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### math question

Posted: Mon Apr 20, 2009 6:37 am
So - i was doing sum comparison tests between my library and my calculator - and i was doing summations. and i accidentally did the sum of x^3, from x = 1 to x = -5. I expected an error (how can you count up from 1 to -5 right?) but instead i got the result of 100. So I did some testing: on my TI-89 and TI-nspireCAS (both computer algebra systems) i got the result of 100. On my TI-84 and Casio (regular calculators) I got an error, and in Maxima (another computer algebra system) i got 0.

Does anyone know how to get that result? or is this some strangeness that TI has simply added to their calculators?

### Re: math question

Posted: Tue Apr 21, 2009 8:17 am
TI-92 solves it like that:

sum(x³,x,a,b)=
-a^4/4 +a³/2 -a²/4 +b²(b+1)²/4

for a=1 and b=-5, you get 100

that sum simply works because TI is able to solve the general case ### Re: math question

Posted: Tue Apr 21, 2009 9:05 pm
What do you eat Kram?
What makes you know all this stuff? ### Re: math question

Posted: Tue Apr 21, 2009 10:57 pm
You are a machine Kram. ### Re: math question

Posted: Wed Apr 22, 2009 3:46 am
Thats simple integration rules if I understand that right... nothing too complicated ### Re: math question

Posted: Wed Apr 22, 2009 3:51 am
IF  ### Re: math question

Posted: Wed Apr 22, 2009 4:53 am
Kram1032 wrote:TI-92 solves it like that:

sum(x³,x,a,b)=
-a^4/4 +a³/2 -a²/4 +b²(b+1)²/4

for a=1 and b=-5, you get 100

that sum simply works because TI is able to solve the general case I did some more work, and found out most of it is different conventions - but TI' s formula is actually different, that case was just coincidence!

for: sum(x³,x,a,b)
if a > b > 0 , then it performs sum(x³,x,(a-1)*-1,b)
if a > 0 > b, then is does sum(x³,x,a,|b-1|)
if 0 > a > b, the result is sum(x³,x,a,|b+1|)

and the expansion of x^3 that it gives dose not work the same as the just stated algorithum for most input took me a little to find out the pattern Im just going to make it throw an error ### Re: math question

Posted: Wed Apr 22, 2009 9:34 am
why would you do that?
If the general case can be solved, you can even solve it for floats and complex numbers....

@"How do you do":
simply type:

Code: Select all

``2nd 4 x ^ 3 , x , a , b ) ENTER``
gives me, after putting it on one ratio,

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``````-(a+b)*(a-b-1)*(a²-+b*(b+1))
____________________________
4``````
then I say:

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``2nd (-) 2nd k a=1 and b=-5``
and my TI returns

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``100``
That's all the magic. I love it^^
We did NOT learn (yet, if we ever will), how to solve such sums generally. It's just what the TI tells me.

As said, interestingly, it's even possible to "count" from, say, -φ to π*i.... that gives:

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``````4*π^4-4*π²-(√5+3)*(3*√5+7)   π³
__________________________-_____ * i
16                2``````
or as floats:

Code: Select all

``17.3988036807-15.5031383402*i``

### Re: math question

Posted: Wed Apr 22, 2009 10:07 am
Kram, you're also randomly janitor at your school ?

### Re: math question

Posted: Wed Apr 22, 2009 10:13 am
nope... gotta think about the real result though... hum... 3? 5? Way too high maths to me... gotta go back to generalise spheres to any dimension. Not so much headache^^

(Btw: Ever tried to PROOF that 1+1=2?)

### Re: math question

Posted: Wed Apr 22, 2009 10:22 am   ### Re: math question

Posted: Thu Apr 23, 2009 3:09 am
Again, that simply a misnomer because of the equation I selected.

Try to do the summation on x*ln(x), it is a summation that can not be generalized, and you will find the ti can not sum it if you specify either of the limits as complex, you will not get a result.

However, it also disproves my previous algorithm since you cant take the log of a negative number and the TI still gives a result back to square one!!

(btw, yes i have! other fun ones, proving 5^(1/2) is irrational, and deriving the direct formula for the nth Fibonacci number are some of my favorites!)

### Re: math question

Posted: Thu Apr 23, 2009 5:26 am
n^(1/k) (n,k= positive integers) is proofen via contradictions, right?

you try to proof, that n^(1/k)=a/b with a,b being positive ints again and somehow you get a contradiction or a limitation (if you assume n=c^k with c again being a positive integer, it'll work as n^(1/k)=c/1 but else, you simply can't proof, that n^(1/k) can be written as a/b, due to a contradiction, which automatically tells you the opposite...)

I read a poem or two which proofed that 2^(1/2) is irrational xD

### Re: math question

Posted: Thu Apr 23, 2009 7:24 am
Kram1032 wrote:n^(1/k) (n,k= positive integers) is proofen via contradictions, right?

you try to proof, that n^(1/k)=a/b with a,b being positive ints again and somehow you get a contradiction or a limitation (if you assume n=c^k with c again being a positive integer, it'll work as n^(1/k)=c/1 but else, you simply can't proof, that n^(1/k) can be written as a/b, due to a contradiction, which automatically tells you the opposite...)

I read a poem or two which proofed that 2^(1/2) is irrational xD
the easy one is proof via contradiction yes - but the most important part is that if n^(1/l)=a/b, that a/b is an irreducible fraction - although if you dont you eventually conclude that the fraction is infinitely reducible , which is also patently absurd and satisfies proof by contradiction ### Re: math question

Posted: Thu Apr 23, 2009 7:45 am
ah, yes, forgot that point^^