2
$\begingroup$

I recently started using mathematica for simple tasks like simplifying etc.

Table of values

First of all let me say if this breaks any rules I apologize since I'm a novice in using this and I'll remove the question.

I want to use the table of values above to find a GENERAL FORMULA that represents the values.

What I want to achieve is a function in terms of '$V$' such that for any value of '$V$' I enter I will get the corresponding '$SS$' value.

In simple terms I'm looking for an equation such that when '$V$' is equal to '6' say, then I can input this into an equation which would probably look something like $D_{5}$ [Since it's the sum of the first 5 terms]= equation in terms of $V$ = $SS$ value.

But I need this equation to hold for an value of $V$ in the entire table.

The formula can also use the $E$ value I just thought it would be easier to study with respect to $V$.

I know there's the FindFormula operator but i'm not sure how to incoperate this with the table to solve such a problem.

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ The first step would be plotting the data. But with only 4 data points there's not much you can do for which you could be confident in future predictions. You might consider the following: v = {6, 9, 12, 15}; ss = {29, 84, 178, 320}; ListLogLogPlot[Transpose[{v, ss}]] followed by lm = LinearModelFit[Log[Transpose[{v, ss}]], x, x]. $\endgroup$
    – JimB
    Commented Sep 11, 2020 at 3:07
  • $\begingroup$ What is the meanings of rows and colums in your table? $\endgroup$
    – Mustafa Kösem
    Commented Sep 11, 2020 at 4:31
  • $\begingroup$ @JimB If I were to calculate more $SS$ values(Which I can) Is it possible to get a better more suited general formula? And how would I do such. Thanks $\endgroup$
    – Jason Moore
    Commented Sep 11, 2020 at 17:00
  • $\begingroup$ @MustafaKösem The Rows and Columns correspond to a vertex and edge of a random graph in which I used a method to find the SS values. I am currently doing some research in different nanostar calculations. $\endgroup$
    – Jason Moore
    Commented Sep 11, 2020 at 17:00

1 Answer 1

Reset to default
1
$\begingroup$ 
f[n_]:=2*(3n+6-1)(n+2)+Total[9Table[Binomial[n,j]+n,{j,0,n}] ]
f[0]
f[1]
f[2]
f[3]

29
84
178
320

Also here is the hole Table:

Table[Flatten[Riffle[Table[{3n+6-1,3n+6-1},n+2],9Table[Binomial[n,j]+n,{j,0,n} ]]],{n,0,3}]

{{5,5,9,5,5},{8,8,18,8,8,18,8,8},{11,11,27,11,11,36,11,11,27,11,11},{14,14,36,14,14,54,14,14,54,14,14,36,14,14}}

Improve this answer
$\endgroup$
6
  • $\begingroup$ Can you please elaborate on how you arrived at 2*(3n+6-1)(n+2)? This seems to be the crux of the formula. I'm inquiring because if i calculate different nanostar graphs my table rows and columns will change and the number of values etc. Is there a blueprint that I can use to change accordingly? Thank you. $\endgroup$
    – Jason Moore
    Commented Sep 11, 2020 at 17:05
  • 1
    $\begingroup$ That formula is the easy part... it just calculates the n-1 values for the initial values of V={6,9,12,15...} but it works for 0,1,2,3.... What values exactly do you need to know? The formula works for any n, and so does the table. $\endgroup$
    – ZaMoC
    Commented Sep 11, 2020 at 17:28
  • 1
    $\begingroup$ @JasonMoore the above formula generates the SS values that you want {29,84,178,320,528,838,1322,2124,3532,6122,11046....}. It doesn't really matter if you use {0,1,2,3,....) or 6+3n for n={0,1,2,3} (which is the V column) or 5+3n (which is the E column). The tricky part is to calculate the values {...,36,...,54,.....,54,,,,36} which come from Binomial[n,k]+n $\endgroup$
    – ZaMoC
    Commented Sep 11, 2020 at 18:16
  • $\begingroup$ Can you indulge me a bit further if possible. What changes would I have to make to this table [imgur.com/FodLUQF] to have the same type of outputs. Yes , the values are indeed different , however the $V$ and $E$ columns follow the same $n-1$ pattern as before but the above formula will not hold for this table. Can you please advise me on what tools specifically that I would have to master , or look at , to be able to manipulate the code to any such table? Thank you @J42161217 $\endgroup$
    – Jason Moore
    Commented Sep 15, 2020 at 20:26
  • $\begingroup$ [Imgur]imgur.com/gallery/ffNlsdQ Backup link of image incase it's taken down. $\endgroup$
    – Jason Moore
    Commented Sep 15, 2020 at 20:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.