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Math Help - [SOLVED] Limit of Truncated Taylor Series to Power.

  1. #1
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    [SOLVED] Limit of Truncated Taylor Series to Power.

    I am trying to figure out a limit, but am getting a bit stuck.
    <br />
\lim_{n\to \infty} \left(1 + a\frac{t}{n} + a^2\frac{t^2}{2n^2} + a^3\frac{t^3}{3!n^3} + a^4\frac{t^4}{4!n^4}\right)^n<br />
    Obviously,
    <br />
\lim_{n\to \infty} \left(1 + a\frac{t}{n}\right)^n = e^{at}<br />
    but I am having trouble extending this.
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  2. #2
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    Quote Originally Posted by lvleph View Post
    I am trying to figure out a limit, but am getting a bit stuck.
    <br />
\lim_{n\to \infty} \left(1 + a\frac{t}{n} + a^2\frac{t^2}{2n^2} + a^3\frac{t^3}{3!n^3} + a^4\frac{t^4}{4!n^4}\right)^n<br />
    Obviously,
    <br />
\lim_{n\to \infty} \left(1 + a\frac{t}{n}\right)^n = e^{at}<br />
    but I am having trouble extending this.
    Squeeze rule: \bigl(1 + a\tfrac{t}{n}\bigr)^n < \bigl(1 + a\tfrac{t}{n} + a^2\tfrac{t^2}{2!n^2} + a^3\tfrac{t^3}{3!n^3} + a^4\tfrac{t^4}{4!n^4}\bigr)^n < \bigl(e^{at/n}\bigr)^n. The outer elements both tend to e^{at}.
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  3. #3
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    And we know the RHS holds since the Truncated Taylor Series is an approximation to e^{at/n}?
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  4. #4
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    Quote Originally Posted by lvleph View Post
    And we know the RHS holds since the Truncated Taylor Series is an approximation to e^{at/n}?
    I was assuming that the truncated series is less than e^{at/n}. But that is only true if at > 0 (which I was taking for granted). If at < 0 then that argument doesn't seem to work.
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  5. #5
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    Yeah, I was thinking the same thing, but in my problem it is true.
    Thank you.
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