Maclaurian' series example "f(x)=tanx" give me answer
surely you can just look it up.
a derivation can be found at
http://www.petervis.com/mathematics/maclaurin_series/maclaurin_series_tanx.html
To find the Maclaurin series for the function f(x) = tan(x), we can use the formula for the Maclaurin series expansion:
f(x) = f(0) + f'(0)x + (f''(0)/2!)x^2 + (f'''(0)/3!)x^3 + ...
First, let's calculate the derivatives of the function f(x) = tan(x) at x = 0:
f(x) = tan(x)
f'(x) = sec^2(x)
f''(x) = 2 * sec^2(x) * tan(x)
f'''(x) = 2 * sec^2(x) * sec^2(x) + 4 * sec^4(x) * tan(x)
Now, let's evaluate these derivatives at x = 0:
f(0) = tan(0) = 0
f'(0) = sec^2(0) = 1
f''(0) = 2 * sec^2(0) * tan(0) = 0
f'''(0) = 2 * sec^2(0) * sec^2(0) + 4 * sec^4(0) * tan(0) = 2
Plugging these values into the Maclaurin series formula, we get:
f(x) = 0 + 1x + (0/2!)x^2 + (2/3!)x^3 + ...
Simplifying, we have:
f(x) = x + (x^3/3) + (2x^5/15) + (17x^7/315) + ...
So, the Maclaurin series expansion for f(x) = tan(x) is:
f(x) = x + (x^3/3) + (2x^5/15) + (17x^7/315) + ...
This series will approximate the function tan(x) for values of x near 0.