Polynomials 3: Powers of x 0 + h

Transcription

1 near small binomial Capter 17 Polynomials 3: Powers of + Wile it is easy to compute wit powers of a counting-numerator, it is a lot more difficult to compute wit powers of a decimal-numerator. EXAMPLE 1. Wile it is easy to find tat: it is a lot more difficult to find tat = = But te main issue is tat te result of a repeated-multiplication wit a base tat is a decimal numerator will usually involve a lot more decimals tan are in te base and tan we really want so tat a lot of te work is wasted. EXAMPLE 2. In = te base, 3.14, as only two decimals but te result, , most probably as a lot more decimals tan we want. In tis capter we will investigate a procedure tat will allow us to get only te number of decimals we want. It is based on te fact tat any decimal numerator is always near a counting numerator in te sense tat any decimal numerator is equal to a counting-numerator plus a small numerator EXAMPLE is near 3 because 3.14 = and 0.14 is small We will tus investigate te powers of te binomial +. We will begin by investigating te case in wic te repeated multiplication involves two copies of te binomial and ten te case in wic te repeated multiplication involves tree copies of te binomial. Ten we will develop a procedure for 197

2 198 CHAPTER 17. POLYNOMIALS 3: POWERS OF X 0 + H te cases in wic te repeated multiplication involves at least tree copies of te binomial Te Second Power: ( + ) 2 1. In tis case, te repeated-multiplication procedure is simple enoug. In order to compute te second power of +, we write, keeping in mind tat we want te monomials to appear in order of diminising sizes and since and bot stand for signed numerators: 2 x 2 0 x a. We begin by looking at wat appens in aritmetic wic is tat te multiplication procedure essentially keeps track and respects te sizes but, because of carryovers, only rougly so. EXAMPLE 4. In order to compute 3.2 2, we actually compute ( ) 2 and write since we are dealing wit plain numerators: tat is Te multiplication procedure kept rougly track of te sizes except for wat te carryover caused: All te way to te left are te ones In te middle are te tents All te way to te rigt are te undredts so tat if we want: No decimal, we write = 10 + (...) One decimal, we write = (...)

3 17.1. THE SECOND POWER: (X 0 + H) Two decimals, we write = were + (...) is tere to remind us tat we are ignoring someting too in te tents to matter ere. EXAMPLE 5. In order to compute 2.8 2, we observe tat 2.8 is nearer 3 tan 2 so tat we actually compute (3 0.2) 2 and write since we are now dealing wit signed numerators: tat is ( 0.2) 2 (+3) (+3) ( 0.2) Te multiplication procedure kept rougly track of te sizes except for wat te carryover caused: All te way to te left are te ones In te middle are te tents All te way to te rigt are te undredts so tat if we want: No decimal, we write = 8 + (...) One decimal, we write = (...) Two decimals, we write = 7.84 were + (...) is tere to remind us tat we are ignoring someting, positive or negative, too in te tents to matter ere. b. In algebra, a very frequent case is wen we want a template for te power of any decimal-numerator in te neigborood of a given. In oter words, we do not want yet to commit ourselves to ow far te decimal-numerator is from te given and we use to represent ow far te decimal-numerator is from te given. Of course, wen, ultimately, we replace by some in te tents number, tere remains te possibility tat a carryover will mess up te result a little bit. Tis toug is someting tat we will not deal wit ere. (But see te Epilogue.)

4 200 CHAPTER 17. POLYNOMIALS 3: POWERS OF X 0 + H EXAMPLE 6. In order to get a template for te second power of any decimal-numerator near 3, bot above 3 and below 3, we write: Anoter, muc more fruitful to get te above template is to go back to te definition of multiplication in terms of te area of a rectangle so tat ( + ) 2 is te area of a + by + square: + + ( +) 2 Wat we ten do is to enlarge te sides of a by square by but, for te sake of clarity, we will construct te enlarged square one step at a time: i. We start wit x 2 0 as te area of a square wit side : 2 ii. We now enlarge te sides of te square by in eac dimension wic adds two + by rectangles:

5 17.1. THE SECOND POWER: (X 0 + H) iii. We complete te enlarged square by adding one by square: 2 2 EXAMPLE 7. In order to get a template to get te second power of any decimalnumerator near 3, bot above 3 and below 3, we visualize te above picture and see in our mind tat we need te area of: i. te original square: 3 2 ii. te two rectangular strips: 2 3 iii. te little square: 2 so tat we ave te template: (3 ) 2 = Tis second approac as tree major advantages over te first one: i. Te terms in te sum are clearly in order of diminising size: Since is in te ones and is in te tents, bot dimensions of te initial square are in te ones so tat x 2 0 is in te ones, one dimension of te rectangles is in te ones but te oter dimension is in te tents so tat 2 is in te tents, bot dimensions of te little square are in te tents so tat 2 is in te undredts. ii. Wen we will need formulas for ( + ) 3, ( + ) 4, etc, not only will repeated multiplication get out of and but, as we sall see, we will never

6 202 CHAPTER 17. POLYNOMIALS 3: POWERS OF X 0 + H need more tan te first few monomials in te result. iii. If all we need is a particular monomial in te result, wic is often te case, we can get it from te picture witout aving to do te wole repeated multiplication. EXAMPLE 8. If, for watever reason, we need te monomial in (3 ) 2, we visualize te two rectangular strips and we write: 2 3 THEOREM 9 (Addition Formula for Quadratics). ( + ) 2 = x Te Tird Power: ( + ) 3 For te sake of brevity we omit te investigation of wat appens in aritmetic. 1. Te repeated-multiplication procedure already begins to be painful: First we must multiply two copies of + : x x Ten, we must multiply x by te tird copy of + x x x x x x Anoter, muc more fruitful approac to te addition formula is to go back to te definition of multiplication in terms of te area/volume of a rectangle so tat ( + ) 3 is te volume of a + by + by + cube: Wat we ten do is to enlarge te tree sides of a by cube by but, for te sake of clarity, we will construct te enlarged cube one step at a time: i. We draw te initial cube wit volume x 3 0 :

7 17.2. THE THIRD POWER: (X 0 + H) ii. We draw te tree slabs wit volume 3x 2 0 : iii. We glue te tree slabs wit volume 3x 2 0 onto wat we already glued: iv. We draw te tree rods wit volume 3 2 :

8 204 CHAPTER 17. POLYNOMIALS 3: POWERS OF X 0 + H v. We glue te tree rods wit volume 3 2 onto wat we already glued: vi. We draw te little finising cube wit volume 3 : We glue te little finising cube wit volume 3 onto wat we already glued: Tis approac as tree major advantages over te repeated-multiplication: i. Te terms in te sum are in order of diminising size. Since is in te ones and is in te tents, all tree dimensions of te initial cube are in te ones so tat x 3 0 is in te ones, two dimensions of te slabs are in te ones but te tird dimension is in te tents so tat, if is in te tents, ten 3x 2 0 is in te tents, one dimensions of te square rods is in te ones so tat, if is in te tents, ten 3 2 is in te undredts. all tree dimensions of te little cube are in te tents so tat, if is in te tents, ten 3 is in te tousandts.

9 17.3. HIGHER POWERS: (X 0 + H) N WHEN N > ii. If all we need is a particular one of te terms, wic will often be te case, we can get it from te picture witout aving to do te wole multiplication. iii. Later on, wen we sall need formulas for ( + ) 4, etc, not only will repeated multiplication get out of and but, as we sall see, we will never need more tan te first few monomials of te result. THEOREM 10 (ADDITION FORMULA for CUBICS). pattern ( + ) 3 = x x Higer Powers: ( + ) n wen n > 3 Here of course: Repeated-multiplication is of course going to be ever more painful We cannot make pictures because we would need to be able to draw in more tan 3 dimensions. So, we need to find a procedure. 1. We begin by looking for a pattern in wat we ave so far. In order to see better wat we are doing, we will not let anyting go witout saying. a. Wen te exponent is 3, we ad: ( + ) 3 = x x = x x x = & 3 & 3 & = 1 & 3 & 3 & 1 Looking at te factors and te coefficients separately, we get te following: Te factors are In oter words, starting wit 3 copies of we get te oters by replacing one by one te copies of by copies of. Te coefficients are Here we cannot see te pattern, b. Wen te exponent is 2, we ave ( + ) 2 = x = x x = & 2 &

10 206 CHAPTER 17. POLYNOMIALS 3: POWERS OF X 0 + H = 1 & 2 & 1 Looking at te factors and te coefficients separately, we get te following: Te factors are In oter words, starting wit we get te oters by replacing one by one te copies of by copies of. Te coefficients are Here again we cannot see te pattern. c. Wen te exponent is 1, we ave ( + ) 1 = + = x = x = & = 1 & 1 Looking at te factors and te coefficients separately, we get te following: Te factors are In oter words, starting wit we get te oters by replacing te one copy of by a copy of. Te coefficients are 1 1 Here we cannot see te pattern, 2. Putting everyting togeter, toug, Te procedure for finding te powers seems to be in all cases: i. Make as many copies of as wat te exponent n in ( + ) n indicates ii. Make as many copies plus 1 of wat te exponent n in ( + ) n indicates iii. Starting wit te second copy, replace one by one te copies of by copies of In order to see a pattern for te coefficients, we write tem starting wit exponent 1 and ending wit exponent 3:

11 17.3. HIGHER POWERS: (X 0 + H) N WHEN N > Te way tings are arranged, we see tat we get eac entry in wat is called te PASCAL TRIANGLE by adding its two parent-entries tat is te two entries just above it. Tus, te next line in te PASCAL TRIANGLE would be: Proving tat all tis is indeed te case would involve more work tan we are willing to do ere and so we will take te following for granted: THEOREM 11 (BINOMIAL THEOREM). Te addition formula for a binomial of degree n is: ( + ) n = x n n 1 xn n(n 1) x n n(n 1)(n 2) According to te BINOMIAL THEOREM, + ( + ) 0 = 0 0 x n n(n 1)(n 2) (1) n 0 n wic is of course as it sould be. Moreover, since te coefficient 1 goes witout saying, tis means tat te very first line in te PASCAL TRIANGLE is 1 so tat te complete PASCAL TRIANGLE is: = 1 PASCAL TRIANGLE parent-entries n := 0 1 n := n := n := n := n := n := Te numerators in te second slanted row (bold-faced) are te coefficients of te +1 powers wic sows tat te coefficient of te +1 power in x n 0 is n. We ceck tat te tird slanted row are te coefficients of te +2 powers wic sows tat te coefficient of te +2 power in x n n(n 1) 0 is. 2

12 208 CHAPTER 17. POLYNOMIALS 3: POWERS OF X 0 + H constant approximation affine approximation quadratic approximation Etc 17.4 Approximations Fortunately, most of te time we only need te very first few terms of te addition formulas. 1. Very often, we will need only te constant approximation of ( + ) n wic is just x n 0. Indeed, very often will be small enoug tat we will not need to consider any of te monomials tat involve it and we will write: ( + ) n = x n 0 + (...) EXAMPLE 9. Te constant approximation of is 16 7 and we write = (...) More generally, te constant approximation of (16 + ) 7 is 16 7 and we write (16 + ) 7 = (...) 2. Wen te constant approximation is too crude, we will often use te affine approximation of ( + ) n wic is x n 0 + n. Indeed, wile may not be small enoug not to matter, te oter powers, 2, 3 etc being smaller tan can often still be ignored and we will ten write ( + ) n = x n 0 + nxn (...) EXAMPLE 10. Te affine approximation of is and we write = (...) More generally, te affine approximation of (16 + ) 7 is (...) and we write (16 + ) 7 = (...) 3. And finally we will also use te quadratic approximation of ( + ) n wic is x n 0 + n + n(n 1) 2 2 wen we will need more precision tat te affine approximation will be able to give us and we will ten write ( + ) n = x n 0 + n + n(n 1) (...) EXAMPLE 11. Te quadratic approximation of is and we write = (...) More generally, te quadratic approximation of (16+) 7 is and we write (16 + ) 7 = (...)