;; Assignment
>> v = 10;
>> v
>> v = [2 3 4];
>> v

;; Reason for a string:
>> '3'
>> 3
>> '3'+45
>> '{.'+0 

>> disp('P(1,2)=10');
>> P=[1 2 3; 4 5 6; 7 8 9];
>> i=2;
>> j=1;
>> disp(strcat('P(',num2str(i),',',num2str(j),')=',num2str(P(i,j))));
;; P(2,1)=4;

>> [a b c]   
>> a=1;
>> b=2;
>> c=3;
>> [a b c]
>> s = ['a' 'b' 'c']
>> s(1)
>> s(2)

>> f='2*x';
>> eval(f);  
>> x = 2;
>> eval(f);

;; Vectors of numbers
>> linspace(0,9.34,7)

;; Booleans
>> 123 == 123
>> 123 == 405
>> 123 ~= 405
>> 123 < 405
>> 123 <= 405
>> 405 <= 405
>> 123 > 405

>> 3 < 4 | 6 < 5
>> 3 < 4 & 6 < 5
>> 3 < 4 & 5 < 6
>> 1 < 2 & 2 < 3 & 3 < 4

>> if 3 < 4 disp('1'); else disp('2'); end
>> if 3 < 4
>>   disp('1');
>> else
>>   disp('2');
>> end

;; Assume the name of the vector is v
>> if v(1) > v(2) v=[v(2) v(1)]; end

;; Plots
>> x = linspace(0,2*pi,100);
>> plot(x,sin(x));
>> hold on;
>> plot(x,cos(x),'color',[1 .5 0]);

>> x = '1:1:m'
>> m = 10;
>> sum(eval(x));
>> m = 20;
>> sum(eval(x));

;; Sums
>> m = input('>');
>> x = 1:1:m;
>> s1 = sum(x);
>> s2 = sum(x.^2);

;; A population model
;; 650000000 in 2006, grows at 1% annually
;; 650000000(1.01) -> 2007
;; 650000000(1.01)(1.01) -> 2008
;; 650000000(1.01)^(t-2006) -> year t
;; Use .^ if t is a vector
>> 650000000(1.01).^(t-2006)
;; When is (1.01)^(t-2006) = 2?
;; When is (t-2006)*log(1.01) = log(2)?
;; When is (t-2006) = log(2)/log(1.01)?
;; So t = 2006 + log(2)/log(1.01) = 2006 + 69 <- doubles every 69 years