Can a^x be negative infinity


by coki2000
Tags: infinity, negative
coki2000
coki2000 is offline
#1
Sep16-10, 12:57 PM
P: 91
I think about on this and found a result, [tex]e^{i\pi k}=e^{-i\pi k}=-1\Rightarrow i\pi k=ln(-1)[/tex]
a>1 and k=1,2,3...

[tex]y=(-a)^{i\alpha }\Rightarrow lny=i\alpha ln(-a)=i\alpha (ln(-1)+lna)=i\alpha (i\pi k+lna)\Rightarrow y=e^{-\alpha \pi k}e^{i\alpha \ln{a}}[/tex]

Then

[tex]\alpha \ln{a}=-\pi \Rightarrow \alpha =\frac{-\pi }{\ln{a}}\Rightarrow e^{-i\pi }e^{\frac{{\pi }^2 k}{\ln{a}}}=-e^{\frac{{\pi }^2 k}{\ln{a}}} [/tex]

If I choose k as infinity, I get a exponential expression as negative infinity.
Is it right? Please explain to me. Thanks
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coki2000
coki2000 is offline
#2
Sep17-10, 05:06 AM
P: 91
Sorry, k would be k=2n-1=1,3,5,7...
jackmell
jackmell is offline
#3
Sep17-10, 07:56 AM
P: 1,666
Not sure what you're doing, but I'd write:
[tex]a^x=-k[/tex]

[tex]e^{x\log(a)}=-k[/tex]

[tex]x\log(a)=\log(-k)[/tex]

[tex]x=\frac{\log(-k)}{\log(a)}=\frac{\ln(k)+i(\pi+2 n \pi)}{\log(a)}[/tex]

and as [itex]k\to\infty[/itex], [itex]x\to \infty+i(\pi+2n\pi)/\log(a)[/itex]


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