Calculate c(sub0), c(sub1), c(sub2), c(sub3), and c(sub4) for the power series the sum from n=0 to infinity of c(subn)x^n that represents f(x)=tanx. Use the first two nonzero terms of the series to approximate the tangent of 1/4 radian. Compare your approximation with the actual value of the tangent of 1/4 radian as shown on a calculator. Your help is very much appreciated!
Usually, the efficient way to compute Taylor series is by using the Taylor series of the known standard functions like sin(x), cos(x) etc., instead of using the definition of the Taylor series in terms of the derivatives of the function.
In this case, you can write:
tan(x) = sin(x)/cos(x)
If we put:
tan(x) = c1 x + c3 x^3 + c5 x^5 + ...
(note that because tan(x) is n odd function, it can only have odd powers of x in the Taylor expansion), then we can find the ck from:
sin(x) = [c1 x + c3 x^3 + c5 x^5...] * cos(x)
if we substitute for sin(x) and cos(x) their respective series expansions and equate equal coefficients of x:
x - x^3/3! + x^5/5! - ... =
[c1 x + c3 x^3 + c5 x^5...] *
[1 - x^2/2! + x^4/4! - ...]
The coefficient of x on the r.h.s. is c1 and this has to be 1, so c1 = 1.
Equating the coefficient of x^3 on both sides gives:
-1/6 = -1/2 + c3 --->
c3 = 1/3
Equating the coefficient of x^5 on both sides gives:
1/120 = 1/24 -1/6 + c5 ---->
c5 = 2/15
So, we have:
tan(x) = x + x/3 + 2/15 x^5 + ...
To find the coefficients c(sub0), c(sub1), c(sub2), c(sub3), and c(sub4) for the power series representing f(x) = tan(x), we can use the Taylor series expansion formula for the tangent function.
The Taylor series expansion for the tangent function is:
tan(x) = x + (x^3)/3 + (2x^5)/15 + (17x^7)/315 + ...
Each term in the series is obtained by taking derivatives of the function and evaluating them at x = 0. We can see that the coefficient of the n-th term in the series is given by:
c(subn) = f^(n)(0) / n!
where f^(n)(0) represents the n-th derivative of f(x) evaluated at x = 0, and n! is the factorial of n.
Let's calculate the coefficients c(sub0), c(sub1), c(sub2), c(sub3), and c(sub4) step by step:
c(sub0) = f(0) / 0! = tan(0) / 0! = 0 / 1 = 0
c(sub1) = f'(0) / 1! = sec^2(0) / 1 = 1
To find c(sub2), we need to calculate the second derivative:
f''(x) = d^2/dx^2(tan(x)) = d/dx(sec^2(x)) = 2sec^2(x)tan(x)
c(sub2) = f''(0) / 2! = 2sec^2(0)tan(0) / 2 = 0
To find c(sub3), we need to calculate the third derivative:
f'''(x) = d^3/dx^3(tan(x)) = d/dx(2sec^2(x)tan(x)) = 2sec^2(x)sec^2(x) + 2sec^4(x)
c(sub3) = f'''(0) / 3! = (2sec^4(0) + 2sec^4(0)) / 6 = 4 / 6 = 2 / 3
To find c(sub4), we need to calculate the fourth derivative:
f''''(x) = d^4/dx^4(tan(x)) = d/dx(2sec^2(x)sec^2(x) + 2sec^4(x)) = 4sec^4(x)sec^2(x) + 8sec^2(x)sec^2(x)tan^2(x)
c(sub4) = f''''(0) / 4! = (4sec^4(0) + 8sec^4(0)tan^2(0)) / 24 = 4 / 24 = 1 / 6
Now we have the coefficients:
c(sub0) = 0
c(sub1) = 1
c(sub2) = 0
c(sub3) = 2 / 3
c(sub4) = 1 / 6
To approximate the tangent of 1/4 radian using the first two nonzero terms of the series, we can use the formula:
Approximation = c(sub0) + c(sub1)x
Substituting the values of the coefficients, we have:
Approximation = 0 + 1x = x
So, the approximation of the tangent of 1/4 radian using the first two nonzero terms is simply x.
Now let's compare this approximation with the actual value of the tangent of 1/4 radian as shown on a calculator:
Using a calculator, the tangent of 1/4 radian is approximately 0.2474.
Comparing with the approximation, which is x, we can substitute x = 1/4 in the approximation:
Approximation = 1/4 = 0.25
As we can see, the approximation of the tangent of 1/4 radian using the first two nonzero terms of the series (0.25) is a relatively close approximation to the actual value (0.2474) obtained from the calculator.