Amritam Gamaya

Karma: 18
• The ques­tion is ill founded. You can in fact re­cover all of the in­for­ma­tion about log x from its Tay­lor se­ries. I think TurnTrout is con­fused maybe be­cause the Tay­lor se­ries only con­verges on a cer­tain in­ter­val, not globally? I’ll an­swer the ques­tion as­sum­ing that’s the con­fu­sion.

If you know all the deriva­tives of log x at x=1, but you know noth­ing else about log x, then you can find a Tay­lor se­ries that con­verges on (0,2). But, given the Tay­lor se­ries, you now also know all the deriva­tives at x=1.9. Writ­ing a Tay­lor se­ries cen­tered at 1.9, you get a se­ries that con­verges on (0,3.8). Con­tin­u­ing in this fash­ion, you can find all val­ues of log x, for all pos­i­tive real in­puts, us­ing only the deriva­tives at x=1. You just need mul­ti­ple “steps.”

That said, there is a fun­da­men­tal limi­ta­tion. Con­sider the func­tions f(x) = 1/​x and g(x) = {1/​x if x > 0, 1 + 1/​x if x < 0}. For x > 0, f(x) = g(x), but for x<0 they are not equal. Clearly both func­tions are in­finitely differ­en­tiable,
but just be­cause you know all the deriva­tives of f at x=1, doesn’t mean you can de­ter­mine it’s value at x=-1.

Okay, so Tay­lor se­ries al­low you to probe all val­ues of a func­tion, but it might take mul­ti­ple steps, and sin­gu­lar­i­ties cause real un­fix­able prob­lems. The cor­rect way to think about this is that func­tions aren’t just differ­en­tiable or not, they are in­finitely differ­en­tiable *on a set*. For ex­am­ple, 1/​x is smooth on (-in­finity,0) union (0,in­finity), which is a set with two con­nected com­po­nents. The Tay­lor se­ries al­lows you to probe all of the val­ues on any in­di­vi­d­ual con­nected com­po­nent, but it very ob­vi­ously can’t tell you any­thing about other con­nected com­po­nents.

As for why it some­times takes mul­ti­ple “steps,” like for log x: for rea­sons, the Tay­lor se­ries has to con­verge on a sym­met­ric in­ter­val. For log x cen­tered at x=1, it sim­ply can’t con­verge at 3 with­out also con­verg­ing at −1, which is ob­vi­ously im­pos­si­ble since it’s out­side the con­nected com­po­nent where log x is differ­en­tiable. The Tay­lor se­ries con­verges on the largest in­ter­val where it can pos­si­bly con­verge, but it still tells you the val­ues el­se­where (in the con­nected com­po­nent) if you’re will­ing to work slightly harder.

Every­thing I said is true for an­a­lytic func­tions. There is still the is­sue of in­finitely differ­en­tiable non-an­a­lytic func­tions as de­scribed here. Log x is not an ex­am­ple of such a func­tion, log x is an­a­lytic. Th­ese coun­terex­am­ples are much more sub­tle, but it has to do with the fact that the er­ror in an n-th deriva­tive ap­prox­i­ma­tion de­cays like O(x^n), so even a Tay­lor se­ries al­lows for er­rors like O(e^x) be­cause ex­po­nen­tial de­cay beats any polyno­mial.

• The idea that the land­lord has to make pay­ments to the bank (mort­gages and debt pay­ments) is os­ten­si­bly fixed by low­ered in­ter­est rates and bailouts, same as any other busi­ness who has lost their rev­enue stream.

The ques­tion is whether land­lords are con­sid­ered es­sen­tial or non-es­sen­tial busi­nesses. If they are es­sen­tial, like gro­cery stores, then we need to make sure they con­tinue to func­tion. If they are non-es­sen­tial, like bar­bers, then they should go into sta­sis and take gov­ern­ment money/​loans to ser­vice any over­head they ab­solutely can’t shut off (like the bar­bers should be do­ing). The con­fu­sion is be­cause shelter is es­sen­tial, but land­lords don’t ac­tu­ally provide shelter (they only steal shelter from those who can’t pay). Land­lords can go into sta­sis like a bar­ber and yet keep pro­vid­ing their es­sen­tial goods like a gro­cery store. This logic proves too much, since I’ve never un­der­stood why we per­mit land­lords to charge so much money with­out pro­vid­ing any ser­vices that I can dis­cern, so I re­treat to Ch­ester­ton’s Fence.

The Solu­tion is Inaction

24 Mar 2020 18:01 UTC
11 points