- #1
newton1
- 152
- 0
1/(1+x^2)=1-x^2 + x^4-x^6+...+(-1)^n(x^2n)+... -1 > x > 1
how to get this??...
how to get this??...
Originally posted by Newton1
1/(1+x^2)=1-x^2 + x^4-x^6+...+(-1)^n(x^2n)+... -1 > x > 1
how to get this??...
Maclaurin's series is a special case of the Taylor series, which is a mathematical representation of a function as an infinite sum of its derivatives at a single point. Maclaurin's series specifically refers to a Taylor series centered at x = 0.
To find the Maclaurin series of a function, you need to calculate the derivatives of the function at x = 0. Then, substitute these values into the general formula for a Taylor series, which is f(x) = f(0) + f'(0)x + (f''(0)/2!)x^2 + (f'''(0)/3!)x^3 + ...
Maclaurin's series is useful in mathematics and engineering because it allows us to approximate complicated functions with a simpler polynomial. This can make calculations and analysis easier and more efficient.
The main difference between Maclaurin's series and Taylor series is the center point at which the series is centered. Maclaurin's series is centered at x = 0, while Taylor series can be centered at any point in the function's domain.
No, Maclaurin's series can only be used for functions that are infinitely differentiable at x = 0. This means that the function must have derivatives of all orders at that point. Otherwise, the series will not converge and cannot be used to approximate the function.