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A root of sin(x)+x^2=1 using newton method

  • 04-12-2010 3:55pm
    #1
    Ste234
    Registered Users, Registered Users 2 Posts: 65 ✭✭


    I don't know what I'm doing wrong here.
    We have to get to 8 decimel places a root of sin(x)+x^2=1
    using 1 as an initial guess.
    I've done it and I get 0.99.... as an answer when the answer according to the solutions and wolframalpha.com is 0.636....

    Anyone know what I may be doing wrong?
    I've used the formula x=x-(f(x)/f'(x))


Comments

  • #2
    Eliot Rosewater
    Registered Users, Registered Users 2 Posts: 3,745 ✭✭✭Eliot Rosewater


    I got the Newton Raphson formula as
    [LATEX]\displaystyle x-\frac{f(x)}{f'(x)}[/LATEX]
    [LATEX]\displaystyle x-\frac{sin(x)+x^2-1}{2x+cos(x)}[/LATEX]

    Letting x=1

    [LATEX]\displaystyle 1-\frac{sin(1)+1^2-1}{2(1) + cos(1)}=0.66875...[/LATEX]

    According to Wolfram Alpha. http://www.wolframalpha.com/input/?i=1-%28sin+1%29%2F%282%2Bcos+1%29


  • #3
    bnt
    Registered Users, Registered Users 2 Posts: 13,073 ✭✭✭✭bnt


    If you're doing this on a calculator, it must be set to Radians. Catches a few people out, that does. Calculus on trig. functions is only valid in Radians, never Degrees. :cool:

    You are the type of what the age is searching for, and what it is afraid it has found. I am so glad that you have never done anything, never carved a statue, or painted a picture, or produced anything outside of yourself! Life has been your art. You have set yourself to music. Your days are your sonnets.

    ―Oscar Wilde predicting Social Media, in The Picture of Dorian Gray



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