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# Note for FMS 100 at KU

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This 5 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at Kansas taught by a professor in Fall. Since its upload, it has received 18 views.

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Date Created: 02/06/15

Why 6 1 Compound interest Suppose that the Bank of Awesome o ers 100 annual interest on their savings accounts and you open an account with 1 on January rst Assuming that you make no deposits or withdrawals how much money will be in your account at the beginning of the next year The answer depends on how frequently interest is computed compounded77 If it is compounded annually iiei computed on December 31 every year then the bank looks on December 31 sees that you have 1 in your account and gives you 1 of interest so that you have a total of 2 1f interest is compounded semiannually then the bank rst looks at your account on June 30 sees one dollar and awards you half of the annual interest 150 for a total of 1150 Then it looks again on December 31 sees 1150 and again awards you half of your annual interest But now June s interest also earns interest so you get 175 instead of 150 giving you a total of 225 If interest is compounded monthly the bank gives you of your annual interest at the end of each month multiplying your balance by 1 each time Thus at the end of the year you have 1 1 12 m 261 If interest is compounded daily your balance is multiplied by 1 K15 365 1 365 m 2171 1f interest is compounded hourly you have 1 1 W160me 525600 times leaving you with 1 1 If compounded every minute you get 1 1 m 1 If as at most real banks interest is compounded constantly you get the limit of this sequence 7 lim lt81lt1 l 1 naoo n 7 lim 1 m 2718281828459045w n naoo The irrational number is referred to simply as e supposedly short for Euler pronounced Oiler i 2 Population growth The polygeminus grew bacterium reproduces on average once every hour Sup pose that you have a culture with one million of these bacteria How many will you have in an hour s time We can model the situation by assuming that the bacteria are all watching the room clock and reproduce instantaneously every hour on the hour In this case there are two million bacteria after an hour A slightly better model is that every half hour each bacterium ips a coin and reproduces if it comes up heads In this case there will be approximately 15 million bacteria after 30 minutes But at the hour the 5 million new bacteria will get a chance to reproduce as well and so there will be 225 million bacteria after an hour If the bacteria each roll a die every ten minutes and reproduce on a six there will be 1 6 million bacteria after an hour If the bacteria are constantly trying to reproduce and succeeding randomly 1 71 an average of once per hour there will be lim lt1 7 million or 5 million naoo n bacteria 3 Formulas related to ex 1 n 1 7 CE 1 7156 We de nede lirn lt1 7 1 Thus 5 lim 1 7 J lim 1 7 1 quotH00 n naoo n naoo n But setting m 3 we also have a 1 lim 1 lim naoo naoo A gt V lim naoo A gt ngm r t V 3 R lim mace A gt 31H V 3 a x n This formula lim 1 7 ex is very useful ngtoo The exercises below are not to be turned in The rst one is quite difficult but should be possible for at least a few of you and thinking about it may help you do the second Exercise Convince yourself of the approximation 1 3 m 55 for small numbers 3 and large numbers Z Can you use limits to make it precise Exercise Show that if you put P dollars in the bank at annual interest rate r for t years you end up with P5 dollars 4 The derivative of 695 Why is 5 its own derivative We compute xh 7 d x e e 7e hm all h7gt0 h x eh71 5 hm h7gt0 h eh7 Thus we need to show that lim 1 You can check with your A calculator by plugging in very small values of 1 that this limit is very close to 1 But unlike in the case of a polynomial limit it s really unclear that the limit should be even a rational number let alone an integer like 1 You would get very similar behavior if you used something like 271829 instead of 5 but that limit is an irrational number slightly larger than 1 Thus a more precise argument is desirable The argument below is technically correct but requires the use of some theory that won t be covered until late in Math 116 Can you tell where the details are omitted1 Recall from above that eh lim 1 Thus using the binomial n theorem h hm 1 7 7 1 e 7 n7gtoo n hm hm h7gt0 h7gt0 h hm lt1n wamp2W 3mgt 1 n7gtoo n 2 n 6 n hm h7gt0 h h h nme Lym 2 lt gtltquotgtlt 2gt 3 lim h7gt0 h 7 1 1 nn71n72 h i 1 h fwfwu 1 eh 7 1 Thus lequot ex h gt511ex 5 lnx Another approach l m following Thomas and Finney Calculus below 1I7ve made the unsupported assumption that the limit as n approaches in nity of the in nite sum exists It does although we haven7t learned how to prove it yet In an appendix h not yet written 171 try to convince you that that sum is actually 7 7 d5 7 dy 7 1 lfy 7 lnx then x 7 ey and so Ty 7 ey Thus E 7 87 l E Probably because the argument to compute the derivative of e above is so technical many texts take a different approach here and de ne ln as not as the inverse function of em but in the following manner De nition lnx is the unique function such that the derivative of lnx is i and ln1 0 Such a function exists for a gt 1 set fa equal to the area bounded by the curves as 1 x a y 0 and y 1 for 0 lt a lt 1 set fa to the negative of 1 this area and is unique by the following argument Suppose f and 9 both satisfy the given conditions Then we have f7g f7g i7 0 Thusf7gmustbea constant function but 7 91 0 so f 7 9 must be zero Thus f 9 Now we use a sequence of very slick arguments to show that lnx is a loga rithm Theorem 1 ln ax i for any positive number a Proof By the chain rule d7ln ax mx 1 Theorem 2 For any positive number a there exists a number Ca such that lnax lnx Ca for all positive 95 Proof We have ilnaac 7 lnx l 7 l 0 so lnax 7 lnx must be a do as as constant function of x We call this constant Ca 1 Theorem 3 Ca lna Proof Set as 1 Then lna lnax Ca lnx Ca ln1 C 11 1 Theorem 4 lnab lna lnb for all positive a and b Proof We have lnax lna lnx Plug in x b D Theorem 5 ln lna 7 lnb for all positive a and b Proof We have lnax lna lnx Plug in x D Theorem 6 lnxquot nlnx for any real number n Proof By the chain rule lnxn h 5271 n n nx nln Thus we have lnx nln x C for some number C Plugging in x 1 gives us ln1 nln1 C or 0 0 C so C 0 and lnx nlnx 1 Theorem 7 lnaquot nlna for all positive numbers a and all real numbers n Proof We have lnx nlnxl Plug in x at 1 De nition Let e be the number satisfying lne 1 Exercise Convince yourself that e exists and is unique Do not attempt to ngtocgt 1 n convince yourself that it equals lim lt1 7 l n Theorem 8 lnx is the logarithm base e that is lne z a for all real numbers a Proof We have lne z alne a1 at 1 Now since ex is the inverse function of lnx and we de ned an as i we have ex e by reversing the argument given at the beginning of this section lfy ex then x lny so L i and y ext We have not shown that the number e de ned here is the number e arising from compound interest this is quite dif cult One can show using advanced techniques from Math 116 that both are equaltothein nitesumllu I will attempt to convince you of this in the appendix to be written laterl

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