Sum of alternating series using four-digit chopping arithmetic

[drawar]

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!

 

[haruspex]

I suggest working backwards from some value of n, and considering each term as a fraction of the one to the left.

Spoiler

E.g. for the target series, one term is -1/2n times the one to the left. So you could start with the max n that gives a nonzero value in four decimal digits. Then add 1, to represent the term to the left, then multiply by -1/(2n-2) to get it relative to the next term to the left, and so on. I.e the sum is
1 +(-1/2)(1+(-1/4)(1+(-1/6)(1+...

 

[drawar]

I suggest working backwards from some value of n, and considering each term as a fraction of the one to the left.

Spoiler

E.g. for the target series, one term is -1/2n times the one to the left. So you could start with the max n that gives a nonzero value in four decimal digits. Then add 1, to represent the term to the left, then multiply by -1/(2n-2) to get it relative to the next term to the left, and so on. I.e the sum is
1 +(-1/2)(1+(-1/4)(1+(-1/6)(1+...

So you could start with the max n that gives a nonzero value in four decimal digits

Oh, that solved the infinity problem, but how do we come up with such n? I was given a function to simulate the 4-digit chopping arithmetic in MATLAB:

function y=fl(x)
if x==0.0,
y=0.0;
return
end
c=floor(log10(abs(x)))-3;
y=fix(x*10^(-c))*10^c;
return

As for the sum, should it be:
[itex] - \frac{1}{2}\left( {1 - \frac{1}{{12}}\left( {1 - \frac{1}{{30}}\left( {1 - \frac{1}{{56}}\left( {1 - ...} \right)} \right)} \right)} \right)[/itex]?

Thanks!

 

[Joffan]

If possible for a given series, I would suggest calculating the the series in pairs, positive and negative, so that you don't lose precision in the subtraction.

This method is possible and somewhat useful for this particular series but the terms decrease too fast to show this to its best advantage.

 

[drawar]

If possible for a given series, I would suggest calculating the the series in pairs, positive and negative, so that you don't lose precision in the subtraction.

This method is possible and somewhat useful for this particular series but the terms decrease too fast to show this to its best advantage.

Thanks but I don't think that method gives a better accuracy than nested polynomials. For example, to calculate [itex]y = a + bx + c{x^2} + d{x^3}[/itex], one has to do 6 multiplications and 3 additions (assuming that the powers of x are calculated separately), while nesting it to [itex]y = a + x(b + x(c + dx))[/itex] requires only 3 multiplications and 3 additions. Less computations mean less roundoff errors and thus increasing the accuracy.

 

[haruspex]

Oh, that solved the infinity problem, but how do we come up with such n? I was given a function to simulate the 4-digit chopping arithmetic in MATLAB:

As for the sum, should it be:
[itex] - \frac{1}{2}\left( {1 - \frac{1}{{12}}\left( {1 - \frac{1}{{30}}\left( {1 - \frac{1}{{56}}\left( {1 - ...} \right)} \right)} \right)} \right)[/itex]?

What's the ratio between consecutive terms? That will tell what to put in the nesting. You want to start with the largest n for which the ratio doesn't truncate to zero in four digits.

 

[drawar]

What's the ratio between consecutive terms? That will tell what to put in the nesting. You want to start with the largest n for which the ratio doesn't truncate to zero in four digits.

Ah thanks, I've figure it out already. n is 86 and the sum is 0.5403. Actually the sum is approximated by Maclaurin's series to be cos(1). :)