unimatix
  • unimatix
For delta x infinitesimal find an infinitesimal (epsilon) such that delta y = dy + epsilon delta x 2sqrt(x)
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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unimatix
  • unimatix
|dw:1441743136199:dw|
unimatix
  • unimatix
|dw:1441743200586:dw|
anonymous
  • anonymous
Can you please use the equation button for this I do not understand the writing.

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unimatix
  • unimatix
sure
Jhannybean
  • Jhannybean
Looks like the definition of a derivative
unimatix
  • unimatix
\[\Delta y = dy + \epsilon \Delta x\]
unimatix
  • unimatix
And I think I've found dy and delta y. Which are \[\Delta y = \frac{ 2 \Delta x }{ \sqrt{x+ \Delta x} + \sqrt{x} }\] \[dy = \frac{ \Delta x }{ \sqrt{x} }\]
Jhannybean
  • Jhannybean
Ahh... this relates to the increment theorem... vaguely remember learning about this x_x
unimatix
  • unimatix
Yep, that sounds right
Jhannybean
  • Jhannybean
Im just giving a shot at this,...
Jhannybean
  • Jhannybean
\(y=2\sqrt{x}\) right?
unimatix
  • unimatix
Yep!
Jhannybean
  • Jhannybean
whew, ok.
Jhannybean
  • Jhannybean
we're ultimately trying to put it in the form of \[\frac{\Delta y}{\Delta x} = f'(x) + \epsilon\] right?...
unimatix
  • unimatix
\[\Delta y = dy+ \epsilon \Delta x\]
unimatix
  • unimatix
I've just got to find epsilon.
unimatix
  • unimatix
\[\Delta y = \frac{ 1 }{ 2\sqrt{x} } \Delta x + \epsilon \Delta x\]
unimatix
  • unimatix
\[\frac{ \Delta y }{ \Delta x } = \frac{ 1 }{ 2\sqrt{x}} + \epsilon \]
unimatix
  • unimatix
\[\epsilon = \frac{ \Delta y }{ \Delta x } - \frac{ 1 }{ 2\sqrt{x} }\]
unimatix
  • unimatix
Maybe substitute back in the Delta y from earlier.
Jhannybean
  • Jhannybean
Oh instead of finding "\(\epsilon\)" can't we take the limit? Just throwing that out there, i might be wrong
anonymous
  • anonymous
ok for y = dy + epsilon delta x what does this even mean :O
unimatix
  • unimatix
I'm sure there is a perfectly good explanation for that...
beginnersmind
  • beginnersmind
@Halmos Maybe it's meant to be? |dw:1441745412913:dw|
Jhannybean
  • Jhannybean
\[y=2\sqrt{x}\]\[y+\Delta y = 2\sqrt{x+\Delta x}\]\[y+\Delta y -y = 2\sqrt{x+\Delta x}-2\sqrt{x}\]\[\Delta y =2\sqrt{x+\Delta x}-2\sqrt{x}\]hmm.....
beginnersmind
  • beginnersmind
IDK, @unimatix you need to check the problem statement. Either that or your text uses some very non-standard terminology.
anonymous
  • anonymous
if you wanna compare this equation to Increment Epsilon Form then you would got \(\epsilon =\dfrac{y}{\Delta x} - \dfrac{dy}{\Delta x} \\\epsilon =\dfrac{dy}{dx}-\dfrac{\Delta y}{\Delta x} \) ermmm first ewuation which is in ur question it does not make sense how would you interpret it ?
unimatix
  • unimatix
@beginnersmind I double checked the problem. I'm using https://www.math.wisc.edu/~keisler/calc.html section 2.2 Lots of people seem to like the book online so idk. It is a bit old.
Jhannybean
  • Jhannybean
\[\Delta y =\frac {2\sqrt{x+\Delta x}-2\sqrt{x}}{1} \cdot \frac{2\sqrt{x+\Delta x}+2\sqrt{x}}{2\sqrt{x+\Delta x}+2\sqrt{x}}\]\[\Delta y = \frac{4(x+\Delta x)-4x}{2\sqrt{x+\Delta x}+2\sqrt{x}}\]\[\Delta y = \frac{4\Delta x}{2\sqrt{x+\Delta x} +2\sqrt{x}}\]\[\frac{\Delta y}{\Delta x}= \frac{4}{2\sqrt{x+\Delta x}+2\sqrt{x}}\]
anonymous
  • anonymous
which problem is it ?
anonymous
  • anonymous
http://prntscr.com/8e5m1s
anonymous
  • anonymous
-..-
beginnersmind
  • beginnersmind
Oh, right, I think I read parts of this book a few years ago. It definitely uses an unusual approach :) "This is a calculus textbook at the college Freshman level based on Abraham Robinson's infinitesimals, which date from 1960. Robinson's modern infinitesimal approach puts the intuitive ideas of the founders of the calculus on a mathematically sound footing, and is easier for beginners to understand than the more common approach via limits. "
beginnersmind
  • beginnersmind
It introduces definitions for dy and dx at the start so you can do algebra with them, beyond the kind of chain rule cancellation you do in "normal" calculus classes.
Jhannybean
  • Jhannybean
Taking the limit of the function...\[\lim_{\Delta x\rightarrow 0} \frac{\Delta y}{\Delta x} = \frac{4}{2\sqrt{x+\Delta x}+2\sqrt{x}} = \frac{4}{4\sqrt{x}} =\frac{1}{\sqrt{x}}\]So...... \[\frac{dy}{dx} = \frac{1}{\sqrt{x}}\]
beginnersmind
  • beginnersmind
Note the title of chapter 1: "Real and Hyperreal numbers"
unimatix
  • unimatix
I didn't find Chapter 1 that bad.
Jhannybean
  • Jhannybean
Okay, but how do we go about solving the problem?
anonymous
  • anonymous
\(\Delta x \approx dx \) and we already know \(dy=f'(x) dx\) Increment theorem would be this cuz of the assumption \(\Delta x = dx \) \( \Delta y =dy+\epsilon dx \) that make since now :D
beginnersmind
  • beginnersmind
@unimatrix It's not that it's bad. It's just that if someone didn't read the book they literally won't know what dz and dx means in this problem.
Jhannybean
  • Jhannybean
dz is using partials?
beginnersmind
  • beginnersmind
Chapter 1 defines them in a specific way, that's slightly different from what you'd see in a normal calculus book or course.
Jhannybean
  • Jhannybean
I love how they use \(d\) for derivative. That makes me happy.

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