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Math Help - Two Analysis II Problems

  1. #1
    Junior Member JoanF's Avatar
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    Two Analysis II Problems

    Hi!
    I need some help in these exercices:

    Exercice 1:
    1.1 - Prove that tan(x+y)= [tan(x)+tan(y)] / [1−tan(x).tan(y)] and then use this to prove that arctg(x) + arctg(y) = arctg ( [x+y] / [1-xy] ) indicating any possible restriction of arguments.

    1.2 - Conclude that: pi / 4 = arctg (1 / 2) + arctg (1 / 3) and pi / 4 = 4.arctg (1 / 5) - arctg (1 / 239)

    1.3 - Use this last equation and Taylor's
    polynomials of arctg(x) to prove that: pi = 3.14159...



    Exercice 2:
    2.1 - Considering a function such that f''(x) + f(x) = 0 for all x belonging to lR, and that f(0) = 0, f'(0) = 0. Prove that all derivatives exist.

    2.2 - Calculate Taylor's polynomial of this fucntion at the point 0 and the respective rest.

    2.3 - Conclude that any function satisfying these conditions is necessarily null.



    I hope someone can help me!


    Thank you very much
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    This is analysis II? seems like it is more calc 3 or something
    Quote Originally Posted by JoanF View Post
    Hi!
    I need some help in these exercices:

    Exercice 1:
    1.1 - Prove that tan(x+y)= [tan(x)+tan(y)] / [1−tan(x).tan(y)]
    try using the fact that tan(x + y) = sin(x + y)/cos(x + y), which you know the addition formulas for. expand them, and simplify and change everything to tangents.

    and then use this to prove that arctg(x) + arctg(y) = arctg ( [x+y] / [1-xy] ) indicating any possible restriction of arguments.
    plug in arctan(x) for x and arctan(y) for y in the previous formula. you should be able to take it from there.

    1.2 - Conclude that: pi / 4 = arctg (1 / 2) + arctg (1 / 3) and pi / 4 = 4.arctg (1 / 5) - arctg (1 / 239)
    just plug in the relevant values for x and y in the equation from the previous part

    1.3 - Use this last equation and Taylor's
    polynomials of arctg(x) to prove that: pi = 3.14159...
    we know that \frac {\pi}4 = \arctan (1/2) + \arctan (1/3). and we can write \arctan x + \arctan y as a single arctangent function. use the Taylor series of this function. plug in the relevant x and y values to get \frac {\pi}4. you can estimate this using the partial sums up to 5 or 6 decimal places accuracy. then, multiply that answer by 4

    Exercice 2:
    2.1 - Considering a function such that f''(x) + f(x) = 0 for all x belonging to lR, and that f(0) = 0, f'(0) = 0. Prove that all derivatives exist.

    2.2 - Calculate Taylor's polynomial of this fucntion at the point 0 and the respective rest.

    2.3 - Conclude that any function satisfying these conditions is necessarily null.
    did you do differential equations? begin by solving the differential equaiton. you will perhaps know what to do after that
    Last edited by Jhevon; April 13th 2009 at 02:38 PM.
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