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xEnOnn

  • 3 years ago

How do I get the Maclaurin series of (2k)/(3^k) ? I tried to make it in the form of a geometry series but still couldn't get anywhere. Thanks.

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  1. yuki
    • 3 years ago
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    The McLaurin series is just the Taylor series when the center is at 0. so do you know how to find the Taylor series of this?

  2. xEnOnn
    • 3 years ago
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    hmm.. The thing is I don't know how I can find the Taylor series of this equation. :(

  3. xEnOnn
    • 3 years ago
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    The usual equations that I could find to turn it into a Taylor Series is of a geometry series form where I can make it in the form of 1/(1-x) become making it a taylor series. In this one, there is no way I could make it in that form. That's why I don't know how to convert it to either the maclaurin or taylor series. Any help on this would be great. Thanks!

  4. yuki
    • 3 years ago
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    \[f(0) + f'(0)x + {f"(0)x^{2}}/{2!} + ...\] so this is the formula, right ? are you having trouble with the finding derivatives?

  5. xEnOnn
    • 3 years ago
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    Actually, I am trying to find something like: \[\frac{2k}{3^{k}} = \sum_{n=0}^{\infty} a(n)\] Then find a(n) in terms of n.

  6. yuki
    • 3 years ago
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    Exactly, that is why you do the Taylor series. Once you expand it, you may see a pattern in the Taylor Series which could give you the nth term.

  7. xEnOnn
    • 3 years ago
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    So I cannot just convert all equations to summations easily like how I could manipulate equations that can turn into the geometry series format and then instantly get its summation series?

  8. yuki
    • 3 years ago
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    For the easy ones, yes. For example the reason you know why 1/(1-x) = 1+x+x^2+... is because the geometric series adds up to a sum (1-x^(n-1))/(1-x) when you are adding n terms. When |x| < 1, what happens is that the x^(n-1) term almost equals to 0 when n is arbitrarily big. so it ends up being 1/(1-x). What's interesting is that when you do the Taylor expansion of 1/(1-x), you will get the same result 1+x+x^2+... so if there is an unusual function, you have to just find out what the pattern of the expansion is on each term and find a formula a(n) for it. Once you have done that, then all you have to do is to add it up which is as same as putting it inside the sigma.

  9. yuki
    • 3 years ago
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    If you want to trust my calculation, then I got the first 5 derivatives for you. So that the Taylor expansion looks like this 2x/3^x = \[f(0) + f'(0) x + f"(0) x^{2}/2! + f ^{(3)}(0)x^{3}/3!+...\] = \[0 + 2/3x + -4\ln3/3x^{2}/2! + 2\ln^{2}x^{3}/3! + 0 + -2\ln^{4}/3x^{5}/5! + 4\ln^{5}3/3x^{6}/6! + ...\] I hope this helps.

  10. yuki
    • 3 years ago
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    woops, I did not type some 3s. it should be \[\ln^{2}3\] and \[\ln^{5}3\]

  11. xEnOnn
    • 3 years ago
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    wow...thanks!! I will try with this method and come back to here again. thank you so much, yuki!

  12. 3lroy
    • 3 years ago
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    is there no manual way of calculating the 2nd and 3 derivatives ....i think if you try that with the taylor series exp

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