anonymous
  • anonymous
I'm trying to find what the summation from 0 to infinity of ((e^-4)4^x)/4! converges to (trying to show it is a valid pdf). Any suggestions on how to approach this would be welcome.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Jhannybean
  • Jhannybean
\[\huge \sum_{0}^{\infty} \frac{4^x(e^{-4})}{4!}\] Is it this? haha
anonymous
  • anonymous
haha yes
jdoe0001
  • jdoe0001
hmmm dunno that one :(

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amistre64
  • amistre64
does the sequence limit to zero?
amistre64
  • amistre64
e^(-4)/4! is just some constant value ...
anonymous
  • anonymous
I believe it does, since the factorial grows faster than the exponential
anonymous
  • anonymous
oh sorry it's over x!
amistre64
  • amistre64
4^x is the only thing that varies .. the constant factors out ...
anonymous
  • anonymous
you're right, without that it wouldn't converge
amistre64
  • amistre64
.... can you pic the question? avoids alot of error
anonymous
  • anonymous
will try, h/o
anonymous
  • anonymous
\[p(x) = \frac{e^{-4}4^x}{x!}\]
amistre64
  • amistre64
pdf, probability density function ...
anonymous
  • anonymous
yeah, pmf not pdf. sorry, the pic was coming out really grainy
anonymous
  • anonymous
\[x = 0,1,2,...\]
amistre64
  • amistre64
we want the sumto equal ...1?
anonymous
  • anonymous
yes
anonymous
  • anonymous
i know i can pull out the constant
amistre64
  • amistre64
then it behooves us to consider that 4^x/x! = e^(4)
amistre64
  • amistre64
sum of ...
amistre64
  • amistre64
do you recall the taylor series for e^x ?
anonymous
  • anonymous
i don't i can look it up, that may be the prod in the right direction i am looking for
amistre64
  • amistre64
its a thought, not sure how well it will turn out tho
anonymous
  • anonymous
ah, yes. that looks like it exactly
anonymous
  • anonymous
\[e^x = \sum \frac{x^n}{n!}\]
amistre64
  • amistre64
im so clever it scares me ....
anonymous
  • anonymous
haha i liked how you looked at it in terms of solving for what would give us e^4
anonymous
  • anonymous
i appreciate the help
amistre64
  • amistre64
youre welcome :)

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