richpu
(Fangqing Pu)
1
Consider the following code:
x =.
s =.
x = 1 - (2 s)^0.5
\[Alpha] = 0.5
n = 4
A = 1 - x^n - \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = 0\), \(\(-1\) + n\)]
\*FractionBox[\(
\(\*SuperscriptBox[\(x\), \(k\)]\)[1 -
\*SuperscriptBox[\(x\), \(\(-k\) + n\)]]\), \(\(-k\) + n\)]\)
p = 1 - x
b = \[Alpha] A*(1 - x)^2/(A + 1 - x^n)
data = Table[Table[{s, f}, {s, 0, 1, 0.01}], {f, {p, b}}];
ListLinePlot[data, PlotLegends -> {price, ad}]
In Mathematica, I want to show both lines of p (for price) and b (for ad), but why line of b is missing?
danl
(Daniel Lichtblau)
2
Please post actual code (not an image), and a question to go with it.
x = 1 - (2 s)^0.5;
[Alpha] = 0.5;
n = 4;
A = 1 - x^n - Sum[(x^k*(1 - x^(-k + n)))/(-k + n), {k, 1, n - 1}];
p = 1 - x;
b = [Alpha]A(1 - x)^2/(A + 1 - x^n);
data = Table[{s, p}, {s, 0, 1, 0.01}];
data2 = Table[{s, b}, {s, 0.0001, 1, 0.01}];
ListLinePlot[{data, data2}, PlotLegends → {price, ad}]
trounev
(Alexander Trounev)
4
Thank You! There are two errors - square brackets in Sum and zero initial value s, after fix it works
x =.
s =.
x = 1 - (2 s)^0.5
\[Alpha] = 0.5
n = 4
A = 1 - x^n - \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = 0\), \(\(-1\) + n\)]
\*FractionBox[\(
\*SuperscriptBox[\(x\), \(k\)] \((1 -
\*SuperscriptBox[\(x\), \(\(-k\) + n\)])\)\), \(\(-k\) + n\)]\)
p = 1 - x
b = \[Alpha] A*(1 - x)^2/(A + 1 - x^n)
data = Table[Table[{s, f}, {s, 0.0001, 1, 0.01}], {f, {p, b}}];
ListLinePlot[data, PlotLegends -> {price, ad}]
