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MathSofiya

  • 2 years ago

Oh I think I get the idea of stripping out a term now. Simply by taking out an \[a_1\] changes where the sum starts? Yes? \[\sum_{n=1}^\infty a_n=a_1+\sum_{n=2}^{\infty} a_n\]

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  1. MathSofiya
    • 2 years ago
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    so when I have \[\sum_{n=1}^{\infty}a_nnx^n=a_1+\sum_{n=2}^{\infty}a_nnx^n\] but it's not n=0? Why?

  2. MathSofiya
    • 2 years ago
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    I'm trying to start at n=0

  3. MathSofiya
    • 2 years ago
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    \[\sum_{n=1}^{\infty}a_nnx^n \rightarrow \sum_{n=0}^{\infty}a_nnx^n\]

  4. MathSofiya
    • 2 years ago
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    I looked at example 1 a. http://tutorial.math.lamar.edu/Classes/CalcII/Series_Basics.aspx but I'm not trying to do an index shift :C

  5. Freyes
    • 2 years ago
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    Well do you need to start at 0? because if you want n = 0 you can just make it at the bottom n = 0

  6. MathSofiya
    • 2 years ago
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    why though?

  7. MathSofiya
    • 2 years ago
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    why is that all I need to do?

  8. Freyes
    • 2 years ago
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    \[\sum_{n =0}^{\infty}\]

  9. Freyes
    • 2 years ago
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    because in your equation it would work for not getting an answer anyway

  10. Freyes
    • 2 years ago
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    because multiplying by 0 just leaves the term out basically, so you can have it as n=0 but you won't get an actual term out.

  11. Freyes
    • 2 years ago
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    if you are trying to get a term out there have to be different conditions met in your equation otherwise it won't work that way. But it looks fine for getting a term out if it's not n = 0.

  12. MathSofiya
    • 2 years ago
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    by that do you mean that \[\sum_{n=1}^{\infty}a_nnx^n=a_1+\sum_{n=0}^{\infty}a_nnx^n\] It leaves the \[a_1\]out?

  13. Freyes
    • 2 years ago
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    no

  14. MathSofiya
    • 2 years ago
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    :(

  15. Freyes
    • 2 years ago
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    it's right the left part

  16. Freyes
    • 2 years ago
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    but if you made that n = 0

  17. Freyes
    • 2 years ago
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    on the left and n= 1 on the right

  18. Freyes
    • 2 years ago
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    it would be \[\alpha _{0}*0*x^0\]

  19. MathSofiya
    • 2 years ago
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    OMG! I think something just clicked!

  20. Freyes
    • 2 years ago
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    Yeah, you see where im going with it

  21. MathSofiya
    • 2 years ago
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    no wait, just a second

  22. Freyes
    • 2 years ago
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    You can have it start at 0.. but it doesnt change anything if you start at 1 in this case

  23. MathSofiya
    • 2 years ago
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    \[\sum_{n=0}^{\infty}a_nnx^n=0\]

  24. Freyes
    • 2 years ago
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    it's like putting 0 + \[\sum_{n=1}^{\infty} (insert rest here)\]

  25. Freyes
    • 2 years ago
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    for n = 0

  26. Freyes
    • 2 years ago
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    I guess if you want to start at 0, it has to be fixed at 0 and you can't change that value for it so it's usually left out of summations and n starts at 1. There are situations where you can have n start at 0 but in you equation those conditions for having a different term at n = 0 is not met.

  27. MathSofiya
    • 2 years ago
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    I'm a very visual learner. Where you said: "on the left and n= 1 on the right" \[\sum_{n=1}^{\infty}a_nnx^n=a_1+\sum_{n=0}^{\infty}a_nnx^n\] ^^^Is wrong I know, but would you mind writing what you mean?

  28. MathSofiya
    • 2 years ago
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    I'm honestly trying hard to get this...I think we're almost there...sorry =/

  29. Freyes
    • 2 years ago
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    \[\sum_{n=0}^{\infty}a _{n}nx ^{n} = 0 + \sum_{n=1}^{\infty}a _{n}nx ^{n}\]

  30. Freyes
    • 2 years ago
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    using the rule that 0 multiplied by anything is 0

  31. MathSofiya
    • 2 years ago
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    Thank you, finally!

  32. Freyes
    • 2 years ago
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    Lol it's ok, glad i could help

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