Today, I was working on a problem in the class which said:

if f(x)=ax^2+b, x<=1 and 1/x, x>1; then find the values of "a" and "b" such that f'(1) be defined. I checked it to be continuous at x=1 first and so I got a+b=1. Secondly, I wanted to get another equation, so I checked the differentiability of the function by the definition of derivation at x=1, so:

[> f := x-> piecewise(x <= 1, a*x^2+b, 1 < x, 1/x);
> limit((f(x)-f(1))/(x-1), x = 1, right);

I have a set E of linear equations in variables x[1],...,x[8] (which I created with X := Matrix(8,1,symbol=x)). I now call

> solve(E);

I get back something like

x[1] = x[2]-x[3], x[2] = x[2], x[3] = 5, ...

I would rather like to get

x[1] = t[1] - 5, x[2] = t[1], x[3] = 5, ...

back.

How to do this? I do want to use the solve function (not LinearSolve et.c.).

I'm supposed to do the following. I tried to ask my teacher how I could go about doing it and she told me to just figure it out. I am supposed to find the linear and quadratic approximations to the function f at x=a. Then, I'm supposed to sketch the graph of the function and its lienar and quadratic approximations.

The linear approximation is:

P₁(x)= f(a) + fprime(a)*(x-a)

The quadratic approximation is:

P₂(x)= f(a) + fprime(a...

I have that problem that I am trying to solve the following equation explicit:

TY = RootOf(6.750969101*10^26*_Z*1000^(9/59)*800^(41/59)-1.015701668*10^27*1000^(50/59)-1.189275761*10^27*800^(50/59)+1.103397224*10^26*_Z*1000^(32/59)*800^(18/59)-7.457053114*10^23*3^(1/5)*1000^(41/59)*800^(50/59)*_Z^(4/5)*sqrt(12500)*4^(19/20)*(800^(9/59)+1000^(9/59))^(1/50)*(702.222720*1000^(9/59)+976.508160*800^(9/59))^(4/5)-7.457053114*10^23*3^(1/5)*1000^(50/59)*800^(41/59)*_Z^(4/5)*sqrt(12500...

Can maple in any way simplify this any further?

-(1/4)*m*omega*sin((1/2)*omega*T)*(2*cos((1/2)*omega*T)^2*xa^2+2*cos((1/2)*omega*T)^2*xb^2-xa^2-2*xa*xb-xb^2)/((cos((1/2)*omega*T)^2-1)*cos((1/2)*omega*T))

I've tried simplify, radnormal, combine but am wondering why maple does not convert for example the denominator in sin(omega*T) 

How do enter and use the notation of a step function (u(t-3)-u(t-4) in maple to use in calculations? thanks

Hi,

I wondered if you would be able to help.

I've set up a for-loop and while-loop that substitutes values of j and k into a given function

I want to get the answer for every right for example right[1], right[2]..but this code give the answer for right[1]nd right[6] only..

 P := array([[1, 4], [2, 3], [3, 2], [4, 1], [6, 5], [6, 1], [6, 2], [5, 3]]);

j := 1; k := j+2; Fold := 0; Fnew := 1; counter := 0;

 for...

hello,i have a problem i have to program analytic solution of heat equation of a rod with two heat sources if you can help me by anything please d'ont hesitate, thank you.

Hi, how I can find the variable Z in this equation? thanks.

I know how to export to .png or .jpg but I'm not quite pleased with the quality of the images.

Is there a way to export high quality images? i.e. no sharp edges of the function in the plot

I have define the following ODE system

restart;
with(LinearAlgebra);
dC1 := diff(C1(t), t) = -2*a1*C1(t)+b1*C2(t);
dC2 := diff(C2(t), t) = 2*a1*C1(t)-(b1+a1)*C2(t)+2*b1*C3(t);
dC3 := diff(C3(t), t) = a1*C2(t)-(2*b1+a2)*C2(t)+b2*O1(t);
dO1 := diff(O1(t), t) = a2*C3(t)-b2*O1(t);
syst := dC1, dC2, dC3, dO1;ics := C1(0) = 1, C2(0) = 0, C3(0) = 0, O1(0) = 0;
func := {C1(t), C2(t), C3(t), O1(t)};
sol := dsolve([syst, ics], func);

After evaluating...

Solve Differential Equation :(3x-y)*dy/dx=2x

Good afternoon Sir.

I request you to help me out in writing Maple procedures to execute, Numerical methods such as

1. Bisection ethod
2. False position method
3. Fixed point iteration
4. Newto-Raphson method
5. Generalized Newton's method
6. Mullers method
7. Graffe's method.

With thanks & regards

M.Anand

My attempt:

RealDomain[solve]({x^2+y^2+z^2 = 3, x+y+z = 3}, {x,y,z});

             {x = -RootOf(_Z^2+(z-3)*_Z+z^2-3*z+3)-z+3, y = RootOf(_Z^2+(z-3)*_Z+z^2-3*z+3), z = z}

In fact, the system in the real domain has a unique solution x = 1, y = 1, z = 1. It is easy to find by hand, noting that the plane  x + y + z = 3  is tangent to the sphere