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Limit of $\frac{x}{x^2-1}$ Question
$$\lim_{x \rightarrow 1-}\frac{x}{x^2-1}$$
My attempt
$$\lim_{x \rightarrow 1-}\frac{x}{x^2-1}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{1}{x^2-1}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{x^2(\frac{1}{x^2})}{x^2(1-\frac{1}{x^2})}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{\frac{1}{x^2}}{1-\frac{1}{x^2}}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{\frac{1}{x^2}}{\frac{x^2-1}{x^2}}=\lim_{x \rightarrow 1-}\frac{x}{x^2-1}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{1}{x^2-1}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{x^2(\frac{1}{x^2})}{x^2(1-\frac{1}{x^2})}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{\frac{1}{x^2}}{1-\frac{1}{x^2}}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{1}{x^2}\cdot\frac{x^2}{x^2-1}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{1}{x^2-1}$$
As you can see I'm going in circles. Can anyone give me a hint on how to start on this problem?
|
$$\lim_{x\to1^-}\frac x{x^2-1}$$
$$=\lim_{x\to1^-}\frac x{(x+1)(x-1)}$$
$$=\lim_{x\to1^-}\frac x{x+1}\cdot\frac1{x-1}$$
$$=\lim_{x\to1^-}\frac x{x+1}\cdot\lim_{x\to1^-}\frac1{x-1}$$
$$=\frac12\lim_{x\to1^-}\frac1{x-1}$$
$$\to-\infty$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Partial sums of harmonic series I've been given the following problem.
Prove that
$$\lim_{N\to\infty}\sum_{i=1}^N\frac1i=\infty.$$
In other words I need to prove that the partial sum of the harmonic series diverges. I know the integral test works in this case, but does anybody know of any other methods for showing this?
|
There are many ways to show this.
\begin{align}
& 1 + \frac 1 2 + \frac 1 3 + \frac 1 4 + \frac 1 5 + \cdots \\[15pt]
= {} & 1 + \frac 1 3 + \frac 1 5 + \frac 1 7 + \cdots \\[8pt]
& \phantom{1} + \frac 1 2 + \frac 1 4 + \frac 1 6 + \cdots \\[15pt]
> {} & \phantom{{} + {}} \frac 1 2 + \frac 1 4 + \frac 1 6 + \cdots \\[8pt]
& {} + \frac 1 2 + \frac 1 4 + \frac 1 6 + \cdots
\quad (\text{This “}{>}\text{'' is true if the sum is finite.)} \\[15pt]
= {} & 1 + \frac 1 2 + \frac 1 3 + \frac 1 4 + \frac 1 5 + \cdots
\end{align}
Here's a more frequently seen way:
\begin{align}
& 1 + \left(\frac 1 2\right) + \left(\frac 1 3 + \frac 1 4\right) + \left( \frac 1 5 + \cdots + \frac 1 8 \right) + \left( \frac 1 9 + \cdots + \frac 1 {16} \right) + \cdots \\[10pt]
\ge {} & 1 + \left(\frac 1 2\right) + \left(\frac 1 4 + \frac 1 4\right) + \left( \frac 1 8 + \cdots + \frac 1 8 \right) + \left( \frac 1 {16} + \cdots + \frac 1 {16} \right) + \cdots \\[10pt]
= {} & 1 + \frac 1 2 + \frac 1 2 + \frac 1 2 + \frac 1 2 + \cdots = \infty.
\end{align}
|
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|
Some Pythagorean triples proofs I'm studying about number theory by myself. So I'm sorry if the question seems to be unclear.
I just want to know how to prove that one member of a Pythagorean triple is always divisible by 5 and that the area of an integer-sided right-angled triangle is always divisible by 6.
I searched for similar questions but didn't find an understandable proof for me.
|
We are considering integer solutions to
$$
a^2+b^2=c^2
$$
Let us consider the equation $a^2+b^2=c^2$ modulo $2$:
$$
\begin{array}{c|cc}
+&0^2&1^2\\
\hline
0^2&0^2&1^2\\
1^2&1^2&0^2
\end{array}
$$
we see that in all four cases at least one of the three numbers is zero modulo $2$. So at least one in the triple is divisible by $2$.
Modulo $3$ we have $1^2=2^2$ so the same table applies. At least one is divisible by $3$.
Modulo $4$ we have $0^2=2^2$ and $1^2=3^2$ so again the same table applies. At least one is divisible by $4$.
Modulo $5$ we have $1^2=4^2=1$ and $2^2=3^2=4$ and so
$$
\begin{array}{c|ccc}
+&0^2&1^2&2^2\\
\hline
0^2&0^2&1^2&2^2\\
1^2&1^2&\times&0^2\\
2^2&2^2&0^2&\times
\end{array}
$$
where $\times$ indicates that result cannot be a square, since $1^2+1^2=2$ and $2^2+2^2=3$ are not values of squares modulo $5$. They are quadratic non-residues mod $5$.
All $7$ cases that are actually possible contain at least one zero. So at least one of the three numbers is divisible by $5$.
Finally, regarding the area being divisible by $6$, we see from the formulas for generating the triples
$$
(a,b,c)=(m^2-n^2,2mn,m^2+n^2)
$$
that the area must be
$$
T=\tfrac12 ab=(m^2-n^2)mn
$$
and if neither $m$ nor $n$ is divisible by $2$ we have $m=n=1$ modulo $2$ and therefore $m^2-n^2$ must be divisible by $2$.
If neither $m$ nor $n$ is divisible by $3$, we see that either $m,n\in\{1,2\}$ modulo $3$ so that $m^2-n^2=0$ modulo $3$. So indeed the area is divisible by both $2$ and $3$, and hence by $6$.
|
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|
Sum of the series $\binom{n}{0}-\binom{n-1}{1}+\binom{n-2}{2}-\binom{n-3}{3}+..........$
The sum of the series $$\binom{n}{0}-\binom{n-1}{1}+\binom{n-2}{2}-\binom{n-3}{3}+..........$$
$\bf{My\; Try::}$ We can write it as $\displaystyle \binom{n}{0} = $ Coefficient of $x^0$ in $(1+x)^n$
Similarly $\displaystyle \binom{n-1}{1} = $ Coefficient of $x^1$ in $(1+x)^{n-1}$
Similarly $\displaystyle \binom{n-2}{2} = $ Coefficient of $x^2$ in $(1+x)^{n-2}$
Now, how can I solve it after that, Help Required, Thanks
|
[Imported from a duplicate question]
Chebyshev polynomials of the second kind have the following representation:
$$ U_n(x)=\sum_{r\geq 0}\binom{n-r}{r}(-1)^r (2x)^{n-2r} \tag{1}$$
hence the wanted sum is just $U_n\left(\frac{1}{2}\right)$, and since $\frac{1}{2}=\cos\frac{\pi}{3}$,
$$ U_n\left(\frac{1}{2}\right) = \frac{\sin((n+1)\pi/3)}{\sin(\pi/3)}.\tag{2} $$
|
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|
Describe all natural numbers $n$ for which $3^{n}-2^{n}$ is divisible by $5$. I need to describe all natural numbers $n$ for which $3^{n}-2^{n}$ is divisible by $5$.
After testing out some $n \in \mathbb{N}$, I came to the conclusion that $3^{n} - 2^{n}$ is divisible by $5$ iff $n$ is even - i.e., if $n$ is of the form $2k$ for $k \in \mathbb{N}$.
Here is my attempt so far:
$(\implies)$:
To show that if $n$ is even, then $3^{n}-2^{n}$ is divisible by $5$, we prove that $\forall k \in N$, $3^{2k}-2^{2k} \equiv \,0 \mod 5$ by induction on $k$:
*
*Basis Step: For $k = 1$, $3^{2(1)}-2^{2(1)}=3^{2}-2^{2}=9-4=5 \equiv\, 0 \mod 5 $.
*Suppose true for $\mathbf{k=m}$: $3^{2m}-2^{2m}\equiv \, 0 \mod 5 \, \implies \, 3^{2m}-2^{2m}=5l$, $l \in \mathbb{Z}$.
*Show true for $\mathbf{k = m+1}$: Consider $3^{2(m+1)}-2^{2(m+1)}\\ = 3^{2m+2}-2^{2m+2}\\ = 3^{2m}\cdot 3^{2}- 2^{2m}\cdot 2^{2}\\ = 9\cdot 3^{2m}-4 \cdot 2^{2m} \\= 5 \cdot 3^{2m} + 4\cdot 3^{2m} - 4 \cdot 2^{2m} \\ = 5(3^{2m}) + 4(3^{2m}-2^{2m}) \\ = 5(3^{2m})+4(5l)\, \text{(by the induction hypothesis)} \\ = 5(3^{2m} + 4l) \, \text{which is divisible by}\, 5.$
So, by induction, then the statement $3^{2k}-2^{2k} \equiv \, 0 \mod 5$ holds $\forall k \geq 1 \, \implies \, $ the statement $3^{n}-2^{n} \equiv \, 0 \mod 5$ holds $\forall n = 2k$, $k \geq 1$.
$(\Longleftarrow)$:
To show that $3^{n}-2^{n}$ divisible by $5$ $\implies$ $n$ even, we will show that $n$ odd $\implies$ $3^{n}-2^{n}$ is not divisible by $5$. Now, suppose the proposition is false. I.e., assume $\exists n \in \mathbb{N}$ for which $n$ is odd, but $3^{n}-2^{n}$ is divisible by $5$.
Since $n$ is odd, it is of the form $2k+1$, $k \in \mathbb{N}$.
So, we have that $3^{2k+1}-2^{2k+1} \equiv \, 0 \mod 5 \, \implies \, 3^{2k+1} - 2^{2k+1} = 5l$, $l \in \mathbb{N}$.
So, $\displaystyle \frac{3^{2k+1}}{5} - \frac{2^{2k+1}}{5} = l$.
At this point, I got stuck. I think my $(\implies)$ direction is fine, but I definitely need help on my $(\Longleftarrow)$ direction.
Could somebody please help me complete this proof?
Thank you.
|
$$3^{2k+1}-2^{2k+1}= 3(9^k)-2(4^k) \equiv 3(-1)^k-2(-1)^k=(-1)^k \mod 5$$
$$3^{2k}-2^{2k}= (9^k)-(4^k) \equiv (-1)^k-(-1)^k=0 \mod 5$$
|
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|
Probability and Conditional Probability Could someone please provide a worked solution to this question?
A box contains 5 amber, 7 blue and 9 green balls. Six of the balls are removed from the box at random and without replacement.
Find the probability that
(a) three out of the six balls are blue;
(b) four of the balls are blue and two are green;
(c) the second ball to be selected is amber, given that the final ball to be selected is blue.
I was wondering if there is a combinatorial argument for these types of problems, because multiplying fractions here seems a little time consuming and messy.
My answers:
(a) $\frac{7}{21}\times\frac{6}{20}\times\frac{5}{19}\times\frac{14}{18}\times\frac{13}{17}\times\frac{12}{16}\times\frac{6!}{(3!)^2}$
(b) $\frac{7}{21}\times\frac{6}{20}\times\frac{5}{19}\times\frac{4}{18}\times\frac{9}{17}\times\frac{8}{16}\times\frac{6!}{(4!)(2!)}$
Any help very much appreciated, as always.
|
Well, for such combinatorical questions there are typically multiple different "counting schemes" to get to the same answer. Here is the thought process I would have to arrive at those. Hope it helps:
a) There are $\binom{5+7+9}{6}$ ways to choose the $6$ balls. There are $\binom{7}{3}$ ways to choose $3$ blue balls and $\binom{5+9}{3}$ ways to choose $3$ non-blue balls. So the Answer is
$\begin{align}\frac{\binom{5+9}{3}\binom{7}{3}}{\binom{5+7+9}{6}}\end{align}$
b) very similar: $\binom{7}{4}$ ways to choose $4$ blue balls. $\binom{9}{2}$ ways to choose $2$ green balls. So the answer is
$\begin{align}\frac{\binom{7}{4}\binom{9}{2}}{\binom{5+7+9}{6}}\end{align}$
Generally: It is always about the "binomial coefficient" $\binom{n}{k}=\frac{n!}{k!(n-k)!}$ which is the number of ways to select $k$ objects (without replacement) out of a total of $n$ objects.
c) Here, order is important, so you can't use binomial coefficients in the same way. The best way here is to multiply fractions for each step like this:
$P(\text{last is blue}) = \frac{7}{5+7+9}$.
$P(\text{second is amber and last is blue}) = \frac{5}{5+7+9} \cdot\frac{7}{4+7+9}$
$P(\text{second is amber}|\text{last is blue}) = \frac{P(\text{second is amber and last is blue})}{P(\text{last is blue})} = \frac{5\cdot7\cdot21}{7\cdot21\cdot20}$
|
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|
Prove $a {a+b \choose b}$ divides the lowest common multiple of $b+1, b+2, ..., b+a$ Prove that for $a, b \in \Bbb N^*$ we have :
$$a {a+b \choose b} | LCM(b+1, ..., b+a)$$
Is there a simple proof of this result that doesn't involve computing an integral? I know a trick exploiting $\int_{0}^{1} {x^n(1-x)^mdx}$ but I am looking for a more "obvious" proof.
|
First note that
$$a\binom{a+b}{b}=\frac{(a+b)!}{(a-1)!b!}$$
To prove that $a {a+b \choose b} | LCM(b+1, ..., b+a)$ it is enough to prove that for each prime $p$ we have
$$p^n | LCM(b+1, ..., b+a)$$
where $n$ is the power of $p$ in $a\binom{a+b}{b}$.
We know by Legendre formula that
$$n=\sum_{k}\lfloor \frac{a+b}{p^k} \rfloor -\lfloor \frac{b}{p^k} \rfloor-\lfloor \frac{a-1}{p^k} \rfloor$$
Now for each $k$ we have
$$\lfloor \frac{b}{p^k} \rfloor+\lfloor \frac{a-1}{p^k} \rfloor \leq \frac{b}{p^k} + \frac{a-1}{p^k} \leq \frac{a+b}{p^k} $$
We also have
$$\frac{a}{p^k} \leq \lfloor \frac{a-1}{p^k} \rfloor +1\\
\frac{b}{p^k} <\lfloor \frac{b}{p^k} \rfloor +1$$
Combining, we get
$$\lfloor \frac{b}{p^k} \rfloor+\lfloor \frac{a-1}{p^k} \rfloor \leq \frac{a+b}{p^k} < \lfloor \frac{b}{p^k} \rfloor+\lfloor \frac{a-1}{p^k} \rfloor +2$$
which implies
$$\lfloor \frac{a+b}{p^k} \rfloor -\lfloor \frac{b}{p^k} \rfloor-\lfloor \frac{a-1}{p^k} \rfloor \in \{ 0,1 \} $$
Now, let $K_0$ be the largest $k$ such that
$$\lfloor \frac{a+b}{p^k} \rfloor -\lfloor \frac{b}{p^k} \rfloor-\lfloor \frac{a-1}{p^k} \rfloor =1 $$
Show that this implies that $p^{K_0}$ has a multiple between $b+1$ and $b+a$.
This shows that $p^{K_0}$ divides $LCM(b+1,..,b+a)$. Now, since all terms in the sum are 0 or 1, it follows that $n \leq K_0$ and hence $p^n | p^{K_0}$.
|
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|
"Smart" solution for $7^{100}+11^{100} \pmod{13}$? I have the following exercise: "Find $7^{100}+11^{100} \pmod{13}$".
My long and boring solution is the following:
*
*I've found that $7^k \equiv 7^{k \pmod{12}} \pmod{13}$ by calculating $7^k \pmod{13} = 7 \cdot (7^{k-1} \pmod{13}) \pmod{13}$ from $k=1$ until the remainder starts to repeat itself.
*The same for $11^k \equiv 11^{k \pmod{12}} \pmod{13}$
*Using the result from steps 1 and 2, $7^{100}+11^{100} \equiv 7^4+11^4 \equiv 9 + 3 \equiv 12 \pmod{13}$
Both steps 1 and 2 require a lot of calculations and I suspect that I've missed some clever trick to avoid them and to solve the exercise in a "smart" way.
Is there a way to improve my solution?
|
Thank to the guys at the comment section, the improved solution is the following:
$$ 7^{100} + 11^{100}$$
$$= 7^{8 \cdot (13-1) + 4} + 11^{8 \cdot (13-1) + 4}$$
$$ \equiv 7^4 + 11^4$$
$$ \equiv (7^2)^2 + (-2)^4 $$
$$\equiv 10^2 + 16$$
$$\equiv 9 + 3 = 12 \pmod{13} $$
|
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|
How many tangent half-angle formulas are there? \begin{align}
\tan \frac{\alpha+\beta} 2 & = \frac{\sin\alpha+\sin\beta}{\cos\alpha + \cos\beta} \tag 1 \\[10pt]
\tan \left( \frac \pi 4 \pm \frac \alpha 2 \right) & = \sec\alpha \pm \tan\alpha \tag 2 \\[10pt]
\frac{1 + i\tan\frac\alpha2}{1-i\tan\frac\alpha2} & = e^{i\alpha} \tag 3 \\[10pt]
\tan\frac\alpha2\cdot\tan\frac\beta2 = \tan\frac\gamma2 & \text{ if and/or only if } \left|\frac{\cos\alpha+\cos\beta}{1+\cos\alpha\cos\beta}\right| = \left|\cos\gamma\right| \tag 4
\end{align}
All of these relate tangents of half angles to trigonometric functions of whole angles.
Are there other tangent half-angle formulas essentially different from these? And how does one decide whether they're essentially different? (I'm not sure I would count the case of $(1)$ in which $\alpha=0$ as essentially different. One could make a case that $(1)$ is not essentially different from $(2)$. We're all accustomed to the mapping $(x,y)\mapsto \dfrac{x+y}{1-xy}$ in connection with tangents, but $(4)$ relates tangents to $(x,y)\mapsto\dfrac{x+y}{1+xy}.$)
|
For $\alpha+\beta+\gamma=\pi$,
\begin{align}
\tan\tfrac\alpha2\,\tan\tfrac\beta2\tan\tfrac\gamma2
&=
\frac{\cos\alpha+\cos\beta+\cos\gamma-1}{\sin\alpha+\sin\beta+\sin\gamma}.
\end{align}
Edit
Some more
\begin{align}
\tan\tfrac\alpha2\,\tan\tfrac\beta2\tan\tfrac\gamma2
&=
\frac{2\sin\alpha\sin\beta\sin\gamma}{(\sin\alpha+\sin\beta+\sin\gamma)^2}
\\
&=
\frac{\sin2\alpha+\sin2\beta+\sin2\gamma}{2(\sin\alpha+\sin\beta+\sin\gamma)^2}
\end{align}
\begin{align}
\cot\tfrac\alpha2+\cot\tfrac\beta2+\cot\tfrac\gamma2-(\tan\tfrac\alpha2+\tan\tfrac\beta2+\tan\tfrac\gamma2)
&=
2(\cot\alpha+\cot\beta+\cot\gamma)
\end{align}
|
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|
Given that $\sqrt[3]{1+9x} ≈ 1+3x+ax^2+bx^3$ for small values of $x$, find the values of the coefficients $a$ and $b$ Given that $\sqrt[3]{1+9x}\approx 1+3x+ax^2+bx^3$ for small values of $x$, find the values of the coefficients $a$ and $b$.
A question from CIE Maths Paper 33 O/N 2015. The answer is
$$a=-9$$
$$b=45$$
The mark sheet says:
State correct unsimplified $x^2$ or $x^3$ and then obtain the answers above
Any answers with explanation would be helpful.
|
The binomial series
$$
(1+x)^\alpha=\sum\binom\alpha k x^k
$$
gives in your case
\begin{align}
(1+9x)^{\frac13}&=1+\frac13·9x+\frac12·\frac13·(\frac13-1)·(9x)^2\\&\qquad+\frac16·\frac13·(\frac13-1)·(\frac13-2)·(9x)^3\\&\qquad+\frac1{24}·\frac13·(\frac13-1)·(\frac13-2)·(\frac13-3)·(9x)^4+...\\
&=1+3x-9x^2+45x^3-270x^4+…
\end{align}
You could also compute the third power on both sides
$$
1+9x=(1+3x)^3+3(1+3x)^2x^2(a+bx)+O(x^4)\\
=1+9x+27x^2+27x^3+3x^2(a+(6a+b)x)+O(x^4)\\
=1+9x+3(9+a)x^2+3(9+6a+b)x^3+O(x^4)
$$
|
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|
Real values of $k$ for which $x^2+(k+1)x+k^2$ has one root double the other
For what real values of $k$ does $x^2+(k+1)x+k^2$ have one root double the other?
For a start, I found the range of $k$ which endows this equation with real roots:
$$-\frac13\le k\le1$$
|
Let $a$ and $b$ be the roots. Since
$$
(x-a)(x-b)=x^2+(k+1)x+k^2\\
x^2-(a+b)x+ab=x^2+(k+1)x+k^2
$$
it follows (by comparing corresponding coefficients) that
$$
\begin{cases}
a+b=-(k+1) & (1)\\
ab=k^2 & (2)
\end{cases}
$$
If $a=2b$, then
$$
\begin{cases}
3b=-(k+1) & (3)\\
2b^2=k^2 & (4)
\end{cases}
$$
Flugging (3) inside (4) gives
$$
\sqrt{2}\left|-\frac{k+1}{3}\right|=|k|
$$
so $k=\frac{3}{7}\sqrt{2}\pm\frac{2}{7}$.
|
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|
A circle inside an ellipse Consider an ellipse with semi-axes $a$ and $b$, taller than it is wide with a small circle of radius $r$ inside. Assume the circle falls to the lowest point possible while staying inside the ellipse.
If $2r\le a-c$ then the circle and ellipse will meet at a single point at the bottom. If $2r>a-c$ the circle and ellipse will intersect at two points on the opposite side, leaving a space between the bottom of the circle and the bottom of the ellipse. For this case, given the radius of the circle and the dimensions of the ellipse how do I calculate the distance $d$ between the bottom of the circle and the bottom of the ellipse?
|
Never underestimate the power of cartesian coordinates
Consider the problem of finding the intersection points between ellipse and circle of radius $r$ centered in $(0,y_c)$, which leads to the system of equations
$$\begin{cases}\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1\\
x^2 + (y-y_c)^2 = r^2.\end{cases} $$
By substitution you obtain the following quadratic equation.
$$ y^2\left(\frac{1}{b^2} -\frac{1}{a^2} \right) + 2y\left(\frac{y_c}{a^2}\right) + \frac{r^2-y_c^2-a^2}{a^2}=0. $$
Since you want the two curves to intersect at only two points with the same ordinate, than this equation must have discriminant equal to $0$, which gives you
\begin{eqnarray}
\left(\frac{y_c}{a^2}\right)^2-\left(\frac{1}{b^2} -\frac{1}{a^2} \right)\left(\frac{r^2-y_c^2-a^2}{a^2}\right) = 0. \tag{1}\label{discr}
\end{eqnarray}
Now you just need to solve the quadratic equation \eqref{discr} with respect to the unknown $y_c$. This rapidly gives
$$y_c^2 = \frac{(b^2-a^2)(a^2-r^2)}{a^2}.$$
The desired distance is
$$w = b-|y_c| - r,$$
that is
$$ w = b - \frac{\sqrt{(b^2 - a^2)(a^2-r^2)}}{a} -r,$$
as, of course, in the other proposed solutions.
|
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"url": "https://math.stackexchange.com/questions/1972994",
"timestamp": "2023-03-29T00:00:00",
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|
Find values of $a$ such that $\frac{x^2-x}{1-ax}$ attains all real values
Find values of $a$ such that
$$\frac{x^2-x}{1-ax}$$
attains all real values.
I first tried to find the range of the above by equating the above to some $y$. Solving I get $x^2+x(ay-1)-y=0$. How do I proceed?
|
Suppose $f(x)=\frac{x^2-x}{1-ax}=k$ for some fixed constants $a,k$. Solving for $x$ we get
$$x^2-x=k(1-ax)=k-kax$$
$$x^2+(ka-1)x-k=0$$
$$x=\frac{1-ka\pm\sqrt{(ka-1)^2+4k}}2$$
For a chosen $a$, $f$ attains all real numbers if and only if the discriminant $(ka-1)^2+4k$ is non-negative for all $k$. Expanding:
$$(ak)^2-2ka+1+4k\ge0$$
$$a^2k^2+(4-2a)k+1\ge0$$
Using the discriminant a second time we find
$$(4-2a)^2-4(a^2)(1)\le0$$
$$16-16a+4a^2-4a^2\le0$$
$$16\le16a$$
$$a\ge1$$
If $a=1$, $f(x)$ behaves like $-x$. However, it has a removable singularity at $x=1$, which means it does not attain a value of $-1$. Therefore $a=1$ is excluded, and $\frac{x^2-x}{1-ax}$ attains all real values when $a>1$.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Show that if $x + \frac1x = 1$, then $x^5 + \frac1{x^5} = 1$. Suppose $x + \frac{1}{x} = 1$.
Without first working out what $x$ is, show that
$x^5 + \frac{1}{x^5} = 1$ as well.
|
Since $x + \frac{1}{x} = 1$ then
\begin{align}
1 &= 1^5 = \left( x + \frac{1}{x} \right)^{5} \\
&= \left( x^5 + \frac{1}{x^5} \right) + 5 \, \left( x^3 + \frac{1}{x^3} \right) + 10 \, \left( x + \frac{1}{x} \right) \\
&= \left( x^5 + \frac{1}{x^5} \right) + 5 \, \left(x + \frac{1}{x} \right)^3 - 15 \, \left( x + \frac{1}{x} \right) + 10 \, \left( x + \frac{1}{x} \right) \\
1 &= \left( x^5 + \frac{1}{x^5} \right) + 5 \, \left(x + \frac{1}{x} \right)^3 - 5 \, \left( x + \frac{1}{x} \right) \\
1 &= x^5 + \frac{1}{x^5}
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1976629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Prove that if $2b^2+1 = 3^b$ where $b$ is a positive integer, then $b \leq 2$
Prove that if $2b^2+1 = 3^b$ where $b$ is a positive integer, then $b \leq 2$.
Is there some way we can transform the equation in order to get the inequality? We have $2b^2 = 3^b-1$.
|
Given: $ \qquad \displaystyle 2b^2+1 = 3^b \tag 1$
We see immediately $b=1$ and $b=2$ give equalities.
What if $b$ grows over $2$, so $b=2+c$? Let's first denote $ß=\lg3 $ and we know, $ß >1$ . With this we make the ansatz:
$$ 2(2+c)^2+1 = 3^{2+c} \tag 2$$
and develop by expanding and rearranging
$$ \begin{array}{rll}
2(2+c)^2+1 &= 9\cdot 3^c \\
9+8c+2c^2 &= 9\cdot (1+ßc + ß^2c^2/2! &+ ß^3c^3/3! + ... ) \\
1+8/9c+2/9c^2 &= 1+ßc + ß^2c^2/2 &+ ß^3c^3/3! + ...\\
\text{ getting }&\\
(8/9-ß)c+1/18(4-9ß^2)c^2 &= &\phantom+ ß^3c^3/3! + ... & \qquad (3)\\
\end{array}$$
*
*The rhs is positive with $c>0$.
*Because $ß>1$ the parentheses on the lhs are negative. So for all $c>0$ the lhs is negative.
So for all $c>0$ in (3) (and thus $b>2$ in (1)) the lhs is smaller than the rhs and the equality (1) does never hold.
|
{
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|
How to solve $\sin(x) + 2\sqrt{2}\cos x =3$ How to solve $\sin(x) + 2\sqrt{2}\cos x=3$ ?
What is general method for doing these kind of questions?
Thanks
|
If you make the following substitutions:
$$
\begin{cases}
X=\cos x\\
Y=\sin x
\end{cases}
$$
then your equations (remembering that $\sin^2 x+\cos^2x=1$) becomes
$$
\begin{cases}
X^2+Y^2=1\\
Y+2\sqrt{2}X=3
\end{cases}
$$
which is the intersection between a circle and a straight line.
Solving it:
$$
\begin{cases}
Y=3-2\sqrt{2}X\\
X^2+(3-2\sqrt{2}X)^2=1\\
\end{cases}
$$
you obtain the equation
$$
X^2+9-12\sqrt{2}X+8X^2=1,
$$
which is
$$
\begin{align*}
9X^2-12\sqrt{2}X+8&=0,\\
(3X-2\sqrt{2})^2&=0,
\end{align*}
$$
which yelds the solution
$$
\begin{cases}
X=\dfrac{2\sqrt{2}}{3}\\
Y=3-2\sqrt{2}\dfrac{2\sqrt{2}}{3}=3-\dfrac{8}{3}=\dfrac{1}{3}
\end{cases}
$$
Substituting back:
$$
\begin{cases}
\cos x=\dfrac{2\sqrt{2}}{3}\\
\sin x=\dfrac{1}{3}
\end{cases}
$$
Or:
$$
x=\arctan\dfrac{1}{2\sqrt2}+2k\pi,\quad\text{for $k\in\mathbf{Z}$}.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1980638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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|
Given $P(x)=x^{4}-4x^{3}+12x^{2}-24x+24,$ then $P(x)=|P(x)|$ for all real $x$ $$P(x)=x^{4}-4x^{3}+12x^{2}-24x+24$$
I'd like to prove that $P(x)=|P(x)|$. I don't know where to begin. What would be the first step?
|
As the other answers pointed out, you need to prove that the polynomial is positive for all values of $x$. A very common way to do it is to express it as a sum of squares.
$$P(x)=x^4−4x^3+12x^2−24^x+24$$
The problems here are the two terms where X is at an odd power: $-4x^3$ and $-24x$
Trying to fit them into squares is almost automatic:
\begin{align}
P(x) & = x^4−4x^3+12x^2−24x+24 \\
& = (x^4 - 4x^3 + 4x^2) + 8X^2 - 24X + 24 \\
& = x^2(x-2)^2 + 6(x^2 - 4X + 4) + 2x^2 \\
& = x^2(x-2)^2 + 6(x-2)^2 + 2x^2
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1981110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 10,
"answer_id": 6
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|
Beta function with different integral limits The beta function is
$$B(x,y) = 2\int_{0}^{\frac{\pi}{2}} \cos^{2x-1}\theta \sin^{2y-1}\theta d\theta$$
for positive values of $x$ and $y$.
How can this integral be used to evaluate other limits of integration?
For example,
$$ \int_{0}^{\pi} \sqrt{\sin\theta} \, d\theta$$
|
Consider:
\begin{align}
\int_{0}^{\pi} \sqrt{\sin\theta} \, d\theta &= \int_{0}^{\pi/2} \sqrt{\sin\theta} \, d\theta + \int_{\pi/2}^{\pi} \sqrt{\sin\theta} \, d\theta \\
&= \int_{0}^{\pi/2} \sqrt{\sin\theta} \, d\theta + \int_{0}^{\pi/2} \sqrt{\sin(\theta + \pi/2)} \, d\theta \\
&= \int_{0}^{\pi/2} \sqrt{\sin\theta} \, d\theta + \int_{0}^{\pi/2} \sqrt{\cos\theta} \, d\theta \\
&= \frac{1}{2} \, \left( B\left(\frac{3}{4}, \frac{1}{2} \right) + B\left( \frac{1}{2} , \frac{3}{4}\right) \right) \\
&= B\left(\frac{3}{4}, \frac{1}{2} \right) = \frac{4\pi}{s},
\end{align}
where $s$ is the lemniscate constant.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Finding an unknown coefficient of a polynomial given a factor Q:Find the value of $a$ given that $x^2+1$ is a factor of $x^4-3x^3+3x^2+ax+2$
No idea where to start, I was going to use the factor theorem but it didn't work out.
Question from year 10 Cambridge maths textbook
|
\begin{align}
x^4-3x^3+3x^2+ax+2 &= (x^4 + x^2) - 3 \, (x^3 + x) + 2 \, (x^2 + 1) + (a+3) \, x \\
&= (x^2 + 1) \, \left( x^2 - 3 \, x + 2 + \frac{(a+3) \, x}{x^2 + 1} \right)
\end{align}
In order for this to be factored then it is required that $a = -3$.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Does this trigonometric pattern continue for all primes $p=6m+1$? (Revised from its original form.) Consider primes $p=6m+1$, and $p>13$:
$p=19=3\times\color{blue}{6}+1$:
Let $\beta = 2\pi/19.\,$ A root of $x^3+x^2-\color{blue}{6}x-7=0$ is $$x=2\big(\cos(2\beta)+\cos(3\beta)+\cos(5\beta)\big)$$
Note that $2+3=5$, or the last multiplier is the sum of the first two.
$p=31=3\times\color{blue}{10}+1$
Let $\beta = 2\pi/31.\,$ A root of $x^3+x^2-\color{blue}{10}x-8=0$ is $$x=2\big(\cos(2\beta)+\cos(4\beta)+\cos(8\beta)+\cos(16\beta)+\cos(30\beta)\big)$$
Similarly, $2+4+8+16=30$. Note that $\cos(30\beta) =\cos(32\beta)$.
$p=37=3\times\color{blue}{12}+1$
Let $\beta = 2\pi/37.\,$ A root of $x^3+x^2-\color{blue}{12}x+11=0$ is $$x=2\big(\cos(2\beta)+\cos(9\beta)+\cos(12\beta)+\cos(15\beta)+\cos(16\beta)+\cos(54\beta)\big)$$
Again, $2+9+12+15+16=54$.
$p=43=3\times\color{blue}{14}+1$
Let $\beta = 2\pi/43.\,$ A root of $x^3+x^2-\color{blue}{14}x+8=0$ is $$x=2\sum_{k=1}^7\cos\big(2^k\beta)$$
But alternatively,
$$x=2\big(\cos(\beta)+\cos(4\beta)+\cos(11\beta)+\cos(16\beta)+\cos(21\beta)+\cos(35\beta)+\cos(88\beta)\big)$$
and, $1+4+11+16+21+35=88$.
Question: Is it true that argument multipliers of the cubic roots obey $\displaystyle\sum_{k=1}^{m-1}a_k=a_m$ for any $p=6m+1$?
|
This situation can be summarized as follows (I take the case p = 19 as an example):
Let $\zeta$ be a $19$-th root of unity then we can form the field of cyclotomic numbers $\mathbb{Q}(\zeta)$. The galois group of this extension (of $\mathbb{Q}$) is the cyclic group $C_{18}$ of order $18$. In order to construct a sub-extension $K$ of dimension $3$, lets chose a subgroup $H$ of $G$ of index $3$, i.e. the cyclic subgroup of order $18/3 = 6$. The orbit of $\zeta^2$ gives $\{\zeta^2, \zeta^3, \zeta^5, \ldots \text{(complex conjugates)}\}$. The sum $x$ of these elements is invariant under $H$ so belongs to $K$, so it's minimal polynomial is cubic : $x^3 + x^2 -6x-7$.The same situation reproduces for the other values of $p$, so the problem rather concerns the powers of a generator in orbits of cosets of subgroups in the automorphism group of the cyclic group $C_p$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\lim ( \sqrt{n^2+n}-n) = \frac{1}{2}$ Here's what I have so far:
Given $\epsilon > 0$, we want to find N such that $\sqrt{n^2+n}-n < \epsilon$ for all $n>N$. And so:
$( \sqrt{n^2+n}-n-\frac{1}{2}) \cdot \frac{\sqrt{n^2+n}-(n+\frac{1}{2})}{\sqrt{n^2+n}-(n+\frac{1}{2})}$
$= \frac{(n^2+n)-(n+\frac{1}{2})}{\sqrt{n^2+n} + (n+\frac{1}{2})}$
And I'm not sure how to go on from here. Help would be appreciated.
|
For each $\varepsilon>0$ you want to find $N$ such that, for $n>N$,
$$
\left|\sqrt{n^2+n}-n-\frac{1}{2}\right|<\varepsilon
$$
that is,
$$
-\varepsilon+\frac{1}{2}<\sqrt{n^2+n}-n<\varepsilon+\frac{1}{2}
$$
It is not restrictive to assume $0<\varepsilon<1/2$.
The inequality $\sqrt{n^2+n}-n>A$, for $A<1/2$, is the same as $\sqrt{n^2+n}>n+A$, and squaring gives
$$
n>2An+A^2
$$
that is $n>A^2/(1-2A)$. In the case of $A=(1-2\varepsilon)/2$ we have
$$
n>\frac{(1-2\varepsilon)^2}{8\varepsilon}
$$
The inequality $\sqrt{n^2+n}-n<B$, with $B>1/2$, is the same as $\sqrt{n^2+n}<n+B$ and, squaring,
$$
n<2Bn+B^2
$$
which is true for every integer $n$.
|
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"url": "https://math.stackexchange.com/questions/1988705",
"timestamp": "2023-03-29T00:00:00",
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|
Find all values of $ r $ such that $ a_n = r^n $ for $ n \in \mathbb{Z}_{\ge2} $ in a recurrence relation I am trying to solve the following problem:
Given the following recurrence relation:
\begin{equation}
a_n = 6a_{n-1} - 8a_{n-2} \ \text{ for } \ n \in \mathbb{Z}_{\ge2}
\end{equation}
Find all values of $ r $ so that when $ a_0 = 1 $ and $ a_1 = r $ then $ a_n = r^n \ \ \forall n \in \mathbb{N} $
My attempt:
\begin{align}
\text{By distinct roots theorem,} \\ \\
\text{Characteristic equation: }
& t^2 - 6t + 8 = 0 \\
& (t-3)^2 + 8 - 9 = 0 \\
& (t-3)^2 = 1 \\
& t = 4 \text{ or } t = 2
\end{align}
\begin{align}
\text{Let } a_0 = 1 = C(4^0) + D(2^0) \text{ where C, D } \in \mathbb{R}
\end{align}
\begin{align}
1 = C + D \\
C = 1 - D
\end{align}
\begin{align}
\text{Let } a_1 = r &= C(4) + D(2) \\
r &= 4C + 2D \\
r &= 4(1-D) + 2D \quad \text{(By substitution)} \\
r &= 4 - 4D + 2D \\
r &= 4 - 2D
\end{align}
It seems that I must have gone wrong somewhere as $ r < 0 $ for $ D > 2$. Hence the answer for $ a_n $ would be very strange (i.e. $ a_n < 0 $ when $ r = -4 $). Could anyone please assist me?
|
Both roots give solutions: $r=4$ for $a_n = 4^n$ ($C=0, D=1$) and $r=2$ for $a_n = 2^n$ ($C=1, D=0$).
|
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|
The sequence $x_{n+m}\le \frac{x_n+x_{n+1}+\cdots+x_{n+m-1}}{m}$. I have problem
Let $m \ge 2 -$ fixed positive integer. The sequence of non-negative real numbers $\{x_n\}_{n=1}^{\infty}$ is that for all $n\in \mathbb N$
$$x_{n+m}\le \frac{x_n+x_{n+1}+\cdots+x_{n+m-1}}{m}$$
Prove that this sequence has a limit.
Here is the solution of this problem.
Q.:Why the sequence is not monotone?
|
Note that, given the unilateral z-transform
$$
A(z) = \sum\limits_{0\, \leqslant \,n} {a_{\,n} \,z^{\,n} } \quad \left| {\;a_{\,n < 0} = 0} \right.
$$
then the Moving Window Sum has a z-transform:
$$
\begin{gathered}
W(A(z),h) = \sum\limits_{0\, \leqslant \,n} {\left( {a_{\,n - h + 1} + \cdots + a_{\,n - 1} + a_{\,n} } \right)\,z^{\,n} } = \hfill \\
= \left( {1 + z + \cdots + z^{\,h - 1} } \right)A(z) = \frac{{1 - z^{\,h} }}
{{1 - z}}A(z) \hfill \\
\end{gathered}
$$
Now, what you have defined is a sequence which is
moving-window average of itself. So, if we put
$$
\left\{ \begin{gathered}
x_{\,n} = 0\quad \left| {\,n < 0} \right. \hfill \\
x_{\,0} = 1 \hfill \\
x_{\,n} = \frac{1}
{h}\left( {x_{\,n - h} + x_{\,n - h + 1} + \cdots + x_{\,n - 1} } \right)\quad \left| {\;1 \leqslant h} \right. \hfill \\
\end{gathered} \right. \tag{1}
$$
we get
$$
\begin{gathered}
X_{\,h} (z) = \sum\limits_{0\, \leqslant \,n} {x_{\,n} \,z^{\,n} } = 1 + \frac{1}
{h}\,\sum\limits_{1\, \leqslant \,n} {\left( {x_{\,n - h + 1} + \cdots + x_{\,n - 1} } \right)\,z^{\,n} } \quad \Rightarrow \hfill \\
\Rightarrow \quad X_{\,h} (z) = 1 + \frac{z}
{h}\,\frac{{1 - z^{\,h} }}
{{1 - z}}X_{\,h} (z)\quad \Rightarrow \quad \left( {1 - \frac{{z\left( {1 - z^{\,h} } \right)}}
{{h\left( {1 - z} \right)}}} \right)X_{\,h} (z) = 1\quad \Rightarrow \hfill \\
\Rightarrow \quad X_{\,h} (z) = \frac{{h\left( {1 - z} \right)}}
{{\left( {h - \left( {h + 1} \right)z + z^{\,h + 1} } \right)}} = \frac{h}
{{\left( {h + \left( {h - 1} \right)z + \cdots + 2\,z^{\,h - 2} + z^{\,h - 1} } \right)\left( {1 - z} \right)}} \hfill \\
\end{gathered} \tag{2}
$$
$X_h(z)$ is a meromorphic function with a single real pole in $z=1$,
which has a well defined Taylor expansion.
The Final Value theorem tells us that the $x$ sequence is limited
$$
\mathop {\lim }\limits_{z\; \to \,1} \left( {1 - z} \right)X_{\,h} (z) = x_\infty = \frac{h}
{{\left( {h + \left( {h - 1} \right) + \cdots + 2\, + 1} \right)}} = \frac{2}
{{\left( {h + 1} \right)}}
$$
Example with $h=3$
$$
x = \left\{ {1,\;\frac{1}
{3},\;\frac{4}
{9},\,\frac{{16}}
{{27}},\,\frac{{37}}
{{81}},\,\frac{{121}}
{{243}},\, \cdots } \right\}
$$
where $3^n x_n$ is the OEIS sequence A103770.
and the Taylor series of $X_3(z)$ effectively corresponds to the $x$ sequence.
As also demonstrated in the OEIS article, the $x$ sequence converges to $1/2$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove $\int_{0}^{1} \frac{\sin^{-1}(x)}{x} dx = \frac{\pi}{2}\ln2$ I stumbled upon the interesting definite integral
\begin{equation}
\int\limits_0^1 \frac{\sin^{-1}(x)}{x} dx = \frac{\pi}{2}\ln2
\end{equation}
Here is my proof of this result.
Let $u=\sin^{-1}(x)$ then integrate by parts,
\begin{align}
\int \frac{\sin^{-1}(x)}{x} dx &= \int u \cot(u) du \\
&= u \ln\sin(u) - \int \ln\sin(u) du
\tag{1}
\label{eq:20161030-1}
\end{align}
\begin{align}
\int \ln\sin(u) du &= \int \ln\left(\frac{\mathrm{e}^{iu} - \mathrm{e}^{-iu}}{i2} \right) du \\
&= \int \ln\left(\mathrm{e}^{iu} - \mathrm{e}^{-iu} \right) du \,- \int \ln(i2) du \\
&= \int \ln\left(1 - \mathrm{e}^{-i2u} \right) du + \int \ln\mathrm{e}^{iu} du \,-\, u\ln(i2) \\
&= \int \ln\left(1 - \mathrm{e}^{-i2u} \right) du + \frac{i}{2}u^{2} -u\ln2 \,-\, ui\frac{\pi}{2}
\tag{2}
\label{eq:20161030-2}
\end{align}
To evaluate the integral above, let $y=\mathrm{e}^{-i2u}$
\begin{equation}
\int \ln\left(1 - \mathrm{e}^{-i2u} \right) du = \frac{i}{2} \int \frac{\ln(1-y)}{y} dy
= -\frac{i}{2} \operatorname{Li}_{2}(y) = -\frac{i}{2} \operatorname{Li}_{2}\mathrm{e}^{-i2u}
\tag{3}
\label{eq:20161030-3}
\end{equation}
Now we substitute equation \eqref{eq:20161030-3} into equation \eqref{eq:20161030-2}, then substitute that result into equation \eqref{eq:20161030-1}, switch variables back to (x), and apply limits,
\begin{align}
\int\limits_{0}^{1} \frac{\sin^{-1}(x)}{x} dx
&= \sin^{-1}(x)\ln(x) + \sin^{-1}(x)\left(\ln2 + i\frac{\pi}{2}\right) \\
&- \frac{i}{2}[\sin^{-1}(x)]^{2} + \frac{i}{2} \operatorname{Li}_{2}\mathrm{e}^{-i2\sin^{-1}(x)} \Big|_0^1 \\
&= \frac{\pi}{2}\ln2
\end{align}
I would be interested in seeing other solutions.
|
\begin{array}{r}
\displaystyle \int_{0}^{1} \frac{\arcsin x}{x} d x=& \int_{0}^{1} \arcsin x d(\ln x)
\stackrel{IBP}{=} \displaystyle -\int_{0}^{1} \frac{\ln x}{\sqrt{1-x^{2}}} d x
\end{array}
Let $\displaystyle I(a)=\int_{0}^{1} \frac{x^{a}}{\sqrt{1-x^{2}}} d x$ and $x=\sin \theta$, then our integral becomes $$
\begin{aligned}
I(a) &=\int_{0}^{\frac{\pi}{2}} \sin ^{2\left(\frac{a+1}{2}\right)-1} \theta \cos ^{2\left(\frac{1}{2}\right)-1} \theta d \theta \\
&=\frac{1}{2} B\left(\frac{a+1}{2}, \frac{1}{2}\right) \\
&=\frac{\sqrt{\pi}}{2} \frac{\Gamma\left(\frac{a+1}{2}\right)}{\Gamma\left(\frac{a}{2}+1\right)}
\end{aligned}
$$
Using logarithmic differentiation yields
$$
\frac{I^{\prime}(a)}{I(a)}=\frac{1}{2}\left[\psi\left(\frac{a+1}{2}\right)-\psi\left(\frac{a}{2}+1\right)\right]
$$
Putting $a=0$ gives
$$
\begin{aligned}
I^{\prime}(0) &=\frac{I(0)}{2} \cdot\left(\psi\left(\frac{1}{2}\right)-\psi(1)\right)=\frac{\pi}{4}(-\ln 4)=-\frac{\pi}{2} \ln 2
\end{aligned}
$$
Now we can conclude that
$$\displaystyle \int_{0}^{1} \frac{\arcsin x}{x} dx = -\int_{0}^{1} \frac{\ln x}{\sqrt{1-x^{2}}} dx =-I^{\prime}(0)= \frac{\pi}{2} \ln 2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1992462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 3
}
|
Show that $x^2+x+4$ is irreducible over $\mathbb Z_{11}$.
Show that $x^2+x+4$ is irreducible over $\mathbb Z_{11}$.
The problem can be solved by putting every element from $\mathbb{Z}_{11}$ and checking if there is any root.
$0^2 + 0 + 4 = 4 \\
1^2 + 1 + 4 = 6 \\
2^2 + 2 + 4 = 10 \\
3^2 + 3 + 4 = 5 \\
4^2 + 4 + 4 = 2 \\
5^2 + 5 + 4 = 1 \\
6^2 + 6 + 4 = 2 \\
7^2 + 7 + 4 = 5 \\
8^2 + 8 + 4 = 10 \\
9^2 + 9 + 4 = 6 \\
10^2 + 10 + 4 = 4 $
But this process is very long one. I am seeking a short method by which I can solve such problem.
Now suppose that it is reducible, then the factors should be linear. So how can we solve the problem based on this kind of observation? Any help will be great. Thanks.
|
By completing the square we get
$$x^2+x+4=(x+6)^2+1.$$
So the original question reduces to the question if $-1$ is a square in $\mathbb Z_{11}$. This question can be answered with no using the first supplement for the law of quadratic reciprocity since $11 \equiv 3 \pmod 4$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1993023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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|
Prove that $x_n = 1 - \sum\limits^{n}_{i=1} \frac{1}{n+i}$ converges
Let $(x_n)^{\infty}_{n=1}$ be the sequence $\displaystyle{x_n = 1 - \sum^{n}_{i=1} \frac{1}{n+i}}$. Prove that $(x_n)^{\infty}_{n=1}$ converges.
I know the sequence is decreasing and bounded below, hence will converge (Monotone Convergence Theorem), but I am having trouble actually proving that the sequence is decreasing.
|
If
$x_n = 1 - \sum^{n}_{i=1} \frac{1}{n+i}
$,
then
$\begin{array}\\
x_{n+1}-x_n
&=\left(1 - \sum^{n+1}_{i=1} \frac{1}{n+1+i}\right)-\left(1 - \sum^{n}_{i=1} \frac{1}{n+i}\right)\\
&=\sum^{n}_{i=1} \frac{1}{n+i}-\sum^{n+1}_{i=1} \frac{1}{n+1+i}\\
&=\sum^{n}_{i=1} \frac{1}{n+i}-\sum^{n+2}_{i=2} \frac{1}{n+i}\\
&=\frac1{n+1}-\frac1{2n+1}-\frac1{2n+2}\\
&=\frac1{2n+2}-\frac1{2n+1}\\
&=\frac{(2n+1)-(2n+2)}{(2n+2)(2n+1)}\\
&=\frac{-1}{(2n+2)(2n+1)}\\
&< 0\\
\end{array}
$
Therefore
$x_{n+1}<x_n$.
Convergence is also easily proved
by comparison with
$\frac1{n^2}$
or,
more easily,
$\frac1{n(n+1)}$.
Explicitly,
$x_{n+1}-x_n
=\frac1{2n+2}-\frac1{2n+1}
>\frac1{2n+2}-\frac1{2n}
=\frac12(\frac1{n+1}-\frac1{n})
$
and this telescopes
so that,
if $m > n$,
$\begin{array}\\
x_m-x_n
&=\sum_{k=n}^{m-1} (x_{k+1}-x_k)\\
&>\sum_{k=n}^{m-1} (\frac12(\frac1{k+1}-\frac1{k}))\\
&=\frac12(\frac1{m}-\frac1{n})\\
&>-\frac1{2n}\\
\end{array}
$
so that,
if $m > n$,
$-\frac1{2n}
<x_m-x_n
< 0
$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1994150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Find a shortest distance from a point to an instersection of two implicit surfaces I have a problem where I need to find a point on a curve that is defined by two implicit surfaces. This point must be at the shortest distance from a reference point that is located somewhere in space.
These two surfaces are a cone-like and a sphere-like. The actual equations are quite complex therefore I will not present them here. However, it is important to mention that unlike in the example I provide below, the intersection curve of these surfaces does not fall in to one plane.
implicit formulation for a sphere: $x^2+y^2+z^2=1$
implicit formulation for a cone: $\frac{x^2+y^2}{0.5}=(z+1)^2$
Explanation how to derive an implicit equation of an intersection of any two 3d surfaces would help a lot as well.
EDIT: line -> curve
EDIT: Adding actual case:
The two surfaces are:
Ellipsoid -> $f_2(x,y,z)=\sqrt{\frac{Ex^2}{D}+Fxy+\frac{Dy^2}{E}+Gz^2}-\sqrt{DE}$
Stretched cone -> $f_1(x,y,z)=\frac{(x+y)-(A+B)}{2}+\sqrt{\left(\frac{(x-y)-(A-B)}{2}\right)^2+Cz^2}$
Where $A, B, C, D, E, F, G$ are the shape parameters. Let's assume that these parameters are:
$A=1$, $B=1$, $C=1$, $D=8$, $E=6$, $F=-1$, $G=1$
However they are not limited to those values, it could pretty much be anything (with some restrictions).
It results in two surfaces looking like this:
Note that the z=0 is a plane of symmetry
|
$$
\left \{
\begin{align*}
x^2+y^2+z^2 &= 1 \\
2(x^2+y^2) &= (z+1)^{2}
\end{align*}
\right.$$
The cone here has a nice symmetry, we can eliminate $x$ and $y$ first:
\begin{align*}
2(1-z^2) &= (z+1)^2 \\
3z^2+2z-1 &=0 \\
(z+1)(3z-1) &= 0 \\
z &= -1 \quad \text{or} \quad \frac{1}{3}
\end{align*}
When $z=-1$, $x^2+y^2=0 \implies (x,y)=(0,0)$
When $z=\dfrac{1}{3}$, $$x^2+y^2=\frac{8}{9}$$
Let the reference point be $(X,Y,Z)$, then you need to minimize
$$d=
\sqrt{
\left( X-\frac{2\sqrt{2}}{3} \cos t \right)^2+
\left( Y-\frac{2\sqrt{2}}{3} \sin t \right)^2+
\left( Z-\frac{1}{3} \right)^2}=
\sqrt{
X^2+Y^2-
\frac{4\sqrt{2}}{3} \left( X\cos t+Y\sin t \right)+\frac{8}{9}+
\left( Z-\frac{1}{3} \right)^2}$$
If $X^2+Y^2\ne 0$, choose $t$ such that
$$X\cos t+Y\sin t=\sqrt{X^2+Y^2}$$
Hence minimal value is attained at
$$
\left(
\frac{2\sqrt{2}X}{3\sqrt{X^2+Y^2}},
\frac{2\sqrt{2}Y}{3\sqrt{X^2+Y^2}},
\frac{1}{3}
\right)$$
See more on cone-sphere intersection.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1994233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Prove: $\log_{2}{3} < \log_{3}{6}$ How should I prove that $\log_{2}{3} < \log_{3}{6}$?
I tried something like this: $2^{\log_{2}{3}}< 2^{\log_{3}{6}}$, $3<6^{\log_{3}{2}}$, $\log_{6}{3}<\log_{6}{6^{\log_{3}{2}}}=\log_{3}{2}$, $\frac{1}{\log_{3}{6}}< \log_{3}{2}$. $\frac{1}{1+\log_{3}{2}}<\log_{3}{2}$, but still nothing.
|
This is essentially the same answer as barak manos's and Djura Marinkov's, mostly just presented in a different fashion, as one long string of self-explanatory equalities and inequalities:
$$\begin{align}
5\log_23&=\log_2243\\
&\lt\log_2256\\
&=8\\
&=5+\log_327\\
&\lt5+\log_332\\
&=5\log_33+5\log_32\\
&=5\log_36
\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1994320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
}
|
Hardcore integral with absolute value Is it possible to solve this integral?
$$\int_0^1\int_0^1\frac{{(y-y_1-\frac{(x-x_1)(y_2-y_1)}{x_2-x_1})(y-y_1-\frac{(x-x_1)(y_3-y_1)}{x_3-x_1})(y-y_3-\frac{(x-x_3)(y_2-y_3)}{x_2-x_3})}}{|{(y-y_1-\frac{(x-x_1)(y_2-y_1)}{x_2-x_1})(y-y_1-\frac{(x-x_1)(y_3-y_1)}{x_3-x_1})(y-y_3-\frac{(x-x_3)(y_2-y_3)}{x_2-x_3})}|}dxdy$$
$x_1,x_2,x_3,y_1,y_2,y_3\in[0,1]$
It would be like adding red area and subtracting green area
|
Label the triangle vertices $v_{1} = (x_{1}, y_{1})$, $v_{2} = (x_{2}, y_{2})$, $v_{3} = (x_{3}, y_{3})$, counterclockwise (say), so that the line $\ell_{ij} = \overline{v_{i} v_{j}}$ divides the square into two regions; call the "right-hand" region (not containing the central triangle) $A_{ij}$. Let $a_{ij}$ denote the area of $A_{ij}$, and $a_{0}$ the area of the central triangle.
Up to a sign, the integral is equal to
$$
4a_{0} - 3 + 2(a_{12} + a_{23} + a_{31}).
$$
In a bit more detail, the function
$$
f_{ij}(x, y) = y - y_{i} - \frac{y_{j} - y_{i}}{x_{j} - x_{i}}(x - x_{i})
$$
is (i) only defined if $x_{i} \neq x_{j}$, i.e., if $v_{i}$ and $v_{j}$ do not lie on a vertical line; (ii) positive above the line and negative below (rather than positive to the right of the oriented segment and negative to the left). As long as no two of the vertices lie on a vertical line, however, the integral
$$
\int_{0}^{1} \int_{0}^{1}
\frac{f_{12}(x, y) f_{23}(x, y) f_{31}(x, y)}
{|f_{12}(x, y) f_{23}(x, y) f_{31}(x, y)|}\, dx\, dy
$$
is equal in absolute value to the expression above.
To prove this, label the areas of regions as shown, with $a_{i}$ abutting $v_{i}$ and $b_{i}$ opposite the central triangle from $v_{i}$. We have
$$
\left.
\begin{aligned}
a_{12} &= a_{1} + a_{2} + b_{3} \\
a_{23} &= a_{2} + a_{3} + b_{1} \\
a_{31} &= a_{3} + a_{1} + b_{2}
\end{aligned}\right\}
\tag{1a}
$$
and
$$
1 = a_{0} + a_{1} + a_{2} + a_{3} + b_{1} + b_{2} + b_{3}.
\tag{1b}
$$
Rearranging,
\begin{align*}
2 - 2a_{0} &= 2(a_{1} + a_{2} + a_{3}) + 2(b_{1} + b_{2} + b_{3}), \\
a_{12} + a_{23} + a_{31} &= 2(a_{1} + a_{2} + a_{3}) + b_{1} + b_{2} + b_{3}.
\end{align*}
Subtracting the second from the first,
$$
2 - 2a_{0} - (a_{12} + a_{23} + a_{31}) = b_{1} + b_{2} + b_{3}.
\tag{2}
$$
The blue area minus green area is equal to
\begin{align*}
a_{0} + a_{1} + a_{2} + a_{3} - (b_{1} + b_{2} + b_{3})
&= 1 - 2(b_{1} + b_{2} + b_{3}) \\
&= 1 - 2\bigl[2 - 2a_{0} - (a_{12} + a_{23} + a_{31})\bigr] \\
&= 4a_{0} - 3 + 2(a_{12} + a_{23} + a_{31}).
\end{align*}
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1994675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Partition Proving Let $A=\{2,6,10,14,\ldots\}$ be the set of integers that are twice an odd number.
Prove that, for every positive integer $n$, the number of partitions of $n$ in which no odd number appears more than once is equal to the number of partitions of $n$ containing no element of $A$.
For example, for $n=6$, the partitions of the first type are
$$6,~~~5+1,~~~4+2,~~~3+2+1,~~~2+2+2,$$
and the partitions of the second type are
$$5+1,~~~4+1+1,~~~3+3,~~~3+1+1+1,~~~1+1+1+1+1+1,$$
and there are $5$ of each type.
On first thought, my mind is blank. I simply do not know how to approach this problem. Solutions are greatly appreciated. Thanks in advance!
|
Let $f(n)$ be the partitions of $n$ with at most one occurrence of each odd number, and $F(z)=\sum_{n} f(n)z^n$.
Then $$\begin{align}F(z) &= (1+z)(1+z^2+z^4+\cdots)(1+z^3)(1+z^4+z^8+\cdots)\cdots \\&=\frac{(1+z)(1+z^3)(1+z^5)\cdots}{(1-z^2)(1-z^4)(1-z^6)\cdots}
\end{align}$$
Now pair off the terms $\frac{1+z^{2k+1}}{1-z^{4k+2}}=\frac{1}{1-z^{2k+1}}$, and you get:
$$\prod_{j\notin A}\frac{1}{1-z^j}$$
And this expands to give you the generating function for counting partitions of $n$ with no instance from $A$.
|
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"url": "https://math.stackexchange.com/questions/1995288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Calculating limit of $\lim_{x\to\infty}\dfrac{\sqrt{x+1}-2\sqrt{x+2}+\sqrt{x}}{\sqrt{x+2}-2\sqrt{x}+\sqrt{x-4}}$ As the title says we want to calculate:
$$\lim_{x\to\infty}\dfrac{\sqrt{x+1}-2\sqrt{x+2}+\sqrt{x}}{\sqrt{x+2}-2\sqrt{x}+\sqrt{x-4}}$$
By multiplying nominator and denominator in their conjugates
$=\lim_{x\to\infty}\dfrac{(\sqrt{x+2}+2\sqrt{x}+\sqrt{x-4})(x+1+x+2\sqrt{x(x+1)}-4(x+2))}{(\sqrt{x+1}+2\sqrt{x+2}+\sqrt{x})(x+2+x-4+2\sqrt{(x+2)(x-4)})-4x)}$
$=\lim_{x\to\infty}\dfrac{(\sqrt{x+2}+2\sqrt{x}+\sqrt{x-4})(-2x-7+2\sqrt{x^2+x})}{(\sqrt{x+1}+2\sqrt{x+2}+\sqrt{x})(-2x-2+2\sqrt{x^2-2x-8})}$
I think now we can take $$2x\approx2\sqrt{x^2+x}\approx2\sqrt{x^2-2x-8}\\[2ex]
\sqrt{x}\approx\sqrt{x+1}\approx\sqrt{x+2}\approx\sqrt{x-4}$$ as $x$ goes to infinity. Hence the limit of above fraction would be $\dfrac{7}{2}$, but wolframalpha gives me $\dfrac{3}{2}$ as the limit of the above fraction.
What am I doing wrong?
|
You can write the fraction as $$\frac{\sqrt{x+1}-2\sqrt{x+2}+\sqrt{x}}{\sqrt{x+2}-2\sqrt{x}+\sqrt{x-4}}=\frac{(\sqrt{x+1}-\sqrt{x+2})+(\sqrt{x}-\sqrt{x+2})}{(\sqrt{x+2}-\sqrt{x})+(\sqrt{x-4}-\sqrt{x})}=\frac{-\frac{1}{\sqrt{x+2}+\sqrt{x+1}}-\frac{2}{\sqrt{x}+\sqrt{x+2}}}{\frac{2}{\sqrt{x+2}+\sqrt{x}}-\frac{4}{\sqrt{x-2}+\sqrt{x}}}=\frac{-\frac{\sqrt{x}}{\sqrt{x+2}+\sqrt{x+1}}-\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{x+2}}}{\frac{2\sqrt{x}}{\sqrt{x+2}+\sqrt{x}}-\frac{4\sqrt{x}}{\sqrt{x-2}+\sqrt{x}}},$$ and as $x\to\infty$, this converges to $$\frac{-\frac{1}{2}-\frac{2}{2}}{\frac{2}{2}-\frac{4}{2}}=\frac{3}{2}.$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1998813",
"timestamp": "2023-03-29T00:00:00",
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|
Proving $∠CAD = 90◦$ In a cyclic quadrilateral $ABCD$, let the diagonals $AC$ and $BD$ intersect at
$X$. Let the circumcircles of triangles $AXD$ and $BXC$ intersect again at $Y$.
If $X$ is the incentre of triangle $ABY$ , show that $∠CAD = 90^{\circ}$.
|
First we show that point $Y$ lies on the edges $CD$. Look at quadrilateral $XCYD$. We will prove that $\angle \, CYD = 180^{\circ}$.
*
*$\angle\, XDY = \angle \, XAY = \alpha$ as inscribed in a circle.
*$\angle \, BAC = \angle \, BAX = \angle \, XAY = \alpha$ since $AC$ passes through the incenter $X$ of triangle $ABY$ and therefore $AC$ is the interior angle bisector of angle $\angle \, BAY$.
*$\angle \, XDC = \angle \, BDC = \angle \, BAC = \alpha$ as inscribed in a circle.
*Hence $\angle \, XDC = \angle \, XDY = \alpha$.
*Analogously, $\angle \, XCD = \angle \, XCY = \beta$.
*In triangle $CDX$ angles sum up to $180^{\circ}$ so $$180^{\circ} = \angle \, CXD + \angle \, XCD + \angle \, XDC = \angle \, CXD + \alpha + \beta$$ and thus $\angle \, CXD = 180^{\circ} - \alpha - \beta$.
*In quadrilateral $XCDY$ angles sum up to $360^{\circ}$ so $$360^{\circ} = \angle \, CXD + \angle \, XCY + \angle \, XDY + \angle \, CYD = \angle \, CXD + \alpha + \beta + \angle \, CYD = $$ $$= 180^{\circ} + \angle \, CYD $$ therefore $\angle \, CYD = 180^{\circ}$ and thus $Y$ lies on $CD$.
Now, we are ready to prove that $\angle \, CAD = 90^{\circ}$.
*$\angle \, AXD = \angle \, BXC$ since $X$ is the intersection point of $AC$ and $BD$.
*$\angle \, AYD = \angle \, AXD$ and $\angle \, BYC = \angle \, BXC$ as inscribed in corresponding circles. Hence $\angle \, AYD = \angle \, BYC$
*$\angle \, XYA = \angle \, XYB$ since $YX$ is angle bisector of $\angle \, AYC$. Hence
$$\angle \, XYD = \angle \, XYA + \angle \, AYD = \angle \, XYB + \angle \, BYC = \angle \, XYC$$
*But $180^{\circ} = \angle \, XYD + \angle \, XYC = 2 \, \angle \, XYD $ so $\angle \, XYD = 90^{\circ}$.
*As quadrilateral $AXYD$ is inscribed in a circle, $180^{\circ} = \angle \, XAD + \angle \, XYD = \angle \, XAD + 90^{\circ}$ so $\angle \, XAD = 90^{\circ}$.
*$X \in AC$ so $\angle \, CAD = \angle \, XAD = 90^{\circ}$.
|
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"url": "https://math.stackexchange.com/questions/2002089",
"timestamp": "2023-03-29T00:00:00",
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|
simplify $\sqrt{(45+4\sqrt{41})^3}-\sqrt{(45-4\sqrt{41})^3}$ simplify $\sqrt{(45+4\sqrt{41})^3}-\sqrt{(45-4\sqrt{41})^3}$.
1.$90^{\frac{3}{2}}$
2.$106\sqrt{41}$
3.$4\sqrt{41}$
4.$504$
5.$508$
My attempt:I do like this but I didn't get any of those five.
$\sqrt{(45+4\sqrt{41})^3}-\sqrt{(45-4\sqrt{41})^3}={\sqrt{45+4\sqrt{41}}}^3-\sqrt{45-4\sqrt{41}}^3$
$=(\sqrt{45+4\sqrt{41}}-\sqrt{45-4\sqrt{41}})(45+4\sqrt{41}+45-4\sqrt{41}+(\sqrt{45+4\sqrt{41}})(\sqrt{45-4\sqrt{41}})$
Now I do the nested radicals formula and I get $254\sqrt{41}$ which is none of those where did I mistaked?
|
Is this the nested radical formula?
$\sqrt{45 \pm 4\sqrt{41}} = a \pm b\sqrt{41}$
$45 \pm 4\sqrt{41} = (a^2 + 41b^2) \pm 2ab\sqrt{41}$
$a^2 + 41b^ = 45; 2ab = 4 \implies a=2;b = 1$
So $\sqrt{45 \pm 4\sqrt{41}} = |2 \pm \sqrt{41}|= \pm 2 + \sqrt{41}$
Plugging that into: $=(\sqrt{45+4\sqrt{41}}-\sqrt{45-4\sqrt{41}})(45+4\sqrt{41}+45-4\sqrt{41}+(\sqrt{45+4\sqrt{41}})(\sqrt{45-4\sqrt{41}}))$
We get $(2+\sqrt{41} - (-2+\sqrt{41}))(90+(2 + \sqrt{41})(-2+\sqrt{41}))=$
$4(90 + (-4 + 41)) = 508$
===
It would have been easier not to do all the expanding:
$(2 + \sqrt{41})^3 - (-2+\sqrt{41})^3=$
$(2^3 + 3*2^2*\sqrt{41} + 3*2*\sqrt{41}^2+\sqrt{41}^3)- (-2^3 + 3*2^2*\sqrt{41} - 3*2*\sqrt{41}^2+\sqrt{41}^3)=$
$2(8 + 6*41) = 508$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate the integral $\int \:\frac{\sqrt{x^2+2\cdot\:\:x+2}}{x}dx$ $$\int \:\frac{\sqrt{x^2+2\cdot\:\:x+2}}{x}dx$$
I have tried to form a square above i also tried to get the x below under the root
but got nothing
|
Rationalise the numerator.
$\displaystyle \int \frac{\sqrt{x^2+2x+2}}{x} \, \text{d}x = \int \frac{x^2+2x+2}{x\sqrt{x^2+2x+2}}\, \text{d}x$
Split apart as follows:
$\displaystyle \int \frac{x^2+2x+2}{x\sqrt{x^2+2x+2}}\, \text{d}x = \int \frac{x+1}{\sqrt{x^2+2x+2}} \, \text{d}x + \int \frac{\text{d}x}{\sqrt{x^2+2x+2}} + \int \frac{2\,\text{d}x}{x\sqrt{x^2+2x+2}}$
The first is an application of the reverse chain rule, the second is an application of the inverse hyperbolic sine formula, and the third requires the substitution $\displaystyle u = \frac{2}{x}$.
$\displaystyle \int \frac{\sqrt{x^2+2x+2}}{x} \, \text{d}x = \sqrt{x^2+2x+2} + \log{\left(x+1+\sqrt{x^2+2x+2} \right)} - \int \frac{\frac{4 \text{d}u}{u^2}}{\frac{2}{u} \sqrt{\frac{4}{u^2}+\frac{4}{u}+2}} $
$\displaystyle \int \frac{\frac{4 \text{d}u}{u^2}}{\frac{2}{u} \sqrt{\frac{4}{u^2}+\frac{4}{u}+2}} = \int \frac{\sqrt{2}\,\text{d}u}{\sqrt{u^2+2u+2}} = \sqrt{2} \log{\left( u+1+\sqrt{u^2+2u+2} \right)}$
$\displaystyle \int \frac{\sqrt{x^2+2x+2}}{x} \, \text{d}x = \sqrt{x^2+2x+2} + \log{\left(x+1+\sqrt{x^2+2x+2} \right)} - \sqrt{2} \log{\left( \frac{2}{x}+1+\sqrt{\frac{4}{x^2}+\frac{4}{x}+2} \right)} $
$\displaystyle = \sqrt{x^2+2x+2} + \log{\left(x+1+\sqrt{x^2+2x+2} \right)} +\sqrt{2}\log{x} - \sqrt{2}\log{\left( x+2+\sqrt{2}\sqrt{x^2+2x+2} \right)}$
|
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|
Find and prove the limit $\lim_{n \to \infty}\frac{2^{n+2}+3^{n+3}}{2^n+3^n}$ I'm trying to find and prove the limit $$\lim_{n \to \infty}\frac{2^{n+2}+3^{n+3}}{2^n+3^n}$$
which according to WolframAlpha happens to be $27$.
The furthest I could get is
$$\frac{2^{n+2}+3^{n+3}}{2^n+3^n} = 4\left(1 + \frac{3^{n}}{2^n+3^n}\right)$$
|
$$\frac{2^{n+2}+3^{n+3}}{2^n+3^n}=\frac{(4)2^{n}+(27)3^{n}}{2^n+3^n}=\frac{(4)2^{n}/3^n+(27)1}{2^n/3^n+1}$$
so the limit is $27$
|
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|
Infimum of complex expression on the unit disc in the complex plane Suppose $|\lambda|<1, Re\lambda < 0, Im\lambda > 0$. What is the largest lower bound independent of $\lambda$ on the expression
$$\frac{1+|\lambda|^2}{2}+Re\lambda - \frac{2\pi}{n+2}Im\lambda$$ where $n$ is a positive integer? If you like, stipulate that $n$ be as large as you need for the problem to make sense, though make it as small of a lower bound on $n$ as you can, if that makes sense.
|
This is equivalent to the bivariate real optimization problem:
\begin{array}{ll}
\text{minimize} & \frac{1+x^2+y^2}{2}+x-\frac{2\pi}{n+2} y \\
\text{subject to} & x^2 + y^2 \leq 1
\end{array}
To solve, we build the Lagrangian
$$L(x,y,z) = \frac{1+x^2+y^2}{2}+x-\frac{2\pi}{n+2} y - z(1 - x^2 - y^2)$$
where $z \geq 0$ is the Lagrange multiplier. The optimality conditions are
$$x + 1 + 2zx = 0 \quad y - \frac{2\pi}{n+2} + 2zy = 0 \quad x^2+y^2\leq 1$$
Rearranging,
$$x = \frac{-1}{1+2z} \quad y = \frac{2\pi}{(1+2z)(n+2)}$$
Note that if we were to guess that $z=0$---its smallest legal value---then we'd see that $x=1$ and $x^2+y^2>1$. Therefore it must be the case that $z>0$, and complementary slackness tells us this implies $x^2+y^2=1$. So we solve:
$$x^2 + y^2 = \frac{1}{(1+2z)^2} \left( 1 + \frac{4\pi^2}{(n+2)^2}\right) = 1$$
which leads us to
$$\frac{1}{1+2z} = (1+(4\pi^2)/(n+2)^2)^{-1/2}$$
Substitution yields
$$x = \frac{-1}{(1+(4\pi^2)/(n+2)^2)^{1/2}}, \quad y = \frac{2\pi}{(n+2)(1+(4\pi^2)/(n+2)^2)^{1/2}}$$
Then bring this back into the expression: Since $x^2+y^2=1$, that first term is equal to 1, and we have
$$1 - \frac{1+(4\pi^2)/(n+2)^2}{(1+(4\pi^2)/(n+2)^2)^{1/2}} = 1 - (1+(4\pi^2)/(n+2)^2)^{1/2}$$
|
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|
Induction divisibility question Q. Prove by induction that $2^{3n-1} + 5(3^n)$ is divisible by $11$ for any even number $n$, where $n$ is an element of natural numbers.
What is have so far:
(base case): $p(2) = 77$, $77/11 = 7$. so base case holds
$p(k) = 2^{3k-1} +5(3^k) $
$p(k+2) = 2^{3k+5} + 5(3^{k+2}) $
$p(k+2) = 2^{3k+1}2^4 + 5(3^{k})(3^2) $
$p(k+2) = 2^{3k+1+(1-1)}2^4 + 5(3^{k})3^2 $
$p(k+2) = 2^{3k-1}2^6 +5(3^{k})3^2$
I am new to induction and I don't know how to continue.
A point in the right direction would be greatly appreciated, thank you.
|
You want to relate $p(k+2)$ back to $p(k)$, and you've gotten most of the way there (though you could've gone straight from $p(k+2)=2^{3k+5}+5(3^{k+2})$ to $p(k+2)=2^{3k-1}2^6+5(3^k)(3^2)$).
It would be nice if instead of $2^{3k-1}2^6+5(3^k)(3^2)$ you had something like $2^{3k-1}2^6+5(3^k)(2^6)$ because then you could just factor this as $2^6\cdot p(k)$, which is divisible by $11$. If we try to re-write $p(k+2)$ to get those terms we want, we get
\begin{align*}
p(k+2)&=2^{3k-1}2^6+5(3^k)(3^2)\\
&=2^{3k-1}2^6+5(3^k)(2^6)-5(3^k)(2^6)+5(3^k)(3^2)\\
&=2^6\cdot p(k) +5(3^k)(3^2-2^6).
\end{align*}
We know $p(k)$ is divisible by $11$, so this sum is divisible by $11$ if and only if the second term $5(3^k)(3^2-2^6)$ is divisible by $11$. Simplify the powers and see what you get.
|
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|
Solving simultaneous equations involving a quadratic I have the question
Solve the simultaneous equation pair
$$x^2 + y^2 = 25\tag1$$
$$2x - y = 5\tag2$$
I have found the value of $y$ from the second equation which is $2x-5$ and substituted this into the first equations $y$ value.
I get $x^2 + (2x -5)^2 = 25$
When I expand the brackets I get the equation
$$x^2+4x^2-20 =0\tag3$$
However, when I checked the solutions the equation should simplify to $x^2 - 4x = 0$ and I do not understand how this is achieved.
|
$$
\begin{cases}
\text{x}^2+\text{y}^2=25\\
2\text{x}-\text{y}=5
\end{cases}\space\Longleftrightarrow\space
\begin{cases}
\text{x}^2+\left(2\text{x}-5\right)^2=25\\
\text{y}=2\text{x}-5
\end{cases}
$$
Solving:
$$\text{x}^2+\left(2\text{x}-5\right)^2=25$$
Gives us $x=0$ or $x=4$
|
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|
Show that $b = ln(\sqrt[3]{\frac{18}{27-k}})$ for $9 = \frac{k}{3-2e^{-b3}}$ I did:
$$9 = \frac{k}{3-2e^{-b3}} \Leftrightarrow \frac{k}{9} = 3-2e^{b3} \Leftrightarrow \frac{k}{9}-\frac{3}{1} = -2e^{-3b} \Leftrightarrow \frac{\frac{k-27}{9}}{\frac{-2}{1}} \Leftrightarrow -\frac{k-27}{18} = \frac{1}{e^{2b}} \Leftrightarrow -\frac{18}{k-27} = e^{3b} \Leftrightarrow \sqrt[3]{-\frac{18}{k-27}} = e^b \Leftrightarrow b = ln{(\sqrt[3]{-\frac{18}{k-27}})}$$
But my book says that the solution if $b = ln(\sqrt[3]{\frac{18}{27-k}})$
It seems I changed the sign somewhere, but I cannot see where.
What did I do wrong?
|
you are there.
$$
\frac{1}{a-b} = -\frac{1}{b-a}
$$
|
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|
How to find x and y coordinates based on the given distance? The problem says: Find the point (coordinates $(x,y)=~?$) which is symmetrical to the point $(4,-2) $ considering the given equation $y=2x-3$
I have found the perpendicular line-slope $y=-~\frac{1}{2}x$ and the
intersection point which is shown in the graph $\left(\frac{6}{5},\frac{-3}{5}\right)$
I'm somehow unable to find the $x$ and $y$ I have found the distance based on the point and point distance equation:$$d(p_1,p_2)= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{\left(\frac{6}{5}-4\right)^2+\left(-\frac{3}{5}+2\right)^2}= \frac{7\sqrt{5}}{5}$$
$$d_1=d_2=d(p_1,p_2)$$
So now I know the distance, the min. distance to the unknown point is the same.
What is the easiest way to find the symmetrical $x$ and $y$ coordinates?
(This is a high school problem)
|
I think you're over-thinking it. When you have the intersect point and another point, you just duplicate the difference to get the point on the other side.
$(\frac{6}{5}, -\frac{3}{5}) - (4, -2)$ is $(-\frac{14}{5}, \frac{7}{5})$
That is what you need to add to the point to get to your intersect point, so either add that do your intersect point, or double it and add it to the initial point to get the point on the other side:
$(-\frac{8}{5}, \frac{4}{5})$
Another option would just be to calculate the x difference and double it to get the opposite x coordinate and pop that into your perpendicular line equation.
Full:
Initial equation: $y = 2x - 3$
Point: $(4, -2)$
So any equation perpendicular to the initial one will have slope $-\frac{1}{2}$ as you suggest. You might have already found the full equation for the line to find your point, but plug in the $y$ value for the given point of -2 and solve for b:
$y = -\frac{1}{2}x$ + b
$b=y +\frac{1}{2}x$
$b=-2 + \frac{1}{2}4$
$b= 0$
So the formula for the perpendicular line passing through point $(4, -2)$ is $y = -\frac{1}{2}x$
Solving to find the intersection point to get the $x$ you set both equations to be equal and solve for x:
$2x - 3 = -\frac{1}{2}x$
$\frac{5x}{2} = 3$
$5x = 6$
$x = \frac{6}{5}$
Plugging that into either equation gives your intersect point:
$(\frac{6}{5},-\frac{3}{5})$
Ok, so now how do you find a point the same distance on the perpendicular line passing through $(4, -2)$? To get from $(4, -2)$ to $(\frac{6}{5}, -\frac{3}{5})$ you have to move $\frac{6}{5} - 4$ in the x direction and $-\frac{3}{5} - -2$ in the y direction, or $(-\frac{14}{5}, \frac{7}{5})$
Add that to the intersect point and you get the point on the opposite side, $(-\frac{8}{5}, \frac{4}{5})$. Plug the x value into your perpendicular equation:
$y = -\frac{1}{2}x$
$y = (-\frac{1}{2}) (-\frac{8}{5}$)
$y = \frac{8}{10}$
$y = \frac{4}{5}$
|
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|
What is the mistake? $$1=1$$
$$\Rightarrow\frac{-1}{1}=\frac{1}{-1}$$
$$\Rightarrow \sqrt{\frac{-1}{1}}=\sqrt{\frac{1}{-1}}$$
$$\Rightarrow\frac{i}{1}=\frac{1}{i}$$
$$\Rightarrow\frac{i}{2}=\frac{1}{2i}$$
$$\Rightarrow\frac{i}{2}+\frac{3}{2i} = \frac{1}{2i} +\frac{3}{2i}$$
$$\Rightarrow i(\frac{i}{2}+\frac{3}{2i} ) = i(\frac{1}{2i} +\frac{3}{2i})$$
$$\Rightarrow\frac{-1}{2}=\frac{1}{2}$$
$$\Rightarrow1=2$$
What is wrong in this?
|
The trick is assuming that
$$\sqrt{\frac{-1}{1}} = \sqrt{\frac{1}{-1}}\Rightarrow
\frac{\sqrt{-1}}{\sqrt{1}} = \frac{\sqrt{1}}{\sqrt{-1}}$$
which is not true because by definition of (complex) square root $-1=\sqrt{-1}\cdot \sqrt{-1} \not=\sqrt{1}\cdot \sqrt{1}=1$.
On the other hand, note that if $a,b,c,d$ are real POSITIVE numbers then
$$\sqrt{\frac{a}{b}}=\sqrt{\frac{c}{d}}\Leftrightarrow
\frac{\sqrt{a}}{\sqrt{b}} = \frac{\sqrt{c}}{\sqrt{d}}.$$
|
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|
Maximum value of algebraic expression
If $2a+3b+4c=100$ and $a,b,c\in \mathbb{N}\;,$ Then $\max(2a^2+5b^2+8c^2)$
$\bf{My\; Try::}$ We can write it as $\displaystyle \frac{a}{50}+\frac{b}{\frac{100}{3}}+\frac{c}{25} = 1$
So it represent a plane which cut $x,y,z$ axis at $\displaystyle A(50,0,0)$ and $\displaystyle B\left(0,\frac{100}{3},0\right)$ and $C(0,0,25)$
Now how can i solve it after that, Help required, Thanks
|
First note that $2a+3b + 4c=100 \implies b \equiv 0 \pmod 2$ and $2a+c \equiv 1 \pmod 3$.
As $b$ is even, we may simplify a bit by letting $b = 2k$, then $a+3k+2c = 50$ and we maximize $V = a^2+10k^2+4c^2$.
As $50^2 = a^2+9k^2+4c^2 + 2(3ak+6kc+2ca)$, we have
$$V= k^2+50^2 - 2(3ak+6kc+2ca) = (k-(3a+6c))^2+50^2 - 4ca - (3a+6c)^2$$
So for any choice of $a, c$, setting $k=3a+6c$ is clearly the best choice.
We are left to minimise $4ca + (3a+6c)^2$, which is clearly increasing in $c, a$ with more sensitivity on $c$. The first choice for minimum is thus $(a, c)=(1, 1)$ but this fails $\pmod 3$, the next best choice is $(2, 1)$ which also doesn't work, but $(a, c) = (1, 2)$ works. So does $(3, 1)$ but this has a slightly higher value, so no further investigation is needed.
Thus the maximum you seek is when $a=1, b=2k=30, c=2$.
|
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|
Find the value of $\lim\limits_{n \to \infty}n\left(\left(\int_0^1\frac{1}{1+x^n}\,\mathrm{d}x\right)^n-\frac{1}{2}\right)$
Find the value of the limit $$\lim_{n \to \infty}n\left(\left(\int_0^1 \frac{1}{1+x^n}\,\mathrm{d}x\right)^n-\frac{1}{2}\right)$$
I can't solve the integral $\int_0^1 \mathrm{\frac{1}{1+x^n}}\,\mathrm{d}x$. But maybe the questions doesn't require solving the integral.
Apparently the $\lim_{n \to \infty}(\int_0^1 \mathrm{\frac{1}{1+x^n}}\,\mathrm{d}x)^n$ should be $\frac{1}{2}$ for the question to make sense. That's all I know.
|
Hint. If one knows the digamma function $\psi=\Gamma'/\Gamma$, one may write, as $n \to \infty$,
$$
\begin{align}
\int_0^1\frac1{1+x^n}\:dx&=\int_0^1\frac{1-x^n}{1-x^{2n}}\:dx
\\\\&\stackrel{u=x^{2n}}=\frac1{2n}\int_0^1\frac{(1-u^{1/2})u^{1/(2n)-1}}{1-u}\:du
\\\\&=\frac1{2n}\left[\psi\left(\frac1{2n}+\frac12 \right)-\psi\left(\frac1{2n} \right) \right]
\\\\&=1-\frac{\ln 2}{n}+\frac{\pi ^2-6 \ln^2 2}{12n^2}+O\left(\frac{1}{n^3}\right)
\end{align}
$$ giving, as $n \to \infty$,
$$
n\ln \left(1-\frac{\ln 2}{n}+\frac{\pi ^2-6 \ln^2 2}{12n^2}+O\left(\frac{1}{n^3}\right) \right)=-\ln 2+\frac{\pi ^2-6 \ln^2 2}{24n}+O\left(\frac{1}{n^2}\right)
$$
that is
$$
n\left[\left(\int_0^1\frac1{1+x^n}\:dx \right)^n-\frac12\right]\to \frac{\pi ^2-6 \ln^2 2}{24}.
$$
|
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|
another trig limit without L'Hospital? $$\lim_{x\to \pi/3 }\dfrac{1-2\cos\left(x\right)}{{\pi}-3x}$$
Here's what I tried:
${\pi}-3x=y$,
$\dfrac{{\pi}-y}{3}=x$
$$\lim_{y\to\ 0} \dfrac{1-2\cos\left(\frac{{\pi}-y}{3}\right)}{y}$$
...
|
Here is a proof without L'Hôpital's rule or derivatives. The two basic trigonometric limits are
\begin{align*}
\lim_{\theta\to 0} \frac{\sin\theta}{\theta} &= 1\\
\lim_{\theta\to 0} \frac{1-\cos\theta}{\theta} &= 0
\end{align*}
The first can be shown geometrically, and the second by an algebraic manipulation of the first.
You got to:
$$\lim_{y\to\ 0} \dfrac{1-2\cos\left(\frac{{\pi}-y}{3}\right)}{y}$$
I would expand the cosine term:
\begin{align*}
\cos\left(\frac{\pi}{3} - \frac{y}{3}\right)
&= \cos\left(\frac{\pi}{3}\right)\cos\left(\frac{y}3\right) + \sin\left(\frac{\pi}{3}\right)\sin\left(\frac{y}3\right) \\
&= \frac{1}{2} \cos\left(\frac{y}3\right) + \frac{\sqrt{3}}{2}\sin\left(\frac{y}3\right)
\end{align*}
So
\begin{align*}
\dfrac{1-2\cos\left(\frac{{\pi}-y}{3}\right)}{y}
&= \frac{1-\cos(y/3)-\sqrt{3}\sin(y/3)}{y} \\
&= \frac{1-\cos(y/3)}{y} - \sqrt{3}\frac{\sin(y/3)}{y}
\end{align*}
Let $u =y/3$, so that $y=3u$. We get
\begin{align*}
\frac{1-\cos(y/3)}{y} - \sqrt{3}\frac{\sin(y/3)}{y}
&= \frac{1-\cos u}{3u} - \sqrt{3}\frac{\sin(u)}{3u}
\\&= \frac{1}{3} \cdot \frac{1-\cos u}{u} - \frac{1}{\sqrt{3}} \frac{\sin u}{u}
\end{align*}
As $y\to 0$, $u\to 0$, so the expression on the right tends to
$$
\frac{1}{3} \cdot 0 - \frac{1}{\sqrt{3}}\cdot 1 = -\frac{1}{\sqrt{3}}
$$
|
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|
Given ${[1+ {(1+ x)}^{1/2}]×\tan(x) = \left[1+ {(1- x)}^{1/2}\right]}$, find $\sin 4 x$. If
$${[1+ {(1+ x)}^{1/2}]×\tan(x) = \left[1+ {(1- x)}^{1/2}\right]}$$
then find the value of $\sin(4x)$.
The options given are:
a) $x$
b) $4x$
c) $2x$
I tried applying many trigo identities but none of them is working and the radicals are posing a big problem...need help...Thanks!!
|
Let: $$t = \tan(x) = \frac{1+\sqrt{1-x}}{1+\sqrt{1+x}}$$
Using the double-angle sine and half-angle tangent formulas:
$$
\sin(4x) \;=\; 2\;\sin(2x)\;\cos(2x) \;=\; 2 \cdot \frac{2 t}{1 + t^2} \cdot \frac{1-t^2}{1+t^2} \;=\; \frac{4t(1-t^2)}{(1+t^2)^2}
$$
Substituting back $t$ in terms of $x$ and simplifying:
$$
\require{cancel}
\begin{align}
\sin(4x) & =\; \frac{2 x (\sqrt{1-x^2}+2 \sqrt{1-x}+2 \sqrt{1+x}+3)}{(\sqrt{1-x}+\sqrt{1+x}+2)^2} \\
& =\; \frac{2 x (\sqrt{1-x^2}+2 \sqrt{1-x}+2 \sqrt{1+x}+3)}{(1-x) + (1+x) + 4 + 2 \sqrt{(1-x)(1+x)} + 4 \sqrt{1-x} + 4 \sqrt{1+x} } \\
& =\; \frac{\cancel{2} x \bcancel{(\sqrt{1-x^2}+2 \sqrt{1-x}+2 \sqrt{1+x}+3)}}{\cancel{2}\bcancel{(3 + \sqrt{1-x^2} + \sqrt{1-x}+\sqrt{1+x})} } \\
& =\; x
\end{align}
$$
|
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|
How many squares are of the form $x^3+7x^2-x$?
How many squares are of the form $x^3+7x^2-x$, where $x$ is a positive integer, that are not multiples of $16$?
Since $x^3+7x^2-x = x(x^2+7x-1)$, we need both $x$ and $x^2+7x-1$ to be perfect squares and so let $x = k^2$, for some $k \in \mathbb{Z}^+$ . Then we need $(k^2)^2+7k^2-1 = k^4+7k^2-1$ to be a perfect square that is not a multiple of $16$. How do we do this?
|
@ovi Why both $x$ and $ x^2+7x-1$ must be a square? Because they are coprime.
From $x^2+7x-1=k^2$ we get $x^2+7x-1-k^2=0$ and the discriminant is $53+4k^2 = m^2 $. From here $53=(m-2k)(m +2k)$ and $53$ is a prime number. It follows $m+2k = 53, m-2k = 1$ and $m=27, k=13$. Therefore $x^2+7x-170=0$ and, from here, $x=10$ but $10$ is not a perfect square.
|
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Diophantine equation $x^4+5y^4=z^4$ I am trying to find all positive integer solution $(x,y,z)$ of equation $x^4+5y^4=z^4$.
Here I fould: $(x,y,z)=(1,2,3)$ and $(d,2d,3d)$. I try to prove if $(x,y)=1$ then $(1,2,3)$ is the unique solution of equation. Could anyone help me for this question?
|
I experimentally searched the possibility of more principal solutions for this equation.
$x^4 +5y^4=z^4$
$z^4≡ x^4\mod y^4$⇒ $x^4<y^4$ or $x<y$
$z^4≡ x^4\mod 5$ ⇒ $x^4 < 5$; the only possible number for x is $1$, that is $x=1$⇒ $(y=2)$ and $z=3$ are the principal solutions.
Also we may write:
$(z^2+x^2)(z+x)(z-x)=5y^4$
Following cases can be considered:
a: $z-x =2$, that is z and x are two consecutive odd or even numbers. One solution is already found i.e. $(z=3)$ and $(x =1)$ and $y=2$. I checked by Python up to $z=10^6$ and got no more solutions.
b: $z-x=1$, that is x and z are two consecutive numbers. This can not be acceptable because $z>y>x$ i.e. the next consecutive number to x can only be y.
c: $z-x=5$, and $(z^2+x^2)(z+x)=y^4$; I checked this up to $z=10^6$ and I got no result.
c:$x+z=5$ that means:we have :$(x=1, z=4), (x=2, z=3), (x=3, z=2) and finally(x=4, z=1)$ which non of them give integer solution for y.
d:$z^2+x^2=5$ , the possible values are $(x=1)$ and$( z=2)$ which do not give an integer solution for y.
e: $z^2-x^2=5$ or $z^2+x^2=y^4$
I checked it up to $ z=10^6$ and could not found any integer solution for y.
Conclusion: equation $x^4 +5y^4=z^4$ is one case of general equation of form $x^4 +Dy^4=z^4$ in which $D=5$ . Lagrange showed that this type of equation depending on certain values of D can have integer solutions in limited numbers.The equation $x^4 +5y^4=z^4$ is one of these equations that have only one set of principal solutions $(x=1, y=2, z=3)$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Taylor expansion for $\frac{1}{x^2 + 2x + 2}$ I'm trying to find the Taylor expansion of the function:
$$ f(x) = \frac{1}{x^2 + 2x + 2} $$
about the point $ x = 0 $. I have worked out the terms up to the fourth derivative, which was very tedious. I found:
$$ f(x) = \frac{1}{2} - \frac{1}{2} x + \frac{1}{4} x^2 + 0 x^3 - \frac{1}{8} x^4 + O \left( x^5 \right) $$
I notice powers of two in the denominator, but I'm not sure of the pattern (and to calculate the next term to confirm would entail another tedious product rule).
Any ideas? Thanks!
|
$$ f(x) = \frac{1}{x^2 + 2x + 2} = a_0 + a_1 x + a_2 x^2 + \cdots $$
\begin{align*}
&& (x^2 + 2x + 2)f(x) &= 1 \quad ({}+ 0 x + 0x^2 + 0x^3 + \cdots) \\
&\implies& 2a_0 &= 1 \\
&& 2a_1 + 2a_0 &= 0 \\
&& 2a_2 + 2a_1 + a_0 &= 0 \\
&& 2a_3 + 2a_2 + a_1 &= 0 \\
&& 2a_4 + 2a_3 + a_2 &= 0 \\
&& 2a_5 + 2a_4 + a_3 &= 0 \\
&&& \vdots \\
&\implies& a_0 &= \frac{1}{2} \\
&& a_1 &= \frac{-1}{2} \\
&& a_2 &= \frac{1}{4} \\
&& a_3 &= \frac{1}{2}\left(\frac{2}{4} + \frac{-1}{2}\right) = 0 \\
&&& \vdots \\
&& a_{k \geq 3} &= \frac{1}{2}\left(2 \cdot a_{k-1} + a_{k-2}\right) \\
&&& \vdots \\
\end{align*}
|
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|
A limit without invoking L'Hopital: $\lim_{x \to 0} \frac{x \cos x - \sin x}{x^2}$ The following limit
$$\ell=\lim_{x \rightarrow 0} \frac{x \cos x - \sin x}{x^2}$$
is a nice candidate for L'Hopital's Rule. This was given at a school before L'Hopital's Rule was covered. I wonder how we can skip the rule and use basic limits such as:
$$\lim_{x \rightarrow 0} \frac{\sin x}{x} \quad , \quad \lim_{x \rightarrow 0} \frac{\cos x -1}{x^2}$$
|
The function $\frac{x\cos(x)-\sin(x)}{x^2}$ is odd, so we only need to look at $0\lt x\lt\frac\pi2$. As shown in this answer, $0\le\sin(x)\le x\le\tan(x)$. Furthermore, $x\cos(x)-\sin(x)=(x-\tan(x))\cos(x)\le0$. So we have
$$
\begin{align}
0
&\ge\color{#C00000}{\frac{x\cos(x)-\sin(x)}{x^2}}\\
&\ge\frac{\sin(x)(\cos(x)-1)}{x^2}\\
&=-\frac{\sin(x)}x\frac{1-\cos^2(x)}{x(1+\cos(x))}\\
&=-\frac{\sin^3(x)}{x^2(1+\cos(x))}
\end{align}
$$
By the Squeeze Theorem, we have
$$
\lim_{x\to0}\frac{x\cos(x)-\sin(x)}{x^2}=0
$$
|
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|
Arithmetic proof of absolute value function of complex numbers I am looking for the arithmetic proof that:
$ |z| = \sqrt{(x^2 + y^2)} $ where $ z = x + i y $
Previously I assumed squaring a function then square rooting it would be analogous to the absolute value function (modulus) but it seems not to be the case in the complex domain. Consider the following simple counter example:
Let $ z = \cos(x) + i \times \sin (x) $
$|z| = |\cos(x) + i \times \sin(x)| = \sqrt{((\cos(x) + i \times \sin(x))^2} $
$ = \sqrt{2 \times i \times \cos(x) \times \sin(x) + \cos^2(x) +i^2 \times \sin^2(x) }$
$ = \sqrt{2i \times \cos(x)\sin(x) + \cos^2(x) - \sin^2(x) } \neq \cos^2(x) + \sin^2(x)$
Why is that?
|
If $|x|=\sqrt {x^2} $ for $x\in \mathbb R $ is not a definition but a result.
$\mathbb R\subset \mathbb C$ and $|x|$ is still defined to be $|x|=|x+0i|=\sqrt {x^2+0^2}=\sqrt{x^2} $.
The definition is $|z|=\sqrt {\text {Re} (z)^2+\text {Im}(z)^2} $ whether $z$ is purely real, complex, or purely imaginary.
In general $|z|\ne \sqrt{z^2} $ unless $\text {Im}(z)=0$ in which case $z=\text {Re}(z) $. So that holds if and only if $z$ is real.
=========
Note: $\sqrt {(a+bi)^2}=\sqrt {(a^2-b^2)+2abi} \ne \sqrt {a^2+b^2} $ UNLESS $2ab=0$ and $b^2=-b^2$. This happens if and only if $b=0$.
So $|z| = |a+bi|=\sqrt {a^2+b^2}=\sqrt {(a+bi)^2}=\sqrt {z^2} $ if and only if $b=0$ if and only if $z \in \mathbb R $.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
What will be the 49th derivative? Let be the function $f=(x^3+3x) \cdot sin(x)$
$f'(x)=\left(3x^2+3\right)\sin\left(x\right)+\left(x^3+3x\right)\cos\left(x\right)$
$f''(x)=\left(3x-x^3\right)\sin\left(x\right)+\left(6x^2+6\right)\cos\left(x\right)$
$f'''(x)=\left(-9x^2-3\right)\sin\left(x\right)+\left(15x-x^3\right)\cos\left(x\right)$
$f''''(x)=\left(x^3-33x\right)\sin\left(x\right)+\left(12-12x^2\right)\cos\left(x\right)$
I still can't the exact pattern in the results, I would appreciate any help.
|
So the general Leibniz rule somewhat is similar to the binomial theorem. Let $u,v$ be n-th differentiable functions then $(uv)^{(n)}=\sum_{k=0}^{n}{n \choose k}u^{(k)}v^{(n-k)}$ where the exponents denote the $kth$ derivative.
When now note that the fourth derivative and higher of $x^3+3x$ will be zero so we only need to consider the cases where the derivative taken is less than fourth one of $x^3+3x$ . For the $sin(x)$ we use the general formula that $(sin(x))^{(n)}=sin(\frac{n\pi}{2}+x)$.
Thus the forty-nith derivative will be.
$((x^3+3x)sin(x))^{(49)}={49 \choose 46}6sin(23\pi+x)+{49 \choose 47}6xsin(\frac{47\pi}{2}+x)+{49 \choose 48}(3x^2+3)sin(24\pi+x)+(x^3+3x)sin(\frac{49\pi}{2}+x)$
Which wil simplfy to
$(x^3 + 3 x) cos(x) + 49 (3 x^2 + 3) sin(x) - 110544 sin(x) - 7056 x cos(x)$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Given $a^2+b^2=2$ prove $a+b\le2$
*
*Given $a^2+b^2=2$ prove $a+b\le2$
*Given $a+b=2$ prove $a^4+b^4\ge2$
I was trying to prove these using the fact that we know $a^2+b^2\ge2ab$ but not sure where to start.
|
By $1$-d version of polarization identity $(a+b)^2 + (a-b)^2 = 2(a^2+b^2)$,
*
*If $a^2 + b^2 = 2$, then
$$(a+b)^2 = 2(a^2+b^2) - (a-b)^2\le 2(a^2+b^2) = 4 \quad\implies\quad a+b \le |a+b| \le 2$$
*If $a + b = 2$, then
$$
8(a^4+b^4) = 4((a^2+b^2)^2 + (a^2-b^2)^2)\\
\ge 4(a^2+b^2)^2 = (2(a^2+b^2))^2 = ((a+b)^2 + (a-b)^2)^2\\
\ge (a+b)^4 = 16\\
{\large\Downarrow}\\
a^4+b^4 \ge 2
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How is $2(\cos^2 x - \sin^2 x)(\sin x - \cos x) = -2(\cos x - \sin x)^2(\cos x + \sin x)$? I was reading the answer of a proof and in it the answer converted this:
$$2(\cos^2(x) - \sin^2(x))(\sin(x) - \cos(x))$$
to this
$$-2(\cos(x) - \sin(x))^2(\cos(x) + \sin(x))$$
can someone explain what happened and how did it end up like this?
|
We just use $a^2-b^2=(a-b)(a+b)$:
\begin{equation}
\begin{split}
2(\cos^2x - \sin^2 x)(\sin x - \cos x)&=-2(\cos^2 x - \sin^2 x)(\cos x - \sin x)\\
&=-2(\cos x - \sin x)(\cos x + \sin x)(\cos x - \sin x)\\&=-2(\cos x - \sin x)^2(\cos x + \sin x)
\end{split}
\end{equation}
|
{
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|
Problem with proof of derivate of $arcsecx$ Look at this proof for derivative of $arcsecx $
$$y=arcsecx \Rightarrow secy=x$$
We have:
$$(secy) \prime =secytany$$
We take derivations from both parts with respect to $x $
$$y \prime = \frac{1}{secytany} = \frac{1}{secy \sqrt{sec^2y-1} } \Rightarrow y \prime = \frac{1}{x \sqrt{x^2-1} } $$
While the correct derivative is:
$$(arcsecx) \prime = \frac{1}{ |x| \sqrt{x^2-1} } $$
What is the problem with the proof above?
|
Hint:
$$
\tan y= \frac{\sin y}{\cos y}=\frac{\sqrt{1-\cos^2 y}}{\cos y}=\sec y \sqrt{1-\frac{1}{\sec ^2 y}} =
$$
$$
= \sec y \sqrt{\frac{\sec^2 y -1}{\sec^2 y}}=\frac{\sec y}{\color{red}{|\sec y|}}\sqrt{\sec^2 y-1}=\color{red}{\mbox{sign}(\sec y)}\sqrt{\sec^2 y-1}
$$
In simple words: the problem is that $\sqrt {a^2}=|a| $
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
A nice pattern for the regularized beta function $I(\alpha^2,\frac{1}{4},\frac{1}{2})=\frac{1}{2^n}\ $? In this post, the problem was given integer/rational $N$, to solve for algebraic number $z$ in the equation,
$$\begin{aligned}\frac{1}{N}
&=I\left(z^2;\ a,b\right)\\[1.5mm]
&= \frac{B\left(z^2;\ a,b\right)}{B\left(a,b\right)}
\end{aligned} $$
using the beta function $B(a,b)$, incomplete beta $B(z;a,b)$ and regularized beta $I(z;a,b)$. It seems for some $a,b,$ there can be a pattern for the solutions.
Given the equation for $n>1$,
$$I\left(x_1^2;\ \tfrac14,\tfrac12\right)=\frac{1}{2^n}\tag1$$
First define $\color{blue}{\gamma = u+\sqrt{-1+u^2}},$ with fundamental unit $u=1+\sqrt{2}.\ $ Then,
For $n=2$:
$$x_1=x_2-\sqrt{1+x_2^2},\quad \color{blue}{x_2 =\gamma}$$
For $n=3$:
$$x_1=x_2-\sqrt{1+x_2^2},\quad \color{green}{x_2 = x_3+\sqrt{-1+x_3^2},\quad x_3=x_4+\sqrt{1+x_4^2},}\quad \color{blue}{x_4 = \gamma}$$
For $n=4$:
$$x_1=x_2-\sqrt{1+x_2^2},\quad x_2 = x_3+\sqrt{-1+x_3^2},\quad x_3=x_4+\sqrt{1+x_4^2},\\ \quad \color{green}{x_4 = x_5+\sqrt{-1+x_5^2},\quad x_5=x_6+\sqrt{1+x_6^2},}\quad \color{blue}{x_6 = \gamma}$$
and so on, where we add the same two nested layers (in green) each time and so end with even index $x_m$.
Questions:
*
*Does this pattern really hold for all $2^n$ and $n>1?$ Why the regularity?
*What is the integral associated with $(1)$ similar to the ones in the post cited above?
|
We have
$$
B\left(z;\ \tfrac14,\tfrac12\right)=4\sqrt[4]{z}\, _2F_1\left(\frac{1}{2},\frac{1}{4};\frac{5}{4};z\right).
$$
By formula 2.1.15 from Erdelyi, "Higher transcendental functions", vol.I
$$
_2F_1\left(\frac{1}{2},\frac{1}{4};\frac{5}{4};z\right)=\sqrt{\frac{1}{1-z}} \, _2F_1\left(\frac{1}{2},\frac{1}{4};\frac{5}{4};-\frac{4 z}{(1-z)^2}\right).
$$
Since $z$ and $-\frac{4 z}{(1-z)^2}$ have different signs when $z$ is real we need to apply this formula one more time
$$
\, _2F_1\left(\frac{1}{2},\frac{1}{4};\frac{5}{4};z\right)=\frac{1}{2}\sqrt[4]{\frac{16 (1-z)^2}{(z+1)^4}} \, _2F_1\left(\frac{1}{2},\frac{1}{4};\frac{5}{4};\frac{16 z (1-z)^2}{(z+1)^4}\right).
$$
In terms of incomplete Beta function one gets
$$
B\left(z;\ \tfrac14,\tfrac12\right)=\frac12 B\left(\tfrac{16 z (1-z)^2}{(z+1)^4};\ \tfrac14,\tfrac12\right).
$$
I think this formula answers the question 1. From this formula one can work out the recursion for the argument.
|
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|
parts of cubic equation Below there is a solved question from a book .
I could not understand how they got $S_{n+3}$ = $S_{n+1}$ - $S_n$
|
All the exercise is centered onto Newton's Identities.
In particular, concerning your question, we have that putting
$$
x^3 - x + 1 = e_0 x^3 - e_1 x^2 + e_2 x - e_3 = 0
$$
then
$$
\begin{gathered}
e_0 = 1 \hfill \\
e_1 = 0 = S_1 \hfill \\
e_2 = - 1 = \frac{1}
{2}\left( {e_1 S_1 - S_2 } \right) = - \frac{1}
{2}S_2 \hfill \\
e_3 = - 1 = \frac{1}
{3}\left( {e_2 S_1 - e_1 S_2 + S_3 } \right) = \frac{1}
{3}\left( { + 2 + S_3 } \right) \hfill \\
e_4 = 0 = \frac{1}
{4}\left( {e_3 S_1 - e_2 S_2 + e_1 S_3 - S_4 } \right) = \frac{1}
{4}\left( {e_3 S_1 - e_2 S_2 + e_1 S_3 - S_4 } \right) \hfill \\
\vdots \hfill \\
e_{n + 3} = 0 = \frac{{\left( { - 1} \right)^{n + 3} }}
{{n + 3}}\left( {e_3 S_n - e_2 S_{n + 1} + e_1 S_{n + 2} - S_{n + 3} } \right) = \frac{{\left( { - 1} \right)^{n + 3} }}
{{n + 3}}\left( { - S_n + S_{n + 1} - S_{n + 3} } \right) \hfill \\
\end{gathered}
$$
|
{
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|
how to show $a_{n}=[\frac{(2n)!!}{(2n-1)!!}]^2 \frac{1}{2n+1}$ converges? Question: $\displaystyle{a_{n} =
\left[{\left(2n\right)!! \over \left(2n - 1\right)!!}\,\right]^{2}
{1 \over 2n + 1}\,,\quad\mbox{prove}\ a_{n}}$ converges.
My thought: I want to prove {$a_{n}$} is an increasing sequence and it has an upper bound. I've figured out $$a_{n} = (\frac{2\cdot 2 }{1\cdot 3})\cdot(\frac{4\cdot4}{3\cdot5})\dots (\frac{(2n-2)(2n-2)}{(2n-3)(2n-1)})(\frac{(2n)^2}{2n-1})(\frac{1}{2n+1})$$ And, $$\frac{a_{n+1}}{a_{n}}=\frac{(2n)^2}{(2n-1)(2n+1)}\frac{(2n+2)^2}{(2n+1)}\frac{(2n-1)}{(2n)^2}(\frac{2n+1}{2n+3}) = \frac{(2n+2)^2}{(2n+1)(2n+3)} \gt 1$$ So it is increasing.
My problem: I'm stuck at proving it has a upper bound. Maybe there are some inequalities that can be used here?
In addition, I've searched this question and find this sequence converges to $\pi/2$ and it is known as Wallis Formula.But here, I only want to show it converges and I'm not going to find its limit.
Thanks for your time!
|
Consider the sequence $(b_n)$ defined by
$$b_n = \frac{2n+2}{2n+1}\cdot a_n.$$
Then $(b_n)$ is decreasing
\begin{align}
\frac{b_{n+1}}{b_n} &= \frac{\frac{2n+4}{2n+3}}{\frac{2n+2}{2n+1}}\cdot\frac{a_{n+1}}{a_n} \\
&= \frac{(2n+4)(2n+1)}{(2n+2)(2n+3)}\cdot\frac{(2n+2)^2}{(2n+1)(2n+3)} \\
&= \frac{(2n+4)(2n+2)}{(2n+3)^2} \\
&= \frac{(2n+3)^2-1}{(2n+3)^2} \\
&< 1
\end{align}
and since $a_n > 0$ and $\frac{2n+2}{2n+1} > 1$ for all $n\geqslant 1$, we have
$$a_1 \leqslant a_n < b_n \leqslant b_1$$
for all $n \geqslant 1$, showing the boundedness of $(a_n)$.
|
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|
Evalute $\int\frac{x^5}{(36x^2+1)^{3/2}}dx$ by trig sub? I'm stuck trying to evaluate the indefinite integral $\int\frac{x^5}{(36x^2+1)^{3/2}}dx$. It looks like it might be solvable by trig substitution, where $tan^2\theta+1=sec^2\theta$. That strategy seemed to payoff until I eliminated the square root. When I've worked these problems before, I usually end up performing a u-substitution after the trig sub, but I'm not sure where to do that or even if this is the right strategy for this integral.
$\int\frac{x^5}{(36x^2+1)^{3/2}}dx=\int\frac{x^5}{((6x)^2+1)^{3/2}}dx$
$x=\frac{1}{6}\tan\theta, dx=\frac{1}{6}\sec^2(\theta) d\theta$
$\frac{1}{6^6}\int\frac{\tan^5(\theta)}{(\tan^2(\theta)+1)^{3/2}}*\sec^2(\theta)d\theta$
$\frac{1}{6^6}\int\frac{\tan^5(\theta)}{(\sec^2\theta)^{3/2}}*\sec^2(\theta)d\theta$
$\frac{1}{6^6}\int\frac{\tan^5(\theta)}{\sec(\theta)}d\theta$
$\frac{1}{6^6}\int\frac{\sin^5(\theta)}{\cos^5(\theta)}*\cos(\theta)d\theta$
$\frac{1}{6^6}\int\frac{\sin^5(\theta)}{\cos^4(\theta)}d\theta$
|
You are almost there.
$$\int\frac{\sin^5(\theta)}{\cos^4(\theta)}d\theta=\int\frac{\sin(\theta)(1-\cos^2(\theta))^2}{\cos^4(\theta)}d\theta=-\int\frac{1-2u^2+u^4}{u^4}du\\
=-\int u^{-4}-2u^2+1 du$$
|
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|
How to solve system of equations involving square roots How to solve the following system of equations? I've tried some basic techniques like adding/substracting and squaring but with no effect.
$$
\left\{
\begin{array}{c}
\sqrt{1 + x_1} + \sqrt{1 + x_2} + \sqrt{1 + x_3} + \sqrt{1 + x_4} = 2\sqrt{5} \\
\sqrt{1 - x_1} + \sqrt{1 - x_2} + \sqrt{1 - x_3} + \sqrt{1 - x_4} = 2\sqrt{3}
\end{array}
\right.
$$
|
Let $\mathbf{r}_k=
\begin{pmatrix}
\sqrt{1+x_{k}} \\ \sqrt{1-x_{k}}
\end{pmatrix}$, then $\displaystyle \sum_{k=1}^{4} \mathbf{r}_k=
\begin{pmatrix}
2\sqrt{5} \\ 2\sqrt{3}
\end{pmatrix}$
Now,
\begin{align*}
\left| \sum_{k=1}^{4} \mathbf{r}_k \right| &=2\sqrt{5+3} \\
&= 4\sqrt{2} \\
\sum_{k=1}^{4} |\mathbf{r}_k| &= \sum_{k=1}^{4} \sqrt{2} \\
&= 4\sqrt{2}
\end{align*}
Considering the inequality
$$\left| \sum_{k=1}^{4} \mathbf{r}_k \right| \le
\sum_{k=1}^{4} |\mathbf{r}_k|$$
in which equality holds if and only if all $\mathbf{r}_{k}$ are equal.
That is
$$4
\begin{pmatrix}
\sqrt{1+x_k} \\ \sqrt{1-x_k}
\end{pmatrix}=
\begin{pmatrix}
2\sqrt{5} \\ 2\sqrt{3}
\end{pmatrix}$$
Hence $$\fbox{$x_1=x_2=x_3=x_4=\frac{1}{4}$}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2051270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
lim 1/(x-3) (epsilon delta) I want to prove that $\lim_{x\to 1}$ $\frac{1}{x-3}=-\frac{1}{2}$
So I started:
$|\frac{1}{x-3} + \frac{1}{2}|$=$\frac{|x-1|}{2|x-3|}$<$\epsilon$
My problem is to bound $\frac{1}{|x-3|}$ from $|x-1|$
Thanks, for any help
|
Start by restricting $|x-1| \le 1$. Then $-1 \le x-1 \le 1$ gives
$$
x-3=(x-1)-2 \\
-1-2 \le x-3 \le 1-2 \\
-3 \le x-3 \le -1 \\
|x-3| \ge 1.
$$
Then, whenever $|x-1| \le 1$,
$$
\left|\frac{1}{x-3}+\frac{1}{2}\right|=\left|\frac{x-1}{x-3}\right|
\le |x-1|.
$$
Therefore, for any $\epsilon > 0$,
$$
\left|\frac{1}{x-3}+\frac{1}{2}\right| < \epsilon
\mbox{ whenever } |x-1| < \delta=\min(\epsilon,1).
$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
What is the probability of drawing the last red ball in the fourth draw? There are 8 Blue(B) and 2 Red(R) balls in a bag. Each time one ball is drawn and replaced by Blue one. What is the probability of drawing the last red ball in the fourth draw ?
I made 3 cases :
A). BBRR = $(\frac{8}{10})^2\cdot \frac{2}{10}\cdot \frac{1}{10}$
B). RBBR = $\frac{2}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \frac{1}{10}$
C). BRBR = $\frac{8}{10}\cdot \frac{2}{10}\cdot \frac{9}{10}.\frac{1}{10}$
Hence, Total Probability :-
=> $P(A) + P(B) + P(C)$
=> ${ (\frac{8}{10})^2\cdot \frac{2}{10}\cdot \frac{1}{10} }$ + $\frac{2}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \frac{1}{10}$ + $\frac{8}{10}\cdot \frac{2}{10}\cdot \frac{1}{10}$
Have I made it Right ?
|
Just to generalize a little bit
Assuming $ N $ = Total Number of draws
Notes:
*
*Probability to draw the first red ball is always 2/10
*Probability to draw the second red ball is always 1/10
If $ i \in (1,N-1) $ is the draw you get the first red ball at, then
*
*Probability to draw a blue ball for $ 1 \le j < i $ is 8/10
*Probability to draw a blue ball for $ i < j < N-1 $ is 9/10
Then you can define a function for computing the probability of a specific scenario
$$ P(i;N) = \left(\frac{8}{10}\right)^{i-1} \left(\frac{9}{10}\right)^{N-1-i} \left(\frac{2}{10}\right) \left(\frac{1}{10}\right) $$
Then you can compute the sum over all the possible scenario
$$ P_{final} = \sum_{i=1}^{N-1}P(i;N) $$
|
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|
Find the solution for $x$ with $0 \le x \lt 13$ so $13 \mid 3x^2+9x+7 $. Find the solution for $x$ with $0 \le x \lt 13$ so $13 \mid 3x^2+9x+7 $.
I got this question in my discrete mathematics class.
I don't really get the idea how 13 can divide $3x^2+9x+7 $
All of my friend told me to try all the number between 0 and 12, but I know that's not how to solve this question.
Can anyone help me?
Thanks.
|
The discriminant of the equation $3x^2+9x+7=0$ equals $81-4 \cdot 9 \cdot 7=-3=10=7^2 \mod 13$. Taking into account that $3^{-1}=9 \mod 13$ we have
$$
x_1=\frac{-9+7}{6}=\frac{-2}{6}=-\frac{-1}{3}=-9=4 \mod 13,
$$
$$
x_2=\frac{-9-7}{6}=\frac{-16}{6}=\frac{-8}{3}=-8 \cdot 9=5 \cdot 9=45=6 \mod 13.
$$
Thus $x=4$ and $x=6.$
|
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|
Solution of Logarithmic Inequalities: $\log_{0.5}(\log_{5}(x^2-4))>\log_{0.5}1$
If $\log_{0.5}(\log_5 (x^2-4)) >\log_{0.5}1$ then x lies in the interval:
(a) $(-3,-\sqrt{5})\cup(\sqrt{5}, 3)$
(a) $(-3,-\sqrt{5})\cup(\sqrt{5}, 3\sqrt{5})$
(c) $(\sqrt{5}, 3\sqrt{5})$
(d) $\phi $
I have solved quite a few logarithmic inequalities and I am familiar with the procedure. I was just not able to get the answer that matches with the given options in the question.
My Working:
$\log_{0.5}(\log_{5}(x^2-4))>\log_{0.5}1 $
$\log_{5}(x^2-4)<1$
$\log_{5}(x^2-4)<\log_5(5)$
$x^2-4<5$
$x^2<9$
$\therefore -3<x<3$
Now the second equation comes from the fact that
$x^2-4>0$
$\therefore x<-2, x>2$
The intervals formed are $(-3,-2)\cup(2,3)$
|
First of all, $\log_{0.5}(\log_5 (x^2-4))$ will remain defined iff
$\log_5 (x^2-4)>0\iff x^2-4>5^0=1\iff x^2>5\iff$ either $x>\sqrt5$ or $x<-\sqrt5\ \ \ \ (1)$
$$\log_{0.5}(\log_5(x^2-4))>\log_{0.5}1\iff\dfrac{\log_e(\log(x^2-5))}{\log_e(0.5)}>\dfrac{\log1}{\log_e(0.5)}=0$$
Now as $0.5=\dfrac12=2^{-1},\log_e(0.5)=\log_e(2^{-1})=-\log_e2$
$\implies0>\dfrac{\log_e(\log_5(x^2-4))}{\log_e(0.5)}=-\dfrac{\log_e(\log_5(x^2-4))}{\log_e2}$
$\implies\log_e(\log_5(x^2-4))<0\iff\log_5(x^2-4)<e^0=1$
$\iff x^2-4<5^1=5\iff x^2<5+4\iff-3<x<3 \ \ \ \ (2)$
What is the intersection of $(1),(2)?$
|
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|
Prove the inequality $\left|\frac{m}{n}-\frac{1+\sqrt{5}}{2}\right|<\frac{1}{mn}$
Prove that the inequality $$\left|\frac{m}{n}-\frac{1+\sqrt{5}}{2}\right|<\frac{1}{mn}$$ holds for positive integers $m, n$ if and only if $m$ and $n$ where $m > n$ are two successive terms of the Fibonacci sequence.
I thought about using the explicit formula for the Fibonacci sequence, which is $$F_n = \dfrac{1}{\sqrt{5}}\left(\left(\dfrac{1+\sqrt{5}}{2}\right)^n-\left(\dfrac{1-\sqrt{5}}{2}\right)^n\right),$$ but this seems to get computational. Is there an easier way to think about this?
|
$\varphi = \frac{1+\sqrt{5}}{2}$ is the root of $x^2-x-1=0$. In fact $x^2-x-1=\left(x - \frac{1+\sqrt{5}}{2}\right) \cdot \left(x - \frac{1-\sqrt{5}}{2}\right)$. Taking $x=\frac{F_{n+1}}{F_n}$, we have:
$$\left|\left(\frac{F_{n+1}}{F_n}\right)^2-\frac{F_{n+1}}{F_n}-1\right|=\left|\frac{F_{n+1}}{F_n} - \frac{1+\sqrt{5}}{2}\right| \cdot \left|\frac{F_{n+1}}{F_n} - \frac{1-\sqrt{5}}{2}\right| \Leftrightarrow $$
$$\left|\frac{F_{n+1}^2-F_{n+1}F_n-F_n^2}{F_n^2}\right|=\left|\frac{F_{n+1}}{F_n} - \frac{1+\sqrt{5}}{2}\right| \cdot \left|\frac{F_{n+1}}{F_n} + \frac{\sqrt{5}-1}{2}\right| \Leftrightarrow $$
Using Will's hint (a particular case of this):
$$\frac{1}{F_n^2}=\left|\frac{F_{n+1}}{F_n} - \frac{1+\sqrt{5}}{2}\right| \cdot \left(\frac{F_{n+1}}{F_n} + \frac{\sqrt{5}-1}{2}\right) \Rightarrow$$
$$\frac{1}{F_n^2}>\left|\frac{F_{n+1}}{F_n} - \frac{1+\sqrt{5}}{2}\right| \cdot \frac{F_{n+1}}{F_n}$$
or
$$\left|\frac{F_{n+1}}{F_n} - \frac{1+\sqrt{5}}{2}\right| < \frac{1}{F_n \cdot F_{n+1}}$$
Now, let's assume $\left|\frac{m}{n} - \frac{1+\sqrt{5}}{2}\right| < \frac{1}{mn}, m>n$ then, using the same polynomial:
$$\left|\left(\frac{m}{n}\right)^2-\frac{m}{n}-1\right|=\left|\frac{m}{n} - \frac{1+\sqrt{5}}{2}\right| \cdot \left|\frac{m}{n} - \frac{1-\sqrt{5}}{2}\right| < $$
$$\frac{1}{mn}\cdot \left(\frac{m}{n} + \frac{\sqrt{5}-1}{2}\right)$$
or
$$\left|m^2-mn-n^2\right|< \frac{n}{m} \cdot \left(\frac{m}{n} + \frac{\sqrt{5}-1}{2}\right)<\frac{n}{m} \cdot \left(\frac{m}{n} + 1\right)<2$$
which means $\left|m^2-mn-n^2\right|=0$ or $\left|m^2-mn-n^2\right|=1$. The first one, being related to $x^2-x-1=0$, has no integer solutions other than $m=n=0$ which is not $m>n$. Fibonacci numbers satisfy the latter.
Are Fibonacci the only numbers satisfying $\left|m^2-mn-n^2\right|=1$? Well, $1$ and $0$ do and if we assume $m=n+q$ we have:
$$|m^2-mn-n^2|=|(n+q)^2 - (n+q)n - n^2|=|n^2+2nq+q^2 - n^2 - nq - n^2|=$$
$$|q^2+nq-n^2|=|n^2-nq-q^2|=1$$
clearly $n>q>0$, otherwise $q^2+nq-n^2 \geq n^2 + n^2 - n^2\geq 1$ (equality in fact is possible only for $n=q=1$). Next we take $n=q+q_1$ and so on, leading to $m>n>q>q_1>...>q_p > 0$, i.e. it's a finite process. $q_p=1$ otherwise it can be "split" further (e.g. $q_p=2$ leads to $q_{p+1}=1$ and $|2^2-2\cdot 1 -1^1|=1$). Eventually, this leads to Fibonacci numbers only.
|
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|
what is the value of $ x$? If $\log_{2}3^4\cdot\log_{3}4^5\cdot\log_{4}5^6\cdot....\log_{63} {64}^{65}=x!$, what is the value of $ x$?
I've tried
$$\log_2 3^4\cdot\log_3 4^5\cdot\log_4 5^6\cdot.... \log_{63} 64^{65}$$
$$=\dfrac{4\log 3}{\log 2}\cdot\dfrac{5\log 4}{\log 3}\cdot\dfrac{6\log 5}{\log 4}\cdots\dfrac{65\log 64}{\log 63}$$
don't know how to solve futher steps, please help.
Thanks
|
Write $\;\;\;\;\;\;\;\dfrac{4\log 3}{\log 2}\cdot\dfrac{5\log 4}{\log 3}\cdot\dfrac{6\log 5}{\log 4}\cdots\dfrac{65\log 64}{\log 63}\;\;\;\;\;$ as
$$=\dfrac{4\log 3}{\log 2}\cdot\dfrac{5\log 4}{\log 3}\cdot\dfrac{6\log 5}{\log 4}\cdots\dfrac{65\cdot 6\log 2}{\log 63}$$
$\{\because \log64=\log2^6=6\log2\}$
Every logarithmic part cancels out, and we have
$$=4\cdot 5 \cdot 6 \cdots \color{red}{6}\cdot 65$$
$$=\color{red}{2\cdot 3} \cdot 4\cdot 5\cdot 6 \cdots 65$$
$$=65!$$
Hence $x=65$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the linear combination of a vector that is in a span So say we have
Span S =
$
\begin{Bmatrix}
\begin{bmatrix}
1 \\
0 \\
1 \\
\end{bmatrix}
\begin{bmatrix}
-1 \\
1 \\
1 \\
\end{bmatrix}
\begin{bmatrix}
1 \\
1 \\
3 \\
\end{bmatrix}
\end{Bmatrix}
$
I know that
$
\begin{bmatrix}
-1 \\
4 \\
7 \\
\end{bmatrix}
$
is in the span because the reduced row echelon form of [A v] is consistent. Now
how do I find what exactly the linear combination that makes the vector?
|
You can read the solutions in the reduced row echelon form of the augmented matrix:
\begin{align}
\begin{bmatrix}1&-1&1&-1\\0&1&1&4\\1&1&3&7\end{bmatrix}\rightsquigarrow\begin{bmatrix}1&-1&1&-1\\0&1&1&4\\0&2&2&8\end{bmatrix}\rightsquigarrow\begin{bmatrix}1&-1&1&-1\\0&1&1&4\\0&0&0&0\end{bmatrix}\rightsquigarrow\begin{bmatrix}1&0&2&3\\0&1&1&4\\0&0&0&0\end{bmatrix}
\end{align}
The last form says the coefficients $x,y,z$ of a linear combination satisfy the relations:
$$x=-2z+3,\quad y= -z+4,\quad\text{or, in matrix form}\quad
\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}3\\4\\0\end{bmatrix}-z\begin{bmatrix}2\\1\\-1\end{bmatrix}.$$
There remains to choose the value of $z$ at your convenience.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the last Digit of $237^{1002}$? I looked at alot of examples online and alot of videos on how to find the last digit But the thing with their videos/examples was that the base wasn't a huge number. What I mean by that is you can actually do the calculations in your head. But let's say we are dealing with a $3$ digit base Number... then how would I find the last digit.
Q: $237^{1002}$
EDIT: UNIVERSITY LEVEL QUESTION.
It would be more appreciated if you can help answer in different ways.
Since the Last digit is 7 -->
*
*$7^1 = 7$
*$7^2 = 49 = 9$
*$7^3 = 343 = 3$
*$7^4 = 2401 = 1$
$.......$
$........$
*$7^9 = 40353607 = 7$
*$7^{10} = 282475249 = 9$
Notice the Pattern of the last digit. $7,9,3,1,7,9,3,1...$The last digit repeats in pattern that is 4 digits long.
*
*Remainder is 1 --> 7
*Remainder is 2 --> 9
*Remainder is 3 --> 3
*Remainder is 0 --> 1
So, $237/4 = 59$ with the remainder of $1$ which refers to $7$. So the last digit has to be $7$.
|
Finding the last digit of ${237}^{1002}$ is same as finding the last digit of $ 7^{1002}$.
We see that
$7^1 = 7$;
$7^2 = 9$;
$7^3 = 3$;
$7^4 = 1$;
$\dots$
Then comes the repetition of $7$, $9$, $3$ and $1$ as the powers of $7$ increase.
Since our power is even, our choices reduce to last digit of $7^2$ and $7^4$ which are $9$ and $1$ respectively.
Since $1002$ is a multiple of $2$ but not a multiple of $4$,
our choice of power $1002$ of $7$ is same as power $2$ of $7$, which is $9$.
Therefore, $7^{1002} = 9$ (last digit)
|
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|
How many committees of $7$ people can be formed in a class of $14$ boys and $10$ girls if the committee is to contain at least $4$ boys? In a class of $14$ boys and $10$ girls, a committee of $7$ is to be formed. How many committees are possible.
(a) if anybody can serve in the committee?
(b) if the committee is to have exactly $4$ boys?
(c) if the committee is to contain at least $4$ boys?
|
Case 1-
C(24,7)
= $\frac{24!}{7!\times17!}$
Case 2-
C(14,4) $\times$ C(10,3)
= $\frac{14!}{4!\times10!} \times \frac{14!}{4!\times10!}$
Case 3-
At least 4 boys, so may be 5 or 6 or 7.
C(14,4) $\times$ C(10,3) + C(14,5) $\times$ C(10,2) + C(14,6) $\times$ C(10,1) + C(14,4)
= $\frac{14!}{4!\times10!} \times \frac{10!}{3!\times7!} + \frac{14!}{5!\times9!} \times \frac{10!}{2!\times8!} + \frac{14!}{6!\times8!} \times \frac{10!}{1!\times9!} + \frac{14!}{7!\times7!}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Is there an alternative form/identity for $\prod_{k=1}^{m} \cos(2^kx)$? I am working with a family of integrals that involve a product of cosines of the form $\prod_{k=1}^{m} \cos(2^kx)$. Is there a formula or identity for simplifying $\prod_{k=1}^{m} \cos(2^kx)$ to a polynomial number of terms in $x$? Or $\cos x$, $\sin x$?
|
$$\cos x\cdot \cos (2x)\cdot\cos(4x)\cdot\cos(8x)...\cos(2^mx)=P$$
Multiply both sides by $\sin x$, so
$$\frac{1}{2}\sin 2x\cdot \cos (2x)\cdot\cos(4x)\cdot\cos(8x)...\cos(2^mx)=P\sin x$$
$$\frac{1}{2^2}\sin 4x\cdot\cos(4x)\cdot\cos(8x)...\cos(2^mx)=P\sin x$$
$$\frac{1}{2^3}\sin 8x\cdot\cos(8x)...\cos(2^mx)=P\sin x$$
keep doing that and get,
$$\frac{1}{2^{m+1}}\sin(2^{m+1}x)=P\cdot \sin x \Rightarrow P=\frac{\sin(2^{m+1}x)}{2^{m+1}\sin x}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Computing $\lim_{x\to0} \frac 8 {x^8} \left[ 1 - \cos\frac{x^2} 2 - \cos\frac{x^2}4 + \cos\frac{x^2}2\cos\frac{x^2}4 \right]$ without using L'Hospital We have to find the following limit.
$$\lim_{x\to0} \frac 8 {x^8} \left[ 1 - \cos\frac{x^2} 2 - \cos\frac{x^2}4 + \cos\frac{x^2}2\cos\frac{x^2}4 \right]$$
In this I thought to use Lhopital . But using that it will become too long . Is there ny short method .
|
HINT:
$$1-a-b+ab=(1-a)(1-b)$$
$$\lim_{h\to0}\frac{1-\cos2h}{h^2}=2\lim_{h\to0}\left(\frac{\sin h}h\right)^2 = \text{?}$$
|
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|
Do not use series expansion or L' Hospital's rule: $f(x)=\frac{\sin 3x+A\sin 2x+B\sin x}{x^5}$ In the following function
$$f(x)=\frac{\sin 3x+A\sin 2x+B\sin x}{x^5},(x\neq 0)$$
is continuous at $x= 0$. Find $A$ and $B$. Also find $f(0)$.
I first thought of using L hospital. But my sir told me to
'Do not use series expansion or L' Hospital's rule.�'
Now how can I proceed?
|
Use $\sin2x=2\sin x\cos x$ and $\sin3x=3\sin x-4\sin^3x$, so your function can be written
$$
\frac{\sin x}{x}\frac{3-4\sin^2x+2A\cos x+B}{x^4}
$$
In order for the limit to be finite, you need that the numerator in the second fraction has limit $0$, so
$$
3+2A+B=0
$$
so $B=-2A-3$. Thus you get, removing $\frac{\sin x}{x}$, that does not contribute to the limit,
$$
\frac{-4\sin^2x+2A\cos x-2A}{x^4}
$$
Set $x=2y$, so the function becomes
$$
\frac{-2\sin^22y+A\cos2y-A}{8y^4}=\frac{-8\sin^2y\cos^2y-2A\sin^2y}{8y^4}=
-\frac{\sin^2y}{y^2}\frac{4\cos^2y+A}{4y^2}
$$
and we need $A=-4$, so the function is
$$
\frac{\sin^4y}{y^4}
$$
which has limit $1$.
Thus $B=5$ and the value to give to the function to ensure continuity is $1$.
Just a confirmation with Taylor expansion:
\begin{align}
\sin 3x+A\sin 2x+B\sin x
&= 3x-\frac{(3x)^3}{6}+\frac{(3x)^5}{120}\\
&\qquad +2Ax-A\frac{(2x)^3}{6}+A\frac{(2x)^5}{120}\\
&\qquad +Bx-B\frac{x^3}{6}+B\frac{x^5}{120}+o(x^5)
\end{align}
Thus we need
$$
\begin{cases}
3+2A+B=0 \\
-27-8A-B=0
\end{cases}
$$
so $A=-4$ and $B=5$. The coefficient of $x^5$ is
$$
\frac{1}{120}(3^5+2^5A+B)=1
$$
|
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|
Partition with minimum size of parts In how many ways can we partition a number $N$? The answer is given by the partition function $p(N)$. So for instance, the partitions of the number $4$ are:
\begin{equation}
4; ~~~ 3+1;~~~ 2+2;~~~ 2+1+1;~~~1+1+1+1
\end{equation}
and hence $p(4) = 5$.
How is this modified if I require each part to be at least of some size $k$? So for instance, if I required that each part must be at least of size $k=2$, the valid partitions of $4$ would be reduced to
\begin{equation}
4; ~~~ 2+2.
\end{equation}
How is this formalized? And is there a systematic way of enumerating all possibilities?
|
Another technique is based upon generating functions. We encode the summands $j$ as powers of $x$ and the coefficient of $x^j$ provides the number of contributions of the summand $j$.
In the example with $N=4$ the number of partitions $p(4)$ is
\begin{align*}
p(4)&=[x^4](x^0+x^1+x^2+x^3+x^4)(x^0+x^2+x^4)(x^0+x^3)(x^0+x^4)\\
&=[x^4](1+x+x^2+x^3+x^4)(1+x^2+x^4)(1+x^3)(1+x^4) \tag{1}
\end{align*}
Here we use the coefficient of operator $[x^j]$ to denote the coefficient of $x^j$ in a series. There are four factors in (1) indicating that the numbers $1,2,3$ and $4$ are the building blocks of a partition of $4$. The first factor in (1) tells us we can use the number $1$ zero or once or twice or up to four times, whereas e.g. the number $3$ can only be used zero or one time.
Now we want to analyse restricted partitions having smallest summand $k=2$. This means we are looking for
\begin{align*}
[x^4]&(1+x^2+x^4)(1+x^3)(1+x^4)\tag{2}\\
&\qquad=[x^4](1+x^2+x^4)(1+x^3+x^4)\tag{3}\\
&\qquad=[x^4](1+x^2+x^3+2x^4)\tag{4}\\
&\qquad=2
\end{align*}
and we are finished.
Comment:
*
*In (2) we take contributions only from $k=2,3$ and $4$.
*In (3) we multiply the two right-most factors out and skip thereby all summands with powers greater then $4$ since they do nothing contribute to $[x^4]$.
*In (4) we multiply out and again skip summands with powers greater than four. We observe, the coefficient of $x^4$ is $2$ and we are done.
|
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|
Compute $\frac{p(12)+p(-8)}{10}$ where $p(x)=x^4+ax^3+bx^2+cx+d$ I have got an olympiad problem which is as follow:
Compute $\frac{p(12)+p(-8)}{10}$ where $p(x)=x^4+ax^3+bx^2+cx+d$ and $p(1)=10$, $p(2)=20$, $p(3)=30$.
I have been told that answer is $1984$.
I thought applying the values and getting a relation between $a,b,c,d$ from $p(1), p(2), p(3)$ will suffice but the problem is that I will end up with with 4 variables and three equations.
I m a newbie to polynomials and I don't think I m gonna get the answer. So, please help me in this. Thanks.
|
We can write $p(x)-10x=0\forall x=1,2,3$
So using factor theorem $p(x)-10x = 0$ has three factors $(x-1)\;,(x-2)\;,(x-3).$
Given $p(x)$ is a $4^{th}$ degree equation with leading coefficients $=1$
So let $x=r$ be the $4^{th}$ root of above equation
So $$p(x)-10x = (x-1)(x-2)(x-3)(x-r)$$
So $$p(12) = 11\cdot 10\cdot 9\cdot (12-r)+10\cdot 12$$
and $$p(-8) = 11\cdot 10 \cdot 9\cdot (8+r)-10\cdot 8$$
So $$\frac{p(12)+p(-8)}{10} = \frac{9\cdot 10 \cdot 11 \cdot 20+10\cdot 4}{10} = $$
|
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|
The sum of all integers from $1$ to $p$ is divisible by $p$ and all prime numbers before $p$. What can $p$ be?
The summation of all integers from $1$ to $p$ is divisible by $p$ and all prime numbers before $p$. Find all possible solutions for $p$, with proof. Here $p$ is a prime number.
This is a preparation problem for the upcoming BdMO. I can't find a way to start. Any hint will be helpful.
|
Let $p_n$ be the $n$-th prime.
By Betrand's Postulate, $p_n < 2^n$ for $n > 1$ and therefore $S_n:=\dfrac{p_n(p_n+1)}{2} < 2^{2n-1} + 2^{n-1} < 2^{2n} = 4^n$.
But $5 < p_n$ for $n > 3$ so $P_n:=\displaystyle\prod_{i=1}^n p_i > 5^{n-1}$ for $n > 3$.
Since $P_n$ must divide $S_n$ then $P_n \leq S_n$, so either $5^{n-1} < 4^n$ and $n > 3$ or $n\leq 3$.
The first case means that $n < 8$ since $5^7 > 4^8$ and thus $5^{n-1} > 4^n$ for $n\geq 8$.
$p_1 = 2$ and $S_1 = 3$ but $2 \leq 2$ and $2\nmid S_1$.
$p_2 = 3$ and $S_2 = 6$ is divisible by $P_2 = 6$. So $p=3$ is a solution.
$p_3 = 5$ and $S_3 = 15$ but $2 < 5$ and $2\nmid S_3$.
$p_4 = 7$ and $S_4 = 28$ but $3 < 7$ and $3\nmid S_4$.
$p_5 = 11$ and $S_5 = 66$ but $7 < 11$ and $7\nmid S_5$.
$p_6 = 13$ and $S_6 = 91$ but $3 < 13$ and $3\nmid S_6$.
$p_7 = 17$ and $S_7 = 153$ but $2 < 17$ and $2\nmid S_7$.
The only solution for $p$ is $3$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Why $x^5y\equiv y^5x\pmod {240}$ if $x,y$ have similar parity? Let $x,y$ be any 2 integers of similar parity. Why do we have :$$x^5y\equiv y^5x\pmod {240}$$
Any hint anyone?
|
Factorising we get $xy(x-y)(x+y)(x^2 + y^2) \equiv 0 \mod 2^4\cdot 3\cdot 5$
Because $x, y$ have the same parity then, if $x,y$ even all factors are divisible by 2, else if $x,y$ odd all but $xy$ are divisible by 2 and $(x^2 - y^2)$ is divisible by 8. Therefore $xy(x-y)(x+y)(x^2 + y^2) \equiv 0 \mod 2^4$
Using Fermat's little theorem, we have $x^5 \equiv x \mod 5$ and $y^5 \equiv y \mod 5$ therefore $x^5y\equiv y^5x\pmod {5}$ also $x^3 \equiv x \mod 3$ and $y^3 \equiv y \mod 3$ therefore $x^5 \equiv x \mod 3$ and $y^5 \equiv y \mod 3$
|
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"timestamp": "2023-03-29T00:00:00",
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|
volume of a truncated cone that is not a frustum I know the formula for a conical frustum is
$$\frac{\pi h}{3}\left( r^2+rR+R^2 \right) $$
What would the formula be for the area of a truncated right circular cone be where the top is not parallel to the base.
With the plane truncating the cone at an angle $\theta$ from parallel and $\theta <.5 \pi$.
|
A plane that meets one nappe of a right circular cone in an ellipse defines an oblique cone with an elliptical base. If the plane lies at distance $d$ from the cone's vertex, and if the base has semi-axes $a$ and $b$, the volume is $\frac{1}{3} \pi abd$.
If the cone has vertex half-angle $\phi$, the cutting plane crosses the cone axis at distance $h$ from the vertex, and the cutting plane makes angle $\theta$ with the flat base, with $0 \leq \theta < \frac{\pi}{2} - \phi$, then the volume of the truncated cone is
$$
\frac{\pi}{3}\, \frac{h^{3} \cot\phi}{(\cot^{2} \phi - \tan^{2} \theta)^{3/2}}.
$$
(If $\theta \geq \frac{\pi}{2} - \phi$, the volume is infinite.)
For brevity, let $k = \cot\phi$ denote the slope of the cone generator and $m = \tan\theta$ denote the slope of the cutting plane.
In Cartesian coordinates with the vertex at the origin and the cone opening around the positive $z$-axis, the cone and cutting plane have respective equations
\begin{align*}
z^{2} &= k^{2}(x^{2} + y^{2}),
\tag{1a} \\
z &= mx + h.
\tag{1b}
\end{align*}
The plane and cone meet where $k^{2}(x^{2} + y^{2}) = z^{2} = (mx + h)^{2}$, or
$$
(k^{2} - m^{2}) x^{2} - 2mhx - h^{2} + k^{2} y^{2} = 0.
\tag{2}
$$
(This is the equation of the elliptical "shadow" of the base, which does not directly give the shape of the base.)
In the longitudinal plane $y = 0$ (shown), the cone and cutting plane meet when
$$
(k^{2} - m^{2}) x^{2} - 2mhx - h^{2} = 0.
$$
The quadratic formula gives the roots
$$
x_{\pm} = \frac{h(m \pm k)}{k^{2} - m^{2}}.
\tag{3}
$$
The semi-major axis is the distance between the corresponding points on the cone,
$$
a = \tfrac{1}{2} \sqrt{1 + m^{2}}(x_{+} - x_{-})
= \sqrt{1 + m^{2}}\, \frac{hk}{k^{2} - m^{2}}
= \sec\theta\, \frac{hk}{k^{2} - m^{2}}.
\tag{4a}
$$
The semi-minor axis of the slant base is the semi-minor axis of the ellipse (2), namely the positive value of $y$ in equation (2) when
$$
x = \frac{x_{-} + x_{+}}{2} = \frac{hm}{k^{2} - m^{2}}.
$$
For this $x$, we have $mx + h = \dfrac{hk^{2}}{k^{2} - m^{2}}$, so
\begin{align*}
y^{2} &= \frac{1}{k^{2}} \bigl[(mx + h)^{2} - k^{2} x^{2}\bigr] \\
&= \frac{1}{k^{2}} \left[\frac{h^{2} k^{4}}{(k^{2} - m^{2})^{2}} - \frac{k^{2} h^{2} m^{2}}{(k^{2} - m^{2})^{2}}\right] \\
&= \frac{h^{2}}{k^{2} - m^{2}}.
\end{align*}
The semi-minor axis is therefore
$$
b = \frac{h}{\sqrt{k^{2} - m^{2}}}.
\tag{4b}
$$
Trigonometry shows the distance from the vertex to the cutting plane is $d = h\cos\theta$. Combining with equations (4a) and (4b), the volume of the slant cone is
$$
\frac{\pi}{3} abd
= \frac{\pi}{3} \left(\sec\theta\, \frac{hk}{k^{2} - m^{2}}\right) \frac{h}{\sqrt{k^{2} - m^{2}}}\, (h\cos\theta)
= \frac{\pi}{3}\, \frac{h^{3} k}{(k^{2} - m^{2})^{3/2}},
$$
as claimed.
|
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|
What is the least positive integer that is divisorous?
We say that a positive integer $N$ is divisorous if the ones digits of the positive divisors of $N$ include all of the base-ten digits from $0$ to $9$. What is the least positive integer that is divisorous?
I didn't see an easy way of finding the least such $N$. We know that $N$ must contain a factor of $2$ and $5$, but how do we find the least such $N$?
|
We first see that such an $N$ must be a multiple of $2$ since otherwise the units digits of the divisors would never be even. We also see that $N$ must be a multiple of $5$ since any number which has a units digit of $5$ must be a multiple of $5$. Thus $N$ is a multiple of $10$ and so we may write $$N = 2 \cdot 5k$$ for some positive integer $k$. Now we take two cases:
Case $1$: $3 \mid N$
In this case we prove that the minimal $N = 2 \cdot 3^3 \cdot 5$. If the exponent of $3$ is reduced, then the new prime $7$ must be added to the prime factorization of $N$ and so $3^2 \nmid N$ otherwise $N$ would be larger. But then we have the minimal such $N$ to be $2 \cdot 3 \cdot 5 \cdot 7$, which doesn't have a divisor with a units digit of $9$. Thus the exponent of $3$ cannot be reduced and so $N$ is minimal in this case.
Case $2$: $3 \nmid N$
In this case we prove that $N > 2 \cdot 3^3 \cdot 5$. We must add a new prime greater than or equal to $7$ into the prime factorization of $2 \cdot 5$. We also must add at least one other prime since otherwise we will have exactly $8$ divisors and it can't be $2$ since we can't get a units digit of $8$. Thus we have $N \geq 2 \cdot 5^2 \cdot 7$, which means that $N > 2 \cdot 3^3 \cdot 5$ since $7 \cdot 5 > 3^3$.
|
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|
Proving the inequality I need to prove this:
$\binom{n}{2}a^2 + \binom{m}{2}b^2
\leq
\binom{n+m}{2} \left( \frac{n\cdot a + m\cdot b}{n+m} \right) ^2$
I tried this (WLOG $n\leq m$):
$\binom{n}{2}a^2 + \binom{m}{2}b^2 = \frac{n(n-1)}{2}a^2 + \frac{m(m-1)}{2}b^2
\\
\binom{n+m}{2} \left( \frac{n\cdot a + m\cdot b}{n+m} \right) ^2
= \frac{n+m-1}{2}\frac{(na+mb)^2}{n+m}$
How would you do it?
|
A full expanding gives $(a-b)^2+2(m+n)ab\geq0$ and the rest for you.
|
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|
Finding the splitting field of the polynomial $x^5+2x^4+5x^2+x+4$ over $F_{11}$. I am working on the following problem:
Find the splitting field of the polynomial $f(x)=x^5+2x^4+5x^2+x+4$ over $F_{11}$.
What I have done:
So far I found that $-2$ is the only root of $f(x)$ in $F_{11}$. I have also factored $f(x)$ as $f(x)=(x+2)(x^2+5x+1)(x^2-5x+2)$ over $F_{11}$.
Apparently the splitting field of $f(x)$ over $F_{11}$ is $F_{11^4}$. Can someone help me understand how one can conclude that.
|
We are considering the field $F_{11}(\alpha,\beta)$ were $\alpha^2 + 5\alpha +1=0$ and $\beta^2 - 5\beta +2=0$. We are going to show that $$F_{11}(\alpha,\beta)=F_{11}(\alpha+\beta)$$
In fact summing the equations we get $$(\alpha + \beta)(\alpha - \beta + 5)=1$$ which means that these elements are inverse one another, thus adding $(\alpha + \beta)$ to the base field we add its inverse $(\alpha - \beta + 5)$ as well, and also their sum and their difference and thus both $\alpha$ and $\beta$. We proved the claim. Now notice that $(\alpha + \beta) = \frac{\sqrt{21}+\sqrt{17}}{2}$ simply solving the equations of degree two. It has minimum polynomial $x^4 -19x^2+1$. It can be factorized in $F_{11}$ as $x^4 -19x^2+1=(x^2+5)(x^2-2)$ thus $\alpha + \beta$ is a solution of one of them, let's call it $p(x)$. So $$F_{11}(\alpha,\beta)=F_{11}(\alpha+\beta)\simeq\frac{F_{11}[x]}{(p(x))}$$
|
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|
$x + \frac1y = y + \frac1z = z + \frac1x$, then value of $xyz$ is? If $x,y,z$ are distinct positive numbers, such that $$x + \frac1y = y + \frac1z = z + \frac1x $$ then value of $xyz$ is?
$$A)\ 4\quad B)\ 3\quad C)\ 2\quad D)\ 1$$
My attempt:
1.I equaled the equation to '$k$'. Using the AM-GM inequality, I found that $k>2$ (the equality does not hold because all are distinct). However, this, I couldn't put to much use.
2. For the next attempt, I substituted the values of $x$ and $y$ in terms of $z$ and $k$ in $xyz$. What I am getting is $xyz=zk^2 - k - z$. That's the farthest I could do..Please, help.
|
By multiplying the first equation by $yz$, the second by $xz$, and the equation connecting the first and third expression by $xy$, we get
$$xyz+z = y^2z+y$$
$$xyz+x = z^2x+z$$
$$xyz+y = x^2y+x$$
Adding these three, cancelling $x+y+z$ from both sides, and dividing by $xyz$ gives $$\frac{x}{z}+\frac{y}{x}+\frac{z}{y} = 3$$
But by AM-GM, the LHS is $\ge 3$ and so we must have equality in AM-GM, meaning $\frac{x}{z} = \frac{y}{x} = \frac{z}{y}$, and so $x^2 = yz, y^2 = xz, z^2 = xy$. Dividing the first by the second gives $\frac{x^2}{y^2} = \frac{y}{x}$ and so $x^3 = y^3 \implies x=y$. Similarly, $y=z$. This contradicts that $x,y,z$ are distinct, so the question is flawed.
|
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|
finding value of $ \int\frac{1}{x^3\sqrt{(1+x)^3}}dx$ finding value of $\displaystyle \int\frac{1}{x^3\sqrt{(1+x)^3}}dx$
Substituting $x=\tan^2 \theta\;, dx = 2\tan \theta \sec^2 \theta$
integral is $=\displaystyle \int \frac{2\tan \theta \sec^2 \theta }{\tan^6 \theta \cdot \sec^3 \theta}d\theta= 2\int\frac{\cos^6 \theta}{\sin^5 \theta}d\theta $
wan,t be able to proceed after, could some help me
|
$$
\left.\begin{aligned}
\int \frac{\cos ^6 \theta}{\sin ^5 \theta}&=-\frac{1}{4} \int \cos ^5 \theta d\left(\frac{1}{\sin ^4 \theta}\right) \\
& =-\frac{\cos ^5 \theta}{4 \sin ^4 \theta}-\frac{5}{4} \int \frac{\cos ^4 \theta \sin ^2 d \theta}{\sin ^4 \theta} \\
& =-\frac{\cos ^5 \theta}{4 \sin ^4 \theta}-\frac{5}{4} \int \frac{\cos ^4 \theta}{\sin ^3 \theta} d \theta \\
& =-\frac{\cos ^5 \theta}{4 \sin ^4 \theta}+\frac{5}{8} \int \cos ^3 \theta d\left(\frac{1}{\sin ^2 \theta}\right) \\
& =-\frac{\cos ^5 \theta}{4 \sin ^4 \theta}+\frac{5 \cos ^3 \theta}{8 \sin ^2 \theta}+\frac{15}{8} \int \frac{\cos ^2 \theta}{\sin \theta} d \theta \\
& =-\frac{\cos ^5 \theta}{4 \sin ^4 \theta}+\frac{5 \cos ^3 \theta}{8 \sin ^2 \theta}+\frac{15}{8} \int \frac{1-\sin \theta}{\sin \theta} d \theta \\
& =-\frac{\cos ^5 \theta}{4 \sin ^4 \theta}+\frac{5 \cos ^3 \theta}{8 \sin ^2 \theta}-\frac{15}{8}[\ln |\csc \theta+\cot \theta|-\cos \theta ]+C\\
& = \frac{1}{4 x^2 \sqrt{1+x}}+\frac{5}{8 x} \sqrt{1+x}-\frac{15}{8} \ln \left|\frac{\sqrt{1+x}+1}{\sqrt{x}}\right|-\frac{1}{\sqrt{1+x}}+C
\end{aligned}\right]
$$
$$
\begin{aligned}
I=&-\frac{1}{2 x^2 \sqrt{1+x}}+\frac{5}{4 x \sqrt{1+x}}-\frac{15}{4} \ln \left|\frac{\sqrt{1+x}+1}{\sqrt{x}}\right| -\frac{1}{2 \sqrt{1+x}}+C
\end{aligned}
$$
|
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|
Prime factor $>250000$ for $1002004008016032$ I need to find a prime factor, $p$, of $1002004008016032$ such that $p \gt 250000$. Now this is a very large number and I know it would be stupid of me to factorize such a large number into its prime factors . Any help how I can solve this ?
|
$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^2+\cdots+xy^{n-2}+y^{n-1})$
Taking $x=10^3,y=2$, we get
$1002004008016032=x^5+x^4y+x^3y^2+x^2y^3+xy^4+y^5=\frac{x^6-y^6}{x-y}$
Also,
$\frac{x^6-y^6}{x-y} = (x+y)(x^2+xy+y^2)(x^2-xy+y^2)$
=$1002\times1002004\times998004 = 1002\times16\times250501\times249501$
Hence, the prime number , $p\gt250000$ is $250501$
|
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|
prove $\sum_{k=0}^{3n+2}\frac{(-1)^k}{6n+5-k}\binom{6n+5-k}{k}=\frac{1}{6n+5}$ I want to
prove that $$\sum_{k=0}^{3n+2}\frac{(-1)^k}{6n+5-k}\binom{6n+5-k}{k}=\frac{1}{6n+5}$$
Let $\displaystyle f_n(x)=\sum_{k=0}^{\lfloor (n+1)/2\rfloor}\binom{n-k+1}{k}x^{n-k}$,
then we have a recurrence relation
$$f_n(x)=xf_{n-1}(x)+xf_{n-2}(x) \quad ; \quad f_1(x)=x+1,\quad f_0(x)=1$$
Hence, $$\sum_{k=0}^{3n+2}\frac{(-1)^k}{6n+5-k}\binom{6n+5-k}{k}=\int_{-1}^0f_{6n+4}(x)dx$$
What will I do from here? Thank you.
|
Chebyshev polynomials of the second kind have the following closed form:
$$ U_n(x) = \sum_{r=0}^{\lfloor n/2\rfloor}\binom{n-r}{r}(-1)^r (2x)^{n-2r} \tag{1} $$
hence:
$$ x^{n-1}\,U_n(x/2) = \sum_{r=0}^{\lfloor n/2\rfloor}\binom{n-r}{r}(-1)^r (x)^{2n-2r-1} \tag{2} $$
and:
$$ S_n=\sum_{r=0}^{\lfloor n/2\rfloor}\binom{n-r}{r}\frac{(-1)^r}{n-r} = 2\int_{0}^{1}x^{n-1} U_n(x/2)\,dx. \tag{3} $$
By substituting $x=2\cos\theta$, the last integral turns into an elementary integral and we get:
$$\boxed{ \sum_{r=0}^{\lfloor n/2\rfloor}\binom{n-r}{r}\frac{(-1)^r}{n-r} = \color{red}{\frac{1}{n}}}\tag{4}$$
for any $n\equiv \pm1\pmod{6}$. As an alternative, we may use the generating function for Chebyshev polynomials of the second kind, by replacing $x$ with $x/2$, then $t$ with $xt$. $S_n$ then becomes the coefficient of $t^n$ in a simple analytic function.
|
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|
Integral of $\sqrt{2+t^2}$ How can I solve the integral of
$$ \int \sqrt{2+t^2}\,dt$$
I obtained it by calculating the length of the arc $a(t)=(t\cos t,t \sin t,t)$ with $t \in [0,2\pi]$.
I try with substitution with the function $\sinh$ but i don't manage to resolve it.
|
One way is to integrate by parts:
$$
\int_0^{2\pi} \underbrace{\sqrt{t^2 + 2}}_u \, \underbrace{dt}_{dv}
= \underbrace{\sqrt{t^2 + 2}}_u \cdot \underbrace{t}_v \bigg|_0^{2\pi}
- \int_0^{2\pi} \underbrace{t}_v \cdot \underbrace{\frac t{\sqrt{t^2+2}} \, dt}_{du}
$$
Continuing without the labels:
\begin{align*}
I &= \int_0^{2\pi} \sqrt{t^2 + 2} \, dt\\[0.3cm]
&= t\sqrt{t^2 + 2} \bigg|_0^{2\pi} - \int_0^{2\pi} \frac{t^2}{\sqrt{t^2+2}} \, dt\\[0.3cm]
&= 2\pi \sqrt{4\pi^2+2} - \int_0^{2\pi} \frac{t^2 \color{red}{+2-2}}{\sqrt{t^2+2}} \, dt\\[0.3cm]
&= 2\pi \sqrt{4\pi^2+2} - \left[\int_0^{2\pi} \frac{t^2 + 2}{\sqrt{t^2+2}} \, dt - \int_0^{2\pi} \frac 2{\sqrt{t^2+2}} \, dt\right]\\[0.3cm]
&= 2\pi \sqrt{4\pi^2+2} - \underbrace{\int_0^{2\pi} \sqrt{t^2+2} \, dt}_{\text{this is $I$}} + \int_0^{2\pi} \frac 2{\sqrt{t^2+2}} \, dt\\[0.3cm]
&= 2\pi \sqrt{4\pi^2+2} - I + 2 \sinh^{-1}\left(\frac t{\sqrt2}\right)\bigg|_0^{2\pi}\\[0.3cm]
&= 2\pi \sqrt{4\pi^2+2} - I + 2\sinh^{-1}(\pi\sqrt2)
\end{align*}
Therefore
$$ I = 2\pi \sqrt{4\pi^2+2} - I + 2\sinh^{-1}(\pi\sqrt2). $$
Solve for $I$ to get
$$ I = \pi \sqrt{4\pi^2+2} + \sinh^{-1}(\pi\sqrt2). $$
Admittedly this does require knowledge of that inverse hyperbolic sine integral (although there are other ways to evaluate that one), which may make this method not that helpful, but here it is for the sake of variety.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate the Volume integral $\iiint\left(x^2+y^2+z^2\right)\mathbb dv$. Suppose $a>0$ and $S = \{(x,y,z) \in \mathbb R^3 : x^2+y^2+z^2=a^2\}$ then
MY FIRST APPROACH: $$\int\int\int\left(x^2+y^2+z^2\right)\mathbb dv=a^2\int\int\int \mathbb dv=a^2.\frac{4}{3}\pi a^3=\frac 4 3\pi a^5$$.
MY SECOND APPROACH:If i use spherical coordinate then $dV = r^2 \sin \theta dr d\theta d\phi.$ and $\left(x^2+y^2+z^2\right)=r^2$ now $$\int\int\int\left(x^2+y^2+z^2\right)\mathbb dv=\iint d\Omega \int_0^a (r^4) dr =\frac 4 5\pi a^5$$
where $\iint d\Omega = \int_0^\pi \sin(\theta)d\theta \int_0^{2\pi} d\phi = 4\pi.$
My question is that why such discrepancy in the answer$?$ and where did i commit mistake in my first approach.
(Note:2nd answer is correct but what's the problem with 1st$?$)
|
Note
$$S = \left\{(x,y,z) \in \mathbb R^3 : x^2+y^2+z^2\le a^2\right\}=\left\{(\theta,\phi,\rho) : 0\le \theta\le 2\pi\, ,\,0\le \phi\le\pi\,\,, \rho^2=a^2\right\}$$
set
\begin{cases}
x=\rho\sin\phi\cos \theta\\
y=\rho\sin\phi\sin \theta\\
z=\rho\cos\phi
\end{cases}
we have
$$\left|\frac{\partial(x,y,z)}{\partial(\rho,\theta,\phi)}\right|=\rho^2\sin\phi$$
thus
$$\int\int\int\left(x^2+y^2+z^2\right)\mathbb {dx\,dy\,dz}=\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{a} \rho^4 \sin\phi\,\mathbb{d\rho\,d\phi\,d\theta}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that given fraction is power of two I do not really know where to start with the following:
Prove that
$$\frac{\displaystyle{ \prod_{k = 1}^{2n - 1} k^{\min(k, 2n - k)}}}{\displaystyle{\prod_{k = 1}^{n - 1}(2k + 1)^{2n - 2k - 1}}}$$ is a power of two.
Could you give me a hint which helps to solve this problem?
|
Much calculus there :
$$\begin{aligned}\frac{\displaystyle{ \prod_{k = 1}^{2n - 1} k^{\min(k, 2n - k)}}}{\displaystyle{\prod_{k = 1}^{n - 1}(2k + 1)^{2n - 2k - 1}}}
&= \frac{\displaystyle{ \prod_{k = 1}^{n} k^{\min(k, 2n - k)}} \times \prod_{k = n + 1}^{2n - 1} k^{\min(k, 2n - k)}}{\displaystyle{\prod_{k = 1}^{n - 1}(2k + 1)^{2n - 2k - 1}}}
\\ &= \frac{\displaystyle{ \prod_{k = 1}^{n} k^k} \times \prod_{k = n + 1}^{2n - 1} k^{2n - k}}{\displaystyle{\prod_{k = 1}^{n - 1}(2k + 1)^{2n - 2k - 1}}}
\\ &= \frac{\displaystyle{ \prod_{k = 1}^{n} k^k} \times \prod_{k = n + 1}^{2n - 1} k^{2n} \times \prod_{k = 1}^{n - 1}(2k + 1)^{2k + 1}}{\displaystyle{\prod_{k = 1}^{n - 1}(2k + 1)^{2n}} \prod_{k = n + 1}^{2n - 1} k^{k}}
\\ &= \frac{\displaystyle{ \left(\prod_{k = 1}^{n} k^k \right)^2} \times \prod_{k = n + 1}^{2n - 1} k^{2n} \times \prod_{k = 1}^{n - 1}(2k + 1)^{2k + 1}}{\displaystyle{\prod_{k = 1}^{n - 1}(2k + 1)^{2n}} \prod_{k = 1}^{2n - 1} k^{k}}
\\ &= \frac{\displaystyle{ \left(\prod_{k = 1}^{n} k^k \right)^2} \times \prod_{k = n + 1}^{2n - 1} k^{2n}}{\displaystyle{\prod_{k = 1}^{n - 1}(2k + 1)^{2n}} \prod_{k = 1}^{n - 1} (2k)^{2k}}
\\ &= \frac{\displaystyle{}\left(\prod_{k = 1}^{2n - 1} k\right)^{2n}}{\displaystyle{}\left(\prod_{k = 1}^{n - 1}(2k + 1)\right)^{2n}} \frac{\displaystyle{ \left(\prod_{k = 1}^{n} k^k \right)^2}}{\displaystyle{} \prod_{k = 1}^{n} k^{2n} \times \prod_{k = 1}^{n - 1} (2k)^{2k}}
\\ &= \frac{\displaystyle{ \left(\prod_{k = 1}^{n} k^k \right)^2} \times \prod_{k = 1}^{n - 1}(2k)^{2n}}{\displaystyle{}\prod_{k = 1}^{n} k^{2n} \times \prod_{k = 1}^{n - 1} (2k)^{2k}}
\\ &= \frac{\displaystyle{ \prod_{k = 1}^{n} k^{2k}} \times 2^{2n(n-1)}}{\displaystyle{}n^{2n} \times \prod_{k = 1}^{n - 1} (2k)^{2k}}
\\ &= \frac{\displaystyle{ } 2^{2n(n-1)}}{\displaystyle{} \prod_{k = 1}^{n - 1} 2^{2k}}
\\ &= 2^{n(n-1)}
\end{aligned}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Additivity of $\arctan(\frac{x+y}{1-xy})$ Show in two different way that,
$$\arctan\left({\frac{x+y}{1-xy}}\right) = \arctan\left(x\right) + \arctan\left(y\right)
$$
The first way I know derive from the addition formula of tangent
$$ \frac{ \tan\left(x\right)+ \tan\left(y\right)}{1- \tan(x) \tan(y)} = \tan\left(x+y \right).
$$
|
Differentiating,
$$ \begin{align}
\frac{d}{dx} \arctan{\left( \frac{x+y}{1-xy} \right)} &= \frac{1}{1+(x+y)^2/(1-xy)^2} \left( \frac{1}{1-xy} - \frac{-(x+y)y}{(1-xy)^2} \right) \\
&= \frac{ 1-xy + xy+y^2 }{(1-xy)^2 + (x+y)^2} \\
&= \frac{ 1+y^2 }{1-2xy+x^2y^2+x^2+2xy+y^2} \\
&= \frac{ 1+y^2 }{1+x^2+y^2+x^2 y^2} \\
&= \frac{ 1+y^2 }{(1+y^2)(1+x^2)} = \frac{1}{1+x^2},
\end{align} $$
and then integrating gives
$$ \arctan{\left( \frac{x+y}{1-xy} \right)} = \arctan{x}+C(y). $$
Putting $x=0$ gives $C(y)=\arctan{y}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Verify $\cos{x}<\left(\frac{\sin{x}}{x}\right)^3$ for $0Verify by Mean Value Theory or otherwise that $\cos{x}<\left(\frac{\sin{x}}{x}\right)^3$ for $0<x<\pi/2$.
I am unable to solve the problem. Please give me a solution of the problem.
|
For $x > 0$ the well-known representation as alternating series
give the estimates
$$
\sin x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!} > x - \frac{x^3}{6} \, ,\\
\cos x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!} < 1 - \frac{x^2}{2} + \frac{x^4}{24} \, .
$$
For $0 < x < \pi/2$ in particular $x^2 < 6$ holds and therefore
$$
\left( \frac{\sin x}{x} \right )^3 - \cos x
> \left( 1 - \frac{x^2}{6} \right )^3 - \left( 1 - \frac{x^2}{2} + \frac{x^4}{24} \right ) \\
= \frac{x^4}{24}- \frac{x^6}{216} =
\frac{x^4}{24} \left( 1 - \frac{x^2}{9} \right) > 0 \, .
$$
|
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|
Incorrect solution in limits $$
\lim_{n\to \infty}\left(\frac{1}{n^4}{+3^{\frac{2}{2+n}}}\right)^{n}
$$
So i re-write it like:
$\lim_{n\to \infty}e^{n\ln{\frac{1}{n^4}\ln3^{\frac{2}{2+n}}}}$ $=$ $e^{\frac{2n}{2+n}\ln{\frac{1}{n^4}\ln3}}=e^{{2}\ln{\frac{1}{n^4}\ln3}}$
So here, $\ln{\frac{1}{n^4}}$ give us minus infinity, but i think that somewhere i lost a way, so i need some hints
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Put $n=\frac{1}{y}$ .The limit reduces to
$L=\lim_{y\to 0}\left(y^4+3^{\frac{2y}{2y+1}}\right)^{\frac{1}{y}}$
Take log on both sides
$\log_{e}L=\lim_{y\to 0} \log_{e}\left(y^4+3^{\frac{2y}{2y+1}}\right)^{\frac{1}{y}}$
$\log_{e}L=\lim_{y\to 0}\frac{1}{y} \log_{e}\left(y^4+3^{\frac{2y}{2y+1}}\right)$
$\log_{e}L=\lim_{y\to 0}\frac{ \log_{e}\left(1+y^4+3^{\frac{2y}{2y+1}}-1\right)}{y^4+3^{\frac{2y}{2y+1}}-1}\frac{y^4+3^{\frac{2y}{2y+1}}-1}{y}$
$\log_{e}L=\lim_{y\to 0}\frac{ \log_{e}\left(1+y^4+3^{\frac{2y}{2y+1}}-1\right)}{y^4+3^{\frac{2y}{2y+1}}-1}\left(y^3+\frac{3^{\frac{2y}{2y+1}}-1}{\frac{2y}{2y+1}}\frac{2}{2y+1}\right)=1\left(2\log_{e}3\right)=\log_{e}9$
As $y\to 0$ it is obvious that $y^4+3^{\frac{2y}{2y+1}}-1\to 0$.
So here the standard result $\lim_{x\to 0}\frac{\log_{e}(1+x)}{x}=1$ has been used.
Also as $y\to 0$ it is obvious that $\frac{2y}{2y+1}\to 0$
so the standard limit $\lim_{x\to 0}\frac{a^x-1}{x}=\log_{e}a$ can be used.
Hence,$L=9$
|
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|
Find the limit of $6^n(2-x_n)$ where $x_n=\sqrt[3]{6+\sqrt[3]{6+\dots+\sqrt[3]{6}}}$ with $n$ roots
Let $x_n=\sqrt[3]{6+\sqrt[3]{6+\dots+\sqrt[3]{6}}}$ where the expression in the RHS has $n$ roots.
Find the following limit: $\lim \limits_{n\to \infty}6^n(2-x_n)$
My approach: I had two approaches. The first one was the following: I showed that $x_n$ is increasing with upper bound which is equal 2 then by Weierstrass theorem its convergent with limit $2$. But we cannot deduce from that the limit of $6^n(2-x_n)$ is zero because the last expression is uncertainty.
The second one was that $x_{n+1}^3=6+x_n$ then $x_{n+1}^3-8=x_n-2$. From the last equation we get: $(x_{n+1}-2)(x_{n+1}^2+2x_{n+1}+4)=x_n-2$. I tried work out with this but I have stuck.
Would be very grateful for hints or solutions.
|
Lets look at the sequence
$$a_n=6^n(8-x_n^3)=6^n(2-x_n)(4+2x_n+x_n^2)$$
Now $$\lim_{n\to\infty}6^n(2-x_n)(4+2x_n+x_n^2)=\lim_{n\to\infty}12\cdot 6^n(2-x_n)$$ so lets instead look at the $\lim a_n$ and whatever the limit is just multiply it by $12$.
We have that
$$\frac{a_{n+1}}{a_n}=\frac{6^{n+1}(8-x_{n+1}^2)}{6^n(8-x_n^2)}=\frac{6(2-x_n)}{(2-x_n)(x_n^2+2x_n+4)}=\frac{6}{x_n^2+2x_n+4}$$
Since $$x_n^2+2x_n+4\geq6^{2/3}+2\cdot 6^{1/3}+4\geq 10$$ since it's increasing.
From this we can see that each $a_n$ is decreasing by $\frac{6}{10}$ at least so we have that
$$a_0(\frac{6}{10})^n\geq a_n> 0 $$ by squeeze theorem we have that $a_n\to 0$ so the original limit tends to $0$.A side note I picked $a_n$ to make calculations easier the same could be done with the original limit.
|
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|
Proof convergence of $\sum\limits_{n = 1}^\infty\frac {n!} {n^n}$ To prove the convergence of $\sum\limits_{n = 1}^\infty\frac {n!} {n^n}$ I used that $\lim\limits_{n\to\infty}|\frac {a_{n+1}} {a_n}|$ has to be $<1$:
$$\lim\limits_{n\to\infty}|\frac {a_{n+1}} {a_n}|$$
$$=\lim\limits_{n\to\infty}|\frac {\frac {(n+1)!} {(n+1)^{n+1}}} {\frac {n!} {n^n}}|$$
$$=\lim\limits_{n\to\infty}|\frac {(n+1)!\cdot n^n} {n!\cdot (n+1)^{n+1}}|$$
$$=\lim\limits_{n\to\infty}|\frac {(n+1)\cdot n^n} {(n+1)^{n+1}}|$$
$$=\lim\limits_{n\to\infty}|\frac {n^n} {(n+1)^n}|$$
$$=\lim\limits_{n\to\infty}|{(\frac {n} {n+1})}^n|=\frac 1 e$$
Okay... problem solved after thankful advice
|
We can also use the Comparison Test: $$0\le \frac{n!}{n^n}=\frac{1\cdot 2\cdot 3\cdot\ldots\cdot n}{n\cdot n\cdot n\cdot\ldots\cdot n}\le \frac{1\cdot 2\cdot n\cdot\ldots \cdot n}{n\cdot n\cdot n\cdot\ldots\cdot n}=\frac{2}{n^2},$$ and $\displaystyle \sum_{n=1}^{+\infty}\frac{2}{n^2}$is convergent, so the given series is convergent.
|
{
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|
Proving a sequence is convergent by convergence of odd and even subsequences? I want to use this method to prove convergence of: $a_{n}=\sqrt{2-a_{n-1}}$, $a_{0}=\frac{2}{3}$.
Here is my attempt at proof:
It can be proven inductively that $0< a_{n}<2$ for all $n$.
I want to show that each of the odd and even subsequences are monotone.
$a_{2n+3}-a_{2n+1}=\sqrt{2-a_{2n+2}}-\sqrt{2-a_{2n}}=\frac{a_{2n}-a_{2n+2}}{\sqrt{2-a_{2n+2}}+\sqrt{2-a_{2n}}}=\frac{a_{2n}-\sqrt{2-\sqrt{2-a_{2n+1}}}}{\sqrt{2-a_{2n+2}}+\sqrt{2-a_{2n}}}$
I'm stuck as I'm not able to show that $(a_{2n+1})_{n\geq0}$ is decreasing.
Thank's in advance.
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If $0<x<1$ then $\sqrt {1-x}>1-x/(2-x)$ (by squaring both sides and calculating). This deserves more publicity. We will use this in (2).(ii). below.
If $a_n$ converges to $L$ then $0\leq L=\sqrt {2-L}\implies (L^2=2-L\land 0\leq L)\implies L=1.$ So it is natural to let $a_n=1+b_n.$
We have $0<|b_n|<1.$ (Equivalent to $0<a_n<2$).
(1). If $1>b_n>0$ then $1+b_{n+1}=\sqrt {2-(1+b_n)}=\sqrt {1-b_n}<1$, which implies $b_{n+1}<0.$..... If $0>b_n>-1$ then $1+b_{n+1}=\sqrt {1-b_n}>1$, which implies $b_{n+1}>0.$
(2).(i). If $b_n<0$ then $b_{n+1}>0$ so $$1+2|b_{n+1}|=1+2b_{n+1}<(1+b_{n+1})^2=1-b_n=1+|b_n|$$ which implies $0<|b_{n+1}|<|b_n|/2.$
(2). (ii). If $b_n>0$ then $b_{n+1}<0$ so $$1-|b_{n+1}|=1+b_{n+1}=\sqrt {1-b_n}\;> 1-b_n/(2-b_n)$$ which implies $|b_{n+1}|<b_n/(2-b_n)<b_n=|b_n|.$
Applying all this, we have:
(3).(i). If $b_n<0$ we have $b_{n+1}>0$ so $|b_{n+2}|<|b_{n+1}|<|b_n|/2.$
(3).(ii). If $b_n>0$ we have $b_{n+1}<0$ so $|b_{n+2}|<|b_{n+1}|/2<|b_n|/2.$
|
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|
How to calculate approximation of integral by developing integrant into power series? How to calculate approximation $I = \int_{-1}^{1} e^{-x^{2}} $ by developing integrant into power series, and how many terms of series is needed, so error is smaller than $\varepsilon = 10^{-3}$ ?
We can rewrite this $I = \int_{-1}^{1} e^{-x^{2}} $ as
$I = \int_{-1}^{1} \sum_{n=0}^{\infty}\frac{(-x)^{2n}}{n!}$
and when we integrate it we get
$2\sum_{n=0}^{\infty} \frac{1}{n!(2n+1)}$ .
How to calculate from here approximation and how many terms of series are needed, so error is smaller than $\varepsilon = 10^{-3}$ ?
|
Since
$$
e^{-x^2}=\sum_{n=0}^\infty\dfrac{(-x^2)^n}{n!}=\sum_{n=0}^\infty\dfrac{(-1)^n}{n!}x^{2n},
$$
we have
$$
I=\int_{-1}^1e^{-x^2}\,dx=2\int_0^1e^{-x^2}\,dx=2\sum_{n=0}^\infty\frac{(-1)^n}{n!(2n+1)}\equiv \sum_{n=0}^\infty(-1)^nb_n
$$
with
$$
b_n=\dfrac{2}{n!(2n+1)}.
$$
Since the sequence $\{b_n\}$ is positive and decreasing, if we set
$$
s_n=\sum_{k=0}^n(-1)^kb_k, \quad r_n=\sum_{k=n+1}^\infty(-1)^kb_k,
$$
then
\begin{eqnarray}
|r_n|&=& \left|\sum_{k=n+1}^\infty (-1)^kb_k\right|=\left|\sum_{k=0}^\infty (-1)^{n+1+k}b_{n+1+k}\right|=\left|\sum_{k=0}^\infty (-1)^kb_{n+1+k}\right|\\
&=&\left|b_{n+1}-b_{n+2}+b_{n+3}-b_{n+4}+b_{n+5}-\ldots\right|\\
&=&b_{n+1}-b_{n+2}+b_{n+3}-b_{n+4}+b_{n+5}-\ldots\\
&=&b_{n+1}-(b_{n+2}-b_{n+3})-(b_{n+4}-b_{n+5})-\ldots\\
&\le&b_{n+1}.
\end{eqnarray}
Hence, if we approximate $I$ by $s_n$, then the error is
$$
|I-s_n|=|r_n|\le b_{n+1}
$$
We just have to find $n$ such that $b_n\le 10^{-3}$. We have
$$
\begin{array}{l|c|c|c|c|c|c|c}
n& 0&1&2&3&4&5&6\\
\hline
b_n&2&\dfrac23&\dfrac15&\dfrac{1}{21}&\dfrac{1}{108}&\dfrac{1}{660}&\dfrac{1}{4680}
\end{array}
$$
We see that
$$
|I-s_5|\le |r_6|\le b_6=\dfrac{1}{4680}<\dfrac{1}{1000},
$$
therefore we can approximate $I$ with $s_5$. Thus
$$
I\approx s_5=2-\dfrac23+\dfrac15-\dfrac{1}{21}+\dfrac{1}{108}-\dfrac{1}{660}=\dfrac{31049}{20790}=1.493458393.
$$
|
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|
Easy way to decompose $\frac{1}{X^{3}\cdot (X-2)^{3}}$ into partial fractions? Is there any easy way or shortcut to decompose $$\frac{1}{X^{3}\cdot (X-2)^{3}}$$ into partial fractions ? because dealing with the usual way of replacing and giving values to $X$ is too clumsy in this case , so I was wondering if there is any quick way to find the decomposition of this rational function ?
|
We may consider that:
$$ \frac{1}{t-1}-\frac{1}{t+1} = \frac{2}{(t-1)(t+1)}\tag{1} $$
so:
$$\begin{eqnarray*} \frac{8}{(t-1)^3 (t+1)^3} &=& \frac{1}{(t-1)^3}-\frac{3}{(t-1)^2(t+1)}+\frac{3}{(t-1)(t+1)^2}-\frac{1}{(t+1)^3}\\&=&\frac{1}{(t-1)^3}-\frac{1}{(t+1)^3}+\frac{3}{2}\frac{2}{(t-1)(t+1)}\left(\frac{1}{t+1}-\frac{1}{t-1}\right)\\&=&\frac{1}{(t-1)^3}-\frac{1}{(t+1)^3}-\frac{3}{2}\left(\frac{1}{t-1}-\frac{1}{t+1}\right)^2\\&=&\frac{1}{(t-1)^3}-\frac{1}{(t+1)^3}-\frac{3/2}{(t-1)^2}-\frac{3/2}{(t+1)^2}+\frac{3}{(t-1)(t+1)}\\&=&\frac{1}{(t-1)^3}-\frac{1}{(t+1)^3}-\frac{3/2}{(t-1)^2}-\frac{3/2}{(t+1)^2}+\frac{3/2}{t-1}-\frac{3/2}{t+1}\tag{2}\end{eqnarray*}$$
and now it is enough to replace $t$ with $X-1$.
Alternative approach. Assuming that
$$\frac{1}{(1-t)^3(1+t)^3}=\frac{A}{(1-t)^3}+\frac{B}{(1-t)^2}+\frac{C}{1-t}+\frac{D}{1+t}+\frac{E}{(1+t)^2}+\frac{F}{(1+t)^3}\tag{3}$$
we have $A=F,B=E$ and $C=D$ since the LHS is an even function.
So, it is enough to find $A,B,C$. If we set $g(t)=\frac{1}{(1-t^2)^3}$, we have:
$$ A = \lim_{t\to 1}(1-t)^3 g(t) = \lim_{t\to 1}\frac{1}{(1+t)^3}=\frac{1}{8}\tag{4}$$
and $A+B+C+D+E+F = g(0)=1$, so it is enough to find $B$, for instance through:
$$ B = \lim_{t\to 1}(1-t)^2\left(g(t)-\frac{1}{8(1-t)^3}\right) = \lim_{t\to 1}\frac{7+4t+t^2}{8(1+t)^3}=\frac{3}{16}.\tag{5}$$
|
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|
Alternative form of Eisenstein integers I just recently got into number theory and algebra so my knowledge is very limited. I stumbled upon the Eisenstein integers $\mathbb{Z}[\omega]$. They are of the form $\varepsilon = a + b \omega$, where $\omega = e^{2 \pi i/3} = -1/2 + i \sqrt{3}/2$.
Now my problem:
I want to show that you can write every Eisenstein integer in the form of $\varepsilon = \frac{a + bi \sqrt{3}}{2}$ where $a \equiv b \bmod 2.$
For now I got: $\omega = -1/2 + i \sqrt{3}/2 \Rightarrow \varepsilon = a + b \omega = a - b/2 + i \sqrt{3}b/2$ then $a = b + 2k \Rightarrow b = a - 2k \Rightarrow \varepsilon = a - a/2 + k + i \sqrt{3}b/2 = \frac{a + bi\sqrt{3}}{2} + k$.
Now I still have to deal with the superfluous $k \in \mathbb{Z}$. How is the argumentation to get rid of it?
|
I think this excessive emphasis on $\omega$ often serves to confuse students. Don't get me wrong, $\omega$ is important. $$\omega = -\frac{1}{2} + \frac{\sqrt{-3}}{2},$$ and $N(\omega) = 1$, which means that $\omega$ is a unit. And furthermore $\omega$ is a complex cubic root of 1.
Let's review the norm function for numbers in $\mathbb{Q}(\sqrt{-3})$: $$N\left(\frac{a}{2} + \frac{b \sqrt{-3}}{2}\right) = \frac{a^2}{4} + \frac{3b^2}{4}.$$ An algebraic number in $\mathbb{Q}(\sqrt{-3})$ is also an algebraic integer if its norm is an integer from $\mathbb{Z}$.
If both $a$ and $b$ are even, then $a^2$ and $3b^2$ are both multiples of 4, so that when you divide them by 4, you still have integers.
If both $a$ and $b$ are odd, then $a^2 \equiv 1 \pmod 4$ and $b^2 \equiv 1 \pmod 4$ as well. But then $3b^2 \equiv 3 \pmod 4$, so that $a^2 + 3b^2 \equiv 0 \pmod 4$. Then, when you divide $a^2 + 3b^2$ by 4, you still have an integer.
Now, how to convert $$m + n \omega = \frac{a}{2} + \frac{b \sqrt{-3}}{2},$$ with $m$ and $n$ both integers from $\mathbb{Z}$?
Well, $a = 2m - n$ and $b = n$.
If both $m$ and $n$ are even, we have no problem, as $a$ and $b$ are as well.
If both $m$ and $n$ are odd, $a$ and $b$ are as well.
If $m$ is odd but $n$ is even, we have $2m$ even and $2m - n$ is still even, so $a$ and $b$ are both even as desired.
If $m$ is even but $n$ is odd, $2m - n$ is odd, so $a$ and $b$ are both odd as desired.
For exercises, I suggest the following: convert the numbers $-\omega$, $-2 + \omega$, $3 - \omega$ and $4 + \omega$ to the $a$ and $b$ form. Also, compute their norms. Bonus: figure out the conversion for $m + n \omega^2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2097658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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|
Proof of an inequality using Cauchy's inequality: $\sum_{k=1}^n \left(p_k+\frac{1}{p_k}\right)^2\geq n^3+2n+\frac{1}{n}$ Given that $p_k> 0$ and $p_1+p_2+\cdots+p_n=1$, prove that
\begin{equation}
\sum_{k=1}^n \left(p_k+\frac{1}{p_k}\right)^2\geq
n^3+2n+\frac{1}{n}.
\end{equation}
I believe that Cauchy's inequality should be used at some point but I haven't figured out how. Expanding the square on the left-hand side gives $2n$ immediately, but I have problem producing the cubic term and $\frac{1}{n}$. Could anyone please offer some insight? It is ok to use any method, not necessarily Cauchy's inequality.
|
The question is from chapter 1 of J. Michael Steele's book ``The Cauchy-Schwarz Master Class''. I didn't know that the book has solutions at the back. Below is the author's solution, which only employs Cauchy's inequality.
First, we expand the square.
\begin{equation}
\sum_{k=1}^n \left(p_k+\frac{1}{p_k}\right)^2 =
\sum_{k=1}^n p_k^2 + 2n + \sum_{k=1}^n \frac{1}{p_k^2}.
\end{equation}
Consider the first sum. Notice that
\begin{equation}
\left(\sum_{k=1}^n p_k^2 \right)
\left(\sum_{k=1}^n 1\right) \geq
\left(\sum_{k=1}^n p_k\right)^2=1,
\end{equation}
from which we obtain
\begin{equation}
\sum_{k=1}^n p_k^2 \geq \frac{1}{n}.
\end{equation}
Now, consider the second sum, for which we first consider the sum $\sum_{k=1}^n\frac{1}{p_k}$. Observe that
\begin{equation}
\sum_{k=1}^n 1 = \sum_{k=1}^n \sqrt{p_k}\frac{1}{\sqrt{p_k}} \leq
\left(\sum_{k=1}^n p_k\right)^{\frac{1}{2}}
\left(\sum_{k=1}^n \frac{1}{p_k} \right)^{\frac{1}{2}}.
\end{equation}
It follows that
\begin{equation}
\sum_{k=1}^n \frac{1}{p_k} \geq n^2.
\end{equation}
Next, we can obtain a lower bound for $\sum_{k=1}^n\frac{1}{p_k^2}$. By Cauchy's inequality,
\begin{equation}
\sum_{k=1}^n\frac{1}{p_k} = \sum_{k=1}^n\frac{1}{p_k} 1 \leq
\left(\sum_{k=1}^n\frac{1}{p_k^2}\right)^{\frac{1}{2}}
\left(\sum_{k=1}^n 1^2\right)^{\frac{1}{2}}.
\end{equation}
Hence,
\begin{equation}
\left(\sum_{k=1}^n\frac{1}{p_k^2}\right)^{\frac{1}{2}}
\left(\sum_{k=1}^n 1^2\right)^{\frac{1}{2}} \geq
\sum_{k=1}^n\frac{1}{p_k} \geq n^2.
\end{equation}
Squaring both sides, we obtain
\begin{equation}
\left(\sum_{k=1}^n\frac{1}{p_k^2}\right) n \geq n^4.
\end{equation}
Therefore,
\begin{equation}
\sum_{k=1}^n\frac{1}{p_k^2} \geq n^3.
\end{equation}
The conclusion that
\begin{equation}
\sum_{k=1}^n \left(p_k+\frac{1}{p_k}\right)^2 \geq
n^3 + 2n + \frac{1}{n}
\end{equation}
follows.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2101222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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|
Finding $\lim_{x\rightarrow 0}\lfloor \frac{2017 \sin x}{x} \rfloor +\lfloor \frac{2017 \tan x}{x} \rfloor $ Finding value of $\displaystyle \lim_{x\rightarrow 0}\bigg\lfloor \frac{2017 \sin x}{x}\bigg \rfloor +\bigg\lfloor \frac{2017 \tan x}{x}\bigg \rfloor,$ where $\lfloor x \rfloor $ is floor function of $x$
Attempt as we know $\sin x< x < \tan x$
wan,t be able to go further, could some help me, thanks
|
Yes, the key is that $\sin x<x<\tan x$ for $x\to0^{+}$, which implies $\displaystyle \frac{\sin x}{x}<1<\frac{\tan x}{x}$. Since after division both expressions are even functions, this double inequality is true both as $x\to0^{+}$ and as $x\to0^{-}$. We also know that both $\displaystyle \lim_{x\to0}\frac{\sin x}{x}=1$ and $\displaystyle \lim_{x\to0}\frac{\tan x}{x}=1$, but due to the inequalities, $\sin(x)/x$ approaches $1$ from below, while $\tan(x)/x$ approaches $1$ from above.
Since $\displaystyle \frac{\sin x}{x}<1$ and $\displaystyle \lim_{x\to0}\frac{\sin x}{x}=1$, for $x$ sufficiently close to $0$ we have:
$$1-\frac{1}{2017}<\frac{\sin x}{x}<1 \implies 2016<\frac{2017\sin x}{x}<2017 \implies \left\lfloor\frac{2017\sin x}{x}\right\rfloor=2016.$$
Similarly, since $\displaystyle \frac{\tan x}{x}>1$ and $\displaystyle \lim_{x\to0}\frac{\tan x}{x}=1$, for $x$ sufficiently close to $0$ we have:
$$1<\frac{\tan x}{x}<1+\frac{1}{2017} \implies 2017<\frac{2017\tan x}{x}<2018 \implies \left\lfloor\frac{2017\tan x}{x}\right\rfloor=2017.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2103227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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|
Intersection of 3 circles on equilateral triangle given the difference in their radii Imagine we are given three intersecting circles centered on the vertices of an equilateral triangle of side length $1$, with $(0,0)$ arbitrarily placed at the bottom left corner. The circles have radii $r$, $r+a$, and $r+b$ respectively (going clockwise again arbitrarily), where $r$ is unknown but $a$ and $b$ are given. For sake of ease we can assume $0<a<b$ and that $a$ and $b$ are "safe," aka a solution is possible. What is the intersection $(x,y)$ of the three circles (in terms of $a$ and $b$) inside the triangle?
I apologize for the crude nature of the diagram, but it was the best I could come up with using MS Paint. Additionally, I'm 99.9% there's a unique solution as long as we restrict our solution set to points inside the triangle, but this may not be the case. Thank you so much for any help.
Edit: if anybody has any bright ideas in this direction, the solution doesn't need to be exact; a close approximation with nicer algebra would be much appreciated.
|
Consider a general point $(x,y)$ inside the triangle,
and let the distance of $(x,y)$ from the vertices $(0,0),$
$\left(\frac12,\frac{\sqrt3}2\right),$ and $(1,0)$ be
$r_1$, $r_2,$ and $r_3$ respectively.
(That is, each of $r_1$, $r_2,$ and $r_3$ is a function of $(x,y)$.)
We want to find a point $(x,y)$ such that
$r_1 = r$, $r_2 = r+a,$ and $r_3 = r+b,$ if such a point exists,
where $r,$ $a,$ and $b$ are given.
Among other things, $(x,y)$ must satisfy the condition $r_3 - r_1 = b.$
This condition defines one branch of
a hyperbola
with foci at $(0,0)$ and $(1,0)$, semi-major axis $\frac b2,$
and eccentricity $\frac 1b.$
In polar coordinates $(\rho,\theta)$, the equation of this hyperbola is
$$
\rho = \frac{\frac b2\left(\left(\frac 1b\right)^2 - 1\right)}
{1+\frac 1b\cos\theta}
= \frac{1 - b^2}{2(b + \cos\theta)}
$$
But $(x,y)$ must also satisfy $r_2 - r_1 = a,$ which puts it on a hyperbola
with foci at $(0,0)$ and $\left(\frac12,\frac{\sqrt3}2\right),$
whose equation is
$$
\rho = \frac{1 - a^2}{2\left(a + \cos\left(\theta - \frac\pi3\right)\right)}
$$
Since the desired point $(x,y)$ must satisfy $r_3 - r_1 = b$ and
$r_2 - r_1 = a$ simultaneously, we have
$$
\frac{1 - b^2}{2(b + \cos\theta)}
= \frac{1 - a^2}{2\left(a + \cos\left(\theta - \frac\pi3\right)\right)}.
$$
Cross-multiply, use the fact that
$\cos\left(\theta-\frac\pi3\right) =
\frac{\sqrt3}2 \sin\theta + \frac12 \cos\theta,$
and collect everything except the $\sin\theta$ terms together,
and we have
$$
\frac{\sqrt3}2(1 - b^2)\sin\theta
= \left(\frac12 - a^2 + b^2\right) \cos\theta + b(1 - a^2) - a(1 - b^2).\tag1
$$
Let $u=\cos\theta$ (so that $\sin^2\theta = 1-u^2$);
let $n=\frac34(1-b^2)^2,$ $m=\frac12 - a^2 + b^2,$
and $k=b(1 - a^2) - a(1 - b^2)$;
rewrite Equation $1$ in those terms; and square both sides. The result is
$$
n(1-u^2) = (mu +k)^2,
$$
which is equivalent to
$$
(m^2 + n)u^2 + 2mku + k^2 - n = 0.
$$
Solving for $u$ via the quadratic formula,
\begin{align}
u &= \frac{-2mk \pm \sqrt{4m^2k^2 - 4(m^2k^2+k^2n-m^2n-n^2)}}{2(m^2+n)} \\
&= \frac{-mk \pm \sqrt{m^2n+n^2-k^2n}}{m^2+n} \\
\end{align}
The conditions of the problem require $0\leq\theta\leq\frac\pi3$,
therefore $\cos\theta \geq \frac12.$ Moreover, $m>0$ and $k>0.$
It follows that only the $+$ case of the $\pm$ sign can possibly
lead to a solution to the original problem.
Wolfram Alpha indicates that
$m^2n+n^2-k^2n$ is positive when $0 < a < b < 1$,
so we don't need to check the sign before taking the square root.
So to find $(x,y)$ given $a$ and $b,$ we set $n,$ $m,$ and $k$ as
described above, then compute the following quantities in the sequence shown:
\begin{align}
u &= \frac{-mk + \sqrt{m^2n+n^2-k^2n}}{m^2+n}, \\
\rho &= \frac{1 - b^2}{2(b + u)}, \\
x &= \rho u, \\
y &= \rho \sqrt{1 - u^2}.
\end{align}
Done!
|
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"url": "https://math.stackexchange.com/questions/2103984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
If $f(g(x)) = 4x^2-8x$ and $f(x)=x^2-4$, then what's the value of $g(x)$? I'm a little stuck with this simple function:
If ${f(g(x)) = 4x^2-8x}$ and ${f(x)=x^2-4}$, then what's the value of ${g(x)?}$
Any tips?
|
\begin{align}
f(g(x))&=4x^2-8x\\
g(x)^2-4&=4x^2-8x\\
g(x)^2&=4x^2-8x+4\\
&=4(x^2-2x+1)\\
&=4(x-1)^2\\
&=[\ \!2(x-1)\ \!]^2\\
\implies g(x)&=\pm2(x-1)
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2104899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
}
|
Infinite sum including logarithm I would like to calculate the following sum:
$$\sum_{n=1}^{\infty}\ln \left( \frac{n^2+2n+1}{n^2+2n} \right)$$
I do know that it converges but I have gone that far:
\begin{align}
& \sum_{n=1}^{\infty}\ln \left( \frac{n^2+2n+1}{n^2+2n} \right) \Longleftrightarrow \sum_{n=1}^\infty \ln \left( \frac{(n+1)^2}{n(n+2)} \right) \\[10pt]
= {} & \ln \left( \frac{4}{3} \right)+\ln \left( \frac{9}{8} \right)+\ln \left( \frac{16}{15} \right)+\cdots+\ln \left( \frac{n}{n-1} \right)\\[10pt]
= {} & \ln \left( \frac{4}{3}\frac{9}{8}\frac{16}{15} \cdots \frac{n}{(n-1)} \right)=\ln (n)
\end{align}
which diverges as $n\to \infty.$
It looked like telescoping in the beginning but now I am confused. Where have I gone wrong?
Thanks.
|
$\begin{array}\\
\sum_{n=1}^{m}\ln \left( \frac{n^2+2n+1}{n^2+2n} \right)
&=\sum_{n=1}^{m}\ln \left( \frac{(n+1)^2}{n(n+2)} \right)\\
&=\sum_{n=1}^{m}( (2\ln(n+1)-\ln(n)-\ln(n+2))\\
&=2\sum_{n=1}^{m}\ln(n+1)-\sum_{n=1}^{m}\ln(n)-\sum_{n=1}^{m}\ln(n+2)\\
&=2\sum_{n=2}^{m+1}\ln(n)-\sum_{n=1}^{m}\ln(n)-\sum_{n=3}^{m+2}\ln(n)\\
&=2(\ln(2)+\ln(m+1)+\sum_{n=3}^{m}\ln(n))-(\ln(1)+\ln(2)+\sum_{n=3}^{m}\ln(n))-(\ln(m+1)+\ln(m+2)+\sum_{n=3}^{m}\ln(n))\\
&=2(\ln(2)+\ln(m+1))-(\ln(2))-(\ln(m+1)+\ln(m+2))\\
&=\ln(2)+\ln(m+1)-\ln(m+2)\\
&=\ln(2)+\ln(1-\frac{1}{m+2})\\
&\to \ln(2)
\qquad\text{since } \ln(1-\frac{1}{m+2}) \to 0
\text{ as } m \to \infty\\
\end{array}
$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.