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\listoverridecount0\ls2}{\listoverride\listid3\listoverridecount0\ls3}}{\info{\title Assignment 23}{\author Jim Wilson}{\operator Jim Wilson}{\creatim\yr2000\mo11\dy21\hr12\min12}{\revtim\yr2000\mo11\dy21\hr13\min36}{\version1}{\edmins41}{\nofpages2}
{\nofwords504}{\nofchars2875}{\*\company Mathematics Education}{\nofcharsws3530}{\vern27}}\widowctrl\ftnbj\aenddoc\formshade\viewkind1\viewscale100\pgbrdrhead\pgbrdrfoot \fet0\sectd \linex0\endnhere\sectdefaultcl {\*\pnseclvl1
\pnucrm\pnstart1\pnindent720\pnhang{\pntxta .}}{\*\pnseclvl2\pnucltr\pnstart1\pnindent720\pnhang{\pntxta .}}{\*\pnseclvl3\pndec\pnstart1\pnindent720\pnhang{\pntxta .}}{\*\pnseclvl4\pnlcltr\pnstart1\pnindent720\pnhang{\pntxta )}}{\*\pnseclvl5
\pndec\pnstart1\pnindent720\pnhang{\pntxtb (}{\pntxta )}}{\*\pnseclvl6\pnlcltr\pnstart1\pnindent720\pnhang{\pntxtb (}{\pntxta )}}{\*\pnseclvl7\pnlcrm\pnstart1\pnindent720\pnhang{\pntxtb (}{\pntxta )}}{\*\pnseclvl8\pnlcltr\pnstart1\pnindent720\pnhang
{\pntxtb (}{\pntxta )}}{\*\pnseclvl9\pnlcrm\pnstart1\pnindent720\pnhang{\pntxtb (}{\pntxta )}}\pard\plain \widctlpar\adjustright \loch\af4\hich\af4\dbch\f4\cgrid {\hich\af4\dbch\af4\loch\f4 Assignment 23
\par 
\par {\pntext\pard\plain\dbch\f4 \hich\af4\dbch\af4\loch\f4 1)\tab}}\pard \fi-360\li360\widctlpar\jclisttab\tx360{\*\pn \pnlvlbody\ilvl0\ls1\pnrnot0\pndec\pnstart1\pnindent360\pnhang{\pntxta )}}\ls1\adjustright {\hich\af4\dbch\af4\loch\f4 
$1,000,000 today or double your money everyday for a month?\hich\af4\dbch\af4\loch\f4  
\par }\pard \widctlpar{\*\pn \pnlvlcont\ilvl0\ls0\pnrnot0\pndec }\adjustright {
\par }\pard \li360\widctlpar{\*\pn \pnlvlcont\ilvl0\ls0\pnrnot0\pndec }\adjustright {\hich\af4\dbch\af4\loch\f4 
I think I saw this problem for the first in elementary school.  A few calculations made it clear that my choice should be to take a doubling penny everyday for a month over the million dollars.  The neat thing about this problem is that until the last few
\hich\af4\dbch\af4\loch\f4  d\hich\af4\dbch\af4\loch\f4 ays of the month, the "earnings\hich\af4\dbch\af4\loch\f4 " are not that great.  It takes over half the month to even make a thousand dollars\hich\af4\dbch\af4\loch\f4 , yet amazingly 
\hich\af4\dbch\af4\loch\f4 enough\hich\af4\dbch\af4\loch\f4 ,\hich\af4\dbch\af4\loch\f4 
 after just 30 days the total sum exceeds $10 million. I think that almost anyone you asked on the street, however, would opt for the million dollars up front, because not many people understand the \hich\af4\dbch\af4\loch\f4 po}{
\hich\af4\dbch\af4\loch\f4 wer}{\hich\af4\dbch\af4\loch\f4  of exponential growt\hich\af4\dbch\af4\loch\f4 h.  I call it expon\hich\af4\dbch\af4\loch\f4 ential because the numbers match that of \hich\af4\dbch\af4\loch\f4 .0\hich\af4\dbch\af4\loch\f4 
2 to th\hich\af4\dbch\af4\loch\f4 e nth power after the first day, with n equal to one less than the number of days passed.}{
\par 
\par {\pntext\pard\plain\dbch\f4 \hich\af4\dbch\af4\loch\f4 2)\tab}}\pard \fi-360\li360\widctlpar\jclisttab\tx360{\*\pn \pnlvlbody\ilvl0\ls1\pnrnot0\pndec\pnstart1\pnindent360\pnhang{\pntxta )}}\ls1\adjustright {\hich\af4\dbch\af4\loch\f4 Th
\hich\af4\dbch\af4\loch\f4 e limit of h(x) as x approaches the following where h(x)=f(x)/g(x).  Let f(x)=x+2 and g(x)=x^2\hich\af4\dbch\af4\loch\f4 -\hich\af4\dbch\af4\loch\f4 4
\par {\pntext\pard\plain\dbch\f4 \hich\af4\dbch\af4\loch\f4 a)\tab}}\pard \fi-360\li720\widctlpar\jclisttab\tx720{\*\pn \pnlvlbody\ilvl0\ls3\pnrnot0\pnlcltr\pnstart1\pnindent720\pnhang{\pntxta )}}\ls3\adjustright {\hich\af4\dbch\af4\loch\f4 
Let x approach positive integer 14.  \hich\af4\dbch\af4\loch\f4 T\hich\af4\dbch\af4\loch\f4 he\hich\af4\dbch\af4\loch\f4  function increases to 1 except where it is undefined as x approaches 14.
\par {\pntext\pard\plain\dbch\f4 \hich\af4\dbch\af4\loch\f4 b)\tab}}\pard \fi-360\li720\widctlpar\jclisttab\tx720{\*\pn \pnlvlbody\ilvl0\ls3\pnrnot0\pnlcltr\pnstart1\pnindent720\pnhang{\pntxta )}}\ls3\adjustright {\hich\af4\dbch\af4\loch\f4 
Let x approach negative integer -50, \hich\af4\dbch\af4\loch\f4 assuming you start at x=0, \hich\af4\dbch\af4\loch\f4 the function \hich\af4\dbch\af4\loch\f4 in\hich\af4\dbch\af4\loch\f4 creas\hich\af4\dbch\af4\loch\f4 es, approaching
\hich\af4\dbch\af4\loch\f4  \hich\af4\dbch\af4\loch\f4  zero\hich\af4\dbch\af4\loch\f4 , as it started \hich\af4\dbch\af4\loch\f4 with a\hich\af4\dbch\af4\loch\f4  \hich\af4\dbch\af4\loch\f4 negative term.
\par {\pntext\pard\plain\dbch\f4 \hich\af4\dbch\af4\loch\f4 c)\tab}}\pard \fi-360\li720\widctlpar\jclisttab\tx720{\*\pn \pnlvlbody\ilvl0\ls3\pnrnot0\pnlcltr\pnstart1\pnindent720\pnhang{\pntxta )}}\ls3\adjustright {\hich\af4\dbch\af4\loch\f4 A
\hich\af4\dbch\af4\loch\f4 s x approaches 2 (h(x) unde\hich\af4\dbch\af4\loch\f4 fined at 2), h(x) is increasing.  However, it is only increasing from x=-2, the other value undefined for h(x).\hich\af4\dbch\af4\loch\f4 
  As x approaches -2, the function decreases.
\par {\pntext\pard\plain\dbch\f4 \hich\af4\dbch\af4\loch\f4 d)\tab}}\pard \fi-360\li720\widctlpar\jclisttab\tx720{\*\pn \pnlvlbody\ilvl0\ls3\pnrnot0\pnlcltr\pnstart1\pnindent720\pnhang{\pntxta )}}\ls3\adjustright {\hich\af4\dbch\af4\loch\f4 A
\hich\af4\dbch\af4\loch\f4 s \hich\af4\dbch\af4\loch\f4 x approaches positive infinity, \hich\af4\dbch\af4\loch\f4 h(x) approaches zero.
\par {\pntext\pard\plain\dbch\f4 \hich\af4\dbch\af4\loch\f4 e)\tab}}\pard \fi-360\li720\widctlpar\jclisttab\tx720{\*\pn \pnlvlbody\ilvl0\ls3\pnrnot0\pnlcltr\pnstart1\pnindent720\pnhang{\pntxta )}}\ls3\adjustright {\hich\af4\dbch\af4\loch\f4 
As x approaches negative infinity, h(x) also approaches zero.
\par }\pard \li360\widctlpar\adjustright {
\par \hich\af4\dbch\af4\loch\f4 This function w\hich\af4\dbch\af4\loch\f4 as intersesting because of the changes that \hich\af4\dbch\af4\loch\f4 occurred\hich\af4\dbch\af4\loch\f4  \hich\af4\dbch\af4\loch\f4 
near the undefined points.  It was hard for me to decipher increasing and decreasing b\hich\af4\dbch\af4\loch\f4 ecause I wasn't sure how to think about approaches to\loch\af4\dbch\af4\hich\f4 \u8230\'c9\loch\f4 
that is I wasn't sure if I should approach the value from x=0 or from x=a value less than the given, depending on \hich\af4\dbch\af4\loch\f4 the\hich\af4\dbch\af4\loch\f4  \hich\af4\dbch\af4\loch\f4 circumstance.  Depending on the perspective, 
\hich\af4\dbch\af4\loch\f4 the\hich\af4\dbch\af4\loch\f4  \hich\af4\dbch\af4\loch\f4 increasing/decreasing opinions might be different.
\par {\pntext\pard\plain\dbch\f4 \hich\af4\dbch\af4\loch\f4 3)\tab}}\pard \fi-360\li360\widctlpar\jclisttab\tx360{\*\pn \pnlvlbody\ilvl0\ls1\pnrnot0\pndec\pnstart1\pnindent360\pnhang{\pntxta )}}\ls1\adjustright {\hich\af4\dbch\af4\loch\f4 Fibonnaci sequence

\par {\pntext\pard\plain\dbch\f4 \hich\af4\dbch\af4\loch\f4 a)\tab}}\pard \fi-360\li720\widctlpar\jclisttab\tx720{\*\pn \pnlvlbody\ilvl0\ls2\pnrnot0\pnlcltr\pnstart1\pnindent720\pnhang{\pntxta )}}\ls2\adjustright {\hich\af4\dbch\af4\loch\f4 A
\hich\af4\dbch\af4\loch\f4 s n increases in \hich\af4\dbch\af4\loch\f4 the\hich\af4\dbch\af4\loch\f4  \hich\af4\dbch\af4\loch\f4 Fibonnaci sequence f\hich\af4\dbch\af4\loch\f4 (n)=f(n-1)+f(n-2), the ratio of f(n)/f(n+1\hich\af4\dbch\af4\loch\f4 )
\hich\af4\dbch\af4\loch\f4  approaches .6180\hich\af4\dbch\af4\loch\f4 3399\hich\af4\dbch\af4\loch\f4  after n=\hich\af4\dbch\af4\loch\f4 21\hich\af4\dbch\af4\loch\f4 .  \hich\af4\dbch\af4\loch\f4 For n<\hich\af4\dbch\af4\loch\f4 21
\hich\af4\dbch\af4\loch\f4 , the ratio decreases then increases and repeats \hich\af4\dbch\af4\loch\f4 the\hich\af4\dbch\af4\loch\f4  \hich\af4\dbch\af4\loch\f4 cycle.\hich\af4\dbch\af4\loch\f4   This says to me that this sequences of ratios is an 
\hich\af4\dbch\af4\loch\f4 alternating\hich\af4\dbch\af4\loch\f4  sequence of terms.  The amount added or \hich\af4\dbch\af4\loch\f4 subtracted\hich\af4\dbch\af4\loch\f4  from \hich\af4\dbch\af4\loch\f4 the\hich\af4\dbch\af4\loch\f4  
\hich\af4\dbch\af4\loch\f4 previous\hich\af4\dbch\af4\loch\f4  term decreases as n increases, so the terms eventually limit to\hich\af4\dbch\af4\loch\f4 .6180339\hich\af4\dbch\af4\loch\f4 9 after term n=22.\hich\af4\dbch\af4\loch\f4   When the ratio 
\hich\af4\dbch\af4\loch\f4 equation\hich\af4\dbch\af4\loch\f4  is changed, that is divide by every second term, third term, etc., \hich\af4\dbch\af4\loch\f4 the\hich\af4\dbch\af4\loch\f4  \hich\af4\dbch\af4\loch\f4 
sequence of new ratios followed our established pattern.  The ratios increase and decrease until about the 21}{\super \hich\af4\dbch\af4\loch\f4 st}{\hich\af4\dbch\af4\loch\f4  term then the sequence limits to a specific limit.
\par {\pntext\pard\plain\dbch\f4 \hich\af4\dbch\af4\loch\f4 b)\tab}}\pard \fi-360\li720\widctlpar\jclisttab\tx720{\*\pn \pnlvlbody\ilvl0\ls2\pnrnot0\pnlcltr\pnstart1\pnindent720\pnhang{\pntxta )}}\ls2\adjustright {\hich\af4\dbch\af4\loch\f4 
I explored this idea of the limit of the seque\hich\af4\dbch\af4\loch\f4 n\hich\af4\dbch\af4\loch\f4 ce of ratios using different values for f(0) and f(1).  I found that the limit of the sequence\hich\af4\dbch\af4\loch\f4 
 was the same as the Fibonnaci sequence is the same no matter the f(0) and f(1) values, but as these terms values increase, the rate at which \hich\af4\dbch\af4\loch\f4 the\hich\af4\dbch\af4\loch\f4  \hich\af4\dbch\af4\loch\f4 sequence limits to 
\hich\af4\dbch\af4\loch\f4 .61803399 increases.  \hich\af4\dbch\af4\loch\f4 But for the numbers I tried, it always took until at least the 18}{\super \hich\af4\dbch\af4\loch\f4 th}{\hich\af4\dbch\af4\loch\f4  \hich\af4\dbch\af4\loch\f4 
term to limit to this number.
\par }\pard \widctlpar\adjustright {
\par {\pntext\pard\plain\dbch\f4 \hich\af4\dbch\af4\loch\f4 4)\tab}}\pard \fi-360\li360\widctlpar\jclisttab\tx360{\*\pn \pnlvlbody\ilvl0\ls1\pnrnot0\pndec\pnstart1\pnindent360\pnhang{\pntxta )}}\ls1\adjustright {\hich\af4\dbch\af4\loch\f4 
Place four numbers in a row:
\par }\pard \li360\widctlpar\adjustright {\hich\af4\dbch\af4\loch\f4 
Looking at the sequence of numbers taking for the given absolute value manipulations does lead to a row of four zeros.  For all of my trials, the most rows I had before reaching a row of zeros was 6.  That is the 7}{\super \hich\af4\dbch\af4\loch\f4 th}{
\hich\af4\dbch\af4\loch\f4  row has 4 zeros.\hich\af4\dbch\af4\loch\f4   Its cool to see that no matter the numbers you start with that the outcome will always be zero because the last non-zero row has four of the same values.
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