MaplePrimes - answers and comments on Question, Why has my sine graph gone weird?

upload code, not pictures

Axel Vogt — Thu, 06 Dec 2012 02:12:12 Z

Please upload code, not pictures. It is more easy to copy than to type, no?

It is smooth, but very rapidly oscillating. I guess.

You may try to avoid the sclaing factor in front. And increase Digits to 15, at least.

Well ...

Preben Alsholm — Thu, 06 Dec 2012 02:24:01 Z

Well, are you serious? There are over 2000 periods of sin.
You could raise numpoints in effect forcing plot to paint the whole thing red (or whatever color).
restart;
phi:=-223000+2*(118470000-3300*t)^(5/8);
h:=(1/(3590-t))^(1/4)*sin(phi);
evalf((eval(phi,t=0)-eval(phi,t=3600))/2/Pi);
plot(h,t=0..3589,numpoints=5000);

Nothing prevents you from using numpoints=10000 if you are still not satisfied!

Absolutely correct.

User7843 — Thu, 06 Dec 2012 04:59:29 Z

Yes! I should have uploaded the code, it is easier to type copy it, sorry.

Yes, the curve is smooth and rapidly oscillating. I took your advice and tried it without the small number in-front, but this didn't seem to make any (noticable) difference. Thanks for the input, however!

Finally, would I just stick "Digits := 15" after "restart;", and before the other lines of code?

cannot resist

acer — Thu, 06 Dec 2012 03:00:46 Z

restart:
f:=-8.7754e-19*(1/(3590-t)^(1/4)):
plot([f,-f],t=2600..3600,filled=true,color=red,view=-6e-19..6e-19);

acer

A steep learning curve...

User7843 — Thu, 06 Dec 2012 04:51:46 Z

Firstly, thanks a lot. I am quite new to Maple and so very ignorant at the moment. I wasn't using Maple commands, I was just plotting with a right-click. Commands seem to be the way forward in this case, where for example, the numpoints are presumably at the core of the issue, and something which can be changed so easily.

I have looked up the commands you chose to use, and I understand all of it, except the evalf line. What does this line do in this particular case? I tried deleting it, and the graph plots (seemingly) fine.

By the way, it's 33000, not 3300 (sorry, I really should have put the code up!), which gives a lot more sineage.

Number of periods

Preben Alsholm — Thu, 06 Dec 2012 05:27:30 Z

@User7843 The line

evalf((eval(phi,t=0)-eval(phi,t=3600))/2/Pi);

gives the number of periods of the sin function. The result is approximately 2262 on the interval I was using.
Try removing evalf. That will show what it does.
I'm using the fact that phi is a monotone (decreasing) function of t.