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TenaciousT

  • 2 years ago

4x^2 tan^−1(9x^3) could someone show me how to represent this as a maclaurin series

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  1. BrushingUpMyMath
    • 2 years ago
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    First you find the MacLaurin series for tan^-1(u), then you substitute u=9x^3 in each term, then you multiply each term by 4x^2. To deal with tan^-1(u), deal with its derivative, which is 1/(1+u^2), then when you're done, integrate each term. So, how to find the Series for 1/(1+u^2)? One trick is to substitute t = - (u^2), so it becomes 1/(1 - t). You should recognize this as the solution to the simple power series: 1 + t + t^2 + t^3 + … Putting back the u’s in place of the t’s we get: 1 – u^2 + u^4 – u^6 + … Integrating each term gives: tan^-1(u) = u – u^3/3 + u^5/5 – u^7/7 + … Substituting back for u=9x^3 gives: tan^-1(9x^3) = 9x^3 – 9^3(x^9)/3 + 9^5(x^15)/5 - 9^7(x^21)/7 + … And finally multiplying by 4x^2, we get the final aswer: 4x^2 tan^-1(9x^3) = (4)(9)(x^5) – (4)(9^3)(x^11)/3 + (4)(9^5)(x^17)5 – (4)(9^7)(x^23)/7 + … Better check the math – I might have messed up in detail, but the general approach should be correct.

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