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How to calculate for Probability -- Summarized

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  1. Feb 26, 2017 #1
    1. The problem statement, all variables and given/known data
    if δ=7.7 %; Pm=0.2349 and
    if δ=30.8 %; Pm=0.9180.

    2. Relevant equations

    1.jpg

    3. The attempt at a solution
    We can calculate value of c if we know value of Pm and δ? Please help me.
    Thanks You!
     
  2. jcsd
  3. Feb 26, 2017 #2

    Ray Vickson

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    If your formula is really correct as written, then for given ##P_m## and ##\delta## you just have a 12th-degree polynomial in the parameter ##c##, so you can find the value of ##c## using a numerical polynomial root-solver.
     
  4. Feb 26, 2017 #3

    Simon Bridge

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    Sure, you have an equation of form: ##y = \sum_{i=2}^{12}a_ix^i## and you want to solve for ##x## given ##\{a_i\}## and ##y##.
    I'd use Newton-Raphson since I have a computer right in front of me.
     
  5. Feb 28, 2017 #4
    Thanks you Ray Vickson and Simon Bridge for your help
    Newton Raphson's Formula,
    $$ x_{n + 1} = x_n + \frac{f ( x_n )}{f ′ ( x_n)} $$

    How i get f(x) from this form $$ y=Σ^{12}_{i=2} a_i x^i $$.
    Please help me!
    Thanks all
     
  6. Mar 1, 2017 #5

    Simon Bridge

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    f(x) is the equation you are trying to evaluate the roots for... that is, it is the equation that has to satisfy f(x)=0
    Can you rearrange the equation you have so it looks like that?
     
  7. Mar 1, 2017 #6

    Ray Vickson

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    You have the wrong formula; it should be
    $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)},$$
    with a "-" sign on the right, not a "+" sign.

    In your case, ##f(x) = \sum_{i=2}^{12} \delta^i (1-\delta)^{12-i} x^i - P_m## (writing ##x## instead of ##c##).
     
  8. Mar 1, 2017 #7
    Thank you very much for your kindness!
    Yes, Simon Bridge. I can't rearrange equation. Now i got from Ray Vickson.
    Now i will try to solve.
     
  9. Mar 1, 2017 #8
    I can't solve.
    I take ## P_M=0.2349 ## and ## δ=7.7%=0.077.##

    ##f(x)= [(δ^2(1-δ)^{10}x^2)+(δ^3(1-δ)^{9}x^3)+(δ^4(1-δ)^{8}x^4)+(δ^5(1-δ)^{7}x^5)+(δ^6(1-δ)^{6}x^6)+(δ^7(1-δ)^{5}x^7)+(δ^8(1-δ)^{4}x^8)+(δ^9(1-δ)^{3}x^9)+(δ^{10}(1-δ)^{2}x^{10})+(δ^{11}(1-δ)^{1}x^{11})+(δ^{12}(1-δ)^{0}x^{12})]-0.2349;##

    ##f(x)= ((0.0027*(x^2))+((2.219e-4)*x^3)+((1.851e-5)*x^4)+((1.544e-6)*x^5)+((1.288e-7)*x^6)+((1.075e-8)*x^7)+((8.968e-10)*x^8)+((7.482e-11)*x^9)+((6.241e-12)*x^{10})+((5.207e-13)*x^{11})+((4.344e-14)*x^{12}))-0.2349;##

    ##f'(x)=(0.0053*(x))+((6.659e-4)*x^2)+((7.406e-5)*x^3)+((7.723e-6)*x^4)+((7.732e-7)*x^5)+((7.525e-8)*x^6)+((7.175e-9)*x^7)+((6.733e-10)*x^8)+((6.241e-11)*x^9)+((5.727e-12)*x^{10})+((5.212e-13)*x^{11});##

    Please help me.
     
  10. Mar 1, 2017 #9

    LCKurtz

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    Entering your formula for ##f(x)## in Maple using ##P=.2349,~\delta = .077## and solving for roots ##x## gives ##6.411427145## and a negative root also. Entering ##P = .9180,~\delta =.308## gives ##2.850506026## plus a negative root. Does that help you?
     
    Last edited: Mar 1, 2017
  11. Mar 1, 2017 #10
    Really Thank you very much.
    Finally i got. But need to long time insert. Look my solving in Maple
    pf.jpg
    If you have any idea, please advice me and help me. Thanks you all.
     
  12. Mar 1, 2017 #11

    LCKurtz

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    It looks like you have figured out how to use Newton's method with Maple. Good job. Of course, if you are going to use Maple in the first place, you don't have to program Newton's method, you can just let Maple solve it for you in two steps like this:
    > restart;
    > f := (delta, P) -> sum(delta^i*(1-delta)^(12-i)*x^i, i = 2 .. 12) - P;
    > fsolve(f(.077, .2349));

    and you can put any other pair of numbers in that last step.
     
  13. Mar 1, 2017 #12
    Thanks you LCKurtz.
    Thanks you all!
     
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