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Material Type: Notes; Professor: Howle; Class: Calculus I-Honors; Subject: MATHEMATICS; University: Texas Tech University; Term: Fall 2007;
Typology: Study notes
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Example: approximate (^3).^198 using differentials.
Let f (x) = (^) x^1 , so f ′(x) = −x−^2.
Recall
f (x 0 + ∆x) ≈ f (x 0 ) + f ′(x 0 )dx
Let x 0 = 4 and ∆x = dx = − 0. 02
f (3.98) ≈ f (4) + f ′(4)(− 0 .02) = 14 + − 16 1 (− 0 .02) = 0 .25 + 0. 00125 = 0. 25125
Compare to calculator value:
f (3.98) ≈ 0. 2512562814
If x is the measured value and x + ∆x represents the exact value, then ∆x is the error in measurement. The difference between f (x + ∆x) and f (x) is called the propagated error at x:
∆f = f (x + ∆x) − f (x)
The relative error is ∆ff ≈ dff.
The percentage error is 100( ∆ff )%.
The Newton-Raphson method uses tangent lines to estimate roots (zeros) of equations.
Theorem: Newton-Raphson Method: To approximate f (x) = 0, start with a preliminary estimate (or guess) x 0 , and generate a sequence of estimates x 1 , x 2 , x 3 ,... using the formula:
xn+1 = xn − (^) ff ′^ ((xxn) n)^
, f ′(xn) 6 = 0
Either this sequence of approximations will approach a limit that is a root of the equation, or the sequence does not have a limit.
Let f (x) be a function on an interval I containing number c.
Extreme Value Theorem: A function f has both an absolute maximum and an absolute minimum on any closed, bounded interval [a, b] where it is continuous.
Note: if f is discontinuous or the interval is not both closed and bounded, we cannot conclude that f has an absolute max and min.
To find the absolute extrema of a continuous function f on [a, b]: