Looking for a special equation...

[pianoman2011]

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!

 

[Deleted member 8382]

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

 

[Trevor]

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

 

[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. :crackup:
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

 

[Morgan]

Any of you guys want to tackle my taxes? :crackup:

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

 

[pianoman2011]

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

 

[Crazy Jay]

I love how math causes you guys to fight JK

 

[Deleted member 8382]

Any of you guys want to tackle my taxes? :crackup:

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

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:

Obviously, no logical operators are allowed:

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

 

[Morgan]

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

 

[pianoman2011]

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

 

[Deleted member 8382]

haven't you done limits at school?

 

[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

 

[Deleted member 8382]

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)

 

[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

 

[Deleted member 8382]

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?

 

[BShanahan14rulz]

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 :scowl: 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.