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This is a problem which I should apparently be solving with Maclaurin series, but I failed to do so.

So I attempted it with binomial series, with 5 terms and an error less than the requirement in 2., however I cannot seem to find my way around this problem with the Maclaurin series expansion.

Using the Maclaurin series;

$F(t)=\int_{0}^{t}\frac{t^4}{(\sqrt{4+t^2})^3}dt$

  1. Find the interval of convergence for F(t).

  2. Approximation for F(1/2) with an error that's less than 5.10^-4

I don't really need someone to put time in a step-by-step guide, not that I wouldn't appreciate it, but perhaps being pointed out an obvious rewriting I should make would also be very helpful.

Thanks a lot for your time.

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HINT:

The McLaurin expansión of $$F(x)=\int_0^x \frac{t^4}{(\sqrt{4+t^2})^3}\,dt$$

is given by: $$\begin{array}{rcl} F(x)&=&\sum_{k=0}^{\infty} \frac{F^{(k)}(0)}{k!}x^k\\ &=&F(0)+F'(0)x+\frac{F''(0)}{2}x^2+\ldots\\ &=&F'(0)x+\frac{F''(0)}{2}x^2+\ldots\\ \end{array}$$

Note that $F(0)=0$ because of the integration interval, and $$F'(x)=\frac{x^4}{(\sqrt{4+x^2})^3}$$

So, to calculate the covnergence radius $R$, you need to calculate the derivatives of $F(x)$, and evaluate them in $x=0$. For that, you'll probably need to find a recurrence formula for the derivatives, so you can use the ratio test.

Then, if $\frac{1}{2}\in(-R,R)$, you can use the McLaurin expansión to approximate $F(\frac{1}{2})$, finding the $k$ terms needed to get an error less tan $5\cdot10^{-4}$ using the remainder of the McLaurin series.

Good luck!

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