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janhardo

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@acer  Thanks Some extra examples...

janhardo 695
May 24 2020
0 0

@acer 

Thanks

Some extra examples makes it more clear.

@Carl Love Thanks  ...

janhardo 695
May 24 2020
0 0

@Carl Love 

Thanks 

@Kitonum Thanks It was asked t...

janhardo 695
May 24 2020
0 0

@Kitonum 

Thanks 

It was asked the do this by two pair of nested loops
Those nested loops do have more uses in other tasks i see and it is to get used of them

Maybe i see also your array specify in this book examples?

What makes it easier  ..to come up with this?   

@acer  Thanks Could not come up o...

janhardo 695
May 24 2020
0 0

@acer 

Thanks

Could not come up on this array

The elements are:

A[1,1]=1^1 , A[1,2]=1^2,   A[1,3]=1^3  A[4,1]= 1^4 

A[2,1]=2^1, A[2,2]=2^2,   A[2,3]=2^3  A[2,3]=2^4  

........................

De outer loop is for from top to bottom en inner loop from left to right

 

excersise set 2 task 1...

janhardo 695
May 23 2020
0 0

@Carl Love 


 

> 

 

 

> 

 

for I from I1 to I2 do statements end do
I stands for the varying index
I1 is the  starting value for the index
I2 is the end value for the index  

 

Array definition : 1-dimensional

> 

restart;
> 

A:=Array(0..6);# 1-dimensional array

Array(%id = 18446746044449226870)

 (1)
> 

A[1]; # a array element

0

 (2)
> 

A[0];

0

 (3)
> 

 

beware the array can start with a 0 or 1 ( depends on  varying index i  )

 Looking at  task (a) ..there are 6 numbers , so the index value  i  goes from 0 ..5  

a general formulae for the sequenze 1,3,9,...    

 

 
   
   
 

 

 

> 

restart;
> 

 

LET

(a)

> 

X :=Array(0..5);

Array(%id = 18446746044449226870)

 (7)
> 

for i from 0 to 5 do
      X[i] := 3^i:
od;

1

  

3

  

9

  

27

  

81

  

243

 (8)
> 

X[1];

3

 (9)
> 

X;

X

 (10)

the one-dimensional  array is conceptual the same as a list

Question : when should i use  for i a 0 or 1 ?  ..

> 

 

(1b) A sequenze  -1,1, ..." = (-1)^(n)" for  n= 0,...

> 

G:=Array(1..12);

G := Array(1..12, {(1) = 0, (2) = 0, (3) = 0, (4) = 0, (5) = 0, (6) = 0, (7) = 0, (8) = 0, (9) = 0, (10) = 0, (11) = 0, (12) = 0})

 (11)
> 

for i from 1 to 12 do
      G[i] := (-1)^i:
od;

-1

  

1

  

-1

  

1

  

-1

  

1

  

-1

  

1

  

-1

  

1

  

-1

  

1

 (12)
> 

G[2];

1

 (13)
> 

G;

G

 (14)

1(c)  this sequenze can be expressed in a general term1,4,7,10,...  3i+1  ( i = 0 ,...n )  

> 

A:=array(0..6):
> 

for i from 0 to 6 do
      A[i] := 3*i+1:
od;

1

  

4

  

7

  

10

  

13

  

16

  

19

 (15)
> 

A[0];

1

 (16)
> 

 


 

Download betounes_ex_set2_1.mw

Overhaul task 1 c...

janhardo 695
May 23 2020
0 0

@Carl Love 

Thanks

I made a overhaul of task 1c and get stuck for drawing cross-hatching one symmetry part of the graph. 

Also i used a piecewise function A__right , but don't understand it  

All integrals ( symbolic, numeric (Int and int ) give the same answer

excersise_set_1 task5 ( last one)...

janhardo 695
May 22 2020
0 0
 

  Don't know yet why the lines are not perpendicular 

 

  


 

> 

 

 

      

> 

restart;
> 

f:=x->x^5-2*x^3-4*x^2+x+4;

proc (x) options operator, arrow; x^5-2*x^3-4*x^2+x+4 end proc

 (1)
> 

p:= plot(f(x),x=-1..2,color=magenta,thickness=2,tickmarks=[3,3],gridlines= false);

 
> 

with(plots,display);

[display]

 (2)
> 

a:=-0.4;b:=2;c:=-1;

-.4

  

2

  

-1

 (3)
> 

#f(a)+m*(x-a);
> 

f1:=D(f);

proc (x) options operator, arrow; 5*x^4-6*x^2-8*x+1 end proc

 (4)
> 

mtan:=D(f)(a);

3.3680

 (5)
> 

tangentline:=plot(f(a)+mtan*(x-a),x=c..b,thickness=2,color=black,gridlines= false):
> 

display({tangentline,p});

 
> 

 

Now a normalline drawing in inflection point tangent line  

> 

mnor:=-1/mtan;

-.2969121140

 (6)
> 

mtan*mnor;

-1.000000000

 (7)
> 

normalline :=plot(f(a)-mnor*(x-a),x=c..b,thickness=2,color=black,gridlines= false):

> 

 

> 

``
> 

display({tangentline,normalline,p});

 
> 

 

 


Download betounes_ex_set_1_andereaanpak_1_opg_5.mw

@fyi Thanks That's in one command,...

janhardo 695
May 22 2020
0 0

@fyi

Thanks

That's in one command, but like it more do in steps like in this example , because its about the how to come on this answer as you doing it by hand .
If maple can come up with a answer that's fantastic.

 

@acerThanks   ...

janhardo 695
May 22 2020
0 0

@acer

Thanks  

Its also not possible to add a enclosed section in a worksheet and posted here ? ..
 

exercise set_1 task 4...

janhardo 695
May 22 2020
0 0


 

> 

 

> 

 

Reading the task this critical points are stationary points ( function derative is zero) + inflection point
lets graph.

> 

restart; with(plots):
> 

f1:= 4*sin(x^3)-x^2;intvx:= x =-2..2;

4*sin(x^3)-x^2

  

x = -2 .. 2

 (1)
> 

plot(f1, intvx);

 
> 

 

Easier to work in Maple with deratives is with the D operator seems

> 

f2:=D(f1);

4*D(sin(x^3))-2*D(x)*x

 (2)
> 

 

Seems to be no easy derative to get with the D operator, then with the diff ( )
diff as expression or function input
lets try the function input.

 

> 

restart;
> 

f3:=x->4*sin(x^3)-x^2;

proc (x) options operator, arrow; 4*sin(x^3)-x^2 end proc

 (3)
> 

diff(f3(x),x);

12*x^2*cos(x^3)-2*x

 (4)
> 

Diff(f3(x),x):%=value(%);

Diff(4*sin(x^3)-x^2, x) = 12*x^2*cos(x^3)-2*x

 (5)
> 

# Diff(f1,x):% = value(%);
> 

 

go back to (4)


 

Download exercise_set_1_task4.mw

 

@Carl Love ThanksWhat is asked in t...

janhardo 695
May 21 2020
0 0

@Carl Love 

Thanks

Some questions

1) 

plots:-display(plots,plot);

 (never seen before) is it a existing command ?

2) int(f-g)     

Why subtraction ?

------------------------------------------------------

I go to the next exercise set 1_3

 

@janhardo  There was een example i...

janhardo 695
May 20 2020
0 0

@janhardo 

There was een example in the book enclosed

https://drive.google.com/open?id=1_FZN4k3_tZn8DSse1uKPnG9d1SVtfFPo

 

@Carl Love  I got this now: f := ...

janhardo 695
May 20 2020
0 0

@Carl Love 

I got this now:

f := x^sin(x);
                                sin(x)
                          f := x     
g := x^cos(x);
                                cos(x)
                          g := x     
evalf(Int(abs(f-g), x= 1..15));
                          50.57916030
Its a powerful command, but that's probably not the solution asked by the author of the exercises.?  

 

@vv  Its a difficult one - inters...

janhardo 695
May 20 2020
0 0

@vv 

Its a difficult one

- intersections points of the two graphs :
How to do this ?: with a expression or in function notation
- calculate areas between intersection points

 

@Carl Love  Thanks For the worked...

janhardo 695
May 20 2020
0 0

@Carl Love 

Thanks

For the worked out answers : my initial approach  was find interceptions with the x-axis and integrate the graph areas one by one.
With int (piecewise ,.. )   its be done faster and easier.

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