f is continuous at 0, f' is continuous at 0, f'' is continuous at 0...PeteSampras said:Homework Statement
Is it possible obtain a Taylor serie at x0=0?Homework Equations
[tex]f(x)= (\frac{x^4}{x^5+1})^{1/2} [/tex][/B]
The Attempt at a Solution
I think that it is not possible , since f' is not differenciable at x=0, since f' have the factor
PeteSampras said:[tex](\frac{x^4}{x^5+1})^{-1/2} [/tex]
but, for example wolfram yield a solution f approx x2
http://www.wolframalpha.com/widget/widgetPopup.jsp?p=v&id=f9476968629e1163bd4a3ba839d60925&title=Taylor Series&theme=blue&i0=sqrt(x^4/(x^5+1))&i1=x&i2=0&i3=10&podSelect=&showAssumptions=1&showWarnings=1
I think that it is not possible , since f' is not differenciable at x=0, since f' have the factor
(x4/x5+1)−1/2
A Taylor Series is a mathematical representation of a function as an infinite sum of terms that are calculated from the values of the function's derivatives at a single point. It is used to approximate functions in the form of polynomials.
To obtain a Taylor Series at x0 = 0, we first need to find the derivatives of the function at x0 = 0. Then, we plug in the values of these derivatives into the formula for a Taylor Series, which is f(x) = f(x0) + f'(x0)(x-x0) + (f''(x0)/2!)(x-x0)^2 + (f'''(x0)/3!)(x-x0)^3 + ...
The purpose of obtaining a Taylor Series at x0 = 0 is to approximate a function at a particular point by using a polynomial. This can be useful in situations where the function is difficult to calculate, but its derivatives are easier to find.
A Maclaurin Series is a special case of a Taylor Series, where x0 = 0. This means that a Maclaurin Series is simply a Taylor Series obtained at x0 = 0. They are essentially the same, but the term "Maclaurin Series" is usually used when x0 = 0 is explicitly stated.
The accuracy of a Taylor Series approximation at x0 = 0 depends on the number of terms used in the series. The more terms that are included, the more accurate the approximation will be. However, it is important to keep in mind that the Taylor Series is only an approximation and will never be an exact representation of the function.