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Volume of a tetrahedron whose 4 faces are congruent. Suppose that I have a tetrahedron such that all four faces consist of congruent triangles, says with the lengths $a,b$ and $c$ for each side. Is there a beautiful method to compute its volume?
PS. The reason for me tagging calculus and linear algebra is that I figured that the technique used to calculate such a problem may come from this areas.
|
The volume of a tetrahedron satisfies
$$36V^2 = a^2 b^2 c^2\left(\;1+2\cos\alpha\cos\beta\cos\gamma-\cos^2\alpha-\cos^2\beta-\cos^2\gamma\;\right) \tag{1}$$
where $a$, $b$, $c$ are lengths of edges coinciding at a vertex, and $\alpha$, $\beta$, $\gamma$ are the angles between those edges ($\alpha$ between $b$ and $c$, etc).
For the tetrahedron in question, we see that $a$, $b$, $c$ and $\alpha$, $\beta$, $\gamma$ are also elements of a triangle. Since $\alpha+\beta+\gamma = 180^\circ$, the trig factor of $(1)$ reduces, and we have
$$36 V^2 = a^2 b^2 c^2 \cdot 4 \cos \alpha \cos \beta \cos \gamma \quad\to\quad 9V^2 = a^2b^2c^2\cos\alpha\cos\beta\cos\gamma \tag{2}$$
which, by the Law of Cosines, we can write as
$$72V^2 = (-a^2+b^2+c^2)(a^2-b^2+c^2)(a^2+b^2-c^2) \tag{3}$$
Another way to get at this result is with the Pseudo-Heron Formula for volume. (See my note, "Heron-like Hedronometric Results for Tetrahedral Volume" (PDF).) If $W$, $X$, $Y$, $Z$ are the face-areas of a tetrahedron, and $H$, $J$, $K$ are the pseudoface-areas (see below) then
$$81V^4 = \left|\;\begin{array}{ccc}
H^2 & X Y - W Z & Z X - W Y \\
X Y - W Z & J^2 & Y Z - W X \\
Z X - W Y & Y Z - W X & K^2
\end{array}\;\right| \tag{4}$$
In an equihedral tetrahedron ($W=X=Y=Z$), this reduces to
$$81V^4 = \left|\;\begin{array}{ccc}
H^2 & 0 & 0 \\
0 & J^2 & 0 \\
0 & 0 & K^2
\end{array}\;\right| = H^2 J^2 K^2 \quad\to\quad 9 V^2 = H J K \tag{5}$$
A pseudoface of a tetrahedron is the quadrilateral shadow of the figure in a plane parallel to two opposite edges. If edges $a$, $b$, $c$ meet at a vertex, and have respective opposite edges $d$, $e$, $f$, then our $H$, $J$, $K$ are related to respective edge-pairs $(a,d)$, $(b,e)$, $(c,f)$, and we have, for instance,
$$16 H^2 = 4 a^2 d^2 - \left(\; b^2 - c^2 + e^2 - f^2 \;\right)^2 \tag{6}$$
In the tetrahedron in question, $a=d$, $b=e$, $c=f$, so that
$$\begin{align}
4 H^2 = a^4 - \left( b^2 - c^2 \right)^2 &= \left(\phantom{-}a^2 - b^2 + c^2 \right)\left(\phantom{-}a^2 + b^2 - c^2 \right)\\
4 J^2 \;\;\phantom{= a^4 - \left( b^2 - c^2 \right)^2} &= \left(\phantom{-}a^2+b^2-c^2\right)\left(-a^2+b^2+c^2\right) \\
4 K^2 \;\phantom{= a^4 - \left( b^2 - c^2 \right)^2} &= \left(
-a^2+b^2+c^2\right)\left(\phantom{-}a^2-b^2+c^2\right)
\end{align}\tag{7}$$
and then $(3)$ follows from $(5)$.
Edited to add some hedronometric context to @Calum's answer ...
*
*The rectangular faces of the tetrahedron's bounding cuboid are exactly the figure's pseudofaces. Indeed, we can deduce that the pseudofaces must be rectangles: each corresponding pseudofacial projection of a tetrahedron with three pairs of opposite congruent edges is necessarily a quadrilateral with two pairs of opposite congruent edges (hence, a parallelogram) and a pair of congruent diagonals (hence, a rectangle).
*Specifically, the rectangular pseudoface $H$ (the projection into a plane parallel to the $a$ edges) has congruent diagonals $a$, and edges $b^\prime := \sqrt{b^2-h^2}$ and $c^\prime := \sqrt{c^2-h^2}$, where $h$ is distance between planes containing the $a$ edges (that is, $h$ is the corresponding "height" of the cuboid, what I call a pseudoaltitude of the tetrahedron). Since $a^2 = (b^\prime)^2 + (c^\prime)^2$ in the rectangle, we have that $h^2=\left(-a^2+b^2+c^2\right)/2$, whence $b^\prime = \sqrt{\left(a^2-b^2+c^2\right)/2\;}$ and $c^\prime=\sqrt{\left(a^2+b^2-c^2\right)/2\;}$. As $H = b^\prime c^\prime$, we reconfirm equation $(7)$.
*As shown in @Calum's cuboid figure, each face of the given tetrahedron is the "hypotenuse-face" of a right-corner tetrahedron whose "leg-faces" are the three half-pseudofaces. (I hadn't really noticed this before!) By de Gua's Theorem , $$\left(\frac12H\right)^2+\left(\frac12J\right)^2+\left(\frac12K\right)^2 = W^2 \quad\to\quad H^2 + J^2 + K^2 = 4 W^2$$ This is consistent with the Sum of Squares identity that holds for any tetrahedron: $$H^2 + J^2 + K^2 = W^2 + X^2 + Y^2 + Z^2$$
*@Calum expresses the tetrahedron's volume in terms of pseudoaltitudes, which I tend to label $h$, $j$, $k$, instead of $x$, $y$, $z$; my $(5)$ expresses the volume in terms of pseudoface areas $H$, $J$, $K$. As it happens, any tetrahedron's volume is given by each pseudoface-pseudoaltitude pair: $$3 V = Hh = Jj = Kk$$ This gives us a bridging relation between the two formulas: $$27 V^3 = H J K \cdot h j k$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Calculate $A^5 - 27A^3 + 65A^2$, where $A$ is the matrix defined below. If $A=\begin{bmatrix} 0 & 0 & 1 \\
3 & 1 & 0 \\
-2&1&4\end{bmatrix}$, find $A^5 - 27A^3 + 65A^2$
$$A=\begin{bmatrix} 0 & 0 & 1 \\
3 & 1 & 0 \\
-2&1&4\end{bmatrix}$$
Let $\lambda$ be its eigenvalue, then
$$(A-\lambda I) = \begin{bmatrix} 0-\lambda & 0 & 1 \\
3 & 1-\lambda & 0 \\
-2&1&4-\lambda\end{bmatrix}$$
$$|A-\lambda I| = -(\lambda)^3 + 5(\lambda)^2 - 6(\lambda) +5$$
Using Cayley-Hamilton theorem
$$A^3-5^2+6A-5=0$$
How do I use this find $A^5 - 27A^3 + 65A^2$?
|
Doing long division:
$$A^5-27A^3+65A^2=(A^2+5A-8)(\underbrace{A^3-5A^2+6A-5}_{=0})+73A-40.$$
Hence it is:
$$A^5-27A^3+65A^2=73A-40I=\begin{pmatrix}-40&0&73\\ 219&33&0\\-146&73&252\end{pmatrix}.$$
|
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|
positive integer solutions to $x^3+y^3=3^z$ I am seeking all positive integer solutions to the equation $x^3+y^3=3^z$.
After doing number crunching, I think there are no solutions. But I am unable to prove it.
Attempt
If $x$ and $y$ have common divisor $d$, we have $d^3(m^3+n^3)=3^z$. So $d$ must be a power of $3$, and we are back to where we started. So we assume $x$ and $y$ are coprime.
Testing the parity, we have sum of 2 cubes to be odd. WLOG, we can assume $x$ is even and $y$ is odd.
Trying mod $3$, we have $x+y=0 \pmod 3$. Since $x$ and $y$ are coprime, $x$ and $y$ must be congruent to $1$ and $-1$ or vice-versa.
If I assume $x=3m+1$ and $y=3n-1$, expand out and simplify, I get $27(m^3+n^3)+27(m^2-n^2)+9(m+n)=3^z$. If I assume $z \geq 3$, this gives $(m^3+n^3)+(m^2-n^2)+\frac{m+n}{3}=3^{z-3}$. But I don't see how to proceed.
I also tried mod $9$ but didn't get anywhere, it didn't cut down the possibilities by much.
I also tried letting $y=x+r$. Then
\begin{align*}
x^3+y^3 &= x^3+(x+r)^3 \\
&= x^3 + (x^3+3x^2r+3xr^2+r^3) \\
&= 2x^3+3x^2r+3xr^2+r^3 \\
&= 3^z
\end{align*}
Then $3\mid 2x^3+3x^2r+3xr^2+r^3$, and $3\mid 3x^2r+3xr^2$, so this implies $3 \mid 2x^3+r^3$. But this doesn't yield any contradiction.
Can anyone supply a proof? Or if my hypothesis is wrong, how to derive all the integer solutions?
Thank you.
|
$27(m^3+n^3) +27(m^2-n^2) + 9(m+n) = 3^z$
$(m+n)(3m^2 +3n^2-3mn+3m-3n+1) = 3^{z-2}$
since $z \ne 2$, we have $3(m^2 +n^2-mn+m-n)+1$ divides a power of $3$ or is equal to $1$, both of which is not possible.
|
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Elementary proof that $24^n$ is the sum of 3 perfect squares Proof that $24^n$ is the sum of 3 perfect squares for every $n >= 1$ . I found this problem in a math contest for 5th graders http://www.mategl.com/atasamente%20pentru%20site/OLM%202014%20Prahova%20(5-12).rar , so obviously a very very elementary proof is required (definitely not induction, very little algebra).
It is easy to figure out that $24^1 = 2^2 + 2^2 + 2^4 $ and $24^2 = 2^6 + 2^8 +2^8$ but I could not see the pattern from here. I also thought about $24^1 = 2^3 + 2^4 $ and $24^2 = 2^6 + 2^9 $ but I can't see a pattern.
|
\begin{align}
24^{2n + 2} &= 24^2 * 24^{2n} = (2^6 + 2^8 + 2^8)* 24^{2n} = (8* 24^n)^2 + (16*24^n)^2 + (16*24^n)^2 \\
24^{2n+1} &= 24 * 24^{2n} = ( 2^2 + 2^2 + 2^4 ) * 24^{2n} = (2*24^n)^2 + (2*24^n)^2 + (4*24^n)^2\\
\end{align}
P.S: Thus the OP was the right track, if they would have computed it for $24^3$ and $24^4$, they would have been able to observe a pattern.
|
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|
Probability of trials without replacement using conditional probability
A bag contains $5$ red marbles and $3$ black marbles. Three marbles are drawn one by one without replacement. What is the probability that at least one of the three marbles drawn be black, if the first marble is red?
Attempt:1
$$
E:\text{first marble is red}\\
F:\text{at least one of the marbles drawn be black}\\
F':\text{none of the marbles drawn be black}\\
n(S)=8*7*6\\
n(F'\cap E)=n(R_1R_2R_3)=5*4*3\\
n(E)=n(\{(R_1R_2R_3),(R_1B_2R_3),(R_1R_2B_3),(R_1B_2B_3)\})\\=5*4*3+5*3*4+5*4*3+5*3*2=60+60+60+30=210\\
\begin{align}
&P(\text{at least 1 of the 3 marbles black|1$^\text{st}$ marble red})
=P(F|E)\\&=1-P(F'|E)=1-\frac{60}{210}=1-\frac{2}{7}=\color{red}{\frac{5}{7}}
\end{align}
$$
Attempt:2
$$
\begin{align}
&P(\text{at least 1 of the 3 marbles black|1$^\text{st}$ marble red})
=P(F|E)\\&=1-P(F'|E)=1-P(R_1R_2R_3)\\&=1-P(R_1)P(R_2|R_1)P(R_3|R_1R_2)=1-\frac{5}{8}.\frac{\frac{5*4}{8*7}}{\frac{5}{8}}.\frac{\frac{5*4*3}{8*7*6}}{\frac{5*4}{8*7}}=1-\frac{5}{28}\\&=\color{blue}{\frac{23}{28}}
\end{align}
$$
OR
$$
\begin{align}
&P(\text{at least 1 of the 3 marbles black|1$^\text{st}$ marble red})=P(F|E)\\&=1-P(F'|E)=1-P(R_1R_2R_3)=1-\frac{5*4*3}{8*7*6}=1-\frac{5}{28}=\color{blue}{\frac{23}{28}}
\end{align}
$$
Attempt:3
$$
\begin{align}
&P(\text{at least 1 of the 3 marbles black|1$^\text{st}$ marble red})=P(F|E)\\&=P(R_1B_2R_3)+P(R_1R_2B_3)+P(R_1B_2B_3)\\&=\frac{5*3*4}{8*7*6}+\frac{5*4*3}{8*7*6}+\frac{5*3*2}{8*7*6}=\frac{10+10+5}{4*7*2}=\color{green}{\frac{25}{56}}
\end{align}
$$
The solution of the above problem is stated to be $\frac{25}{56}$ as in my Attempt 3, but why am I not getting the right solution in Attempt $1$ and Attempt $2$ ?
|
I use the elementary events for the answer. Firstly I use your definition of events:
$E:\text{first marble is red}\\
F:\text{at least one of the marbles drawn be black}\\$
You have to find the probability that you draw at least 3 black marbles given the first marble is red. This is $P(F|E)$
In total we have these 8 cases:
$$rrr, rrb, rbr, brr, bbr, brb, rbb, bbb$$
Now we list the cases where at least 1 black marble is drawn and the first marble is red.
$$rrb, rbr, rbb$$
The probability is $$P(F\cap E)=\frac{5}{8}\cdot \frac{4}{7}\cdot \frac{3}{6}+\frac{5}{8}\cdot \frac{3}{7}\cdot \frac{4}{6}+\frac{5}{8}\cdot \frac{3}{7}\cdot \frac{2}{7}=\frac{25}{56}$$
Now we calculate P(A). The events are $rrr, rrb, rbr, rbb$ Therefore
$$P(E)=\frac{5}{8}\cdot \frac{4}{7}\cdot \frac{3}{6}+\frac{5}{8}\cdot \frac{4}{7}\cdot \frac{3}{6}+\frac{5}{8}\cdot \frac{3}{7}\cdot \frac{4}{6}+\frac{5}{8}\cdot \frac{3}{7}\cdot \frac{2}{6}=\frac{5}{8}$$
Now we apply the Bayes theorem
$$P(F|E)=\frac{P(F\cap E)}{P(E)}=\frac{\frac{25}{56}}{\frac{5}{8}}=\frac{25}{56}\cdot \frac{8}{5}=\frac{5}{56}\cdot \frac{8}{1}=\frac{5}{7}\cdot \frac{1}{1}=\boxed{\frac57}$$
$$ \begin{align} &P(\text{at least 1 of the 3 marbles
black|1$^\text{st}$ marble
red})=P(F|E)\\&=1-P(F'|E)=1-P(R_1R_2R_3)=1-\frac{5*4*3}{8*7*6}=1-\frac{5}{28}=\color{blue}{\frac{23}{28}}
\end{align} $$
Here you miss the Bayes theorem as well.
$$P(F'|E)=\frac{P(F'\cap E)}{P(E)}$$
$P(F'\cap E)=\frac{5\cdot 4\cdot 3}{8\cdot 7\cdot 6}$
And $P(E)=\frac58$ Thus
$$P(F|E)=1-P(F'|E)=1-\frac{P(F'\cap E)}{P(E)}=1-\frac{\frac{5\cdot 4\cdot 3}{8\cdot 7\cdot 6}}{\frac58}=1-\frac27=\boxed{\frac57}$$
Many ways lead to Rome.
|
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|
Why can't $\frac{1}{\sin(x)\cos(x)}$ be expressed in the form $\frac{A}{\sin x}+\frac{B}{\cos x}$ I've tried expressing $\frac{1}{\sin(x)\cos(x)}$ as partial fractions:
$$\frac{1}{\sin(x)\cos(x)} = \frac{A}{\sin x}+\frac{B}{\cos x}
\implies 1=A\cos x + B\sin x$$
I let $x=\frac{\pi}{2}$, getting $A=-1$. Then I let $x=\pi$, getting $B=1$. This means that $1=\cos x-\sin x$ which is obviously wrong most of the time.
But why is it wrong? If the denominator had something like $(x^2-1)$ I still can get a correct answer for different values I substitute.
|
\begin{align}
\sin \sin x + \cos x \cos x &= 1 \\
\dfrac{\sin x \sin x}{\sin x \cos x} + \dfrac{\cos x \cos x}{\sin x \cos x}
&= \dfrac{1}{\sin x \cos x} \\
\dfrac{\sin x}{\cos x} + \dfrac{\cos x}{\sin x}
&= \dfrac{1}{\sin x \cos x} \\
\end{align}
|
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|
Prove $\sum_{i=1}^{n}i^3=\frac{1}{4} n^2(n+1)^2$ (induction) Problem
Prove $$\sum_{i=1}^{n}i^3=\frac{1}{4} n^2(n+1)^2$$
Attempt to solve
I would try to prove this with induction. We have sum and the sum as function of $p(n)$. Now i try to prove that the sum equals $p(n)$ with induction. $p(n)=\frac{1}{4}n^2(n+1)^2$
$$ 1^3+2^3+\cdots + \ n^3 = \frac{1}{4}n^2(n+1)^2 $$
Base case
$$ p(1)=\frac{1}{4}\cdot 1^2 (1+1)^2 = \frac{2^2}{4} = \frac{4}{4} = 1 $$
Induction step
Assume that $n \ge 1$ and $p(n)$ is valid.
$$ 1^3+2^3+ \dots + n^3 = \frac{1}{4} n^2(n+1)^2$$
substitute $n=n+1$
$$ 1^3+2^3 + \dots + n^3 + (n+1)^3 = \frac{1}{4} (n+1)^2 ((n+1)+1)^2 $$
$$ 1^3+2^3 + \dots + n^3 + (n+1)^3 = \frac{1}{4} (n+1)^2 (n+2)^2 $$
we know that
$$ 1^3+2^3+ \dots + n^3 = \frac{1}{4} n^2(n+1)^2$$
so we have
$$ \frac{1}{4}n^2(n+1)^2 + (n+1)^3 = \frac{1}{4}n^2(n+1)^2(n+2)^2 || \cdot 4 $$
$$ n^2(n+1)^2+4(n+1)^3=n^2(n+1)^2(n+2)^2 $$
$$ n^2(n+1)^2+4(n+1)^2(n+1)=n^2(n+1)^2(n+2)^2 || : (n+1)^2$$
$$ n^2+4(n+1)=n^2(n+2)^2 $$
$$ n^2+4n+4 = n^2(n^2+4n+4)$$
$$ n^2+4n +4 \neq n^2(n^2+4n+4) $$
Problem is there is extra $n^2$ on right side compared to left side which makes equation not true. Is there computational error or is the problem more fundamental perhaps i don't understand induction proofs well enough?
|
As @packsciences said, your mistake is after "so we have". The correct way to do it is:
$$\frac{1}{4}n^2(n+1)^2 + (n+1)^3 = (n+1)^2 (\frac{1}{4}n^2 + n+1) = \frac{1}{4}(n+1)^2 (n^2 + 4n + 4) = \frac{1}{4} (n+1)^2 (n+2)^2$$
|
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|
The limit of $\frac{n^3-3}{2n^2+n-1}$ I have to find the limit of the sequence above.
Firstly, I tried to multiply out $n^3$, as it has the largest exponent.
$$\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} =
\lim_{n\to\infty}\frac{n^3(1-\frac{3}{n^3})}{n^3(\frac{2}{n} + \frac{1}{n^2} - \frac{1}{n^3})} =
\lim_{n\to\infty}\frac{1-\frac{3}{n^3}}{\frac{2}{n} + \frac{1}{n^2} - \frac{1}{n^3}}$$
$$
\begin{align}
\lim_{n\to\infty}1-\frac{3}{n^3} = 1 \\[1ex]
\lim_{n\to\infty}\frac{2}{n} + \frac{1}{n^2} - \frac{1}{n^3} = 0 \\[1ex]
\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} = \frac{1}{0}
\end{align}
$$
Then, after realizing $\frac{1}{0}$ might not be a plausible limit, I tried to multiply out the variable with the largest exponent in both the dividend and the divisor.
$$\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} = \lim_{n\to\infty}\frac{n^3(1 - \frac{3}{n^3})}{n^2(2 + \frac{1}{n} - \frac{1}{n^2})} =
\lim_{n\to\infty}n\cdot\frac{1 - \frac{3}{n^3}}{2 + \frac{1}{n} - \frac{1}{n^2}}$$
$$
\begin{align}
\lim_{n\to\infty}1-\frac{3}{n^3} = 1 \\
\lim_{n\to\infty}2 + \frac{1}{n} - \frac{1}{n^2} = 2 \\
\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} = \frac{1}{2} \\
\lim_{n\to\infty}n = \infty
\end{align}
$$
So, my questions about this problem:
*
*Could $\frac{1}{0}$ be a valid limit?
*Does $\infty\cdot\frac{1}{2}$
equal to $\infty$?
*In conclusion, what is the limit of the sequence
above? $\infty?$
Thank you!
|
Could $\frac{1}{0}$ be a valid limit?
You should be asking instead :"Is the following true?" $$\lim_{n \rightarrow \infty} \frac{f(n)}{g(n)} = 0$$
where $\lim_{n \rightarrow \infty} f(n) = K $, where $K$ is finite and $\lim_{n \rightarrow \infty} g(n) = +\infty$. The answer is yes. This is one of the limit rules.
Does $\infty \frac{1}{2}$ equal to $\infty$?
You should be asking instead:"Is the following true?" $$\lim_{n \rightarrow \infty} K f(n)= \infty$$
where $K$ is finite and $\lim_{n \rightarrow \infty} f(n) = +\infty$. The answer is yes. This is one of the limit rules.
In conclusion, what is the limit of the sequence above?
Following the same steps as you did, in both approaches, I have nothing to add. The limit is $\infty$.
|
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Proof by induction, $1/2 + ... + n/2^n < 2$ So I'm having trouble with proving this homework question by induction.
$$
\frac{1}{2^1} + \frac{2}{2^2} + ... +\frac{n-1}{2^{n-1}} + \frac{n}{2^n} <2
$$
I know how to prove that the series converges to 2 (using things like the ratio method), but actually using induction is where I get confused.
Base case is easy, n=1.
$$
\frac{1}{2^1}<2
$$
Induction case we assume that
$$
\frac{1}{2^1} + \frac{2}{2^2} + ... +\frac{k-1}{2^{k-1}} + \frac{k}{2^k} <2
$$
Then we get to fun old induction. How do I show that
$$
\frac{1}{2^1} + \frac{2}{2^2} + ... +\frac{k-1}{2^{k-1}} + \frac{k}{2^k} + \frac{k+1}{2^{k+1}} <2 ?
$$
|
Denote $$S_n=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\cdots+\frac{n}{2^n}.\tag1$$Then $$\frac{1}{2}S_n=\frac{1}{2^2}+\frac{2}{2^3}+\frac{3}{2^4}+\cdots+\frac{n}{2^{n+1}}.\tag2$$Thus by $(1)-(2)$, we obtain $$\frac{1}{2}S_n=\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots+\frac{1}{2^n}\right)-\frac{n}{2^{n+1}}< 1-\frac{n}{2^{n+1}}<1,$$ which implies $$S_n < 2.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Fourier series of $f(x)=\sin^2x\cos^2x$ at $(-\pi,\pi)$ Find Fourier series of $f(x)=\sin^2 x \cos^2 x$ at $(-\pi,\pi)$
$f(x)$ is even so we only have to evaluate $a_0,a_n$
$$
a_0 = \frac{1}{\pi}
\int_{-\pi}^{\pi} \sin^2 x \cos^2 x dx
= \frac{1}{4\pi}
\int_{-\pi}^{\pi}\sin^2(2x)
= \frac{1}{4\pi}
\int_{-\pi}^{\pi}\frac{1-\cos 2x}{2} dx
= \frac{1}{4}
$$
and
$$
\begin{split}
a_n &= \frac{1}{\pi}
\int_{-\pi}^{\pi}\sin^2 x \cos^2 x \cos(nx)dx\\
&= \frac{1}{4\pi}
\int_{-\pi}^{\pi}\frac{\cos(nx)-\cos(2x)\cos(nx)}{2}dx\\
&=\frac{-1}{16\pi}\int_{-\pi}^{\pi}(\cos((n+2)x)+\cos((n-2)x))dx\\
&=0
\end{split}
$$
So $f(x)\approx\frac{1}{8}$
Is it correct?
|
Simply write
$$\sin^2x\cos^2x=\dfrac14\sin^22x=\dfrac18-\dfrac18\cos4x$$
|
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|
Number of ways to pay (generating functions) I've just started learning generating functions.
*
*Let $a_n$ be the number of ways in which you can pay $n$ dollars using 1 and 2 dollar bills. Find the generating function for $(a_0, a_1, a_2, \ldots)$ and general term $a_n$.
*Prove that the number of ways to pay $n$ dollars using only 1, 2, and 3 dollar bills is equal to closest integer of $\frac{(n+3)^2}{12}$?
What I've got so far:
Let $f(x) = \frac{1}{1-x}$ be the generating function for $(1, 1, 1, ...)$. $1 + x^2 + x^4 + \ldots$ is the generating function for $(1, 0, 1, 0, 1, \ldots)$ which is equal to $f(x^2) = \frac{1}{1-x^2} = \frac{1}{(1-x)(1+x)}$.
$$
\begin{align}
& (1 + x + x^2 + \ldots)(1 + x^2 + x^4 + \ldots) \\
&= \frac{1}{(1+x)(1-x)^2} \\
&= \frac{1}{4} \cdot \frac{1}{1+x} + \frac{1}{4} \cdot \frac{1}{1-x} + \frac{1}{2} \cdot \frac{1}{(1-x)^2} \\
&= \frac{1}{4} \Sigma_{n \geq 0}{-1 \choose n}x^n + \frac{1}{4} \Sigma_{n \geq 0}x^n + \frac{1}{2} \Sigma_{n \geq 0}{2 + n - 1 \choose n}x^n \\
&= \Sigma_{n \geq 0}\Bigl((-1)^n \frac{1}{4} x^n + \frac{1}{4} x^n + \frac{1}{2}{n + 1 \choose n} x^n \Bigr) \\
&= \Sigma_{n \geq 0}\Bigl((-1)^n \frac{1}{4} x^n + \frac{1}{4} x^n + \frac{1 \cdot 2}{2 \cdot 2}(n + 1) x^n \Bigr) \\
&= \Sigma_{n \geq 0} \frac{1}{4} \Bigl((-1)^n x^n + x^n + 2(n + 1) x^n \Bigr)
\end{align}
$$
I don't know is this is correct nor how to proceed.
I guest that we use similar method to solve problem (2)?
|
we have with $n_1$ 1 dollar bills and $n_2$ 2 dollar bills we have: $n_1+2n_2 = n$. which has a solution for every $n_1$ such that $n-n_1$ is even.
So if $n$ is odd then there are $(n+1)/2$ possible ways of paying.
So if $n$ is even then there are $(n/2)+1$ possible ways of paying.
Hence problem 1 is done.
Let the solution to problem 1) be $a_n$.
So for the problem 2) we have $n_1+2n_2+3n_3 = n$.
Let the solutions to problem 2) be $b_n$.
Then we have $b_n = \sum_{i,3 | (n-i)} a_i$.
|
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|
Find $\{ (x,y) \in \mathbb{R}^2 : x^2+y^2=3\}\cap\{(a,b) \in \mathbb{R}^2 : 2a^2 + 3 b^2=6\}$
Find $A \cap B$, where
$A=\{ (x,y) \in \mathbb{R}^2 : x^2+y^2=3\},\quad B=\{(a,b) \in \mathbb{R}^2 : 2a^2 + 3 b^2=6\}$
I know that $(\sqrt{3},0)$ and $(-\sqrt{3},0)\in A \cap B$ so $A\cap B\neq\emptyset$, but how can I find all of elements of $A \cap B$ ?
|
An.option:
Ellipse $2x^2+3y^2=6$, or
$3(x^2+y^2) =6+x^2$.
With $d^2$ the (distance)$^2$ from the origin:
$d^2:= x^2+y^2= 2 +(1/3)x^2 \le 3$,
since $|x| \le √3$ for points on the ellipse.
$d^2 = 3$ , for $x = ^{+}_{-}√3$, else $d^2<3$, i.e
only $2$ points on the circle with $r^2=3$.
|
{
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|
In $\triangle ABC$, $\frac{\cot A+\cot B}{\cot(A/2)+\cot(B/2)}+\frac{\cot B+\cot C}{\cot(B/2)+\cot(C/2)}+\frac{\cot C+\cot A}{\cot(C/2)+\cot(A/2)}=1$ In $\Delta ABC$ , prove $$\frac{\cot A+\cot B}{\cot(A/2)+\cot(B/2)}+\frac{\cot B+\cot C}{\cot(B/2)+\cot(C/2)}+\frac{\cot C+\cot A}{\cot(C/2)+\cot(A/2)}=1$$
I tried to take the $LHS $ in terms of $\sin$ and $\cos$ but ended up getting a huge equation.
I also tried to convert $\cot$ to $\; \dfrac {1}{\tan}\;$ but was still unable to solve the problem.
|
Hint: Note that $$\begin{align}\frac{\cot(A)+\cot(B)}{\cot\left(\frac{A}{2}\right)+\cot\left(\frac{B}{2}\right)}&=\frac{\sin(B)\cos(A)+\sin(A)\cos(B)}{4\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\Bigg(\cos\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)+\cos\left(\frac{B}{2}\right)\sin\left(\frac{A}{2}\right)\Bigg)}
\\&=\frac{\sin(C)}{4\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\cos\left(\frac{C}{2}\right)}\,.\end{align}$$
Similarly,
$$\frac{\cot(B)+\cot(C)}{\cot\left(\frac{B}{2}\right)+\cot\left(\frac{C}{2}\right)}=\frac{\sin(A)}{4\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\cos\left(\frac{C}{2}\right)}$$
and
$$\frac{\cot(C)+\cot(A)}{\cot\left(\frac{C}{2}\right)+\cot\left(\frac{A}{2}\right)}=\frac{\sin(B)}{4\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\cos\left(\frac{C}{2}\right)}\,.$$
Thus, the required identity is equivalent to
$$\sin(A)+\sin(B)+\sin(C)=4\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\cos\left(\frac{C}{2}\right)\,.$$
|
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|
Solving the factorial equation $(n + 4)! = 90(n + 2)!$ Solve the equation below:
$(n+4)!
= 90
(n+2)!$
I did this:
$(n+4)(n+3)(n+2)!
= 90
(n+2)!$
$n^2+7n+12+90=0$
$n^2+7n+102=0$
Is there anymore to this?
|
Well, you know that $$\frac{(n+4)!}{(n+2)!} = (n+4)\cdot(n+3) = n^2+7n+12=90$$$$n^2+7n-78=0$$The solutions to this quadratic are $n=6$ and $n=-13$, so we take $n=6$ to be our answer.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the coefficient of $x^9$ in the expansion of $(1+x)(1+x^2)(1+x^3)...(1+x^{100})$ The coefficient of $x^9$ in the expansion of $(1+x)(1+x^2)(1+x^3)...(1+x^{100})$.
I tried the following concept, how to sum 9 using 1-9 only without repetition.
1:9, 2:1+8,3:7+2, 4: 6+3, 5: 5+4,
6:1+2+6 ,7:1+3+5, 8:2+3+4
The answer is 8.
How will it be solved using Binomial theorem.
|
Yes we can count directly the cases that is
*
*$x^9$
*$x\cdot x^8$
*$x\cdot x^2\cdot x^6$
*$x\cdot x^3\cdot x^5$
*$x^2\cdot x^7$
*$x^2\cdot x^3\cdot x^4$
*$x^3\cdot x^6$
*$x^4\cdot x^5$
to obtain $8$.
|
{
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|
Problem with simple inverse Laplace transform I have the function:
$$F(s) = \frac{s^4+3s^3+2s^2+4s+4}{(s+3)(s^2+1)}$$
and I have to make inverse Laplace. I tried to collect $s^3$ from the first and second element of the numerator in order to obtain:
$$\frac{2s^2+4s+4}{(s+3)(s^2+1)} + \frac{s^3}{s^2+1}$$
for the first, I calculated residues and for the second, I made a polynomial division. I obtained
$$\frac{1}{s+3} + \frac{1}{s^2+1} + s - \frac{s}{s^2+1}$$
When I calculated the inverse I obtained:
$$e^{-3t}H(t) + \sin(t)H(t) + \delta_0' - \cos(t)H(t)$$
But, in the solution there isn't $\cos(t)H(t)$
|
Because you have a mistake when computing the Residues, you should get
$$F(s) = \frac{s^4+3s^3+2s^2+4s+4}{(s+3)(s^2+1)} = \frac{1}{s+3} + \frac{1}{s^2+1} + s \tag{1} $$
which answers your question on why you have an extra $\cos(t)H(t)$ appearing. To see why you have done a calculation mistake, take a look here:
$$\frac{1}{s+3} + \frac{1}{s^2+1} + s = \frac{s^2+1+s+3+s(s+3)(s^2+1)}{(s+3)(s^2+1)}$$
which upon arranging terms you get equation $(1)$.
|
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|
Value of 'a' for which $f(x)=\frac{x^3}{3} +\frac{x^2}{2} +ax+b $ is injective Let $f(x)=\frac{x^3}{3} +\frac{x^2}{2} +ax+b \forall x \in R$.
Find the least value of a for which $f(x)$ is injective function.
My approach , for $f(x)$ to be injective either f(x) should be increasing or decreasing function.'
$Y=f'(x)=x^2+x+a$ [Increasing function]
If $Y>0$, then $x^2+x+a>0$, it is possible when $1-4a<0$ or $a>\frac{1}{4}$
If $Y<0$, then $x^2+x+a<0$, it is possible when $1-4a \ge 0$ or $a \le \frac{1}{4}$
The options are (A) $\frac{1}{4}$ which is the correct option, the other given options are (B) $1$, (C) $\frac{1}{2}$, (D) $\frac{1}{8}$
Though my optin is correct but i have one doubt when the equality=$0$, then Y=$0$ hence it is neither increasing nor decreasing function.
|
Hint: you need $f^\prime$ to be non-negative for all $x$.
|
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|
Finding $\lim_{x\to-\infty}\sqrt{x^2-5x+1}-x$ results in loss of information Let $f(x) = \sqrt{x^2-5x+1}-x$
Find $\lim_{x\to\infty}f(x)$
$$\lim_{x\to\infty} \sqrt{x^2-5x+1}-x$$
$$\lim_{x\to\infty} \dfrac{x^2-5x+1-x^2}{\sqrt{x^2-5x+1}+x}$$
$$\lim_{x\to\infty} \dfrac{-5x+1}{\sqrt{x^2-5x+1}+x}$$
$$\lim_{x\to\infty} \dfrac{\dfrac{-5x+1}{x}}{\dfrac{\sqrt{x^2-5x+1}+x}{x}}$$
$$\lim_{x\to\infty} \dfrac{-5+\dfrac{1}{x}}{\sqrt{\dfrac{x^2}{x^2}-\dfrac{5}{x}+\dfrac{1}{x^2}}+1}$$
From here, I know that $\lim_{x\to\infty} \dfrac{1}{x} = 0$, $\lim_{x\to\infty} \dfrac{5}{x} = 0$ and $\lim_{x\to\infty} \dfrac{1}{x^2} = 0$
$$\lim_{x\to\infty} \dfrac{-5}{\sqrt{1}+1}$$
$$\lim_{x\to\infty} \dfrac{-5}{2}$$
$$\lim_{x\to\infty} f(x) = \dfrac{-5}{2}$$
Everything up to here seems fine. The issue is when I try to find $\lim_{x\to-\infty} f(x)$
I also know that $\lim_{x\to-\infty} \dfrac{1}{x} = 0$, $\lim_{x\to-\infty} \dfrac{5}{x} = 0$ and $\lim_{x\to-\infty} \dfrac{1}{x^2} = 0$
This would make me conclude that $\lim_{x\to-\infty}f(x) = \dfrac{-5}{2}$.
However, this is not the case because $\lim_{x\to-\infty}f(x) = \infty$
Desmos view of $f(x)$
Why am I arriving to the wrong answer and how can I algebraically prove that the answer is $\infty$?
|
hint
At $-\infty$, $x$ becomes negative, thus
$$\sqrt{x^2}=|x|=-x$$
$$\frac{\sqrt{x^2-5x+1}}{x}=\frac{\sqrt{x^2-5x+1}}{-(-x)}=$$
$$-\sqrt{\frac{x^2-5x+1}{x^2}}.$$
the denominator goes to zero.
|
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|
Cubic Diophantine with two variables The question is : prove that $y^2 = x^3+(x+4)^2$ have no solutions in positive integers $x,y$.
I tried to play with the equation and get to $ x^3 = (y+x+4)(y-x-4)$, if $p|x$ then $p^3 | x^3$ and $ p | y+x+4$ or $ p | y-x-4$ , assume that $p$ divides both then we get that $ p | 2x-8$ and since $p|x$ we get that $ p | 8$ meaning $p=2$, so if we assume that $x$ is odd then $ p^3 | y+x+4$ or $p^3| y-x-4$ but not both and so we can write $y+x+4 = b^3 $ and $ y-x-4= a^3$, but that did not help me solving the problem.
|
Based on your argument, we can split into two cases:
*
*$y + x + 4$ and $y - x - 4$ are relatively prime.
In this case, as you said, we can assume that $y + x + 4 = b^3$ and $y - x - 4 = a^3$. Then we have $x = ab$. Hence, $b^3 - a^3 = 2x + 8 = 2ab + 8$. This means that $b > a$ and have the same parity, so $b - a \geq 2$.
However, this means that $2ab + 8 = b^3 - a^3 = (b-a)(b^2 + ba + a^2) \geq 2(b^2 + ba + a^2)$. Therefore, $a^2 + b^2 \leq 4$. A quick check show that this is impossible.
*$y + x + 4$ and $y - x - 4$ are not relatively prime. By your argument this means that the greatest common divisor is a power of 2. There are two cases:
*
*$y + x + 4 = 2b^3$ and $y - x - 4 = 4a^3$; or
*$y + x + 4 = 4b^3$ and $y - x - 4 = 2a^3$.
For the first case, $x = 2ab$ and $4ab + 8 = 2b^3 - 4a^3$, which implies $2ab + 4 = b^3 - 2a^3$, so $b = 2b'$ is even, which implies $2ab' + 2 = 4b'^3 - a^3$, which implies $a = 2a'$ is even, which implies $2a’b' + 1 = 2b'^3 - 4a'^3$, contradiction. The second case is similar.
|
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|
Laurent expansion of $f(x)=\frac{1}{(z-1)(z-2)^2}$
I am trying to find the Laurent expansion of $f(x)=\frac{1}{(z-1)(z-2)^2}$ when $0<|z-1|<1$.
I thought that if
\begin{align}
f(z)&=\frac{1}{z-1}-\frac{1}{z-2}+\frac{1}{(z-2)^2} \\
&=\frac{1}{z-1}+\frac{1}{2-z}+\frac{d}{dz}\left(\frac{1}{2-z}\right) \\
&=\frac{1}{z-1}+\frac{1}{1-(z-1)}+\frac{d}{dz}\left(\frac{1}{1-(z-1)}\right) \\
&=\frac{1}{z-1}+\sum_{n=0}^{\infty}(z-1)^n+\frac{d}{dz}\left(\sum_{n=0}^{\infty}(z-1)^n\right) \\
&=\frac{1}{z-1}+\sum_{n=0}^{\infty}\left((z-1)^n+n(z-1)^{n-1}\right) \\
&=\frac{1}{z-1}+\sum_{n=0}^{\infty}(z-1)^n\left(1+n(z-1)^{-1}\right) \\
\end{align}
But this does not agree with the series generated by wolfram. What about this method is incorrect?
alternative approach suggested by David
\begin{align}
f(z)&=\frac{1}{z-1}\left(\frac{1}{(z-2)^2}\right) \\
&=\frac{1}{z-1}\frac{d}{dz}\left(-\frac{1}{z-2}\right) \\
&=\frac{1}{z-1}\frac{d}{dz}\left(\frac{1}{1-(z-1)}\right) \\
&=\frac{1}{z-1}\frac{d}{dz}\left(\sum_{n=1}^{\infty}(z-1)^n\right) \\
&=\frac{1}{z-1}\left(\sum_{n=1}^{\infty}n(z-1)^{n-1}\right) \\
&=\sum_{n=1}^{\infty}n(z-1)^{n-2}
\end{align}
|
Use this one: let $z-1=w$ to make it easier then with $0<|w|<1$
$$\dfrac{1}{1-w}=\sum_{n\geq0}^\infty w^n$$
differentiating shows
$$\dfrac{1}{(1-w)^2}=\sum_{n\geq1}^\infty nw^{n-1}$$
therefore
$$\dfrac{1}{w(w-1)^2}=\sum_{n\geq1}^\infty nw^{n-2}$$
|
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|
If $\sum\limits_{\text{cyc}}x=\frac\pi2$, then $2\sqrt{\sum\limits_{\text{cyc}}\tan x}\le\sum\limits_{\text{cyc}}\frac{\sqrt{\tan x}}{\cos x}$
Let $x$, $y$, $z$ be positive real numbers such that $x+y+z=\frac{\pi}{2}$. Then,
$$2\sqrt{\tan x+\tan y+\tan z}\leq \frac{\sqrt{\tan x}}{\cos x}+\frac{\sqrt{\tan y}}{\cos y} +\frac{\sqrt{\tan z}}{\cos z}$$
My try:
I have tried to use the formula of a triangle (if we have $a+b+c=\pi$).
So I make the following substitution :
$$x=0.5a \qquad y=0.5b \qquad z=0.5c$$
And now the idea is to use the formula :
$$\cos(A)^2=\frac{a^2+b^2-c^2}{2ab}$$
But if we make this substitution :
$$u=\frac{a}{b} \qquad v=\frac{b}{c} \qquad w=\frac{c}{a}$$
We get :
$$\cos(A)^2=\frac{u^2+1-\frac{1}{v^2}}{2\frac{u}{v}}$$
Furthermore we know with the condition that we have :
$$\tan a \tan b \tan c=\tan a+\tan b+\tan c$$
And $$uvw=1$$
So, in conclusion, we have two conditions on one inequality, but I think it's hard to isolate each variable.
|
Let $\tan{x}=a$, $\tan{y}=b$ and $\tan{z}=c$.
Thus, $a$, $b$ and $c$ are positives, $ab+ac+bc=1$ and we need to prove that
$$\sum_{cyc}\sqrt{a(1+a^2)}\geq2\sqrt{a+b+c}.$$
Indeed, by C-S and Schur we obtain:
$$\sum_{cyc}\sqrt{a(1+a^2)}=\sum_{cyc}\sqrt{a(a^2+ab+ac+bc)}=\sum_{cyc}\sqrt{a(a+b)(a+c)}=$$
$$=\sum_{cyc}\sqrt{a^2(a+b+c)+abc}=\sqrt{\sum_{cyc}\left(a^2(a+b+c)+abc+2\sqrt{(a^2(a+b+c)+abc)(b^2(a+b+c)+abc)}\right)}\geq$$
$$\geq\sqrt{\sum_{cyc}\left(a^2(a+b+c)+abc+2(ab(a+b+c)+abc)\right)}=$$
$$=\sqrt{\sum_{cyc}(a^3+3a^2b+3a^2c+5abc)}\geq\sqrt{\sum_{cyc}(4a^2b+4a^2c+4abc)}=$$
$$=2\sqrt{(a+b+c)(ab+ac+bc)}=2\sqrt{a+b+c}$$ and we are done!
|
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|
Application of Fermat's little theorem to check divisibility Using Fermat's little theorem to prove: $(i)19\mid 2^{2^{6k+2}}+3$, where $k=0,1,2.....$$(ii)13\mid 2^{70}+3^{70}$My Approach: I couldn't think of how to go with $(i)$ but i tried $(ii)$ to show $2^{70} \equiv 0\pmod {13}$ and $3^{70} \equiv 0\pmod {13}$.Since,
$$2^{12} \equiv 1 \pmod {13}\Rightarrow (2^{12})^5\equiv 1 \pmod {13}\Rightarrow2^{60} \equiv 1 \pmod {13}$$Again, $$2^4 \equiv 3 \pmod {13}\Rightarrow2^8.2^2 \equiv 10\pmod {13}$$Using both result: $2^{70} \equiv 10\pmod {13}$I failed again to show that. Any hints or solution will be appreciated.Thanks in advance.
|
We have
$$2^{6k+1}\equiv 8^{2k}\cdot 2\equiv 2\pmod{9} $$
from which
$$2^{6k+2}\equiv 4\pmod {18} $$
hence by Fermat's little theorem
$$2^{2^{6k+2}}\equiv 2^4\equiv -3\pmod {19} $$
For the second $2^4\equiv 3\pmod {13} $ and $2^{12}\equiv 1\pmod {13}$ by Fermat little theoren hence
$$2^{70}+3^{70}\equiv 2^{70}+2^{280}\equiv 2^{10}+2^{4}\equiv 9\cdot 4+3\equiv 0\pmod {13} $$
|
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|
Limit of $\ \lim_{x \to \frac{\pi}{4}} \frac{\tan x-1}{x-\pi/4} $ without l'hopital's Find limit of
$$\ \lim_{x \to \frac{\pi}{4}} \frac{\tan x-1}{x-\pi/4} $$
I started by defining $\ x = y + \frac{\pi}{4} $
$$\ \lim_{x \to \frac{\pi}{4}} \frac{\tan x-1}{x-\pi/4} = \lim_{y \to 0} \frac{\tan (y+ \pi/4)-1}{y}= ? $$
This is exactly the same question where the solution suggested also used $\ x = y + \pi/4 $ yet I couldn't understand this equality: $$\ \tan(x) = \tan(y + \pi/4) = \frac{\tan y -1 }{1- \tan y} $$ and then how is that exactly equal to limit of $$\ \lim_{y \to 0}\frac{1}{y} \left( 1-\frac{\tan y +1}{1 - \tan y}\right)$$
|
$$\frac{\tan\left(y+\dfrac\pi4\right)-1}y=\frac{\dfrac{\tan y+\tan\dfrac \pi4}{1-\tan y\tan \dfrac \pi4}-1}y=2\frac{\tan y}{y}\frac1{1-\tan y}\to 2$$
|
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|
Proving a convexity inequality
Given $f: \mathbb{R} \to \mathbb{R}$ convex, show that:
$$ \frac{2}{3}\left(f\left(\frac{x+y}{2}\right) + f\left(\frac{z+y}{2}\right) + f\left(\frac{x+z}{2}\right)\right) \leq f\left(\frac{x+y+z}{3}\right) + \frac{f(x) + f(y) + f(z)}{3}.$$
I have tried some ideas, such as transforming it into
$$ f\left(\frac{x+y}{2}\right) + f\left(\frac{z+y}{2}\right) + f\left(\frac{x+z}{2}\right) - 3f\left(\frac{x+y+z}{3}\right)\\
\leq f(x) + f(y) + f(z) - f\left(\frac{x+y}{2}\right) - f\left(\frac{z+y}{2}\right) - f\left(\frac{x+z}{2}\right) $$ (which graphically seemed intuitive) and using that such an $f$ lies above its tangents, but did not succeed… Ideas are welcome :)
|
Let $x\geq y\geq z$.
Consider two cases.
*
*$x\geq y\geq\frac{x+y+z}{3}\geq z$.
From here we see that $2y\geq x+z$ and easy to check that
$$\left(\frac{x+y+z}{3},\frac{x+y+z}{3},\frac{x+y+z}{3},z\right)\succ\left(\frac{x+z}{2},\frac{x+z}{2},\frac{y+z}{2},\frac{y+z}{2}\right),$$ which by Karamata gives
$$3f\left(\frac{x+y+z}{3}\right)+f(z)\geq2f\left(\frac{x+z}{2}\right)+2f\left(\frac{y+z}{2}\right).$$
Thus, it's enough to prove that
$$f(x)+f(y)\geq2f\left(\frac{x+y}{2}\right),$$ which is Jensen.
*$x\geq \frac{x+y+z}{3}\geq y\geq z$.
This case is similar. Try to kill it by yourself.
|
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|
Find $n$ such that there are $11$ non-negative integral solutions to $12x+13y =n$
What should be the value of $n$, so that $12x+13y = n$ has 11 non-negative integer solutions?
As it is a Diophantine equation, so we check whether the solution exists? it exists if $gcd(12,13)|n$ that is $1|n$ and hence integer solutions to $12x+13y = n$ exist.
To find the solutions explicitly, we write $1 $ as the linear combinations of $12$ and $13$ that is $1 = 12(1) + 13(-1)$, next we write $n = 12(n) + 13(-n)$ so that the integer solutions to $12x+13y = n$ is $x = n + t(12)$ and $y = -n + t(13)$. To check for non-negative integral solutions, $x \geq 0$ and $y \geq 0$ so that $t \geq -\frac{n^2}{12}$ and $t \geq \frac{n^2}{13}$ so $t \geq \frac{n^2}{13}$. But now I am struggling how to relate this to the 11 number of non-negative integral solutions?.Any help?
|
Note that from the following hint
$$-1\cdot 12+1\cdot 13=1 \implies (-1+k \cdot 13)\cdot 12+(1-k\cdot 12)\cdot 13=1$$
we have
$$-n\cdot 12+n\cdot 13=n \implies (-n+k \cdot 13)\cdot 12+(n-k\cdot 12)\cdot 13=1$$
and we need
*
*$-n+k \cdot 13\ge 0$
*$n-k\cdot 12\ge 0$
that is
$$\frac n{13}\le k\le\frac n{12}$$
which leads to $n=10\cdot 12\cdot13$ with
$$12x+13y=10\cdot 12\cdot13$$
which has the following solutions
$$x=13k \quad y=120-12k \quad k=0,1,2,\ldots,10$$
|
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|
If $\frac{\cos x}{\cos y}+\frac{\sin x}{\sin y}=-1$, then what's the value of $4\left(\frac{\cos^3y}{\cos x}+\frac{\sin^3y}{\sin x}\right)$? I was trying to solve this problem:
If
$$\frac{\cos x}{\cos y}+\frac{\sin x}{\sin y}=-1$$
then what is the value of
$$S=4\left(\frac{\cos^3y}{\cos x}+\frac{\sin^3y}{\sin x}\right)$$
I tried to solve this problem using a change of variables
$\cos^3y=a\cos x$ and $\sin^3y=b \sin x$.
So:
$$\frac{S}{4}=a+b$$
$$\frac{\cos^6y}{a^2}+\frac{\sin^6y}{b^2}=1$$
and
$$\frac{\cos^2y}{a} + \frac{\sin^2y}{b}=-1$$
I tried to manipulate these equations, but it gives no result
|
I don't know if this helps, but this is what I've discovered so far
The first equation can be written as
$$\cos x \sin y + \sin x \cos y = -\cos y \sin y$$
or using the double angle formula
$$\sin (x+y) = -\cos y \sin y$$
By the triple angle formulas for sine and cosine, we have that
$$\sin^3(y)=\frac{3\sin(y)-\sin(3y)}{4}, \, \cos^3(y) = \frac{3\cos(y) + \cos(3y)}{4}$$
Plugging this in and simplifying the fractions, we get
$$S=\frac{3\cos(y)\sin(x) + \cos(3y)\sin(x) +3\sin(y)\cos (x)-\sin(3y)\cos(x)}{\cos x \sin x}$$
Rearranged, this is
$$S=\frac{3\cos(y)\sin(x) +3\sin(y)\cos (x) + \cos(3y)\sin(x) -\sin(3y)\cos(x)}{\cos x \sin x}$$
And using the sum and difference of angle formulas for sine, we get
$$S=\frac{3\sin(x+y) + \sin(x-3y)}{\cos x \sin x}$$
EDIT: Using the graphing software desmos and plugging in test values for $x,y$ that satisfy the first equation, we get $\fbox{4 }$. I guess I'll leave it up to someone smarter to prove why is works, instead of just showing that it does.
|
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|
Differential equation $y''+y=\tan(x)$ Given the following differential equation:
\begin{align}
y''+y=\tan(x)
\end{align}
I tried this way:
\begin{align}
y_H(x)=A\cos(x)+ B\sin(x)\\ A,B\in \Re \\
y_p(x)=A(x)\cos(x)+B(x)\sin(x)
\end{align}
I continue with the Wronski method
\begin{align}
W(F)= \begin{bmatrix}
\cos(x) & \sin(x) \\
-\sin(x) & \cos(x)
\end{bmatrix}\\\\
det \ W(F) = 1 \neq 0
\end{align}
So is invertible
\begin{align}
W(F)^{-1}= \begin{bmatrix}
\cos(x) & -\sin(x)\\
\sin(x) & \cos(x)
\end{bmatrix}\\\\
\end{align}
Now since:
\begin{align}
W(F) \cdot\begin{bmatrix}
A'(x) \\ B'(x)
\end{bmatrix}= \begin{bmatrix}
0 \\ b(x)
\end{bmatrix}\\\\
\begin{bmatrix}
A'(x) \\ B'(x)
\end{bmatrix}=
\begin{bmatrix}
\cos(x) & -\sin(x) \\
\sin(x) & \cos(x)
\end{bmatrix}
\cdot
\begin{bmatrix}
0 \\ \tan(x)
\end{bmatrix}\\\\
B'(x)= \cos(x) \tan(x)=\sin(x) \\
B(x) = -\cos(x)\\\\
A'(x) = -\sin(x)\tan(x)\\
A(x) = -\int \sin(x) \tan(x) dx = \cos(x)\tan(x) - \int \frac{1}{\cos(x)} dx= \\
= \sin(x) - \frac{1}{2} \log{\frac{1+sin(x)}{1-\sin(x)}}\\\\
y_p(x)=A(x)\cos(x) + B(x)\sin(x) = \\\\
= \sin(x)\cos(x) - \frac{1}{2}\cos(x) \log{\frac{1+sin(x)}{1-\sin(x)}} +
-\cos(x)\sin(x)= \\\\ - \frac{1}{2}\cos(x) \log{\frac{1+sin(x)}{1-\sin(x)}}
\end{align}
So now the linear combination of the two solution should be
\begin{align}
y(x)= A\cos(x) + B\sin(x) - \frac{1}{2}\cos(x) \log{\frac{1+sin(x)}{1-\sin(x)}}
\end{align}
but the correct solution is
\begin{align}
y(x)= A\cos(x) +B\sin(x) + \cos(x) \log\left(\frac{\cos(\frac{x}{2})-\sin(\frac{x}{2})}{cos(\frac{x}{2})+sin(\frac{x}{2})}\right)
\end{align}
Is my solution wrong? If yes where I did the mistake?
Is my solution right? How can I get the other form? What steps I need to do?
|
Your solution is correct just note that $(\sin(x/2)\pm \cos+x/2))^2=1\pm \sin(x)$. Now use the log property that $\log(a^b)=b\log(a)$ and you are done. And if you are confused with the discrepancy of plus and minus sign before the log expression we have $\log(\frac{1}{x})=-\log(x)$
|
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|
Integrate $\int \sqrt{\frac{1+x}{1-x}}dx$ using $x = \cos(u) $ I have to integrate $\int \sqrt{\frac{1+x}{1-x}}$ using $x = \cos(u)\Rightarrow dx = -\sin(u) du$
My attempts:
$$\int \sqrt{\frac{1+x}{1-x}}dx=\int \sqrt{\frac{(1+x)(1-x)}{(1-x)(1-x)}}dx=\int \frac{\sqrt{1-x^2}}{1-x}dx=\int \frac{\sqrt{1-x^2}}{1-x}dx$$
$$\int \frac{\sqrt{\sin^2(u)}}{1-\cos(u)}( -\sin(u))du=\int \frac{-\sin(u)^2}{1-\cos(u)}du=\int \frac{-(1 - \cos^2(u)) }{1-\cos(u)}du$$ , someone could help because i block here because i don't get the same resultat with this website
|
Try this-
$$\int\sqrt{\frac{1+x}{1-x}}dx=\int\frac{1+x}{\sqrt{1-x^2}}dx=\int\frac{1}{\sqrt{1-x^2}}+\frac{x}{\sqrt{1-x^2}}dx$$
The first integral is $\arcsin x$ and the second is evaluated by substituting $u=x^2\implies du=2xdx$
|
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|
Finding value of $\int^{2\pi}_{0}\frac{\sin^2(x)}{a-b\cos x}dx$ without contour Integration
Finding value of $\displaystyle \int^{2\pi}_{0}\frac{\sin^2(x)}{a-b\cos x}dx\;\;, a>b>0$ without contour Integration
Try: Let $\displaystyle I =\int^{2\pi}_{0}\frac{\sin^2 x}{a-b\cos x}dx=2\int^{\pi}_{0}\frac{\sin^2 x}{a-b\cos x}dx$
Using Integration by parts
$$ I =\frac{2}{b}\bigg[\sin (x)\cdot \ln|a-b\cos x|\bigg]\bigg|^{\pi}_{0}-\frac{2}{b}\int^{\pi}_{0}\cos x\cdot \ln|a-b\cos x|dx$$
$$I=-\frac{2}{b}\int^{\pi}_{0}\cos x\cdot \ln(a-b\cos x)dx$$
Could some help me to solve it, Thanks in advance
|
Your integral is equal to $$2\int_{0}^{\pi}\frac{\sin^2x}{a-b\cos x} \, dx$$ and we can express the integrand as $$\frac{\cos x} {b} +\frac{a}{b^2}+\frac{b^2-a^2}{b^2(a-b\cos x)} $$ and hence the integral in question is $$2\cdot 0+\frac{2\pi a} {b^2}+\frac{2(b^2-a^2)}{b^2}\int_{0}^{\pi}\frac{dx}{a-b\cos x} $$ The last integral can be evaluated using a tricky substitution $$(a-b\cos x) (a+b\cos y) =a^2-b^2$$ to get $\pi/\sqrt{a^2-b^2}$ and hence the desired answer is $$\frac{2\pi a} {b^2}-\frac{2\pi\sqrt{a^2-b^2}}{b^2}=\frac{2\pi}{a+\sqrt{a^2-b^2}}$$
|
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|
Approximation of square roots Recently, I've seen a YouTube video where they approximate square roots real quick.
They use this approximation :
$$\sqrt{x} \approx \lfloor \sqrt{x} \rfloor+\dfrac{x-(\lfloor \sqrt{x} \rfloor)^2}{2\lfloor \sqrt{x} \rfloor}$$
I want to know the math behind this approximation. Can someone help me out?
|
Let, $n \in N $
Suppose, we want to evaluate $\sqrt{n^2+c}$
$$\implies \sqrt{n^2(1+\dfrac{c}{n^2})}$$
$$\implies n \sqrt{1+\dfrac{c}{n^2}}$$
Now, we shall use the binomial expansion for $\sqrt{1+\dfrac{c}{n^2}}$
We know, $\sqrt{1+\dfrac{c}{n^2}}=1+\dfrac{c}{2n^2}-\dfrac{c^2}{8n^4}+\cdots$
So, $n \sqrt{1+\dfrac{c}{n^2}}=n+\dfrac{c}{2n}-\dfrac{c^2}{8n^3}+ \cdots$
Therefore, we can say, $\sqrt{n^2+c}=n+\dfrac{c}{2n}+\cdots$
From there we get the approximation, $\sqrt{x} \approx \lfloor x \rfloor + \dfrac{x-(\lfloor \sqrt{x} \rfloor)^2}{2\lfloor \sqrt{x} \rfloor}$
|
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|
How evaluate $ \int_0^\infty \frac{\ln^2\;\left|\;\tan\left(\frac{ax}{2}-\frac{\pi}{4}\right)\;\right|}{1+x^2} \;dx$ How evaluate (without using Complex analysis)
$$ \int_0^\infty \frac{\ln^2\;\left|\;\tan\left(\frac{ax}{2}-\frac{\pi}{4}\right)\;\right|}{1+x^2}\; dx\quad (a\gt0)$$
My Attempt:
I used the expansion of the following functions: $$ \ln\left|2\sin(x)\right| \text{ and }\ln\left|2\cos(x)\right| $$
To get the following expansion: $$ \ln\;\left|\;\tan\left(\frac{ax}{2}-\frac{\pi}{4}\right)\;\right|=2\sum_{n=1}^\infty (-1)^n \frac{\sin[(2n-1)ax]}{2n-1} $$
Then I expressed the square of the logarithmic function as follows:$$ \ln^2\;\left|\;\tan\left(\frac{ax}{2}-\frac{\pi}{4}\right)\;\right|=4\sum_{m=1}^\infty\sum_{n=1}^\infty (-1)^{m+n} \frac{\sin[(2m-1)ax]\sin[(2n-1)ax]}{(2m-1)(2n-1)} $$
And used the formula of the product of two sines, then integrated the following well known integral under the summation sign:
$$ \int_0^\infty \frac{\cos(bx)}{1+x^2} \;dx$$
And expressed the final result in terms of $\; \arctan\;\left(\;e^{-a}\;\right)\; $
but this was not the right answer according to the one evaluated by Complex analysis, which is
$$ \frac{{\pi}^3}{8}-2\pi\; \arctan^2\;\left(\;e^{-a}\;\right) $$
Any hint for another method or idea?
|
Using OP's series expansion, we have
\begin{align*}
I
&= 2 \sum_{m, n \geq 0} \frac{(-1)^{m+n}}{(2m+1)(2n+1)} \int_{0}^{\infty} \frac{\sin[(2m+1)ax]\sin[(2n+1)ax]}{1+x^2} \, dx \\
&= \pi \sum_{m, n \geq 0} \frac{(-1)^{m+n}}{(2m+1)(2n+1)} \left( e^{-a|2m-2n|} - e^{-a(2m+2n+2)} \right).
\end{align*}
We will split the sum into several parts and analyze them separately.
*
*It is straightforward that
$$ \sum_{m, n \geq 0} \frac{(-1)^{m+n}}{(2m+1)(2n+1)} e^{-a(2m+2n+2)} = \arctan^2(e^{-a}). $$
*Using the identity $\frac{1}{(2m+1)(2n+1)} = \frac{1}{(2m-2n)(2n+1)} + \frac{1}{(2n-2m)(2m+1)}$, we have
\begin{align*}
\sum_{m, n \geq 0} \frac{(-1)^{m+n}}{(2m+1)(2n+1)} e^{-a|2m-2n|}
&= \sum_{m = 0} \frac{1}{(2m+1)^2} + \sum_{\substack{m, n \geq 0 \\ m \neq n }} \frac{(-1)^{m-n}}{(m-n)(2n+1)} e^{-a|2m-2n|} \\
&= \frac{\pi^2}{8} + \sum_{n \geq 0} \frac{1}{2n+1} \sum_{\substack{k \geq -n \\ k \neq 0}} \frac{(-1)^k}{k} e^{-2a|k|}.
\end{align*}
*The last sum can be simplified further: using the fact that $\frac{(-1)^k}{k}e^{-2a|k|}$ is an odd function of $k$,
\begin{align*}
\sum_{n \geq 0} \frac{1}{2n+1} \sum_{\substack{k \geq -n \\ k \neq 0}} \frac{(-1)^k}{k} e^{-2a|k|}
&= \sum_{n \geq 0} \frac{1}{2n+1} \sum_{k = n+1}^{\infty} \frac{(-1)^k}{k} e^{-2a|k|} \\
&= \sum_{k = 1}^{\infty} \left( \sum_{n=0}^{k-1} \frac{1}{2n+1} \right) \frac{(-1)^k}{k} e^{-2a|k|}.
\end{align*}
Symmetrizing the inner sum, we find that
\begin{align*}
\sum_{n=0}^{k-1} \frac{1}{2n+1}
= \frac{1}{2} \sum_{\substack{i, j \geq 0 \\ i+j = k-1}} \left( \frac{1}{2i+1} + \frac{1}{2j+1} \right)
= \sum_{\substack{i, j \geq 0 \\ i+j = k-1}} \frac{k}{(2i+1)(2j+1)}.
\end{align*}
Plugging this back,
\begin{align*}
\sum_{n \geq 0} \frac{1}{2n+1} \sum_{\substack{k \geq -n \\ k \neq 0}} \frac{(-1)^k}{k} e^{-2a|k|}
&= \sum_{i, j \geq 0} \frac{1}{(2i+1)(2j+1)} (-1)^{i+j+1} e^{-a(2i+2j+2)} \\
&= - \arctan^2(e^{-a}).
\end{align*}
Combining altogether, we obtain the desired answer.
|
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|
limit of $\sqrt{(x^2+1)/(x^3+1)}$ as $x$ approaches negative infinity $\lim_{x \rightarrow - \infty } \sqrt{ \frac{x^2+1}{x^3+1} } $
My teacher says that no limit exists, but Wolfram Alpha says the limit is 0. I'm confused.
Any helps are welcome.
|
The answer depends on the underlying domain of the expression. To see this, you may write
$$\sqrt{ \frac{x^2+1}{x^3+1} } \stackrel{x<-1}{=} \sqrt{-1} \cdot \frac{1}{\sqrt{|x|}} \cdot \sqrt{ \frac{1+\frac{1}{x^2}}{1+\frac{1}{x^3}} } = \begin{cases} \mbox{not defined} & \mbox{ in } \mathbb{R} \\ \pm i \cdot \frac{1}{\sqrt{|x|}} \cdot \sqrt{ \frac{1+\frac{1}{x^2}}{1+\frac{1}{x^3}} } \stackrel{x \to -\infty}{\longrightarrow} 0 & \mbox{ in } \mathbb{C}\end{cases}$$
|
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|
Find the values of $p$ and $q$
If $p^3+p=q^2+q$ where $p$ and $q$ are prime numbers,
Find all the solutions (p, q)
I tried to solve this exercise using that:
$p^2 = -1(\text{mod} \, q)$ and $q = -1(\text{mod} \, p)$;
So: $q+1=ap$ and $p^2+1=bq$, where $b$ and $q$ integers.
Then I tried to solve a quadratic equation, but I could not finish the problem
|
Clearly $p\neq q$, and because $p$ and $q$ are prime and
$$p(p^2+1)=p^3+p=q^2+q=q(q+1),$$
we must have $p\mid q+1$ and $q\mid p^2+1$. Write
$$q+1=ap\qquad\text{ and }\qquad p^2+1=bq,$$
to find that $p^2-abp+b+1=0$. In particular $b+1\equiv0\pmod{p}$, say $b=cp-1$, but then
$$p^2+1=(cp-1)q=(cp-1)(ap-1)=acp^2-(a+c)p+1.$$
Note that $a$, $b$ and $c$ are positive integers, and that $a>1$ as otherwise $p=q+1$ which implies that $(p,q)=(3,2)$ which is not a solution. The equation above simplifies to
$$(ac-1)p=a+c,$$
and as $p\geq2$ clearly we cannot have $a,c\geq2$. Hence $c=1$ and so
$$p^2+1=bq=(cp-1)q=(p-1)q.$$
In particular $p-1\mid p^2+1$. As $p-1\mid p^2-1$ it follows that $p-1=2$, so $p=3$ and hence $q=5$.
|
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|
Probability of getting a particular sum from n m-sided dice c# I am trying to generate a probability of getting a specific number x from n dice, with no guarantee of them having the same number of sides. (eg, 1d6 + 2d10)
Does there exist a mathematical formula for this?
Does there exist a formula for getting above x?
|
The real trick here is to use generating functions.
The way they work is by assigning the probabilities to a polynomial, with each term having coefficient equal to its probability of being rolled, and power the value of the roll.
Then, one simply looks at the coefficient of the term with power equivalent to the desired sum for the probability.
For example, let's take your situation of 1 fair 6 sided dice and 2 fair ten sided die. The polynomial we write is $$(\frac16x+\frac16x^2+\frac16x^3+\frac16x^4+\frac16x^5+\frac16x^6)\cdot(\frac1{10}x+\frac1{10}x^2+\frac1{10}x^3+\frac1{10}x^4+\frac1{10}x^5+\frac1{10}x^6+\frac1{10}x^7+\frac1{10}x^8+\frac1{10}x^9+\frac1{10}x^{10})^2$$$$\frac1{600}\cdot(x+x^2+x^3+x^4+x^5+x^6)\cdot(x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9+x^{10})^2$$
Let's find the coefficient of $x^{15}$. Using Wolfram Alpha, I found the coefficient to be 0.085. This tells us the probability of the sum totalling 15 is $\color{red}{0.085}$
|
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|
Probability of at least once out of the three tosses Suppose a pair of fair dice is tossed 3 times. Let $X$ be the sum of the outcomes at each toss. Find the probability of getting a sum of 7 (i.e., $X=7$) at least once out of the three tosses.
My attempt:
The sample space is $6^3=216.$ There are 15 combinations that gives a sum of $7$.
$$P(X=7)=\frac{15}{216}$$
I am stuck here. If a pair of fair dice is tossed 3 times, should the sample space be 36 or 216? Am I on the right path?
|
let X = getting total of 7 then sample space S={(1,6)(6,1)(2,5)(5,2)(3,4)(4,3)}=6 times
Total number of outcomes =
{(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)
(2,1).....................(2,6)
(3,1).....................(3,6)
(4,1)
(5,1)
(6,1),,,,,,,,,,,,,,,,,,,,,(6,6)}=6x6=36 outcomes
P(getting total 7)=p(x)=6/36=1/6
q=5/6
getting 7 atleast once
p(x>=1)=1-p(x<1)
=1-p(x=0)
=1- ncx p^x q^n-x , n=3 p =1/6 q=5/6
=1- [3c0 (1/6)^0 (5/6)^3-0]
=0.42
|
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"timestamp": "2023-03-29T00:00:00",
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|
Using Vieta's formula to find the sum of the roots for a given cubic equation. Vieta's formula states that, if a cubic equation has three different roots, the following is true:
$$\begin{eqnarray*}
x_1 + x_2 + x_3 &=& -b/a\\
x_1x_2 + x_1x_3 + x_2x_3 &=& c/a \\
x_1x_2x_3 &=& -d/a
\end{eqnarray*}$$
Then, how is the following calculated?
$x_1^3$ + $x_3^3$ + $x_2^3$
|
Use that $$(a+b+c)^3=a^3+b^3+c^3+3ab(a+b)+3ac(a+c)+3bc(b+c)+6abc$$
$$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
If $(a-b+5)x^2 + (2b+1)y^2 = b^2-2b-11$ is the equation of a unit circle, then find the sum of the possible values of $ab$. If
$$(a-b+5)x^2 + (2b+1)y^2 = b^2-2b-11$$
is the equation of the unit circle, then what is the sum of the values that $ab$ can take?
|
Here's an extended hint ...
The standard equation for an origin-centered unit circle is $x^2+y^2=1$. More generally, such a circle has the equation
$$p x^2 + p y^2 = p$$
for any non-zero $p$. Therefore, for $$(a-b+5)x^2+(2b+1)y^2=b^2-2b-11 \tag{1}$$
to be the equation of a unit circle, the coefficients of $x^2$ and $y^2$, and the expression on the right-hand side, must all be equal (and non-zero).
$$a-b+5=2b+1=b^2-2b-11 \quad(\neq 0)\tag{2}$$
The last two parts of $(2)$ allow us to solve for $b$:
$$2b+1=b^2-2b-11 \quad\to\quad b^2-4b-12=0 \tag{3}$$
The first two parts of $(2)$ then tell us what $a$ is:
$$a-b+5=2b+1 \quad\to\quad a=3b-4 \tag{4}$$
The rest of the exercise is left to the reader.
|
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|
minimum value of $\bigg||z_{1}|-|z_{2}|\bigg|$
If $z_{1}\;,z_{2}$ are two complex number $(|z_{1}|\neq |z_{2}|)$ satisfying
$\bigg||z_{1}|-4\bigg|+\bigg||z_{2}|-4\bigg|=|z_{1}|+|z_{2}|$
$=\bigg||z_{1}|-3\bigg|+\bigg||z_{2}|-3\bigg|.$Then minimum of $\bigg||z_{1}|-|z_{2}|\bigg|$
Try: Let $|z_{1}|=a$ and $|z_{2}|=b$ and $a,b\geq 0$ and $a\neq b$
So $$|a-4|+|b-4|=|a|+|b| = |a-3|+|b-3|.$$
Let $A(0,0)$ and $B(3,3)$ and $C(4,4)$ and $P(a,b)$
Then $PA=PB=PC.$ and we have to find $|a-b|$
so i did not understand how can i conclude it,
could someone please explain me
Thanks
|
From the equality
$$|a-4|+|b-4|=a+b = |a-3|+|b-3|$$
we have that $a,b\ge 3$ is not possible.
So, assume $a\ge 3, b\le 3$. Then
$$a+b = a-3+3-b=a-b$$ which is only possible if $b=0.$ In such a case
$$|a-4|+4=a =a-3+3=a.$$ This is only possible if $a\ge 4.$
Now assume $a,b\le 3.$ In such a case
$$4-a+4-b=a+b=3-a+3-b.$$ But there is no solution.
So, $a\ge 4$ and $b=0.$
Because of the symmetry we can have $b\ge 4$ and $a=0.$
|
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|
Find $c$ and $n$ such that $\frac{x^3 \arctan x}{x^4 + \cos x +3} \sim cx^n$ as $x \to 0$ Where is my mistake in the below:
$$\frac{1}{c} \lim_{x \to 0} \frac{x^3 \arctan x}{x^{4+n} + x^n \cos x + 3x^n} = \frac{1}{c} \lim_{x \to 0} \frac{x^4}{x^{4+n}+x^n \cos x + 3x^n} \\ =\frac{1}{c} \lim_{x \to 0} \frac{1}{x^n + x^{n-4} \cos x + 3x^{n-4}}$$
Looking at it now it feels like it is the step that I am about to do, because it appears as though the denominator becomes smaller and smaller going to infinity. However, my reasoning was that since it is given that there can be an equivalence between $cx^n$ and our function, i.e. the limit is finite and goes to $1$ as $x \to 0$, then it also must be true that if we invert the fraction, then the limit is still finite and is $1/1=1$ as $x \to 0$. So, next step:
$$c \lim_{x \to 0} \left(x^n + x^{n-4}\cos x + 3x^{n-4} \right)$$
$$c \lim_{x \to 0} (x^{n-4}(\cos x -1))=1$$
$$-c \lim_{x \to 0} \left(x^{n-4} \frac{x^2}{2}\right) = 1$$
We need $x^{n-4} = x^{-2}$ $\implies n=2$
$$-c \frac{1}{2} =1 \implies c=-2$$
|
It should be obvious that $c\neq 0$ otherwise the definition of $\sim$ is in trouble. Next we can note that the given condition implies $$\lim_{x\to 0}\frac{cx^n(3+\cos x+x^4)}{x^3\arctan x} =1$$ Using the limit $\lim_{x\to 0} (1/x)\arctan x=1$ we can see that the above condition is equivalent to $$4c\lim_{x\to 0}x^{n-4}=1$$ Now it should be obvious that $n=4$ as $n<4$ makes the limit infinite and $n>4$ makes the limit $0$ and then $c=1/4$.
|
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|
Induction proof: $\frac{1}{2^1} + \frac{2}{2^2} + \frac{3}{2^3} +...+\frac{n}{2^n}$ $<2$ Prove by induction the following.
$$\frac{1}{2^1} + \frac{2}{2^2} + \frac{3}{2^3} +\dots+\frac{n}{2^n}<2.$$
Caveat: The $<$ will be hard to work with directly. Instead, the equation above can be written in the form,
$$\frac{1}{2^1} + \frac{2}{2^2} + \frac{3}{2^3} +\dots+\frac{n}{2^n}=2-\text{something}.$$
First, find the "something" and then use that form of the equation to prove the assertion.
I can't seem to figure out the form that this equation can be written as. Also, once I find the form how would I do the proof.
I understand it involves using a Basic Step and an Induction Step
|
Show that
$$\frac{1}{2^1} + \frac{2}{2^2} + \frac{3}{2^3} +\dots+\frac{n}{2^n}\leq 2-\frac{An+B}{2^n}$$
for some $A,B\geq 0$. Then the base case is satisfied if
$$\frac{1}{2}\leq 2-\frac{A+B}{2}$$
that is $A+B\leq 3$.
The induction step works if for all $n\geq 1$,
$$2-\frac{An+B}{2^n}+\frac{n+1}{2^{n+1}}\leq2-\frac{A(n+1)+B}{2^{n+1}},$$
that is
$$n+1\leq (2An+2B)-(A(n+1)+B)=An+B-A.$$
It follows that $A=1$ and $B=2$ and we may conclude that:
For all $n\geq 1$,
$$\frac{1}{2^1} + \frac{2}{2^2} + \frac{3}{2^3} +\dots+\frac{n}{2^n}\leq 2-\frac{n+2}{2^n}$$
P.S. Looking back to the the induction proof steps it is easy to realize that the above inequality is actually an equality!!
|
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|
Proving that $\sum_{n=1}^\infty \frac{\sin\left(n\frac{\pi}{3}\right)}{(2n+1)^2}=\frac{G}{\sqrt 3} -\frac{\pi^2}{24}$ Trying to show using a different approach that $\int_0^1 \frac{\sqrt x \ln x}{x^2-x+1}dx =\frac{\pi^2\sqrt 3}{9}-\frac{8}{3}G\, $ I have stumbled upon this series: $$\sum_{n=1}^\infty \frac{\sin\left(n\frac{\pi}{3}\right)}{(2n+1)^2}$$ The linked proof relies upon this trigamma identity. Now by rewriting the integral as:
$$I=\int_0^1 \frac{\sqrt{x}\ln x}{x^2-2\cos\left(\frac{\pi}{3}\right)x+1}dx$$
And using that: $$\frac{\sin t}{x^2-2x\cos t+1}=\frac{1}{2i}\left(\frac{e^{it}}{1-xe^{it}}-\frac{e^{-it}}{1-xe^{-it}}\right)=\Im \left(\frac{e^{it}}{1-xe^{it}}\right)=$$
$$=\sum_{n=0}^{\infty} \Im\left(x^n e^{i(n+1)t}\right)=\sum_{n=0}^\infty x^n\sin((n+1)t)$$
$$I=\frac{1}{\sin \left(\frac{\pi}{3}\right)}\sum_{n=0}^\infty \sin\left(\frac{\pi}{3} (n+1) \right)\int_0^1 x^{n+1/2} \ln x dx$$
$$\text{Since} \ \int_0^1 x^p \ln x dx= -\frac{1}{(p+1)^2}$$
$$I=-\frac{2}{\sqrt 3} \sum_{n=0}^\infty \frac{\sin\left((n+1)\frac{\pi}{3}\right)}{(n+1+1/2)^2}=-\frac{8}{\sqrt 3}\sum_{n=1}^\infty \frac{\sin\left(n\frac{\pi}{3}\right)}{(2n+1)^2} $$
And well by using the previous link we can deduce that the series equals to $\frac{G}{\sqrt 3} -\frac{\pi^2}{24}$, where $G$ is Catalan's constant.
I thought this might be a coefficient of some Fourier series, or taking the imaginary part of $\left(\sum_{n=1}^\infty \frac{e^{i\frac{n\pi}{3}}}{(2n+1)^2}\right)$, but I was not that lucky afterwards.
Is there a way to show the result without relying on that trigamma identity? Another approach to the integral would of course be enough.
|
This is a
major revision
of the last few lines.
The conclusion is that
$\sum_{n=1}^\infty \frac{\sin\left(n\frac{\pi}{3}\right)}{(2n+1)^2}
=\frac{\sqrt{3}}{72}\psi^{(1)}(\frac56)-\frac{\sqrt{3}\pi^2}{144}
$
where
$\psi^{(1)}$
is a polygamma function
(reference below).
$\begin{array}\\
\sum_{n=1}^\infty \frac{\sin\left(n\frac{\pi}{3}\right)}{(2n+1)^2}
&=\sum_{n=1}^\infty \left(\frac{\sin\left((3n-2)\frac{\pi}{3}\right)}{(2(3n-2)+1)^2}+\frac{\sin\left((3n-1)\frac{\pi}{3}\right)}{(2(3n-1)+1)^2}+\frac{\sin\left(3n\frac{\pi}{3}\right)}{(2(3n)+1)^2}\right)\\
&=\sum_{n=1}^\infty \left(\frac{\sin\left(-2\frac{\pi}{3}\right)}{(6n-3)^2}+\frac{\sin\left(-\frac{\pi}{3}\right)}{(6n-1)^2}+\frac{\sin\left(n\pi\right)}{(6n+1)^2}\right)\\
&=\sum_{n=1}^\infty \left(-\frac{\sqrt{3}/2}{(6n-3)^2}+\frac{\sqrt{3}/2}{(6n-1)^2}\right)\\
&=\frac{\sqrt{3}}{2}\sum_{n=1}^\infty \left(-\frac1{(6n-3)^2}+\frac1{(6n-1)^2}\right)\\
&=\frac{\sqrt{3}}{2}\sum_{n=1}^\infty \left(\frac1{(6n-1)^2}-\frac1{(6n-3)^2}\right)\\
&=\frac{\sqrt{3}}{2}\left(\sum_{n=1}^\infty \frac1{(6n-1)^2}-\frac19\sum_{n=1}^\infty\frac1{(2n-1)^2}\right)\\
&=\frac{\sqrt{3}}{2}\left(\frac1{36}\sum_{n=1}^\infty \frac1{(n-1/6)^2}-\frac19(1-\frac14)\zeta(2)\right)\\
&=\frac{\sqrt{3}}{2}\left(\frac1{36}\sum_{n=0}^\infty \frac1{(n+5/6)^2}-\frac19\frac34\frac{\pi^2}{6}\right)\\
&=\frac{\sqrt{3}}{72}\psi^{(1)}(\frac56)-\frac{\sqrt{3}\pi^2}{144}\\
\end{array}
$
$\psi^{(1)}(z)$
is a polygamma function:
https://en.wikipedia.org/wiki/Polygamma_function
$\psi^{(m)}(z)
=(-1)^{m+1}m!\sum_{n=0}^{\infty} \dfrac1{(z+n)^{m+1}}
$
so that
$\psi^{(1)}(z)
=\sum_{n=0}^{\infty} \dfrac1{(z+n)^2}
$
Also
$\begin{array}\\
\zeta(m)
&=\sum_{n=1}^{\infty} \dfrac1{n^m}\\
&=\sum_{n=1}^{\infty} \dfrac1{(2n-1)^m}+\sum_{n=1}^{\infty} \dfrac1{(2n)^m}\\
&=\sum_{n=1}^{\infty} \dfrac1{(2n-1)^m}+\dfrac1{2^m}\sum_{n=1}^{\infty} \dfrac1{n^m}\\
&=\sum_{n=1}^{\infty} \dfrac1{(2n-1)^m}+\dfrac1{2^m}\zeta(m)\\
\end{array}
$
so
$\zeta(m)(1-2^{-m})
=\sum_{n=1}^{\infty} \dfrac1{(2n-1)^m}
$.
|
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|
Finding the centroid of a tetrahedron I have four points to form a tetahedron
$$A=(0,-\frac{1}{2},-\frac{1}{4}\sqrt{\frac{1}{2}})
\\B=(0,\frac{1}{2},-\frac{1}{4}\sqrt{\frac{1}{2}})
\\C=(-\frac{1}{2},0,\frac{1}{4}\sqrt{\frac{1}{2}})
\\D=(\frac{1}{2},0,\frac{1}{4}\sqrt{\frac{1}{2}})$$
it looks like this :
I'm asked to find the centroid of the tetrahedron formed by these points.
I looked up in the internet and there is an easy way to find it:
$$G=(\frac{x_1+x_2+x_3+x_4}{4}+\frac{y_1+y_2+y_3+y_4}{4}+\frac{z_1+z_2+z_3+z_4}{4})$$
I tried to check it with the triple intregrals. In order to do it I find the plane at which A, D and B lie, and that would be my first integration limit and the plane formed by A,B and C.
The planes :
$$\overline{AB}=(0,1,0)\\
\overline{AD}=(\frac{1}{2},\frac{1}{2},\frac{1}{2}\sqrt{\frac{1}{2}})$$
And after takind the cross product of both, and then using $a(x-x_0)+b(y-y_0)+c(z-z_0)$ to obtain the plane I get $$\frac{1}{2}\sqrt{\frac{1}{2}}x-\frac{1}{2}z=\frac{1}{8}\sqrt{\frac{1}{2}}$$ and the other one containing A,B,C is $$ \frac{1}{2}\sqrt{\frac{1}{2}}x+\frac{1}{2}z=\frac{1}{8}\sqrt{\frac{1}{2}}$$
I divide it into this two planes, and then see how it looks like in the $x,y$ plane
And I get the integrals:
$$ \int_{0}^{\frac{1}{2}}\int_{-x+\frac{1}{2}}^{\frac{1}{2}}\int_{\frac{1}{4}\sqrt{\frac{1}{2}}}^{-\frac{1}{4}\sqrt{\frac{1}{2}}+x\sqrt{\frac{1}{2}}} zdz dy dx dx+\int_{\frac{1}{2}}^{0}\int_{0}^{-x+\frac{1}{2}}\int_{\frac{1}{4}\sqrt{\frac{1}{2}}}^{-\frac{1}{4}\sqrt{\frac{1}{2}}+x\sqrt{\frac{1}{2}}}z dz dy dx $$
and the other part
$$ \int_{0}^{\frac{1}{2}}\int_{x+\frac{1}{2}}^{\frac{1}{2}}\int_{\frac{1}{4}\sqrt{\frac{1}{2}}}^{\frac{1}{4}\sqrt{\frac{1}{2}}\sqrt{\frac{1}{2}}} zdz dy dx dx+\int_{\frac{1}{2}}^{0}\int_{0}^{-x-\frac{1}{2}}\int_{\frac{1}{4}\sqrt{\frac{1}{2}}}^{\frac{1}{4}\sqrt{\frac{1}{2}}-x\sqrt{\frac{1}{2}}}z dz dy dx $$
I don't get 0.
I can't see why. Is the formula I use not valid?
|
Few days ago, using vectog calculus, I have proved an analogous formula for centroid of a triangle. To appply this method here, two things are needed to check.
*
*The centroid is a common point of four lines, each of them joins a vertex with the centroid of the opposite face.
*The centroid divides this face in the ratio 3:1.
With this knowledge our task is trivial.
|
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|
how is that expression is generated
I have no idea that how is red arrowed mammoth term is generated from the yellow arrowed term
Please explain
|
The author starts by substituting $t$ for $e^x - 1$, hence
$$ \sin (e^x - 1) = \sin t $$
Now the power series of the sine function is plugged in (yellow arrowed line on the right)
$$ \sin (e^x - 1) = \sin t = t -\frac{t^3}{3!} + \frac{t^5}{5!} \mp \cdots $$
Now we resubstitute $e^x -1$ for $t$, at the same time we use the power series expansion of the exponential function $$ e^x = 1 + x + \frac{x^2}2 + \frac{x^3}{6} + \cdots $$
or
$$ e^x - 1 = x + \frac{x^2}2 + \frac{x^3}{6} + \cdots $$
giving
$$ \sin (e^x - 1) = \left(x + \frac{x^2}2 + \frac{x^3}{6} + \cdots\right) - \frac 16\left(x + \frac{x^2}2 + \frac{x^3}{6} + \cdots\right)^3 + \frac 1{120}\left(x + \frac{x^2}2 + \frac{x^3}{6} + \cdots\right)^5 \mp \cdots $$
which is the red arrowed expression.
|
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"timestamp": "2023-03-29T00:00:00",
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|
how to find subdifferential of a function $x^2+ |x-1|+|x-2|$ Given a function
$f(x) = x^2+ |x-1|+|x-2| $
find it's subdifferential.
My approach to solving this was to divide the answer into 5 parts:
*
*For |x-1|>1 and |x-2|>2
$f(x) = x^2+ x-1+x-2$ and $f'(x) = 2x+2$
*For |x-1|<1 and |x-2|<2
$f(x) = x^2-(x-1)-(x-2)$ and $f'(x) = 2x-2$
*For |x-1|>1 and |x-2|<2
$f(x) = x^2+(x-1)-(x-2)$ and $f'(x) = 2x$
*For |x-1|<1 and |x-2|>2
$f(x) = x^2-(x-1)+(x-2)$ and $f'(x) = 2x$
*For |x-1|=1 and |x-2|=2
$ f(x) = x^2$ and $f'(x) = 2x$
Does this look right? Is this the correct approach?
|
First, I have to admit I am not familiar with a "subdifferential". It look to me like you are just taking the derivative. Second, dividing with things like "|x- 1|> 1 and |x- 2|> 2" is confusing and probably not what you want. Instead divide the real line into three intervals: $x\le 1$, $1< x\le 2$, and $x> 2$.
For $x\le 1$ both x-1 and x-2 are negative: $x^2+ |x- 1|+ |x- 2|= x^2- (x- 1)- (x- 2)= x^2- 2x+ 3$. The derivative is $2x- 2$.
For $1< x\le 3$ x-1 is positive but x-2 is still negative:$x^2+ |x- 1|+ |x- 2|= x^2+ (x- 1)- (x- 2)= x^2+ 10$. The derivative is $2x$.
For $x> 2$ both x-1 and x-2 are positive: $x^2+ |x- 1|+ |x- 2|= x^2+ (x- 1)+ (x- 2)= x^2+ 2x- 3$. The derivative is $2x+ 2$.
|
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|
Find Derivative of $y=\tan^{-1}\left(\sqrt{\frac{x+1}{x-1}}\right)$ for $|x|>1$?
Question. If $y=\tan^{-1}\left(\sqrt{\frac{x+1}{x-1}}\right)$ for $|x|>1$ then $\frac{d}{dx}=?$
Answer: $\displaystyle\frac{dy}{dx}=-\frac{1}{2|x|\sqrt{x^2-1}}$
My 1st attempt- I followed the simple method and started by taking darivative of tan inverse and the following with chain rule and i got my answer as ($-\frac{1}{2x\sqrt{x^2-1}}$), which is not same as the above correct answer. 2nd method is that you can substitute $x=\sec\left(\theta\right)$ and while solving in last step we will get $\sec^{-1}\left(\theta\right)$ whose derivative contains $\left|x\right|$, but still i searched and don't know why its derivative has $\left|x\right|$
Here's my attempt stepwise
$\displaystyle\frac{dy}{dx}=\frac{1}{1+\left(\sqrt{\frac{x+1}{x-1}}\right)^2}\cdot\frac{1}{2\sqrt{\frac{x+1}{x-1}}}\cdot\frac{\left(x-1\right)-\left(x+1\right)}{\left(x-1\right)^2}$
$\displaystyle=\frac{\left(x-1\right)}{\left(x-1\right)+\left(x+1\right)}\cdot\frac{1\sqrt{x-1}}{2\sqrt{x+1}}\cdot-\frac{2}{\left(x-1\right)^2}$
$\displaystyle=-\frac{1}{2x}\cdot\frac{\left(x-1\right)\sqrt{x-1}}{\left(x-1\right)^2}\cdot\frac{1}{\sqrt{x+1}}$
$\displaystyle=-\frac{1}{2x\sqrt{x-1}\sqrt{x+1}}$
$\displaystyle=-\frac{1}{2x\sqrt{x^2-1}}$
Can you tell what i am doing wrong in my 1st attempt?
|
By the chain rule,
$$\left(\arctan\sqrt{\frac{x+1}{x-1}}\right)'=-\frac2{(x-1)^22\sqrt{\dfrac{x+1}{x-1}}\left(1+\dfrac{x+1}{x-1}\right)}=-\frac1{2x(x-1)\sqrt{\dfrac{x+1}{x-1}}}
\\=-\frac1{2x\text{ sgn}(x-1)\sqrt{x^2-1}}.$$
The claim follows from
$$\text{ sgn}(x-1)=\text{ sgn}(x)$$ (when $|x|>1$).
|
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|
Find the remainder $R$ of $(1^2+1)(2^2+1)...(p^2+1)$ divided by $p$
Find the remainder $R$ of $(1^2+1)(2^2+1)...(p^2+1)$ divided by $p$, with $p$ being a prime number greater than $3$.
For $p \equiv 1 \mod 4$, there exists an integer $j$ such that $p\mid j^2+1$ (since $-1$ is a quadratic residue of $p$), therefore $R=0$.
For $p \equiv 3 \mod 4$, how can we find $R$? Does $R$ depend on the value of $p$?
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If $p=2$ or $p\equiv 1\pmod{4}$, then $R$ is obviously $0$ because $-1$ is a quadratic residue modulo $p$. We assume from now on that $p\equiv 3\pmod{4}$.
Let $f(x)$ denote the polynomial $x^p-x$ over the Galois field $K=\operatorname{GF}(p)$. Note that $f$ factorizes into
$$f(x)=\prod_{t\in K}(x-t).$$
Let $E$ be the extension $K[X]/(X^2+1)$ of $K$, which is just the Galois field $\operatorname{GF}(p^2)$. (This is where we use the assumption that $p\equiv 3\pmod{4}$, since $X^2+1$ is irreducible over $K$.) Write $i$ for the image of $X$ under the canonical projection $K[X]\to K[X]/(X^2+1)=E$.
Now, we have
$$R=\prod_{k=1}^p(k^2+1)=\prod_{k=1}^p(i-k)(-i-k)=\left(\prod_{k=1}^p(i-k)\right)\left(\prod_{k=1}^p(-i-k)\right).$$
By the definition of $f$, we get
$$R=f(i)f(-i)=\left(i^p-i\right)\left((-i)^p-(-i)\right)=\left(i^{p-1}-1\right)\left((-i)^{p-1}-1\right).$$
Since $\frac{p-1}{2}$ is odd, we have
$$R=\left((-1)^{\frac{p-1}{2}}-1\right)\left((-1)^{\frac{p-1}{2}}-1\right)=\big((-1)-1\big)^2=4.$$
This is an equality in $K$. Translating this result back to $\Bbb{Z}$, we conclude that
$$R=\begin{cases}
0&\text{if}\ p=2 \vee p\equiv 1\pmod{4},\\
1&\text{if}\ p=3,\\
4&\text{if}\ p>3 \wedge p\equiv 3\pmod{4}.
\end{cases}$$
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"timestamp": "2023-03-29T00:00:00",
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|
find the sum to n term of $\frac{1}{1\cdot2\cdot3} + \frac{3}{2\cdot3\cdot4} + \frac{5}{3\cdot4\cdot5} + \frac{7}{4\cdot5\cdot6 } + ... $ $$\frac{1}{1\cdot2\cdot3} + \frac{3}{2\cdot3\cdot4} + \frac{5}{3\cdot4\cdot5} + \frac{7}{4\cdot5\cdot6 } + ... $$
$$=\sum \limits_{k=1}^{n} \frac{2k-1}{k\cdot(k+1)\cdot(k+2)}$$ $$= \sum \limits_{k=1}^{n} - \frac{1}{2}\cdot k + \sum \limits_{k=1}^{n} \frac{3}{k+1} - \sum \limits_{k=1}^{n}\frac{5}{2\cdot(k+2)} $$
I do not know how to get a telescoping series from here to cancel terms.
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When terms are in A.P in denominator, we use difference
of last and first terms in product to distribute the terms
of denominator over numerator, to change into
telescoping series.
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Divisibility 1,2,3,4,5,6,7,8,9,&10 Tried:
Seems the ten-digit number ends with $240$ or $640$ or $840$ (Is not true, there are more ways the number could end)
$8325971640,$
$8365971240,$
$8317956240,$
$8291357640,$
$8325971640,$
$8235971640,$
$1357689240,$
$1283579640,$
$1783659240,$
$1563729840,$
$1763529840,$
$1653729840,$
$7165239840,$
$7195236840,$
$2165937840,$
$9283579640$
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The number which are divisible by $8$ is also divisible by $2$ and $4$.
The number which are divisible by $9$ is also divisible by $3$.
The number of which is divisible by $6$ is also divisible by $2$ and $3$.
The number which are divisible by $10$ is also divisible by $2$ and $5$.
Also also the number we expect is divisible by $11$ and $7$.
So the number is in the form $=P×2^{3i}×3^{2j}×5^k×7^m×11^n$,
Where $i$, $j$, $k$, $m$, & $n$ are positive any positive integer and $P$ is any positive integer integer.Using this condition we will produce a required ten digit number.
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Trigonometric Identities: Given $\tan(2a)=2$ and $\frac{3\pi}{2}Given $\tan(2a)=2$ and $\frac{3\pi}{2}<a<2\pi$ find value of $\tan(a)$
I first found the values of $\cos(2a)$ and $\sin(2a)$ and then used the half angle formula.
$$\tan(a)=\tan\frac{2a}{2}=\frac{1-\cos(2a)}{\sin(2a)}
\implies\frac{5}{2\sqrt 5}\left(1-\frac{\sqrt{5}}5\right)$$
I then simplified that to $\frac{\sqrt{5}}{2}-1$
Then because it is in fourth quadrant I multiplied by negative 1 and got:
$$1-\frac{\sqrt{5}}{2}$$
Am I doing something wrong, or am I correct?
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Just use the double-angle identity.
$$\tan(2a) = \frac{2\tan{a}}{1-\tan^2(a)}$$
$$\implies \frac{2\tan{a}}{1-\tan^2(a)} = 2$$
$$\implies 2\tan{a} = 2-2\tan^2{a} \implies \tan{a} = 1-\tan^2{a} \implies \tan^2{a}+\tan{a}-1 = 0$$
Set $t = \tan{a}$ and solve for $t$.
$$t^2+t-1 = 0$$
$$t = \frac{-1\pm\sqrt{5}}{2}$$
Plug in $\tan{a}$.
$$\tan{a} = \frac{-1\pm\sqrt{5}}{2}$$
But $\frac{3\pi}{2} < a < 2\pi$, so $\tan{a} < 0$.
$$\tan{a} = \frac{-1-\sqrt{5}}{2}$$
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Prove $\sum^{k}_{i=0}{F(i)} + 1 = F(k+2)$ without induction I want to prove that $\sum^{k}_{i=0}{F(i)} + 1 = F(k+2)$, where $F(0) = 0$, $F(1) = 1$ and for all $n \geq 2$, $F(n) = F(n-1) + F(n-2)$ without using induction.
I want to prove it without induction because I want to implement the equation within another induction proof about Fibonacci numbers. Having 2 induction proofs together would be messy.
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Use the general term formula: $$F_n=\frac{\phi^n-\psi^n}{\sqrt{5}}, \ \ \text{where} \ \ \phi=\frac{1+\sqrt{5}}{2}, \ \ \psi=\frac{1-\sqrt{5}}{2}.$$
Use the sum of geometric progression:
$$\begin{align}\sum^{n}_{i=0}{F(i)} + 1 &= \frac{\phi^0-\psi^0}{\sqrt{5}}+\frac{\phi^1-\psi^1}{\sqrt{5}}+\cdots + \frac{\phi^n-\psi^n}{\sqrt{5}}+1=\\
&=\frac{1}{\sqrt{5}}([\phi^1+\phi^2+\cdots +\phi^n]-[\psi^1+\psi^2+\cdots +\psi^n])+1=\\
&=\frac{1}{\sqrt{5}}\cdot \left(\frac{\phi(\phi^n-1)}{\phi-1}-\frac{\psi(\psi^n-1)}{\psi-1}\right)+1=\\
&=\frac{1}{\sqrt{5}}\cdot \left(\frac{\phi(\phi^n-1)}{-\psi}-\frac{\psi(\psi^n-1)}{-\phi}\right)+1=\\
&=\frac{1}{\sqrt{5}}\cdot \frac{\phi^2(\phi^n-1)-\psi^2(\psi^n-1)}{-\psi\cdot \phi}+1=\\
&=\frac{\phi^{n+2}-\psi^{n+2}}{\sqrt{5}}+ \frac{-\phi^2+\psi^2}{\sqrt{5}}+1=\\
&=\frac{\phi^{n+2}-\psi^{n+2}}{\sqrt{5}}=\\
&=F(n+2).\end{align}$$
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Find $a$, $b$, $c$, $d$ such that $(ax+b)^2(x+c) = 4x^3 + dx^2 + 55x - 100$ How would I find $a$, $b$, $c$, and $d$ in :
$$(ax+b)^2(x+c) = 4x^3 + dx^2 + 55x - 100$$
I have already worked out that $a=2$, but I may be wrong. I am not sure how to find the values of the other letters.
Thanks in advance!
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Hint: $$(ax+b)^2(x+c)={a}^{2}c{x}^{2}+{a}^{2}{x}^{3}+2\,abcx+2\,ab{x}^{2}+{b}^{2}c+x{b}^{2}=a^2x^3+x^2(a^2c+2ab)+x(2abc+b^2)+b^2c$$
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Solve the equation $2x^3+x^2-7x-6=0$ given that the difference of two roots is $3$ Q:Solve the equation $2x^3+x^2-7x-6=0$ given that the difference of two roots is $3$.My book solve it leting the roots of the equation be $\alpha,\alpha+3,\beta$ then find the equation whose roots are $\alpha-3,\alpha,\beta-3$.And i know how to find the equation whose roots are diminished by $3$ and they get it $2x^3+19x^2+53x+36=0.$
Hence $(x+1)$ is a common factor of $2x^3+x^2-7x-6=0$ and $2x^3+19x^2+53x+36=0.$And they showed all the roots are: $-1,2,\frac{-3}{2}$Now my Question is "Is there exist any easier way to solve it?" Because in this process i need a lot of work in order to find the new equation and find GCD/HCF of these two equation.Any hints or solution will be appreciated.
Thanks in advance.
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$$2x^3+x^2-7x-6$$ and $$2x^3+19x^2+53x+36$$
have a common root, which is such that
$$x^2-7x-6=19x^2+53x+36.$$
By solving the quadratic, this root is one of $-1$ or $-\dfrac73$. Then $-1$ fits and by long division you reduce to
$$2x^2-x-6=0.$$
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How many $4$-element subsets of $S = \{a, b, c, d, e, f, g, h, i\}$ contain at least one of $a$ and $b$? Consider the set $S = \{a,b,c,d,e,f,g,h,i,j\}$. How many $4$-element subsets of $S$ contain at least one of $a$ and $b$?
So my thought is :
1) number of $4$-element subsets in the set $S = 210$
2) $4$-element subsets of $S$ contain exactly one of $a$ and $b$ $= \dfrac{2!}{1!1!} \cdot \dfrac{8!}{3!5!} = 112$
3) $4$-element subsets of $S$ contain both $a$ and $b$ $= C(8,2) = \dfrac{8 \cdot 7}{2 \cdot 1} = 28$
How many $4$-element subsets of $S$ contain at least one of $a$ and $b$?
\begin{align*}
& = 210 - (112+28)\\
& = 70
\end{align*}
Is it right?
thanks!
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You have correctly calculated that the number of four-element subsets of the ten-element set $S$ is
$$\binom{10}{4} = 210$$
that the number of subsets of $S$ that contain exactly one of $a$ or $b$ is
$$\binom{2}{1}\binom{8}{3} = 112$$
and that the number of subsets of $S$ that contain both $a$ and $b$ is
$$\binom{2}{2}\binom{8}{2} = 28$$
However, you have drawn the wrong conclusion. The number of subsets that contain at least one of $a$ and $b$ is the sum of the number that contain exactly one of $a$ or $b$ and the number that contain both $a$ and $b$. Hence, the number of subsets that contain at least one of $a$ and $b$ is
$$\binom{2}{1}\binom{8}{3} + \binom{2}{2}\binom{8}{2} = 140$$
Alternatively, we could subtract the number of subsets of $S$ that contain neither $a$ nor $b$ from the total. The number of subsets of $S$ that contain neither $a$ nor $b$ is
$$\binom{2}{0}\binom{8}{4} = 70$$
Subtracting this from the number of four-element subsets of $S$ yields
$$\binom{10}{4} - \binom{2}{0}\binom{8}{4} = 140$$
as before.
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If $\tan(A-B)=1$ and $\sec(A+B)=2/\sqrt{3}$, find the minimum positive value of $B$
If $\tan(A-B)=1$ and $\sec(A+B)=2/\sqrt{3}$, find the minimum positive value of $B$
I am using $\tan(A-B)=1$, so $A-B=n\pi+\pi/4$ and $A+B = 2n\pi\pm\pi/6$. Solving these I am getting $B =7\pi/24$ and $A =37\pi/24$.
The book I am refering to has marked the answer as $19\pi/24$.
Am I doing something wrong?
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Your answer is correct. For all $n \in \mathbb{Z}$:
$$\tan(A-B) = 1 \implies A-B = \frac{\pi}{4}+\pi n$$
Recall that secant, like cosine, is an even function. (Which you've apparently referred to as well, after the format editing.)
$$\sec(-\theta) = \sec(\theta)$$
$$\sec(A+B) = \frac{2}{\sqrt 3} \implies \cos(A+B) = \frac{\sqrt 3}{2} \implies \pm(A+B) = \frac{\pi}{6}+2\pi n$$
Therefore, there are $2$ cases, neither of which yields the book's answer.
Case $1$: With $+(A+B)$, by subtracting the two equations to eliminate $A$, the following equation is obtained.
$$-2B = \frac{\pi}{4}+\pi n-\bigg(\frac{\pi}{6}+2\pi n\bigg)$$
$$-2B = \frac{\pi}{4}+\pi n-\frac{\pi}{6}-2\pi n$$
$$-2B = \frac{\pi}{12}-\pi n$$
$$B = \frac{\pi n}{2}-\frac{\pi}{12}$$
The minimum here is $B = \frac{11\pi}{24}$.
Case $2$: With $-(A+B)$, by adding the two equations to eliminate $A$, the following equation is obtained.
$$-2B = \frac{\pi}{4}+\pi n+\bigg(\frac{\pi}{6}+2\pi n\bigg)$$
$$-2B = \frac{\pi}{4}+\pi n+\frac{\pi}{6}+2\pi n$$
$$-2B = \frac{5\pi}{12}+3\pi n$$
$$B = -\frac{3\pi n}{2}-\frac{5\pi}{24}$$
The minimum here is $B = \frac{7\pi}{24}$.
Thus, the minimum positive value of $B$, as you correctly found, is $\frac{7\pi}{24}$.
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Find diagonalizable decomposition of $A$.
Let $A=\begin{pmatrix}1&2\\-2&-3\end{pmatrix}.$ Find $P$ and $D$, such that: $$A=PDP^{-1}$$
First let's calculate the characteristic polynomial for $A$: $$P_a(x)=\det(A-xI_2)=(x+1)^2.$$
So the eigenvalues of $A$ are $\lambda=-1$ with $a(\lambda)=2$ where $a(\lambda)$ is the algebraic multiplicity of $\lambda$.
So now let's find $V_\lambda.$
For that we have:
$$(A-\lambda I_2)(x, y)=(0,0)\implies y=-x$$
So we have:
$$V_\lambda=\{\alpha(1,-1)\mid \alpha\in\mathbb{R}\}$$
$$\dim_{\mathbb{R}}V_\lambda=1=g(\lambda)$$
where $g(\lambda)$ is the geometrically multiplicity of $\lambda.$Now:
$$D=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$$
How do we find $P$?
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In order to solve for the eigenvectors, you simply put them back in the equation. Suppose that $A$ is
$$ A = \begin{pmatrix} 1& 2 \\ -2 & -3 \end{pmatrix} \tag{1}$$
then if solved for our eigenvalues by the equation
$$ \det(A- \lambda I) =0 \tag{2} $$
it gave us $\lambda_{1}, \lambda_{2}$ which you below $\lambda_{1},\lambda_{2} =-1$ in order to get $P$ we do the following.
$$ A -(-1)I x=0 \tag{3}$$
this becomes the following equation
$$ A = \begin{pmatrix} 1& 2 \\ -2 & -3 \end{pmatrix} + \begin{pmatrix} 1& 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2& 2 \\ -2 & -2 \end{pmatrix} \tag{4}$$
then apply $x$
$$ \begin{pmatrix} 2& 2 \\ -2 & -2 \end{pmatrix}\begin{pmatrix} x_{1} \\x_{2} \end{pmatrix} = \begin{pmatrix} 0 \\0 \end{pmatrix} \tag{5}$$
$$ 2x_{1} + 2x_{2} = 0 \\ -2x_{1} - 2x_{2} = 0 \tag{6}$$
then you see they're equal and opposite
$$ v_{1} = \begin{pmatrix} 1 \\ -1\end{pmatrix} $$
which would mean $v_{2}$ nearly the same
we would need to normalize them of course..let's normalize them
$$ q_{1} =\frac{v_{1}}{\|v_{1}\|} = \frac{v_{1}}{\sqrt{2}} \tag{7}$$
if you use python. You will see
import numpy as np
A = np.matrix([[1,2],[-2,-3]])
w,v = np.linalg.eig(A)
v
Out[2]:
matrix([[ 0.70710678, -0.70710678],
[-0.70710678, 0.70710678]])
just demonstrating this
test = 1/np.sqrt(2)
test
Out[4]: 0.7071067811865475
Note that diagonalizable means linearly independent.
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|
Combinatorial proof of $\binom{3n}{3} =3\binom{n}{3} +6n\binom{n}{2} +n^3$?
Give a combinatorial proof of the following identity: $$\binom{3n}{3} =3\binom{n}{3} +6n\binom{n}{2} +n^3.$$
I've been working on this proof for hours, however I'm not able to show LHS = RHS-
I completely understand binomial theorem and few combinatorial proofs but not able to succeed this one.
Help would be appreciated.
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Notice that the LHS is:
$$\frac{3n \times (3n-1) \times (3n-2)}{3!} = \frac{n \times (3n-1) \times (3n-2)}{2}$$
$$\to \frac{9n^3-9n^2+2n}{2}$$
For the RHS:
$$3 \times \frac{n\times(n-1)\times(n-2)}{3!} + 6n \times \frac{n\times(n-1)}{2!} + \frac{2n^3}{2}$$
$$\Longrightarrow \frac{n\times(n-1)\times(n-2)}{2} + \frac{6n^2 \times(n-1)}{2} + \frac{2n^3}{2}$$
$$\to \frac{9n^3-9n^2+2n}{2}$$
Hence, RHS=LHS.
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Find the value of $(\frac{\alpha}{\alpha +1})^3+(\frac{\beta}{\beta +1})^3+(\frac{\gamma}{\gamma +1})^3$ Find the value of
$(\frac{\alpha}{\alpha +1})^3+(\frac{\beta}{\beta +1})^3+(\frac{\gamma}{\gamma +1})^3$ where $\alpha,\beta,\gamma$ are roots of the equation $x^3+2x^2+3x+3=0$.
I tried to use the formula which is wrong
$(\frac{\alpha}{\alpha +1})^3+(\frac{\beta}{\beta +1})^3+(\frac{\gamma}{\gamma +1})^3=((\frac{\alpha}{\alpha +1})+(\frac{\beta}{\beta +1})+(\frac{\gamma}{\gamma +1}))^3-3((\frac{\alpha}{\alpha +1})(\frac{\beta}{\beta +1})(\frac{\gamma}{\gamma +1}))$
And then I broke the terms to get
$(\frac{13}{5})^3-\frac{36}{5}$
but this is $\ne 44$, which should be the answer.
I know where I was wrong, the formula is $(a+b+c)^3=a^3+b^3+c^3-3(a+b)(b+c)(c+a)$
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$$\alpha+\beta+\gamma=-2,$$ $$\alpha\beta+\alpha\gamma+\beta\gamma=3$$ and
$$\alpha\beta\gamma=-3.$$
Thus, $$\sum_{cyc}\frac{\alpha}{\alpha+1}=\frac{\sum\limits_{cyc}\alpha(\beta+1)(\gamma+1)}{\prod\limits_{cyc}(\alpha+1)}=\frac{\sum\limits_{cyc}(\alpha\beta\gamma+2\alpha\beta+\alpha)}{\alpha\beta\gamma+\alpha\beta+\alpha\gamma+\beta\gamma}=\frac{-9+6-2}{-3+3-2+1}=5,$$
$$\sum_{cyc}\frac{\alpha\beta}{(\alpha+1)(\beta+1)}=\frac{\sum\limits_{cyc}\alpha\beta(\gamma+1)}{\prod\limits_{cyc}(\alpha+1)}=\frac{-9+3}{-1}=6$$ and
$$\prod_{cyc}\frac{\alpha}{\alpha+1}=\frac{\alpha\beta\gamma}{\prod\limits_{cyc}(\alpha+1)}=\frac{-3}{-1}=3.$$
Id est, $$\sum_{cyc}\left(\frac{\alpha}{\alpha+1}\right)^3=\left(\sum_{cyc}\frac{\alpha}{\alpha+1}\right)^3-3\sum_{cyc}\frac{\alpha}{\alpha+1}\sum_{cyc}\frac{\alpha\beta}{(\alpha+1)(\beta+1)}+3\prod_{cyc}\frac{\alpha}{\alpha+1}=$$
$$=5^3-3\cdot5\cdot6+3\cdot3=44.$$
|
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|
Is $x(1 - 2x) \le \frac{1}{8}$ and further, is $x(1 - ax) \le \frac{1}{4a}$ It is clear that $x(1-x) \le \frac{1}{4}$
Does it likewise follow that $x(1-2x) \le \frac{1}{8}$?
Here's my reasoning:
(1) For $x < \frac{1}{4}$, $x(1-2x) < \frac{1}{8}$
(2) For $\frac{1}{4} < x < \frac{1}{2}$, $x(1-2x) < \frac{1}{8}$
(3) For $\frac{1}{2} < x$, $x(1-2x) < 0$
Further, can this be generalized to $x(1-ax) \le \frac{1}{4a}$
Since:
(1) For $x < \frac{1}{2a}$, $x(1-ax) < \left(\frac{1}{2a}\right)\left(\frac{1}{2}\right) = \frac{1}{4a}$
(2) For $\frac{1}{2a} < x < \frac{1}{a}$, $x(1-ax) < \left(\frac{1}{2a}\right)\left(\frac{1}{2}\right) = \frac{1}{4a}$
(3) For $\frac{1}{a} < x$, $x(1-ax) < 0$
Are both of these observations correct? Is only one correct? Is there an exception that I am missing?
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Without any derivative, just high-school theory of quadratic equations:
A quadratic polynomial (with real coefficients) has a global extremum at the arithmetic mean of its roots. Further, this extremum is a maximum if its leading coefficient is negative, a minimum if the leading coefficient is positive.
So here, the extremum is attained at $\dfrac1{2a}$ and the leading coefficient is $-a$. This extremum is a maximum if $a>0$, a minimum if $a<0$ and it is equal to
$$\frac1{2a}\biggl(1-a\,\frac1{2a}\biggr)=\frac1{4a}.$$
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What is the biggest possible sum $|X_1-X_2|+|X_2-X_3|+\cdots+|X_{n-1}-X_n|$ where $X_1,X_2,\cdots,X_n$ are first $n$ positive integers? What is the biggest possible sum $|X_{1}-X_{2}|+|X_{2}-X_{3}|+\cdots+|X_{n-1}-X_{n}|$ where $X_{1},X_{2},\cdots,X_{n}$ are first $n$ positive integers?
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Be greedy. $n$ should go next to $1,2$, then $n-1$ next to $1$ and so on. There are other arrangements that give the same sum, but none better. For $n=12$ it gives $7,5,9,3,11,1,12,2,10,4,8,6$ for $2+4+6+8+10+11+10+8+6+4+2$ and for $n=13$ it gives $8,5,10,3,12,1,13,2,11,4,9,6,7$ for $3+5+7+9+11+12+11+9+7+5+3+1$ The patterns in the elements of the sums should be clear.
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|
Determine that the function $f(x)=\sqrt{x^2-x-6} \text{ in } x_0=3$ is continous with the $\varepsilon$-$\delta$-definition of limit/criterion
[Proof-verification] Determining whether the function $f(x)=\sqrt{x^2-x-6} \text{ in } x_0=3$; $x_0\in \mathbb{R}$ is continous $\color{red}{\text{ in }x_0}$ or not with the
$\varepsilon$-$\delta$-definition of limit/criterion:
$$f(x)=\sqrt{x^2-x-6} \text{ in } x_0=3$$
Proof:
Let $\varepsilon>0$ and ${\mid x-x_0\mid}<\delta \iff {\mid x-3\mid}<\delta$
\begin{align}
{\mid f(x)-f(x_0)\mid}&= {\mid\sqrt{x^2-x-6}-\sqrt{3^2-3-6}\mid}\\
& ={\mid\sqrt{x^2-x-6}\mid}\\
& ={\mid\sqrt{x+2} \cdot \sqrt{x-3}\mid}\\
& <\sqrt{x+2}\cdot \sqrt{x+2}\\
& = x+2\\
& < x-3+5 \\
& <\delta+5 =: \varepsilon \\
& \iff \delta = \varepsilon -5
\end{align}
$\implies$ the function is continous in $x_0=3 \qquad \qquad \qquad \qquad \qquad_\blacksquare $
Is this proof correct?
|
First of all, the function $f$ is not defined on $\mathbb{R}$ but for
$$
x\in(-\infty,-2]\cup[3,+\infty).
$$
So one can only talk about its continuity of $f$ at $x=3$ from the right.
For $x>3$, you are right to get
$$
|f(x)-f(3)|=\sqrt{x-3}\sqrt{x+2}.
$$
Note that you don't need the absolute value for the square root terms.
But then you made a mistake: the $\delta$ you get must be positive.
The term $\sqrt{x-3}$ should not be dropped and it would give you the desired $\delta$.
Consider instead for $0<x-3<1$ the inequality
$$
\sqrt{x-3}\sqrt{x+2}\leq 6\sqrt{x-3}\le\varepsilon.
$$
|
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|
Find the inverse Laplace of $ \ \frac{2}{(s-1)^3(s-2)^2}$. Find the inverse Laplace of $ \ \frac{2}{(s-1)^3(s-2)^2}$.
Answer:
To do this we have to make partial fractions as follows:
$ \frac{2}{(s-1)^3(s-2)^2}=\frac{A}{S-1}+\frac{B}{(s-1)^2}+\frac{C}{(s-1)}+\frac{J}{(s-1)^3}+\frac{K}{(s-2)^2}+\frac{L}{s-2}$
Am I right so far?
Does there is any easy way?
|
You have one too many $\frac {1}{(s-1)}$ terms.
$\frac {2}{(s-1)^3(s-2)^2} = \frac {A}{(s-1)} + \frac {B}{(s-1)^2} + \frac {C}{(s-1)^3} + \frac {D}{(s-2)} + \frac {E}{(s-2)^2}$
Traditionally, you would then multiply through to clear out the denominators.
$2 = A(s-1)^2(s-2)^2 + B(s-1)(s-2)^2 + C(s-2)^2 + D(s-1)^3(s-2) + E(s-1)^3$
Here is a trick though.
$\lim_\limits{s\to 1} \frac {2(s-1)^3}{(s-1)^3(s-2)^2} = C $
and
$\lim_\limits{s\to 2} \frac {2(s-2)^2}{(s-1)^3(s-2)^2} = E$
But, you might find it easier to do this in pieces.
If you start with
$\frac {2}{(s-1)^3(s-2)^2} = \frac {As^2+Bs+C}{(s-1)^3} + \frac {Ds+E}{(s-2)^2}$
(Note that this will not have the same values for $A,B,C,D,E$ as the first approach)
Then there is less multiplying up front. But, you might need to do more decomposition down the line.
|
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|
What am I doing wrong solving this system of equations? $$\begin{cases}
2x_1+5x_2-8x_3=8\\
4x_1+3x_2-9x_3=9\\
2x_1+3x_2-5x_3=7\\
x_1+8x_2-7x_3=12
\end{cases}$$
From my elementary row operations, I get that it has no solution. (Row operations are to be read from top to bottom.)
$$\left[\begin{array}{ccc|c}
2 & 5 & -8 & 8 \\
4 & 3 & -9 & 9 \\
2 & 3 & -5 & 7 \\
1 & 8 & -7 & 12
\end{array}\right]
\overset{\overset{\large{R_1\to R_1-R_3}}{{R_2\to R_2-2R_3}}}{\overset{R_3\to R_3-2R_4}{\large\longrightarrow}}
\left[\begin{array}{ccc|c}
0 & 2 & -3 & 1 \\
0 & -3 & 1 & -5 \\
0 & -13 & 9 & -17 \\
1 & 8 & -7 & 12
\end{array}\right]
\overset{\overset{\large{R_3\,\leftrightarrow\, R_4}}{R_2\,\leftrightarrow\, R_3}}{\overset{R_1\,\leftrightarrow\,R_2}{\large\longrightarrow}}
\left[\begin{array}{ccc|c}
1 & 8 & -7 & 12 \\
0 & 2 & -3 & 1 \\
0 & -3 & 1 & -5 \\
0 & -13 & 9 & -17
\end{array}\right]$$
$$\overset{R_4\to R_4-R_3}{\large\longrightarrow}
\left[\begin{array}{ccc|c}
1 & 8 & -7 & 12 \\
0 & 2 & -3 & 1 \\
0 & -3 & 1 & -5 \\
0 & 10 & 8 & -12
\end{array}\right]
\overset{\overset{\large{R_3\to R_3+R_2}}{R_4\to R_4-5R_2}}{\large\longrightarrow}
\left[\begin{array}{ccc|c}
1 & 8 & -7 & 12 \\
0 & 2 & -3 & 1 \\
0 & -1 & -2 & -4 \\
0 & 0 & 23 & -17
\end{array}\right]
\overset{\overset{\large{R_2\to R_2+2R_3}}{R_3\to-R_3}}{\large\longrightarrow}$$
$$\left[\begin{array}{ccc|c}
1 & 8 & -7 & 12 \\
0 & 0 & -7 & -7 \\
0 & 1 & 2 & 4 \\
0 & 0 & 23 & -17 \\
\end{array}\right]
\overset{R_2\,\leftrightarrow\,R_3}{\large\longrightarrow}
\left[\begin{array}{ccc|c}
1 & 8 & -7 & 12 \\
0 & 1 & 2 & 4 \\
0 & 0 & -7 & -7 \\
0 & 0 & 23 & -17 \\
\end{array}\right]$$
However, the answer in the book $(3, 2, 1)$ fits the system.
Was there an arithmetical mistake, or do I misunderstand something fundamentally?
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Hint: Try inputing the solution $(3,2,1)$ into every step. That will allow you to identify the step where you went wrong.
|
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|
Let ${x_n}=2^{n}a_n$, and $a_{n+1}=\sqrt{\frac{1-\sqrt{1-a_n^2}}{2}}, a_0=1$, how to prove ${x_n}$ converges? How to show $x_n=2^na_n$ converges, where $a_{n+1}=\sqrt{\frac{1-\sqrt{1-a_n^2}}{2}}$
The question originated from Professor David McKinnon.
Attempt:
I did prove it is increasing but failed to show it is bounded above. The sequence should converge to $\frac{\pi}{2}$, if I'm not mistaking.
In addition, ${a_n}$ is a decreasing sequence, and it is bounded below by 0, so I tried to use the fact that ${a_n}$ is a Cauchy to prove that ${x_n}$ is also a Cauchy.
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Since $$\frac{a_{n+1}^2}{a_n^2} = \frac{1 - \sqrt{1-a_n^2}}{2a_n^2} = \frac{1}{2(1+\sqrt{1-a_n^2})} \le \frac{1}{2}$$
We have
$$a_n^2 = a_0^2 \prod_{k=1}^n \frac{a_k^2}{a_{k-1}^2} \le 2^{-n}
\quad\implies\quad
\sum_{k=0}^\infty a_k^2 \le \sum_{k=0}^\infty 2^{-k} = 2$$
Since
$$\frac{x_{n+1}^2}{x_n^2} = \frac{4a_{n+1}^2}{a_n^2}
= \frac{2}{1+\sqrt{1-a_n^2}}
= 1 + \frac{1-\sqrt{1-a_n^2}}{1+\sqrt{1-a_n^2}}
= 1 + \frac{a_n^2}{(1+\sqrt{1-a_n^2})^2}$$
We find
$$1 \le \frac{x_{n+1}^2}{x_n^2} \le 1 + a_n^2 \le e^{a_n^2}$$
LHS tells us the sequence $(x_n)$ is increasing while RHS tells us it is bounded from above.
$$x_n^2 = x_0^2 \prod_{k=1}^n \frac{x_k^2}{x_{k-1}^2}
\le \prod_{k=1}^n e^{a_{k-1}^2} = e^{\sum_{k=0}^{n-1} a_k^2}
< e^{\sum_{k=0}^\infty a_k^2} = e^2$$
As a result, sequence $(x_n)$ converges to some number $\le e$.
|
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|
Need help showing that $\zeta (x)$= ${1\over 1-2^{1-x}}$ $\eta (x)$ This question is I am working on is the extension of the domain of zeta function.$\zeta(x)$=$1+\frac{1}{2^x}+\frac{1}{3^x}+\frac{1}{4^x}+\frac{1}{5^x}+\frac{1}{6^x}+ \cdots$$\eta(x)$=$1-\frac{1}{2^x}+\frac{1}{3^x}-\frac{1}{4^x}+\frac{1}{5^x}-\frac{1}{6^x}+ \cdots$Note: The zeta $\zeta(x)$ function consists of positive terms only and the eta $\eta(x)$ function converges for all positive values for $x$ even when $x$ is less than $0$ where $0\lt x \lt 1$. Also in the equation $\zeta (x)$= ${1\over 1-2^{1-x}}$ $\eta (x)$, the left hand side is only defined when $x\gt 1$ because that is the domain of the zeta series. The right hand side can be used to define the zeta function for $0\lt x \lt 1$ and the right hand side totally depends on $\eta(x)$ in the function because it is already defined for all positive values for $x$. Here is my attempt so far:$({1\over 1-2^{1-x}})$ $\eta (x)$$\qquad\qquad\qquad=({1\over 1-2^{1-x}})\cdot 1 -({1\over 1-2^{1-x}})(\frac{1}{2^x})+({1\over 1-2^{1-x}})(\frac{1}{3^x})-({1\over 1-2^{1-x}})(\frac{1}{4^x})+({1\over 1-2^{1-x}})(\frac{1}{5^x})-({1\over 1-2^{1-x}})(\frac{1}{6^x})+\cdots$$\qquad\qquad\qquad=(\frac{2^x}{2^x-2})-(\frac{1}{2^x-2})$$+$[$({2/3})^{x}\over{2^x-2}$]$-$($2^{-x}\over{2^x-2}$)$+$[$({2/5})^{x}\over{2^x-2}$]$-$($3^{-x}\over{2^x-2}$)$+\cdots$By adding each pair of terms from above we get(note: I attempted to input the info but its too difficult to do due to the complexity from combining the terms but I have the next step written down and after that I am stuck showing it). Need help showing that $\zeta (x)$= ${1\over 1-2^{1-x}}$ $\eta (x)$. Getting cancellations is hard
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An idea for you to develop:
$$\zeta(s)=\sum_{n=1}^\infty\frac1{n^s}=\sum_{n=1}^\infty\frac1{(2n)^s}+\sum_{n=1}^\infty\frac1{(2n-1)^s}=\frac1{2^s}\zeta(s)+\sum_{n=1}^\infty\frac1{(2n-1)^s}\implies\ldots$$
Observe you're going to get lots of "poles"...which in fact are removable singularities.
|
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|
Using the Well Ordering Principle to Show that n^3 - n is divisible by 6. A Guide On Well-Ordering Principle
MIT OCW Guide on WOP
I read the following two sources and I think I understand the structure of proofs involving WOP, but I am stuck on the open-ended part where you prove P(n-1) for the following problem:
*
*Use the well-ordering principle to show that n^3 - n is divisible by 6.
Any help would be greatly appreciated!
Note: After factoring out (n-1)^3 - (n-1) it took the form of n^3 - 3n^2 + 2n and the only difference between that and n^3 - n is "3n^2 - 3n" which I proved was divisible by 6 via induction. Is there a better way to solve this problem?
Edit: Show that $n^3-n$ is divisible by $6$ using induction, post is useful however I not supposed to use induction for the main part of the proof.
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Define $f(n) = n^3-n$.
Since $f(-n) = -f(n)$, we only have to prove that $f(n)$ is divisible by $6$ for $n \ge 0$. So our domain is the well-ordered set of natural numbers and we always have $f(n) \ge 0$.
We note that both $f(0)$ and $f(1)$ are divisible by $6$.
Let $k \gt 1$ be the least number such that $f(k)$ is not divisible by $6$, so we employ Euclidean division and write
$\tag 1 f(k) = k^3 - k = 6q + r \quad \text{ with } q \ge 0 \text{ and } 1 \le r \le 5$
Plugging $k-1$ into $f$ we get
$\tag 2 f(k-1) = (k-1)[(k-1)^2-1]$
$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = (k-1) \,(k^2 -2k)$
$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = k^3 -3k^2 + 2k$
$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = (k^3 - k) + k + 2k - 3k^2$
$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = 6q + r + 3k\,(1+k)$
Now both $2$ and $3$ must divide $3k\,(1+k)$, so we write that
$\tag 3 f(k-1) = 6q + r + 6q^{'} = 6(q+q^{'}) + r$
So $f(k-1)$ is also not divisible by $6$, a contradiction.
|
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|
Parametrization of a rotated ellipse I need to parametrize an ellipse centered at the origian in cartesian coordinates given by
$a x^2 + b xy + c y^2 = 1$
Now I dug up from old notes that this can be brought into normal form through a rotation $\tan 2 \theta = \frac {b}{a-c}$, which eliminates the $xy$ term in rotated coordinates $x' = x \cos \theta + y \sin \theta, y = - x \sin \theta + y \cos \theta$ and we get
$\left( \frac{a + c + \sqrt{(a-c)^2+b^2}}{2}\right) x'^2 + \left( \frac{a + c - \sqrt{(a-c)^2+b^2}}{2}\right)y'^2 = 1$
which can be expressed as a normal equation $\frac{x'^2}{\alpha^2} + \frac{y'^2}{\beta^2} = 1$ if $4ac > b^2$ and both of the coefficients above have positive signs (which we may assume). This is then parametrized in the standard way by
$x' = \alpha \cos t, y' = \beta \sin t$ for $0\le t < 2\pi$
in the rotated coordinates and by $x = x' \cos \theta - y' sin \theta, y = x' sin \theta + y' \cos \theta$ in the original coordinates. Now while the above works, it's been pointed out to me that I am making things more complicated than necessary. Are there better ways of achieving this? I was thinking of diagonalizing the matrix
$(x, y)' \begin{pmatrix} a & b/2 \\ b/2 & c \end{pmatrix} (x,y)=1$
but I dont think that would make things easier. Is there a direct parametrization of the ellipse in terms of $a,b,c$?
|
Rewriting equation is quicker : $4a^2x^2 + 4abxy + 4acy^2 = 4a$ (multiplying by $4a$) ; $(2ax + by)^2 - b^2y^2 + 4acy^2 = 4a$ (it is a trick removing the $xy$ term) ; $(2ax + by)^2 + (4ac - b^2)y^2 = 4a$, then you can compare with equation $X^2/p + Y^2/q = 1$ of a conic (ellipse or hyperbola) for which you know a parametrization.
|
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How can I find a polynomial that fits a given table?
The first difference is 1, 4, 7, 10, 13, 16
The second difference is 3, 3, 3, 3, 3, 3
Since the second difference is constant this would be quadratic and I would have
$\frac{3}{2}n^{2}$
So now I will take the differences between the original sequence and the values of $\frac{3}{2}n^{2}$ for n = 1, 2, ...
3 - 3/2, 4 - 6, 8 - 27/2, 15 - 24, 25 - 75/2
Which is 3/2, -2, -11/2, -9, -25/2
so the difference between each of those terms are
$-3 \frac{1}{2}, -3 \frac{1}{2}, -3 \frac{1}{2}, -3 \frac{1}{2}, -3 \frac{1}{2}$
So then I have $\frac{3}{2} n^{2} -3.5n+5$ but when I plug in zero i get 5, and I should get 3, so what did I do wrong?
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Let $\Delta_k$ be the $k^{\text{th}}$ finite diffetence. then
$$P(x)=P(x_0)+$$
$$\sum_{k=1}^n\frac{\Delta_k}{k!}(x-x_0)(x-x_1)...(x-x_{k-1})=$$
$$3+1.(x-x_0)+\frac 32(x-x_0)(x-x_1)=$$
$$3+(x-0)+\frac 32(x-0)(x-1)=$$
$$3+x+\frac 32x^2-\frac 32x=$$
$$\frac 32x^2-\frac 12x+3.$$
|
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Explanation for binomial sums $\sum_{n=0}^{\infty} \binom{-4}{n-1} (-1)^{n-1} x^n = \sum_{n=0}^{\infty} \binom{-4}{n} (-1) x^{n+1}$ I was looking at some negative binomial coefficient problems and I stumbled upon this explanation
$$\sum_{n=0}^{\infty} \binom{n+2}{3} x^n = \sum_{n=0}^{\infty} \binom{n+2}{n-1} x^n= \sum_{n=0}^{\infty} \binom{-4}{n-1} (-1)^{n-1} x^n = \sum_{n=0}^{\infty} \binom{-4}{n} (-1) x^{n+1} $$
I was wondering how the author arrived at the conclusion that
$$\sum_{n=0}^{\infty} \binom{n+2}{n-1} x^n = \sum_{n=0}^{\infty} \binom{-4}{n-1} (-1)^{n-1} x^n = \sum_{n=0}^{\infty} \binom{-4}{n} (-1) x^{n+1}$$
|
Index transformation (replace $n$ by $n+1$) gives
$$ \sum_{n=0}^\infty \binom{-4}{n-1}(-1)^{n-1} x^n
= \sum_{n=-1}^\infty \binom{-4}{n}(-1)^n x^{n+1} $$
Now, on the right hand side, the term for $n=-1$, namely
$$ \binom{-4}{-1} (-1)^{-1} x^0 = 0 $$
is zero due to $\binom{-4}{-1} = 0$. Hence
$$ \sum_{n=0}^\infty \binom{-4}{n-1}(-1)^{n-1} x^n
= \sum_{n=-1}^\infty \binom{-4}{n}(-1)^n x^{n+1}
= \sum_{n=0}^\infty \binom{-4}{n}(-1)^n x^{n+1} $$
|
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For any $2$ x $2$ matrix $A$, does there always exist a $2$ x $2$ matrix $B$ such that det($A+B$) = det($A$) + det($B$)? For each invertible $2$ x $2$ matrix $A$, does there exist an invertible $2$ x $2$ matrix $B$ such that the following conditions hold?
(1) $A + B$ is invertible
(2) det($A+B$) = det($A$) + det($B$)
I know that for $2$ x $2$ matrices det($A+B$) = det($A$) + det($B$) + tr($A$)tr($B$) - tr($AB$). So this means tr($A$)tr($B$) = tr($AB$). Right now I am having trouble proving that there exists a $B$ that satisfies this equation as well as condition (1).
|
Since we aretalking of $2\times2$ matrices, its slightly easier to write down explicitly.
So $\det(A+B)=\det(A)+\det(B)$ happens when $(a_{11}+b_{11})(a_{22}+b_{22})-(a_{12}+b_{12})(a_{21}+b_{21})=(a_{11}a_{22}-a_{12}a_{21})+(b_{11}b_{22}-b_{12}b_{21})$
$\implies a_{11}b_{22}+b_{11}a_{22}-a_{12}b_{21}-b_{12}a_{21}=0=\det\begin{bmatrix}a_{11}\,a_{12}\\b_{21}\,b_{22}\end{bmatrix}+\det\begin{bmatrix}b_{11}\,b_{12}\\a_{21}\,a_{22}\end{bmatrix}$
Choosing $b_{11} = -a_{21}, b_{12} = -a_{22}, b_{21} = a_{11}$ and $b_{22} = a_{12}$ we get our matrix B.
The key point here is to choose $b_{ij}$ such that the two determinants are zero. Part 1 follows by considering matrix B such that $\det(A+B)\neq0$
$\rule{17cm}{1pt}$
Example for $\det(A+B)\neq0$ consider the matrix $A=\begin{bmatrix}4\: 5 \\7\: 9\end{bmatrix}$. We can choose matrix B as $\begin{bmatrix}-7 -9 \\4 \:\: 5\end{bmatrix}$.
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Find the general solution to the ODE $x\frac{dy}{dx}=y-\frac{1}{y}$ I have been working through an ODE finding the general solution and following the modulus through the equation has left me with four general solutions, as shown below. Online ODE solvers, however, have only calculated the two (positive and negative), on the right hand side of the page. Am I incorrect with my use of the modulus operator after I integrated each side then?
$$\begin{align*}x\frac{dy}{dx}&=y-\frac{1}{y}\quad\quad(-1<y<1)\\
\int\frac{1}{y-\frac{1}{y}}\ dy&=\int\frac{1}{x}\ dx\\
\int\frac{y}{y^2-1}\ dy&=\int\frac{1}{x}\ dx\\
\frac{1}{2}\ln{|y^2-1|}&=\ln{|x|}+c\\
\sqrt{|y^2-1|}&=A|x|\ \text{, where $A=e^c$}\\
|y^2-1|&=A^2x^2\\
\implies y^2-1&=A^2x^2&1-y^2&=A^2x^2\\
y^2&=A^2x^2+1&y^2&=1-A^2x^2\\
\\
\therefore y&=\pm\sqrt{Kx^2+1}&y&=\pm\sqrt{1-Kx^2}\ \text{, where $K=A^2$}\\
\end{align*}$$
Thanks :)
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Another approach is this. Rewrite the DE as
$$
\frac{yy'}{y^2-1}=\frac 1x
$$
Which is the same as
$$
\frac 12 \frac{d}{dx}\ln(y^2-1)=\frac 1x
$$
Upon integration, one gets
$$
\ln\dfrac{y^2(x)-1}{y^2(x_0)-1}=2\ln\frac{x}{x_0}
$$
Hence, one obtains
$$
y^2(x)=1+[y^2(x_0)-1]\left(\frac{x}{x_0}\right)^2
$$
|
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|
$1+\frac {1}{4}(1+\frac {1}{4}) +\frac {1}{9}(1+\frac {1}{4} +\frac {1}{9})+....$ Show that $$1+\frac {1}{4} \bigg(1+\frac {1}{4}\bigg) +\frac {1}{9} \bigg(1+\frac {1}{4} +\frac {1}{9}\bigg)+.....$$
converges.
Can you find the exact value of the sum.
My effort:
I have proved the convergence with comparing to $$\bigg(\sum _1^\infty \frac {1}{n^2}\bigg)^2$$
I have not figure out the exact sum.
Any suggestions??
|
$$
2S = \sum_{i\leq j} \frac{1}{i^{2}j^{2}} + \sum_{i\geq j} \frac{1}{i^{2}j^{2}} = \left(\sum_{n\geq 1}\frac{1}{n^{2}}\right)^{2} + \sum_{n\geq 1}\frac{1}{n^{4}} = \frac{\pi^{4}}{36} + \frac{\pi^{4}}{90}
$$
|
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|
Binomial Theorem with Three Terms $(x^2 + 2 + \frac{1}{x} )^7$
Find the coefficient of $x^8$
Ive tried to combine the $x$ terms and then use the general term of the binomial theorem twice but this does seem to be working.
Does anyone have a method of solving this questions and others similar efficiently?
Thanks.
|
The multinomial theorem can come to the rescue:
$$
(a+b+c)^n=\sum_{i+j+k=n}\binom{n}{i,j,k}a^ib^jc^k
$$
where $\dbinom{n}{i,j,k} = \dfrac{n!}{i! \, j! \, k!}$.
Here $n=7$, $a=x^2$, $b=2$, $c=x^{-1}$. How can we get $a^ic^k=x^8$? We need
$$
2i-k=8,\qquad i+k\le 7
$$
Hence $k=2i-8$ and $3i-8\le 7$, so $i\ge4$ and $i\le 5$. Hence we have the cases
*
*$i=4$, $k=0$, $j=3$;
*$i=5$, $k=2$, $j=0$.
Thus the coefficient is
$$
2^3\binom{7}{4,3,0}+\binom{7}{5,0,2}=
8\frac{7!}{4!\,3!\,0!}+\frac{7!}{5!\,0!\,2!}=8\cdot35+21=301
$$
|
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|
Laurent series of $ \frac{z-12}{z^2 + z - 6}$ for $|z-1|>4$
How do you find the Laurent series for $f(z) = \dfrac{z-12}{z^2 + z - 6}$ valid for $|z-1|>4$?
I know that $f(z) = \dfrac{z-12}{z^2 + z - 6} = \dfrac{-2}{z-2} + \dfrac{3}{z+3}$
It is easy for me to extract a series for $\dfrac{3}{z+3}$, but have no idea how to do it for $\dfrac{-2}{z-2}$.
Please help? Thank you!
|
Hint. Starting from your partial fraction decomposition, we have that
$$\frac{z-12}{z^2 + z - 6} = -\frac{2}{u-1} + \frac{3}{u+4}=-\frac{2/u}{1-1/u} + \frac{3/u}{1+4/u}$$
where $u=z-1$. Now note that $1/|u|<4/|u|<1$ when $|z-1|>4$.
|
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|
Prove that $\sum^n_{k=0}\frac{(-1)^k}{k+x}\binom{n}{k}=\frac{n!}{x(x+1)\cdots(x+n)}$. Given the following formula
$$
\sum^n_{k=0}\frac{(-1)^k}{k+x}\binom{n}{k}\,.
$$
How can I show that this is equal to
$$
\frac{n!}{x(x+1)\cdots(x+n)}\,?
$$
|
Induction step:
$$\begin{align}
\sum_{k=0}^{n+1}&\frac{(-1)^k}{x+k}\binom{n+1}k=\frac1x+\frac{(-1)^{n+1}}{x+n+1}+\sum_{k=1}^{n}\frac{(-1)^k}{x+k}\left[\binom nk+\binom n{k-1}\right]
\\&=\frac{n!}{x(x+1)\cdots(x+n)}+\frac{(-1)^{n+1}}{x+n+1}+\sum_{k=1}^{n}\frac{(-1)^k}{x+k}\binom{n}{k-1}
\\&=\frac{n!}{x(x+1)\cdots(x+n)}+\frac{(-1)^{n+1}}{x+n+1}-\sum_{k=0}^{n-1}\frac{(-1)^k}{(x+1)+k}\binom{n}{k}
\\&=\frac{n!}{x(x+1)\cdots(x+n)}+\frac{(-1)^{n+1}}{x+n+1}-\frac{n!}{(x+1)(x+2)\cdots(x+n+1)}+\frac{(-1)^n}{x+n+1}
\\&=\frac{n!(x+n+1)-n!x}{x(x+1)\cdots(x+n+1)}=\frac{(n+1)!}{x(x+1)\cdots(x+n+1)}
\end{align}$$
|
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|
how to determine the bounds of this integral?
Let $ G:= \{ (x,y) \in \mathbb{R}^2 \mid x^2 + 4y^2 >1, x^2+y^2 <4 \}$
I want to determine $ \int_G x^2+y^2 d(x,y)$.
So, its the area between a circle with radius 2 und an ellipse with semi-axis 1 in $x$-direction and semi-axis $\frac{1}{2}$ in $y$-direction, right?
How can I determine the bounds? How can I use polar coordinates to transform this?
|
Let
$$ x = r \cos \theta, \quad y = r \sin \theta. $$
Then since $x^2+y^2 < 4,$ we see that $r < 2$, and since $x^2+4y^2 > 1$, we see that
$$ r^2\cos^2\theta+4r^2\sin^2\theta = r^2+3r^2\sin^2\theta > 1 \quad \iff \quad r > \frac{1}{\sqrt{1+3\sin^2\theta}}. $$
Thus
\begin{align}
\int_G x^2+y^2 \, \mathrm d(x,y) &= \int_0^{2\pi}\int_{\frac{1}{\sqrt{1+3\sin^2\theta}}}^2 r^2 \cdot r \, \mathrm dr \, \mathrm d\theta.\\
\end{align}
Can you proceed?
|
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|
Simplfying expression efficiently What would be the best way to simplify such an expression? Simply foiling out the expression? could I take out $\frac{b-a}{4}$ and simplifying?
$$\frac{3}{4}(b-a)\left( \frac{2a}{3} +\frac
{b}{3} \right)^2 + \left( \frac{b-a}{4} \right)b^2$$ I know that this is equivalent to $\frac{b^3-a^3}{3}$
|
Well, first we notice that there's a common factor of $\frac{b - a}{4}$ in there, so we'll pull that out, leaving
$$\left(\frac{2a+b}{3}\right)^2 + b^2.$$
Expanding that bracket out, we obtain
$$\frac{4a^2 + 4ab + b^2}{3} + b^2$$
Combining those fractions, that's
$$\frac{4a^2+4ab+4b^2}{3}$$
Pulling out the $\frac{4}{3}$ (which, in particular, cancels out the factor of $\frac{1}{4}$ that we pulled out in the first step, so our pulled-out bit is now $\frac{b-a}{3}$), we're left with
$a^2 + ab + b^2.$
At this point, there's not much left to do besides combining the two parts again, giving our full expression as being equal to
$$\frac{(b-a)(a^2+ab+b^2)}{3}.$$
Multiplying through that top thing, we obtain
$$\frac{b^3-a^3}{3},$$
as expected.
|
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|
Is $1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{x} < \log_2 x$ If $x \ge 5$, is $1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{x} < \log_2 x$
I believe the answer is yes.
Here is my thinking:
(1) $\log_2{5} > 2.32 > 2.284 > 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}$
(2) Assume up to $x$ that $\log_2 x > \sum\limits_{i \le x}\frac{1}{i}$
(3) $(x+1)^{x+1} > {{x+1}\choose{x+1}}x^{x+1} + {{x+1}\choose{x}}x^x = x^{x+1} + (x+1)x^x = 2x^{x+1} + x^x > 2(x^{x+1})$
(4) $(x+1)\log_2\left(\frac{x+1}{x}\right) > 1$
(5) $\log_2(x+1) - \log_2(x) > \frac{1}{x+1}$
(6) $\log_2(x+1) - \log_2(x) + \log_2(x) = \log_2(x+1) > \sum\limits_{i \le x+1}\frac{1}{i}$
|
Yes because $H_n-\log n$ is bounded by $1$ and $\log_2n$ is a multiple of $\log n$.
|
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|
Find the value of $a$
Find $a$ for which $f(x) = \left(\frac{\sqrt{a+4}}{1-a} -1\right)x^5-3x+\ln5 \;$ decreases for all $x$ with $a\neq 1$ and $a\geq -4$.
My try:
For $f(x)$ to decrease
$$5x^4\left(\frac{\sqrt{a+4}}{1-a} -1\right) -3 <0\implies x^4\left(\frac{\sqrt{a+4}}{1-a} -1\right) < 3/5 $$
How can I proceed further?
|
HINT
We have
$$f(x) = \left(\frac{\sqrt{a+4}}{1-a} -1\right)x^5-3x+\ln5 \implies f'(x) = 5\left(\frac{\sqrt{a+4}}{1-a} -1\right)x^4-3\le 0$$
then we need
$$\left(\frac{\sqrt{a+4}}{1-a} -1\right)x^4- \frac35 \le 0 $$
that is true for all $x$ when
$$\frac{\sqrt{a+4}}{1-a} -1\le0 \iff \frac{\sqrt{a+4}+a-1}{1-a}\le0$$
|
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|
Closed form for $K(n)=[0;\overline{1,2,3,...,n}]$ I just started playing around with fairly simple periodic continued fractions, and I have a question. The fractions can be represented "linearly": for $n\in\Bbb N$,
$$K(n)=[0;\overline{1,2,3,...,n}]$$
I am seeking for a closed form for $K(n)$. I found the first few.
$n=1$:
$$K(1)=\frac1{1+K(1)}$$
$$\Rightarrow K(1)=\frac{-1\pm\sqrt{5}}2$$
$n=2$:
$$K(2)=\frac1{1+\frac1{2+K(2)}}$$
$$\Rightarrow K(2)=-1\pm\sqrt{3}$$
$n=3$:
$$K(3)=\frac1{1+\frac1{2+\frac1{3+K(3)}}}$$
$$\Rightarrow K(3)=\frac{-4\pm\sqrt{37}}3$$
$n=4$:
$$K(4)=\frac1{1+\frac1{2+\frac1{3+\frac1{4+K(4)}}}}$$
$$\Rightarrow K(4)=\frac{-9\pm2\sqrt{39}}5$$
As you may be able to tell, these results are all found by simplifying the fraction until one has a quadratic in $K(n)$, at which point the quadratic formula may be applied.
I would be surprised if there wasn't a closed form expression for $K(n)$, as they can all be found the same way. I've failed to recognize any numerical patterns in the results, however.
So, I have two questions:
$1)$: How does one express $K(n)$ in the $\operatorname{K}_{i=i_1}^\infty \frac{a_i}{b_i}$ notation? I was thinking something like
$$K(n)=\operatorname{K}_{i\geq0}\frac1{1+\operatorname{mod}(i,n)}$$
$2)$: What is a closed form for $K(n)$?
Thanks.
Update:
I'm pretty sure that all the $\pm$ signs in the beginning of the question should be changed to a $+$ sign.
|
Following up on Daniel Schepler's comment. Let $$P_n(x) = \frac{1}{1 + \frac{1}{2 + \ddots \frac{1}{n+x}}}.$$ This is basically the RHS of the recurrence equation for $K(n)$. Then:
\begin{align*}
P_1(x) &= \frac{1}{x+1} \\
P_2(x) &= \frac{x+2}{x+3} \\
P_3(x) &= \frac{2x+7}{3x+10} \\
P_4(x) &= \frac{7x+30}{10x+43} \\
P_5(x) &= \frac{30x+157}{43x+225} \\
P_6(x) &= \frac{157x+972}{225x+1393}.
\end{align*}
Note that $P_n(x) = P_{n-1}\left( \frac{1}{x+n}\right)$. Therefore, if $P_{n-1}(x) = \frac{ax+b}{cx+d}$, then
\begin{align*}P_n(x) &= \frac{\frac{a}{x+n} + b}{\frac{c}{x+n} + d} \\
&= \frac{bx + (a+bn)}{dx + (c+dn)}
\end{align*}
Thus in general, we may write $$P_n(x) = \frac{a_n x + a_{n+1}}{b_n x + b_{n+1}}$$
where $a$ and $b$ satisfy the recurrence $a_{n+1} = a_{n-1} + n a_n$ and likewise for $b$, with the initial conditions $a_1 = 0, a_2 = b_1 = b_2 = 1$. This recurrence gives the OEIS sequences linked by Jean-Claude Arbaut. $K(n)$ is a solution to $x - P_n(x) = 0$, or a root of the quadratic $$b_n x^2 + (b_{n+1} - a_n) x - a_{n+1} = 0.$$
|
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|
Calculate limit of \Gamma function Show that
$$\lim _{x \to \infty} \log \left( \frac{ \sqrt{x} \Gamma\left(\frac{x}{2}\right) } {\Gamma \left( \frac{x+1}{2}\right)} \right) = \frac{1}{2} \log(2),$$ where $\Gamma$ is the Gamma function.
I reduced this problem to calculate the limit:
$$\lim_{x \to \infty} \frac{B\left(\frac{x}{2},\frac{x}{2}\right)}{B\left(\frac{x+1}{2},\frac{x+1}{2}\right)} = 2, $$ where $B$ is the Beta function, but I don't know if this is useful or if there is other way to calculate this limit.
Any help will be very appreciated, thanks!
|
Old topic, but there is a much easier way to prove it.
First of all, the following relationship exists between Beta and Gamma functions:
$$B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$
So we have:
$$\Gamma \left( \frac{x+1}{2}\right) = \Gamma \left( \frac{x}{2}+\frac{1}{2}\right) = \frac{\Gamma(\frac{x}{2})\Gamma(\frac{1}{2})}{B(\frac{x}{2},\frac{1}{2})}$$
On the other hand, Stirling's approximation gives the equation below for large x:
$$B(\frac{x}{2},\frac{1}{2})=\Gamma(\frac{1}{2})(\frac{x}{2})^{(-\frac{1}{2})}=\Gamma(\frac{1}{2})\sqrt{\frac{2}{x}}$$
By substituting all in the original formula, we have:
$$\log \left( \frac{ \sqrt{x} \Gamma\left(\frac{x}{2}\right) } {\Gamma \left( \frac{x+1}{2}\right)} \right)=\log \left( \frac{\frac{\sqrt{x} \Gamma\left(\frac{x}{2}\right)}{1}} {\frac{\Gamma(\frac{x}{2})\Gamma(\frac{1}{2})}{\Gamma(\frac{1}{2})\sqrt{\frac{2}{x}}}} \right)=\log(\sqrt{x \frac{2}{x}})=\log(\sqrt2)=\frac{1}{2}\log(2)$$
|
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|
System of 3 equations problem
If
$$x+y+z=1\\x^2+y^2+z^2={3\over2}\\x^3+y^3+z^3=1$$
Then how much is
$$x^4+y^4+z^4$$
So what I'm pretty sure about, this shouldn't be solved for $x$,$y$, and $z$, but I should try to make $x^4+y^4+z^4$ from these equations.
I did this:
$(x^2+y^2+z^2)^2=x^4+y^4+z^4+2x^2y^2+2x^2z^2+2y^2z^2$
And from here I can see:
$x^4+y^4+z^4=1-2(x^2y^2+x^2z^2+y^2z^2)$
But I'm missing $(x^2y^2+x^2z^2+y^2z^2)$
I tried finding $(x+y+z)^2$ and other combinations of squaring, tried multiplying some of these, but still couldn't find what I needed.
I'm in high school so try and do this for my level of knowledge, with at least a similar way to mine
|
Alternativley, working with trinomials is cumbersome, so make them binomials:
$$\begin{cases}x+y+z=1\\x^2+y^2+z^2={3\over2}\\x^3+y^3+z^3=1\end{cases} \Rightarrow \begin{cases}x+y=1-z\\x^2+y^2={3\over2}-z^2\\x^3+y^3=1-z^3\end{cases}.$$
$(1)^2-(2)$:
$$2xy=2z^2-2z-\frac12$$
$(1)^3-(1)$:
$$3xy(x+y)=3z(z-1) \Rightarrow 3xy(1-z)=3z(z-1) \Rightarrow xy=-z$$
So:
$$-2z=2z^2-2z-\frac12 \Rightarrow z=\pm \frac12.$$
$(1)^4$:
$$(x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4 \Rightarrow \\
x^4+y^4=(x+y)^4-4xy(x^2+y^2)-6(xy)^2 \Rightarrow \\
x^4+y^4+z^4=(1-z)^4-4(-z)(\frac32-z^2)-6(-z)^2+z^4.$$
Plugging $z=\frac12$:
$$x^4+y^4+z^4=\frac1{16}+\frac52-\frac32+\frac1{16}=\frac98.$$
Plugging $z=-\frac12$:
$$x^4+y^4+z^4=\frac{81}{16}-\frac52-\frac32+\frac1{16}=\frac98.$$
|
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|
Evaluating $\liminf_{n\to\infty}n\{n\sqrt2\}$
How can we evaluate $$\liminf_{n\to\infty}n\{n\sqrt2\},$$where $\{\cdot\}$ denotes the fractional part of $\cdot$?
The first thing came to my mind is Pell's equation $x^2-2y^2=1$.
Knowing that $\sqrt2$ has a continued fraction $[1;2,2,2\cdots]$, I tried to estimate the limit by using $\sqrt2$'s convergents' denominators. It seems like the limit approximately equals to $0.36$.
|
For $n > 0$, let $m = \lfloor n\sqrt{2}\rfloor$. Since $\sqrt{2} \not\in \mathbb{Q}$,
$$n\sqrt{2} > m \implies 2n^2 > m^2 \implies 2n^2 \ge m^2 + 1 \implies \sqrt{2}n \ge \sqrt{m^2+1}$$
This implies
$$n\{n\sqrt{2}\} \ge n(\sqrt{m^2+1} - m) \ge \frac{1}{\sqrt{2}}\sqrt{m^2+1}(\sqrt{m^2+1}-m)\\ = \frac{1}{\sqrt{2}}\frac{\sqrt{m^2+1}}{\sqrt{m^2+1}+m} \ge \frac{1}{2\sqrt{2}}$$
As a result, $\displaystyle\;\liminf_{n\to\infty}\, n\{n\sqrt{2}\} \ge \frac{1}{2\sqrt{2}}$.
For the other direction, consider following pair of sequences of integers $(m_k), (n_k)$ defined by
$$m_k + n_k \sqrt{2} = (1 + \sqrt{2})^{2k+1}\quad\text{ for } k \in \mathbb{N}$$
It is easy to check they are increasing and $m_k^2 - 2n_k^2 = -1$. As a result,
$$\liminf_{n\to\infty}\, n\{n\sqrt{2}\}
\le \liminf_{k\to\infty}\, n_k\{n_k\sqrt{2}\}
= \liminf_{k\to\infty}\frac{1}{\sqrt{2}}\frac{\sqrt{m_k^2+1}}{\sqrt{m_k^2+1} + m_k}\\
= \lim_{m\to\infty} \frac{1}{\sqrt{2}}\frac{\sqrt{m^2+1}}{\sqrt{m^2+1}+m}
= \frac{1}{2\sqrt{2}}$$
Combine these, we get $\displaystyle\;\liminf_{n\to\infty}\, n\{n\sqrt{2}\} = \frac{1}{2\sqrt{2}}$
|
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|
Trying to find a modulo for $2^{24}$ I was trying to figure out which modulo $n$ would make $2^{24}$ congruent to $1 \bmod n$. One answer is $241$, and I wonder if there is a good way to find it.
I tried to use Fermat's little theorem or Chinese Remainder theorem, but $24$ is not a prime and $241$ is not composite.
|
We want to find factors of
\begin{align}(2^{24}-1)&=(2^{12}-1)(2^{12}+1)\\
&= (2^6-1)(2^6+1)(2^{12}+1)
\end{align}
Hence we can for example, pick our $n$ to be $2^6-1, 2^6+1$ or $2^{12}+1$. Notice that $241$ is a factor of $2^{12}+1$.
Also note that $$2^{24}=(2^3)^8=(7+1)^8$$
Hence we can pick $n=7$.
Also $$2^{24}=(2^8)^3=((2^8-1)+1)^3$$
Hence, we can pick $n=2^8-1$ or its factors and so on.
|
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Find all natural numbers $n$, such that polynomial $n^7+n^6+n^5+1$ would have exactly 3 divisors.
Find all natural numbers $n$, such that polynomial $$n^7+n^6+n^5+1$$ would have exactly 3 divisors.
What I found out is that this polynomial must be a square of prime number, otherwise it will have more than 3 divisors. Also, n must be an even number, otherwise this polynomial would be divisible by 4 (if it's an odd number, then it would be possible and would fit conditions only and only if n=1). My guts are telling me that somehow I have to prove that it's impossible with even numbers too, though I have no idea how.
|
You can factorize this polynomial like this: $$(n+1)(n^2+1)(n^4-n+1)$$ so this one has almost always at least 3 divisors.
Actualy, for $n>1$ we have $n+1>2$, $n^2+1>2$ and $n^4-n+1>2$ so it could be only $n=1$ which works.
How I got this this factorization:
\begin{eqnarray} (n^7+n^5)+(n^6+1) &=& n^5(n^2+1) + (n^2+1)(n^4-n^2+1)\\
&=& (n^2+1)(n^5+n^4-n^2+1)\\
&=& (n^2+1)\Big(n^4(n+1)-(n-1)(n+1)\Big)\\
&=&(n+1)(n^2+1)(n^4-n+1)
\end{eqnarray}
|
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|
Prove $\int_0^1\frac{\log(t^2-t+1)}{t^2-t}\mathrm dt=\frac{\pi^2}9$ I am in the middle of proving that
$$\sum_{k\geq1}\frac1{k^2{2k\choose k}}=\frac{\pi^2}{18}$$
And I have reduced the series to
$$\sum_{k\geq1}\frac1{k^2{2k\choose k}}=\frac12\int_0^1\frac{\log(t^2-t+1)}{t^2-t}\mathrm dt$$
But this integral is giving me issues. I broke up the integral
$$\int_0^1\frac{\log(t^2-t+1)}{t^2-t}\mathrm dt=\int_0^1\frac{\log(t^2-t+1)}{t-1}\mathrm dt-\int_0^1\frac{\log(t^2-t+1)}t\mathrm dt$$
I preformed the substitution $t-1=u$ on the first integral, then split it up:
$$\int_0^1\frac{\log(t^2-t+1)}{t-1}\mathrm dt=\int_{-1}^0\frac{\log(2u+i\sqrt3+1)}u\mathrm du+\int_{-1}^0\frac{\log(2u-i\sqrt3+1)}u\mathrm du-2\log2\int_{-1}^0\frac{\mathrm du}u$$
But the last term diverges, but I don't know what I did wrong. In any case, I would be surprised if there wasn't an easier way to go about this. Any suggestions? Thanks.
|
Another Approach is to employ Feynman's Trick:
Let
$$I(x) = \int_{0}^{1} \frac{\ln\left| x^2\left(t^2 - t\right) + 1\right|}{t^2 - t}\:dt$$
Note $I = I(1)$ and $I(0) = 0$
Thus
\begin{align}
I'(x) &= \int_{0}^{1} \frac{2x\left(t^2 - t\right)}{\left(x^2\left(t^2 - t\right) + 1\right)\left( t^2 - t\right)}\:dt = \frac{2}{x}\int_{0}^{1} \frac{1}{\left(t - \frac{1}{2}\right)^2 + \frac{4 - x^2}{4x^2}}\:dt\\
&= \frac{4}{x}\int_{0}^{\frac{1}{2}} \frac{1}{t^2 + \frac{4 - x^2}{4x^2}}\:dt = \frac{8}{\sqrt{4 - x^2}}\arctan\left(\frac{x}{\sqrt{4 -x^2}} \right)
\end{align}
We now integrate to solve $I(x)$
$$I(x) = \int\frac{8}{\sqrt{4 - x^2}}\arctan\left(\frac{x}{\sqrt{4 -x^2}} \right) \:dx = 4\left[\arctan\left( \frac{x}{\sqrt{4 - x^2}}\right) \right]^2 + C $$
Where $C$ is a constant of integration. As $I(0) = 0$ we find $C = 0$ and so:
$$I(x) = 4\left[\arctan\left( \frac{x}{\sqrt{4 - x^2}}\right) \right]^2$$
And finally
$$ I = I(1) = 4\left[\arctan\left( \frac{1}{\sqrt{3}}\right) \right]^2 = \frac{\pi^2}{9}$$
|
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|
Factorising $X^{16}- X$ over $\mathbb F_4$. I need to factorise $X^{16}- X$ over $\mathbb F_4$. How might I go about this? I have factorised over $\mathbb F_2$ and I know the quadratic must split but I'm not sure about the quartic and octic. Is there any theory that can help me here or is bruteforce the only way?
Here is the factorisation over $\mathbb F_2$:
$$X(X+1)(X^2+X+1)(X^4+X^3+X^2+X+1)(X^8+X^7+X^5+X^4+X^3+X+1)$$
|
Let $q:=4$. You are factorizing $X^{q^2}-X$ over $\mathbb{F}_q$. Thus, you should expect that $$X^{q^2}-X=(X^q-X)\,f_1(X)\,f_2(X)\,\cdots\,f_k(X)$$
where $f_1(X),f_2(X),\ldots,f_k(X)\in\mathbb{F}_q[X]$ are all the irreducible monic quadratic polynomials over $\mathbb{F}_q$ (whence $k=\dfrac{q^2-q}{2}$, which equals $6$ when $q=4$). You are right that the quadratic polynomial $X^2+X+1$ factors into linear terms over $\mathbb{F}_q=\mathbb{F}_4$, namely,
$$X^q-X=X^4-X=X(X+1)(X^2+X+1)=X(X+1)(X+t)(X+t+1)\,,$$
where $t$ is an element of $\mathbb{F}_4\setminus\mathbb{F}_2$.
Therefore, you have to find $6$ polynomials of the form
$$X^2+aX+b\,,$$
with $a,b\in\mathbb{F}_4=\{0,1,t,t+1\}$, that are not divisible by $X$, $X+1$, $X+t$, or $X+t+1$. If $a=0$, then note that every element of $\mathbb{F}_4$ is a square ($0=0^2$, $1=1^2$, $t=(t+1)^2$, and $t+1=t^2$), so $b=c^2$ for some $c\in\mathbb{F}_4$, and $X^2+aX+b=(X+c)^2$. Therefore, $a\neq 0$.
If $a=1$, then $b\in\{t,t+1\}$, and both choices work. If $a=t$, then $b\in\{1,t\}$ and both choices work. If $a=t+1$, then $b\in\{1,t+1\}$ and both choices work. This shows that
$$\begin{align}X^{16}-X&=X(X+1)(X+t)\big(X+(t+1)\big)\,(X^2+X+t)\,\big(X^2+X+(t+1)\big)\\&\phantom{abcf}(X^2+tX+1)\,(X^2+tX+t)\,\big(X^2+(t+1)\,X+1\big)\,\big(X^2+(t+1)\,X+(t+1)\big)\,.\end{align}$$
I cannot offer a good way to directly deal with the octic $X^8+X^7+X^5+X^4+X^3+X+1$, except from the work above. However, there is an easy way to factor the quartic polynomial $X^4+X^3+X^2+X+1$. Let $Y:=X+\dfrac{1}{X}$. Then,
$$\begin{align}X^4+X^3+X^2+X+1&=X^2\,(Y^2+Y+1)=X^2(Y+t)(Y+t+1)\\&=(X^2+tX+1)\,\big(X^2+(t+1)X+1\big)\,.\end{align}$$
|
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|
Proving a three variables inequality Given that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$ show that:$$(a+1)(b+1)(c+1)\ge 64$$
My attempt: First I tried expanding the LHS getting that$$abc+ab+bc+ca+a+b+c \ge 63$$
I applied Cauchy-Schwarz on $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ getting that $a+b+c\ge9$.
Then I also tried to manipulate the first condition and got $abc=ab+bc+ca$, then I applied AM-GM on $a+b+c$ getting the following$$a+b+c\ge3\sqrt[3]{abc}$$
Finally I substituted $ab+bc+ca=abc$ in my initial expression, getting:$$2abc+(a+b+c)\ge63$$The last thing I tought about was that I have both $a+b+c\ge9$ and $a+b+c\ge3\sqrt[3]{abc}$ so if I somehow related them I would have $\sqrt[3]{abc} \ge 3 \rightarrow abc\ge27$ and with this conditions the problem would follow by summing, but the direction of the inequality is not allowing me to do as intended...
I'm stuck here, have tried lot of other things but nothing really worked, also partial help is appreciated!
|
For positive variables we need to prove that
$$\ln\left((1+a)(1+b)(1+c)\right)\geq\ln64$$ or
$$\sum_{cyc}(\ln(1+a)-2\ln2)\geq0$$ or
$$\sum_{cyc}\left(\ln(1+a)-2\ln2+\frac{9}{4}\left(\frac{1}{a}-\frac{1}{3}\right)\right)\geq0,$$
which is true because
$$f(a)=\ln(1+a)-2\ln2+\frac{9}{4}\left(\frac{1}{a}-\frac{1}{3}\right)\geq0$$ for all $a>0$.
Indeed, $$f'(a)=\frac{(a-3)(4a+3)}{4a^2(a+1)},$$ which gives $a_{min}=3$ and since $f(3)=0$, we are done!
|
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|
Different answers with $\sec(x) = 2\csc(x)$ My son and I were solving this last night and we get different answers depending on which identities we use. The question also did specify $0 \leqslant x < 2\pi$
Here's our work:
$$\sec x = 2 \csc x$$
$$\frac 1 {\cos x} = \frac 2 {\sin x}$$
cross multiply:
$$2 \cos x = \sin x$$
and square both sides (I think this introduces a problem?)
$$4 \cos^2 x = \sin^2 x$$
Now we used the identity $\sin^2 x + \cos^2 x = 1$
Let's replace $\sin x$:
$$4 \cos^2 x = 1 - \cos^2 x$$
$$5 \cos^2 x = 1$$
$$\cos^2 x = \frac 1 5$$
$$\cos x = ±\sqrt{\frac 1 5}$$
$$\cos^{-1}\left(±\sqrt \frac 1 5\right) = 1.10, 2.03$$
That gave us two answers within the range requested.
But let's replace $\cos x$ instead:
$$4 \cos^2 x = \sin^2 x$$
$$4 (1 - \sin^2 x) = \sin^2 x$$
$$4 - 4 \sin^2 x = \sin^2 x$$
$$4 = 5 \sin^2 x$$
$$\frac 4 5 = \sin^2 x$$
$$±\sqrt \frac 4 5 = \sin x$$
$$\sin^{-1}\left(±\sqrt \frac 4 5\right) = x = 1.1, -1.1$$
Two answers, but we can throw out the negative one because it is not within the range specified.
Then we used the $\tan x$ identity (which is what we should have done to begin with since squaring obviously seems to introduce invalid answers):
$$\tan x = \frac {\sin x} {\cos x}$$
$$2 \cos x = \sin x$$
$$2 = \sin x / \cos x$$
$$2 = \tan x$$
$$\tan^{-1} 2 = 1.1$$
So now I assume $1.1$ is the right answer. But where did $-1.1$ and $2.03$ come from?
They don't show up in the graphs:
AH! But they do show up in the squared version, which I now understand is where the extra answers came from:
What is the fundamental mistake here? How would one use the squaring method, and then at the end know which solution(s) to throw out as a side effect?
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1) $a^2 = b^2$ => $a = b$ or $a = - b$
2) $a = b$ => (square both sides) $a^2 = b^2$
The idea here is that in the first case you have that $a = -b$, but in the second case (which is your case), your $a^2 = b^2$ inevitably adds the $a = -b$ solutions to your total, which obviously are wrong since your original equation is $a = b$
The proper way to solve this problem is to do this:
$2\cos(x) - \sin(x) = 0$
$\cos(x)*\frac{2}{\sqrt{1^2+2^2}} - \sin(x)*\frac{1}{\sqrt{1^2+2^2}} = 0$
$\cos(x)*\cos(\arccos(\frac{2}{\sqrt{5}})) - \sin(x)*\sin(\arcsin(\frac{1}{\sqrt{5}})) = 0$
$\cos(x + \arccos(\frac{2}{\sqrt{5}})) = 0$
I guess you can take it from here
|
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|
Find limit $\lim\limits _{x\rightarrow \infty }\left(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} -\sqrt{2x^{4}}\right)$ $\displaystyle \lim\limits _{x\rightarrow \infty }\left(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}}\displaystyle -\sqrt{2x^{4}}\right)$
$\displaystyle \lim\limits _{x\rightarrow \infty }\left(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} -\sqrt{2x^{4}}\right)$$\displaystyle =\displaystyle \lim\limits _{x\rightarrow \infty }\dfrac{\left(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} -\sqrt{2x^{4}}\right)\left(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} +\sqrt{2x^{4} \ }\right)}{\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} +\sqrt{2x^{4} \ }} =$
$\displaystyle =\lim\limits _{x\rightarrow \infty }\dfrac{x^{2}\sqrt{x^{4} +1} -x^{4}}{\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} +\sqrt{2x^{4} \ }}$
What is the next step should be? Please help!
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Use the trick twice for the numerator to obtain
$$\dfrac{x^{2}\sqrt{x^{4} +1} -x^{4}}{\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} +\sqrt{2x^{4} \ }}=\dfrac{x^{2}\sqrt{x^{4} +1} -x^{4}}{\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} +\sqrt{2x^{4} \ }}\cdot \dfrac{x^{2}\sqrt{x^{4} +1} +x^{4}}{x^{2}\sqrt{x^{4} +1} +x^{4}}=$$
$$=\dfrac{x^4}{\left(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} +\sqrt{2x^{4} \ }\right)\left(x^{2}\sqrt{x^{4} +1} +x^{4}\right)}\sim\frac{1}{4\sqrt 2 x^2}\to 0$$
|
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|
Prove: $\sum _{n=1}^{\infty} \left(\frac{n^{2} + 1}{n^{2} +n +1}\right)^{n^2} < \infty$ without L'Hôpital's Given $\sum _{n=1}^{\infty} \left(\frac{n^{2} + 1}{n^{2} +n +1}\right)^{n^2}$, prove that it converges.
I tried to use the Ratio test.
I got a terrible algebraic expression: $$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \left(\frac{n^4 +2n^3 +3n^2 + 2n +2}{n^4 + 4n^3 + 7n^2 + 6n +3}\right)\cdot \left(\frac{n^2 + 2n + 2}{n^2 + 3n + 3}\right)^{2n +1}$$
Now problem is I'm not allowed to use L'Hoptial and I don't really know what to do with this disaster.
|
By Taylor's expansion we have
$$\left(\frac{n^{2} + 1}{n^{2} +n +1}\right)^{n^2}=\left(1-\frac{n}{n^{2} +n +1}\right)^{n^2}=e^{n^2 \log\left(1-\frac{n}{n^{2} +n +1}\right)}=e^{n^2 \left(\frac{-n}{n^{2} +n +1}+O(1/n^2)\right)}\sim \frac c{e^n}$$
and then refer to limit comparison test.
|
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|
Proof verification for $\lim_{n\to\infty}\frac{1}{n}(1+\sqrt2+\dots + \sqrt{n}) = +\infty$
Show that:
$$
\lim_{n\to\infty}\frac{1}{n}(1+\sqrt2+\dots + \sqrt{n}) = +\infty
$$
I've tried the following way. Consider the following sum:
$$
\sqrt n + \sqrt{n-1} + \dots + \sqrt{n-\frac{n}{2}} + \dots + \sqrt{2} + 1
$$
Now if we take only $n\over 2$ terms of the sum we obtain that:
$$
\sqrt n + \sqrt{n-1} + \dots > {n \over 2} \sqrt{n\over 2}
$$
Let:
$$
x_n = {1 \over n}(1 + \sqrt{2} + \dots + \sqrt{n}),\ \ n\in \Bbb N
$$
Using the above we have that:
$$
x_n > {1\over n} {n\over 2}\sqrt{n\over 2} = {1\over 2}
\sqrt{n \over 2}
$$
Now taking the limit for RHS its obvious that:
$$\lim_{n\to\infty}{1\over2}\sqrt{n\over2} = +\infty
$$
Which implies:
$$
\lim_{n\to \infty}x_n = + \infty
$$
Have I done it the right way? Also i would appreciate alternative ways of showing that limit. Thanks!
|
Just to give an alternative, note first that the sequence is increasing:
$$\begin{align}
{1\over n+1}(1+\sqrt2+\cdots+\sqrt n+\sqrt{n+1})-{1\over n}(1+\sqrt2+\cdots\sqrt n)
&={\sqrt{n+1}\over n+1}-{1+\sqrt2+\cdots+\sqrt n\over n(n+1)}\\
&\gt{\sqrt{n+1}\over n+1}-{n\sqrt n\over n(n+1)}\\
&={\sqrt{n+1}-\sqrt n\over n(n+1)}
\end{align}$$
Now consider
$$\begin{align}
{1\over n^2}(1+\sqrt2+\cdots+\sqrt{n^2})
&\gt{1\over n^2}(1+1+1+2+2+2+2+2+3+\cdots+(n-1)+n)\\
&={1\over n^2}(3\cdot1+5\cdot2+7\cdot3+\cdots+(2n-1)(n-1)+n)\\
&\gt{2\over n^2}(1^2+2^2+3^2+\cdots+(n-1)^2)\\
&={2\over n^2}\cdot{(n-1)n(2n-1)\over6}\\
&={(n-1)(2n-1)\over3n}\\
&\to\infty
\end{align}$$
|
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|
Ideals generated by two elements in $\mathbb{Z}[x]$ Consider the following ideal $(2+x,x^2+5)$ in $\mathbb{Z}[x]$. Then I showed that $(2+x,x^2+5)=(9,2+x)$.
But I am not able to do the same for ideals $(1-4x,x^2+5)$ and $(1+2x,x^2+5)$.
Is there some method?
EDIT: My goal is to show that these ideals can be written as $(c,ax+b)$ for some appropriate integers $a,b,c$.
Would be very grateful for help!
|
Note that $\langle 1-4x,x^2+5\rangle=\langle x+20,81\rangle$. This is because
$$x^2+5=(x-20)\cdot(x+20)+5\cdot 81\,,$$
$$1-4x=(-4)\cdot (x+20)+1\cdot 81\,,$$
$$x+20=4\cdot(x^2+5)+x\cdot(1-4x)\,,$$
and
$$81=16\cdot(x^2+5)+(1+4x)\cdot(1-4x)\,.$$
Similarly, $\langle 1-2x,x^2+5\rangle=\langle x+10,21\rangle$. This is because
$$x^2+5=(x-10)\cdot(x+10)+5\cdot 21\,,$$
$$1-2x=(-2)\cdot(x+10)+1\cdot 21\,,$$
$$x+10=2\cdot(x^2+5)+x\cdot(1-2x)\,,$$
and
$$21=4\cdot(x^2+5)+(1+2x)\cdot(1-2x)\,.$$
|
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|
How to calculate the following integrals? How the calculate the following integrals? Therein $D$ is a constant.
$$(1)\;\;\int_{0}^{2\pi}\frac{1}{1-D\cdot\cos\theta} d\theta$$
and
$$(2)\;\;\int_{0}^{2\pi}\frac{1}{1+D\cdot\cos\theta} d\theta$$
|
$$I_2=\int_0^{2\pi}\frac{d\theta}{1+D\cos(\theta)}=\int_0^{\pi}\frac{d\theta}{1+D\cos(\theta)}+\int_{\pi}^{2\pi}\frac{d\theta}{1+D\cos(\theta)}$$ Let $\theta=\tan(\frac{x}{2})$
$$I_2=\int_0^{\infty}\frac{1}{1+D\frac{1-x^2}{1+x^2}}\frac{2dx}{1+x^2}+\int_{-\infty}^0\frac{1}{1+D\frac{1-x^2}{1+x^2}}\frac{2dx}{1+x^2}=2\int_{-\infty}^{\infty}\frac{dx}{1+x^2+D-Dx^2}=2\int_{-\infty}^\infty \frac{dx}{(1-D)x^2+(1+D)}=\frac{2}{1-D}\int_{-\infty}^{\infty}\frac{dx}{x^2+\frac{1+D}{1-D}}=\frac{2\pi}{\sqrt{1-D^2}}$$ Hopefully all my computation is correct, the other integral can be done in the same way. Just use Weierstrass Substitution.
|
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|
Finding matrix $A^n,$ when $\lim_{n \to \infty}$
Finding $\lim_{n\rightarrow \infty}\begin{pmatrix}
1 & \frac{x}{n}\\ \\
-\frac{x}{n} & 1
\end{pmatrix}^n$ for all $x\in \mathbb{R}$
Try:
Let $$ A = \begin{pmatrix}1&\frac{x}{n}\\\\-\frac{x}{n}&1\end{pmatrix}.$$
Then $$ A^2 = \begin{pmatrix}1-\frac{x^2}{n^2}&\frac{2x}{n}\\\\-\frac{2x}{n}&1-\frac{x^2}{n^2}\end{pmatrix}$$
And then $$A^3 = \begin{pmatrix}1-3\frac{x^2}{n^2}&\frac{3x}{n}-\frac{x^3}{n^3}\\\\-3\frac{x}{n}+\frac{x^3}{n^3}&1-\frac{x^2}{n^2}\end{pmatrix}$$
So by using same way and taking $\lim_{n\rightarrow \infty}A^n = \begin{pmatrix}1&0\\\\0&1\end{pmatrix}$
But answer given as $$\begin{pmatrix}\cos x &\sin x\\\\-\sin x&\cos x\end{pmatrix}$$
Could some help me where I am missing and also explain how to solve it?
|
You seem to be arguing that the terms proportional to $x$, $x^2$, $x^3$, etc. vanish in the limit $n \to \infty$. But the problem with this argument is that the numerical coefficients in front of these terms also grow as $n \to \infty$, and diverge without bound. To see this, define
$$
A_n = \begin{bmatrix}1 & x/n\\ -x/n & 1 \end{bmatrix}^n. $$
Here are the terms in the upper-left entry for the first few values of $n$:
\begin{align}
(A_1)_{12} &= \frac{x}{n} = x \\
(A_2)_{12} &= \frac{2x}{n} = x\\
(A_3)_{12} &= \frac{3x}{n} - \frac{x^3}{n^3} = x - \frac{x^3}{27} \\
(A_4)_{12} &= \frac{4 x}{n} - \frac{4 x^3}{n^3} = x - \frac{x^3}{16} \\
(A_5)_{12} &= \frac{5 x}{n} - \frac{10 x^3}{n^3} + \frac{x^5}{n^5} = x - \frac{10 x^3}{125} + \frac{x^5}{3125}\\
\end{align}
Just looking at the term proportional to $x$ in this sequence, it seems pretty likely that this will not vanish as $n \to \infty$.
|
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|
maximum value of $x_{1}+x_{2}+x_{3}+\cdots+x_{10}$
Consider the following quantities $x_{1},x_{2},x_{3},\cdots \cdots ,x_{10}$
and $-1\leq x_{1},x_{2},x_{3},\cdots \cdots ,x_{10}\leq 1.$ and
$x^3_{1}+x^3_{2}+\cdots+x^{3}_{10}=0.$
Then maximum of $x_{1}+x_{2}+x_{3}+\cdots\cdots+x_{10}$
Try : Let $x_{i} = \sin(a_{i})$ for $i = 1$ to $10$ and $\displaystyle a_{i}\in \bigg[-\frac{\pi}{2},\frac{\pi}{2}\bigg]$
So we have given
$\sin^3(a_{1})+\sin^3(a_{2})+\cdots \cdots+\sin^3(a_{10})=0$
and we have to find maximum value of
$\sin(a_{1})+\sin(a_{2})+\cdots+\sin(a_{10}).$
I did not understand how to calculate it.
Could some help me, thanks
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Let us transform the variables using $f(t)=\sqrt[3]t$, to equivalently maximise $ \sum f(x_k)$ subject to $\sum x_k = 0$ and $x_k \in [-1, 1]$, where index $k\in \mathbb N, 1\leqslant k\leqslant 10$. WLOG using symmetry we may assume the $x_k$ are in non-descending order.
Noticing $f(x)$ is convex for $x< 0$, if the first $n$ among the $x_k$ are the negatives, Karamata's Inequality gives: $$\sum_{k\leqslant n} f(x_k)\leqslant (n-1)f(-1) + f\left(\sum_{k\leqslant n}x_k +n-1\right)$$
Hence we may replace the first $n$ (i.e. all negative) terms with $n-1$ copies of $-1$ and some number, say $p^3 = \sum_{k\leqslant n}x_k +n-1$.
Now the remaining $10-n$ terms are non-negative, a domain where $f$ is concave, so again by Karamata or Jensen, we may replace them all by their arithmetic mean (say $q^3$) for a sum which is at least as high.
Now consider the numbers $(p^3, q^3)$, which obviously obey $-1 \leqslant p$ and $q>0$.
$\bullet \quad$ If $p^3+q^3 \geqslant 0$, note $f(p^3)+f(q^3)\leqslant 2f(\frac12p^3+\frac12q^3) \iff p+q\leqslant 2\sqrt[3]{\frac{p^3+q^3}2} \iff 3(p-q)^2(p+q)\geqslant 0$, so we may replace both by a non-negative number.
$\bullet \quad$ OTOH, if $p^3+q^3< 0$, then $f(p^3)+f(q^3) \leqslant f(-1)+f(1+p^3+q^3) \iff p+q \leqslant -1+\sqrt[3]{1+p^3+q^3}\\ \iff -3(1+p)(1+q)(p+q)\geqslant 0$, so we may replace $(p^3, q^3)$ with $(-1, 1+p^3+q^3)$ for at lease as high a sum. Note that $1+p^3+q^3\geqslant 0$
In both cases, we are left with copies of $-1$ and non-negative numbers, which after another application of Jensen, leave us with $x_k \in \{-1, a\}$ for some $a>0$.
Thus we have for some natural $m \in [10]$ and some real $a \in (0, 1)$,
$$\sum_k f(x_k) \leqslant mf(-1) + (10-m)f(a) = -m+(10-m)\sqrt[3]a$$
subject to $(10-m)a=m \implies a = \frac m {10-m}$.
So we are left to find the natural $m$ that maximises $-m+(10-m)\sqrt[3]{\frac{m}{10-m}} = \sqrt[3]{m(10-m)^2}-m$.
Using calculus (extending $m$ to reals and checking both the integers around the unimodal maximum of $m=\frac{10}9$) or an easy enumeration, it is found $m=1$ is optimal choice, for a maximum of $9^{2/3}-1=3\sqrt[3]3-1$.
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|
Derivative of piecewise function with $\sin\frac{1}{x}$ term I was going through my calculus book, and I am not sure I understand this part
$f(x) = \begin{cases} \frac{x^2}{4}+x^4\sin(\frac{1}{x}) &\text{if $x\neq0$ } \\ 0 &\text{if $x=0$ } \end{cases}$
$ f'(x) = \begin{cases} \frac{x}{2}-x^2\cos(\frac{1}{x})+4x^3\sin(\frac{1}{x}) &\text{if $x\neq0$ } \\ 0 &\text{if $x=0$ } \end{cases}$
$ f''(x) = \begin{cases} \frac{1}{2}+12x^2\sin(\frac{1}{x})-\sin(\frac{1}{x})-6x\cos(\frac{1}{x}) &\text{if $x\neq0$ } \\ \frac{1}{2} &\text{if $x=0$ } \end{cases}$
So, I know that when I have piecewise function, I need to look at left and right limit, but I don't see why second part in second derivative is $\frac{1}{2}$, or rather why does the $\sin\frac{1}{x}$ term go to 0?
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If $x\neq0$, then\begin{align}\frac{f'(x)-f'(0)}x&=\frac{\frac x2-x^2\cos\left(\frac1x\right)+4x^3\sin\left(\frac1x\right)}x\\&=\frac12-x\cos\left(\frac1x\right)+4x^2\sin\left(\frac1x\right)\end{align}and therefore\begin{align}f''(0)&=\lim_{x\to0}\frac12-x\cos\left(\frac1x\right)+4x^2\sin\left(\frac1x\right)\\&=\frac12.\end{align}
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|
Find the geometric locus $z \in \mathbb C$ so that $\frac{z+2}{z(z+1)}\in \mathbb R$
Find the geometric locus of the set of $z \in \mathbb C$ so that
$$\frac{z+2}{z(z+1)}\in \mathbb R$$
Source: IME (Military Engineering Institute, Brazil, entrance exam, 1974)
My attempt: With the notation $z=a+bi$, the solution provided in a book is either the line $b=0$ or the circle $(a+2)^2+b^2=2$ but I could not find it, or not able to recognize this locus set from the algebraic development I did (or the solution or the statement has some mistake).
Hints and solutions are welcomed.
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A little Partial Fraction Decomposition before rationalization of the denominator will reduce calculation.
$$\dfrac{z+2}{z(z+1)}=\dfrac1{z+1}+\dfrac{2(z+1-z)}{z(z+1)}=\dfrac2z-\dfrac1{z+1}$$
Now the imaginary part of $\dfrac2z$ is $-\dfrac{2y}{x^2+y^2}$
and that of $\dfrac1{z+1}$ is $-\dfrac y{(x+1)^2+y^2}$
Finally, $\dfrac{z+2}{z(z+1)}$ will be real if $-\dfrac{2y}{x^2+y^2}=-\dfrac y{(x+1)^2+y^2}$
$\iff0=y(2x^2+2y^2+4x+2-x^2-y^2)=y\{(x+2)^2+y^2-2\}$
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Floor equation $\lfloor 3x-x^2 \rfloor = \lfloor x^2 + 1/2 \rfloor$
Solve the equation:
$$\left \lfloor 3x-x^2 \right \rfloor = \left \lfloor x^2 + 1/2 \right \rfloor$$
In the solution it writes
We notice that $x^{2}+\frac{1}{2}> 0$ therfore $\left \lfloor x^2 - 1/2 \right \rfloor \geq 0$ . From there $\left \lfloor 3x-x^2 \right \rfloor = n \geq 0$.
So far all of this I understand but than it writes:
But $3x-x^2 \leq \frac{9}{4}$ and $\left \lfloor 3x-x^2 \right \rfloor< 3$.
I dont understand this last part how did they get $3x-x^2 \leq \frac{9}{4}$ ?
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If we put $-x^2+3x$ into translated form we get:
$$-(x^2-3x)=-\left[\left(x-\frac 3 2\right)^2-\left(\frac32\right)^2\right]=-\left(x-\frac 3 2\right)^2+\frac94$$
We can now see that this is $x\mapsto x^2$ that has been dilated by a factor of -1 (graphically: reflected about the $x$ axis) and translated $\frac 3 2$ units right and $\frac 9 4$ units up. (In simple terms, the graph is open downwards and the vertex is at $\left(\frac 3 2, \frac 9 4\right)$.) Hence, for all $x$, $-x^2+3x\le \frac 9 4$.
Edit
I think there's a typo in the first part of your presented solution (see coffeemath's comment): it should be:
... therefore $\left\lfloor x^2+\frac 1 2\right\rfloor\ge0$. From there...
In any case, I think the solution is directing you to to identify potential cases to investigate: this equation can only hold for floors of 0, 1 and 2.
For example, investigating the floor 0 case:
$$\begin{align}
\left\lfloor x^2+\frac 1 2\right\rfloor = 0&\Rightarrow 0\le x^2+\frac 1 2<1\\
x^2+\frac 1 2\ge 0&\Rightarrow x\in\mathbb R\\
x^2+\frac 1 2 < 1&\Rightarrow
x\in\left]-\frac{\sqrt 2}{2},\frac{\sqrt2}{2}\right[\\\\
\left\lfloor-x^2+3x\right\rfloor=0&\Rightarrow0\le-x^2+3x<1\\
-x^2+3x\ge0&\Rightarrow x\in\left[0, 3\right]\\
-x^2+3x<1&\Rightarrow x\in\left]-\infty,\frac{3-\sqrt 5}{2}\right[\;\;\bigcup\;\;\left]\frac{3+\sqrt5}{2},\infty\right[
\end{align}$$
Taking the intersection of all these sets gives us part of the solution to the original problem:
$$x\in\left[0, \frac{3-\sqrt 5}{2}\right[$$
Similarly, investigating the floor 1 case gives us that the equation holds for $x\in\left[\frac{\sqrt 2}{2}, 1\right[$, and the floor 2 case gives us that the equation holds for $x\in\left[\frac{\sqrt6}{2},\frac{\sqrt{10}}{2}\right[$.
The solution set to the problem is thus:
$$\left[0, \frac{3-\sqrt 5}{2}\right[\;\;\bigcup\;\;\left[\frac{\sqrt 2}{2}, 1\right[\;\;\bigcup\;\;\left[\frac{\sqrt6}{2},\frac{\sqrt{10}}{2}\right[$$
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|
With $z\in \mathbb C$ find the maximum value for $\lvert z\rvert$ such that $\lvert z+\frac{1}{z}\rvert=1$
With $z\in \mathbb C$ find the maximum value for |z| such that
$$\left\lvert z+\frac{1}{z}\right\rvert=1.$$
Source: List of problems for math-contest training.
My attempt: it is easy to see that the given condition is equivalent
$$\lvert z^2+1\rvert=\lvert z\rvert$$
and if $z=a+bi$,
\begin{align*}
\lvert z\rvert&=\lvert a^2-b^2+1+2ab i\rvert=\sqrt{(a^2-b^2+1)^2+4 a^2b^2}\\
&=\sqrt{(a^2-b^2)^2+2(a^2-b^2)+1+4a^2b^2}\\
&=\sqrt{(a^2+b^2)^2+2(a^2-b^2)+1}\\
&=\sqrt{\lvert z\rvert^4+2(a^2-b^2)+1}
\end{align*}
I think it is not leading to something useful... the approach I followed is probably not useful.
Hints and answers are welcomed.
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Here's a solution using the dual problem. Assume that the solution of the problem occurs at $z = re^{i\theta}.$ Consider
$$
F_r(t) = |re^{it}+r^{-1}e^{-it}|^2 = r^2 + r^{-2} + e^{2it}+e^{-2it} = r^2+r^{-2}+2\cos 2t.
$$ If $z= re^{i\theta}$ is the maximizer, we claim that $F_r'(\theta)=0$. That is, given $|z|=r$, the angle $\theta$ should be chosen so that $F_r(t)$ is either maximized or minimized. To see this, assume to the contrary that $F_r'(\theta)\neq 0$. Then, there is small $u$ (we allow $u$ to have negative sign) such that $$F_r(\theta-u)<1=F(\theta)<F_r(\theta+u).$$ If it is the case that $r\leq 1$, then by choosing slightly large $r'$, we can make $F_{r'}(\theta+u)=1$. This contradicts maximality of $r$. If $r>1$, then by choosing slightly large $r'$, we can make $F_{r'}(\theta-u) = 1$. This also leads to a contradiction. Now it follows that $\cos(2\theta)$ should be either $-1$ or $1$. However, we can easily see that $\cos(2\theta)$ cannot be $1$. Thus, from $\cos(2\theta)=-1$, we have
$$
1=F_r(\theta)=r^2 +r^{-2} -2 = (r-r^{-1})^2,
$$or
$$
r-1/r = \pm 1.
$$ This gives
$$
r= \frac{\pm 1+ \sqrt{5}}{2}.
$$ Therefore, the larger one $r=\frac{ 1+ \sqrt{5}}{2}$ is the solution.
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Solving $\Bigl\{\begin{smallmatrix}x+\frac{3x-y}{x^2+y^2}=3\\y-\frac{x+3y}{x^2+y^2}=0\end{smallmatrix}$ in $\mathbb R$
$$\begin{cases}
x+\dfrac{3x-y}{x^2+y^2}=3 \\
y-\dfrac{x+3y}{x^2+y^2}=0
\end{cases}$$
Solve in the set of real numbers.
The furthest I have got is summing the equations, and I got
$$x^3+(y-3)x^2+(y^2+2)x+y^3-3y^2-4y=0.$$
But I have no idea how to solve this problem. How can I solve it?
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$$\displaystyle x+\frac{3x-y}{x^2+y^2}=3\cdots (1)$$
$$\displaystyle y-\frac{x+3y}{x^2+y^2}=0\cdots\cdots(2)\times i $$
Now adding these two equations
and substituting $z=x+iy$ and $\bar{z}=x-iy$
and $$|z|^2=x^2+y^2$$
So we have $$z+\frac{3-i}{z}=3\Rightarrow z^2-3z+(3-i)=0$$
On solving that equation we have
$$z=2+i\;\;,1-i$$
So we get $$(x,y)=(2,1)\;\;,(1,-1)$$
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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.