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## LINEAR ALGEBRA

by: Cassidy Grimes

15

0

5

# LINEAR ALGEBRA MATH 700

Cassidy Grimes

GPA 3.51

Staff

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COURSE
PROF.
Staff
TYPE
Class Notes
PAGES
5
WORDS
KARMA
25 ?

## Popular in Mathematics (M)

This 5 page Class Notes was uploaded by Cassidy Grimes on Monday October 26, 2015. The Class Notes belongs to MATH 700 at University of South Carolina - Columbia taught by Staff in Fall. Since its upload, it has received 15 views. For similar materials see /class/229540/math-700-university-of-south-carolina-columbia in Mathematics (M) at University of South Carolina - Columbia.

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Date Created: 10/26/15
CHANGE OF BASES AND SUMS OF POWERS OF INTEGERS RALPH HOWARD DEPARTMENT OF MATHEMATICS UNIVERSITY OF SOUTH CAROLINA COLUMBIA7 SICA 292087 USA HOWARDQMATH SC EDU 1 SOME BASES OF THE POLYNOMIALS OF DEGREE n Let 7 be the vector space of real polynomials of degree 3 71 That is 73 a0a1anzna0an E R We have seen that dim 73 n I and that it has the usual basis U1xxn coming from the powers of x In di erent problems there di erent bases of 73 that are better adapted to the problem For example let a E R then the ordered basis A 1xia 03 2 3 quotquot 71 has the property that if f E 73 is expressed in this basis 96 0 11 m ZakT k0 then the coordinate a0 an are given by ak fka where f0 is the k th derivative of f and f0 f be de nition Problem 1 Derive this formula for ak by taking the k th derivative of both sides of 11 and then setting z a Note this is exactly the usual derivation of the coef cients in a Taylor series that you know and love from calculus D 1 2 RALPH HOWARD In our terminology the coordinate vector of the vector f E 73 is f a f a fa lleA fn71a fna which is just the list of the values of the derivatives of f at z 1 So in a context where one is working with the derivatives of polynomials at the point z a the basis A is the natural one to use 2 FORMULAS FOR SUMS OF POWERS We now give a basis of 73 were it is easy to derive summation formulas77 the precise meaning of this will be cleared up below and then by expessing the usual basis 1x 2 z in terms of this basis we can derive formulas for sums of powers This is a theme we will see repeatedly during the term Make a problem easier by changing to a nicer basis Set Soz 1 andforlgkgnset Skz z 71z 7 2 7 k 1 This has k factors and so has degree k For small values of k we have 5095 51W 5295 959c 1 53z z 71z 7 2 S4z z 71z 7 2x 7 3 Let S 50z81x8nx Then S is an ordered basis of 73 We now de ne the change of basis matrices between the usual basis LI and the basis 8 by k Skz Zakixi i0 CHANGE OF BASES AND SUMS OF POWERS OF INTEGERS 3 and k 21 M Z innSm i0 So for example 53z z 71 7 2 x3 7 3 2x 133 ag z 131 1301 which implies 133 1 132 73 131 2 130 0 Likewise 3 53W 352 51i basss 53252 531S1 53050 yielding bgg 1 bgg 3 bgl 1 bgo 0 Problem 2 Show that AB BA n1 on 73 D On 736 the matrices A am and B are 7 by 7 matrices given explicitly by 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 71 1 0 0 0 0 Aakil 0 2 73 1 0 0 0 0 76 11 76 1 0 0 0 24 750 35 710 1 0 0 7120 274 7225 85 715 1 and 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 0 Bbkil 0 1 3 1 0 0 0 0 1 7 6 1 0 0 0 1 15 25 10 1 0 0 1 31 90 65 15 1 ln matrix notation this means Soz 1 0 0 0 0 0 0 1 51z 0 1 0 0 0 0 0 z 52z 0 71 1 0 0 0 0 2 53z 0 2 73 1 0 0 0 3 S4z 0 76 11 76 1 0 0 4 S5z 0 24 750 35 710 1 0 5 56z 0 7120 274 7225 85 715 1 6 4 RALPH HOWARD and 1 1 0 0 0 0 0 0 SO c 0 1 0 0 0 0 0 81 2 0 1 1 0 0 0 0 52 3 0 1 3 1 0 0 0 Sg 4 0 1 7 6 1 0 0 S4 m5 0 1 15 25 10 1 0 S5 6 0 1 31 90 65 15 1 56 Problem 3 Check that the the values for a and bki are correct for 239 k S 4 There are some obvious patterns in these matrices Problem 4 Show the following 1 1 bkk 1f01 all k 2 2 ak0bk00fork21 3 The signs in the matrix A have a chess board pattern That is 71 kaki Z LIMA 4916164 Not easy use that AB I bki 2 0 for all 239 k Hard 6 bkl 1 for k 2 1 Hard D U rb From the point of View of summation formulas what makes the Skz nice is the relation 1 51495 mlt5kd 1 Sk195 which holds for k 2 0 Problem 5 Verify this formula D The standard trick with telescoping series shows that Siam Sk1 SIX 39 SAN ltsk1uv 17 Sk10 Sk1N 1 k1 At the last step we have used Sk10 0 Problem 6 Verify this D In summation notation this is 22 sky k i 15mm 1 CHANGE OF BASES AND SUMS OF POWERS OF INTEGERS 5 Now we can give formulas for sums of powers of integers Using equa tions 21 and 22 we have bkifsz rluv l 1 For small values of k this gives N 1 E j b11ESZN b1051ltN j0 N 1 1 12 bzzgsst 1 bzlgsluv 1 52050N 1 7390 N 1 1 1 E jg 533153N 1 532 S1N 1 b31 SON 1 70 b3080N 1 739 Problem 7 Use these to derive the familiar formulas for Zioj 2L0 jZ and Ziojg D

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