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11-03-2010, 08:29 PM #33
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pianoman2011
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Re: Looking for a special equation...

I finally translated the equation into something my TI-84 can understand! Pretty much exactly what you said Trevor...

f(x) = ( abs[x] / x ) * ( (abs[x] + .5 ) - ( fPart( (abs[x] + .5) / 1)

Essentially, my calculator does not have a button for normal modulo operation. I'll give you some examples.

Ex1. "(x+4)%1" is actually "1*fPart((x+4)/1)" in my calculator
Ex2. "(x+5.5)%3" is actually "3*fPart((x+5.5)/3)" in my calculator

Thanks for finding the equation Trevor!

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11-07-2010, 01:02 AM #34
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Hallucynogenyc
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Re: Looking for a special equation...

IMO this isn't neat. You're just taking profit of the % function, which implies rounding by itself.

You want to separate the decimals from the integer number, and you say you don't want to use the the round() function because it is already given in the calculator... but you use the fPart function, which instead of separating the Integer part directly, returns the decimals and you then subtract them from the original number to get the Integer part. Not neat, IMO.

Anyway, I reopen the question: Give me that same function without using the operators |x|, x%y, nor any other function. Obviously, no logical operators are allowed: I myself am smart enough to look for the source code of the round() function in a calculator
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11-07-2010, 01:43 AM #35
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Re: Looking for a special equation...

Quote:
 Originally Posted by Hallucynogenyc I myself am smart enough to look for the source code of the round() function in a calculator
As far as I know it's probably just memory manipulation, not math.

-Trevor
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11-07-2010, 03:03 AM #36
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Re: Looking for a special equation...

Quote:
 Originally Posted by Vaporizer Integers by definition are whole numbers. No decimals or fractions. I don't see how this is gonna be accurate for anything but estimating. If you used this definition and the TI is using a true integer 5.9 + 6.1 = 11 not 12. I'll stick with the >5 rule lol
I have to quote myself where I was wrong in how I worded my statement.
Integers, by definition, are WHOLE numbers WITHOUT fractions or decimals.
Now how can you round a whole number? You can't, it's whole.
So the number you are working with is not even an integer to start with.
Integer Definition
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11-07-2010, 03:21 AM #37
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Re: Looking for a special equation...

Any of you guys want to tackle my taxes?

I'm not the smartest of people but I know I'm not thickest. However, you guys just made me feel DUMB! I found the introduction to Calculus a real ball ache and we have more to come this year!

I'm so glad I hang around a place that can make me feel stoopid. Kind of good to get grounded oc*****nally.

Glad all got worked out. (Mostly at this point anyway)

On a seperate point...

Where the hell have you been Hallucynogenyc?

M
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11-07-2010, 05:45 AM #38
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Re: Looking for a special equation...

Quote:
 Give me that same function without using the operators |x|, x%y, nor any other function.
Now that's a hell of a challenge.

-Jakob
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11-07-2010, 06:52 AM #39
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Crazy Jay
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Re: Looking for a special equation...

I love how math causes you guys to fight JK

11-07-2010, 02:31 PM #40
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Re: Looking for a special equation...

Quote:
 Originally Posted by Morgan Any of you guys want to tackle my taxes? I'm not the smartest of people but I know I'm not thickest. However, you guys just made me feel DUMB! I found the introduction to Calculus a real ball ache and we have more to come this year! I'm so glad I hang around a place that can make me feel stoopid. Kind of good to get grounded oc*****nally. Glad all got worked out. (Mostly at this point anyway) On a seperate point... Where the hell have you been Hallucynogenyc? M
Well, I just know how to defend myself a bit on maths hehe About me, read the post I made on the "missing or absent members thread", on the "other" section. Thanks for your interest

Quote:
 Originally Posted by twhite828 As far as I know it's probably just memory manipulation, not math. -Trevor
And that's why that sentece you quoted was preceded by:
Quote:
 Obviously, no logical operators are allowed:

Quote:
 Originally Posted by pianoman2011 Now that's a hell of a challenge. -Jakob
Well, I'm going to share the idea I've been working with all the time.

lim x->oo (2a^x) = 0 if -0.5<=a<=0.5, oo otherwise
lim x->oo (1/1+(2a^x))...

Continue yourself
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11-07-2010, 02:59 PM #41
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Re: Looking for a special equation...

Quote:
 Originally Posted by Hallucynogenyc Well, I just know how to defend myself a bit on maths hehe About me, read the post I made on the "missing or absent members thread", on the "other" section. Thanks for your interest
... Will do...

M
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11-07-2010, 03:40 PM #42
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pianoman2011
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Re: Looking for a special equation...

Quote:
 Well, I'm going to share the idea I've been working with all the time. lim x->oo (2a^x) = 0 if -0.5<=a<=0.5, oo otherwise lim x->oo (1/1+(2a^x))... Continue yourself
Damn Hallucinogen! That is a little over my head right now, I might have to write it down and work over it for a little while.

-Jakob
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11-08-2010, 01:10 AM #43
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Hallucynogenyc
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Re: Looking for a special equation...

haven't you done limits at school?
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11-08-2010, 04:23 AM #44
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pianoman2011
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Re: Looking for a special equation...

Yeah, but only briefly and I didn't fully understand them. And the "a" variable in your equation throws me off a little

-Jakob
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11-08-2010, 08:09 PM #45
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Hallucynogenyc
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Re: Looking for a special equation...

Quote:
 Originally Posted by pianoman2011 Yeah, but only briefly and I didn't fully understand them. And the "a" variable in your equation throws me off a little -Jakob
a is a number. lim x->oo f(a,x) only means that you need to substitute the x for oo in f(a,x). All we need the term limit is because in fact, oo by itself doesn't exist, but forget that now.

First you need to know a couple of things:

1.-oo = infinite
2.-Anything divided by oo is 0.
3.-Anything multiplied by oo is oo.
4.-Anything elevated to oo is oo, (0 or any value between 0 and 1)^oo=0 and 1^oo = 1.
5.-Anything added to oo is oo.

(I know all of this is garbage when two limits interact, but it's all he needs to know for now)
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11-08-2010, 08:22 PM #46
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pianoman2011
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Re: Looking for a special equation...

Ohhh okay, I didn't realize that "a" was a constant. And yea I have no idea about two limits interacting, that's kinda what threw me off.

(0 or any value between 0 and 1)^oo=0

Didn't know about that though! When I think about it, it makes sense.

-Jakob
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 473nm Oceanus 5mW RPL Style: <5mW
 532nm Rayfoss 200mW Adjustable Focus Pointer: 150-190mW
 532nm BudgetGadget 50mW Keychain Pointer: 60mW
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11-08-2010, 09:26 PM #47
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Hallucynogenyc
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Re: Looking for a special equation...

Quote:
 Originally Posted by pianoman2011 Ohhh okay, I didn't realize that "a" was a constant. And yea I have no idea about two limits interacting, that's kinda what threw me off. (0 or any value between 0 and 1)^oo=0 Didn't know about that though! When I think about it, it makes sense. -Jakob
Well, two limits interacting just means something like

calculate: lim x->oo 2x/(3x+1). if you just use what I told ya, it gives you oo/oo, which has an undetermined solution. However, you'll learn that all of that can mostly still be solved

So... did you get what I meant with my equation?
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11-12-2010, 09:42 PM #48
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Re: Looking for a special equation...

lim x->oo (2a^x) = 0 if -0.5<=a<=0.5, oo otherwise
lim x->oo (1/1+(2a^x))=1 if -0.5<=a<=0.5, else oo

because lim x->oo (1/1) = 1, regardless of x, and limit of 2a^x is 0.

...? I failed Calc 2 twice, and had to drop out... Don't think they even offer calc at the comm. college Limits are all very interesting and everything, but simplifying integrals is where I fell on my face. That, and I couldn't understand my professor, my teaching assistant, or my tutor.

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