Recent content by hereiscassie

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    Taylor Series Expansion for f(t) and G(x) with Convergence Analysis"

    Thank you so much! you helped me a lot!:)
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    Taylor Series Expansion for f(t) and G(x) with Convergence Analysis"

    oh right! so after that i get x^2 < 1 and then x will equal -1 and 1 right? and then the IOC would be -1 < x < 1 but then how would i test these boundaries using AST?? do I use the sum i had with these?
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    Taylor Series Expansion for f(t) and G(x) with Convergence Analysis"

    I fixed part b and i got: 4x - (4x^3)/3 + (4x^5)/5 - (4x^7)/7 +...+ [(-1)^n](4)(x^(2n +1))/(2n + 1) +... and I am still slightly confused in part c... I did the ratio test and i can't get past \stackrel{lim}{n \rightarrow \infty} \left|\frac{(-1)x^2(2n +1)}{2n +3}| srry, its supposed...
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    Taylor Series Expansion for f(t) and G(x) with Convergence Analysis"

    Let f be the function given by f(t) = 4/ (1 + t^2) and G be the function given by G(x) = {Integral from 0 to x} f(t)dt . (a) Find the first four nonzero terms and the general term for the power series expansion of f(t) about t = 0. (b) Find the first four nonzero terms and the general term...
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    Doing a MacLaurin Series and more

    Doing a MacLaurin Series and more! The function f is defined by f(x) = 1/(1+x^3). The MacLaurin series for f is given by 1 - x^3 + x^6 - x^9 +...+ (-1)^n(x^3n) +... which converges to f(x) for -1 < x < 1. a) Find the first three nonzero terms and the general term for the MacLaurin series...
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    Taylor Series Help: ƒ(x) = e^(x/2), g(x) = ((e^(x/2)) - 1)/x

    good! there was a part c for this too. it said: c) for the function g in part b, find g'(2) and use it to show that \sum_{n= 1}^\infty \frac{n}{4(n+1)!} = 1/4 i don't know how to start it, or what to do. help please?