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drawar

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## Homework Statement

Let [itex]a_{n}[/itex] be an alternating series whose terms are decreasing in magnitude. How to compute the sum as precisely as possible using four-digit chopping arithmetic? In particular, apply the method to compute [itex]\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}}}{{(2n)!}}}[/itex] and specify how the term [itex]\frac{{{{( - 1)}^n}}}{{(2n)!}}[/itex] is obtained.Here is an excerpt from my textbook regarding the definition of finite digit chopping arithmetic:

Any positive real number [itex]y[/itex] can be written in decimal floating point form: [itex]y=0.{d_1}{d_2}{d_3}...{d_k}{d_{k + 1}}{d_{k + 2}}... \times {10^n}[/itex] where each digit [itex]d_{k}[/itex] satises [itex]0 \leq[/itex] [itex]d_{k}[/itex] [itex]\leq 9[/itex] and [itex]d_{1}[/itex] [itex]\neq 0[/itex]. To put [itex]y[/itex] to a [itex]k[/itex]-digit decimal floating point number, the first method is chopping (a.k.a. rounding towards [itex]0[/itex] or truncation): [itex]fl(y)=0.{d_1}{d_2}{d_3}...{d_k} \times {10^n}[/itex]

Here, [itex]fl(y)[/itex] denotes the floating form of [itex]y[/itex].

## Homework Equations

## The Attempt at a Solution

I was told that calculating by nesting is more precise than calculating term by term, but how to calculate the sum of a series going to infinity? Any hint given is much appreciated, thanks!

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