How do I know when to factor out a variable?

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  • #1
billllib
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Homework Statement:
How do I move v to one side of the equation from gamma?
Relevant Equations:
gamma = \frac 1 {\sqrt {1 - \frac {v^2} {c^2}}}
Take the equation for gamma. I know how to get from step 1 to gamma. How would I do that if I have not have seen the instructions? There are many ways to factor step 1 without getting the same values of

## gamma = \frac 1 {\sqrt {1 - \frac {v^2} {c^2}}} ##

step 1 =

Speaking of factoring and gamma How do I move v to one side of the equation from gamma?
 

Answers and Replies

  • #2
haruspex
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step 1 =

I don’t see any connection between your question and that image.
 
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  • #3
billllib
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Step 1 is how I derive gamma. Its the information I start with.
 
  • #4
willem2
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If you derive equations for distance, time or speed in a moving frame, gamma appears in all of them.
Can you tell us exactly what was the equation you derived?
 
  • #5
Step 1 is how I derive gamma. Its the information I start with.

It looks from your working like you're doing the light clock derivation, so I assume ##D## is the hypotenuse and ##L## is the vertical distance between the mirrors. How can you write ##x## in terms of ##\Delta t_{B}##? Do you have a clear diagram to work from?
 
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  • #6
billllib
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I am not so much as asking what the steps are. I am more curious the history of gamma. How do you know where to move the variables when factoring in math? Is there a pattern when moving variables and factoring? Because there seems to be many different ways to move variables and factor. I know the rule that if you do one thing to one side you do it to the other side. But there are many ways to move variables and factor and you can get very different results even when trying to get a certain a variables.

The second part of the question I am having trouble. I have all the variables in gamma except v and solving for v. I guess I could also say they are many different ways to move variables and factor for the gamma equation. How do you know the method without being shown a pattern? Is there a pattern to factoring and moving variables?

If it is not clear what I am asking I could try to explain again.
 
  • #7
haruspex
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I am not so much as asking what the steps are. I am more curious the history of gamma. How do you know where to move the variables when factoring in math? Is there a pattern when moving variables and factoring? Because there seems to be many different ways to move variables and factor. I know the rule that if you do one thing to one side you do it to the other side. But there are many ways to move variables and factor and you can get very different results even when trying to get a certain a variables.

The second part of the question I am having trouble. I have all the variables in gamma except v and solving for v. I guess I could also say they are many different ways to move variables and factor for the gamma equation. How do you know the method without being shown a pattern? Is there a pattern to factoring and moving variables?

If it is not clear what I am asking I could try to explain again.
Well, it still isn't clear to me.
You write that you know how to get from step 1 to gamma, but you have not posted those steps. So I see no example of the "moving variables and factoring" that you ask about.
 
  • #8
billllib
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Okay let's say we already derived gamma.
I have all the variables except v. Each step I take is a branch on the tree. The original is the base a circle in the middle. and the goal is to get v to one side. Of course there are different branches then the branches listed below.

Origin
##gamma = \frac 1 {\sqrt {1 - \frac {v^2} {c^2}}}##
branch 1

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

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

etc...



branch 2
## gamma {\sqrt \frac {v^2} {c^2} } = \frac {\sqrt { \frac {v^2} {c^2}}} {\sqrt {1 - \frac {v^2} {c^2}}}##
## gamma {\sqrt \frac {v^2} {c^2} } = \frac 1 {\sqrt {1} }##


etc...

How do I know which branch to take? Different branches give different results then v = ... . So is there a pattern or a method I can follow or just trial and error?

Also I know how to get from step 1 to gamma it is my notes I just don't feel like writing it. What makes you think I don't? What I don't know is how to get v to one side.

Hence the questions.

Thanks for the help.
 
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  • #9
haruspex
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Okay let's say we already derived gamma.
I have all the variables except v. Each step I take is a branch on the tree. The original is the base a circle in the middle. and the goal is to get v to one side. Of course there are different branches then the branches listed below.

Origin
##gamma = \frac 1 {\sqrt {1 - \frac {v^2} {c^2}}}##
branch 1

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

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

etc...



branch 2
## gamma {\sqrt \frac {v^2} {c^2} } = \frac {\sqrt { \frac {v^2} {c^2}}} {\sqrt {1 - \frac {v^2} {c^2}}}##
## gamma {\sqrt \frac {v^2} {c^2} } = \frac 1 {\sqrt {1} }##


etc...

How do I know which branch to take? Different branches give different results then v = ... . So is there a pattern or a method I can follow or just trial and error?

Also I know how to get from step 1 to gamma it is my notes I just don't feel like writing it. What makes you think I don't? What I don't know is how to get v to one side.

Hence the questions.

Thanks for the help.
Neither of your branches makes algebraic sense, so I wouldn't pick either.
Branch 1 goes wrong at the first step, branch 2 at the second.

You have ##\gamma=\frac 1{\sqrt{1-\frac{v^2}{c^2}}}##. Think through the steps that equation tells you to take to calculate γ from v, then undo them in reverse order.
So first is inversion:
##\frac 1{\gamma}=\sqrt{1-\frac{v^2}{c^2}}##
Next is that square root:
##\frac 1{\gamma^2}=1-\frac{v^2}{c^2}##
Now the subtraction from 1:
##1-\frac 1{\gamma^2}=\frac{v^2}{c^2}##
Now the division by c2:
##c^2(1-\frac 1{\gamma^2})=v^2##
And finally undo the squaring of v.

The last two steps can be done in the other order if you think of ##\frac{v^2}{c^2}## as ##(\frac vc)^2##.
 
  • #10
billllib
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I did make a mistake in branch 1 it should be just c not c^2. Why does branch 2 not make sense besides the forgotten minus sign. How do you know which branch to take? Is it just trial and error? Let's call your math branch 3. First step, how do you get 1/y? I get y/1.
 
  • #11
haruspex
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I did make a mistake in branch 1 it should be just c not c^2
No, still wrong. You are now saying:
##\gamma = \frac 1 {\sqrt {1 - \frac {v^2} {c^2}}}##
branch 1
##\gamma (c) = \frac 1 {\sqrt {1 - \frac {(v^2) (c^2)} {(c^2)} }}##
Let's take that in smaller steps:
##\gamma = \frac 1 {\sqrt {1 - \frac {v^2} {c^2}}}##
## = \frac c {c\sqrt {1 - \frac {v^2} {c^2}}}##
## = \frac c {\sqrt {c^2(1 - \frac {v^2} {c^2})}}##
## = \frac c {\sqrt {c^2 - {v^2} }}##
##\frac{\gamma}c= \frac 1{\sqrt {c^2 - {v^2} }}##
branch 2
## \gamma {\sqrt \frac {v^2} {c^2} } = \frac {\sqrt { \frac {v^2} {c^2}}} {\sqrt {1 - \frac {v^2} {c^2}}}##
## \gamma {\sqrt \frac {v^2} {c^2} } = \frac 1 {\sqrt {1} }##
I am baffled as to how you think that step is correct. You are in effect saying ##\sqrt {1 - \frac {v^2} {c^2}}=\sqrt {\frac {v^2} {c^2}}##.

I am forced to conclude you need to revise basic algebraic principles.
 
  • #12
billllib
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## = \frac c {c\sqrt {1 - \frac {v^2} {c^2} } }## Also I don't understand this example.

I think it should be

## = \frac 1{c\sqrt {1 - \frac {v^2} {c^2} } }##

Okay a quick test is to reverse the process. Do you know where I can get the rules of algebra as a quick refresher? The more basic rules of algebra I know but the removal of the square root switching sides is confusing me.
 
  • #13
haruspex
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## = \frac c {c\sqrt {1 - \frac {v^2} {c^2} } }## Also I don't understand this example. I think it should be
## = \frac 1{c\sqrt {1 - \frac {v^2} {c^2} } }##
My step was ## = \frac 1 {\sqrt {1 - \frac {v^2} {c^2} } } = \frac c {c\sqrt {1 - \frac {v^2} {c^2} } }##
Given a fraction ##\frac ab## you can multiply top and bottom by the same thing:
##\frac ab =\frac {ac}{bc}##
That's effectively rule 3 at https://algebrarules.com/.
Next, I took the c at the bottom inside the square root. I can break that into two steps:
##c=\sqrt{c^2}##
##\sqrt x\sqrt y=\sqrt{xy}##
Plugging in c2 for x and ## {1 - \frac {v^2} {c^2} }## for y gives:
## c\sqrt {1 - \frac {v^2} {c^2} } = \sqrt {c^2 (1-\frac {v^2} {c^2}) }= \sqrt {c^2 - c^2\frac {v^2} {c^2} }= \sqrt {c^2 - {v^2} }##
 
  • #14
billllib
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## \frac 1 { \frac {1} {c^2} } { gamma} =\frac 1 {\sqrt {1 - \frac {1} {v^2} }} ## Would this work according to the rules of algebra?


Sorry for the many questions.
 
  • #15
haruspex
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## \frac 1 { \frac {1} {c^2} } { gamma} =\frac 1 {\sqrt {1 - \frac {1} {v^2} }} ## Would this work according to the rules of algebra?
Certainly not.
It's no good making stabs in the dark like this. To manipulate algebra every step needs to be justified by a standard rule. What rule do you think would justify that?
 
  • #16
billllib
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I justify this by moving ##c## up from the 3rd fraction to the second fraction .

## gamma = { \frac 1 {c} } { \frac 1 {\sqrt {1 - \frac {1} {v^2} }} }##


Then I put the c into the square root to get the regular gamma. Or would that get me 2 c^2? inside the square root?
I think it would.
Which rule am I breaking now when moving ##c^2## to the left of the equal sign?
 
  • #17
haruspex
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I justify this by moving ##c## up from the 3rd fraction to the second fraction .

## gamma = { \frac 1 {c} } { \frac 1 {\sqrt {1 - \frac {1} {v^2} }} }##


Then I put the c into the square root to get the regular gamma. Or would that get me 2 c^2? inside the square root?
I think it would.
Which rule am I breaking now?
I thought your starting point was ##\gamma=\frac 1{\sqrt{1-\frac{v^2}{c^2}}}##
How do you get from there to
## \gamma = { \frac 1 {c} } { \frac 1 {\sqrt {1 - \frac {1} {v^2} }} }##?
 
  • #18
billllib
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I put the ##c^2## where the ##v^2## is, I am referencing the original equation. Then I put ## \frac 1 { \frac {1} {c^2} } = ## because the ##c^2## was on the bottom with ##v^2##.
 
  • #20
billllib
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Because of the rule ## \frac a a,## cancels out. Or in this case ##c^2##instead of ##a##
 
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  • #21
haruspex
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Because of the rule ## \frac a a,## cancels out. Or in this case ##c^2##instead of ##a##
But you do not have ##\frac{c^2}{c^2}## anywhere.
In post #18 you described two steps. Please post the first step, i.e. what you had before the step and what you have after.
 
  • #22
billllib
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step 1
## \gamma=\frac 1{\sqrt{1-\frac{v^2}{c^2}}} ##

step 2 which I left out

## \gamma=\frac 1{\sqrt{1-\frac { {v^2} {c^2} } {c^2} }} ##


step 3
original This is incorrect
## \gamma = { \frac 1 {c} } { \frac 1 {\sqrt {1 - \frac {v^2} {1}}} }##

what I should have wrote
##
\frac 1 { \frac {1} {c^2} } { gamma} = { \frac 1 {\sqrt {1 - \frac {v^2} {1} }} }##

what step 4 would have been if I followed the origin but should be ignored due to the screw up
## \frac 1 { \frac {1} {c^2} } { gamma} = { \frac 1 {\sqrt {1 - \frac {v^2} {1} }} }##
 
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  • #23
haruspex
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step 1
## \gamma=\frac 1{\sqrt{1-\frac{v^2}{c^2}}} ##
step 2 which I left out
## \gamma=\frac 1{\sqrt{1-\frac { {v^2} {c^2} } {c^2} }} ##
No, you can't do that. Why do you think you can just put an extra c2 in there? That clearly makes it different in value.
Is ##\frac 49## equal to ##\frac{4\times 9}9##?
 
  • #24
billllib
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I think because of the rule ##c^2/c^2 ## to elimate the ## { \sqrt c^2}## and move it to the other side of the equal sign.
 
  • #25
haruspex
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I think because of the rule ##c^2/c^2 ## to elimate the ## { \sqrt c^2}## and move it to the other side of the equal sign.
##c^2/c^2 ## is not a rule. If you mean the rule that you can multiply numerator and denominator by the same then that would give you
## \frac 1{\sqrt{1-\frac{v^2}{c^2}}} = \frac 1{\sqrt{1-\frac { {v^2} {c^2} } {c^2c^2} }} ##
## = \frac 1{\sqrt{1-\frac { {v^2} {c^2} } {c^4} }} ##
Note the fourth power in the denominator.
 
  • #26
billllib
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## c = \frac a b## I was taught to go ## ca = \frac a {ba}##

then

##ca = \frac 1 b ##

I could be remembering wrong. Did I go wrong somewhere?

I am just applying this rule to gamma.

Thanks for the help.
 
  • #27
haruspex
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## c = \frac a b## I was taught to go ## ca = \frac a {ba}##
then
##ca = \frac 1 b ##
I could be remembering wrong. Did I go wrong somewhere?
Sorry, but that is all wrong.
To make it clear, I am going to break it down into the tiniest steps and describe why each is valid.

It is always valid to do the same operation to both sides of the equation.
If ## c = \frac a b## then you can multiply both sides of the equation by b to get
## bc =b \times\frac a b##.
The RHS there can be written instead as ##b\times(a\div b)##, and dividing by something is the same as multiplying by its inverse:
##bc=b\times(a\times\frac 1 b)##.
Multiplication is "commutative", which means ##x\times y=y\times x##, so ##bc=b\times(\frac 1 b\times a)##.
It is also "associative", which means ##x\times(y\times z)=(x\times y)\times z##, so
##bc=(b\times\frac 1 b)\times a##.
Of course, ##x\times \frac 1x=1##, so:
##bc=1\times a=a##.

Now, you don't want to have to go through that every time, so you need to get used to going directly from ## c = \frac a b## to ##bc=a##.
This is analogous to going from ##c=a-b## to ##c+b=a##. In both cases, where b moves across to the other side of the equals sign it does the opposite of what it did before. If it subtracted it now adds; if it divided it now multiplies.

Going back to what you posted, ## ca = \frac a {ba}##, what you have done there is to multiply the LHS by a but divide the RHS by a. Do you see that?
 
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  • #28
billllib
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But even if I try to go ## c = \frac a b## then ## \frac c a = \frac a {a b} ## then ## \frac c a = b ## then ## \frac {ca} a = a b ## then

## c = ab## So how do I move ##"a"## to the other side? Would it be ## cb-a = 1 ?##

This is assuming I didn't make any errors. I could have easily made a careless mistake.

This is correct?
 
  • #29
SammyS
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But even if I try to go ## c = \frac a b## then ## \frac c a = \frac a {a b} ## then ## \frac c a = b ## then ## \frac {ca} a = a b ## then

## c = ab## So how do I move ##"a"## to the other side? Would it be ## cb-a = 1 ?##

This is assuming I didn't make any errors. I could have easily made a careless mistake.

This is correct?
Not correct.

First error:

You have done the following correctly.
But even if I try to go ## c = \frac a b## then ## \frac c a = \frac a {a b} ##

But following this you should have:
then ##\frac c a=\frac 1 b ##

and the rest also has errors.

Added in Edit: Let's address the rest of that post.
So how do I move "a" to the other side? Would it be ##cb−a=1##?
There seems to be no way for that to follow from your equation, ##c a=b ## .

Let's suppose that you intended it to follow from ## c = \frac a b ## .

Multiplying Both Sides of the equation by ##b## and simplifying gives, ##c b =a ##. It looks like I moved the ##b## from the denominator on the right to be on the main line on the left. Indeed, I didn't move anything. It just looks that way.

To get rid of the ##a## on the right, you can do one of two things. Either subtract a from Both Sides, or else divide Both Sides by ##a## (same as multiplying Both Sides by ##\frac 1 a ##). Subtracting gives you a zero on the right side, whereas dividing gives you ##1## on the right side,

This whole idea of moving quantities from one part of an equation to another is a bad idea in general. None of the replies in this thread give algebraic rules for moving variables. Only once does @haruspex even mention the word "move". In Post #27, he does give the mathematical explanation for a variable disappearing from one side but appearing on the other. More importantly, he points out a number of other mathematical properties to use in solving an equation for a specific variable.

I intentionally made this be an addition to an existing post rather than burying @haruspex's latest post (#32). Seems it took you a long time to notice, or at least respond to his Post#9. Study that post and Post #27.

Try to answer your own question before asking us.

What's the justification for the next step in Post #9 ? You tell us.
 
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  • #30
haruspex
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## \frac c a = \frac a {a b} ## then ## \frac c a = b ##
By what rule?
So how do I move ##"a"## to the other side? Would it be ## cb-a = 1 ?##
Again, by what rule.
Stop making wild guesses and justify every step you make by appeal to a rule. If you will not make the effort to do that then we will make no progress.
 
  • #31
billllib
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In post #9 ,the first step, what algebra rule is that?
 
  • #32
haruspex
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In post #9 ,the first step, what algebra rule is that?
You mean this one?
##\gamma=\frac 1{\sqrt{1-\frac{v^2}{c^2}}}##
##\frac 1{\gamma}=\sqrt{1-\frac{v^2}{c^2}}##
The rule there is "do the same to both sides". In this case, the thing I did to both sides was to invert, i.e. turned a fraction like ##\frac xy## into ##\frac yx##.
 

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