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Math Help - tayor series

  1. #1
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    tayor series

    Let f(x) = log(1 + x).

    Let
    f : R--->R be infinitely differentiable. Prove or disprove the following two statements.
    (i) ‘The Taylor series of
    f always converges for at least one point.’

    (ii) ‘The Taylor series of
    f always converges to the function for at least two points.’

    I don't understand this question so can you please help?

    thanks
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  2. #2
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    Quote Originally Posted by charikaar View Post
    Let f(x) = log(1 + x).

    Let
    f : R--->R be infinitely differentiable. Prove or disprove the following two statements.
    (i) ‘The Taylor series of
    f always converges for at least one point.’

    (ii) ‘The Taylor series of
    f always converges to the function for at least two points.’

    I don't understand this question so can you please help?

    thanks
    Try finding the Taylor series first...

    Assume that f(x) can be expressed as a polynomial.

    Then \ln{(1 + x)} = a + bx + cx^2 + dx^3 + \dots.

    By letting x = 0 we see that a = 0.


    Take the derivative of both sides

    (1 + x)^{-1} = b + 2cx + 3dx^2 + 4ex^3 + \dots.

    By letting x = 0 we see that b = 1.


    Take the derivative of both sides

    -(1 + x)^{-2} = 2c + 3!dx + 4\cdot 3 ex^2 + 5\cdot 4 fx^3 + \dots.

    By letting x = 0 we see that 2c = -1 and so c = -\frac{1}{2}.


    Take the derivative of both sides

    2(1 + x)^{-3} = 3!d + 4! ex + 5 \cdot 4 \cdot 3 fx^2 + 6\cdot 5 \cdot 4 gx^3 + \dots.

    By letting x = 0 we see that 3!d = 2, so d = \frac{2}{3!} = \frac{1}{3}.


    Take the derivative of both sides

    -3!(1 + x)^{-4} = 4!e + 5!fx + \dots

    By letting x = 0 we see that 4!e = -3! and so d = -\frac{1}{4}.


    You should be able to see that if we were to keep going we would have

    \ln{(1 + x)} = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \frac{1}{4}x^4 + \dots

     = \sum_{n = 1}^{\infty}\frac{(-1)^{n + 1}x^n}{n}.


    Can you find for which values of x that this series will converge for?

    Hint: A geometric series converges for |r| < 1...
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  3. #3
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    Quote Originally Posted by charikaar View Post
    Let f(x) = log(1 + x).

    Let
    f : R--->R be infinitely differentiable. Prove or disprove the following two statements.
    (i) ‘The Taylor series of
    f always converges for at least one point.’

    (ii) ‘The Taylor series of
    f always converges to the function for at least two points.’

    I don't understand this question so can you please help?

    thanks
    What, exactly, is the question? You first say "Let f(x) = log(1 + x)" and then say "Let f : R--->R be infinitely differentiable". Which is it? log(1+ x) is not defined for x\le -1. If that is the function in question, then it is easy to show, perhaps by actually calculating its Taylor's series about some x= a, as Prove It does, that its Taylor's series converges for all x in it radius of convergence.

    If the problem is to determine whether the Taylor's series of any infinitely differentiable function converges to the function value in at least one x value, then look at the general formula for the Taylor's series, at x= a,
    \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n
    and see what happens if x= a.

    If the problem is to determine whether the Taylor's series of any infinitely differentiable function converges to the function value in at least two x values, then that is much harder.

    Look at the function
    f(x)= e^{-\frac{1}{x^2}} is x\ne 0
    f(0)= 0

    It's not too difficult to show that all derivatives exist at 0: its derivatives, of all orders, for x not 0, are e^{-\frac{1}{x^2}} over a polynomial in x. As x goes to 0, the exponential "dominates" and so this goes to 0. Taking the nth derivative to be 0 at x= 0 gives us an infinitely differentiable function. But its derivative is 0 at x= 0 for all orders of derivative so its Taylor's series at x= 0 is just
    \sum_{n= 0}^\infty \frac{0}{n!}x^n= 0+ 0+ 0+ \cdot\cdot\cdot.

    That clearly converges, to 0, for all x but f(x)\ne 0 except at x= 0.
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  4. #4
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    Thank you very much for your replies.

    Please ignore Let f(x) = log(1 + x).
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