Edit: looks like was already raised by Dacyn and answered to my satisfaction by Robert_AIZI. Correctly applying the fundamental theorem of calculus will indeed prevent that troublesome zero from appearing in the RHS in the first place, which seems much preferable to dealing with it later.
My real analysis might be a bit rusty, but I think defining I as the definite integral breaks the magic trick.
I mean, in the last line of the ‘proof’, 1+I+I2+… gets applied to the zero function.
Any definitive integral of the zero function is zero, so you end up with f(x)=0, which is much less impressive.
More generally, asking the question Op(f)=0 for any invertable linear operator Op is likely to set yourself up for disappointment. Since the trick relies on inverting an operator, we might want to use a non-linear operator.
I(f(x)):=∫x0f(x)dx+C where C is some global constant might be better. (This might affect the radius of convergence of that Taylor series, do not use for production yet!)
This should result in… uhm… C+(Cx+C)+(Cx22+Cx+C)+…?
Which is a lot more work to reorder than the original convention used in the ‘proof’ where all the indefinite integrals of the zero function are conveniently assumed to be the same constant, and all other indefinite integrals conveniently have integration constants of zero.
Even if we sed s/C//ϵ and proclaim that ϵ should be small (e.g. compared to x) and we are only interested in the leading order terms, this would not work. What one would have to motivate is throwing everything but the leading power of x out for every In evaluation, then later meticulously track these lower order terms in the sum to arrive at the Taylor series of the exponential.
Edit: looks like was already raised by Dacyn and answered to my satisfaction by Robert_AIZI. Correctly applying the fundamental theorem of calculus will indeed prevent that troublesome zero from appearing in the RHS in the first place, which seems much preferable to dealing with it later.
My real analysis might be a bit rusty, but I think defining I as the definite integral breaks the magic trick.I mean, in the last line of the ‘proof’,1+I+I2+…gets applied to the zero function.Any definitive integral of the zero function is zero, so you end up with f(x)=0, which is much less impressive.More generally, asking the question Op(f)=0 for any invertable linear operator Op is likely to set yourself up for disappointment. Since the trick relies on inverting an operator, we might want to use a non-linear operator.I(f(x)):=∫x0f(x)dx+C
where C is some global constant might be better. (This might affect the radius of convergence of that Taylor series, do not use for production yet!)This should result in… uhm…C+(Cx+C)+(Cx22+Cx+C)+…?Which is a lot more work to reorder than the original convention used in the ‘proof’ where all the indefinite integrals of the zero function are conveniently assumed to be the same constant, and all other indefinite integrals conveniently have integration constants of zero.Even if we sed s/C//ϵ and proclaim thatϵshould be small (e.g. compared to x) and we are only interested in the leading order terms, this would not work. What one would have to motivate is throwing everything but the leading power of x out for everyInevaluation, then later meticulously track these lower order terms in the sum to arrive at the Taylor series of the exponential.