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unimatix

  • one year ago

For delta x infinitesimal find an infinitesimal (epsilon) such that delta y = dy + epsilon delta x 2sqrt(x)

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  1. unimatix
    • one year ago
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    |dw:1441743136199:dw|

  2. unimatix
    • one year ago
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    |dw:1441743200586:dw|

  3. anonymous
    • one year ago
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    Can you please use the equation button for this I do not understand the writing.

  4. unimatix
    • one year ago
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    sure

  5. Jhannybean
    • one year ago
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    Looks like the definition of a derivative

  6. unimatix
    • one year ago
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    \[\Delta y = dy + \epsilon \Delta x\]

  7. unimatix
    • one year ago
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    And I think I've found dy and delta y. Which are \[\Delta y = \frac{ 2 \Delta x }{ \sqrt{x+ \Delta x} + \sqrt{x} }\] \[dy = \frac{ \Delta x }{ \sqrt{x} }\]

  8. Jhannybean
    • one year ago
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    Ahh... this relates to the increment theorem... vaguely remember learning about this x_x

  9. unimatix
    • one year ago
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    Yep, that sounds right

  10. Jhannybean
    • one year ago
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    Im just giving a shot at this,...

  11. Jhannybean
    • one year ago
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    \(y=2\sqrt{x}\) right?

  12. unimatix
    • one year ago
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    Yep!

  13. Jhannybean
    • one year ago
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    whew, ok.

  14. Jhannybean
    • one year ago
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    we're ultimately trying to put it in the form of \[\frac{\Delta y}{\Delta x} = f'(x) + \epsilon\] right?...

  15. unimatix
    • one year ago
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    \[\Delta y = dy+ \epsilon \Delta x\]

  16. unimatix
    • one year ago
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    I've just got to find epsilon.

  17. unimatix
    • one year ago
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    \[\Delta y = \frac{ 1 }{ 2\sqrt{x} } \Delta x + \epsilon \Delta x\]

  18. unimatix
    • one year ago
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    \[\frac{ \Delta y }{ \Delta x } = \frac{ 1 }{ 2\sqrt{x}} + \epsilon \]

  19. unimatix
    • one year ago
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    \[\epsilon = \frac{ \Delta y }{ \Delta x } - \frac{ 1 }{ 2\sqrt{x} }\]

  20. unimatix
    • one year ago
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    Maybe substitute back in the Delta y from earlier.

  21. Jhannybean
    • one year ago
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    Oh instead of finding "\(\epsilon\)" can't we take the limit? Just throwing that out there, i might be wrong

  22. anonymous
    • one year ago
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    ok for y = dy + epsilon delta x what does this even mean :O

  23. unimatix
    • one year ago
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    I'm sure there is a perfectly good explanation for that...

  24. beginnersmind
    • one year ago
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    @Halmos Maybe it's meant to be? |dw:1441745412913:dw|

  25. Jhannybean
    • one year ago
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    \[y=2\sqrt{x}\]\[y+\Delta y = 2\sqrt{x+\Delta x}\]\[y+\Delta y -y = 2\sqrt{x+\Delta x}-2\sqrt{x}\]\[\Delta y =2\sqrt{x+\Delta x}-2\sqrt{x}\]hmm.....

  26. beginnersmind
    • one year ago
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    IDK, @unimatix you need to check the problem statement. Either that or your text uses some very non-standard terminology.

  27. anonymous
    • one year ago
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    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 ?

  28. unimatix
    • one year ago
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    @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.

  29. Jhannybean
    • one year ago
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    \[\Delta y =\frac {2\sqrt{x+\Delta x}-2\sqrt{x}}{1} \cdot \frac{2\sqrt{x+\Delta x}+2\sqrt{x}}{2\sqrt{x+\Delta x}+2\sqrt{x}}\]\[\Delta y = \frac{4(x+\Delta x)-4x}{2\sqrt{x+\Delta x}+2\sqrt{x}}\]\[\Delta y = \frac{4\Delta x}{2\sqrt{x+\Delta x} +2\sqrt{x}}\]\[\frac{\Delta y}{\Delta x}= \frac{4}{2\sqrt{x+\Delta x}+2\sqrt{x}}\]

  30. anonymous
    • one year ago
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    which problem is it ?

  31. anonymous
    • one year ago
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    http://prntscr.com/8e5m1s

  32. anonymous
    • one year ago
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    -..-

  33. beginnersmind
    • one year ago
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    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. "

  34. beginnersmind
    • one year ago
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    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.

  35. Jhannybean
    • one year ago
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    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...... \[\frac{dy}{dx} = \frac{1}{\sqrt{x}}\]

  36. beginnersmind
    • one year ago
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    Note the title of chapter 1: "Real and Hyperreal numbers"

  37. unimatix
    • one year ago
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    I didn't find Chapter 1 that bad.

  38. Jhannybean
    • one year ago
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    Okay, but how do we go about solving the problem?

  39. anonymous
    • one year ago
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    \(\Delta x \approx dx \) 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

  40. beginnersmind
    • one year ago
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    @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.

  41. Jhannybean
    • one year ago
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    dz is using partials?

  42. beginnersmind
    • one year ago
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    Chapter 1 defines them in a specific way, that's slightly different from what you'd see in a normal calculus book or course.

  43. Jhannybean
    • one year ago
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    I love how they use \(d\) for derivative. That makes me happy.

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