Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Principal roots in the derivation of the quadratic formula The following is an abbreviation of a common derivation of the quadratic formula:
$$
ax^2+bx+c=0\\
\vdots\\
\sqrt{(x+\frac{b}{2a})^2}=\sqrt{\frac{b^2-4ac}{4a^2}}\\
x+\frac{b}{2a}=\frac{\pm\sqrt{b^2-4ac}}{2a}\\
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\
$$
Following the square root extraction, the equation should have three plus-minus signs:
$$
\pm(x+\frac{b}{2a})=\frac{\pm\sqrt{b^2-4ac}}{\pm2a}\\
$$
But instead there's only one. Is it because the signs cancel each other out, or are moved from one expression to the other (through multiplication in $-1$ or $\frac{-1}{-1}$) in six of the eight permutations that the three plus-minus signs allow; or is it a matter of choosing a principal root for some arcane reason beyond the scope of elementary algebra (eg. branch points)?
| Actually, there're cyclic properties for higher order equations.
For cubic $x^{3}+3px=2q$,
$$x_{k}=e^{\frac{2i\pi k}{3}} \sqrt[3]{q+\sqrt{p^{3}+q^{2}}}+
e^{\frac{4i\pi k}{3}} \sqrt[3]{q-\sqrt{p^{3}+q^{2}}}$$ for $k=0,1,2$.
We have same set of roots by writting $x_{k}$ as
$$e^{\frac{4i\pi k}{3}} \sqrt[3]{q+\sqrt{p^{3}+q^{2}}}+
e^{\frac{2i\pi k}{3}} \sqrt[3]{q-\sqrt{p^{3}+q^{2}}}$$
For biquadratic $x^{4}-2ax^{2}+b=0$,
$$x=\pm \sqrt{a \pm \sqrt{a^{2}-b}}=
\pm \sqrt{\frac{a+\sqrt{b}}{2}} \pm \sqrt{\frac{a-\sqrt{b}}{2}}$$
Four combinations of $\pm$ signs giving four distinct roots.
Also de Moivre's quintic $x^{5}+5ax^{3}+5a^{2}x=2b$,
$$x_{k}=e^{\frac{2i\pi k}{5}} \sqrt[5]{b+\sqrt{a^{5}+b^{2}}}+
e^{\frac{8i\pi k}{5}} \sqrt[5]{b-\sqrt{a^{5}+b^{2}}}$$ for $k=0,1,2,3,4$.
It also gives same set of roots by writing $x_{k}$ as
$$x_{k}=e^{\frac{4i\pi k}{5}} \sqrt[5]{b+\sqrt{a^{5}+b^{2}}}+
e^{\frac{6i\pi k}{5}} \sqrt[5]{b-\sqrt{a^{5}+b^{2}}}$$
Last but not least, there are $n$ roots for $n$ degree polynomial equation. Redundancy of root (don't mess up with multiple roots) means you've done an extra cycle, but make sure you don't miss any.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1654541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Prove $e^{z+w}=e^ze^w$ for some $z,w\in\mathbb{C}$ by power series. Prove that
$$e^{z+w} = \sum_{k=0}^\infty \frac{z^k}{k!}\sum_{m=0}^\infty \frac{w^m}{m!}=e^ze^w$$
The proof provided is:
\begin{align}
\sum_{n=0}^\infty \frac{(z+w)^n}{n!} &= \sum_{n=0}^\infty \frac{1}{n!} \sum_{k=0}^n \frac{n!}{k!(n-k)!} z^{n-k}w^k \\
&= \sum_{n=0}^\infty \sum_{k=0}^n \frac{z^{n-k}w^k}{k!(n-k)!} \tag{2} \\
&= \sum_{k=0}^\infty \frac{w^k}{k!}\sum_{n=0}^\infty \frac{z^{n-k}}{(n-k)!}\tag{3} \\
&= \sum_{k=0}^\infty \frac{w^k}{k!}\sum_{m=0}^\infty \frac{z^{m}}{m!} & (m=n-k)\tag{4}
\end{align}
Now from line $(2)$ to line $(3)$ why does the upper limit of the $k$ summation change to infinity? On the line $(4)$ why is the bottom limit of the inner sum not $m=-k$ as opposed to $m=0$?
| Note that for $k \ge n$ it is $\frac{1}{(n - k)!} = 0$.
Or go the other way around:
$\begin{align}
\mathrm{e}^w \cdot \mathrm{e}^z
&= \sum_{r \ge 0} \frac{w^r}{r!} \cdot \sum_{s \ge 0} \frac{z^s}{s!} \\
&= \sum_{n \ge 0} \sum_{0 \le k \le n} \frac{w^k z^{n - k}}{k! (n - k)!} \\
&= \sum_{n \ge 0} \frac{1}{n!}
\sum_{0 \le k \le n} \frac{n!}{k! (n - k)!} w^k z^{n - k} \\
&= \sum_{n \ge 0} \frac{1}{n!}
\sum_{0 \le k \le n} \binom{n}{k} w^k z^{n - k} \\
&= \sum_{n \ge 0} \frac{1}{n!} (w + z)^n \\
&= \mathrm{e}^{w + z}
\end{align}$
Here we use first the Cauchy product of series (i.e., gather all terms with exponent of $w$ and $z$ adding up to $n$), and later the binomial theorem.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\int_0^\infty \frac{x^{k-1} + x^{-k-1}}{x^a + x^{-a}}dx = \frac{\pi}{a \cos(\frac{\pi k}{2a})}$. I need help in proving this identity
$$\int_0^\infty \frac{x^{k-1} + x^{-k-1}}{x^a + x^{-a}}dx = \frac{\pi}{a \cos(\frac{\pi k}{2a})}$$
for $0<k<a.$ It might be done using residues, but I don't know which contour to choose.
| From this post, we know that
$$ \int_0^\infty\frac{u^{\alpha-1}}{1+u^\beta}dx=\frac{\pi}{\beta\sin(\frac{\pi\alpha}{\beta})}.$$
So
\begin{eqnarray}
\int_0^\infty \frac{x^{k-1} + x^{-k-1}}{x^a + x^{-a}}dx &=&
\int_0^\infty \frac{x^{a+k-1}}{1 + x^{2a}}dx+\int_0^\infty \frac{x^{a-k-1}}{1 + x^{2a}}dx\\
&=&\frac{\pi}{2a\sin(\frac{(a+k-1)\pi}{2a})}+\frac{\pi}{2a\sin(\frac{(a-k-1)\pi}{2a})}\\
&=& \frac{\pi}{a \cos(\frac{\pi k}{2a})}
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1657473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Proof of $n(n^2+5)$ is divisible by 6 for all integer $n \ge 1$ by mathematical induction
Prove the following statement by mathematical induction:
$n(n^2+5)$ is divisible by 6 for all integer $n \ge 1$
My attempt:
Let the given statement be p(n).
(1) $1(1^2+5)$=6 Hence, p(1) is true.
(2) Suppose for all integer $k \ge 1$, p(k) is true.
That is, $k(k^2+5)$ is divisible by 6
We must show that p(k+1) is true.
$(k+1)((k+1)^2+5)$=$k^3+3k^2+3k+1+5(k+1)$
=$k^3+3k^2+8k+6$
=$k(k^2+5)+3k^2+3k+6$
I'm stuck on this step. I feel I have to show $3k^2+3k+6$ is divisible by 6. But, how can I show $3k^2+3k+6$ is divisible by 6?
| Hint:
let $$f(n)=n(n^2+5),n\geq1$$ then
$$f(n+1)=(n+1)(n^2+2n+6)=n^3+2n^2+6n+n^2+2n+6=n^3+3n^2+8n+6=$$
$$=n^3+5n+3n^2+3n+6=3(n^2+n)+6+n(n^2+5)=f(n)+6\left(\frac{n(n+1)}{2}+1\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1658054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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More elegant way of verifying the solution of $y'=\frac{-(x+2)+\sqrt{x^2+4x+4y}}{2}$? The question asks to verify that for any value of $c$, $y=c^2+cx+2c+1$ satisfies the solution.
$$y'=\frac{-(x+2)+\sqrt{x^2+4x+4y}}{2}\tag{1}$$
I am aware we can just prove this by direct substitution.
However, I think it's no coincidence that the author specifically put the DE in the form of a solution to a quadratic equation. $(1)$ can be interpreted as a solution of the quadractic equation $$(y')^2+(x+2)y'+(1-y)=0$$
Which can be factored to $$(y'-c)(y'+c+x+2)=0\tag{2}$$
The first factor gives $y'=c$, which comes from the solution we were already given $(y=c^2+cx+2c+1)$. The second factor we just added when we constructed the quadratic equation, and it's not necessarily a solution of $(1)$.
However, I don't think we can conclude that $y=c^2+cx+2c+1$ is a solution of $(1)$ just because $y'=c$ is a solution of $(2)$. Is there any way to complete the question from here, or is another approach neccesary?
| Since you're given $y = c^2+cx+2c+1$, we can first verify that $y^{\prime} = c$ by taking the first derivative.
Next we can arrange $y = c^2+cx+2c+1$ into a standard form polynomial of $c$.
$$y = c^2+cx+2c+1 \implies 0 = c^2+cx+2c+1 - y = (1)c^2+(x+2)c+(1 - y)$$
Now using the quadratic formula we find the roots of our polynomial to solve for $c$. Our goal isn't really to solve $c$, but rather the form that results is quite familiar.
$$c = \frac{-(x+2) \pm \sqrt{(x^2+2)-4(1)(1-y)}}{2(1)} = \frac{-(x+2) \pm \sqrt{x^2+4x+4y}}{2}$$
Since we know from our derivative and the given differential equation that $c = y^{\prime} = \frac{-(x+2) + \sqrt{x^2+4x+4y}}{2}$ and solving the candidate solution for $c$ tells us that $c = \frac{-(x+2) \pm \sqrt{x^2+4x+4y}}{2}$, we simply take the solution for $c$ to be the branch where we add the radical to show that our candidate solution satisfies the differential equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1660741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Integrate $\int \frac{\mathrm dx}{(x-2)(\sqrt{x^2-4x+3})}$
Integrate $$\int \frac{\mathrm dx}{(x-2)(\sqrt{x^2-4x+3})}$$
$$\int \frac{\mathrm dx}{(x-2)(\sqrt{x^2-4x+3})}=\int \frac{\mathrm dx}{(x-2)(\sqrt{(x-2)^2-(1)^2})}=\operatorname{arcsec}(x-2)+\rm C$$
Wolfram writes that the answer is $-\arctan\left(\frac{1}{\sqrt{x^2-4x+3}}\right)+\rm C\;.$
| $$\begin{align}
\int\frac{\mathrm{d}x}{\left(x-2\right)\sqrt{x^{2}-4x+3}}
&=\int\frac{\mathrm{d}x}{\left(x-2\right)\sqrt{\left(x-1\right)\left(x-3\right)}}\\
&=\int\frac{\mathrm{d}t}{t\sqrt{t^{2}-1}};~~~\small{\left[x=t+2\right]}\\
&=\int\frac{1}{t^{2}}\cdot\frac{t\,\mathrm{d}t}{\sqrt{t^{2}-1}}\\
&=\int\frac{\mathrm{d}u}{1+u^{2}};~~~\small{\left[\sqrt{t^{2}-1}=u\right]}\\
&=-\int\frac{\left(-u^{-2}\right)\,\mathrm{d}u}{u^{-2}+1}\\
&=-\int\frac{\mathrm{d}w}{w^{2}+1};~~~\small{\left[u^{-1}=w\right]}\\
&=-\arctan{\left(w\right)}+\color{grey}{constant}\\
&=-\arctan{\left(\frac{1}{u}\right)}+\color{grey}{constant}\\
&=-\arctan{\left(\frac{1}{\sqrt{t^{2}-1}}\right)}+\color{grey}{constant}\\
&=-\arctan{\left(\frac{1}{\sqrt{\left(x-2\right)^{2}-1}}\right)}+\color{grey}{constant}.\blacksquare\\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1661343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Fraction involving Surds Fraction involving Surds. Can anyone please show me the working out?
$$
\frac{(\sqrt{6}-1)}{\sqrt{3}} + \frac{(\sqrt{6}+2)}{2\sqrt{3}}
$$
I did this and it was incorrect:
$$
2\sqrt{3}(\sqrt{6}-1) + \frac{\sqrt{3}(\sqrt{6} + 2)}{\sqrt{3}\times 2\sqrt{3}}
$$
$$
2\sqrt{18} - 2\sqrt{3} + \sqrt{18} + \frac{2\sqrt{3}}{\sqrt{3}\times 2\sqrt{3}}
$$
Thanks!
Edit: This is my answer, which I now understand from my math teacher:
$$
\frac{(\sqrt{6}-1)}{\sqrt{3}} + \frac{(\sqrt{6}+2)}{2\sqrt{3}}
$$
$$
\frac{2sqrt{3}(\sqrt{6}-1)}{\sqrt{3}} + \frac{sqrt3(\sqrt{6}+2)}{2\sqrt{3}}
$$
$$
2\sqrt{18} - 2\sqrt{3} + sqrt{18} + 2\sqrt{3}\6
| $\begin{array}\\
\frac{(\sqrt{6}-1)}{\sqrt{3}} + \frac{(\sqrt{6}+2)}{2\sqrt{3}}
&=\frac{2(\sqrt{6}-1)}{2\sqrt{3}} + \frac{(\sqrt{6}+2)}{2\sqrt{3}}
\qquad\text{put the fractions over a common denominator}\\
&=\frac{2(\sqrt{6}-1)+(\sqrt{6}+2)}{2\sqrt{3}}
\qquad\text{add the numerators}\\
&=\frac{2\sqrt{6}-2+\sqrt{6}+2}{2\sqrt{3}}
\qquad\text{use the distributive law to get rid of parentheses}\\
&=\frac{3\sqrt{6}}{2\sqrt{3}}
\qquad\text{simple algebra}\\
&=\frac{3\sqrt{2\cdot 3}}{2\sqrt{3}}
\qquad\text{setup to get rid of }\sqrt{3}\\
&=\frac{3\sqrt{2}\sqrt{ 3}}{2\sqrt{3}}
\qquad\text{use }\sqrt{xy} = \sqrt{x}\sqrt{y}\\
&=\frac{3\sqrt{2}}{2}
\qquad\text{and its gone}\\
\end{array}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1663819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Exponential equation with same, unknown bases I have the following equation:
$(x-3)^{(x^2-x)} = (x-3)^2$
The book says solutions are: $x_1 = -1, x_2 = 2, x_3 = 3, x_4 = 4$
I was only able to get -1, 2, 4 by doing this:
$(x-3)^{(x^2-x)} = (x-3)^2 / \log$
$(x^2-x) \cdot \log(x-3) = 2\log(x-3)$ (exponents out)
$(x^2-x) \cdot \log(x-3) - 2\log(x-3) = 0$
$\log(x-3)(x^2-x - 2) = 0$
$\log(x-3) = 0 \; \lor \; x^2-x-2=0 $
$x-3=1 \; \lor \; x^2-x-2=0 $
$x_1 = 4, \; x_2 = -1, \; x_3 = 2$
The explanation in the book says only this:
$x-3 = 1 \; \lor \; (x-3 = 0 \; \land \; x^2-x \ne 0) \; \lor \; (x-3 \ne 0 \; \land \; x^2-x=2)$
What exactly am I missing here? I don't quite undersand the book's explanation though.
| If $a^m=a^n$
either $m=n$
or $a=0, mn>0$
or $a=1$
or $a=-1,m-n=$ some even integer
By taking logarithm, we implicitly assume $a\ne0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1664114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Compare five ways of solving cubic equation by iterations (nested expressions) Say we have a depressed cubic equation in the general form:
$$x^3-bx-c=0$$
There are basically five ways of solving it by iterations. Let's consider them in no particular order (the names are my own).
1) The continued fraction / nested radical method:
$$x=\sqrt{b+\frac{c}{x}}=\sqrt{b+\cfrac{c}{\sqrt{b+\cfrac{c}{x}}}}=\sqrt{b+\cfrac{c}{\sqrt{b+\cfrac{c}{\sqrt{b+\cfrac{c}{x}}}}}}$$
Take some $x_0$ and solve the recurrence:
$$x_n=\sqrt{b+\frac{c}{x_{n-1}}}$$
2) The continued fraction / nested square method:
$$x=-\cfrac{c}{b-x^2}=-\cfrac{c}{b-\cfrac{c^2}{(b-x^2)^2}}=-\cfrac{c}{b-\cfrac{c^2}{\left( b-\cfrac{c^2}{(b-x^2)^2} \right)^2}}$$
3) The branching continued fraction method:
$$x=\cfrac{b}{x}+\cfrac{c}{x^2}=\cfrac{b}{\cfrac{b}{x}+\cfrac{c}{x^2}}+\cfrac{c}{\left( \cfrac{b}{x}+\cfrac{c}{x^2} \right)^2}$$
It can also be expressed in several different ways, like this one:
$$x=\cfrac{c+bx}{x^2}=\cfrac{c+b\cfrac{c+bx}{x^2}}{\cfrac{(c+bx)^2}{x^4}}$$
But anyway, for some $x_0$:
$$x_n=\cfrac{b}{x_{n-1}}+\cfrac{c}{x_{n-1}^2}$$
4) Nested cubic roots method:
$$x=\sqrt[3]{c+bx}=\sqrt[3]{c+b\sqrt[3]{c+bx}}=\sqrt[3]{c+b\sqrt[3]{c+b\sqrt[3]{c+bx}}}$$
5) And nested cubes method:
$$x=\frac{1}{b} \left(-c+x^3 \right)=\frac{1}{b} \left(-c+\frac{1}{b^3} \left(-c+x^3 \right)^3 \right)$$
They all work in different ways, so I wonder, what are general rules of using one method or the other? How to get all three roots by using them?
For example we have the equation with three real roots $(-3.65617,0.42297,3.23319)$:
$$x^3-12x+5=0$$
Let's take $x_0=1$. Then methods 1, 3 and 4 will converge to the third root, while methods 2 and 5 will converge to the second root.
Changing $x_0$ to other values will not lead to the first root, but some methods will not converge (mainly, method 3).
| Let me try two general methods and have look if these show up in your list:
$$
f(x) = x^3 - bx - c = 0
$$
Newton-Raphson-Iteration
$$
x_{n+1}
= x_n - \frac{f(x_n)}{f'(x_n)}
= x_n - \frac{x_n^3 - b x_n - c}{3x_n^2 - b}
$$
Fixed point iteration:
$$
x^3 - bx - c = 0 \iff \\
x^3 - c = b x
$$
If $b \ne 0$ we can write this as
$$
x^3 - bx - c = 0 \iff \\
F(x) = \frac{x^3-c}{b} = x
$$
which leads to the iteration
$$
x_{n+1} = \frac{x_n^3-c}{b}
$$
The Netwon-Raphson I have not spotted, but the fixed point iteration is your number 5 (nested cubes).
Newton-Raphson has quadratic convergence, so it is fast.
Fixed point iteration features attractive $\lvert F(x^*) \rvert < 1$ and not attractive fixed points $x^*$. In the latter case one could try to iterate $x = F^{-1}(x)$, because $(F^{-1})'(x) = 1 / F'(F^{-1}(x))$.
The properties of these iterations result from the properties of their general method.
Note that the different roots (up to three), need different start values for the iteration.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1669163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve system of simultaneous equations in $3$ variables: $x+y+xy=19$, $y+z+yz=11$, $z+x+zx=14$ Solve the following equation system:
$$x+y+xy=19$$
$$y+z+yz=11$$
$$z+x+zx=14$$
I've tried substituting, adding, subtracting, multiplying... Nothing works. Could anyone drop me a few hints without actually solving it? Thanks!
| Add $1$ to both sides of all the equations. To get
\begin{align*}
(x+1)(y+1) & = 20\\
(y+1)(z+1) & = 12\\
(z+1)(x+1) & = 15\\
\end{align*}
Now let $u=x+1,v=y+1,w=z+1$. And you have
\begin{align*}
uv&=20\\
vw&=12\\
wu&=15
\end{align*}
From this you can get
$$(uvw)=\pm 60.$$
Now use the above equations to compute $u=\pm 5$ and so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1669228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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$1 + {1 \over 3} - {1 \over 2} + {1 \over 5} + {1 \over 7} - {1 \over 4} + {1 \over 9} + {1 \over 11} - {1 \over 6} + +-...$ conditionally convergent $$1 + {1 \over 3} - {1 \over 2} + {1 \over 5} + {1 \over 7} - {1 \over 4} + {1 \over 9} + {1 \over 11} - {1 \over 6} + +-...$$
I want to show first that $S_{3n}$, $S_{3n+1}$ and $S_{3n+2}$ converges to the same limit, I show
$$S_{3n} = (1 + {1 \over 3} - {1 \over 2}) + ({1 \over 5} + {1 \over 7} - {1 \over 4}) + ({1 \over 9} + {1 \over 11} - {1 \over 6}) + ... + ({1 \over 4n-3} + {1 \over 4n-1} - {1 \over 2n})$$
but how to proceed from here?
| $$\frac{1}{4n-3}+\frac{1}{4n-1}-\frac{1}{2n}=\frac{1}{4n-3}+\frac{1}{4n-1}-\frac{1}{4n}-\frac{1}{4n}=\frac{1}{4n-3}-\frac{1}{4n-2}+\frac{1}{4n-1}-\frac{1}{4n}+\frac{1}{2}\left(\frac{1}{2n-1}-\frac{1}{2n}\right)$$ so $$\lim _{n\to \infty}S=\frac{3}{2}\lim _{n\to \infty}\sum _n\left(\frac{1}{2n-1}-\frac{1}{2n}\right)$$ and $$\log (x+1)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}..$$ so $$\log(2)=\log(1+1)=\sum \left(\frac{1}{2n-1}-\frac{1}{2n}\right)$$ so $S=\frac{3}{2}\log (2)$ Hope it will help
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1669982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Simplify the boolean function below by using algebra laws. I've been stuck on this question for some time, if anyone happens to solve it please explain step by step.
$$(A +B ) \times ( A' + C ) \times ( B + C )$$
| $(A +B ) \times ( A' + C ) \times ( B + C ) = [(A+B)\times A' + (A+B)\times C]\times (B+C) $
=$[A\times A'+B \times A' + A \times C + B \times C]\times (B+C) $
$A \times A' =0$
So, $B \times B \times A' +B \times A' \times C + A \times B \times C +A \times C \times C + B \times B \times C + B \times C \times C$
=$A' \times B + A' \times B \times C+ A \times B \times C+A \times C +B \times C = A' \times B + A \times C + B \times C$
write $B \times C$ as $A \times B \times C + A' \times B \times C$
=$ A' \times B + A \times C + A \times B \times C + A' \times B \times C$
=$A' \times B + A' \times B \times C+ A \times C + A \times B \times C$
=$A' \times B + A \times C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1671698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Using the $\epsilon - \delta$ definition for limit show that $\lim_{(x,y)\to (0,0)} \frac{3x^3+2y^3}{x^2+\sin^2(xy)+y^2}=0$. Using the $\epsilon - \delta$ definition for limit show that $\lim_{(x,y)\to (0,0)} \frac{3x^3+2y^3}{x^2+\sin^2(xy)+y^2}=0$.
My answer: $\frac{3x^3+2y^3}{x^2+\sin^2(xy)+y^2} \leq \frac{3x^3+2y^3}{x^2+y^2} \leq 3\frac{x^3+y^3}{x^2+y^2}$.
I somehow need need to relate $x^3+y^3$ to $x^2+y^2$ so that I can bound the fraction. But I cant seem to arrive at a relation that would work. Help please!
| Suppose that $(x,y)\ne (0,0)$. Then
$$\small \left|\frac{3x^3+2y^3}{x^2+\sin^2(xy)+y^2}\right|\le \left|\frac{3x^3+2y^3}{x^2+y^2}\right|\le 3|x|\frac{x^2}{x^2+y^2}+2|y|\frac{y^2}{x^2+y^2}\le 3|x|+2|y|\le 5\sqrt{x^2+y^2}.$$
Now finding a suitable $\delta$ will be straightforward.
| {
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"url": "https://math.stackexchange.com/questions/1673363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Determining whether there are solutions to the cubic polynomial equation $x^3 - x = k - k^3$ other than $x = -k$ for a given parameter $k$ Let $k$ be a real parameter, and consider the equation
$$x^3 - x = k - k^3 .$$
Obviously, $x=-k$ is a solution. Is it the only one? How to prove it?
| Write it as
$$
x^3+k^3=x+k
$$
so that you get
$$
(x+k)(x^2-kx+k^2)=x+k
$$
If $x=-k$, you have a solution, otherwise
$$
x^2-kx+k^2-1=0
$$
and the roots are
$$
\frac{k\pm\sqrt{4-3k^2}}{2}
$$
so you two other roots when $4-3k^2>0$, just one more if $4-3k^2=0$.
Well, in the case $4-3k^2$ one of the roots could be the same as $-k$. This happens precisely when $-k$ is a root of $x^2-kx+k^2-1=0$, that is, when
$$
k^2+k^2+k^2-1=0
$$
that is, $k=1/\sqrt{3}$ or $k=-1/\sqrt{3}$.
So we have
*
*three solutions for $-2/\sqrt{3}<k<2/\sqrt{3}$, but $k\ne1/\sqrt{3}$ and $k\ne-1/\sqrt{3}$;
*two solutions for $k=\pm2/\sqrt{3}$ or $k=\pm1/\sqrt{3}$;
*one solution for $|k|>2/\sqrt{3}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Riemann integral of $ f(x) = \begin{cases} \frac{1}{n}, & \text{if $x=\frac{1}{n}, \ \ n=1, 2,3 ,\cdots$ } \\ 0, & \text{other where} \end{cases}$ Let $f:[0, 1]\to\mathbb{R}$ and $ f(x) =
\begin{cases}
\frac{1}{n}, & \text{if $x=\frac{1}{n}, \ \ n=1, 2,3 ,\cdots$ } \\
0, & \text{other where}
\end{cases}$
I want to find $\int_{0}^{1} f\, dx.$ I think the answer is $0$.
Any ideas or insight would be greatly appreciated
| For any $\epsilon > 0$, choose $n > \dfrac{2}{\epsilon}$, define a partition $P_n = \{0,\dfrac{1}{n}, 1-\dfrac{1}{n}, 1\}$, we have: $U(f,P_n)-L(f,P_n)= \dfrac{1}{n}\left(\dfrac{1}{n}-0\right)+0\left(1-\dfrac{1}{n}-\dfrac{1}{n}\right)+1\left(1-\left(1-\dfrac{1}{n}\right)\right)- 0= \dfrac{1}{n}+\dfrac{1}{n^2}< \dfrac{2}{n}< \epsilon$. Thus $f$ is Riemann integrable on $[0,1]\Rightarrow \displaystyle \int_{0}^1 f(x)dx = \displaystyle \lim_{n\to \infty} U(f,P_n) = \displaystyle \lim_{n\to \infty} \left(\dfrac{1}{n}+\dfrac{1}{n^2} \right)= 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A series of natural numbers is divided into groups A series of natural numbers is divided into groups: $$(1),(2,3,4), (5,6,7,8,9,10,11), (12,13,......,26)$$ and so on.
The sum of numbers in $n^{th}$ group is $a.2^{2n-1}-(2n+b)2^{2n-1}+n+1$. Then find $a+b$
The main problem I am facing here is to find first term of $n^{th}$ group. Series for first term is like $1,2,5,12,27,...$ In this series I observed that difference of difference is in G.P. but don't know how to use this information.
| The leading term of $n$-th group is $2^{n}-n$ so the last term is $2^{n+1}-n-2$.
Hence, the sum is
\begin{align*}
& \quad \frac{1}{2} (2^{n+1}-n-2+2^{n}-n)(2^{n+1}-n-2-2^{n}+n+1) \\
&= (3\cdot 2^{n-1}-n-1)(2^{n}-1) \\
&= 3\cdot 2^{2n-1}-3\cdot 2^{n-1}-n\cdot 2^{n}+n+1 \\
&= 3\cdot 2^{2n-1}-(2n+3) 2^{n-1}+n+1
\end{align*}
| {
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If a prime $p>2$ is expressible as $p=a^2+b^2$, then $4\mid (p-1)$ Show that if $p\in\mathbb{P}, p>2$ is a sum of squares, i.e $p=a^2+b^2$, then $p\equiv 1\pmod{4}$.
Have established that the remainder can't be either of $0,2$ for obvious reasons. So we must consider $1$ or $3$. Attempting assume by contradiction that $4\mid (p-3)$:
$$4k+3= a^2+b^2\Longleftrightarrow 4k + 2ab +3 = (a+b)^2$$
I can't find a contradiction, yet when I think about a prime divided by 4 giving the remainder of 3, for example 7, there is no way to express it as a sum of squares.
Another attempt at the same approach is $4k+3 = a^2+b^2$ so we have
$$4k-a^2=b^2-3$$
I believe I should somehow conclude one of the summands is not a square of an integer. How to proceed?
| First of all, recall that if $n$ is a perfect square then $n\equiv0,1\pmod4$:
*
*$k\equiv0\pmod4 \implies k^2\equiv0^2\equiv0\pmod4$
*$k\equiv1\pmod4 \implies k^2\equiv1^2\equiv1\pmod4$
*$k\equiv2\pmod4 \implies k^2\equiv2^2\equiv0\pmod4$
*$k\equiv3\pmod4 \implies k^2\equiv3^2\equiv1\pmod4$
Therefore, $a^2+b^2\equiv0,1,2\pmod4$.
*
*If $a^2+b^2\equiv0\pmod4$ then $a^2+b^2$ is divisible by $4$
*If $a^2+b^2\equiv2\pmod4$ then $a^2+b^2$ is divisible by $2$
Therefore, if $a^2+b^2$ is prime then $a^2+b^2\equiv1\pmod4$.
| {
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Condensation of an expression There is an expression: $(h+2p)^2(x+2y)^2 - 3p^2(x+2y)^2-3y^2(h+2p)^2+9p^2y^2$
Is there a way to simplify this to the form $l^2 + 4lk + k^2$? I tried to substitute $l = (h+2p)(x+2y)$ and $k= 3py$ and this takes care of the expressions at the two ends, but the expressions in the middle are stumping me.
Alternatively, one may try expressing it in the form $(l+2k)^2 - 3lk$, but I am unable to do this as well. However, I suspect that there is some relationship as $-3lk$ bears resemblance to $- 3p^2(x+2y)^2-3y^2(h+2p)^2$ if you factor out the $-3$. Is it possible to use Fibonnacci Identity in this endeavor?
EDIT: The original expression is $(h^2 + 4ph + p^2)(x^2 + 4xy + y^2)$
I converted this to $$((h+2p)^2-3p^2)((x+2y)^2-3y^2)$$
EDIT: I tried an alternative approach. I multiplied out this expression:
$$(h^2 + 4ph + p^2)(x^2 + 4xy + y^2)$$
and got the following:
$$h^2x^2 + p^2x^2 + h^2y^2 + p^2y^2 + 4hpx^2 + 4xyh^2 + 4xyp^2 +4hpy^2 + 16hpxy$$
(I'm leaving out some steps) This can be factored into :
$$(hx + py)^2+4(px+hy)(hx+py)+(px+hy)^2 +12hpxy$$
This would have been perfect except for the pesky $12hpxy$.
|
Answer:
$$(h x-p y)^2 + 4(h x-p y)(h y+p x+4 p y) +(h y+p x+4 p y)^2 = (h^2 + 4p h + p^2)(x^2 + 4x y + y^2) $$
How I found this result:
Rewrite
$$(h^2 + 4p h + p^2)(x^2 + 4x y + y^2) = ((h+2p)^2-3p^2)((x+2y)^2-3y^2) = (A^2-3B^2)(C^2-3D^2)$$
where $A=h+2p$, $B=p$, $C=x+2y$ and $D=y$. Now, we have
$$(A^2-3B^2)(C^2-3D^2) = (AC+3BD)^2 - 3(AD + BC)^2 = \lbrace[AC+3BD-2(AD+BC)]+2(AD+BC)\rbrace^2 - 3(AD + BC)^2 $$
Let $l=AC+3BD-2AD-2BC$ and $k=AD + BC$. Now, $(l+2k)^2-3k^2=l^2 +4lk + k^2 $, which is what we wanted. Finally, substitute $A$, $B$, $C$ and $D$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrating $\int \frac{\sin x-\cos x}{(\sin x+\cos x)\sqrt{(\sin x \cos x + \sin^2x\cos^2x)}}\,dx$ I came across a question today...
Integrate $\int \dfrac{\sin x-\cos x}{(\sin x+\cos x)\sqrt{(\sin x \cos x + \sin^2x\cos^2x)}}\,dx$
How to do it? I tried
1. to take $\sin x \cos x =t$ but no result
2. to convert the thing in the square root into $\sin x +\cos x$ so that I could take $\sin x + \cos x = t$ but then something I got is $\int\frac{-2}{t|t+1|\sqrt{t-1}}\,dt$. Now I don't know how to get past through it.
| Notice,
\begin{align}
& \qquad \int\frac{\sin x-\cos x}{(\sin x+\cos x)\sqrt{\sin x\cos x+\sin^2 x\cos^2x}}\ dx\\
& \quad=\int\frac{\sin x-\cos x}{(\sin x+\cos x)\sqrt{\frac{(\sin x+\cos x)^4-1}{4}}}\ dx\\
&\quad=2\int\frac{(\sin x-\cos x)dx}{(\sin x +\cos x)\sqrt{(\sin x+\cos x)^4-1}}
\end{align}
let $\sin x+\cos x=t\implies (\cos x-\sin x)\ dx=dt$,
$$=-2\int\frac{dt}{t\sqrt{t^4-1}} \qquad \qquad\qquad\qquad\qquad$$
let $t^4-1=u^2\implies 4t^3\ dt=2u\ du$,
\begin{align}
&=-2\int\frac{udu}{2u(u^2+1)}\\
&=-\int\frac{du}{1+u^2}\\
&=-\tan^{-1}(u)+C\\
&=-\tan^{-1}\left(\sqrt{t^4-1}\right)+C\\
&=-\tan^{-1}\left(2\sqrt{\sin x\cos x+\sin^2x\cos^2 x}\right)+C\\
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Minimum of $\frac{x}{1+y^2}+\frac{y}{1+x^2}$ on $x,y\ge 0$, $x+y=2$ let $x,y\ge 0$, and such $x+y=2$ find the minimum
$$\dfrac{x}{1+y^2}+\dfrac{y}{1+x^2}$$
I think $x=y=1$ is minimum of the value $1$,How can I prove?
| Hint:
The minimum is indeed when $x=y=1$. Using AM-GM or Cauchy Schwarz inequality, it is sufficient to show $xy^2+yx^2\le 2$.
Addendum:
$$\begin{align}\frac{x}{1+y^2}+\frac{x(1+y^2)}4 \ge x, &\quad \frac{y}{1+x^2}+\frac{y(1+x^2)}4 \ge y \\ \implies \frac{x}{1+y^2}+\frac{y}{1+x^2} &\ge \frac32-\frac{xy^2+yx^2}4 \\&= \frac32-\frac{x+y}4xy \\&\ge \frac32-\frac{x+y}4\frac{(x+y)^2}4=1\end{align} $$
| {
"language": "en",
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Solve for $x$ if $4^{\frac{x}{y} + \frac{y}{x}}$ $=$ $32$ and $\log_3(x+y)+\log_3(x-y)=1$
Question:
Solve for $x$ if $4^{\frac{x}{y} + \frac{y}{x}}$ $= 32$ and $\log_3(x+y)+\log_3(x-y)=1$
My attempt: With the first equation
$$4^{\frac{x}{y} + \frac{y}{x}} = 32$$
$$2^{2(\frac{x}{y} + \frac{y}{x})} = 2^5$$
$$ 2(\frac{x}{y} + \frac{y}{x}) = 5 $$
$$ \frac{x}{y} + \frac{y}{x} = \frac{5}{2} $$
$$ \frac{x^2 + y^2}{xy} = \frac{5}{2} $$
Now with the second equation
$$\log_3(x+y)+\log_3(x-y)=1$$
$$\log_3((x+y)(x-y)) = 1 $$
$$ \log_3(x^2-y^2) = 1 $$
$$ x^2-y^2 = 3$$
Now I have 2 equations:
$$ \frac{x^2 + y^2}{xy} = \frac{5}{2} $$
$$ x^2-y^2 = 3$$
Now I am stuck..
| The set $$ x^2 - y^2 = 3 $$ in the plane is a hyperbola. You can draw it. What kind of set is
$$ x^2 -\frac{5}{2}xy + y^2 = 0, $$ or
$$ 2x^2 -5xy + 2y^2 = 0? $$
The second version factors...............
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $p_n=q_{n-1}=\frac{\alpha^n-\beta^n}{\alpha-\beta}$
If $\frac{p_n}{q_n}$ be the nth convergent of the simple continued fraction $$\cfrac{1}{a+\cfrac{1}{a+\cfrac{1}{a+\ddots}}}$$ show that $$p_n=q_{n-1}=\frac{\alpha^n-\beta^n}{\alpha-\beta}$$ where, $\alpha$, $\beta$
are the roots of the equation $x^2-ax-1=0$
Attempt
I can solve the 1st part of the problem.
We have $p_n=ap_{n-1}+p_{n-2}$ and $q_n=aq_{n-1}+q_{n-2}$.
Also $\frac{p_n}{q_n}=\frac{1}{a+\frac{p_{n-1}}{q_{n-1}}}$. Equation the numerator we get $p_n=q_{n-1}$.
Now how to get the 2nd part of the problem.
| As you have already noted, we have $p_n = a p_{n - 1} + p_{n - 2}$ and $q_n = a q_{n - 1} + q_{n - 2}$. Moreover (as Daniel Fischer wrote in his comment already) we start with $p_0 = 0$, $p_1 = 1$. Let $\alpha, \beta$ be the roots of $X^2 - aX - 1$. The rest is induction:
For $n = 0$, we have $\frac{\alpha^0 - \beta^0}{\alpha - \beta} = 0 = p_0$.
For $n = 1$, we have $\frac{\alpha - \beta}{\alpha - \beta} = 1 = p_1$.
Let now $n > 1$. We have $p_n = a p_{n - 1} + p_{n - 2} = a \frac{\alpha^{n - 1} - \beta^{n - 1}}{\alpha - \beta} + \frac{\alpha^{n - 2} - \beta^{n - 2}}{\alpha - \beta}$ by the induction hypothesis. So $$(\alpha - \beta) p_n = a \alpha^{n - 1} + \alpha^{n - 2} - a\beta^{n - 1} - \beta^{n - 2}$$
By assumption we have $\alpha^2 - a \alpha - 1 = 0$ and so $a \alpha = \alpha^{2} - 1$ and similarly for $\beta$. Plugging this in the above equation we get: $$(\alpha - \beta) p_n = (\alpha^2 - 1) \alpha^{n - 2} + \alpha^{n - 2} - (\beta^2 - 1) \beta^{n - 2} - \beta^{n - 2} = \alpha^n - \beta^n$$
and hence $p_n = \frac{\alpha^n - \beta^n}{\alpha - \beta}$, which was to prove.
You might be interested to know that for $a = 1$ one can use the continued fraction $$1 + \frac{1}{1 + \frac{1}{1 + \cdots}}$$to approximate the golden ratio $\phi = \frac{1 + \sqrt{5}}{2}$ (which is in fact a root of $X^2 - X - 1$).
En effet, by staring at the fraction it is clear that it is bounded form below by $0$ and from above by $2$. Moreover, it is monotonically decreasing after some point, hence the fraction converges. We have $\frac{p_n}{q_n} = 1 + \frac{1}{\frac{p_{n - 1}}{q_{n - 1}}}$ and taking the limit $x := \lim_{n \rightarrow \infty} \frac{p_n}{q_n}$ we get $x = 1 + \frac{1}{x}$ or equivalently $x^2 - x - 1 = 0$. Solving this we get $x = \frac{1 + \sqrt{5}}{2}$ (for the limit must be positive).
| {
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"timestamp": "2023-03-29T00:00:00",
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Recovering a quadratic polynomial from three values using calculus I'm asked to solve this using calculus:
Let $$ f(x) = ax^2 + bx +c .$$ If $ f(1) = 3 $, $f(2) = 7$, $f(3) = 13$, then find $a$, $b$, and $f(0)$.
I know I can solve this using solving three equations simultaneously. And I can also solve this using Gauss Jordan or Gaussian elimination method by writing the augmented matrix. But I'm wondering is there any other method to solve this.
Solving by any method it turns out that $a = b = c = 1$.
| A solution that deliberately doesn't use any calculus...
Given:
\begin{align}
f(1) \colon\quad \phantom{4}a+\phantom{2}b+c &= 3 \tag{1}\\
f(2) \colon\quad 4a+2b+c&=7 \tag{2}\\
f(3) \colon\quad 9a+3b+c&=13 \tag{3}
\end{align}
Subtract $(1)$ from $(2)$ and $(3)$, respectively:
\begin{align}
3a+\phantom{2}b &= 4 \tag{A}\\
8a+2b &= 10 \tag{B}
\end{align}
Subtract $2(A)$ from $(B)$: $2a=2$, or $a=1$.
Plug $a$ into $(A)$: $b=4-3a=4-3=1$.
Plug $a$ and $b$ into $(1)$: $c=3-a-b=3-1-1= 1$.
Hence, $f(x)=x^2+x+1$ and $f(0)=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Mistake in proof for sequence of cubes I know there are thousands of proofs for this to have a look at, but I started one myself in a slightly different way than what is easily found when googling. To me, the proof seems like it should work, but the result is not what it should be
$$\begin{align}
\sum_{k=0}^n k^3 & = \sum_{k=0}^n (n-k)^3 \\
& = \sum_{k=0}^n (n^3 -3n^2k + 3nk^2 - k^3) \\
& = (n+1)n^3 - 3n^2\frac{n(n+1)}{2} + 3n\frac{n(2n+1)(n+1)}{6} - \sum_{k=0}^n k^3\\
\Rightarrow 2 \sum_{k=0}^n k^3 & = \frac{2n^3(n+1)}{2} - \frac{3n^3(n+1)}{2} + \frac{(2n^3 + n^2)(n+1)}{2} \\
\Rightarrow \sum_{k=0}^n k^3 & = \frac{(n^3 + n^2)(n+1)}{4}
\end{align}$$
I realise it might be easier or something like that when considering a sum starting from $k=1$ (that is what everyone seems to be doing in any case), but I thought it should work like this as well. Did I make any mistake throughout my proof or forget something? I assume it is provable like this...
| You are correct. Just take $n^2$ common from the first parenthesis.
$$\frac{(n^3 + n^2)(n+1)}{4} = \frac{n^2(n+1)^2}{4} = \left(\frac{n(n+1)}{2}\right)^2$$
| {
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Prove using the formal definition of a limit that How would I go about proving this limit?
$\lim_{x\to\infty}\frac{1}{x^4+x^2+5}=0$
so far I have:
$|f(x) -L|< ϵ$ wherever $x > N$
$|\frac{1}{x^4+x^2+5} - 0| < ϵ $ wherever $x > N$
$|\frac{1}{x^4+x^2+5}| < ϵ ,\; x > ∞$, assuming, $x > 0$
taking the absolute value $\frac{1}{x^4+x^2+5} < ϵ$
${x^4+x^2+5}$ > $\frac{1}{ϵ}$
i am not sure whether this is correct, but what would be the next step?
| We have: $x^4+x^2+5 > x^4 \Rightarrow \dfrac{1}{x^4+x^2+5} < \dfrac{1}{x^4}$. Thus you need $\dfrac{1}{x^4} < \epsilon \Rightarrow x^4 > \epsilon^{-1} \Rightarrow x > \epsilon^{-\frac{1}{4}}$. This means for a given $\epsilon > 0$, choose $N$ a natural number such that $N \geq 1+ \epsilon^{-\frac{1}{4}}$, then whenever $x > N \Rightarrow |f(x)| < \epsilon \Rightarrow \displaystyle \lim_{x \to \infty} f(x) = 0$.
| {
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Spotting that $\,x^8 + x^7 + 1\,$ is reducible. I saw a puzzle recently that came down to spotting that $\,x^8 + x^7 + 1\,$ is reducible — namely, we have
$$x^8 + x^7 + 1 = \left(1 + x + x^2\right)\left(1-x+x^3-x^4+x^6\right)$$
After playing around a bit, I noticed that this is a special case of the fact that
$$\left.\Phi_m\left(x\right) \quad \big| \quad x^{mk} \big[\Phi_m\left(x\right) - 1\big] + 1\right.$$
for each $k$, which holds since
$$x^{mk} \big[\Phi_m\left(x\right) - 1\big] + 1 = x^{m k} \Phi_m\left(x\right) + \left(1 - x^{m k}\right)$$
and of course both terms on the RHS are divisible by $\,\Phi_m\left(x\right)$. In the cases $\,m = 1, 2, 3\,$ this recovers the statements:
*
*$x-1\;$ divides $\;x^{k+1} - 1$
*$x+1\;$ divides $\;x^{2k+1} + 1$
*$x^2 + x + 1\;$ divides $\;x^{3k+2} + x^{3k+1} + 1$
Of course the first two are very well-known. Is it simply the case that many people also know the third statement off the tops of their heads? Or is there some other way of spotting that $x^8 + x^7 + 1$ is reducible?
| The trick is that $8 \equiv 2 \pmod 3$ and $7 \equiv 1 \pmod 3.$ That means that, if we take a cube root of unity, say
$$ \omega = \frac{-1 + i \sqrt 3}{2}, $$
we get
$$ \omega^8 + \omega^7 + 1 = \omega^2 + \omega + 1 = 0. $$
The same holds for
$$ \bar{\omega } = \omega^2. $$
That means that
$$ (x - \omega)(x - \bar{\omega}) = x^2 + x + 1 $$
divides the original polynomial,
as the quadratic is the minimal polynomial for $\omega.$
As you noticed, the same thing can be done with other roots of unity. If we had
$$ x^{1347} + x^{949} + x^{216} + x^{98} + 1, $$
the exponents are $2,4,1,3 \pmod 5,$ and the displayed polynomial is divisible by
$$ x^4 + x^3 + x^2 + x + 1 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1691337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Find $\lim_{(x,y) \to (0,0)} \frac{\tan(x^3+y^3)}{\sin(x^2+y^2)}$
Find $$\lim_{(x,y) \to (0,0)} \frac{\tan(x^3+y^3)}{\sin(x^2+y^2)}$$
What do you guys think about this approach, and are there any faster and easier approaches?
My approach:
*
*$\frac{\tan(x^3+y^3)(x^2+y^2)}{\sin(x^2+y^2)(x^2+y^2)}$
2.$\frac{\tan(x^3+y^3)}{(x^2+y^2)}$
*$\frac{\sin(x^3+y^3)}{\cos(x^3+y^3)\rightarrow 1(x^2+y^2)}$
*$\frac{\sin(x^3+y^3)}{(x^2+y^2)}$
*Polar: $rcos^3\phi + rsin^3\phi = 0 + 0 = 0$
Thus,
$$\lim_{(x,y) \to (0,0)} \frac{\tan(x^3+y^3)}{\sin(x^2+y^2)} = 0$$
| Note that
$$ \frac{\tan(x^3 + y^3)}{\sin(x^2 + y^2)} = \frac{1}{\cos(x^3+y^3)} \frac{\sin(x^3 + y^3)}{\sin(x^2 + y^2)} = \underbrace{\frac{1}{\cos(x^3+y^3)}}_{\text{(1)}}
\underbrace{\frac{\sin(x^3 + y^3)}{x^3 + y^3}}_{\text{(2)}}
\underbrace{\frac{x^2 + y^2}{\sin(x^2 + y^2)}}_{\text{(3)}}
\underbrace{\frac{x^3 + y^3}{x^2 + y^2}}_{\text{(4)}}.$$
The first term in the product is continuous at $(x,y) = (0,0)$. The second term is of the form $g(p(x,y))$ where
$$ g(z) = \begin{cases} \frac{\sin(z)}{z} & z \neq 0 \\ 1 & z = 0 \end{cases} $$
is continuous and $p(x,y) \rightarrow 0$ as $(x,y) \rightarrow (0,0)$ and so tends to $g(0)$. The third term is also of the form $g(p(x,y))$ for a different $g$ and $p$. The last term can be dealt easily using polar coordinates.
| {
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"url": "https://math.stackexchange.com/questions/1691538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How are these equal? How are both sides of these equations equal to one another?
$(\sqrt[4]{x}-\sqrt[4]{y})\cdot(\sqrt{x^{3}}+\sqrt{y^{3}})\cdot\frac{\sqrt[4]{x}+\sqrt[4]{y}}{x-\sqrt{xy}+y}=
(\sqrt[4]{x}-\sqrt[4]{y})\cdot(\sqrt[4]{x}+\sqrt[4]{y})\cdot\frac{\sqrt{x^{3}}+\sqrt{y^{3}}}{x-\sqrt{xy}+y}$
I copied this verbatim from a calculus study guide (I'm trying to teach myself), and I'm confused as to how $(\sqrt{x^{3}}+\sqrt{y^{3}})$ switches places with $\sqrt[4]{x}+\sqrt[4]{y}$
| $(\sqrt[4]{x}-\sqrt[4]{y})\cdot(\sqrt{x^{3}}+\sqrt{y^{3}})\cdot\frac{\sqrt[4]{x}+\sqrt[4]{y}}{x-\sqrt{xy}+y}=
(\sqrt[4]{x}-\sqrt[4]{y})\cdot(\sqrt[4]{x}+\sqrt[4]{y})\cdot\frac{\sqrt{x^{3}}+\sqrt{y^{3}}}{x-\sqrt{xy}+y}$
The $\sqrt{x^{3}}+\sqrt{y^{3}}$ and$\sqrt[4]{x}+\sqrt[4]{y}$ are swapped. That's all.
$$a\times\frac{b}{c}=b\times\frac{a}{c}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Quick way to solve $16x \equiv -9 \pmod {3^5}$ Quick way to solve $16x \equiv -9=234 \pmod {3^5}$
Any hints?
I tried all the numbers from $1$ to $10$ but none work.
| Hint:
If $16x \equiv -9 \bmod {3^5}$, then $16x \equiv -9 \bmod {3^2}$ and so $x \equiv 0 \bmod {3^2}$.
Let $x=9y$. Then we can cancel $9$ in $16x \equiv -9 \bmod {3^5}$ and get $16y \equiv -1 \bmod {3^3}$, which is simpler.
| {
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Prove that $7 | (3^{2n + 1} + 2^{n +2})$ Prove that $7 | (3^{2n + 1} + 2^{n +2})$
So far I have:
Base case: n = 1
$ = (3^{2(1) + 1} + 2^{(1) +2})$
$ = (3^{3} + 2^{3})$
$ = (35)$ which divides 7
Inductive Step:
$ = (3^{2(n +1) + 1} + 2^{(n +1) +2})$
$ = (3^{2n + 3} + 2^{n + 3})$
$ 7j = (3^{2n + 3} + 2^{n + 3})$
-I'm a bit stuck as to how to proceed, any help or correcitons are appreciated.
- I believe I understand where we need to go now but i'm not sure how to get there. It seems obvious that any n plugged into this formula would be divisible by 7 given that $3^1 + 2^2 = 7$ Since you have that $3^1+2^2$ nested inside the given formula it seems easily understood that any sum with any positive n would also be divisible by 7. I just don't see the steps in between to actually prove it.
| Let $7|3^{2k+1}+2^{k+2}$.
To prove that is divisible by $7|3^{2k+3}+2^{k+3}$, it suffices to prove that the difference is divided by
$$\left((3^{2k+3}+2^{k+3})-(3^{2k+1}+2^{k+2}) \right)=\left((3^{2k+1+2}+2^{k+2+1})-(3^{2k+1}+2^{k+2}) \right)=$$
$$=\left(9 \cdot 3^{2k+1}+2\cdot 2^{k+2}-3^{2k+1}-2^{k+2} \right)=$$
$$=8 \cdot 3^{2k+1}+2^{k+2}=7 \cdot 3^{2k+1} +\left(3^{2k+1}+2^{k+2} \right)$$
Obviously, both terms are divisible by 7, the second induction hypothesis
| {
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Simple method to solve a geometry question for junior high school student Rencently, my sister asked me a geometry question that came from her mock examination, please see the following graph.
Here,
*
*$\angle DOE=45°$
*the length of $DE$ is constant, and $DE=1$. Namely, $OD,OE$ are changeable.
*$\triangle DEF$ is equilateral triangle.
Q: What is the maximum length of $OF$?
My solution
Denote $OD,OE,\angle ODE$ as $x,y,\theta$, respectively.
Via sine theorem
$$
\begin{cases}
ED^{2} = OE^{2} + OD^{2} - 2OE \times OD\cos \angle EOD \\[6pt]
\cos \theta = \dfrac{EO^{2} + ED^{2} - OD^{2}}{2 EO \times ED}
\end{cases}
$$
$$
\begin{align}
1^{2} &= x^{2} + y^{2} - 2xy\cos 45^{\circ} \\
&= x^{2} + y^{2} - \sqrt{2} xy
\end{align}
$$
$$
\implies
\begin{cases}
\color{red}{xy} = \dfrac{x^{2} + y^{2} - 1}{\sqrt{2}} \color{red}{\leq}
\dfrac{x^{2} + y^{2}}{2} \implies x^{2} + y^{2} \color{red}{\leq}
2 + \sqrt{2} \\[6pt]
\cos \theta = \dfrac{x^{2} + 1 - y^{2}}{2x}
\end{cases}
$$
Via cosine theorem
$$
\frac{y}{\sin \theta} = \frac{DE}{\sin \angle EOD} = \frac{1}{\sin
45^{\circ}} \implies \sin \theta = \frac{y}{\sqrt{2}}
$$
$$\begin{align}
OF^{2} &= EO^{2} + EF^{2} - 2EO \times EF\cos \angle OEF \\
&= x^{2} + 1^{2} - 2x\cos(\theta + 60^{\circ}) \\
&= x^{2} + 1 - 2x(\cos \theta \cos 60^{\circ} - \sin \theta \sin
60^{\circ}) \\
&= x^{2} + 1 - 2x\left(\frac{x^{2} + 1 - y^{2}}{2x} \frac{1}{2} -
\frac{y}{\sqrt{2}} \frac{\sqrt{3}}{2}\right) \\
&= \frac{x^{2} + y^{2} + 1}{2} + \frac{\sqrt{3} xy}{\sqrt{2}} \\
&= \frac{x^{2} + y^{2} + 1}{2} + \frac{\sqrt{3}}{\sqrt{2}}
\frac{x^{2} + y^{2} - 1}{\sqrt{2}} \\
&= \frac{(\sqrt{3} + 1)(x^{2} + y^{2})}{2} + \frac{1 - \sqrt{3}}{2}
\\
&\color{red}{\leq} \frac{(\sqrt{3} + 1)(2 + \sqrt{2})}{2} + \frac{1 -
\sqrt{3}}{2} = \frac{1}{2}(3 + \sqrt{3} + \sqrt{2} + \sqrt{6})
\end{align}
$$
However, for junior high school student, she doesn't learn the following formulae:
*
*sine theorem
*cosine therem
*$\cos(x+y)=\cos x \cos y-\sin x \sin y$
*fundamental inequality $x y\leq \frac{x^2+y^2}{2}$
Question
*
*Is there other simple/elegant method to solve this geometry question?
Update
Thanks for MXYMXY's hint
Here, the line $O'F$ pass the center of the circle. Namely, $O'D=OF$
In Rt $\triangle O'OF$, the inequality $O'F>OF$ holds.
|
Let $FX$ be the median of the equilateral triangle such that the length of $FX =$ constant $= [m]$, say
In $\triangle OXF$, $t \le s + [m]$ by triangle inequality.
$t_{max}$ occurs when the triangle is degenerated into a straight line OXF.
When that occurs, $\angle OXD$ has to be right-angled because (1) $OXF$ is a straight line; and (2) $FX$ is the median of the equilateral triangle.
The rod $DE$ should be placed in such a way that (1) $OX$ is the angle bisector of $\angle DOE$; (2) $\triangle ODE$ is isosceles with OD = OE; and (3) it should be perpendicular to $OX$.
The required maximum length $= OX + [m]$ can be easily found and detail is therefore skipped.
Added:-
The required maximum $= OX + [m] = \dfrac {\dfrac {1}{2}}{\tan 22.5^0} +[\dfrac {\sqrt 3}{2}]$
| {
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"timestamp": "2023-03-29T00:00:00",
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Integration using Euler $\int \frac{\sqrt{x^2+2x-1} }x\,dx$ I've just tried to use the Euler's formula for my integral, but I can't get the correct answer. So if anyone could help me I would really appreciate that.
This is my integral:
$$\int\frac{\sqrt{x^2+2x-1} }{x}\,dx$$
P.S. The ingral must be solven using Euler's formula
This is where I've got stuck:
I started with this substitution:
$$\sqrt{x^2+2x-1} = -x + t$$
After derivating I get $dx= t^2 + 2x -1 /2(t+1)^2$.
After immpleneting it into my integral, I get to this point
$$\int\frac{(t^2+2t-1)(t^2+2t-1)}{(t^2+1)2(t+1)^2}\,dt$$ I don't have any idea what I should do next (thought to do another substitution but don't know what to substitute).
| Before applying any substitutions, rewrite the integral as
\begin{align}
I=&\int\frac{\sqrt{x^2+2x-1} }{x}\,dx\\
= &\ \sqrt{x^2+2x-1} +\int \frac1{\sqrt{x^2+2x-1} }dx
- \int \frac1{x\sqrt{x^2+2x-1} }dx
\end{align}
Then, with the Euler substitution $\sqrt{x^2+2x-1} = t-x $
\begin{align}
&\int \frac1{\sqrt{x^2+2x-1} }dx=\int \frac1{1+t}dt =\ln|1+t|\\
&\int \frac1{x\sqrt{x^2+2x-1} }dx=\int \frac2{1+t^2}dt = 2\tan^{-1}t\\
\end{align}
As a result
\begin{align}
I= \sqrt{x^2+2x-1} +\ln\left|1+x+ \sqrt{x^2+2x-1}\right|\\
- 2\tan^{-1}(x+ \sqrt{x^2+2x-1})\\
\end{align}
| {
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Computing $\int_0^{2\pi}\sqrt{1+\sin x}dx$ Problem Evaluate the following integral
$$\int_0^{2*\pi}\sqrt{1+\sin x}dx$$
Attempted solution
Note that
\begin{align*}
& \int_{0}^{2\pi}\sqrt{1+\sin x}dx\\
= & \int_{0}^{\pi}\sqrt{1+\sin x}dx+\int_{\pi}^{2\pi}\sqrt{1+\sin x}dx\\
= & \int_{0}^{\pi}\sqrt{\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)^{2}}dx+\int_{0}^{\pi}\sqrt{1+\sin\left(x-\pi\right)}d\left(x-\pi\right)\\
= & \int_{0}^{\pi}\sqrt{\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)^{2}}dx+\int_{0}^{\pi}\sqrt{1-\sin x}dx\\
= & \int_{0}^{\pi}\sqrt{\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)^{2}}dx+\int_{0}^{\pi}\sqrt{\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)^{2}}dx\\
= & \int_{0}^{\pi}\left|\cos\frac{x}{2}+\sin\frac{x}{2}\right|dx+\int_{0}^{\pi}\left|\cos\frac{x}{2}-\sin\frac{x}{2}\right|dx\\
= & \int_{0}^{\frac{\pi}{2}}\cos\frac{x}{2}+\sin\frac{x}{2}dx+\int_{\frac{\pi}{2}}^{\pi}\cos\frac{x}{2}+\sin\frac{x}{2}+\int_{0}^{\frac{\pi}{2}}\cos\frac{x}{2}-\sin\frac{x}{2}dx+\int_{\frac{\pi}{2}}^{\pi}\sin\frac{x}{2}-\cos\frac{x}{2}dx\\
= & \int_{0}^{\frac{\pi}{2}}2\cos\frac{x}{2}dx+\int_{\frac{\pi}{2}}^{\pi}\sin\frac{x}{2}dx\\
= & \left.4\sin\frac{x}{2}\right|_{0}^{\frac{\pi}{2}}-\left.4\cos\frac{x}{2}\right|_{\frac{\pi}{2}}^{\pi}=\boxed{4\sqrt{2}}.
\end{align*}
Question I don't know if I solved this integral in the best way possible. I really appreciate if someone can offer some alternative solution.
| A formula for $\sin x+\sin y$ may help:
$$1+\sin x = \sin\frac{\pi}{2}+\sin x = 2\left(\sin\left(\frac{\pi}{4}+\frac{x}{2}\right)\right)^2$$
and now it should be easier
| {
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"timestamp": "2023-03-29T00:00:00",
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An ellipse $x^2+4y^2=4$ is rotated anticlockwise through a right angle in its own plane about its center An ellipse $x^2+4y^2=4$ is rotated anticlockwise through a right angle in its own plane about its center.If the locus of the point of intersection of a tangent to ellipse in its original position with the tangent at the same point of the ellipse in its new position is given by the curve
$(x^2+y^2)^2=\lambda(x^2+y^2)+\mu xy$ where $\mu$ and $\lambda$ are positive integers.Find the value of $\lambda+\mu.$
The equation of the ellipse $\frac{x^2}{4}+\frac{y^2}{1}=1$,when it is rotated anticlockwise through a right angle in its own plane about its center,then its equation becomes $\frac{x^2}{1}+\frac{y^2}{4}=1$.
Let the point of tangency be $(2\cos\theta,\sin\theta)$
So the tangent to the first ellipse $\frac{x^2}{4}+\frac{y^2}{1}=1$ is $\frac{x\cos\theta}{2}+\frac{y\sin\theta}{1}=1$
and the tangent to the second ellipse $\frac{x^2}{1}+\frac{y^2}{4}=1$ is $\frac{2x\cos\theta}{1}+\frac{y\sin\theta}{4}=1$
Eliminating $\theta $ from the two equations gives
$16x^2+4y^2=25x^2y^2$
But i am not correct,i dont know where i am wrong.
| If we let the point of tangency be $(2\cos\theta,\sin\theta)$, then this point moves to $(-\sin\theta,2\cos\theta)$ by the rotation, at which we have to consider the tangent line of the ellipse in its new position.
So, we have the following two lines :
$$\frac{\cos\theta}{2}x+\frac{\sin\theta}{1}y=1$$
$$\frac{-\sin\theta}{1}x+\frac{\cos\theta}{2}y=1$$
Representing $\cos\theta,\sin\theta$ by $x,y$ and using $\cos^2\theta+\sin^2\theta=1$ will give you the answer.
| {
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computing the value of $\sin^2(\frac{\pi}{10}$) The question asks me to prove the following formula: $$\cos5\theta=\cos\theta(16\sin^4\theta-12\sin^2\theta+1)$$ which is pretty straightforward to do using De Moivre's theorem. They further ask me to show that the exact value for $\sin^2\frac{\pi}{10}=\frac{3-\sqrt{5}}{8}$.
I proceed by simply solving the quadratic equation $$16\sin^4\frac{\pi}{10}-12\sin^2\frac{\pi}{10}+1=0$$ However, this yields two answers, namely $\frac{3\pm \sqrt{5}}{8}$. I don't know how to eliminate the answer where it's a plus and the justification from the book is
Since $\sin^2(\frac{\pi}{10})\lt \sin^2(\frac{3\pi}{10})$
I don't understand how that relates to the problem.
| Say you were asked to find $\sin^2\left( \frac{3\pi}{10} \right) $.
You would proceed by saying that $$\cos \left( \frac{15\pi}{10} \right) = \cos \left( \frac{3\pi}{2} \right) = 0 \\
16\sin^4\frac{3\pi}{10}-12\sin^2\frac{3\pi}{10}+1=0$$
which is exactly the same quadratic equation you have solved when finding
$\sin \left( \frac{3\pi}{10} \right)$. Since you know that
$\sin \left( \frac{\pi}{10} \right) < \sin \left( \frac{3\pi}{10} \right)$ you know that the smaller root of the quadratic is $\sin \left( \frac{\pi}{10} \right)$.
And by the way, you can actually find the square root of that:
$$\sin \left( \frac{\pi}{10} \right) = \sqrt\frac{{3-\sqrt{5}}}{8}
= \frac{\sqrt{5}-1}{4}$$
| {
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Inequality $abdc$ $\leq$ $3$ $a+b+c+d=6$
and
$a^2+b^2+c^2+d^2=12$.
and $a,b,c,d$ are reals.
Prove: $abcd$ $\leq$ $3$ without Lagrange multipliers, complex numbers or convexity help.
Using Cauchy–Schwarz inequality I found: $a,b,c,d \in [0,3]$.
How solve inequality?
| Let $a=1+x$, $b=1+y$, $c=1+z$ and $d=1+t$.
Hence, $x+y+z+t=2$, $\sum\limits_{cyc}(1+x)^2=12$, which gives $x^2+y^2+z^2+t^2=4$ and
$xy+xz+yz+xt+yt+zt=0$.
By the way, $0=(xy+xz+yz+xt+yt+zt)^2=$
$=x^2y^2+x^2z^2+y^2z^2+x^2t^2+y^2t^2+z^2t^2+2\sum\limits_{cyc}x^2(yz+yt+zt)+6xyzt$.
Id est, $abcd=(1+x)(1+y)(1+z)(1+t)=3+\sum\limits_{cyc}xyz+xyzt=$
$=3+\frac{1}{2}\sum\limits_{cyc}x\sum\limits_{cyc}xyz+xyzt=3+\frac{1}{2}\left(\sum\limits_{cyc}x^2(yz+yt+xt)+4xyzt\right)+xyzt=$
$=3+\frac{1}{2}\sum\limits_{cyc}x^2(yz+yt+xt)+3xyzt=$
$=3+\frac{1}{4}\left(-(x^2y^2+x^2z^2+y^2z^2+x^2t^2+y^2t^2+z^2t^2)-6xyzt\right)+3xyzt=$
$=3-\frac{1}{4}(x^2y^2+x^2z^2+y^2z^2+x^2t^2+y^2t^2+z^2t^2)+\frac{3}{2}xyzt\leq$
$\leq3-\frac{3}{2}|xyzt|+\frac{3}{2}xyzt\leq3$.
| {
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Find all solutions to the equation $x^2 + y^2 + xy = (xy)^2$ Can anyone help me find the number of solutions to the equation:
$$ x^2 + y^2 + xy = (xy)^2 $$
Let me give a brief account of what I've tried to proceed with:
Case 1: One of $x$ and $y$ is odd. This results into a contradiction where the parity of LHS and RHS differs.
Case 2: Both of $x$ and $y$ are even. I have shown that no solution apart from $x=y=0$ exists and I've proved that using Infinite Descent.
Now I'm stuck in case $3$ where both $x$ and $y$ are odd.
| First, write it as $(x^2-1)(y^2-1)=xy+1$.
In the narrow ranges where $|x|\leq 1$ or $|y|\leq 1$, you can solve yourself.
Then, when $x>1$ and $x>1$ you have $(x^2-1)(y^2-1)>=2(x^2+y^2-2)=2(x-y)^2+4xy-4\geq 4xy-4$.
But $xy\geq 4$, so $4xy -4> xy +3\cdot 4 -4 \geq xy-1$.
The case when $x<-1$ and $y<-1$ is the same.
There are no cases when $x<-1$ and $y>1$ because then $xy+1<0$.
Finally, if $x=0$, $y=0$. If $x=-1$ then $y=1$, and if $x=1,$ $y=-1$.
The key is that in most cases, it is "obvious" that $(x^2-1)(y^2-1)$ is a lot bigger than $xy$.
Another approach. Let $d=xy$. Then assume $d\neq 0$ and you have:
$$x^2+\frac{d^2}{x^2}=d^2-d$$
or
$$x^4-(d^2-d)x^2 + d^2=0$$
So by the quadratic formula:
$$x^2=\frac{d^2-d \pm \sqrt{d^2(d-3)(d+1)}}{2}$$
When is $(d-3)(d+1)=(d-1)^2-4$ a perfect square? The only case of perfect squares that differ by four is $0$ and $4$. That means $d=3$ or $d=-1$.
The only cases left to handle then are $d=0$ and $d=3$.
| {
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} |
Find the sum of the areas of regions $X$ and $Y$ Right triangle $ABC$ is inscribed in a circle with $AC = 6$, $BC = 8$ and $AB=10$. $AC$ and $CB$ are semi-circles. Find the sum of the areas of regions $X$ and $Y$.
This is not so obvious to me. I started off with the formula for area which is A=$\pi r^2$ and since each semi-circle, we are going to have divide A by $2$. However, how would I find $r^2$ or the radius?
Any ideas would help.
| The inscribed triangle is right-angled, therefore its hypotenuse $AB$ is a diameter (by the converse to Thales' Theorem) and the large circle $C$ has radius $5$.
Let the white region between $X$ and the triangle be $X'$, and the white region between $Y$ and the triangle be $Y'$.
Then $$A(X') + A(Y') = \frac{1}{2}A(C) - A(triangle) = \frac{1}{2}(5^2\pi)-\frac{1}{2}(6\times8) = \frac{25}{2}\pi-24 $$
Finally the smaller semi-circles (that is $X\cup X'$ and $Y\cup Y'$) have area $ \frac{1}{2}(3^2\pi) $ and $ \frac{1}{2}(4^2\pi) $ respectively, so we obtain:
$$ A(X) + A(Y) = \frac{1}{2}(3^2\pi) + \frac{1}{2}(4^2\pi) - (A(X') + A(Y')) = \frac{9}{2}\pi + 8\pi - (\frac{25}{2}\pi-24) = 24 $$
| {
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Divide an equilateral triangle into at least $100$ regions? A piece of paper has the shape of an equilateral triangle. What is the minimum number of straight lines parallel to its sides that divide the triangle into at least $100$ regions?
The answer said it is $16$. But. But. Don't know how to solve it. Thank you.
| Let us start with one line, parallel to each side of the triangle. Consider the following diagram:
$$n=3$$
By drawing 3 lines, we split it up into 7 segments.
Now what we can do is split up the inner triangle with the same logic:
$$ n = 6 $$
Note that this splits up the inner triangle into another 7 parts. In addition to this, it also adds 6 other segments into the triangle, between the first and seconds sets of lines.
This means that with $6$ lines, we have $7 + 7 + 6 = 20 $ segments
It is obvious that if we add another 3 lines, one each parallel to the sides that it will split up the inner triangle into another 7 segments.
$$n=9$$
In addition to this 7, we see again, it adds another 6 segments, next to the 6 created after the previous step. And another 6 is also added, inside 1st set of lines but outside the second set.
Therefore, here we have $20 + 7 + 6 + 6 =39$ segments.
At this point we can put together a formula for this. Let $n$ be the number of lines divided by 3 and $f(n)$ be the number of segments. (We will work in sets of 3 as it is easier. Then we have $$f(1) = 7,\quad f(2)=20\quad f(3) = 39$$
For each addition of 1 in $n$, the inner triangle is split into 7. We also see an increase in multiples of 6.
We know that when $n=2$, we have $7 + 7 + 6$ segments.
When $n=3$, we have $7+7+7+6+6+6$ segments.
We can see a pattern here.
Thus, we can make the following equation:
$$ f(n) = 7n + 6n + 6(n-1) + 6(n-2) + ... 6,\quad n>0$$
$$\implies f(n) = 7n + 6n + 6n +... + 6n -6 - 12- 18 -...-6(n-1)$$
The $-6-12...$ bit can be written as $$-1(6 + 12 + 18 + ... + n-1) = -3(1+2+3+...+(n-1) = -3\left(\frac{(n-1)((n-1)+1)}{2}\right) = \frac{-2}{3}n(n-1)$$
So, now we have $$f(n) = 7n + 6n(n-1) -\frac{2}{3}n(n-1) = 7n + 6n^2 - 6n -\frac{2}{3}n + \frac{2}{3} = 6n^2 + \frac{1}{3}n + \frac{2}{3}$$
Now we want to investigate the case with 100 segments.
Let $f(n) = 100$
Then $$ 100 = 6n^2 + \frac{1}{3}n + \frac{2}{3} \implies n \approx 4.04$$
Therefore, we need at least $n=5$ for 100 segments (as $ n$ must be a whole number).
Therefore, we need $ 3(5) = 15$ lines for at least 100 segments in the triangle.
| {
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Dividing polynomials A polynomial $f(x)$ with real coefficients leaves the remainder $15$ when divided by $x-3$, and the remainder $2x+1$ when divided by $(x-1)^2 $. Then the reminder when $f(x)$ is divided by $(x-3)(x-1)^2 $ is
$a)$ $2x^2 -2x+3$
$b)$ $6x-3$
$c)$ $x^2 + 2x$
$d)$ $3x+6$
I have tried using the rule $ f(x) = p(x)q(x) + r(x) $ and then later trying to substitute the zeros of the roots in the same expression.
That is if we substitute x=3 in the equation written just above we get the result that $ f(3) = 15 $ since the first term tends to zero as 3 is the zero for $p(x)$.
Two , from the following logic I tried checking the options to see whether they satisfy the conditions $f(3) = 15$ and $f(1) = 3$.
Where I am getting stuck is that options 1 , 2 and 3 all of them satisfy the conditions mentioned above.
The correct answer mentioned is $ (a) $.
| Suppose $f=p.(x-3)(x-1)^2+r$ and $r=q_1(x-3)+r_1$ and $r=q_2(x-1)^2+r_2$
Now $$f=p(x-3)(x-1)^2+r=(x-3)(p(x-1)^2+q_1)+r_1$$ and $$f=p(x-3)(x-1)^2+r=(x-1)^2(p(x-3)+q_2)+r_2$$
Thus we should have $$r_1=15 \quad *$$ and $$r_2=2x+1 \quad **$$
Because the reminder of $f$ by $x-3$ and $(x-1)^2$ respectively should be $15$ and $2x+1$ .
The first three options satisfied first equation, but we see
$2x^2-2x+3=2(x-1)^2+2x+1$
$6x-3=0.(x-1)^2+6x-3$
$x^2+2x=(x-1)^2+4x-1$
So the answer is a !
| {
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Prove $\cos(2\theta) + \cos\left(2 \left(\frac{\pi}{3} + \theta\right)\right) +\cos\left(2 \left(\frac{2\pi}{3} + \theta\right)\right) = 0$ It's been a while since I've done trig proofs. I know that
$$\cos(2\theta) + \cos\left(2 \left(\frac{\pi}{3} + \theta\right)\right) + \cos\left(2 \left(\frac{2\pi}{3} + \theta\right)\right) = 0$$
is true. This can be easily seen by plugging in values, for example where $\theta=0$ we get
$$1 -\frac{1}{2} -\frac{1}{2} = 0$$
and the same can be seen for any $\theta$. Is there a straight forward way to prove this statement is true.
| By a formula that is well known,
$$cos(A)+cos(B)=2cos((A+B)/2)cos((A-B)/2)$$
Let $A = 2\theta $, $B=4\pi/3 + 2\theta$. Then by the above,
$$
cos(2\theta)+cos(4\pi/3 + 2\theta)=2cos(2\theta+2\pi/3)cos(2\pi/3) = -cos(2\theta +2\pi/3)
$$
The first formula used above must have been rote learnt or at least encountered in high school. It's use gives a 2 line proof as I have shown.
| {
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Finding the points on the curve where the gradient of the tangent is equal to 2. So, my question is as following.
"Find all the points on the curve $y= 2x^3 + 3x^2 - 10x +3$ where the gradient of the tangent is 2."
I used $\frac{dy}{dx}$ in order to get the formula for the gradient function.
$\frac{dy}{dx} = 6x^2 + 6x - 10$ (or when divided by 2 it's $3x^2 + 3x - 5$)
How do I continue? When I set the derived equation equal to 2, I do not get the answer given in the book.
Any input would be appreciated!
| Why have you divided by 2?
I would have thought that the answer would be something like:
$dy/dx = 6x^2 + 6x - 10$
Then set this equal to 2:
$6x^2 + 6x - 10=2$
$\implies 6x^2 + 6x - 12=0 $
$\implies 2x^2 + 2x - 4=0 $
$\implies (2x+4)(x-1)=0 $
$\implies x=-2$ or $x=1$
Then substitute these values of $x$ in to your given equation for $y$ to give the points:
$y(-2)=2(-2)^3+3(-2)^2-10(-2)+3=2(-8)+3(4)+20+3=-16+12+23=19$
and $y(1)=2(1)^3+3(1)^2-10(1)+3=2(1)+3(1)-10+3=2+3-10+3=-2$.
| {
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How to calculate $\left( 1+\tan 5^\circ\right) \left( 1+\tan 10^\circ\right)\left( 1+\tan 15^\circ\right)\cdots\left( 1+\tan 40^\circ\right)$ I curious practical solution.(Step by step)
$\left( 1+\tan 5^\circ\right) \left( 1+\tan 10^\circ\right)\left( 1+\tan 15^\circ\right)\cdots\left( 1+\tan 40^\circ\right)$
Answer is $16$.
| If $\displaystyle A+B = \frac{\pi}{4}\;,$ Then $(1+\tan A)(1+\tan B) = 2$
So $(1+\tan 5^0)(1+\tan 40^0)=2$ and $(1+\tan 10^0)(1+\tan 35^0)=2$
and $(1+\tan 15^0)(1+\tan 30^0)=2$ and $(1+\tan 20^0)(1+\tan 35^0)=2$
So $$(1+\tan 5^0)(1+\tan 40^0)(1+\tan 10^0)(1+\tan 35^0)(1+\tan 15^0)(1+\tan 30^0)(1+\tan 20^0)(1+\tan 35^0)=2^4=16$$
| {
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How do I demonstrate that a polynomial of degree $2$ divides one of degree $n$? Let $f$ and $g$ the polynomials
$$f(x) = (x+1)^{2n-1}+(-1)^n(x+2)^{n+1}\qquad\text{and}\qquad g(x) = x^2 + 3x + 3$$
How do I demonstrate that $g$ divides $f$? I tried finding the roots of $g$ then testing them for $f$, it didn't work ...
I also wrote g as $(x+1)^2 + x + 2$ but I couldn't get anywhere.
| For $f_n(x) = (x+1)^{2n-1}+(-1)^n(x+2)^{n+1}$ with $g(x) = x^2+3x+3=(x+1)^2+x+2$, first consider $n=1,2$:
$$f_1(x) = x+1-x^2-4x-4 = -x^2-3x-3\\
f_2(x) = (x+1)^3+(x+2)^3=(2x+3)\left((x+1)^2-(x+1)(x+2)+(x+2)^2\right)\\
=(2x+3)(-(x+1)+(x+2)^2)=(2x+3)(x^2+3x+3)$$
So we have that $g(x)\mid f_1(x),\ g(x)\mid f_2(x)$. It appears that a useful way to approach the question is to show that we end up with something either of the form $(x+1)^2+(x+2)$ or $(x+2)^2-(x+1)$. To this end, with $y=x+1\to g(y)=y^2+y+1$, consider that we have
$$f_n(y) = y\left(y^2\right)^{n-1}+(-1)^n(y+1)^2(y+1)^{n-1}\\
=(-1)^n(y^2+y+1)(y+1)^{n-1}+y\left((y^2)^{n-1}+(-1)^n(y+1)^{n-1}\right)$$
For even $n$, $(y^2)+(y+1)$ is always a factor of $(y^2)^{n-1}+(-1)^n(y+1)^{n-1}$ (just as $a+b\mid a^{2n-1}+b^{2n-1}$). Now the only remaining issue is to determine whether $(y^2)+(y+1)$ is a factor of $f_n(x)$ for odd $n$. But we see that for odd $n=2m+1$, $(y^2)^{n-1}+(-1)^n(y+1)^{n-1}$ is a difference of squares, and therefore factors as
$$\left((y^2)^m-(y+1)^m\right)\left((y^2)^m+(y+1)^m\right)$$
If $m$ is odd, we are done; if $m$ is even, we can split the left-hand term as a difference of squares again; we can repeat as needed until we have a sum with odd outer exponents. Therefore $g(y)\mid f_n(y)$ and thus $g(x)\mid f_n(x)$.
| {
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Proving that $\cos\frac{2\pi}{13}+\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13}=\frac{\sqrt{13}-1}{4}$ How can I prove that:
$\cos\frac{2\pi}{13}+\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13}=\frac{\sqrt{13}-1}{4}$
Without using complex numbers?
I tried to raise by 2 and to multipy by 2, and got:
$2y^2=3+3\cos\frac{4\pi}{13}+2\cos\frac{10\pi}{13}+\cos\frac{12\pi}{13}+ 2\cos\frac{14\pi}{13}+2y$
But I'm stuck from here.
Thanks.
| Let $t = \frac{\pi}{13}$
then you know you want $$\cos(2t)+\cos(6t)+\cos(8t)$$
and if you call that $x$, square it, then you should have $$x^2 = \cos^2(2t)+\cos^2(6t)+\cos^2(8t)[\cos(2t)+\cos(4t)+\cos(6t)+\cos(8t)+\cos(10t)+\cos(12t)]$$
Note that,
$$[\cos(2t)+\cos(4t)+\cos(6t)+\cos(8t)+\cos(10t)+\cos(12t)] = -0.5$$
Hence,
$$x^2 = \cos^2(2t)+\cos^2(6t)+\cos^2(8t) - 0.5 $$
By the law of cosines,
$$x^2 = \frac{1}{2}(\cos(4t)+\cos(12t)+\cos(16t) + 3) - 0.5 $$
Hence,
$$2x^2 = \cos(4t)+\cos(12t)+\cos(16t) + 2 $$
and as we know $$x = \cos(2t) + \cos(6t) + \cos(8t)$$
Add $$2x^2 + x = \frac{-1}{2} +2 = \frac{3}{2}$$
Then we have,
$$4x^2 + 2x -3 = 0$$
Use Quadratic formula, then you will get two solutions, but you know the answer is positive.
| {
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Given a Poisson process $N(t)$ with rate $2$, find $P( N(3) = 4 | N(1) = 1 )$.
Given a Poisson process $N(t)$ with rate $2$, find
a. $P( N(3) = 4 | N(1) = 1 )$
b. $P( N(1) = 1 | N(3) = 4 )$
Using the method, I have
$$\frac{P( N(3) = 4 \cap N(1) = 1 )}{P( N(1) = 1)}$$
$$P( N(1) = 1) = 2 / \exp(-2)$$
Then because $P( N(3) = 4 \cap N(1) = 1 )$ is independent, therefore
$$\frac{P( N(3) = 4 ) P( N(1) = 1 )}{P( N(1) = 1)} = P( N(3) = 4 )$$
Am I correct in saying this for the first a?
| That's false, they are not independent.
Notice that
$$\{N(3)= 4, N(1) = 1\}\iff\{N(1) = 1,N(3)-N(1) = 3\}$$
and
$N(1)$ is independent of $N(3)-N(1)$. Hence
\begin{align*}
P(N(3) = 4|N(1) = 1) &= \frac{P(N(3) = 4,N(1) = 1)}{P(1) = 1)}\\
&=\frac{P(N(3)-N(1) = 3, N(1) = 1)}{P(N(1) = 1)}\\
&=\frac{P(N(3)-N(1) = 3)P(N(1) = 1)}{P(N(1) = 1)}\\
&=P(N(3)-N(1) = 3)
\end{align*}
| {
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Find all functions $f: \mathbb N \rightarrow \mathbb N$ such that $f(n!)=f(n)!$ Find all functions $f: \mathbb N \rightarrow \mathbb N$ (where $\mathbb N$ is the set of positive integers) such that $f(n!)=f(n)!$ for all positive integers $n$ and such that $m-n$ divides $f(m)-f(n)$ for all distinct positive integers $m,n$.
My work so far:
$f(1!)=f(1)=f(1)!$ and $f(2!)=f(2)=2$. Then $f(1)=1$ or $f(1)=2$ or $f(2)=1$ or $f(2)=2$.
Case 1: $f(1)=f(2)=1$. I proved $f(n) \equiv 1$
Case 2: $f(1)=f(2)=2$. I proved $f(n) \equiv 2$
Case 3: $f(1)=2$ and $f(2)=1$. I proved that this case is impossible.
Case 4: $f(1)=1$ and $f(2)=2$. I need help here.
| [Sangchul Lee's proof below is substantially nicer than this one.]
Assume $f(1) = 1, f(2) = 2$.
It is a fact that $4 \mid f(6)-f(2) = f(3)! - 2$, so $f(3)! \equiv 2 \pmod{4}$.
Therefore $f(3) = 2$ or $3$.
Also $2 \mid f(3)-f(1) = f(3)-1$, so $f(3)$ is odd; hence $f(3) = 3$.
The following five theorems are to be viewed jointly as an inductive step.
Lemma 1: $$f(r)! \equiv (r-1)! \pmod{r!-(r-1)!}$$
Proof:
Generally, $$n \mid f(r)! - f(r!-n)$$
so $$f(r!) \equiv f(r!-n) \pmod{n}$$
so $$f(r)! \equiv f(r!-n) \pmod{n}$$
Letting $n = r!-(r-1)!$, obtain $$f(r)! \equiv f((r-1)!) = f(r-1)! \pmod{r!-(r-1)!}$$
Inductively, $f(r-1) = r-1$, so $$f(r)! \equiv (r-1)! \pmod{r!-(r-1)!}$$
Lemma 2: $$f(r) \equiv r \pmod{2}$$
This is immediate from the fact that $2 \mid f(r)-f(r-2)$, and inductively $f(r-2) = r-2$, so $$f(r) \equiv r \pmod{2}$$
Lemma 3: $$f(r) \geq r$$
Indeed, if $f(r) < r-1$, then $f(r)! \leq (r-2)!$; but $(r-2)! < r! - (r-1)!$ and so the statement of Lemma 1 that $f(r)! \equiv (r-1)! \pmod{r!-(r-1)!}$ is precisely the statement that $f(r)! = (r-1)!$, which is precisely the statement that $f(r) = r-1$. This is a contradiction.
Also, by Lemma 2, $f(r) \not = r-1$; so $f(r) \geq r$.
Lemma 4: $f(r) < 2(r-1)$.
We have by Lemma 1 $$f(r)! \equiv (r-1)! \pmod{r!-(r-1)!}$$
But if $f(r) \geq 2(r-1)$ then $(r-1)! (r-1) \mid f(r)!$, so $f(r)! \equiv 0 \pmod{(r-1)! (r-1)}$.
Theorem 5: If $r > 5$, have $f(r) = r$.
Let $p$ be any prime less than $r$.
Then $$r - (r-p) \mid f(r)-f(r-p)$$
so (inductively) $$p \mid f(r) - r$$
Therefore $f(r) \equiv r \pmod{p}$ for any prime $p < r$.
By the Chinese remainder theorem, this fixes the value of $f(r)$ modulo $\prod_{p < r} p$.
But $\prod_{p<r} p > 2(r-1)$ for $r > 5$, by Bertrand's postulate and induction.
Indeed, there is a prime between $\frac{r}{2}$ and $r$, so $$\prod_{p < r} p \geq \left( \prod_{p < r/2} p \right) \times \frac{r}{2} > 2 \left(\frac{r}{2} - 1 \right) \times \frac{r}{2} = (r-2) \times \frac{r}{2}$$
which, for $r>5$, is bigger than $2(r-1)$.
Therefore, since $f(r) < 2(r-1)$, this fixes the value of $f(r) = r$.
We still have the base cases $r=4$ and $r=5$ to deal with.
*
*$r=4$ gives, by Lemmata 2, 3 and 4, $f(4) = 4$.
*$r=5$ gives $f(5) = 5$ or $7$; but $f(5) \equiv 5 \pmod{3}$ and so it can't be $7$.
This completes the five base cases $r=1, \dots, 5$, and hence the proof.
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How does it equal -1/12? So all my friends keep telling me that if you add up all the numbers from 1 to infinity, (1+2+3+4...) then the answer is -1/12. They showed me this proof with infinite sums, but I didn't understand it, so all I'm asking is does it really equal -1/12?
| It doesn't. there's a function called the reimann zeta function. It has a complicated definition but $\zeta (s) $ will equal $1/1^s + 1/2^s + 1/3^s+... $ IF that expression has a value. If that expression does not have a value $\zeta (s) $ will have a different value.
As it turns out $\zeta (-1) = -1/12$. Now IF 1+2+3+... converged (it doesn't obviously but if it did) then it would have to be that 1+2+3+... = $\zeta (-1)=-1/12$. But 1+2+3+... DOESN'T converge so this is utterly irrelevant and meaningless.
Here's an analogy. It's a different result but it's a similar idea:
Let $N_x = 1 + x + x^2 + x^3 + ....$. To cut to the chase, if $-1 < x < 1$ then $N_x = \frac 1{1 - x}$. This is because $(1 - x)(1 + x + x^2 + x^3 + ...) = (1 + x + x^2 + x^3 + ...) - (x + x^2 + x^3 + x^4 + ...) = 1$.
So, for example $N_{1/2} = 1 + 1/2 + 1/4 + 1/8 + .... = 2 = \frac 1{1 - 1/2}$. And $N_{-1/2} = 1 - 1/2 + 1/4 - 1/8 + 1/16 - 1/32 + ... = \frac 1{1-(-1/2)} = \frac 2 3$.
Neat, huh?
But does that mean $N_{-1} = 1 -1 + 1 - 1 + 1 - 1 + .... = \frac 1{1-(-1)} = 1/2$? Or that $N_{2} = 1 + 2^2 + 2^3 + 2^4 + .... = \frac 1{1-2} = -1$?
Obviously not. Why not? Well, because when we said $(1 - x)(1 + x + x^2 + x^3 + ...) = (1 + x + x^2 + x^3 + ...) - (x + x^2 + x^3 + x^4 + ...) = 1$, we were assuming $(1 + x + x^2 + x^3 + ...)$ converges to a meaningful value. It does converge to a meaningful value if $-1 < x < 1$ and if so then everything we said was true. But if $|x| \ge 1$ then $(1 + x + x^2 + x^3 + ...)$ doesn't converge to a meaningful value and nothing we said makes any sense.
So it's the same thing with $1 + 2 + 3 + 4 + ....$. IF $1 + 2^{-s} + 3^{-s} + ....$ equals anything than that thing equals $\zeta(s)$. But $1 + 2^1 + 3^1 + ...$ doesn't equal anything. So it doesn't equal $\zeta(-1) = -1/12$.
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If $\tan(\frac{\pi}{4}+\frac{y}{2})=\tan^3(\frac{\pi}{4}+\frac{x}{2})$,then prove that $\sin y=\frac{\sin x(3+\sin^2 x)}{(1+3\sin^2 x)}$ If $\tan(\frac{\pi}{4}+\frac{y}{2})=\tan^3(\frac{\pi}{4}+\frac{x}{2})$,then prove that $\sin y=\frac{\sin x(3+\sin^2 x)}{(1+3\sin^2 x)}$
$\tan(\frac{\pi}{4}+\frac{y}{2})=\tan^3(\frac{\pi}{4}+\frac{x}{2})$
$\frac{1+\tan\frac{y}{2}}{1-\tan\frac{y}{2}}=(\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}})^3$
Applying componendo and dividendo rule on both sides,
$\tan\frac{y}{2}=\frac{\tan\frac{x}{2}(3+\tan^2\frac{x}{2})}{(1+3\tan^2\frac{x}{2})}$
I am stuck here.
| Now using Weierstrass substitution, $$\cos\left(\dfrac\pi2+A\right)=\dfrac{1-\tan^2\left(\dfrac\pi4+\dfrac A2\right)}{1+\tan^2\left(\dfrac\pi4+\dfrac A2\right)}$$
As $\sin A=-\cos\left(\dfrac\pi2+A\right)$ applying Componendo and Dividendo
$$\tan^2\left(\dfrac\pi4+\dfrac A2\right)=\dfrac{1+\sin A}{1-\sin A}$$
Replace the values of $\tan^2\left(\dfrac\pi4+\dfrac y2\right),\tan^2\left(\dfrac\pi4+\dfrac x2\right)$ in $$\tan^2\left(\dfrac\pi4+\dfrac y2\right)=\tan^6\left(\dfrac\pi4+\dfrac x2\right)=\left\{\tan^2\left(\dfrac\pi4+\dfrac x2\right)\right\}^3$$
and apply Componendo and Dividendo
| {
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Prove that $n^4-n^2$ is divisible by $8$ if $n$ is an odd positive integer.
Prove that $n^4-n^2$ is divisible by $8$ if $n$ is an odd positive integer.
I'm supposed to use proof by induction, but I failed at it miserably. So far I have this:
$$(n^4) - (n^2) = (n^2)((n^2)-1) = n(n-1)n(n+1)$$
Let $n = 2k + 1$ because its always odd. Then
$$\begin{align}
&(2k+1)(2k)(2k+1)(2k+2)
\\\implies\quad &(2k+1)(2k)(2k+1)2(k+1)
\\\implies\quad &(2k+1)2(k)(2k+1)2(k+1)
\\\implies\quad &4(2k+1)(k)(2k+1)(k+1)
\\\implies\quad &4((2k+1)^2)(k)(k+1)
\\\implies\quad &4((2k+1)^2)((k^2)+k)
\end{align}$$
Now $(2k+1)^2$ is odd and $k^2 + k$ is always even because
$$\begin{cases}
\mathrm{odd}^2 + \mathrm{odd} = \mathrm{even} \\
\mathrm{even}^2 + \mathrm{even} = \mathrm{even}
\end{cases}$$
So that means I can take out a $2$ and write $((k^2)+k) = 2X$.
$$\begin{align}
&4(2X)((2k+1)^2)
\\\implies\quad &8X((2k+1)^2)
\end{align}$$
and because there is an $8$ I can say its always divisible by $8$, no matter what odd integer $k$ is? How can this be done using proof by induction?
| You can do it more simply. Your statement that if $n$ is odd you can express it as $2k+1$ is correct. Then write $n^4-n^2=(2k+1)(2k)(2k+1)(2k+2)=(2k+1)^2(2k)(2k+2)$ Now if $k$ is odd, $2k+2$ has two factors of $2$-it is divisible by $4$. If $k$ is even, $2k$ is divisible by $4$. Either way, $2k(2k+2)$ is divisible by $8$. Another approach is just to check the odd residues $\bmod 8$ Induction was well covered by Zelzy
| {
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Show that the triangle is equilateral triangle If a triangle $ABC$ has the equality $$h_a\cdot\sqrt{3} +\frac{a}{2}= b + c$$ $h_a$ is the height from $A$, then show that the triangle is equilateral.
Using sine rule, I tried to show that bring equality to a form of showing that $A=B=C=\frac{\pi}{3}$ but I managed.
Does anyone have a solution?
Thank you very much!
| Considering the area of the triangle, we have
$$\frac 12ah_a=\frac 12bc\sin\angle{BAC},$$
i.e.
$$\sin\angle{BAC}=\frac{ah_a}{bc}=\frac{a(2b+2c-a)}{2\sqrt 3\ bc}\tag1$$
Using $(1)$ and the law of cosines,
$$\left(\frac{b^2+c^2-a^2}{2bc}\right)^2=\cos^2\angle{BAC}=1-\left(\frac{a(2b+2c-a)}{2\sqrt 3\ bc}\right)^2,$$
i.e.
$$12b^2c^2-a^2(2b+2c-a)^2-3(b^2+c^2-a^2)^2=0$$
Expanding the LHS,
$$4a^4-4a^3b-4a^3c-2a^2b^2+8a^2bc-2a^2c^2+3b^4-6b^2c^2+3c^4=0$$
Making two groups,
$$a^2(4a^2-4ab-4ac-2b^2+8bc-2c^2)+3(b^2-c^2)^2=0$$
then
$$a^2(4a^2-4ab-4ac+b^2-3b^2+2bc+6bc+c^2-3c^2)+3(b^2-c^2)^2=0$$
and rearranging
$$a^2(4a^2+b^2+c^2-4ab-4ac+2bc-3b^2+6bc-3c^2)+3(b^2-c^2)^2=0$$
to have
$$a^2((2a-b-c)^2-3(b-c)^2)+3(b^2-c^2)^2=0$$
then
$$a^2(2a-b-c)^2=3a^2(b-c)^2-3(b^2-c^2)^2$$
$$a^2(2a-b-c)^2=3(b-c)^2(a^2-(b+c)^2)$$
$$a^2(2a-b-c)^2=3(b-c)^2(a-b-c)(a+b+c)$$
Now, the LHS is non-negative, and the RHS is non-positive as $a-b-c\lt 0$, from which we have to have
$$a^2(2a-b-c)^2=3(b-c)^2(a-b-c)(a+b+c)=0,$$
i.e.
$$2a-b-c=b-c=0\quad\Rightarrow\quad a=b=c$$
as desired.
| {
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Prove that $\int_1^{\frac{1+z}{1-z}} \frac{d^n}{dz^n} (z^2-1)^n= -\frac{2(z-1)^n}{(n+1)! }\frac{d^n}{dz^n} (\frac{z}{z-1})^{n+1} $
Prove that
$$\int_1^{\frac{1+z}{1-z}} \frac{d^n}{dz^n} (z^2-1)^n= -\frac{2(z-1)^n}{(n+1)! }\frac{d^n}{dz^n} (\frac{z}{z-1})^{n+1} $$
Here are some attempts.
$\frac{d^n}{dz^n} (z^2-1)^n=2^n n! L_n(z),$ where $L_n(z)$ is the Legendre polynomials. I tried to use following formula for Legendre polynomials
$$\frac{x^2-1}{n} \frac{d}{dx}L_n(x)=xL_n(x)-L_{n-1}(x).$$
But I get tripped up on it.
| Since $\frac{d^n}{dz^n} (z^2-1)^n$ is Rodrigues' formula of Legendre polynomials, we use the following formula from https://en.wikipedia.org/wiki/Legendre_polynomials.
$L(z)=\sum_{i=0}^n {{n \choose i} {-n-1 \choose i}(\frac{1-x}{2})^{i}}.$
Using the property ${n \choose i}=(-1)^i{-n+i-1 \choose i}$ and leting $k:=n-i$ we have
$L_n(x)=\sum_{k=0}^n {{n \choose k} {2n-k \choose n}(\frac{x-1}{2})^{n-k}}. $
Thus we have
$\int_1^{\frac{1+z}{1-z}} \frac{d^n}{dz^n} (z^2-1)^n= \int_1^{\frac{1+z}{1-z}}L_n(z) dx=\int^{\frac {z+1}{1-z}}_{1} \sum_{k=0}^n {{n \choose k} {2n-k \choose n}(\frac{x-1}{2})^{n-k}} dx=
\\= \left.\sum_{k=0}^n {n \choose k} {2n-k \choose n}\frac{(x-1)^{n-k+1}}{2^{n-k}(n-k+1)}\right|_1 ^{\frac {z+1}{1-z}}=
\\=\frac{1}{n+1} \sum_{k=0}^n {n+1 \choose k} {2n-k \choose n}(1-\frac{z+1}{1-z})^{n-k}\frac{1}{2^{n-k+1}}-\frac{1}{n+1} \sum_{k=0}^n {n+1 \choose k} {2n-k \choose n}(0)^{n-k+1}\frac{1}{2^{n-k}}=
\\=\frac{1}{n+1} \sum_{k=0}^n {n+1 \choose k} {2n-k \choose n}\frac{(2z)^{n-k+1}}{2^{n-k}(1-z)^{n-k+1}}=
\\=2\frac{(z-1)^{n} (-1)^n}{n+1} \sum_{k=0}^n {n+1 \choose k} {2n-k \choose n}z^{n+1-k}(1-z)^{-2n+k-1}.$
Appling the General Leibniz Rule for product differentiation for $\frac{d^n}{dz^n}\left(\frac{z}{z-1}\right)^{n+1}$ we have
$-\frac{2(z-1)^{n}}{(n+1)!} \frac{d^n}{dz^n}\left(\frac{z}{z-1}\right)^{n+1}=
=-\frac{2(z-1)^{n}}{(n+1)!} \sum_{k=0}^n {n \choose k} \frac{d^k }{dz^k}z^{n+1}\frac{d^{n-k} }{dz^{n-k}}\frac{1}{(z-1)^{n+1}}=\\
-\frac{2(z-1)^{n}}{n+1} \sum_{k=0}^n {n+1 \choose k} {2n-k \choose n}\frac{(n+1-k)!}{(n+1)!}\frac{d^k z^{n+1}}{dz^k}\frac{n!}{(2n+k)!}\frac{d^{n-k} (z-1)^{-(n+1)}}{dz^{n-k}}=
\\ 2\frac{(z-1)^{n} (-1)^n}{n+1} \sum_{k=0}^n {n+1 \choose k} {2n-k \choose n}z^{n+1-k}(1-z)^{-2n+k-1}.$
There is another proof. But it uses a little-known identity.
| {
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Evaluation of $ \lim\limits_{x\rightarrow \infty}x\left[\ln \left(e\left(1+\frac{1}{x}\right)^{1-x}\right)\right]$
Evaluation of $\displaystyle \lim_{x\rightarrow \infty}x\left[\ln \left(e\left(1+\frac{1}{x}\right)^{1-x}\right)\right]$
$\bf{My\; Try::}$ Let $\displaystyle l=\displaystyle \lim_{x\rightarrow \infty}x\left[\ln \left(e\left(1+\frac{1}{x}\right)^{1-x}\right)\right]=\lim_{x\rightarrow \infty}x\left[1+(x-1)\ln\left(1+\frac{1}{x}\right)\right]$
So we get $$l=\lim_{x\rightarrow \infty}x\left[1+(1-x)\left(\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3}-\frac{1}{4x^4}-.........\infty\right)\right]$$
So we get $$l=\lim_{x\rightarrow \infty}x\left[1-1+\frac{1}{2x}-\frac{1}{3x^2}+\frac{1}{4x^3}+\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3}.....\right] = \frac{3}{2}$$
My Question is How can we solve it without using series expansion,
If yes then plz explain here, Thanks
| Let $y = 1/x$ and apply L'Hospital's rule:
$$\begin{align}l &= \lim_{y \to 0}\frac{1 + (1 - 1/y)\ln(1+y)}{y}\\ &= \lim_{y \to 0}\frac{1 - \frac{\ln(1+y)}{y}}{y} + \lim_{y \to 0}\frac{\ln(1+y)}{y} \\ &= \lim_{y \to 0}\frac{\ln(1+y)(1+y)-y}{y^2(1+y)}+ \lim_{y \to 0}\frac{1}{1+y} \\ &= \lim_{y \to 0}\frac{1 + \ln(1+y) - 1}{2y + 3y^2}{}+1 \\ &= \lim_{y \to 0}\frac{1 }{(1+y)(2 + 6y)}{}+1\\ &= \frac{1}{2} + 1 \\ &= \frac{3}{2}\end{align}$$
| {
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In how many ways can a positive integer $n$ be expressed as a summation of positive integers less than $n$ For example if I take $n=5$, then I can express it in the following ways:
*
*$1+1+1+1+1$
*$2+3$
*$3+2$
*$1+4$
*$4+1$
*$1+1+3$
*$1+3+1$
*$3+1+1$
*$2+2+1$
*$2+1+2$
*$1+2+2$
Please note that the order of terms in the expression of summation also counts if the terms are distinct.
| Take the string $ \large 1{\boxed.} 1\boxed. 1\boxed. 1\boxed. 1$
In the $\;(n-1)\;$ boxes, either put a $+\;$ or a comma.
$1,\;\;1,\;\;1+1+1$ e.g. would represent $1+1+3$
Since you have $2$ choices for each box, # of compositions = $2^{n-1}$
but as you have specified positive integers less than $n$, ans = $2^{n-1} - 1 = 2^4 -1 = 15$
| {
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How do I show that $\frac a{1 - a^2} + \frac b{1 - b^2} + \frac c{1 - c^2} \ge \frac {3 \sqrt 3}2$ For $0 \lt a, b, c \lt 1$, if $ab + bc + ca = 1$, show that $$\frac a{1 - a^2} + \frac b{1 - b^2} + \frac c{1 - c^2} \ge \frac {3 \sqrt 3}2.$$
I want to use trigonometric substitution:
For the angles $A, B, C$ of any acute triangle, $$\tan A + \tan B + \tan C = \tan A \tan B \tan C,$$ $$\frac 1{\tan A \tan B} + \frac 1{\tan B \tan C} + \frac 1{\tan C \tan A} = 1.$$
Also, $\tan A, \tan B, \tan C \gt 0$. So I substitute $a, b, c$ for $\frac 1{\tan A}, \frac 1{\tan B}, \frac 1{\tan C}$ respectively. Then the inequality in question becomes $$\frac {\tan A}{1 - \tan^2 A} + \frac {\tan B}{1 - \tan^2 B} + \frac {\tan C}{1 - \tan^2 C} \le -\frac {3 \sqrt 3}2.$$
Here $A, B, C \not = \frac {\pi}4$ since $a, b, c \not = 1$.
By the trigonometric identity $\tan 2A = \frac {2 \tan A}{1 - \tan^2 A}$, we have
$$\tan 2A + \tan 2B + \tan 2C \le -3 \sqrt 3,$$
where $0 \lt A, B, C \lt \frac {\pi}2$, $A, B, C \not = \frac {\pi}4$, and $A + B + C = \pi$.
How do I proceed?
Edit: The restriction $a, b, c \lt 1$ was added after the question had received some answers, thanks to Michael Rozenberg, who pointed out this mistake.
| We can prove $S=\sum\tan A=\prod\tan A$
Using AM GM inequality if $\tan A,\tan B,\tan C\ge0,$ $$\dfrac{\sum\tan A}3\ge\sqrt[3]{\prod\tan A}$$
$$\iff\dfrac S3\ge\sqrt[3]S\implies\left(\dfrac S3\right)^3\ge S\iff S^2\ge27$$ as $S\ne0$
Can you take it from here?
| {
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Maclaurin series of $x^3/(e^x-1)$ how would i taylor expand $f(x)=\frac{x^3}{e^x-1}$ around $x=0$?
I was thinking of writing
$\frac{x^3}{e^x-1}\approx\frac{x^3}{1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\dots}$
$~~~~~~~~= \frac{1}{\frac{1}{x^3}+\frac{1}{x^2}+\frac{1}{2x}+\frac{1}{6}+\frac{x}{24}+\frac{x^2}{120}+\dots}$
$~~~~~~~~=6\bigg(\frac{1}{1+(\frac{6}{x^3}+\frac{6}{x^2}+\frac{3}{x}+\frac{x}{4}+\frac{x^2}{20}+\dots)}\bigg)$
And then use the geometric series $\frac{1}{1+x}=1-x+x^2-x^3+\dots$ but that didn't get ge the right answer..
| If you assume it has a Taylor expansion, you can write:
$$\frac{x^3}{e^x-1} = a_0+a_1x+a_2x^2+a_3x^3+\ldots$$
Rearrange:
$$x^3 = \left( a_0+a_1x+a_2x^2+a_3x^3+\ldots \right)(e^x-1) $$
Use the series of $e^x$:
$$x^3 = \left( a_0+a_1x+a_2x^2+a_3x^3+\ldots \right)\left( x + \frac{x^2}{2}+ \frac{x^3}{6}+ \frac{x^4}{24} + \ldots \right) $$
Expand and group in powers of $x$:
$$x^3 = a_0x+ \left( \frac{a_0}{2}+a_1 \right)x^2 + \left( \frac{a_0}{6}+ \frac{a_1}{2}+a_2 \right)x^3 + \left( \frac{a_0}{24}+ \frac{a_1}{6}+\frac{a_2}{2} + a_3 \right)x^4 + \ldots$$
Identifying the coefficients of the corresponding powers of $x$ then allows you to find the coefficients $a_0,a_1,a_2,a_3,\ldots$ recursively.
$$\left\{
\begin{array}{l}
a_0 = 0 \\
\frac{a_0}{2}+a_1 = 0 \\
\frac{a_0}{6}+ \frac{a_1}{2}+a_2 = 1 \\
\frac{a_0}{24}+ \frac{a_1}{6}+\frac{a_2}{2} + a_3 = 0 \\
\cdots
\end{array}
\right. \Rightarrow \left\{
\begin{array}{l}
a_0 = 0 \\
a_1 = 0 \\
a_2 = 1 \\
a_3 = -\frac{1}{2} \\
\cdots
\end{array}
\right.$$
Tedious and perhaps not that elegant, but it works!
$$\frac{x^3}{e^x-1} = x^2-\frac{x^3}{2}+\frac{x^4}{12}+\ldots$$
| {
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Are these metrics? I want to find if the below functions are metrics. I have worked through each of the three conditions, but am stuck on the positivity of $f(a, b)$ (first condition-see below) and the triangle inequality of $g(a, b)$ (third condition-see below).
Edit: Have now solves these two parts, see the answers for positivity of $f(a, b)$ and the triangle inequality of $g(a, b)$
Please could you help me on these two points?
$f(a, b)=a^2+ab+b^2$
*
*Initially I thought this would not satisfy positivity. But for $f$
to be negative, $ab$ must be negative, and greater in modulus value
to $a^2+b^2$. So for this to work, $a$ and $b$ must be opposite sign
and less than $1$. I have tried small opposite signed values of $a$
and $b$ but have not found $f(a, b)$ to be positivity. Not sure.
*Symmetry. $f(b, a)=b^2+ba+a^2=a^2+ab+b^2$ so this is satisfied
*Triangle inequality. Is $f(a, b)+f(b, c) \geq f(a, c)$ satisfied? Let $X=f(a, b)+f(b, c)-f(a, c)=a^2+b^2+ab+c^2+b^2+cb-a^2-c^2-ac=b^2+ab+cb-ac$. Assume $X \geq 0 $ for all $a, b, c \in \mathbb{R}\implies b^2 +ab+cb \geq ac$. My initial thought is that if we take $b$ to be very small and $a$ and $c$ to be comparatively larger, then we may arrive at a contradiction. Take $b=0.01, a=100, c=100 \implies 0.0001+2 \geq 10000$. Contradiction. Triangle inequality does not hold.
$g(a, b)= \frac{|a^3-b^3|}{2}$
*
*Positivity satisfied since $|x| \geq 0 \forall x \in \mathbb{R}$ and $2 \geq 0$
*$g(b, a)=\frac{|b^3-a^3|}{2}=\frac{|-(-b^3+a^3)|}{2}=\frac{|-1||a^3-b^3|}{2}=\frac{|a^3-b^3|}{2}=g(a, b)$. So symmetry is satisfied.
*Must consider the sign of: $X=g(a, b)+g(b, c)-g(a, c)=\frac{|a^3-b^3|}{2}+\frac{|b^3-c^3|}{2}-\frac{|a^3-c^3|}{2}$ For the triangle inequality to be satisfied $X$ must be positive. I tried to find some counter-examples, such as take $a=100, c=0 \implies \frac{|100^3-b^3|}{2}+\frac{|b^3|}{2}-\frac{|100^3|}{2}$. But if $b$ is positive then the whole thing is positive. If it is negative , then the first two terms are positive and larger than the third. From this reasoning, I believe $X$ is always positive and the triangle inequality holds. Not sure.
| $f(a,b)=a^2+ab+b^2 \geq \frac {1}{4} a^2 + ab + b^2=(\frac {1}{2} a + b)^2 \geq 0$.
This shows that your $f$ satisfies non-negativity condition.
| {
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How can I solve this line & plane intersect question and verify the given answer? Find an equation for the plane that passes through the point $(3,2,1)$ and contains the line of intersection of the planes with equations $x+y+z=3$ and $x+2y+3z=6$.
The given answer from the key is: $-x+2y+5z=6$.
| Let $ p_1 = x+y+z -3$ and let $p_2 = x+2y+3z-6$. So, now, the given plane equations are $p_1 = 0$ and $p_2 = 0$.
Think about the equation $kp_1 + p_2 = 0$. This is certainly a plane. Also, it contains the intersection of the given two planes, because any point that satisfies both $p_1 = 0$ and $p_2=0$ also satisfies $kp_1 + p_2 = 0$.
So, by varying $k$, we can produce an entire family of planes that contain the line of intersection. Now we just have to choose the right $k$ to get a specific plane of this family that contains the point $(3,2,1)$.
Now
\begin{align}
&\;(3,2,1) \text{ lies on the plane } kp_1 + p_2 = 0 \\
\Rightarrow\;& (3,2,1) \text{ satisfies the equation } k(x+y+z -3) + (x+2y+3z-6) =0 \\
\Rightarrow\;& k(3+2+1-3) + (3 +4 +3 -6) = 0 \\
\Rightarrow\;& 3k + 4 = 0 \\
\Rightarrow\;& k = -4/3 \\
\end{align}
Substitute $k= -4/3$ in $kp_1 + p_2 = 0$. The equation of the desired plane is:
$$
-\tfrac43(x+y+z -3) + (x+2y+3z-6)=0
$$
$$
\text{i.e. }-4(x+y+z -3) + 3(x+2y+3z-6)=0
$$
$$
\text{i.e. }-x + 2y +5z = 6
$$
| {
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Apolar polynomial of a polynomial of degree 2 Suppose $P(z)=3z^2-2(a+b+c)z+ab+bc+ca,$ where $a, b, c$ are any complex numbers lying on or within the unit circle $|z|=1.$ I am trying to find an apolar polynomial $Q(z)$ (in terms of a,b,c) to $P(z)$ such that both the zeros of $Q(z)$ lie within a unit distance from $a.$ Could any one guide me in this direction?
| I'm going to charge in
and see how far I can get.
It turns out that
I got stopped,
but I will show how far I got anyway.
The zeros of
$P(z)=3z^2-2(a+b+c)z+ab+bc+ca,
$
(I am assuming that
the second term is $z$, not $z^2$)
are
$\begin{array}\\
r_1, r_2
&=\dfrac{2(a+b+c)\pm \sqrt{4(a+b+c)^2-12(ab+bc+ca)}}{6}\\
&=\dfrac{2(a+b+c)\pm \sqrt{4(a^2+b^2+c^2+2ab+2ac+2bc)-12(ab+bc+ca)}}{6}\\
&=\dfrac{2(a+b+c)\pm \sqrt{4(a^2+b^2+c^2-ab-ac-bc)}}{6}\\
&=\dfrac{(a+b+c)\pm \sqrt{a^2+b^2+c^2-ab-ac-bc}}{3}\\
\end{array}
$
Now
$\begin{array}\\
2(a^2+b^2+c^2-ab-ac-bc)
&=a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2\\
&=(a-b)^2+(b-c)^2+(c-a)^2\\
\end{array}
$
so
$r_1, r_2
=\dfrac{(a+b+c)\pm \sqrt{(a-b)^2+(b-c)^2+(c-a)^2)}/\sqrt{2}}{3}
$.
Choosing $a$ as the point
to work with,
$a_1
=a-r_1
=\dfrac{(2a-b-c)- \sqrt{(a-b)^2+(b-c)^2+(c-a)^2)}/\sqrt{2}}{3}
$
and
$a_2
=a-r_2
=\dfrac{(2a-b-c)+ \sqrt{(a-b)^2+(b-c)^2+(c-a)^2)}/\sqrt{2}}{3}
$,
and we want
$|a_1|<1
$
and
$|a_2|<1
$.
Choosing $a_1$,
this means
$|(2a-b-c)+ \sqrt{(a-b)^2+(b-c)^2+(c-a)^2)}/\sqrt{2}|
< 3
$.
The fact that
$a, b,$ and $c$
are complex
leaves me unsure of what to do
at this point.
So,
I'll stop and hope that
someone else can make use of
what I have done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1740212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
solve $x^2 -4x +13 \equiv 0 \pmod{81}$? How do I solve $x^2 -4x +13 \equiv o \pmod{81}$ ?
I know that this is the same as $x^2 -4x +13 \equiv x^2 + 2x + 1 \equiv (x +1)^2\equiv 0\pmod{3^4}$
but why is $x \equiv -1\pmod{3}$ the only solution of the congruence $f(x) \equiv 0\pmod{3}$ ? (why is $x \equiv 2\pmod{3}$ not a solution?)
How do I use hensel's lemma to find/ prove that there are no solutions to this problem?
| $$x^2-4 x+13\equiv 0\pmod {81}\iff81|(x^2-4 x+13)=(x-2)^2+9\iff$$ $$\iff 9|((x-2)^2/9+1\iff \exists y\;(\;[x-2=3 y]\land [9|y^2+1]\;)\implies$$ $$\implies \exists y\;(3|y^2+1)\iff \exists y\;(y^2\equiv 2\pmod 3\;).$$ Which can't happen because if $y\equiv 0\pmod 3$ then $y^2\equiv 0\pmod 3,$ and if $y\equiv \pm 1\pmod 3$ then $y^2\equiv 1\not \equiv 2\pmod 3.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1741007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How do I show that $\frac {a^2}b + \frac {b^2}c + \frac {c^2}d + \frac {d^2}a \ge 4$ for $a^2 + b^2 + c^2 + d^2 = 4$? Let $a, b, c, d$ be positive real numbers such that $a^2 + b^2 + c^2 + d^2 = 4$, show that $$\frac {a^2}b + \frac {b^2}c + \frac {c^2}d + \frac {d^2}a \ge 4.$$
My try:
$$\frac {a^2}b + \frac {b^2}c + \frac {c^2}d + \frac {d^2}a \ge a + b + c + d,$$
yet $$a + b + c + d \le \sqrt{4(a^2 + b^2 + c^2 + d^2)} = 4.$$
Thus, direct application of Cauchy-Schwarz inequality is too weak. I tried other methods but with no significant progress:
$$(\frac {a^2}b + \frac {b^2}c + \frac {c^2}d + \frac {d^2}a)^2 \ge \frac {(a^{4/3} + b^{4/3} + c^{4/3})^3}{a^2 + b^2 + c^2 + d^2} = \frac {(a^{4/3} + b^{4/3} + c^{4/3})^3}4.$$
I also observed that
$$(\frac {a^2}b + \frac {b^2}c + \frac {c^2}d + \frac {d^2}a) + (\frac {a^2}c + \frac {b^2}d + \frac {c^2}a + \frac {d^2}b) + (\frac {a^2}d + \frac {b^2}a + \frac {c^2}b + \frac {d^2}c) + (\frac {a^2}a + \frac {b^2}b + \frac {c^2}c + \frac {d^2}d) = 4 (\frac 1a + \frac 1b + \frac 1c + \frac 1d) \ge 16,$$
since $$\frac 1a + \frac 1b + \frac 1c + \frac 1d \ge \sqrt{\frac {(1 + 1 + 1 + 1)^3}{a^2 + b^2 + c^2 + d^2}} = 4.$$
Now my work might seem stupid or off-topic here, but I provide it here because I wish any of these attempts will lead to a solution. Any hints will be appreciated.
| This may also be shown using Jensen's inequality with $x \mapsto \frac{1}{\sqrt{x}}$ and weights $\frac{a^2}{4}$, $\frac{b^2}{4}$, $\frac{c^2}{4}$, $\frac{d^2}{4}$, which by the given assumption sum to one.
\begin{align*}
\frac {a^2}b + \frac {b^2}c + \frac {c^2}d + \frac {d^2}a &=
4 \left( \frac{a^2}{4}\frac{1}{\sqrt{b^2}} + \frac{b^2}{4}\frac{1}{\sqrt{c^2}} + \frac{c^2}{4}\frac{1}{\sqrt{d^2}}+ \frac{d^2}{4}\frac{1}{\sqrt{a^2}} \right) \\
&\geq 4 \left( \frac{1}{\sqrt{ \frac{a^2}{4} b^2 + \frac{b^2}{4} c^2 + \frac{c^2}{4} d^2 + \frac{d^2}{4} a^2} } \right)
\end{align*}
So it suffices to show that
\begin{equation}
\frac{a^2}{4} b^2 + \frac{b^2}{4} c^2 + \frac{c^2}{4} d^2 + \frac{d^2}{4} a^2 \leq 1
\end{equation}
This is the same ending point as the other existing answers, which factor and use AM-GM to finish.
$$\frac{a^2}{4} b^2 + \frac{b^2}{4} c^2 + \frac{c^2}{4} d^2 + \frac{d^2}{4} a^2 = \frac{1}{4} (a^2+c^2)(b^2+d^2) \leq \frac{1}{4^2}(a^2+b^2+c^2+d^2)^2 = 1 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1743834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 2
} |
Proving a trigonometric identity:
Prove that $\sin \frac{{2\pi }}{7} + \sin \frac{{4\pi }}{7} + \sin
\frac{{8\pi }}{7} = \frac{{\sqrt 7 }}{2}$.
I have tried to square both side and got ${\sin ^2}\frac{{2\pi }}{7} + {\sin ^2}\frac{{4\pi }}{7} + {\sin ^2}\frac{{8\pi }}{7} = \frac{7}{4}$. But I cannot proceed further. Any help would be appreciated.
| Starting from ${\sin ^2}\frac{{2\pi }}{7} + {\sin ^2}\frac{{4\pi }}{7} + {\sin ^2}\frac{{8\pi }}{7} = \frac{7}{4}$, we only need to prove
$$-\cos\frac{4\pi}{7}-\cos\frac{8\pi}{7}-\cos\frac{16\pi}{7}=\frac12,$$
or equivalently
$$\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7}=\frac{1}{2}.$$
Multiplying $2\sin\frac{\pi}7$ on both sides,
$$\sin\frac{2\pi}{7}+\left(\sin\frac{4\pi}{7}-\sin\frac{2\pi}{7}\right)+\left(\sin\frac{6\pi}{7}-\sin\frac{4\pi}{7}\right)=\sin\frac\pi7.$$
This is obviously true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1745060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Simplifying radicals inside radicals: $\sqrt{24+8\sqrt{5}}$ Simplify: $\sqrt{24+8\sqrt{5}}$
I removed the common factor 4 out of the square root to obtain $2\sqrt{6+2\sqrt{5}}$, but the answer key says it is $2+2\sqrt{5}$.
Am I missing out on some general rule here?
| To see whether $\sqrt {24+8\sqrt 5 }=A+B\sqrt 5$ with rational $A, B,$ suppose $$24+8\sqrt 5=(A+B\sqrt 5)^2=(A^2+5 B^2)+2 A B\sqrt 5.$$ Since $\sqrt 5$ is not rational, this requires $$24=A^2+5B^2 \;\text {and }\; 2 A B=8.$$ From $2 A B=8$ we have $B=4/A.$ (Note that we cannot have $A=0$ else $24+8\sqrt 5=5 B^2$ is rational, implying $\sqrt 5$ is rational.)
Substituting $4/A$ for $B$ in $24=A^2+5 B^2,$ and multiplying through by $A^2,$ and re-arranging, we get $$(A^2)^2-24 A^2+80=0.$$ This is quadratic in $A^2,$ with solutions $$A^2\in \{4, 20\}.$$ But $A^2=20$ implies $A$ is not rational. This leaves $A^2=4,$ so $A=\pm 2.$ Since we have $B=4/A,$ this implies $(A,B)=(2,2)$ or $(A,B)=(-2,-2).$ Since $A+B\sqrt 5>0,$ we have $A=B=2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
} |
Prove that the identity is true for all natural numbers for the identity:
$$\frac{n!}{x(x+1)(x+2)...(x+n)} = \frac{A_0}{x+0} + \frac{A_1}{x+1}+...+ \frac{A_n}{x+n}$$
prove
$$A_k= (-1)^kC(n,k)$$
I think this might work by induction, but i am not able to arrive at a final answer. please help!!!
| Multiply both sides by $x+k$
$$\frac{n!}{x(x+1)(x+2)\cdots(x+k-1)(x+k+1)\cdots(x+n)} = \frac{A_0 (x+k)}{x+0} + \frac{A_1 (x+k)}{x+1}+...+A_k+\cdots+ \frac{A_n (x+k)}{x+n}$$
and make the specification $x=-k$. Then
$$
\frac{n!}{(-k)(-k+1)(-k+2)\cdots(-1)\cdot 1 \cdots (n-k)} = \frac{n!}{(-1)^k k!\, (n-k)!}=(-1)^k \binom{n}{k} =A_k.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluate: $\csc^2\left(\frac{\pi}{9}\right)+\csc^2\left(\frac{2\pi}{9}\right)+\csc^2\left(\frac{4\pi}{9}\right)$
$$\csc^2\left(\frac{\pi}{9}\right)+\csc^2\left(\frac{2\pi}{9}\right)+\csc^2\left(\frac{4\pi}{9}\right) \;=\; \text{???}$$
$\bf{My\; Try::}$ Let $\displaystyle \frac{\pi}{9} = \theta\;,$ Then $9\theta = \pi\Rightarrow 6\theta = \pi-3\theta$
So $\sin (6\theta) = -\sin (3\theta)\Rightarrow 2\sin 3 \theta \cdot \cos 3 \theta+\sin 3 \theta =0$
So we get $\displaystyle \sin 3 \theta \cdot \left[2\cos 3 \theta +1\right] =0$
So we get $\sin 3\theta = 0$ or $2\cos 3 \theta+1=0$
Now i did not understand how can I convert into $\sin^2$ form, Help me
Thanks
| As a partial answer, finding the trig value for $\sin\left(\frac{\pi}{9}\right)$ is discussed here: http://mathworld.wolfram.com/TrigonometryAnglesPi9.html
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1749306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Implicit Euler method and explicit Euler method I wanna know what is the difference between explicit Euler's method and implicit Euler's method. And is the local truncation error for both of them is $O(h)$ and the coefficient of the $O(h)$ term is $h/2$?
| The error of both explicit and implicit Euler are $O(h)$. So
$$f(x-h) = f(x) - h f'(x) + \frac{h^2}{2} f''(x) - \frac{h^3}{6} f'''(x) + \cdots$$
and
$$f(x+h) = f(x) + h f'(x) + \frac{h^2}{2} f''(x) + \frac{h^3}{6} f'''(x) + \cdots$$
So the backward Euler is
$$f(x) - f(x-h) = h f'(x) - \frac{h^2}{2} f''(x) + \frac{h^3}{6} f'''(x) - \cdots$$
$$f'(x) = \frac{f(x) - f(x-h)}{h} + \frac{h}{2} f''(x) - \frac{h^2}{6} f'''(x) + \cdots$$
the backward Euler is first order accurate
$$f'(x) = \frac{f(x) - f(x-h)}{h} + O(h)$$
And the forward Euler is
$$f(x+h) - f(x) = h f'(x) + \frac{h^2}{2} f''(x) + \frac{h^3}{6} f'''(x) + \cdots$$
the forward Euler is first order accurate
$$f'(x) = \frac{f(x+h) - f(x)}{h} + O(h)$$
We can do a central difference and find
$$f(x+h) - f(x-h) = (h f'(x) + \frac{h^2}{2} f''(x) + \frac{h^3}{6} f'''(x) + \cdots) - (- h f'(x) + \frac{h^2}{2} f''(x) - \frac{h^3}{6} f'''(x) + \cdots)$$
$$f(x+h) - f(x-h) = (h f'(x) + \frac{h^2}{2} f''(x) + \frac{h^3}{6} f'''(x) + \cdots) + (h f'(x) - \frac{h^2}{2} f''(x) + \frac{h^3}{6} f'''(x) - \cdots)$$
$$f(x+h) - f(x-h) = 2 h f'(x) + \frac{h^3}{6} f'''(x) + \cdots$$
$$f'(x) = \frac{f(x+h) - f(x-h)}{2h} - \frac{h^2}{12} f'''(x) + \cdots$$
Therefore, the central difference is second order accurate.
$$f'(x) = \frac{f(x+h) - f(x-h)}{2h} + O(h^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1750238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Jordan Canonical form of $T(f(x))=f'''(x)+2f(x)$ For each linear operator T, find a Jordan canonical form $J$ of $T$ and a canonical basis $\beta $ for $T$
$T$ is the linear operatonr acting on $P_2( \mathbb{R})$
I guess I could throw in random polynomials until one gets the generelized eigenvectors magically. I am sure some here can do that no problem. uhm, this is how I have learned it and still interested in other ways as long at it is methodical.
here we go, first find the matrix representation of $T$. Basis for $P_3(\mathbb{R})= \{ 1,x,x^2,x^3\}$
$$
\begin{aligned}
T(1)&=0+2 &= 2+0x+0x^2+0x^3
\\ T(x)&=0+2x &= 0+2x+0x^2+0x^3
\\ T(x^2)&=2+2x^2 &= 2+0x+2x^2+0x^3
\\ T(x^3)&=6x+2x^3 &= 0+6x+0+2x^3
\end{aligned}$$
making $$[T]_\beta = \begin{bmatrix}
2 & 0 & 2 & 0
\\ 0 & 2& 0& 6
\\ 0 & 0& 2& 0
\\ 0 &0& 0& 2
\end{bmatrix} $$
It is upper triangular ofcourse so the char poly is $(\lambda-2)^4$, with e-vals $\lambda=1$ with algebreic multiplicity of 4.
we need to find e-vectors
$$ (T-2I)=\begin{pmatrix} 0& 0& 2& 0
\\ 0& 0& 0&6
\\ 0& 0& 0 &0
\\ 0& 0& 0& 0 \end{pmatrix}$$
$\dim(E_{\lambda_1}=2)$. there are 2 missing e-vectors that are generilized to make into jordan form aka $\dim(K_{\lambda_1})$
We need dot diagram for row 1 $$ \begin{matrix} * & * \end{matrix} $$
row2 $$\begin{aligned}
r_2 &=rank((T-\lambda_1 I )^{2-1})-rank((T-\lambda I)^2)
\\ &=rank((T-\lambda I))-rank((T-\lambda I)^2)
\\ &=2-rank \begin{pmatrix} 0& 0& 2& 0
\\ 0 & 0& 0& 6
\\ 0 &0& 0& 0
\\ 0 &0& 0& 0 \end{pmatrix}
\begin{pmatrix} 0& 0& 2& 0
\\ 0 & 0& 0& 6
\\ 0 &0& 0& 0
\\ 0 &0& 0& 0 \end{pmatrix}
\\&=2-0
\end{aligned}$$
So the Dot diagram for $T_1$ is $$ \begin{matrix}
* (T-I\lambda)v_1\ & * (T-I\lambda)v_2
\\*v_1 &*v_2
\end{matrix}$$
So, picked $(1,0,-1,0)=(T-I\lambda)v_1$ as end gen e-vector with $v_1=(0,0,1/2,1/6)$ it was found by straight inverse also $(T-I\lambda)v_2=(0,1,0,-1)$ but could not find $v_2$ I dont think it exists. there is some trickery to get our $v_1,v_2$ that works but I can't duplicate it at this time.
Do know from the dot product that the jorad form is
$$J=\begin{pmatrix} 2 & 1& 0 &0
\\ 0 &2& 0 &0
\\ 0 &0 &2 &1
\\ 0 &0& 0& 2 \end{pmatrix} $$
having trouble finding $Q$ s.t $ J=Q^{-1}AQ$
| I know it is snotty to answer your own question.But, It has been an hour or 2 since the question was ask . Turns out the ordering is important that is why it is called an ordered basis.
Basis for $N((t-I2)^2)$ is
$$\left \{ \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}
, \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}
, \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}
, \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} \right \} $$
Select a non e-vector $(0,0,1,0)$ as $v_1$
$$\begin{aligned}
(T-2I)(v_1)
&= \begin{pmatrix} 0 & 0 & 2 & 0
\\ 0& 0& 0& 6
\\ 0 &0& 0& 0
\\ 0&0&0&0
\end{pmatrix}
\begin{pmatrix}0 \\ 0\\1\\0 \end{pmatrix}
=\begin{pmatrix} 2 \\ 0\\0\\0 \end{pmatrix}
\end{aligned} $$
We will pick another that $(0,0,0,1)$
$$\begin{aligned}
(T-2I)(v_2)
&= \begin{pmatrix} 0 & 0 & 2 & 0
\\ 0& 0& 0& 6
\\ 0 &0& 0& 0
\\ 0&0&0&0
\end{pmatrix}
\begin{pmatrix}0 \\ 0\\0\\1 \end{pmatrix}
=\begin{pmatrix} 0 \\ 6\\0\\0 \end{pmatrix}
\end{aligned} $$
dot diagram is lookin like
$$\begin{matrix}
* \begin{pmatrix} 2 \\ 0\\0\\0 \end{pmatrix}
& * \begin{pmatrix} 0 \\ 6\\0\\0 \end{pmatrix}
\\ * \begin{pmatrix}0 \\ 0\\1\\0 \end{pmatrix}
& * \begin{pmatrix}0 \\ 0\\0\\1 \end{pmatrix} \end{matrix} $$
was selecting the ordered basis to be $\{ 2,6x,x^2,x^3 \}$ and keeped getting a matrix representation that was not quite jordan form but following the dot diagram closely
consider the ordere basis of $P_3(\mathbb{R})=\{ 2,x^2,6x,x^3\}$ and finding $[T]$ like in the question but with our new basis we do get the Jordan form
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1751378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that there are two values to the third side,one of which is $m$ times the other. Let $1<m<3$. In $\triangle ABC$, if $2b=(m+1)a$ and $\cos A=\frac{1}{2}\sqrt{\frac{(m-1)(m+3)}{m}}$, prove that there are two values to the third side, one of which is $m$ times the other.
$\frac{b}{a}=\frac{m+1}{2}>1$
We need to prove that $c$ has two values, one of which is $m$ times the other.
Applying cosine law,
$$c^2=a^2+b^2-2ab\cos C$$
I am stuck here as $\cos C$ is not given. Is there any other method possible for solving this question?
| By the law of cosines,
$$a^2=b^2+c^2-2bc\cos A=\left(\frac{(m+1)a}{2}\right)^2+c^2-2\cdot\frac{(m+1)a}{2}\cdot c\cdot \frac 12\sqrt{\frac{(m-1)(m+3)}{m}}$$
from which we have
$$\frac{(m-1)(m+3)}{4}a^2=-c^2+\frac{(m+1)ac}{2}\sqrt{\frac{(m-1)(m+3)}{m}}$$
Now dividing the both sides by $a^2$ and letting $x=c/a$ give
$$\frac{(m-1)(m+3)}{4}=-x^2+\frac{(m+1)x}{2}\sqrt{\frac{(m-1)(m+3)}{m}},$$
i.e.
$$x^2-\frac{(m+1)}{2}x\sqrt{\frac{(m-1)(m+3)}{m}}+\frac{(m-1)(m+3)}{4}=0$$
and so
$$\begin{align}\frac ca&=x\\&=\frac 12\left(\frac{(m+1)}{2}\sqrt{\frac{(m-1)(m+3)}{m}}\pm\frac{(m-1)}{2}\sqrt{\frac{(m-1)(m+3)}{m}}\right)\\&=\frac 14\sqrt{\frac{(m-1)(m+3)}{m}},\quad\frac m4\sqrt{\frac{(m-1)(m+3)}{m}}\end{align}$$
so, we have
$$c=\frac a4\sqrt{\frac{(m-1)(m+3)}{m}},\quad\frac {ma}4\sqrt{\frac{(m-1)(m+3)}{m}}$$
from which the claim follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1755123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to write the following polynomial in $(1-\frac{x}{a}) (1-\frac{x}{b}) (1-\frac{x}{c}) (1-\frac{x}{d})$? I was given the following problem:
Write the polynomial $f(x) = \frac{1}{24} \displaystyle \prod_{i \mathop = 1}^4 (x-i)$ in the form $(1-\frac{x}{a}) (1-\frac{x}{b}) (1-\frac{x}{c}) (1-\frac{x}{d})$
So far I have done this:
$f(x) = \frac{1}{24} \displaystyle \prod_{i \mathop = 1}^4 (x-i) = \frac{1}{24}(x-1)(x-2)(x-3)(x-4)
= \frac{1}{24}(1-x)(2-x)(3-x)(4-x)
= (\frac{1-x}{1}) (\frac{2-x}{2}) (\frac{3-x}{3}) (\frac{4-x}{4})
= ({1-x}) (1-\frac{x}{2}) (1-\frac{x}{3}) (1-\frac{x}{4})$
EDIT: Seems like I got it, thanks for all the help!
| hint: $x-i = i\left(\dfrac{x}{i}-1\right)=-i\left(1-\dfrac{x}{i}\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1757075",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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} |
Evaluate $\int \frac{1-x}{(1+x)\sqrt{x+x^2+x^3}}dx$ Evaluate $$\int \frac{1-x}{(1+x)\sqrt{x+x^2+x^3}}dx$$ i used substitution $x=\tan^2 y$ so $dx=2\tan y \sec^2 y dy$ so the integral becomes
$$I=\int\frac{2\cos 2y\: \tan y\: \sec^2 y \:dy}{\sqrt{\tan^2 y+\tan^4 y+\tan^6 y}}=\int\frac{2\cos 2y \:\sec^2 y\: dy}{\sqrt{1+\tan^2 y+\tan^4 y}}$$ I was stuck here
| This is the kind of solution that comes after the seventh cup of tea:
One can note that
$$
(1+x)^4=(1+x^2)^2+4(x+x^2+x^3),
$$
and so your integral equals
$$
\int\frac{(1-x)(1+x)^3}{\bigl((1+x^2)^2+4(x+x^2+x^3)\bigr)\sqrt{x+x^2+x^3}}\,dx
$$
or
$$
\int
\frac{1}{1+\Bigl(\frac{2\sqrt{x+x^2+x^3}}{1+x^2}\Bigr)^2}
\frac{(1-x)(1+x)^3}{(1+x^2)^2\sqrt{x+x^2+x^3}}
\,dx.
$$
Since
$$
D\frac{2\sqrt{x+x^2+x^3}}{1+x^2}=\frac{(1-x)(1+x)^3}{(1+x^2)^2\sqrt{x+x^2+x^3}}
$$
we finally find that
$$\int\frac{1-x}{(1+x)\sqrt{x+x^2+x^3}}\,dx=\arctan\biggl(\frac{2\sqrt{x+x^2+x^3}}{1+x^2}\biggr)+C.$$
| {
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Show that the curve $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ form an ellipse If the definition of an ellipse is the set of points $(x,y)$ such that given two focus points $F_1, F_2$ the sum of the distances from $(x,y)$ to each focus point is constant, how can one show that the curve $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1, \quad 0 < b \leq a \quad$ forms an ellipse?
The methods that I know of are to either derive the formula by considering the foci $(-c,0), (c,0)$ and the constant distance $2k$, or to set $(x,y) = (a\cos{v}, b\sin{v})$. Is there some other, relatively easy way to show that the points satisfying the equation form an ellipse?
I am asking because in a book I am reading the author states that a "direct calculation" shows that the curve indeed forms an ellipse, but I do not understand what kind of calculation this might be.
| You can write a point on the curve as $(a\cos t,b\sin t)$ for a unique $t\in[0,2\pi)$. Consider the distanc from $(c,0)$, where $c=\sqrt{a^2-b^2}$:
\begin{align}
\sqrt{(a\cos t-c)^2+(b\sin t)^2}
&=\sqrt{a^2\cos^2t-2ac\cos t+a^2-b^2+b^2\sin^2t} \\
&=\sqrt{a^2\cos^2t-2ac\cos t+a^2-b^2+b^2\sin^2t} \\
&=\sqrt{c^2\cos^2t-2ac\cos t+a^2} \\
&=|c\cos t-a| \\
&=a-c\cos t
\end{align}
Similarly, the distance from $(-c,0)$ is
$$
a+c\cos t
$$
so the sum is $2a$.
| {
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How to diagonalize a matrix $A$ and then compute $A^{10}$?
Diagonalize the following matrix $$A = \begin{bmatrix} -2 & -8 & -12\\ 1 & 4 & 4\\ 0 & 0 & 1\end{bmatrix}$$ and compute $A^{10}$.
I have found that the eigenvalues of $A$ are $\lambda _1 = 0$, $\lambda _2 = 1$, $\lambda _1 = 2$. Then, I found the following eigenvectors who form my matrix $P$.
$$v_1=\left (\begin{matrix}1\\ 1\\ 1\end{matrix} \right ), \qquad v_2=\left (\begin{matrix}-1/4\\ 0\\ -1/2\end{matrix} \right ), \qquad v_3=\left (\begin{matrix}0\\ -1/4\\ 0\end{matrix} \right )$$
At the end my matrix
$$P=\left ( \begin{matrix}
1 & 1 &1 \\
-\frac{1}{4}& 0 &-\frac{1}{2} \\
0 &-\frac{1}{4} & 0
\end{matrix} \right )$$
In linear algebra, two $n$-by-$n$ matrices $A$ and $B$ are called similar if $B=P^{-1}AP$. I get that
$$B=\left ( \begin{matrix}
0 & 0 & 0\\
0 &1 &0 \\
0& 0 & 2
\end{matrix} \right )$$
So is this all good? How can I use that to get value of $A^{10}$?
| We have $x^{10} \equiv 511 x^2 - 510 x \bmod x(x-1)(x-2)$ and so $A^{10} = 511 A^2 - 510 A$ by the Cayley–Hamilton theorem.
| {
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Evaluating the rational integral $\int \frac{x^2+3}{x^6(x^2+1)}dx $
Evaluate $$\int \frac{x^2+3}{x^6(x^2+1)}dx .$$
I am unable to break into partial fractions so I don't think it is the way to go. Neither is $x=\tan \theta$ substitution. Please give some hints. Thanks.
| Both methods work just fine here:
(1) Partial fractions One can apply the Method of Partial Fractions to any rational expression, and since the denominator of the integrand is already factored into a product of irreducible polynomials (over $\Bbb R$), we can proceed directly to writing the decomposition:
$$\frac{x^2 + 3}{x^6 (x^2 + 1)} = \frac{A}{x^6} + \frac{B}{x^5} + \frac{C}{x^4} + \frac{D}{x^3} + \frac{E}{x^2} + \frac{F}{x} + \frac{G}{x^2 + 1} .$$
In fact, since the integrand is even, only even terms appear in the decomposition: $$\frac{x^2 + 3}{x^6 (x^2 + 1)} = \frac{A}{x^6} + \frac{C}{x^4} + \frac{E}{x^2} + \frac{G}{x^2 + 1} .$$
(2) Trigonometric substitution This is probably less efficient than partial fractions here, but it's not a bad option: The usual substitution $x = \tan \theta, dx = \sec^2 \theta\, d\theta$, transforms the integral to $$\int \frac{\tan^2 \theta + 3}{\tan^6 \theta},$$ which can be handled in several ways. One approach is to write the numerator as $\sec^2 \theta + 2$ and splitting the integral as $$\int \frac{\sec^2 \theta \,d\theta}{\tan^6 \theta} + 2\int \cot^4 \theta \,d\theta .$$ The first can be handled with substitution and the second with the usual trick of writing $\cot^2 = \csc^2 - 1$ and applying the elementary integral $\int \csc^2 \theta = - \cot \theta + C$.
| {
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Find the range of the function $y = \sqrt{x^2 + 1}− x$? I have a function $y = \sqrt{x^2 + 1} − x$, where the Domain is $(−\infty,+\infty)$.
Explanation for the domain
I need to make sure the domain of the function does not include values of $x$
that will make the square root negative.
This means that I need:
$x^2+1\ge0$
the discriminant is negative and $x \in\mathbb R$. In fact, $x^2\ge 0$, $\forall x\in\mathbb R$ and $1 > 0$.
I wish, if it's possible, to explain the value of the range with an algebraic demonstration. I am using an example to solve the function for $x$ using $y$ as parameter:
$\sqrt{x^2 + 1} − x = y$
$\sqrt{x^2 + 1} = y + x$
irrational equation, therefore:
$$
\begin{cases}
& y + x >= 0\\
& x^2 + 1 =(y+x)^2
\end{cases}
$$
$$
\begin{cases}
& x >= -y\\
& x^2 + 1 =y^2 + x^2 + 2yx
\end{cases}
$$
$$
\begin{cases}
& x >= -y\\
& 1 =y^2 + 2yx
\end{cases}
$$
$$
\begin{cases}
& x >= -y\\
& x = \frac{1 - y^2}{2y}
\end{cases}
$$
now I need to find for which values is $x >= -y$, therefore:
$ \frac{1 - y^2}{2y} < 0$
solving the numerator and denominator
numerator
$1 - y^2 > 0$
$y^2 -1 < 0$
solve for $-1 < y < +1$
denominator
$2y > 0$
solve for $y > 0$
combine the tow solutions the inequality has occurred when
$-1 < y < 0$ or $y> 1$
| Hint: Expand the fraction with $\sqrt{x^2+1}+x$ to see that
$$y=\frac{1}{\sqrt{x^2+1}+x}$$
| {
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"question_score": "2",
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Find the number of solutions of $\sin x+2 \sin 2x- \sin 3x=3$ In $(0 \:\:\pi)$Find the number of solutions of $$\sin x+2 \sin 2x- \sin 3x=3$$
The equation can be written as
$$\sin x+4 \sin x \cos x=3+\sin 3x$$ i.e.
$$\sin x(1+4\cos x)=3+\sin 3x$$ i.e.,
$$\sin x(1+4\cos x)=3+\sin x(3-4\sin^2 x)=3+\sin x(4\cos ^2x-1)$$ so
$$\sin x(4\cos^2 x-4\cos x-2)=-3$$ Any hint from here?
| Here $$\sin x-\sin 3x+2\sin 2x = -2\cos(2x)\sin x+2\sin 2x\leq |2\cos 2x\sin x+2\sin 2x|$$
So $$\sin x-\sin 3x+2\sin 2x\leq 2|\sin 2x+\cos 2x|\leq 2\sqrt{2}|\sin (2x+45^\circ)|\leq 2\sqrt{2}<3$$
So no solution in $x\in (0,\pi)$
| {
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How to integrate the function $\ f(x)=\arctan(\sqrt{\sqrt{x}-1}) $ How to integrate the function $\ f(x)=\arctan(\sqrt{\sqrt{x}-1}) $
I tried by $\ u=\arctan(\sqrt{\sqrt{x}-1}) $ and $\ dv=dx $, but wolfram alpha doesn't agree.
My solution is $\arctan(\sqrt{\sqrt{x}-1}) \cdot x -\frac{1}{2}\sqrt{x}-\frac{1}{2}\ln(\sqrt{x}-1)$
| Integrate by parts:
$\int f \: \text{dg} = fg -\int g \: \text{df} $
where
$f = \arctan(\sqrt{\sqrt{x}-1})$
$\text{dg} = \text{dx}$
$\text{df} = \frac{1}{4x \sqrt{\sqrt{x}-1}} \text{dx}$
$g=x$
This gives
$$\int \arctan(\sqrt{\sqrt{x}-1})\: \text{dx} = x \arctan(\sqrt{\sqrt{x}-1})
- \int \frac{1}{4x \sqrt{\sqrt{x}-1}}\: \text{dx}$$
Focusing on the remaining integral, substitute $u = \sqrt{x}$.
$$
- \frac{1}{2} \int \frac{u}{\sqrt{u-1}} \: \text{du}$$
substitute $s = u-1$
$$
- \frac{1}{2} \int \sqrt{s} + \frac{1}{\sqrt{s}} \: \text{ds}$$
Now it should be straight forward to finish.
| {
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Show that $\frac{d^n}{dx^n}\left[ \frac{1}{1+x^2} \right] = \frac{(-1)^nn!}{(x^2+1)^{\frac{n+1}{2}}}\sin[(n+1)\arctan x]$ We have, $$\frac{1}{1+x^2}=\frac{A}{x-i}+\frac{B}{x+i} $$
From this we get $A=\frac{1}{2i}, B=-\frac{1}{2i}$.
So,
$$\frac{1}{1+x^2}=\frac{1}{2i}\left[\frac{1}{x-i} + \frac{1}{x+i} \right] $$
Now,
\begin{align*}\left(\frac{1}{1+x^2}\right)^{(n)}&=\frac{1}{2i}\left[\left(\frac{1}{x-i} \right)^{(n)} + \left( \frac{1}{x+i}\right)^{(n)} \right] \\
&= \frac{1}{2i}\left[ \frac{(-1)^{n}n!}{(x-i)^{n+1}} - \frac{(-1)^{n}n!}{(x+i)^{n+1}}\right] \\
&= \frac{(-1)^{n}n!}{2i} \left( \frac{(x+i)^{n+1}-(x-i)^{n+1}}{(x^2+1)^{n+1}}\right)
\end{align*}
Then we have, $$ (x+i)^{n+1}=(x^2+1)^{\frac{n+1}{2}}\left(\cos((n+1)\cdot \mathrm{arccot}x) + i \sin((n+1)\mathrm{arccot}x) ) \right)$$
$$ (x-i)^{n+1}=(x^2+1)^{\frac{n+1}{2}}\left(\cos((n+1)\cdot \mathrm{arccot}(-x)) + i \sin((n+1)\mathrm{arccot}(-x)) ) \right)$$
But this is where I'm stuck, because we are not winning $\arctan$, maybe it is some elementary trygonometry identity that I'm missing.
| HINT:
Let $x=r\cos y,1=r\sin y\implies\cot y=x\implies y=\text{arccot}(x)$
See Are $\mathrm{arccot}(x)$ and $\arctan(1/x)$ the same function?
$$(x-i)^m=r^m(\cos y-i\sin y)^m=(1+x^2)^{m/2}\left\{\cos(-y)+i\sin(-y)\right\}^m$$
Using De Moivre' Theorem,
$$\dfrac1{(x+i)^m}=\dfrac{(x-i)^m}{(x^2+1)^m}=\dfrac{(1+x^2)^{m/2}(\cos my-i\sin my)}{(x^2+1)^m}$$
Can you take it from here?
| {
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Rational solutions of $y^2 = x^3 - x$ I believe that the only rational solutions of $$y^2 = x^3 - x$$ are the obvious ones $(-1,0)$, $(0,0)$, $(1,0)$, and that this was proved by Fermat using the method of descent.
Can anyone outline a proof, either Fermat’s proof or a different one; or, failing that, point to a suitable reference, preferably available online?
| For the sake of any future readers, I thought I would extend coffeemath’s excellent answer a little to give a self-contained proof, since it’s quite easy.
Primitive Pythagorean triples
We briefly recall the classification of primitive Pythagorean triples. Let $x^2+y^2=z^2$ with $x$, $y$ coprime natural numbers. An odd square is congruent to $1$ modulo $4$, and an even square is congruent to $0$, so the sum of two odd squares can never be a square. Therefore one of $x$ and $y$ must be even. Since they are coprime, it follows that the other is odd. So suppose $x=2k$ for some $k$, and $y, z$ are odd. Then we have $4k^2+y^2=z^2$, hence $$k^2=(\frac{z+y}{2})(\frac{z-y}{2}).$$It follows (since they are coprime) that $\frac{z+y}{2}$ and $\frac{z-y}{2}$ are both square, so let $a^2=\frac{z-y}{2}$ and $b^2=\frac{z+y}{2}$. Then $$x=2ab,\qquad y=b^2-a^2,\qquad z=a^2+b^2$$ with $a<b$ and $a,b$ coprime. Note that, since $y$ and $z$ are odd, one of $a,b$ is odd and the other even.
The main proof
We want to show that the equation $$x^4+4y^4=z^2$$ has no non-trivial integer solutions. The strategy of the proof – Fermat’s method of descent – is to show that, if it does, then there is another triple of integers $p,q,r$ with $p^4+4q^4=r^2$ having $0<r<z$.
If $x,y$ have a common prime factor $w$ then $(\frac x w)^4+4(\frac y w)^4$ = $(\frac{z}{w^2})^2$, and we are done. Otherwise $x,y$ are coprime. By the classification above we have that $x$ is odd and $$2y^2=2ab,\qquad x^2=b^2-a^2,\qquad z^2=a^2+b^2$$ for $a<b$ with $a,b$ coprime. Since $y^2=ab$, it follows that $a$ and $b$ are both square, say $a=c^2$ and $b=d^2$. Therefore $x^2=d^4-c^4$, which we rewrite as $$x^2+c^4=d^4.$$Using the classification again, recalling that $x$ is odd, we have$$c^2=2ef,\qquad x=f^2-e^2,\qquad d^2=e^2+f^2$$ for some coprime $e,f$. Since $c^2=2ef$ we know that one of $e,f$ is a square and the other is twice a square, so $e,f$ are equal in some order to $p^2,2q^2$ for some $p,q$, and since $d^2=e^2+f^2$ it follows that $$p^4+4q^4=d^2$$ where $d<d^2=b<z$, as required.
| {
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"timestamp": "2023-03-29T00:00:00",
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Compute the minimum value of $a^n + b^n + c^n$ subject to $a^2 + b^2 + c^2 = 1 $
Assume that $a,b,c$ are non-negative real numbers and $n$ is a natural number $n \ge 3$. What is $f(n)=$ the minimum value of $a^n + b^n + c^n$ ?
I find ;
$$f(3) = \frac{1}{\sqrt{3}}\qquad f(4) = \frac{1}{3}$$
then I guess
$$f(n) = \left(\frac{1}{\sqrt{3}} \right)^n \times 3 $$
Is that true?
| Your conjecture is true. First, $f(n)=a^n+b^n+(1-(a^2+b^2))^{\frac{n}{2}}$, calculate its derivative w.r.t. $a$, one obtains it has the minima $a=\sqrt{\frac{1-b^2}{2}}$. Inserting this into $f(n)$, we obtain $f(n)=2(\frac{1-b^2}{2})^{\frac{n}{2}}+b^n$, now calculate its derivative w.r.t. $b$, you obtain it has the minima $b=\frac{1}{\sqrt{3}}$. Now one concludes that $a=b=c=\frac{1}{\sqrt{3}}$ and $\min f(n)=3^{\frac{2-n}{2}}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is $11^2+12^2+13^2+14^2+15^2+16^2=1111$ special? Is this pure coincidence or is this a special case of some well-known number-theoretic result?
If the latter is true, is there some notable generalization?
EDIT: Thanks to the interesting answers below, a follow-up question is now on MathOverflow: https://mathoverflow.net/questions/239033/repdigit-numbers-which-are-sum-of-consecutive-squares?noredirect=1#comment591447_239033
| The reason is that $$11^2+12^2+13^2+14^2+15^2+16^2=600+420+91$$ in which the summands are such that successively give $1,1,1,1$ when making the sum. In fact,
$$(10+1)^2+(10+2)^2+(10+3)^2+(10+4)^2+(10+5)^2+(10+6)^2=6\cdot10^2+20(1+2+3+4+5+6)+(1^2+2^2+3^2+4^2+5^2+6^2)=6\cdot10^2+20(\frac{6\cdot7}{2})+\frac{6(6+1)(2\cdot6+1)}{6}=600+420+91$$
I don't know if it is generalizable.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the numerical value of $\sin 10^\circ \sin 50^\circ \sin 70^\circ$. Prerequisite
This problem is found in "Trigonometry" by I. M. Gelfand [in English].
It is asked in the section "Double the angle". So, assume that I know the sin/cos angle additions [i.e.: $\sin(A + B) = \sin A \cos B + \cos A \sin B$, etc.] as well as everything learned prior.
I've check other sources and they say to use Morrie's Law, however I have not actually learned it in the book.
Problem
Find the numerical value of $\sin 10^\circ \sin 50^\circ \sin 70^\circ$.
Hint: If the value of the given expression is $M$, find $M \cos 10^\circ$.
| $$\begin{eqnarray*}\sin(A)\sin(B)\sin(C) &=& \frac{1}{2}\left(\cos(A-B)-\cos(A+B)\right)\sin C\\&=&\frac{1}{4}\left(\sin(C+A-B)+\sin(C-A+B)-\sin(C+A+B)-\sin(C-A-B)\right)\end{eqnarray*}$$
hence:
$$\begin{eqnarray*} \sin(10^\circ)\sin(50^\circ)\sin(70^\circ)&=&\frac{1}{4}\left(\sin(110^\circ)+\sin(30^\circ)+\sin(-10^\circ)-\sin(130^\circ)\right) \\&=&\frac{1}{4}\left(\cos(20^\circ)+\sin(30^\circ)-\sin(10^\circ)-\cos(40^\circ)\right)\end{eqnarray*}$$
but:
$$ \cos(20^\circ)-\cos(40^\circ) = 2\sin(30^\circ)\sin(10^\circ) = \sin(10^\circ)$$
hence:
$$ \sin(10^\circ)\sin(50^\circ)\sin(70^\circ) = \color{red}{\frac{1}{8}}.$$
| {
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"question_score": "4",
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How to prove that $\frac{1}{x^4+y^4} e^{-\frac{1}{x^2+y^2}}\to0$ when $(x,y)\to (0,0) $?
How can I calculate the limit
$$
\lim_{(x,y)\to (0,0)} \frac{1}{x^4+y^4} e^{-\frac{1}{x^2+y^2}}
$$
and show that it is zero?
When switching to polar coordinates, I get:
$$
\lim_{(x,y)\to (0,0)} \frac{1}{r^4 (\cos^4 \theta+\sin^4 \theta) } e^{-\frac{1}{r^2}}
$$
but I have no idea how to show the term $\frac{1}{r^4 (\cos^4 \theta+\sin^4 \theta) }$ is bounded.
| You need to show that $\cos^4\theta+\sin^4\theta$ never gets too small. A nice way to do this is to show
$$\begin{align}
\cos^4\theta+\sin^4\theta
&=(\cos^2\theta+\sin^2\theta)^2-2\cos^2\theta\sin^2\theta\\
&=1-{1\over2}(2\sin\theta\cos\theta)^2\\
&=1-{1\over2}\sin^22\theta\\
&\ge1-{1\over2}\\
&={1\over2}
\end{align}$$
| {
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max of $e$ with $a+b+c+d+e=8$ and $a^2+b^2+c^2+d^2+e^2=16$ Given that a,b,c,d,e are real number such that:
$\begin{cases}
a+b+c+d+e=8\\
a^2+b^2+c^2+d^2+e^2=16
\end{cases}$
determine the maximun value of $e$.
I started like that :
$\max(e)=\max(8-a-b-c-d)=\min(a+b+c+d)$
but i don't kow how to carry on
can you help please ?
| The geometry is the same as in 3 dimensions, when taking the plane section $x+y+z = c$ of a sphere $x^2+y^2+z^2=C$. Extrema of $z$ are when $x=y$. The only remaining thing to understand is why the answer for the parameters given in the problem does not involve square roots.
Let $n$, which equals $4$ in the posted problem, be the number of variables other than the last one, which I will continue to call "$e$".
Max/min $e$ in $(n+1)$ dimensions occur at the two solutions of
$na+e=c$ and $na^2+e^2=C$, so $(c-e)^2 = n(C - e^2)$ or $$(n+1)e^2 - 2ce + (c^2 - nC) = 0 .$$
To get a problem with extremal $e$ rational in the parameters,
$c^2 - (n+1)(c^2 - nC^2) = n((n+1)C - c^2)$ must be a perfect square.
There is no special perfect square expression that appears in the algebra, and the problem was just based on a choice of small integer values that lead to a square.
| {
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express the value of an expression as a common factor In the following problem, by adding $0.141414$..., $0.414141..., 0.151515...$, and $0.515151...$, I get $1.111....$ Then the expression becomes square root ($11 \times 1.1111$....). My answer is $11 \times \sqrt{0.1010101...}$, which is different from the answer sheet. Any help is highly appreciated.
What is the value of the following expression? Express your answer as a common fraction.
| Notice that $0.\overline{14}+0.\overline{41}+0.\overline{15}+0.\overline{51}=1.\overline{2} = 1+\frac{2}{9}$.
Now $\sqrt{11(1+\frac{2}{9})}=\sqrt{\frac{99}{9}+\frac{22}{9}}=\sqrt{\frac{121}{9}}=\frac{11}{3}=3\frac{2}{3}$
| {
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Two real numbers, $x$ and $y$, satisfy the condition $x + y = 2 $. Show $xy(x^2+y^2) \leq 2$
Question: Two real numbers, $x$ and $y$, satisfy the condition $x + y = 2 $.
Show $xy(x^2+y^2) \leq 2$
What I have attempted:
Consider $$x+y=2$$
$$ \Leftrightarrow (x+y)^2 = 2^2 $$
$$ \Leftrightarrow (x+y)^2 = 4 $$
Notice that $$ x^2 + y^2 = (x+y)^2 - 2xy $$
But $(x+y)^2 = 4$
$$ \Longrightarrow x^2 + y^2 = 4-2xy $$
Now we have this inequality; $$xy(x^2+y^2) \leq 2$$
$$\Leftrightarrow xy(4-2xy) \leq 2 $$
$$ \Leftrightarrow 2xy(2-xy) \leq 2 $$
$$ \Leftrightarrow 2(2xy-x^2y^2) \leq 2 $$
Now I am stuck, how should I continue also is there another way to approach this question?
| Another way:
Let $x=r\cos \theta$; $y=r\sin\theta$. It is clear that we need only consider the case when $0 \le \theta \le \frac \pi 2$.
Since $x+y=2$ then $0\le r \le 2^\frac12$
(i.e., the longest distence from $(0,0)$ to the line $x+y=2$ when $0 \le \theta \le \frac \pi 2$).
Therefore, $$xy(x^2+y^2)=\frac12r^4\sin 2\theta \le \frac12 (2^\frac12)^4=2.$$
| {
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Conic Sections: Hyperbola (Finding the Locus) This is a multipart question so bear with me until I get to the part where I am stuck on.
$H$: $xy=c^2$ is a hyperbola
(i) Show that $H$ can be represented by the parametric equations $x=ct$ , $y= \frac{c}{t}$.
If we take $y= \frac{c}{t}$ and rearrange it to $t= \frac{c}{y}$ and subbing this into $x=ct$
$$ x = c(\frac{c}{y}) $$
$$ \therefore xy = c^2 $$
(ii) Find the gradient of the normal to $H$ at the point $T$ with the coordinates $(ct, \frac{c}{t})$
As $xy = c^2$
$$ \Leftrightarrow y = c^2 x^{-1} $$
$$ \Leftrightarrow y' = -\frac{c^2}{x^2} $$
$$ \Leftrightarrow y'_{x=ct} = \frac{-1}{t^2} $$
Hence gradient of the normal is $m = t^2$
The normal to H at the point $T$ meets $H$ again at $P(cp , \frac{c}{p})$
(iiI) By finding the gradient of the line TP and comparing it with the gradient from (ii), find a relationship between $t$ and $p$
So repeating steps of (ii) again
$xy = c^2$
$$ \Leftrightarrow y = c^2 x^{-1} $$
$$ \Leftrightarrow y' = -\frac{c^2}{x^2} $$
$$ \Leftrightarrow y'_{x=cp} = \frac{-1}{p^2} $$
Hence gradient of the normal is $$m = p^2$
Comparing gradients
$$ p^2 = t^2 $$
$$ \Leftrightarrow |p| = |t| $$ Now do we assume $t>0$ and $p>0$? which gives us
$$ \Leftrightarrow p = t $$ as our relationship.
Now this is where I am mostly stuck on
(iv) Hence or otherwise by finding the parametric equations of $M$, the midpoint of $TP$ in terms of $t$, show that the coordinates of $M$ lie on the curve $4x^3y^3 + c^2(x^2-y^2)^2 = 0$
Note: This can also be stated as "Find the Locus of M"
So basically what I have done is to find the normal line of TP
Which is given as
$$ (y-y_1)=m(x-x_1) $$
Because they want everything in terms of $t$ , let $p=t$
$$ (y-\frac{c}{t}) = t^2(x-ct) $$
$$ \Leftrightarrow (y-\frac{c}{t}) = t^2x - ct^3 $$
$$ \Leftrightarrow y = t^2x - ct^3 + \frac{c}{t} $$
Now this is stuck how should I continue?
| (iii)
You have an error here.
That the normal to H at the point $T$ meets $H$ again at $P(cp , \frac{c}{p})$ means that the normal to $H$ at the point $T$ is the line $TP$.
The gradient of the line $TP$ is given by
$$\frac{c/p-c/t}{cp-ct}=\frac{-(p-t)}{(p-t)pt}=-\frac{1}{pt}$$
Since this is equal to $t^2$, we have
$$-\frac{1}{pt}=t^2\quad\Rightarrow\quad pt^3=-1$$
(iv)
Here, you have to consider the coordinates of $M$, not the equation of the line $TP$.
Let $M(X,Y)$.
Then, using $p=-1/t^3$,
$$X=\frac{cp+ct}{2}=\frac{-c/t^3+ct}{2}=\frac{c(t^4-1)}{2t^3}$$
$$Y=\frac{c/p+c/t}{2}=\frac{-ct^3+c/t}{2}=\frac{c(1-t^4)}{2t}$$
Then,
$$\begin{align}&4X^3Y^3+c^2(X^2-Y^2)^2\\&=4\left(\frac{c(t^4-1)}{2t^3}\right)^3\left(\frac{c(1-t^4)}{2t}\right)^3+c^2\left(\left(\frac{c(t^4-1)}{2t^3}\right)^2-\left(\frac{c(1-t^4)}{2t}\right)^2\right)^2\\&=\frac{-c^6(t^4-1)^6}{16t^{12}}+\frac{c^2\cdot c^4(1-t^4)^6}{16t^{12}}\\&=0\end{align}$$
| {
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sum of all positive integral values of $a\;,$ for which equation $\lfloor x \rfloor ^3+x-a=0$ has solution
The sum of all positive integral values of $a\;,$ Where $a\in \left[1,1500\right]$
for which the equation $\lfloor x \rfloor ^3+x-a=0$ has solution, Where $\lfloor x \rfloor $ Represent floor of $x$
$\bf{My\; Try::}$ Given $\lfloor x \rfloor ^3+x-a=0\Rightarrow x=\underbrace{a-\lfloor x \rfloor^3}_{\bf{integer\; quantity}}$
So Here $x$ must be an $\bf{Integer\; quantity.}$
Now How can I solve after that, Help Required, Thanks
| $$\lfloor x \rfloor ^3+x-a=0$$
$a\in \mathbb Z \Rightarrow x \in Z$
$$x^3+x=a$$
$$x(x^2+1)=a$$
$a\in[1;1500]$
Then $2\le x\le11$
If $x=2$ then $a=10$
If $x=3$ then $a=3\cdot10=30$
...
If $x=11$ then $a=11\cdot(11^2+1)=1342$
| {
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Determine the value of c that makes the blue area above y = c equal to the blue area below y = c.
Determine the value of c that makes the blue area above y = c equal to the blue area below y = c.
edit: I'm kind of stuck on this problem, not sure what steps to do so that I can find the equal areas.
edit2: The answer looks right so far, thanks to everyone who helped out!
| Let $x=a$ be the first intersection and $x=b$ be the second.
The areas are equal, so $\displaystyle\int_0^a c-(8x-27x^3)dx=\int_a^b 8x-27x^3-cdx$. Carrying out the integration, we have:
$$cx-4x^2+\frac{27}{4}x^4\Bigg|_0^a= 4x^2-\frac{27}{4}x^4-cx\Bigg|_a^b$$
$$ca-4a^2+\frac{27}{4}a^4= 4b^2-\frac{27}{4}b^4-cb-4a^2+\frac{27}{4}a^4+ca$$
$$0= 4b^2-\frac{27}{4}b^4-cb=b(4b-\frac{27}{4}b^3-c)$$
Thus, $c=4b-\dfrac{27}{4}b^3$
Now, we also recall that $f(b)=8b-27b^3=c$
We set the two equations equal giving:
$$4b-\dfrac{27}{4}b^3=8b-27b^3$$
$$0=4b-\dfrac{81}{4}b^3$$
By normal methods, we find that $b=\dfrac{4}{9}$. We substitute this back into our equation relating $b$ and $c$ to find that: $$f(\dfrac{4}{9})=8\dfrac{4}{9}-27(\dfrac{4}{9})^3=\dfrac{32}{27}=c$$
| {
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Determine the composition of the functions $f(x)=4x+3$ and $g(x)=-5x^2+1$ Answer:
\begin{align*}
(f \circ g)(x) & = f(g(x))\\
& = 4(-5x^2+1)+3\\
& = -20x^2+8+3\\
& = -20x^2+11
\end{align*}
\begin{align*}
(g \circ f)(x) & = g(f(x))\\
& = -5(4x+3)^2+1\\
& = -5[(4x+3)(4x+3)]+1\\
& = -5(16x^2+24x+9)+1\\
& = -80x^2-19x-45+1\\
& = -80x^2-19x-44
\end{align*}
Is my working out and answer correct?
| You have the right idea, except your algebra is off. For the first one
$$4(-5x^2+1)+3=-20x^2+4+3$$
$$=-20x^2+7$$
For the second one:
$$-5(16x^2+24x+9)+1=-80x^2-120x-45+1$$
$$=-80x^2-120x-44$$
| {
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Hottest and coldest points on a heated circular plate (use Lagrange multipliers) A circular plate given by the relationship $x^2 + y^2 \leq 1$ is heated according to the spatial temperature function $T(x,y) = 2x^2 + y^2-y$. Find the hottest and coldest point on the plate using Lagrange multipliers.
I tried to solve it but I got only one point (0,1) instead of 4 points .. I want to know how can I get 4 points ?? Is there any method to solve this problem?
| I can provide a solution without using Lagrange multipliers.
$2x^2+y^2-y=2x^2+(y-\frac{1}{2})^2-\frac{1}{4}$
The coldest point is $\left(0,\frac{1}{2}\right)$, where the temperature is $-\frac{1}{4} $
The hottest point must be somewhere along the circle $x^2+y^2=1$, since the further away the a point is from $\left(0,\frac{1}{2}\right)$, the hotter it is.
Let's use polar coordinates: $x=\cos\theta$, $y=\sin\theta$
We are then trying to maximize $2\cos^2\theta+\sin^2\theta-\sin\theta$
The derivative of this expression is $-\sin(2\theta)-\cos\theta$
The derivative equals zero at the following values of $\theta$: $\frac{\pi}{2}$, $\frac{3\pi}{2}$, $\frac{7\pi}{6}$, $\frac{11\pi}{6}$
Plugging these values of $\theta$ into the original equation, we can see that the hottest points are $\left(\pm\frac{\sqrt{3}}{2},-\frac{1}{2}\right)$ with temperature $\frac{9}{4}$.
| {
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Fake proofs using matrices Having gone through the 16-page-list of questions tagged fake-proofs, and going though both the relevant MSE Question and Wikipedia page, I didn't find a single fake proof that involved matrices.
So the question (or challange) here is: what are some fake proof using matrices?
In particular, the fake proof should use a property, an operation, ..., specific to matrices (or at least not present in $\mathbb{R}$ or $\mathbb{C}$), e.g.
*
*Noncommutativity
*(Non-)existence of an inverse
*Matrix sizes
*Operations as $\det$, $\text{trace}$, ...
*Eigenvalues and diagonalization
*Matrix decompositions and normal forms
*...
Note: It does not matter if the result being "proven" is correct or not. The fallacy in the proof itself is what matters.
Examples:
*
*Proof that 1 = 0
Proof: it is a well-known fact that $(x+y)^2 = x^2 + 2xy + y^2$.
Now let
$$x = \begin{pmatrix}0 & 1\\ 0 & 0 \end{pmatrix},\;\;y = \begin{pmatrix}1 & 0\\ 0 & 0 \end{pmatrix}.$$
On the one hand, we have that
$$ (x+y)^2 = \begin{pmatrix}1 & 1\\ 0 & 0 \end{pmatrix}^2 = \begin{pmatrix}1 & 1\\ 0 & 0 \end{pmatrix},$$
on the other hand we have
$$x^2 + 2xy + y^2 = \begin{pmatrix}0 & 0\\ 0 & 0 \end{pmatrix} + 2\begin{pmatrix}0 & 0\\ 0 & 0 \end{pmatrix} + \begin{pmatrix}1 & 0\\ 0 & 0 \end{pmatrix} = \begin{pmatrix}1 & 0\\ 0 & 0 \end{pmatrix}.$$
Since two matrices are equal if and only if all their entries are equal, we conclude that $1 = 0$.
The mistake here is that $x$ and $y$ do not commute. Thus $(x+y)^2 = x^2 + xy + yx + y^2 \neq x^2 + 2xy + y^2$.
*Proof that 2 = 0
Proof: We know that $\det (AB) = \det (BA)$, since $$\det (AB) = (\det A) (\det B) = (\det B) (\det A) = \det (BA).$$
Now consider the matrices
$$ A = \begin{pmatrix}1 & 0 & 1\\ 0 & 1 & 0 \end{pmatrix}, \; \; B = \begin{pmatrix}1 & 0\\ 0 & 1\\ 1 & 0 \end{pmatrix}.$$
We have that
$$AB = \begin{pmatrix}2 & 0\\ 0 & 1 \end{pmatrix}, \;\; BA = \begin{pmatrix}1 & 0 & 1\\ 0 & 1 & 0 \\1 & 0 & 1\end{pmatrix}.$$
Hence $\det (AB) = 2$ and $\det (BA) = 0$, therefore $2 = 0$.
The mistake here is that $\det$ is defined for square matrices only, and thus $\det AB = \det BA$ only holds in general if $A$ and $B$ are square.
| To prove: $0=1$.
Certain identities get funky when we pass over to infinite-order matrices. We see such matrices, for example, in representations of operators in quantum mechanics.
Everyone knows that
$Tr(AB-BA)=Tr(AB)-Tr(BA)=0$.
So let
$A_{i,j}=\delta_{i,j-1}, B_{i,j}=A_{j,i}$
Here $\delta$ is the Kronecker delta function, and $i$ and $j$ run through all natural mumbers.
Then $AB-BA$ has a 1 in the upper left position and zeroes everywhere else. The sum of diagonal elments is 1 and yet "$Tr(AB-BA)=0$".
$Tr(AB-BA)=0$ is guaranteed only for finite matrices or, among infinite matrices, when the traces of $AB$ and $BA$ converge absolutely.
| {
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Chain rule to differentiate $\sin ^2\frac{x}{2}$ I have this equation $$\sin ^2(\frac{x}{2})$$
Using the chain rule $
M'(N(x)).N'(x)$:
$$\begin{align*}
&M= (\sin \frac{x}{2})^2 \\
&N= \frac{x}{2}\end{align*}$$
That makes
$$2\sin \frac{x}{2}*\frac{1}{2}$$ or $$\sin \frac{x}{2}$$
Going again, we should have
$$\frac{1}{2} \cos \frac{x}{2}$$ or $$\frac{\cos \frac{x}{2}}{2}$$
Yet solution is giving me
$$(\sin \frac{x}{2})(\cos \frac{x}{2})$$
Why...?
| Write down
$$\begin{cases}h(x)=x^2\\g(x)=\sin x\\k(x)=\frac x2\end{cases}\;\;\implies \sin^2\frac x2=h\circ g\circ k(x)=h(g(k(x)))\implies$$
$$\left(\sin^2\frac x2\right)'=h'(g(k(x))\cdot g'(k(x))\cdot k'(x)=2\sin\frac x2\cdot\cos\frac x2\cdot\frac12=\frac12\sin x$$
using the trigonometric identity
$$\sin2\alpha=2\sin\alpha\cos\alpha$$
| {
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Is there a simple, intuitive way to see that $f(x)=x-\sqrt{x^2-1}<1$ if $x>1$ Is there a simple intuitive way to show that $f(x)=x-\sqrt{x^2-1}<1$ if $x>1$?
I sense it could be done more simple than this:
1 - take the derivative
$f'(x)=1-\frac{x}{\sqrt{x^2-1}}<0$ if $x>1$ so the slope of $f(x)$ in $(1,\infty)$ is negative.
2 - conclude
Since $f(1)=1$ and the derivative is negative we have showed that $f(x)=x-\sqrt{x^2-1}<1$ if $x>1$
Question: is there a simple, intuitive way to see that $f(x)=x-\sqrt{x^2-1}<1$ if $x>1$?
| For $x\gt 1$ one has $2x\gt 2$ thus $2x-2\gt 0$. This means
$x^2-1\gt x^2-1-(2x-2)$$
But the r.h.s of the previous inequality is $(x-1)^2$ so we have $(x^2-1)\gt (x-1)^2$.
Now take the square root and bear in mind that $\sqrt{(x-1)^2}=x-1$ because $x\gt 1$ we get
$\sqrt{x^2-1}\gt x-1$ and so
$$-\sqrt{x^2-1}\lt -(x-1)$$
Now add $x$ to get the inequality expected
$$x-\sqrt{x^2-1}\lt x-x+1=1$$
| {
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Integrate $\int e^x\frac{1+\sin x}{1+\cos x}dx$ Integrate
$$\int e^x\cdot\frac{1+\sin x}{1+\cos x}\,dx$$
My try;
First step:
I let
$$\frac{1+\sin x}{1+\cos x} = u$$
$$e^x = v$$
and then I applied integration by parts:
$$\frac{1+\sin x}{1+\cos x}=u \implies du=\frac{\sin x+\cos x+1}{(1+\cos x)^2}dx$$
$$v = e^x \implies dv = e^x$$
and
$$\int e^x\cdot\frac{1+\sin x}{1+\cos x}\,dx=e^x\cdot\frac{1+\sin x}{1+\cos x}-\int e^x\cdot\frac{\sin x+\cos x+1}{(1+\cos x)^2}dx$$
| Use $\sin x=2\sin\frac{x}{2}\cos\frac{x}{2},1+\cos x=2\cos^2\frac{x}{2}$. The integrand becomes $e^x\tan\frac{x}{2}+\frac{1}{2}e^x\sec^2\frac{x}{2}$.
Since the derivative of $\tan\frac{x}{2}$ is $\frac{1}{2}\sec^2\frac{x}{2}$ we can integrate immediately to get $e^x\tan\frac{x}{2}+C$.
| {
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Prove $\int_{0}^{\infty}\frac{2x}{x^8+2x^4+1}dx=\frac{\pi}{4}$ $$\int_{0}^{\infty}\frac{2x}{x^8+2x^4+1}dx=\frac{\pi}{4}$$
$u=x^4$ $\rightarrow$ $du=4x^3dx$
$x \rightarrow \infty$, $u\rightarrow \infty$
$x\rightarrow 0$, $u\rightarrow 0$
$$=\int_{0}^{\infty}\frac{2x}{u^2+2u+1}\cdot\frac{du}{4x^3}$$
$$=\frac{1}{2}\int_{0}^{\infty}\frac{1}{u^2+2u+1}\cdot\frac{du}{x^2}$$
$$=\frac{1}{2}\int_{0}^{\infty}\frac{1}{u^2+2u+1}\cdot\frac{du}
{\sqrt{u}}$$
Convert to partial fractions
$$\int_{0}^{\infty}\frac{1}{\sqrt{u}}-\frac{2}{u+1}+\frac{1}{(u+1)^2}du$$
$$=\left.2\sqrt{u}\right|_{0}^{\infty}-\left.2\ln(1+x)\right|_{0}^{\infty} -\left.\frac{1}{1+u}\right|_{0}^{\infty}$$
Where did I went wrong during my calculation?
| At first - substitution:
$$I=\int\limits_0^\infty{2x\,dx\over x^8+2x^4+1} = \int\limits_0^\infty{d(x^2)\over x^8+2x^4+1} = \int\limits_0^\infty{dy\over (y^2+1)^2}.$$
And then - by parts:
$$I = \int\limits_0^\infty{1+y^2-y^2\over (y^2+1)^2}\,dy = \int\limits_0^\infty{dy\over y^2+1} + {1\over2}\int\limits_0^\infty y\,d{1\over y^2+1}\,dy$$
$$ = \int\limits_0^\infty {dy\over y^2+1} + {1\over2}{y\over y^2+1}\biggr|_0^\infty - {1\over2}\int\limits_0^\infty {dy\over y^2+1} = {1\over2}\arctan y\biggr|_0^\infty = {\pi\over4}.$$
| {
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Need help with solving the expression $(x + y)^2 = (x + 1)(y - 1)$ for $x$ and $y$ Solve for $x$ and $y$ in
$(x+y)^2=(x+1)(y-1)$. I've figured out that
$$x^2+2xy+y^2=xy-x+y-1$$ but I'm not sure how to proceed. I'm somewhat familiar with solving quadratic equations by completing the square, could that help somehow? I'd appreciate any tips.
Thanks -Kurns
| The only solution is $x=-1,y=1$.
If $x+1,y-1$ are both positive, then AM-GM implies their product is at most the square of their AM which is $\frac{1}{4}(x+y)^2<(x+y)^2$. If they are both negative, then $-x-1,-y+1$ are both positive and so $(x+1)(y-1)=(-x-1)(-y+1)<(-x-y)^2=(x+y)^2$. If one of $x+1,y-1$ is positive and the other negative, then their product is negative. In none of these cases can we have equality.
If one of $x+1,y-1$ is zero, then $x+y$ must be zero and hence both $x+1,y-1$ must be zero.
| {
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How to solve this recurrence relation $ f(n) = \frac{10+7f(n-1)}{2+f(n-1)}$ I had a problem in which I ended up getting the following recurrence relation:
$$\begin{align}
&f(1) = 7\\
&f(n) = \dfrac{10+7f(n-1)}{2+f(n-1)}
\end{align}$$
I haven't solved much recurrence relations, and I am unaware of any general procedure. The only thing I was able to do till now was to get this relation. I have tried putting $n$ as $n+1$, but it soon gets unsolvable.
In a more crude form, $$f(n) = 5+ \dfrac{1}{\frac{1}{f(n-1)}+\frac{1}{2}}$$
| Let we state it in a slightly different way. Let $(a_1,b_1)=(7,1)$ and:
$$\begin{pmatrix} a_{n+1} \\ b_{n+1}\end{pmatrix} = \begin{pmatrix}7 & 10 \\ 1 & 2\end{pmatrix}\begin{pmatrix} a_{n} \\ b_{n}\end{pmatrix} = M \begin{pmatrix} a_{n} \\ b_{n}\end{pmatrix}.$$
The eigenvalues of $M$ are $\frac{9\pm\sqrt{65}}{2}$. By the Cayley-Hamilton theorem, the characteristic polynomial of the sequence
$$ \left\{\begin{pmatrix} a_{n+1} \\ b_{n+1}\end{pmatrix}\right\}_{n\geq 1} $$
is the same as the characteristic polynomial of $M$. Component-wise, we have:
$$ a_n = A_+ \left(\frac{9+\sqrt{65}}{2}\right)^n + A_- \left(\frac{9-\sqrt{65}}{2}\right)^n, $$
$$ b_n = B_+ \left(\frac{9+\sqrt{65}}{2}\right)^n + B_- \left(\frac{9-\sqrt{65}}{2}\right)^n, $$
where the constants $A_{\pm},B_{\pm}$ can be found by computing the first values of $(a_i,b_i)$, then solving a linear system. In our case $(a_1,b_1)=(7,1)$ and $(a_2,b_2)=(59,9)$ give:
$$ \frac{a_n}{b_n}=\frac{(5+\sqrt{65})(9+\sqrt{65})^n-(5-\sqrt{65})(9-\sqrt{65})^n}{2((9+\sqrt{65})^n-(9-\sqrt{65})^n)}$$
but the LHS is exactly $f(n)$. That also gives us:
$$ \lim_{n\to +\infty} f(n) = \color{red}{\frac{5+\sqrt{65}}{2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1809236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Let $M=\big( \begin{smallmatrix} A & B \\ C & D \end{smallmatrix} \big)$. Prove $\det(M)=\det(A)\cdot \det(D-C·A^{-1}·B)$. Let (shown in matrix blocks) $M=\big( \begin{smallmatrix} A & B \\ C & D \end{smallmatrix} \big)$ a square matrix such that $A$ is invertible and $D$ is a square matrix. I have to prove that
$$\det(M)=\det(A)\cdot \det(D-C·A^{-1}·B)$$
I also have an indication to use it:
Consider previously these cases:
\begin{align*}
A_1 &= \begin{pmatrix} A & 0 \\ 0 & I \end{pmatrix},\\
A_2 &= \begin{pmatrix} I & 0 \\ 0 & D \end{pmatrix},\\
A_3 &= A_1·A_2,\\
A_4 &= \begin{pmatrix} I & A^{-1}·B \\ 0 & I \end{pmatrix},\text{ and}\\
A_5 &= A_3\cdot A_4.
\end{align*}
The problem is difficult in general for me and I don't even know how to use the indication.
| The indication shows that
$\det\begin{pmatrix}A&B\\0&D\end{pmatrix}=\det(A)\det(D)$.
Similarly $\det\begin{pmatrix}A&0\\C&D\end{pmatrix}=\det(A)\det(D)$.
Now $\begin{pmatrix}I&0\\-CA^{-1}&I\end{pmatrix}\begin{pmatrix}A&B\\C&D\end{pmatrix}=\begin{pmatrix}A&B\\0&D-CA^{-1}B\end{pmatrix}$
It follows from the indication that $\det(I)\det(I)\det\begin{pmatrix}A&B\\C&D\end{pmatrix}=\det(A)\det(D-CA^{-1}B)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1811433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Calculating $\lim \limits_{x \to 0} \frac{\cos x\sin x-x}{\sin^3 x}$ using taylor series I am given the following limit, and need to calculate it using taylor series:
$$\lim \limits_{x \to 0} \frac{\cos{x}\sin{x}-x}{\sin^3{x}}$$
$$\sin{x}=x-\frac{x^3}{3!}+\frac{x^5}{5!}...$$
$$\cos{x}=1-\frac{x^2}{2!}+\frac{x^4}{4!}...$$
The problem I'm facing is how many terms to take when plugging in each taylor series into the original limit. I read in another thread that the lowest exponent in the denominator that I'll get is $x^3$, which means the largest degree I should have in the numerator is also $x^3$. Doing this I get:
$$\frac{\left(1-\frac{x^2}{2}\right)\left(x\right)-x}{x^3+...}= \frac{\frac{x^3}{2}}{x^3+...}$$
Of course I get the limit is $-\frac12$ which is incorrect. What did I do wrong?
Please, I'd prefer hints and a correction rather than a full blown answer.
| You did not expand each element up to the same order in their Taylor series :
$$\cos (x)=1-\frac{x^2}{2!}+o(x^3)$$
$$\sin(x)=x-\frac{x^3}{3!}+o(x^3)$$
$$(\sin(x))^3=x^3+o(x^3)$$
Now we have : $$\lim \limits_{x \to 0} \frac{(\cos x)(\sin x)-x}{(\sin x)^3}=\lim \limits_{x \to 0} \frac{x-\frac{x^3}{3!}-\frac{x^3}{2!}-x+o(x^3)}{x^3+o(x^3)}=\lim \limits_{x \to 0} \frac{-\frac{2x^3}{3}+o(x^3)}{x^3+o(x^3)}=\lim \limits_{x \to 0} \frac{-2x^3}{3x^3}=-\frac{2}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find $f'(x)$ at given value of x Find $f'(x)$ at the given value of x
$f(x)=\sqrt{x+2}$
Find $f'(7)$
My question for this one is do I approach this question by trying to find the derivative of the initial equation and then once I have found the derivative do I simply plug in 7 for x and solve?
So once I start solving I would get $\frac{\sqrt{x+h+2}-\sqrt{x+2}}{h}$
$\lim_{h\to0} \frac{\sqrt{x+h+2}-\sqrt{x+2}}{h} = \lim_{h\to0} \frac{\sqrt{x+h+2}-\sqrt{x+2}}{h} \cdot \frac{\sqrt{x+h+2}+\sqrt{x+2}}{\sqrt{x+h+2}+\sqrt{x+2}}$
From that point I would get
$\frac{(x+h+2-x-2)}{h\sqrt{x+h+2}+\sqrt{x+2}}$
| $$f'(x)=\frac{1}{2\sqrt{x+2}}$$
$$f'(7)=\frac{1}{2\sqrt{7+2}}=\frac{1}{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
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