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Calculus Gymnasiematematik(high school math)


→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:

Calculus Gymnasiematematik(high school math) matematik 4

Integration by parts

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