- #1

jostpuur

- 2,112

- 18

Do there exist numbers [itex]A,B,C[/itex] such that

[tex]

x\log(x) = Ax^B + O(x^C)\quad\quad\textrm{as}\; x\to 0^+

[/tex]

and such that

[tex]

1\leq C

[/tex]

?

The approximation is trivial if [itex]C < 1[/itex], because then [itex]x^{1-C}\log(x)[/itex] would approach zero, and [itex]A[/itex] and [itex]B[/itex] could be chosen to be almost anything (only [itex]C<B[/itex] needed). But if [itex]1\leq C[/itex], then the approximation could have some content. Obviously conditions

[tex]

A < 0 < B < 1

[/tex]

should hold, because [itex]x\log(x) < 0[/itex] when [itex]0<x<1[/itex], and [itex]D_x(x\log(x))\to \infty[/itex].

update:

I see these numbers do not exist, because if they did, then also [itex]\log(x)[/itex] could be approximated with some [itex]\alpha x^{\beta}[/itex] where [itex]\beta <0[/itex].

[tex]

x\log(x) = Ax^B + O(x^C)\quad\quad\textrm{as}\; x\to 0^+

[/tex]

and such that

[tex]

1\leq C

[/tex]

?

The approximation is trivial if [itex]C < 1[/itex], because then [itex]x^{1-C}\log(x)[/itex] would approach zero, and [itex]A[/itex] and [itex]B[/itex] could be chosen to be almost anything (only [itex]C<B[/itex] needed). But if [itex]1\leq C[/itex], then the approximation could have some content. Obviously conditions

[tex]

A < 0 < B < 1

[/tex]

should hold, because [itex]x\log(x) < 0[/itex] when [itex]0<x<1[/itex], and [itex]D_x(x\log(x))\to \infty[/itex].

update:

I see these numbers do not exist, because if they did, then also [itex]\log(x)[/itex] could be approximated with some [itex]\alpha x^{\beta}[/itex] where [itex]\beta <0[/itex].

Last edited: