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Madi S.
Software Engineer
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Web Development
TutorMe
Question:

Within the context of a Client-Server architecture, explain the difference between stateful and stateless communication protocols, and provide an example of each.

Madi S.
Answer:

A stateful protocol is one where the server retains state information about the client across every request made by the client, and vice versa. State information may be any information regarding the session or status of the client/server, or any such information. A TCP connection is an be an example of such a protocol. On the other hand, a stateless protocol is one where the server does not maintain any state information about the client across multiple requests. Requests are seen as independent of one another, and the state of the request only lives for the duration of the request. Once a request is complete, the server forgets about the state of that request and does not share it across other requests. The HTTP protocol may be an example of such.

Discrete Math
TutorMe
Question:

Consider the following logical equivalence relation: $$ \neg (p \rightarrow (q \wedge r) ) \equiv (p \wedge \neg q) \vee (p \wedge \neg r) $$. Show that this equivalence relation holds by means of algebraic manipulation.

Madi S.
Answer:

Firstly, note the following relevant logical equivalence laws: $$ {\displaystyle \neg (p\wedge q)\equiv \neg p\vee \neg q} $$ ... De Morgean's Law 1 $$ {\displaystyle \neg (p\vee q)\equiv \neg p\wedge \neg q} $$ ... De Morgean's Law 2 $$ {\displaystyle p\wedge (q\vee r)\equiv (p\wedge q)\vee (p\wedge r)} $$ ... Distributive Law $$ {\displaystyle \neg (\neg p)\equiv p} $$ ... Double Negation Law $$ {\displaystyle p\implies q\equiv \neg p\vee q} $$ ... Implication Law Now, taking the LHS of the given equivalence relation, we proceed with algebraic manipulation and have that: $$ \neg (p \rightarrow (q \wedge r) ) $$ $$ \equiv $$ $$ \neg (\neg p \vee (q \wedge r)) $$ ... By the Implication Law $$ \equiv $$ $$ \neg (\neg p) \wedge \neg (q \wedge r) $$ ... By De Morgan's Law 2 $$ \equiv $$ $$ p \wedge \neg (q \wedge r) $$ ... By the Double Negation Law $$ \equiv $$ $$ p \wedge (\neg q \vee \neg r) $$ ... By De Morgan's Law 1 $$ \equiv $$ $$ (p \wedge \neg q) \vee (p \wedge \neg r) $$ ... By the Distributive Law Hence, by means of algebraic manipulation, the two statements are equivalent, and therefor the equivalence relation holds as required.

Calculus
TutorMe
Question:

Consider the function $$ f(x)= \frac{1}{x-1} $$. Determine the limit of $$ f(x) $$ as $$ x $$ approaches 1, if it exists. I.e. determine $$ \lim_{x\to1} f(x) $$ if it exists. Furthermore, state by referencing relevant limit theory, why the limit exists or does not exist.

Madi S.
Answer:

Note that we are interested in knowing how $$ f(x) $$ behaves $$ \textbf{around} $$ the point $$ x=1 $$, and that we are not interested in solving $$ f(1) $$, which is in fact undefined. From limit theory, we have that for any function $$ g(x) $$, $$ \lim_{x\to a} g(x) = L $$ $$ \textbf{if and only if} $$ $$ \lim_{x\to a^-} g(x) = \lim_{x\to a^+} g(x) = L $$. $$ \cdots \mathbf{(A)} $$ We will now determine the limit of $$ f(x) $$ as $$ {x\to1} $$ by investigating the behavior of the two one-sided limits of $$ f(x) $$. Firstly, note that taking the limit as $$ x $$ approaches $$ 1 $$ from the left, we have that: $$ \lim_{x\to1^-} f(x) = \lim_{x\to1^-} (\frac{1}{x-1}) = -\infty $$. $$ \cdots \mathbf{(B)} $$ Note that as $$ x $$ approaches $$ 1 $$, the denominator becomes an infinitely $$ \textbf{negative} $$ small value. This can further be seen graphically from the graph of $$ f(x) $$, and by plugging in values of $$ x < 1 $$ into $$ f(x) $$, e.g. $$ 0.9, 0.99, 0.999, etc$$. Similarly, taking the limit as $$ x $$ approaches $$ 1 $$ from the right, we have that: $$ \lim_{x\to1^+} f(x) = \lim_{x\to1^+} (\frac{1}{x-1}) = \infty $$. $$ \cdots \mathbf{(C)} $$ Again, note that as $$ x $$ approaches $$ 1 $$, the denominator becomes an infinitely $$ \textbf{positive} $$ small value. This can further be seen graphically from the graph of $$ f(x) $$, and by plugging in values of $$ x > 1 $$ into $$ f(x) $$, e.g. $$ 1.1, 1.01, 1.001, etc$$. Hence from $$ \mathbf{(B)} $$ and $$ \mathbf{(C)} $$ above, we have that $$ \lim_{x\to1^-} f(x) \neq \lim_{x\to1^+} f(x) $$, and thus by the theorem in $$ \mathbf{(A)} $$ we have that the limit does not exist, i.e. $$ \lim_{x\to1} f(x) $$ $$ \textbf{does not exist} $$.

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