# having difficulty using a for loop to solve an equation

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## having difficulty using a for loop to solve an equation

 here is my task: Create two arrays x and y, whose entries are defined as x  = i, y  = j i = 1, 2, 3, 4, 5. y=5 − 2j, j = −2, −1, 0, 1, 2. Then, compute the cubed sum of these two vectors in two different ways. First, use a for loop to construct the vector z as z  = (x + y )^ 3 , i = 1, 2, 3, 4, 5. my code: %Vectorization clc; clear; %x=i;                %array created per spec %i=[1 2 3 4 5];      %array created per spec y=5-2j;             %array created per spec j=[-2 -1 0 1 2];    %array created per spec z=(y).^3;           %formula to test for loop for m=j   t=z end   the result: t =   65 - 142i t =   65 - 142i t =   65 - 142i t =   65 - 142i t =   65 - 142i I know the answer should be t= (the cube for the elements in j) I'm new to octave but I've always needed adult supervision with loops.  What am I missing?
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## Re: having difficulty using a for loop to solve an equation

 "i" and "j" are sqrt(-1). Use something else "ii" and "jj" is what I normally do.Dmitri.On Mon, Jun 19, 2017 at 6:47 PM, naryana.shankara wrote:here is my task: Create two arrays x and y, whose entries are defined as x  = i, y  = j"i" and "j" are sqrt(-1). Use something else. My choise :  "ii" and "jj" Dmitri.--​ ​ _______________________________________________ Help-octave mailing list [hidden email] https://lists.gnu.org/mailman/listinfo/help-octave
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## Re: having difficulty using a for loop to solve an equation

 In reply to this post by naryana.shankara got rid of i and j my code looks like this:%Vectorizationclc;clear;a=[1 2 3 4 5];      %array created per specx=a;                %array created per specb=[-2 -1 0 1 2];    %array created per specy=5-2.*b;           %array created per spec  z=(y+x).^3;         %formula to be solved in loopfor m=a  t=zend and the output is:t =   1000    729    512    343    216t =   1000    729    512    343    216t =   1000    729    512    343    216t =   1000    729    512    343    216t =   1000    729    512    343    216>>I was expecting my result to be something along the lines of:t =   1000    729    512    343    216>>ort=1000t=729t=512t=343t=216>>Did I leave something out or forget to add something?Thank you for your helpOn Mon, Jun 19, 2017 at 5:47 PM, naryana.shankara wrote:here is my task: Create two arrays x and y, whose entries are defined as x  = i, y  = j i = 1, 2, 3, 4, 5. y=5 − 2j, j = −2, −1, 0, 1, 2. Then, compute the cubed sum of these two vectors in two different ways. First, use a for loop to construct the vector z as z  = (x + y )^ 3 , i = 1, 2, 3, 4, 5. my code: %Vectorization clc; clear; %x=i;                %array created per spec %i=[1 2 3 4 5];      %array created per spec y=5-2j;             %array created per spec j=[-2 -1 0 1 2];    %array created per spec z=(y).^3;           %formula to test for loop for m=j   t=z end the result: t =   65 - 142i t =   65 - 142i t =   65 - 142i t =   65 - 142i t =   65 - 142i I know the answer should be t= (the cube for the elements in j) I'm new to octave but I've always needed adult supervision with loops.  What am I missing? -- View this message in context: http://octave.1599824.n4.nabble.com/having-difficulty-using-a-for-loop-to-solve-an-equation-tp4683785.html Sent from the Octave - General mailing list archive at Nabble.com. _______________________________________________ Help-octave mailing list [hidden email] https://lists.gnu.org/mailman/listinfo/help-octave _______________________________________________ Help-octave mailing list [hidden email] https://lists.gnu.org/mailman/listinfo/help-octave
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## Re: having difficulty using a for loop to solve an equation

 In reply to this post by naryana.shankara On Tue, Jun 20, 2017 at 1:47 AM, naryana.shankara wrote:here is my task: Create two arrays x and y, whose entries are defined as x  = i, y  = j i = 1, 2, 3, 4, 5. y=5 − 2j, j = −2, −1, 0, 1, 2. Then, compute the cubed sum of these two vectors in two different ways. First, use a for loop to construct the vector z as z  = (x + y )^ 3 , i = 1, 2, 3, 4, 5. my code: %Vectorization clc; clear; %x=i;                %array created per spec %i=[1 2 3 4 5];      %array created per spec y=5-2j;             %array created per spec j=[-2 -1 0 1 2];    %array created per specFirst of all, I would say it is an ill-formed question.In the introduction, two arrays are defined, however the position is also defined, especially for the y-array.So in a formal sense only z1 and z2 are defined.When declaring your array, first define the index then use it.j=-2:2;y=5-2*j;Likewise for x. _______________________________________________ Help-octave mailing list [hidden email] https://lists.gnu.org/mailman/listinfo/help-octave