An Accidental Hint at Euler's Equation

I had this thought at lunch today. It's not a sophisticated treatment of the subject, but I found it amusing.

We know that when you differentiate a sinusoid twice, you get back a scaled and negated version of the original. Here the scaling factor is unity.

y=sin(t)y¨=sin(t)=y y = sin(t) \\ \ddot{y} = -sin(t) = -y

We also know that double-differentiating an exponential gets us back a scaled (but not negated) version of the original. Here again, the scaling factor is unity.

y=ety¨=et=y y = e^t \\ \ddot{y} = e^t = y

These two behaviors are tantalizingly similar. So how might we get the exponential to behave like the sinusoid with respect to double-differentiation? Maybe we could throw in the square root of negative one.

y=ejty¨=j2ejt=ejt=y y = e^{j t} \\ \ddot{y} = j^2 e^{j t} = -e^{j t} = -y

Making the exponent complex makes the exponential behave like a sinusoid with respect to double-differentiation. Hence, we've stumbled onto something like Euler's equation (shown below for reference).

ejt=cost+jsint e^{j t } = cos \, t + j \, sin \, t

Note by Steven Chase
1 year ago

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Good Lord. Thanks sir for posting these.

Md Zuhair - 1 year ago

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Glad you liked it

Steven Chase - 1 year ago

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Very interesting. Great work.

Krishna Venkatraman - 8 months, 2 weeks ago

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Thanks

Steven Chase - 8 months, 2 weeks ago

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