Understanding Engineering Mathematics

14 Analysis for Engineers – Limits, Sequences, Iteration, Series and All That The topics considered in this chapter are often re ...

evaluation and properties of limits continuity slope of a curve and theory of differentiation infinite sequences iteration and ...

However, we will show that it cannot in fact be represented by a decimal with a finite number of places – i.e. it isnota rationa ...

On my calculator √ 2 = 1 .41421356237. My calculator squares this to 1.9999999998, so it is clearly not the square root of 2. ...

Problem 14.2 Evaluatef.x/=x^2 Y 2 xY3forx= 1 .1, 1.01, 1.001, 1.0001. What do you thinkf.x/approaches asxgets closer and closer ...

The table of values suggests that asx→1,f(x)→2 (the arrow represents ‘tends to’). However, of course, f( 1 )= 12 − 1 1 − 1 = 0 0 ...

The properties of limits are fairly well what we might expect. Thus, if lim x→a f(x)=b lim x→a g(x)=c then L1. lim x→a kf (x)=kb ...

and again you can see we get this directly from lim x→ 1 [ x+ 1 √ x^2 + 2 x+ 3 ] = 2 √ 6 Note that it wouldnotbe correct to writ ...

q B A P 1 Figure 14.4 From Figure 14.4 we see that ifPAis an arc of a unit circle subtending an angleθ at the centre (173 ➤ ), t ...

So asx→∞, ex xn →∞–i.e.exis ‘stronger’ thanxn; for any positive integern.Sofor example: xn ex =xne−x→0asx→∞ In fact, it is not d ...

A functionf(x)such that lim x→a f(x) =f(a) is said to bediscontinuous atx=a. For example,f(x)= x+ 3 (x− 1 )(x+ 5 ) is discontinu ...

This is a rather sophisticated problem, but is worth working through because it brings together a number of the key subtleties o ...

The functiong(x)is continuousand indeed is equivalent to g(x)= 1 x− 1 forx> 2 Exercise on 14.4 Consider the following functio ...

curve is f(a+h)−f(a) a+h−a = f(a+h)−f(a) h Ash→0, the chord approaches the tangent to the curve atx=a, so the slope of the curve ...

arithmetic operations such as addition and multiplication. Fortunately, we can express cosx(xin radians!) in terms of just such ...

Exercise on 14.6 Adapt the argument of this section to show that the series 1 + 1 √ 2 + 1 √ 3 + 1 √ 4 +···+ 1 √ r +··· does not ...

(iii) In this caseun= 1 +(n− 1 )3 is pretty obvious. (iv) In this case it is perhaps not quite so easy to spot a pattern for an ...

l n Figure 14.9A sequence approaching a limitl. If a sequence has a finite limit, we say itconverges, if not, itdiverges. Graphi ...

wherexn is thenth approximation. Starting with a first approximation x 1 we find another by: x 2 =F(x 1 ) and another x 3 =F(x 2 ...

14.9 Infinite series An infinite series (105 ➤ ) is one of the form S=u 1 +u 2 +u 3 +···+un+··· i.e. the sum of the terms of an ...