## 490 Reputation

10 years, 94 days
Maple is to me difficult. The first version I bought was Maple9, and it was more than 15 years ago. But, I couldn't use it, feeling it too difficult. But, three years ago, I thought Maple might be helpful to my study, and since then, I have continued to learn Maple. As I got able to read the Maple help, I think that I could get to use maple better now than before. But, I feel that I am a beginner yet.

## Adding a term, then subtracting the term...

Maple 2016

Hello people in mapleprimes,

I have an equation called aa in the following.

alias(`&delta;x`=dx, `&Delta;x`=Dx,`&Delta;y`=Dy,`&Delta;z`=Dz):
aa:=Dz=f(x+Dx,y+Dy)-f(x,y);

As for modification of this expression, I ask your favor to teach me.

Then, I want to change this aa to D[1](f)(x+theta__1*Dx,y+Dy)*Dx+D[2](f)(x,y+theta__2*Dy)*Dy.

But, to do so, I have to split aa into the one including -f(x,y+Dy)+f(x,y+Dy) between two terms

of aa.

But, as maple cancels these terms, I can't do so. How can I insert two terms, then

obtain the expression  D[1](f)(x+theta__1*Dx,y+Dy)*Dx+D[2](f)(x,y+theta__2*Dy)*Dy?

taro

## limit of multivariate function...

Maple 2016

Hello people in mapleprimes,

I have a question. I hope someone give an answer to me.

rho:=sqrt((x-a)^2+(y-b)^2);

limit(x-a+y-2,rho=0);

brings an error

Error, invalid input: limit expects its 2nd argument, p, to be of type Or(name = algebraic, set(name = algebraic)), but received ((x-a)^2+(y-b)^2)^(1/2) = 0

Isn't other way than the following?

limit(x-a+y-2,{x=a,y=b});

Best wishes.

taro

## Notation of diff and ' that is diff(f(x)...

Maple 2016

#Hello people in Mapleprimes,

#After

restart;interface(typesetting=extended);

diff(f(x),x);

#shows f'(x).

#But,

diff(f(t1),t1);

#shows just df(t1)/dt1, not f'(t).

#Can't I show f'(t1) not df(t1)/dt1?

#Is this a way peculiar to Maple?

#I hope someone could give me some hints.

#Best wishes.

#taro

## Fundamental theorem of calculus...

Maple 2016

Hello people in mapleprime,

Though I wrote the title as Fundamental theorem of calculus,

what I am considering is just how to continue the chain of codes in calculation.

restart;

#I defined F__0 as

F__0:=x->Int(f(t),t=a..x);

#Then, the difference between a primitive function of f(x), F(x), and F__0 is no more than a constant C, so I write.

bb:=F(x)-F__0(x)=C;

#Then, substituting "a" into equation "bb", I obtain the value of F(a)

bb1:=subs(x=a,bb):cc:=simplify(%);

#Then, I substituted the value of C in "bb1" into "bb,"  obtaining the following "cc1."

cc1:=subs(isolate(cc,C),bb);

#And, then, I isolated the term of Int(f(t),t=a..x)  in cc1,

dd:=isolate(cc1,Int(f(t), t = a .. x));

#And, then, I substitute x=b into the outcome of dd, and obtain the final equation.

subs(x=b,dd);

Surely, with the above code, I could get the fundamental theorem. But, it looks in a little roundabout way.

So, I thought I would ask here about whether there aren't any better ways to do the fundamentally the same thing or

hints to improve the above code.

taro

## variable raised to -2 and over variable ...

Maple 2015

Hello people in the mapleprimes,

I have a question, so I hope someone give me answers to it.

I calculated for the solution of the follwing differential equation.

restart
b:=diff(y(x),x)+a*y(x)=f(x);#where a and f(x) is not specified.
dsolve(b,y(x));

subs({f(x)=exp(x),a=2},%);where f(x) and a are specified.

c:=value(%);

The solution of the above was

y(x) = (1/3)*exp(x)+_C1/(exp(x))^2,  (A)

where please note that the second term takes

the form of fraction _C/(exp(x))^2.

On the other hand, next I calculated the following differential equation where f(x) and a are specified from the start.

restart
d:=diff(y(x),x)+2*y(x)=exp(x);

dsolve(b,y(x));

Then,

y(x) = (1/3)*exp(x)+exp(-2*x)*_C1  (B)

was the obtained solution.

Each (A) and (B) are the same substantially mathematically. But, for Maple, the variable powered to minus brabra

is not the same as one over variable powered to brabra, so that (A) and (B) takes different forms, and maple will see them

different with each other.

Surely, with algsubs, algsubs(_C1/(exp(x))^2=exp(-2*x)*_C1,c) transforms (A) to (B).

But, I want to know whether there are some other ways than that  to modify (A) to (B).

If there are any good ways for it, I will be happy if you teach them to me.

taro

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