In mathematics, the Maclaurin collection is defined because the extended collection of precise capabilities. on this series, the approximated cost of the given feature may be determined as the sum of the derivatives of any function. when the function expands to zero as opposed to other values a = 0.
The formulation utilized by the Maclaurin collection calculator for computing a series expansion for any characteristic is:
$$ Σ^∞_{n=0} \frac{f^n (0)} {n!} x^n $$
Example:
Calculate Maclaurin expansion of \( e^y \) up to \( n = 4 \)?
Solution:
Given function \( f(y) = e^y \) and order point \( n = 0 \) to \( 4 \)
Maclaurin equation for the function is:
$$ f(y) = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!} y^k $$
$$ f(y) \approx \sum_{k=0}^{4} \frac{f^{(k)}(0)}{k!} y^k $$
So, calculate the spinoff and compare them on the given point to get the end result into the given formulation.
$$ f^0(y) = f(y) = e^y $$
Evaluate function:
$$ f(0) = e^0 = 1 $$
Take the first derivative:
$$ f^1(y) = [f^0(y)]' = e^y $$
$$ f^1(0) = e^0 = 1 $$
Second Derivative:
$$ f^2(y) = [f^1(y)]' = e^y $$
$$ f^2(0) = e^0 = 1 $$
Third Derivative:
$$ f^3(y) = [f^2(y)]' = e^y $$
$$ f^3(0) = e^0 = 1 $$
Fourth Derivative:
$$ f^4(y) = [f^3(y)]' = e^y $$
$$ f^4(0) = e^0 = 1 $$
as a result, replacement the values of derivatives within the components:
$$ f(y) \approx 1 + \frac{1}{1!} y + \frac{1}{2!} y^2 + \frac{1}{3!} y^3 + \frac{1}{4!} y^4 $$
$$ f(y) \approx 1 + y + \frac{y^2}{2} + \frac{y^3}{6} + \frac{y^4}{24} $$
$$ e^y \approx 1 + y + \frac{y^2}{2} + \frac{y^3}{6} + \frac{y^4}{24} $$
Maclaurin calculator reveals the strength collection extensions for any function by means of following those guidelines:
