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ilovemath7 Group Title

If I need to find a closed solution for the summation of x*a^x from x = 0 to n, what do I do? For starters, I've decided to do an integral from 0 to b of the same function, but now I have 2 variables! ah! Help please :D

  • 2 years ago
  • 2 years ago

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  1. ilovemath7 Group Title
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    The only hint I was given was to take the integral of the function I was trying to take the summation of. But, now I have\[\int\limits_{0}^{b}x*a^x da\]

    • 2 years ago
  2. ilovemath7 Group Title
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    I can't seem to do a u-sub that will work for me.

    • 2 years ago
  3. ilovemath7 Group Title
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    I want to assume that x is a constant and pull it outside, but I don't know if I can do that.

    • 2 years ago
  4. extremity Group Title
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    you can, since youre differentiating with respect to a.

    • 2 years ago
  5. ilovemath7 Group Title
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    Because S = (0)a^(0) + (1)a^(1)...+(n)a^(n); you see why I think it's a constant?

    • 2 years ago
  6. ilovemath7 Group Title
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    Yes! But what happens to the other x?

    • 2 years ago
  7. extremity Group Title
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    wel, the other x is a constant too right?

    • 2 years ago
  8. extremity Group Title
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    whats the integral of a^x , if x is constant?

    • 2 years ago
  9. Jemurray3 Group Title
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    I have a wonderful solution to this, I believe.

    • 2 years ago
  10. ilovemath7 Group Title
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    Bring it :D

    • 2 years ago
  11. ilovemath7 Group Title
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    hm, you are right extremity

    • 2 years ago
  12. ilovemath7 Group Title
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    \[\int\limits_{0}^{b}a^x da\]?

    • 2 years ago
  13. Jemurray3 Group Title
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    \[ \sum_{x=0}^n xa^{\lambda x} = \sum_{x=0}^n \frac{d}{d\lambda} \frac{a^{\lambda x}}{\ln(a)} = \frac{1}{\ln(a)} \frac{d}{d\lambda} \sum_{x=0}^n a^{\lambda x}\] \[ = \frac{1}{\ln(a)} \frac{d}{d\lambda} \frac{1- a^{\lambda (n+1)}}{1- a^\lambda}\]

    • 2 years ago
  14. ilovemath7 Group Title
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    \[a^{x+1} /(x+1)\]

    • 2 years ago
  15. ilovemath7 Group Title
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    woah.

    • 2 years ago
  16. ilovemath7 Group Title
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    where'd you get lambda?

    • 2 years ago
  17. Jemurray3 Group Title
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    just made it up. Set it to 1 when you're all done.

    • 2 years ago
  18. ilovemath7 Group Title
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    it's to differentiate x from the exponent?

    • 2 years ago
  19. Jemurray3 Group Title
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    yep. I may have made a typo, but I'm getting \[ \frac{a(n+1) -a^n - n}{(1-a)^2} \] as a final result. Assuming of course that a does not equal 1.

    • 2 years ago
  20. Jemurray3 Group Title
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    Oops, I did make a typo

    • 2 years ago
  21. Jemurray3 Group Title
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    should be \[ \frac{na^{n+2} - na^{n+1} - a^{n+1} + a}{(1-a)^2} \]

    • 2 years ago
  22. Jemurray3 Group Title
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    And wolfram alpha confirms. Awesome.

    • 2 years ago
  23. Jemurray3 Group Title
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    By the way, if a = 1, obviously the sum is just \[ \sum_{x=0}^n x = \frac{n(n+1)}{2}\]

    • 2 years ago
  24. ilovemath7 Group Title
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    lol

    • 2 years ago
  25. ilovemath7 Group Title
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    Thanks! I am going back over this to make sure I understand. Thanks for all your help!

    • 2 years ago
  26. Jemurray3 Group Title
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    No problem. Be careful with your derivatives. And @TuringTest / @Gravion should recognize this technique.... ;)

    • 2 years ago
  27. ilovemath7 Group Title
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    :D

    • 2 years ago
  28. ilovemath7 Group Title
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    Are you still on?

    • 2 years ago
  29. ilovemath7 Group Title
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    I have a question: what if I need to use the hint (i.e., that's what they want us to do). I kinda see what you are doing, but I can move on from the integral from 0 to infinity of x*a^x

    • 2 years ago
  30. ilovemath7 Group Title
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    the integral doesn't converge, so...

    • 2 years ago
  31. Jemurray3 Group Title
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    That is the hint. I integrated, then differentiated.

    • 2 years ago
  32. Jemurray3 Group Title
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    And it should converge if a is sufficiently small.

    • 2 years ago
  33. ilovemath7 Group Title
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    hm...

    • 2 years ago
  34. Jemurray3 Group Title
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    If |a| < 1, that reduces to \[ \frac{a}{(1-a)^2} \]

    • 2 years ago
  35. ilovemath7 Group Title
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    I'm sorry, I'm thrown off by the ln(a). I can't get it in my own work...

    • 2 years ago
  36. ilovemath7 Group Title
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    I'm getting a^(x+1)/ (x+1) though, which I can use on the limits of intergration...but then I get x[infinity - a] which doesn't make sense!

    • 2 years ago
  37. Jemurray3 Group Title
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    Did you follow my reasoning up until the part where you take the derivative?

    • 2 years ago
  38. ilovemath7 Group Title
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    um, actually no. I don't understand d/dlambda

    • 2 years ago
  39. ilovemath7 Group Title
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    you said you made lamda up---to differentiate from the other x (I guess) but then how do you also do a derivative for it?

    • 2 years ago
  40. ilovemath7 Group Title
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    wait, the "integrate then differentiate" thing is what I am supposed to do (that's the hint they gave" but I dunno why I can't understand this!

    • 2 years ago
  41. Jemurray3 Group Title
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    I invented a parameter to put in the problem, differentiated and integrated with respect to said parameter, and at the end of the day, after everything was done, I set that parameter to 1.

    • 2 years ago
  42. Jemurray3 Group Title
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    It's not a trivial series to sum, what class are you doing this for?

    • 2 years ago
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