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I was watching Carl Bender's lecture 2 on YouTube and I stumbled across a 'fundamentally hard problem', namely the Schrodinger equation

$y''(x)+Q(x)y=0$ (eq.1)

Carl mentions that this is a 'fundamentally hard problem' because the most natural way to proceed with solving (eq.1) reduces your calculations to solving that same original problem! In other words, there is nothing you can do to solve (eq.1). He does this to motivate the power of asymptotic methods by sacrificing the notion of equal signs (=) and introducing a new notion (~).

My question goes beyond this, in particular, are there asymptotic problems which are fundamentally hard? If so, what notion from asymptotics must we sacrife in order to make progress?

Andy
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  • The main way this happens in my experience is when you want to make an approximation but you have no well-defined "extreme case" to use to develop an approximation scheme. In this case the solution to your problem just is what it is, and cannot be accurately represented in a "simple" manner. Of course it can be estimated piecewise if it is, say, continuous, but it requires many pieces which drifts into the realm of numerical analysis rather than asymptotics. Hopefully this ramble made sense. – Ian Jan 06 '20 at 14:07
  • Fundamentally difficult asymptotic problems are numerous. Check out the wiki for the Riesz function. – user321120 Jan 06 '20 at 15:09

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