# Taylor series

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How do I find the Taylor series for this function:

(1/x) Integral(x to 0) (e^t - e^-t)/2t dt.

...up to 4 terms? And the approximate value at x = 1.

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#### Solution Preview

Solution. As e^t=1+t+t^2/2!+t^3/3!+t^4/4!+t^5/5!+...+t^n/n!+...,

and

e^(-t)=1-t+t^2/2!-t^3/3!+t^4/4!-t^5/5!+...+(-t)^n/n!+...,

we have

e^t-e^(-t)=2t+2t^3/3!+2t^5/5!+2t^7/7!+...+2t^(2n-1)/(2n-1)!+...

Hence,

[e^t-e^(-t)]/(2t)=1+t^2/3!+t^4/5!+t^6/7!+...+t^(2n-2)/(2n-1)!+... ...

#### Solution Summary

This provides an example of finding a Taylor series.

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