Y = (Sin[2 [Pi]t]) (1 + (1/5) Sin[6 [Pi]t] + (1/10) Sin[8 [Pi]t])
I'm trying to plot this function in Mathematica, however when I run Plot[Y, {t, -15, 15}] nothing will show up on the graph.
I think the problem is I'm entering the syntax wrong for all the trig functions?
Sorry I'm new to Mathematica. Thank you for any help! Joe
During evaluation of In[116]:= SetDelayed::write: Tag Times in (Sin[6.28319 sin^2 x] (1+1/5 Sin[18.8496 sin^2 x]+1/10 Sin[25.1327 sin^2 x]))[t_] is Protected. >>
Out[116]= $Failed
– Joe Mar 09 '13 at 23:07In[120]:= Y[t_] := (Sin[ 2 [Pi]t]) (1 + (1/5) Sin[6 [Pi]t] + (1/10) Sin[8 [Pi]t])
During evaluation of In[120]:= SetDelayed::write: Tag Times in (Sin[6.28319 sin^2 x] (1+1/5 Sin[18.8496 sin^2 x]+1/10 Sin[25.1327 sin^2 x]))[t_] is Protected. >>
Out[120]= $Failed
– Joe Mar 09 '13 at 23:19In[2]:= Y[ t_] := (Sin[ 2 [Pi]t]) (1 + (1/5) Sin[6 [Pi]t] + (1/10) Sin[8 [Pi]t])
In[4]:= Plot[Y[t_], {t, -15, 15}]
Out[4]= a blank graph
– Joe Mar 09 '13 at 23:25Clear[x,t,Y]Then make your $Y$ function again, using the delayed evaluationY[t_] := (your function in t)(Thetshould turn green in your expression.) Then do your plot again:Plot[ Y[t] , {t , -15, 15} ]. – Alexander Gruber Mar 09 '13 at 23:25_aftertin yourPlot._is used in conjunction with:=to denote a delayed evaluation, meaning that it won't evaluate what's on the RHS of the:=untilYis called with an argument. When you puttinY[t]in thePlot, you're actually fixing at, so there's no need for_. (You're calling a function, not making a function.) – Alexander Gruber Mar 09 '13 at 23:28