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janhardo

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@tomleslie Thanks!That look' s&...

janhardo 700
June 12 2020
0 0

@tomleslie 

Thanks!

That look' s great : the tread-riser patern is clearly to see for the Riemann sum cases.

But i you are this far in the programming: the final step to a procedure and to make it general is not far anymore ?
(Although it is not asked in the task)

Note: it is not about using maple in-built command, because it is exercise in programming.
Although it could be handy to use them.

It is another programming style you used then in the book i showed, but that's good..


 

@Carl Love ThanksIndeed complicated...

janhardo 700
June 12 2020
0 0

@Carl Love 

Thanks

Its simple idea : a tread-riser, but i can't figure it out to plot  

@acer  Thanks Its all about clear...

janhardo 700
June 11 2020
0 0

@acer 

Thanks

Its all about clearity
For this Riemann sum (right hand rule )example it is drawing 3 plots ( 2 verical lines:interval and the tread-riser outline sements )

Well, how to get the points for the plot( seq( points), i=1..10) ?  
 

@acer ThanksOn the the frontpage is...

janhardo 700
June 11 2020
0 0

@acer 
Thanks

On the the frontpage is written: An introduction to Programming using Maple

There is lot help in the book how Maplecommands can be used in this book.
Indeed it is not for learning how to use maple , their must be some base understanding for finding you way in Maple

Also as the authors note : Often the code for doing a given task can be made clearer by using arrays even though the task can be done without them.

Ofcourse clarity comes at the expense of using storage space in memory.
Examples shows that it is often necessary to sacrify this storage space by creating arrays, because the values of xi,yi,i=0..,N , are needed not only needed in calcullation  of sums ( example) , but also at other places in the code. 

Is memory these days a issue on high end computer ?

It could be translate it to moderner seq and map and other techniques? in maple as i have seen in another your example of task solutions. 
I will analyze how it can be done on the "modern" maple way

 

@Carl Love ThanksThis is a procedur...

janhardo 700
June 11 2020
0 0

@Carl Love 

Thanks

This is a procedure you made, but that's not needed in this task ( even before the procedure is introduced in the book )
Its too difficult for me now in the learning process to understand the procedure.

.
I enclosed the worksheet where i am struggling to try to getting to draw the magenta lines for the Riemann rightsum rule

Above is the Riemann lefthand rule drawed with the tread-riser outline 

Note: the tread-riser  is made out of 3 points connected with linesegments
There is no i +1:  y -value possible yet in the code what i needed to get a point for the tread riser step 

The undersum (rectangle under graph) starts at X0 and the uppersum(rectangle above graph)  by X1 for calculating the sum.  

There are 3 plots : a vertical line at a and a vertical line at b and the tread riser step    


 

> 

restart;with(plots):

 task 1 (a)

 

> 

N:=10: # code op blz 47 parabool (x^2+1)
> 

dX:=(b-a)/N: #

> 

X:=a: Y:=f(X): s:=0:
> 

for i from 1 to N do
> 

   s:= s+Y*dX; # geen i,is f(x).dx gesommeerd

> 

   X:= X+dX; Y:=f(X); #geen i
> 

end do:
> 

 

> 

 

Wat doe de code hierboven ? : de loop telt  y.dx  (f(x).dx)  op  in een som s
s= s+ y0.dx = y0.dx
s = s+y1.dx = y0dx + y1.dx

s=  .........

s= s+ yN-1.dx = y0.dx+y1.dx+ ....+yN-1.dx (alg )

> 

 

task 1(b)Via Array

> 

X:=Array(0..N); Y:=Array(0..N);

`Array(0..10, {(1) = 0, (2) = 0, (3) = 0, (4) = 0, (5) = 0, (6) = 0, (7) = 0, (8) = 0, (9) = 0, (10) = 0})`

  

`Array(0..10, {(1) = 0, (2) = 0, (3) = 0, (4) = 0, (5) = 0, (6) = 0, (7) = 0, (8) = 0, (9) = 0, (10) = 0})`

 (1)
> 

dX:=(b-a)/N:
> 

X[0]:=a: Y[0]:=f(X[0]):
> 

s:=0:
> 

for i from 1 to N do
> 

  s:= s+Y[i-1]*dX; #linkersom

> 

  X[i]:= X[i-1]+dX; Y[i]:=f(X[i]);
> 

end do:
> 

 

> 

 

xi en yi worden opgeslagen in een array  en word ook een som gemaakt s als in het voorgaand vb

task 2

> 

 

vb 3.2 : riemann som met graphics (linkersom)

> 

with(plots,display);

[display]

 (2)

Before we can do any calculations, we need a function  f  to work with, so we choose the following polynomial function and use the interval [0, 1]:

 

> 

a:=0;b:=1;

0

  

1

 (3)
> 

 

> 

f:=x->x^2-x^4+1;

proc (x) options operator, arrow; x^2-x^4+1 end proc

 (4)

 

The staircase pattern consists of "treads" and "risers" and we will produce the plots of these as we do the calculations.

The initial computations are done with N =10. The partition points x are stored in an array X and the values y = f (x) are stored in an array Y. The tread/riser plots are stored in an array called tr.

 

> 

N:=10;

10

 (5)
> 

X:=Array(0..N); Y:=Array(0..N);tr:=Array(1..N);

`Array(0..10, {(1) = 0, (2) = 0, (3) = 0, (4) = 0, (5) = 0, (6) = 0, (7) = 0, (8) = 0, (9) = 0, (10) = 0})`

  

`Array(0..10, {(1) = 0, (2) = 0, (3) = 0, (4) = 0, (5) = 0, (6) = 0, (7) = 0, (8) = 0, (9) = 0, (10) = 0})`

  

Array(%id = 18446745233214237262)

 (6)
> 

 

Next we calculate Delta*x = (b - a) / N, assign values to x[0],  y[0] = f (x[0]), initialize the summation variable s, and plot the left-hand vertical part of the outline, which is stored in ver0.

 

> 

dX:=evalf((b-a)/N);X[0]:=a;Y[0]:=f(X[0]);s:=0;

.1000000000

  

0

  

1

  

0

 (7)
> 

ver0:=plot([[X[0],0],[X[0],Y[0]]],color=magenta): #verticale lijn bij x0 = a (intervalgrens)

> 

 

Now we can do the main calculation of the left-hand rule Riemann sum and produce the remaining parts of the outline with the following do loop:

 

> 

for i from 1 to N do
> 

  s:=s+Y[i-1]*dX:
> 

  X[i]:=X[i-1]+dX; Y[i]:=f(X[i]):
> 

  tr[i]:=plot([[X[i-1],Y[i-1]],[X[i],Y[i-1]],[X[i],Y[i]]],color=magenta): # plot van trapje is 2 lijnstukkken ( 3 punten)

> 

end do:
> 

verN:=plot([[X[N],0],[X[N],Y[N-1]]],color=magenta): # verticale lijn bij x= b (interval grens)
> 

hor:=plot(0,x=a..b,color=magenta):
> 

p:=plot(f(x),x=a..b,color=black,thickness=2):
> 

 

> 

display({seq(tr[i],i=1..N),hor,ver0,verN,p},scaling=constrained,axes=none);

 
> 

 

> 

 

Opgave 2(i) rechterhandregel (via bovensom : interval van x1 ..N
Bovensom van x1..N
Ondersom van x0 ..N-1

 

> 

with(plots,display);

[display]

 (8)

Before we can do any calculations, we need a function  f  to work with, so we choose the following polynomial function and use the interval [0, 1]:

 

> 

a:=0;b:=1;

0

  

1

 (9)
> 

 

> 

f:=x->x^2-x^4+1;

proc (x) options operator, arrow; x^2-x^4+1 end proc

 (10)

 

The staircase pattern consists of "treads" and "risers" and we will produce the plots of these as we do the calculations.

The initial computations are done with N =10. The partition points x are stored in an array X and the values y = f (x) are stored in an array Y. The tread/riser plots are stored in an array called tr.

 

> 

N:=10;

10

 (11)
> 

X:=Array(0..N); Y:=Array(0..N);tr:=Array(1..N);

`Array(0..10, {(1) = 0, (2) = 0, (3) = 0, (4) = 0, (5) = 0, (6) = 0, (7) = 0, (8) = 0, (9) = 0, (10) = 0})`

  

`Array(0..10, {(1) = 0, (2) = 0, (3) = 0, (4) = 0, (5) = 0, (6) = 0, (7) = 0, (8) = 0, (9) = 0, (10) = 0})`

  

Array(%id = 18446745233214192806)

 (12)
> 

 

Next we calculate Delta*x = (b - a) / N, assign values to x[0],  y[0] = f (x[0]), initialize the summation variable s, and plot the left-hand vertical part of the outline, which is stored in ver0.

 

> 

dX:=evalf((b-a)/N);X[0]:=a;Y[0]:=f(X[0]);s:=0;

.1000000000

  

0

  

1

  

0

 (13)
> 

ver0:=plot([[X[0],0],[X[0],Y[1]]],color=magenta): #verticale lijn bij x0 = a (intervalgrens) #aangepast
> 

 

> 

NULL# rechter riemann som

 

Now we can do the main calculation of the left-hand rule Riemann sum and produce the remaining parts of the outline with the following do loop:

 

> 

for i from 1 to N do
s:= s+Y[i-1]*dX;  
X[i]:= X[i-1]+dX; Y[i]:=f(X[i]);
tr[i]:=plot([[X[i-1],Y[i]],[X[i],Y[i]],[X[i],Y[i-1]]],color=magenta): # plot van trapje is 2 lijnstukkken ( 3 punten)
end do:
> 

verN:=plot([[X[N],0],[X[N],Y[N]]],color=magenta): # verticale lijn bij x= b (interval grens)
> 

hor:=plot(0,x=a..b,color=magenta):
> 

p:=plot(f(x),x=a..b,color=black,thickness=2):
> 

display({seq(tr[i],i=1..N),hor,ver0,verN,p},scaling=constrained,axes=none);

 
> 

? riemann sum
> 

 


 

Download exc_set_3_task_1_a_b.mw

 

 

@Carl Love  As a pdf . exc_set_2_...

janhardo 700
June 09 2020
0 0

@Carl Love 

As a pdf .

exc_set_2_task_8_info_bijgevoegd_en_vraag.pdf

exc set 2 task 8 about vector function...

janhardo 700
June 09 2020
0 0

@janhardo 

This question is curve handled by a vector-valued function

exc_set_2_task_8_info_bijgevoegd_en_vraag.mw

exc set 2 task 8 vector-valued function...

janhardo 700
June 08 2020
0 0

@janhardo 

Don't know how to start with this asked vector-valued function for a Curve in this task
Could be too difficult for me ?

Note: some background information how to handle Curves

exc_set_2_task_8_info_bijgevoegd_en_vraag.mw 

@Carl Love  Thanks The parametric...

janhardo 700
June 08 2020
0 0

@Carl Love 

Thanks

The parametric description i can understand with calculus, but with vector-Calculus i should do need more experience.    

@Carl Love  Thanks I will look at...

janhardo 700
June 08 2020
0 0

@Carl Love 

Thanks

I will look at it

@Carl Love  Thanks That awesome p...

janhardo 700
June 08 2020
0 0

@Carl Love 

Thanks

That awesome power of Maple is is then versatility to program on different ways ?
   

@Carl Love ThanksYes, too difficult...

janhardo 700
June 08 2020
0 0

@Carl Love 
Thanks

Yes, too difficult, although it are 2 position vectors ( velocity, accelaration) moving along the Curve and that is rather easy to imagine.
The length of the folium is a lineintegral ?

@Carl Love  Thanks I got all sort...

janhardo 700
June 08 2020
0 0

@Carl Love 

Thanks

I got all sorts of lists , its about list-building with high level commands as i understand it.

@Carl Love  Thanks Maybe understa...

janhardo 700
June 07 2020
0 0

@Carl Love 

Thanks

Maybe understandable for me in a easy example.?
The book examples are less demanding for advanced programming ? 

@Carl Love ThanksThe book text : Ma...

janhardo 700
June 07 2020
0 0

 @Carl Love 

Thanks

I see a example : folium of Descartes made by a vector function 

 

 

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