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Math Help - Appreciate Some Guidance, Help, Steps on this question

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    Appreciate Some Guidance, Help, Steps on this question

    Use Maclaurin’s series:
    8.1. to find the first 4 terms of the expansion of Sinh(2x), then
    8.2. use this expansion to evaluate Sinh(2x) when x = 0.5, to 4 decimal places
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    Re: Appreciate Some Guidance, Help, Steps on this question

    Hello, jono1966!

    \text{8.1 Use Maclaurin’s series to find the first 4 terms of the expansion of }\sinh(2x)

    \text}8.2. Use this expansion to evaluate }\sinh(2x)\text{ when }x = \frac{1}{2}\text{, to 4 decimal places.}

    The Maclaurin series begins: . \sinh(2x) \;=\;(2x) + \frac{(2x)^3}{3!} + \frac{(2x)^5}{5!} + \frac{(2x)^7}{7!} + \cdots

    If x = \tfrac{1}{2}, we have: . \sinh(1) \;=\;1 + \frac{1^3}{3!} + \frac{1^5}{5!} + \frac{1^7}{7!} \;=\; 1.175198413

    Therefore: . \sinh(1) \:\approx\:1.1752

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    Re: Appreciate Some Guidance, Help, Steps on this question

    Hi did you get a reply for this question please ?

    Use Maclaurin’s series:
    8.1. to find the first 4 terms of the expansion of Sinh(2x), then
    8.2. use this expansion to evaluate Sinh(2x) when x = 0.5, to 4 decimal places

    I have the same question and am struggling thanks for any advice
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    Re: Appreciate Some Guidance, Help, Steps on this question

    Quote Originally Posted by dean54321 View Post
    Hi did you get a reply for this question please ?

    Use Maclaurin’s series:
    8.1. to find the first 4 terms of the expansion of Sinh(2x), then
    8.2. use this expansion to evaluate Sinh(2x) when x = 0.5, to 4 decimal places

    I have the same question and am struggling thanks for any advice
    post #2 is a reply ...
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    Re: Appreciate Some Guidance, Help, Steps on this question

    I need to use the workings below for 8.1 could you give me any advice please. This is so the assessment criteria is covered.

    Question 8 – There are standard differential coefficients for both Cosh and Sinh, or you may wish to use their exponential definitions to successively differentiate. It may help to use the exponential definitions when evaluating f(0), f’(0), f’’(0) etc.

    I have tried with this so far but am struggling from here thanks.

    f"(x)=2^2sinh(2x) so f"(0)=0
    f"'(x)=2^3cosh(2x), f"'(0)=2^3
    fv(x)=2^4sinh(2x), fv(0)=0



    Thanks very much
    Last edited by dean54321; April 20th 2015 at 11:26 AM.
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    Re: Appreciate Some Guidance, Help, Steps on this question

    There are standard differential coefficients for both Cosh and Sinh, or you may wish to use their exponential definitions to successively differentiate. It may help to use the exponential definitions when evaluating f(0), f’(0), f’’(0) etc.
    \sinh(x) = \frac{1}{2}(e^x - e^{-x})

    f(x) = \frac{1}{2}(e^x - e^{-x}) \text{ ... } f(0) = 0

    f'(x) = \frac{1}{2}(e^x + e^{-x}) \text{ ... } f'(0) = 1

    f''(x) = \frac{1}{2}(e^x - e^{-x}) \text{ ... } f''(0) = 0

    f'''(x) = \frac{1}{2}(e^x + e^{-x}) \text{ ... } f'''(0) = 1

    f^{iv}(x) = \frac{1}{2}(e^x - e^{-x}) \text{ ... } f^{iv}(0) = 0

    see a pattern?

    the general Maclaurin series is ...
    f(x) = f(0) + f'(0)x + \frac{f''(0) x^2}{2!} + \frac{f'''(0) x^3}{3!} + ...


    \sinh(x) = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!} + ...

    now find the composite function ...

    \sinh(2x) = ?
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    Re: Appreciate Some Guidance, Help, Steps on this question

    Hi

    yes i can see a pattern is the following correct

    Sinh (2x) = 2 x 1/2 (ex - e-x)

    f (x) = 2 x 1/2 (ex - e-x)..... f (0) = 0
    fi (x) = 2 x 1/2 (ex + e-x)..... f (0) = 2
    fii (x) = 2 x 1/2 (ex - e-x)..... f (0) = 0
    fiii (x) = 2 x 1/2 (ex + e-x)..... f (0) = 2
    fiv (x) = 2 x 1/2 (ex - e-x)....... f (0) = 0

    putting these values into maclaurins series

    Sinh (2x) = 2x + 2x^3/3! + 2 x^5/5! + 2x^7/7! + 2x^9/9! .........
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    Re: Appreciate Some Guidance, Help, Steps on this question

    No, it is not ... please read the last instruction I gave you regarding finding the composite function.
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    Re: Appreciate Some Guidance, Help, Steps on this question

    Sorry I don't get what the composite function is.
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    Re: Appreciate Some Guidance, Help, Steps on this question

    f(x) = \sinh(x) = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!} + ...

    the following are various examples of composite functions of f[g(x)] = \sinh[g(x)] ...

    g(x) = 2x ...

    f(2x) = \sinh(2x) = 2x + \frac{(2x)^3}{3!} + \frac{(2x)^5}{5!} + \frac{(2x)^7}{7!} + ...

    g(x) = x+1 ...

    f(x+1) = \sinh(x+1) = (x+1) + \frac{(x+1)^3}{3!} + \frac{(x+1)^5}{5!} + \frac{(x+1)^7}{7!} + ...

    g(x) = x^2 ...

    f(x^2) = \sinh(x^2) = x^2 + \frac{(x^2)^3}{3!} + \frac{(x^2)^5}{5!} + \frac{(x^2)^7}{7!} + ...

    g(x) = 3 ...

    f(3) = \sinh(3) = 3 + \frac{(3)^3}{3!} + \frac{(3)^5}{5!} + \frac{(3)^7}{7!} + ...

    ... if you still don't understand, then I recommend you research composite functions.
    Last edited by skeeter; April 21st 2015 at 11:27 AM.
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