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anonymous

  • one year ago

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.

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  1. Jhannybean
    • one year ago
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    \[\huge \sum_{0}^{\infty} \frac{4^x(e^{-4})}{4!}\] Is it this? haha

  2. anonymous
    • one year ago
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    haha yes

  3. jdoe0001
    • one year ago
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    hmmm dunno that one :(

  4. amistre64
    • one year ago
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    does the sequence limit to zero?

  5. amistre64
    • one year ago
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    e^(-4)/4! is just some constant value ...

  6. anonymous
    • one year ago
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    I believe it does, since the factorial grows faster than the exponential

  7. anonymous
    • one year ago
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    oh sorry it's over x!

  8. amistre64
    • one year ago
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    4^x is the only thing that varies .. the constant factors out ...

  9. anonymous
    • one year ago
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    you're right, without that it wouldn't converge

  10. amistre64
    • one year ago
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    .... can you pic the question? avoids alot of error

  11. anonymous
    • one year ago
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    will try, h/o

  12. anonymous
    • one year ago
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    \[p(x) = \frac{e^{-4}4^x}{x!}\]

  13. amistre64
    • one year ago
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    pdf, probability density function ...

  14. anonymous
    • one year ago
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    yeah, pmf not pdf. sorry, the pic was coming out really grainy

  15. anonymous
    • one year ago
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    \[x = 0,1,2,...\]

  16. amistre64
    • one year ago
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    we want the sumto equal ...1?

  17. anonymous
    • one year ago
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    yes

  18. anonymous
    • one year ago
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    i know i can pull out the constant

  19. amistre64
    • one year ago
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    then it behooves us to consider that 4^x/x! = e^(4)

  20. amistre64
    • one year ago
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    sum of ...

  21. amistre64
    • one year ago
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    do you recall the taylor series for e^x ?

  22. anonymous
    • one year ago
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    i don't i can look it up, that may be the prod in the right direction i am looking for

  23. amistre64
    • one year ago
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    its a thought, not sure how well it will turn out tho

  24. anonymous
    • one year ago
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    ah, yes. that looks like it exactly

  25. anonymous
    • one year ago
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    \[e^x = \sum \frac{x^n}{n!}\]

  26. amistre64
    • one year ago
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    im so clever it scares me ....

  27. anonymous
    • one year ago
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    haha i liked how you looked at it in terms of solving for what would give us e^4

  28. anonymous
    • one year ago
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    i appreciate the help

  29. amistre64
    • one year ago
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    youre welcome :)

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