I finished "Batman Arkham city" on PS3. I have already played "Batman Arkham asylum" in last year and now I am playing the second time on PC.
One of the thing which I was charmed is cool motions in the battles of Batman.
Of course the player has to accurately push the buttons which enable the motion. However, the cool motions or combos will succeed by pushing buttons gently in these games.
The timing at which I (feeling impatiently) find myself pushing the buttons is slightly overquicker than the timing at which nice players recognize. From the result, the number of times pushing the buttons are too many.
Cool motions or combos are needed appropriate pushing operation times and accurate timings. Action games are quite difficult, but much exciting for me.
2013/11/23
2013/11/20
real number system 8 (nested intervals)
We shall prove that a contracting closed interval becomes a point. It is definitely no doubt. But it is only enabled by infinite operations and is in the base of the real number system.
For any bounded closed interval sequences $I_n=[a_n,b_n], (-\infty<a_n\leq b_n<\infty)$ , if $I_n\supset I_{n+1}, (n\in\mathbb{N})$ and $|a_n-b_n|\rightarrow 0$ as $n\rightarrow \infty$, there exists a real number $c$ such that
\[ \bigcap_{n=1}^{\infty}I_n = c \]
In other words, if the real number sequence $a_1,a_2,\cdots $ is monotonic increasing bounded above, the bounded real number sequence $b_1,b_2,\cdots $ is monotonic decreasing bounded below and $a_n\leq b_n$ for any $n$ , there is a real number $c$ such that $a_n\leq c\leq b_n$ . Furthermore, if $|a_n-b_n|\rightarrow 0$ as $n\rightarrow \infty$, $c$ is a point. That is to say, $\lim_{n\rightarrow\infty}a_n=\lim_{n\rightarrow\infty}b_n=c$ .
In real number system, as a bounded monotone real number sequence has a limit number, we put $\lim_{n\rightarrow\infty}a_n=a,\quad \lim_{n\rightarrow\infty}b_n=b$ . Hence, for any $n$ ,
\[ a_n\leq a\leq b\leq b_n \]
and
\[ I_n\supset [a,b] \]
It means $\bigcap_{n=1}^{\infty}I_n\ne \phi$ . There is a real number $c\in [a,b]$ .
As $a_n\leq c\leq b_n$ for any $n$ , $|a_n-c|\leq b_n-a_n$ . Therefore $a=c$, because $|a_n-b_n|\rightarrow 0$ . Similarly $b=c$ .
It is called the method of nested intervals. By the method we are able to get a root of some kinds of equations computationally.
For any bounded closed interval sequences $I_n=[a_n,b_n], (-\infty<a_n\leq b_n<\infty)$ , if $I_n\supset I_{n+1}, (n\in\mathbb{N})$ and $|a_n-b_n|\rightarrow 0$ as $n\rightarrow \infty$, there exists a real number $c$ such that
\[ \bigcap_{n=1}^{\infty}I_n = c \]
In other words, if the real number sequence $a_1,a_2,\cdots $ is monotonic increasing bounded above, the bounded real number sequence $b_1,b_2,\cdots $ is monotonic decreasing bounded below and $a_n\leq b_n$ for any $n$ , there is a real number $c$ such that $a_n\leq c\leq b_n$ . Furthermore, if $|a_n-b_n|\rightarrow 0$ as $n\rightarrow \infty$, $c$ is a point. That is to say, $\lim_{n\rightarrow\infty}a_n=\lim_{n\rightarrow\infty}b_n=c$ .
In real number system, as a bounded monotone real number sequence has a limit number, we put $\lim_{n\rightarrow\infty}a_n=a,\quad \lim_{n\rightarrow\infty}b_n=b$ . Hence, for any $n$ ,
\[ a_n\leq a\leq b\leq b_n \]
and
\[ I_n\supset [a,b] \]
It means $\bigcap_{n=1}^{\infty}I_n\ne \phi$ . There is a real number $c\in [a,b]$ .
As $a_n\leq c\leq b_n$ for any $n$ , $|a_n-c|\leq b_n-a_n$ . Therefore $a=c$, because $|a_n-b_n|\rightarrow 0$ . Similarly $b=c$ .
It is called the method of nested intervals. By the method we are able to get a root of some kinds of equations computationally.
2013/11/17
intervals and segments
An interval or a segment in $\mathbb{R}$ means the set of every points $x$ between $a\in \mathbb{R}$ and $b\in \mathbb{R}$ , ($a<b$).
We shall define intervals and segments in $\mathbb{R}$ accuratelly.
If the set does not include both endpoints, it is called the open interval or the segment. We write the open interval or the segment $(a,b)$ .
The interval or the closed interval which includes both endpoints is expressed $[a,b]$ .Furthermore, it is easy for you to understand the left half open interval $(a,b]$ and the right half open interval $[a,b)$ .
As $\pm\infty$ are not real numbers, the following intervals are impossible.
\[ [-\infty,a),\quad [-\infty,a],\quad (a,\infty],\quad [a,\infty],\quad [-\infty,\infty] \]
The intervals $(-\infty,a), (a,\infty), (-\infty,\infty)$ are regarded as open and $(-\infty,a], [a,\infty)$ are regarded as half open.
We shall define intervals and segments in $\mathbb{R}$ accuratelly.
If the set does not include both endpoints, it is called the open interval or the segment. We write the open interval or the segment $(a,b)$ .
The interval or the closed interval which includes both endpoints is expressed $[a,b]$ .Furthermore, it is easy for you to understand the left half open interval $(a,b]$ and the right half open interval $[a,b)$ .
As $\pm\infty$ are not real numbers, the following intervals are impossible.
\[ [-\infty,a),\quad [-\infty,a],\quad (a,\infty],\quad [a,\infty],\quad [-\infty,\infty] \]
The intervals $(-\infty,a), (a,\infty), (-\infty,\infty)$ are regarded as open and $(-\infty,a], [a,\infty)$ are regarded as half open.
2013/11/08
epsilon-delta proofs 8
In a metric space, we are able to estimate the distance between elements with the distance function. Therefore, without a real numerical sequence explicitly we can define the "limit".
We write $\lim_{x\rightarrow c}f(x)=y$ when $f(x)$ approaches $y$ as $x$ approaches $c$. Suppose that the set $A$ including $x$ and the set $B$ onto which $f(x)$ maps are both metric spaces. Hence, there are the distance function $d_A(p,q)$ of $A$ and $d_B(p,q)$ of $B$.
$\lim_{x\rightarrow c}f(x)=y$ means that for any $\epsilon>0$ , there exists a $\delta>0$ such that if
\[ d_A(x,c)<\delta \]
then
\[ d_B(f(x),y)<\epsilon \]
Preceding definition was as follow.
$a_n$ approches $a$, when for any $\epsilon >0$, there is a $N>0$ such that if $n\geq N$,
\[ |a_n-a|< \epsilon \]
Let us equate $d_A(x,c)$ with $n\geq N$ and $d_B(f(x),y)$ with $|a_n-a|$ . Only $n\geq N$ is not a distance function. Please understand carefully that $d_B(f(x),y)<\epsilon$ is true for "every" $x$ which satisfies $d_A(x,c)<\delta$ .
We write $\lim_{x\rightarrow c}f(x)=y$ when $f(x)$ approaches $y$ as $x$ approaches $c$. Suppose that the set $A$ including $x$ and the set $B$ onto which $f(x)$ maps are both metric spaces. Hence, there are the distance function $d_A(p,q)$ of $A$ and $d_B(p,q)$ of $B$.
$\lim_{x\rightarrow c}f(x)=y$ means that for any $\epsilon>0$ , there exists a $\delta>0$ such that if
\[ d_A(x,c)<\delta \]
then
\[ d_B(f(x),y)<\epsilon \]
Preceding definition was as follow.
$a_n$ approches $a$, when for any $\epsilon >0$, there is a $N>0$ such that if $n\geq N$,
\[ |a_n-a|< \epsilon \]
Let us equate $d_A(x,c)$ with $n\geq N$ and $d_B(f(x),y)$ with $|a_n-a|$ . Only $n\geq N$ is not a distance function. Please understand carefully that $d_B(f(x),y)<\epsilon$ is true for "every" $x$ which satisfies $d_A(x,c)<\delta$ .