janhardo

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11 years, 66 days

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These are replies submitted by janhardo

@tomleslie 

Thanks!

For the extensive answer.

Was it in the Riemann sum example that a function what is passed to the procedure was already defined as a function with a arrow operator.

X:= Array(0..N, [seq(j, j=a..b, (b-a)/N)]): # x values stored in array 
 Y:= Array(0..N, f ~ (X)):   # y -values stored in array
 

Now in this example of Polygonal approximation the function is defined as a expression , so this must first defined as a function in order to use the ~ elementwise operator.

xv := Array ()  # x- values stored in array 
yv:=unapply( f, indets(f, 'name')[]) ~ (xv);   y-values stored in a list ?  ..no as a array as showed the ouput 

I noticed the difference between '[ ]' and [ ] 

Think i understood it now for the f part 
In the book the functions are given as expressions. 

return display ( ) .. to make sure there will be a plot ?

I try the attached functions in the book example...but noticed the procedure definition by p,L := approxL ( ):

Why is that ?

 

 

 

 

 

 

@tomleslie 
Thanks

First there is a series x-values stored in Array xv 

Its a difficult one to decipher. yv:=unapply( f, indets(f, 'name')[])~(xv);  What is the idea behind this statement    

yv := xv^3 - xv*sin(xv)  outside the procedure call 
unapply( ) should make a function from the expression in  ( ) 

Problem to check the outcome from statements in procedure , beause it are local variables ?

Complicated this.

  

 

 

 

@vv 

Thanks

Yes, you do have absolutely right, its my lack of knowledge of how to handling series precisely. 
Thought that with some basic calculus, partial deratives , etc i could follow it.
But no.

Don't know if the calculus textbook has this sort of example i posted here ? 
I do have seen some examples ..but complicated

 

Could not find a solution strategy .


 

restart;

the following alternating series converges to the value shown( the sum of the series)

 

S:= Sum((-1)^k/((k + 2)*k!), k = 0 .. infinity): S = value(S);

Show this by considering the integral

Sum((-1)^k/((k+2)*factorial(k)), k = 0 .. infinity) = 1-2/exp(1)

(1)

S:=S;

Sum((-1)^k/((k+2)*factorial(k)), k = 0 .. infinity)

(2)

IN:= Int(x*exp(-x),x): IN = value(IN);

Int(x*exp(-x), x) = -(x+1)*exp(-x)

(3)

 

Are not standard integrals

F:= Int(x*exp(-x), x): F = value(F);

Int(x*exp(-x), x) = -(x+1)*exp(-x)

(4)

 

Sx:= convert(F, FPS); #maclaurin series for F

Sum((-1)^(k+2)*(k+1)*x^(k+2)/factorial(k+2), k = 0 .. infinity)

(5)

 

from Sx to S ?


 

Download forum_excset3_task_3b.mw

 

 

@vv 
Thanks

Can you elaborate what the meant here  : You can also compute int(int(1/(1-x),x),x) + c1*x+c2; and adjust the constants such that f(0)=f'(0)=0.

@vv 

Thanks

Using the right commands and if i  look at the integral : i should never came on this one
Seems to be a double integral trick  ? 

                      -int(1/(x - 1), [x = 0 .. y, y = 0 .. z])


But don't get answer  :  ?  ..yes with value( )   

@Carl Love 

Thanks

Using this limit gives the radius of convergence for any algebraic function (not complex).
This limit outcome can be interval , real number axis  or one real number ?

Now to find a example of this.

Integrating and differentiating a powerseries makes now sense after have seen a example 
Not adding a constant of integration to a powerseries makes that it not good fit with a function as it seems in a seen example. 
 

 

@janhardo 

i bought years back some used calculus books : thomas calculus, early trancedentals and another calculus book

There is lot to find there in those books about powerseries, so i  want to dive further with Maple in the powerseries...  

@Carl Love 

Thanks

There is a lot to say more about this converge of powerseries  
-  d'Alembert 
-  Cauchy 
 

A dutch example https://www.hhofstede.nl/modules/machtreeksen.htm

A question here  : Why is such convergence area  in fact always symmmetrical with middle a ( waarom is zo'n convergentiegebied.... )

Well this goes too far for me to investigate 
  

@acer 

Thanks 

Some handy properties of powerseries: 

  •  by summing or multiplication of two powerseries around x = a you get a new powerseries around x = a  with as radius of convergence the smallest of the original convergence radiusses.
  •  by differentation or integration of a powerseries the radius of convergense keeps preserved. 

@Carl Love 

Thanks

With this clear example i can try to answer  other  series tasks     

 

@janhardo 

The posts are crossing
Note: its my interest in series that i zoom in on the task in the programming book how series can be related with eachother.

@vv 

Thanks

I don't know what the solutions strategy is yet ?


 

restart;

S:=Sum((-1)^(n+1)/(n*(n+1)),n = 1 .. infinity)=sum((-1)^(n+1)/(n*(n+1)),n = 1 .. infinity);

Sum((-1)^(n+1)/(n*(n+1)), n = 1 .. infinity) = -1+2*ln(2)

(1)

f := x -> Sum(x^(n+1)/(n*(n+1)),n = 1 .. infinity)= sum(x^(n+1)/(n*(n+1)),n = 1 .. infinity);

proc (x) options operator, arrow; Sum(x^(n+1)/(n*(n+1)), n = 1 .. infinity) = sum(x^(n+1)/(n*(n+1)), n = 1 .. infinity) end proc

(2)

f(-1);

Sum((-1)^(n+1)/(n*(n+1)), n = 1 .. infinity) = -1+2*ln(2)

(3)

diff(f(x),x$1);

Sum(x^(n+1)/(x*n), n = 1 .. infinity) = sum(x^(n+1)/(x*n), n = 1 .. infinity)

(4)

diff(f(x),x$2);

Sum(x^(n+1)*(n+1)/(x^2*n)-x^(n+1)/(x^2*n), n = 1 .. infinity) = sum(x^(n+1)*(n+1)/(x^2*n)-x^(n+1)/(x^2*n), n = 1 .. infinity)

(5)

simplify(%); # f''(x)

Sum(x^(n-1), n = 1 .. infinity) = sum(x^(n-1), n = 1 .. infinity)

(6)

S=value(Sum(x^(n - 1), n = 1 .. infinity));

(Sum((-1)^(n+1)/(n*(n+1)), n = 1 .. infinity) = -1+2*ln(2)) = sum(x^(n-1), n = 1 .. infinity)

(7)

Limit(x^(n - 1),n=infinity);

Limit(x^(n-1), n = infinity)

(8)

value(%);

limit(x^(n-1), n = infinity)

(9)

 

Its a geometrical series : "(x[n+1])/(x[n];)= x    " how to recognize this  ?
There is no closed value to get for sum(x^(n-1), n = 1 .. infinity)  ?

 

I lost oversight  two times differentiate and two times integrating


 

Download uitwerkingvraagbeatwioording_forum.mw

   

@vv 

Thanks 

As i understand it correct : S series is included in F series. 
But there is a given series S and how  to know at forehand that series F is related with series S ?

@Carl Love 

Thanks 

I am puzzling on the first book example for a given sum  how this to get  by  a integral.
Yes, there is studymaterial about power series. 

Seems to be useful infinite series handling, probably in further programming tasks? 

 

    

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