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JenniferSmart1

  • 3 years ago

I did this problem a month ago and I somewhat forgot how to do it.

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  1. JenniferSmart1
    • 3 years ago
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    the third line where n=2. How did we determine that it can be written as... \[\frac{C_0}{1\cdot2\cdot3\cdot4}\]

  2. saifoo.khan
    • 3 years ago
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    Is that Taylor series or something? D:

  3. JenniferSmart1
    • 3 years ago
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    no not really the original function is \[C_{n+2}=-\frac{C_n}{(n+2)(n+1)}\]

  4. saifoo.khan
    • 3 years ago
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    Sorry. This goes over my dumb head. :'( Math gods like @satellite73 @amistre64 @asnaseer can help you with problems like these. :/

  5. JenniferSmart1
    • 3 years ago
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    I'm sure @UnkleRhaukus can help...where is he?

  6. saifoo.khan
    • 3 years ago
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    @eseidl @eSpeX can you guys solve this? Last time I saw Rhau in chem section.

  7. eseidl
    • 3 years ago
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    Am I missing something? Isn't it just:\[C_{n+2}=-\frac{c_n}{(n+2)(n+1)}\]Input n=2\[C_{2+2}=-\frac{c_2}{(2+2)(2+1)}\]So,\[C_4=-\frac{c_2}{4*3}\]But we know\[c_2=-\frac{c_0}{1*2}\]From the table where n=0.Thus:\[C_4=-\frac{c_0}{4*3*2*1}=-\frac{c_0}{4!}\]

  8. JenniferSmart1
    • 3 years ago
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    yeah I think you're right, Thanks!

  9. eseidl
    • 3 years ago
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    lol, yeah it just seemed too straight forward to call in the "math gods" (which I definitely am not one of lol)

  10. JenniferSmart1
    • 3 years ago
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    It is straight forward I know...LOL and Turing has taught me how to do this...I just forgot

  11. JenniferSmart1
    • 3 years ago
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    I guess doing math and watching The Walking Dead can prevent one from doing simple math ;P

  12. UnkleRhaukus
    • 3 years ago
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    @eseidl all you missing is the negative signs cancel \[n=2\qquad C_4=\frac{-C_2}{4\cdot3}=-\frac{-\tfrac{C_0}{2\cdot1}}{4\cdot3}=\frac{C_0}{4\cdot3\cdot 2\cdot1}=\frac{C_0}{4!}\],

  13. eseidl
    • 3 years ago
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    @UnkleRhaukus oooops, yeah good call :)

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