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I wish to evaluate the expression

knowing that

where a is a constant.  It is not hard to see, assuming enough differentiability,  that the expression evaluates to

I know how to do this when all the derivatives are expressed in terms of the diff() operator.  Here it is:

eq := diff(u(x,t),t) = a^2*diff(u(x,t),x,x);
expr := diff(u(x,t),t,t);
eval['recurse'](expr,[eq]);

However, I would prefer to do the computations when all derivatives are expressed in terms of the D operator but cannot get that to work.  What is the trick?

I googled everywhere for this and most results just tell me what diff and D does...

 

Basically I have a function, let's say

 

f:= x -> x^2

How do I turn the derivative of f into a function?

 

I tried

 

fprime := x -> diff(x^2,x)

 

But tihs just shows me diff(x^2,x), instead of x -> 2x

guys , i have a metric and i want to define a componenets of a tensor and then obtain its covariant derivative with respect to a metric, what is your idea ?

N_1=-A(r)^1/2 , A_2=A_3=A_4=0 , what is D_[nu] N_1 =?

 in general i want to define N[1]=-A(r)^(1/2) and N[2] = N[3]= N[3] = N[4] = 0 And define F[mu, nu] = 2*(D_[mu] N[nu]-D_[nu] N[mu]) And define Omega[mu, nu] = 2*(D_[mu] N[nu]+D_[nu] N[mu]) and compute expression F_[alpha, beta] F_[~alpha`, ~beta ] And N_[alpha] N_[~beta`] F_[ ~alpha, ~lambda ] Omega_[beta, lambda])

i have problem with this how to difine this tensorial terms and how to compute them.

Covariant.mw

 

thxxxx

How to find the nth derivative of (logx)/x  and (e^x)logx by using leibenitz theorem....?

 

 

 

 

When I take the derivative of this function wrt p, I am getting this:

 

Why the program gives , instead of only

Sorry for the format, I just copy and paste.

Thanks,

 

 

 

related topic is here

Suppose I have 2 differential equations in vector form, and I want to solve them using dsolve. I am not able to figure the syntax for what I would do for scalar ODE to initial its derivative at t=0, which is D(x)(0)=some_value, but do the same when x is a vector.

Here is an example:

restart;
x := t-> <x1(t),x2(t)>;
eq:=diff~(x(t),t$2) =~ <sin(t),t>;
ic1:=x(0)=~0;

So far so good. Now I wanted to also make initial conditions for derivative at zero to be some value. Only syntax I know is using D(x)(0)=some_value. But this works for scalar ODE. When I tried

ic2:=D(x)(0)=~0;

I got

This does not work:

ic2:=diff~(x)(0)=~0;

any help on the correct syntax to use? I am using Maple 2015

 

Hi, 


     I've been playing around with the Physics package, and I'm confused on evaluaing derivatives of explicit funcitons of the coordinates. This code below doesnt behave as I would think. I'm trying to define z as a function of X[mu]*X[mu], and take diff(z, X[mu]). You can see that each method d_, diff,  disagree and none are satisfactory ansers. (Maple 2015, Windows 8.1 64-bit, Intel i5 Haswell) 

# Declare coordinates for 2 dimensions, flat space

restart:
with(Physics):
Setup(mathematicalnotation = true, dimension = 2):
Coordinates(X):

# Method 1: Using Define and various differential operators
Define(z):
z :=sqrt(R^2-X[mu]*X[mu]);
d_[mu](z(X));
d_[1](z(X));
diff(z, x1);  #This one is correct
diff(z, X[mu]); # off by 2

# Method #2: Using functions
# Off by a factor of 2
z2 := mu -> sqrt(R^2-X[mu]*X[mu]);
diff(z2(mu), X[mu]); # off by 2

 PhysicsDiffBug.mw

In the following, the diff operator calcuates the derivative correctly, but the D operator doesn't.  A bug?

restart;

f := x -> a[1][2]*x;    # the double index on a[][] is intended

proc (x) options operator, arrow; a[1][2]*x end proc

 

diff(f(x), x);

a[1][2]

 

D(f)(x);

(D(f))(x)

 


Here is a worksheet containing the commands above in case you want to try it yourself: mw.mw

Hello everyone!

I'm pretty new with Maple. I think I've understood the way Maple handles differentiation fairly well, but upon a specific request from my PhD tutor I have to perform a task which is giving me a hard time. 

My question is: is in any way possible to express a derivative of a function or expression in terms of the function itself?
I'll try to explain myself with an example: let f(x) and f'(x) be the function and its first derivative:

Instead of expressing f'(x) in the way shown, I'd like to express it as a function of f(x), such as in the following:

I would apply the same process to the higher order derivatives, if possible.

A huge thank you to whoever will help me!

When I take the derivative of abs(x), I use the chain rule and get this

When I ask Maple to differentiate abs(x), I get this:

I read the help file on "signum", and I expected this to work, but it does not.

 

 

How can I represent signum in normal calculus syntax when working the derivative of functions involving abs(x)?

 

 

 

Determine using determinants the range of values of a (if any) such that
f(x,y,z)=4x^2+y^2+2z^2+2axy-4xz+2yz
has a minimum at (0,0,0).

From the theory, I understand that if the matrix corresponding to the coefficients of the function is positive definite, the function has a local min at the point. But, how do I get the range of values of a such that f is a min? Is this equivalent to finding a such that det(A) > 0?

 

2.

Now modify the function to also involve a parameter b: g(x,y,z)=bx^2+2axy+by^2+4xz-2a^2yz+2bz^2. We determine conditions on a and b such that g has a minimum at (0,0,0).
By plotting each determinant (using implicitplot perhaps, we can identify the region in the (a,b) plane where g has a local minimum.

Which region corresponds to a local minimum?

Now determine region(s) in the (a,b) plane where g has a local maximum.

I don't understand this part at all..

I have the function:   f(x,y) = 1/(sqrt(2*Pi)) * e-1/2(x^2+y^2)

I need to take the total derivative of this function in maple, but I don't know the syntax.

I've plotted the graph for this max function. Is there any way I can find the points of discontinuity in general and then use that to compute the derivatives at points where it exists?

Hi Everyone,

I have an expression that contains a second order derivative: EXPR:=ay''(x) + bz'(x) + cf'(x)+... The variable y  obeys an ordinary differential equation, y''(x) = f(y,x). I would like to replace the second order deriavtive in my expression with f(y,x). So far I have tried applyrule([y''(x)=f(y,x)],EXPR), subs(y''(x)=f(y,x),EXPR) and algsubs(y''(x)=f(y,x),EXPR) and nothing seems to work. Any helpful suggestions?

 

Hello,

 

could you help me solve this error ? I don't understand what it means.

 


> eq3:=diff(x(t),t,t)+Gamma*diff(x(t),t)+omega[0]^2*(x(t)-(diff(x(t),t,t)+Gamma*diff(x(t),t)+omega[0]^2*x(t)+omega[0]^2*X[0])/omega[0]^2) = -omega[0]^2*X[0]:
> dsolve(eq3);
Warning, it is required that the numerator of the given ODE depends on the highest derivative. Returning NULL.

 

Thanks.

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