janhardo

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These are replies submitted by janhardo

@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? 

 

    

Is there a connection between of a function with his:   derative, integral and serie ?

 

@Carl Love 

Thanks

Yes, i understand it that you mention it for completeness.

Complex analyse is not that easy as it is for real numbers.

  
 
 

@acer 

Thanks, indeed this question has nothing to do with complex numbers and know nothing about complex functions

 

@Carl Love 

Thanks

I do have  some basic understanding of complex numbers, but complex functions( analyse) i did not yet studied.
 

 

@Carl Love 

Thanks
I did 

convert(ln(1+x),FPS); and get 

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

But value(sum((-1)^k*x^(k + 1)/(k + 1), k = 0 .. infinity)); gives back again 

 

So f(x)= ln(1+x)   gives a serie, but this serie gives not the function back
Domain is < 0 to ? from ln(1+x)  and serie start at x = 0
It must be functions what their domain ranges from 0 ...n  to get from a serie a function back then?   

  

Don't know how to handle this vectorvalued function task for 4 points with tangentlines (or velocity lines?) and accelleration lines 

There was a task made  with a parametric curve also with 3 points and tangentlines, but the vectornotation differs
I tried task 8(i) for curve see post

I choose 4 points from t = 0 ,1/4 , 1/2, 3/4  (they must no too close and interesting )

blz35.pdf

blz36.pdf

hfdts2_blz37.pdf

  

@vv 

Thanks

Is it possible to get a function from a FPS serie ?

Expanding the integrand in a Maclaurin series ( as book tip ) : the integrand is here : ln(1+x).

 

 

@janhardo 

I must use this form for the vectorfunction in task 8

s(t):=[t^5-t+1,2*t^6-t^2-t+1,t=-1..1.2], labels=[x,y];


 

restart; with(plots): with(DEtools): with(VectorCalculus):

r(t):=[cos(t),sin(t),t=0..2*Pi], labels=[x,y];

[cos(t), sin(t), t = 0 .. 2*Pi], labels = [x, y]

(1)

plot(r(t)):

s(t):=[t^5-t+1,2*t^6-t^2-t+1,t=-1..1.2], labels=[x,y];

[t^5-t+1, 2*t^6-t^2-t+1, t = -1 .. 1.2], labels = [x, y]

(2)

 

plot(s(t));

 

========================================================

standaard parametrisch eplot ..nu via vector

V:=<cos(t),sin(t)>;

Vector(2, {(1) = cos(t), (2) = sin(t)})

(3)

 

2D curve defined parametrically by a free vector, graphed with the SpaceCurve command

 

SpaceCurve(V,t=0..2*Pi):

 

R:=<t^5-t+1,2*t^6-t^2-t+1>;

Vector(2, {(1) = t^5-t+1, (2) = 2*t^6-t^2-t+1})

(4)

SpaceCurve(R,t=-1..1.2);

 

 

===========================================================

R:= PositionVector([cos(t),sin(t)]);

Vector(2, {(1) = cos(t), (2) = sin(t)})

(5)

PlotPositionVector(R,t=0..2*Pi,curveoptions=[color= red],tangent=true): # is niet goed bij vraagstuk 8 exc set 2

 

PositionVector graphed by PlotPositionVector command

 


 

Download info__parametrisch_en_vectorfunctie.mw

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