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Today I read that the ten most important British inventions are:
→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
Q.E.D.
If the primitive function of an integrand can be found it is always best to take advantage of the fundamental theorem of calculus.
In order to be able to determine integrals whose indefinte integrals(primitive functions) cannot be found immediately some of the following motions can be useful:
∫f'(g(x))g'(x)dx = f((g(x)) + C
Then perform the substitution u = g(x). If we differentiate this we get du = g'(x) dx.
Substitution in the integral above yields the following:
∫ f(g(x)) g'(x) dx = ∫f ‘(u) du = f(u) + C
substituting back to g(x) and we have the answer: f((g(x)) + C.
Example: Determine ∫x/(x2+1)dx
This integral can be dealt with tby using the substitution
u = x2 + 1.
Then du = 2x dx → x dx = du/2.
Substitution transforms the integral to: ∫du/(2u) = ln|u| + C.
√Substituting back gives us the answer: 0,5·ln|x2 +1| + C = ln√(x2 + 1).
A good way of illustrating probabilities is to use so-called Venn-diagrams. In effect this means representing the probability of an event with circles.
Mutually excluding events can be represented by two separate non-overlapping ciecles.
P(A) + P(B) = P(A U B)
Simultanously events can be illustrated by partially overlapping circles. The overlap then represents the instance of both events happening simultanously.
A Diophantine equation is an equation in which only integers are allowed as coefficients. Also the solutions must be integers. This can be written as ax + by = c. This is a linear diophantine equation.
For non-linar diophantine equations there is no general solution formula available.
Ex. 8x + 7y = 148.
has the solutions x=1 and y = 20.
It can be showed that in order to give integer-solutions 148 must be divisible by the Greatest Common divisor of 8 and 7 in this case 1.
Speaking at the Matematikbromötet on Stockholm Nov 12th. 2012
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