Even sums can be odd! - Solution

Graphs of Even Functions and their Integrals with t0=0

y=1

From the plot it appears that the integral is an odd function because if the integral graph is reflected in the y axis it results in an upside down version of itself (i.e. a reflection in the t-axis of the original plot).

y=|t| using |t|=(t*t)^0.5

From the plot it appears that the integral is an odd function because if the integral graph is reflected in the y axis it results in an upside down version of itself (i.e. a reflection in the t-axis of the original plot).

y=cos(t)

From the plot it appears that the integral is an odd function.

y=cosh(t) using cosh(t)=0.5*(e^t+e^(-1*t))

The Maclaurin series for a general function, f(t), defined around t=0 and with all its derivatives defined at t=0 is given by:

If a function is even then it will only have even powers of t in its Maclaurin series, i.e. f'(0) and f'''(0) and f(n)(0) for n odd, must all be 0.

So, for an even function the Maclaurin series is:

Integrating above we get

Providing t0=0.

As all the terms in the above are even powers of t then the integral is even, providing t0=0. For non-zero t0=0, no such conclusion can be made as the lower bound of the integration can result in a non-zero constant term, meaning that the integral is no longer even.

An example of the integral where t0 is non zero is given below.

y=cosh(t) using cosh(t)=0.5*(e^t+e^(-1*t)) and integral with t0=-5

The integral is not an odd function in this case.

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