Looking for a special equation...

[Meatball]

A piecewise function would be best IMO, but depending on the domain you're wanting to work with, it could be one heck of one function to write out. Not much use on a calculator consequently.

I thought about modifying Newton's method a bit, in order to find y values that aren't zeros, but then I ran into taking limits of derivatives of functions that I don't even have yet.

I gotta take a nap now. Thanks guys.

 

[pianoman2011]

Cool story bro, I've taken calculus at the collegiate level too. More of it at a higher level. And I'm sure a whole lot people here have more and higher-level calculus under their belt than I do.

Perhaps your problem is that you focus on showboating rather than on your studies.

-Trevor

I don't see what your problem is. I wasn't showboating, I was just letting you know that I'm not a complete idiot, because that's how you made me seem when you first responded to me. I'm not doing this for an assignment. I'm not doing this to show off any knowledge. Quite the contrary... I posed a question because I didn't know the answer. So maybe you can vent and pick on people somewhere else. Honestly IDGAF if you've taken calc at a "collegiate level", cause that collegiate-level knowledge isn't helping to answer my question one bit.

if you don't want to use custom functions, then... can limits be used? can a piecewise function be used?

A limit function would be fine, I'm pretty sure those could go into my calculator. I've just been wrapping my head around this recently and I can't come up with anything! :X

I'm still wondering what is so hard abt the > than 5 rule.......
I'm no rocket scientist......I didn't fail math or programming languages either.
I'm at a quandary why to have it calculate the number when you can plug it in already rounded in your head or round the result in your head :thinking:

This is like the "duh" factor.............am I missing something?

Edit: Ok, do you realize the error factor when rounding via multiple times and used in an equation.......?
My last post here.......he's gotten responses in 4 formats.......enough is enough.
I may need to call the Troll Overlord in on this............lmao

The point of knowing the answer to this question isn't to be able to figure out how to round numbers... it's more of a "how can I mathematically do that" kinda' thing. And I'd love you to call in TJ if he's good at math!

A piecewise function would be best IMO, but depending on the domain you're wanting to work with, it could be one heck of one function to write out. Not much use on a calculator consequently.

I thought about modifying Newton's method a bit, in order to find y values that aren't zeros, but then I ran into taking limits of derivatives of functions that I don't even have yet.

I gotta take a nap now. Thanks guys.

Thanks for the info. I think I'll look up what a piecewise function is and try to figure some stuff out. Maybe I'll be able to figure something out.

-Jakob

 

[Trevor]

I'm in calculus, way above rounding numbers...

Nopeeeee college calculus at my local college!

If that's not showboating, I don't know what is. It just appears you have something to prove.

f(x) = ( x + 0.5 ) - ( ( x + 0.5 ) % 1 );

...but if you could do that you'd probably have round(); anyway.

-Trevor

 

[pianoman2011]

If that's not showboating, I don't know what is. It just appears you have something to prove.

f(x) = ( x + 0.5 ) - ( ( x + 0.5 ) % 1 );

...but if you could do that you'd probably have round(); anyway.

-Trevor

When I look at those quotes, it does look a little showboaty. The first one was just to say that my question had nothing to do with any of my homework assignments. The second one was just to let you know that I'm not in highschool calc

That equation you posted up there looks like it has a lot of potential to be the one I'm looking for. I think if I throw in an "abs()" or two I might find my answer. Thanks!

-Jakob

Edit: I actually take back what I said about abs(). I think I need to turn that into a step function somehow. Bringing me back to the math class I took a few years ago... time to dig through my notes on step-functions!

 

[Trevor]

f(x) = ( |x| / x ) * ( ( |x| + 0.5 ) - ( ( |x| + 0.5 ) % 1 ) );

May not be optimal (speaking in CPU cycles), but that should do it.

Also, just busting your balls... don't worry about it.

-Trevor

 

[Jimmymcjimthejim]

If that's not showboating, I don't know what is. It just appears you have something to prove.

I don't enjoy making posts like this, but seriously Trevor, ever since you entered this thread you've adopted an arrogance within your posts. The posts you quoted are only responses to your vulgar assumptions, and the fact that a student is curious enough to expand his knowledge of calculus is a blessing. Take it somewhere else man.

 

[Trevor]

I don't enjoy making posts like this, but seriously Trevor, ever since you entered this thread you've adopted an arrogance within your posts. The posts you quoted are only responses to your vulgar assumptions, and the fact that a student is curious enough to expand his knowledge of calculus is a blessing. Take it somewhere else man.

A question such as this one is oddly specific and screamed "homework problem" to me... therefore I wasn't going to be particularly kind. I was proven incorrect. Forgive the assumption. :tinfoil:

*avoids making comment about how rounding isn't calculus and attempts to leave quietly*

-Trevor

 

[pianoman2011]

f(x) = ( |x| / x ) * ( ( |x| + 0.5 ) - ( ( |x| + 0.5 ) % 1 ) );

May not be optimal (speaking in CPU cycles), but that should do it.

Also, just busting your balls... don't worry about it.

-Trevor

Thanks man, not worrying anymore

So I tried your equation with x=3.55

f(x) = (|x|/x)*((|x|+0.5)-((|x|+0.5)%1))
I'm assuming ^ means times .01
f(3.55) = (3.55/3.55)*(3.55+0.5)-((3.55+.5)*.01)

f(3.55) = 1*4.05-(4.05*.01)

f(3.55) = 4.05-.0405

f(3.55) = 4.0095

It came close to 4! But not quite... Trying to figure out what didn't work there.

I don't enjoy making posts like this, but seriously Trevor, ever since you entered this thread you've adopted an arrogance within your posts. The posts you quoted are only responses to your vulgar assumptions, and the fact that a student is curious enough to expand his knowledge of calculus is a blessing. Take it somewhere else man.

Thanks for defending me jimmy, I do believe me and Trevor are now ok I just hope my question doesn't turn into one big fight :undecided:

-Jakob

 

[pianoman2011]

A question such as this one is oddly specific and screamed "homework problem" to me... therefore I wasn't going to be particularly kind. I was proven incorrect. Forgive the assumption. :tinfoil:

*avoids making comment about how rounding isn't calculus and attempts to leave quietly*

-Trevor

Don't leave, you were on a roll!

-Jakob

 

[Trevor]

The % means essentially 'the remainder of.'

So, you're dividing ( |x| + 0.5 ) by 1 and taking the remainder, and subtracting it from ( |x| + 0.5 ), effectively rounding. You then multiply by ( |x| / x ) to get the original sign back.

@above: Thanks.

-Trevor

 

[pianoman2011]

Figured out what I can directly type into my calculator to give me the answer! It's the integer function:

[MATH]
[NUM]
[5]

^^That's what you press on a TI-84 to get to the function.

So int(x+0.5) gives me that answer.
Example: int(4.7+.5)=5

Now I'm just wondering how the int() function works :evil:

-Jakob

 

[pianoman2011]

The % means essentially 'the remainder of.'

So, you're dividing ( |x| + 0.5 ) by 1 and taking the remainder, and subtracting it from ( |x| + 0.5 ), effectively rounding. You then multiply by ( |x| / x ) to get the original sign back.

@above: Thanks.

-Trevor

Ohh ok now I just have to figure out how to put the "remainder of" in terms my calculator can recognize!

-Jakob

 

[Trevor]

Now I'm just wondering how the int() function works :evil:

-Jakob

It's low level enough (programmatically speaking) to where I'd bet it's just memory manipulation, not a pure (mathematically speaking) formula.

I'd start by checking out how the "float" (or "double" too) datatype works.

EDIT: [2nd] + [0] -> [R] -> scroll down a bit. round() is there on my TI-83. I thought you were trying to avoid built-in functions.

EDIT 2: round(X, 0) gives me a nice step function.

-Trevor

 

[pianoman2011]

It's low level enough (programmatically speaking) to where I'd bet it's just memory manipulation, not a pure (mathematically speaking) formula.

I'd start by checking out how the "float" (or "double" too) datatype works.

EDIT: [2nd] + [0] -> [R] -> scroll down a bit. round() is there on my TI-83. I thought you were trying to avoid built-in functions.

-Trevor

Yeah, avoiding built-in functions was one of the things I was attempting I think I'm gonna write down all of the information/equations I got out of this thread, look up how the float and integer functions work, and then try to work out an equation that doesn't use "int(x)".

I'll let ya' know when I come to a solution! Thanks.

-Jakob

 

[Jimmymcjimthejim]

Forgive the assumption. :tinfoil:

Need there be more to be said, you are forgiven :yh:
Wish I could help you guys out, but I'm too tired to think right now. Would've expected the solution to be a little bit simpler, but that's just a matter of taste. Night' folks.

 

[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