Let's find the Maclaurin Series of tanhx up to powers of x^5(adsbygoogle = window.adsbygoogle || []).push({});

Yeah! Good idea!

I know

Right, f(x) = tanh

[tex]f'(x) = sech^{2}(x)[/tex]

[tex]f''(x) = -2sech^{2}(x)tanh(x)[/tex]

[tex]f''(x) = 4sech^{2}(x)tanh^{2}(x) - 2sech^{4}(x)[/tex]

giving f(0) = 0, f'(0) = 1, f''(0) = 0 f'''(-2)

but according to my text book apparently f'(0) = -1, how can this be. Especially as f'''(0) = -2, not just 2.

Thanks

Thomas

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# Maclaurin Series of Tanh(x)

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