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# Solving an equation of two variables where one var varies, and plot

Posted 10 years ago
 Hello All I am trying to solve for p below. \sum{k=3}^{n/2} {p^k(1-p)^{n-k}\binom{n}{k}} = \sum{k=2}^{n/2} {p^k(1-p)^{n-k}\cdot\left(\binom{n}{k} - \binom{n/2}{k}\cdot2^k\right)} How can I tell wolfram alpha to vary n from say 4 to 100 and solve for p? I would like a graphical display of n v p also. Thanks J
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Posted 10 years ago
 Thank you for your helpYes i am getting to the point where an investment in Mathematica seems appropriate.Much appreciated.
Posted 10 years ago
 I know you want to solve this in Wolfram Alpha but all I can offer is one way to do it in Mathematica: solutions = p /. Table[Flatten[N[Solve[Sum[Binomial[n, k]*p^k*(1 - p)^(n - k), {k, 3, n/2}] == Sum[p^k*(1 - p)^(n - k)*(Binomial[n, k] - Binomial[n/2, k]*2^k), {k, 2, n/2}] && 0 < p < 1, p, Reals]]], {n, 6, 100}] ListPlot[Table[{n, solutions[[n - 5]]}, {n, 6, 100}]] with output {0.272727, 0.166667, 0.105573, 0.074773, 0.0556628, 0.0430781, \ 0.0343294, 0.0280016, 0.0232764, 0.0196548, 0.0168176, 0.0145537, \ 0.0127181, 0.0112094, 0.0099541, 0.00889856, 0.00800249, 0.00723531, \ 0.00657342, 0.0059984, 0.00549568, 0.00505362, 0.00466285, \ 0.00431572, 0.00400596, 0.00372841, 0.00347873, 0.00325332, \ 0.00304914, 0.0028636, 0.00269449, 0.00253993, 0.0023983, 0.0022682, \ 0.00214841, 0.00203786, 0.00193563, 0.0018409, 0.00175297, \ 0.00167118, 0.00159499, 0.0015239, 0.00145745, 0.00139526, \ 0.00133697, 0.00128225, 0.00123083, 0.00118244, 0.00113684, \ 0.00109384, 0.00105323, 0.00101484, 0.00097851, 0.000944098, \ 0.00091147, 0.000880505, 0.000851092, 0.000823128, 0.00079652, \ 0.000771181, 0.000747033, 0.000724002, 0.00070202, 0.000681023, \ 0.000660955, 0.000641761, 0.000623391, 0.000605799, 0.000588941, \ 0.000572778, 0.00055727, 0.000542384, 0.000528087, 0.000514347, \ 0.000501137, 0.000488429, 0.000476199, 0.000464422, 0.000453077, \ 0.000442143, 0.0004316, 0.000421429, 0.000411614, 0.000402137, \ 0.000392985, 0.000384141, 0.000375592, 0.000367326, 0.000359329, \ 0.000351591, 0.0003441, 0.000336846, 0.000329819, 0.00032301, \ 0.000316409} 
Posted 10 years ago
 You'll need to have n greater than 5 to get something on the left-hand side (unless you define it to be 0 in that case). Also, I assume that you'll only want solutions for p that are between 0 and 1 (and, of course, not a complex number).Have you already solved the equation for a fixed n and just want to see if it is possible to put things in a loop?