Hi dr. Peyam! Thanks for the class! A doubt: if I had chosen the ε of the δ3 as ε/(2*|g(xo)|), could I separate this exercise it into two cases, one where g is different from 0 and another where g is equal to 0 ? Thanks for your attention!
i this f∘g is cont by the following proof (1) f is cont. ∀ε>0 ∃ρ: ∀y, |y-b|< ρ ⇒|f(y)-f(b)|0 ∃ δ >0: ∀x, |x-a|< δ ⇒|g(x)-g(a)|0 be given, set y=g(x) and b=g(a). Then we can choose ρ= ρ(ε) satisfies (1) and δ= δ{ρ(ε)}= δ∘ρ(ε) satisfies (2), thus ∀ε>0 ∃ρ,δ ∀x, |x-a|< δ ⇒|g(x)-g(a)|
If f(x):R to R is continuous, and g,h:R to R, g(x)=f(x+t) and h(x)=f(tx) [ t€R+ ]. Then, are g(x) and h(x) continuous or uniformly continuous or discontinuous?
I love this sort of theorems, thanks
Thanks for the great video> Appreciate it doctor!
Really great work Sir !!
Hi dr. Peyam! Thanks for the class!
A doubt:
if I had chosen the ε of the δ3 as ε/(2*|g(xo)|), could I separate this exercise it into two cases, one where g is different from 0 and another where g is equal to 0 ?
Thanks for your attention!
Amazing proof! Great job, from a mathematics major in Canada.
Thank you 😊
Is this a real analysis lesson, due to involving the sequence definition of continuity?
Yeah
@@drpeyam Thank you. I've just started self-studying real analysis. I will definitely be watching this video in the near future.
I think this is also taught in calc 2
Do f and g have to have the same domain for this proof to apply?
No it’s continuous on the intersection of their domains
Thanks must go to Dr .Peyam for the eloquent proof in his great video ❤
The video sound is pretty good, beyond my imagination
Dr sir, please suggest a textbook to read along with 😢
Analysis by Ross
@@drpeyam Thank you very much sir
and the proof I spent two hours trying to figure out is done in 11 minutes
Great video sir
Is f∘g continuous?
Yeah that too
i this f∘g is cont by the following proof
(1) f is cont. ∀ε>0 ∃ρ: ∀y, |y-b|< ρ ⇒|f(y)-f(b)|0 ∃ δ >0: ∀x, |x-a|< δ ⇒|g(x)-g(a)|0 be given, set y=g(x) and b=g(a).
Then we can choose ρ= ρ(ε) satisfies (1)
and δ= δ{ρ(ε)}= δ∘ρ(ε) satisfies (2), thus
∀ε>0 ∃ρ,δ ∀x, |x-a|< δ ⇒|g(x)-g(a)|
Tbh I dislike that during proofs people "choose ε=1". Literary ε is a small number "given by the enemy", as my teacher would say...
It’s just for simplicity. You could fix any epsilon you like (such as 0.00001) and do the proof with that instead of 1
😂take epsilon=c , c>0
「上記のギフトのいずれかを選択できます」、
If f(x):R to R is continuous, and
g,h:R to R, g(x)=f(x+t) and h(x)=f(tx)
[ t€R+ ]. Then, are g(x) and h(x) continuous or uniformly continuous or discontinuous?
They are continuous
Neat!
Hello