I've been on this question a week now and still no conclusive answer! What I need is a function that produces the inequalities that determine a triangle given the 3 points and then using a 4th point, prints true if the 4th point satisfies 2 or 3 or the inequalities and prints false if it only satisfies 1 or none of the inequalities. I need to have this solved by tonight so any quick help would be greatly appreciated!

Consider the curve defined by f(x, y) = 3+2x+y+2x^2+2xy+3y^2 = 0.Locally on the curve we can view y as a function of x, i.e. y = y(x).Compute formulas for the first and second derivative of y with resoect to x.

2. Give the Maple command(s) to compute \frac{\partial^8 f}{\partial^5 x \partial^3 y} for f(x, y) = e^{2x+ cos(y)}.

Ok so my problem is that I need to write a function which produces the three inequalities defining a triangle given the 3 points that form its vertices, then with the 4th point supplied to the function I need the function to print false if the 4th point satisfies less than 2 of these inequalities and true if it satisfies more than 2 of these inequalities. Can anybody help me with this? On a previous post I obtained part of the function that produces the triangle, but the 2 inequalities part was not solved.

Thanks for your time

How to show the code in file with extension .lib and edit it?

Hi,

I'm new to these forums. I'm using Maple 17. What am I missing about the odd behaviour exhibited below? (the code can be copy pasted into Maple)

Suppose I define m,n,p,q as integers and x as real, then define the function h(x,m,n,p,q) below.

assume(m::integer, n::integer, p::integer, q::integer, x::real);
h:=(x,m,n,p,q)->-(1/8*I)*(exp((2*I)*Pi*x)-1)^2*(exp(-(2*I)*Pi*x*(n-q+2))-exp(-(2*I)*Pi*x*(n-q+1)))/(Pi^3*x*(m-x)*(p-x));

If I do:

int(h(x, m, n, p, q), x = -infinity .. infinity)

It says it's 0, but that's not true. The integral is not always 0 but depends on m,n,p,q ... and even Maple acknowledges this. If I do:

int(h(x, 1, 1, 1, 1), x = -infinity .. infinity)

I get -(1/2*I)/Pi ... so clearly not 0.

Also, if I do:

int(h(x, m, n, m, n), x = -infinity .. infinity)

I get -(1/2*I)/(m*Pi) ... again, not 0.

What am I missing? How can I correct this and obtain the analytic expression for int(h(x, m, n, p, q), x = -infinity .. infinity)?

Trying without the assume() causes Maple to run into issues not knowing anything about m,n,p,q.

[ Edit: I finally solved the integral on paper, see my post below if you're curious. The Maple inconsistency and wrong result explained above are still there though ]

Any help would be greatly appreciated.

Regards,

Alex.

I am not able to find way to do this very basic and common operation.

I use worksheet mode, and many times I'd like to split/divide a large execution group I've build of some code to 2 execution groups at some place. i.e. I'd like to point my mouse at a line and say divide here. Here is an example:

I see only the options Insert->Execution group-> After Cursor or Before cursor. Both of which do not do what I want. I want to divide it at that point.  So what I end up doing is to make a new execution group manually (using the Insert command), then go back and cut and paste the code I want in the new group.

I hope there is an option to do this. I do these sorts of things all the time when using Mathematica, which has Divide cell, Merge cells and other options. A cell in Mathematica is similar to execution group in Maple. Are there other options to maniuplate execution groups other insert before/after cursor that I might have missed? I am using 17.02 on windows 7.

Please note, I only use worksheet mode.

I am stuck with an IVP which is

eq1:=diff(y(x),x$2)+2/x*(diff(y(x),x))+y^M=0;

ic:=y(0)=a,D(y)(0)=0;

its quite easy to find the series solution of the ode 

dsolve({eq1, ic}, y(x), series);

y(x)=a-(1/6)*exp(M*ln(a))*x^2+(1/120)*(exp(M*ln(a)))^2*M*x^4/a;

But I am facing problem when I try to solve it numerically,

dsolve(subs(a=1,M=3,{eq1,ic}),numeric);

THanks

Hello

Am trying to convert a vector projection to matrix type output.

Attached is a sample worksheet.

Transformation_matri.mw 

Consider:

> restart:
> int(exp(I*x)/x,x=-infinity..infinity,CauchyPrincipalValue)
                                                                                                   Pi I

> int(exp(I*3*x)/x,x=-infinity..infinity,CauchyPrincipalValue)
                                                                                                   Pi I

> int(exp(I*k*x)/x,x=-infinity..infinity,CauchyPrincipalValue) assuming k::positive;
                                                                                                  2 I Pi

I would expect the last line to also give me Pi I

Mathematica gives a Pi I via:

FullSimplify[Integrate[Exp[I k x]/x, {x, -\[Infinity], \[Infinity]}, PrincipalValue->True, Assumptions->{k>0}],  Assumptions->{k>0}]

Any idea what this could be?

thanks.

intEq:= 2*cosh(x)-sinh(x)-(2-x)= 1 + int( (2-x+t)* phi(t),t=0..x):
intsolve(intEq,phi(x));

I'd like to extract the (1/2) out, so that it shows as (1/2)*exp(x/2) *  (cosh(x/2) - etc....)

Any recommended way to do it? My attempts did not work well. I am newbie in Maple. Thanks.

I need to make a graph that has dates 1/14/2000 on the x axis and intrest rates on the y axis but when i import my spread sheet from excell the dates keep getting converted into lager numbers like 37000 for some reason and i cannot figure out how to stop this odd conversion. Has anyone encountered this?? i tried looking up dates in maple hepl but cant find much on this topic. Hope i explained what is happening well enough. Graph_One_b.mwGraph_One_b.mw

I have to solve a complex equation whose solutions, I'm looking for, are real numbers, I chose a command 'solve ' to solve this problem but the program shows me this message 'warning solutions May Have Lost'. How I can resolve this problem?

this is the equation and the message :

solve({((abs((((1-(nc-I*kc))/(1+(nc-I*kc)))+(((nc-I*kc)-(0.518-I*0.023))/((nc-I*kc)+(0.518-I*0.023)))*exp(-I*4*3.14*h*(nc-I*kc)/lamda))/(1+((1-(nc-I*kc))/(1+(nc-I*kc)))*(((nc-I*kc)-(0.518-I*0.023))/((nc-I*kc)+(0.518-I*0.023)))*exp(-I*4*3.14*2500*(nc-I*kc)/581))))^2)*100=43.12,((abs((((1-(nc-I*kc))/(1+(nc-I*kc)))+(((nc-I*kc)-(0.176-I*0.023))/((nc-I*kc)+(0.176-I*0.023)))*exp(-I*4*3.14*h*(nc-I*kc)/lamda))/(1+((1-(nc-I*kc))/(1+(nc-I*kc)))*(((nc-I*kc)-(0.176-I*0.023))/((nc-I*kc)+(0.176-I*0.023)))*exp(-I*4*3.14*2500*(nc-I*kc)/720))))^2)*100=44.37},{kc,nc});
{kc,nc};

Warning, solutions may have been lost            {kc, nc}

So I need a help .

Hi,

I have a system of equations for which I want to get a solution around an initial guess, unfortunatly I cannot succed it. Here is the system and the initial guess

solve({(-(((1/((0.005+0.009)^2+(0.072+1*(Ladssec+Ll))*(0.072+1*(Laqssec+Ll))))*((0.005+0.009)*sin(-0.785)+(0.072+1*(Ladssec+Ll))*cos(-0.785)))*((Ladssec*(-id0+(0.953/Lfd)+(0.849/L1d)))+Laqssec*id0)-((1/((0.005+0.009)^2+(0.072+1*(Ladssec+Ll))*(0.072+1*(Laqssec+Ll))))*((0.072+1*(Laqssec+Ll))*sin(-0.785)-(0.005+0.009)*cos(-0.785)))*((Laqssec*(-iq0+(0.468/L1q)))+Ladssec*iq0))/(2*2.5))=-1.7858
,(-((((0.005+0.009)/((0.005+0.009)^2+(0.072+1*(Ladssec+Ll))*(0.072+1*(Laqssec+Ll))))*(Ladssec/Lfd))*((Ladssec*(-id0+(0.953/Lfd)+(0.849/L1d)))+Laqssec*id0)-(((0.072+1*(Laqssec+Ll))*Ladssec)/(((0.005+0.009)^2+(0.072+1*(Ladssec+Ll))*(0.072+1*(Laqssec+Ll)))*Lfd))*((Laqssec*(-iq0+(0.468/L1q)))+Ladssec*iq0)+((Ladssec*iq0)/Lfd)) /(2*2.5))=0.1036
,(-((-((0.005+0.009)/((0.005+0.009)^2+(0.072+1*(Ladssec+Ll))*(0.072+1*(Laqssec+Ll))))*(Ladssec/L1d))*((Ladssec*(-id0+(0.953/Lfd)+(0.849/L1d)))+Laqssec*id0)-(((0.072+1*(Laqssec+Ll))*Ladssec)/(((0.005+0.009)^2+(0.072+1*(Ladssec+Ll))*(0.072+1*(Laqssec+Ll)))*L1d))*((Laqssec*(-iq0+(0.468/L1q)))+Ladssec*iq0)+((Ladssec*iq0)/L1d)) /(2*2.5))=0.2279
,(-((((0.072+1*(Ladssec+Ll))*Laqssec)/(((0.005+0.009)^2+(0.072+1*(Ladssec+Ll))*(0.072+1*(Laqssec+Ll)))*L1q))*((Ladssec*(-id0+(0.953/Lfd)+(0.849/L1d)))+Laqssec*id0)-(-((0.005+0.009)/((0.005+0.009)^2+(0.072+1*(Ladssec+Ll))*(0.072+1*(Laqssec+Ll))))*(Laqssec/L1q))*((Laqssec*(-iq0+(0.468/L1q)))+Ladssec*iq0)-((Laqssec*id0)/L1q)) /(2*2.5))=-2.3313

,(-(1*Rfd*((1/((0.005+0.009)^2+(0.072+1*(Ladssec+Ll))*(0.072+1*(Laqssec+Ll))))*((0.072+1*(Laqssec+Ll))*sin(-0.785)-(0.005+0.009)*cos(-0.785)))*Ladssec)/(Lfd))=0.0023
,(-(1*Rfd*(1-(Ladssec/Lfd)+((((0.072+1*(Laqssec+Ll))*Ladssec)/(((0.005+0.009)^2+(0.072+1*(Ladssec+Ll))*(0.072+1*(Laqssec+Ll)))*Lfd)) *Ladssec)))/(Lfd))=-0.0054
,(-(1*Rfd*(-(Ladssec/Lfd)+((((0.072+1*(Laqssec+Ll))*Ladssec)/(((0.005+0.009)^2+(0.072+1*(Ladssec+Ll))*(0.072+1*(Laqssec+Ll)))*L1d)) *Ladssec)))/(Lfd))= 0.0018
,(-(1*Rfd*(-((0.005+0.009)/((0.005+0.009)^2+(0.072+1*(Ladssec+Ll))*(0.072+1*(Laqssec+Ll))))*(Laqssec/L1q))*Ladssec)/(Lfd))=5.8369e-04

,(-(1*R1d*((1/((0.005+0.009)^2+(0.072+1*(Ladssec+Ll))*(0.072+1*(Laqssec+Ll))))*((0.072+1*(Laqssec+Ll))*sin(-0.785)-(0.005+0.009)*cos(-0.785)))*Ladssec)/(L1d))=0.1303
,(-(1*R1d*(-(Ladssec/Lfd)+((((0.072+1*(Laqssec+Ll))*Ladssec)/(((0.005+0.009)^2+(0.072+1*(Ladssec+Ll))*(0.072+1*(Laqssec+Ll)))*Lfd))*Ladssec)))/(L1d))=0.1157
,(-(1*R1d*(1-(Ladssec/Lfd)+((((0.072+1*(Laqssec+Ll))*Ladssec)/(((0.005+0.009)^2+(0.072+1*(Ladssec+Ll))*(0.072+1*(Laqssec+Ll)))*L1d)) *Ladssec)))/(L1d))= -0.3192
,(-(1*R1d*(-((0.005+0.009)/((0.005+0.009)^2+(0.072+1*(Ladssec+Ll))*(0.072+1*(Laqssec+Ll))))*(Laqssec/L1q))*Ladssec)/(L1d))=0.0333

,(-(1*R1q*((1/((0.005+0.009)^2+(0.072+1*(Ladssec+Ll))*(0.072+1*(Laqssec+Ll))))*((0.005+0.009)*sin(-0.785)+(0.072+1*(Ladssec+Ll))*cos(-0.785))) *Laqssec)/(L1q))=-0.1323
,(-(1*R1q*(((0.005+0.009)/((0.005+0.009)^2+(0.072+1*(Ladssec+Ll))*(0.072+1*(Laqssec+Ll))))*(Ladssec/Lfd))*Laqssec)/(L1q))=-0.0044
,(-(1*R1q*(-((0.005+0.009)/((0.005+0.009)^2+(0.072+1*(Ladssec+Ll))*(0.072+1*(Laqssec+Ll))))*(Ladssec/L1d))*Laqssec)/(L1q))= 0.0051
,(-(1*R1q*(1-(Laqssec/L1q)+((((0.072+1*(Ladssec+Ll))*Laqssec)/(((0.005+0.009)^2+(0.072+1*(Ladssec+Ll))*(0.072+1*(Laqssec+Ll)))*L1q))*Laqssec)))/(L1q))=-0.2349}, {L1d, L1q, Ladssec, Laqssec, Lfd, R1d, R1q, Rfd},useassumptions)

Lfd ~=0.233, R1d~=0.0848, L1d~=0.2, L1q~=-0.09, Ladssec~=0.099, Laqssec~=-0.1, Rfd~=0.00173, R1q~=0.0129

1. 1.    The bending moment M, in a simply supported beam is represented by the formula: (5marks)

 M = w x/2 (l-x)

Where: w = uniformly distributed load in KN per m

            x = distance from the left hand end of the beam in m 

            l = span of the beam in m 

If  = 4KN per m and  = 5m, calculate the value of  where  = 12KNm

Any thoughts on this guys?