cant solve stuff without a formula. So i tried a couple things. .

First I tried to Plot it out, which got me part of the way there but did not give me the full graph. http://www.wolframalpha.com/input/?i=plot+1%2F40%2C+1000%2F1360%2C10000%2F3480+

So next i went to a buddy of mine who is more a nerd then i am who has a program on his computer that given a set of numbers will find a formula that can generate more in that set. The program is called Eureqa and the formula it generated that fit the first 3 numbers in the set was 38.574093 + 1.426013*x - 0.00010458704*x*x

Plugging 20 000 000 into the equation got a value of -41806295701

and here is your graph of the formula via wolfram http://www.wolframalpha.com/input/?i=plot+38.574093+%2B+1.426013*x+-+0.00010458704*x*x

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Eureqa found a slightly better fitting equation. (100546.73 + 4841.2861*x)/(2633.7004 + x)but it yields a pretty drastically different result for 20 000 000. the problem is that with such a small data set to work there are multiple 'solutions' that could fit. – Mickey Slater Jun 13 '11 at 17:58
up vote 4 down vote

As Phwd pointed out, the key word in Wolfram|Alpha is "fit", so
fit {{1,40},{1000,1360},{10000,3480}}
will give you some least square fits to a linear, log and periodic (sinusoidal) functions. But none of these work very well.

However, if you want to be more specific, in this case W|A actually accepts the normal Mathematica input (this is not always the case). Since the data looks logarithmic + corrections, I tried
Fit[{{1,40},{1000,1360},{10000,3480}}, {1, x, x^2, Log[x]}, x] to find the least squares fit. The result was
39.8988 + 0.101156 x + 8.141317224831925*^-6 x^2 + 175.282 Log[x]

(Aside: you could also try taking the log of the x values first, then a quadratic fit).

As expected, with 4 free parameters and 3 data points, we get a very good fit! enter image description here

The extrapolation up to x=20,000,000 should not be trusted (but I find 3.25855*10^9).

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That notation doesn't work for me, it gives an output of {(Fit[data, funs, vars] | 40 Fit[data, funs, vars] 1000 Fit[data, funs, vars] | 1360 Fit[data, funs, vars] 10000 Fit[data, funs, vars] | 3480 Fit[data, funs, vars]), {1, x, x^2, log(x)}, x} – endolith Apr 7 '12 at 4:28
@endolith: I'm not quite sure what you're talking about... – Simon Apr 7 '12 at 5:29
http://www.wolframalpha.com/input/?i=[Fit{{1%2C40}%2C{1000%2C1360}%2C{10000%2C3‌​480}}%2C+{1%2C+x%2C+x^2%2C+Log[x]}%2C+x] – endolith Apr 7 '12 at 15:04
1  
@endolith: Sorry about that, there was a problem with escaping the square bracket in the link text, which meant the bracket was in the was in the wrong place -- for some reason this broke W|A's natural language input processing and it returned half-baked nonsense. It's fixed now. – Simon Apr 8 '12 at 0:00
up vote 2 down vote

You are looking for regression analysis. So first you would need to understand according to your data what you would expect.

Is it a wave, is it exponential, quadratic? This type of information leads to better results. On first inspection, one could see a linear regression will not do.

Linear fit http://www.wolframalpha.com/input/?i=linear+fit+{1%2C+40}%2C{1000%2C+1360}%2C{10000%2C+3480}

So the next step (within Wolfram's limitation) is a quadratic, which fits but only because there are so little points.

-0.000108587 x^2+1.43002 x+38.5701

Which agrees with what @Mickey is saying

Quadratic fit http://www.wolframalpha.com/input/?i=quadratic+fit+{1%2C+40}%2C{1000%2C+1360}%2C{10000%2C+3480}

The same could be achieved for cubic (i.e. no x^3),

-0.000108587 x^2+1.43002 x+38.5701

Exponential (exponential fit) and Logarithmic (log fit) do not work well.

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