for loop to input "n" number of var "x"

[albisher]

hi i am new use having problems when i want to enter variables

let .. x1, x2, x3, .... xn
are the variables needed to be input by user

if i want to deal using for loop

i do not know how to do it!
i am usin 89-titanum in programmin numerical methods for engineering

ok ... i have it like this
user will input n,a,b
then calculater find h
and then

for i,0,n
clrio
a+h*i->z
z->x[i]
f(z)->fx[i]
disp "xi=",x[i]
disp "f(xi)=",fx[i]
pause
1+i->i
endfor


but it is giving ERROR:
index out of range

it is reterning me to the line
z->x[i]

any ideas ? pls help ;>

[IAmACalculator]

Since you're doing [font="courier new"]For I,0,N

[albisher]

then how do i need to get it!!

is there any good explenation about the way i deal with arrays?

what i understand ... my error because i start from 0

but when i changed program .. for i,1,n
still there is another error with same line

it is saying : Dimention

[leofox]

You have to make sure the list you are using has a size. If the size is 0 and you are trying to store to element 1, it will say err:dimension.
You have to enter the 68K equivalent of N->dim(X

[bukwirm]

In TI-BASIC list indexes start at 1, not 0. As for the "Dimension" error, make sure the list exists and has at least 1 element.

For more details about TI lists/matrices, see my signature, then ask if you still have specific questions.

[albisher]

ok, let me explain it more deeply.

i need to fill

x0, x1, x2, ..... xn

so i chose to use for loop

as a simple thing of what i am doing

input "n=", n @nujmber of segments
input "a=", a @1st element value
input "b=", b @last element value

(b-a)/n -> h @ step size between each element

for i,1,n+1

a+h*(i-1)->z
z->x[i]
1+i->i

endfor

..........

will this be good way of thinking in ti-89 titanium basic programming lang?

[IAmACalculator]

[albisher]

:>

finally i unerstod it hehehe thanks friends

input "n=", n
newList(n)

input "a=", a
input "b=", b

(b-a)/n -> h

for i,1,n+1
a+h*(i-1)->z
z->x[i]
endfor


this willll helllllp meeee looooot
;]

i am getting better each day even that is so slow but .. i am being good

[DarkerLine]

newList(n)->x.

Actually, {}->x will work too in your case, since storing to an element one past the end of the list will append that element to the end. For example {1,2} -> x: 3 -> x[3] will make x = {1,2,3}.