lemelinm

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18 years, 50 days

 

 

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Mario Lemelin
Maple 14.00 Win 7 64 bits
Maple 14.00 Ubuntu 10,04 64 bits
messagerie : mario.lemelin@cgocable.ca téléphone :  (819) 376-0987

MaplePrimes Activity


These are answers submitted by lemelinm

each time I want ot define a function  :=  I get that stupid ' align='absmiddle'> instead.

 

I mean to write

f:=(x,y) -> cos(x,y)

f:= (x,y) -> cos(x*y)

D[1$2,2$3](f)(x,y)

Mario

I think it's time to stop, take a shower and continue tomorow! :-)

 

 

I have learn a lot in the way you have done it. Since I want to show it to students,  I will try to do it without "taylor(qx*fx - px, x=4, 9)".  Do you know (or someone else) a simple application of the use of the Padé Approximant?

Thanks!

I have tried your suggestion.  Of course, this was my mistake about  collecting x^k.

But even then, I still unable to reproduce the answer of Maple.  PLease help me solve this one as an example and if as a bonus you have a simple application, I will be very thankfull.

mario.lemelin@cgocable.ca

Thanks Alejandro

 

mario.lemelin@cgocable.ca

I saw them but I don't have accès to those journals.  Do you know a website?

mario.lemelin@cgocable.ca

Hi,

 

I am willing to help you under somes conditions:

1) I am sure that the file 7853_Assingment 3.mw is not your work!  True or false?

2) Do you understand what is happening in the for loop?

3) does the range is [0,3]?

4) do you know why equation (5) and (6) behave like this?

5)  Give me YOUR file for 3 and 4 in your homework and then I will help you!

 

mario.lemelin@cgocable.ca

Set:

f:=x -> 0.25*x^4+.............

Always do a plot of the function before doing anything

plot(f(x),x=1..2)

You can see that there is something happening between 1.1 and 1.6.  Take a closer look.

You can see here the advantage of ploting.

By writing fsolve(..  I guess that you want to solve f(x)=0

 

fsolve(f(x)=0,x=1.6..3)  but is the answer in the domain you selected?

"compare your result with solve(f(x)=0,x) wich give two reals roots and two complex roots"

1)

I guess you know how to find the maximum and the minimum.  So you see that there is at least one point of inflexion in the interval [1,2]

To find it, you do the second derivative test and search the point where it's equal to 0

fsolve(D[1,1](f)(x)=0,x=1.2..1.6)

Here you know how to find that point.  Now it's your job to ask your teacher the meaning of the second derivative test and why it should be equal to 0

About the reflection, I guess again that you mean inflexion!

Hope it help...

mario.lemelin@cgocable.ca

Suppose you have

g(x) -> (3*x-2)/(x-1)

a)

Limit(g(x),x=infinity,right)

Limit(g(x),x=infinity,left)

b)

The same thing for the singularity.  Suppose it's a then

Limit(g(x),x=a,left):%=value(%)

same for right

c)

If you have

f:=x -> -x^2+2*x+8

plot(f(x),x=-2..4)

maximize(f(x),x=-2..4)

you will obtain the value of f(x) but not the point x.  This is a question of the firts derative test to find the maximum, suppose it's b then f(b) will give you 9.

d)

this is the test of the second derivative

 

Hope this will help!

 

By the way, you could use a package that will help you investigate futher more.

with(Student:-Calculus1);

mario.lemelin@cgocable.ca

With the words "Maple Videos"+Maplesoft

You have to add Maplesoft because they are  many things with the name Maple+Videos+Maplesoft

mario.lemelin@cgocable.ca

Sorry for the first two line:

f:=x->exp(x);g:=x->2*cos(x)

 

Mario Lemelin

mario.lemelin@cgocable.ca

Usually, I ask to plot before doing anything else.  In this case, I do:

f:=x->exp(x)

g:=x->2*cos(x)

plot([f(x),g(x)],x=-3..3)

Then the plot reveile what is happening

so now I can do:

fsolve(f(x)=g(x),x=-3..-1)

and

fsolve(f(x)=g(x),x=0..1)

 

Mario Lemelin

mario.lemelin@cgocable.ca

In Maple 11, I write:

int(exp(x),x)

and receive

                          exp(x)

 

There is no factor of lnexp(x)squared.  Since I cannot reproduce, I am not able to say.  But is this what you were trying to do?

 

 

 

Mario Lemelin

mario.lemelin@cgocable.ca

The FuctionAdvisor is very interesting.  But it only work with known functions as you can see with:

FunctionAdvisor(known_functions)

but not for a contruction like x^2*ln(x).

If it could be algorithmic, I would be very interested to know how if this is not too complicated for no too nasty nesting.

 

Mario Lemelin

mario.lemelin@cgocable.ca

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