Applications on Recurrence Relations
Published On :2021-05-04 00:20:00
Ex (1): The number of bacteria in a colony doubles every hour. If a colony begin with five bacteria, how many will be present in n hours?
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{a_n} = 2;{a_{n - 1}};Where;;{a_n} = # ;;of;bactiria;at;time;n,;with;intial;condintion;{a_0} = 5\
Rightarrow {a_n} = 2;{a_{n - 1}};,;;{a_0} = 5\\
Observe;that;;;;{a_1} = 2;{a_0} = 2 * 5 = 10\
{a_2} = 2;{a_1} = 2 * 10 = 20;\
{a_3} = 2;{a_2} = 2 * 20 = 40
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Exleft( 2 right):;{a_n} = 2;{a_{n - 1}} - ;{a_{n - 2}};,;;;;{a_0} = 0;,{a_1} = 3\\
1 - Find;{a_2},;{a_3},;;{a_4};.\\
Rightarrow ;;{a_2} = 2;{a_1} - {a_0} = 2left( 3 right) - 0 = 6\
;{a_3} = 2;{a_2} - {a_1} = 2left( 6 right) - 3 = 9\
;{a_4} = 2;{a_3} - {a_2} = 2left( 9 right) - 6 = 12\\
2 - Determine;whetther;;{a_n} = 3n,;is;a;solution;for;this;recurrence;relation;.\\
Rightarrow plug;in;{a_n} = 3n;,;in;the;recurrence;relation;,\
;{a_n} = 2;{a_{n - 1}} - ;{a_{n - 2}}\
3n;does;it;equal;2left( {3n - 3} right) - left( {3n - 6} right)\
left( {6n - 6} right) - left( {3n - 6} right) = 3n\
Thus;;{a_n} = 3n;;is;a;solution;of;the;recurrence;relation.\\
3 - Is;{a_n} = {2^n},;;a;solution;for;this;recurrence;relation;?;\\
Rightarrow ;;{a_n} = 2;{a_{n - 1}} - ;{a_{n - 2}}\
{a_n} = {2^n} Rightarrow {2^n}does;it;equal;2 * {2^{n - 1}} - {2^{n - 2}}\
{2^n} - {2^{n - 2}}\
{2^n}; ne {2^n} - {2^{n - 2}}\
Thus;;{a_n} = {2^n};;is;a;solution;of;the;recurrence;relation.
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