What is the maclaurin series for x/sinx and the maclaurin series for 2/sinx..
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Start out with the Maclaurin series for sin(x):
sin(x) = x - x^3/3! + x^5/5! - x^7/7! + ... .
Now, note that x/sin(x) is even as:
-x/sin(-x) = -x/[-sin(x)] = x/sin(x).
Thus, only even-powered terms will appear in the series.
So, for some A, B, C, D, ..., you can write:
x/sin(x) = A + Bx^2 + Cx^4 + Dx^6 + ...
==> x = sin(x)(A + Bx^2 + Cx^4 + Dx^6 + ...)
==> x = (x - x^3/3! + x^5/5! - x^7/7!)(A + Bx^2 + Cx^4 + Dx^6 + ...)
==> x = Ax + (B - A/3!)x^3 + (A/5! - B/3! + C)x^5 + (D - C/3! + B/5! - A/7!)x^7 + ... .
(i) By comparing x coefficients: A = 1
(ii) By comparing x^3 coefficients: B - A/3! = 0 ==> B = A/3! = 1/6.
(iii) By comparing x^5 coefficients: A/5! - B/3! + C = 0 ==> C = B/3! - A/5! = 7/360
(iv) By comparing x^7 coefficients: D = C/3! - B/5! + A/7! = 31/15120.
Therefore:
x/sin(x) = 1 + (1/6)x^2 + (7/36)x^4 + (31/15120)x^6 + ... .
The second problem can be done in a similar fashion.
I hope this helps!
sin(x) = x - x^3/3! + x^5/5! - x^7/7! + ... .
Now, note that x/sin(x) is even as:
-x/sin(-x) = -x/[-sin(x)] = x/sin(x).
Thus, only even-powered terms will appear in the series.
So, for some A, B, C, D, ..., you can write:
x/sin(x) = A + Bx^2 + Cx^4 + Dx^6 + ...
==> x = sin(x)(A + Bx^2 + Cx^4 + Dx^6 + ...)
==> x = (x - x^3/3! + x^5/5! - x^7/7!)(A + Bx^2 + Cx^4 + Dx^6 + ...)
==> x = Ax + (B - A/3!)x^3 + (A/5! - B/3! + C)x^5 + (D - C/3! + B/5! - A/7!)x^7 + ... .
(i) By comparing x coefficients: A = 1
(ii) By comparing x^3 coefficients: B - A/3! = 0 ==> B = A/3! = 1/6.
(iii) By comparing x^5 coefficients: A/5! - B/3! + C = 0 ==> C = B/3! - A/5! = 7/360
(iv) By comparing x^7 coefficients: D = C/3! - B/5! + A/7! = 31/15120.
Therefore:
x/sin(x) = 1 + (1/6)x^2 + (7/36)x^4 + (31/15120)x^6 + ... .
The second problem can be done in a similar fashion.
I hope this helps!
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In order to compute a maclaurin series, you need to evaluate the expression and all of its derivatives at x=0. If you cannot do this directly, then limits will suffice -- compute the limit as x->0 of x/sin(x).
L'Hopital's rule: lim x->0 = 1/cos(x), which comes out to 1. You're not done, you only have the first term in the series. You still need to differentiate the expression (which is easy enough, you use the product rule with x*csc(x) and that will give you what you need), and if you can't directly evaluate later expressions, then use limits again. If any of the limits do not exist, then the expression has no MacLaurin series. This first one should have a MacLaurin series with a radius of convergence of pi, if I remember correctly. The second one does not have a MacLaurin series because the limit as x->0 of 2/sin(x) does not exist. However, the second one does have a Taylor series about 0
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L'Hopital's rule: lim x->0 = 1/cos(x), which comes out to 1. You're not done, you only have the first term in the series. You still need to differentiate the expression (which is easy enough, you use the product rule with x*csc(x) and that will give you what you need), and if you can't directly evaluate later expressions, then use limits again. If any of the limits do not exist, then the expression has no MacLaurin series. This first one should have a MacLaurin series with a radius of convergence of pi, if I remember correctly. The second one does not have a MacLaurin series because the limit as x->0 of 2/sin(x) does not exist. However, the second one does have a Taylor series about 0
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1 + x^2/6 + (7 x^4)/360 + (31 x^6)/15120 + (127 x^8)/604800 +
(73 x^10)/3421440
(73 x^10)/3421440
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hi
1
keywords: help,series,for,will,please,sinx,be,What,maclaurin,What will be maclaurin series for x/sinx? please help :(