This should be very interesting... I assume this is in response to the poll on whether there should be more postings on issues related to maths?

--Robin Kay--
P.S: TSE discountinued due to lack of interest, see my web site.

Whaddaya mean lack of interest? It got a news item on TICALC.ORG, for Pete's sake!!!

If you scroll down in the discussions on that poll, you'll see a suggestion for a math list.

This is awesome! Now we math geeks can actually talk about math on the calculator, which was designed for math in the first place. I'm going over there as soon as I'm done typing this up, and I urge others to do the same thing. Help promote more math, over games, on the TI line of calculators. For a long time now, everyone in my school could plau tetris, mario, and password-lock their calculator, but couldn't find the solver on it. That irritated me, now I can see that not everyone in the TI community just gets the calculators for the games :)

Good luck to all, and visit my website.

Wolf

I completely agree, there should be more math programs for the calculator than games!!! if the math programs were as fun... nevermind :Þ one thing I would like to know is why the calculator gives you a wrong answer when you enter a odd powered root with an even powered radicand such as for the cubed root of 64, it gives you 3.9999999999999... and so on... well, if any of you figure this out, tell me...

Um, my calculator gives me 4 for that one, and still, 3.9r=4 so technically, it's not wrong...

Are you by any chance in approximate mode? Which calculator is this?

OK, my TI-86 gives for 3xroot(64) 3.9999999999999
I get the same thing for 64^(1/3), but if I use 10^((log 64)/3) I get 4.0000000000001
Now if I use e^((ln 64)/3) I do get 4

I know it has to do with the transcendental algorithms that the calulator uses internally. But why that I don't know. What surprised me was using the Log and natural log I got the better answers instead of them being the same.
Now on my TI-92 10^((log(64))/3) gave me 3.999... answer. Otherwise the same as the TI-86.

Speaking of the solver... it's amazed me that no teacher has ever mentioned it in any of my math-related classes, yet it is one of the most useful features of some of the calculators. And the polynomial solver on the 85/86? That's not been metnioned either.

What?
My math teacher actually taught us all how to use the solver! But maybe its just cause its Calculus. What math are you in?

I'm replying to the comments on the sovler, and our teacher tells us that the fastest and the best way to solve equations, those that are simple, for the solver, but also the ones for simult and poly. She actually recommends that we use those, and I'm a sophomore in Algebra 2, and next year I'm taking calculus. Most everyone in the class has a TI-86, and one kid has an 82, and my teacher says that as long as there are so many people, might as well spend time to go over the functions in class after a quiz if we have time.

Keep smiling and visit my website for customized programs just for you!

Wolf.

Stop promoting your web site. It sucks.

Good idea. I have some questions. Will this relate to all intresting math, or just intresting math that a calculator can do? If one wanted to put something cool in the newsletter, how would they do it? Email ticalc I guess.

- The former, but most interesting math is calc-related.. ;)

- ask@ticalc.org

--BlueCalx

Speaking of math, does anyone have or know of a place where I can get some 3d raycasting equations/algorithms? I'd be very appreciative.

Glad to see that at least somewhat more of an emphasize is being placed on the math capabilities of calculators.

http://www.bol.ucla.edu/~permadi/raycast

--Robin Kay--

looks great, thx.

What exactly are you planing?

planning

I can't spell well

Well,i guess this is related to math..
this problem has stumped me for a while and i need someone to tell me the fault in it:
-1/1=1/-1
i^2=-1
i^2/1=1/i^2
squareroot each side and you get:
i/1=1/i
now cross multiply and you get:
1=-1
How is this correct, or what is the fault in this? Note i do not know the answer, please help.

>-1/1=1/-1
>i^2=-1
>i^2/1=1/i^2

ok up to here
since i^2=-1 and -1/1=-1 when you sqrt -1 you get i
>squareroot each side and you get:
>i/1=1/i
no you would get i=i
that solves it (i think) someone tell me if im wrong
>now cross multiply and you get:
>1=-1
>How is this correct, or what is the fault in this? >Note i do not know the answer, please help.

No, you keep it in the same form, square root the whole 2 sides, for example if you squareroot each side of 3/4, it would be SqRoot(3) /SqRoot(4)
So i^2/1=1/i^2 would be:
SqRoot(i^2)/SqRoot(1) = SqRoot(1)/SqRoot(i^2)
so i/1 =1/i, so cross multiply you get -1=1

Look at this again,

i^2/1 = 1/i^2
-1 = -1
i/1 = i
i/i = -i
(-i)^2 = i^2
i^2/1 = i^2
1/i^2 = (-i)^2

Now consider

(-1)^2 = 1^2
1 = 1
The square root of both sides must remain equal.
i^2/1 = 1/i^2
-1 = -1

-1 is not equal to 1
Just as
1/i is not equal to i/1
That is
-i is not equal to i

extraneous solution that arose when you changed the degree of the eqation?

1=i^4
1/i^2=i^4/i^2=i^2
sqrt(i^2)=i
therefore sqrt(1/i^2)=i, not 1/i.

This is basically the same argument as saying you should reduce 1/-1 before square rooting. I'm going to look into this more though, perhaps DeMoivre's theorem will yield an answer.

yeah that's right

-1/1=1/-1
i^2=-1
i^2/1=1/i^2
squareroot each side and you get:
i/1=1/i which is false.
i^2/1=1/i^2
the correct squareroot of each side is:
i=i

That is why i/1=1/i when you cross multiply and you get -1=1 which is also false.

No, yall are just showing me other ways to get around it, but every step i did had no fault in it, of course i know you can work it out other ways to prove -1 does not equal 1.

Look the squareroot of a number has both a + and - value. You cannot say 1=1, take the sqrt(1)=sqrt(1) and say -1=1. i^2/1=1/i^2 taking the sqrt of both sides as i/1=1/i is false.

It is doing the same as saying sqrt(1)=sqrt(1) is -1=1

The sqrt(i^2/1)=sqrt(1/i^2) is i=i which is true. i has a sqrt -i sign value and because it is sign value must be treated as such, just like a -1.

In doing algebra both sides of the equation must remain equal. Or you are doing it wrong. It is that simple.

but then there are other things such as i^4=1 right?
well take the fourth root of each side, you get i=1
how can you explain that?
Im thinking its just an exception for i...

1 has four fourth roots, 1, -1, i, and -i. Your calculator gives only one solution if you use 4xv1, which is 1.

So in this case -1 is the correct root to use and you get i=-1

That thing that looks like a 'v' was supposed to be an nth root symbol. I guess it didn't come out right.

Each nonzero number, including i²/1=1/i²=-1, has exactly 2 unequal square roots. The principal value of the square root function is always r*e^(i*theta), where -pi/2 < theta <= pi/2. (The real part is positive, or the real part is 0 and the imaginary part is "positive")

Therefore, when dealing with negative or complex numbers, sqr(ab) is not always equal to sqr(a)sqr(b), but +/- sqr(a)sqr(b). A similar statement holds true for sqr(a/b).

Look at this:

i/1 = i

1/i = -i

Of course 1 = -1 is not equal.

simply i/1 is not equal to 1/i

It is a difference in sign. i has a sign value. That is why it i and not 1.