My tutor showed me something today, and I still can't completely wrap my head around why it makes sense. Consider the following integral equation:(adsbygoogle = window.adsbygoogle || []).push({});

##\int f(x) = f(x) - 1##

Then:

##\int (f(x)) - f(x) = -1 \implies f(x) \left( \int (\text{Id}) - 1\right) = -1##

so we get the geometric series

##f(x) = \frac{1}{1 - \int(Id)} = 1 + \int + \int\int + \int\int\int + ... = 1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3 + ... = e^x##

which satisfies our original equation if we choose the right constant of integration. Similarly, take the differential equation (using D as the differential operator):

##Df = f - 1 \implies f\left(D(\text{Id}) - 1\right) = -1##

##f = 1 + D + (D \circ D) + (D \circ D \circ D) + ... = 1##

which also satisfies our original equation.

Is this mathematically valid, or was he pulling my leg? And if so, could someone help me make sense of why this is valid?

Trying this on some other differential equations seems to show that this doesn't always work -- I was wondering if anyone could point me to an article that says more about the idea behind this.

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# Treatment of Integral as an Operator?

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