Math Is Fun Forum

cadjeff

Member

Registered: 2007-05-08

Posts: 26

expansions and approximations

my knowledge on this subject is limited, for anyone who is into this, i'd very much appreciate any pointers/explanation on this.

thank you,

cad

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: expansions and approximations

I'll do the second one.

(i) Can't be bothered to draw it (good start ). But:
- f(x) is a straight line along y=1
- f_1(x) is a straight line from (0,0) to (1,1)
- f_2(x) is a curve from (0,0) to (1,1), going steeply up at first and then getting more gradual (as if f_1(x) was being pulled towards (0,1).
- f_3(x) is similar to f_2(x), but the pull towards (0,1) is stronger this time and so the curve is more apparent.

(ii)

This clearly converges to 0 as n --> ∞ (and so its square root also does).

I'm guessing the definition of convergence in the least squares sense in the scan has a typo, and that it should say f(x) - f_n(x) inside the inner brackets. As it is, the expression inside the brackets is 0 and so everything would trivially converge in that sense.

If this is the case, then the above working shows that f_n --> f in the least squares sense.

(iii)

This is true regardless if the value of n, since 0^(1/n) is always 0.

Hence x^(1/n) does not converge uniformly.

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Why did the vector cross the road?
It wanted to be normal.