Problem 23. Use the geometric series, = 1+x+r² + x³ + •. to obtain a series for and use this to obtain the series Eo ", %3D 1 1 E(-1)"; arctan a = r - -.. D 2n +1 0 Use the series above to obtain the series =Eo(-1)"z+1

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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2:21
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< RealAnalysis-ISBN-fix...
(b) Let s(x, N) = E-0 2 rn+1 and c(x, N) = En=o 2n r2n
and use a computer algebra system to plot these for -4n <a < 4n, N =
1, 2, 5, 10, 15. Describe what is happening to the series as N becomes
larger.
Problem 23. Use the geometric series, = 1+x+x²+x³ + • . .= E ",
to obtain a series for and use this to obtain the series
1
arctan x = x -
(-1)"
2n+1.
5*
2n +1*
n=0
Use the series above to obtain the series = E.(-1)"
The series for arctangent was known by James Gregory (1638-1675) and it is
sometimes referred to as "Gregory's series." Leibniz independently discovered
1 =1-+ - by examining the area of a circle. Though it gives us
a means for approximating a to any desired accuracy, the series converges too
slowly to be of any practical use. For example, if we compute the sum of the
first 1000 terms we get
/1000
4E(-1)";
2n + 1
2 3.142591654
n=0
which only approximates a to two decimal places.
Newton knew of these results and the general scheme of using series to
compute areas under curves. These results motivated Newton to provide a
series approximation for a as well, which, hopefully, would converge faster. We
will use modern terminology to streamline Newton's ideas. First notice that
1 = L-0 V1 - x² dx as this integral gives the area of one quarter of the unit
circle. The trick now is to find series that represents VI– a².
To this end we start with the binomial theorem
(a + b)N = £ (, )ax-",
where
N!
n! (N – n)!
N (N – 1) (N – 2).. · (N – n+1)
n!
II"-0 (N – j)
n!
Unfortunately, we now have a small problem with our notation which will
be a source of confusion later if we don't fix it. So we will pause to address this
matter. We will come back to the binomial expansion afterward.
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Transcribed Image Text:2:21 ( Safari < RealAnalysis-ISBN-fix... (b) Let s(x, N) = E-0 2 rn+1 and c(x, N) = En=o 2n r2n and use a computer algebra system to plot these for -4n <a < 4n, N = 1, 2, 5, 10, 15. Describe what is happening to the series as N becomes larger. Problem 23. Use the geometric series, = 1+x+x²+x³ + • . .= E ", to obtain a series for and use this to obtain the series 1 arctan x = x - (-1)" 2n+1. 5* 2n +1* n=0 Use the series above to obtain the series = E.(-1)" The series for arctangent was known by James Gregory (1638-1675) and it is sometimes referred to as "Gregory's series." Leibniz independently discovered 1 =1-+ - by examining the area of a circle. Though it gives us a means for approximating a to any desired accuracy, the series converges too slowly to be of any practical use. For example, if we compute the sum of the first 1000 terms we get /1000 4E(-1)"; 2n + 1 2 3.142591654 n=0 which only approximates a to two decimal places. Newton knew of these results and the general scheme of using series to compute areas under curves. These results motivated Newton to provide a series approximation for a as well, which, hopefully, would converge faster. We will use modern terminology to streamline Newton's ideas. First notice that 1 = L-0 V1 - x² dx as this integral gives the area of one quarter of the unit circle. The trick now is to find series that represents VI– a². To this end we start with the binomial theorem (a + b)N = £ (, )ax-", where N! n! (N – n)! N (N – 1) (N – 2).. · (N – n+1) n! II"-0 (N – j) n! Unfortunately, we now have a small problem with our notation which will be a source of confusion later if we don't fix it. So we will pause to address this matter. We will come back to the binomial expansion afterward. Next Dashboard Calendar To Do Notifications Inbox
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