Maclaurin series approximation. To approximate ∫(arctan(x) - x)/x^2 dx on [0, 1/2]?
the Maclaurin series of arctan(x) (and, from that, the Maclaurin series of the integrand) is used. An alternating series for the (exact) value S of the definite integral results. An approximation to S is obtained by using the minimum number of terms that, by the Alternating Series Test, guarantee an absolute error less than 0.001. What is the approximation?
can someone provide step by step. thanks a lot.
Update:ans is -37/960
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arctan(x) = integral (1/(1+x^2) dx) = integral ( SUM ( (-1)^(n) (x^2)^n ) dx ) n = 0 to infinity
= SUM ( (-1)^n (x^(2n+1)/(2n+1) )
Therefore
(arctan(x) - x) / x^2 = SUM ( (-1)^n x^(2n-1)/(2n+1) ) n = 1 to infinity
And integral (arctan(x) - x)/x^2 ) = SUM ( (-1)^n x^2n/[2n(2n+1)] ) n = 1 to infinity
The truncation error after truncating the series at n = N is bounded by
SUM ( (1/2)^2n (-1)^n / (2n(2n+1) ) from n = N+1 , to infinity
The absolute value of the n-th term is 1/(2n(2n+1)) = (1/(2n) - 1/(2n+1) )
So an upper bound for the error terms is given by (1/2)^(2(N+1) ) / 2(N+1) = (1/2)^(2N+3) / (N+1)
Since we want the maximum error to be less than 0.001, we have (1/2)^(2N+3) / (N+1) < 0.001
Trying some values of N, we find that the smallest N this happens is N = 3
(1/2)^(9) / (4) < .001, therefore we can trucate the series after N = 3, and the approximation is
P(x) = - x^2 / 6 + x^4 /20 - x^6 /42
The exact integral is F(x) = {- arctan(x) / x - 0.5 ln (1+x^2)} | 0 to x
= - arctan(x) / x - 0.5 ln (1 + x^2) - ( -1 + 0) = 1 - arctan(x)/x - 0.5 ln(1+x^2)
The exact value of the integral at x = 1/2 equals
F(1/2) = 1 - 2 arctan(1/2) - 0.5 ln (1 + 1/4) = -0.0388669936587
The approximate value is
P(1/2) = - (1/2)^2 / 6 + (1/2)^4 /20 - (1/2)^6 /42
= -0.038913690476
The absolute difference is | F(1/2) - P(1/2) | = | -0.03886699 + 0.03891369 | = 0.0000467
which is less than 0.001 as expected.
Start with the geometric series 1/(1 - x) = Σ(n = 0 to â) x^n.
Let x = -t^2:
1/(1 + t^2) = Σ(n = 0 to â) (-1)^n x^(2n).
Integrate both sides from 0 to x:
arctan x = Σ(n = 0 to â) (-1)^n x^(2n+1)/(2n+1)
.............= x + Σ(n = 1 to â) (-1)^n x^(2n+1)/(2n+1).
So, arctan(x) - x = Σ(n = 1 to â) (-1)^n x^(2n+1)/(2n+1)
==> (arctan(x) - x)/x^2 = Σ(n = 1 to â) (-1)^n x^(2n-1)/(2n+1).
Hence, integrating both sides from 0 to 1/2 yields
â«(x = 0 to 1/2) (arctan(x) - x) dx/x^2
= Σ(n = 1 to â) (-1)^n x^(2n)/(2n (2n+1)) {for x = 0 to 1/2}
= Σ(n = 1 to â) (-1)^n (1/2)^(2n)/(2n (2n+1))
= Σ(n = 1 to â) (-1)^n / (2^(2n+1) * n(2n+1)).
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Since this is an alternating series, we look for the term in the series where
|(-1)^n / (2^(2n+1) * n(2n+1))| < 0.001
==> 1 / (2^(2n+1) * n(2n+1)) < 0.001
==> 2^(2n+1) * n(2n+1) > 1000
==> n = 3 or higher.
So we can truncate this series after the n = 2 term for the desired accuracy:
â«(x = 0 to 1/2) (arctan(x) - x) dx/x^2
â Σ(n = 1 to 2) (-1)^n / (2^(2n+1) * n(2n+1))
= -1/(2^3 * 1 * 3) + 1/(2^5 * 2 * 5)
= -37/960.
I hope this helps!