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anonymous

  • 5 years ago

1/3x^ = 2x+1

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  1. anonymous
    • 5 years ago
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    What's your x (on the left-hand side) to the power of?

  2. anonymous
    • 5 years ago
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    2

  3. anonymous
    • 5 years ago
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    You need to remove any x's you have from the denominator. So, multiply everything through by 3x^2 (I said 3x^2 instead of just x^2 because we'd end up multiplying through by 3 anyway - may as well get it over and done with). So,

  4. anonymous
    • 5 years ago
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    \[\frac{1}{3x^2}=2x+1 \rightarrow 1=3x^2(2x+1)=6x^3+3x^2\]

  5. anonymous
    • 5 years ago
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    he may have meant (1/3)x^2...

  6. anonymous
    • 5 years ago
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    Is that what you meant, mac?

  7. anonymous
    • 5 years ago
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    that would sound more likely to be the homework assigned to an algebra class

  8. anonymous
    • 5 years ago
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    Because solving 6x^3+3x^2=1 algebraically is not nice.

  9. anonymous
    • 5 years ago
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    ^not at all...

  10. anonymous
    • 5 years ago
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    \[\frac{1}{3}x^2=2x+1 \rightarrow x^2=6x+3\] after multiplying through by 3. You have a quadratic to solve:\[x^2-6x-3=0\]

  11. anonymous
    • 5 years ago
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    ergh... that still wont give integer roots

  12. anonymous
    • 5 years ago
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    You can complete the square or use the quadratic formula.

  13. anonymous
    • 5 years ago
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    Can't help that...

  14. anonymous
    • 5 years ago
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    \[x^2-6x-3=0 \rightarrow x=3 \pm 2\sqrt{3}\]

  15. anonymous
    • 5 years ago
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    as for the 6xcubed i would use long division... or rather synthetic... is that the best way?

  16. anonymous
    • 5 years ago
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    Do you mean you would use long division or synthetic after finding a factor?

  17. anonymous
    • 5 years ago
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    i would use them to check if i can guess a factor lol

  18. anonymous
    • 5 years ago
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    Yes

  19. anonymous
    • 5 years ago
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    i wonder where mac went...

  20. anonymous
    • 5 years ago
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    The method for solving cubics is a pain, so I usually try to find one solution and, like you said, use long/synthetic division on the cubic and go from there. Else, numerical solution, like Newton-Raphson method.

  21. anonymous
    • 5 years ago
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    Mac's lurking...

  22. anonymous
    • 5 years ago
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    newton-raphson method? whats that?

  23. anonymous
    • 5 years ago
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    It's a method for finding solutions to equations when you can't do so algebraically (or can't be bothered). The only condition is that the function you're looking at has to be continuous (because you use calculus).

  24. anonymous
    • 5 years ago
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    im curious, as the derivative cant tell you solutions... can it??

  25. anonymous
    • 5 years ago
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    You're right, the derivative itself won't give you the solutions (unless you have a multiple root), but the method takes your guess, then uses calculus to find the equation of the tangent at the function value of your guess for x, then seeks out where the tangent cuts the x-axis. That point - where it cuts the x-axis - is closer to the root than your initial guess (usually...there are cases where it won't work). That new x-value is your new 'guess' which goes through the process again to yield a better 'guess' for x...you can keep going until you get to a level of accuracy you want.

  26. anonymous
    • 5 years ago
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    ... so this will only work with f(x) with a degree 3 or more so that f'(x) is at least quadratic... and you find the tangent, then find the zero of that equation, and then use that in the original and find the tangent and use the zero of that equation and so on?

  27. anonymous
    • 5 years ago
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    The method will will work with any function that has at least one derivative. I'll try and find some information for you. It's a very useful tool to have, mainly because you end up having at your disposal a means to solve many practical problems that either don't have algebraic solutions, or where the solutions are a pain to get at and you don't need your answer to be exact.

  28. anonymous
    • 5 years ago
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    but then you will never get the exact zero, you will only get closer and closer to it? as the zero of a tangent line can never be equal to the value put into the original function for that tangent line unless you have a multiple root case ( as you said above)

  29. anonymous
    • 5 years ago
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    Check out the image - this is what's going on. http://en.wikipedia.org/wiki/Newton's_method

  30. anonymous
    • 5 years ago
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    thank you man

  31. anonymous
    • 5 years ago
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    You're welcome ;)

  32. anonymous
    • 5 years ago
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    what the heckkk that is freaking genius. damn newton -.-

  33. anonymous
    • 5 years ago
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    Yeah, I think it's his best thing outside of the calculus and Laws of Motion.

  34. anonymous
    • 5 years ago
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    when it comes to physics laws of that sort, id always put kepler over newton... idk why newton got all the fame for those, keplers were way better -.-

  35. anonymous
    • 5 years ago
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    Try it with something like finding the square root of 2 - it will find it pretty damn quick. \[x=\sqrt{2} \rightarrow f(x)=x^2-2=0 \rightarrow f'(x)=2x\]Then\[x_{n+1}=x_n-\frac{f(x)}{f'(x)} \rightarrow x_{n+1}=x_n-\frac{x^2_n-2}{2x_n}\]

  36. anonymous
    • 5 years ago
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    i didnt understand the second half of that lol

  37. anonymous
    • 5 years ago
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    I'm going to do an example.

  38. anonymous
    • 5 years ago
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    You have to find something to start with, so we make an educated guess for sqrt(2). Since \[1<2<4 \rightarrow \sqrt{1}<\sqrt{2}<\sqrt{4} \iff 1 <\sqrt{2}<2\] as sqrt function is monotonic increasing (i.e. the direction of the inequality is preserved). So we have for an initial guess for x,\[x \approx \frac{3}{2}\]Then, a better 'guess' for x will be,\[x_2 = \frac{3}{2}-\frac{\left( \frac{3}{2} \right)^2-2}{2\frac{3}{2}} = \frac{17}{12} \approx 1.416667\]and the square root of 2 is about \[1.414213562\]

  39. anonymous
    • 5 years ago
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    You can see after one trial, it's pretty close.

  40. anonymous
    • 5 years ago
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    The closer you are to your answer in the first place, the faster it will converge. You could then plug 17/12 into the formula to get a better approximation, and continue.

  41. anonymous
    • 5 years ago
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    Next one would be 577/408 ~= 1.414215686

  42. anonymous
    • 5 years ago
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    dang, the formula, has you divide the tangent line function by the tangent line slope, why?

  43. anonymous
    • 5 years ago
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    Well, going back to the sqrt of 2, when you start out, you don't know what the sqrt of 2 is, but you know it should be something, so call that something, x. Then x =sqrt(2). If I square both sides and subtract 2 from both sides, I get x^2-2=0 So, finding x = sqrt(2) amounts to finding where on the x-axis x^2-2 will cut it. So I plot my parabola, and imagine the setup that's in that Wikipedia pic. Since the method finds where the graph cuts the x-axis, it will find that x where x^2-2=0; that is, the x that came from x=sqrt(2).

  44. anonymous
    • 5 years ago
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    yeah i get that, but i dont get how subtracting the tangent line formula divided by the slope of the tangent gets you the intersection of the tangent line?

  45. anonymous
    • 5 years ago
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    The reason for why we divide the function by the derivative is due to the derivation. The derivation can be done in either a geometrical setting (using that picture, say) or with Taylor series expansion on an arbitrary function. I'm sure proofs would be online. It's a pretty old theorem ;)

  46. anonymous
    • 5 years ago
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    oh wait i get it...

  47. anonymous
    • 5 years ago
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    You know, you can derive it yourself by analyzing each step in that pic. and doing what the mathematics would dictate in that situation.

  48. anonymous
    • 5 years ago
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    your taking y=mx+b and doing 0=mx+b so mx=-b so x=-b/m and you take that, which is the distance from the x value to the zero, and subtract it from your original point which will show you where on the x axis it is...

  49. anonymous
    • 5 years ago
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    nope that made no sense... ill figure this out, gimmy a sec

  50. anonymous
    • 5 years ago
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    You're half there.

  51. anonymous
    • 5 years ago
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    It's difficult to explain online. They need to put a drawing feature in this thing.

  52. anonymous
    • 5 years ago
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    lol they do, like tutorvista

  53. anonymous
    • 5 years ago
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    I think mac must think his thread's been hijacked.

  54. anonymous
    • 5 years ago
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    i think he left

  55. anonymous
    • 5 years ago
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    so youre taking x and subtracting y/y'(x)

  56. anonymous
    • 5 years ago
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    i still dont get why ><

  57. anonymous
    • 5 years ago
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    Hang on, I'll see if I can put a small proof together.

  58. anonymous
    • 5 years ago
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    how is the y value divided by the slope at that point, equal to the difference btween x and the intersection of the tangent line of x

  59. anonymous
    • 5 years ago
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    no you dont need to make a proof of it i just need to understand how y/y' is the difference between x and the zero of y'

  60. anonymous
    • 5 years ago
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    Wait, if I prove it using the geometrical route, you'll see...I'm just going to see if it's feasible (versus finding something online).

  61. anonymous
    • 5 years ago
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    ok :/

  62. anonymous
    • 5 years ago
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    Alright, writing it up isn't hard, but you'll see what's going on if you do a drawing like the one seen in Wikipedia.

  63. anonymous
    • 5 years ago
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    And go along with it...

  64. anonymous
    • 5 years ago
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    i can understand how it all works, but how does the drawing explain how the formula works?

  65. anonymous
    • 5 years ago
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    (i already made a drawring btw lol)

  66. anonymous
    • 5 years ago
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    Suppose x_1 is close to a root of the equation f(x)=0. Then the equation of the tangent at x=x_1 is\[y-y_1=m(x-x_1)\]that is,\[y-f(x_1)=f'(x_1)(x-x_1)\]This tangent crosses the x-axis at a point where x = x_2 and y = 0. Hence,\[0-f(x_1)=f'(x_1)(x_2-x_1) \rightarrow x_2=x_1-\frac{f(x_1)}{f'(x_1)}\]

  67. anonymous
    • 5 years ago
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    Then repeat...

  68. anonymous
    • 5 years ago
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    oh wow that makes sense...

  69. anonymous
    • 5 years ago
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    Happy now?

  70. anonymous
    • 5 years ago
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    yes :D thanks for spending time to teach me this stuff :O

  71. anonymous
    • 5 years ago
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    No worries - it's good you want to know :)

  72. anonymous
    • 5 years ago
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    its incredible that you can put up with teaching me like that, even alot of teachers would loose it pretty soon

  73. anonymous
    • 5 years ago
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    i honestly am really curious as to where mac went...

  74. anonymous
    • 5 years ago
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    he stopped observing a while ago even

  75. anonymous
    • 5 years ago
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    You learnt quickly - it wasn't a problem. You would have learnt quicker if the interface was easier (but it's not the designers' fault - where's the holography already?).

  76. anonymous
    • 5 years ago
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    Yeah, he probably got his answer and was all, "WTF is going on?!"

  77. anonymous
    • 5 years ago
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    whats after differential equations?

  78. anonymous
    • 5 years ago
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    Eh?

  79. anonymous
    • 5 years ago
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    like course wise, you take calc1 calc2 calc3 diff eq then what?

  80. anonymous
    • 5 years ago
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    Have no idea man, not American ;D

  81. anonymous
    • 5 years ago
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    but you studied in europe, so the course order must be nearly identicle, especially in advanced mathematics, so instead i should probably ask, what did you do (courses) for you degree and btw, you have a masters? phD?

  82. anonymous
    • 5 years ago
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    your degree*

  83. anonymous
    • 5 years ago
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    Partial differential equations is usually end of the line for calculus. Then there's other stuff like Lebesgue integration (overtakes Riemann integral), and other stuff in algebra and geometry - the further you go, the more the areas start to link up. E.g. differential geometry, algebraic topology, etc. But for calculus, numerical and algebraic methods for solving partial differential equations is usually end of line. If you know how to solve (methods to solve) nth order, non-linear, inhomogeneous partial differential equations with non-constant coefficients, you're done!

  84. anonymous
    • 5 years ago
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    Since i did not understand any of nth order, non-linear, inhomogeneous partial differential equations with non-constant coefficients i guess im not done :D

  85. anonymous
    • 5 years ago
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    lol, keep going, though!

  86. anonymous
    • 5 years ago
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    do you know all that is taught of mathematics to the extent that research is your only option?

  87. anonymous
    • 5 years ago
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    No, you can do whatever you want with it. I teach as well. You can use it in industry and government, as in modelling for things...haven't really sold it, have I?

  88. anonymous
    • 5 years ago
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    well not in that sense, i meant like if you wanted to know more on math, your only option would be to study unsolved theorems or unproved research or do research on your own

  89. anonymous
    • 5 years ago
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    anddd man i need help on something: \[\int\limits_{}^{} (\sin x - \cos x) \div (\sin x)\]

  90. anonymous
    • 5 years ago
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    nevermind i got it

  91. anonymous
    • 5 years ago
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    Yes, you're right on your last point.

  92. anonymous
    • 5 years ago
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    thats pretty crazy, that you know everything O.o

  93. anonymous
    • 5 years ago
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    I don't know everything..!

  94. anonymous
    • 5 years ago
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    you know everything in math...

  95. anonymous
    • 5 years ago
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    No I don't ;)

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