Can somebody explain where the insanity in this problem has gone to?

e^(pi*i)+1=0
e^(pi*i)=-1
e^(2*pi*i)=1 (square both sides)
2*pi*i=0 (natural log of both sides)

2*pi*i=0? Help me solve my dilemma!

Are you feeling okay?

You gotta think about the complex coordinate system in polar coordinates, kinda hard to explain in a post like this. r*e^(theta*i), where r is the radius and theta is the angle. In this case, we are pointing to -1. If it were e^(2*pi*i), we would point to 1, and e^(1/2*pi*i) points to i.

It has to do with taylor polynomials, you can use them to derive that e^(pi*i)=-1.

Anyway, if that confuses you try these,

a=b
0=b-a
0/(b-a)=(b-a)/(b-a)
0=1

x^2 = x*x
x^2 = x+x+...x (x x's)
now, take the derivative..
2x = 1+1+...1 (x 1's)
2x = x
2x/x = x/x
2 = 1

I hope that clears things up for you.

Man, you're 'problems aren't even difficult.

1)Dividing (b-a)/(b-a) isn't allowed in ordinairy maths because (b-a) = 0. Or, in other words, 0/0 is quite a mistake.

2)x^2^=sum(i=1..x,xi)
but d(x^2,x)<>d(sum(i=1..x,xi),x)

So, you're less interesting than you probably hoped for.

Greetings

On a second thought, when you square things it can add roots, at least if there is a polynomial. And, since ln(-1) exists as i, maybe that is why it gets screwed up, since you tried to square it before taking the natural log.

e^(pi*i)+1=0
e^(pi*i)=-1
e^(2*pi*i)=1 (square both sides)

instead of this last line we can say

e^(pi*i)*e^(pi*i)=-1*-1

So

ln(e^(pi*i)*e^(pi*i)) = ln(-1*-1)

2ln(e^(pi*i)) = 2ln(-1) (since ln(a^2) = 2ln(a))

And, ln(-1) clearly does not exist in the rational field, but this is equivalent to what you are trying to do. Maybe you just shouldn't mix all these complex numbers with normal ones.

Puzzles resolved (sort of):


exp(pi*i) = -1 ==>(implies) 2pi*i = 0

This is wrong because in general, exp(pi*i*(2n+1)) = -1 (for n an integer)
So if you use this more general expression, you end up with:

4*pi*i*n = 0

for n=0, the beginning and ending equations agree. Why it doesn't work for all n still puzzles me. See below for another puzzle which is basically the same thing.


a=b
0=b-a
0/(b-a)=(b-a)/(b-a)
0=1

This one is easy. b-a = 0, and you can't divide by zero.


x^2 = x*x
x^2 = x+x+...x (x x's)
now, take the derivative..
2x = 1+1+...1 (x 1's)
2x = x
2x/x = x/x
2 = 1

I think the problem here is that the sum is only true for integer x and doesn't work when x is real. Since differentiation isn't defined on the integers, it doesn't make sense to differentiate a function on the integers.


exp(i*2*pi*n) = 1 (for n an integer)
e = e*1
= e*exp(i*2*pi*n)
= exp(1+i*2*pi*n)
= (exp(1+i*2*pi*n))^(1+i*2*pi*n)
= exp((1+i*2*pi*n)^2)
= exp(1+i*4*pi*n-4*pi^2*n^2)
= exp(1+i*4*pi*n)*exp(-4*pi^2*n^2)
= e*exp(-4*pi^2*n^2)
!=(not equal to) e (for n != 0)

Evil. I think where we use exp(a)^b = exp(a*b) is where things go wrong, in this puzzle and the first one.
To the guy who said we shouldn't mix "normal" numbers with complex ones: every real number is a complex number, i.e. a = a + i*0, but you probably knew that. Real numbers are "unreal" just as imaginary numbers are. Think about how you can get infinite precision on the reals. You obviously can't do that to anything in real life. The reals just have properties that make them closer to common sense than imaginary numbers, but they're still completely abstract and exist only in the mind.

Sorry for the long off-topic post, but I'm new here, so it's not my fault. Where is this mythical math forum?

jk

I'm sure nobody cared then and even fewer people care now, but I have the definitive answer to the puzzle.

Like I said above, if you remember the fact that the complex exponential is periodic, you get this:

exp(2*pi*i*n) = 1 ==> 2*pi*i*n = 0
(the 4 was a goof on my part)

It seems like this is only valid for n = 0, but then I remembered an obscure fact about the logarithm: it's multi-valued. In general,

ln(z) = ln(r) + i*(theta + 2*pi*n)

Where z is a complex number, r is its modulus, theta is its argument, and n is an integer. This is fairly obvious if you write z as a phasor (z = r*exp(theta*i)) before taking its logrithm. If you don't know, the modulus is like the radius and the argument is like the polar angle in polar coordinates. So for z = 1 + 0*i,

ln(z) = ln(1) + i*(0 + 2*pi*n) = 2*pi*n*i

and the paradox is resolved.

I just noticed that this is offtopic. You should post this in a math forum.

You just noticed, eh?

Hehe yeah well, I couldn't think of any other way to phrase that. I mean "oh and by the way, that was offtopic" might have sounded funny. I foresaw a comment such as yours, though.

OH MY GOD an actual MATH-RELATED post on a calculator site!! Bravo!

Look at the type of article it is attached to, sherlock.

Dude, your web page is cool but weird

YES!!!!!!!!!!!!!!!!

I GOT A TI-89!!!!!!!!! NOW I CAN TAKE ADVANTAGE OF THIS!!!!!!!!!!!!!!!!!!!!!!!!!!

of course... I'll have to learn asm all over, or maybe I'll join the dark side of C

What's wrong with C? Why learn assembly for a single processor when you can learn C for which there is a compiler for practically every computer system on the face of the earth?

Cause assembly is cool. If a person codes assembly, it tends to run a little faster than a computer converting C into assembly.

That's why people learn assembly.

>>it tends to run a little faster than a computer converting C into assembly.

Right, but this is only true if you are an advanced programmer in asm because the compilers and linkers become better and better. And since he wants to start it would be better when he starts with C, that's much easier.

Yeah? Is there one for the TI-83+?

Assembly is fun.

Also most of the sofware written for the calc is so diesigned for the calc that I do not see it being useful on other platforms.

-Samuel

It depends. A purely numeric program (that doesn't use the estack) should be portable.

You should try TIGCC (at http://tigcc.ticalc.org, or just click the Web Page link above) instead, it has a few more features:
- Global and static variables are allowed in RAM programs with TIGCC.
- Programs can be automatically compressed with TIGCC. This also allows RAM programs of up to 64 KB of uncompressed size rather than just 24 KB.
- TIGCC is given with a static library called TIGCCLIB which contains functions like most ANSI stdio.h and stdlib.h functions, grayscale and sprite routines and many more.
- Programs written with TIGCC also run on AMS versions lower than 2.04 without having to install a memory resident emulation program.
- Programs written with TIGCC also work on AMS 1, while SDK programs often fail to work even with an emulation program because they use ROM_CALLs available on AMS 2 only. SDK programs even sometimes use ROM_CALLs only introduced in AMS 2.04 or 2.05.
- TIGCC is a port of GCC 3, so you get GCC specific features such as GNU C extensions with it.
- TIGCC supports third-party static libraries, and a few of those have already been written, like ExtGraph by Thomas Nussbaumer.
- TIGCC also allows you to program in assembly (not just inline assembly), and you even have the choice between 2 different assemblers.
- TIGCC also allows you to write kernel-based programs, including dynamic libraries and programs which use them (even though I do not recommend to use this feature).

Hehe listing all its features, eh? I think it's inherently obvious that TI-GCC is the best compiler around, even without that.

Not all its features. Only those the SDK doesn't have.

The only good thing about the TI-SDK is the Developer Guide TI offers, it contains a lot of material about the calculators' architecture.

> - Programs can be automatically compressed with TIGCC. This also allows RAM programs of up to 64 KB of uncompressed size rather than just 24 KB
Hey, but in SDK there´s no size limit at all!.

>>Hey, but in SDK there´s no size limit at all!.

Wrong! There is a size limit.
With the TI-GCC you can only develop RAM-Apps, not Flash-Apps. You also can develop RAM-Apps with the TI-SDK, but the size is limited to 24k(TI-GCC->64k!) due their anti-piracy mechanism. Also Flash-Apps have a size limit, but it's at 4M, wich is about five times the Flash-ROM available on a TI-89/92+.

Yes, and you cannot get the FlashApps signed, so currently you cannot actually develop a FlashApp which works on a real calculator, not even with the SDK. Unsigned FlashApps only work on TI's simulator.

Which, is one more reason (if i may) why I think that this version 1.0 of FS should still be called a beta :)

You can, actually, but to do that the app needs to be submitted in one of TI's app contests (until the freeware key is released).

In an e-mail TI written to me, they already have released the freeware key with the version 1.0.
But I didn't find anything, so I think it isn't true.