MHB What Are the Properties of the Maps Defined in the Content?

  • Thread starter Thread starter mathmari
  • Start date Start date
  • Tags Tags
    Properties
mathmari
Gold Member
MHB
Messages
4,984
Reaction score
7
Hey! :o

Let $a\in \mathbb{R}$. We define the map $\text{cost}_a:\mathbb{R}\rightarrow \mathbb{R}$, $x\mapsto a$. We define also $-f:=(-1)f$ for a map $f:\mathbb{R}\rightarrow \mathbb{R}$.

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a map and $\lambda\in \mathbb{R}$.

Show that:

  1. for $a,b\in \mathbb{R}$ it holds that $\text{cost}_a+\text{cost}_b=\text{cost}_{a+b}$.
  2. for $a\in \mathbb{R}$ it holds that $\lambda\text{cost}_a=\text{cost}_{\lambda a}$.
  3. $-(f+g)=(-f)+(-g)$.
  4. $f+f=2f$.
  5. $f+(-f)=\text{cost}_0$.

Could you give me a hint how we could these? Aren't all of these trivial? (Wondering)
 
Physics news on Phys.org
Hey mathmari!

I guess the first one can be shown as follows:
$$\DeclareMathOperator{\cost}{cost}
\forall a,b\in\mathbb R,\,\forall x\in\mathbb R : (\cost_a+\cost_b)(x)=\cost_a(x)+\cost_b(x)=a+b=\cost_{a+b}(x)$$
Therefore:
$$\forall a,b\in\mathbb R : \cost_a+\cost_b=\cost_{a+b}$$
(Thinking)

And yes, it does look rather trivial. (Tauri)
 
Klaas van Aarsen said:
I guess the first one can be shown as follows:
$$\DeclareMathOperator{\cost}{cost}
\forall a,b\in\mathbb R,\,\forall x\in\mathbb R : (\cost_a+\cost_b)(x)=\cost_a(x)+\cost_b(x)=a+b=\cost_{a+b}(x)$$
Therefore:
$$\forall a,b\in\mathbb R : \cost_a+\cost_b=\cost_{a+b}$$
(Thinking)

Ahh ok! (Malthe)

In the same way we could show also the other ones, or not? (Wondering)

$$2. \ \ \ \DeclareMathOperator{\cost}{cost}
\forall a\in\mathbb R,\,\forall x\in\mathbb R : \lambda \cost_a(x)=\lambda a=\cost_{\lambda a}(x)$$
Therefore:
$$\forall a\in\mathbb R : \lambda\cost_a=\cost_{\lambda a}$$

$$3. \ \ \ \forall x\in\mathbb R : -(f+g)(x)=(-1)(f+g)(x)=(-1)(f(x)+g(x))=(-1)f(x)+(-1)g(x)=(-f)(x)+(-g)(x)$$
Therefore:
$$-(f+g)=(-f)+(-g)$$

$$4. \ \ \ \forall x\in\mathbb R : (f+f)(x)=f(x)+f(x)=2f(x)$$
Therefore:
$$f+f=2f$$

$$5. \ \ \ \forall x\in\mathbb R : (f+(-f))(x)=(f+(-1)f)(x)=f(x)+(-1)f(x)=(1-1)f(x)=0\cdot f(x)=0=\text{cost}_0(x)$$
Therefore:
$$f+(-f)=\text{cost}_0$$ Is everything correct? Could we improve something? (Wondering)
 
Looks all good to me. (Nod)
 
Klaas van Aarsen said:
Looks all good to me. (Nod)

Great! Thank you! (Yes)
 
For original Zeta function, ζ(s)=1+1/2^s+1/3^s+1/4^s+... =1+e^(-slog2)+e^(-slog3)+e^(-slog4)+... , Re(s)>1 Riemann extended the Zeta function to the region where s≠1 using analytical extension. New Zeta function is in the form of contour integration, which appears simple but is actually more inconvenient to analyze than the original Zeta function. The original Zeta function already contains all the information about the distribution of prime numbers. So we only handle with original Zeta...

Similar threads

Replies
9
Views
2K
Replies
3
Views
3K
Replies
6
Views
3K
Replies
1
Views
2K
Replies
3
Views
2K
Replies
3
Views
2K
Replies
9
Views
2K
Replies
2
Views
1K
Back
Top