UNIT 1: ANALYTICAL METHODS FOR ENGINEERS

Transcription

1 UNIT : NLYTICL METHODS FOR ENGINEERS Unit code: /0/0 QCF Level: Credit value: OUTCOME TUTORIL LGEBRIC METHODS Unit content Be able to analyse and model engineering situations and solve problems using algebraic methods lgebraic methods: polynomial division; quotients and remainders; use of factor and remainder theorem; rules of order for partial fractions (including linear, repeated and quadratic factors); reduction of algebraic fractions to partial fractions. Eponential, trigonometric and hyperbolic functions: the nature of algebraic functions; relationship between eponential and logarithmic functions; reduction of eponential laws to linear form; solution of equations involving eponential and logarithmic epressions; relationship between trigonometric and hyperbolic identities; solution of equations involving hyperbolic functions. rithmetic and geometric: notation for sequences; arithmetic and geometric progressions; the limit of a sequence; sigma notation; the sum of a series; arithmetic and geometric series; Pascal s triangle and the binomial theorem. Power series: epressing variables as power series functions and use series to find approimate values e.g. eponential series, Maclaurin s series, binomial series. You should judge your progress by completing the self assessment eercises. This is a useful web site that eplains partial fractions D.J.Dunn

2 . PRTIL FRCTIONS Partial fractions are a useful way of simplifying polynomial epressions. This is done so that they are easier to integrate or epand. The degree of factor refers to the highest power of in any factor (terms inside the brackets). FIRST DEGREE LINER FCTORS In the following epression, we have factors containing and 0 so they are linear and first order. (Remember 0 is and not normally written and is normally written as ). f() a b c Every time two linear factors are multiplied the powers increase by so if we have three factors on the bottom line, the denominator has a highest power of. So long as the numerator f() has a lower power then the epression is eactly the same as the sum of three partial fractions such that: f() B C a b c a b c The method of solution is shown in the following eamples. WORKED EXMPLE No. Find the partial fractions for B B - Multiply through by ( ) The relationship must be valid for any value of so conveniently put = B(0) Hence = 8/7 7 7 Now start again and this time multiply through by ( + ) and make = - 0 B B Hence B = /7 7 In the last eample we could use the following method to obtain the result. To find cover the factor ( ) and put = in the remainder. () 8 7 To find B cover the factor ( +) and put = - in the remainder. (-) B - 7 When we have three polynomials on the bottom line we need three partial fractions but we can still use the cover up method.

3 WORKED EXMPLE No. Find the partial fractions for Cover ( ) and put = () () Cover ( + ) put = - B 8 Cover ( ) put = 0 C 8 B hence C When the numerator has a higher degree than the denominator, we can use long division to reduce the epression until the remainder meets this condition. REMINDER THEOREM When we divide a number by another and the result is not eact, it leaves a remainder. For eample: remainder or in other words. In this case is the divisor, is the quotient and is the remainder. We can say that = + In the same way when we divide polynomials we can say: Divisor quotient + remainder polynomial In particular if the divisor is ( + a) and the polynomial is f () then f () ( + a) quotient + remainder. If = -a then f (a) = (a a) quotient + remainder. f (a) = remainder This gives an easy way of finding the remainder when a polynomial is divided by ( a) THE FCTOR THEOREM If the remainder is zero, then the polynomial will factorise.

4 WORKED EXMPLE No. - Divide and find the remainder by the remainder theorem. The first term of the resulting polynomial must be = Multiply ( + ) ( ) = + The reminder is found by subtracting Now divide again (- - +) ( +) The result must be (- ) = - ( + ) (-) = - - The reminder is found by subtracting Now divide again (- + ) ( +) The result must be (-) = - ( + ) (-) = - - The reminder is found by subtracting The whole process is usually shown like this The result of the division is the polynomial - and the remainder is The final remainder is. This may be found using the remainder theorem as follows. The divisor is + use = - f( ) ( )( ( ) ( ) ) 0(( ) ( ) )

5 WORKED EXMPLE No. Find the partial fractions for the epression Multiply out the brackets 8 If we now divide out using long division we get: 8 8 remainder 7 remainder hence 8 8 The remainder now has a lower degree than the denominator so we can change this to partial fractions. B 8 8 () () 7 B hence

6 REPETED FCTORS If we have two or more identical factors, they can be written as the factor to a suitable power. This is called a repeated factor. I can be shown that the partial fractions take the following form. a b a b b b The method of solution is shown in the net eample. B C D WORKED EXMPLE No. Find the partial fractions for the epression B C D The partial factors take the form of : can be solved by the cover up method. However, when we multiply both sides by ( +) and put = -, only D will remain so D may be solved by the cover up method. D 7 - B C We now have: 7 Multiply both sides by ( )(+) and: B C 7 B C 0 7 Multiply out in full B B B C C C The coefficients of must add up to zero so /7 + B = 0 B = -/7 0 It is clear that we did not need to multiply out to obtain this result as the coefficients of are clear without doing this. To find C we can equate the coefficients of and it is clear that /7 + C = 0 and C = -/7 but this is not so obvious without multiplying out. nother way is to equate coefficients of o (i.e. the constants) B C 0 C C = - /

7 QUDRTIC (SECOND DEGREE) FCTORS Second degree factors are brackets that contain in them. These should be factorised if possible to change them into first degree factors. If they are not easily factorised it can be shown that for each quadratic factor there is a partial fraction in the following form. f() C D a b a c repeated quadratic factor is dealt with in the same way as before hence: f() C D E F a c a b a c The method of solution is outlined in the following eample. WORKED EXMPLE No. Find the partial fractions for the epression The partial factors take the form of : and B are found by the cover up method. () () B 7 B C D C D 7 C D 7 Equate coefficients of C C (multiply it out in full if you need convincing) Equate coefficients of o (i.e. the constant terms) D D D /77 /

8 SERIES EXPNSION The following eample shows how partial fractions help us to epand an epression. WORKED EXMPLE No.7 Simplify the following epression so that it is a series in ascending orders of. B Use the cover up method to find putting = -. B C B C Equate the coefficients of 0 B B - Equate the constants. C C - (.) Rearrange into the required form. Note this can now be epanded to form:

9 SELF SSESSMENT EXERCISE No. Find the partial fractions for the following epressions nswers ( 7)