Tag Archives: oxford代写

计算机科学的逻辑|Logic and Proof代写 

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计算机科学的逻辑|Logic and Proof代写 
问题 1.

Consider the proposition
$$
A=(\neg P \supset Q) \supset(\neg R \supset S) .
$$
First, we eliminate $\supset$ using the fact that $(\neg B \vee C)$ is equivalent to $(B \supset C)$. We get
$$
(\neg(\neg \neg P \vee Q)) \vee(\neg \neg R \vee S) .
$$

证明 .

Then, we put this proposition in NNF. We obtain
$$
(\neg P \wedge \neg Q) \vee(R \vee S)
$$
Using distributivity we obtain
$$
(\neg P \vee R \vee S) \wedge(\neg Q \vee R \vee S)
$$


英国论文代写Viking Essay为您提供作业代写代考服务

Oxford COURSE NOTES ：

Show that the infinite sequent $\Gamma \rightarrow \Delta$ where
$$
\Gamma=
$$
and
$$
\left.\Delta=<\left(P_{1} \supset Q\right)\right\rangle $$ is falsifiable. (ii) Prove that for every $i>0$, the sequent $\Gamma \rightarrow \Delta^{\prime}$, where $\Gamma$ is as above and $\Delta^{\prime}=<\left(P_{0} \supset P_{i}\right)>$ is provable.





线性代数|Linear Algebra代写 

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线性代数|Linear Algebra代写 
问题 1.

If $t$ is a transformation represented by
$$
\operatorname{Rep}{B, D}(t)=\left(\begin{array}{ll} 1 & 0 \ 1 & 1 \end{array}\right){B, D} \quad \text { so that } \vec{v}=\left(\begin{array}{c}
v_{1} \
v_{2}
\end{array}\right){B} \mapsto\left(\begin{array}{c} v{1} \
v_{1}+v_{2}
\end{array}\right)_{D}=t(\vec{v})
$$

证明 .

then the scalar multiple map $5 t$ acts in this way.
$$
\vec{v}=\left(\begin{array}{l}
v_{1} \
v_{2}
\end{array}\right){B} \longmapsto\left(\begin{array}{c} 5 v{1} \
5 v_{1}+5 v_{2}
\end{array}\right){D}=5 \cdot t(\vec{v}) $$ Therefore, this is the matrix representing $5 t$. $$ \operatorname{Rep}{B, D}(5 t)=\left(\begin{array}{ll}
5 & 0 \
5 & 5
\end{array}\right)_{B, D}
$$


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Oxford COURSE NOTES ：

The elementary reduction matrices are obtained from identity matrices with one Gaussian operation. We denote them:
(1) $I \stackrel{k \rho_{i}}{\longrightarrow} M_{i}(k)$ for $k \neq 0$;
(2) $I \stackrel{\rho_{i} \leftrightarrow \rho_{j}}{\longrightarrow} P_{i, j}$ for $i \neq j$;
(3) $I \stackrel{k \rho_{i}+\rho_{j}}{\longrightarrow} C_{i, j}(k)$ for $i \neq j$.





λ微积分和类型|Lambda Calculus and Types代写 

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λ微积分和类型|Lambda Calculus and Types代写
问题 1.

In this subsection we work with the Church variant of $\lambda_{\rightarrow}^{0}$ having one atomic type 0 , rather than with $\lambda_{\rightarrow}^{\mathbb{A}}$, having an arbitrary set of atomic types. We will write $\pi=\pi^{0}$. The reader is encouraged to investigate which results do generalize to $\mathbb{T}^{\mathbb{A}}$.
Let $\mathcal{M}={\mathcal{M}(A)}_{A \in \mathbb{}}$ be a family of non-empty sets indexed by types $A \in \mathbb{T}$.
(i) $\mathcal{M}$ is called a type structure for $\boldsymbol{\lambda}_{\rightarrow}^{0}$ if
$$
\mathcal{M}(A \rightarrow B) \subseteq \mathcal{M}(B)^{\mathcal{M}(A)}
$$
Here $X^{Y}$ denotes the collection of set-theoretic functions
$$
{f \mid f: Y \rightarrow X} .
$$

证明 .

Let $X$ be a set. The full type structure $\mathcal{M}$ over the ground set $X$ defined in was specified by
$$
\begin{aligned}
\mathcal{M}(0) & \triangleq X \
\mathcal{M}(A \rightarrow B) & \triangleq \mathcal{M}(B)^{\mathcal{M}(A)}, \quad \text { for all } A, B \in \mathbb{\pi} .
\end{aligned}
$$


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Oxford COURSE NOTES ：

Let $\mathcal{M}$ be a typed applicative structure. A layered non-empty subfamily of $\mathcal{M}$ is a family $\Delta={\Delta(A)}_{A \in \pi}$ of sets, such that the following holds
$$
\forall A \in \Pi . \emptyset \neq \Delta(A) \subseteq \mathcal{M}(A) .
$$
$\Delta$ is called closed under application if
$$
f \in \Delta(A \rightarrow B), g \in \Delta(A) \Rightarrow f g \in \Delta(B) .
$$
$\Delta$ is called extensional if
$$
\forall A, B \in \pi \forall f, g \in \Delta(A \rightarrow B) \cdot[[\forall a \in \Delta(A) . f a=g a] \Rightarrow f=g] .
$$





离散数学|Discrete Mathematics代写 

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离散数学|Discrete Mathematics代写
问题 1.

Unfortumately, the answer is yes. The smallest such number is $341=11 \cdot 31$. This is not a prime, but it satisfies
$$
341 \mid 2^{340}-1 .
$$
(How do we know that this divisibility relation holds without extensive computation? We can use Fermat’s Theorem. It is sufficient to argue that both 11 and 31 are divisors of $2^{340}-1$, since then so is their product, 11 and 31 being different primes. By Fermat’s Theorem,
$$
11 \mid 2^{10}-1
$$

证明 .

Next we invoke the result : It implies that
$$
2^{10}-1 \mid 2^{340}-1
$$
Hence
$$
11 \mid 2^{340}-1
$$
For 31, we don’t need Fermat’s Theorem, but only again:
$$
31=2^{5}-1 \mid 2^{340}-1
$$


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Oxford COURSE NOTES ：

4.3.2. For $n=1$ and $n=2$, if we require that $L_{n}$ be of the given form, then we get
$$
L_{1}=1=a+b, \quad L_{2}=3=a \frac{1+\sqrt{5}}{2}+b \frac{1-\sqrt{5}}{2}
$$
Solving for $a$ and $b$, we get
$$
a=\frac{1+\sqrt{5}}{2}, \quad b=\frac{1-\sqrt{5}}{2}
$$
Then
$$
L_{n}=\left(\frac{1+\sqrt{5}}{2}\right)^{n}+\left(\frac{1-\sqrt{5}}{2}\right)^{n}
$$





连续数学|Continuous Mathematics代写 

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连续数学|Continuous Mathematics代写 
问题 1.

The connection between the homotopy methods and Newton’s method is deeper than may be seen at first glance. Let us start with the homotopy
$$
h(t, x)=f(x)-e^{-t} f\left(x_{0}\right)
$$
In this equation, $t$ will run from 0 to $\infty$. We seek a curve or path, $x=x(t)$, on which
$$
0=h(t, x(t))=f(x(t))-e^{-t} f\left(x_{0}\right)
$$

证明 .

As usual, differentiation with respect to $t$ will lead to a differential equation describing the path:
$$
\begin{aligned}
0 &=f^{\prime}(x(t)) x^{\prime}(t)+e^{-t} f\left(x_{0}\right) \
&=f^{\prime}(x(t)) x^{\prime}(t)+f(x(t)) \
x^{\prime}(t) &=-\left[f^{\prime}(x(t))\right]^{-1} f(x(t))
\end{aligned}
$$


英国论文代写Viking Essay为您提供作业代写代考服务

Oxford COURSE NOTES ：

On the other hand,
$$
(P A){\imath \jmath}=\sum{k=1}^{n} P_{\imath k} a_{k \jmath}=\sum_{k=1}^{n} \delta_{p_{1} k} a_{k j}=a_{p_{\imath} j}
$$
Thus we have proved, for all pairs $(i, j)$, that
$$
(P A){i j}=(L U){i j}
$$





计算机网络|Computer Networks代写 

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计算博弈论|Computational Game Theory代写 
static int
deliverSWP(SwpState state, Msg *frame)
{
SwpHdr hdr;
char *hbuf;
hbuf = msgStripHdr(frame, HLEN);
load_swp_hdr(&hdr, hbuf)
if (hdr->Flags & FLAG_ACK_VALID)
{
/* received an acknowledgment---do SENDER side */
if (swpInWindow(hdr.AckNum, state->LAR + 1,
state->LFS))
{
do
{
struct sendQ_slot *slot;
slot = &state->sendQ[++state->LAR % SWS];
evCancel(slot->timeout);
msgDestroy(&slot->msg);
semSignal(&state->sendWindowNotFull);
} while (state->LAR != hdr.AckNum);
} }
if (hdr.Flags & FLAG_HAS_DATA)
{
struct recvQ_slot *slot;
/* received data packet---do RECEIVER side */
slot = &state->recvQ[hdr.SeqNum % RWS];
if (!swpInWindow(hdr.SeqNum, state->NFE,
state->NFE + RWS - 1))
{
/* drop the message */
return SUCCESS;
}
msgSaveCopy(&slot->msg, frame);
slot->received = TRUE;
if (
英国论文代写Viking Essay为您提供作业代写代考服务

Oxford COURSE NOTES ：

static bool
swpInWindow(SwpSeqno seqno, SwpSeqno min, SwpSeqno max)
{
SwpSeqno pos, maxpos;
pos = seqno - min; /* pos *should* be in range [0..MAX)*/
maxpos = max - min + 1; /* maxpos is in range [0..MAX]*/
return pos < maxpos;
}





计算博弈论|Computational Game Theory代写 

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计算博弈论|Computational Game Theory代写 
问题 1.

$$
\max _{p}(p \cdot v-e(p)) .
$$
Denote the solution of this problem by $p^{*}(v)$. Assuming that $e$ is differentiable, the first-order condition for this problem implies that

证明 .

$$
v=e^{\prime}\left(p^{}(v)\right) \text { for all } v . $$


英国论文代写Viking Essay为您提供作业代写代考服务

Oxford COURSE NOTES ：

Proof. I first prove part a. By the definition of Nash equilibrium we have $U_{2}\left(\alpha_{1}^{}, \alpha_{2}^{}\right) \geq U_{2}\left(\alpha_{1}^{}, \alpha_{2}\right)$ for every mixed strategy $\alpha_{2}$ of player 2 or, since $U_{2}=-U_{1}$, $U_{1}\left(\alpha_{1}^{}, \alpha_{2}^{}\right) \leq U_{1}\left(\alpha_{1}^{}, \alpha_{2}\right)$ for every mixed strategy $\alpha_{2}$ of player 2
Hence
$$
U_{1}\left(\alpha_{1}^{}, \alpha_{2}^{}\right)=\min {\alpha{2}} U_{1}\left(\alpha_{1}^{*}, \alpha_{2}\right)
$$





计算机科学和哲学的数学研究|Mathematics for Computer Science and Philosophy代写 

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计算机科学和哲学的数学研究|Mathematics for Computer Science and Philosophy代写 
问题 1.

Here $\alpha \vee \beta$ depends on the same assumptions on which $\alpha$, or $\beta$, depends. So IV is used as follows:
$x \quad(n) \quad \alpha$
$x \quad(m) \quad \alpha \vee \beta \quad \operatorname{I\vee n}$
Or alternatively:
$x \quad(n) \quad \beta$
$x \quad(m) \quad \alpha \vee \beta \quad \operatorname{IV} n$

证明 .

The right number $n$ indicates the line of $\alpha$, or $\beta$. The left numbers, instead, are the same as those of $\alpha$, or $\beta$. For example, IV is used in the following derivation of $\sim p \vee q$ from $p \supset q$ :
1 (1) $p \supset q \quad \mathrm{~A}$
$2 \quad$ (2) $\sim(\sim p \vee q) \quad \mathrm{A}$
$3 \quad$ (3) $p$ A


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Oxford COURSE NOTES ：

you may start as follows:
1 (1) $p \wedge q \quad \mathrm{~A}$
2 (2) q $r \mathrm{~A}$
The second step is to write $\alpha$ at the end of the derivation, for $\alpha$ must occur as the last formula. Thus, in our example you may write $p \wedge r$ at the end of the sequence:
$1 \quad$ (1) $p \wedge q \quad \mathrm{~A}$
2 (2) $q \supset r \quad \mathrm{~A}$
() $p \wedge r$





概率与计算|Probability and Computing代写 

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概率与计算|Probability and Computing代写
问题 1.

$$
\phi_{0 j}^{2}=\frac{\pi_{j}\left(1-\pi_{j}\right)}{\zeta_{j} \xi_{1 j} \xi_{2 j}}
$$
Therefore, the test statistic can be expressed as
$$
X_{A(g)}^{2}=\frac{T^{2}}{V\left(T \mid H_{0}\right)}=\frac{\left.\sum_{j} \mid \tilde{w}{j}\left(p{1 j}-p_{2 j}\right)\right]^{2}}{\sum_{j} \tilde{w}{j}^{2} \sigma{0 j}^{2}}=\frac{\tilde{T}^{2}}{V\left(\tilde{T} \mid H_{0}\right)}
$$

证明 .

$$
F(x) \leq\left|y^{}\right||x| $$ and $$ -F(x)=F(-x) \leq p(-x)=\left|y^{}\right||x|
$$
that is,
$$
|F(x)| \leq\left|y^{}\right||x| . $$ This shows that $F:=x^{} \in X^{}$ and $\left|x^{}\right| \leq\left|y^{}\right|$. Since the reversed inequality is trivial for any linear extension of $y^{}$, the theorem is proved in the case of real scalars.


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Oxford COURSE NOTES ：

using the weights
$$
\tilde{w}{j}=\frac{1}{\phi{0 j}^{2} g^{\prime}\left(\pi_{j}\right)} .
$$Under $H_{0}: \pi_{1 j}=\pi_{2 j}=\pi_{j}$, then $\tilde{T} \approx \mathcal{N}\left(0, \sigma_{\bar{T}{0}}^{2}\right)$, where $E(\tilde{T})=0$ and $$ \begin{aligned} V\left(\tilde{T} \mid H{0}\right) &=\sigma_{T_{0}}^{2}=\sum_{j}^{2} \tilde{w}^{2} \sigma_{0 j}^{2}=\sum_{j} \tilde{w}{j}^{2} \frac{\phi{0 j}^{2}}{N} \
&=\frac{1}{N} \sum_{j j}\left(\frac{1}{\phi_{0 j}^{2} g^{\prime}\left(\pi_{j}\right)^{2}}\right)=\frac{\phi_{\widetilde{T}_{0}}^{2}}{N}
\end{aligned}
$$





类别、证明和过程|Categories, Proofs and Processes代写 

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类别、证明和过程|Categories, Proofs and Processes代写 
constructor TtdRecordList.Create(aElementSize : integer);
begin
inherited Create;
{save the actual element size}
FActElemSize := aElementSize;
{round the actual size to the nearest 4 bytes}
FElementSize := ((aElementSize + 3) shr 2) shl 2;
{calculate the maximum number of elements}
{$IFDEF Delphi1}
FMaxElemCount := 65535 div FElementSize;
{$ELSE}
FMaxElemCount := MaxInt div FElementSize;
{$ENDIF}
end;
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Oxford COURSE NOTES ：

function TtdRecordList.Remove(aItem : pointer;
aCompare : TtdCompareFunc) : integer;
begin
Result := IndexOf(aItem, aCompare);
if (Result<>tdc_ItemNotPresent) then
Delete(Result);
end;
function TtdRecordList.IndexOf(aItem : pointer;
aCompare : TtdCompareFunc) : integer;
var
ElementPtr : PAnsiChar;
i : integer;
begin
ElementPtr := FArray;
for i := 0 to pred(Count) do begin
if (aCompare(aItem, ElementPtr) = 0) then begin
Result := i;
Exit;
end;
inc(ElementPtr, FElementSize);
end;
Result