Administrivia Syllabus and other course information posted on the internet: links on blackboard & at www.dankalman.net. Check assignment sheet for reading assignments and eercises. Be sure to read about polynomials, including roots, factorization, long division and remainders. If you haven t done so already, get and install Freemat and start working on the tutorial. 1 Matrinomials Lectures 1+ O Outline Basics: polynomials and roots Horner s Form and Quick Computation Products and Sums of Roots Reverse Polynomials Sums of Reciprocal Roots Long Division and Remainders Palindromials 1
Reminder: What s a polynomial? 7 5 14 15 5 Polynomials have ROOTS 0 1 7 5 1 1) 1)(5 ( 1 7 5 Roots are related to Factors Complete factorization ).4. )(.4. 1)( ( 10 16 10 16 1) ( 1 7 5 i i 6
Comple Number Review Etension of the real numbers created by inventing a nonreal number 1. Any combination of the form with a and b real numbers is defined to be a comple number. These form a number system with addition, subtraction, and multiplication defined using the usual rules of algebra. Eg (+7i)+(4-i) = 7 + 4i; ( + 7i)(4 i) = 1 9i + 8i 1i (FOIL) = 1 + 19i + 1 (because i = 1) = + 19i. There is even division. Eg: 7 7 4 4 4 7 4 9 7 9 7 4 4 4 5 5 5. Historically, there has been a progression of different number systems developed as various types of numbers were recognized and defined: whole numbers, fractions (rational numbers), negative numbers, and irrational numbers (giving us the reals). The comple number system is just another step in this progression. 9 Fundamental Theorem of Algebra Refers to any polynomial of the form where the coefficients,,, are comple numbers and 0. This includes the case that the coefficients are real numbers. The theorem says that such a polynomial can always be epressed in this form: where the roots,,, are comple numbers, not necessarily all distinct. This is an eistence theorem: there must eist n comple roots (again not necessarily distinct). 10 Finding them is another matter.
Solutions by Radicals For quadratics, cubics, and quartics all roots epressible from coefficients using addition, subtraction, multiplication, division, and square-, cube-, and fourth-roots. Familiar quadratic formula: Similar formula for roots of 0: Even more complicated formula(s) for quartics. Abel-Galois-Ruffini Theorem: No such formulas eist for quintics and higher degree equations. For eample, it can be shown that the quintic 15 5 has at least one root that is not epressible in terms of radicals. This proves that there is no general formula using radicals for all solutions of all quintics. 1 Wikipedia Ecerpt 14 4
Finding Roots is Hard But we can find out some things easily The sum of the roots is 11/5 The average of the roots is 11/0 The sum of the reciprocals of the roots is 7/ We ll get back to roots in a bit First let s look at computation Can you compute p() in your head? 17 Horner s Form Standard descending form Horner form Also referred to as partially factored or nested form 18 5
Derivation of Horner Form 1 Quick Evaluation Compute p(): (((5 11) + 6) + 7) Answer = 7 Compute p(): (((5 11) + 6) + 7) Answer = 180 Compute p(/5): (((5 # 11)# + 6)# + 7)# Answer = /15? 6
Getting Back to our Roots Coefficients and combinations of roots 5 Product of Roots ±Constant term / highest degree coefficient Eample: (The ± sign is + because degree is even) Product of the roots is /5 6 7
For our eample Key Idea of Proof Say the roots are r, s, t, and u. p() = 5( r)( s)( t)( u) Multiply this out to find the constant term 5rstu = - Note constant term is 5(- r)(- s)(- t)(- u), so for odd degree we get an etra sign 9 Sum of Roots nd highest degree coefficient divided by highest degree coefficient Eample: Sum of the roots is Average of the roots is 11/5 11/0 0 8
A System To Remember Specified sum and product of two unknowns For eample: this system 5 8. Solution: and y are the roots of the quadratic equation 580. Reason: 58iff 5 and 8. The unknowns in any system of the form must be given by. Eercises 1. Prove the sum of the roots result. Prove this: For any polynomial p, the average of the roots of p is equal to the average of the roots of the derivative p. 4 9
Reverse Polynomial Consider the polynomial p( ) 7 11 The reverse polynomial is 6 Rev p( ) 1 7 11 5 1 5 1 Question: How are the roots of Rev p related to the roots of p? 4 1 5 4 5 6 Answer: Roots of reverse polynomial are reciprocals of roots of the original. 7 Sum of Reciprocal Roots 1 st degree coefficient divided by the constant coefficient Eample: Sum of the reciprocal roots is 7/ Average of the reciprocal roots is 7/1 8 10
Eercises 1. Prove this: For any polynomial p of degree n with nonzero constant term, Rev p() = n p(1/).. Use 1 to prove that the roots of Rev p() are the reciprocals of the roots of p().. Use to prove: if p() = a n n + + a 1 + a 0 and a 0 0, then the sum of the reciprocals of the roots of p is a 1 / a 0 41 Polynomial Long Division Eample: ( 5 + 6) (-) 4 11
Polynomial Long Division Redo the eample working from the constants upward 45 Polynomial Long Division Another eample: ( 5 + 6) (-1) (Do it on white board / scratch paper) Answer -6 4 5 Similar to the long division 1 to find. Alternate form of answer: -6 (1 + + + ) = -6 + /(-1) Part in parentheses is a power series for the rational function 1/(1-). Similar to mied fraction form of answer to a division problem: fraction part = remainder / divisor 46 1
Amazing Application Start with p() = + Find the derivative Reverse both (Do it on whiteboard/scratch paper) Do a long division problem of the reversed p() into the reversed p (), working from the constants forward (Do it on whiteboard/scratch paper) (Do it on whiteboard/scratch paper) Answer: + + 6 + 8 + 18 4 The coefficients have an astonishing interpretation: sums of powers of roots 49 Checking the answer p() = + = (-)( 1) Roots are, 1, and -1 Sum of roots = Sum of squares of roots = 6 Sum of cubes = 8 Sum of fourth powers = 18 Etc. 50 1
Proof Hints Rev p() = n p(1/) Logarithmic Derivative: f / f = (ln f ) If f () = ( r) ( s) ( t) then 1 1 1 (ln f ( ))' r s t Geometric Series: 1 1 1/ a 1 4 and 1 a 1 / a 5 Palindromials p() = reverse p() Recall: if p(r) = 0 (r0) then [rev p](1/r) = 0. So for palindromials, whenever r is a nonzero root, so is 1/r. Eample: 4 + 7 - + 7 + 1 1 and -1 are not roots, so roots come in reciprocal pairs Must factor as (-r)(-1/r)(-s)(-1/s) Rewrite: ( u+ 1) ( v+ 1) where u = r+1/r and v = s + 1/s 54 14
Matching Coefficients ( u + 1) ( v + 1) = 4 + 7 - + 7 + 1 u + v = -7 and uv + = - Two unknowns. Sum = -7, product = -4 They are the roots of + 7 4 = 0 u and v are given by 7 65 Our factorization is 7 65 7 65 1 1 57 Solve for 65 1 7 Use quadratic formula on each factor Roots from first factor are 1 7 65 98 14 4 Remaining roots are 1 4 7 65 1 7 4 65 1 0 65 7 65 98 14 65 98 14 65 58 15
General Reduction Method p() = a 6 + b 5 + c 4 + d + c + b + a p()= (a + b + c + d + c/ + b/ + a/ ) p()/ =a( +1/ ) + b( +1/ ) + c(+1/)+d We want roots of a( +1/ ) + b( +1/ ) + c(+1/)+d Almost a polynomial in u = (+1/). u = + + 1/ +1/ = u u = + + / + 1/ = + u + 1/ +1/ = u u Leads to a cubic polynomial in u: a(u u) + b(u ) + c(u)+ d Solve for u, and then substitute in u = (+1/) and solve for. Note u + 1 = 0. 61 Eample: 1 Let 1 Substitute, using and 60 0 1 If then so 10, giving. Roots obtained from 1and 1similarly. 6 16
Eample Make the standard reduction It s another palindromial! Reduce again Solve with quadratic formula Find u: so Solve for u 65 Solve for We have found 4 values for u We know + 1/ = u Solve u + 1 = 0 with quadratic formula for each known u value That gives 8 roots Here is one: 66 17