hello; I am asking for the difference between > Digits:=200; > > eq:=-x^3+6*x^2-6*x-1: > fsolve(eq=0,x)[1]; > > solve(eq=0,x)[1]; > evalf(%); The result of evalf(%) above contains a complex part. No matter how large I make Digits to be. the result of fsolve contains only real part. The question is, would you consider this equation to have complex or real roots? thanks, Nasser

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10/14/2005 10:04:07 PM

In article <rXV3f.2604$tV6.1659@newssvr27.news.prodigy.net>, Nasser Abbasi <nma@12000.org> wrote: |>I am asking for the difference between |>> Digits:=200; |>> eq:=-x^3+6*x^2-6*x-1: |>> fsolve(eq=0,x)[1]; |>> solve(eq=0,x)[1]; |>> evalf(%); |>The result of evalf(%) above contains a complex part. No matter how |>large I make Digits to be. |>the result of fsolve contains only real part. Yes, the solution of a cubic with rational coefficients and three irrational real roots (the "casus irreducibilis") using radicals must involve complex numbers: the solutions are not in a radical extension of Q contained in the real line. Do a floating-point computation involving I and roundoff error is likely to give you a nonzero imaginary part in the result. But Maple does know how to get real solutions involving trigonometric functions: > map(simplify @ evalc,[solve(eq,x)]); 1/2 1/2 1/2 1/2 [2 2 cos(%1) + 2, -2 cos(%1) + 2 - 3 2 sin(%1), 1/2 1/2 1/2 -2 cos(%1) + 2 + 3 2 sin(%1)] 1/2 23 %1 := 1/3 arctan(-----) 3 |>The question is, would you consider this equation to have complex or |>real roots? Real, of course. You can also use Sturm's Theorem to count the number of real roots. > sturm(sturmseq(eq,x),x,-infinity,infinity); 3 Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada

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10/14/2005 10:27:06 PM

Tukey (inventor of FFT) thinks that an approximate solution of the exact problem is often more useful than the exact solution of an approximate problem. I find it hard to argue which one is more important or useful. Once you believe in one of them, your belief will lead your research style to either algorithm-centered or model-construction-centered. Anybody wants to elaborate on either of these two views? b83503104@yahoo.com wrote: > Tukey (inventor of FFT) thinks that an approximate solution of the > exact problem is often more useful than the exact solution of an > approximate ...

Hi, I have a problem with solve, det and sub in matlab symbolic. I'm trying to solve : det(K) = 0 . K depends on 'alpha', symbolic unknowns. Theory predicts six solutions, and I found them. But when I try to check the results using subs(det(K),'solutions',solutions), 4 solutions give effectively det(K) = 0, while the others don't nullify the determinant. I'm very annoyed by this. Dou you have any idea about the origines of this and where is the bug? Thanks a lot! Sig Ponge wrote: > Theory predicts six solutions, and I found them. > > But when I t...

Hello; (This is a general CAS question I think, but my example uses Maple. I am sure one can find such an 'issue' in other CAS systems) Is it normal/OK behaviour for things to work as follows: Given some equation, ask solve to solve it for some unknown. Now, if I call simplify() first on the equation then call solve() on the result, then I get an answer/solution otherwise I do not or I get some other warning/error. This means solve() could not find a solution in the present form of the equation or expression (even though a solution does exist if the form 'l...

Dear Group members, I ran into a case where Mathematica Solve yields different solutions depending on whether I solve it analytically (and then plug in some numbers) or numerically. I'm solving an algebraic equation, with some square roots. Mathematica does NOT complain about branch cut or anything else when solving the problem analytically. Based on physical reasoning that created the equation in the 1st place the purely numerical solution is the correct one. I attached a 3-line pdf version of the notebook that demonstrates this issue. Thank you for any suggestions! https://docs.goo...

Hi, I use solve comman to solve an equation(see below). I got the result, however, it somehow is a string (10000/17) rather than a numerical number(588.23). What can I do to make it numerical number? Thanks. I appreciate your help in adavnce !!! ------------------------- p=50 wQF=40 a=0.2 w=0.2 ub=800 lb=400 v=20 syms x y f=1/(ub-lb); %pdf of a uniform distribution F=(x-lb)/(ub-lb); % cdf F(x) FQF=(p-wQF)*(1+a)*(1-int(f,lb,(1+a)*y))-(wQF-v)*(1-w)*int(f,lb,(1-w)*y); qQF=solve(FQF,y) Result: qQF= 10000/17 ...

I'm starting with Matlab and I don't know what to do to solve numerically x+e*sin(x)=omega*t where e, omega and t are calculated earlier in the program. Many thanks, tania In article <eefb68a.-1@webx.raydaftYaTP>, Tania <tania.re@wanadoo.fr> wrote: > I'm starting with Matlab and I don't know what to do to solve > numerically > > x+e*sin(x)=omega*t > > where e, omega and t are calculated earlier in the program. > > Many thanks, > > tania ------ Hello Tania, Do you have the Symbolic Math Toolbox? If so, try that with 'so...

These are part of our solutions not shown here , if the solution you want isn't in the list, do not give up, just contact us and we will find it to you email: lsms9[at] yahoo.com it's ls...@yahoo.com we have a lot of solutions manual over 700 solutions manual solution manual Math solution manual physics solution manual Calculus solution manual Dynamics solution manual mechanics solution manual electromagnetics solution manual Thermodynamics solution manual Circuits =2E. =2E. in low cast price .. Are you looking for solution manuals, solution manual, solutions I have solution m...

Well, just for the records: Maple 9 is unable to execute this: fsolve(min(x+51580800.0,5652700000.0,x+51580800.0+.8585358682e13/x^(1/2)) = 4368842661., x, 1684100000.0 .. 5652700000.0); It gives up with the following error message: Error, (in PiecewiseTools:-Convert) unable to convert On the other hand, Maple V and Maple 6 does it well. I'm trying to convert an old Maple V worksheet to Maple 9 and this work is not being easy! Black boxes, black boxes ... Greetings, Humberto. Humberto Jose Bortolossi wrote: > fsolve(min(x+51580800.0,5652700000.0,x+51580800.0+.8...

Dear all, I have a matrix defined as: Matrix[{\[Alpha]_, \[Beta]_, \[Gamma]_}, \[Theta]_] := {{\[Alpha]^2 \ (1 - Cos[\[Theta]]) + Cos[\[Theta]], \[Alpha] \[Beta] (1 - Cos[\[Theta]]) - \[Gamma] Sin[\[Theta]], \[A lpha] \[Gamma] (1 - Cos[\[Theta]]) + \[Beta] Sin[\[Theta]]}, {\[Alpha] \[Beta] (1 \ - Cos[\[Theta]]) + \[Gamma] Sin[\[Theta]], \[Beta]^2 (1 - Cos[\[Theta]]) + Cos[\[Theta]], \[Beta] \[Gamma] (1 - Cos[\[Theta]]) - \[Alpha] Sin[\[Theta]]}, {\[Alpha] \[Gamma] \ (1 - Cos[\[Theta]]) - \[Beta] Sin[\[Theta]], \[Beta] \[Gamma] (1 - Cos[...

Hi all, I am working in beam physics.I am getting used to mathematica now.But feel little uncomfortable doing few things. Now I solved a differential (simulataneous ) equation in {X''(z),Y''(z),X(z),Y(z)} with four intial conditons and getting the answer for X & Y in interpolating function.I plot it for required z and its all fine and great and FAST!! Question 1: Now I want to find the minimum of this solution X(z).How do i use the function FindMinimum for this interpolating function? Question 2: In the long run I want to find the value of the intial ...

please describe the details of shooting method to solve schrodinger equation for an electron confined in a 1D infinite-Quantum well with width d.thank youe very much. Define vector x first, with length n and solve for y(x).Set y(1)=0;y(2)=1; and vary E until you obtain y(n)==0 or abs(y(n))<1e-6. Simply "integrate" the SE to obtain y(3) from y(2) and y(1) and so on. But you better use FEMLAB if you have that. /Per reza mohseni wrote: > > > please describe the details of shooting method to solve schrodinger > equation for an electron confined in a 1D infinite-Qu...

Hello, Problem: Need to calculate coefficients B(n) in equation (1), i.e. solutions of the equation (where P(n) and M(n) are known variables). eq(1): sum[P(n)/(1-(M(n)/B(n)))]=1, where n ranges from 0 to N. For each increasing n, coefficient B takes one value between an increasing interval, e.g. below are the possible values B can assume, in agreement with index n: n=1 then 0<B1<0.5 n=2 then 0.5<B2<0.7 n=3 then 0.7<B3<1 In sum it's the equivalent of solving (A*B1 + A*B2 + A*B3 + ... + AnBn=1, where A is a constant) for all coefficients B. With the added particular...

hello, I am now trying to solve a cubic equation like this: Z^3 + A*Z^2 +B*Z +C = 0 while A, B and C have some value. I want to solve the equation. but when I use the command: solve('Z^3 + A*Z^2 +B*Z +C = 0','Z') it only gives out simbolic solution, while i need the numerical solution. Is there anyway to get the numerical solution? thanks Done/ roots.m will help Zhuoyang Lian wrote: > > > hello, I am now trying to solve a cubic equation like this: > Z^3 + A*Z^2 +B*Z +C = 0 > while A, B and C have some value. I want to solve the equation. but > when I use the...

Hello, Problem: Need to calculate coefficients B(n) in equation (1), i.e. solutions of the equation (where P(n) and M(n) are known variables). eq(1): sum[P(n)/(1-(M(n)/B(n)))]=1, where n ranges from 0 to N. For each increasing n, coefficient B takes one value between an increasing interval, e.g. below are the possible values B can assume, in agreement with index n: n=1 then 0<B1<0.5 n=2 then 0.5<B2<0.7 n=3 then 0.7<B3<1 In sum it's the equivalent of solving (A*B1 + A*B2 + A*B3 + ... + AnBn=1, where A is a constant) for all coefficients B. With the added particula...

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