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'Strange' trigonometric roots of $x^5-4x^4+2x^3+5x^2-2x-1$ - could someone explain? This quintic equation has $5$ real roots:
$$x^5-4x^4+2x^3+5x^2-2x-1=0 \tag{1}$$
The roots are, from left to right:
$$x_1=\frac{\cos \frac{19}{22} \pi}{\cos \frac{1}{22} \pi}$$
$$x_2=\frac{\cos \frac{9}{22} \pi}{\cos \frac{19}{22} \pi}$$
$$x_3=\frac{\cos \frac{7}{22} \pi}{\cos \frac{5}{22} \pi}$$
$$x_4=\frac{\cos \frac{1}{22} \pi}{\cos \frac{7}{22} \pi}$$
$$x_5=\frac{\cos \frac{5}{22} \pi}{\cos \frac{9}{22} \pi}$$
I found these roots numerically, using ISC. The equation was found on Wikipedia in a different form (for $x-4/5$), no solutions were provided.
I can derive this equation for each root individually. But I don't see how do all five roots 'fit' together.
Why ${1,5,7,9,19}$? Why there is no $3/22, 13/22$ or $17/22$ in any of the arguments?
In any case, I would be grateful for the explanation for how these roots all fit together.
| We just have to find the minimal polynomial of
$$ \alpha = \frac{\cos\frac{19\pi}{22}}{\cos\frac{\pi}{22}}=-\frac{\cos\frac{3\pi}{22}}{\cos\frac{\pi}{22}}=3-4\cos^2\frac{\pi}{22}=1-2\cos\frac{\pi}{11} \tag{1}$$
then find the conjugate roots. But it is well-known that the minimal polynomial of $\cos\frac{2\pi}{m}$ has degree $\frac{\varphi(m)}{2}$ (i.e. $5$ in our case) and the algebraic conjugates of $\cos\frac{2\pi}{m}$ are $\cos\frac{2\pi k}{m}$ with $\gcd(k,m)=1$, so the claim is straightforward.
| {
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Prove: $\frac{r_a}{bc} + \frac{r_b}{ca} + \frac{r_c}{ab} = \frac{1}{r} - \frac{1}{2R}$, for circumradius R, inradius $r$, and exradii $r_x$
In $\triangle ABC$, prove:
$$\frac{r_a}{bc} + \frac{r_b}{ca} + \frac{r_c}{ab} = \frac{1}{r} - \frac{1}{2R}$$
for circumradius $R$, inradius $r$, and exradii $r_a$, $r_b$, $r_c$ in the standard arrangement.
It is known that $r_a = \sqrt{\dfrac{s\left(s-b\right)\left(s-c\right)}{s-a}}$, where $s = \dfrac12\left(a+b+c\right)$ is the semiperimeter of $\triangle ABC$. Similar formulas exist for $r_b$, $r_c$ and $r$. But how does $R$ connect with all of this?
| We will use the $\sum_{cyc}$ notation for cyclic sums:
$$
\sum_{cyc}f(a,b,c) = f(a,b,c)+f(b,c,a)+f(c,a,b)
$$
for any function $f$. Also, let $\Delta$ be the area of triangle $ABC$.
We have:
$$ \frac{r_a}{bc}=\frac{\Delta}{(s-a)bc}=\frac{\frac{1}{2}\sin(A)}{s-a}=\frac{a}{4R(s-a)}\tag{1}$$
hence we have to prove that:
$$ 2+\sum_{cyc} \frac{a}{s-a} = \frac{4R}{r} \tag{2}$$
that is equivalent to:
$$ 2r+\sum_{cyc}\frac{a}{\cot\left(\frac{A}{2}\right)} = 2R\tag{3}$$
or to:
$$ 2r+\sum_{cyc}2R(1-\cos(A)) = 2R\tag{4}$$
that follows from Carnot's theorem.
| {
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When is $\sqrt{x/y^2}$ equal to $\sqrt{x}/y$? The solution to the quadratics is given by
$r = -\dfrac{b}{2a}\pm\sqrt{\dfrac{b^2-4ac}{4a^2}}$, which is shortened to
$r = -\dfrac{b}{2a}\pm\dfrac{\sqrt{b^2-4ac}}{2a}$, but I'm wondering how if this is justified, given that $4a^2$ can be negative if $a \in \mathbb{C}$, and $\sqrt{\dfrac{x}{y}} \neq \dfrac{\sqrt{x}}{\sqrt{y}}$ if $x$ and $y$ are negative, but given that we have $a^2$, is this justified?
Is $\sqrt{x/y^2} = \sqrt{x}/y$? Is $r = -\dfrac{b}{2a}\pm\dfrac{\sqrt{b^2-4ac}}{2a}$ always true?
| That's a good question.
But for the sake of the quadratic equation we don't need $\sqrt{x/y^2} = \sqrt{x}/y$. We just need $\sqrt{x/y^2} = \pm \sqrt{x}/y$.
If $y^2 = c \ne 0$ then $y = \{x_1,x_2\}$ where $x_1 = -x_2$. This is true for all $c$, real or complex. Arbitrarily declare one of them, say $x_1$, to be the $\sqrt{c}$ then we can always have $\sqrt{y^2} = \pm y$. Always.
(If $y^2 = 0 \implies y = 0$ and $0 = \pm 0$.)
If $y^2 = c = r*e^{i\theta} \in \mathbb C$ then $y = \{\sqrt{r}e^{i(\theta/2},\sqrt{r}e^{i\theta/2 + \pi}\}$ = $\pm \sqrt{r} e^{i\theta/2}$
So $\sqrt{x/y^2} = \pm \sqrt{x}/y$. So equation is good even for complex numbers.
| {
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Positive integers $a$ and $b$ are such that $a+b=a/b + b/a$. Find $a^2+ b^2$. This is PRE RMO 2015 question. Positive integers $a$ and $b$ are such that $$a+b=a/b + b/a$$ Find $$a^2+ b^2$$
My try:-
Given that
$$a+b=a/b + b/a$$On simplification we get
$$a^2 b+ b^2 a= a^2 + b^2$$
But in my book the given answer is 2. Please tell me how it is possible.
| You know that $a^2b+b^2a=a^2+b^2$. However, you haven't used that $a,b$ are positive integers. Rewrite the expression as $a^2(b-1)+b^2(a-1)=0$. Then, since $a,b > 0$, we have $a^2 >0$ and $b^2>0$. Also, $b-1\geq 0$ and $a-1\geq 0$. So $a^2(b-1)+b^2(a-1)$ is positive or zero, where it is zero if $b-1= 0$ and $a-1= 0$.
Hence $a=b=1$ and hence $a^2+b^2=1^2+1^2=2$.
| {
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Why can't we use the law of cosines to prove Fermat's Last Theorem? In investigating approaches to Fermat's Last Theorem I came across the following and I can't figure out where I am going wrong. Any input would be greatly appreciated.
We want to show that $a^n + b^n = c^n$ cannot hold for odd $n>1$ and pairwise relatively prime $a$, $b$, and $c$. Assuming by way of contradiction that we have $a^n + b^n = c^n$ we must have $a$, $b$, and $c$ forming the sides of a triangle since $(a+b)^n > c^n$ so $a+b>c$. Therefore the law of cosines can apply and we can write:
$$c^2 = a^2+b^2 - 2ab{\cos{C}}$$
where $C$ is the angle opposite to side $c$. If we add and subtract $2ab$ on the right-hand side we get
$$c^2 = {(a+b)}^2 -2ab(\cos{C}+1)$$
Now, $a+b$ and $c$ share a common factor since $(a+b) | (a^n+b^n)$ for odd $n$ and $c^n = a^n+b^n$. (Here $x | y$ means as usual, "$x$ divides $y$").Therefore, they share the same factor with $2ab(\cos{C}+1)$.
Now, $\cos{C} + 1$ must be a rational number since $a$, $b$, and $c$ are all integers. So let $\cos{C} +1 = \frac{r}{s}$ where $r$ and $s$ are integers and $(r,s)=1$. (i.e. $\frac{r}{s}$ is a reduced fraction). (Here, $(r,s)$ means as usual the greatest common divisor of $r$ and $s$.)
Now assuming $a$, $b$, and $c$ are relatively prime we must have $(ab) |s$ for otherwise $c$ and $2ab$ would share a common factor. Even moreso we must have $ab=s$ since otherwise $\frac{2abr}{s}$ would not be an integer. (Since $c - a - b$ is even, we don't need $2 | s$). So we can write:
$$\cos{C}+1 = \frac{r}{ab}$$ or equivalently $$\cos{C} = \frac{r - ab}{ab}$$
Now we had from the law of cosines:
$$c^2 = a^2+b^2 - 2ab{\cos{C}}$$
so making the substitution $\cos{C} = \frac{r - ab}{ab}$ we get
$$c^2 = a^2 + b^2 - 2r + 2ab$$
If we subtract $a^2$ to both sides and factor out the $b$ on the right-hand side, we get:
$$c^2 - a^2 = b(b + 2a) - 2r$$
Now, $(c - a) | (c^2 - a^2)$ and also $(c-a) | (c^n - a^n)$. Then we must have $((c-a),b) >1$ since $b^n = c^n - a^n$. From the equation above, we must therefore also have $(b,2r) > 1$. Similarly we can show that we must have $(a,2r) > 1$.
However, both of these conclusions are problematic since $r$ was initially assumed to be relatively prime to $s = ab$. The only other option is that $a$ and $b$ are both even, but this is also problematic since $a$ and $b$ are assumed to be relatively prime.
Thus we cannot have $a^n + b^n = c^n$ for odd $n>1$ and pairwise relatively prime $a$, $b$, and $c$.
I'm sure someone has thought of this approach before so where am I going wrong?
| As Maik points out in the accepted answer we don't necessary need $ab=s$ in order to ensure that $a$, $b$, and $c$ are pairwise relatively prime. We do however require that $(c,\dfrac{2abr}{s})>1$ since $(c, (a+b) )>1$.
Now we need $s|ab$ because otherwise we would not get an integer value for $\dfrac{2abr}{s}$. Also we cannot have $(c,2ab)>1$ because this would imply that $a$, $b$, and $c$ share a common factor. (As noted in some of the comments to my original question we could have $c$ being even if $a$ and $b$ are both odd, but then $4|c$, $4|{(a+b)}^2$ so we would need $4|2ab$ implying that either $a$ or $b$ is even, a contradiction.)
What I forgot was that we can still have $(c,r)>1$ and thus avoid any contradictions with $a$, $b$ and $c$ being pairwise relatively prime.
| {
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Using determinants to find a recursive sequence I am trying to compute a three diagonal determinant in order to find the recursive relation.
Let $\Delta_{n}$=$\begin{vmatrix}
11 & 3 & 0 & 0 & \dots & 0\\
13 & 11 & 3 & 0 & \dots & 0\\
0 & 13 & 11 & 3 & \dots & 0\\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots\\
0 & 0 & 0 & 0 & \dots 11 & 3\\
0 & 0 & 0 & 0 & \dots 13 & 11\\
\end{vmatrix}$
Once I have expanded $\Delta_{n+2}$ along the first row I obtain:
$\Delta_{n+2}$=$11\Delta_{n+1}$ - 3$\begin{vmatrix}
13 & 3 & 0 & 0 & \dots & 0\\
0 & 11 & 3 & 0 & \dots & 0\\
0 & 0 & 11 & 3 & \dots & 0\\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots\\
0 & 0 & 0 & 0 & \dots 11 & 3\\
0 & 0 & 0 & 0 & \dots 13 & 11\\
\end{vmatrix}$
Expanding the above matrix along the first row I obtain:
$\Delta_{n+2}$=$11\Delta_{n+1}$ - 3(13)$\Delta{n} + 0 + 0 + \dots$
I hope that I am on the right track with my matrix expansions. Any comments are appreciated.
| Set
$$A_n(a,b,c)=\left( \begin{matrix}
a & b & 0 & \cdots & \cdots & 0 \\
c & a & b & \cdots & \cdots & 0 \\
0 & c & a & b & \cdots & 0 \\
\vdots & \ddots & \ddots & \ddots & \ddots & \vdots \\
0 & \cdots & \cdots & c & a & b \\
0 & 0 & \cdots & 0 & c & a \\
\end{matrix} \right)_{n\times n}
$$
then
$$|A_{n+2}(a,b,c)|=a|A_{n+1}(a,b,c)|-bc\,|A_{n}(a,b,c)|$$
Indeed we can show
$$|A_n(a,b,c)|=\frac{1}{\sqrt{a^2-4bc}}\left[\left(\frac{a+\sqrt{a^2-4bc}}{2}\right)^{n+1}-\left(\frac{a-\sqrt{a^2-4bc}}{2}\right)^{n+1}\right]$$
Hint
set $x_{n}=|A_n(a,b,c)|$ then
$$x_{n+2}=a\,{x_{n+1}}-bc\,{x_{n}}$$
where
$x_1=a$ and $x_0=1$
we have
$$\lambda_1=\frac{a+\sqrt{a^2-4bc}}{2}$$
and
$$\lambda_2=\frac{a-\sqrt{a^2-4bc}}{2}$$
$...$
| {
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Exact value of $\cos^2(\frac{\pi}{8})+\sin^2(\frac{15\pi}{8})?$ I tried separating it into $\cos^2(\frac{\pi}{8})+\sin^2(\frac{\pi}{8}+\frac{14\pi}{8})?$ and using the angle sum identity but it didn't help.
| Notice that for all $x$ we have $\sin\left(2\pi-x\right)=\sin(-x)=-\sin(x)$ due to $2\pi$-periodicity of $\sin$ and the fact it is an odd function.
Applying, have
$$\sin\left(\frac{15\pi}{8}\right)^2=\left(\sin\left(2\pi-\frac{\pi}8\right)\right)^2=\left(\sin\left(-\frac{\pi}8\right)\right)^2=\sin\left(\frac{\pi}8\right)^2;$$
returning back to the original expression we get
$$\cos\left(\frac{\pi}8\right)^2+\sin\left(\frac{15\pi}8\right)^2=\cos\left(\frac{\pi}8\right)^2+\sin\left(\frac{\pi}8\right)^2,$$
which by the fundamental trigonometric identity $\sin(x)^2+\cos(x)^2=1$ easily gives
$$\therefore\cos\left(\frac{\pi}8\right)^2+\sin\left(\frac{15\pi}8\right)^2=\boxed{1}\,.$$
| {
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Find the covaraince of the number of hearts drawn and the number of clubs drawn from a deck of cards.
Two cards are drawn without replacement from a pack of cards. The random
variable $X$ measures the number of heart cards drawn, and the random
variable $Y$ measures the number of club cards drawn. Find the covariance
and correlation of $X$ and $Y$:
I did the following:
Y/X 0 1 2
0 (26/52)*(25/51) (13/52)*(26*51) (13/52)*(12/51)
1 (13/52)*(26*51) (13/52)*(13/51) 0
2 (13/52)*(12/51) 0 0
| Now sum up the rows and the columns to get the marginal distribution. For instance $P(Y=0)= (26/52)\cdot (25/51) +2\cdot (13/52)\cdot (26\cdot 51) +(13/52)\cdot (12/51)$
With these probabilities you can calculate the expected value of $X$ and $Y$.
$E(Y)=\mu_y=0\cdot P(Y=0)+1\cdot P(Y=1)+2\cdot P(Y=2)$. Similar calculation for $\mu_x$
The covariance is
$$Cov(X,Y)=\left[\sum_{x=0}^2 \sum_{y=0}^2 x\cdot y\cdot P(X=x,Y=y)\right]-\mu_x\cdot \mu_y$$
The first four summands of the brackets are
$0\cdot 0\cdot P(X=0, Y=0)+0\cdot 1\cdot P(X=0, Y=0)+0\cdot 2\cdot P(X=0, Y=2)+1\cdot 0\cdot P(X=1, Y=0)+\ldots$
$=0\cdot 0\cdot (26/52)\cdot (25/51)+0\cdot 1\cdot 2 \cdot (13/52)\cdot (26/ 51)+0\cdot 2 \cdot (13/52)\cdot (12/51) +1\cdot 0\cdot 2\cdot (13/52)\cdot (26/ 51)+\ldots $
The correlation coefficient is
$Corr(X,Y)=\frac{Cov(X,Y)}{\sqrt{Var(X)\cdot Var(Y)}}$
with $Var(X)=\sum_{x=0}^2 P(X=x)\cdot (x-\mu_x)^2$
| {
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Coefficient of $x^{41}$ in $(x^5 + x^6 + x^7 + x^8 + x^9)^5$
What is the coefficient of coefficient of $x^{41}$ in $(x^5 + x^6 + x^7 + x^8 + x^9)^5$?
Using summation of G.P., this is equivalent to finding the coefficient of $x^{41}$ in
$$\left(x^5 \left(\frac{1-x^5}{1-x}\right)\right)^5$$
and thus finding coefficient of $x^{16}$ in $(\frac{1-x^5}{1-x})^5$. How to proceed after this?
| In other words, how many ways are there to write
$$ 41 = a + b+c+d+e $$
where $a,b,c,d,e$ are integers between $5$ and $9$, inclusive.
The highest sum we can make is $9\cdot 5=45$, so to get $41$ instead we need to remove $4$ units from among the 5 variables. Fortunately, even removing all $4$ of them from the same variable still leaves $5$, which is a valid value. So what we're counting is the same as
How many ways are there to write $4$ as an (ordered) sum of $5$ non-negative integers?
This is a standard combinatorial problem; by the stars-and-bars formula, the answer is $$\binom{4+5-1}{4} = 70$$
| {
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Prove that: $\sum\limits_{n=1}^\infty \frac{n^2}{2^n}=6$. Prove that: $$\sum\limits_{n=1}^\infty \dfrac{n^2}{2^n}=6.$$ I am trying to find the sum of this infinite series. Got stuck. Any help will be highly appreciated.
| Just for variety, reindexing sums and using the geometric series formula:
$$\begin{align}
\sum_{n=1}^{\infty}\frac{n^2}{2^n}
&=\sum_{n=1}^{\infty}\frac{\sum_{k=1}^{n}(2k-1)}{2^n}\\
&=\sum_{k=1}^{\infty}(2k-1)\sum_{n=k}^{\infty}\frac{1}{2^n}\\
&=\sum_{k=1}^{\infty}(2k-1)\frac{(1/2)^k}{1-1/2}\\
&=2\sum_{k=1}^{\infty}\frac{(2k-1)}{2^k}\\
&=2\sum_{k=1}^{\infty}\frac{k}{2^{k-1}}-2\sum_{k=1}^{\infty}\frac{1}{2^k}\\
&=2\sum_{k=1}^{\infty}\frac{\sum_{j=1}^k1}{2^{k-1}}-2\frac{1/2}{1-1/2}\\
&=2\sum_{j=1}^{\infty}\sum_{k=j}^{\infty}\frac{1}{2^{k-1}}-2\\
&=2\sum_{j=1}^{\infty}\frac{1/2^{j-1}}{1-1/2}-2\\
&=4\sum_{j=1}^{\infty}\frac{1}{2^{j-1}}-2\\
&=4\frac{1}{1-1/2}-2\\
&=6
\end{align}$$
| {
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Prove that $\cos (5A) = 16 \cos^5 (A) - 20 \cos^3 (A) + 5 \cos (A)$
Prove the given trigonometric identity
$$\cos (5A) = 16 \cos^5 (A) - 20 \cos^3 (A) + 5 \cos (A)$$
My attempt
L.H.S.$=\cos5A$
$$\cos(A+4A)$$
$$\cos A\cos4A-\sin A\sin4A$$
Now how should I move further?
| One approach is to use De Moivre's Theorem along with the Binomial Theorem.
Consider $(\cos A+i\sin A)^5$. By De Moivre's Theorem, we get $\cos(5A)+i\sin(5A)$. Notice that $\cos(5A)$ is the real portion of the complex expression.
On the other hand, we can use the Binomial Theorem to expand $(\cos A+i\sin A)^5$.
$$(\cos A+i\sin A)^5=(\cos A)^5+5(\cos A)^4(i\sin A)+10(\cos A)^3(i\sin A)^2+10(\cos A)^2(i\sin A)^3+5(\cos A)(i\sin A)^4+(i\sin A)^5$$
Selecting only the real terms from above and equating with $\cos(5A)$, we have
$$\cos(5A)=\cos^5A-10\cos^3A\sin^2A+5\cos A\sin^4A\ .$$
Using the Pythagorean Identity $\sin^2A=1-\cos^2A$, we get
\begin{align}
\cos(5A)&=\cos^5A-10\cos^3A(1-\cos^2A)+5\cos A(1-\cos^2A)^2\\
&=\cos^5A-10\cos^3A+10\cos^5A+5\cos A-10\cos^3A+5\cos^5A\\
&=16\cos^5A-20\cos^3A+5\cos A\ .
\end{align}
| {
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Simplifying $\sqrt[5]{1+g+g^3}=\frac {\sqrt{1+g^2}}{\sqrt[10]{5}}$ and similar ones I saw that Ramanujan simplified many radicals such as:
For $g^5=2$
$$\sqrt[5]{1+g+g^3}=\frac {\sqrt{1+g^2}}{\sqrt[10]{5}}\tag{1}$$
For $g^4=5$
$$\frac {\sqrt[5]{3+2g}-\sqrt[5]{4-4g}}{\sqrt[5]{3+2g}+\sqrt[5]{4-4g}}=2+g+g^2+g^3\tag{2}$$
For $g^5=2$
$$\frac {\sqrt{g+3}+\sqrt{5g-5}}{\sqrt{g+3}-\sqrt{5g-5}}=g^2+g\tag{3}$$
For $g^5=3$
$$\sqrt[3]{2-g^3}=\frac {1+g-g^2}{\sqrt[3]{5}}\tag{4}$$
For $g^5=3$
$$\frac {\sqrt{g^2+1}+\sqrt{5g-5}}{\sqrt{g^2+1}-\sqrt{5g-5}}=g^3+g^2+g+\frac {1}{g}\tag{5}$$
And the list goes on and on... My question is: How do we prove such equations and is there a general method that can be used?
After giving it some thought, I wondered if you can use componendo et dividendo to prove it. Where given the fraction $\frac {a}{b}=\frac {c}{d}$, we have $\frac {a+b}{a-b}=\frac {c+d}{c-d}$. But I don't think we can use that method to find other examples of $(1)$ or $(4)$.
| Each equation can be manipulated until it is an equality of polynomials in $g$. For example, putting (1) to the power of $10$, it is equivalent to
$$
5(1+g+g^3)^2=(1+g^2)^5.
$$
(note that the operation can be reversed because both sides of the original equation are positive).
A polynomial in $g$ can be simplified using $g^{k+5}=2g^k$, until only exponents less than $5$ remain. At this point the expressions should be identical (if they are not, and the coefficients are rational, then equality does not hold). In this case the LHS is
$$\begin{eqnarray*}
5(1+g+g^3)^2
&=&5(1+g^2+g^6+2g+2g^3+2g^4)\\
&=&5(1+4g+g^2+2g^3+2g^4)
\end{eqnarray*}$$
and the RHS is
$$\begin{eqnarray*}
(1+g^2)^5
&=&1+5g^2+10g^4+10g^6+5g^8+g^{10}\\
&=&1+5g^2+10g^4+20g+10g^3+4\\
&=&5(1+4g+g^2+2g^3+2g^4)
\end{eqnarray*}$$
as required.
| {
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Arc of curve of function - Find the minimum length
This is the arc of the curve:
$$y = x^3 + x^2 - \frac{29x}{2} + 1 \\ t > 0 \\ x \in [t,t+1]$$
Find $t$ for which the length of the arc of the curve is minimum.
Should I use $ \int {\sqrt{1 + [f'(x)]^2}} dx$ ?
Thank you very, but very much!
| doing André Nicolas say, we have:
$$f'(x)=3x^2+2x-\frac{29}{2}$$
and
$$g(t)=\int_t^{t+1}{\sqrt{1 + \left(3x^2 + 2x - \frac{29}{2}\right)^2}} dx$$ $$g(t)=\int_0^{t+1}{\sqrt{1 + \left(3x^2 + 2x - \frac{29}{2}\right)^2}} dx-\int_0^{t}{\sqrt{1 + \left(3x^2 + 2x - \frac{29}{2}\right)^2}} dx$$
now we calculate $g'(t)$:
$$g'(t)=\sqrt{1 + \left(3(t+1)^2 + 2(t+1) - \frac{29}{2}\right)^2}-\sqrt{1 + \left(3t^2 + 2t - \frac{29}{2}\right)^2}=0$$
$$1 + \left(3(t+1)^2 + 2(t+1) - \frac{29}{2}\right)^2-1-\left(3t^2 + 2t - \frac{29}{2}\right)^2=0$$
$$\left(3t^2+2t-\frac{29}{2}+6t+5\right)^2-\left(3t^2+2t-\frac{29}{2}\right)^2=0$$
$$2(6t+5)(3t^2+2t-\frac{29}{2})+(6t+5)^2=0$$
if $t$ distinc to $-\frac{5}{6}$, we factorizing $(6t+5)$:
$$6t^2+4t-29+6t+5=0$$
finally we have:
$$6t^2+10t-24=0$$
wich have the solutions: $t=-\frac{16}{12}$ and $t=3$
Finally the result is:
$$t=3$$
| {
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Prove that $\cos\frac {2\pi}{7}+ \cos\frac {4\pi}{7}+ \cos\frac {8\pi}{7}=-\frac{1}{2}$
Prove that
$$\cos\frac {2\pi}{7}+ \cos\frac {4\pi}{7}+ \cos\frac {8\pi}{7}=-\frac{1}{2}$$
My attempt
\begin{align}
\text{LHS}&=\cos\frac{2\pi}7+\cos\frac{4\pi}7+\cos\frac{8\pi}7\\
&=-2\cos\frac{4\pi}7\cos\frac\pi7+2\cos^2\frac{4\pi}7-1\\
&=-2\cos\frac{4\pi}7\left(\cos\frac\pi7-\cos\frac{4\pi}7\right)-1
\end{align}
Now, please help me to complete the proof.
| Consider the polynomial $x^6+x^5+\dots+x+1$. This has roots at $e^{i2n\pi/7} = cos(2n\pi/7)+ isin(2n\pi/7)$ for $n = 1,2,\dots,7$.
Since $cos(x) = cos(2\pi - x)$ the sum you want is just half of the real part of the sum of the roots of this polynomial. Now by Vieta's formula the sum of the roots is $-1$, hence the sum is $-1/2$.
| {
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Laurent expansion - Faster technique I'm currently preparing for an exam in complex analysis. There is a type of exercise, where I need to compute Laurent expansions about different places. However, my current technique does consume quite a lot of my available time, so my question is if anyone knows a faster technique.
I will show my technique on an example. Here's the task:
Compute all isolated singularities of the following function and specify their type. Also give the two terms with the smallest degree of the Laurent expansion about those singularities and their residuum.
$$f(z) = \frac{z}{(z - 1)^2(z - 2)}$$
And now to my technique. The function has singularities where $(z - 1)^2(z - 2) = 0$, this is at $z_1 = 1$ and $z_2 = 2$. Thus, the singularities are also isolated.
$$\lim_{z \to 1} |f(z)| = \frac{1}{\lim_{z \to 1} |(z - 1)^2(z - 2)|} = \infty\\
\lim_{z \to 2} |f(z)| = \frac{2}{\lim_{z \to 2} |(z - 1)^2(z - 2)|} = \infty$$
So, both singularities are poles.
For a Laurent expansion we would like the function to only consist of addends, so we do a partial fraction decomposition.
$$\frac{z}{(z - 1)^2(z - 2)} = \frac{A}{z - 1} + \frac{B}{(z - 1)^2} + \frac{C}{z - 2} \Leftrightarrow\\
z = A(z - 1)(z - 2) + B(z - 2) + C(z - 1)^2 \Leftrightarrow\\
z = Az^2 - 3Az + 2A + Bz - 2B + Cz^2 - 2Cz + C \Leftrightarrow\\
z = z^2(A + C) + z(-3A + B - 2C) + 1(2A - 2B + C)$$
Okay, now we solve the equation system.
$$\begin{pmatrix}
1 & 0 & 1 & 0\\
-3 & 1 &-2 & 1\\
2 &-2 & 1 & 0\\
\end{pmatrix}
\Leftrightarrow
\begin{pmatrix}
1 & 0 & 1 & 0\\
0 & 1 & 1 & 1\\
0 &-2 & 1 & 0\\
\end{pmatrix}
\Leftrightarrow
\begin{pmatrix}
1 & 0 & 1 & 0\\
0 & 1 & 1 & 1\\
0 & 0 & 1 & 2\\
\end{pmatrix}
\Leftrightarrow\\
\begin{pmatrix}
1 & 0 & 0 & -2\\
0 & 1 & 1 & 1\\
0 & 0 & 1 & 2\\
\end{pmatrix}
\Leftrightarrow
\begin{pmatrix}
1 & 0 & 0 & -2\\
0 & 1 & 0 & -1\\
0 & 0 & 1 & 2\\
\end{pmatrix}$$
With that we follow:
$$f(z) = -\frac{2}{z - 1} - \frac{1}{(z - 1)^2} + \frac{2}{z - 2}$$
Let's first compute the Laurent expansion about $z_1 = 1$, we substitute with $w = z - 1 \Leftrightarrow z = w + 1$.
$$f(z) = -w^{-2} - 2w^{-1} + \frac{2}{z - 2}$$
Now we need to adjust the last term in a form like $a_j w^j$. We do so by using the geometric series.
$$\frac{2}{z - 2} = \frac{2}{w + 1 - 2} = 2 \cdot \frac{1}{w - 1} = -2 \cdot \frac{1}{1 - w}$$
As we expand about $z_1 = 1$, thus $|w| = |z - 1| < |1|$, the standard geometric series can be used.
$$-2 \cdot \frac{1}{1 - w} = -2 \sum\limits_{n = 0}^\infty w^n$$
Back to the function, here's the Laurent expansion about $z_1 = 1$.
$$f(z) = -w^{-2} - 2w^{-1} + -2 - 2w - 2w^2 - \ldots$$
The first two terms are $-w^{-2}$ and $-2w^{-1}$. The residue is the factor of $w^{-1}$, which is $-2$.
Last we would need to expand around $z_2 = 2$ but this works analogously, so I will not write it down here. There I brought $\frac{-2}{z - 1}$ and $\frac{-1}{(z - 1)^2}$ into a form with $w$ and then used the geometric series again. However, the later term is difficult because it's to the power of $2$. But it can be seen as the derivative of a similar geometric series. As we have compact convergence for the geometric series in this case, we can extract the derivative and so on.
Okay, so that was my technique. Very time consuming is the step where I need to compute the partial fraction decomposition. And after that, where I need to compute the exact series representation of nearly every addend of $f$. And this process repeats for every singularity.
I wonder if there are faster ways as I only need the two terms with the smallest degree.
Thanks for your help :)
| Here are some hints for faster calculation.
Task: Compute all isolated singularities of the following function and specify their type.
\begin{align*}
f(z)=\frac{z}{(z-1)^2(z-2)}\tag{1}
\end{align*}
Answer: From the representation (1) of $f(z)$ we see it has exactly two isolated singularities. One of them is a pole of second order at $z=1$ and the the other is a simple pole at $z=2$.
Note: Since you can deduce the isolated singularities directly from $f$ there is effectively nothing to calculate.
Task: Also give the two terms with the smallest degree of the Laurent expansion about those singularities and their residuum.
Answer: We start with the partial fraction decomposition
\begin{align*}
f(z)=\frac{z}{(z-1)^2(z-2)}=\frac{A}{z-1}+\frac{B}{(z-1)^2}+\frac{C}{z-2}\tag{2}
\end{align*}
We can calculate the coefficients of terms with simple poles easily.
Answer: To obtain $C$ we multiply (2) with $z-2$ and put $z=2$ in the remaining expression.
\begin{align*}
C=\left.f(z)(z-2)\right|_{z=2}=\left.\frac{z}{(z-1)^2}\right|_{z=2}=2
\end{align*}
Similarly we obtain $B$ by multplication with $(z-1)^2$ and putting $z=1$ in the remaining expression.
\begin{align*}
B=\left.f(z)(z-1)^2\right|_{z=1}=\left.\frac{z}{z-2}\right|_{z=1}=-1
\end{align*}
Putting $B$ and $C$ in (2) and isolating $A$ we obtain
\begin{align*}
A&=[z^0]\left(\frac{z}{(z-1)^2(z-2)}-\frac{1}{(z-1)^2}-\frac{2}{z-2}\right)\tag{3}\\
&=[z^0]\left(-\frac{1}{(z-1)^2}-\frac{1}{1-\frac{z}{2}}\right)\\
&=-1-1\\
&=-2
\end{align*}
Note: In order to calculate (3) we observe that the LHS is the constant $A$, so the RHS has also to be a constant. Let's consider the series expansion of the RHS around $z=0$ and let's check for the constant terms of the series.
The first term $\frac{z}{(z-1)^2(z-2)}$ is $z$ multiplied by the product of two binomial series, so there is no contribution to the constant term of the RHS, since the resulting series starts with $z$ having smallest power $1$.
The second term $[z^0]\frac{1}{(z-1)^2}=1$ since the binomial series starts with $1$ and also the the third term has $[z^0]\frac{1}{1-\frac{z}{2}}=1$ since the binomial series start with $1$.
We conclude the contributions to the constant term is
\begin{align*}
A=0-1-1=-2
\end{align*}
as shown in (3).
Answer: We observe the partial fraction decomposition is
\begin{align*}
f(z)=\frac{z}{(z-1)^2(z-2)}=-\frac{2}{z-1}-\frac{1}{(z-1)^2}+\frac{2}{z-2}\tag{4}
\end{align*}
and we are nearly finished. Note that the task is not to specifiy the Laurent series
\begin{align*}
\sum_{n=-\infty}^{\infty}a_n z^n
\end{align*}
We only need to find the two terms with smallest degree of the Laurent series. But, if you look at the presentation (4) the first two terms are already the Laurent expansion of $f$ around $z=1$ while the third part is a Taylor series around $z=1$ and does nothing contribute to $a_{-2}$ or $a_{-1}$.
On the other hand, the third part is already the Laurent expansion around $z=2$ while the first two parts are just a Taylor series around $z=2$ and contribute nothing to the principal part. Here are the coefficients of the two terms with smallest degree $a_{-1}=2$ and $a_{0}$.
Therefore we can answer:
Answer: From the partial fraction decomposition (4) we see the two terms with the smallest degree around the poles are
\begin{align*}
&z=1\qquad\rightarrow\qquad a_{-2}=-1,& & a_{-1}=-2\\
&z=2\qquad\rightarrow\qquad a_{-1}=2,& & a_{0}=\ldots
\end{align*}
It remains the calculation of $a_0$ when expanding $f$ around $z=2$.
| {
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If $ax^3+bx+c=0$, $bx^3+cx+a=0$, $cx^3+ax+b=0$ have a common root for distinct non-zero $a$, $b$, $c$, then at least one has three real roots
Let $a$, $b$, $c$ be distinct nonzero real numbers. If the equations
$$\begin{align}
E_1: ax^3+bx+c=0 \\
E_2: bx^3+cx+a=0 \\
E_3: cx^3+ax+b=0
\end{align}$$
have a common root, prove that at least one of these equations has three real roots (not necessarily distinct).
This question was a math olympiad problem.Below was my attempt but I've got confused:
I know that the discriminant of the equation $x^3+Ax+B=0$ is given by $-4A^3-27B^2$, so it suffices to prove that the discriminant of at one least of the equations is non-negative.
It's easy to prove that the common root $r$ must be real. Now assume on the contrary that all three equations had a pair of complex conjugate roots. Let $\alpha, \bar{\alpha}, \beta, \bar{\beta}, \gamma, \bar{\gamma}$ be the complex conjugate roots of $E_1, E_2$ and $E_3$ respectively.
Then by viete theorem, we have $$\alpha \bar{\alpha} r=|\alpha|^2r=\frac{-c}{a},$$
$$|\beta|^2r=\frac{-a}{b},$$
$$|\gamma|^2r=\frac{-b}{c}.$$
Multiplying the three equations gives $$|\alpha \beta \gamma|^2r^3=-1.$$
Hence $r<0.$ Hence $-c/a<0$ or $c/a>0.$ Similarly $a/b>0$ and $b/c>0.$ This implies that $a,b,c$ all all positive or or negative.
By symmetry, we can assume $a>b>c>0$, but then none of the discriminant could be non-negative. So what's wrong with the argument and how can I complete the proof? Thanks in advance!
| Let $r$ be the common root, then
$$(a+b+c)(r^3+r+1)=0 \tag{$E_1+E_2+E_3$}$$
Case I:
If $r^3+r+1=0$, then $(ar^3+br+c)-a(r^3+r+1)=0$
$$\implies (a-b)r=(c-a)$$
By symmetry, $$r=\frac{c-a}{a-b}=\frac{a-b}{b-c}=\frac{b-c}{c-a}$$
$$c=\frac{a+b \pm i(a-b)\sqrt{3}}{2}$$
Or equivalently,
$$r=\frac{-1 \pm i\sqrt{3}}{2}$$
that contradicts with $r^3+r+1=0$.
Case II:
If $a+b+c=0$, then $r=1$.
For $x\neq 1$,
$$\frac{ax^3+bx+c}{x-1}=ax^2+ax-c=0$$
$$\Delta_{1}=a^2+4ac$$
Similarly,
$$\Delta_{2}=b^2+4ba$$
$$\Delta_{3}=c^2+4cb$$
Since $a+b+c=0$, either two of $a$, $b$ and $c$ are of equal sign.
That is either one of $ab$, $bc$ and $ca$ is positive.
At least one of the $\Delta_{1}$, $\Delta_{2}$ and $\Delta_{3}$ is positive.
Further points to be noticed
For $r \ne 1$ and $a+b+c=0$,
$$\frac{r^3-1}{r-1}=-\frac{b}{a}=-\frac{c}{b}=-\frac{a}{c} \implies a=b\omega=c\omega^2 \land \omega^3=1$$
which does not give distinct, non-zero and real values of $a$, $b$ and $c$.
Also considering the matrix equation:
$$
\begin{pmatrix}
a & b & c \\
b & c & a \\
c & a & b
\end{pmatrix}
\begin{pmatrix}
r^3 \\ r \\ 1
\end{pmatrix}=
\begin{pmatrix}
0 \\ 0 \\ 0
\end{pmatrix}$$
The eigenvalues are $\lambda_1=a+b+c$ and
$\lambda_{2,3}=\pm \sqrt{\frac{(b-c)^2+(c-a)^2+(a-b)^2}{2}}$.
For the matrix is singular,
$$\lambda_1 \lambda_2 \lambda_3=0$$
The eigenvector corresponding to $\lambda_1=a+b+c$ is
$$\vec{v_1}=(1,1,1) \implies r^3=r=1$$
For $a=b=c$, $$\lambda_{2,3}=0 \implies r^3+r+1=0$$
| {
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Integrate $\int \frac{(1-x)^2 \cdot e^x}{(1+x^2)^2}dx$ Integrate using integrating by parts :
$$\int \frac{(1-x)^2 \cdot e^x}{(1+x^2)^2}dx$$
My attempt :
$$I=\int \frac{(1-x)^2 \cdot e^x}{(1+x^2)^2}dx$$
$$I=\int \frac{e^x}{(1+x^2)^2}\cdot(1-x)^2dx$$
$$I=\frac{e^x}{(1+x^2)^2}\cdot\frac{(1-x)^3}{-3}-\int \frac{(1-x)^3}{-3}\cdot\frac{(1+x^2)^2e^x-e^x\cdot2(1+x^2)\cdot2x}{(1+x^2)^4}dx$$
Now it is little bit confusing. How can I simplify this ? Is there another method to do it differently ?
| Just if you look for another way to do it.
Because of the square in denominator and the exponential in numerator, you could have assumed that the result is
$$\int \frac{(1-x)^2 }{(1+x^2)^2}e^x\,dx= \frac {P_n(x)}{1+x^2}e^x$$ where $P_n(x)$ is a polynomial of degree $n$.
Differentiate both sides
$$\frac{(1-x)^2 }{(1+x^2)^2}e^x=\frac{ \left(x^2+1\right) P'_n(x)+(1-x)^2 P_n(x)}{\left(1+x^2\right)^2}e^x$$ that is to say $$(1-x)^2=\left(x^2+1\right) P'_n(x)+(1-x)^2 P_n(x)$$ Comparing the degrees : $2$ for the lhs and $n+2$ in the rhs makes $n=0$; so $P_n(x)$ is just a constant. This implies that $P'_n(x)=0$ and what is left is $$(1-x)^2=(1-x)^2 P_0(x)$$ which makes $P_0(x)=1$.
| {
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Solving $\int\cos(x)\ln\left(\frac{9}{6-\sin(x)}\right)\,dx $ I have:
$$\int\cos(x)\ln\left(\frac{9}{6-\sin(x)}\right)\,dx $$
I've tried by parts but without results, i don't know how to start, any tips?
| $$\int \cos { \left( x \right) } \ln { \left( \frac { 9 }{ 6-\sin { \left( x \right) } } \right) dx=\int { \ln { \left( \frac { 9 }{ 6-\sin { \left( x \right) } } \right) d\left( \sin { \left( x \right) } \right) = } } } \\ =\sin { \left( x \right) \ln { \left( \frac { 9 }{ 6-\sin { \left( x \right) } } \right) -\frac { 1 }{ 9 } \int { \left( 6-\sin { \left( x \right) } \right) \sin { \left( x \right) } dx } = } } \\ =\sin { \left( x \right) \ln { \left( \frac { 9 }{ 6-\sin { \left( x \right) } } \right) -\frac { 2 }{ 3 } \int { \sin { \left( x \right) dx+\frac { 1 }{ 9 } \int { \sin ^{ 2 }{ x } dx } } } = } } \\ =\sin { \left( x \right) \ln { \left( \frac { 9 }{ 6-\sin { \left( x \right) } } \right) -\frac { 2 }{ 3 } \int { \sin { \left( x \right) dx+\frac { 1 }{ 18 } \int { \left( 1-\cos { \left( 2x \right) } \right) } dx } } = } } \\ =\sin { \left( x \right) \ln { \left( \frac { 9 }{ 6-\sin { \left( x \right) } } \right) +\frac { 2 }{ 3 } \cos { \left( x \right) } +\frac { 1 }{ 18 } \left( x-\frac { \sin { \left( 2x \right) } }{ 2 } \right) } +C } $$
| {
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How to find eigenvectors given complex eigenvalues I am given that: $\vec{x}' = A\vec{x}$, where $A=\begin{pmatrix} -3 & 0 & 2 \\ 1 & -1 & 0 \\ -2 & -1 & 0 \end{pmatrix}$
I want to find the general solution of this in terms of real-valued functions.
Computing $\det(A-\lambda I)=0$, I found the characteristic equation to be $(\lambda +2)(\lambda^2 +2\lambda+3)=0.$ So $\lambda_1=-2,\lambda_{2,3} = -1\pm\sqrt{2}i$.
I'm having trouble dealing with $\lambda_{2,3}$. Can anyone carry me through the computations?
Thanks.
| Let's write $\lambda_2=-1+i\sqrt{2}$, $\lambda_3=-1-i\sqrt{2}$.
We have
$$A-\lambda_2 I=\begin{pmatrix} -2-i\sqrt{2} & 0 & 2 \\ 1 & -i\sqrt{2} & 0 \\ -2 & -1 & 1-i\sqrt{2} \end{pmatrix} $$
Applying row operations, we obtain
\begin{align*}
A-\lambda_2 I&\to \begin{pmatrix} 1 & 0 & \frac{-2+i\sqrt{2}}{3} \\ 1 & -i\sqrt{2} & 0 \\ -2 & -1 & 1-i\sqrt{2} \end{pmatrix} \\
&\to \begin{pmatrix} 1 & 0 & \frac{-2+i\sqrt{2}}{3} \\ 0 & -i\sqrt{2} & \frac{2-i\sqrt{2}}{3} \\ 0 & -1 & \frac{-1-i\sqrt{2}}{3} \end{pmatrix} \\
&\to \begin{pmatrix} 1 & 0 & \frac{-2+i\sqrt{2}}{3} \\ 0 & 1 & \frac{1+i\sqrt{2}}{3} \\ 0 & -1 & \frac{-1-i\sqrt{2}}{3} \end{pmatrix} \\
&\to \begin{pmatrix} 1 & 0 & \frac{-2+i\sqrt{2}}{3} \\ 0 & 1 & \frac{1+i\sqrt{2}}{3} \\ 0 & 0 & 0 \end{pmatrix}.
\end{align*}
From this, we can see that a vector $x=(x_1,x_2,x_3)^T$ is mapped to $0$ under $A-\lambda_2 I$ if
\begin{align*}
&x_1+\frac{-2+i\sqrt{2}}{3}x_3=0 \\
&x_2+\frac{1+i\sqrt{2}}{3}x_3=0,
\end{align*}
or,
\begin{align*}
x_1&=\frac{2-i\sqrt{2}}{3}x_3,\\
x_2&=\frac{-1-i\sqrt{2}}{3}x_3.
\end{align*}
Therefore, the eigenspace of $A$ associated with eigenvalue $\lambda_2$ is spanned by
$$x=\begin{pmatrix} \frac{2-i\sqrt{2}}{3} \\ \frac{-1-i\sqrt{2}}{3} \\1\end{pmatrix},$$
i.e., $x$ is an eigenvector with eigenvalue $\lambda_2$. This gives your solution for $\lambda_2$. To find the eigenvector associated with $\lambda_3$, essentially do the same as what I did. If you require further explanation of what I did, please let me know and I will edit.
| {
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Monotonicity of the sequence $(a_n)$, where $a_n=\left ( 1+\frac{1}{n} \right )^n$ Define $a_n=\left ( 1+\frac{1}{n} \right )^n$ for $n\geq 1$. I want to show that it is increasing. First, we have
$$\frac{a_{n+1}}{a_n}=\left ( \frac{1+\frac{1}{n+1}}{1+\frac{1}{n}} \right )^n\left ( 1+\frac{1}{n+1} \right )=\left ( 1-\frac{1}{(n+1)^2} \right )^n\left ( \frac{n+2}{n+1} \right )$$
Using the Bernoulli inequality, we see that, for all $n\geq 1$,
$$\left ( 1-\frac{1}{(n+1)^2} \right )^n\geq 1-\frac{n}{(n+1)^2}=\frac{n^2+n+1}{n^2+2n+1}.$$
How do you then show that
$$\left ( 1-\frac{n}{(n+1)^2} \right )\left ( \frac{n+2}{n+1} \right )>1?$$
Edit: This is not a duplicate question; the question is not to show about the existence of the definition of Euler number $e$. The question is about showing that it is increasing, the way I have shown that I have been stuck with, not other ways. It seems that the question is too easy that I have been too tired to think at this late.
| It can be proved using function. We have $f(x) = \left(1+\dfrac{1}{x}\right)^x, x \in [1,\infty)$ and taking log: $\log f(x) = x\log(x+1) - x\log x\implies f'(x) = f(x) \left(\log(x+1)+ \dfrac{x}{x+1} - \log x - 1\right)= f(x)\left(\log(x+1) - \log x - \dfrac{1}{x+1}\right)= f(x)g(x)\implies g'(x) = \dfrac{1}{x+1}-\dfrac{1}{x}+\dfrac{1}{(1+x)^2}= \dfrac{x(x+1)-(x+1)^2+x}{x(1+x)^2}= -\dfrac{1}{x(1+x)^2} < 0\implies g(x) > \displaystyle \lim_{x \to \infty} g(x)=0\implies f'(x) > 0\implies f(n+1) > f(n)$.
| {
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I need help with a limit without using L'Hopital I need to do the following limit without using L'Hopital and I have not been able, please help
$$\lim\limits_{x \to 3} \left(\frac{x-1}{2x-4}\right)^{\frac{1}{x-3}}$$
| Since $\lim _{ x\rightarrow 0 }{ { \left( 1+x \right) }^{ \frac { 1 }{ x } } } =e$ (or $\\ \lim _{ x\rightarrow \infty }{ { \left( 1+\frac { 1 }{ x } \right) }^{ x } } =e\\ $)
$$\lim _{ x\to 3 } \left( \frac { x-1 }{ 2x-4 } \right) ^{ \frac { 1 }{ x-3 } }=\lim _{ x\to 3 } \left( 1+\frac { 3-x }{ 2x-4 } \right) ^{ \frac { 1 }{ x-3 } }=\\ =\lim _{ x\to 3 }{ \left[ \left( 1+\frac { 1 }{ \frac { 2x-4 }{ 3-x } } \right) ^{ \frac { 2x-4 }{ 3-x } } \right] } ^{ \frac { 3-x }{ 2x-4 } \frac { 1 }{ x-3 } }=\lim _{ x\to 3 }{ { e }^{ -\frac { 1 }{ 2x-4 } } } =\frac { 1 }{ \sqrt { e } } $$
You can solve it with other way let say $x-3=z,$ then $$\lim _{ z\rightarrow 0 }{ \left( 1+\frac { -z }{ 2+2z } \right) ^{ \frac { 1 }{ z } } } =\lim _{ z\rightarrow 0 }{ { \left( \left( 1+\frac { 1 }{ -\frac { 2+2z }{ z } } \right) ^{ -\frac { 2+2z }{ z } } \right) }^{ -\frac { z }{ 2+2z } \frac { 1 }{ z } } } =\lim _{ z\rightarrow 0 }{ { e }^{ -\frac { 1 }{ 2z+2 } } } =\frac { 1 }{ \sqrt { e } } $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The cubic equation $x^3-5x^2+6x-3 = 0$ has solutions $\alpha$, $\beta$ and $\gamma$. The cubic equation $x^3-5x^2+6x-3 = 0$ has solutions $\alpha$, $\beta$ and $\gamma$.
Find the value of $$\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}$$
| Since $$\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} = \frac{\alpha^2 \beta^2 + \beta^2 \gamma^2 + \alpha^2\gamma^2}{\alpha^2 \beta^2 \gamma^2}$$
You know that for a cubic $x^3 + ax^2 + bx + c$ the sum of roots are given by $\alpha + \beta + \gamma = -a = 5$, the sum of the alternate pairs $\alpha\beta + \beta\gamma + \alpha \gamma = b = 6$ and the product $\alpha \beta \gamma = -c = 3$.
So squaring the second equation gives us $$\alpha^2 \beta^2 + \beta^2 \gamma^2 + \alpha^2 \gamma^2 + 2\alpha\beta\gamma(\alpha + \beta +\gamma) = 6^2$$
This yields $\alpha^2\beta^2 + \beta^2 \gamma^2 + \alpha^2 \gamma^2 = 36 -2(3)(5) = 6$ and hence $$\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} = \frac{6}{3^2} = \frac{6}{9} = \color{red}{\frac{2}{3}}$$
since squaring the third equation gives us $\alpha^2\beta^2\gamma^2 = 3^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove this inequality with trigonometry $9\cos^2{x}-10\cos{x}\sin{y}-8\cos{y}\sin{x}+17\ge 1$ let$x,y\in R$,show that $$9\cos^2{x}-10\cos{x}\sin{y}-8\cos{y}\sin{x}+17\ge 1$$
Maybe use Cauchy-Schwarz inequality can solve it?and I can't
Adit it:I think the right hand can replace constant $9$ ,and it's best
| Let $u=9\cos^2x-10\cos x\sin y-8\cos y\sin x$
As $-\sqrt{a^2+b^2}\le-(a\cos v+b\sin v)\le\sqrt{a^2+b^2}$
$\implies-\sqrt{(10\cos x)^2+(8\sin x)^2}\le-(10\cos x\sin y+8\cos y\sin x)\le\sqrt{(10\cos x)^2+(8\sin x)^2}$
$u\ge9\cos^2x-\sqrt{(10\cos x)^2+(8\sin x)^2}=(\sqrt{9\cos^2x+16}-1)^2-17$
Now $3\le\sqrt{16}-1\le\sqrt{9\cos^2x+16}-1\le\sqrt{9+16}-1=4$
$\implies3^2\le(\sqrt{9\cos^2x+16}-1)^2\le4^2$
$\cdots$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding GCD of $95$ and $39$ My Algebra instructor gave us this problem which is to find gcd of $95$ and $39$ and express it as $95x+39y$. Also are $x$, $y$ unique?
Now they are relatively prime so GCD is $1$. I have no clue how to find $x$ and $y$ and if they are unique or not.
Thanks!
| Although you could use the extended Euclidean algorithm to calculate that value, sometimes it's simpler to use Euler's theorem and exponentiation by squaring.
I don't want to solve your problem for you, so I will use similar numbers $94$ and $51$ instead.
More precisely, say we want to find $x$ and $y$ such that $94x+51y = \gcd(94,51) = 1$. This implies that $94x=1 \pmod{51}$, thus
\begin{align}
x & = 94^{\phi(51)-1} &\pmod{51} \\
& = 94^{(17-1)(3-1)-1} &\pmod{51} \\
& = 94^{31} &\pmod{51} \\
& = 94^{30+1} &\pmod{51} \\
& = 94\cdot(94^{15})^2 &\pmod{51} \\
& = 94\cdot(94\cdot(94\cdot(94\cdot (94\cdot1^2)^2)^2)^2)^2 &\pmod{51} \\
& = 94\cdot(94\cdot(94\cdot(94\cdot (94\cdot1^2)^2)^2)^2)^2 &\pmod{51} \\
& = 19 &\pmod{51}
\end{align}
That means we can pick any $x = 19 \pmod{51}$ to get a solution. For example when $x = 19$ then $94\cdot 19 + 51y = 1$ gets us $y = \frac{1-94\cdot 19}{51} = -35$. On the other hand if $x = 19+51 = 70$ then $y=-129$.
I hope this helps $\ddot\smile$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the eigenvalues of a symmetric matrix
Find the eigenvalues of a $3 \times 3$ symmetric matrix with $1$ on the main diagonal and $\frac{1}{\sqrt 3}$ off the main diagonal.
Since each row on addition give the same value, one of the three eigenvalue is $1+\frac{2}{\sqrt 3}$.
Is there an easy way to find the other two values without using the formula $\det(A-\lambda I_3) = 0$.
| You might want to take advantage that your $3 \times 3 $ matrix is a circulant matrix of the form:
$$ \begin{bmatrix} c_0 & c_{n-1} & c_{n-2} & \cdots & c_1 \\ c_1 & c_0 & c_{n-1} & \cdots & c_{2}\\ \vdots & c_1& c_0& \ddots & \vdots \\ c_{n-2} & &\ddots & \ddots& c_{n-1}\\ c_{n-1} & c_{n-2} &\cdots & c_1 & c_{0} \end{bmatrix}$$ and the eigenvalues can be computed through the formula:
$$ \lambda_j = c_0 + c_1 \omega_j +c_2 \omega_j^2 + \cdots+ c_{n-1}\omega_j^{n-1}, \quad j = 0,1,\ldots,n-1,\tag 1$$
where $\omega_j = \exp\left(\frac{2 \pi i j}{n}\right)$ are the $n-$th roots of unity.
Concretely, $n = 3$ and we have $\omega_0 = 1, \omega_1 = -\frac{1}{2} + \frac{i\sqrt 3}{2}, \omega_2 = -\frac{1}{2} - \frac{i\sqrt 3}{2}.$
According to $(1)$:
$$\lambda_0 = 1 + \frac{1}{\sqrt 3} \cdot 1^1 + \frac{1}{\sqrt 3}\cdot 1^2 = 1 +\frac{2}{\sqrt 3}. $$
$$\lambda_1 = 1 + \frac{1}{\sqrt 3}\cdot \left(-\frac 12 +\frac{i\sqrt 3}{2}\right) + \frac{1}{\sqrt 3}\cdot \left(-\frac 12 +\frac{i\sqrt 3}{2}\right)^2 = 1-\frac{1}{\sqrt 3}.$$
Similarly, $\lambda_2 = 1 - \frac{1}{\sqrt 3}.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that if $ a,b,c > 0 $, then $ [(1 + a) (1 + b) (1 + c)]^{7} > 7^{7} (a^{4} b^{4} c^{4}) $.
Problem. Prove that if $ a,b,c > 0 $, then $ [(1 + a) (1 + b) (1 + c)]^{7} > 7^{7} (a^{4} b^{4} c^{4}) $.
I don’t know how to solve this problem... What I can think of is to just simplify this inequality:
$$
\left[ \frac{(1 + a) (1 + b) (1 + c)}{7} \right]^{7} > a^{4} b^{4} c^{4}.
$$
How can I proceed with solving this problem?
Note: This is a question of sequence and series, specifically AM-GM-HM inequality...
| Notice
$$\frac{(1+a)(1+b)(1+c)-1}{7}=\frac{a+b+c+ab+bc+ca+abc}{7} \ge \sqrt[7]{a^4b^4c^4}$$
Because of $\text{AM-GM}$.
We conclude
$$\frac{(1+a)(1+b)(1+c)}{7}>\frac{(1+a)(1+b)(1+c)-1}{7} \ge\sqrt[7]{a^4b^4c^4}$$
This is equivalent to
$$
\left[ \frac{(1 + a) (1 + b) (1 + c)}{7} \right]^{7} > a^{4} b^{4} c^{4}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1876539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Landau symbol and taylor series $$\lim_{x\rightarrow \infty} \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt x=\frac{1}{2}$$
I saw the calculation $$\lim_{x\rightarrow \infty} \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt x=$$...$$=\sqrt x (\sqrt{1+\sqrt{1/x+\sqrt{1/x^3}}}-1)= $$ $$\sqrt x(1+\frac{1}{2}\sqrt{1/x+\sqrt{1/x^3}}+ \color{red}{\mathcal o \left(1/x+\sqrt{1/x^3}\right)}-1)=...$$
Main question: Now wouldn't it have to be $\mathcal o(\sqrt{1/x+\sqrt{1/x^3}})$ instead? So I could write $\mathcal o(\sqrt{1/x+\sqrt{1/x^3}})$ or $\mathcal O(1/x+\sqrt{1/x^3})$?
Are there other ways to calculate the limit?
| $$
\begin{align}
\lim_{x\to\infty}\left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)
&=\lim_{x\to\infty}\left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}\\
&=\lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}\\
&=\lim_{x\to\infty}\frac{\sqrt{1+\sqrt{\frac1x}}}{\sqrt{1+\sqrt{\frac1x+\sqrt{\frac1{x^3}}}}+1}\\
&=\frac12
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Some questions about highly composite numbers A highly composite number is a natural number $n\ge 1$, such that $t(m)<t(n)$ for all $m$ with $1\le m<n$ , where $t(n)$ is the number of divisors of $n$.
The link shows that the prime factorization of such a number $n$ contains non-increasing exponents and contains all primes upto the largest prime factor of $n$.
It is also claimed that for every $n>36$ the last exponent is always $1$, in other words, if $p$ is the largest prime factor of a highly composite number $n>36$, then $p^2$ does not divide $n$.
How can I prove this ?
The largest highly composite number not divisible by $9$, seems to be $$1680=2^4\cdot 3\cdot 5\cdot 7$$
Is this true, and how can I prove it ?
Finally
Given the sequence of the exponents, is there an efficient method to determine whether the sequence corresponds to a highly composite number ?
It is clear that it is sufficient to test the numbers corresponding to a non-increasing sequence and that the length is bounded by the smallest primorial exceeding the given number $n$. But for large $n$, many such sequences have to be checked.
Just for the sake of curiousity : What is the largest known highly composite number ?
| Answer to question 1:
Suppose we have $n = 2^{a_1} 3^{a_2} \ldots p_k^{a_k}$ with $a_1 \ge a_2 \ge \ldots \ge a_k \ge 2$, $n > 36$ and $n$ is highly composite. I claim that $a_1 \ge 3$. Suppose $a_1 \le 2$; if $k=2$, then $n \le 2^2 3^2 = 36$, whereas if $k \ge 3$, then $t(\frac{4n}{5}) = \frac{a_1+3}{a_1+1} \cdot \frac{a_3}{a_3+1}\cdot t(n) \ge \frac{5}{3}\cdot\frac{2}{3}\cdot t(n) > t(n)$, so $n$ is not highly composite.
But then $t(\frac{p_{k+1} n}{2 p_k}) = 2 \cdot \frac{a_1}{a_1+1} \cdot \frac{a_k}{a_k+1}\cdot t(n) \ge 2 \cdot \frac{3}{4} \cdot \frac{2}{3} \cdot t(n) = t(n)$, while $\frac{p_{k+1} n}{2 p_k} < n$ by Bertrand's Postulate.
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplify limit problem $\lim\limits_{x\to 0}\frac{\sqrt{2+x^2}-\sqrt{2+x^2}}{x^2}$ This problem is from Calculus: A complete course 8 ed.(Adams/Essex).
Problem no. 35 chapt 1.2.
Evaluate the limit or explain why it does not exist.
$\displaystyle\lim_{x\to 0}\frac{\sqrt{2+x^2}-\sqrt{2+x^2}}{x^2}$
So I multiplied the fraction with the conjugate of the numerator and ended up with this. I see that $x^2$ can be divided from both numerator and denominator. But after that I am not sure how I should continue.
$\displaystyle\lim_{x\to 0}\frac {2x^2}{x^2(\sqrt{2+x^2} + \sqrt{2-x^2})}$
The answer states that next step is
$\displaystyle\frac {2}{\sqrt{2} + \sqrt{2}}$
But I do not see how it got there. I tried looking up square root laws and how simplify but could not see anything that was useful for me.
| $\displaystyle\quad\lim_{x\to 0}\frac {2x^2}{x^2(\sqrt{2+x^2} + \sqrt{2-x^2})}$
$\displaystyle=\lim_{x\to 0}\frac {2}{(\sqrt{2+x^2} + \sqrt{2-x^2})}$
$\displaystyle=\frac {2}{(\sqrt{2+0^2} + \sqrt{2-0^2})}$
$\displaystyle=\frac {2}{(\sqrt{2} + \sqrt{2})}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Limit of $\frac{5^x}{6^x}$ at infinity How should I calculate the limit of the given function?
Should I differentiate or should I account for changes intuitively?
Please provide a general answer accounting for both:
*
*when numerator is smaller and
*when numerator is greater$$\lim_{x\rightarrow\infty}\frac{5^x}{6^x}$$
| Note that $\bigl(\frac{5}{6} \bigr)^x$ is a decreasing function of $x$, because $\bigl(\frac{5}{6} \bigr)^x = e^{x \log (\frac{5}{6})}$, and as $x$ increases, $x \log(\frac{5}{6})$ decreases because $\log(\frac{5}{6}) < 0$, and hence $e^{x \log(\frac{5}{6})}$ is a decreasing function of $x$ for $x>0$ because the exponent then tends to $-\infty$. Now, because $\frac{5}{6}$ is a positive quantity, it follows that $\bigl(\frac{5}{6}\bigr)^x$ is bounded below by zero.
Hence, $\bigl(\frac{5}{6}\bigr)^x$ is a decreasing sequence which is bounded below. By completeness, $\bigl(\frac{5}{6}\bigr)^x$ is a convergent sequence, say to some $C$. Now, because $C = \displaystyle\lim_{x \to \infty} \biggl(\frac{5}{6}\biggr)^x$, it follows that:
$$
\frac{5}{6} C = \displaystyle\lim_{x \to \infty} \biggl(\frac{5}{6}\biggr)^{x+1} = \displaystyle\lim_{x \to \infty} \biggl(\frac{5}{6}\biggr)^x = C
$$
because if $x+1 \to \infty$, then $x \to \infty$ and vice versa. Hence $\frac{C}{6} = 0$ and $C=0$. Which is to say, $\displaystyle\lim_{x \to \infty} \biggl(\frac{5}{6}\biggr)^x=0$.
There was nothing special about $\frac{5}{6}$ here, any proper fraction would do.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\sqrt[2]{2} \cdot \sqrt[4]{4}\cdot \sqrt[8]{8}\cdot \dots$ Evaluate: $$\lim_{n\to \infty }\sqrt[2]{2}\cdot \sqrt[4]{4}\cdot \sqrt[8]{8}\cdot \dots \cdot\sqrt[2^n]{2^n}$$
My attempt:First solve when $n$ is not infinity then put infinity in.
$$2^{\frac{1}{2}}\cdot 4^{\frac{1}{4}}\cdot \dots\cdot (2^n)^{\frac{1}{2^n}}$$
$$=2^{\frac{1}{2}}\cdot 2^{\frac{2}{4}}\cdot \dots\cdot 2^{\frac{n}{2^n}}$$
Now calculate the sum of the powers:
$$\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\dots+\frac{n}{2^n}$$
$$=\frac{2^{n-1}+2\cdot2^{n-2}+3\cdot2^{n-3}+\dots+n\cdot2^0}{2^n}$$
Now calculate the numerator:
$$2^0+2^1+2^2+\dots+2^{n-1}=2^n-1$$
$$+$$
$$2^0+2^1+\dots+2^{n-2}=2^{n-1}-1$$
$$+$$
$$2^0+2^1+\dots+2^{n-3}=2^{n-2}-1$$
$$+$$
$$\vdots$$
$$+$$
$$2^0=2^1-1$$
$$=2^1+2^2+2^3+\dots+2^n-n=2^{n+1}-n-1$$
Now put the numerator on the fraction:
$$\frac{2^{n+1}-n-1}{2^n}=2-\frac{n}{2^n}-\frac{1}{2^n}$$
Now we can easily find $\lim_{n \to \infty}\frac{1}{2^n}=0$
Then we just have to find $\lim_{n \to \infty }\frac{n}{2^n}$, that by graphing will easily give us the answer zero.
That gives the total answer is $4$.
But now they are two problems:
1.I cannot find $\lim_{n \to \infty }\frac{n}{2^n}$ without graghing.
2.My answer is too long.
Now I want you to help me with these problems.Thanks.
| I believe this is a "known" representation of the key summation:
$$\sum_{n=1}^{\infty} \frac{n}{2^n} \;=\; \frac{1}{2} + \left( \frac{1}{4} + \frac{1}{4} \right) + \left( \frac{1}{8} + \frac{1}{8} + \frac{1}{8} \right) + \cdots \;=\; 2 $$
| {
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"timestamp": "2023-03-29T00:00:00",
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$a_1 + a_2 + \dots a_n = 1$ find min of $a_1^2 +\frac{a_2^2}{2} + \dots + \frac{a_n^2}n.$ Given $n$ numbers $a_1$, $\cdots$ and such that $a_1$, $a_2$, $\cdots$, $a_n > 0$ and their sum is $1$, I want to find the minimum value of
$$a_1^2 + \frac{a_2^2}{2} + \cdots + \frac{a_n^2}{n}.$$
I have tried using weighted AM-GM inequality, like this:
$$\frac{a_1^2 + \frac{a_2^2}{2} + \cdots + \frac{a_n^2}{n}}{a_1 + a_2 + \cdots + a_n } \geqslant \frac{a_1^{a_1} \cdots a_n^{a_n}}{2^{a_2} \cdots n^{a_n}}$$
but was unable to make progress on the right hand side.
Is there a better way to apply AM-GM inequality? Or is there some different way altogether to solve this?
| Consider the continuous function $f(a_1,a_2\dots,a_n)=a_1^2+a_2^2/2+\dots +a_n^2/n$. We will minimize it in the compact region defined by constraints $a_i\geq 0$ and $a_1+a_2+\dots + a_n=1$.
The idea is that since the region is compact the minimum must exist, and we will find necessary conditions for the point at which the minimum is achieved.
Suppose that $a_1+a_k=z$, then $a_1^2+\frac{a_k^2}{k}$ must be maximized with respect to $a_1+a_k=z$, when does this happen? we want to maximize $a_1^2+\frac{(z-a_1)^2}{k}$,the derivative of this function is $2a_1-\frac{2(z-a_1)}{k}$. So the minimum is reached when $(k+1)a_1=z$ or $a_1=\frac{z}{k+1}$, or when $ka_1=a_k$.
We conclude the maximum is reached at:
$a_1,2a_1,3a_1\dots na_1$, which clearly adds up to $1$ when $a_1=\frac{2}{n(n+1)}$.
| {
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solve $\tan{x} = \tan{3x}$ I'm asked to solve $\tan{x} = \tan{3x}$
Here's my attempt:
$$\tan{x} = \tan{3x}$$
$$\tan{x} = \tan{(x + 2x)}$$
$$\tan{x} = \frac{\tan{x} + \tan{2x}}{1-\tan{x}\tan{2x}}$$
Recall the identity:
$$\tan{2x} = \frac{2\tan{x}}{1-\tan^2{x}}$$
So then we have:
$$\tan{x} = \frac{\tan{x} + \frac{2\tan{x}}{1-\tan^2{x}}}{1-\tan{x}\frac{2\tan{x}}{1-\tan^2{x}}}$$
$$\tan{x} - \tan^2{(x)} \cdot \frac{2\tan{x}}{1-\tan^2{x}} = \tan{x} + \frac{2\tan{x}}{1-\tan^2{x}}$$
$$-\tan^2{(x)} \cdot \frac{2\tan{x}}{1-\tan^2{x}} = \frac{2\tan{x}}{1-\tan^2{x}}$$
$$-\tan^2{x} \cdot \frac{2\tan{x}}{1-\tan^2{x}} \cdot \frac{1-\tan^2{x}}{2\tan{x}} = 1$$
$$\tan^2{x} = -1$$
This does obviously not compute. Why is my way wrong and how can I go about solving it?
| You can't obviously cancel in the last (or one before the last, in fact) step, thus you must have
$$\frac{2\tan x}{1-\tan^2x}=0\iff \tan x=0\iff x=k\pi\;,\;\;k\in\Bbb Z.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\lim_{n\to\infty}\frac{n^2+5n^3}{2n^3+3\sqrt{4+n^6}}$ $$\frac{n^2+5n^3}{2n^3+3\sqrt{4+n^6}}$$
$$=\frac{\frac{1}{n}+5}{2+\frac{3}{n^3}\sqrt{4+n^6}}$$
$$=\frac{\frac{1}{n}+5}{2+3\sqrt{\frac{4+n^6}{n^9}}}$$
$$=\frac{0+5}{2+0}$$
$$=\frac{5}{2}$$
...but the given answer is $1$. Where did I make a mistake??
| The error is that you used $n^3=\sqrt{n^9}$. But...
| {
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Help simplify using quadratic formula $\dfrac{4\pm\sqrt{28}}{2}=2\pm\sqrt7$
My Question is how did $\dfrac{4\pm\sqrt{28}}{2}$ become simplified as $2\pm\sqrt7$
Can you help me by explaining the steps clearly
:) Many Thanks
| $\dfrac{4\pm\sqrt{28}}{2}=$
$\dfrac{4\pm\sqrt{4\cdot7}}{2}=$
$\dfrac{4\pm\sqrt{4}\cdot\sqrt{7}}{2}=$
$\dfrac{4\pm2\cdot\sqrt{7}}{2}=$
$\dfrac{2\pm1\cdot\sqrt{7}}{1}=$
$2\pm\sqrt{7}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1888902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How does $\pi - \arctan(\frac{8}{x}) - \arctan(\frac{13}{20-x}) = \arctan(\frac{x}{8}) + \arctan(\frac{20-x}{13})$? How does $$\pi - \arctan(\frac{8}{x}) - \arctan(\frac{13}{20-x}) = \arctan(\frac{x}{8}) + \arctan(\frac{20-x}{13})$$
| $$\frac{\pi}{2} - \arctan\left(\frac{8}{x}\right) + \frac{\pi}{2} - \arctan\left(\frac{13}{20-x}\right) = \arctan\left(\frac{x}{8}\right) + \arctan\left(\frac{20-x}{13}\right)$$
$$\mbox{arccot} \left(\frac{8}{x}\right) +\mbox{arccot}\left(\frac{13}{20-x}\right)=\arctan\left(\frac{x}{8}\right) + \arctan\left(\frac{20-x}{13}\right)$$
$$\color{red}{\boxed{ \color{black}{\mbox{arccot} \left(\frac{1}{t}\right) =\arctan t}}} $$
$\color{red}{\text{This is valid only when}\qquad t \gt 0}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1890218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solution of inequality Question:
$x$ and $y$ are different integers and both $x$ and $y$ are greater than $0$. If $x^2-y^2<8$ and $x+y>3$, what is the maximum value of $x$?
My solution:
I used plugged in number method, and found for $x=4$ and $y=3$, $4^2-3^2=7$ and $4+3=7$, satisfies both equations. but, if $x=5$ and $y=4$, values doesn't satisfy first equation. hence, maximum possible value of $x= 4$.
Is there any other easy way than this to solve this problem?
| There is no maximum value of $y$:
Let $y$ be any integer with $y\ge3$, and let $x=y-1$.
Then $x^2-y^2<0<8$, and $x+y=2y-1\ge5>3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1891156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
factors of $(x+y)^p-x^p-y^p$ Assume $x,y\in\mathbb{N}$ and $p$ - prime, $p\ge 5$.
$5xy(x+y)(x^2+xy+y^2)$ divides $(x+y)^5-x^5-y^5$ (in fact, there is equality).
$7xy(x+y)(x^2+xy+y^2)^2$ divides $(x+y)^7-x^7-y^7$ (there is equality also).
$11xy(x+y)(x^2+xy+y^2)$ divides $(x+y)^{11}-x^{11}-y^{11}$ but $11xy(x+y)(x^2+xy+y^2)^2$ does not.
It is easy to prove that $pxy(x+y)$ divides $(x+y)^p-x^p-y^p$.
Does $pxy(x+y)(x^2+xy+y^2)$ divide $(x+y)^p-x^p-y^p$ always?
Is there a characterisation of $p$'s for which $pxy(x+y)(x^2+xy+y^2)^2$ divides $(x+y)^p-x^p-y^p$ ?
Edit:
For $p=17$ there is no divisibility by $(x^2+xy+y^2)^2$, similarily as for $p=11$.
| By dehomogenizing the polynomials, it suffices to work with $f_p (x) = (x+1)^{p} - x^p - 1$. Let $g(x) = x^2 + x + 1$ and for any polynomial $f$, let $\text{ord}_g (f)$ denote the maximum power of $g$ that divides $f$. We claim $\text{ord}_g (f_p) = 1$ if $p \equiv 5 \pmod{6}$ and $\text{ord}_g (f_p) = 2$ if $p\equiv 1\pmod{6}$.
Let $\omega = (-1+\sqrt{-3})/2$ be the root of $g(x)$ with negative real part. We know $\omega^3 = 1$ and $\omega+1$ is a root of $x^2 - x + 1$, so $(\omega+1)^3 = -1$.
If $p \equiv 5 \pmod{6}$, writing $p=6k+5$ we find $$f_p (\omega) = (\omega + 1)^{6k+5} - \omega^{6k+5} - 1 = -(\omega+1)^2 - \omega^2 - 1 = -2(\omega^2 + \omega + 1) = 0.$$ Thus, $\omega$ is a root of $f_p$ and it follows $\overline{\omega}$ also is, hence $g$ divides $f_p$. But $\omega$ is not a root of $f_p'(x) = p(x+1)^{p-1} - px^{p-1}$ since $$f_p'(\omega) = p(\omega+1)^{6k+4} - \omega^{6k+4}) = p(-(\omega+1) - \omega) = -p(2\omega+1) \neq 0.$$ Thus, $g$ does not divide $f_p'$ and it follows $\text{ord}_g (f_p) = 1$.
However, if $p\equiv 1 \pmod{6}$, writing $p=6k+1$ we find $$f_p (\omega) = (\omega+1)^{6k+1} - \omega^{6k+1} - 1 = (\omega+1) - \omega - 1 = 0$$ and $$f_p'(\omega) = p((\omega+1)^{6k} - \omega^{6k}) = 0,$$ hence $g^2$ divides $f_p$. But $f_p ''(x) = p(p-1)(x+1)^{p-2} - p(p-1)x^{p-2}$ and $$f_p ''(\omega) = p(p-1)((\omega+1)^{6k-1} - \omega^{6k-1}) = p(p-1)\left(\frac{1}{\omega+1} - \frac{1}{\omega} \right) = p(p-1) \frac{-1}{\omega^2 + \omega} = p(p-1) \neq 0.$$ So $g^3$ does not divide $f_p$ and we conclude $\text{ord}_g (f_p) = 2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to determine $x, y, z$ coordinates of the third vertex of a 3D triangle? Given a line $AB$ such that $A(0,0,0)$ and $B(4,7,9)$. How can I obtain a point $C(x,y,z)$ of $\Delta ABC$ with $AB$, $AC$, and $BC$ known?
Any help will be appreciated!
| We are given two fixed points $A$ and $B$ in three space and two positive numbers $a$ and $b$ which together with $c=|AB|$ satisfy the triangle inequalities. We are looking for a point $C$ such that $|AB| = c, \,\, |BC|=a$ and $|CA| = b$.
Assume we have found one such point $C$. Let $H \in AB$ be the (unique) point on $AB$ such that $CH$ is perpendicular to $AB$. Since we are in three space, it turns out that all points $C$ that satisfy the conditions $|BC|=a$ and $|CA| = b$ form a circle lying on the plane through $H$ perpendicular to line $AB$. Moreover, the center of the circle is $H$ and its radius is $|CH|$. WE want to find a parametrization of that circle.
As already mentioned, point $H \in AB$ is such that $CH$ is perpendicular to $AB$, i.e. $CH$ is the altitude of triangle $ABC$ through the vertex $C$. Then $H$ splits the segment $AB$ into two segments $AH$ and $BH$. Let $|AH| = c_b$ and $|BH| = c_a$. By construction $c_a+c_b = c.$ Furthermore, by Pythagoras' theorem $$b^2 - c_b^2 = |CH|^2 = a^2-c_a^2.$$ Thus, we have a system of 2 equations for the unknown variables $c_a,c_b$:
\begin{align}
c_b^2-c_a^2 &= b^2-a^2\\
c_b+c_a &= c
\end{align}
and when we solve it we obtain
\begin{align}
c_b &= \frac{b^2-a^2+c^2}{2c}\\
c_a &= \frac{a^2-b^2+c^2}{2c}
\end{align}
Furthermore, we can find the length of the altitude
$$|CH| = \sqrt{b^2-c_b^2} = \frac{\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}}{2c}$$
Next, observe that the coordinate system is with origin $A$. Let $\overrightarrow{e}_1 = (1,0,0)$. Then clearly $\overrightarrow{e}_1$ and $\overrightarrow{AB}$ are linearly independent. Therefore vector $\overrightarrow{e}_1 \times \overrightarrow{AB}$ is perpendicular to vector $\overrightarrow{AB}$ and vector $\overrightarrow{AB} \times(\overrightarrow{AB} \times \overrightarrow{e}_1)$ is perpendicular to both vectors $\overrightarrow{AB}$ and $\overrightarrow{e}_1 \times \overrightarrow{AB}$. Furthermore, the vectors $$\frac{\overrightarrow{e}_1 \times \overrightarrow{AB}}{|\overrightarrow{e}_1 \times \overrightarrow{AB}|} \,\,\, \text{ and }\,\,\, \frac{\overrightarrow{AB} \times(\overrightarrow{AB} \times \overrightarrow{e}_1)}{|\overrightarrow{AB} \times(\overrightarrow{AB} \times \overrightarrow{e}_1)|}$$ are of unit length and are perpendicular to each other, as well as each of them is perpendicular to $\overrightarrow{AB}$. Therefore, the circle we are looking for can be parametrized as
$$\overrightarrow{AC} = \overrightarrow{AH} + |CH| \left(\cos{\theta} \, \frac{\overrightarrow{AB} \times(\overrightarrow{AB} \times \overrightarrow{e}_1)}{|\overrightarrow{AB} \times(\overrightarrow{AB} \times \overrightarrow{e}_1)|} + \sin{\theta} \, \frac{\overrightarrow{e}_1 \times \overrightarrow{AB}}{|\overrightarrow{e}_1 \times \overrightarrow{AB}|}\right).$$ But $$\overrightarrow{AH} = c_b\frac{\overrightarrow{AB}}{|AB|} = \frac{c_b}{c}\overrightarrow{AB}.$$ Thus
$$\overrightarrow{AC} = \frac{c_b}{c}\overrightarrow{AB} + \sqrt{c^2-c_b^2} \left(\cos{\theta} \, \frac{\overrightarrow{AB} \times(\overrightarrow{AB} \times \overrightarrow{e}_1)}{|\overrightarrow{AB} \times(\overrightarrow{AB} \times \overrightarrow{e}_1)|} + \sin{\theta} \, \frac{\overrightarrow{e}_1 \times \overrightarrow{AB}}{|\overrightarrow{e}_1 \times \overrightarrow{AB}|}\right).$$ We know everything in this equation -- we have vectors $\overrightarrow{AB}$ and $\overrightarrow{e_1}$, as well as
\begin{align}
c_b &= \frac{b^2-a^2+c^2}{2c}\\
\sqrt{b^2-c_b^2} &= \frac{\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}}{2c}.
\end{align}
My brief computations show that in numbers
$$
\begin{pmatrix}
C_1\\ C_2 \\ C_3
\end{pmatrix} = \frac{c_b}{c}\begin{pmatrix}
4\\ 7\\ 9
\end{pmatrix} + \sqrt{c^2-c_b^2} \left(\frac{\cos{\theta}}{\sqrt{4749}} \, \begin{pmatrix}
-65\\ 14\\ 18
\end{pmatrix} + \frac{\sin{\theta}}{\sqrt{130}} \, \begin{pmatrix}
0\\ -9\\ 7
\end{pmatrix}\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1893667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find the fourier series of a piecewise continuous function
Consider the function
$f(x) =
\begin{cases}
0 & & x \in [-\pi, 0) \\
1 & & x \in [0, \frac{\pi}{2})\\
0 & & x \in [\frac{\pi}{2}, \pi)\\
\end{cases}$
Calculate its fourier series
I am unsure if this is correct but i have:
$a_0=\frac{1}{2\pi} \int_{-\pi}^{\pi}f(x)dx=\frac{1}{2\pi} \int_{0}^{\frac{\pi}{2}}1dx=\frac{1}{2\pi}[x]_{0}^{\frac{\pi}{2}}=\frac{1}{4}$
$a_n=\frac{1}{\pi} \int_{-\pi}^{\pi}f(x)\cos(nx)dx=\frac{1}{\pi} \int_{0}^{\frac{\pi}{2}}\cos(nx)dx=\frac{1}{\pi}[\frac{1}{n}\sin(nx)]_{0}^{\frac{\pi}{2}}=\frac{1}{n\pi}$
$b_n=\frac{1}{\pi} \int_{-\pi}^{\pi}f(x)\sin(nx)dx=\frac{1}{\pi} \int_{0}^{\frac{\pi}{2}}\sin(nx)dx=\frac{1}{\pi}[\frac{1}{n}(-\cos(nx))]_{0}^{\frac{\pi}{2}}=\frac{-1}{n\pi}$
So we have $f(x)=\frac{1}{4}+\sum_{n=0}^{\infty}[\frac{1}{n\pi} \cos(nx) - \frac{1}{\pi} \sin(nx)]$
Is this correct?
| It should be $$\\ { a }_{ 0 }=\frac { 1 }{ 2 } \\ { a }_{ n }=\frac { \sin { \left( \frac { \pi n }{ 2 } \right) } }{ n\pi } \\ { b }_{ n }=\frac { 1-\cos { \left( \frac { \pi n }{ 2 } \right) } }{ n\pi } \\ \\ \\ $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1894416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Showing a series is convergent I am trying to show
$$ \sum_{n \geq 0} \frac{ 3^n }{\sqrt{2^n + 10^n}} $$
converges. I try to compare as follows
$$ \frac{ 3^n }{\sqrt{2^n + 10^n}} < \frac{ 3^n }{\sqrt{10^n} } = \frac{3^n}{ \sqrt{10}^n } = \left( \frac{ 3 }{ \sqrt{10} } \right)^n $$
Since $\sqrt{10} > 3$, then the $\sum \left( \frac{ 3 }{ \sqrt{10} } \right)^n $ converges and thus the result follows by the comparison test. Is there another method we could have used to show convergence?
| There are other ways forward. Let's invoke the ratio test on the summand $a_n=\frac{3^n}{\sqrt{2^n+10^n}}$. Proceeding we have
$$\begin{align}
\frac{a_{n+1}}{a_n}&=\frac{3^{n+1}}{\sqrt{2^{n+1}+10^{n+1}}}\,\frac{\sqrt{2^n+10^n}}{3^n}\\\\
&=3\sqrt{\frac{2^n+10^n}{2^{n+1}+10^{n+1}}}\\\\
&=\frac{3}{\sqrt{10}}\sqrt{\frac{1+\left(\frac15\right)^n}{1+\left(\frac15\right)^{n+1}}}\\\\
&\to \frac{3}{\sqrt{10}}\,\,\text{as}\,\,n\to \infty\\\\
&<1
\end{align}$$
Therefore, the ratio test guarantees that the series converges.
We can use the root test with similar success. Proceeding here, we have
$$\begin{align}
\sqrt[n]{|a_n|}&=\frac{3}{\sqrt{10}}\sqrt[n]{\frac{1}{\sqrt{1+\left(\frac15\right)^n}}}\\\\
&\to \frac{3}{\sqrt{10}}\,\,\text{as}\,\,n\to \infty\\\\
&<1
\end{align}$$
Therefore, the root test guarantees that the series converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1895675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
How prove this inequality $a^2m_{b}m_{c}+b^2m_{c}m_{a}+c^2m_{a}m_{b}\ge\frac{1}{4}(a^2+b^2+c^2)^2$ If $a,b,$ and $c$ are the side lengths of a triangle, $m_a, m_b,$ and $m_c$ are the lengths of the medians, prove that
$$a^2m_{b}m_{c}+b^2m_{c}m_{a}+c^2m_{a}m_{b}\ge\dfrac{1}{4}(a^2+b^2+c^2)^2$$
$m_a^2 = \dfrac{2b^2+2c^2-a^2}{4}$
but I can't seem to arrive at the desired result.
| We need to prove that
$$\sum\limits_{cyc}m_a^2bc\geq\frac{4}{9}\left(\sum\limits_{cyc}m_a^2\right)^2$$ or
$$\sum\limits_{cyc}(2b^2+2c^2-a^2)bc\geq(a^2+b^2+c^2)^2$$ or
$$\sum\limits_{cyc}(-a^4+2a^3b+2a^3c-2a^2b^2-a^2bc)\geq0$$ or
$$\sum\limits_{cyc}(-2a^4+4a^3b+4a^3c-4a^2b^2-2a^2bc)\geq0$$ or
$$\sum\limits_{cyc}(-2a^4+a^3b+a^3c+2a^3b+2a^3c-4a^2b^2+a^3b+a^3c-2a^2bc)\geq0$$ or
$$\sum\limits_{cyc}(a-b)^2(-a^2-b^2-ab+2ab+ac+bc)\geq0$$ or
$$\sum\limits_{cyc}(a-b)^2(c^2-a^2-b^2+2ab-c^2+ac+bc-ab)\geq0$$ or
$$\sum\limits_{cyc}(a-b)^2(c+a-b)(c+b-a)+\sum\limits_{cyc}(a-b)^2(c-a)(b-c)\geq0$$ or
$$\sum\limits_{cyc}(a-b)^2(c+a-b)(c+b-a)+(a-b)(b-c)(c-a)\sum\limits_{cyc}(a-b)\geq0$$ or
$$\sum\limits_{cyc}(a-b)^2(c+a-b)(c+b-a)\geq0$$
Done!
| {
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"url": "https://math.stackexchange.com/questions/1895897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Question about Spivak's chapter 5 problem 3 part 3 Question:
Find a $\delta$ such that $|f(x)-L|<\epsilon$ for all x satisfying $0<|x-a|<\delta$
Relating equations to the problem;
$$f(x)=x^4+\frac{1}{x} \\a=1, L=2 $$
rewriting the equation as the definition of the limits
$$|x^4+\frac{1}{x}-2|<\epsilon$$ if and only if $$0<|x-1|<\delta$$
As x approaching closer and closer to a $x^4$ approach to 1 and $\frac{1}{x}$ approach to 1 ,realizing it's the addition of limits rewriting the definition as;
$$|x^4+\frac{1}{x}-(1+1)|<\epsilon$$ now I got
$|x^4-1|<\frac{\epsilon}{2}$ if $0<|x-a|<\delta_1$ ,and I got $|\frac{1}{x}-1|<\frac{\epsilon}{2}$ if $0<|x-a|<\delta_2$
starting with $x^4$ factoring I got
$$|(x-1)(x^3+x^2(1)+1^2(x)+1^3)|<\frac{\epsilon}{2}$$
choose $\delta_1=min(1)$ , and I got $0<|x-1|<1$, using triangle inequality $|x|-|1|\le|x-1|<1$, setting $|x|-|1|<1$, and I got $|x|<2$
rewriting the expression $(x^3+x^2(1)+(1^2)(x)+1^3)\le|x^3|+|x^2||1|+|1^2||x|+|1^3|$, sub |x| in the equation I got $|2^3|+|2^2||1|+|1^2||2|+|1^3|=15$, and I just need to choose that $\delta_1= min(1, \frac{\epsilon}{30})$
for the next limit of $\frac{1}{x}$, I know I can choose $\delta_2$ such that
if $$0<|x-1|<\delta_2$$ then $$|\frac{1}{x}-1|<min(\frac{|1|}{2},\frac{\epsilon|1^2|}{4})$$
now choose $\delta=min(\delta_1,\delta_2)$ so I got $\delta=min(1,\frac{1}{2},\frac{\epsilon}{4},\frac{\epsilon}{30})$
but when I looked at the solution it is $\delta=min(1,\frac{1}{2},\frac{\epsilon}{4}, \frac{\epsilon}{8\times2\times2})$
So it is $\frac{\epsilon}{32}$ rather than $\frac{\epsilon}{30}$, could anyone check my math and see where I did wrong, and furthermore, in his final answer Spivak ignore $\frac{\epsilon}{4}$, and he got $\delta=min(\frac{1}{2},\frac{\epsilon}{32})$, can anyone give an explanation on why that is so? Is it because $\frac{\epsilon}{4}$ is bigger than $\frac{\epsilon}{32}$, and it did not matter since $\frac{\epsilon}{32}$ is closer to 1?
|
We can actually achieve a broader bound for $\delta$ through straightforward analysis. To that end, we proceed.
Note that we have
$$\bbox[5px,border:2px solid #C0A000]{x^4+\frac1x-2= (x-1)\left(x^3+x^2+x+1-\frac{1}{x}\right)}$$
Now, suppose we restrict $x$ such that $0<|x-1|<\frac12$. Then, $\frac12 <x<\frac32$. The supremum of $x^3+x^2+x+1-\frac1x$ on $(1/2,3/2)$ is $\frac{179}{24}$. Hence, $\left|x^4+\frac1x\right
|< \frac{179}{24}|x-1|$
Therefore, for all $\epsilon>0$, and $\delta =\min\left(\frac12,\frac{24\epsilon}{179}\right)$,
$$\left|x^4+\frac1x-2\right
|<\epsilon$$
whenever $0<|x-1|<\delta$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1896339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
Ellipse and hyperbola have the same foci I am trying to solve the following problem:
An ellipse and a hyperbola have the same foci, $A$ and $B$, and intersect at four points. The ellipse has major axis $50$, and minor axis $40$. The hyperbola has conjugate axis of length $20$. Let $P$ be a point on both the hyperbola and ellipse. What is $PA \times PB$?
So I say the center of the ellipse is at $(0,0)$ and the equation of the ellipse is $$\frac{x^2}{25^2}+\frac{y^2}{20^2}=1$$
I calculate that the foci of the ellipse are located at $(15,0)$ and $(-15,0)$.
The general equation of a hyperbola is: $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=\pm 1 \quad \cdots \cdots (*)$$ since the length of the conjugate axis is $20$, we can say
$$2a = 20 \implies a = 10$$
Since $a$ is $10$ we search for $b$ with the condition that the hyperbola formed from $(*)$ will have foci at $(15,0)$ and $(-15,0)$.
I get $$b = 5 \sqrt{5}$$
Now plugging into $(*)$ the values I have for $a$ and $b$ and get,
$$\frac{y^2}{125}-\frac{x^2}{100}=-1$$
for the equation of the hyperbola. Now we need one intersection point and for that I used Mathematica and get $$P = \left ( \frac{50}{3}, \frac{20 \sqrt{5}}{3} \right )$$
The whole line of reasoning leads to the following diagram:
With $A$ and $B$ the foci and $P$ one of the points of intersection. I used the distance formula to get the length of $PA$ and $PB$ and got $15$ and $35$ as seen in the diagram. $$15 \times 35 = 525$$
Of course, this is not the answer given, which is $500$. Where did I go wrong?
Thanks to all for their nice solutions.
| Here's a geometric solution.
For any point $P$ on the ellipse, the sum of distances $PA+PB$ to the foci $A$ and $B$ is constant and equals the length of the major axis. So $PA+PB=50$.
Also, for an ellipse the distance $f$ from the center of the ellipse to one of the foci satisfies the equation $f^2+b^2=a^2$, where $2a$ and $2b$ are the lengths of the major and minor axes respectively. So $f^2=25^2-20^2=225$ and therefore $f=15$.
For a hyperbola the distance $f_H$ from the center to one of the foci satisfies the equation $f_H^2=a^2+b^2$, where now $2a$ and $2b$ stand for the lengths of the transverse and conjugate axes respectively. Since the ellipse and hyperbola share the same foci, we know $f_H=f=15$. But we are given that $2b=20$, so $a^2=15^2-10^2=125$ and therefore $a=5\sqrt 5$.
Finally, for a hyperbola the difference of distances $PA-PB$ is constant and equals the length of the transverse axis, i.e., $PA-PB=10\sqrt 5$.
We now have two equations in two unknowns: $PA+PB=50$, and $PA-PB=10\sqrt 5$. Solve these and obtain $PA\cdot PB=500$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1896801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
$\log_{4n} 40\sqrt{3} = \log_{3n} 45$, find $n^3$. $\log_{4n} 40\sqrt{3} = \log_{3n} 45$, find $n^3$.
I can't seem to find any identities to help me in this problem. Any hints or answers? Thanks in advance!
| set $\log_{4n} 40\sqrt{3} = \log_{3n} 45=t$, thus
\begin{cases}
4^t n^t=40\sqrt{3}\\
3^t n^t=45
\end{cases}
we have
$$\left(\frac{4}{3}\right)^t=\frac{8\sqrt{3}}{9}=\frac{8}{3\sqrt 3}\implies t=\frac 32$$
finally
$$8n^{\frac 32}=40\sqrt 3$$
then
$$n^3=75$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1897276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve a system of quadratic equations in two variables for integral solutions. Let's say I have two equations like:
$x^2 - y^2 = 6$ or
$x^2 + y^2 = 6$
What is the best way to solve these sort of equations for finding only positive integral solutions?
| General Case:
These equations belong to the conic sections:
$$
Ax^2 + Bxy + Cy^2 +Dx + Ey + F = 0
$$
If $x$ and $y$ are real values you end up with curves like hyperbolas and circles.
If you restrict the unknowns to integer values, these equations are called Diophantine equations.
An example is Pell's equation:
$$
x^2-ny^2=1
$$
which is equivalent to the type of your first equation.
You might restrict its general solutions to positive ones.
For the second type of equations one ends up with Pythgorean triples:
$$
a^2 + b^2 = c^2
$$
Your Case:
For your given equations there seem to be no solutions according to WA (link and link).
First Equation:
Looking for a solution of
$$
x^2 - y^2 = 6 \\
$$
Case $x \le y$: For $x$ and $y = x + k$ for $k \ge 0$ we have
$$
6 = x^2 - (x + k)^2 = -k^2 - 2kx \le 0
$$
so no solution for $x \le y$.
Case $x > y$: For $x = y + k$ and $k > 0$ we have
$$
6 = (y + k)^2 - y^2 = k^2 + 2ky = f(k,y)
$$
So we simply try out the first few positive integer values for $k$:
$$
f(1, y) = 1 + 2y \ne 6
$$
because the left hand side is odd and the right hand side even.
$$
f(2, y) = 4 + 4 y \ge 8 > 6
$$
As $f$ is increasing with $k$, we have $f(k,y) > 6$ for $k \ge 2$, so no solution here either.
Second Equation:
Looking for a solution of
$$
x^2 + y^2 = 6 \\
$$
We can assume $x \ge y$, otherwise we rename $x$ into $y$ and vice versa.
So we have $x = y + k$ and $k \ge 0$ and get
$$
6 = (y + k)^2 + y^2 = 2y^2 + k^2 + 2k = g(k, y)
$$
We try out the first few non-negative integer values for $k$:
$$
g(0, y) = 2y^2 = 6 \iff y = \pm \sqrt{3} \\
$$
which is no integer.
$$
g(1, y) = 2y^2 + 3 = 6 \iff y^2 = 3/2
$$
which is no integer as well.
$$
g(2, y) = 2 y^2 + 8 \ge 10 > 6
$$
as $g$ is increasing with $k$, it will not get better, so no solution here either.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How can one find the factorization $a^4 + 2a^3 + 3a^2 + 2a + 1 = (a^2 + a + 1)^2$ from scratch? I want to prove $a^4 + 2a^3 + 3a^2 + 2a + 1$ always be perfect square for $a \in \mathbb{N}$. Using GeoGebra I found $a^4 + 2a^3 + 3a ² + 2a + 1 = (a^2 + a + 1)^2$ which is trivial to verify.
Since this equation has no real roots, it's not so easy to decompose it through division by $a-a_0$ where $a_0$ is a root. I don't want to use complex numbers if possible. How do find this factorization from scratch / without using a computer?
| The coefficients $1,2,3,2,1$ are suggestive of the following:
$$\begin{array}{c} & 1a^4 & + & 2a^3 & + & 3a^2 & + & 2a & + & 1 \\ \hline
= & a^4 & + & a^3 & + & a^2 \\ & & + & a^3 & + & a^2 & + & a \\ & & & & + & a^2 & + & a & + & 1 \end{array} $$
$$\begin{array}{l} = & a^2(a^2+a+1)+a(a^2+a+1)+1(a^2+a+1) \\ = & (a^2+a+1)^2. \end{array}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate $\sqrt [3] {7+5\sqrt {2}}+\sqrt [3] {7-5\sqrt {2}}$ i need help.
I tried but I could not(real numbers).
| Let $a$ and $b$ be the two terms of the sum. Note that they are both real numbers, and $a^3 + b^3 = 14$ while $ab = -1$.
$$a^3+b^3 = (a+b)(a^2-ab+b^2) = (a+b)((a+b)^2-3ab)$$
Write $x=a+b$ , so $14 = x(x^2+3)$ or $x^3 + 3x -14 =0$. This equation has only one real solution. Solve it and you are done.
| {
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Short proof for the determinant of a $4$ by $4$ matrix
Prove that $\det \begin{bmatrix}x&y&z&t\\-y&x&-t&z\\-z&t&x&-y\\-t&-z&y&x\end{bmatrix} = (x^2+y^2+z^2+t^2)^2$
I'm looking for an elegant proof that doesn't involve bruteforce.
Since the answer is given, I'm thinking we can argue that the determinant here is a homogeneous polynomial $P(x,y,z,t)$ with degree $4$, that is invariant under $x\to -x$ and permutations of $x,y,z,t$.
As a result, $P(x,y,z,t) = \lambda (x^4+y^4+z^4+t^4) + \delta (x^2y^2+x^2z^2 + x^2t^2+y^2z^2 + y^2t^2 + z^2t^2)$
$\lambda$ and $\delta$ can be found by computing $P(0,0,0,1)$ or some such.
The problem is, it doesn't look easy to prove that $P$ doesn't change under permutation of $x,y,z,t$, neither that it's invariant when the variables are negated.
Can you suggest another short proof, or prove the two claims above ?
| Hint:
$$
\begin{pmatrix}
x & y & z & t \\
-y & x & -t & z \\
-z & t & x & -y \\
-t & -z & y & x
\end{pmatrix}
\begin{pmatrix}
x & -y & -z & -t \\
y & x & t & -z \\
z & -t & x & y \\
t & z & -y & x
\end{pmatrix} = (x^2+y^2+z^2+t^2)
\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}
$$
| {
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Find $\sum_{r=2}^{100}\frac{3^r(2-2r)}{(r+1)(r+2)}$ The question of course seems to be summation of a telescopic series, but despite many attempts am not able to reduce it ...
$$\sum_{r=2}^{100}\frac{3^r(2-2r)}{(r+1)(r+2)}$$ is equal to ??
I've tried converting $2$'s to $(3-1)$ to reduce the expression, but that too didnt work out.
Edit:
I meant that the question seems to be one of 'telescopic method', but its not necessary to do it that way.
For confirmation or as a hint the answer is:
$$\frac{3}{2}-\frac{3^{101}}{101(102)}$$
Other might be useful information is that in this case,
$$\sum_{r=2}^{100}...=\sum_{r=1}^{100}...$$
| With fraction decomposition we get $\frac{(2-2r)}{(r+1)(r+2)}=\frac{4}{r+1}-\frac{6}{r+2}$ (as mentioned by claude-leibovici), so
\begin{align}
\sum_{r=2}^{100}\frac{3^r(2-2r)}{(r+1)(r+2)} &= \sum_{r=2}^{100} 3^r \cdot (\frac{4}{r+1}-\frac{6}{r+2})\\
&= \sum_{r=2}^{100} \frac{3^r \cdot 4}{r+1}-\frac{3^r \cdot 6}{r+2} \\
&= \sum_{r=2}^{100} \frac{3^r \cdot 4}{r+1}-\frac{3^{r+1} \cdot 2}{r+2} \\
\end{align}
I am sure frome here you se the telescopic series.
| {
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Why we use $2\cos\theta=\sqrt3$ to find $x^{72}+x^{66}$ given that $x+\frac1x=\sqrt3$? $x + \frac{1}{x} = \sqrt 3$ then find the value of $x^{72}+x^{66}$.
My teacher said that in this case, where $x + \frac{1}{x}$ is equal to any value which is less than $2$, like $1$ or $\sqrt{3}$, just put this value in equal of $2 \cos \theta$ and after you will get value of $\theta$ after that divide the value $180$ by that theta value and after getting the result put that result in equation $x^n + 1=0$, as "$n$".
Please explain what is basic concept behind this.
| If
$$x+\dfrac{1}{x} = 2\cos\theta$$
then
$$x^{n}+\dfrac{1}{x^n} = 2\cos(n\theta)$$
A way to prove this is with a concept known as mathematical induction, which you may not be ready to do/understand yet. If you wanted to read up on it, attempt it, it would be a little easier to prove the equivalent: If $x^2+1 = 2x\cos\theta$ then $x^N+1 = x^{2n}+1 = 2x^n\cos(n\theta)$.
$$x^{72}+x^{66} = x^{66}(x^6+1)$$
So we are interested in the case where $N=6$.
If $x+\dfrac{1}{x}=\sqrt{3}$ then we solve
$$2\cos(\theta) = \sqrt{3} \iff \cos(\theta) = \dfrac{\sqrt{3}}{2}$$
So $\theta = 30^o$ (there are other possibilities). I suspect the smallest acute angle is what is desired (I really don't understand the necessity of the whole $180^o$ thing, I thought my previous answer was neater and succinct).
$\dfrac{180^o}{30^o} = 6 = N = 2n$.
Since $x^{2}+1 = 2\cos\theta$ implies $x^{2n}+1 = 2x^n\cos(n\theta)$ we have
$$x^6 + 1 = 2x^3\cos(3\cdot 30^o) = 2x^3\cdot\cos(90^o) = 2x^3(0)=0$$
$$x^{72}+x^{66} = x^{66}(x^6+1) = x^{66}(0)=0$$
| {
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Find any rational approximation of $a = \frac{\sin(\sqrt{2})}{\sqrt{2}}$ I need to find any ratinal approximation of $a = \frac{\sin(\sqrt{2})}{\sqrt{2}} $ with accuracy $d = \frac{1}{500}$.
I generally know how to do these type of tasks (using Taylor expansion), but I got problem with this one.
I defined function $f(x) = \begin{cases} 1 \ \mbox{dla} \ x=0 \\ \frac{\sin x}{x} \ \mbox{dla} \ x \neq 0 \end{cases}$, to handle problem with dividng by zero in $f(0)$.
I calculate first and second derivative:
$f'(x) = \frac{x^2 \cdot \cos x - \sin x}{x^2}$
$f''(x) = \frac{x^3 \cdot \sin x - 2x^2 \cdot \cos x + \sin x }{x^4}$
But well, what now? I have no idea how to use the Lagrange remainder theorem here.
| $$\frac{1}{x}\sin x=\frac{1}{x}(x-\frac{x^3}{3!}+\frac{x^5}{5!}...)=1-\frac{x^2}{3!}+\frac{x^4}{5!}...$$
hence
$$=$$
| {
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Finding the roots of a 3rd degree polynomial I found this question from an old math olympiad questionnaire.
Let A, B, and C be the roots of $$x^3-4x-8=0$$ Find the numerical value of the expression $$\frac{A+2}{A-2}+\frac{B+2}{B-2}+\frac{C+2}{C-2}$$
I tried using Rational Roots Theorem. And then, I realized this equation involves complex numbers. Afterwhich, I tried various manipulations. But everytime I do manipulate I either go back from where I started or I went off too far and probably did something wrong in my calculations. I also tried tried doing something like "let u be x squared" and then tried manipulations there. I did found a value for x which are 0,0,+/-2 sqrt of 3, and I know these are wrong, so I gave up.
I do not know how to continue or what else to try. Can anyone help?
| \begin{align*}
\frac{A+2}{A-2}+\frac{B+2}{B-2}+\frac{C+2}{C-2} &= \frac{A-2+4}{A-2}+\frac{B-2+4}{B-2}+\frac{C-2+4}{C-2} \\
&=3 +\frac{4}{A-2}+\frac{4}{B-2}+\frac{4}{C-2}
\end{align*}
The equation with $\alpha - 2, \beta -2, \gamma - 2$ as roots (where $\alpha, \beta, \gamma$ are the roots of the given equation) is
$(y+2)^3 - 4(y+2) - 8 = y^3 + 6y^2 + 8y - 8 = 0$ and we want the sum of the reciprocals of the roots of this equation. This is
$\frac{8}{8} = 1$. Thus
\begin{align*}
\frac{A+2}{A-2}+\frac{B+2}{B-2}+\frac{C+2}{C-2} &= 3+4 = 7
\end{align*}
| {
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Is there any solution such that.. Is there any positive integer solutions of $x$, $y$, $z$ such that both $$\frac{x^2+y^2+z^2}{x+y+z}$$ and $$\frac{x^2+y^2+z^2}{xyz}$$ are integers?
Edit: Forgot to mention that $x$, $y$, $z$ are distinct positive integers.
| Special case
Well, if $x = y = z$ then your equations become
$$\frac{3x^2}{3x} = x$$
$$\frac{3x^2}{x^3} = \frac{3}{x}$$
If they are coupled, then you notice that $x = 1$ is the simplest result. There is also $x = 3$: $3/x$ is an integer if and only if $x = 1$ or $x = 3$.
Unless you count negative integers too, in that case you have also $x = y = z = -1$ and $x = y = z = -3$
| {
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Show that $\sum^{6}_{i=1} a_{i}=\frac{15}{2}$ and $ \sum^{6}_{i=1} a^{2}_{i}=\frac{45}{4} \implies \prod_{i=1}^{6} a_{i} \leq \frac{5}{2}$ Let $a_{i}$, $1 \leq i \leq 6,$ be real numbers such that
$\displaystyle\hspace{1.2 in}\sum^{6}_{i=1} a_{i}=\frac{15}{2}\;\;$ and $\;\;\displaystyle\sum^{6}_{i=1} a^{2}_{i}=\frac{45}{4}$.
Prove that $\hspace{.15 in}\displaystyle\prod_{i=1}^{6} a_{i} \leq \frac{5}{2} $.
I was thinking if I consider the first summation and extended it, it going be pretty long which $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}= \frac{15}{2}$ and the second one like $a^{2}_{1}+a^{2}_{2}+a^{2}_{3}+a^{2}_{4}+a^{2}_{5}+a^{2}_{6}= \frac{45}{4}$. But I do not think this is the shortest of doing that, I am wondering if someone would be able to give me a hint so I can think better than this. Thank you
| Partial answer: Consider the polynomial $$P\left(x\right)=x^{6}+\sum_{m=1}^{6}b_{m}x^{m-1}
$$ where $b_{m}\in\mathbb{R}
$ and assume that $a_{i},\, i=1,\dots,6
$ are the roots of this polynomial. By the Laguerre-Samuleson inequality we have that $$b_{1}=-\sum_{i=1}^{6}a_{i}=-\frac{15}{2}
$$ $$b_{2}=\sum_{i=1}^{6}a_{i}^{2}=\frac{45}{4}
$$ and $$-\frac{b_{1}}{2}+\frac{\sqrt{6b_{2}-b_{1}^{2}}}{2}\leq a_{i}\leq-\frac{b_{1}}{6}+\frac{\sqrt{6b_{2}-b_{1}^{2}}}{2}\sqrt{5}\tag{1}
$$ ($(1)$ and the other inequalities obviously hold for every polynomial of degree $n$) so in our case we have $$0\leq a_{i}\leq\frac{5}{2}
$$ but this not ensure the inequality. So we have to recall the Brunk's inequalities and Boyd-Hawkins inequalities. Let $$s=\frac{\sqrt{6b_{2}-b_{1}^{2}}}{2}.
$$ If we assume that $$a_{1}\geq a_{2}\geq\dots\geq a_{6}
$$ then $$-\frac{b_{1}}{6}+\frac{s}{\sqrt{5}}\leq a_{1}\leq-\frac{b_{1}}{6}+s\sqrt{5},\tag{2}
$$ $$-\frac{b_{1}}{6}-s\sqrt{5}\leq a_{6}\leq-\frac{b_{1}}{6}-\frac{s}{\sqrt{5}}\tag{3}
$$ and $$-\frac{b_{1}}{6}-s\sqrt{\frac{k-1}{6-k+1}}\leq a_{k}\leq-\frac{b_{1}}{6}+s\sqrt{\frac{6-k}{k}},\, k=2,\dots,5.\tag{4}
$$ It is important to note that the equality holds in the RHS of $(2)$ if and only if the equality holds in the RHS of $(3)$ and this is equivalent to $$a_{1}=-\frac{b_{1}}{6}+s\sqrt{5}
$$ $$a_{2}=\dots=a_{6}=-\frac{b_{1}}{6}-\frac{s}{\sqrt{5}}.
$$ In this case the product is exactly $5/2$. In a similar way it is possible to prove that the equality in the RHS of $(4)$ holds if and only if$$a_{1}=\dots=a_{k}
$$ $$a_{k+1}=\dots=a_{6}
$$ hence $$a_{1}=\dots=a_{k}=-\frac{b_{1}}{6}+s\sqrt{\frac{6-k}{k}}
$$ $$a_{k+1}=\dots=a_{6}=-\frac{b_{1}}{6}-s\sqrt{\frac{k}{6-k}}
$$ and again if we maximize a root necessary the product is less or equal to $5/2$. This argument not consider all the possible cases but maybe it may be helpful for a complete proof.
| {
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Probability density of $Y = bX+c$ is $f_Y(x) = \frac{1}{b}f_X(\frac{x-c}{b})$ for $b>0$ We know that if $Y = bX+c$ where $b>0$, its density function with respect to $f_X$ (density of $X$) is: $f_Y(x) = \frac{1}{b}f_X(\frac{x-c}{b})$. Now how can I get the density of $Y$ when $b<0$?
Following the same approach as for case $b>0$, I got ($F_Y$ is the cumulative distribution function of $Y$):
$$F_Y(x) = P(Y\leq x) = P(bX+c\leq x) = (\text{since $b<0$}) =\\
P(X \geq \frac{x-c}{b}) = 1 - P(X< \frac{x-c}{b})=1 - P(X\leq \frac{x-c}{b})=\\
1 - \int_{-\infty}^{\frac{x-c}{b}}f_X(\xi)d\xi=1 - \int_{-\infty}^{x}\frac{1}{b}f_X(\frac{t-c}{b})dt$$
If only I could write the last expression as the integral of something, then I could get $f_Y$.
Update:
As pointed out in the comments, the last part should be $1-\int_{\infty}^x \frac{1}{b}f_X(\frac{t-c}{b})dt$.
Solution: I think I got it. We have $$P(X\geq \frac{x-c}{b}) = \int_{\frac{x-c}{b}}^\infty f_X(\xi)d\xi = \text{(using $\xi = (t-c)/b$ and reversing the limits)}=\int_{-\infty}^x -\frac{1}{b} f_X(\frac{t-c}{b})dt$$
So $f_Y(x) = -\frac{1}{b} f_X(\frac{x-c}{b})$. Note: the expression is positive since $b<0$.
| $$P(X\geq \frac{x-c}{b}) = \int_{\frac{x-c}{b}}^\infty f_X(\xi)d\xi = \text{(using $\xi = (t-c)/b$ and reversing the limits)}=\int_{-\infty}^x -\frac{1}{b} f_X(\frac{t-c}{b})dt$$
So $f_Y(x) = -\frac{1}{b} f_X(\frac{x-c}{b})$.
Note: the expression is positive since $b<0$.
| {
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Prove that $n^3+3n^2+3n+1=2(1+2+...+n)(n+1)+(n+1)^2$ I'm trying to prove that $n^3+3n^2+3n+1=2(1+2+...+n)(n+1)+(n+1)^2$, which is part of a larger proof. We can write:
$n^3+2n^2+n=2(1+2+...+n)(n+1)$
$n^3+2n^2+n=(1+2+...+n)(2n+2)$
$n^3+2n^2+n=(2n+4n+6n+...+2n^2)+(2+4+6+...+n)$
I have no idea how to proceed, though.
| You can simplify by $n+1$ to get
$$(n+1)^2=2(1+2+\cdots n)+(n+1).$$
Then by induction,
$$(n+1)^2-n^2=2n+1=2(1+2+\cdots n)-2(1+2+\cdots n-1)+(n+1)-n.$$
| {
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Inequality with $a+b+c=1$ Let $a,b,c\in\mathbb{R^+}$ such that $a+b+c=1$. Prove that $$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+3(ab+bc+ca)\geq\frac{11}{2}.$$
I am trying to resolve this problem but actually i found some issues:
$$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=3+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{9}{2}$$
I only use Nesbitt's inequality. Then we only need that $3(ab+bc+ca)\geq 1$, but it is not correct, because $3(ab+bc+ca)\leq 1$.
Maybe some ideas will be more that grateful.
| $$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+3(ab+bc+ca)\geq \frac{11}{2}$$
Using $a+b+c=1$, this is equivalent to resolve:
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+3(ab+bc+ca)\geq \frac{5}{2}$$
But we can use MA-MG in the expression:
$$\frac{2}{3}\cdot \frac{a}{b+c}+\frac{3}{2}(ab+ca)=\frac{2a^2}{3a(b+c)}+\frac{3a(b+c)}{2}\geq \sqrt{\frac{2a^2}{3a(b+c)}\cdot \frac{3a(b+c)}{2}}=2a$$
Similar to the above expression:
$$\frac{2}{3}\cdot \frac{b}{c+a}+\frac{3}{2}(ab+bc)\geq 2b$$
$$\frac{2}{3}\cdot \frac{c}{a+b}+\frac{3}{2}(ca+bc)\geq 2c$$
Then we need to prove that
$$\frac{1}{3}\left \lbrace \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right\rbrace \geq \frac{1}{2}$$
And this is Nesbitt's inequality.
| {
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Help simplifying the expression $\bigg(-1+{n+1 \choose 2}\bigg)+{n+1 \choose 2}+\bigg(1+{n+1 \choose 2}\bigg)+\cdots+\bigg(n-2+{n+1 \choose 2}\bigg)$ I need help simplifying the following sum
If $n \geq 2$ then simplify in terms of $n$ -
$$s_n =\bigg(-1+{n+1 \choose 2}\bigg)+{n+1 \choose 2}+\bigg(1+{n+1 \choose 2}\bigg)+\cdots+\bigg(n-2+{n+1 \choose 2}\bigg)$$
Here is my work I am not sure if I made any mistakes:
I know that the terms ${n+1 \choose 2}$ occur exactly $n-times$ in the above expression so after regrouping terms I get
\begin{align}
s_n &=n{n+1 \choose 2}+(-1+0+1+2+\cdots+ (n-2))\\
& = -1+n{n+1 \choose 2} + (1+2+\cdots+ (n-2))\\
& = -1+n{n+1 \choose 2}+{n-1 \choose 2}\\
& = -1+{n^3+2n^2-3n+2 \above 1.5pt 2}\\
& ={n(n^2+2n-3) \above 1.5pt 2}\\
& ={n(n+3)(n-1) \above 1.5 pt 2}
\end{align}
The sequence is $s_n =5,18,42,80,135,210,308\ldots$ and we can compare that to this sequence A212343. The motivation here are the sums of the following sets $\{2,3\}$, $\{5,6,7\}$, $\{9,10,11,12\}$ and so on.
|
The calculation is correct. Using the summation symbol and a slightly different calculation
we obtain
\begin{align*}
s_n&=\sum_{j=-1}^{n-2}\left(j+\binom{n+1}{2}\right)\\
&=\sum_{j=1}^n\left(j-2+\binom{n+1}{2}\right)\tag{1}\\
&=\left(\sum_{j=1}^nj\right)-2n+n\binom{n+1}{2}\\
&=\frac{1}{2}n(n+1)-2n+n\frac{1}{2}(n+1)n\tag{2}\\
&=\frac{1}{2}n\left(n^2+2n-3\right)
\end{align*}
in accordance with your result.
Comment:
*
*In (1) we shift the index $j$ to start from $1$.
*In (2) we use the summation formula $\sum_{j=1}^nj=\frac{1}{2}n(n+1)$.
| {
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"source": "stackexchange",
"question_score": "2",
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Determine the coefficient of $xy$ in the expansion of $(x+y+2)^7$ The problem: Determine the coefficient of $xy$ in the expansion of $(x+y+2)^7$
My approach:
We can rewrite the equation substituting $x+y =j$
$$(x+y+2)^7=(j+2)^7$$
This is simpler because we know the coefficients thanks to the formula:
$$(a+b)^{n}=\sum _{{k=0}}^{n}{n \choose k}a^{{n-k}}b^{{k}} $$
With $n=7$ we have $1,7,21,35,35,21,7,1$ as coefficients and the expansion looks like this:
$$j^7+14\cdot j^6+(21\cdot 2^2) j^5+(35\cdot 2^3) j^4+(35\cdot 2^4) j^3+(21\cdot 2^5) j^2+(7\cdot 26)j+2^7$$
Now the only time that $xy$ appears is in the expansion of $j^2$ therefore we have:
$$(21\cdot 2^5) j^2=(21\cdot 2^5) (x+y)^2= (21\cdot 2^5)(x^2+2xy+y^2)=21\cdot 2^5x^2+21\cdot 2^6xy+21\cdot 2^5y^2$$
The coefficient is $21\cdot 2^6$
Is this correct? Is there a simpler proof?
| Yes to both questions.
We can write:
$$(x+y+2)^7=\sum_{p+q+r=7}\binom7{p,q,r}2^rx^py^q$$
To get $xy$ we make $p=q=1$, so the coefficient of $xy$ is:
$$\binom7{5,1,1}2^5=\frac{7!}{5!1!1!}\cdot 2^5=42\cdot 2^5=21\cdot2^6$$
| {
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Find the range of change number $c=\frac{\text{Im } z}{\text{Re } z}$ if $|z-3-4i|\le 1$ I am working on the following problem:
Find the range of change of the numbers $c=\frac{\text{Im }z}{\text{Re }z}$ if $|z-3-4i|\le 1$
This is what I did:
Letting $z=x+iy$, I can write from $|z-3-4i|\le 1$ $$(x-3)^2+(y-4)^2\le 1$$ From that it can be written $$2\le x\le 4$$ and $$3\le y\le 5$$ Therefore $$\frac{3}{4}\le c\le\frac{5}{2}$$
Is there anything wrong with my work? Any help would be greatly appreciated.
| We have a circle centered at $(3,4)$ of radius $1.$
construct the line between the origin and the center of the circle.
Let $\theta$ be the angle that this line forms with the x axis.
$\tan \theta = \frac 43$
Construct the line between the origin and the tangents.
Let $\phi$ be the angles between each tangent and the line to the center of the circle.
$\sin \phi = \frac 15\\
\cos \phi = \sqrt {1-\sin^2\phi} = \frac {\sqrt {24}}{5}\\
\tan \phi = \frac {\sqrt 6}{12}$
$c = [\tan (\theta - \phi), \tan (\theta + \phi)]$
$\tan (\theta + \phi) = \frac {tan \theta + \tan \phi}{1-\tan\theta \tan\phi} = \frac {48+3\sqrt{6}}{36 - 4\sqrt {6}} = \frac {6+\sqrt 6}{4}\\
\tan (\theta - \phi) = \frac {tan \theta - \tan \phi}{1+\tan\theta \tan\phi} = \frac {48-3\sqrt 6}{36+4\sqrt 6}= \frac {6-\sqrt 6}{4}$
| {
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} |
inequality for positive number $a+b+c=3$ Let $a,b,c$ be positive numbers such that $a+b+c=3$. Prove that
$$\frac{2-\sqrt{a}}{\sqrt{c+3a}}+\frac{2-\sqrt{b}}{\sqrt{a+3b}}+\frac{2-\sqrt{c}}{\sqrt{b+3c}}\:\ge \:\frac{3}{2}.$$
| By AM-GM and C-S we obtain:
$$\sum_{cyc}\frac{2-\sqrt{a}}{\sqrt{c+3a}}=\sum_{cyc}\frac{2(4-2\sqrt{a})}{2\sqrt{4(c+3a)}}\geq\sum_{cyc}\frac{2(4-a-1)}{4+c+3a}=$$
$$=\sum_{cyc}\frac{6(b+c)}{4(a+b+c)+3c+9a}=\sum_{cyc}\frac{6(b+c)}{13a+4b+7c}=$$
$$=\sum_{cyc}\frac{6(b+c)^2}{(b+c)(13a+4b+7c)}\geq\frac{6\left(\sum\limits_{cyc}(b+c)\right)^2}{\sum\limits_{cyc}(b+c)(13a+4b+7c)}=$$
$$=\frac{24(a+b+c)^2}{\sum\limits_{cyc}(11a^2+37ab)}\geq\frac{24(a+b+c)^2}{\sum\limits_{cyc}(16a^2+32ab)}=\frac{3}{2}.$$
Done!
| {
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If we know the GCD and LCM, how to find all unordered pairs? If we know that $\gcd(a,b)=12$ and $\operatorname{lcm}(a,b)=360$, how can we find all unordered pairs $a,b$?
The answer in the back of the book I got this from gives $(12,360), (24,180), (36,120), (60,72)$ as the solutions.
| If $\gcd(a,b) = 12$, then $12 \mid a$ and $12 \mid b$. Hence $\dfrac{a}{12}$ and $\dfrac{b}{12}$ are integers and
$\gcd\left( \dfrac{a}{12}, \dfrac{b}{12} \right) = 1$
. Hence...
\begin{align}
ab &= \gcd(a,b) \times \operatorname{lcm}(a,b) \\
ab &= 12 \times 360 \\
\dfrac{a}{12} \dfrac{b}{12} &= \dfrac{360}{12} \\
\dfrac{a}{12} \dfrac{b}{12} &= 30 \\
\end{align}
So, if you are looking for unordered pairs,
$\left(\dfrac{a}{12}, \dfrac{b}{12} \right)
\in \{ (1,30), (2,15), (3,10), (5,6)\}$
Note that the condition $\gcd\left( \dfrac{a}{12}, \dfrac{b}{12} \right) = 1$ has been satisfied in each case.
So,
$(a,b) \in \{ (12,360), (24,180), (36,120), (60, 72)\}$
| {
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What is this number: $x = \sqrt{1+2^1\sqrt{1+2^2\sqrt{1+2^3\sqrt{1+...}}}}\approx 4.14031...$? I was watching this Mathologer video about Ramanujan's nested radical identity
$$
3 = \sqrt{1 + 2\sqrt{1+3\sqrt{1+4\sqrt{1+...}}}}\, ,
$$
and I decided to look at this for different sets of "radical coefficients". Using powers of $2$ yields the quantity
$$
x = \sqrt{1+2^1\sqrt{1+2^2\sqrt{1+2^3\sqrt{1+...}}}}\, .
$$
I can't figure out much about the nature of the number $x$. I can't even prove that it exists, let alone anything analytically about its value. Nevertheless, by looking at the truncation of this expression at different levels in Mathematica, it seems to be the case that:
1) The expression converges relatively quickly.
2) $x \approx 4.14031456214125981180937290...$
I tried feeding this number into an Inverse Symbolic Calculator... to no avail, although $x$ does match about 8 digits of
$$
\frac{5}{4}\, \frac{\sqrt{3} + \sqrt{5}\ln(5)}{\ln(5)}\, .
$$
Can anyone tell me anything else about this quantity?
Edited to add:
I note that the apparently simpler Nested Radical Constant
$$
\sqrt{1 + \sqrt{2 + \sqrt{3 + ...}}}
$$
has no closed form, so we clearly have no reason to expect my quantity $x$ to have a closed form either...
Edit 2:
There was an answer that someone posted, and then deleted for some reason, which gave a method for getting a lower bound on this quantity. A variation of that method leads to:
\begin{align}
x &= \sqrt{1+2^1\sqrt{1+2^2\sqrt{1+2^3\sqrt{1+...}}}}\\
&> \sqrt{1+2^1\sqrt{0+2^2\sqrt{0+2^3\sqrt{0+...}}}}\\
&= \sqrt{1+2^1 \times 2^{2/2}\times 2^{3/4}\times 2^{4/8}\times 2^{5/16}...}\\
&= \sqrt{1 + 2^{\sum_{m=0}^\infty \frac{m+1}{2^m}}}\\
&= \sqrt{1 + 2^4} = \sqrt{17} \approx 4.1231...
\end{align}
This is apparently a relatively tight lower bound.
| What does
$$
x = \sqrt{1+2^1\sqrt{1+2^2\sqrt{1+2^3\sqrt{1+...}}}}\, .
$$
mean? I interpret it as the limit of this sequence:
\begin{align}
x_1 &= \sqrt{1 + 2^1} \\
x_2 &= \sqrt{1 + 2^1 \sqrt{1 + 2^2}} \\
x_3 &= \sqrt{1 + 2^1 \sqrt{1 + 2^2 \sqrt{1 + 2^3}}} \\
x_4 &= \sqrt{1 + 2^1 \sqrt{1 + 2^2 \sqrt{1 + 2^3 \sqrt{1 + 2^4}}}} \\
\end{align}
Comparing $x_{n+1}$ to $x_n$:
\begin{align}
x_{n+1}
&= \sqrt{1 + 2^1 \sqrt{1 + 2^2 \sqrt{1 + 2^3 \dotsb \sqrt{1 + 2^n \sqrt{1 + 2^{n+1}}}}}} \\
&> \sqrt{1 + 2^1 \sqrt{1 + 2^2 \sqrt{1 + 2^3 \dotsb \sqrt{1 + 2^n \sqrt{1}}}}} = x_n \\
\end{align}
so the sequence is strictly increasing.
I have not found a majorant sequence yet. First numerical results seem to suggest convergence.
One can calculate $x_n$ via the sequence
\begin{align}
y_{n, 0} &= 1 \\
y_{n, k+1} &= \sqrt{1 + 2^{n-k} y_{n, k}} \quad (k \in \{ 0, \dotsc, n-1 \} ) \\
x_n &= y_{n, n}
\end{align}
This allows to interpret the evaluation of the term for $x_n$ as a series of iterations against the functions
$$
r_{n-k}(x) = \sqrt{1 + 2^{n-k} x}
$$
which varies for $d = n-k$ from $n$ down to $1$.
The images below show the iterations for $x_3$ and $x_8$:
In this view the fixed points of the $r_d$ would influence the iteration.
$$
x_1^* = 1 + \sqrt{2} = 2.4142\dotsc \\
x_2^* = 2 + \sqrt{5} = 4.2361\dotsc \\
$$
| {
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Elegant way to find the roots of a Quartic I need to find the roots of this quartic: $y=x^4+6x^3+11x^2+6x-24024$ in order to be able to solve my problem.
I'm wondering if there's a nice easy way to find its roots (such as substituting $x$ with something). I do know the Rational Root Theorem, but I feel like listing out the factors of $24024$ would take too long.
| You may notice that
$$ x^4+6x^3+11x^2+6x-24024 = (x^2+3x+1)^2-24025 = (x^2+3x+1)^2-155^2 $$
by looking for the square of a second degree polynomial that matches the coefficients pattern $1,6,11$. It goes so smooth that the identity $a^2-b^2=(a-b)(a+b)$ also provides a factorization:
$$\boxed{ x^4+6x^3+11x^2+6x-24024 = \color{blue}{(x-11)(x+14)}\color{green}{(x^2+3x+156)}.}$$
The fact that $24024$ is exceptionally close to an integer square and $x^4+6x^3+11x^2+6x+1$ is a palyndromic polynomial (hence $x^2$ times a quadratic polynomial in the variable $x+\frac{1}{x}$) and a square leads to an alternative, equivalent solution.
| {
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Why $4a^3+27b^2<0 \iff x^3+ax+b=0$ has exactly three solutions? According to Exc. 8 Sec. 4.3 of the book Advanced Calculus by Fitzpatrick,
For numbers $a$ and $b$, prove that the following equation has exactly three solutions if and only $4a^3+27b^2<0$: $$ x^3+ax+b=0, \ \ \ \ \ x \ \text{in} \ \mathbb R.$$
Long time struggling, I can't prove neither directions. The book is pretty rudimentary and discussions like here are beyond the scope of the book.
Let $f(x) = x^3 + ax + b$. So $f'(x) = 3x^2 + a$. Set $3x^2 + a = 0$, so that at the points $\sqrt{-a/3}$ and $-\sqrt{-a/3}$, $f'(x) = 0$. There exists exactly two maximum/minimum due to the existence of $3$ solutions. These $2$ max/min must be at $x = \sqrt{-a/3}$ and $x = -\sqrt{-a/3}$. Note that must $a<0$. Also must $f(-\sqrt{-a/3}) > 0$ and $f(\sqrt{-a/3})<0$. Putting $\pm \sqrt{-a/3}$ into $f(x)$, I fail to conclude $4a^3 + 27b^2 < 0$. As, $$-(-a/3)^{3/2}-a(-a/3)^{1/2}+b>0 \\ (-a/3)^{3/2}+a(-a/3)^{1/2}+b<0$$ results in $a<0$ and no more thing!
Please help!
| For the general cubic, the theory goes like this:
Let $p(x)=x^3+ax^2+bx+c$
Then $\dfrac{dp}{dx}=3x^2+2ax+b$
We need the two solutions of $\dfrac{dp}{dx}=0$, namely $r_1, r_2$, and that $p(r_1)p(r_2)\lt0$, i.e. the two stationary points are on either side of the $x$-axis.
Solving for the stationary points, $\dfrac{-2a\pm\sqrt{4a^2-12b}}{6}=\dfrac{-a\pm\sqrt{a^2-3b}}{3}$
A few preliminary results:
$r_1r_2=\dfrac b 3\qquad;\qquad r_i^2=\dfrac{2a^2-3b\pm2a\sqrt{a^2-3b}}{9}$
Putting it all together, we require:
$(r_1(r_1^2+ar_1+b)+c)(r_2(r_2^2+ar_2+b)+c)\lt0$
If we let $a=0$ we get Jack's result.
| {
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Show that $\gcd(a+b,a-b)=1\iff\gcd(a^2+b^2,a^2-b^2)=1.$
Problem. Let $a$ and $b$ be two integers. Show that
$$\gcd(a+b,a-b)=1\iff\gcd(a^2+b^2,a^2-b^2)=1.$$
I am really clueless for how to approach this problem. First let's try $(\Rightarrow)$. We can write
$$x(a+b)+y(a-b)=1.$$
Then I tired to use the identity $a^2+b^2=(a+b)^2-2ab$ but it doesn't lead anywhere.
| Let $p$ be a prime number distinct of $2$ suppose that $p$ divides $a+b$ and $p$ divides $a-b$, then $p$ divides $(a+b)(a-b)=a^2-b^2$ and $p$ divides $(a-b)^2+(a+b)^2=2(a^2+b^2)$ and thus $p$ divides $a^2+b^2$.
If $a+b$ and $a-b$ are even, $a$ and $b$ have the same parity. This implies that $a^2$ and $b^2$ have the same parity and $a^2+b^2, a^2-b^2$ are even. Thus $gcd(a^2+b^2,a^2-b^2)=1$ implies that $gcd(a+b,a-b)=1$.
Let $p$ be a prime, suppose that $p$ divides $a^2-b^2$ and $p$ divides $a^2+b^2$, $p$ divides $(a^2-b^2)+(a^2+b^2)=2a^2$, it divides also $2b^2$, thus $p$ divides $2a$ and $2b$.
It divides $a^2-b^2=(a+b)(a-b)$, if $p$ divides $a+b$, it divides $(a+b) -2b=a-b$ if $p$ divides $a-b$, it divides $(a-b)+2b=a+b$. Thus $p$ divides $a+b$ and $p$ divides $a-b$, we deduce that $gcd(a+b,a-b)=1$ implies that $gcd(a^2+b^2,a^2-b^2)=1$.
| {
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Sketching a graph using completing the square For the question $f(X) = X^2 + 4KX + (3+11K)$, where $K$ is a constant.
given that $K = 1$,
Sketch the graph $y=f(X)$, showing the coordinates of any point at which the graph crosses a coordinate axis.
)"
I get
... $(x+2)^2 - 4^2 + (3 + 11)$ which is equivalent to $(x+2)^2 + 10$
and i get $(-2, 10)$ for the minimum coordinates, however i do not get how the $-2$ value was achieved for the first coordinate value.
| The reason why we perform the completing square trick is to change the original to the form $y = (x+h)^2 + k$ so that we can deduce the minimum of y.
Since the squared term is always positive, the minimum value of it is 0. This occurs when x = -h. In your case, h = 2.
Answering part
Performing completing square, we have $y = x^2 + 4x + \4 - \4 +(3+11) = (x+2)^2 - 4 + (3+11)= (x+2)^2 + 10$.
The -2 comes directly from setting $x + 2 = 0$.
| {
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How is this property of real numbers proven?
Prove that $\dfrac{x^2}{(x − 1)^2} + \dfrac{y^2}{(y − 1)^2} + \dfrac{z^2}{(z − 1)^2} ≥ 1$ for all real numbers $x, y, z$, each different from $1$ and satisfying $xyz = 1$.
How do I prove this?
| Since,$xyz=1$,we have any two of x,y,z is negative or all must be positive and in both case all three are non zero.
CASE 1:
If all x,y,z are positive I hope that you may easily prove it.
CASE 2:
if x,y is negative then we have $$x>x-1$$
But $$ x^2≤(x-1)^2.$$since if we assume $x=\frac{1}{2} $ then$x^2=(x-1)^2$
Therefore $$ 0<\frac{x^2}{(x-1)^2}<1$$
Similarly it follows for y,
But for z we have $$\frac{z^2}{(z-1)^2}≤1$$if $z≤\frac{1}{2}$,else z>1
therfore totally we have $$ \frac{x^2}{(x-1)^2}+
\frac{y^2}{(x-1)^2}
+
\frac{z^2}{(z-1)^2}
≥1 $$
hence proved.
| {
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"timestamp": "2023-03-29T00:00:00",
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Compute $\sum\limits_{n=1}^\infty \frac{5+4n-1}{3^{2n+1}}$ I have troubles finding the limit of the following series: $\sum_{n=1}^\infty \frac{5+4n-1}{3^{2n+1}}$
So far I figured it'd easier to split the sum into:
$\sum_{n=1}^\infty \frac{5}{3^{2n+1}} \sum_{n=1}^\infty \frac{4n-1}{3^{2n+1}}$
= $\sum_{n=1}^\infty 5 \cdot\frac{1}{3^{2n+1}} +\sum_{n=1}^\infty 4n-1 \cdot \frac{1}{3^{2n+1}}$
And with $\sum_{n=1}^\infty \frac{1}{w^n} = \frac{1}{w-1}$ you get the following terms:
$5\cdot \frac{1}{3^{2n+1}-1} + 4n-1\cdot \frac{1}{3^{2n+1}-1}$
which is bascially a sequence, but im stuck right here.. help is very appreciated!
| depend on the geometric series , we can get the following to compute the sum
$${\frac {x^2}{1-x^2}}=\sum _{n=1}^{\infty }x^{2n}\quad {\text{ for }}|x|<1\!$$
$${\displaystyle {\frac {x^2}{(1-x^2)^{2}}}=\sum _{n=1}^{\infty }nx^{2n}\quad {\text{ for }}|x|<1\!}$$
| {
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Prove $\sum_{k=0}^{n-1} {n-1 \choose k}cos(\frac{2\pi k}{n}+\frac{2\pi}{n})(-1)^k \neq 0$ Let $f(x)$ be a polynomial such that $f(N)=cos((N-1)\frac{2\pi}{n})$ and $N\in\{1,2,3,...n\}$. I am interested in finding the degree of $f(x)$. Using the method of successive differences, I found the $n-1$ difference to be $$D_{n-1}=\sum_{k=0}^{n-1} {n-1 \choose k}cos(\frac{2\pi k}{n}+\frac{2\pi}{n})(-1)^k$$
I am attempting to prove $D_{n-1}\neq 0$ for $n=1,2,3,...$.
So far I have attempted to take the limit as $n\rightarrow\infty$ and checked $n$ values up to $30$. Neither of these have been useful.
| Let's abbreviate $\omega_n = e^{2\pi i/n}$, so that $D_{n-1}$ is the real part of $$A_{n-1} = \sum_{k=0}^{n-1} \binom{n-1}{k} \omega_n^{k+1} (-1)^k = \omega_n \sum_{k=0}^{n-1} \binom{n-1}{k} (-\omega_n)^k.$$
Now by the binomial theorem, this is just $\omega_n (1-\omega_n)^{n-1}$.
For any number $t$ we have $(1 + e^{it})^2 = (2 + 2 \cos t) e^{it} = 4 \cos^2 (t/2) e^{it}$, so if $-\pi \le t \le \pi$ you get $1 + e^{it} = 2 \cos(t/2) e^{it/2} $. In particular, $-\omega_n = e^{i(2\pi/n-\pi)}$, so
$$ 1 - \omega_n = 2 \sin \frac{\pi}{n} \exp\left[ \left(\frac{1}{n} - \frac{1}{2} \right) \pi i \right]. $$
Thus,
\begin{align}
A_{n-1}
&= \left (2 \sin \frac{\pi}{n} \right)^{n-1} \exp\left[ \frac{2\pi i}{n} + (n-1) \left( \frac{1}{n} - \frac{1}{2} \right) \pi i \right] \\
&= \left (2 \sin \frac{\pi}{n} \right)^{n-1} \exp\left[ \left( \frac{3-n}{2} + \frac{1}{n} \right) \pi i \right]. \\
\end{align}
Finally,
$$ D_{n-1} = \operatorname{Re} A_{n-1} = \left (2 \sin \frac{\pi}{n} \right)^{n-1} \cos \left[ \left( \frac{3-n}{2} + \frac{1}{n} \right) \pi \right] $$
which is always nonzero.
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Prove $\sqrt{3} + \sqrt{5}$ is irrational If it is assumed that $\sqrt{3}$ is known to be irrational (not the case for $\sqrt{5}$), then prove that $\sqrt{3}+\sqrt{5}$ is irrational.
My approach:
Assume that $\sqrt{3}+\sqrt{5}$ is rational. Then there exist coprime integers $p$ and $q$ so that $\frac{p}{q}$ is rational and $\frac{p}{q}=\sqrt{3}+\sqrt{5}$. Thus $(\sqrt{3}+\sqrt{5})^2=2(4+\sqrt{15})=\frac{p^2}{q^2}$, which implies that $p^2$ is even, so $p$ is also even. Let $p:=2m$ for some integer $m$, then $p^2=4m^2$. Thus $q^2(4+\sqrt{15})=2m^2$, which implies that $q^2$ is even, and so $q$ is even. But this contradicts that $p$ and $q$ are coprime, and we arrive at a contradiction.
My way of proving this does not use the fact that $\sqrt{3}$ is
irrational. Please let me know if my proof is correct, and how to use
the above mentioned fact.
| This is from here:
Prove that $\sqrt{2}+\sqrt{3}$ is irrational.
More generally,
suppose
$r
=\sqrt{a}+\sqrt{b}
$
is rational,
where
$a$ and $b$
are positive integers.
Then
$r(\sqrt{a}-\sqrt{b})
=a-b
$
so
$\sqrt{a}-\sqrt{b}
=\dfrac{a-b}{r}
$
is also rational.
Adding and subtracting these,
$\sqrt{a}$ and $\sqrt{b}$
are rational.
Therefore,
if either or both
of
$\sqrt{a}$ and $\sqrt{b}$
are irrational,
then
$\sqrt{a}+\sqrt{b}$
is irrational
(and similarly for
$\sqrt{a}-\sqrt{b}$).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Does this exhibit a parameterization of $S^1 \times S^1$? Guillemin and Pollack asks:
Explicitly exhibit enough parameterizations to cover $S^1 \times S^1 \subset R^4$.
My solution is given by $8$ parameterizations (closely following their example of parameterizing a circle):
$$f_1(x,z) = (x, \sqrt{1 - x^2}, z, \sqrt{1 - z^2})$$
$$f_2(x,z) = (x, -\sqrt{1 - x^2}, z, \sqrt{1 - z^2})$$
$$f_3(x,z) = (x, \sqrt{1 - x^2}, z, -\sqrt{1 - z^2})$$
$$f_4(x,z) = (x, -\sqrt{1 - x^2}, z, -\sqrt{1 - z^2})$$
$$f_5(y,w) = (y, \sqrt{1 - y^2}, w, \sqrt{1 - w^2})$$
$$f_6(y,w) = (y, -\sqrt{1 - y^2}, w, \sqrt{1 - w^2})$$
$$f_7(y,w) = (y, \sqrt{1 - y^2}, w, -\sqrt{1 - w^2})$$
$$f_8(y,w) = (y, -\sqrt{1 - y^2}, w, -\sqrt{1 - w^2})$$
I believe these are all I would need In terms of parameterizations but I am unsure, can anyone confirm or deny this?
| As pointed out by John Hughes and Andrew Hwang I have not considered every parameterization needed and made some mistakes in the existing parameterization. I have included an additional parameterizations and fixed the existing ones:
$$f_1(x,z) = (x, \sqrt{1 - x^2}, z, \sqrt{1 - z^2})$$
$$f_2(x,z) = (x, -\sqrt{1 - x^2}, z, \sqrt{1 - z^2})$$
$$f_3(x,z) = (x, \sqrt{1 - x^2}, z, -\sqrt{1 - z^2})$$
$$f_4(x,z) = (x, -\sqrt{1 - x^2}, z, -\sqrt{1 - z^2})$$
$$f_5(y,w) = (\sqrt{1 - y^2}, y, \sqrt{1 - w^2}, w)$$
$$f_6(y,w) = (-\sqrt{1 - y^2}, y, \sqrt{1 - w^2}, w)$$
$$f_7(y,w) = (\sqrt{1 - y^2}, y, -\sqrt{1 - w^2}, w)$$
$$f_8(y,w) = (-\sqrt{1 - y^2}, y, -\sqrt{1 - w^2}, w)$$
$$f_9(x,w) = (x, \sqrt{1 - x^2}, w, \sqrt{1 - w^2})$$
$$f_{10}(x,w) = (x, -\sqrt{1 - x^2}, w, \sqrt{1 - w^2})$$
$$f_{11}(x,w) = (x, \sqrt{1 - x^2}, w, -\sqrt{1 - w^2})$$
$$f_{12}(x,w) = (x, -\sqrt{1 - x^2}, w, -\sqrt{1 - w^2})$$
$$f_{13}(y,z) = (\sqrt{1 - y^2}, y, \sqrt{1 - z^2}, z)$$
$$f_{14}(y,z) = (-\sqrt{1 - y^2}, y, \sqrt{1 - z^2}, z)$$
$$f_{15}(y,z) = (\sqrt{1 - y^2}, y, -\sqrt{1 - z^2}, z)$$
$$f_{16}(y,z) = (-\sqrt{1 - y^2}, y, -\sqrt{1 - z^2}, z)$$
Note: $f_9$, $f_{10}$ ... are unnecessary, as an appropriate variable substitution can be made.
| {
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"url": "https://math.stackexchange.com/questions/1941519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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I conjecture this inequality $\sqrt[4]{\frac{(xy+yz+xz)(x^2+y^2+z^2)}{9}}\ge\sqrt[3]{\frac{(x+y)(y+z)(z+x)}{8}}$ Let $x,y,z>0$,prove or disprove
$$\sqrt[4]{\dfrac{(xy+yz+xz)(x^2+y^2+z^2)}{9}}\ge\sqrt[3]{\dfrac{(x+y)(y+z)(z+x)}{8}}$$
I tried many times ,use
$$x^2+y^2+z^2\ge\dfrac{1}{3}(x+y+z)^2$$
and kown $$9(x+y)(y+z)(z+x)\ge 8(x+y+z)(xy+yz+xz)$$
| This inequality is nice. And without any advanced techniques/theorems, it is also hard.
The technique that Michael used in his solution was developed more than ten years ago (credit goes to Nguyen Anh Cuong who was, to the best of my knowledge, the first to introduce the idea, under the name "ABC method"). The general idea is the following.
Suppose that we need to prove $F(p,q,r)\ge 0$ where $p=x+y+z,q=xy+yz+zx,r=xyz$ and $x,y,z\ge 0$. If $f(r):= F(p,q,r)$ is a concave function (w.r.t. $r$) then it suffices to prove the inequality for $(x-y)(y-z)(z-x)=0$ or $xyz=0$.
Let me introduce another nice and "simple" (not quite!) solution.
We will show that, for any $x,y,z\ge 0$:
$$\sqrt[4]{\frac{(xy+yz+xz)(x^2+y^2+z^2)}{9}}\ge \sqrt[4]{\frac{(x+y)(y+z)(z+x)(x+y+z)}{24}} \ge \sqrt[3]{\frac{(x+y)(y+z)(z+x)}{8}}$$
Although these new inequalities are tighter, they are much easier to solve.
Indeed:
*
*The left-hand side inequality is equivalent to
$$\boxed{8(xy+yz+xz)(x^2+y^2+z^2) \ge 3(x+y)(y+z)(z+x)(x+y+z)}$$
(boxed, because I believe this is an important lemma that Olympiad students should keep in mind). Denote $A=xy(x^2+y^2)+yz(y^2+z^2)+zx(z^2+x^2),B=x^2y^2+y^2z^2+z^2x^2$ and $C=xyz(x+y+z)$ then the above inequality becomes $$8(A+C) \ge 3(A+2B+4C),$$
which is true because $A\ge 2B\ge 2C$.
*The right-hand side inequality is equivalent to $$8(x+y+z)^3\ge 27(x+y)(y+z)(z+x),$$ which is just AM-GM: $(a+b+c)^3 \ge 27abc$ where $a=x+y,b=y+z,c=z+x$.
We are done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $\int_0^{\infty}\frac{\ln (x)}{(x^2+1)(x^3+1)}\ dx=-\frac{37}{432}\pi^2$ with real method I came across the following integral:
$$\large{\int_0^\infty \frac{\ln (x)}{(x^2+1)(x^3+1)}\ dx=-\frac{37}{432}\pi^2}$$
I know it could be solved with resuide method, and I want to know if there are some real methods can sove it?
Meanwhile,I remember a similar integral:
$$\large{\int_0^\infty \frac{1}{(x^2+1)(x^a+1)}\ dx=\frac{\pi}{4}}$$
And I want to know the following one:
$${\color{red}{\large{\int_0^\infty \frac{\ln x}{(x^2+1)(x^a+1)}\ dx = \huge{?}}}}$$
Using the Mathematica I got the follow result.
Could you suggest some ideas how to prove this? Any hints will be appreciated.
| A real-analytic technique may be to exploit
$$ I = \int_0^1 \frac{1-x^3}{(1+x^2)(1+x^3)}\log(x)\,dx $$
then expand $f(x)=\frac{1-x^3}{(1+x^2)(1+x^3)}$ as a Taylor series around $x=0$,
$$ f(x) = \sum_{n\geq 0}\left(1-x^2-2 x^3+x^4+2 x^5+x^6-2 x^7-x^8 + x^{10} \right) x^{12n} $$
and exploit
$$ \int_0^1 x^k\log(x)\,dx = -\frac{1}{(k+1)^2} $$
to convert $I$ into a combination of Dirichlet $L$-functions $L(s,\chi)$ where $s=2$ and $\chi$ is a Dirichlet character $\!\!\pmod{12}$, or a combination of trigamma functions evaluated at multiples of $\frac{1}{12}$: the reflection formula for $\psi'$ is helpful, since it gives
$$ \sum_{n\geq 0}\left(\frac{1}{(12n+1)^2}+\frac{1}{(12n+11)^2}\right) = \frac{2+\sqrt{3}}{36}\,\pi^2,$$
$$ \sum_{n\geq 0}\left(\frac{1}{(12n+3)^2}+\frac{1}{(12n+9)^2}\right)=\frac{1}{72}\,\pi^2,$$
$$ \sum_{n\geq 0}\left(\frac{1}{(12n+4)^2}+\frac{1}{(12n+8)^2}\right)=\frac{1}{108}\,\pi^2,$$
$$ \sum_{n\geq 0}\left(\frac{1}{(12n+5)^2}+\frac{1}{(12n+7)^2}\right)=\frac{2-\sqrt{3}}{36}\,\pi^2,$$
$$ \sum_{n\geq 0}\frac{1}{(12n+6)^2}=\frac{1}{288}\pi^2.$$
To deduce $I=\color{red}{-\frac{37}{432}\pi^2}$ is now just a matter of simple algebra.
| {
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"timestamp": "2023-03-29T00:00:00",
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$(2a)^2 +b^2=c^{4}a^2d^2 +4d$ integer solutions I'm trying to find possible values for the Diophantine equation:
$$
(2a)^2 +b^2=c^{4}a^2d^2 +4d \tag{1}\\
$$
With:
$$
\gcd(a,d)=1 \tag{2}
$$
Where $a,b,c,d $ all integers $ >0$. Specifically i'm interested in possible values for $d$.
For instance $d=1$ has solutions because for $d=1$ condition $(2)$ is automatically satisfied for every $a$.
But $d=2$ has no solutions (i think, see proof below).
I'm hoping to prove that there are no solutions for $d \neq 1$.
I have two questions:
1) Does anyone know how to determine all possible values for $d$ in equation above?
2) I tried below to prove some statements myself ($d\neq 2,3 \nmid d, 4 \nmid d$ and $a,b,d$ have no common prime divisors $>2$). But i'm not shure they are correct. So if you can verify some of the statements below please let me know.
$(*1) \enspace d \neq 2$
$$
d=2 \implies \text{ everything must be divisible by } 4 \implies\\
a^2 +{{b^2}\over{4}}=c^{4}a^2 +2 \tag{3}\\
$$
$a$ must be odd because of $(2)$. If $c$ is odd then ${{b^2}\over{4}}$ must be even. But then ${{b^2}\over{4}}$ must be divisible by $4$ because it's a perfect square. Then if we take $(3) \pmod 4$ we get the contradiction : $1+0 \equiv 1+2 \pmod 4 $. So assume $c$ is even and ${{b^2}\over{4}}$ is odd. Rewriting $(3)$ shows that we can then isolate a term that must be divisible by $3$: ${{b^2}\over{4}}=(c^{4}-1)a^2 +2=
(c^{2}+1)(c-1)(c+1)a^2 +2 \implies {b^2\over{4}} \equiv 2 \pmod 3$ . Contradiction because $2$ is not in the quadratic residues $\pmod 3$ .
$(*2) \enspace 3 \nmid d$
Also $3$ can not divide $d$ : This would mean: $ 3 \mid (2a)^2 + b^2 \implies 3 \mid a^2 \land 3 \mid b^2 $ (see for example here). This is a contradiction with equation $(2)$.
$(*3) \enspace a,b,d $ 'coprime' for divisors $>2$
From $(1)$ and $(2)$ we see that any prime $p > 2$ that divides $d$ can not divide $a$ and therefore can also not divide $b$ because $d$ divides the right hand side of $(1)$ . So $\gcd(p,b)=1$ for prime $p>2$ with $p \mid d$.
Also a prime $p' > 2$ that divides $a$ can not divide $d$ and therefore can't divide $b$. So $\gcd(p',b)=1$.
$(*4) \enspace 4 \nmid d$
$$
d= \text{ even } \implies a= \text{ odd and
everything is divisible by } 4 \implies\\
a^2 +{{b^2}\over{4}}=c^4a^2{{d^{2}}\over{4}} +d \\
a^2 +b'^2=c^4a^2d'^2 +2d' \\
d'= \text{ even } \implies \text{ right side divisible by } 4 \implies \text{ left side two odd squares }\\
\implies 1+1\equiv 0 \pmod 4 \implies \text{ contradiction }\\
$$
We conclude that $4 \nmid d$.
| If $p$ is a prime dividing $d$, we need $4 a^2 + b^2 \equiv 0 \mod p$. Since $a$ can't be divisible by $p$, that says $-4$ is a square mod $p$. The primes with this property are $2$ and those $\equiv 1 \mod 4$.
| {
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rational solutions of $x^4+x^2y^2+y^4=x^2$ I am trying to find rational solutions of $x^4+x^2y^2+y^4=x^2$, except for $(\pm 1,0), (0,0) $. I guess these three solutions are the only ones, but I failed to prove it. It seems that $x^4+x^2y^2+y^4=(x^2+xy+y^2)(x^2-xy+y^2) $ may be useful, but I do not know how to proceed.
| HINT.-A better way could be the following:$$(x^2+y^2)^2=x^4+x^2y^2+y^4+x^2y^2=x^2+x^2y^2$$ Then by Pythagorean triples you do have two possibilities
$$\begin{cases}1)\space\space x=2xy\space\space\text{ and }\space\space xy=x^2-y^2\\2)\space\space x=x^2-y^2\space\space\text{ and }\space\space xy=2xy\end{cases}$$ Can you end now?
| {
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Let $a, b \in \mathbb{R}^+$. Prove that $\frac{2a^2+3b^2}{2a^3+3b^3} + \frac{2b^2+3a^2}{2b^3 + 3a^3} \le \frac{4}{a+b}$.
Let $$a, b \in \mathbb{R}^+$$ Prove that $$\frac{2a^2+3b^2}{2a^3+3b^3} + \frac{2b^2+3a^2}{2b^3 + 3a^3} \le \frac{4}{a+b}$$
We can make the denominators common on the LHS by AM-GM but the problem is there are 2 terms and the power is 3, so I am unable to simplify it. It gives
$$5(a^2+b^2)(a+b) \le 8 \cdot 6^{3/2} \cdot (ab)^{3/2}$$
I am not able to proceed. Please help.
Thanks.
| Sometimes the most elementary and the most straightforward is the best.
It is easy to factorize $(RHS- LHS)$ as $(a-b)^2\times(something)$, isn't it?
$$\left(\frac{2}{a+b} - \frac{2a^2+3b^2}{2a^3+3b^3} \right) +\left(\frac{2}{a+b} - \frac{2b^2+3a^2}{2b^3 + 3a^3} \right) \ge 0$$
To make it more elegant:
\begin{align}
\frac{2}{a+b} - \frac{2a^2+3b^2}{2a^3+3b^3} &= \frac{2a^3+3b^3-2a^2b-3ab^2}{(a+b)(2a^3+3b^3)} \\
&=\frac{(a-b)(2a^2-3b^2)}{(a+b)(2a^3+3b^3)} \\
&=\frac{(a-b)(2a^2-2b^2)}{(a+b)(2a^3+3b^3)} - \frac{(a-b)b^2}{(a+b)(2a^3+3b^3)}\\
&\ge - \frac{(a-b)b^2}{(a+b)(2a^3+3b^3)}.
\end{align}
Similarly,
$$\frac{2}{a+b} - \frac{2b^2+3a^2}{2b^3+3a^3} \ge - \frac{(b-a)a^2}{(a+b)(2b^3+3a^3)}.$$
Taking the sum of the two inequalities, it remains to prove
$$- \frac{(a-b)b^2}{2a^3+3b^3} - \frac{(b-a)a^2}{2b^3+3a^3} \ge 0$$
or equivalently
$$(a-b)(2a^5+3a^2b^3-2b^5-3a^3b^2) \ge 0,$$
which is true from the following factorization (again):
$$2a^5+3a^2b^3-2b^5-3a^3b^2 = 2(a^5-b^5) + 3(a^2b^3 - a^3b^2) = (a-b)(2a^4+2a^3b-a^2b^2+2ab^3+2b^4).$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $m$ and $n$ so that the given function has the range $[-3, 5]$ Find $m$ and $n$ real numbers so that $f(x) = \frac{3x^2 + mx + n}{x^2 + 1}$ takes all and only the values from the interval $[-3, 5]$.
I started by solving the following double inequality:
$$-3 \leq \frac{3x^2 + mx + n}{x^2 + 1} \leq 5$$
From the left inequality I got $6x^2 + mx + n + 3 \geq 0$. The coefficient of $x^2$ is positive so the discriminant $\delta$ has to be less than or equal to $0$.
$\delta = m^2 - 24n - 72 \leq 0$
From the right inequality I got $2x^2 -mx + 5 - n \geq 0$. By using the same technique as the one used to solve the left inequality I got:
$\delta = m^2 - 40 + 8n \leq 0$
By combining these two resulted inequalities I got $m \leq 48$, that is $m \in [-4\sqrt3, 4\sqrt3]$.
Now I need to find the values of $m$ and $n$ so that $f$ is surjective, because it must take all the values in the given interval.
I don't know yet how to proceed, so I would appreciate any help from you guys!
Thank you!
| Here's another way. Consider $\large y=\dfrac{3x^2+mx+n}{x^2+1}$. Rearranging, we have:
$$x^2(y-3)+-mx+(y-n)=0$$
Now, since $x$ has to be real (since domain of $y$ is $\mathbb{R}$), so we see that:
$$m^2 - 4(y-3)(y-n) \ge 0 \\
\implies 4y^2-4(n+3)y+12n-m^2 \le 0 $$
Since $y$ is surjective, so the entire range of $y$ must be exactly $[-3,5]$, so our equation should be equivalent to $(y+3)(y-5)\le 0$, or:
$$4y^2-8y-60 \le 0$$
So we just compare the coefficients to get the two equations:
$$-4(n+3)=-8 \\
12n-m^2= -60$$
Which gives:
$$ m=\pm 4\sqrt 3 \text{ and } n= -1$$
| {
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"url": "https://math.stackexchange.com/questions/1948350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Inequality based on triangle: $\frac{3}{2}\le\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2$
If $a,b,c$ are sides of a triangle, prove that ${3\over2 }\le {{a\over b+c}} + {{b\over c+a}}+{{c\over a+b}} \lt 2$ .
| The left inequality.
$\sum\limits_{cyc}\frac{a}{b+c}-\frac{3}{2}=\sum\limits_{cyc}\left(\frac{a}{b+c}-\frac{1}{2}\right)=\sum\limits_{cyc}\frac{a-b-(c-a)}{2(b+c)}=\sum\limits_{cyc}\frac{(a-b)^2}{2(a+c)(b+c)}\geq0$.
The right inequality.
$\sum\limits_{cyc}\frac{a}{b+c}<\sum\limits_{cyc}\frac{a+a}{a+b+c}=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1950188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Divisibilty of $n^3+3$ and $n^5+5$ by a prime. Given that $n^3+3$ and $n^5+5$ are divisible by a prime $p$. How many values can $p$ take, with $n$ an integer?
The best I could do is $p=2$ for odd n, and I showed that $p$ cannot be equal to $3$. I know by subtracting the first equation to second, we will get
$n^5-n^3+2=n^3(n-1)(n+1)+2$. The expression is clearly not divisible by $3$ which means $p$ cannot take value of $3$.
My only question is: Are there any values that $p$ can take? I've tried to solve it various ways, none worked.
If you can help me, I greatly appreciate it, thank you
| Hint $\ {\rm mod}\ p\!:\ \begin{align}\color{#0a0}{n^3\equiv -3}\\\color{#c00}{n^5\equiv -5}\end{align}\Rightarrow\, (\color{#0a0}{-3})^5 \equiv(\color{#0a0}{n^3})^5\equiv(\color{#c00}{n^5})^3\equiv(\color{#c00}{-5})^3\Rightarrow\, 0\equiv 3^5\!-5^3\equiv 2\cdot 59 $
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Help in solving exponential equation Solve the following equation:
$$\frac{8^x + 27^x}{12^x + 18^x} = \frac{7}{6}$$
All I managed to do is rewrite the given equation in a simpler form:
$$\frac{4^x}{6^x + 9^x} + \frac{9^x}{6^x + 4^x} = \frac{7}{6}$$
I don't know what should be done next.
Thank you in advance!
| Hint:
$$\frac{8^x + 27^x}{12^x + 18^x} = \frac{7}{6}$$
$$6\cdot8^x-7\cdot12^x-7\cdot18^x+6\cdot27^x=0$$
$$6\cdot2^{3x}-7\cdot2^{2x}\cdot3^x-7\cdot2^x\cdot3^{2x}+6\cdot3^{3x}=0$$
Let $\left(\frac23\right)^x=t$
Then
$$6\cdot t^{3}-7\cdot t^2-7\cdot t+6=0$$
$$t\in \{-1;\frac23;\frac32\}$$
$$x=\pm1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1955076",
"timestamp": "2023-03-29T00:00:00",
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To find the locus of a point
A point moves such that the sum of its distances from the coordinate axes is equal to its distance from the circle $x^2+y^2=4$.
I tried and got this equation for the locus:
$$x+y=\sqrt{x^2+y^2}-2$$
Should I replace $x+y$ with $|x|+|y|$ because distances are absolute? In that case, how will I represent $y$ as an explicit function of $x$?
Thanks in advance.
| Let the point of the locus be $(a,b)$.
It's equivalent to two circles are touching
$$
\left \{
\begin{array}{rcl}
(x-a)^2+(y-b)^2 &=& (|a|+|b|)^2 \\
x^2+y^2 &=& 4
\end{array} \right.$$
$$
\left \{
\begin{array}{rcl}
x^2+y^2-2ax-2by &=& 2|a b| \\
x^2+y^2 &=& 4
\end{array} \right.$$
Eliminating quadratic terms:
\begin{align*}
ax+by &=2-|ab| \\
y &= \frac{2-|ab|-ax}{b} \\
x^2+\left( \frac{2-|ab|-ax}{b} \right)^2 &=4 \\
b^2x^2+[a^2x^2-2a(2-|ab|)x+(2-|ab|)^2] &= 4b^2 \\
(a^2+b^2)x^2-2a(2-|ab|)x+(2-|ab|)^2-4b^2 &=0 \\
\Delta &= 0 \\
a^2(2-|ab|)^2-(a^2+b^2)[(2-|ab|)^2-4b^2] &= 0 \\
4b^2(a^2+b^2)-b^2(2-|ab|)^2 &= 0 \\
4(a^2+b^2)-(2-|ab|)^2 &= 0 \\
4(a^2+b^2)-a^2b^2+4|ab|-4 &= 0 \\
\end{align*}
Note that distance from circle is either maximum or minimum.
Considering the minimal distance only:
If maximal distance is also included:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1956301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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} |
Limit of $\frac{x}{x^2-1}$ Question
$$\lim_{x \rightarrow 1-}\frac{x}{x^2-1}$$
My attempt
$$\lim_{x \rightarrow 1-}\frac{x}{x^2-1}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{1}{x^2-1}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{x^2(\frac{1}{x^2})}{x^2(1-\frac{1}{x^2})}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{\frac{1}{x^2}}{1-\frac{1}{x^2}}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{\frac{1}{x^2}}{\frac{x^2-1}{x^2}}=\lim_{x \rightarrow 1-}\frac{x}{x^2-1}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{1}{x^2-1}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{x^2(\frac{1}{x^2})}{x^2(1-\frac{1}{x^2})}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{\frac{1}{x^2}}{1-\frac{1}{x^2}}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{1}{x^2}\cdot\frac{x^2}{x^2-1}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{1}{x^2-1}$$
As you can see I'm going in circles. Can anyone give me a hint on how to start on this problem?
| I think one can prove that if f(x0) is finite, then
$$
\lim_{x->x_0}f(x)g(x) = f(x_0) \lim_{x->x_0}g(x)
$$
One can thus avoid partial fractions by substitution
$$
\lim_{x \rightarrow 1^-} \frac{x}{x^2 - 1} =
\lim_{\epsilon \rightarrow 0^+} \frac{1 - \epsilon}{(1 - \epsilon)^2 - 1} =
\lim_{\epsilon \rightarrow 0^+} \frac{1}{\epsilon}\frac{1 - \epsilon}{\epsilon - 2} =
\lim_{\epsilon \rightarrow 0^+} -\frac{1}{2\epsilon} = - \infty
$$
Or even a direct evaluation
$$
\lim_{x \rightarrow 1^-} \frac{x}{x^2 - 1} =
\lim_{x \rightarrow 1^-} \frac{x}{x + 1}\frac{1}{x - 1} =
0.5 \lim_{x \rightarrow 1^-} \frac{1}{x - 1} = - \infty
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1956525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Find the $n$th derivative of $f(x)=(\sqrt{x^2-1}+\sqrt{x-1})^2$ In this case $n=16$ and the point is $x=1$.
I know $\sqrt{x}$ is not differentiable in zero, but this function actually has left derivatives.
Also, I know that $f(x)=(\sqrt{x^2-1}+\sqrt{x-1})^2=x^2+x-2+2\sqrt{(x^2-1)(x-1)}$, so it is enogth to compute the derivative of $2\sqrt{(x^2-1)(x-1)}$.
Also, I tried to find the $n$th derivative of $\sqrt{g(x)}$ but its expresion was too complicated.
| For convenience take $x = 1+t$, so $x=1$ corresponds to $t=0$.
Thus
$$ \eqalign{f(x) &= \left(\sqrt{(1+t)^2-1} + \sqrt{t}\right)^2\cr
&= t^2 + 3t + 2 + 2 \sqrt{t^2+2t}\sqrt{t}\cr
&= t^2 + 3t + 2 + 2 t \sqrt{t+2}\cr
&= t^2 + 3t + 2 + 2 \sum_{k=0}^\infty \dfrac{\sqrt{2} (-1)^k (2k)! t^{k+1}}{8^k (k!)^2 (1-2k)}
}$$
The $16$'th derivative at $t=0$ comes from the $k=15$ term:
$$ \eqalign{\dfrac{d^{16}}{dt^{16}} \dfrac{2\sqrt{2} (-1)^{15} 30! \; t^{16}}{8^{15} (15!)^2 (-29)} &= \dfrac{2\sqrt{2} (-1)^{15} 30! 16!}{8^{15} (15!)^2 (-29)} \cr &= \frac{213458046676875}{33554432}
\sqrt{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1957533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Incorrect method to find a tilted asymptote Suppose I want to find the slanted asymptote for the graph of $\displaystyle y=\frac{x^2+x-6}{x+2}$.
Using division, we have $\displaystyle y=x-1-\frac{4}{x+2};\;$ so $y=x-1$ is the slanted asymptote.
I would like to find out, though, what is wrong with the following incorrect way of finding the asymptote:
$\displaystyle y=\frac{x^2+x-6}{x+2}=\frac{x+1-\frac{6}{x}}{1+\frac{2}{x}}\approx\frac{x+1}{1}=x+1$, so $y=x+1$ is the slanted asymptote.
| $\frac{y}{x+1} \;\to\; 1$ indeed, but:
$$y - (x+1) = \frac{x^2 + x - 6 - x^2 - 3 x - 3}{x+2} = \frac{-2 x - 9}{x+2} \quad \to \quad -2$$
which is why the slope of $x+1$ is correct, but its intercept is off by $-2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1959652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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} |
Properties of $\log^2(x)$ I have been trying to figure out the solution to a logarithm problem, and keep running into the equation $\\10\log^2_2(x) = x$. In so doing I've been trying to simplify $\log_2^2(x)$. What I have is
$\log_2^2(x) = \log_2(x)\log_2(x) = \log_2(x^{\log_2(x)})$
which as far as I can tell is no easier to work with.
Am I missing something? Or is there a completely different alternative way to look at this problem? Thanks!
| This is a fun one! We can solve it using the Lambert W Function, the inverse of the function
$$f(x) = xe^x$$
$$10\log_2^2(x) = x$$
$$\sqrt{10}\log_2(x) = \sqrt{x}$$
$$2^{\log_2(x)\sqrt{10}} = 2^{\sqrt{x}}$$
$$x^{\sqrt{10}} = 2^{\sqrt{x}}$$
$$x = 2^{\frac{\sqrt{x}}{\sqrt{10}}}$$
$$x = e^{\frac{\sqrt{x}}{\sqrt{10}}\log 2}$$
$$(\sqrt{x})^2 = e^{\frac{\log 2}{\sqrt{10}}\sqrt{x}}$$
$$\sqrt{x} = e^{\frac{\log 2}{2\sqrt{10}}\sqrt{x}}$$
Let $c = \frac{\log 2}{2\sqrt{10}}$:
$$\sqrt{x} = e^{c\sqrt{x}}$$
$$-c\sqrt{x}e^{-c\sqrt{x}} = -c$$
$$-c\sqrt{x} = W(-c)$$
$$\sqrt{x} = \frac{W(-c)}{-c}$$
$$x = \frac{(W(-c))^2}{c^2}$$
$$x = \frac{\left(W\left(-\frac{\log 2}{2\sqrt{10}}\right)\right)^2}{\left(\frac{\log 2}{2\sqrt{10}}\right)^2}$$
$$x =\left(W\left(-\frac{\log 2}{2\sqrt{10}}\right)\right)^2\frac{\log^2 2}{40}$$
There are actually $3$ solutions for $x$. One results in taking this function as is. The other results in using $-\sqrt{10}$ instead of $\sqrt{10}$. The last results in taking $\sqrt{10}$ and using the second branch $W_{-1}$ of the Lambert W function (since there are multiple solutions to $xe^x = y$ for all $y<0$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1960298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the equation of a circle I have the question "Find the equations of the following circles (in some cases more than one circle is possible).
A circle passes through the points $(1,4)$, $(7,5)$ and $(1,8)$"
I know that to find the equation of a circle I use $(X-a)^2 + (Y-b)^2 = r^2$,
However, I know you do this for one set of points I'm not sure what to do for three set of points and also the radius is not given.
The final answer should be $2X^2 + 2Y^2 -15X - 24Y + 77 = 0$.
| The triangle with vertices at
$$A(1;4),\quad B(7;5),\quad C(1;8) $$
has side lengths $AB=\sqrt{37},\, AC=4,\, BC=3\sqrt{5}$ by the Pythagorean theorem and area
$$[ABC]=\frac{1}{2}\left|1\cdot 5+7\cdot 8+1\cdot 4-4\cdot 7-5\cdot 1-8\cdot 1\right|=12 $$
by the shoelace formula. It follows that its circumradius is given by
$$R=\frac{abc}{4[ABC]} = \frac{12\sqrt{185}}{48} = \frac{1}{4}\sqrt{185}. $$
The barycentric coordinates of the circumcenter are straightforward to compute through $a^2,b^2,c^2$, hence we know both the centre
$$ O\left(3+\frac{3}{4};6\right)$$
and the radius of our circumcircle. Its equation readily follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1962587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the following Definite Integral $\int_{-1}^{1}(\{x+1\}\{x^2+2\})+(\{x^2+3\}\{x^3+4\})dx$, where $\{.\}$ is the fractional part of $x$. Integrate: $$\int_{-1}^{1}(\{x+1\}\{x^2+2\})+(\{x^2+3\}\{x^3+4\})dx$$
My attempt: We know that within the interval $[0,1)$, the following is true:
$$\{x+1\}=x$$
$$\{x^2+2\}=x^2$$
$$\{x^2+3\}=x^2$$
$$\{x^3+4\}=x^3$$
Therefore $$I_1=\int_{0}^{1}x^3+x^5=5/12.$$
Similarily on the interval $[-1,0)$ we can write:
$$\{x+1\}=x+1$$
$$\{x^2+2\}=x^2$$
$$\{x^2+3\}=x^2$$
$$\{x^3+4\}=x^3+1$$
Therefore, $$I_2=\int_{-1}^{0}x^3+2x^2+x^5=2/3-5/12.$$ Hence $$\int_{-1}^{1}(\{x+1\}\{x^2+2\})+(\{x^2+3\}\{x^3+4\})dx=2/3.$$ However the answer is $-2/3.$ I think I've made a mistake in the second transformations. Regardless, please explain where am I going wrong.
| If $x\in [-1,1]$, then $x^2,x^3,x^4\in [0,1]$. Now, when we add at least 1 to $y$, where $y\in [-1,1]$ its fractional part will be positive. Thus, you have a positive expression under the integral, so result cannot be negative.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1963753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
The range and domain of $o(x)=3+\sqrt{16-(x-3)^2}$ What is the domain and range of $$o(x)=3+\sqrt{16-(x-3)^2}\tag1$$
For the domain, I know that the expression under the radical has to be larger than or equal to $0$. So therefore, I get this:$$16-(x-3)^2\geq 0\\(7-x)(x+1)\geq 0\\\therefore x\in [-1,7]$$
But for the range, I thought of the function as a square root function; because it is in the form $y=\sqrt x+b$. Since the square root of something cannot be less than $0$, I thought the range was $[3,+\infty)$. Which is wrong (according to the book).
The range is supposedly $[3,7]$.
| You are right that $\sqrt{x}+b $ can not be less than $b $ so $3 +\sqrt {16- (x-3)^2} $ can't be less than $3$.
But note: $\sqrt {a - x} $ can not ever be more than $\sqrt {a} $. So $b \le b +\sqrt {x -a} \le b + \sqrt a$.
So $16 - (x-3)^2 \le 16$ so $0 \le \sqrt {16 - (x-3)^2}\le 4$ so $3 \le 3+\sqrt {16 - (x-3)^2}\le 7$ so
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1964574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Calculating the integral $\int\limits_{0}^{2\pi} \frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}$ I wanted to calculate $$\int\limits_{0}^{2\pi} \frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}$$
So I solved the indefinite integral first (by substitution):
$$\int\frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}=\frac{1}{b^2}\int\frac{d \theta}{\cos^2\theta \left(\frac{a^2}{b^2} \tan^2\theta+1 \right)} =\left[u=\frac{a}{b}\tan\theta, du=\frac{a}{b\cos^2\theta} d\theta \right ]\\=\frac{1}{b^2}\int\frac{b}{a\left(u^2+1 \right)}du=\frac{1}{ab}\int\frac{du}{u^2+1}=\frac{1}{ab} \arctan \left(\frac{a}{b}\tan\theta \right )+C$$
Then:
$$\int\limits_{0}^{2\pi} \frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}=\frac{1}{ab} \arctan \left(\frac{a}{b}\tan (2\pi) \right )-\frac{1}{ab} \arctan \left(\frac{a}{b}\tan 0 \right )=0$$
Which is incorrect (the answer should be $2\pi/ab$ for $a>0,b>0$).
On the one hand, the substitution is correct, as well as the indefinite integral itself (according to Wolfram it is indeed $\frac{1}{ab} \arctan \left(\frac{a}{b}\tan\theta \right )$ ), but on the other hand I can see that had I put the limits during the substitution I'd get $\int\limits_{0}^{0} \dots = 0$ because for $\theta = 0 \to u=0$ and for $\theta = 2\pi \to u=0$.
Why is there a problem and how can I get the correct answer?
Edit: Here is Wolfram's answer:
Wolfram is correct because $$\frac{a^2 b^2}{2}\int\limits_{0}^{2\pi} \frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}$$ is the area of an ellipse (defined by $x=a\cos t , y=b\sin t$), that is $$\frac{a^2 b^2}{2}\int\limits_{0}^{2\pi} \frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}=\pi ab$$
| Let $a,b>0, $ by symmetry,
$$
\begin{aligned}
\int_{0}^{2 \pi} \frac{d \theta}{a^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta} &=4 \int_{0}^{\frac{\pi}{2}} \frac{d \theta}{a^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta} \\
&=4 \int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} \theta d \theta}{a^{2} \tan ^{2} \theta+b^{2}} \\
&=4 \int_{0}^{\infty} \frac{d t}{a^{2} t+b^{2}} \\
&=\frac{4}{a b}\left[\tan ^{-1}\left(\frac{a t}{b}\right)\right]_{0}^{\infty} \\
&=\frac{2 \pi}{a b}
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1965164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
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Then minimum value of $\frac{\sec^4 \alpha}{\tan^2 \beta}+\frac{\sec^4 \beta}{\tan^2 \alpha}$
If $\displaystyle \alpha, \beta \in \left(0,\frac{\pi}{2}\right),$ Then minimum value of $$\frac{\sec^4 \alpha}{\tan^2 \beta}+\frac{\sec^4 \beta}{\tan^2 \alpha}$$
$\bf{My\; Try::}$ I have tried it using Cauchy-Schwarz Inequality
$$K=\frac{\sec^4 \alpha}{\tan^2 \beta}+\frac{\sec^4 \beta}{\tan^2 \alpha}\geq \frac{(\sec^2 \alpha+\sec^2 \beta)^2}{\tan^2 \alpha+\tan^2 \beta} = \frac{(2+\tan^2 \alpha+\tan^2 \beta)^2}{\tan^2 \alpha+\tan^2 \beta}$$
So $$K\geq \frac{4+(\tan^2 \alpha+\tan^2 \beta)^2+4\cdot (\tan^2 \alpha+\tan^2 \beta)}{\tan^2 \alpha+\tan^2 \beta} $$
$$K\geq 4+(\tan^2 \alpha+\tan^2 \beta)+\frac{4}{\tan^2 \alpha+\tan^2 \beta}\geq 4+4 = 8$$
My question is can we solve it some simple way, If yes Then plz explain me, Thanks
| Let $a = \frac{\sec^2\alpha}{\tan \beta}$ and $b = \frac{\sec^2\beta}{\tan\alpha}$. Then we have:
$$a^2 + b^2\geq 2ab = \frac{2\sec^2\alpha\sec^2\beta}{\tan \beta\tan\alpha} = \frac{2}{\sin\alpha\cos\alpha\sin\beta\cos\beta} = \frac{8}{\sin(2\alpha)\sin(2\beta)}\geq 8.$$
Both equality happens when $\alpha = \beta = \pi/4.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1966853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Limit calculation using Taylor series $$\lim_{x \to 0} \frac{xe^{-2x}-xe^{2x}}{\sqrt{1+x^{2}}-\sqrt{1-x^2}}$$
Any tip about denominator, for numerator i am trying to plug in the value $-2x$ and $2x$ in pre-computed series of $e^x$. I have to calculate the limit using taylor series.
| $$
\lim_{x \to 0} \frac{xe^{-2x}-xe^{2x}}{\sqrt{1+x^{2}}-\sqrt{1-x^2}}
$$
Use Taylor expansions:
$$
e^{-2x}=1-2x+O(x^2)\\
e^{2x}=1+2x+O(x^2)\\
\sqrt{1+x^2}=1+\frac{1}{2}x^2+O(x^4)\\
\sqrt{1-x^2}=1-\frac{1}{2}x^2+O(x^4)
$$
Then,
$$
\lim_{x \to 0} \frac{xe^{-2x}-xe^{2x}}{\sqrt{1+x^{2}}-\sqrt{1-x^2}}=\lim_{x \to 0} \frac{x(1-2x+O(x^2))-x(1+2x+O(x^2))}{(1+\frac{1}{2}x^2+O(x^4))-(1-\frac{1}{2}x^2+O(x^4))}\to\\
\lim_{x \to 0} \frac{-4x^2+O(x^3)}{x^2+O(x^4)}=\lim_{x \to 0} \frac{-4+O(x)}{1+O(x^2)}=-4
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1967315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Definite integration involving Trigonometry. $$ \frac{\int_0^{\pi/2} (\sin x)^{2½+1} \, dx}{\int_0^{\pi/2} (\sin x)^{2½-1} \, dx}$$
I'm puzzled.
How to proceed...
| We have
$$\sin^{\sqrt{2}+1}x=\sin^{\sqrt{2}-1}x\sin^2x=\sin^{\sqrt{2}-1}x-\sin^{\sqrt{2}-1}x\cos^2x.$$
Now use integration by parts
$$\int f'g=fg-\int fg'$$
with
$$f(x)=\frac{1}{\sqrt{2}}\sin^{\sqrt{2}}x,\quad g(x)=\cos x,$$
it follows that
$$\begin{aligned}&\int_0^{\frac{\pi}{2}}\sin^{\sqrt{2}-1}x\cos^2x\\
=&\frac{1}{\sqrt{2}}\sin^{\sqrt{2}}x\cos x|_{x=0}^{\frac{\pi}{2}}+\frac{1}{\sqrt{2}}\int_0^{\frac{\pi}{2}}\sin^{\sqrt{2}}x\sin xdx\\
=&\frac{1}{\sqrt{2}}\int_0^{\frac{\pi}{2}}\sin^{\sqrt{2}+1}xdx
\end{aligned}$$
Therefore
$$\begin{aligned}\int_0^{\frac{\pi}{2}}\sin^{\sqrt{2}+1}xdx&=\int_0^{\frac{\pi}{2}}\sin^{\sqrt{2}-1}xdx-\int_0^{\frac{\pi}{2}}\sin^{\sqrt{2}-1}x\cos^2xdx\\
&=\int_0^{\frac{\pi}{2}}\sin^{\sqrt{2}-1}xdx-\frac{1}{\sqrt{2}}\int_0^{\frac{\pi}{2}}\sin^{\sqrt{2}+1}xdx,
\end{aligned}$$
which implies
$$\frac{\sqrt{2}+1}{\sqrt{2}}\int_0^{\frac{\pi}{2}}\sin^{\sqrt{2}+1}xdx=\int_0^{\frac{\pi}{2}}\sin^{\sqrt{2}-1}xdx.$$
Hence
$$\frac{\int_0^{\frac{\pi}{2}}\sin^{\sqrt{2}+1}xdx}{\int_0^{\frac{\pi}{2}}\sin^{\sqrt{2}-1}xdx}=\frac{\sqrt{2}}{\sqrt{2}+1}=\sqrt{2}(\sqrt{2}-1)=2-\sqrt{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1967806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $\sum^{n}_{k=1}(k+2)(k+4) = \frac{2n^{3} + 21n^{2} + 67n}{6}$ When doing induction should you always try to put your final answer as the "desired " form? For example if: $$\sum^{n}_{k=1}(k+2)(k+4) = \frac{2n^{3} + 21n^{2} + 67n}{6}$$ we ought to give the final answer as $$\frac{2(k+1)^{3} + 21(k+1)^{2} + 67(k+1)}{6}?$$
I just expanded both the $\text{LHS}_{k+1}$ and the $\text{RHS}_{k+1}$ to show they were equal after the induction. Like this:
Show that $$\sum^{n}_{k=1}(k+2)(k+4) = \frac{2n^{3} + 21n^{2} + 67n}{6}$$ for all integers $n \geq 1$.
For $n = 1$,
$$\sum^{1}_{k=1}(k+2)(k+4) = 15$$
and
$$\frac{2(1)^{3} + 21(1)^{2} + 67(1)}{6} = 15$$
Assume that it is true for some integer $n = k$, thus $$\sum^{k}_{k=1}(k+2)(k+4) = \frac{2k^{3} + 21k^{2} + 67k}{6}$$ so the $\text{LHS}_{k+1}$ $$\sum^{k+1}_{k=1}(k+2)(k+4) = \sum^{k}_{k=1}(k+2)(k+4) + (k+3)(k+5)$$ $$= \frac{2k^{3} + 21k^{2} + 67k}{6} + \frac{6(k+3)(k+5)}{6}$$ $$=\frac{2k^{3} + 27k^{2} + 115k + 90}{6}$$ Now the $\text{RHS}_{k+1}$ $$\frac{2(k+1)^{3} + 21(k+1)^{2}+ 67(k+1)}{6} = \frac{2k^{3} + 27k^{2} + 115k + 90}{6}$$ Thus $\text{LHS}_{k+1} = \text{RHS}_{k+1}$ Q.E.D.
| $(k+2)(k+4)=k^2+6k+8$
$$\sum^{n}_{k=1}{(k+2)(k+4)}=\sum^{n}_{k=1}{k^2}+6\sum^{n}_{k=1}{k}+\sum^{n}_{k=1}{8}=\frac{n(n+1)(2n+1)}{6}+6 \cdot \frac{n(n+1)}{2}+8n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1968267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Let $n \in \mathbb{Z}^+$, prove the identity $ \sum_{k=1}^{n-1} \binom {n} {k} \frac{kn^{n-k}}{k+1}=\frac{n(n^{n}-1)}{n+1}$ Let $n \in \mathbb{Z}^+$, prove the identity $$ \sum_{k=1}^{n-1} \binom {n} {k} \frac{kn^{n-k}}{k+1}=\frac{n(n^{n}-1)}{n+1}$$
First of all $$ \sum_{k=1}^{n-1} \binom {n} {k} \frac{kn^{n-k}}{k+1}=n^{n}\Bigg(\sum_{k=1}^{n-1} \binom {n} {k} \frac{kn^{-k}}{k+1} \Bigg)$$
$$=n^n\Bigg(\sum_{k=1}^{n-1} \binom {n}{k} \bigg(1-\frac{1}{k+1}\bigg)\bigg(\frac{1}{n^k}\bigg)\Bigg)$$
$$=n^n \sum_{k=1}^{n-1} \binom {n} {k} \frac{1}{n^k}-n^n \sum_{k=1}^{n-1} \binom {n}{k} \bigg( \frac{1}{k+1}\bigg)\bigg(\frac{1}{n^k} \bigg)$$
We have for the first sum $$(1+\frac{1}{x})^n = \sum_{k=0}^n \binom{n}{k}\frac{1}{x^k}.$$
For the second sum
$$(1+x)^n = \sum_{k=0}^n \binom{n}{k}x^k.$$
Integrating both sides from $0$ to $x$, we see that
$$\frac{(1+x)^{n+1}-1}{n+1} = \sum_{k=0}^n \binom{n}{k}\frac{x^{k+1}}{k+1}.$$
Putting $x=1$ yields $$\frac{2^{n+1}-1}{n+1}=\sum_{k=0}^n \binom{n}{k}\frac{1}{k+1}$$
Here where I have stopped. I could not get them similar for what I have. Would someone help me out !
| There is a straightforward direct computational argument:
$$\begin{align*}
\sum_{k=1}^{n-1}\binom{n}k\frac{kn^{n-k}}{k+1}&=\sum_{k=1}^{n-1}\binom{n}k\left(1-\frac{1}{k+1}\right)n^{n-k}\\
&=\sum_{k=1}^{n-1}\binom{n}kn^{n-k}-\sum_{k=1}^{n-1}\binom{n}k\frac{n^{n-k}}{k+1}\\
&=(n+1)^n-n^n-1-\frac{1}{n+1}\sum_{k=1}^{n-1}\binom{n+1}{k+1}n^{n-k}\\
&=(n+1)^n-n^n-1-\frac{1}{n+1}\sum_{k=2}^n\binom{n+1}kn^{n+1-k}\\
&=(n+1)^n-n^n-1-\frac{1}{n+1}\left((n+1)^{n+1}-n^{n+1}-(n+1)n^n-1\right)\\
&=-1+\frac{n^{n+1}+1}{n+1}\\
&=\frac{n(n^n-1)}{n+1}
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1968896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
How to evaluate the given determinant
Question Statement:-
Show that $$\begin{vmatrix}
(b+c)^2 & c^2 & b^2 \\
c^2 & (c+a)^2 & a^2 \\
b^2 & a^2 & (a+b)^2 \\
\end{vmatrix}=2(ab+bc+ca)^3$$
Attempt at a Solution:-
1st attempt(which was in vain):-
LHS:-$$\begin{align}\begin{vmatrix}
(b+c)^2 & c^2 & b^2 \\
c^2 & (c+a)^2 & a^2 \\
b^2 & a^2 & (a+b)^2 \\
\end{vmatrix}=\begin{vmatrix}
2bc & c^2 & b^2 \\
(c^2-a^2)-(c+a)^2 & (c+a)^2 & a^2 \\
(b^2-a^2)-(b
+a)^2 & a^2 & (a+b)^2 \\
\end{vmatrix}\left[\begin{array}{11}
C_1\rightarrow C_1-C_2-C_3\end{array}\right]
=\begin{vmatrix}
2bc & c^2 & b^2 \\
-2(a^2+ac) & (c+a)^2 & a^2 \\
-2(a^2+ab) & a^2 & (a+b)^2 \\
\end{vmatrix}=
2\begin{vmatrix}
bc & c^2 & b^2 \\
-(a^2+ac) & (c+a)^2 & a^2 \\
-(a^2+ab) & a^2 & (a+b)^2 \\
\end{vmatrix}=
\dfrac{2}{bc}\begin{vmatrix}
b^2c^2 & c^2 & b^2 \\
-bc(a^2+ac) & (c+a)^2 & a^2 \\
-bc(a^2+ab) & a^2 & (a+b)^2 \\
\end{vmatrix}\left[C_1\rightarrow bc\cdot C_1\right]=
\dfrac{2}{bc}\begin{vmatrix}
b^2c^2-b^2c^2 & c^2 & b^2 \\
-bc(a^2+ac)-b^2(c+a)^2 & (c+a)^2 & a^2 \\
-bc(a^2+ab)-a^2b^2 & a^2 & (a+b)^2 \\
\end{vmatrix}[C_1\rightarrow C_1-b^2C_2]=
\dfrac{2}{bc}\begin{vmatrix}0 & c^2 & b^2 \\
-b(a+c)(ab+bc+ac) & (c+a)^2 & a^2 \\
-ab(ab+bc+ac) & a^2 & (a+b)^2 \\
\end{vmatrix}\\
=-2\left(\dfrac{ab+bc+ac}{c}\right)\begin{vmatrix}
0 & c^2 & b^2 \\
a+c & (c+a)^2 & a^2 \\
a & a^2 & (a+b)^2 \\
\end{vmatrix}
\end{align}$$
I was pretty much stuck after this so I tried another approach.
2nd Attempt:-
$$\begin{vmatrix}
(b+c)^2 & c^2 & b^2 \\
c^2 & (c+a)^2 & a^2 \\
b^2 & a^2 & (a+b)^2 \\
\end{vmatrix}=
(abc)^4\begin{vmatrix}
\left(\dfrac{1}{b}+\dfrac{1}{c}\right)^2 & \dfrac{1}{b^2} & \dfrac{1}{c^2} \\
\dfrac{1}{a^2} & \left(\dfrac{1}{a}+\dfrac{1}{c}\right)^2 & \dfrac{1}{c^2} \\
\dfrac{1}{a^2} & \dfrac{1}{b^2} & \left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2 \\
\end{vmatrix}\left[\begin{array}{11}
R_1\rightarrow \dfrac{R_1}{b^2c^2} \\
R_2\rightarrow \dfrac{R_2}{a^2c^2} \\
R_2\rightarrow \dfrac{R_3}{a^2b^2 }\end{array}\right]$$
And after starting along the route that I have shown in the second attempt I figured it was much more useless than the previous one. So, I thought that the Mathematics Stack Exchange is the only route left. So, your help is very much needed.
| Let us continue with your valiant first attempt (note that you incorrectly missed out a factor of $2$, from your third stage to the fourth stage)
$$-\color{red}{2}\left(\dfrac{ab+bc+ac}{c}\right)\begin{vmatrix}
0 & c^2 & b^2 \\
a+c & (c+a)^2 & a^2 \\
a & a^2 & (a+b)^2 \\\end{vmatrix}$$
Using a brute force approach to evaluate the determinant of the $3\times 3$ matrix (aided by the fact that the top left entry is $0$), the determinant evaluates to
$$0-c^2((a+b)^2(a+c)-a^2)+b^2(a^2(a+c)-(c+a)^2a)$$
which, after a fair amount of algebraic manipulation, simplifies to
$$-c(a^2c^2+b^2c^2+a^2b^2+2a^2bc+2abc^2+2ab^2c)=-c(ab+bc+ac)^2$$
We thus have the desired result
$$-2\left(\frac{ab+bc+ac}{c}\right)\times-c(ab+bc+ac)^2=2(ab+bc+ac)^3$$
A less brutal approach is to make use of the Matrix determinant lemma see here.
The matrix, denote by $M$, can be expressed as
$$M=vv^T-A=-(A+(-vv^T))$$
where $v= \left( \begin{array}{c}
b+c \\
a+c \\
a+b \end{array} \right) $ and $A= (ab+bc+ca)\left( \begin{array}{ccc}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0 \end{array} \right) $
Invoking the Matrix determinant lemma, we have
$$\det(M)=-(1-v^TA^{-1}v)\det(A)\tag{Eq.1}$$
Now, it is straightforward to show that
$$\det(A)=2(ab+bc+ac)^3$$ and $$A^{-1}=\frac{1}{2(ab+bc+ca)}\left( \begin{array}{ccc}
-1 & 1 & 1 \\
1 & -1 & 1 \\
1 & 1 & -1 \end{array} \right)$$
Leading to $v^TA^{-1}v=2$
Substituting all the values into Eq.$1$, we have
$$\det(M)=-(1-2)\times2(ab+bc+ac)^3=2(ab+bc+ca)^3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1969441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $ \int^{\frac{91\pi}{6}}_0 |\cos(x)| \, \mathrm{d}x$ I am trying to evaluate the exact value of the following definite integral:
\begin{align} \int^{\frac{91\pi}{6}}_0 |\cos(x)| \, \mathrm{d}x \end{align}
Since $ \int^{\frac{\pi}{2}}_0 \cos(x) \, \mathrm{d}x$ has symmetry, I did the following:
\begin{align}
\frac{91\pi}{6} \cdot \frac{2}{\pi} = \frac{91}{3}
\end{align}
\begin{align}
\int^{\frac{\pi}{2}}_0 \cos(x) \, \mathrm{d}x = [\sin(x)]^{\frac{\pi}{2}}_0 = 1
\end{align}
\begin{align}
\therefore \text{Bounded Area } = 1 \cdot \frac{91}{3} = \frac{91}{3}
\end{align}
Apparently, the correct answer is $ \frac{61}{2} $ which is very close to my answer. I cannot understand why my answer is wrong. Could someone please advise me?
| Observe that
$$\int_0^\pi|\cos x|dx=\left.2\int_0^{\pi/2}\cos x dx=2\sin x\right|_0^{\pi/2}=2\implies$$
$$\int_0^{91\pi/6=15\pi+\frac\pi6}|\cos x|dx=\left.2\cdot15+\int_\pi^{\frac{7\pi}6}(-\cos x)dx=30-\sin x\right|_\pi^{\frac{7\pi}6}=30+\frac12=\frac{61}2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1971315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
find the distance between two skew lines Given the following parametric find the distance between the two lines:
$x=2+t\;,\;\; y=1+6t\;,\;\; z=2t$ and
$x=1+2s\;,\;\; y=6+14s\;,\;\; z=-1+5s$
I tried using a point on the first line called P and entered $s=0$ and $s=2$ to get q and r. I then took the cross product of qr and qp then used the distance formula but this didn't seem to work.
| You can write a point on the first line as $P = (2,1,0) + t(1,6,2)$ and a point on the second line as $Q = (1,6,-1) + s(2,14,5)$. If you take the difference $P - Q$ you get $(1,-5,1) + t(1,6,2) + s(-2,-14,-5)$.
So your problem is equivalent to finding the distance between the plane given by $(1,-5,1) + t(1,6,2) + s(-2,-14,-5)$ and the origin. The first step here is to find a normal vector $\mathbf{n}$ to the plane, you can use $\mathbf{n} = (1,6,2) \times (-2,-14,-5) = (-2,1,-2)$. Now take any vector $\mathbf{v}$ in the plane (for example, $\mathbf{v} = (1,-5,1))$ you can compute the distance as:
$$D = \frac{|\mathbf{v} \cdot \mathbf{n}|}{||\mathbf{n}||} = \frac{|-2 - 5 - 2|}{\sqrt{4 + 1 + 4}} = \frac{9}{3} = 3$$
In general, if your lines are given by $\mathbf{p} + t\mathbf{v}$ and $\mathbf{q} + s\mathbf{w}$, you can compute:
$$D = \frac{|(\mathbf{p} - \mathbf{q}) \cdot \mathbf{n}|}{||\mathbf{n}||}; \qquad \qquad \mathbf{n} = \mathbf{v} \times \mathbf{w}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1973152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Sum of all real values of $x$ in trigonometric equation of cotangent The sum of all real values of $x$ in $\left[0,2\pi\right]$ which satisfy the equation $3\cot^2 x+8\cot x+3=0$
on solving $\displaystyle \cot x = \frac{-8\pm \sqrt{64-36}}{6} = \frac{-8\pm 2\sqrt{7}}{6} = -\frac{4}{3}\pm \frac{\sqrt{7}}{3}$
I want to go further could some help me with this, Thanks
| Check
$$\frac{-8 \pm \sqrt{28}}{6} = \frac{-8 \pm \sqrt{4 \times 7}}{6}= \frac{-4}{3} \pm \frac{\sqrt{7}}{3}$$
"The sum of all real values of x...which satisfy the equation" haven't you found the solutions to the $\cot$ quadratic? So what is the sum of these (two) values?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1974125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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