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May 19

predicting results from a trend

I have this graph (http://dota2.cyborgmatt.com/prizetracker/international2014) that I want to extend to see the predicted results at any day. I've been out of school too long to know how to do this. I would love an explanation of what to do with this set of data. I'm nearly sure it's quadratic in nature. Oh and the daily tracker is good enough, don't need to use the hourly tracker. Thanks Bluefist ^talk 05:37, 19 May 2014 (UTC)[reply]

Do you have a set of data or do you only have the graph? The graph looks like

y_{i}\approx {\sqrt {a(x_{i}-b)}}

, so if you have

(x_{i},y_{i})

you should compute

(x_{i},y_{i}^{2})

and make a regression analysis to find reasonable values for a and b. Bo Jacoby (talk) 17:58, 19 May 2014 (UTC).[reply]

If you hover over the points it tells you the data in a popup. Also I don't know how to do any of that. Bluefist ^talk 16:19, 20 May 2014 (UTC)[reply]

The data are these:

Subtract y₀ from the y-column.

Then divide by y₁₀−y₀

 0 0.000000
 1 0.250462
 2 0.419510
 3 0.529392
 4 0.638706
 5 0.729377
 6 0.794454
 7 0.854983
 8 0.915841
 9 0.962325
10 1.000000

Square the y-values, and multiply by 10

 0  0.00000
 1  0.62731
 2  1.75988
 3  2.80256
 4  4.07946
 5  5.31991
 6  6.31157
 7  7.30996
 8  8.38765
 9  9.26069
10 10.00000

The two columns are now reasonably equal. $x\approx 10\left({\frac {y-y_{0}}{y_{10}-y_{0}}}\right)^{2}$ and so $y\approx y_{0}+(y_{10}-y_{0}){\sqrt {\frac {x}{10}}}$ . This formula does what you requested: the predicted results (y) at any day (x). Note that y₀ = 1600000 and y₁₀ = 5920247. Bo Jacoby (talk) 22:30, 20 May 2014 (UTC).[reply]

Logarithmic operation

Is there any logarithmic operation of adding 1?? We know that (defining the logarithmic operation of an operation as the operation that relates the binary logarithms of the numbers the way the original operation relates the original numbers):

The logarithmic operation of doubling is adding 1.
The logarithmic operation of squaring is doubling.
The logarithmic operation of raising a number to the power of its binary logarithm is squaring.

But what operation is there that can be the logarithmic operation of adding 1?? Georgia guy (talk) 15:36, 19 May 2014 (UTC)[reply]

See List of logarithmic identities, especially the summation/subtraction section of that article. The formal answer to you question is "raising 2 to the power of the number, adding 1 to the result, taking the base-2 logarithm of the result", but the identities for the logarithm of a sum aren't as pretty as the identities for products, quotients or powers. Icek (talk) 18:34, 19 May 2014 (UTC)[reply]

\log(x+1)=\log(x)+\log(1+{\frac {1}{x}})

. "The logarithmic operation of adding 1 is adding the logarithm of 1+1/x". Bo Jacoby (talk) 19:29, 19 May 2014 (UTC).[reply]

But you have to add that x = 2^y, if y is the number for which the logarithmic operation shall be performed. Icek (talk) 17:49, 20 May 2014 (UTC)[reply]