Re: finding a linear approximation function L(x)

On 3 sep, 21:51, The World Wide Wade <aderamey.a...@xxxxxxxxxxx>
wrote:
In article
<10809be9-0403-463a-9f85-f649aa82b...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,





 conrad <con...@xxxxxxxxxx> wrote:
Suppose I want to find the value of sqrt(99.8)

I can either find a linear approximation
of the form:
L(x) = f(a) + f'(a)(x - a)

That is, the linearization of f at a

or I can use the differential form:
Given a dx and dx = h and given an x
Then dy = f'(x)*dx
f(x + h) ~= f(x) + dy

The latter form is simple to compute.
My trouble with finding a linear
approximation of the first form is
in determining a suitable value for 'a'

How should this be done?

--
conrad

Difficult to believe you find the latter simple and the former not.
They are exactly the same method. I prefer the notation of the former
because it doesn't contain any wacky differentials.- Masquer le texte des messages précédents -

- Afficher le texte des messages précédents -

Bonsoir,

There is a known way to build approximations :
considering Taylor/exponential development
exp(h*D) * f = f(x +h) , D = d/dx , Id identity
exp(D)~ Id +h*D*(Id + h/2*D) ~ Id +h*D*exp(h/2*D)
{ with given convergence constraints to neglect
D^n terms since some n ).
Or f(x +h) ~ f(x) + h*f{(x +h/2)

In your case sqrt(99.8) = sqrt(100-0.2), h =-.1
and sqrt(99.8) ~ sqrt(100) -0.2*1/(2*sqrt(99.9))
~ 10 - 0.1/sqrt(100) ~10 -0.1 = 9.99
My TI82 gives :9.989994995

Alain
.

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