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Need help proving an interval $\frac {1} {ek} \le \frac {1}{k} (1 - \frac {1}{k} )^{k-1} \le \frac {1}{2k}$ I am trying to proof
$$\frac {1} {ek} \le \frac {1}{k} (1 - \frac {1}{k} )^{k-1} \le \frac {1}{2k} $$
for k>=2
to prove this I first multiply by k getting
$$\frac {1} {e} \le \left(1 - \frac {1}{k} \right)^{k-1} \le \frac {1}{2} $$
then use case $k=2$ as a base case
$$\frac {1} {e} <= \frac {1}{2} <= \frac {1}{2} $$
which is good, then assumed
$$\frac {1} {e} <= (1 - \frac {1}{k} )^{k-1} <= \frac {1}{2} $$
to be true for any k>2 and try to prove for k+1
so I sustitute k+1 on k getting
$$\frac {1} {e} <= (1 - \frac {1}{k+1} )^{k} <= \frac {1}{2} $$
which equals
$$\frac {1} {e} <= (\frac {k}{k+1} )^{k} <= \frac {1}{2} $$
so I am trying to get $$ (\frac {k}{k+1} )^{k} $$ to any of the original formulas to finish the prove but I have been unsuccesful. I have devoted a lot of time to it and dont see the solution, if anyone does thanks in aadvance.
if you see any other choice that is easy to prove this pls let me know because I dont have to do it by induction
|
Use the well-known inequality proved here: https://math.stackexchange.com/a/1161287/148510.
For $0 < x < 1$,
$$1-x \leqslant -\ln x \leqslant \frac{1-x}{x}.$$
Then with $x = 1-1/k$, we have
$$\ln(1 -1/k) \geqslant \frac{-\frac1{k}}{1-\frac1{k}}=\frac{-1}{k-1}.$$
Hence,
$$(k-1)\ln(1 -1/k) \geqslant -1 \implies \left(1 - \frac1{k}\right)^{k-1}\geqslant e^{-1}.$$
For the other inequality, an application of the Bernoulli inequality yields
$$\left(\frac{k}{k-1}\right)^{k-1} = \left(1+\frac{1}{k-1}\right)^{k-1}\geqslant 1 + \frac{k-1}{k-1}= 2.$$
Hence,
$$\left(1 - \frac1{k}\right)^{k-1} = \left(\frac{k-1}{k}\right)^{k-1} = \left(\frac{k}{k-1}\right)^{-(k-1)} \leqslant \frac1{2}.$$
|
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|
Find the maximum and minimum of $x^2+2y^2$ if $x^2-xy+2y^2=1$.
Find the maximum and minimum of $x^2+2y^2$ if $x,y\in\mathbb R$ and $$x^2-xy+2y^2=1$$
My attempt:
Clearly, since $x^2-x(y)+(2y^2-1)=0$ and $2y^2-y(x)+(x^2-1)=0$, we have that
$$\Delta_1=y^2-8y^2+4=4-7y^2\ge 0$$
and
$$\Delta_2=x^2-8x^2+8=8-7x^2\ge 0$$
so that $y^2\le \frac{4}{7}$ and $x^2\le \frac{8}{7}$.
Thus $x^2+2y^2\le \frac{8}{7}+\frac{8}{7}=\frac{16}{7}$.
But clearly $x^2+2y^2\neq \frac{16}{7}$, since if $\begin{cases}x^2=\frac{8}{7}\\y^2=\frac{4}{7}\end{cases}$, then $x^2-xy+2y^2\neq 1$
so that equality can't be achieved and we only have that $x^2+2y^2<\frac{16}{7}$.
|
An inequality approach: using $a^2+b^2\geq 2ab$ inequality below, we have
$$
x^2-xy+2y^2=1\implies 1+xy=x^2+2y^2\geq2\sqrt{2}xy\implies xy\leq\frac{1}{2\sqrt{2}-1}\cdot
$$
It follows that
$$
x^2+2y^2=1+xy\leq 1+\frac{1}{2\sqrt{2}-1}=
\boxed{\frac{2\sqrt{2}}{2\sqrt{2}-1}}\cdot
$$
Similarly, using $a^2+b^2\geq -2ab$, we have
$$
1+xy=x^2+2y^2\geq-2\sqrt{2}xy\implies xy\geq-\frac{1}{2\sqrt{2}+1}\cdot
$$
which implies
$$
x^2+2y^2=1+xy\geq 1-\frac{1}{2\sqrt{2}+1}=\boxed{\frac{2\sqrt{2}}{2\sqrt{2}+1}}\cdot
$$
Equality is realised for the max when
$$
y=\frac{1}{\sqrt{-\sqrt{2}+4}},\quad x=\sqrt{2}y;
$$
and for min
$$
y=\frac{1}{\sqrt{\sqrt{2}+4}},\quad x=-\sqrt{2}y\cdot
$$
|
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|
Finding polynomial Coefficients
Let $f(x) = x^5 - x^4 + ax^3 + bx^2 + 8x + 4$
The root will make sure, $f(2) = 0$ Which shows:
$$2^5 - 2^4 + a2^3 + 4b + 16 + 4 = 0$$
$$16 + 8a + 4b + 20 = 0 \implies 8a + 4b = -36 \implies 2a + b = -9$$
It also follows that $f'(2) = 0$ but:
$$f'(x) = 5x^4 - 4x^3 + 3ax^2 + 2bx + 8$$
$$f'(2) = 80 - 32 + 12a + 4b + 8 = 0$$
$$12a + 4b = -56 \implies -3a - b = 14$$
The system is:
$$2a + b = -9$$
$$-3a - b = 14$$
$$-a = 5 \implies a = -5 \implies b = -19$$
Is this even correct?
Theorem: If $x=a$ is a double root of $f(x)$ then $f(a) = f'(a) = 0$.
Proof:
$$f(x) = (x-a)^2 g(x)$$
$$f'(x) = 2(x-a)g(x) + (x-a)^2g'(x)$$
$$f(a) = 0$$
$$f'(a) = 2(a-a)g(x) + (a-a)^2g(a) = 0$$
|
Your idea is correct, but you made a tiny mistake in solving the system. $a=-5$ is correct, but $b$ should be equal to $1$.
|
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|
Minimizing the product $xy$ subject to a polynomial constraint on $x, y$
Given that $$16y(x^2+1)=25x(y^2+1),$$ where $x,y$ are positive integers, find the smallest possible value of $xy$.
I wrote my expression as a quadratic in $x$ and calculated it in form of $y$. Then $xy$ became in $y$ but it had square roots, so differentiating it and then solving and then substituting didn't seem a feasible method.
|
Let's first note that $x > y > 1.$ Since $(y, y^2+1) = 1,$ we have $y\mid 25x,$ and similarly, $x\mid 16y.$ Suppose that $5\not\mid y,$ so that $y\mid x.$ Writing $x = ky$ yields $ky\mid 16y \Rightarrow k\mid 16,$ and clearly $k\not = 1.$ Hence we now have $16y(k^2y^2+1) = 25ky(y^2+1),$ which after rearranging becomes
$$y^2(16k-25) = 25 - \frac{16}{k}.$$
Since $k \ge 2,$ the LHS is at least $7y^2$ while the right-hand side is less than $25,$ so $y = 1,$ which is impossible. Hence $5\mid y.$
Suppose that $25\not\mid y.$ Now $y = 5y' \Rightarrow y' \mid 5x \Rightarrow y' \mid x.$ Then $x = ky'$ and $ky' \mid 80y' \Rightarrow k\mid 80.$ Since $x > y, k > 5.$ Using $y = 5y'$ and $x = ky'$ in the original equality yields
$$80y'(k^2y'^2 +1) = 25ky'(25y'^2+1)$$
which, after rearrangement, becomes
$$y'^2(16k-125) = 5 - \frac{16}{k}.$$
As $k>5,$ the RHS is positive, which implies that $k\ge 8.$ But then $k = 8$ (otherwise the LHS is bigger than $5$), whence $y' = 1$ gives $y = 5, x = 8.$
Now if $25\mid y,$ then $x \ge 2$ so $xy \ge 50$ is bigger than the product $5\cdot 8 = 40,$ so we're done.
|
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|
Parametrizing intersection of a plane and surface I'm working on…
Parametrize the curve which is the intersection of the plane $2x+4y+z=4$ with the surface $z=x^2+y^2$.
I tried eliminating $z$ by plugging it into the first equation and also tried parametrizing each equation before finding the intersection but was not successful. The answer is supposed to be $x=3\cos(t)-1$, $y=3\sin(t)-2$, $z=14-6\cos(t)-12\sin(t)$.
|
Step 1: As you suggest, eliminate $z$ by plugging it into the first equation:
$$2x + 4y + x^2 + y^2 = 4$$
Now, complete the square. You get the equation of a circle:
$$(x + 1)^2 - 1 + (y + 2)^2 - 4 = 4$$
$$(x + 1)^2 + (y + 2)^2 = 9$$
This is a circle with center $(-1,-2)$ and radius $3$. So $x = -1 + 3\cos(t)$ and $y = -2 + 3\sin(t)$, and $z = 4 - 2x - 4y$.
|
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|
Find the slope of the tangent line to the graph of $f^{-1}$ Given function $f$, find the slope of the line tangent to the graph $f^{-1}$ at the point on the graph $f^{-1}$. $f(x)=\sqrt{5x}$; $(4,\frac{16}{5})$?
Here is what I have thus far:
$f'(x)= \frac{\sqrt{5}}{2\sqrt{x}}$
can it also be $f'(x)= \frac{2.5}{\sqrt{5x}}$?
We then need to plug in for $x$ at $x=\frac{16}{5}$
$f'(x)= \frac{\sqrt{5}}{2\sqrt{\frac{16}{5}}}=\frac{5}{8}$
We then have to reciprocate $\frac{5}{8}$
Let $M$ be the slope of the tangent line $f^{-1}$
$M=\frac{8}{5}$
Are these steps and arithmetic correct?
|
y= f(x) = $\sqrt{5x}$ $\rightarrow$ $y^2$ = $5x$ $\rightarrow$ $\frac{1}{5}y^2$ = x
So $f^{-1}(x)$ = $\frac{1}{5}x^2$ , which is easily verified by taking $f^{-1}(f(x))$= x.
Now:
$(f^{-1})'(x)$ = $\frac{2}{5}x$. The slope of the tangent line of $f^{-1}$ at (4,$\frac{16}{5}$) is $(f^{-1})'(4)$ = $\frac{2}{5}*4$ =$\frac{8}{5}$.
So the tangent line to the curve of $(f^{-1})'(x)$ at ( 4,$\frac{16}{5}$) is y-$\frac{16}{5}$ = $\frac{8}{5}$ (x - 4) $\rightarrow$ y = $\frac{8}{5}x$ -$\frac{16}{5}$.
A graph of this situation is as follows:
This is the point slope form of the tangent line.
|
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|
Find all solutions to $x^{10} = 1 \pmod {377}$ Find all solutions to $x^{10} \equiv 1 \pmod {377}$
I noticed that $x^{10} \equiv 1 \pmod {377}$ can be written as:
$(x^5+1)(x^5-1)\equiv 0 \pmod {377}$
also $377 = 13 \times 29$
Any help would be greatly appreciated
|
Hint: by the Chinese remainder theorem, we can find all solutions to this equation by first finding all solutions to
$$
x^{10} \equiv 1 \pmod{13}\\
x^{10} \equiv 1 \pmod{29}
$$
To find the solutions mod $13$, it is slightly useful to note that $x^{10} = x^{12 - 2} = x^{-2}$.
Full solution: The first equation has solution
$$
x^{10} \equiv 1 \pmod{13} \iff \\
x^{-2} \equiv 1 \pmod{13} \iff \\
x^{2} \equiv 1 \pmod{13} \iff \\
x \equiv \pm 1 \pmod{13}
$$
so, we have two solutions to the first equation. For the second, the same trick looks a little weirder:
$$
x^{10} \equiv 1 \pmod{29} \implies\\
(x^{10})^3 \equiv 1 \pmod{29} \implies\\
x^{30 - 28} \equiv 1 \pmod{29} \implies\\
x^2 \equiv 1 \pmod{29}
$$
so, in order for $x$ to be a solution to the original equation, we must also have $x^2 \equiv 1 \pmod{29}$. So, our full solution is
$$
x \equiv \pm 1 \pmod{29}
$$
and you may verify that both of these satisfy the original equation.
Now, applying the CRT (calculating $[13^{-1}]_{29} = 9$ and $[29^{-1}]_{13} = 9$), we end up with $4$ solutions to the original equation (mod 377). In particular, we have
$$
x \equiv 1 \pmod {377}\\
x \equiv -1 \pmod {377}\\
x \equiv 13[13^{-1}]_{29}(-1) + 29[29^{-1}]_{13}(1) \equiv 144 \pmod {377}\\
x \equiv 13[13^{-1}]_{29}(1) + 29[29^{-1}]_{13}(-1) \equiv -144 \pmod {377}\\
$$
|
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|
Number of solutions of the equation $x_1 + x_2 + x_3 = 9$ How to calculate the number of solutions of the equation $x_1 + x_2 + x_3 = 9$ when $x_1$, $x_2$ and $x_3$ are integers which can only range from 1 to 6.
|
We can find the number of solutions using binomial theorem.
The coefficient of $x^9$ in the following will be the required answer.
$$(x+x^2+\cdots+x^6)^3$$
This above, is a Geometric Progression. Therefore,
$$=\left (\frac{x-x^7}{1-x}\right )^3$$
$$=(x-x^7)^3(1-x)^{-3}$$
Now apply binomial theorem to get the coefficient of $x^9$
$$\left (\binom{3}{0}x^3-\binom{3}{1}x^9+\binom{3}{2}x^{15}-\binom{3}{3}x^{21} \right )\left (\binom{2}{0}+\binom{3}{1}x+\binom{4}{2}x^2+\cdots\infty\right )$$
We can neglect all terms with exponent $>9$
$$\left (\binom{3}{0}x^3-\binom{3}{1}x^9\right )\left (\binom{2}{0}+\binom{3}{1}x+\binom{4}{2}x^2+\cdots+\binom{11}{9}x^9\right )$$
We get the the coefficeient of $x^9$ as
$$\binom{3}{0}\binom{8}{6}-\binom{3}{1}\binom{2}{0}$$
$$=28-3$$
$$=25$$
|
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|
Tough inequality in positive reals numbers. Let $a, b, c$ be positive real. prove that
$$\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\geq\left(1+\frac{a+b+c}{\sqrt[3]{abc}}\right)$$
Thanks
|
The inequality is unchanged if you multiply $a$, $b$, and $c$ by a positive number $k$, so WLOG $abc = 1$. Then the left hand is greater than or equal to
$$1 + \frac{a}{b} + \frac{b}{c} + \frac{c}{a}$$
and $$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \ge a + b + c$$
since
\begin{align}&\frac{a}{b} + \frac{b}{c} + \frac{c}{a}\\
&= \frac{1}{3}\left(\frac{a}{b} + \frac{a}{b} + \frac{b}{c}\right) + \frac{1}{3}\left(\frac{b}{c} + \frac{b}{c} + \frac{c}{a}\right) + \frac{1}{3}\left(\frac{c}{a} + \frac{c}{a} + \frac{a}{b}\right)\\
&\ge\sqrt[3]{\frac{a}{b}\frac{a}{b}\frac{b}{c}} + \sqrt[3]{\frac{b}{c}\frac{b}{c}\frac{c}{a}} + \sqrt[3]{\frac{c}{a}\frac{c}{a}\frac{a}{b}}\\
&= \frac{a}{\sqrt[3]{abc}} + \frac{b}{\sqrt[3]{abc}} + \frac{c}{\sqrt[3]{abc}}\\
&= a + b + c.
\end{align}
|
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|
Solve $a_{n+1} - a_n = n^2$ using generating functions The Full Question
Using the method of generating functions, solve $a_{n+1} - a_n = n^2$ where $a_0 = 1$
My Research
Scanned the website for similar answers, reviewed the following links:
Solve the following recurrences using generating functions.
can we use generating functions to solve the recurrence relation $a_n = a_{n-1} + a_{n-2}$, $a_1=1$, $a_2=2$?
Solving a recurrence equation using generating functions
Using Generating Functions to Solve Recursions
Solve the following recursive relation by using generating functions
Solve a recursion using generating functions?
Solving recurrences using generating functions
None of them would make my question a duplicate and none of them really addressed what I was having difficulty with.
My Work
If we multiply each side by $x^{n+1}$ and sum each of the possible cases $a_0,a_1,\dots$
we have:
$$\sum\limits_{j=0}^{\infty}{a_{n+1}x^{n+1}} - x\sum\limits_{j=0}^{\infty}a_nx^n = \sum\limits_{j=0}^{\infty}n^2x^{n+1}$$
If we let $f(x) = \sum\limits_{j=0}^{\infty}a_nx^n$ our equation transforms into:
$$f(x)-a_0 -xf(x) = \sum\limits_{j=0}^{\infty}n^2x^{n+1}$$
Factoring $f(x)$ and observing the LHS is a well known generating functions and using the fact that $a_0 = 1$, we get:
$$f(x)(1-x) = \frac{x(1+x)}{(1-x)^3} + 1 \implies f(x) = \frac{x(1+x)}{(1-x)^4} + \frac{1}{1-x}$$
We now need the coefficients of $x^n$ of these two functions we are adding. The right function it is well known that is has coefficients of $1$. We need partial fraction decomposition to obtain the coefficients of the left hand side.
$$\frac{x(1+x)}{(1-x)^4}= \frac{A}{1-x}+\frac{B}{(1-x)^2}+\frac{C}{(1-x)^3}+\frac{D}{(1-x)^4}$$
$\implies x(1+x)= A(1-x)^3+B(1-x)^2+C(1-x)+D$
Expanding this, we have
$Ax^3+3Ax^2 - 3Ax + A + Bx^2 -2Bx + B +C -Cx +D$
Which gives us the system of equations:
1) $A+B+C+D = 0$
2) $-3A-2B-C =1$
3) $3A+B = 1$
4) $A = 0$
5) $D = 0$
$3(0)+B=1 \implies B=1$ Using equations 3&4
$-3(0) -2 - C=1\implies C=3$ Using equations Using equations 2&1 and the result above
$0+1+3+0 = 0$ plugging in our results to equation 1
that is obviously not correct. Where did I make my mistake. I feel I'm very close and must have messed up the partial fraction decomposition somewhere, but maybe the mistake is somewhere else. Can anyone find where I went wrong?
|
You made a mistake in solving
$$ x(1+x)= A(1-x)^3+B(1-x)^2+C(1-x)+D$$
A faster way to solve this is the following:
$$x = 1 \Rightarrow D=2$$
Next derivate
$$1+2x=-3A(1-x)^2-2B(1-x)-C$$
plug in $x=1$. Then derivate again, and plug in $x=1$. Continue...
|
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Minimise $x+y$ under the condition $2x+32y \leq 9xy$ where $x,y > 0$ Let $x$ and $y$ be positive real numbers such that $2x+32y\leq 9xy$. Then find the smallest possible value of $x+y$. I tried using AM-GM but it is difficult to find the case of equality. Is Cauchy-Schwarz also helpful? Thanks.
|
Rearranging the inequality yields
\begin{align*}
0 &\le 9xy -2x -32y \\
\frac{64}9 &\le 9xy -2x -32y + \frac{64}9 = \bigg(3 x - \frac{32}3\bigg) \bigg(3 y - \frac23\bigg).
\end{align*}
The AM-GM inequality* now gives
$$
\frac83 \le \sqrt{\bigg(3 x - \frac{32}3\bigg) \bigg(3 y - \frac23\bigg)} \le \frac12 \bigg(3 x - \frac{32}3 + 3 y - \frac23 \bigg) = \frac32(x+y) -\frac{17}3,
$$
which implies $x+y \ge 50/9$. Moreover, the case of equality is $3x-32/3 = 3y-2/3$, which has exactly one point of intersection with the line $x+y=50/9$, namely $x=40/9$, $y=10/9$; one confirms that this point is on the boundary $2x+32y=9xy$ as well. In summary, the smallest possible value of $x+y$ is $50/9$.
*AM-GM applies only when the two factors are both positive, so one must check that points with $0\le x\le 32/9$ and $0\le y\le 2/9$ don't satisfy the necessary inequality.
|
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The $ABCD$ paralelograms sides are $AB,BC,CD,DA$. On these line segments there are points in the same order: $X,Y,Z,V$. The $ABCD$ paralelograms sides are $AB,BC,CD,DA$. On these line segments there are points in the same order: $X,Y,Z,V$. We know, that: $$\frac{AX}{XB}=\frac{BY}{YC}=\frac{CZ}{ZD}=\frac{DV}{VA}=k$$
$k$ is a positive constant what is less then $\frac{1}{2}$. What is the value of $k$, if the area of $XYZV$ is 86% of the area of $ABCD$?
Can you tell me a step by step answer to this question?
|
Let $AB = CD = a$ and $BC = AD = b$. Then we have:
$$\frac{AX}{XB} = k \implies \frac{a}{XB} = k+1 \implies XB = \frac{a}{k+1}$$
From this we obtain that:
$$AX = \frac{ka}{k+1}$$
We get simular results for all other segments. Now check that $XYZV$ is inscribed in $ABCD$ and we have:
$$P_{ABCD} - P_{XBY} - P_{YCZ} - P_{ZDV} - P_{VAX} = P_{XYZV}$$
Now using the condition and the formula for area of a triangle we have:
$$P_{ABCD} - \frac{XB\cdot YB \cdot \sin \angle XBY}{2} - \frac{YC\cdot ZC \cdot \sin \angle YCZ}{2} - \frac{ZD\cdot VD \cdot \sin \angle ZDV}{2} - \frac{VA\cdot XA \cdot \sin \angle VAX}{2} = \frac{86P_{ABCD}}{100}$$
Substitute for all the segments and the recall the formula for area of a paralelogram: $P = ab \sin \alpha$ and you'll get:
$$P_{ABCD} -\frac{2kP_{ABCD}}{(k+1)^2} = \frac{86P_{ABCD}}{100}$$
Divide by $P_{ABCD} \not = 0$ and you'll get:
$$\frac{7}{50} = \frac{2k}{(k+1)^2}$$
$$7(k+1)^2 = 100k$$
$$7k^2-86k + 7 = 0$$
Solbing it you'll get: $k_1 = \frac 17 \left(43 - 30\sqrt{2}\right)$ and $k_2 = \frac 17 \left(43 + 30\sqrt{2}\right)$. Since $k_2 >2$ the only solution is $k_1$
|
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|
Prove that $ab \leq \frac14$ and $ (1+1/a)(1+1/b)\ge 9$ when $a+b=1, a \ge 0, b \ge 0$ Our teacher gave us some identities firstly and said we could use one of them to prove it. The identities are: $$\frac{(a^2+b^2)}{2}≥\left(\frac{(a+b)}{2}\right)^2$$ $$(x+y)^2≥2xy$$ and $$\frac{(x+y)}{2} \ge \sqrt{xy}$$
Here's what I did to solve this exercise:
$$\frac{a+b}2 \ge \sqrt{ab} \implies \frac12 \ge \sqrt{ab} \implies \frac14\ge ab \implies ab\le\frac14$$
$$\left(1+\frac1a\right)\left(1+\frac1b\right) = 1+\frac1b+\frac1a+\frac1{ab} \ge 9 \Rightarrow \frac{a+b+1}{ab}\ge8 \Rightarrow \frac2{ab}\ge8 \Rightarrow 2\le8ab \Rightarrow ab\le\frac14$$
I came to a known point, but I'm not sure that this is the right form. If there is any other more clear proof please show it to me.
|
Subtract the inequality from the equation
$$
1=(a+b)^2=a^2+2ab+b^2\\
0\le(a-b)^2=a^2-2ab+b^2
$$
to get
$$
1\ge4ab\implies\frac14\ge ab\tag{1}
$$
Apply $(1)$ to get
$$
\begin{align}
\left(1+\frac1a\right)\left(1+\frac1b\right)
&=1+\frac1a+\frac1b+\frac1{ab}\\
&=1+\frac{a+b}{ab}+\frac1{ab}\\[3pt]
&=1+\frac2{ab}\\[3pt]
&\ge1+\frac2{1/4}\\[8pt]
&=9\tag{2}
\end{align}
$$
|
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|
Prove $(1 + \frac{1}{n})^n$ is bounded above I've checked similar questions on the site but couldn't find satisfactory solutions or hints.
Also, is there a more general approach to proving whether a given sequence is bounded below or above?
|
By the binomial formula:
\begin{eqnarray*}
\left(1+\frac{1}{n}\right)^n=1+1+\sum_{k=2}^n\binom{n}{k}\cdot\left(\frac{1}{n}\right)^k.
\end{eqnarray*}
Notice that
\begin{eqnarray*}
\binom{n}{k}\cdot\left(\frac{1}{n}\right)^k=\frac{n(n-1)\cdots(n-k+1)}{n^k}\cdot\frac{1}{k!}<\frac{1}{k(k-1)}=\frac{1}{k-1}-\frac{1}{k}.
\end{eqnarray*}
So we get
\begin{eqnarray*}
\left(1+\frac{1}{n}\right)^n<2+\sum_{k=2}^n\left(\frac{1}{k-1}-\frac{1}{k}\right)=3-\frac{1}{n}<3.
\end{eqnarray*}
|
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|
Find the derivative of the function. y = $\sqrt{7x+\sqrt{7x+\sqrt{7x}}} $ This question is really tricky. I am wondering if I am right?
|
$$\dfrac{dy}{dx}=\dfrac{d\left( \sqrt{7x+\sqrt{7x+\sqrt{7x}}}\right) }{dx}\\
=\dfrac{1}{2\sqrt{7x+\sqrt{7x+\sqrt{7x}}}}\dfrac{d\left( 7x+\sqrt{7x+\sqrt{7x}}\right) }{dx}\\
=\dfrac{1}{2\sqrt{7x+\sqrt{7x+\sqrt{7x}}}}\left ( 7+\dfrac{d\left(\sqrt{7x+\sqrt{7x}}\right) }{dx}\right)\\
=\dfrac{1}{2\sqrt{7x+\sqrt{7x+\sqrt{7x}}}}\left ( 7+\left (\dfrac{1}{2\sqrt{7x+\sqrt{7x}}}\dfrac{d\left(7x+\sqrt{7x}\right) }{dx}\right)\right)\\
=\dfrac{1}{2\sqrt{7x+\sqrt{7x+\sqrt{7x}}}}\left ( 7+\left (\dfrac{1}{2\sqrt{7x+\sqrt{7x}}}\left(7+\dfrac{d\left(\sqrt{7x}\right) }{dx}\right)\right)\right)\\
=\dfrac{1}{2\sqrt{7x+\sqrt{7x+\sqrt{7x}}}}\left ( 7+\left (\dfrac{1}{2\sqrt{7x+\sqrt{7x}}}\left(7+\left(\dfrac{1}{2\sqrt{7x}}\dfrac{d\left(7x\right) }{dx}\right)\right)\right)\right)\\
=\dfrac{1}{2\sqrt{7x+\sqrt{7x+\sqrt{7x}}}}\left ( 7+\left (\dfrac{1}{2\sqrt{7x+\sqrt{7x}}}\left(7+\left(\dfrac{7}{2\sqrt{7x}}\right)\right)\right)\right)=\dots$$
|
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|
Fourier Series Expansion, error in coefficients? After reworking the problem many times I keep getting the same (incorrect?) answer.
So the problem as stated is
Find the Fourier expansion of :
$$
f(x) = \begin{cases}
x &\text{ if }0 < x < \pi,\\
2(\pi - x) &\text{ if }\pi < x < 2\pi
\end{cases}
$$
I get the following Fourier coefficients:
\begin{align}
a_0 &= \frac{-π}{4}
\\
a_n &= \frac{ 1 - (-1)^n}{π n^2}
\\
b_n &= \frac{4 + 3(-1)^n}{n}
\end{align}
They are apparently wrong since I'm meant to show that
$$
\frac{π^2}{8} = \sum_{n=0}^{+\infty} \frac{1}{(2n-1)^2}
$$
I've checked the jump discontinuities $ \frac{1}{2(f^+(\pi) + f^-(\pi))}$ and $\frac{1}{2(f^+(2\pi) + f^-(2\pi))}$ so it can only be an error in my coefficients?
|
Consider the shifted variant $f(x)=\pi-|x|$ on $[-π,π]$ as fundamental period.
Then, since it is an even function, $b_n=0$ and
\begin{align}
a_0&=\frac1{\pi}\int_0^\pi(\pi-x)dx\\&=\frac1{2\pi}[-(\pi-x)^2|_0^\pi=\frac{\pi}2
\\
a_n&=\frac{2}{\pi}\int_0^\pi(\pi-x)\cos(nx)dx
\\&=\frac2{n\pi}[(\pi-x)\sin(nx)]_0^\pi+\frac2{n\pi}\int_0^\pi \sin(nx)\\
&=\frac2{n^2\pi}[-\cos(nx)]_0^\pi=\frac{2(1-(-1)^n)}{n^2\pi}
\end{align}
which implies $a_{2n}=0$ and $a_{2n+1}=\frac4{(2n+1)^2\pi}$
Comparing function and series at $x=0$ gives
$$
\pi=f(0)=\frac\pi2+\frac4\pi\sum_{n=0}^\infty\frac1{(2n+1)^2}
$$
which rearranged leads to
$$
\frac{\pi^2}{8}=\sum_{n=0}^\infty\frac1{(2n+1)^2}
$$
|
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|
Which mixed numbers have the property $\sqrt{n + \frac{p}{q}}=n\sqrt{\frac{p}{q}}$? Could I please have help with describing mixed numbers (aka mixed fractions) that have this property:
Show that $\sqrt{9\frac{9}{80}}=9\sqrt{\frac{9}{80}}$ and $\sqrt{4\frac{4}{15}}=4\sqrt{\frac{4}{15}},\;$ where $\sqrt{9\frac{9}{80}} = \sqrt{9+\frac{9}{80}}$ etc.
I can easily show this mathematically but however, which mixed numbers have this property and which don't, is there a rule, etc.
|
HINT
For $x,y\in \mathbb{Z^{+}}$
$$\begin{align}\sqrt{x\dfrac{x}{y}} &= x\sqrt{\dfrac{x}{y}}\\ &\iff \\\sqrt{x+\dfrac{x}{y}}&=\sqrt{\dfrac{x^3}{y}}\\\sqrt{\dfrac{\color{blue}{x(y+1)}}{y}} &=\sqrt{\dfrac{\color{blue}{x^3}}{y}}\end{align}$$
Comparing you get $$\color{blue}{y = x^2-1}$$
That means it works for all $x,y$ that satisfy above equation
Check this for mixed numbers notation
|
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|
Why is $\sqrt{3}=[1;1,2,1,2,\dots]$?
Why is $\sqrt{3}=[1;1,2,1,2,\dots]$ ?
$\displaystyle[1;1,2,1,2,\dots]=1+\frac{1}{[1;2,1,2,\dots]}=1+\frac{1}{1+\frac{1}{2+\frac{1}{[1;2,1,2,\dots]}}}$
If I set $x=[1;2,1,2,\dots]$ then;
$\frac{x+1}{x}=\frac{5x+2}{3x+1}$
with solutions $x_{1,2}=\frac{1\pm\sqrt{3}}{2}$
what did go wrong ?
|
Let's begin by computing $y=[1;2,1,2,\cdots]$. This is $$1+\frac{1}{2+\frac{1}{1+\frac{1}{2+\ddots}}}.$$
Now, $\frac{1}{y-1}$ equals $$2+\frac{1}{1+\frac{1}{2+\ddots}}.$$
Then, $$\frac{1}{\frac{1}{y-1}-2}=y.$$
Simplifying the fractions results in $2y^2-2y-1$. Solving for $y$ results in $\frac{1\pm\sqrt{3}}{2}$.
Since $x=1+\frac{1}{y}$, you get that $x=\pm\sqrt{3}$.
|
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|
Algebraic manipulation with $e^x$ I have these inequalities;
$ 1 + x \le e^x \le \frac{1}{1-x} $ for $|x| <1$
This is what i require;
$ 1 - \frac{|x|}{1-x} \le e^x \le 1 + \frac{|x|}{1-x} $ for $0 < x <1$
This is my attempt;
$ 1 + x \le e^x \le \frac{1}{1-x} $ for $|x| <1$
$\implies x \le e^x - 1 \le \frac{1}{1-x} -1 = \frac{x}{1-x} $ for $|x| <1$
$\implies 1 \le \frac{e^x - 1}{x} \le \frac{1}{1-x} $ for $0 < x <1$
I'm not sure how to continue with my argument.
|
Note that for $x\in(0,1)$
$$
e^x\leq\frac{1}{1-x}=\frac{1-x+x}{1-x}=1+\frac{x}{1-x}=1+\frac{|x|}{1-x}\cdot
$$
This takes care of the the inequality on the right. For the one on the left, (still assuming $x\in(0,1)$)
$$
e^x\geq 1+x>1>1-\underbrace{\frac{|x|}{1-x}}_{+}\cdot
$$
|
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|
Prove or disprove $n \geq 2 ~\rightarrow~ \prod \limits_{i=1}^{n} \left ( 1 - \frac{1}{i^2} \right ) ~=~ \frac{n+1}{2n}$ I am working on one of my HW assignments
$$
\forall n \in \mathbb{Z}, ~
n \geq 2
~\rightarrow~
\prod \limits_{i=1}^{n} \left (
1 - \frac{1}{i^2}
\right )
~=~
\frac{n+1}{2n}
$$
And i am not clear whether it should be proved or disproved.
my main concern is base case n=2.
$$P_l (2) = \prod \limits_{i=1}^{2} \left ( 1 - \frac{1}{i^2} \right ) = \left ( 1 - \frac{1}{1^2} \right ) \cdot \left ( 1 - \frac{1}{2^2} \right ) = 0 \cdot \frac{3}{4} = 0$$
$$P_r (2) = \frac{2+1}{2 \cdot2} = \frac{3}{4}$$
but prove for k+1 works
Induction Hypothesis
$$ \prod \limits_{i=1}^{n} \left (
1 - \frac{1}{i^2}
\right )
~=~
\frac{n+1}{2n} \rightarrow
\prod \limits_{i=1}^{n+1} \left (
1 - \frac{1}{i^2}
\right )
~=~
\frac{(n+1)+1}{2(n+1)}= \frac{(n+2)}{2(n+1)}
$$
$$
\prod \limits_{i=1}^{n+1} \left (
1 - \frac{1}{i^2}
\right )
= \prod \limits_{i=1}^{n} \left (
1 - \frac {1}{i^2}
\right )\cdot \left (1 - \frac {1}{(n+1)^2} \right )
$$
By substitution
$$
= \frac{n+1}{2n} \cdot \left (1 - \frac {1}{(n+1)^2} \right )
$$
$$
= \frac{n+1}{2n} \cdot \frac{n(n+2)}{(n+1)^2}
$$
$$
= \frac{n \cdot (n+1)\cdot (n+2)}{2n\cdot (n+1)\cdot (n+1)} = \frac{(n+2)}{2(n+1)}
$$
Now why my base case isn't working? if it should work for all n>2
Can someone point out what am I doing wrong?
|
\begin{align}
\prod \limits_{i=1}^{n} \Big(1 - \frac{1}{i^2}\Big)&=\prod \limits_{i=1}^{1} \Big(1 - \frac{1}{i^2}\Big) \times \prod \limits_{i=2}^{n} \Big(1 - \frac{1}{i^2}\Big)\\
&=\prod \limits_{i=1}^{1} \Big(1 - \frac{1}{i^2}\Big) \times \frac{\prod_{i=2}^{n}(i^2-1)}{\prod_{i=2}^{n}i^2}\\
&=\prod \limits_{i=1}^{1} \Big(1 - \frac{1}{i^2}\Big) \times \frac{\prod_{i=2}^{n}(i-1)\prod_{i=2}^{n}(i+1)}{(\prod_{i=2}^{n}i)^2}\\
&=\prod \limits_{i=1}^{1} \Big(1 - \frac{1}{i^2}\Big) \times \frac{(\frac1n\prod_{i=1}^{n}i)\Big(\frac{n+1}{2}\times\prod_{i=1}^{n}i\Big)}{(\prod_{i=1}^{n}i)^2}\\
&=\prod \limits_{i=1}^{1} \Big(1 - \frac{1}{i^2}\Big) \times \frac{n+1}{2n}
\end{align}
|
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|
Show by induction: $ \frac{a_{2n+4}}{a_{n+2}}=\frac{a_{2n+2}}{a_{n+1}}+\frac{a_{2n}}{a_{n}}$ I would appreciate if somebody could help me with the following problem:
Q: Sequence $\{a_n\}$; satisfy $a_{n+2}=a_{n+1}+a_{n}, a_1=1,a_2=1$
Show by induction:
$$ \frac{a_{2n+4}}{a_{n+2}}=\frac{a_{2n+2}}{a_{n+1}}+\frac{a_{2n}}{a_{n}}$$
I tried induction .But not getting the result!
|
We are give that $a_{n+2}=a_{n+1}+a_{n}$, $a_1=1$, and $a_2=1$.
Thus, it is easy to see that $a_3=a_2+a_1=1+1=2$, $a_4=a_3+a_2=2+1=3$.
Step 1:
Let's use induction on $n$ to show that for $m\ge1$, $n\ge2$, we have
$$a_{n+m}=a_{n}a_{m+1}+a_{n-1}a_{m}$$
For the first benchmark, let $n=2$. Note that
$$a_{m+2}=a_{m+1}+a_m=a_{m+1}(1)+a_{m}(1)=a_{m+1}(a_2)+a_m(a_1)$$
So, the identity works for $n=2$.
For the second benchmark, let $n=3$. Note that
$$a_{m+3}=a_{m+2}+a_{m+1}=2a_{m+1}+a_m=a_3a_{m+1}+a_2a_m$$
So, the identity works for $n=3$. Let's assume that the identity works for $n$. Then,
$$a_{m+n+1}=a_{m+n}+a_{m+n-1}=(a_{n}a_{m+1}+a_{n-1}a_m)+(a_{n-1}a_{m+1}+a_{n-2}a_m)$$
Collecting terms we have
$$a_{m+n+1}=(a_{n}+a_{n-1})a_{m+1}+(a_{n-1}+a_{n-2})a_m=a_{n+1}a_{m+1}+a_{n}a_{m}$$
and the identity is proven.
Step 2:
Here, note that for $m=n$, we have
$$a_{2n}=a_{n+n}=a_{n}a_{n+1}+a_{n-1}a_{n}= a_{n}(a_{n+1}+a_{n-1})$$
Step 3:
Now, observe that the term $a_{2n+4}$ can be written $a_{2n+4}=a_{2(n+2)}$. Using the new identity, we see that
$$\frac{a_{2n+4}}{a_{n+2}}=\frac{a_{n+2}(a_{n+3}+a_{n+1})}{a_{n+2}}=a_{n+3}+a_{n+1}$$
$$\frac{a_{2n+2}}{a_{n+1}}=\frac{a_{n+1}(a_{n+2}+a_{n})}{a_{n+1}}=a_{n+2}+a_{n+1}$$
$$\frac{a_{2n}}{a_{n}}=\frac{a_{n}(a_{n+1}+a_{n-1})}{a_{n}}=a_{n+1}+a_{n-1}$$
Finally,
$$\frac{a_{2n+2}}{a_{n+1}}+\frac{a_{2n}}{a_{n}}=\left(a_{n+2}+a_{n+1}\right)+\left(a_{n+1}+a_{n-1}\right)=a_{n+3}+a_{n+1}=\frac{a_{2n+4}}{a_{n+2}}$$
And the proof is complete!
|
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|
INMO Problem with even function proof.
Let $n$ be a natural number. Show that
$$\left[ \frac{n}{1} \right ] + \left[ \frac{n}{2} \right ] + \left[ \frac{n}{3} \right ] + \cdots + \left[ \frac{n}{n} \right ] + [\sqrt{n}]$$ is even. Here $[x]$ means the greatest integer less than or equal to $x$.
Let $$f(n) = [\sqrt{n}] + \sum_{k=1}^{n} \left[ \frac{n}{k} \right]$$
$$f(1) = [1] + [1] = 2 \checkmark$$
The statement for $f(n+1)$ is to be proved:
$$f(n+1) = [\sqrt{n+1}] + \sum_{k=1}^{n+1} \left[\frac{n+1}{k} \right]$$
I just need some hints from now?
|
Hint: Prove by induction.
When is $ \lfloor \frac{n}{k} \rfloor \neq \lfloor \frac{n+1}{k} \rfloor $?
When they are not equal, what is the difference of these 2 terms?
Ans: The terms are different if there exists some integer $i$ such that $ \frac{n}{k} < i \leq \frac{ n+1}{k}$. Multiplying throughout by $k$, we get $ n < ik \leq n+1$. Since $n$, $n+1$ are consecutive integers, this then implies that $ik = n+1$, or that $k$ is a divisor of $n+1$.
Thus, this term increases by 1 if $k \mid n+1$. Otherwise, it stays constant.
Hence, letting $ g(n) = \sum_{i=1}^n \lfloor \frac{n}{k} \rfloor$, we get that
$$g(n+1) - g(n) = \text{ number of divisors of } n$$
|
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|
Solving a radical equation for real roots I'm attempting to solve the derivative of my function $f(x)$ for real roots.
$$
\\ \begin{align*}
\\ f(x) &= 3x^2 + 3\arcsin{x}
\\ f^{\prime}(x) &= 6x + \dfrac{3}{\sqrt{1-x^2}}
\\
\\ 0 &= 6x + \dfrac{3}{\sqrt{1-x^2}}
\\
\\ 0 &= 6x \sqrt{1-x^2} + 3
\\ 0 &= \sqrt{36x^2 - 36x^4} + 3
\\ -3 &= \sqrt{36x^2-36x^4}
\\ (-3)^2 &= (\sqrt{36x^2-36x^4})^2
\\ 0 &= -36x^4 + 36x^2 - 9
\\ \frac{0}{9} &= \frac{9(-4x^4 + 4x^2 - 1)}{9}
\\ 0 &= -4x^4 + 4x^2 - 1
\\ \end{align*}
$$
I've considered the rational root theorem at this point, through which I find possible roots to be $x = \{ \pm1, \pm\frac{1}{2}, \pm\frac{1}{4} \}$.
Clearly I've made an error, as the actual root is $x = -\dfrac{1}{\sqrt{2}}$.
|
Your work is correct, $1/\sqrt 2$ is indeed a root of $-4x^4+4x^2-1$. The rational root test just tells you that the only possible rational roots of your polynomial are $\pm 1$, $\pm 1/2$ and $\pm 1/4$, it doesn't tell you anything about possible irrational roots, so there's no contradiction. For example, the rational root test also tells you that the only possible rational roots of $x^2-2$ are $\pm 1$ and $\pm 2$, but that doesn't contradict the fact that the actual roots are $\pm\sqrt 2$.
You can easily solve $-4x^4+4x^2-1=0$ by writing it as $-4(x^2)^2+4x^2-1=0$, solving for $x^2$ and then taking a square root.
|
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|
A quick way to prove the inequality $\frac{\sqrt x+\sqrt y}{2}\le \sqrt{\frac{x+y}{2}}$ Can anyone suggest a quick way to prove this inequality?
$$\frac{\sqrt x+\sqrt y}{2}\le \sqrt{\frac{x+y}{2}}$$
|
Divide by $\sqrt y$ and let $t=\frac xy$ to obtain $\frac{\sqrt t+1}2\leq\sqrt\frac{t+1}2$. Comparing the derivatives of both sides, we see
$$\begin{array}{c}\left(\frac{\sqrt t+1}2\right)'&&\left(\sqrt\frac{t+1}2\right)'\\
\|&&\|\\
\frac1{2\sqrt{4t}}&\leq&\frac1{2\sqrt{2t+2}}&\quad\text{for }t\geq1\\
\frac1{2\sqrt{4t}}&\geq&\frac1{2\sqrt{2t+2}}&\quad\text{for }t\leq1\end{array}$$
so $$\frac{\sqrt t+1}2=-1+\int_1^u\left(\frac{\sqrt u+1}2\right)'du\leq-1+\int_1^u\left(\sqrt\frac{u+1}2\right)'du=\sqrt\frac{t+1}2$$
for all $t\geq0$.
|
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|
Chinese Remainder Theorem for $x\equiv 0 \pmod{y}$ Can anyone solve the following system of congruences using CRT step-wise, without skipping any part?
$$\begin{cases} x\equiv 3 \pmod{7}\\ x\equiv 3 \pmod{13}\\ x\equiv 0 \pmod{12}\end{cases}$$
The answer to this problem $276$ but, I keep getting the answer $264$. What is it that I am doing wrong?
|
$x$ is a multiple of $12$ so $x=12k$. We also have $12k\equiv 3\bmod 7$, $12$ is congruent to $5\bmod 7$ and the inverse of $5\bmod 7$ is $3$ by inspection. So $k\equiv 3\cdot5\cdot k\equiv 3\cdot3\equiv 2\bmod 7$. Therefore $k$ is $7l+2$.
So $x=12(7l+2)=84l+24$.
We now have $84l+24\equiv 3 \bmod 13$. This tells us $84l\equiv 5 \bmod 13\implies 6l\equiv 5 \bmod 13$. The inverse of $6\bmod 13$ is $11$ by inspection, therefore $l\equiv 11\cdot 6 \cdot l\equiv 11\cdot 5\equiv 3 \bmod 13$. Therefore $l=13j+3$
And so $x=84l+24=84(13j+3)+24=1092j+276$.
So indeed the solution is to take $n\equiv 276\bmod 1092$. And by the chinese remainder theorem the solution is unique.
|
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|
How to solve $\int_0^{\infty} \frac{\log(x+\frac{1}{x})}{1+x^2}dx$? Here is my question
$$\int_0^{\infty} \frac{\log(x+\frac{1}{x})}{1+x^2}dx$$
I have tried it by substituting $x$ = $\frac{1}{t}$. I got the answer $0$ but the correct answer is $\pi log(2)$. Any suggestion would be appreciated.
|
You may write
$$
\begin{align}
\int_0^{\infty} \frac{\log(x+\frac{1}{x})}{1+x^2}dx&=\int_0^{\infty} \frac{\log(1+x^2)-\log x}{1+x^2}dx\\\\
&=\int_0^{\infty} \frac{\log(1+x^2)}{1+x^2}dx-\int_0^{\infty} \frac{\log x}{1+x^2}dx.
\end{align}
$$ Clearly, by the change of variable $ x \to \dfrac 1 x$, we get
$$
\int_0^{\infty} \frac{\log x}{1+x^2}dx=-\int_0^{\infty} \frac{\log x}{1+x^2}dx=0.
$$ On the other hand, by the change of variable $$x= \tan \theta, \quad dx =(1+\tan^2 \theta)d\theta, \quad 1+ \tan^2 \theta=\dfrac{1}{\cos^2 \theta},$$ we obtain the classic evaluation:
$$
\int_0^{\infty} \frac{\log(1+x^2)}{1+x^2}dx=-2\int_0^{\pi/2} \log ( \cos \theta) \: d\theta=\pi \log 2.
$$
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"timestamp": "2023-03-29T00:00:00",
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|
Set A = {1,2,3,4,5} Pick randomly one digit and remove it. What is the prob. that we pick an odd digit the 2nd time. The probability that we pick any number for the first time is $\dfrac{1}{5}$
the sample space of sample spaces after the first event is then
{2,3,4,5}
{1,3,4,5}
{1,2,4,5}
{1,2,3,5}
{1,2,3,4}
prob. to pick an odd from the 1st sample space is $\dfrac{1}{2}$
prob. to pick an odd from the 2nd sample space is $\dfrac{3}{4}$
prob. to pick an odd from the 3rd sample space is $\dfrac{1}{2}$
prob. to pick an odd from the 4th sample space is $\dfrac{3}{4}$
prob. to pick an odd from the 5th sample space is $\dfrac{1}{2}$
The final result is:
$\dfrac{1}{5}$ * $\dfrac{1}{2}$ + $\dfrac{1}{5}$ * $\dfrac{3}{4}$ + $\dfrac{1}{5}$ * $\dfrac{1}{2}$ + $\dfrac{1}{5}$ * $\dfrac{3}{4}$ + $\dfrac{1}{5}$ * $\dfrac{1}{2}$ = $\dfrac{3}{5}$
Is this reasoning correct? Are there any simpler ways to solve this problem?
|
There is a simpler way. No digit should have an inherent advantage of being picked second. So since 3 of the 5 digits are odd, there is a $\frac{3}{5}$ chance of the second number picked being odd.
|
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|
Pascals Identity Let there be a group of n boxers and we want to select k people out of it, suppose one of the persons name is ‘Prem’ , so no of ways to choose k people = (combinations in which Prem is present + combinations in which Prem is not present) i.e $ {n-1} \choose {k-1} $ + $ {n-1} \choose k $, this is equal to $ n \choose k $, this was given in book. $ \qquad \qquad \quad $ However what I did was that to assume we know two people among them named ‘Prem’ and ‘Ram’ so now no of combinations = (combinations in which both are present + combination in which Ram is present but not Prem + combination in which Prem is present but not Ram + combination in which none of them are present ) = $ {n-2} \choose {k-2} $ + 2 $ {n-1} \choose {k-1} $ + $ {n-2} \choose {k} $ , this also should be equal to $ n \choose k $ , however when i tried with a value namely $ 6 \choose 4 $ , i didn’t get the two values same, what did i do wrong?
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Short answer: Your middle term should be $2\binom{n-2}{k-1}$.
Explanation: If, from $n$ people—Prem, Ram, and $n-2$ others—you must choose Prem but not Ram, and $k-1$ others, then the $k-1$ are drawn from the last group of $n-2$ others. Likewise if you choose Rem but not Pram, and $k-1$ others. Both these terms are $\binom{n-2}{k-1}$; hence, the middle term should be $2\binom{n-2}{k-1}$.
For $n = 6, k = 4$, we have
$$
\binom{4}{2} + 2\binom{4}{3} + \binom{4}{4} = 6 + 2(4) + 1 = 15 = \binom{6}{4}
$$
Note that your identity can also be derived from the first identity, applied recursively:
$$
\begin{align}
\binom{n}{k} & = \binom{n-1}{k-1} + \binom{n-1}{k} \\
& = \binom{n-2}{k-2} + \binom{n-2}{k-1}
+ \binom{n-2}{k-1} + \binom{n-2}{k} \\
& = \binom{n-2}{k-2} + 2\binom{n-2}{k-1} + \binom{n-2}{k}
\end{align}
$$
|
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|
Finding value of $\sin (15\, ^{\circ})$ with half angle identity The answer I got when trying to solve it was $\sqrt{\frac{1 - \sqrt3}{2} }$ but the book says it's $\sqrt{ \frac{2 - \sqrt3}{2}}$ and I don't know how the two on the top half gets there.
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By the half angle formulas $$\sin\dfrac{\pi}{6}=2\sin\dfrac{\pi}{12}\cos\dfrac{\pi}{12}.$$ Therefore if you substitute $x=\sin\dfrac{\pi}{12},$ you will get the equation $$\dfrac{1}{2}=2x\sqrt{1-x^2}.$$ One way to solve this equation is squaring both sides. But, if you square an equation number of roots will be double. Therefore you should check the roots by backward substitution. $$1=16x^2(1-x^2)$$ $$3=(4x^2-2)^2$$ Since $$0\lt\sin\dfrac{\pi}{12}=x\lt\frac12$$ we have $4x^2-1\lt0.$ Therefore choose $$4x^2-2=-\sqrt3$$ $$x^2=\dfrac{2-\sqrt3}{4}=\dfrac{(\sqrt3-1)^2}{8}$$ Therefore choose $$x=\dfrac{\sqrt3-1}{2\sqrt2}.$$ Hence
$$\sin\dfrac{\pi}{12}=\dfrac{\sqrt6-\sqrt2}{4}$$
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|
Find All Solutions to System of Congruence $$
\begin{cases}
x\equiv 2 \pmod{3}\\
x\equiv 1 \pmod{4}\\
x\equiv 3 \pmod{5}
\end{cases}
$$
$
n_1=3\\
n_2=4\\
n_3=5\\
N=n_1 * n_2 * n_3 =60\\
m_1 = 60/3 = 20\\
m_2 = 60/4 = 15\\
m_3 = 60/5 = 12\\
gcd(20,3)=1=-20*1+3*7\\
gcd(15,4)=1=-15*1+4*4\\
gcd(12,5)=1=12*3-5*7\\
x=-20*2-15*1+12*3\equiv -19 \equiv 41 \pmod{60}\\
$
The above is what I've tried so far. Can someone tell me where I'm going wrong? It's my first time doing this and looking at the explanation and examples for the Chinese Remainder Theorem makes my head want to explode.
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I think you're a bit confused about the recipe used to solve the system of congruences using the Chinese Remainder Theorem. Using your notation, the actual solution is given by
\begin{align*}
x = 2 \cdot m_1 \cdot ({m_1}^{-1} \text{ mod } 3) + 1 \cdot m_2 \cdot ({m_2}^{-1} \text{ mod } 4) + 3 \cdot m_3 \cdot ({m_3}^{-1} \text{ mod } 5) \, .
\end{align*}
Let's see why this answer works. Since $m_1$ is divisible by both $4$ and $5$, the first term is "invisible" when we consider $x$ mod $4$ and mod $5$: it is congruent to $0$ mod $4$ and mod $5$, so it doesn't contribute to the answer of the last two congruences. Thus, we can focus on making this first term satisfy the first congruence. Okay, so far we know the first term should have as factors $2$ (the congruence we want) and $m_1$ (to get rid of the effect mod $4$ and $5$). But now we don't satisfy the first congruence; the $m_1$ throws things off. To fix this, we multiply by the ${m_1}^{-1}$ mod $3$. Then mod $3$ we have
\begin{align*}
x &= 2 \cdot m_1 \cdot ({m_1}^{-1} \text{ mod } 3) + 1 \cdot m_2 \cdot ({m_2}^{-1} \text{ mod } 4) + 3 \cdot m_3 \cdot ({m_3}^{-1} \text{ mod } 5)\\
&\equiv 2 \cdot m_1 \cdot ({m_1}^{-1} \text{ mod } 3) \equiv 2 \pmod{3}
\end{align*}
as desired. Similar reasoning shows that $x$ satisfies the given congruences mod $4$ and $5$.
You have the right idea with your solution, but you're missing some factors. Since you've shown $1 = 20 \cdot (-1) + 3 \cdot 7 \equiv 20 \cdot (-1) \pmod{3}$, then
$$
{m_1}^{-1} \text{ mod } 3 = 20^{-1} \text{ mod } 3 = -1 \equiv 2 \pmod{3} \, .
$$
So the first term should be $2 \cdot 20 \cdot 2$. Can you figure out the other two terms now?
|
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|
How do I know that $\ln(x^2+1)-x \arctan(x)$ is always negative or zero? I did google the function and I can clearly see that it is always negative or zero, but I have no idea how I would have found this on my own. Both the logarithm and the $x\cdot \arctan(x)$ are positive.
What I did do is derive the function but that did not seem to be more obvious.
$\rho' = \frac{x}{1+x^{2}}-\arctan(x)$
This also does not look obvious to me.
$\rho'' = -\frac{2\cdot x^{2}}{(1+x^{2})^{2}}$
This is most definitely negative, thus concave (or concave downwards). Is that enough to deduce anything ?Am I missing something?
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let $$y = \ln(1 + x^2 ) - x \tan^{-1} x .$$ the function is even and $y = 0$ at $x = 0 \text{ and } \lim_{x \to \infty} y = -\infty. $ taking the derivative, you find $$\frac{dy}{dx} = \frac{2x}{1+x^2} - \frac{x}{1+x^2} - \tan^{-1} x =\frac{x}{1+x^2} - \tan^{-1} x$$ we will have established that $y < 0 \text{ for all } x \neq 0$ if we can show that $$f(x) = \frac{x}{1+x^2} - \tan^{-1} x <0, x \neq 0.$$ observe $x = 0$ is the only zero of $f.$ suppose $$x > 0, f'(x) = \frac{1}{1+x^2} - \frac{2x^2}{1+x^2} -\frac 1{1+x^2} = -\frac{2x^2}{1+x^2} < 0$$ together with $f(0) = 0$ proves the claim and we are done.
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|
To Find $a$ such that $2^{1990} \equiv a\pmod {1990}$. To Find $a$ such that $2^{1990} \equiv a\pmod {1990}$.
$1990 = 2 \times 5 \times 199$. Now $a \equiv 0 \pmod {2}$, $a \equiv 4 \pmod{5}$ and $a \equiv 29 \pmod{199}$. Taking first two together we get $a \equiv 4 \pmod {10}$.
$199x \equiv 1\pmod {10}$ has a solution $x =-1$ and $10x \equiv 1\pmod{199}$ has a solution $x = 20$.
Thus using Chinese Remainder Theorem we have,
$a \equiv 4(-1)199 + 29(20)10 \equiv 5004 \equiv 1024\pmod {1990}$.
Thus our $a = 1024$.
Is the solution correct? Is there any better proof?
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Checking your answer of $a=1024=2^{10}$:
$$\begin{align}2^{1990}&\equiv 0\equiv 2^{10}&\pmod 2\\
2^{1990}&\equiv \left(2^{4}\right)^{495}2^{10}\equiv 2^{10}&\pmod{5}\\
2^{1990}&\equiv \left(2^{198}\right)^{10}2^{10}\equiv 2^{10}&\pmod{199}
\end{align}$$
After your edits, your answer is correct.
|
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|
Simplify $\left(\sqrt{\left(\sqrt{2} - \frac{3}{2}\right)^2} - \sqrt[3]{\left(1 - \sqrt{2}\right)^3}\right)^2$ I was trying to solve this square root problem, but I seem not to understand some basics.
Here is the problem.
$$\Bigg(\sqrt{\bigg(\sqrt{2} - \frac{3}{2}\bigg)^2} - \sqrt[3]{\bigg(1 - \sqrt{2}\bigg)^3}\Bigg)^2$$
The solution is as follows:
$$\Bigg(\sqrt{\bigg(\sqrt{2} - \frac{3}{2}\bigg)^2} - \sqrt[3]{\bigg(1 - \sqrt{2}\bigg)^3}\Bigg)^2 = \Bigg(\frac{3}{2} - \sqrt{2} - 1 + \sqrt{2}\Bigg)^2 = \bigg(\frac{1}{2}\bigg)^2 = \frac{1}{4}$$
Now, what I don't understand is how the left part of the problem becomes:
$$\frac{3}{2} - \sqrt{2}$$
Because I thought that $$\sqrt{\bigg(\sqrt{2} - \frac{3}{2}\bigg)^2}$$
equals to $$\bigg(\bigg(\sqrt{2} - \frac{3}{2}\bigg)^2\bigg)^{\frac{1}{2}}$$
Which becomes $$\sqrt{2} - \frac{3}{2}$$
But as you can see I'm wrong.
I think that there is a step involving absolute value that I oversee/don't understand.
So could you please explain by which property or rule of square root is this problem solved?
Thanks in advance
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This is a mistake that might have been avoided by being aware of the numerical values of some of the expressions involved. A very rough approximation for $\sqrt{2}$ is $1.41$. Clearly $\frac{3}{2} = 1.5$. Then $\sqrt{2} - \frac{3}{2} \approx -0.09$, and that squares to $0.0081$, and the square root of that is $0.09$, not $-0.09$.
At no point in this problem is it actually necessary to take the square root of a negative number. But the possibility of going down that path, even if erroneously, would perhaps have alerted you that some things in this problem should be evaluated rather than canceled. Thus, $$\left(\sqrt{2} - \frac{3}{2}\right)^2 = \left(-\frac{3}{2} + \sqrt{2}\right)^2 = \frac{17}{4} - 3 \sqrt{2} \neq -\frac{3}{2} + \sqrt{2}.$$
I'm not suggesting that you always run through problems with crude numerical approximations. But it can be a good way to check you're on the right track.
|
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Hiding 3 coins in a pie, and slicing the pie in 8 equal pieces - probability Okay so I have this question on my maths sheet and it's in the permutations section but I can't get my head round it (probably just being dumb)
I know that $P(n,r) = \frac{n!}{(n-r)!}$
and that $C(n,r) = \frac{n!}{r!(n-r)!}$
And old lady bakes a pie for her grandson, puts three coins into the pie and slices it into 8 equal portions. What is the probability that the grandson will find 2 or more coins in his slice of the pie, assuming that each coin is equally likely to be anywhere within the pie?
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If an event has probability $p$, then the probability that it will occur exactly $k$ times in $n$ trials is
$$\binom{n}{k}p^k(1 - p)^{n - k}$$
Since the pie has been cut into eight pieces of equal size, the probability that a particular coin is in that slice is $p = 1/8$. The number of trials is three since we must check if each coin is in that slice. Hence, the probability that exactly two of the three coins will be found in a particular slice is
$$\binom{3}{2}\left(\frac{1}{8}\right)^2\left(1 - \frac{1}{8}\right)^1$$
The probability that all three of the coins will be found in the same slice is
$$\binom{3}{3}\left(\frac{1}{8}\right)^3\left(1 - \frac{1}{8}\right)^0$$
Thus, the probability that at least two coins will be found in the same slice is
\begin{align*}
\binom{3}{2}\left(\frac{1}{8}\right)^2\left(\frac{7}{8}\right)^1 + \binom{3}{3}\left(\frac{1}{8}\right)^3\left(\frac{7}{8}\right)^0 & = 3 \cdot \frac{1}{64} \cdot \frac{7}{8} + 1 \cdot \frac{1}{512} \cdot 1\\
& = \frac{21}{512} + \frac{1}{512}\\
& = \frac{22}{512}\\
& = \frac{11}{256}
\end{align*}
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How to solve this limit: $\lim_{x\to\infty}\sqrt{x+\sqrt{x}}-\sqrt{x-1}$ Compute the limit $\lim\limits_{x\to+\infty}\sqrt{x+\sqrt{x}}-\sqrt{x-1}$
my attempt:
I tried to multiply top and bottom by the conjugate
$$\begin{align}
\lim_{x\to+\infty}\sqrt{x+\sqrt{x}}-\sqrt{x-1}&=\lim_{x\to+\infty}\left(\sqrt{x+\sqrt{x}}-\sqrt{x-1}\right)\frac{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}\\
&=\lim_{x\to+\infty}\frac{\left(\sqrt{x+\sqrt{x}}\right)^2-\left(\sqrt{x-1}\right)^2}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}\\
&=\lim_{x\to+\infty}\frac{(x+\sqrt{x})-(x-1)}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}\\
&=\lim_{x\to+\infty}\frac{1+\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}
\end{align}$$
But I don't know what I can do after this.
|
You may just observe that, as $x\to \infty$:
$$
\frac{1+\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}\sim \frac{\sqrt{x}\left(1+\frac1{\sqrt{x}}\right)}{\sqrt{x}\left(\sqrt{1+\frac{1}{\sqrt{x}}}\right)+\sqrt{x}\left(\sqrt{1-\frac{1}{x}}\right)}\sim \frac12.
$$
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|
Simplifying trig expression I have
$$\frac{\tan{15^\circ}}{1-\tan{15^\circ}^2}$$
and need to simplify it. The only equation I have that is even close to a match for it is $2\frac{\tan{15^\circ}}{1-\tan{15^\circ}^2}$. But the numerator isn't right with the $\times 2$. The answer should be $\frac{\sqrt{3}}{6}$. What do I do with this?
|
$$
2 \frac{\tan(15^{\circ})}{1-\tan(15 ^{\circ})^2} = \tan(30) = \frac{1}{\sqrt{3}} \tag{1}
$$
because:
$$
\frac{\tan(\alpha) + \tan(\beta)}{1-\tan(\alpha)\tan(\beta)} = \tan(\alpha+\beta) \tag{2}
$$
Following equation (1):
$$
\frac{\tan(15^{\circ})}{1-\tan(15 ^{\circ})^2} = \frac{1}{2}\tan(30) = \frac{1}{2\sqrt{3}}
$$
And finally we employ a very useful method in dealing with roots, multiplying by $1$:
$$
\frac{1}{2\sqrt{3}}=\frac{1}{2\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{6}$$
|
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|
Evaluating a definite integral by definition: $A(x) = \int_{-1}^x (t^2 + 1)\space dt$ I have an area function $A(x)$ defined as
$$A(x) = \int_{-1}^{x} (t^2 + 1)\space dt$$
... and I would like to use the definition of definite integral to evaluate it.
I started this way
$$A(x) = \lim_{n \to \infty} \sum_{k=1}^n { (t^2 + 1) {x - (-1) \over n}}\tag 1$$
I thought that $$\lim_{n \to \infty} \sum_{k=1}^n {x +1 \over n }\tag 2$$ turns into $x + 1$, so that
$$A(x) = \lim_{n \to \infty} \space\sum_{k=1}^n { t^2}{x + 1 \over n} + x + 1\tag 3$$
and saying that $t = k{x + 1 \over n}$ it would become
$$A(x) = \lim_{n \to \infty} \space\sum_{k=1}^n { k^2 (x+1)^3 \over n^3} + x + 1\tag4$$
I'm not sure if I didn't make any mistake until here, and even if I didn't, I don't know what to do now. Maybe I could use
$$\sum_{k = 1}^{n} k^2 = {n(n + 1)(2n + 1) \over 6}\tag 5$$
but in order to do that, I would have to first get rid of $n^2$ in denominator.
Could you please give me a friendly kick in the head to make it correct? Any help will be highly appreciated.
|
Remember that
$$\int_a^b dt \, f(t) = \lim_{n \to \infty} \frac{b-a}{n} \sum_{k=0}^n f \left [ a + (b-a) \frac{k}{n} \right ] $$
Here, $a=-1$ and $b=x$, so that the sum we are interested in is
$$ \begin{align} \int_{-1}^x dt \, (1+t^2) &= \lim_{n \to \infty} \frac{x+1}{n} \sum_{k=0}^n \left [ 1+ \left ( -1 + \frac{x+1}{n} k \right )^2 \right ] \\ &= x+1 + \lim_{n \to \infty} \frac{x+1}{n} \sum_{k=0}^n \left [1 - 2 \frac{x+1}{n} k + \left ( \frac{x+1}{n} \right )^2 k^2 \right ]\\ &= 2 (x+1) + \lim_{n \to \infty} \left [\frac{-2 (x+1)^2}{n^2} \sum_{k=0}^n k + \frac{(x+1)^3}{n^3} \sum_{k=0}^n k^2 \right ]\\ &= 2 (x+1) + \lim_{n \to \infty} \left [\frac{-2 (x+1)^2}{n^2} \frac{n (n+1)}{2} + \frac{(x+1)^3}{n^3} \frac{n (n+1) (2 n+1)}{6} \right] \\ &= 2 (x+1) - (x+1)^2 + \frac13 (x+1)^3 \\ &= \frac13 (x+1) (x^2-x+4)\end{align}$$
You can verify that this is what you get by integrating using the Fundamental theorem.
|
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|
Converge or diverge $\sum^{\infty}_{n=2}\frac{4^n(n!)^2}{(2n-1)!}$ Show if the series converges or diverges.$$\sum^{\infty}_{n=2}\frac{4^n(n!)^2}{(2n-1)!}$$Can someone please help with proving this? (I think it converges)
|
$a_n=\dfrac{4^n(n!)^2}{(2n-1)!} = \dfrac{(2^n\cdot n!)^2}{(2n-1)!} = \dfrac{2\cdot 4\cdots (2n-2)\cdot (2n)}{1\cdot 2\cdot 3\cdots (2n-2)\cdot (2n-1)}\cdot 2\cdot 4\cdots (2n) > \dfrac{2\cdot 4\cdot 6\cdots (2n-2)\cdot (2n)}{1\cdot 2\cdot 3\cdots (2n-2)\cdot (2n-1)}\cdot 1\cdot 3\cdot 5\cdots (2n-1) = \dfrac{(2n-1)!2n}{(2n-1)!}= 2n = b_n \to \text{ divergence}$ by comparison test with $\sum_{n} (2n)$
|
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|
Compute $\lim_{x\rightarrow0}\frac{e^{x^2} - \cos x}{\sin^2 x}$ Find the limit as $x$ approaches $0$ of
$\dfrac{e^{x^2} - \cos x}{\sin^2 x}$
What I tried is as $x$ approaches $0$, $e^{x^2}$ tends to $1$ and so the numerator tends to $1-\cos x$ and after doing some trigonometric simplifications I got the answer as $\frac 12$.
Is this right? Any help would be appreciated.
|
Use Taylor expansion. As $x\rightarrow0$, you have
$$e^{x^2}=1+x^2+O(x^4)$$
$$\cos x=1-\frac12x^2+O(x^4)$$
$$\sin^2x=x^2+O(x^4)$$
Thus
$$\frac{e^{x^2}-\cos x}{\sin^2 x}=\frac{1+x^2-1+\frac12x^2+O(x^4)}{x^2+O(x^4)}=\frac{\frac32+O(x^2)}{1+O(x^2)}\underset{x\rightarrow 0}{\longrightarrow} \frac32$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
When $p=2$ or $p$ prime, with $p=1\pmod{4}$, $x^2\equiv -1\pmod{p}$ is soluble - trouble understanding proof Theorem: When $p=2$ or $p$ prime, with $p=1\pmod{4}$, $x^2\equiv -1\pmod{p}$ is soluble
Proof: When $p=2$, the statement is clear.
Assume $p\equiv 1\pmod{4}$, let $r=\frac{p-1}{2}$ and $x=r!$
Then since $r$ is even $x^2\equiv (1\cdot 2\cdot...\cdot r)((p-1)\cdot(p-2)\cdot...\cdot(p-r))\equiv (p-1)!\equiv -1\pmod{p}$
Thus, when $p\equiv 1\pmod{4}$, the congruence $x^2\equiv -1\pmod{p}$ is soluble.
Point of contention: I understand the general argument
I understand the relation
$(1\cdot 2\cdot...\cdot r)((p-1)\cdot(p-2)\cdot...\cdot(p-r))\equiv (p-1)!\equiv -1\pmod{p}$
But I cant work out how $x^2\equiv(1\cdot 2\cdot...\cdot r)((p-1)\cdot(p-2)\cdot...\cdot(p-r))\pmod{p}$
How is $((\frac{p-1}{2})!)^2\equiv(1\cdot 2\cdot...\cdot r)((p-1)\cdot(p-2)\cdot...\cdot(p-r))\pmod{p}$
|
Suppose $\;p=4k+1\;$ , so that
$$\frac{p-1}2=2k\implies\;\; \text{since}\;\;j=p-j\pmod p\;\;\text{for}\;\;0\le j\le p \,, $$
we get that
$$\frac{p-1}2+1=\frac{p-1}2-1\pmod p\;,\;\;\frac{p-1}2+2=\frac{p-1}2-2\pmod p\,,\ldots$$
so that
$$1\cdot2\cdot\ldots\cdot k\cdot\ldots\cdot\left(\frac{p-1}2-1\right)\cdot\left(\frac{p-1}2\right)\cdot\left(\frac{p-1}2-1\right)\cdot\ldots\cdot k\cdot\ldots \cdot1=\left[\left(\frac{p-1}2\right)!\right]^2$$
If you know some basic group theory things can possibly go easier than the above.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integrating $ \int^{\pi}_{0} \frac{\cos^2 x}{a^2 \cos^2 x + b^2 \sin^2 x} \ dx $ I am trying to show that
$$ \int^{\pi}_{0} \frac{\cos^2 x}{a^2 \cos^2 x + b^2 \sin^2 x} \ dx = \frac{\pi}{a(a+b)} $$ where $ a,b > 0$.
I have tried a few things, but none have worked.
For example, one approach was
$$ \int^{\pi}_{0} \frac{\cos^2 x}{a^2 \cos^2 x + b^2 \sin^2 x} \ dx = \int^{\pi}_{0} \frac{\cos^2 x}{(a^2 - b^2)\cos^2 x + b^2} \ dx$$ and then divide by $ a^2 - b^2 $ throughout. However I am not given that $ a \not= b $, so this approach didn't work.
I also tried splitting up the integral, but then was not sure how to proceed:
$$ \int^{\pi}_{0} \frac{\cos^2 x}{a^2 \cos^2 x + b^2 \sin^2 x} \ dx = \int^{\frac{\pi}{2}}_{0} \frac{\cos^2 x}{a^2 \cos^2 x + b^2 \sin^2 x} \ dx + \int^{\pi}_{\frac{\pi}{2}} \frac{\cos^2 x}{a^2 \cos^2 x + b^2 \sin^2 x} \ dx $$
Substituting $ x \to x + \frac{\pi}{2} $ followed by $ x \to \frac{\pi}{2} -x $ a in the second integral gives
$$ \int^{\frac{\pi}{2}}_{0} \frac{\cos^2 x}{a^2 \cos^2 x + b^2 \sin^2 x} \ dx $$
Hence $$ \int^{\pi}_{0} \frac{\cos^2 x}{a^2 \cos^2 x + b^2 \sin^2 x} \ dx = 2 \int^{\frac{\pi}{2}}_{0} \frac{\cos^2 x}{a^2 \cos^2 x + b^2 \sin^2 x} \ dx $$
Next I thought I could substitute $ u = a^2 \cos^2 x + b^2 \sin^2 x $ which changes my limits from $[0, \frac{\pi}{2}]$ to $[a^2, b^2]$, however I could not complete the substitution as I didn't see how to express $ du = (b^2 - a^2) \sin 2 x \ dx $ or $ \cos^2 x$ in terms of $u$.
|
If we set $x=\arctan t$ we simply have:
$$I(a,b)=\int_{0}^{\pi}\frac{\cos^2 x}{b^2\sin^2 x+a^2\cos^2 x}\,dx = 2\int_{0}^{+\infty}\frac{dt}{(1+t^2)(a^2+b^2 t^2)}$$
that is straightforward to compute through partial fraction decomposition or through the residue theorem.
|
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|
System of equations with radicals: $2\sqrt[4]{\frac{x^4}{3}+4}=1+\sqrt{\frac{3}{2}y^2}$ and $2\sqrt[4]{\frac{y^4}{3}+4} = 1+\sqrt{\frac{3}{2}x^2}$
Solve the system of equations (in $\mathbb R$):
$$\begin{matrix}
2\sqrt[4]{\frac{x^4}{3}+4}=1+\sqrt{\frac{3}{2}y^2}
\\
2\sqrt[4]{\frac{y^4}{3}+4}
=
1+\sqrt{\frac{3}{2}x^2}
\end{matrix}.$$
This is from an older question, which was closed and deleted because of lack of context. (Here is link for users who can see deleted questions.) I will post my solution below - so I hope this time the question will not be closed for the lack of effort.
I found the system not too easy and somewhat interesting. (Of course I might have missed some straightforward way to the solution.) I'd be glad to see some other methods to solve it.
|
Without loss of generality, we can assume that $x,y\ge0$. (Notice that all exponents are even. We can then add signs to get the remaining solutions.)
So our equations are changed to
$$2\sqrt[4]{\frac{x^4}{3}+4}=1+\sqrt{\frac{3}{2}}y\tag{1}$$
$$2\sqrt[4]{\frac{y^4}{3}+4}=1+\sqrt{\frac{3}{2}}x\tag{2}$$
If we subtract the two equations, we get (1)-(2):
$$2\left(\sqrt[4]{\frac{x^4}{3}+4}-\sqrt[4]{\frac{y^4}{3}+4}\right)=y-x.\tag{3}$$
Notice that the function $x\mapsto\sqrt[4]{\frac{x^4}{3}+4}$ is increasing.
So for $x>y$ the LHS is positive and the RHS is negative. Similarly, for $y<x$ the signs of the two expressions are opposite.
So we can only find a real solution for $x=y$, which gives us
$$2\sqrt[4]{\frac{x^4}{3}+4}=1+\sqrt{\frac{3}{2}}x\tag{4}$$
From this we get
$$16\left(\frac{x^4}{3}+4\right)=\left(1+\sqrt{\frac{3}{2}}x\right)^4\tag{5}$$
This is a quartic equation. In theory, this can be done by hand, but it will be very probably quite messy. WolframAlpha returns this. (One of the solution, according to WA, is $\sqrt6$.)
|
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|
Find $dy/dx$ of $(xy^2)+5 = x + 2y^2$ For the solution I got $$\frac{y^2-1}{ 4y-2xy} = dy/dx$$
I just want to know if this is correct. Also it says to evaluate $dy/dx$ at $(1,2)$.
Would the solution to that be $3/4$?
|
Here are the steps
$$ xy^2+5=x+2y^2 $$
$$ \frac{d}{dx}\left[xy^2+5\right]=\frac{d}{dx}\left[x+2y^2\right] $$
$$ y^2\frac{d}{dx}\left[x\right]+x\frac{d}{dx}\left[y^2\right]=1+2\frac{d}{dx}\left[y^2\right] $$
$$ y^2+2xy\frac{d}{dx}[y]=1+4y\frac{d}{dx}[y] $$
$$ y^2-1=\left(4y-2xy\right)\frac{d}{dx}[y] $$
$$ \frac{d}{dx}[y] = \frac{y^2-1}{4y-2xy} $$
So now at $(1, 2)$, we have
$$ \frac{d}{dx}[y] = \frac{2^2-1}{4\cdot 2-2\cdot 1\cdot 2} = \frac{3}{8-4} = \frac34 $$
|
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|
sum of the squares of the reciprocals of the two parts of the focal chord of a parabola
Find the sum of the squares of the reciprocals of the two parts of the focal chord of a parabola.
My attempt:
Let $y^2=4ax$ be a parabola. Let PQ be the focal chord through the focus S$(a, 0)$ of the parabola such that the co-ordinates of P & Q are $(at_1^2, 2at_1)$ & $(at_2^2, 2at_2)$. Then we have, $t_1t_2=-1$. Clearly, if $l=PS$ and $l'=QS$, then $l=\sqrt{(at_1^2-a)^2+(2at_1-0)^2}=\dots=a(t_1^2+1)$ and similarly, $l'=a(t_2^2+1)=a\frac{t_1^2}{1+t_1^2}$ (since, $t_1t_2=-1$).
Then $\frac{1}{l}+\frac{1}{l'}=\dots=\frac{1}{a}$.
Then $\frac{1}{l^2}+\frac{1}{l'^2}=\frac{1}{a}-2\frac{1}{ll'}$
But the answer should be independent of $l$ & $l'$. What to do?
|
Loosely speaking, the "axis chord" has parts $a$ and $\infty$, so the sum of the squares of the reciprocals of the lengths is
$$\frac{1}{a^2}+\frac{1}{\infty^2} = \frac{1}{a^2} + 0 = \frac{1}{a^2}$$
However, the focal chord perpendicular to the axis has parts $2a$ and $2a$, so the corresponding sum is
$$\frac{1}{4a^2} + \frac{1}{4a^2} = \frac{1}{2a^2}$$
The sum is not independent of the choice of chord.
As you have shown, though: The sum of the (un-squared) reciprocals of the lengths is independent of that choice.
|
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|
Find the coefficient of $x^{30}$. Find the coefficient of $x^{30}$ in the given polynomial
$$
\left(1+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9+x^{10}+x^{11}+x^{12}\right)^5
$$
I don't know how to solve problems with such high degree.
|
This turns out to be a detailed explanation of Jack's answer.
$$
\begin{align}
\left(\frac{1-x^{13}}{1-x}\right)^5
&=(1-x^{13})^5(1-x)^{-5}\\
&=\sum_{j=0}^5(-1)^j\binom{5}{j}x^{13j}\sum_{k=0}^\infty(-1)^k\binom{-5}{k}x^k\tag{1}\\
&=\sum_{j=0}^5(-1)^j\binom{5}{j}x^{13j}\sum_{k=0}^\infty\binom{k+4}{k}x^k\tag{2}\\
&=\sum_{m=0}^\infty\sum_{j=0}^5(-1)^j\binom{5}{j}\binom{m-13j+4}{m-13j}x^m\tag{3}\\
&=\sum_{m=0}^{60}\sum_{j=0}^5(-1)^j\binom{5}{j}\binom{m-13j+4}{m-13j}x^m\tag{4}\\
&=\sum_{m=0}^{60}\sum_{j=0}^{\lfloor m/13\rfloor}(-1)^j\binom{5}{j}\binom{m-13j+4}{4}x^m\tag{5}
\end{align}
$$
Explanation:
$(1)$: Binomial Theorem
$(2)$: $\binom{-n}{k}=(-1)^k\binom{n+k-1}{k}$
$(3)$: change variables $k\mapsto m-13j$
$(4)$: if $m\gt60$, then the sum in $j$ is an order $5$ difference of a degree $4$ polynomial
$(5)$: if $k\lt0$, then $\binom{n}{k}=0$; if $0\le k\le n$, then $\binom{n}{k}=\binom{n}{n-k}$
Plugging in $m=30$, we get the coefficient of $x^{30}$ to be
$$
\begin{align}
\sum_{j=0}^2(-1)^j\binom{5}{j}\binom{34-13j}{4}
&=\binom{5}{0}\binom{34}{4}-\binom{5}{1}\binom{21}{4}+\binom{5}{2}\binom{8}{4}\\
&=17151
\end{align}
$$
|
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|
Proving $\left(a+\frac{2}{a}\right)^2+\left(b+\frac{2}{b}\right)^2\ge \frac{81}{2}$ for all positive real $a,b$ such that $a+b=1$ I approached this problem in two different ways, but only one was successful. I'll post the latter as an answer, while here follows the first approach:
I expanded the squares:
$$a^2+\frac{4}{a^2}+4+b^2+\frac{4}{b^2}+4\ge\frac{81}{2} \\ a^2+b^2+\frac{4}{a^2}+\frac{4}{b^2}\ge \frac{65}{2}$$ and then multiplied both sides by $a^2b^2$ to get $$a^2b^2(a^2+b^2)+4(a^2+b^2)\ge\frac{65}{2}a^2b^2 \\ (a^2+b^2)(a^2b^2+4)\ge\frac{65}{2}a^2b^2, $$ which, combining $a+b=1$ and $(a+b)^2=a^2+b^2+2ab$, can be rewritten as $$(1-2ab)(a^2b^2+4)\ge\frac{65}{2}a^2b^2.\tag{$\star$}$$ Setting $c=ab$ makes $(\star)$ an inequality in one variable, but it didn't help me. Is this totally the wrong track? And, I was wondering if there are other approaches besides what I came up with.
|
To begin with, we note that equality occurs when $a=b=1/2$. Thus, assuming WLOG $a>b$ and using $b=1-a$, let us differentiate the LHS to show it is increasing: $$2\left(a+\frac{2}{a}\right)\left(1-\frac{2}{a^2}\right)+2\left(1-a+\frac{2}{1-a}\right)\left(-1+\frac{2}{(1-a)^2}\right)>0 $$ $$ \left(a+\frac{2}{a}\right)\left(1-\frac{2}{a^2}\right)>\left(1-a+\frac{2}{(1-a)}\right)\left(1-\frac{2}{(1-a)^2}\right).\tag{1}$$ Again, we differentiate to show the LHS of $(1)$ is increasing: $$\left(1-\frac{2}{a^2}\right)^2+\frac{4}{a^3}\left(a+\frac{2}{a}\right)>0$$ whereas its RHS is decreasing: $$-\left(1-\frac{2}{(1-a)^2}\right)^2-\frac{4}{(1-a)^3}\left(1-a+\frac{2}{1-a}\right)<0,$$ and we are done.
|
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|
Evaluate $\int_{-\pi}^\pi \! \cos(kx)\cos^n(x) \, \mathrm{d}x$ My question is:
Evaluate $$\int_{-\pi}^\pi \! \cos(kx)\cos^n(x) \, \mathrm{d}x$$ for $k=0,1,...,(n-1)$ and $n \in \mathbb{N}$.
I've tried integration by parts but without much success. Any help/trick for this integral? Thanks!
|
$$\begin{align} 2^n \cos^n x &= \left(e^{ix} + e^{-ix}\right)^n\\
&= e^{inx} + {n \choose 1}e^{i(n-2)x} + {n \choose 2}e^{i(n-4)x}+ \cdots+{n \choose 2}e^{-i(n-4)x}+{n \choose 1}e^{-i(n-2)x} + e^{-inx}\\
&=2\cos nx+ 2{n \choose 1}\cos(n-2)x+2{n \choose 2}\cos(n - 4)x + \cdots
\end{align}$$
multiplying by $\cos kx,$ we have
$$\begin{align}2^n \cos^n x\cos kx &= 2\cos nx\cos kx+ 2{n \choose 1}\cos(n-2)x\cos kx+{n \choose 2}\cos(n - 4)x\cos kx + \cdots \\
&=\cos(n+k)x +\cos(n-k)x+{n\choose 1}\left(\cos(n-2+k)x -\cos(n-2-k)x\right)\\ &+{n\choose 2}\left(\cos(n-4+k)x -\cos(n-4-k)x\right) +\cdots
\end{align}$$
only contribution to $\int_{-\pi}^{\pi} 2^n \cos^n x\cos kx\, dx$ are $2\pi{n \choose j}$ with $j = \frac{n+k}2$ and $-2\pi{n\choose j}$ with $j = \frac{n-k}2$ therefore $$\int_{-\pi}^{\pi} 2^n \cos^n x\cos kx\, dx = 2\pi\left({n\choose {\frac{n+k}2}} - {n\choose {\frac{n-k}2}}\right)$$
|
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|
Complex Numbers Question, IIT JEE [2006]. Please tell me whether I solved it properly? $Q.$The value of $\sum\limits_{k=1}^{10}(\sin{\frac{2k\pi}{11}-i\cos\frac{2k\pi}{11}})$ is-?
I solved it like this-
$\frac{\sum\limits_{k=1}^{10}(\cos{\frac{2k\pi}{11}+i\sin\frac{2k\pi}{11}})}{i}$
If we observe these are the roots of the equation $z^{11}=1$
So $1+z_1+z_2+...+z_{10}=1$ (De Moivre
s Theorem)
So $z_1+z_2+...+z_{10}=-1$
$-1=i^2$
So $\frac{i^2}{i}=i$
|
$$\frac1i\sum_{k=1}^{10}\left(\cos\frac{2k\pi}{11}+i\sin\frac{2k\pi}{11}\right)=-i\sum_{k=1}^{10} \left(e^{\frac{2\pi i}{11}}\right)^k=$$
$$=-ie^{\frac{2\pi i}{11}}\frac{e^{\frac{2\cdot10\pi i}{11}}-1}{e^{\frac{2\pi i}{11}}-1}=-ie^{\frac{2\pi i}{11}}\frac{e^{\frac{-2\pi i}{11}}-1}{e^{\frac{2\pi i}{11}}-1}=-i\frac{e^{\frac{2\pi i}{11}}-1}{1-e^{\frac{2\pi i}{11}}}=i$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Basis of a $2\times2$ matrix How would I find the basis for an arbitrary matrix W such that:
$$
W =\left\{
\begin{pmatrix} a & b \\ c & a +b +c\end{pmatrix} \ \big| \ \ a ,b ,c \in \mathbb{R}
\right\}
$$
|
Hint
$$\begin{pmatrix}a&b\\c&d\end{pmatrix}\in W\iff d=a+b+c\iff\begin{pmatrix}a&b\\c&d\end{pmatrix}=a\begin{pmatrix}1&0\\0&1\end{pmatrix}+b\begin{pmatrix}0&1\\0&1\end{pmatrix}+c\begin{pmatrix}0&0\\1&1\end{pmatrix}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that if $a\neq 1$, then $\sum_{k=0}^{n-1}ka^k = \frac{1-na^{n-1}+(n-1)a^n}{(1-a)^2}$
Need to show that if $a\neq 1$, then
$$\sum_{k=0}^{n-1}ka^k = \frac{1-na^{n-1}+(n-1)a^n}{(1-a)^2}$$
Here is my attempt:
$$\begin{aligned}
S & =\sum_{k=0}^{n-1}ka^k \\
&= \sum_{k=0}^{n}(k-1)(a^{k-1}) \\
\end{aligned}$$
from here we can say:
$$\begin{aligned}
S & =(1-1)(a^{1-1})+(2-1)(a^{2-1})+(3-1)(a^{3-1})+...+(n-1)(a^{n-1})\\
& =(0)(a^{0})+(1)(a^{1})+(2)(a^{2})+(3)(a^{3})+...+(n-1)(a^{n-1})\\
& =a+2a^2+3a^3+(n-1)(a^{n-1})\\
\end{aligned}$$
now compute $(a)S$:
$$\begin{aligned}
(a)S & =(a)(a)+(a)(2a^2)+(a)(n-1)(a^{n-1})\\
& =a^2+2a^3+(n-1)(a^{n-1+1})\\
& =a^2+2a^3+(n-1)(a^{n})\\
\end{aligned}$$
now compute $S-(a)S$:
$$\begin{aligned}
S-(a)S & = a+2a^2+3a^3+(n-1)(a^{n-1})-[a^2+2a^3+(n-1)(a^{n})]\\
& = a+2a^2+3a^3+(n-1)(a^{n-1})-a^2-2a^3-(n-1)(a^{n})\\
& = a+a^2+a^3+(n-1)(a^{n-1})-(n-1)(a^{n})\\
\end{aligned}$$
re-writing above:
$$(1-a)S = a+a^2+a^3+(n-1)(a^{n-1})-(n-1)(a^{n})$$
dividing both-sides by $(1-a)$:
$$S = \frac{a+a^2+a^3+(n-1)(a^{n-1})-(n-1)(a^{n})}{1-a}$$
Am I on the right track so far? If not, please point out where I went wrong.
Also, any examples would be very appreciated.
Update:
A lot of people have provided answers, however, no one has pointed out what is wrong with my current approach? I am looking to learn not to copy any answer. Please consider my "solution" and help direct me. Honestly I don't even want the final solution, I just want help understanding how to answer this. I really appreciate all the effort and time put.
|
let $$\begin{align}1+S_n &= 1 + a + 2a^2 + 3a^3 + \cdots + (n-1)a^{n-1}\\
aS_n &= a + a^2 +2a^3+ \cdots + (n-1)a^n\end{align}$$ subtracting the second equation from the first, we get $$(1-a)S_n + 1 = 1+a^2 + a^3 + \cdots+a^{n-1} - (n-1)a^n $$ rearranging this, we get $$(1-a)S_n+1+a+(n-1)a^n = 1+ a + a^2 + \cdots + a^{n-1} = \frac{1-a^n}{1-a} $$
can you find $S_n$ from the last equation?
|
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|
Generating function for recurrence in two variables Given characteristic polynomial for the recurrence in two variables (say $F(x,y)$)
$$
(y^2-1)^x
$$
and initial values can generating function for $F(x,y)$ be derived?
I know how to do it for a recurrence with one variable but have no idea how to do it in the case of two variables.
PS. The recurrence itself is unknown. Only characteristic polynomial and initial values are given.
EDIT. I can be wrong with the meaning of "characteristic polynomial" for a recurrence with two variables. The idea is that I've got a set of linear recurrences $F(x, y)$ with one variables each of which has the following characteristic polynomial: $F(1,y) = y^2 - 1, F(2,y) = (y^2 - 1)^2, F(3,y) = (y^2 - 1)^3, ...$
So I just generalized it by $F(x,y) = (y^2 - 1)^x$
|
We want $F(z,w)$ a generating function such that
$$F(z,w) = \sum_{x=0}^{\infty} \sum_{y=0}^{\infty} (y^2-1)^x z^xw^y \tag{1}$$
$(y^2-1)^x z^x$ is a power series.
$$ F(z,w) = \sum_{y=0}^{\infty} \frac{w^y}{1 - (y^2-1)z} \tag{2}$$
$$ F(z,w) = -\frac1{z} \sum_{y=0}^{\infty} \frac{w^y}{y^2 - 1 - \frac1{z}} \tag{3}$$
$$ F(z,w) = -\frac1{z} \sum_{y=0}^{\infty} \frac{w^y}{\left(y - \sqrt{1 + \frac1{z}}\right)\left(y + \sqrt{1 + \frac1{z}}\right)} \tag{4}$$
$$ F(z,w) = \frac1{2z\sqrt{1 + \frac1{z}}} \left[ \sum_{y=0}^{\infty} \frac{w^y}{y + \sqrt{1 + \frac1{z}}} - \sum_{y=0}^{\infty} \frac{w^y}{y - \sqrt{1 + \frac1{z}}} \right] \tag{5}$$
Now we need a generating function such that
$$G(w) = \sum_{y=0}^{\infty} \frac{w^y}{y + c} \tag{6}$$
$$\sum_{y=0}^{\infty} w^y = \frac1{1-w} \tag{7}$$
$$w^{c-1}\sum_{y=0}^{\infty} w^y = \sum_{y=0}^{\infty} w^{y+c-1} = \frac{w^{c-1}}{1-w} \tag{8}$$
$$\int\sum_{y=0}^{\infty} w^{y+c-1} dw = \sum_{y=0}^{\infty} \frac{w^{y+c}}{y+c} = \int \frac{w^{c-1}}{1-w} dw \tag{9}$$
$$w^{-c}\sum_{y=0}^{\infty} \frac{w^{y+c}}{y+c} = \sum_{y=0}^{\infty} \frac{w^{y}}{y+c} = w^{-c}\int \frac{w^{c-1}}{1-w} dw \tag{10}$$
From wolfram +c and wolfram -c
$$\int \frac{w^{c-1}}{1-w} dw = \frac{w^c (c w \, _{2}F_{1}(1, c+1, c+2, w)+c+1)}{c (c+1)}+constant \tag{11}$$
Where $_{2}F_{1}$ is a hypergeometric function.
$$ F(z,w) = \frac1{2\,z\,c} \left[ \frac{c w \, _{2}F_{1}(1, c+1, c+2, w)+c+1}{c (c+1)} + \frac{c w \, _{2}F_{1}(1, -c+1, -c+2, w)+c-1}{c (c-1)}\right] \tag{12}$$
Where $c = \sqrt{1 + \frac1{z}}$
Sanity check wolfram
|
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|
Solution verification: Prove by induction that $a_1 = \sqrt{2} , a_{n+1} = \sqrt{2 + a_n} $ is increasing and bounded by $2$ I have the following recursive relation (sequence):
\begin{align}
a_1 = \sqrt{2}, \quad a_{n+1} = \sqrt{2 + a_n}
\end{align}
My Try:
I'm a little skeptical of my manipulations near the end but it looks like it works out.
Base Case:
Let $n=1$ then
\begin{align}
&a_2 = \sqrt{2 + a_1} \\
&a_2 < 2 \\
&\sqrt{2 + \sqrt{2}} < 2
\end{align}
The base case holds.
Induction hypothesis:
Let $n=k$
$$a_1 = \sqrt{2} \quad a_{k+1} = \sqrt{2 + a_k} \quad a_{k+1} = \sqrt{2}$$
Induction Step:
Now we have to prove that $a_{k+2} < 2$. Let $n=k+1$.
\begin{align}
a_{k+2} &= \sqrt{2 + a_{k+1}} \\
\implies a_{k+2} &= \sqrt{2 + \sqrt{2 + a_k}} \\
\end{align}
Now we have to show that $a_{k+2} < 2$.
\begin{align}
a_{k+2} &< 2\\
\sqrt{2 + \sqrt{2 + a_k}} &< 2\\
2 + \sqrt{2 + a_k} &< 4 \\
\sqrt{2 + a_k} &< 2 \\
\end{align}
Q.E.D
Are my steps correct?
Thanks for your time!
|
Your base case and inductive step are both overly complicated.
Base case: We have $a_1 = \sqrt{2}$, which is less than $2$.
Inductive step: Assume we have $a_k < 2$
Now $a_{k+1} = \sqrt{2 + a_k} < \sqrt{2 + \sqrt{2}} < \sqrt{2 + 2} = 2$.
It may also be useful to remark that the sequence is strictly positive, so it is also bounded below.
The error you are making is that your base case proves $a_2$ but it should prove $a_1$ and your inductive step proves $a_{k+2}$ when you should be proving $a_{k+1}$.
Also, I'm a bit confused by your inductive step. There's no need to write the definition of the sequence in the inductive step and I'm confused why you have $a_{k+1} = \sqrt{2}$. Did you mean $a_{k+1} < \sqrt{2}$? Because if so, your inductive step (if you are assuming it for $n = k$) should say $a_k < \sqrt{2}$. Indeed, the statement you are trying to prove is $\{P_k : \text{each } a_k \text{ is bounded by } 2\}$. So in the inductive step you assume $P_n$ which is "$a_n$ is bounded by $2$"
|
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|
How find this minimum Help me!
Let $x,y,z\ge0$ such that: $xy+yz+zx=1$.
Find the minimum value of: $A=\dfrac{1}{x^2+y^2}+\dfrac{1}{y^2+z^2}+\dfrac{1}{z^2+x^2}+\dfrac{5}{2}(x+1)(y+1)(z+1)$
I found minimum value of $A$ is $\dfrac{25}{2}$ iff $(x,y,z)=(1,1,0)$ or any permutation. But I can't solve that.
|
first to prove:
$\dfrac{1}{x^2+y^2}+\dfrac{1}{y^2+z^2}+\dfrac{1}{z^2+x^2} \ge \dfrac{5(1+xyz)}{2}$ ...(1)
$A \ge \dfrac{25}{2} \implies x+y+z+2xyz\ge 2 $...(2)
(2) is easy to prove. But (1) is hard. I have a ugly solution and I think to wait some days to post it if there is no better way.
|
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|
Integral using trigonometric substitution I'd like to ask for feedback on my calculation for this integral:
$$\int{\frac{dx}{2-\cos{x}}}$$
Using half-angle substitution:
$$t = \tan{\frac{x}{2}}$$
$$\cos{x} = \frac{1-t^2}{1+t^2}$$
$$dx = \frac{2dt}{1+t^2}$$
So
$$\int{\frac{dx}{2-\cos{x}}} = \int{\frac{1}{2-\frac{(1-t^2)}{1+t^2}}}\cdot\frac{2dt}{1+t^2}$$
$$= 2\,\,\int{\frac{1}{2(1+t^2)-(1-t^2)}}\,\,dt$$
$$= 2\,\,\int{\frac{1}{3(\frac{1}{3}+t^2)}}\,\,dt$$
$$= \frac{2}{3}\,\,\int{\frac{1}{(\frac{1}{\sqrt(3)^2})+t^2}}\,\,dt$$
Integrating the expression
$$=\frac{2}{3}\frac{1}{\sqrt{3}}\cdot \tan^{-1}{(\frac{t}{\sqrt{3}})}+c$$
$$=\frac{2}{3\sqrt{3}}\cdot \tan^{-1}{(\frac{1}{\sqrt{3}}\tan{(\frac{x}{2})})}+c$$
I believe the last expression should have $\frac{2}{\sqrt{3}}$ and not $\frac{2}{3\sqrt{3}}$, but in my calculation this appears correct. Any feedback would be greatly appreciated!
|
Note also that it is $\tan^{-1}(\sqrt 3t)$, not $\tan^{-1}\left(\frac{t}{\sqrt 3}\right)$.
Setting $t=\frac{1}{\sqrt 3}\tan\theta$ gives you
$$\begin{align}\frac 23\int\frac{1}{\left(\frac{1}{\sqrt 3}\right)^2+t^2}dt&=\frac 23\int\frac{1}{\frac 13+\frac 13\tan^2\theta}\cdot\frac{d\theta}{\sqrt 3\cos^2\theta}\\&=\frac 23\cdot\frac{1}{\frac 13\cdot \sqrt 3}\int d\theta\\&=\frac{2}{\sqrt 3}\tan^{-1}(\sqrt 3 t)+C\\&=\frac{2}{\sqrt 3}\tan^{-1}\left(\sqrt 3\tan\frac x2\right)+C\end{align}$$
|
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|
Find the type of triangle from equation. In triangle $ABC$, the angle($BAC$) is a root of the equation
$$\sqrt{3}\cos x + \sin x = \frac{1}{2}.$$
Then the triangle $ABC$ is
a) obtuse angled
b) right angled
c) acute angled but not equilateral
d) equilateral.
Thanks in advance.
|
$\frac{\sqrt{3}}{2}\cdot \cos x + \frac{1}{2}\sin x = \frac{1}{4}$
$\sin ({x + \frac{\pi}{3}}) = \frac{1}{4}$
|
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|
Calculating the work integral of a vector field. Let F be the vector field
F = $<\dfrac{1}{x} + 2x + y, \ \ \dfrac{1}{y} + x + 1>$ .
Compute the work integral $\int_{c} F dr$ where C is the path
r(t) = (1 + cos t)i + (2 + sin t) j $ \ \ \ \ -\dfrac{\pi}{2} \leq t \leq \dfrac{\pi}{2}$.
I am very confused as to how to approach this problem. I understand that we simply have to parameterize F, find v(t), and dot product both of them, but given these expressions that would be very tedious. Is there a simpler way to approach this problem?
|
Fundamental Theorem for Line Integrals
$$\eqalign{
& W = \int\limits_C {F \bullet dr} \cr
& - - - - Test\;For\;Gradient\;Field \cr
& F = \left( {{1 \over x} + 2x + y\,,\,{1 \over y} + x + 1} \right) = \left( {M\;,\;N} \right) \cr
& F = \nabla f\;\;if\;\;\;{\partial _y}M = {\partial _x}N \cr
& {\partial _y}M = 1 \cr
& {\partial _x}N = 1 \cr
& W = \int\limits_C {F \bullet dr} = \int\limits_C {\nabla f \bullet dr} = f({p_1}) - f({p_0}) \cr
& \nabla f = \left( {{1 \over x} + 2x + y\,,\,{1 \over y} + x + 1} \right) = \left( {{\partial _x}f\;,\;{\partial _y}f} \right) \cr
& - - - - Find\;\;f(x,y) \cr
& \left\{ \matrix{
{\partial _x}f = {1 \over x} + 2x + y \hfill \cr
{\partial _y}f = {1 \over y} + x + 1 \hfill \cr} \right. \cr
& \int {{\partial _x}fdx = \ln (x) + {x^2}} + yx + g(y) = f(x,y) \cr
& {\partial _y}f = x + g'(y) \cr
& g'(y) = {1 \over y} + 1 \cr
& g(y) = \ln (y) + y + c \cr
& f(x,y) = \ln (x) + {x^2} + yx + \ln (y) + y + c \cr
& - - - - Find\;\;{p_1}\;and\;{p_0} \cr
& x = 1 + \cos t \cr
& y = 2 + \sin t \cr
& {{ - \pi } \over 2} \le t \le {\pi \over 2} \cr
& 1 \le x \le 1 \cr
& 1 \le y \le 3 \cr
& {p_0} = (1,1) \cr
& {p_1} = (1,3) \cr
& - - - - Finding\;The\;Work \cr
& W = \int\limits_C {F \bullet dr} = \int\limits_C {\nabla f \bullet dr} = f({p_1}) - f({p_0}) \cr
& W = \left( {\ln (1) + {1^2} + 3 + \ln (3) + 3} \right) - \left( {\ln (1) + {1^2} + 1 + \ln (1) + 1} \right) = \ln (3) + 4 \cr} $$
|
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|
How many planes are there with the desired property? Two points $A$ and $B$ are given, for example $A(2/5/7)$ and $B(4/11/16)$.
The object is to find all planes containing the points $A$ and $B$ with distance
$2$ to the origin.
I tried the hesse-normal-form, but since only one direction-vector is given,
the normal-vector cannot be calculated.
I also tried to add a third point $C(c_1,c_2,c_3)$ and set up the plane
equation, but this also led to nowhere.
Can anyone determine the desired planes and, for each one, an equation ?
|
Let plane $\alpha$ be $p(X)=Ax+By+Cz+D=0$, then $\rho(\alpha,X)=\frac{|p(X)|}{\sqrt{A^2+B^2+C^2}}$.
So we put $A^2+B^2+C^2=1$ and get equations
$D=\pm 2$;
$2A+5B+7C+D=0$;
$4A+11B+16C+D=0$.
$2A+5B+7C+D=0 \Leftrightarrow A=-\frac{5}{2}B-\frac{7}{2}C\mp1$
Plugging this into $4A+11B+16C\pm 2=0$ we get
$-10B-14C\mp 4 +11B+16C\pm 2 =0$.
$B+2C\mp 2=0$, $B=-2(C\mp 1)$, plug it back into $A=-\frac{5}{2}B-\frac{7}{2}C\mp1$ we get $A=5(C\mp 1)-\frac{7}{2}C\mp1=\frac{3}{2}C\mp 6$, and plug them all into $A^2+B^2+C^2=1$:
$(\frac{3}{2}C\mp 6)^2 + (2(C\mp 1))^2+C^2=1$,
$\frac{9}{4}(C^2\mp 8C +16)+4(C^2\mp 2C + 1) + C^2=1$,
$\frac{29}{4}C^2\mp 26C +39=0$
$\emptyset$
And the answer there is no such plane. :)
|
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|
Simplifying Square Roots Frustration Okay, I'm really frustrated with this.
So, when you have $3 \sqrt 5 + 5 \sqrt 5$,
you get $8\sqrt5$, right?
Okay, so what do I do for here:
$\sqrt{11} - 3 \sqrt{11}$
Is it just $-3 \sqrt{11}$ ?
What about for $6 \sqrt 2 + 4 \sqrt{50}$ ?
Do I multiply the $6 \sqrt{2}$ so that I can make what's inside of the square root equal to $50$?
|
*
*$\sqrt{11}-3\sqrt{11}=\color{red}{1}\cdot\sqrt{11}-3\sqrt{11}=-2\sqrt{11}.$
*$6\sqrt 2+4\sqrt{50}=6\sqrt 2+4\color{blue}{\sqrt{2\cdot 5^2}}=6\sqrt{2}+4\cdot \color{blue}{5\sqrt 2}=6\sqrt 2+20\sqrt 2=26\sqrt 2$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Calculate a determinant. Let $a_{1}, \cdots, a_{n}$ and $b$ be real numbers. I like to know the determinant of the matrix
$$\det\begin{pmatrix}
a_{1}+b & b & \cdots & b \\
b & a_{2}+b & \cdots & b \\
\vdots & \vdots & \ddots & \vdots \\
b & b & \cdots & a_{n}+b
\end{pmatrix}=?$$
I guess the answer is $$a_{1} \cdots a_{n}+ \sum_{i=1}^{n} a_{1} \cdots a_{i-1} b a_{i+1}\cdots a_{n} $$ after some direct calculations for $n=2,3$. The question is how to calculate it for general $n$. Thanks!
|
Applying multilinearity and an inductive argument:
$$\begin{vmatrix}
a_{1}+b & b & \cdots & b \\
b & a_{2}+b & \cdots & b \\
\vdots & \vdots & \ddots & \vdots \\
b & b & \cdots & a_{n}+b
\end{vmatrix}=\begin{vmatrix}
a_{1} & b & \cdots & b \\
0 & a_{2}+b & \cdots & b \\
\vdots & \vdots & \ddots & \vdots \\
0 & b & \cdots & a_{n}+b
\end{vmatrix}+\begin{vmatrix}
b & b & \cdots & b \\
b & a_{2}+b & \cdots & b \\
\vdots & \vdots & \ddots & \vdots \\
b & b & \cdots & a_{n}+b
\end{vmatrix}$$
$$=\begin{vmatrix}
a_{1} & 0 & \cdots & b \\
0 & a_{2} & \cdots & b \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & a_{n}+b
\end{vmatrix}+
\begin{vmatrix}
a_{1} & b & \cdots & b \\
b & b & \cdots & b \\
\vdots & \vdots & \ddots & \vdots \\
b & b & \cdots & a_{n}+b
\end{vmatrix}+
\begin{vmatrix}
b & 0 & \cdots & b \\
b & a_{2} & \cdots & b \\
\vdots & \vdots & \ddots & \vdots \\
b & 0 & \cdots & a_{n}+b
\end{vmatrix}+
\begin{vmatrix}
b & b & \cdots & b \\
b & b & \cdots & b \\
\vdots & \vdots & \ddots & \vdots \\
b & b & \cdots & a_{n}+b
\end{vmatrix}=\ldots$$
Note the very last determinant is zero.
For example:
$$\begin{vmatrix}a_1+b&b&b\\
b&a_2+b&b\\
b&b&a_3+b\end{vmatrix}=
\begin{vmatrix}a_1&b&b\\
0&a_2+b&b\\
0&b&a_3+b\end{vmatrix}+
\begin{vmatrix}b&b&b\\
b&a_2+b&b\\
b&b&a_3+b\end{vmatrix}=$$
$$=\begin{vmatrix}a_1&0&b\\
0&a_2&b\\
0&0&a_3+b\end{vmatrix}+
\begin{vmatrix}a_1&b&b\\
0&b&b\\
0&b&a_3+b\end{vmatrix}+
\begin{vmatrix}b&0&b\\
b&a_2&b\\
b&0&a_3+b\end{vmatrix}+
\begin{vmatrix}b&b&b\\
b&b&b\\
b&b&a_3+b\end{vmatrix}=\ldots$$
$$=a_1a_2a_3+b(a_1a_2+a_1a_3+a_2a_3)$$
and you were right.
|
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|
How prove $a + b +c + \frac{1}{abc} \geq 4\sqrt{3}$? Let $a, b, c >0$ and $a^2+b^2+c^2 =1$ How prove $a + b +c + \frac{1}{abc} \geq 4\sqrt{3}$?
|
Let's consider the function
$$f(a,b,c) = a+b+c+\frac{1}{abc}.$$
We want to minimize it.
Then
$$f'_a = 1 - \frac{1}{(abc)a}, f'_a = 1 - \frac{1}{(abc)b}, f'_c = 1 - \frac{1}{(abc)c}$$
Imposing all of them to be $0$, we have:
$$\left\{\begin{array}{l}(abc)a = 1\\(abc)b = 1\\(abc)c = 1\end{array}\right.$$
This means that $a=b=c$. Now, consider the function:
$$f(x) = 3x+\frac{1}{x^3},$$
where $a=b=c=x$.
Since $a^2+b^2+c^2 = 3x^2 = 1$, then $x = \frac{\sqrt{3}}{3} > 0$ and get that the minimum is
$$f\left(\frac{\sqrt{3}}{3}\right) = \sqrt{3}+3\sqrt{3} = 4 \sqrt{3}$$
Addition
It is easy to show that the function $f(a,b,c)$ is convex and thus it has a relative minimum.
|
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|
Proving the existence of a square root of $ -1_{A} $ in a $ 2 $-dimensional unital algebra $ A $ over $ \Bbb{R} $. Suppose that $ A $ is a $ 2 $-dimensional unital algebra over $ \Bbb{R} $ with a basis $ \{ 1_{A},u \} $, and assume that $ A $ does not have any zero divisors. Show that $ A $ contains an element $ b $ such that $ b^{2} = - 1_{A} $.
I’m sure this is straightforward. I’ve tried thinking about it in terms of a bijective map to show the existence of the element and by writing
$$
b = \lambda_{1} 1_{A} + \lambda_{2} u, \qquad \lambda_{1},\lambda_{2} \in \Bbb{R},
$$
and then evaluating $ b^{2} $, but I didn’t get anywhere useful.
Any help appreciated.
|
Not sure how old Frobenio handled this one, but here's how I do it:
Since $A = \text{span}(\{ 1_A, u \})$, we have
$u^2 = \alpha 1_A + \beta u, \;\; \alpha, \beta \in \Bbb R. \tag{1}$
Then
$u^2 - \beta u = \alpha 1_A, \tag{2}$
or, "completing the square",
$u^2 - \beta u + \dfrac{\beta^2}{4} 1_A = \alpha 1_A + \dfrac{\beta^2}{4} 1_A = (\alpha + \dfrac{\beta^2}{4})1_A, \tag{3}$
or
$(u - \dfrac{\beta}{2}1_A)^2 = (\alpha + \dfrac{\beta^2}{4})1_A. \tag{4}$
If
$\alpha + \dfrac{\beta^2}{4} \ge 0, \tag{5}$
there exists $\gamma \in \Bbb R$ such that
$\gamma^2 = \alpha + \dfrac{\beta^2}{4}, \tag{6}$
whence (4) becomes
$(u - \dfrac{\beta}{2}1_A)^2 = \gamma^2 1_A \tag{7}$
or
$(u - \dfrac{\beta}{2}1_A)^2 - \gamma^2 1_A = 0 \tag{8}$
or
$(u -\dfrac{\beta}{2} 1_A - \gamma 1_A)(u - \dfrac{\beta}{2}1_A + \gamma 1_A) = 0 \tag{9}$
or
$(u -(\dfrac{\beta}{2} + \gamma)1_A)(u - (\dfrac{\beta}{2} - \gamma) 1_A) = 0; \tag{10}$
neither factor in (10) is $0$ by the linear independence of the basis $\{1_A, u \}$; thus each is a zero divisor in $A$. If we rule out the existence of zero divisors in $A$, then we must have
$\alpha + \dfrac{\beta^2}{4} < 0; \tag{11}$
in this case we may find $\gamma \in \Bbb R$ such that
$-\gamma^2 = \alpha + \dfrac{\beta^2}{4} \tag{12}$
whence
$(u - \dfrac{\beta}{2} 1_A)^2 = -\gamma^2; \tag{13}$
or
$(\dfrac{u - \dfrac{\beta}{2}1_A}{\gamma})^2 = \dfrac{(u - \dfrac{\beta}{2} 1_A)^2}{\gamma^2} = -1_A, \tag{14}$
as sought. QED.
|
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|
The complex roots of a biquadratic polynom In my recent post I have a problem with the following function: $x^4-4x^2+16$, and what I need is to find the complex roots.
Here is my answer:
First step, I make the substitution $x^2=y$ which involving $y^2-4y^2+16$, with $x^2=2\pm i\sqrt{12}$. Therfore:
$x^4-4x^2+16=(x^2-2-i\sqrt{12})(x^2-2+i\sqrt{12})
=(x+\sqrt{2+i\sqrt{12}})(x+\sqrt{2-i\sqrt{12}})(x-i\sqrt{2+i\sqrt{12}})(x-\sqrt{2-i\sqrt{12}})$
factorized irreducible over $\mathbb{R}\subset\mathbb{C}$.
Here is author's answer:
$x{_1,_2,_3,_4}=\pm(\sqrt{3}\pm i)$.
I realize the problem is at factorization: $x^4-4x^2+16=(x^2+bx+c)(x^2+dx+e)$.
If I pair the complex roots from author's answer I'll obtain:
$x^4-4x^2+16=(x+\sqrt{3}+i)(x+\sqrt{3}-i)(x-\sqrt{3}-i)(x-\sqrt{3}+i)=(x^2+\sqrt{12}x+4)(x^2-\sqrt{12}x+4)$
factorized irreducible over $\mathbb{Q}\subset\mathbb{R}$, but reducible over $\mathbb{C}$.
My final question: Why $x_1{,_2,_3,_4}=\pm\sqrt{2\pm i\sqrt{12}}$ isn't enough to make the assessment that it is the final answer ? How can I get author's answer $x{_1,_2,_3,_4}=\pm(\sqrt{3}\pm i)$ begining from my complex roots found ? Why we obtain 2 different form of complex roots, who's belong in the same field $\mathbb{C}$ when we change factorization ?
|
Note that your answer gives that same result as author's answer, but the more important thing to pay attention with is when you use $\sqrt z$ with $z$ a complex numbers this is not defind so in I would say that your answer is incomplete because you did not find the complex square roots of $2\mp i\sqrt {12}$, and when you did this you will find the same result as author's.
Another way to see things is that any complex number is of the form $x+iy$ where $x,y\in \Bbb R$ which is the canonical form. and your answer does not give the roots of this form.
Maybe we can use that:
$$\begin{align}x^4-4x^2+16&=x^2-2\cdot4x^2+16+4x^2\\
&=(x^2-4)^2+4x^2\\
&=(x^2-2ix-4)(x^2+2ix-4)\\
&=((x-i)^2-3)((x+i)^2-3)\\
&=(x-i-\sqrt3)(x-i+\sqrt3)(x+i-\sqrt3)(x+i+\sqrt3)\end{align}$$
this is how the author did it, but you can simplify your expression, for instance:
$$\pm\sqrt{2\pm i\sqrt{12}}=\mp \sqrt{2+\mp 2i\sqrt 3}=\mp \sqrt{3^2+\mp 2i\sqrt 3 +i^2}=\mp \sqrt{(i\mp\sqrt 3)^2}=\pm(\sqrt{3}\pm i)$$
|
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|
Solving $\int dx {\sqrt{x^2+a}} e^{-A x^2} erf \left( c(x-b) \right)$ I got as far as:
$$\int dx {\sqrt{x^2+a}} e^{-A x^2} erf \left( c(x-b) \right) $$
$$=\frac{2}{\sqrt{\pi}} \int dx \int^{c(x-b)}_0 dy {\sqrt{x^2+a}} e^{-A x^2 - y^2}$$
$$=\frac{-2 c}{\sqrt{\pi}} \int db \int dx {\sqrt{x^2+a}} e^{-A x^2 - c^2 (x-b)^2} $$
$$=\frac{-2 c}{\sqrt{\pi}} \int db e^{-b^2 \left\{ c^2 + \frac{c^4}{A+c^2} \right\} } \int dx {\sqrt{x^2+a}} e^{ - \left(A+c^2 \right) \left( x-\frac{b c^2}{A+c^2} \right)^2} $$
|
Hint:
$\int\sqrt{x^2+a}e^{-Ax^2}\text{erf}(c(x-b))~dx$
$=\dfrac{2}{\sqrt\pi}\int\sum\limits_{n=0}^\infty\dfrac{(-1)^nc^{2n+1}(x-b)^{2n+1}\sqrt{x^2+a}e^{-Ax^2}}{n!(2n+1)}~dx$
$=\dfrac{2}{\sqrt\pi}\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{2n+1}\dfrac{(-1)^{n-k+1}(2n)!b^{2n-k+1}c^{2n+1}x^k\sqrt{x^2+a}e^{-Ax^2}}{n!k!(2n-k+1)!}~dx$
|
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|
Residue $\frac{e^z}{z^3\sin(z)}$ I want to find the residue of $$\frac{e^z}{z^3\sin(z)}$$ and get $$ \frac 1 {3!} \lim_{z \to 0} \left( \frac{d^3}{dz^3} \left(\frac{ze^z}{\sin(z)} \right)\right) = \frac{1}{3}$$
Can anyone confirm this? I tried using the Laurent Series, but I didn't know how to compute it.
Oh, I was able to compute the Laurent Series and confirm it was $\frac{1}{3}$
I don't know how to close this question though!
|
Near zero,
$$ e^z = 1 + z + \frac{z^2}{2} + \frac{z^3}{6} + O(z^4). $$
Also,
$$\sin{z}=z-\frac{1}{6}z^3 + O(z^5),$$
so the denominator is
$$ z^{-4} \left(1-\frac{1}{6}z^2+ O(z^4)\right)^{-1} = \frac{1}{z^4}\left( 1-\frac{z^2}{6} +O(z^4) \right), $$
using the binomial theorem.
Hence the Laurent series has principal part
$$ \frac{e^z}{z^3\sin{z}} = \frac{1}{z^4}\left( 1 + z + \frac{z^2}{2} + \frac{z^3}{6} + O(z^4) \right)\left( 1+\frac{z^2}{6} +O(z^4) \right) \\
= \frac{1}{z^4} + \frac{1}{z^3} + \frac{2}{3z^2} + \frac{1}{3z} + O(1), $$
and the residue is $1/3$.
|
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|
Show that $\dfrac{m - \sqrt[3]{2}}{\sqrt[3]{3}}$ is an algebraic integer. Let $m$ be an integer such that $m \equiv 2 \pmod 3$. Show that the number $$\dfrac{m - \sqrt[3]{2}}{\sqrt[3]{3}}$$ is an algebraic integer.
The usual technique, doing $x = \dfrac{m - \sqrt[3]{2}}{\sqrt[3]{3}}$ and trying to find an algebraic expression in terms of $x$ seems not to work in this case (at least I couldn't do it). Can anyone help me? This is a question from an old exam.
|
$x\sqrt[3]{2} = m - \sqrt[3]{2} \to \sqrt[3]{2} = \dfrac{m}{x+1} \to 2 = \dfrac{m^3}{x^3+3x^2+3x+1} \to 2x^3+6x^2+6x+2-m^3=0$. Thus $x$ is algebraic number.
|
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|
Find the standard matrix representation of the linear transformation T in M2,2 let $T: M_{2,2} \rightarrow M_{2,2}$ be a linear transformation defined by:
$$T \left(\begin{bmatrix}
a & b\\
c & d\\
\end{bmatrix}\right) = \begin{bmatrix}a + b& b + a \\ c - d&d+b\end{bmatrix}
$$
Find the standard matrix for $S$ by using the standard basis of $M_{2,2}$
This is what i've done so far:
$$T \left(\begin{bmatrix}
1 & 0\\
0 & 0\\
\end{bmatrix}\right) = \begin{bmatrix}1& 1 \\ 0&0\end{bmatrix} = x_1
$$
$$T \left(\begin{bmatrix}
0 & 1\\
0 & 0\\
\end{bmatrix}\right) = \begin{bmatrix}1& 1 \\ 0&1\end{bmatrix} = x_2
$$
$$T \left(\begin{bmatrix}
0 & 0\\
1 & 0\\
\end{bmatrix}\right) = \begin{bmatrix}0&0\\ 1&0\end{bmatrix} = x_3
$$
$$T \left(\begin{bmatrix}
0 & 0\\
0 & 1\\
\end{bmatrix}\right) = \begin{bmatrix}0& 0\\ -1&1\end{bmatrix} = x_4
$$
But I'm not too sure where to go from here. All I know that these need to somehow become part of a larger matrix
|
Let $v_1,v_2,v_3,v_4$ denote the standard basis. Note that
$$
T(v_1) = T \left(\begin{bmatrix}
1 & 0\\
0 & 0\\
\end{bmatrix}\right) = \begin{bmatrix}1& 1 \\ 0&0\end{bmatrix} =
1v_1 + 1v_2 +0v_3 + 0v_4\\
T(v_2) = T \left(\begin{bmatrix}
0 & 1\\
0 & 0\\
\end{bmatrix}\right) = \begin{bmatrix}1& 1 \\ 0&1\end{bmatrix} =
1v_1 + 1v_2 + 0v_3 + 1v_4\\
T(v_3) = T \left(\begin{bmatrix}
0 & 0\\
1 & 0\\
\end{bmatrix}\right) = \begin{bmatrix}0&0\\ 1&0\end{bmatrix} =
0v_1 + 0v_2 + 1v_3 + 0v_4\\
T(v_4) = T \left(\begin{bmatrix}
0 & 0\\
0 & 1\\
\end{bmatrix}\right) = \begin{bmatrix}0& 0\\ -1&1\end{bmatrix} =
0v_1 + 0v_2 + (-1)v_3 + 1v_4
$$
It follows that the matrix of the transformation is given by
$$
S =
\pmatrix{
1&1&0&0\\
1&1&0&0\\
0&0&1&-1\\
0&1&0&1
}
$$
|
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|
Find the value of : $\lim_{x\to\infty}\frac{\sqrt{x-1} - \sqrt{x-2}}{\sqrt{x-2} - \sqrt{x-3}}$ I'm trying to solve evaluate this limit
$$\lim_{x\to\infty}\frac{\sqrt{x-1} - \sqrt{x-2}}{\sqrt{x-2} - \sqrt{x-3}}.$$
I've tried to rationalize the denominator but this is what I've got
$$\lim_{x\to\infty}(\sqrt{x-1} - \sqrt{x-2})({\sqrt{x-2} + \sqrt{x-3}})$$
and I don't know how to remove these indeterminate forms $(\infty - \infty)$.
EDIT: without l'Hospital's rule (if possible).
|
\begin{align}
\frac{\sqrt{x-1} - \sqrt{x-2}}{\sqrt{x-2} - \sqrt{x-3}}
&=\frac{\sqrt{x-1} - \sqrt{x-2}}{\sqrt{x-2} - \sqrt{x-3}}\,\,
\frac{\sqrt{x-1} + \sqrt{x-2}}{\sqrt{x-1} + \sqrt{x-2}}\,\,
\frac{\sqrt{x-2} + \sqrt{x-3}}{\sqrt{x-2} + \sqrt{x-3}}\\ \ \\
&=\frac{x-1-(x-2)}{x-2-(x-3)}\,\frac{\sqrt{x-2} + \sqrt{x-3}}{\sqrt{x-1} + \sqrt{x-2}}=\frac{\sqrt{x-2} + \sqrt{x-3}}{\sqrt{x-1} + \sqrt{x-2}}\\ \ \\
&=\frac{\sqrt{1-2/x}+\sqrt{1-3/x}}{\sqrt{1-1/x}+\sqrt{1-2/x}}\to\frac22=1
\end{align}
|
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|
Solve this equation: $\log_3(3-2\cdot3^{x+1})=2+2x$ Solve this equation: $\log_3(3-2\cdot3^{x+1})=2+2x$. I put $(2+2x)^3=3-2\cdot3^{x+1}$. But I don't know how to go on.
|
We have
$$3-2\cdot 3^{x+1} = 3^{2+2x}$$
Setting $3^{x+1}=t$, we obtain
$$3-2t = t^2 \implies t^2+2t-3 =0 \implies (t+3)(t-1) = 0 \implies t \in \{-3,1\}$$
Hence, if we want $x \in \mathbb{R}$, we have
$$3^{x+1} = 1 \implies x+1 =0 \implies x=-1$$
|
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|
How to solve $\int_{-1}^{1} (x^{4/3} + 4x^{1/3}) dx$? I started by integrating it this way:
$$(\frac{3}{7} x^{7/3} + \frac{16}{4}x^{4/3})$$
What is wrong with it?
|
The answer is
$\frac{6}{7}$
Integrating,
$ \int^1_{-1} x^{4/3} 4 x^{1/3} \,dx = (\frac{3}{7}x^{7/3} + 3 x^{4/3})^{1}_{-1} = \frac{3}{7} + 3 + \frac{3}{7} - 3 = \frac{6}{7}. $
In the OP as it reads now, it appears that you increased the power on the second term and multiplied by the new power instead of dividing.
|
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|
If $a+b+c\le 1$ then $3(a+b+c)-(a^2+b^2+c^2-ab-bc-ac)\ge (\sqrt{a}+\sqrt{b}+\sqrt{c})^2$
Let $a,b,c\ge 0$ such that $a+b+c\le 1$, prove that
$$3(a+b+c)-(a^2+b^2+c^2-ab-bc-ac)\ge (\sqrt{a}+\sqrt{b}+\sqrt{c})^2\tag{1}$$
I conjecture:
Let $a_{i}\ge 0$, $i=1,2,\cdots$, $a_{1}+a_{2}+\cdots+a_{n}\le 1$, $n\ge 3$,then
$$n(a_{1}+a_{2}+\cdots+a_{n})-(a^2_{1}+a^2_{2}+\cdots+a^2_{n}-a_{1}a_{2}-a_{2}a_{3}-\cdots-a_{n}a_{1})\ge (\sqrt{a_{1}}+\sqrt{a_{2}}+\cdots+\sqrt{a_{n}})^2$$
This inequality is stronger than Cauchy-Schwarz inequality,because
$$(a^2+b^2+c^2-ab-bc-ac)\ge 0$$
Applying Cauchy-Schwarz inequality
$$3(a+b+c)\ge(\sqrt{a}+\sqrt{b}+\sqrt{c})^2$$
How to prove the required statement (1)?
|
Since $a+b \leqslant 1 $ by Cauchy - Schwarz we get $$\sqrt{a} +\sqrt{b} \leq \sqrt{2}\cdot \sqrt{a+b} \leqslant \sqrt{2} $$
hence $$\frac{(\sqrt{a} +\sqrt{b})^2}{2} \leqslant 1$$
multiplying both sides of the above inequality by $(\sqrt{a} -\sqrt{b} )^2 $ we get $$\frac{(a-b)^2 }{2} \leqslant (\sqrt{a} -\sqrt{b})^2 . $$
Now we have
\begin{align}
3 (a+b+c ) -(\sqrt{a} +\sqrt{b} +\sqrt{c})^2 &= 2a +2b +2c -2\sqrt{ab} -2\sqrt{ac} -2\sqrt{bc} \\
&= (\sqrt{a} -\sqrt{b} )^2 +(\sqrt{a} -\sqrt{c} )^2 +(\sqrt{c} -\sqrt{b} )^2\\
&\geqslant \frac{1}{2} \cdot \left ( (a-b)^2 + (a-c)^2 +(c-b)^2 \right)\\
&= a^2 +b^2 +c^2 -ab -ac -ab
\end{align}
what completes the proof.
|
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|
Finding pole order and calculating residue of: $\frac1{z-\tan z} - \frac1{z}$ Part of the the problem I'm trying to do involves finding the residue of
$\frac{1}{z-\tan z} - \frac{1}{z}$ at z=0
I am not sure of the order of the pole
Here's what I did to find the order of the pole
f(z) = $\frac{1}{z-\tan z} - \frac{1}{z}$
= $\frac{\tan z}{z(z - \tan z)}$
Now $z - \tan z = \frac{1}{3}z^3 + \frac{2}{15}z^5 + ...\quad$
and $\quad z(z - \tan z) = \frac{1}{3}z^4 + \frac{2}{15}z^6 + ... $
So f has pole of order 4 at z=0.
Is this correct and to find the residue I use the formula
$\frac{1}{n!}\lim_{z\to 0} \frac{d^3}{dz^3}(z^4f(z))$ right?
|
You may write
$$
\begin{align}
\frac{1}{z-\tan z} - \frac{1}{z}&=\frac{1}{z-\left(z+\frac{1}{3}z^3 + \frac{2}{15}z^5 + \mathcal{O}(z^7)\right)} - \frac{1}{z}\\\\
&=\frac{1}{-\frac{1}{3}z^3 - \frac{2}{15}z^5 + \mathcal{O}(z^7)} - \frac{1}{z}\\\\
& =-\frac{3}{z^3}\frac{1}{1 + \frac{2}{5}z^2 +\mathcal{O}(z^4)} - \frac{1}{z}\\\\
& =-\frac{3}{z^3}\left(1- \frac{2}{5}z^2 + \mathcal{O}(z^4)\right) - \frac{1}{z}\\\\
& =-\frac{3}{z^3}+ \frac{6}{5z} - \frac{1}{z}+ \mathcal{O}(z)\\\\
& =-\frac{3}{z^3}+ \frac{1}{5z} + \mathcal{O}(z)
\end{align}
$$ giving the value $\dfrac15$ for the desired residue.
|
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|
Proving that $12^n + 2(5^{n-1})$ is a multiple of 7 for $n\geq 1$ by induction
Prove by induction that $12^n + 2(5^{n-1})$ is a multiple of $7$.
Here's where I am right now:
Assume $n= k $ is correct:
$$12^k+2(5^{k-1}) = 7k.$$
Let $n= k+1 $:
$$12^{k+1} + 2(5^k)$$
$$12^k(12) + 2(5^k)$$
Any ideas on how I can proceed from here?
|
You should write $12^k+2\cdot 5^{k-1}=7r$ (using $k$ with two different meanings is wrong).
Then you have $12^k=7r-2\cdot 5^{k-1}$; now
\begin{align}
12^{k+1}+2\cdot 5^k
&=12\cdot 12^k+2\cdot 5^k\\
&=12\cdot(7r-2\cdot 5^{k-1})+2\cdot 5^k\\
&=7\cdot12r-24\cdot5^{k-1}+10\cdot5^{k-1}
\end{align}
and you should be able to conclude.
Remember to check the basis of the induction, that is, the case $n=1$.
Using congruences it is easier, of course: just observe that $12\equiv 5\pmod{7}$, so
$$
12^n+2\cdot 5^{n-1}\equiv 5^n+2\cdot 5^{n-1}\equiv 5^{n-1}(5+2)\equiv 5^{n-1}\cdot7\equiv0\pmod{7}
$$
|
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|
Boolean Algebra - proof without associativity? I would like to prove the following:
$(x\cdot y) + (\overline{x} + \overline{y}) = 1$
without the Associativity Property. I can't seem to do this algebraically (without truth tables).
|
So I cannot use Associative:
$X + (Y + Z) = (X + Y) + Z = (X + Z) + Y = X + Y + Z $
$(x\cdot y) + (\overline{x} + \overline{y})$
Expand minimized terms using complement
$(x\cdot y) + (\overline{x} \cdot (\overline{y} + y) + \overline{y} \cdot (\overline{x} + x))$
$(x\cdot y) + (\overline{x}\cdot\overline{y} + \overline{x}\cdot y + \overline{y} \cdot \overline{x} + \overline{y}\cdot x)$
Remove duplicates with Idempotent.
$(x\cdot y) + (\overline{x}\cdot\overline{y} + \overline{x}\cdot y + \overline{y}\cdot x)$
Extract common terms with Redundancy and Complement.
$x \cdot (y + \overline{y}) + \overline{x} \cdot (\overline{y} + y)$
I may of used Associative by removing brackets. Complement again.
$x + \overline{x}$
$1$
A lot easier using deMorgan's.
Laws and Theorems of Boolean Algebra
|
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|
$f(x)=\sum_{i=0}^{\infty} (x^{2^n})/(1-x^{2^{n+1}})$. Find $f(99)$. $f(x)=\sum_{i=0}^{\infty} (x^{2^n})/(1-x^{2^{n+1}})$. Find $f(99)$.
ATTEMPT: The following series can be re-written as $f(x)=\sum_{i=0}^\infty \left(\frac{1}{1-x^{2^n}}\right) \cdot \left( \frac{1}{1+x^{2^n}}\right)$ and then expanded along.
|
If this problem is $$f(x)=\sum_{n=0}^{\infty}\dfrac{x^{2^n}}{1-x^{2^{n+1}}}$$
Hint:Let $x^{2^n}=t$,then
$$\dfrac{x^{2^n}}{1-x^{2^{n+1}}}=\dfrac{t}{1-t^2}=\dfrac{1}{1-t}-\dfrac{1}{1-t^2}=\dfrac{1}{1-x^{2^n}}-\dfrac{1}{1-x^{2^{n+1}}}$$
so
$$f(x)=\dfrac{1}{1-x}-\lim_{N\to\infty}\dfrac{1}{1-x^{2^{N+1}}}=\dfrac{1}{1-x}$$
|
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|
What's wrong?: Find the infinite sum $S = 1 + \frac{3}{4} + \frac{7}{16} + \frac{15}{64} + \frac{31}{256} + \ldots$ The answer I got by hand is not the same to the one I found using a spreadsheet.
$\displaystyle S = 1 + \frac{3}{4} + \frac{7}{16} + \frac{15}{64} + \frac{31}{256} + \ldots$
$\displaystyle \frac{1}{4}S = \hspace{8.5pt} \frac{1}{4} + \frac{3}{16} + \frac{7}{64} + \frac{15}{256} + \ldots$
$\displaystyle \frac{3}{4}S = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots \qquad \leftarrow S- \frac{1}{4}S$
For the Infinite Sum on the RHS $\displaystyle \left(S = \frac{a}{1-r}\right)$:
$\displaystyle a = 1$
$\displaystyle r = \frac{1}{2}$
Then
$\displaystyle \frac{3}{4}S = \frac{1}{1-\frac{1}{2}}$
$\displaystyle \frac{3}{4}S = 2$
$\displaystyle S = \frac{8}{3}$
Using Excel the answer is $\displaystyle \frac{5}{3}$, but I don't know where is the issue.
Thanks!!
|
We can pattern match this to a difference of geometric series. The sum in question is $$\begin{align} 1 + \frac{3}{4} + \frac{7}{16} + \frac{15}{64} + \frac{31}{256} = 1+\sum_{k=1}^\infty \frac{2^{k+1}-1}{4^{k}} \\ = 1+\sum_{k=1}^\infty \frac{2^{k+1}}{4^{k}}-\sum_{k=1}^\infty \frac{1}{4^{k}}\end{align}$$ Now use geometric series...
|
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|
Limit and Integral problem work verification-2 I have to calculate the following:
$$\large\lim_{x \to \infty}\left(\frac {\displaystyle\int\limits_{x^{2}}^{2x}t^{4}e^{t^{2}}dt}{e^{x}-1-x - \frac{x^2}{2}- \frac{x^3}{6}-\frac{x^4}{24}}\right)$$
My attempt:
Let $F(x)=\displaystyle\int\limits_0^xt^4e^{t^2}dt$. Then,
$$\large\lim_{x\to\infty}\left(\frac{\displaystyle\int\limits_{x^{2}}^{2x}t^{4}e^{t^{2}}dt}{e^{x}-1-x - \frac{x^2}{2}- \frac{x^3}{6}-\frac{x^4}{24}}\right)=\lim_{x \to \infty}\left(\frac {F(2x) - F(x^2)}{e^{x}-1-x - \frac{x^2}{2}- \frac{x^3}{6}-\frac{x^4}{24}}\right)$$
Applying L'Hôpital's rule, we have,
$$\large\begin{align}\lim_{x \to \infty}\left(\frac {32x^4e^{4x^2} - 2x^9e^{x^4}}{e^{x}-1-x - \frac{x^2}{2}- \frac{x^3}{6}}\right) &= \lim_{x \to \infty}(32x^4e^{4x^2-x} - 2x^9e^{x^4-x}) \\&= \lim_{x \to \infty}\bigg(2x^4e^{4x^2-x}(16-x^5e^{x^4-4x^2})\bigg) = -\infty\end{align}$$
Am I right?
|
Hint: Notice that $$\left(\int_{x^2}^{2x}t^4 e^{t^2} dt\right)' = \color{red}2\dot\,(2x)^4 e^{(2x)^2} - \color{red}{2x}\dot\,(x^2)^4 e^{(x^2)^2}$$
and $$e^x = \sum_{k=0}^\infty \frac{x^k}{k!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + O(x^4)$$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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|
Use generating function to find coefficient Use a generating function to find the coefficient of $x^{22}$ in:
$$\frac{1+3x}{(1-x)^8}$$
I know I need to use a binomial expansion on the lower term, but what about the upper term?
|
Recall that $$\frac1{1-x}=\sum_{n=0}^\infty x^n.$$
So $$\frac{\mathsf d^7}{\mathsf dx^7}\left[\frac1{1-x}\right] = \frac{7!}{(1-x)^8}$$
and
$$\begin{align*}
\frac{\mathsf d^7}{\mathsf dx^7}\sum_{n=0}^\infty x^n =\sum_{n=7}^\infty \frac{n!}{(n-7)!}x^{n-7}=\sum_{n=0}^\infty\frac{(n+7)!}{n!}x^{n}
\end{align*}$$
Hence
$$\begin{align*}\frac{1+3x}{(1-x)^8} &= \frac {1+3x}{8!}\frac{\mathsf d^7}{\mathsf dx^7}\left[\frac1{1-x}\right]\\
&= \frac{1+3x}{7!}\sum_{n=0}^\infty\frac{(n+7)!}{n!}x^n\\
&= \sum_{n=0}^\infty\frac{(n+7)!}{n!7!}x^n + \sum_{n=0}^\infty \frac{3(n+7)!}{n!7!}x^{n+1}\\
&= \sum_{n=0}^\infty\binom {n+7}7 x^n + \sum_{n=1}^\infty 3\binom {n+6}7x^n\\
&= 1 + \sum_{n=1}^\infty \left(\binom{n+7}7+3\binom{n+6}7\right)x^n
\end{align*} $$
So the coefficient of $x^{22}$ is
$$\binom {22+7}{7} + 3\binom{22+6}7 = 5112900.$$
|
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|
Solve the equation $\log_{2} x \log_{3} x = \log_{4} x$ Question:
Solve the equations
a)
$$\log_{2} x + \log_{3} x = \log_{4} x$$
b)
$$\log_{2} x \log_{3} x = \log_{4} x$$
Attempted solution:
The general idea I have been working on is to make them into the same base, combine them and take the inverse to get the solution for x.
a)
$$\log_{2} x + \log_{3} x = \log_{4} x \Leftrightarrow$$
$$\log_{2} x + \frac{\log_{2} x}{\log_{2} 3} = \frac{\log_{2} x}{\log_{2} 4} \Leftrightarrow$$
$$\log_{2} x + \frac{\log_{2} x}{\log_{2} 3} = \frac{\log_{2} x}{2} \Leftrightarrow$$
$$\frac{2 \log_{2} 3 \log_{2}x + 2\log_{2} x - \log_{2}3 \log_{2}x}{2 \log_{2}3} = 0 \Leftrightarrow $$
Moving the denominator over and solving for $\log_{2} x$
$$\frac{\log_{2} x (2\log_{2}3 + 1 - \log_{2}3)}{2\log_{2} 3} = 0 \Leftrightarrow$$
$$\log_{2} x = 0 \Rightarrow x = 2^{0} = 1$$
b)
$$\log_{2} x \log_{3} x = \log_{4} x$$
$$\log_{2} x \frac{\log_{2} x}{\log_{2} 3} - \frac{\log_{2} x}{\log_{2} 4} = 0 \Leftrightarrow$$
$$\frac{4 (\log_{2}x)^2 - \log_{2}3 \log_{2}x}{4 \log_{2}3} = 0 \Leftrightarrow $$
$$4 (\log_{2}x)^2 - \log_{2}3 \log_{2}x = 0 \Leftrightarrow $$
Substituting $t = \log_{2} x$ gives:
$$4t^2 - \log_{2}3t = 0 \Rightarrow$$
$$t^2 - \frac{\log_{2}3t}{4} = 0 \Rightarrow$$
$$t \left(t- \frac{\log_{2}3}{4}\right) = 0$$
$$t_{1} = 0 \Rightarrow x = 1$$
$$t_{2} = \frac{\log_{2}3}{4} \Rightarrow x = 2^{\frac{\log_{2}3}{4}}$$
However, the second solution here should be $\sqrt{3}$, so I must have made a mistake somewhere. Any suggestions?
|
I think that you made two mistakes.
The first mistake is in a) when factorizing, instead you typed it should have been $\frac{\log_{2} x(2\log_{2} 3+ 2 -\log_{2} 3)}{2\log_{2} 3}=0$ though it doesn´t affect the answer.
And the other mistake in b) is the one that @Gregory Grant says, once you do that you get to answer $x=\sqrt 3$
|
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|
Show that $ \frac{\cos5x + \cos4x} {1-2\cos3x} = -\cos2x -\cos x $ The Question reads -
$$ \frac{\cos5x + \cos4x} {1-2\cos3x} = -\cos2x -\cos x $$
I tried using the obvious approach by converting $5x , 4x $ and $ 3x$ to either $2x$ or $x$ but all that seemed to do was to further complicate the fraction. Any hints would be much appreciated.
I also tried applying various identities but to no avail.
|
Let's apply the identity $\cos (x+y)+\cos(x-y)=2\cos x \cos y$ to the product
$$(1-2\cos 3x)(-\cos 2x-\cos x)=-\cos 2x -\cos x+2\cos 3x \cos 2x+2\cos 3x \cos x$$
Note that $2\cos 3x \cos 2x=\cos x+\cos 5x$ while $2\cos 3x \cos x=\cos 2x+\cos 4x$
Thus
$$\begin{align}
(1-2\cos 3x)(-\cos 2x-\cos x)&=-\cos 2x -\cos x+2\cos 3x \cos 2x+2\cos 3x \cos x\\\\
&=-\cos 2x -\cos x+(\cos x+\cos 5x)+(\cos 2x +\cos 4x)\\\\
&=\cos 5x + \cos 4x
\end{align}$$
which was to be shown!
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Solving a Diophantine equation3 The Diophantine equation that I have to solve is:
$$343x^2-27y^2=1$$
This question has already been posted by other user but it has not received an answer.
I proved to solve it.
This is my attempt:
substituting for $x^2=u$ and $y^2=v$ the equation becomes a Diophantine linear equation:$$343u-27v=1$$
using Euclid's algorithm to solve the equation, the solutions are:
$$
\left\{
\begin{array}{ll}
u=10+27k \\
v=127+343k\end{array}
\right.
$$
substituting these solutions in first equation we obtain:
$$y^2=127+343\cdot (\frac {x^2-10}{27})$$
$x^2-10$ has to be a multiple of $27$ therefore $$x^2=c27+10 \tag{1}$$ (with $c$ integer)
$(1)$ becomes $$(x-1)(x+1)=9\cdot (3c+1)$$
obtaining two systems:
$$
\left\{
\begin{array}{ll}
x+1=n9 \\
x-1=\frac {3c+1}{n}\end{array}
\right.
$$
and
$$\left\{
\begin{array}{ll}
x-1=n9 \\
x+1=\frac {3c+1}{n}\end{array}
\right.
$$
The solution of the first is $c=3p+1$ (with $p$ integer):
indeed if $x+1\equiv 0\pmod 3$ $c$ has to be $c\equiv 1\pmod 3$ because $c(x-1)-1$ has to be a multiple of $3$.
The solution of the second system is $c=3p+2$: indeed if $x-1\equiv 0\pmod 3$ $c\equiv 2\pmod 3$. How can I can continue and are there other solutions?
|
$343x^2-27y^2=1$
$⇒343x^2≡1 mod 27$
$343≡19 mod 27$
we can find that if $x=8$ we get:
$343 . 8^2 ≡ 1 mod 27$
We can see that x belong to a set its members make an arithmetic progression and have following specification:
$s . 19 ≡ 1 mod 27$; $s = [10, 37, 64, . . a+27k]$
where $a=10$; and $d=27$ is common ratio.
With $k=5257504398$ we get $x=376766$
Also $y^2=343 k$ that gives $y=1342879$
Therefore the parametric solution is:
$x=\sqrt{10+27k}$
$y=\sqrt{343 k}$
This equation probably has more solutions.
|
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|
Prove that $7$ divides $1 + 2^{(2^n)} + 2^{(2^{n+1})}$ by induction Prove that $7$ divides $1 + 2^{(2^n)} + 2^{(2^{n+1})}$ by induction.
I ran into the above problem. The base case $n=1$ gives $21$ which is divisible by $7$.
Now assume it is true for $n$. Then for $n+1$, we have the expression
$$ 1 + 2^{(2^{n+1})} + 2^{(2^{n+2})}$$
which is equal to
$$1 + 2^{(2^n\cdot 2)} + 2^{(2^n \cdot 2^2)}$$
but I do not really see how it helps.
Any ideas?
|
As in Winther's post let $x_n=2^{2^n}$. Then the sequence is
$I_n=1+x_n+x_n^2$, and $x_{n+1}=x_n^2$.
We suppose $I_n\equiv0\pmod 7 \implies x_n+x_n^2\equiv-1 \pmod 7$,
squaring both sides:
$x_n^2+2x_n^3+x_n^4\equiv1 \implies I_{n+1}=1+x_n^2+x_n^4\equiv 2(1-x_n^3)\equiv 0 \pmod 7$
since $x_n^3=2^{3\cdot2^n}\equiv1^{2^n}\equiv 1 \pmod 7$
|
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|
A supposed to be easy calculus problem Find the values of $m$ if the line $y=mx+2$ is a tangent to the curve $x^2-2y^2=1$.
My working:
First we differentiate $x^2-2y^2=1$ with respect to $y$ to get the gradient. We get $y^2=\frac{1}{2}x^2-\frac{1}{2}\implies y=\pm\sqrt{\frac{1}{2}x^2-\frac{1}{2}}$.
We take the positive one for demonstration
$\frac{dy}{dx}=\frac{1}{2}x(\frac{1}{2}x^2-\frac{1}{2})^{-\frac{1}{2}}=\frac{x}{2\sqrt{\frac{1}{2}x^2-\frac{1}{2}}}$
$\implies(1-2m^2)x^2=-2m^2$
Since the tangent touches the curve, we can make $x^2-2(mx+2)^2=1$, we then get $(1-2m^2)x^2=9+8mx$
$\implies(1-2m^2)x^2=-2m^2$ and $(1-2m^2)x^2=9+8mx$ are two equations with two unknowns, then we should be able to find the values of $m$, but I couldn't find any easy way to solve those 2 simultaneous equations. Is there any easier method?
I tried solving $9+8mx=-2m^2$ but we still have two unknowns in one equation?
Also, if we don't use those two simultaneous equations, can we solve this question with a different method?
I am trying to solve WITHOUT implicit differentiation.
Many thanks for the help!
|
Let $(a,b)$ be a point of tangency. We have $2x-4y\frac{dy}{dx}=0$, so the slope of the tangent line at $(a,b)$ (if $b\ne 0$) is $\frac{a}{2b}$.
The tangent line has equation $y-b=(x-a)(a/2b)$. Simplifying , and comparing with $y=mx+2$, we find that $b-a^2/(2b)=2$. It follows that $2b^2-a^2=4b$. Since $a^2-2b^2=1$, we conclude that $b=-1/4$. The rest is routine.
Remark: Note that the tangent line happens to be at a point on the "lower" half of the hyperbola. So taking the positive square root turns out not to be useful.
|
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|
Finding value of a quadratic The polynomial
\begin{equation*}
p(x)= ax^2+bx+c
\end{equation*}
has $1+\sqrt{3}$ as one of it's roots and also $p(2)=-2$. Is there any way to know the value of $a$, $b$ and $c$? I tried but I can form only $2$ equations how can I figure out value of $3$ unknown with $2$ equations so something must be missing.
Also $a$, $b$ and $c$ are rationals.
|
Hint: If one of the roots is $1 + \sqrt{3}$ and $a, b, c$ are rational then what must the other root be?
Solution 1:
The other root must then be $1 - \sqrt{3}$.
You have two options now: You can make a system of three linear equations in three variables and simply solve.
Or, you can simply expand $\alpha[x - (1 - \sqrt{3})][x - (1 + \sqrt{3})] = \alpha(x^2 - 2x - 2)$ and use $p(2) = -2$ to get $\alpha = 1$.
Solution 2:
We only need two equations if we know that $a, b, c$ are rational.
$$(1+\sqrt{3})^2 a + (1+\sqrt{3})b + c = 0$$
$$4a + 2\sqrt{3} a + b + \sqrt{3} b + c = 0$$
$a, b, c$ are rational, so we know that $-2a = b$ because the $\sqrt{3}$ terms must cancel each other.
So, $4a + 2b + c = -2$ gives us $c = -2$.
Substituting back into the above we get $4a + b = 2$, so $a = 1$ and $b = -2$.
Thus the quadratic is $x^2 - 2x - 2$.
|
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|
Finding value of $m$ such that such that the polynomial is factorized A polynomial $2x^2+mxy+3y^2-5y-2$ Find the value of $m$ much that $p(xy)$ can be factorized into two linear factors
|
Putting $m = 7$ you obtain
\begin{align*}
(x+3y+1)(y+2x-2) &= xy+2x^2-2x+3y^2+6xy-6y+y+2x-2 \\
&= 2x^2 + 7xy +3y^2-5y-2
\end{align*}
|
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|
Solving a Radical Equation $5(\sqrt{1-x} + \sqrt{1+x}) = 6x + 8\sqrt{1-x^2}$ (squaring doesn't help) How should I approach this problem:
$$ 5(\sqrt{1-x} + \sqrt{1+x}) = 6x + 8\sqrt{1-x^2} $$
I've tried squaring both sides but to get rid of all the radicals requires turning it into a quartic equation, which I don't know how to solve?
Any tips? Thanks!
|
Here is a really nice method: when looking at terms such as $\sqrt{1-x}$, $\sqrt{1+x}$, and $\sqrt{1-x^2}$, think trigonometry!
Make the trig substitution:
$$x = \cos{\theta}$$
Now:
$$\sqrt{1-x}=\sqrt{2}\sin{\frac{\theta}{2}}$$
$$\sqrt{1+x}=\sqrt{2}\cos{\frac{\theta}{2}}$$
$$\sqrt{1-x^2}=\sin{\theta}$$
The equation with $\theta$ is then:
$$5\left(\sin{\frac{\theta}{2}} + \cos{\frac{\theta}{2}}\right) = 6\cos{\theta} + 8\sin{\theta} $$
Can you take it from here?
Edit: Another trick is used to finish the problem. One of the advantages of trig substitutions is that there are many identities to use. In this case:
$$6\cos{\theta} + 8\sin{\theta} = 10\left(\frac{3}{5}\cos{\theta} + \frac{4}{5}\sin{\theta}\right)$$
Now, if we let $\phi = \sin^{-1}\left(\frac{3}{5}\right)$:
$$6\cos{\theta} + 8\sin{\theta} = 10\left(\sin{\phi}\cos{\theta} + \cos{\phi}\sin{\theta}\right)=10\sin(\phi+\theta)$$
|
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|
What will be the equation of side $BC$.
The equation of two equal sides $AB$ and $AC$ of an isosceles triangle $ABC$ are $x+y=5$ and $7x-y=3$ respectively . What will be the equation of the side $BC$ if the area of the triangle $\triangle ABC$ is $5$ square units.
$a.)x+3y-1=0\\
b.)x-3y+1=0\\
c.)2x-y-5=0\\
\color{green}{d.)x+2y-5=0}\\$
$\quad\\~\\~\\$
Let the slope of the required line be $m$.
I used the slope formula between two lines
$\angle B=\angle C\\~\\\left|{\dfrac{(-1-m)}{(1-m)}}\right|=\left|{\dfrac{(7-m)}{(1+7m)}}\right|\\
\implies m=-3,\dfrac{1}{3}$
But the book is giving right answer as option $d.)$.
By finding $m$ i only found slope not the whole equation.
Also i would like to know if their is clean simple short way , and also using $Area=5$ .
I have studied maths upto $12th$ grade.
Update : by using Geogebra I found that $x-3y+1=0$ fits perfectly
|
So, the coordinate of $A(1,4)$
We can set the coordinates of $B(a,5-a);C(b,7b-3)$
We have $|AB|^2=|AC|^2\implies(a-1)^2+(a-1)^2=(b-1)^2+\{7(b-1)\}^2$
$\implies a-1=\pm(5b-5)=\pm5(b-1)$
$$\triangle ABC=\dfrac12\begin{vmatrix}1 & 4 & 1 \\ a & 5-a & 1 \\ b & 7b-3 &1\end{vmatrix}=4|(a-1)(b-1)|$$
$\implies\triangle ABC=4\cdot5|(b-1)^2|$
$\implies20(b-1)^2=5\iff b-1=\pm\dfrac12$
Can you take it home from here?
|
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|
Get rid of the square roots of the denominator: $\dfrac{1}{\sqrt{7}-2\sqrt{5}+\sqrt{3}}$
How to get rid of the square roots of the denominator: $\dfrac{1}{\sqrt{7}-2\sqrt{5}+\sqrt{3}}$?
*
*I squared the whole denominator, but that didn't help.
*Also I searched for a propriety or identity like $A^2-B^2$, but I didn't see one that could fit.
Any help is appreciated.
|
$$\begin{align}\frac{1}{\sqrt 7-2\sqrt 5+\sqrt 3}&=\frac{1}{\sqrt 7-2\sqrt 5+\sqrt 3}\cdot \frac{\sqrt 7-2\sqrt 5-\sqrt 3}{\sqrt 7-2\sqrt 5-\sqrt 3}\\&=\frac{\sqrt 7-2\sqrt 5-\sqrt 3}{(\sqrt 7-2\sqrt 5)^2-3}\\&=\frac{\sqrt 7-2\sqrt 5-\sqrt 3}{4(6-\sqrt{35})}\cdot\frac{6+\sqrt{35}}{6+\sqrt{35}}\\&=\frac{(\sqrt 7-2\sqrt 5-\sqrt 3)(6+\sqrt{35})}{4}\end{align}$$
|
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|
Decomposition into partial fractions to compute an integral I'm having problems with:
$$\int_{-\infty}^{\infty}\frac{x^4+1}{x^6+1}dx$$
I was thinking: $\frac{x^4+1}{x^6+1}$ is an even function and the interval $(-\infty,\infty)$ is symmetric about 0, we could write the integral like:
$$2\int_{0}^{\infty}\frac{(x^4+1)}{x^6+1}dx$$
And for $\frac{x^4+1}{x^6+1}$, I will use the partial fractions. I will write $x^6+1$ like a sum of cubes $x^6+1=(x^2)^3+1^3$ and use the formula $a^3+b^3=(a+b)(a^2-ab+b^2)$ and in the end we will have
$$x^6+1=(x^2+1)(x^4-x^2+1)$$ But now I'm stuck and do not know how to decompose $x^4-x^2+1$. I was thinking about $(x^2-1)x^2+1$ or $(x^2-\frac{1}{2})^2+\frac{3}{4}$ but I need a form for that like $-(-1+\sqrt3 x-x^2) (1+\sqrt3 x+x^2)$...
A little help here?
|
$$\begin{eqnarray*}\int_{\mathbb{R}}\frac{x^4+1}{x^6+1}\,dx&=&2\int_{\mathbb{R}^+}\frac{x^4+1}{x^6+1}\,dx=2\int_{0}^{1}\frac{x^4+1}{x^6+1}\,dx+2\int_{1}^{+\infty}\frac{x^4+1}{x^6+1}\,dx\\&=&4\int_{0}^{1}\frac{x^4+1}{x^6+1}\,dx=4\int_{0}^{1}\sum_{k\geq 0}(-1)^k\left(x^{6k}+x^{6k+4}\right)\,dx\\&=&4\sum_{k\geq 0}(-1)^k\left(\frac{1}{6k+1}+\frac{1}{6k+5}\right)=4\sum_{k\geq 0}\frac{\chi(k)}{k}\end{eqnarray*}$$
where $\chi(k)$ is a Dirichlet character $\!\!\pmod{12}$, that equals one if $k\in\{1,5\}\pmod{12}$, minus one if $k\in\{7,11\}\pmod{12}$ and zero otherwise. We know in advance that:
$$ \sum_{k\geq 0}\frac{(-1)^k}{2k+1}=\arctan 1=\frac{\pi}{4},$$
$$ \sum_{k\geq 0}\frac{(-1)^k}{6k+3}=\frac{1}{3}\,\arctan 1=\frac{\pi}{12},$$
hence by summing these series we get $\sum_{k\geq 0}(-1)^k\left(\frac{1}{6k+1}+\frac{1}{6k+5}\right)=\frac{\pi}{3}$, so:
$$ \int_{\mathbb{R}}\frac{x^4+1}{x^6+1}\,dx = \color{red}{\frac{4\pi}{3}}.$$
|
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|
How to find $\frac{0}{0}$ limit without L'Hôpital's rule I am having trouble solving this limit. I tried applying L'Hôpital's rule but I got $\frac{0}{0}$.
$$\lim_{x\to0} {\frac{\frac{1}{1+x^3} + \frac{1}{3}\log{\left(1+3x^3\right)}-1}{2\sin{\left(3x^2\right)}-3\arctan{\left(2x^2\right)}}}$$
I would appreciate any hints in the right direction. Thanks in advance for your help.
|
Recall some standard limits (which can be computed without using
l'Hospital's rule ):
\begin{eqnarray*}
\lim_{u\rightarrow 0}\frac{\frac{1}{1+u}-1+u}{u^{2}} &=&1 \\
\lim_{u\rightarrow 0}\frac{\log (1+u)-u}{u^{2}} &=&-\frac{1}{2} \\
\lim_{u\rightarrow 0}\frac{\sin u-u}{u^{3}} &=&-\frac{1}{6} \\
\lim_{u\rightarrow 0}\frac{\arctan u-u}{u^{3}} &=&-\frac{1}{3}.
\end{eqnarray*}
Now one can re-write the original expression using those of the standard
limits above as follows
\begin{eqnarray*}
{\frac{\frac{1}{1+x^{3}}+\frac{1}{3}\log {\left( 1+3x^{3}\right) }-1}{2\sin {%
\left( 3x^{2}\right) }-3\arctan {\left( 2x^{2}\right) }}} &{=}&\frac{\left(
\frac{1}{1+x^{3}}-1+x^{3}\right) +\left( \frac{1}{3}\log {\left(
1+3x^{3}\right) -x}^{3}\right) }{\left( 2\sin {\left( 3x^{2}\right) -6x}%
^{2}\right) -\left( 3\arctan {\left( 2x^{2}\right) -6x}^{2}\right) } \\
&=&\frac{\left( \frac{\frac{1}{1+(x^{3})}-1+(x^{3})}{(x^{3})^{2}}\right)
+3\times \left( \frac{\left( \log {\left( 1+\left( 3x^{3}\right) \right) -}%
\left( {3x}^{3}\right) \right) }{(3x^{3})^{2}}\right) }{2\times 3^{3}\left(
\frac{\sin {\left( 3x^{2}\right) -(3x}^{2})}{(3x^{2})^{3}}\right) -3\times
2^{3}\left( \frac{\arctan {\left( 2x^{2}\right) -(2x}^{2})}{\left(
2x^{2}\right) ^{3}}\right) }
\end{eqnarray*}
It follows that
\begin{equation*}
\lim_{x\rightarrow 0}\left( \frac{\frac{1}{1+x^{3}}+\frac{1}{3}\log {\left(
1+3x^{3}\right) }-1}{2\sin {\left( 3x^{2}\right) }-3\arctan {\left(
2x^{2}\right) }}\right) =\frac{\left( 1\right) +3\times \left( -\frac{1}{2}%
\right) }{2\times 3^{3}\left( -\frac{1}{6}\right) -3\times 2^{3}\left( -%
\frac{1}{3}\right) }=\frac{1}{2}.
\end{equation*}
|
{
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"url": "https://math.stackexchange.com/questions/1302336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Maximum of given expression? Suppose $a,b,c>0$ and further that $a^{2} + b^{2} + c^{2}=2abc + 1 $.
The problem is to find $\max \big(a-2bc\big) \big(b-2ca\big) \big(c-2ab\big) $.
Give me some help. I've tried $X=a-2bc$, $Y=b-2ca$, $Z=c-2ab$
which yields $X^2 + Y^2 + Z^2 = 1-2XYZ$, but $\frac12$ is not the maximum because
$XYZ=0$.
Can someone give me an elegant solution?
|
let $$x=a-2bc,y=b-2ca,z=c-2ab$$
since
$$a^2+b^2+c^2=2abc+1\Longrightarrow x^2+y^2+z^2+2xyz=1$$
so we only find $xyz$ maximu,since
$$1=x^2+y^2+z^2+2xyz\ge 3(x^2y^2z^2)^{\frac{1}{3}}+2xyz\Longrightarrow xyz\le\dfrac{1}{8}$$
so
$$(a-2bc)(b-2ac)(c-2ab)\le\dfrac{1}{8}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1304018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Find value of $xy\sqrt{y^2 - x^2}$ for the given differential equation.
If $(y^3 - 2x^2y)dx + (2xy^2 - x^3)dy = 0 $ , then prove that the value of $xy\sqrt{y^2 - x^2}$ is a constant.
This is what I've tried :
$$ y(y^2 - 2x^2) dx + x(2y^2 - x^2) dy = 0 \\
\cfrac{dy}{dx} = \cfrac{(2x^2 - y^2)y}{(2y^2 - x^2)x} \\
$$
I then tried substituting $y/x = v$
Further, solving it, I got :
$$xy\sqrt{y^2 - x^2} = \cfrac{x^{4/3}}{y^{4/3}}$$
But, I'm not sure how to come with a constant at RHS.
|
From the very beginning, use $y=x\,v$, $y'=x v'+v$ and replace in the original equation. You will get a common factor $x^3$; simplify to get $$x \left(2 v^2-1\right) v'+3 v \left(v^2-1\right)=0$$ which is separable becoming $$\frac 1x \frac{dx}{dv}=-\frac{2 v^2-1}{3 v \left(v^2-1\right)}$$ Use partial fraction decomposition to get $$\frac 1x \frac{dx}{dv}=-\frac{1}{3 v}-\frac{1}{6 (v+1)}-\frac{1}{6 (v-1)}$$ Integrate both sides $$\log(x)+C=-\frac{1}{6} \log \left(1-v^2\right)-\frac{1}{3}\log (v)$$ that is to say $$Cx^6=\frac{1}{v^2(1-v^2)}$$ Make the cross product, replace $v$ by $\frac y x$ to obtain $$C x^2 y^2 (x^2-y^2)=1$$ from which the result arrives.
|
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|
Proof of Different Polynomial Decompositions into Linear Factors From G. Polya "Mathematics and Plausible Reasoning" p. 18.
How do you prove that provided the roots of a polynomial are different from zero,
$$a_0 + a_1x+a_2x^2 + ... + a_nx^n$$
$$\,= a_0\left(1-\frac{x}{\alpha_1}\right)\left(1-\frac{x}{\alpha_2}\right)...\left(1-\frac{x}{\alpha_n}\right)$$
with $\alpha_1, \alpha_2,...\alpha_n$ corresponding to the polynomial roots.
|
\begin{align}
&a_0+a_1x+\ldots+a_nx^n=\\
&\qquad=a_n(x-\alpha_1)\cdot\ldots\cdot(x-\alpha_n)=\\
&\qquad=a_n\alpha_1\frac{x-\alpha_1}{\alpha_1}\cdot\ldots\cdot\alpha_n\frac{x-\alpha_n}{\alpha_n}=\\
&\qquad=a_n\alpha_1\cdot\ldots\cdot\alpha_n\left(\frac{x}{\alpha_1}-1\right)\cdot\ldots\cdot\left(\frac{x}{\alpha_n}-1\right)=\\
&\qquad=(-1)^n\,a_n\alpha_1\cdot\ldots\cdot\alpha_n\left(1-\frac{x}{\alpha_1}\right)\cdot\ldots\cdot\left(1-\frac{x}{\alpha_n}\right)
\end{align}
Setting $x=0$ results in:
$\qquad a_0+a_1x+\ldots+a_nx^n\,=\,(-1)^n\,a_n\alpha_1\cdot\ldots\cdot\alpha_n\left(1-\frac{x}{\alpha_1}\right)\cdot\ldots\cdot\left(1-\frac{x}{\alpha_n}\right)$
simplifying to
$$a_0=(-1)^n\,a_n\alpha_1\cdot\ldots\cdot\alpha_n$$ Therefore:
$\qquad a_0+a_1x+\ldots+a_nx^n \,=\,a_0\left(1-\frac{x}{\alpha_1}\right)\cdot\ldots\cdot\left(1-\frac{x}{\alpha_n}\right)$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Finding kernel and range of a linear transformation We are given:
Find $\ker(T)$, and $\textrm{rng}(T)$, where $T$ is the linear transformation given by
$$T:\mathbb{R^3} \rightarrow \mathbb{R^3}$$
with standard matrix
$$ A = \left[\begin{array}{rrr}
1 & -1 & 3\\
5 & 6 & -4\\
7 & 4 & 2\\
\end{array}\right]\textrm{.}
$$
The kernel can be found in a $2 \times 2$ matrix as follows:
$$ L = \left[\begin{array}{rrr}
a & b\\
c & d\\
\end{array}\right] = (a+d) + (b+c)t
$$
Then to find the kernel of $L$ we set
$$(a+d) + (b+c)t = 0$$
$$d = -a$$
$$c = -b$$
so that the kernel of $L$ is the set of all matrices of the form
$$ A = \left[\begin{array}{rrr}
a & b\\
-b & -a\\
\end{array}\right]
$$
but I do not know how to apply that to this problem.
|
$$
A = \left[\begin{array}{rrr}
1 & -1 & 3\\
5 & 6 & -4\\
7 & 4 & 2\\
\end{array}\right]
$$
Consider a linear map represented as a $m × n$ matrix $A$ .
The kernel of this linear map is the set of solutions to the equation $Ax = 0$
$$
ker(A)=\{x \in R^n|Ax=0\}
$$
$$
det(A)=1(12+16)-(-1)(10+28)+3(20-42)=0
$$
Since $det(A)=0$ , $x\ne0$ and $0$ is a vector here.
$$
\left[\begin{array}{rrr}
1 & -1 & 3\\
5 & 6 & -4\\
7 & 4 & 2\\
\end{array}\right]
\left[\begin{array}{r}
a\\b\\c
\end{array}\right]
=\left[\begin{array}{r}
0\\0\\0
\end{array}\right]
$$
In row-reduced form,
$$
A = \left[\begin{array}{rrr}
1 & 0 & \frac{14}{11}\\
0 & 1 & \frac{-19}{11}\\
0 & 0 & 0\\
\end{array}\right]
$$
$$x=\frac{-14}{11}z$$
$$y=\frac{19}{11}z$$
$$
\left[\begin{array}{r}
a\\b\\c
\end{array}\right]
=\left[\begin{array}{r}
-14\\19\\11
\end{array}\right]z
$$
Similarly for $2×2$ matrix .
|
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"timestamp": "2023-03-29T00:00:00",
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|
How i show this beautiful inequality :$\frac{x^n}{x^m+y^m}+\frac{y^n}{y^m+z^m}+\frac{z^n}{z^m+x^m}\geq \frac{3} {2}(\frac{1}{\sqrt{3}})^{n-m}$? let $m,n$ be integers, show that if $ n>m\geq 0 $ :
$$\frac{x^n}{x^m+y^m}+\frac{y^n}{y^m+z^m}+\frac{z^n}{z^m+x^m}\geq \frac{3}
{2}\left(\frac{1}{\sqrt{3}}\right)^{n-m}$$
where real $x,y,z > 0 $ and $xy + yz + zx = 1$
Thank you for your help .
|
This is a sketch with some details I didn't want to fill in but they'll check out. Let $n>m\geq 0$ be fixed and
$$f(x,y,z) = \frac{x^n}{x^m+y^m}+\frac{y^n}{y^m+z^m}+\frac{z^n}{z^m+x^m}$$
we'll try to find the minimum under $xz+zy+yx=1$ using Lagrange multipliers. Let
$$ g(x,y,z) = xz + zy+ yx$$
Following the idea of the multipliers, we try to solve
$$ \nabla f = \lambda \nabla g$$
We see the first relationship as
$$ \frac{(n-m) x^{n-1}x^m+ n y^m x^{n-1}}{(x^m + y^m)^2}-m\frac{ x^{m-1} z^n }{(z^m + x^m)^2} = \lambda(z+y) $$
It's not so easy to see, but since all 3 equations are the same having $x,y,z$ mixed up, it means we must have $x=y=z= \frac{1}{\sqrt{3}}$ for the three equations to hold with
$$ \lambda = \frac{ n-m}{4} \left ( \frac{1}{\sqrt{3}} \right) ^{n-m-2} $$
Also, you'll see that this value is a minimum by checking the Hessian(or by a clever intuition), use $g(x,y,z)=1$ ,$x,y,z,>0$ and $n >m$. Thus
$$f(x,y,z) \Big | _g \geq \min_{x,y,z} f(x,y,z) \Big | _g = f \left(\frac{1}{\sqrt{3}} , \frac{1}{\sqrt{3}} , \frac{1}{\sqrt{3}} \right)= \frac{3}{2} \left ( \frac{1}{\sqrt{3}} \right)^{n-m} $$
Edit: It occurred to me I should say something about the max of $f(x,y,z) \big| _g$, you'll see it occurs at infinity since we may take $ y = \epsilon \approx 0$ then
$$xz \approx 1 \implies x \approx \frac{1}{z}$$
So as $x \to \infty$ we see that
$$f(x,y,z) \Big |_g = \mathcal{O} (x^{n-m}) \to \infty $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Calculate the sum $\sum_{n=3}^{\infty}\frac{4n-3}{n^3-4n}$ $$\sum_{n=3}^{\infty}\frac{4n-3}{n^3-4n}$$
I think it is related to power series, because it is the topic, but I have no idea how to get there.
Could you give a hint?
|
By partial fractions
$$\begin{align}\sum\limits_{n=3}^\infty\frac{4n-3}{n^3-4n}&=\sum\limits_{n=3}^\infty\frac{5}{8(n-2)}+\frac{3}{4n}-\frac{11}{8(n+2)}
\\&=\sum\limits_{n=1}^4\frac{5}{8n}+\sum\limits_{n=3}^4\frac{3}{4n}+\sum\limits_{n=5}^\infty\frac{5}{8n}+\frac{3}{4n}-\frac{11}{8n}
\\&=\sum\limits_{n=1}^4\frac{5}{8n}+\sum\limits_{n=3}^4\frac{3}{4n}
\\&=\frac{167}{96}
\end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Is it possible to have an $n\times n$ real matrix $A$ such that $A^TA$ has an eigenvalue of $-1$?
Question: Is it possible to have an $n\times n$ real matrix $A$ such that $A^TA$ has an eigenvalue of $-1$?
I can prove that it is not possible for $n=1,2$, but I am not sure for the general case.
Case $n=1$: $a^2v=-v$ $\implies$ $(a^2+1)v=0$ $\implies$ $v=0$.
Case $n=2$: Write $A=\begin{pmatrix}a&b \\ c&d\end{pmatrix}$. Then, $A^T A=\begin{pmatrix}a^2+c^2&ab+cd\\ ab+cd&b^2+d^2\end{pmatrix}$, so
$$
\begin{align}
\det(A^TA+1)
&= \begin{vmatrix}a^2+c^2+1&ab+cd\\ ab+cd&b^2+d^2+1\end{vmatrix}\\
&= 1+a^2+b^2+c^2+d^2+(ad-bc)^2\\
&\neq 0.
\end{align}
$$
Now, can we generalize to all $n$?
|
According to wiki, the matrix $A^TA$ is positive definite for any non-singular $A$. The proof there is easily modified to show that singular $A$ give $A^TA$ positive semi-definite.
|
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|
Calculus I: Find $y'$ equation: $x^4y+\sqrt{xy}=5$ I need to find $y'$ from the equation $x^4y+\sqrt(xy)=5$
Wolffram and Mathway(pro) are giving me way different answers and I can not follow what they are doing and where I am going wrong.
My attempt:
$$x^4y+\sqrt{xy}=5$$
$$x^4y+(xy)^{1/2} = 5$$
$$x^4y'+y(4x^3)+\frac{1}{2}(xy)^\frac{-1}{2}(xy'+y)=0$$
$$y'(x^4+\frac{1}{2}(xy)^{1/2}(x+y))=-4x^3y$$
$$y' = -4x^3y / (x^4+1/2(xy)^{1/2}(x+y))$$
|
Use multiplication rule:
$$(x^4y)'=4x^3y+x^4y'$$
$$(\sqrt{xy})'=\frac{y}{2\sqrt{xy}}+\frac{xy'}{2\sqrt{xy}}$$
$$5'=0$$
We get
$$4x^3y+x^4y'+\frac{y}{2\sqrt{xy}}+\frac{xy'}{2\sqrt{xy}}=0$$
$$x^4y'+\frac{xy'}{2\sqrt{xy}}=-4x^3y-\frac{y}{2\sqrt{xy}}$$
$$y'\bigg(x^4+\frac{x}{2\sqrt{xy}}\bigg)=-4x^3y-\frac{y}{2\sqrt{xy}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1311243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
$(1+i)^6$ in polar form $re^{i\theta}$ I used De Moivre's formula and got
$$
\left(\frac{\sqrt{2}}{2}\right)^6 \times
\cos\left(6 \times \frac{1}{4\pi}\right) +
i\sin\left(6 \times \frac{1}{4\pi}\right) =
\frac{1}{8} e^{\frac{3}{2\pi}}.
$$
But the answer is $8e^{\frac{3}{2\pi}}$, can someone explain where the $8$ is from?
Thanks!
|
$$(1+i)^6=\left(\sqrt{2}e^{\frac{\pi}{4}i}\right)^6=\left(\sqrt{2}\right)^6e^{6\left(\frac{\pi}{4}i\right)}=8e^{\frac{3}{2}\pi i}=8e^{-\frac{\pi}{2}i}=-8i$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Combinatorial identity of $\sum\limits_{k=1}^n \frac{{n \choose k}.{(-1)^{k}}}{k+1}$ I don't know how to deal with this example: Find a closed form of
$$
\sum\limits_{k=1}^n \frac{{n \choose k}.{(-1)^{k}}}{k+1}
$$
|
$\sum_{k=0}^n {n \choose k}x^k=(1+x)^n$
Integrating both sides to get
$\sum_{k=0}^n {n \choose k}\frac{x^{k+1}}{k+1}=\frac{(1+x)^{n+1}}{n+1}+c$
For $x=0$ we have $0=\frac{1}{n+1}+c$, i.e. $c=-\frac{1}{n+1}$
$x\sum_{k=0}^n {n \choose k}\frac{x^{k}}{k+1}=\frac{(1+x)^{n+1}}{n+1}-\frac{1}{n+1}$
For $x=-1$ we have
$-\sum_{k=0}^n {n \choose k}\frac{(-1)^{k}}{k+1}=\frac{(1-1)^{n+1}}{n+1}-\frac{1}{n+1}=-\frac{1}{n+1}$
So $\sum_{k=1}^n {n \choose k}\frac{(-1)^{k}}{k+1}=\frac{1}{n+1}-{n \choose 0}\frac{(-1)^0}{1}=\frac{1}{n+1}-1$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Minimal Polynomial of $\sqrt{2}+\sqrt{3}+\sqrt{5}$ To find the above minimal polynomial, let
$$x=\sqrt{2}+\sqrt{3}+\sqrt{5}$$
$$x^2=10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}$$
Subtracting 10 and squaring gives
$$x^4-20x^2+100=4(31+2\sqrt{60}+2\sqrt{90}+2\sqrt{150})$$
$$x^4-20x^2+100=4(31+4\sqrt{15}+6\sqrt{10}+10\sqrt{6})$$
$$x^4-20x^2-24=40\sqrt{6}+24\sqrt{10}+16\sqrt{15}$$
$$x^4-20x^2-24=8(2\sqrt{6}+2\sqrt{10}+2\sqrt{15})+24\sqrt{6}+8\sqrt{10}$$
$$x^4-20x^2-24=8(x^2-10)+24\sqrt{6}+8\sqrt{10}$$
$$x^4-28x^2-104=24\sqrt{6}+8\sqrt{10}$$
Again, squaring both sides
$$x^8-56x^6+576x^4+5428x^2+10816=4096+765\sqrt{6}$$
But if I square again, I will get a degree 16 polynomial. Mathematica says the minimal polynomial is degree 8, which would make sense since elements of $\mathbb{Q}[\sqrt{2},\sqrt{3},\sqrt{5}]$ look like
$$a+b\sqrt{2}+c\sqrt{3}+d\sqrt{5}+e\sqrt{6}+f\sqrt{10}+g\sqrt{15}+h\sqrt{30}$$
Where am I making mistakes?
|
The 'easy' way to get at the minimal polynomial in this case is to take the product of all the terms $x\pm\sqrt2\pm\sqrt3\pm\sqrt5$; this is, in essence, because the Galois group over $\mathbb{Q}$ in this case is just $(\mathbb{Z}/2\mathbb{Z})^3$, with each of the three copies of $\mathbb{Z}/2\mathbb{Z}$ corresponding to a sign change on one of the three square root terms. The simple approach (take the product of $(x-\alpha)$ over all of the possible $\alpha$ obtained by applying the automorphisms in the Galois group to $\sqrt2+\sqrt3+\sqrt5$) works here because $\sqrt2$, $\sqrt3$ and $\sqrt5$ are 'mutually irreducible' over $\mathbb{Q}$ (in the sense that $\sqrt2\not\in\mathbb{Q}(\sqrt{3}, \sqrt{5})$, etc.) , so we can compose extensions in a clean fashion.
|
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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.