→Taylor-expansion is a method of approximating a function f(x) around a point a with a polynomial of the argument x in the vicinity of a. The polynomial itself consists of the derivatives of the function of various orders.
Tn(x) = f(a) + f ‘ (a) (x-a) + f ”(a) (x-a)2/2! + f(3)(x) (x-a)3/3! + …. + R(x). The Taylor-expansion of the exponential function is:
e<sup>x<\sup> = 1 + x + x2/2 + x3/3! + ……. + xn/n!
To prove the validity of this statement consider the special case of MacLaurin-polynomials were a function is expanded around x=0. Observe the polynomials
p(x) = a +bx + cx2 + dx3
p'(x) = b + 2cx + 3dx2
p”(x) = 2c + 2 *3dx.
x= 0 in each of those equalities and get expressions for the coefficients a,b,c and d. p(0) = a → a=p(0)
p'(0) = b → b = p'(0)
p”(0) = 2c → c = p”(0)/2
p(3) (0) = 2•3 d → d= p(3)(0)/ 2•3
Therefore the given polynomial can be written as
p(x) = p(0) + p'(0) x + p”(0)x2 /2+ p(3)(0)/3 !+ ……..pn(0) Xn\n!
Learn about buoyancy and how to compare apple and pears:
Integration by parts can be regarded as the inverse to the product rule for differentiation. Suppose U(X) and V(x) are two differentiable functions. According to the product rule
dU(x)V(x)/dx = U(x) dV(x)/dx + V(x)dU(x)/dx = U(x) dV(x)/dx+ V(x)dU(x)/dx
Integrating both sides of this equation and transposing terms, we obtain
∫U(x)dV(x)/dx dx = U(x)V(x) – ∫ V(x)dU/dx dx
This is the general formula for integration by parts.
In each application we break up the integrand into a product of two pieces U and V’: where V’ is easier to integrate.
With this method one can differentiate for example lnx.
Let U=lnx then dU/dx = 1/x and dU = dx/x.
and dV = dx → V=x.
∫lnxdx = x lnx – ∫x1/xdx = xlnx – x + C