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anonymous

  • one year ago

Friends please if you can help me with this exercise, this came into my calculus test 1: parameters found for "w" and "n" the equation cos (wx) - x / n has a real solution and calculate how many second solutions has exactly ..... please still do not know how to solve it thanks

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  1. perl
    • one year ago
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    The directions are somewhat unclear. Are you quoting word for word?

  2. perl
    • one year ago
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    Is this the equation : \( \Large \cos (wx) - \frac xn = 0 \)

  3. anonymous
    • one year ago
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    yes that is the equation

  4. anonymous
    • one year ago
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    search parameters of "w" and "n" for the equation has a real solution

  5. anonymous
    • one year ago
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    but also sought to find the exact number of solutions of the equation

  6. perl
    • one year ago
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    That's harder :) We can try using calculus

  7. anonymous
    • one year ago
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    That confused me

  8. anonymous
    • one year ago
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    This exercise was in my calculus test 1

  9. perl
    • one year ago
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    this problem is within the subject of calculus 1

  10. perl
    • one year ago
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    we can see that as n gets large \(\Large \cos (wx) - \frac xn \to \cos(wx) \)

  11. perl
    • one year ago
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    We could modify the equation \( \Large n\cdot \cos (wx) -x = 0 \) The derivative is \( \Large -n \sin(wx)w -1 = 0\) However I don't see how this helps

  12. perl
    • one year ago
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    Again just to be clear you want to know exactly how many real solutions are there to the equation: $$\Large \cos (wx) - \frac xn = 0$$ Does n have to be an integer? Clearly n=0 won't work.

  13. perl
    • one year ago
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    I don't see anything that pops out. @Michele_Laino @ParthKohli

  14. ParthKohli
    • one year ago
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    They call these things transcendental equations, don't they? Graphing is the way to go about these.

  15. perl
    • one year ago
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    yes . not an algebraic equation. i dont see any other way than by checking for sign change, using intermediate value theorem . but this does not give you a nice result

  16. anonymous
    • one year ago
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    theorem of Bolzano or intermediate value theorem that thought but how to use it with the values of "w" and "n"

  17. anonymous
    • one year ago
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    but forget to mention \[w >0,n \ge1\]

  18. perl
    • one year ago
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    we can assume n are integers. https://www.desmos.com/calculator/0pdovzb6kn

  19. anonymous
    • one year ago
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    I think it has infinitely many solutions. As n goes larger, the continuity of the cosine (regarding W values) will ensure the existence of number k= cos (c) for some c in R such that k between (-1,1). Now we can vary n in such a way that x/n is between (-1,1) for any value of x (since n is independent). That is to say, for any x in the domain, there exist some n such that their ratio is between (-1,1). on the other hand, we can change w as we want such that the equality still holds . in short, infinitely many solutions That's just what I think

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