Last time we started on the problem of approximating the
minimum value of the function
x
2
– sin
x
.
We showed that there is a unique value
x
* for which this
function achieves its global minimum on (–
∞
,
∞
), and that
this value is the unique real number satisfying the equation
2
x
– cos
x
= 0.
Let’s estimate this number with Newton’s method.
Trial and error gives
x
* to be about 1/2, so take
x
1
= 0.5.
With
g
(
x
) = 2
x
– cos
x
, we have
g
′
(
x
) = 2 + sin
x
.
x
n
+1
=
x
n
– (2
x
n
– cos
x
n
) / (2 + sin
x
n
).
With
x
1
= 0.5 , we get
x
2
= 0.450626693077243046568,
x
3
= 0.450183647577774742501
,
x
4
= 0.45018361129487381641,
x
5
= 0.45018361129487357304,
x
6
= 0.45018361129487357304,
...
So after just a few iterations, the process stabilizes,
and we’ve found
x
* to twenty significant figures.
Hence the minimum value taken on by
x
2
– sin
x
is about
(
x
4
)
2
– sin
x
4
= –0.232465575.
..
In general, when Newton’s method works, the error at the
n
+1st stage is approximately equal to the square of the
error at the
n
th stage; hence each iteration gives you about
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View Full Documenttwice as many correct digits as the iteration before:
x
1
=
0
.5
, we get
x
2
=
0.450
626693077243046568,
x
3
=
0.4501836
47577774742501,
x
4
=
0.450183611294873
81641,
etc.
Questions on Section 4.6?
Section 4.7: Antiderivatives
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 Fall '11
 Staff
 Calculus, Derivative

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