Re: FindRoot vs Solve
- To: mathgroup at smc.vnet.net
- Subject: [mg22477] Re: [mg22427] FindRoot vs Solve
- From: BobHanlon at aol.com
- Date: Sat, 4 Mar 2000 02:26:42 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
eqn = (28. Cos[phi Degree] - 20. == 0);
Solve[eqn, phi]
Solve::"ifun": "Inverse functions are being used by \!\(Solve\), so some \
solutions may not be found."
{{phi -> -44.4153}, {phi -> 44.4153}}
FindRoot[eqn, {phi, 0.1, 0.1, 90.}]
FindRoot::"regex": "Reached the point \!\({96.71124978216424`}\) which is \
outside the region \!\({\({0.1`, 90.`}\)}\)."
{phi -> 96.7112}
FindRoot[lhs==rhs, {x, xstart, xmin, xmax}] searches for a solution, stopping
the search if x ever gets outside the range xmin to xmax. As indicated by the
FindRoot::regex message, the search went outside of the specified range.
Consequently, the search stopped. The returned value is not the "solution"
merely the value at the time that the search terminated.
You can either try a new starting value
FindRoot[eqn, {phi, 30., 0.1, 90.}]
{phi -> 44.4153}
or eliminate the restriction on the search region.
FindRoot[eqn, {phi, 0.1}]
{phi -> 44.4153}
Bob Hanlon
In a message dated 3/1/2000 2:06:22 AM, mitterma at linz.vai.co.at writes:
>Why is FindRoot not able to solve an equation using "Degree"? Solve can
>do
>it.
>
>In[]=
>
>Solve[28. Cos[phi Degree] - 20. == 0, phi]
>FindRoot[28. Cos[phi1 Degree] - 20. == 0, {phi1, 0.1, 0.1, 90.}]
>
>Out[]=
>
>{{phi->-44.4153},{phi->44.4153}}
>{phi1->96.7112}
>
>Only the following statement gives the right result:
>
>In[]=
>FindRoot[28. Cos[phi2] - 20. == 0, {phi2, 0.1, 0.1, 90.}]
>Out[]=
>{phi2->0.775193}
>
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From: BobHanlon at aol.com
To: mathgroup at smc.vnet.net
by imo16.mx.aol.com (mail_out_v25.3.) id k.77.1e1eb0f (1840);
Wed, 1 Mar 2000 12:22:44 -0500 (EST)
Subject: [mg22477] Re: [mg22405] (no subject)
There is a standard package
Needs["Calculus`VectorAnalysis`"]
sphericalToCartesian =
Thread[Rule[{r, theta, phi},
CoordinatesFromCartesian[{x, y, z}, Spherical]]]
{r -> Sqrt[x^2 + y^2 + z^2],
theta -> ArcCos[z/Sqrt[x^2 + y^2 + z^2]],
phi -> ArcTan[x, y]}
polarToCartesian = Drop[sphericalToCartesian /. z -> 0, {2}]
{r -> Sqrt[x^2 + y^2], phi -> ArcTan[x, y]}
cartesianToSpherical =
Thread[Rule[{x, y, z}, CoordinatesToCartesian[{r, theta, phi}, Spherical]]]
{x -> r*Cos[phi]*Sin[theta], y -> r*Sin[phi]*Sin[theta],
z -> r*Cos[theta]}
cartesianToPolar = Drop[cartesianToSpherical /. theta -> Pi/2, -1]
{x -> r*Cos[phi], y -> r*Sin[phi]}
Bob Hanlon
In a message dated 3/1/2000 1:53:32 AM, MAvalosJr at aol.com writes:
>Anyone know of a program that transforms Polar equations to Cartesian and
>
>visa versa?
>