Is There a Mistake in These Differential Equation Notes?

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Discussion Overview

The discussion revolves around a potential mistake in the manipulation of a differential equation, specifically the transformation of the expression \(\frac{x-y}{2x} = \frac{dy}{dx}\) into different forms. Participants are examining the algebraic steps involved and questioning the validity of certain transformations and simplifications.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions whether the transformation from \(\frac{x-y}{2x}\) to \(\frac{1}{2}(1-\frac{y}{x})\) is valid, suggesting an alternative form of \(\frac{1}{2}(x-\frac{y}{x})\).
  • Another participant provides a simplification of the expression, asserting that \(\frac{x-y}{2x}\) can be expressed as \(\frac{1}{2}(1-\frac{y}{x})\) through algebraic manipulation.
  • Concerns are raised about the introduction of unnecessary factors during simplification, with some arguing that it complicates understanding.
  • Participants discuss the educational background of the original poster, suggesting that they may understand procedural rules without grasping the underlying concepts.
  • Humorous anecdotes about early learning experiences in mathematics are shared, illustrating the varied backgrounds of participants.

Areas of Agreement / Disagreement

There is no consensus on the correctness of the transformations discussed. Participants express differing views on the validity of certain algebraic steps and the necessity of introducing factors during simplification.

Contextual Notes

Some participants highlight the potential confusion caused by introducing extra factors in algebraic manipulations, indicating that this may lead to misunderstandings in the application of mathematical rules.

snowJT
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I was copying notes in class down, now that class just ended, I was reading over the notes, and I saw this line that went from...

THIS:
\frac{x-y}{2x} = \frac{dy}{dx}

To THIS:
\frac{1}{2}(1-\frac{y}{x}) = \frac{dy}{dx}

But I'm wondering... wouldn't It be this?

\frac{1}{2}(x-\frac{y}{x}) = \frac{dy}{dx}

if not.. how come?
 
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Do you think that:
\frac{7-2}{7}=7-\frac{2}{7}??
 
\frac{x-y}{2x} = \frac{1}{2}\left(\frac{x}{x} - \frac{y}{x}\right) = \frac{1}{2}(1-\frac{y}{x})
 
thanks, I see it like that; however, when you try and multiply it back...

\frac{1}{2}(1-\frac{y}{x})= \frac{1}{2}-\frac{y}{2x}

no?
 
That is correct. So?
 
Yes, but you can't multiply it back to get \frac{x}{x} to get \frac{x-y}{2x}
 
What...?
 
well...

I don't think this makes sense... because how can you get x back when you go from this

\frac{1}{2}(1-\frac{y}{x})

to this:

\frac{1}{2}\left(\frac{x}{x} - \frac{y}{x}\right) <-- x over x appears from 1 over 1 which doesn't make sense
 
Last edited:
So you're saying that 2 divided by 2 is not 1?
 
  • #10
no, I'm saying you can't get x when you go backwards
 
  • #11
snowJT said:
no, I'm saying you can't get x when you go backwards

I don't know what your mean. This can be simplified as \frac{1}{2}(1-\frac{y}{x})=\frac{1}{2}-\frac{y}{2x}=\frac{2x-2y}{4x}=\frac{x-y}{2x}

What's wrong with that?
 
  • #12
thanks... that cleared it up well, I see it thanks... there were some other people here beside me at school who wern't getting it too, but now they do too, lol
 
  • #13
@cristo: what's wrong is that you needlessly introduce extra factors of 2 for no reason.
@snowJT: you were taught in elementary school or primary school, or some time before the age of 11 to put things over a common denominator, i.e. that a/b + c/d = (ad+bc)/bd, so use it, but sensibly so that you don't introduce unnecessary factors that are going to cancel out later. You are just adding together fractions. That is something you have been doing for years.
 
  • #14
matt grime said:
@cristo: what's wrong is that you needlessly introduce extra factors of 2 for no reason.

How is that wrong? Clearly I know that the factor of two was included unnecessarily, but since the OP could not see the answer, I expanded out the fraction fully, then used the rule that he would be familiar with, namely that a/b+c/d=(ad+bc)/bd.

I understand your point, but sometimes, when someone doesn't see what is going on, it is *simpler* to follow rules that they will know, and then simplify later. Skipping steps, in this case, may have added confusion.
 
  • #15
"Putting things over a common denominator" should be well known from the age of *insert age when you learn what a fraction is*, and knowing that that if you have denominators 2x and 2 then you need only 2x shuold be clear if you explain that is all you're doing. Of course, multipling the 1/2 in is the real culprit here.
 
  • #16
haha I know this is kinda random, but i remember when I was like 3 and my sister said she was learning algebra at school, and asked me what a+a was, i was convinecd it was b :D
 
  • #17
it's not b
 
  • #18
cristo said:
How is that wrong? Clearly I know that the factor of two was included unnecessarily, but since the OP could not see the answer, I expanded out the fraction fully, then used the rule that he would be familiar with, namely that a/b+c/d=(ad+bc)/bd.

I found it interesting that the OP could understand the algebra when cristo included the redundant factor but could not understand it previously.

It seems to be a case of OP not being able to grasp that 1/2 = x/2x but still being able to use the "method of cross multiplication" to add fractions. I think this implies that he was taught how to do things but not why those procedures work. Sad.
 
  • #19
Gib Z said:
haha I know this is kinda random, but i remember when I was like 3 and my sister said she was learning algebra at school, and asked me what a+a was, i was convinecd it was b :D

Haha that's very funny, everyone knows that it's really a + 1 that equals b. :-p
 
  • #20
uart said:
Haha that's very funny, everyone knows that it's really a + 1 that equals b. :-p

That's actually true if you're talkling in ASCII. :biggrin:
 

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