Explaining the Principle of mathematical induction

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1        Explaining the Principle of mathematical induction:

Formally the principle of a proof by induction can be stated as follows:

A proposition P (n) involving a positive integer n, is true for all positive integral values of n if, P (1), and P (k)  P (k +1) is true.

This can be explained using a staircase as a simple analogy. Image the proposition that a man can climb a given uniform staircase, to prove this statement we need to show two things. These are that the man can get onto the first step and that he is able to climb from one step to an other. Now relating this to the formal principle of induction, the staircase can be considered the general proposition P (n). The first step of the staircase is P (1), the second P (2), the third P (3), and so on.  If we can show that the man can get onto the first step P (1) then we have ironically finished the first step of proving the proposition. The second step would be to prove that he can get from one step to an other formally put  P (k)  P (k +1). If we can show this than it follows that man can get from the first to the second step, second to the third,… n steps. Thus it can be said that P (n) is true for all positive integral values of n.

2        Definition of the derivative function f (x)

The derivative function is a general expression for the gradient of a curve at any given point. It is based on the principle of limiting a cord on a given curve spanned between two points. As the cord gets smaller, the gradient of the cord gets closer and closer to the gradient of the tangent at the given point. This is illustrated in the diagram above. The formula for the derivative from first principles is also given above; this is derived from the gradient formula ∆y/∆x (gradient of tangent to a given point).  The value of h is a very small quantity for it tends towards zero, meaning it almost equals zero.

3 a)         Obtaining the derivative of  f (x) = x2  

  1. Obtaining the derivative of  f (x) = x3                  

4        Proving by induction that the derivative of xn is nxn-1

        

        Step 1                prove true for n = 1

        

        

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                         LHS:

                        From first principles:

        

                        RHS:        

                        nx 1 – 1  =   n0  =   1

                LHS = RHS   .·. Formula holds

        Step 2                Assume true for n = k

                         dy/dx  xk  =   kx k-1

        

        

                        Now we have to prove true for n = k + 1

                        dy/dx  x k + 1   =   (k + 1)x k +1 -1

        

                        RHS:

                        (k + 1)x k +1 -1  =  (k + 1)xk

                                

                        LHS:

                              

                        dy/dx  x k ...

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