janhardo

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11 years, 39 days

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These are replies submitted by janhardo

@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

@Carl Love 

Thanks

I read here : "formal power series can be seen as an extension of polynomials. We speak of “formal” series, because we are not worried about convergence issues."

But it is still not clear how the proven serie has a connection with the serie expension of this integrand function : how are those two series related ?

Th e proof is not yet fully understood, but maybe later? 

 

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