Fraction calculation, can't find mistake

[late347]

Homework Statement

I did some homework and wolfram alpha afterwards gave different result than what I have. Apparently I must have done some error, but I couldn't find error after I doublechecked about two times. So I decided to see if you guys can find an error. And what is the correct answer also. I was pulling my hair out after doing this problem for some time $$ \frac{\frac{1}{s}-s}{\frac{1}{s}-\frac{1}{s-1}} ~~~~~~ $$

Homework Equations

The Attempt at a Solution

here we expanded
$(\frac{1}{s}-\frac{s^2}{s})/(\frac{s-1}{s(s-1)}-\frac{s}{s(s-1)})$here I tried to get the plusses and minuses right, but I'm not sure I succeeded
$[\frac{1}{s}+\frac{-s^2}{s}] / [\frac{s-1}{s(s-1)}+\frac{-s}{s(s-1)}]$

$(\frac{1-s^2}{s}) / [\frac{(s-1)-s}{s(s-1)}]$

$[ \frac{1-s^2}{s}] ~~*~~ [\frac{s(s-1)}{-1} ]$$\frac{(s-1)(s+1)}{s} * \frac{s(s-1}{-1}$$\frac{(s-1)^2(s+1)(s)}{-s}$

$(-1)(s-1)^2(s+1)$

 

[fresh_42]

What is $(1-s^2)$ factored?

 

[SammyS]

Homework Statement

I did some homework and wolfram alpha afterwards gave different result than what I have. Apparently I must have done some error, but I couldn't find error after I doublechecked about two times. So I decided to see if you guys can find an error. And what is the correct answer also. I was pulling my hair out after doing this problem for some time $$ \frac{\frac{1}{s}-s}{\frac{1}{s}-\frac{1}{s-1}} ~~~~~~ $$

Homework Equations

The Attempt at a Solution

here we expanded
$(\frac{1}{s}-\frac{s^2}{s})/(\frac{s-1}{s(s-1)}-\frac{s}{s(s-1)})$here I tried to get the plusses and minuses right, but I'm not sure I succeeded
$[\frac{1}{s}+\frac{-s^2}{s}] / [\frac{s-1}{s(s-1)}+\frac{-s}{s(s-1)}]$

$(\frac{1-s^2}{s}) / [\frac{(s-1)-s}{s(s-1)}]$

$[ \frac{1-s^2}{s}] ~~*~~ [\frac{s(s-1)}{-1} ]$$\frac{(s-1)(s+1)}{s} * \frac{s(s-1}{-1}$$\frac{(s-1)^2(s+1)(s)}{-s}$

$(-1)(s-1)^2(s+1)$

Looks OK.

What's the problem you're having?

 

[late347]

basically $(1-s^2) = (1^2-s^2)$

so that is (1+s) (1-s)

= 1-s +s -s^2
= 1- s^2

 

[fresh_42]

basically $(1-s^2) = (1^2-s^2)$

so that is (1+s) (1-s)

= 1-s +s -s^2
= 1- s^2

Yes.

 

[late347]

Yes.

can you bold print my mistake in quote for example

 

[fresh_42]

You wrote
$[ \frac{1-s^2}{s}] ~~*~~ [\frac{s(s-1)}{-1} ] = \frac{(s-1)(s+1)}{s} * \frac{s(s-1}{-1}$
But $(s-1)(s+1) \neq (1-s)(1+s)$

 

[late347]

I think I got it right this time.

(s-1)(s+1)

= s^2 +s -s -1

$=~~ (+s^2~~-1)$

the correct part said that the factors were as follows

$(1-s) (1+s) = 1 +s -s -s^2$
$= -s^2 +1$

the latter case is the "counter number" to the other number (multiplied by minus one)

 

[SammyS]

Looks OK.

What's the problem you're having?

Oh!

not quite OK !

sign mistake

 

[fresh_42]

I think I got it right this time.

(s-1)(s+1)

= s^2 +s -s -1

$=~~ (+s^2~~-1)$

The other way around. Until $1-s^2$ everything was fine, and by the way, afterwards, too. You simply changed the sign by substituting $1-s^2 = (s-1)(s+1)$ which is wrong by $-1$. You should have used $1-s^2=(1-s)(1+s)$ or $1-s^2=(-1)(s-1)(s+1)$

 

[late347]

Oh!

not quite OK !

sign mistake

looks like the tricky math prob got under your skin at first also ! xD

 

[fresh_42]

looks like the tricky math prob got under your skin at first also ! xD

Oh!

not quite OK !

sign mistake

By the way. Me, too. I only found out as I calculated it on my own. At first sight I assumed a false input in Wolfram.

 

[late347]

ok i will try to post the new answer in full, but it will be tedious because of Latex code

 

[late347]

$\frac{1/s-s}{1/s-[1/(s-1)]}$$\frac{(1-s^2)/s}{(-1)/[s(s-1)]}$$\frac{(1-s)(1+s)}{s}*\frac{s(s-1)}{-1}$$\frac{(1-s)(1+s)(s)(s-1)}{(-1)(s)}$$\frac{(-1)(s-1)(s)(s-1)(s+1)}{-1s}$

$\frac{(s-1)^2(s)(s+1)}{s}$

$(s-1)^2(s+1)$

 

[SammyS]

What does Wolfram give?

Did I somehow miss that in one of these posts?

 

[late347]

What does Wolfram give?

Did I somehow miss that in one of these posts?

I think I did a wrong factorization for the term $(1^2 -s^2)$evidently wolfram alpha did not become fooled by that.

this skewed my result from early on. I tried to use the memory formula for the $(a^2-b^2)$ for finding factors

 

[fresh_42]

What does Wolfram give?

Did I somehow miss that in one of these posts?

No. The OP hasn't said and I haven't done. Far too many parenthesis to type in.

 

[late347]

code for wolfram alpha&& inside these &&

&& (1/s-s)/(1/s-1/(s-1)) &&

 

[late347]

looks like I did the problem correctly as post # 14 will show the same alternative form result as wolfram alpha.

$(1-s)(1-s^2)$

or

$(s-1)^2(s+1)$

 

[fresh_42]

code for wolfram alpha&& inside these &&

&& (1/s-s)/(1/s-1/(s-1)) &&

I wouldn't have relied on this ambiguous inline coding. I prefer to use parenthesis to avoid false interpretations.
Concerning the sign error all of us blundered into: it happens. Nothing to worry about.

 

[late347]

I think modern wolfram alpha shows the input in variable forms like fraction line form of the division as a picture

 

[late347]

I wouldn't have relied on this ambiguous inline coding. I prefer to use parenthesis to avoid false interpretations.
Concerning the sign error all of us blundered into: it happens. Nothing to worry about.

did you get the same result as post 14 from me

 

[fresh_42]

did you get the same result as post 14 from me

Yes.

I think modern wolfram alpha shows the input in variable forms like fraction line form of the division as a picture

This way one can see, whether the input was correct or not. However, one shouldn't get used to it in the sense that "everybody knows what is meant", because it is a total unnecessary source of potential mistakes.

We often read things like "e^t^2/2" where it is not clear without additional knowledge that $e^{\frac{1}{2}t^2}$ is meant and not $\frac{1}{2}e^{t^2}$ or $e^t \,$.

 

[SammyS]

What I get was s³ - s² - s +1

 

[late347]

What I get was s³ - s² - s +1

¨now we have competition between the answers it seems

looks like that comes out at the same factorization so it looks like what I got,