# Factor Equation: (1-v^2/c^2)^3/sqr((1-v^2/c^2)^2)

• eNathan
Sqrt[(-0.36 + 1)/0.] = ComplexInfinityThis is despite the fact that our simplification is correct. Just so you can see that to mathematica I'm not wrong. The difference is subtle, but it's there, and it does make your formulation different in a mathematically meaningful way.You are correct; their simplification is wrong, although this seems to have occurred because the brackets in your original equation were difficult to read.However, your equation does not eliminate any divide by zero error. In fact, it introduces one. You probably haven't noticed that because google doesn't handle division by zero correctly, giving...Sqrt[(-0.36 + 1)/0.] =
eNathan
sqr(sqr((1-v^2/c^2)^2)) * ((1-v^2/c^2) / sqr((1-v^2/c^2)^2))

There is a purpose in this equation, but it is way to long. I know this can be factored becase there are a lot of "like terms" If you want to, change the square roots to ^.5 instead of sqr(x) you can use x^.5

Why dont' you start by replacing 1-v^2/c^2 with some letter, such as x and make your life easier.

eNathan said:
sqr(sqr((1-v^2/c^2)^2)) * ((1-v^2/c^2) / sqr((1-v^2/c^2)^2))

There is a purpose in this equation, but it is way to long. I know this can be factored becase there are a lot of "like terms" If you want to, change the square roots to ^.5 instead of sqr(x) you can use x^.5

$$\sqrt{\frac{\sqrt{\gamma^{2}}\gamma}\sqrt{\gamma^{2}}}}=\sqrt{\gamma},\gamma=1-(\frac{v}{c})^{2}$$

$$\sqrt{1- \frac{v^2}{c^2}}$$

I think, you'll need to check it though.

$$(Sqrt\$$(1-v^2÷/c^2)$$^5\ = (Sqrt\ $$(c-v)$$^5\$$(c+ v)$$^5\)÷/c^(10)\)$$

Zurtex, it does not cancel down to sqr(1- v^2 / c^2)
huan.conchito, are you saying it factors to (sqr((1-v^2/c^2)^5 ?

And sorry but I am not to good at Latex. And I am not quite sure what gamma^2 is

Well, then write your original expression properly; both Zurtex and I found by our best efforts of decoding that what you had written, reduced to $$\sqrt{1-(\frac{v}{c})^{2}}$$
(That's $$\sqrt{\gamma}$$ in my notation)

Maybe we were wrong, but what you've actually written, certainly looks like that.

arildno said:
Well, then write your original expression properly; both Zurtex and I found by our best efforts of decoding that what you had written, reduced to $$\sqrt{1-(\frac{v}{c})^{2}}$$
(That's $$\sqrt{\gamma}$$ in my notation)

Maybe we were wrong, but what you've actually written, certainly looks like that.
Actually I'm lazy and just copied this into mathematica, replaced the sqr with Sqrt[ and the appropriate ) with ] and got it to display it better and simplify it a little.

roflmfao

arildno said:
Well, then write your original expression properly; both Zurtex and I found by our best efforts of decoding that what you had written, reduced to $$\sqrt{1-(\frac{v}{c})^{2}}$$
(That's $$\sqrt{\gamma}$$ in my notation)

Maybe we were wrong, but what you've actually written, certainly looks like that.

Would you like me to mathematically prove to you it cannot be factored to
sqr(1- (v/c)^2 )

?

eNathan said:
Would you like me to mathematically prove to you it cannot be factored to
sqr(1- (v/c)^2 )

?
Would you like me to show you screen shots of me copying and pasting what you've put up into mathematica and it showing you the answer? Please be more clear and less rude to people trying to help you.

eNathan, your notation was at best "confusing", so i dont' think you ought to make posts like that. you didn't even clearly bracket yuour expression or try and set it in latex to make it easier. People have tried to decode it and two people have independently got the same thing, when it really ought to be your job to sort it out.

eNathan said:
Zurtex, it does not cancel down to sqr(1- v^2 / c^2)
huan.conchito, are you saying it factors to (sqr((1-v^2/c^2)^5 ?

And sorry but I am not to good at Latex. And I am not quite sure what gamma^2 is
im sure my answer is right, i did it the reverse too

Zurtex, don't get all emotional on me. I was simply asking you if you want me to show you the math. Yes, I can see why you got your answer, but it is not 100% the same as mines. So I will show you the math right here.
matt grime, I'm sorry if it was too confusing for you to understand. It is a little hard for me to.
understand.huan.conchito I was sure too.

Ok, here is the math, pay close attention and please tell me who is wrong, google or mathematica

With my long equation, when v = .6 and c = 1 the result is 0.8 You can clarify this at

Again, let v = 0.6 and let v = 1. With the factored form -- that is using the equation $sqr(1-v^2/c^2)$ -- the result is 0.8 everything is fine right :) Again, to clarify this go to

But, let's try this equation when v > c

With my equation, when v = 1.6 and c = 1, the result is -1.2489996 You can clarify this at http://www.google.com/search?hl=en&...^2)+/+sqr((1-+1.6+^2/+1.0+^2)^2))&btnG=Search

Again, let v = 1.6 and c = 1. With your equation $sqr(1-v^2/c^2)$ the result is 1.2489996 i to clarify this go to

Now, notice that my equation is capable of resulting in a negative number when v > c, while your "factored" version gives an imaginary number (positive), or in other cases you simply get a division by 0 error.

But with my form of the equation, there is no division by 0 error, and there are no imaginary numbers. You can get both a positive and negative number respectively, while your "factored" version cannot.

Do you see how my equation serves a different purpose than your's?

eNathan said:
Now, notice that my equation is capable of resulting in a negative number when v > c, while your "factored" version gives an imaginary number (positive), or in other cases you simply get a division by 0 error.

But with my form of the equation, there is no division by 0 error, and there are no imaginary numbers. You can get both a positive and negative number respectively, while your "factored" version cannot.

Do you see how my equation serves a different purpose than your's?

You are correct; their simplification is wrong, although this seems to have occurred because the brackets in your original equation were difficult to read.

However, your equation does not eliminate any divide by zero error. In fact, it introduces one. You probably haven't noticed that because google doesn't handle division by zero correctly, giving a result of zero and not undefined.

A good simplification would be:

$$\mathrm{sgn}(1-\tfrac{v^2}{c^2})\sqrt{\lvert 1-\tfrac{v^2}{c^2}\rvert}$$

First of all did you mean:

$$\sqrt{ \sqrt{ \left(1-\frac{v^2}{c^2}\right)^2 } } \frac{1-\frac{v^2}{c^2}}{\sqrt{\left(1-\frac{v^2}{c^2}\right)^2}}$$

Because if you hadn't noticed your way of expressing it is at best messy, also if you think you know how to express it better why have you bothered asking us?

Now, next really depends how you take the square root sign, however you seem to be taking it as a function rather than something that simply undoes the square sign (like we were taking it as), so carrying on:

$$\sqrt{\left|1-\frac{v^2}{c^2}\right|} \frac{1-\frac{v^2}{c^2}}{\left|1-\frac{v^2}{c^2}\right|}$$

$$\sqrt{\left| \frac{1}{1-\frac{v^2}{c^2}}\right|} \left( 1-\frac{v^2}{c^2} \right)$$

$$\sqrt{\left| \frac{c^2}{c^2-v^2}\right|} \left( 1-\frac{v^2}{c^2} \right)$$

$$\frac{|c|}{\sqrt{\left| c^2-v^2\right|}} \left( 1-\frac{v^2}{c^2} \right)$$

$$\frac{|c|}{\sqrt{\left| c^2-v^2\right|}} \left( \frac{c^2 - v^2}{c^2} \right)$$

$$\frac{1}{\sqrt{\left| c^2-v^2\right|}} \left( \frac{c^2 - v^2}{|c|} \right)$$

$$\frac{c^2 - v^2}{|c|\sqrt{\left| c^2-v^2\right|}}$$

$$\text{sign}\left(c^2 - v^2\right)\frac{ \sqrt{ \left| \left(c^2 - v^2\right)^2 \right| }}{|c|\sqrt{\left| c^2-v^2\right|}}$$

$$\text{sign}\left(c^2 - v^2\right) \frac{ \sqrt{ \left| c^2 - v^2 \right| }}{|c|}$$

Spot any mistake in my workings? It's the same as the answer above by the way.

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I still see no unconditional apology from enathan for the complete and utter mess he first posted.
Your disgustingly rude behaviour towards me, matt grime and Zurtex belongs nowhere.

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arildno said:
I still see no unconditional apology from enathan for the complete and utter mess he first posted.
Your disgustingly rude behaviour towards me, matt grime and Zurtex belongs nowhere.

I can't believe I have to quote myself

Zurtex, don't get all emotional on me. I was simply asking you if you want me to show you the math. Yes, I can see why you got your answer, but it is not 100% the same as mines. So I will show you the math right here.
matt grime, I'm sorry if it was too confusing for you to understand. It is a little hard for me to.
understand.huan.conchito I was sure too.

The first think I say to Zurtex is to "not get all emotional on me". This was a joke, but I said it because he though I was being rude to him, when I was merely asking him if he wants to see my results of the math. I tell matt grime that I am sorry if it was confusing. Now, if I am not mistaken "sorry" is somewhat of an apology. And I even tell him that it's a little hard for even me to read the equation. I know my typeset is horrible, and I am not too good at Latex. And can you please tell me where I was rude to you? The only answer you have me was "$$\sqrt{1-(\frac{v}{c})^{2}}$$" I'm sorry if I did not thank you for that? All I did was point out how your version's and my version are not identical. I was simply explaining something to you, and you think I was insulting your mathematica skills?

And master_coda, I will look into your simplified version later I am in school right now hehe

eNathan said:
Would you like me to mathematically prove to you it cannot be factored to
sqr(1- (v/c)^2 )

?
What the f**k do you think this is then?
And no, no one besides you here regards algebraic manipulations as "difficult" or "confusing".
It is trivial; being able to handle this is the most elementary skill a person with even marginal mathematical competence should master.
The sole confusion here is yours, and your inability to present a problem correctly.

PS:
Have you thought of street-sweeping as an alternative career path?
I think you would do well there.

arildno said:
What the f**k do you think this is then?
And no, no one besides you here regards algebraic manipulations as "difficult" or "confusing".
It is trivial; being able to handle this is the most elementary skill a person with even marginal mathematical competence should master.
The sole confusion here is yours, and your inability to present a problem correctly.

PS:
Have you thought of street-sweeping as an alternative career path?
I think you would do well there.

Ok, let me break down the English here. I asked Zurtex if he would like to mathematical show him why those two equations are not the exact same. He asked me if I would like a print screen. And yes, I was right (although, you can't really blame Zurtex for it). Their is no need to say "F**K" here, we are debating math not cursing at each other. You seem to be very mad at me for some reason, and I hope you don't cause this thread to be closed. I am using the equation in a programming language, so I presented it in the form which I would use it.

PS.
street-sweeping
I plan to go to college and be a carpenter of a physicists. One pays more, and one I like doing more.
So stop insulting people here. You are the only one doing this, not me. I think you cannot accept the fact that I did show everyone why those two equations are not identical in post #13. So please just stop running your mouth about how I cause confusion and I insult everybody.

And Zurtex I am going over your math...

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Actually like I pointed out before, it's a matter of how you treat the square root sign, when I see:

$$\sqrt{1 - \cos^2 x} = \sin x$$

I see nothing wrong with it, however if you are using the square root as a function and not something that simply undoes squaring then it should be:

$$\sqrt{1 - \cos^2 x} = |\sin x|$$

Guys, please don't get edgy, eNathan I'm sorry but you didn't really come off as polite to people helping you, saying "Would you like me to mathematically prove to you it cannot be factored to..." just comes off as rude. You don't offer any of your own workings and people are just trying to help you.

Well, I am sorry then. Do you want me to include the words "sorry" and "please" in every statement I say? And yes, I am appreciative for the mathematical help.

Arildno,i think you're overreacting,too.And taking advantage that there isn't any moderator interested in abruptly ending this senseless debate...

Daniel.

Ok, without any further desputes (im assuming) is there any way to factor
$sqr(sqr((1-v^2/c^2)^2)) * ((1-v^2/c^2) / sqr((1-v^2/c^2)^2))$ and get the exact same results? Yes, my equation does use Sqr() as an undo function, and does the same with ^2 squaring numbers. So in shorts, sqr(1- (v/c)^2) will not work. But nice try. Does anybody have a solution?

Yes,use the function \frac{blah,blah,blah} for writing it properly.

Daniel.

$\sqr(\sqr((1- \frac {v^2, c^2} )^2)) * \frac{((1- \frac{v^2, c^2)}, \sqr((1-\frac{v^2, c^2})^2))$

I am not good at latex so I hope that turned out right :P

eNathan said:
Ok, without any further desputes (im assuming) is there any way to factor
$sqr(sqr((1-v^2/c^2)^2)) * ((1-v^2/c^2) / sqr((1-v^2/c^2)^2))$ and get the exact same results? Yes, my equation does use Sqr() as an undo function, and does the same with ^2 squaring numbers. So in shorts, sqr(1- (v/c)^2) will not work. But nice try. Does anybody have a solution?
If the square root is an undo function then the answer is:

$$\sqrt{1- \left(\frac{v}{c} \right)^2}$$

And please stop saying factor! We are simplifying not factoring.

Google does not treat the square root as this.

eNathan please click on the below code to see how it is done:

$$\sqrt{ \sqrt{ \left(1-\frac{v^2}{c^2}\right)^2 } } \frac{1-\frac{v^2}{c^2}}{\sqrt{\left(1-\frac{v^2}{c^2}\right)^2}}$$

If you want to learn LaTeX, highly suggestable on this forum go here and click on the LaTeX and the code will pop up:

eNathan said:
$$\sqr{\sqr{1- \frac{v^2}{c^2} )^2)) * \frac{((1- \frac{v^2, c^2)}, \sqr((1-\frac{v^2, c^2})^2))[/itex] I am not good at latex so I hope that turned out right :P It didn't.The functions are \sqrt{...} and \frac{numerator}{denominator}... Daniel. Ok look http://www.google.com/search?hl=en&q=sqr(+1-+(1.5/1)^2 So you are saying that the "i" means negative? And I think if I try to run those numbers in code, it will give me a "division by zero error" for some reason... The problem is really that a negative times a negatives is a positive. I think it would make more mathematical sense to make the roots of negative numbers negative, which is what I was also thought I am my alg class. When does making it an imaginary positive come in handy? Or express somethign that actually exists? Ok, here is a simpler version of it. To let everybody know, I used the sqr() and ^2 so many times because I wanted to get the positive form of the number. |x| is the same as sqr(x^2) So here it is let [tex]x =1 - \frac{v^2} {c^2}$$

$$\sqrt{|x|} * \frac{x} {|x|}$$

phew, I am finally getting this Latex code thingy

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So u can simplify the sqrt with the denominator and be left with a product.

Daniel.

Ok, with all this said, is ther any way to simplify

$$\sqrt{|x|} * \frac{x} {|x|}$$

Or is that the simplest it can get?

$$|x|=\left(\sqrt{|x|}\right)^{2}$$

So,unless that "x" is 0,u could simplify.

Danie.

dextercioby said:
$$|x|=\left(\sqrt{|x|}\right)^{2}$$

So,unless that "x" is 0,u could simplify.

Danie.

From what it looks like, you just said that $$abs(x)$$ = $$sqr( abs(x) )^2$$

How is that a simplfication of the equation? I am not being mean, I am just woundering ...

he didn't say that was a simplification he said you cuold use it to simplify.

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