Equilibrium? (Help with concept) Finding eqb visually/graphically , Finding eqb algebraically??
what exactly do you mean by "equilibrium"?
Idk just a concept my professor said to go over and practice graphically algebraically
I'm in Calc I if that helps
LIke I said I'm having my midterm tomorrow so just going over stuff and the list said equilibrium which I do not understand. So the best you can explain will help
i've not really heard this term before in the context of calculus do you have an example from your lecture notes or textbook?
equilibria?
oh ok I got it, you just need to solve when x_n+1 = x_n (they can use the variable x, or a, or whatever)
like, for the first example a_n+1 = sqrt(a) you just set a = a_n+1 to get a = sqrt(a) which is only true for a = 1 or 0
it's just algebra, really
the n+1 disappears?
we substitute a for a_n+1 since a = a_n+1
if it makes it less confusing you can just use a new variable like z or something a_n+1 = sqrt(a) becomes z = sqrt(z) which gives the same result, just make sure to report your answer in terms of a_n or whatever variable is already in the problem
okay so what if numbers were involved?
same concept
?
yeah, you just set a_n+1 and a_n equal and solve, regardless of what else is in the equation
can you just give a quick example with numbers ? :S
Just want to note an example down for reference with actuAl numbers
like, the second example you have has a_n+1 = 0.33an + 214.5 you just set a_n+1 and an equal to each other to get a = 0.33a + 214.5 and solve for a
a-214.5=0.33a
I would recommend subtracting 0.33a from both sides and dividing to isolate a
a-0.33a=214.5
good, finish simplifying the left side
6.71E-1a=214.5 a=3.2E2
I would recommend not converting to base 10 notation here
.67 do 214.5/.67=320
like, a-0.33a is just 1 - 0.33 which is 0.67a 0.67a = 214.5 a = 214.5/0.67 = 320.149
yes
NOw how about to graphing an equilibria or whatever :S
uh well since a_n+1 = an you should look for a part of the graph that levels off/flatlines
i think
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