# No solution from RSolve[ ] of two equations?

Posted 10 months ago
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 I know S[n] is Sum[a[k], {k,1,n}] a[1]==1 I want to get the general solution of a[n] and S[n].So I tried this one.But I can't get the answer. What can I do for this problem?
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Posted 10 months ago
 I do not understand what S_k...(*) means. Maybe this approach is helpful: s[n_] := Sum[a[i], {i, 1, n}] ss[n_] := Sum[(n - i + 1) a[i], {i, 1, n}] (* your equation: *) eqn[n_] := (n + 1) s[n + 1] == Log[2, n + 2] + ss[n] nMax = 10; (* as an example *) vars = Table[a[n], {n, 1, nMax + 1}]; equations = Table[eqn[n], {n, 0, nMax}]; (* solutions: *) Solve[equations, vars] Though not rigorous according to this it seems that $$a[n]= -\frac{\log(n) - \log(n+1)}{n \log(2)}$$ is the general solution.
Posted 10 months ago
 Oh. Thank you very much. Your approach is helpful for me.
Posted 10 months ago
 @Park Seong Hyun Welcome to Wolfram Community! Please next time make sure you know the rules: https://wolfr.am/READ-1STThe rules explain how to format your code properly. Posting code Images doesn't help other members to copy your code. Please EDIT your post and make sure code blocks start on a new paragraph and look framed and colored like this. int = Integrate[1/(x^3 - 1), x]; Map[Framed, int, Infinity] You can also embed notebook or attach notebook.
Posted 10 months ago
 Looking at your equations eqn[ n ] (in Henriks notation). expanding them (by hand ) and and forming the difference eqn[ j + 1 ] - eqn [ j ] you arrive at ( j + 2 ) a[ j + 2 ] == Log[ ( j + 3 ) / ( j + 2 ) ] / Log[ 2 ] and therefore a[ j_] := Log[(1 + j)/j]/(j Log[2]) 
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