solns3 - Mathematics 105 Spring 2004 M. Christ 1 Midterm...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Mathematics 105 Spring 2004 M. Christ 1 Midterm Exam Wednesday February 18 Guidelines. The exam will be based on the first three problem sets, on Chapters 1 and 2 of Spivak, and on the lectures. You might be asked to solve problems similar to those in the problem sets, to state definitions, to state theorems, and to give examples (possibly with justification). Mastery of the proofs in the text, especially those discussed in lecture, could be tested; you might be asked to reproduce a proof, or to carry out one step in a proof, or to use theorems, lemmas, and arguments from the text to prove other statements. Solutions to selecta from problem set #3 2-29(b) . Show that if the directional derivative D x f ( a ) exists, then for any t R , D xt f ( a ) = tD x f ( a ). Solution. We are given that lim s f ( a + sx )- f ( a ) s exists, and we are asked to evaluate lim h f ( a + htx )- f ( a ) h . If t = 0 then the latter limit is lim h h = 0 = 0 D x f ( a ), so the desired equation holds. If t 6 = 0 we evaluate the limit by making the substitution h = s/t , where t remains fixed and s is a function of h . Certainly s 0 as h 0 and vice versa, so we obtain lim h f ( a + htx )- f ( a ) h = lim s f ( a + sx )- f ( a ) s/t = t lim s f ( a + sx )- f ( a ) s = tD x f ( a ) . Thus D tx f ( a ) exists and equals tD x f ( a ). 2-29(c) . Suppose that f is differentiable at a . Show that D x f ( a ) exists and equals Df ( a )( x ), and that D x + y f ( a ) = D x f ( a ) + D y f ( a ). Solution. The latter conclusion follows from the former, because Df ( a )( x ) + Df ( a )( y ) = Df ( a )( x + y ) since Df ( a ) is a linear transformation. If x = 0 then D x f ( a ) = 0, as one sees directly from its definition (see the solution for part (b)), while Df ( a )( x ) = 0 as well, so they are equal. Assume now that x 6 = 0. We know from the definition of Df ( a ) that | f ( a + tx )- f ( a )- Df ( a )( tx ) | | tx | tends to zero as t 0. Pulling out the nonvanishing factor of | x | from the denominator doesnt change this, and a little algebraic manipulation allows us to rewrite this fact as lim t f ( a + tx )- f ( a ) t- Df ( a )( x ) = 0 . This means precisely that f ( a + tx )- f ( a ) t Df ( a )( x ) as t 0. Thus D x f ( a ) exists and equals Df ( a )( x )....
View Full Document

This note was uploaded on 05/07/2008 for the course MATH 105 taught by Professor Michaelchrist during the Spring '04 term at University of California, Berkeley.

Page1 / 4

solns3 - Mathematics 105 Spring 2004 M. Christ 1 Midterm...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online