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LW - Explaining a Math Magic Trick by Robert AIZI

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Manage episode 416878688 series 3337129
Вміст надано The Nonlinear Fund. Весь вміст подкастів, включаючи епізоди, графіку та описи подкастів, завантажується та надається безпосередньо компанією The Nonlinear Fund або його партнером по платформі подкастів. Якщо ви вважаєте, що хтось використовує ваш захищений авторським правом твір без вашого дозволу, ви можете виконати процедуру, описану тут https://uk.player.fm/legal.
Link to original article
Welcome to The Nonlinear Library, where we use Text-to-Speech software to convert the best writing from the Rationalist and EA communities into audio. This is: Explaining a Math Magic Trick, published by Robert AIZI on May 5, 2024 on LessWrong. Introduction A recent popular tweet did a "math magic trick", and I want to explain why it works and use that as an excuse to talk about cool math (functional analysis). The tweet in question: This is a cute magic trick, and like any good trick they nonchalantly gloss over the most important step. Did you spot it? Did you notice your confusion? Here's the key question: Why did they switch from a differential equation to an integral equation? If you can use (1x)1=1+x+x2+... when x=, why not use it when x=d/dx? Well, lets try it, writing D for the derivative: f'=f(1D)f=0f=(1+D+D2+...)0f=0+0+0+...f=0 So now you may be disappointed, but relieved: yes, this version fails, but at least it fails-safe, giving you the trivial solution, right? But no, actually (1D)1=1+D+D2+... can fail catastrophically, which we can see if we try a nonhomogeneous equation like f'=f+ex (which you may recall has solution xex): f'=f+ex(1D)f=exf=(1+D+D2+...)exf=ex+ex+ex+...f=? However, the integral version still works. To formalize the original approach: we define the function I (for integral) to take in a function f(x) and produce the function If defined by If(x)=x0f(t)dt. This rigorizes the original trick, elegantly incorporates the initial conditions of the differential equation, and fully generalizes to solving nonhomogeneous versions like f'=f+ex (left as an exercise to the reader, of course). So why does (1D)1=1+D+D2+... fail, but (1I)1=1+I+I2+... works robustly? The answer is functional analysis! Functional Analysis Savvy readers may already be screaming that the trick (1x)1=1+x+x2+... for numbers only holds true for |x|<1, and this is indeed the key to explaining what happens with D and I! But how can we define the "absolute value" of "the derivative function" or "the integral function"? What we're looking for is a norm, a function that generalizes absolute values. A norm is a function x||x|| satisfying these properties: 1. ||x||0 for all x (positivity), and ||x||=0 if and only if x=0 (positive-definite) 2. ||x+y||||x||+||y|| for all x and y (triangle inequality) 3. ||cx||=|c|||x|| for all x and real numbers c, where |c| denotes the usual absolute value (absolute homogeneity) Here's an important example of a norm: fix some compact subset of R, say X=[10,10], and for a continuous function f:XR define ||f||=maxxX|f(x)|, which would commonly be called the L-norm of f. (We may use a maximum here due to the Extreme Value Theorem. In general you would use a supremum instead.) Again I shall leave it to the reader to check that this is a norm. This example takes us halfway to our goal: we can now talk about the "absolute value" of a continuous function that takes in a real number and spits out a real number, but D and I take in functions and spit out functions (what we usually call an operator, so what we need is an operator norm). Put another way, the L-norm is "the largest output of the function", and this will serve as the inspiration for our operator norm. Doing the minimal changes possible, we might try to define ||I||=maxf continuous||If||. There are two problems with this: 1. First, since I is linear, you can make ||If|| arbitrarily large by scaling f by 10x, or 100x, etc. We can fix this by restricting the set of valid f for these purposes, just like how for the L example restricted the inputs of f to the compact set X=[10,10]. Unsurprisingly nice choice of set to restrict to is the "unit ball" of functions, the set of functions with ||f||1. 2. Second, we must bid tearful farewell to the innocent childhood of maxima, and enter the liberating adulthood of suprema. This is necessary since f ranges over the infinite-dimensional vector space of continuous functions, so the Heine-Borel theorem no longer guarant...
  continue reading

1659 епізодів

Artwork
iconПоширити
 
Manage episode 416878688 series 3337129
Вміст надано The Nonlinear Fund. Весь вміст подкастів, включаючи епізоди, графіку та описи подкастів, завантажується та надається безпосередньо компанією The Nonlinear Fund або його партнером по платформі подкастів. Якщо ви вважаєте, що хтось використовує ваш захищений авторським правом твір без вашого дозволу, ви можете виконати процедуру, описану тут https://uk.player.fm/legal.
Link to original article
Welcome to The Nonlinear Library, where we use Text-to-Speech software to convert the best writing from the Rationalist and EA communities into audio. This is: Explaining a Math Magic Trick, published by Robert AIZI on May 5, 2024 on LessWrong. Introduction A recent popular tweet did a "math magic trick", and I want to explain why it works and use that as an excuse to talk about cool math (functional analysis). The tweet in question: This is a cute magic trick, and like any good trick they nonchalantly gloss over the most important step. Did you spot it? Did you notice your confusion? Here's the key question: Why did they switch from a differential equation to an integral equation? If you can use (1x)1=1+x+x2+... when x=, why not use it when x=d/dx? Well, lets try it, writing D for the derivative: f'=f(1D)f=0f=(1+D+D2+...)0f=0+0+0+...f=0 So now you may be disappointed, but relieved: yes, this version fails, but at least it fails-safe, giving you the trivial solution, right? But no, actually (1D)1=1+D+D2+... can fail catastrophically, which we can see if we try a nonhomogeneous equation like f'=f+ex (which you may recall has solution xex): f'=f+ex(1D)f=exf=(1+D+D2+...)exf=ex+ex+ex+...f=? However, the integral version still works. To formalize the original approach: we define the function I (for integral) to take in a function f(x) and produce the function If defined by If(x)=x0f(t)dt. This rigorizes the original trick, elegantly incorporates the initial conditions of the differential equation, and fully generalizes to solving nonhomogeneous versions like f'=f+ex (left as an exercise to the reader, of course). So why does (1D)1=1+D+D2+... fail, but (1I)1=1+I+I2+... works robustly? The answer is functional analysis! Functional Analysis Savvy readers may already be screaming that the trick (1x)1=1+x+x2+... for numbers only holds true for |x|<1, and this is indeed the key to explaining what happens with D and I! But how can we define the "absolute value" of "the derivative function" or "the integral function"? What we're looking for is a norm, a function that generalizes absolute values. A norm is a function x||x|| satisfying these properties: 1. ||x||0 for all x (positivity), and ||x||=0 if and only if x=0 (positive-definite) 2. ||x+y||||x||+||y|| for all x and y (triangle inequality) 3. ||cx||=|c|||x|| for all x and real numbers c, where |c| denotes the usual absolute value (absolute homogeneity) Here's an important example of a norm: fix some compact subset of R, say X=[10,10], and for a continuous function f:XR define ||f||=maxxX|f(x)|, which would commonly be called the L-norm of f. (We may use a maximum here due to the Extreme Value Theorem. In general you would use a supremum instead.) Again I shall leave it to the reader to check that this is a norm. This example takes us halfway to our goal: we can now talk about the "absolute value" of a continuous function that takes in a real number and spits out a real number, but D and I take in functions and spit out functions (what we usually call an operator, so what we need is an operator norm). Put another way, the L-norm is "the largest output of the function", and this will serve as the inspiration for our operator norm. Doing the minimal changes possible, we might try to define ||I||=maxf continuous||If||. There are two problems with this: 1. First, since I is linear, you can make ||If|| arbitrarily large by scaling f by 10x, or 100x, etc. We can fix this by restricting the set of valid f for these purposes, just like how for the L example restricted the inputs of f to the compact set X=[10,10]. Unsurprisingly nice choice of set to restrict to is the "unit ball" of functions, the set of functions with ||f||1. 2. Second, we must bid tearful farewell to the innocent childhood of maxima, and enter the liberating adulthood of suprema. This is necessary since f ranges over the infinite-dimensional vector space of continuous functions, so the Heine-Borel theorem no longer guarant...
  continue reading

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