Mathematics - 2023-01-15

Wolgwang

00:33

@robjohn I see.

63

A: How to show that $\int_0^1 \left(\sqrt[3]{1-x^7} - \sqrt[7]{1-x^3}\right)\;dx = 0$

Let $m, n > 0$. Then observe that $$ \int_{0}^{1} \sqrt[n]{1-x^m} \; dx$$ is the area of the region given by inequalities $$ 0 \leq x \leq 1 \quad \text{and} \quad 0 \leq y \leq \sqrt[n]{1-x^m}.$$ But the last inequality is equivalent to $0 \leq x^m + y^n \leq 1$. Thus $$ \int_{0}^{1} \sqrt[n]{1-...

copper.hat

01:44

@ILikeMathematics sorry, i missed the $f''(x_0) = 0$ completely

Simd

05:12

Is it possible to prove that if a,b >0 then a/b>0?

leslie townes

sure, but the structure of the argument (and whether or not the argument counts as a proof) would depend on what one is assuming in the first place. different sources will formulate that background differently.

it can be proved the axioms for an 'ordered field' (see en.wikipedia.org/wiki/Ordered_field but note even there they offer two sets) if a and b are assumed to be in an ordered field and a/b is interpreted the way that it would be there.

math.stackexchange.com/questions/4356436/… offers a proof from the axioms for an ordered field that if a is in an ordered field and a > 0 then a^{-1} > 0, which is related.

although even the MSE answer references 'trichotomy,' which is buried in consequences of wikipedia's sets of axioms and not expressly stated there. so 'the axioms for an ordered field' sometimes vary even if they describe the exact same notion.

Simd

@leslietownes this is for high school so they won’t know fields

leslie townes

"this is for high school" is the kind of context that it is helpful to mention. and while they may not use the words, i have definitely seen high school level books that do, more or less, axiomatize field arithmetic and deduce boring consequences of the distributive law in the same way that a more advanced book would. although you're right, it's not common.

generally, in hs level or below, i would not axiomatize stuff at this level or ask people to prove anything. you can 'explain' how it makes sense in terms of other given things.

Simd

Thanks. So there is no elementary proof?

@leslietownes how would you do that?

leslie townes

e.g. if they are willing to believe things like positive x positive = positive, positive x negative = negative, which incidentally are not far removed from field axioms, you can explain it that way.

but again, this isn't so much 'proof' as 'explanation.' explaining relative to other things.

for example. if a and b are positive, i think a/b ought to be positive. why? well, a/b can't be 0, because then e.g. the product a * (a/b) = a * 0 = 0 would also be 0, and it's not, and a/b can't be negative, because then b * (a/b) would be a positive times a negative, and hence negative, but the product is a, which has to be positive.

so a/b being positive is the only possibility left.

this basically is the proof you get from the field axioms, the difference is you aren't citing the axioms, but saying things like "if you believe [some fact that happens to be a consequence of the field axioms], then you also have to believe [this consequence]."

the explanation above doesn't explain why a*0 = 0, for example, or why positive * negative = negative. it just relates this issue about a/b being positive to those other things.

Simd

05:30

I like this, thank you

leslie townes

just don't tell them that they're reasoning from the field axioms. :D

a lot of it goes back to "if you believe that the distributive law holds, then you will need believe that a bunch of other stuff holds too." although i'm not sure if the distributive law is pedagogically convincing as a phenomenon worth taking as a starting point.

Simd

Don’t we want b*(a/b) for the proof a/b is not zero as well!

leslie townes

yes. sounds right. if it looks like i'm off by a transposition, the answer is always yes.

Simd

05:45

:)

Koro

Hahn Banach theorem: Suppose that X is a real or complex normed linear space. Let p be a real valued map defined on $X\times X$ such that $p$ is subadditive, and $p(ax)=|a| p(x)$ for any scalar $a$. Suppose that $Y\subset X$ is a subspace, $f$ is a linear functional on $Y$ satisfying $f(y)\le p(y)$ for all y in Y. Then, there exists a linear functional $f'$ on X which is an extension of $f$ such that $f' (x)\le p(x)$ for every x in X.

I understand the proof for X being a real normed space.

For X being a complex space: we write $f(y)=f_1(y)+i f_2(y), y\in Y, f_1=\Re(f), f_2=\Im(f)$. $f_i$'s are real valued.

now, I want to extend $f_1$ to all of X but I can't use Hahn Banach for the real case as $Y$ is not real space.

leslie townes

i agree that these things need a lot of careful thought, but, isn't it though? if you just forget that you can also multiply by non-real complex scalars? maybe handle that separately?

Koro

So I consider the set $Y'=Y$ (equal as sets), then noting that $Y'$ is a vector space over $\mathbb R$.

leslie townes

oh, i get it, you don't like the same notation for Y the real complex vector space and "Y" the real vector space.

Koro

@leslietownes I'm trying to understand the book's argument for the complex case :).

I'm now presenting my understanding to see if I understood it correctly.

leslie townes

06:00

i'm generally against notationally formalizing these things but this is one instance where it might actually help. i have seen sets of notes where people write down 'some obvious thing' to extend to the complex case and for some reason it doesn't work. usually because they goof up paying attention to norms, and not on complex linearity, but it happens.

Koro

$f_1$ is a linear functional on $Y'$, which is by its definition a real normed space so HB extension works on $f_1$.

Now we note that $f(y)=f_1(y)+if_2(y)\implies if(y)=f(iy)=f_1(iy)+if_2(iy)=if_1(y)-f_2(y)$. Comparing real and imaginary parts, we get: $f_2(y)=-f_1(iy)$.

So consider $f'(x):= f_1'(x)-i f_1'(ix)$ which extends $f$ on Y to X.

Then it can be shown that $f'$ is linear functional on complex space X and that f' is dominated by $p$.

Alessandro Codenotti

@Yai0Phah Indeed it is very false, $[0,1]^{\Bbb N}$ embeds into $\ell^2$ for another example

PNDas

08:29

Hi, suppose we are solving a quasilinear PDE with the method of characteristics. And suppose set of characteristic curve forms a region D. On D we can find solution. What happens outside D? Should I say solution doesn't exist? or solution is not found by this method? or solution is not unique?

e.g. solve $u_x+xu_t+tu=0 \,, (x,t)\in \mathbb R\times(0,\infty)$ and $u(0,t)=f(t)\,,t\in[0,1]$

the set of characteristic curves is $\{(x,t):t=\frac{x^2}2+s,s\in[0,1]\}$.

Also by our terminology, characteristic curve lies in xy- plane. Some people also call it base characteristic curve.

robjohn