DOC PREVIEW
HARVARD MATH 1A - Lecture 2: Worksheet

This preview shows page 1 out of 2 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 2 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 2 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

Math 1A: introduction to functions and calculus Oliver Knill, 2011Lecture 2: WorksheetIn this lecture, we want to learn what a function is and get acquaintedwith the most important examples.Trigonometric functionsThe cosine and sine functio ns can be defi ne d geometrically by the co-ordinates (cos(x), sin(x)) of a point on the unit circle. The tangentfunction is defined as tan(x) = sin(x)/ cos(x)).cos(x) = adjacent side/hypothenusesin(x) = opposite side/hypothenusetan(x) = opposite side/adja dc ent sidePythagoras theorem gives us the import ant identitycos2(x) + sin2(x) = 1Define also cot(x) = 1/ tan(x). Less important but sometimes usedare sec(x) = 1/ cos(x), csc(x) = 1/ sin(x).1 Find cos(π/3), sin(π/3).2 Where does c os and sin have roots, place s, where the functionis zero?3 Find tan(3π/2) and cot(3π/2).4 Find cos(3π/2 ) and sin(3π/2).5 Find tan(π/4) and cot(π/4).cosHΦLsinHΦLΦ1cosHxLx2ΠsinHxLx2ΠtanHxLxΠ2-Π2The exponential functionThe function 2xis first of all defined for all integers like 210= 1 024.By taking roots, we can define it for rational numbers like 23/2=81/2=√8 = 2.828.... Since the functio n 2xis monotonone on the setof rationals, we can fill the gaps and define the function 2xfor any x.By taking square roots again and again, we see 21/2, 21/4, 21/8, ... weapproach 20= 1.2xxThere is nothing special about 2 and we can take any positive base aand define the exp onential ax. It satisfies a0= 1 and the rema r kablerule:ax+y= ax· ayIt is sp ectacular because it provides a link between addition and mul-tiplication.We will espe cially consider the exponential exph(x) = (1 + h)x/h,where h is a positive parameter. This is a supercool exponential be-cause it satisfies exph(x + h) = ( 1 + h) exph(x) so that[exph(x + h) − exph(x)]/h = exph(x) .Hold on to that. We will look at this later again. In modern language,we would say that ”the quantum derivative of the quantum exponen-tial is the function itself for any Planck constant h” .For h = 1, we have the function 2xwe have started with. In thelimit h → 0 , we get the important exponential function e xp(x) whichwe also call ex. For x = 1, we get the Euler number e = e1=2.71828....1 What is 2−5?2 Find 21/2.3 Find 271/3.4 Why is A = 23/4smaller than B = 24/5? Take the 20th powerof both numbers .5 Assume h = 2 find


View Full Document

HARVARD MATH 1A - Lecture 2: Worksheet

Download Lecture 2: Worksheet
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Lecture 2: Worksheet and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Lecture 2: Worksheet 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?