numeric solving vs graph  Hewlett Packard
This is a discussion on numeric solving vs graph  Hewlett Packard ; Hello all,
I had a quite simple problem, or so I thought. I needed to find the
first 3 values of t for which dx/dt =0, with x(t) = 4 sin3t e^(t).
I put the equation on the stack, LS+Calc ...

numeric solving vs graph
Hello all,
I had a quite simple problem, or so I thought. I needed to find the
first 3 values of t for which dx/dt =0, with x(t) = 4 sin3t e^(t).
I put the equation on the stack, LS+Calc > DERVX. It's some crazy
formula filling up the display. Functions in ALG don't really make it
look any better, so I leave it as is.
I put 0 on the stack and then =, which gives me dx/dt = 0.
Then I go to LS+S.SLV > SOLVE (the first one). The result after a few
seconds is { }.
Graphing that function, showing f', and getting the roots in Graph
mode reveal the correct answer of t=0.416 for the first, and from that
I can deduce t(n) = 0.416+n*PI/3.
How can I make it work numerically?
Thanks! Hope this was clear enough....
Christoph

Re: numeric solving vs graph
On Sep 20, 11:11*pm, Christoph Koehler
wrote:
> Hello all,
>
> I had a quite simple problem, or so I thought. I needed to find the
> first 3 values of t for which dx/dt =0, with x(t) = 4 sin3t e^(t).
>
> I put the equation on the stack, LS+Calc > DERVX. It's some crazy
> formula filling up the display. Functions in ALG don't really make it
> look any better, so I leave it as is.
> I put 0 on the stack and then =, which gives me dx/dt = 0.
>
> Then I go to LS+S.SLV > SOLVE (the first one). The result after a few
> seconds is { }.
>
> Graphing that function, showing f', and getting the roots in Graph
> mode reveal the correct answer of t=0.416 for the first, and from that
> I can deduce t(n) = 0.416+n*PI/3.
>
> How can I make it work numerically?
>
> Thanks! Hope this was clear enough....
>
> Christoph
I got that dx/dt = (12 cos(3t)  4 sin(3t))/e^t.
The symbolic solver also gave me { }, so I used the numeric solver.
Use STEQ to store level 1 of the stack into variable EQ, then launch
the numeric solver ([RS]7 (NUM.SLV) 1. Solve equation). Supply an
initial guess (I used 1) and I got the root 1.0646... Different
guesses will result in different roots, though this method is somewhat
blind.
If you want the first 3 values where dx/dt = 0, it's probably easier
to use the grapher. Use STEQ, then hold [LS] F1, ERASE DRAW. When in
the graphing environment, move the cursor close to the root, press
FCN, then ROOT. Do this for the first 3 intersections and get t =
1.0646, t = 2.1118, t = 3.1590.
S.C.

Re: numeric solving vs graph
In article
,
sc_usenet@hotmail.com wrote:
> On Sep 20, 11:11*pm, Christoph Koehler
> wrote:
> > Hello all,
> >
> > I had a quite simple problem, or so I thought. I needed to find the
> > first 3 values of t for which dx/dt =0, with x(t) = 4 sin3t e^(t).
> >
> > I put the equation on the stack, LS+Calc > DERVX. It's some crazy
> > formula filling up the display. Functions in ALG don't really make it
> > look any better, so I leave it as is.
> > I put 0 on the stack and then =, which gives me dx/dt = 0.
> >
> > Then I go to LS+S.SLV > SOLVE (the first one). The result after a few
> > seconds is { }.
> >
> > Graphing that function, showing f', and getting the roots in Graph
> > mode reveal the correct answer of t=0.416 for the first, and from that
> > I can deduce t(n) = 0.416+n*PI/3.
> >
> > How can I make it work numerically?
> >
> > Thanks! Hope this was clear enough....
> >
> > Christoph
>
> I got that dx/dt = (12 cos(3t)  4 sin(3t))/e^t.
>
> The symbolic solver also gave me { }, so I used the numeric solver.
> Use STEQ to store level 1 of the stack into variable EQ, then launch
> the numeric solver ([RS]7 (NUM.SLV) 1. Solve equation). Supply an
> initial guess (I used 1) and I got the root 1.0646... Different
> guesses will result in different roots, though this method is somewhat
> blind.
>
> If you want the first 3 values where dx/dt = 0, it's probably easier
> to use the grapher. Use STEQ, then hold [LS] F1, ERASE DRAW. When in
> the graphing environment, move the cursor close to the root, press
> FCN, then ROOT. Do this for the first 3 intersections and get t =
> 1.0646, t = 2.1118, t = 3.1590.
>
> S.C.
Curious. For the same dy/dx as you have, I find different zeroes.
Namely: 0.4163, 1.4635 and 2.5101, to 4 decimal places.

Re: numeric solving vs graph
On Sun, 21 Sep 2008 09:40:12 0500:
> Different guesses will result in different roots,
> though this method is somewhat blind.
An aid to systematic searching for numeric ROOTS over an interval:
http://groups.google.com/group/comp....f2e21bace9dd00
UserRPL program listing:
http://groups.google.com/group/comp....f?dmode=source
[r>] [OFF]

Re: numeric solving vs graph
On Sep 21, 2:13*pm, Virgil wrote:
> In article
> ,
>
>
>
>
>
> *sc_use...@hotmail.com wrote:
> > On Sep 20, 11:11*pm, Christoph Koehler
> > wrote:
> > > Hello all,
>
> > > I had a quite simple problem, or so I thought. I needed to find the
> > > first 3 values of t for which dx/dt =0, with x(t) = 4 sin3t e^(t).
>
> > > I put the equation on the stack, LS+Calc > DERVX. It's some crazy
> > > formula filling up the display. Functions in ALG don't really make it
> > > look any better, so I leave it as is.
> > > I put 0 on the stack and then =, which gives me dx/dt = 0.
>
> > > Then I go to LS+S.SLV > SOLVE (the first one). The result after a few
> > > seconds is { }.
>
> > > Graphing that function, showing f', and getting the roots in Graph
> > > mode reveal the correct answer of t=0.416 for the first, and from that
> > > I can deduce t(n) = 0.416+n*PI/3.
>
> > > How can I make it work numerically?
>
> > > Thanks! Hope this was clear enough....
>
> > > Christoph
>
> > I got that dx/dt = (12 cos(3t)  4 sin(3t))/e^t.
>
> > The symbolic solver also gave me { }, so I used the numeric solver.
> > Use STEQ to store level 1 of the stack into variable EQ, then launch
> > the numeric solver ([RS]7 (NUM.SLV) 1. Solve equation). Supply an
> > initial guess (I used 1) and I got the root 1.0646... Different
> > guesses will result in different roots, though this method is somewhat
> > blind.
>
> > If you want the first 3 values where dx/dt = 0, it's probably easier
> > to use the grapher. Use STEQ, then hold [LS] F1, ERASE DRAW. When in
> > the graphing environment, move the cursor close to the root, press
> > FCN, then ROOT. Do this for the first 3 intersections and get t =
> > 1.0646, t = 2.1118, t = 3.1590.
>
> > S.C.
>
> Curious. For the same dy/dx as you have, I find different zeroes.
>
> Namely: 0.4163, 1.4635 and 2.5107, to 4 decimal places.
That's weird. I just did the same thing again and got your roots. Now
I'm not sure what happened last time.
S.C.

Re: numeric solving vs graph
sc_usenet@hotmail.com wrote:
> I got that dx/dt = (12 cos(3t)  4 sin(3t))/e^t.
>
> The symbolic solver also gave me { }, so I used the numeric solver.
> ...
> If you want the first 3 values where dx/dt = 0, it's probably easier
> to use the grapher.
> ...
> Use STEQ, then hold [LS] F1, ERASE DRAW. When in
> the graphing environment, move the cursor close to the root, press
> FCN, then ROOT.
Well, I don't want to disturb this discussion about various more or less
elegant ways to make the calculator solve the problem, but in this case
there is a much easier and more straightforward way: simply use your own
brain, the calculator almost provided the solution anyway:
> I got that dx/dt = (12 cos(3t)  4 sin(3t))/e^t.
So the rest is trivial: a local minimum or maximum of x(t) requires
dx/dt = 0 => 12 cos(3t)  4 sin(3t) = 0 => t = 1/3 arctan 3 = 0.4163...
Since tan x = tan(x + Pi), simply add multiples of Pi/3 for further
solutions, i.e. t = 0.4163, 1.4635, 2.5107, 3.5579, 4.6051...
It's that easy.
Okay, I didn't check the second derivate. <8)
Maybe this is a good example for combining the benefits of our little
electronic helpers with good old plain manmade math. The former
provides the derivate, the latter gives the final result.
In this case even the correct one. 8)
Dieter

Re: numeric solving vs graph
On Sep 21, 8:07*pm, Dieter wrote:
> Maybe this is a good example for combining the benefits of our little
> electronic helpers with good old plain manmade math. The former
> provides the derivate, the latter gives the final result.
> In this case even the correct one. 8)
>
Exactly  the calculator is (or should be) and extension to the
brain, not a replacement.
I didn't think about that though, since I was more focused on
calculator usage with this question.
S.C.

Re: numeric solving vs graph
In article
,
sc_usenet@hotmail.com wrote:
> On Sep 21, 2:13*pm, Virgil wrote:
> > In article
> > ,
> >
> >
> >
> >
> >
> > *sc_use...@hotmail.com wrote:
> > > On Sep 20, 11:11*pm, Christoph Koehler
> > > wrote:
> > > > Hello all,
> >
> > > > I had a quite simple problem, or so I thought. I needed to find the
> > > > first 3 values of t for which dx/dt =0, with x(t) = 4 sin3t e^(t).
> >
> > > > I put the equation on the stack, LS+Calc > DERVX. It's some crazy
> > > > formula filling up the display. Functions in ALG don't really make it
> > > > look any better, so I leave it as is.
> > > > I put 0 on the stack and then =, which gives me dx/dt = 0.
> >
> > > > Then I go to LS+S.SLV > SOLVE (the first one). The result after a few
> > > > seconds is { }.
> >
> > > > Graphing that function, showing f', and getting the roots in Graph
> > > > mode reveal the correct answer of t=0.416 for the first, and from that
> > > > I can deduce t(n) = 0.416+n*PI/3.
> >
> > > > How can I make it work numerically?
> >
> > > > Thanks! Hope this was clear enough....
> >
> > > > Christoph
> >
> > > I got that dx/dt = (12 cos(3t)  4 sin(3t))/e^t.
> >
> > > The symbolic solver also gave me { }, so I used the numeric solver.
> > > Use STEQ to store level 1 of the stack into variable EQ, then launch
> > > the numeric solver ([RS]7 (NUM.SLV) 1. Solve equation). Supply an
> > > initial guess (I used 1) and I got the root 1.0646... Different
> > > guesses will result in different roots, though this method is somewhat
> > > blind.
> >
> > > If you want the first 3 values where dx/dt = 0, it's probably easier
> > > to use the grapher. Use STEQ, then hold [LS] F1, ERASE DRAW. When in
> > > the graphing environment, move the cursor close to the root, press
> > > FCN, then ROOT. Do this for the first 3 intersections and get t =
> > > 1.0646, t = 2.1118, t = 3.1590.
> >
> > > S.C.
> >
> > Curious. For the same dy/dx as you have, I find different zeroes.
> >
> > Namely: 0.4163, 1.4635 and 2.5107, to 4 decimal places.
>
> That's weird. I just did the same thing again and got your roots. Now
> I'm not sure what happened last time.
>
> S.C.
We'll probably never know.

Re: numeric solving vs graph
On Sep 21, 7:07*pm, Dieter wrote:
> Maybe this is a good example for combining the benefits of our little
> electronic helpers with good old plain manmade math. The former
> provides the derivate, the latter gives the final result.
> In this case even the correct one. 8)
I totally agree.
I can do this by hand. But in a test setting I'd rather concentrate on
the subject matter and not worry about taking derivatives. This is not
for a Calculus class but EE.
Thanks for all the tips
Christoph