+- Tetration Forum (https://tetrationforum.org)
+-- Forum: Tetration and Related Topics (https://tetrationforum.org/forumdisplay.php?fid=1)
+--- Forum: Mathematical and General Discussion (https://tetrationforum.org/forumdisplay.php?fid=3)
+--- Thread: a curious limit (/showthread.php?tid=623)
Pages:
1
2
a curious limit - JmsNxn - 04/14/2011
I'm wondering if the following limit is non-zero; v E R
\( \lim_{h\to\0}h^{1-e^{vi}} \)
and if so, what is it equal to? Thanks
I know it doesn't converge for \( v =\pi (1 + 2k) \,\,\,\,\{k \,\epsilon\, N\} \)
RE: a curious limit - nuninho1980 - 04/14/2011
attention: h -> o is bad but yes h -> 0. because 'o' isn't number and is letter. lol
RE: a curious limit - JmsNxn - 04/14/2011
(04/14/2011, 08:10 PM)nuninho1980 Wrote: attention: h -> o is bad but yes h -> 0. because 'o' isn't number and is letter. lol
no no, I put 0, but latex just designs it to look like o
RE: a curious limit - bo198214 - 04/14/2011
(04/14/2011, 08:01 PM)JmsNxn Wrote: I'm wondering if the following limit is non-zero; v E R
\( \lim_{h\to\0}h^{1-e^{vi}} \)
and if so, what is it equal to? Thanks
I know it doesn't converge for \( v =\pi (1 + 2k) \,\,\,\,\{k \,\epsilon\, N\} \)
The powers with non-integer exponents are not uniquely defined in the complex plane.
In your case you would need to put:
\( \lim_{h\to\0} e^{(1-e^{vi})\log(h)} \)
But then the standard logarithm has a cut on \( (-\infty,0] \), which is quite arbitrary: one could put a cut however one likes. For example \( h \) could spiral around 0, while moving towards 0 and would increase/decrease its imaginary part by \( 2\pi i \) in each round.
I guess it really depends on how \( h \) approaches 0.
RE: a curious limit - JmsNxn - 04/14/2011
let's take the limit from positive (keep it simple first)
so:
\( \lim_{h\to\0^+}\, (1-e^{vi})ln(h)\, =\, f(v) \)
Is there any way of re-expressing this limit?
RE: a curious limit - bo198214 - 04/15/2011
(04/14/2011, 10:55 PM)JmsNxn Wrote: let's take the limit from positive (keep it simple first)
so:
\( \lim_{h\to\0^+}\, (1-e^{vi})ln(h)\, =\, f(v) \)
Is there any way of re-expressing this limit?
But then its not difficult, since \( \ln(h)\to -\infty \) on the reals, the whole limit goes to (complex) \( \infty \) except for \( 1-e^{vi}=0 \), for which the whole limit is 0.
RE: a curious limit - JmsNxn - 04/15/2011
(04/15/2011, 07:14 AM)bo198214 Wrote: But then its not difficult, since \( \ln(h)\to -\infty \) on the reals, the whole limit goes to (complex) \( \infty \) except for \( 1-e^{vi}=0 \), for which the whole limit is 0.
Alright, how about
\( \lim_{h\to\0^{v+}} (1-e^{vi})ln(h) \)
where \( \lim_{h\to\0^{v+}} \) is taken to mean approaching along the \( e^{vi} \) axis.
I think it's the equivalent of:
\( = \lim_{h\to\0^{+}} (1-e^{vi})ln(he^{vi}) \)
\( = \lim_{h\to\0^{+}} (1-e^{vi})(ln(h) + vi) \)
which I guess converges to negative infinity again, except for 1-e^{vi}=0
hmm, seems this is less interesting than I thought.
RE: a curious limit - JmsNxn - 04/16/2011
is there any way of letting h approach zero such that:
\( \lim_{h\to\0} (1-e^{vi})ln(h) = 0 \)?
RE: a curious limit - bo198214 - 04/16/2011
(04/16/2011, 07:22 PM)JmsNxn Wrote: is there any way of letting h approach zero such that:
\( \lim_{h\to\0} (1-e^{vi})ln(h) = 0 \)?
The logarithm of \( z=r e^{i\phi} \) is \( \log( r)+i\phi \).
So regardless how you approach 0, i.e. \( r\to 0 \), you will allways have that \( |\log(z)|=\sqrt{\log( r)^2+\phi^2}\to \infty \).
So the answer is no (except \( 1=e^{vi} \)).
RE: a curious limit - JmsNxn - 04/16/2011
(04/16/2011, 07:41 PM)bo198214 Wrote: The logarithm of \( z=r e^{i\phi} \) is \( \log( r)+i\phi \).
So regardless how you approach 0, i.e. \( r\to 0 \), you will allways have that \( |\log(z)|=\sqrt{\log( r)^2+\phi^2}\to \infty \).
So the answer is no (except \( 1=e^{vi} \)).
that's what I thought