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Math Help - newton's method

  1. #1
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    Unhappy newton's method


    Q1 - Use Newton's method to find the value of x for the function f(x)=2x^2 + 4x - 3 = 0 ; x > 0
    Note : Perform only 3 iterations.




    Q2 - Evaluate the integral ∫ (1 + sinx)^6 cosxdx by proper substitution.




    Q3 - Evaluate the sum ∑ k(k - 2)(k + 2) using the required theorems.


    Note on this sign "∑" on the upper side it is 35 and on the lower side it is k-1....I dont know how to use the tags here coz am new...sorry plz...

    Help me in this
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  2. #2
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    The exact same questions you posted here has been posted elsewhere and already received a response:
    http://www.mathhelpforum.com/math-he...t-hw-help.html
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  3. #3
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    According to Q2

    Quote Originally Posted by Angel Rox View Post


    Q2 - Evaluate the integral ∫ (1 + sinx)^6 cosxdx by proper substitution.
    Substitute z:= sin(x) => z' = cos(x) =>  dx = \frac{dz}{z'}

    => \int (1+z)^6 cos(x) * \frac{dz}{cos(x)} = \int (1+z)^6 = ...
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  4. #4
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    Yeah I saw that response which u have mentioned but i couldnt understand his performing methodology because he has just given some simple hints...so if any one could simplify it a lil bit...I'll be thankful...
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  5. #5
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    Quote Originally Posted by Angel Rox View Post
    Yeah I saw that response which u have mentioned but i couldnt understand his performing methodology because he has just given some simple hints...so if any one could simplify it a lil bit...I'll be thankful...
    What parts of it you did not understand?
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  6. #6
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    I think i can do Q2 by substituting 1+ sinx = u

    But in Q 1 which approximation we should take next?

    and in Q3 he wrote we need to expand to get and then sum these two terms seperately... How we r going to do this??
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  7. #7
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    Q1. Try x_0 = 1 as Grandad mentioned to find x_1, then use x_1 to find x_2,...,then use x_n to find x_{n+1}

    Q3. The lower side should read k=1.

    And what Grandad meant is that once you have expanded it to:
    \sum_{k=1}^{35} (k^3 - 4k)

    Separate it into: \sum_{k=1}^{35} k^3 - 4\sum_{k=1}^{35} k

    Recall that:
    \sum_{k=1}^n k = \frac{n(n+1)}{2}

    \sum_{k=1}^n k^3 = \left(\frac{n(n+1)}{2}\right)^2
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  8. #8
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    Quote Originally Posted by Angel Rox View Post
    I think i can do Q2 by substituting 1+ sinx = u

    But in Q 1 which approximation we should take next?

    and in Q3 he wrote we need to expand to get and then sum these two terms seperately... How we r going to do this??
    And thank Grandad as well. His post was sufficient for your question and my reply was based on it.
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  9. #9
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    Ok I have thanked him too...Thanx to all of u ...
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