Question

The Maclaurin series expansion for the arctangent of x is defined for |x| ≤ 1 as arctan x = ∑ n=0 [infinity] (−1)n ______ 2n +1 x 2n+1 (a) Write out the first 4 terms (n = 0,...,3). (b) Starting with the

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Maxsilver · Answer 1 · 2021-02-27T08:07:10+0000

Answer:

a)
n =0, ((-1)^0)/(2*0+1) x^(2*0+1)= x

n =1, ((-1)^1)/(2*1+1) x^(2*1+1)= -(x^3)/(3)

n =2, ((-1)^2)/(2*2+1) x^(2*2+1)= (x^5)/(5)

n =3, ((-1)^3)/(2*3+1) x^(2*3+1)= -(x^7)/(7)

b) n=0

$arctan(\pi/6) \approx \pi/6 = 0.523599$

The real value for the expression is
$arctan (\pi/6) = 0.482348$

And if we replace into the formula of relative error we got:

$\% error= (|0.523599 -0.482348|)/(0.482348) * 100= 8.55\%$

n =1

$arctan(\pi/6) \approx \pi/6 -((pi/6)^3)/(3) = 0.47576$

$\% error= (|0.47576 -0.482348|)/(0.482348) * 100= 1.37\%$

n =2

$arctan(\pi/6) \approx \pi/6 -((pi/6)^3)/(3) +((pi/6)^5)/(5) = 0.483631$

$\% error= (|0.483631 -0.482348|)/(0.482348) * 100= 0.27\%$

n =3

$arctan(\pi/6) \approx \pi/6 -((pi/6)^3)/(3) +((pi/6)^5)/(5)-((pi/6)^7)/(7) = 0.48209$

$\% error= (|0.48209 -0.482348|)/(0.482348) * 100= 0.05\%$

$\arctan (\pi/6) = 0.48$

Explanation:

Part a

the general term is given by:

a_n = ((-1)^n)/(2n+1) x^(2n+1)

And if we replace n=0,1,2,3 we have the first four terms like this:

n =0, ((-1)^0)/(2*0+1) x^(2*0+1)= x

n =1, ((-1)^1)/(2*1+1) x^(2*1+1)= -(x^3)/(3)

n =2, ((-1)^2)/(2*2+1) x^(2*2+1)= (x^5)/(5)

n =3, ((-1)^3)/(2*3+1) x^(2*3+1)= -(x^7)/(7)

Part b

If we use the approximation
$arctan x \approx x$ we got:

n=0