Wikipedia:Reference desk/Archives/Mathematics/2022 March 21

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March 21

Change of variables formula

In the article Integration by substitution the following change of variables formula is mentioned:

Let φ : [a, b] → I be a differentiable function with a continuous derivative, where I ⊆ R is an interval. Suppose that f : I → R is a continuous function. Then

$\int _{a}^{b}f(\varphi (x))\varphi '(x)\,dx=\int _{\varphi (a)}^{\varphi (b)}f(u)\,du.$

My question is how to apply this is the measure theory sense without $\phi$ being injective ($\phi$ is real valued and real Borel measurable and additional hypothesis may be imposed on it too). Specifically, I want to apply this formula to the case of finding the PDF of a function of a real valued random variable.-- Abdul Muhsy talk 04:01, 21 March 2022 (UTC)

OK I see that now.- Abdul Muhsy talk 08:59, 21 March 2022 (UTC)

Integrands equal when integral is

In the article Integration by substitution, it is mentioned that

$\int _{S}p_{Y}(y)\,dy=\int _{S}p_{X}(\phi ^{-1}(y))\left|{\frac {d\phi ^{-1}}{dy}}\right|\,dy,$ so

$p_{Y}(y)=p_{X}(\phi ^{-1}(y))\left|{\frac {d\phi ^{-1}}{dy}}\right|.$

Can someone explain how the first equality implies the second?- Abdul Muhsy talk 09:01, 21 March 2022 (UTC)

By using the linearity of the integration operator, it follows from this statement:

If $\int _{S}f(y)\,dy=0$ for all $S\subseteq T,$ function $f$ vanishes on $T$.

This assumes that $f$ is continuous and that we are dealing with measurable sets, where each $y\in T$ has some measurable neighbourhood $N\subseteq T.$  --Lambiam 09:36, 21 March 2022 (UTC)

"Distance-constant" continuous injective functions

Suppose we have a metric space $K$ and a continuous injective function $f:\mathbb {R} \rightarrow K$ (treating $\mathbb {R}$ as a metric space with the usual Euclidean metric) satisfying $\forall n\in \mathbb {Z} ^{+},\forall a\in \mathbb {R} ,d_{K}(f(a),f(a-{\frac {1}{n}}))=d_{K}(f(a),f(a+{\frac {1}{n}}))$. Does it follow that $\forall b\in \mathbb {R} ,d_{K}(f(0),f(b))=k*b*d_{K}(f(0),f(1))$ for some constant $k$? My intuition says no, given that as soon as one removes the injectivity constraint, one can immediately take the mapping from the real line to the circle in $\mathbb {R} ^{2}$ (with the Euclidean metric restricted to the subspace) given by $f(a)=(\cos(a),\sin(a))$, but I have yet to construct a counterexample or a proof. GalacticShoe (talk) 22:43, 21 March 2022 (UTC)

By putting $b=1$ in the sought implication we see that it requires $k=1.$ Isn't the corkscrew mapping $f:\mathbb {R} \rightarrow \mathbb {R} ^{3}$ given by $f(a)=(\cos a,\sin a,a)$ a counterexample?  --Lambiam 08:36, 22 March 2022 (UTC)

I think you're right; that's a nice example. There may be more exotic ones as well. I'm thinking take K to be the torus defined as the quotient of R×R under $(x,y)\sim (x+1,y)\sim (x,y+1),\,$. Then define f(r) = (ar, br) where a and b are constants with a/b irrational. --RDBury (talk) 09:15, 22 March 2022 (UTC)

PS. A slightly less exotic, but equivalent version of this is f(x) = (a cos(cx), a sin(cx), b cos(dx), b sin(dx)) mapping R to R⁴, assuming a, b, c, d > 0 and d/c is irrational. Or basically take any curve R → Rⁿ whose generalized curvatures according to the Frenet–Serret formulas are constant. For n=2 this is either a circle or a line and for n=3 it's a circle, line or helix. I don't know if all curves for n>3 are known, but it seems like they would be. --RDBury (talk) 09:46, 22 March 2022 (UTC)

Did you mean (a cos(cx), a sin(cx), b cos(dx), b sin(dx))?  --Lambiam 20:28, 22 March 2022 (UTC)

Yes, corrected. Nice catch. --RDBury (talk) 22:01, 22 March 2022 (UTC)

Thank you so much for your reply! I'll look into seeing if I can find a list of constant-curvature curves in $\mathbb {R} ^{n}$. Just out of curiosity, would you happen to know if a smooth curve satisfying the above property must be of constant generalized curvature? (Also, would you happen to know if there are non-smooth curves satisfying the above property as well?) Many thanks! GalacticShoe (talk) 16:56, 22 March 2022 (UTC)

Ah yes, that works! Thank you so much, you've been a great help. GalacticShoe (talk) 16:50, 22 March 2022 (UTC)