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Problems;

Problems 1-12 below are to be done using a power series with an error estimate. If a hand calculator is available they can be worked with the errors reduced by an additional factor of 1000.

1             Approximate ln (1.2) within 0.01.

2            Approximate arctan (1/10) within 10-7.

3            Approximate e-1/4 within 0.00001.

4            Approximate 09_infinite_series-666.gif within 0.01.

5            Approximate 09_infinite_series-667.gif within 0.0001.

6            Approximate 09_infinite_series-668.gif within 0.001.

7            Approximate 09_infinite_series-669.gif within 0.0001.

8            Approximate 09_infinite_series-670.gif within 0.00001.

9            Approximate 1/(1 - 0.003) within 0.0001.

10            Approximate ln 3 within 0.1 by the method of Example 6. Hint: ln 3 = -ln(1 - x) where x = ⅔.

11            Approximate ln 3 within 0.001 by the method of Example 7.

12            (a) Approximate ln (1½) within 0.00001 by the method of Example 7.

(b) Approximate ln 3 within 0.00002 using the formula ln 3 = ln 2 + ln (½).

In Problems 13-18 find a power series approximation with an error estimate for f(x) valid for -½ ≤ x ≤ ½. Then approximate f(½) within 0.01.

13            09_infinite_series-671.gif

14            09_infinite_series-672.gif

15            09_infinite_series-673.gif.

16            09_infinite_series-674.gif

Hint: x2 = when x = ½.

17           09_infinite_series-675.gif

09_infinite_series-676.gif

19            Using the power series for arctan x at x = 1, show that

09_infinite_series-677.gif

20            Using the power series for 09_infinite_series-678.gif at x = 1, show that

09_infinite_series-679.gif


Last Update: 2006-11-25