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Hint required - Find values of $a$ and $b$ such that the limit equals $\frac{5}{12}$. Find the values of $a$ and $b$ such that $$\lim_{x\to 0} \frac{\sqrt[3]{ax+b}-2}{x}= \frac{5}{12}.$$
Can someone please provide me with a hint on how to approach this question?
| $$\lim_{x\to 0} \frac{\sqrt[3]{ax+b}-2}{x}= \frac{5}{12}$$
An approach without using L'Hôpital's rule:
Since the denominator approaches zero as $x$ approaches zero, the numerator must also approach zero as $x$ approaches zero for the limit to be finite, i.e.,
$$\tag{1} \lim_{x\to0}\sqrt[3]{ax+b}-2=0
\implies b^\frac13 - 2 = 0$$
Now, consider the numerator:
$$\begin{align}\sqrt[3]{ax+b}-2 &= \sqrt[3]{\left(\frac{ax}{b}+1\right)b}-2 \\
&= b^{\frac13} \left(1+\frac{ax}{b}\right)^{\frac13} - 2 \\
\end{align}$$
Using the binomial approximation, i.e., $(1+x)^n \approx1+nx$ for small $x$, we get
$$ b^{\frac13}\left(1+\frac{ax}{3b}\right) - 2 = \underbrace{(b^\frac13-2)}_{0} + \frac{ax}{3b^\frac23} = \frac{ax}{3b^\frac23}$$
Thus,
$$ \tag{2} \lim_{x\to 0} \frac{\sqrt[3]{ax+b}-2}{x} = \lim_{x\to 0} \frac{\frac{ax}{3b^{2/3}}}{x} = \frac{a}{3b^\frac23} = \frac{5}{12}$$
It is now easy to calculate $a$ and $b$ from $(1)$ and $(2)$
| {
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How to calculate $\lim\limits_{n\rightarrow \infty }(n-\sum^n_{k=1}\cos\frac{2k}{n\sqrt{n}}) $? I tried this:
$\lim\limits_{n\rightarrow \infty }(n-\sum^n_{k=1}\cos\frac{2k}{n\sqrt{n}}) $=
= $ \lim\limits_{n\rightarrow \infty }(1-\cos\frac{2}{n\sqrt{n}}) $ + $ \lim\limits_{n\rightarrow \infty }(1-\cos\frac{4}{n\sqrt{n}}) $ + ... + $ \lim\limits_{n\rightarrow \infty }(1-\cos\frac{2n}{n\sqrt{n}}) $ =
= $ \lim\limits_{n\rightarrow \infty }\frac{(1-\cos\frac{2}{n\sqrt{n}})(1+\cos\frac{2}{n\sqrt{n}})}{1+\cos\frac{2}{n\sqrt{n}}} $ + $ \lim\limits_{n\rightarrow \infty }\frac{(1-\cos\frac{4}{n\sqrt{n}})(1+\cos\frac{4}{n\sqrt{n}})}{1+\cos\frac{4}{n\sqrt{n}}} $ + ...+ $ \lim\limits_{n\rightarrow \infty }\frac{(1-\cos\frac{2n}{n\sqrt{n}})(1+\cos\frac{2n}{n\sqrt{n}})}{1+\cos\frac{2n}{n\sqrt{n}}} $ =
= $ \lim\limits_{n\rightarrow \infty }\frac{\sin^2\frac{2}{n\sqrt{n}}}{1+\cos\frac{2}{n\sqrt{n}}} $ + $ \lim\limits_{n\rightarrow \infty }\frac{\sin^2\frac{4}{n\sqrt{n}}}{1+\cos\frac{4}{n\sqrt{n}}} $ + ... + $ \lim\limits_{n\rightarrow \infty }\frac{\sin^2\frac{2n}{n\sqrt{n}}}{1+\cos\frac{2n}{n\sqrt{n}}} $ = ...
, but from here I don't know what to do.
| You face the sum of cosines with angles in arithmetic progression.
So
$$\sum_{k=1}^n \cos \left(\frac{2 k}{n^{3/2}}\right)=\sin \left(\frac{1}{\sqrt{n}}\right) \cos \left(\frac{n+1}{n^{3/2}}\right) \csc
\left(\frac{1}{n^{3/2}}\right)$$
$$S_n=n-\sin \left(\frac{1}{\sqrt{n}}\right) \cos \left(\frac{n+1}{n^{3/2}}\right) \csc\left(\frac{1}{n^{3/2}}\right)$$ Now, using Taylor series
$$S_n=\frac{2}{3}+\frac{13}{15 n}+\frac{4}{315
n^2}+O\left(\frac{1}{n^3}\right)$$
If $n>6$, the relative error is smaller than $0.1$%. If $n>14$, it becomes smaller than $0.01$%.
| {
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find the maximum of $x+\frac{1}{2y}$ under the condition $(2xy-1)^2=(5y+2)(y-2)$ Assume $x,y \in \mathbb{R}^+$ and satisfy $$\left(2xy-1\right)^2 = (5y+2)(y-2)$$, find the maximum of the expression $x+\frac{1}{2y}$.
$\because (5y+2)(y-2) \geq 0$ , $\therefore y\geq 2$ or $y \leq -\frac{2}{5}$, and we have $$2xy = \sqrt{(5y+2)(y-2)}+1$$
then $f(x,y)=x+\frac{1}{2y}=\frac{2xy+1}{2y}=\frac{\sqrt{(5y+2)(y-2)}+2}{2y}=g(y)$,$$\therefore g'(y) = \frac{2 + 2 y - \sqrt{-4 - 8 y + 5 y^2}}{y^2 \sqrt{-4 - 8 y + 5 y^2}}$$
and $2+2y = \sqrt{5y^2-8y-4} \Rightarrow y = 8+6\sqrt{2}$, with the help of graphing calculator ,i know this is the maximum point. $$\therefore f_{\text{max}}(x,y)=g(8+6\sqrt{2})=\frac{3\sqrt{2}}{2}-1$$
Is there any other way to work out it?
| Alternative solution:
Let $z = x + \frac{1}{2y}$.
We have $x = z - \frac{1}{2y}$ and
$2xy = 2yz - 1$.
Then, we have
$$(2yz - 2)^2 = (5y + 2)(y - 2)$$
or
$$(4z^2 - 5)y^2 + (-8z + 8)y + 8 = 0. \tag{1}$$
Since the quadratic equation in $y$ (1) has real roots, its discriminant is non-negative, i.e.
$$\Delta := -64z^2 - 128z + 224 \ge 0$$
which results in
$$- \frac{3}{2}\sqrt 2 - 1\le z \le \frac{3}{2}\sqrt 2 - 1.$$
Also, when $y = 8 + 6\sqrt 2$ and $x = \frac98\sqrt 2 - \frac12$, we have $(2xy - 1)^2 = (5y + 2)(y - 2)$ and
$z = \frac{3}{2}\sqrt 2 - 1$.
Thus, the maximum of $x + \frac{1}{2y}$ is $\frac{3}{2}\sqrt 2 - 1$.
| {
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Discriminant of depressed cubic If the cubic equation $x^3+px+q$ has roots $\alpha , \beta , \gamma $ then we know that $\alpha + \beta + \gamma =0 $, $\alpha \beta + \alpha \gamma + \beta \gamma =p $ and $\alpha \beta \gamma =-q $.
The discriminant is $\Delta = (\alpha - \beta )^2 (\alpha - \gamma )^2 (\beta - \gamma ) ^2 $.
I know the answer should be $\Delta = -4p^3 -27q^2 $ and when I substitute expressions involving roots instead of $ p$ and $q$ in the expression for $\Delta $ and try to verify it, it basically just comes out with a big mess and I haven’t been able to verify it yet. Is there a better way to show the value of the discriminant in terms of $p$ and $q$ ?
|
If the cubic equation $x^3+px+q$ has roots $\alpha , \beta , \gamma $ then we know that $\alpha + \beta + \gamma =0 $, $\alpha \beta + \alpha \gamma + \beta \gamma =p $ and $\alpha \beta \gamma =-q $.
The discriminant is $\Delta = (\alpha - \beta )^2 (\alpha - \gamma )^2 (\beta - \gamma ) ^2 $.
I know the answer should be $\Delta = -4p^3 -27q^2$
For convenience, I will employ changes of variables.
Let $~A = \alpha\beta, ~B = \alpha + \beta \implies \gamma = -B.$
Then,
$(\alpha-\beta)^2 = (B^2 - 4A).$
$[(\alpha - \gamma )(\beta - \gamma )]^2$
$ = [(2\alpha + \beta)(\alpha + 2\beta)]^2$
$ = [2\alpha^2 + 5\alpha\beta + 2\beta^2]^2$
$ = [2B^2 + A]^2 = 4B^4 + 4B^2A + A^2.$
Putting this all together, you have that
$\Delta = (B^2 - 4A)(4B^4 + 4B^2A + A^2)$
$= ~(4B^6 + 4B^4A + B^2A^2)$
$+ ~(-16B^4A - 16B^2A^2 - 4A^3)$
$$=~ 4B^6 - 12B^4A - 15B^2A^2 - 4A^3. \tag1 $$
So, the problem reduces to determining whether (1) above matches $-4p^3 -27q^2.$
$p = A + B(-B) = A - B^2.$
$q = -AB$.
Therefore,
$-4p^3 - 27q^2$
$=~ -4(A-B^2)^3 -27(-AB)^2$
$=~-4[A^3 - 3A^2B^2 + 3AB^4 - B^6] - 27A^2B^2$
$= -4A^3 + 12A^2B^2 - 12AB^4 + 4B^6 - 27A^2B^2$
$$= -4A^3 - 15A^2B^2 - 12AB^4 + 4B^6. \tag2 $$
The expressions in (1) and (2) match.
| {
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Solving the system $a^3 + 15ab^2 = 9$, $\;\frac 35 a^2b + b^3 = \frac 45$ I have problem solving the following system of two cubic equations.
$$\begin{cases}
a^3 + 15ab^2 = 9 \\
\frac 35 a^2b + b^3 = \frac 45
\end{cases}
$$
I don't have any idea how to approach this kind of problem.
I'm looking for solutions included in the set of real numbers (no complex solutions).
Is there any general solution to the problem?
| Starting from
$$\begin{align}
a^3 + 15ab^2 = 9 \\
\frac 35 a^2b + b^3 = \frac 45
\end{align}$$
let $b=ax$ as obviously $b\neq0$:
$$\begin{align}
a^3 + 15x^2a^3 = 9 \\
3 xa^3 + 5x^3a^3 = 4
\end{align}$$
dividing both sides eliminates $a$:
$$
\frac{1+ 15x^2}{3 x + 5x^3} = \frac 94
$$
so you have a cubic in $x$, for which there are explicit solutions.
| {
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Solve for x: $\lfloor x\rfloor \{\sqrt{x}\}=1$ where $\{x\}=x-\lfloor x\rfloor$ Solve for x:
$\lfloor x\rfloor \{\sqrt{x}\}=1$
where $\{x\}=x-\lfloor x\rfloor$
My Attempt:
I took intervals of $x$ as $x\in (2,3)$ so $\sqrt x\in (1,2)$. Due to which $\lfloor x\rfloor=2$ and $\{\sqrt x\}=\sqrt x-\lfloor \sqrt x\rfloor=\sqrt x-1$.
Substituting in the equation
$\lfloor x\rfloor \{\sqrt{x}\}=1$
I got
$2(\sqrt x-1)=1$
$x=\frac{9}{4}$
which satisfies the equation.
But when I apply the same process to $x\in (3,4)$ I end up getting value of $x$ that does not satisfy the given equation. But when I plotted the graph a solution clearly exists between two consecutive integers.
What am I missing
| There is obviously no solution for x < 1. Let k, n be integers with k ≥ 1 and $k^2 ≤ n ≤ k^2 + 2k$, and look for solutions x in the interval [n, n+1); this covers all solutions.
In this interval, the equation is equivalent to $n \cdot (\sqrt x - k) = 1$ or $x = (k + 1 / n)^2$. Since $n ≥ k^2$, $x ≤ (k + 1 / k^2)^2 = k^2 + 2/k + 1/k^4$. For k ≥ 3, the right hand side is less than $k^2 + 1$, so we have $n = k^2$ and $x = (k + 1 / k^2)^2$.
For k = 2, n ≥ 5 is also impossible because in that case $x = (k + 1 / n)^2 ≤ (2 + 1/5)^2 = 121/25 < 5$. For k = 1, $x = (1 + 1 / n)^2$, so if n = 1 then x = 4, if n = 2 then x = 2.25, if n = 3 then x = 16/9, and x = 2.25 is the only solution.
Summary: The solutions are x = 2.25, and $x = (k + 1 / k^2)^2$ for k ≥ 2. To check, let k = 10, $x = 10.01^2$, $\sqrt x = 10.01$, and 100 * 0.01 = 1.
| {
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How to find $\lim_{x\to-\infty} \frac{\sqrt{x^2+2x}}{-x}$? By factorization:
$$\lim_{x\to-\infty} \frac{\sqrt{x^2+2x}}{-x}\tag{1}$$
$$=\lim_{x\to-\infty} \frac{x\sqrt{1+\frac{2}{x}}}{-x}$$
$$=\lim_{x\to-\infty}-\sqrt{1+\frac{2}{x}}$$
If I input $x=-\infty$, the limiting value seems to be $-1$. But according to desmos, the limiting value should be $1$.
By L'Hopital's rule:
$$\lim_{x\to-\infty} \frac{\sqrt{x^2+2x}}{-x}$$
$$=\lim_{x\to-\infty} \frac{\dfrac{x+1}{\sqrt{x^2+2x}}}{-1}$$
$$=-\lim_{x\to-\infty} \dfrac{x+1}{\sqrt{x^2+2x}}$$
$$=-\lim_{x\to-\infty} \frac{1}{\dfrac{x+1}{\sqrt{x^2+2x}}}$$
$$=-\lim_{x\to-\infty} \dfrac{\sqrt{x^2+2x}}{x+1}$$
$$=-\lim_{x\to-\infty} \frac{\dfrac{x+1}{\sqrt{x^2+2x}}}{1}$$
$$=-\lim_{x\to-\infty} \dfrac{x+1}{\sqrt{x^2+2x}}$$
I can't get a determinate form.
My questions:
*
*How do I find $(1)$ using factorization?
*How do I find $(1)$ using L'Hopital's rule?
Related
| This answer is based on @KaviRamaMurthy's comments.
Alternative way to find the limit:
$$\lim_{x\to-\infty} \frac{\sqrt{x^2+2x}}{-x}$$
$$\text{[$\sqrt{x^2+2x}$ is positive. Moreover, $-x$ is also positive. So, $\frac{\sqrt{x^2+2x}}{-x}$ is positive.]}$$
$$=\lim_{x\to-\infty} \frac{\sqrt{x^2+2x}}{|x|}$$
$$=\lim_{x\to-\infty} \frac{\sqrt{x^2+2x}}{\sqrt{x^2}}$$
$$=\sqrt{\lim_{x\to-\infty} \frac{x^2+2x}{x^2}}$$
$$\text{[Recall that $\lim_{x\to a} \sqrt[n]{f(x)}=\sqrt[n]{\lim_{x\to a} f(x)}$]}$$
$$=\sqrt{\lim_{x\to-\infty} 1+\frac{2}{x}}$$
$$=\sqrt{1}$$
$$=1\text{(Ans.)}$$
Let us consider another case where the function is negative. Let us consider $\lim_{x\to-\infty} \frac{\sqrt{x^2+2x}}{x}$.
Another case:
$$\lim_{x\to-\infty} \frac{\sqrt{x^2+2x}}{x}$$
$$=-\lim_{x\to-\infty} \frac{\sqrt{x^2+2x}}{-x}$$
$$=-\lim_{x\to-\infty} \frac{\sqrt{x^2+2x}}{|x|}$$
$$=-\lim_{x\to-\infty} \frac{\sqrt{x^2+2x}}{\sqrt{x^2}}$$
$$=-\sqrt{\lim_{x\to-\infty} \frac{x^2+2x}{x^2}}$$
$$=-\sqrt{\lim_{x\to-\infty} 1+\frac{2}{x}}$$
$$=-\sqrt{1}$$
$$=-1\text{(Ans.)}$$
| {
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Why does the method of inclusion/exclusion give the wrong answer when finding the number of integers b/w 1 and 10 that are not divisible by 2,3 or 5? Let $S=\{1,2,\dots, 10\}$.
METHOD 1: I'm first counting the integers that are divisible by $2, 3$ or $5$ in $S$ and then subtracting from the total as follows:
Let $A, B, C$ be the set of integers that are divisible by $2, 3$ and $5$, respectively.
$A=\{2,4,6,8,10\}, B=\{3, 6, 9\}$ and $C=\{5, 10\}$.
We have $|A\cup B\cup C|=|A|+|B|+|C|-|A\cap B| - |A\cap C| - |B\cap C| + |A\cap B\cap C|=5+3+2-1-1-0=8$ is the number of integers that are divisible by $2, 3$ or $5$, so that $10-8=2$ must be those not divisible by either of $2, 3$ or $5$.
METHOD 2: Here I'm counting the integers not divisible by $2, 3$ or $5$ in $S$ directly as follows:
Let $A, B, C$ be the set of integers that are not divisible by $2, 3$ and $5$, respectively.
$A=\{1,3,5,7,9\}, B=\{1,2,4,5,7,8,10\}$ and $C=\{1,2,3,4,6,7,8,9\}$.
Then $|A\cup B\cup C|=|A|+|B|+|C|-|A\cap B| - |A\cap C| - |B\cap C| + |A\cap B\cap C|=5+7+8-3-4-5+2=10$, which is clearly wrong.
| In the second method you want to get $|A\cap B \cap C|$ not $|A \cup B \cup C|$.
| {
"language": "en",
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The Maclaurin series of $1-(1-\frac{x^2}{2} + \frac{x^4}{24})^{2/3}$ has all coefficients positive It was shown in a previous post that the Maclaurin series of $1 - \cos^{2/3} x$ has positive coefficients. There @Dr. Wolfgang Hintze: has noticed that the truncation $1- \frac{x^2}{2} + \frac{x^4}{24}$ can be substituted for $\cos x$ ( seems to be true for all the truncations). The proof is escaping me.
Thank you for your attention!
$\bf{Added:}$ Thomas Laffey in this paper directs to a proof of the fact that if $a_1$, $\ldots$, $a_n\ge 0$ then $\alpha = \frac{1}{n}$ makes the following series positive:
$$1- (\prod_{i=1}^n (1- a_i x))^{\alpha}$$
Numerical testing suggests that $\alpha = \frac{\sum a_i^2}{(\sum a_i)^2} \ge \frac{1}{n}$ works as well ( see the case $n=2$ tested here). So in our case, instead of $\alpha = \frac{1}{2}$ we can take $\alpha = \frac{2}{3}$. Clearly, this would then be the optimal value. This would be a test case for $n=2$. The result for $\cos x$ used the special properties of the function ( solution of a certain differential equation of second order). Maybe $1- x/2 + x^2/24$ is as general as any quadratic with two positive (distinct) roots.
| Hint
Consider
$$f(a)=1-\left(1-\frac{x^2}{2}+\frac{x^4}{24}\right)^a$$ and use the binomial expansion.
It should write
$$f(a)=\frac a 2 \sum_{n=1}^\infty\frac {P_n(a)}{b_n} x^{2n}$$ You should see it very quickly.
Edit
After your edit and remarks, considering
$$f(k)=1-\big[ (1-ax)(1-b x)\big]^k$$ let $b=a c$ to obtain
$$f(k)=k\sum_{n=1}^\infty \frac{a^n}{n!}\, Q_n(c)\, x^n$$ with
$$Q_1(c)=c+1 \qquad\text{and} \qquad Q_2(c)=(c^2+1)-k (c+1)^2$$ If, as you did, we choose to cancel $Q_2(c)$ that is to say to use $k=\frac{c^2+1}{(c+1)^2}$ we have
$$\color{blue}{Q_n(c)= (n-2)\, 2^{\left\lfloor \frac{n-1}{2}\right\rfloor }\, \frac{c\, (c^2-1)^2}{(c+1)^{n}} \,\,R_n(c)}$$ with
$$\left(
\begin{array}{cc}
n & R_n(c) \\
3 & 1 \\
4 & c^2+3 c+1 \\
5 & \left(c^2+c+1\right) \left(c^2+4 c+1\right) \\
6 & \left(c^2+1\right) \left(c^2+4 c+1\right) \left(3 c^2+8 c+3\right) \\
7 & \left(c^2+3 c+1\right) \left(c^2+4 c+1\right) \left(6 c^4+7 c^3+6 c^2+7 c+6\right)
\end{array}
\right)$$
| {
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"source": "stackexchange",
"question_score": "6",
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fourth-order finite difference for $(a(x)u'(x))'$ Previously I asked here about constructing a symmetric matrix for doing finite difference for $(a(x)u'(x))'$ where the (diffusion) coefficient $a(x)$ is spatially varying. The answer provided there works for getting a second order accurate method. What about getting a fourth-order accurate method? If I follow the same idea and apply the following fourth-order accurate formula for first derivative
$ u'_i = \dfrac{u_{i-1} - 8u_{i-1/2} + 8u_{i+1/2} - u_{i+1}}{6\Delta x}$ (1)
in succession, then I end up getting a formula for $(a u')_i'$ which involves $u_{i-2}, u_{i-3/2}, u_{i-1}, u_{i-1/2}, u_i, u_{i+1/2}, u_{i+1}, u_{i+3/2}, u_2$. It involves the mid point values $u_{i+n/2}$ and it depends on nine neighbouring values of $u_i$, which doesn't sound right for fourth order accurate scheme. For the special case of $a(x)=1$ it also doesn't reduce to the formula
$ u'' = \dfrac{-u_{i-2} + 16u_{i-1} - 30u_i + 16u_{i+1} - u_{i+2}}{12\Delta x^2}$. (2)
So what's wrong with applying (1) in succession? What's the correct approach to get a finite difference formula for $(au')'$ with fourth order accuracy?
Edit 1: After reading this post I managed to derive a symmetric four-order centered difference scheme for $[a(x)u'(x)]'$ by applying
$ u'(x) = \dfrac{u_{i-3/2} - 27u_{i-1/2} + 27u_{i+1/2} - u_{i+3/2}}{24\Delta x} + \mathcal{O}(\Delta x^4)$
twice in succession. The resulting formula for $[a(x)u'(x)]'$ is
$ (au')'_i = \dfrac{1}{576\Delta x^2} \left( a_{i-\frac{3}{2}}u_{i-3}
- 27(a_{i-\frac{3}{2}} + a_{i-\frac{1}{2}})u_{i-2}
+ (27a_{i-\frac{3}{2}} + 729a_{i-\frac{1}{2}}+27a_{i+\frac{1}{2}})u_{i-1}
- (a_{i-\frac{3}{2}} + 729a_{i-\frac{1}{2}} + 729a_{i+\frac{1}{2}} + a_{i+\frac{3}{2}})u_i
+ (27a_{i-\frac{1}{2}} + 729a_{i+\frac{1}{2}}+27a_{i+\frac{3}{2}})u_{i+1}
- 27(a_{i+\frac{1}{2}} + a_{i+\frac{3}{2}})u_{i+2}
+ a_{i+\frac{3}{2}}u_{i+3} \right)$
which involves the mid point values of $a(x)$ and has seven stencils. Is there a formula that involves even less computations / stencils(e.g. five stencils)?
| A bit of work with five-point Lagrange interpolants based on a uniform grid for $a$ and $u$ shows, with
\begin{align}
w_{-2} &:= \tfrac{1}{144}a_{-2} - \tfrac{1}{18} a_{-1} - \tfrac{1}{12}a_0 + \tfrac{1}{18} a_1 - \tfrac{1}{144} a_2, \\
w_{-1} &:= -\tfrac{1}{18}a_{-2} + \tfrac{4}{9} a_{-1} + \tfrac{4}{3} a_0 - \tfrac{4}{9}a_1 + \tfrac{1}{18}a_2 \\
w_0 &:= -\tfrac{5}{2}a_0, \\
w_{1} &:= \tfrac{1}{18}a_{-2} - \tfrac{4}{9}a_{-1} + \tfrac{4}{3}a_0 + \tfrac{4}{9}a_1 - \tfrac{1}{18}a_2, \\
w_{2} &:= -\tfrac{1}{144}a_{-2} + \tfrac{1}{18}a_{-1} - \tfrac{1}{12}a_0 - \tfrac{1}{18}a_1 + \tfrac{1}{144}a_2
\end{align}
that $(au')'(0) = (w_{-2}u_{-2}+ w_{-1}u_{-1} + w_0u_0 + w_1 u_1 + w_2 u_2)/h^2 + O(h^4)$. It typically won't be a symmetric rule.
ADDED EDIT
Since I was thinking in terms of using grid values I started with the Lagrange interpolant for $a$
\begin{align*}
A(x) &= \\
& \quad a_{-2}\,
\frac{x - x_{-1}}{x_{-2} - x_{-1}}
\frac{x - x_{0}}{x_{-2} - x_{0}}
\frac{x - x_{1}}{x_{-2} - x_{1}}
\frac{x - x_{2}}{x_{-2} - x_{2}} \\
&+ a_{-1}\,
\frac{x - x_{-2}}{x_{-1} - x_{-2}}
\frac{x - x_{0}}{x_{-1} - x_{0}}
\frac{x - x_{1}}{x_{-1} - x_{1}}
\frac{x - x_{2}}{x_{-1} - x_{2}} \\
&+ a_{0}\,
\frac{x - x_{-2}}{x_{0} - x_{-2}}
\frac{x - x_{-1}}{x_{0} - x_{-1}}
\frac{x - x_{1}}{x_{0} - x_{1}}
\frac{x - x_{2}}{x_{0} - x_{2}} \\
&+ a_{1}\,
\frac{x - x_{-2}}{x_{1} - x_{-2}}
\frac{x - x_{-1}}{x_{1} - x_{-1}}
\frac{x - x_{0}}{x_{1} - x_{0}}
\frac{x - x_{2}}{x_{1} - x_{2}} \\
&+ a_{2}\,
\frac{x - x_{-2}}{x_{2} - x_{-2}}
\frac{x - x_{-1}}{x_{2} - x_{-1}}
\frac{x - x_{0}}{x_{2} - x_{0}}
\frac{x - x_{1}}{x_{2} - x_{1}}
\end{align*}
and the corresponding $U$ for $u$. Setting $x_{k} = kh$ for a uniform mesh simplifies the expressions and makes the template symmetric about $x=0$. Computing $(AU')'$ and evaluating it at $x=0$ gives the result which can be verified to be $O(h^4)$. The symmetric template is key.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4455880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Showing $\int_{0}^{2022}x^{2}-\lfloor{x}\rfloor\lceil{x}\rceil dx = 674$ It is from the 2022 MIT Integration Bee Question 3 states as follows:
$$\int_{0}^{2022}x^{2}-\lfloor{x}\rfloor\lceil{x}\rceil dx$$
I know that the answer is $674$, but I do not know the process and the steps to derive this solution. Can someone please help?
| To keep things simple, let's limit the integration to the interval $(n, n+1)$. where $n \in \mathbb{Z}$.
$$\int_n^{n+1} x^2 - \lfloor x \rfloor \lceil x \rceil dx $$
$$= \int_n^{n+1} x^2 - n (n + 1) dx $$
$$= \left(\frac{1}{3}x^3 - n (n + 1)x\right)\Bigg|_{n}^{n+1} $$
$$= \left(\frac{1}{3}(n+1)^3 - n (n + 1)^2\right) - \left(\frac{1}{3}n^3 - n^2 (n + 1)\right)$$
$$= \frac{1}{3}n^3 + n^2 + n + \frac{1}{3} - n^3 - 2n^2 - n - \frac{1}{3}n^3 + n^3 + n^2 $$
$$= \frac{1}{3} $$
Gee, sure is convenient that all of those $n$'s cancel out and we're left with a constant.
So all we have to do is add up the integral value $\frac{1}{3}$ over $2022$ of these $1$-unit intervals, to get $2022 / 3 = 674.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4456609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
What is the minimum value of the function $f(x)= \frac{x^2+3x-6}{x^2+3x+6}$? I was trying to use the differentiation method to find the minimum value of the person but it did not give any result, I mean when I differentiated this function and equated to zero for finding the value of $x$ when the function value would be minimum, it gave me an absurd relation like $-6=6$.
How can we solve this? Why is the differentiation method not working here? Please help !!!
$f(x)= \frac{x^2+3x-6}{x^2+3x+6}$
$f'(x)= \frac{(x^2+3x+6)(2x+3)-(x^2+3x-6)(2x+3)}{(x^2+3x+6)^2}$
Then equated this to $0$ to find the value of $x$.
Thanks in advance !!!
| You do not need calculus/derivatives for this. Let $u=x^2+3x+6=\left(x+\frac32\right)^2+\frac{15}{4}$, and note that you would like to minimize $\frac{u-12}{u}=1-\frac{12}{u}$.
$u$ can take values in $[15/4,\infty)$. And $1-\frac{12}{u}$ is an increasing function for positive $u$.
It follows that the minimum occurs when $u=\frac{15}{4}$, which happens when $x=-\frac{3}{2}$. And the minimum value obtained is $1-\frac{12}{u}=1-\frac{12}{15/4}=-\frac{11}{5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4457378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Single Solution for this Recurrence: $a(n)=3^n-a(n-1)+1$ I've solved this recurrence using the iteration method for even and odd values of $n$, but I cannot seem to find a singular explicit function that solves this recurrence for all values of $n$.
The recurrence is
$$a(n) =
\begin{cases}
1, & \text{if $n=0$} \\
3^n-a(n-1)+1, & \text{otherwise}
\end{cases}$$
| Let $A(x)=\sum_{n \ge 0} a_n x^n$ be the ordinary generating function. The recurrence and initial condition imply that
\begin{align}
A(x)
&= a_0 x^0 + \sum_{n \ge 1} a_n x^n \\
&= 1 + \sum_{n \ge 1} (3^n - a_{n-1} + 1) x^n \\
&= 1 + \sum_{n \ge 1} (3x)^n - x \sum_{n \ge 1} a_{n-1} x^{n-1} + \sum_{n \ge 1} x^n \\
&= 1 + \frac{3x}{1-3x} - x A(x) + \frac{x}{1-x},
\end{align}
so
\begin{align}
A(x)
&= \frac{1 + \frac{3x}{1-3x} + \frac{x}{1-x}}{1+x} \\
&= \frac{1 - 3x^2}{(1-x)(1+x)(1-3x)} \\
&= \frac{1/2}{1-x} - \frac{1/4}{1+x} + \frac{3/4}{1-3x} \\
&= \frac{1}{2}\sum_{n\ge 0} x^n - \frac{1}{4}\sum_{n\ge 0} (-x)^n + \frac{3}{4}\sum_{n\ge 0} (3x)^n \\
&= \sum_{n\ge 0} \left(\frac{1}{2} - \frac{1}{4} (-1)^n + \frac{3}{4} 3^n\right)x^n,
\end{align}
which immediately yields
$$a_n = \frac{2-(-1)^n+3^{n+1}}{4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4460339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding the area of a triangle on a unit circle Assume that $0< \theta < \pi$. For three points $A(1,0)$, $B(\cos(\theta),\sin(\theta))$ and $C(\cos(2\theta),\sin(2\theta))$ on a unit circle, the area of triangle $ABC$ is: ???
I drew out the unit circle and tried to get the dimensions. I ended up drawing a triangle underneath the main triangle to try and get the base, but that ended up giving me a square root and I couldn't find a way to get rid of it.
The answer is: $\sin(1-\cos)$
| The three side lengths of $\triangle ABC$ are:
$$BC = AB = \sqrt{(\cos \theta - 1)^2 + (\sin \theta - 0)^2}$$
$$= \sqrt{\cos^2 \theta - 2 \cos \theta + 1 + \sin^2 \theta}$$
$$= \sqrt{2 - 2 \cos \theta}$$
$$AC = \sqrt{(\cos(2\theta) - 1)^2 + (\sin(2\theta)-0)^2}$$
$$= \sqrt{\cos^2(2\theta) - 2\cos(2\theta) + 1 + \sin^2(2\theta)}$$
$$= \sqrt{2 - 2\cos(2\theta)}$$
$$= \sqrt{2 - 2(1 - 2\sin^2 \theta)}$$
$$= \sqrt{4\sin^2\theta}$$
$$= 2 \sin \theta$$
(In that last step, we can assume that $\sin\theta$ is positive because were are given $0 < \theta < \pi$.)
If we cut this isosceles triangle in half by bisecting $\angle B$, we get two right triangles with a hypotenuse of $\sqrt{2 - 2\cos\theta}$ and a base leg of $\sin \theta$. The other leg, the height of the triangle, is:
$$\sqrt{(\sqrt{2 - 2\cos\theta})^2 - \sin^2 \theta}$$
$$= \sqrt{2 - 2\cos\theta - \sin^2 \theta}$$
$$= \sqrt{2 - 2\cos\theta - (1 - \cos^2 \theta)}$$
$$= \sqrt{1 - 2\cos\theta + \cos^2 \theta}$$
$$= \sqrt{(1 - \cos\theta)^2}$$
$$= 1 - \cos\theta$$.
We thus have:
$$\text{area of } \triangle ABC = \frac{1}{2} \text{(base)}\text{(height)}$$
$$=\frac{1}{2} (2 \sin \theta)(1 - \cos \theta)$$
$$=(\sin \theta)(1 - \cos \theta)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4464374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
The complex number $(1+i)$ is root of polynomial $x^3-x^2+2$. Find the other two roots. The complex number $(1+i)$ is root of polynomial $x^3-x^2+2$.
Find the other two roots.
$(1+i)^3 -(1+i)^2+2= (1-i-3+3i)-(1-1+2i) +2= (-2+2i)-(2i) +2= 0$.
The other two roots are found by division.
$$
\require{enclose}
\begin{array}{rll}
x^2 && \hbox{} \\[-3pt]
x-1-i \enclose{longdiv}{x^3 -x^2 + 2}\kern-.2ex \\[-3pt]
\underline{x^3-x^2- i.x^2} && \hbox{} \\[-3pt]
2 +i.x^2
\end{array}
$$
$x^3-x^2+2= (x-1-i)(x^2) +2+i.x^2$
How to pursue by this or some other approach?
|
$x^3-x^2+2= (x-1-i)\,x^2 + 2 + i\,x^2\;$ How to pursue by this or some other approach?
The posted answers cover the other approaches.
OP's approach is also salvageable, but first the polynomial division must be taken to completion:
$$
x^3-x^2+2= \left(x-1-i\right)\left(x^2 + ix - 1 + i\right)
$$
The quotient can be further factored either with the quadratic formula, or directly "by inspection":
$$
x^2 + ix - 1 + i = (x^2-1) + i(x+1)=(x+1)(x-1)+i(x+1)=(x+1)(x-1+i)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4464516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Determine the image of the unit circle $S^1$ by the action of the matrix $e^A$. We have:
$$e^{ \begin{pmatrix}
-5 & 9\\
-4 & 7
\end{pmatrix} }$$
I need to determine the image of the unit circle $S^1$ by the action of the matrix $e^A$.
I think that I know how to calculate $e^A$:
I get the Jordan decomposition:
$$A = \begin{pmatrix}
-5 & 9\\
-4 & 7
\end{pmatrix} =
\begin{pmatrix}
-6 & 1\\
-4 & 0
\end{pmatrix}
\cdot
\begin{pmatrix}
1 & 1\\
0 & 1
\end{pmatrix}
\cdot
\frac{1}{4}
\begin{pmatrix}
0 & -1\\
1 & -6
\end{pmatrix}
$$
With eigenvalues: $\lambda$ = 1, algebraic multiplicity = 2, eigenvecotrs: $\left\{ \begin{pmatrix}
1\\
0
\end{pmatrix}, \begin{pmatrix}
0\\
1
\end{pmatrix} \right\}$
$$ \displaystyle e^A = \sum^{\infty}_{i = 0} \frac{A^i}{i!}$$
$$e^A =
\begin{pmatrix}
-6 & 1\\
-4 & 0
\end{pmatrix}
\cdot
\left(
\begin{pmatrix}
1 & 0\\
0 & 1
\end{pmatrix}
+
\displaystyle \sum^{\infty}_{i = 1} \frac{ \begin{pmatrix}
1 & 1\\
0 & 1
\end{pmatrix}}{i!}
\right)
\cdot
\frac{1}{4}
\begin{pmatrix}
0 & -1\\
1 & -6
\end{pmatrix}$$
$$e^A = \begin{pmatrix}
-5 & 9\\
-4 & 7
\end{pmatrix} =
\begin{pmatrix}
-6 & 1\\
-4 & 0
\end{pmatrix}
\cdot
\begin{pmatrix}
\displaystyle \sum^{\infty}_{i = 1} \frac{1}{i!}& \displaystyle \sum^{\infty}_{i = 1} \frac{2^{i-1}}{i!}\\
0 & \displaystyle \sum^{\infty}_{i = 1} \frac{1}{i!}
\end{pmatrix}
\cdot
\frac{1}{4}
\begin{pmatrix}
0 & -1\\
1 & -6
\end{pmatrix}$$
Where: $$\displaystyle \sum^{\infty}_{i = 1} \frac{2^{i-1}}{i!} = \frac{1}{2} \sum^{\infty}_{i = 1} \frac{2^{i}}{i!} = \frac{1}{2}(e^2 - 1)
$$
So:
$$e^A =
\begin{pmatrix}
-6 & 1\\
-4 & 0
\end{pmatrix}
\cdot
\begin{pmatrix}
e & \displaystyle \frac{e^2}{2} - \displaystyle \frac{1}{2}\\
0 & e
\end{pmatrix}
\cdot
\frac{1}{4}
\begin{pmatrix}
0 & -1\\
1 & -6
\end{pmatrix} =
\begin{pmatrix}
\displaystyle \frac{-3e^2 + e + 3}{4} & \displaystyle \frac{9e^2 - 9}{2}\\
\displaystyle \frac{-e^2 + 1}{2} & 3e^2 + e - 3
\end{pmatrix}
$$
Now, I don't know if I did it correctly up to this point and what I should do next - to operate on my unit circle.
Solution:
Because of @Oscar Lanzi we know that:
$$e^{\begin{pmatrix}-5 & 9\\-4 & 7\end{pmatrix}}=e\begin{pmatrix}-5 & 9\\-4 & 7\end{pmatrix}$$
Then because of that:
Equation of unit circle under linear transformation - can't understand role of inverse matrix
(answer by @Prototank)
We know that the image of unit circle in action of the matrix $A$ is given by: $$65x^{2}-166xy+106y^{2}=1$$
Now we need to scale by $e$ and we get the image of unit circle in action of the matrix $e^A$:
$$65x^{2}-166xy+106y^{2}=e^2$$
| The calculation won't be easy, but standard.
To understand the image of any invertible matrix $T$ acting on the unit circle, first do the SVD decomposition $$T=U\Sigma V = U\begin{pmatrix} \sigma_1 & \\ & \sigma_2\end{pmatrix}V$$ Note that $V$ send the circle back to itself, as it keeps norms. Now it should be clear that the image is just an ellipse with two principal axes being $\sigma_1, \sigma_2$ along the directions of the two column vectors of $U$.
Also to find $U$ and $\sigma_i$, we only have to diagonalize $TT^*$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4467835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
If $x$ and $y$ are positive integers such that $5x+3y=100$, what is the greatest possible value of $xy$? I first wrote $y$ in terms of $x$. in $5x + 3y = 100$, I subtracted $5x$ from both sides to get $3y = 100 - 5x$. Therefore, $y = \frac{100 - 5x}{3}$. I substituted this into $xy$ to get $x(\frac{100 - 5x}{3}$) – You want to find the maximum value of this. This can be simplified as $\frac{-5x^2 + 100x}{3}$. I factored out a $-5$ to get $\frac{-5(x^2 - 20x)}{3}$. Completing the Square, I got $\frac{-5((x - 10)^2 - 100)}{3}$, or $\frac{-5(x - 10)^2 + 500}{3}$. The maximum value of this is when $x = 10$, since it makes $-5(x - 10)^2$ equal $0$ ($0$ is the greatest value because otherwise it would be negative). So my answer was $\boxed{\frac{500}{3}}$, but I'm pretty certain that isn't correct because the product of two positive integers can't be a fraction. Can someone help me out?
~
EDIT: I found a case where $x=11$. Then, the product is $165$. Not sure if that is the maximum, though.
| Since $$3y= 100-5x\implies 3\mid 20-x$$ we have $20-x =3t$ for some integer $t$. Then $x= 20-3t$ and $y= 5t$. So we have $$xy = 5t(20-3t) \leq \max \{20\cdot 8, 15\cdot 11\}$$
(Maximum value of quadratic function is at $10\over 3$ so in the set of integers it is at $3$ or at $4$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4468673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Possible growth rates of a matrix entry with respect to exponentiation Let $A = \begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}$, so $A^n = \begin{pmatrix}1 & n \\ 0 & 1\end{pmatrix}$. Thus, $(A^n)_{1,1} = 1 = \Theta(1)$, and $(A^n)_{1,2} = n = \Theta(n)$.
Given a constant $c$, if $B = \begin{pmatrix} c/2 & c/2 \\ c/2 & c/2\end{pmatrix}$ then $B^n = \begin{pmatrix} c^n/2 & c^n/2 \\ c^n/2 & c^n/2\end{pmatrix}$. Thus, $(B^n)_{1,1} = c^n/2 = \Theta(c^n)$.
We have observed three possible growth rates of a particular entry with respect to the exponent of a matrix, constant, linear, and exponential. What other growth rates are possible? For example, does there exist an $m\times m$ matrix $C$ and indices $i,j$ such that $(C^n)_{i,j} = \Theta(n^2)$?
| Consider $A=I+S,$ where $S$ a matrix with entries $s_{i,i+1}=1$ and $0$ otherwise. If the dimension is equal $m,$ then $$A^n=I+ \sum_{k=1}^{m-1} {n\choose k}S^k$$ Thus $$(A^n)_{1,1+j}={n\choose j}\approx {n^j\over j!}\quad \qquad 1\le j\le m-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4470374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Inequality about $x^p$ for $p\geq 2$ a real number I am stuck with the following inequality
$$g^2(n+2)+2g(n+2)g(n)-g^2(n+1)-2g(n+1)g(n+2)\geq 0,$$
for all integer $n\geq 1$. Here, $g(x)=x^p$ where $p\geq 2$ is a real number. I need help.
| It suffices to prove that
$$(n + 2)^{2p} + 2(n + 2)^p n^p - (n + 1)^{2p} - 2(n + 1)^p(n + 2)^p \ge 0$$
or
$$[(n + 2)^p - (n + 1)^p]^2
- 2(n + 1)^{2p} + 2(n + 2)^p n^p \ge 0$$
or (multiplying both sides by $(n + 1)^{-2p}$)
$$\left(\left(1 + \frac{1}{n + 1}\right)^p - 1\right)^2
- 2 + 2\left(1 - \frac{1}{(n + 1)^2}\right)^p \ge 0.$$
Using Bernoulli inequality, we have
$$\left(1 + \frac{1}{n + 1}\right)^p \ge 1 + \frac{p}{n + 1}$$
and
$$\left(1 - \frac{1}{(n + 1)^2}\right)^p \ge 1 - \frac{p}{(n + 1)^2}.$$
It suffices to prove that
$$\left(\frac{p}{n + 1}\right)^2
- 2 + 2\left(1 - \frac{p}{(n + 1)^2}\right) \ge 0$$
or
$$\frac{p(p - 2)}{(n + 1)^2} \ge 0$$
which is true.
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4471178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the limit $\lim _{x\to 0}\frac{\cos x-1+\frac{x}{2}\cdot \sin x}{\ln ^4(x+1)}$ I need to find $\displaystyle \lim _{x\to 0}\frac{\cos x-1+\frac{x}{2}\cdot \sin x}{\ln ^4\left(x+1\right)}$.
I tried using the following:
\begin{align*}
\ln(1+x)&\approx x,\\
\sin(x)&\approx x-\frac{x^3}{2},\\
\cos(x)&\approx 1-\frac{x^2}{2!}+\frac{x^4}{4!},
\end{align*}
I managed to get
$\displaystyle \lim _{x\to 0}\frac{\cos x-1+\dfrac{x}{2}\cdot \sin x}{\ln ^4\left(x+1\right)}=\lim _{x\to 0}\frac{\left(\dfrac{\left(-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}\right)}{x^{4}}+\dfrac{x^{2}-\frac{x^{4}}{3!}}{2x^{4}}\right)}{\left(\dfrac{\ln\left(1+x\right)-x}{x}+1\right)^{4}}$
but I can't see how to use that $\displaystyle \lim _{x\to 0}\frac{\ln\left(1+x\right)-x}{x}=0$ in here.
Am I missing something? I didn't learn the little-$\mathcal{O}$ notation yet.
Thanks
| You only need to add to your equalities the $\log$
$$\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + \textrm{h.o.t}\\
x \sin x = x^2 - \frac{x^4}{6} + \textrm{h.o.t}\\
\log(1+x) = x + \textrm{h.o.t, so} \\
\log^4(1+x) = x^4 + \textrm{h.o.t}$$
Now calculate carefully and see that
$$\frac{\cos x-1 - \frac{x}{2} \sin x}{\log^4(1+x)} =\ldots = \frac{-\frac{x^4}{24} + \textrm{h.o.t}}{x^4 + \textrm{h.o.t}}$$
Now divide numerator and denominator by $x^4$, and get your result. You were on the right track!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4477665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Solving the integral $\int_{0}^{1} \frac{s \ln{s} - t \ln{t}}{(s^{2} - t^{2})\sqrt{1-s^{2}}}\,\mathrm{d}s$ For $t \in [0,1]$, let
$$f(t) = \int_{0}^{1} \frac{s \ln{s} - t \ln{t}}{(s^{2} - t^{2})\sqrt{1-s^{2}}}\,\mathrm{d}s.$$
Is it possible to evaluate this integral to obtain an explicit expression for $f(t)$? Mathematica and Maple both fail to give an answer.
| Split the integral into two
\begin{align}
I=& \int_{0}^{1} \frac{s \ln{s} - t \ln{t}}{(s^{2} - t^{2})\sqrt{1-s^{2}}}ds=I_1+I_2
\end{align}
where, with $b = \sqrt{1-t^2}$
\begin{align}
I_1= & \int_{0}^{1} \frac{s \ln{s} - s \ln{t}}{(s^{2} - t^{2})\sqrt{1-s^{2}}}ds
\overset{x=\sqrt{1-s^2}}=\frac12 \int_{0}^{1}
\frac{\ln\frac{1-x^2}{1-b^2}}{b^2-x^2}dx\\
=&\ \frac1{4b}\bigg[\text{Li}_2\left(\frac{b-1}{b+1}\right)-\text{Li}_2\left(\frac{b+1}{b-1} \right)\bigg]\\
\\
I_2= & \int_{0}^{1} \dfrac{s \ln{t} - t \ln{t}}{(s^{2} - t^{2})\sqrt{1-s^{2}}}ds\\
=& \int_{0}^{1} \dfrac{\ln{t}}{(s +t)\sqrt{1-s^{2}}}ds=\frac{2\ln t}{\sqrt{1-t^2}}\tanh^{-1}\sqrt{\frac{1-t}{1+t}}\\
\end{align}
Thus
\begin{align}
I=& \frac1{4\sqrt{1-t^2}}\left[ \text{Li}_2\bigg(\frac{\sqrt{1-t^2}-1}{\sqrt{1-t^2} +1}\bigg)-\text{Li}_2\bigg(\frac{\sqrt{1-t^2} +1}{\sqrt{1-t^2}-1} \bigg)
+ 8\ln t\tanh^{-1}\sqrt{\frac{1-t}{1+t}}\right]
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Simplest proof of $|I_n+J_n|=n+1$ Answering another question, I realised that I "know" that the determinant of the sum of an identity matrix $I_n$ and an all-ones square matrix $J_n$ is $n+1$. I.e.
$$|I_n+J_n|=\left|\begin{pmatrix}
1 & 0 & 0 &\cdots & 0 \\
0 & 1 & 0 &\cdots & 0 \\
0 & 0 & 1 &\cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & 1
\end{pmatrix} +\begin{pmatrix}
1 & 1 & 1 & \cdots & 1 \\
1 & 1 & 1 & \cdots & 1 \\
1 & 1 & 1 & \cdots & 1 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 1 & 1 & \cdots & 1
\end{pmatrix}\right| =\left|\begin{matrix}
2 & 1 & 1 & \cdots & 1 \\
1 & 2 & 1 & \cdots & 1 \\
1 & 1 & 2 & \cdots & 1 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 1 & 1 & \cdots & 2
\end{matrix} \right|=n+1.$$
I can prove it by adding and subtracting rows and columns without changing the determinant:
*
*subtracting the first row from the others $
\left|\begin{matrix}
2 & 1 & 1 & \cdots & 1 \\
1 & 2 & 1 & \cdots & 1 \\
1 & 1 & 2 & \cdots & 1 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 1 & 1 & \cdots & 2
\end{matrix} \right|=
\left|\begin{matrix}
2 & 1 & 1 & \cdots & 1 \\
-1 & 1 & 0 & \cdots & 0 \\
-1 & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
-1 & 0 & 0 & \cdots & 1
\end{matrix} \right|$
*subtracting the other rows from the first $\left|\begin{matrix}
2 & 1 & 1 & \cdots & 1 \\
-1 & 1 & 0 & \cdots & 0 \\
-1 & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
-1 & 0 & 0 & \cdots & 1
\end{matrix} \right|=
\left|\begin{matrix}
n+1 & 0 & 0 & \cdots & 0 \\
-1 & 1 & 0 & \cdots & 0 \\
-1 & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
-1 & 0 & 0 & \cdots & 1
\end{matrix} \right|$
*adding the other columns to the first $\left|\begin{matrix}
n+1 & 0 & 0 & \cdots & 0 \\
-1 & 1 & 0 & \cdots & 0 \\
-1 & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
-1 & 0 & 0 & \cdots & 1
\end{matrix} \right|=
\left|\begin{matrix}
n+1 & 0 & 0 & \cdots & 0 \\
0 & 1 & 0 & \cdots & 0 \\
0 & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & 1
\end{matrix} \right|$
*and clearly $\left|\begin{matrix}
n+1 & 0 & 0 & \cdots & 0 \\
0 & 1 & 0 & \cdots & 0 \\
0 & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & 1
\end{matrix} \right| =n+1$
but I feel there should be something more intuitive.
What would be a more intuitive or insightful way to prove this?
| The eigenvalues of $J_n$ are $0$ with geometric multiplicity $n-1$ (because the rank of $J_n$ is one) and $n$ with algebraic multiplicity at least $1$ (an eigenvector is the “all $1$” column vector).
Thus the algebraic multiplicity of $0$ is $n-1$ and of $n$ is $1$. Hence the characteristic polynomial of $J_n$ is
$$
p(x)=\det(J_n-xI_n)=(0-x)^{n-1}(n-x)
$$
For $x=-1$, we get
$$
p(-1)=\det(J_n+I_n)=1^{n-1}(n+1)=n+1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4480703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Real part of $ \quad 1- \frac{2}{\pi} \arctan(r^{\rho}e^{i\rho\theta}).$ To solve the Dirichlet problem using mellin transform, i needed to find the real part of $ \quad 1- \displaystyle\frac{2}{\pi} \arctan(r^{\rho}e^{i\rho\theta}).$
I already know the result will be
\begin{cases}
\quad 1- \displaystyle \frac{1}{\pi}\arctan(\frac{2r^{\rho}\cos(\rho \theta)}{1-r^{2\rho}}) & \text{ si } r\in [0, 1]\\ \quad \\
\quad \displaystyle \frac{1}{\pi}\arctan(\frac{2r^{\rho}\cos(\rho \theta)}{r^{2\rho}-1}) & \text{ si } r\in ]1,+\infty[.\\
\end{cases}
I find it in "Dautray R., Lions J.L., Mathematical analysis and numerical calculation for science and technology."
I want to know how they found it ?
| For a complex number $z \ne \pm \mathrm{i}$, we have
$$\arctan z = \frac{1}{2\mathrm{i}}\ln \frac{\mathrm{i} - z}{\mathrm{i} + z}$$
where $\ln u$ is the principal branch of complex logarithm ($u\ne 0$)
$$\ln u = \ln |u| + \mathrm{i}\arg u, \quad -\pi < \arg u \le \pi.$$
(See, e.g. http://scipp.ucsc.edu/~haber/archives/physics116A10/arc_10.pdf)
Also, if $c, d$ are real numbers with $c^2 + d^2 \ne 0$, we have
$$\arg (c + d\mathrm{i}) =
\left\{\begin{array}{ll}
\arctan \frac{d}{c} & c > 0 \\[5pt]
\arctan \frac{d}{c} + \pi & c < 0, d \ge 0 \\[5pt]
\arctan \frac{d}{c} - \pi & c < 0, d < 0\\[5pt]
\frac{\pi}{2} & c = 0, d > 0\\[5pt]
-\frac{\pi}{2} & c = 0, d < 0
\end{array}
\right.$$
See, e.g. How to figure out the Argument of complex number?
$\phantom{2}$
Consider the case when $r > 0$ and $\rho > 0$.
Let $a = r^\rho \cos \rho\theta$
and $b = r^\rho \sin \rho\theta$.
Then $a^2 + b^2 = r^{2\rho}$.
(1) If $0 < r < 1$, we have
$$\mathrm{Re} [\arctan (a + b\mathrm{i})]
= \mathrm{Re} \left(\frac{1}{2\mathrm{i}}\ln \frac{1 - a^2 - b^2 + 2a \mathrm{i}}{a^2 + (1 + b)^2}\right) = \frac12 \arctan \frac{2a}{1 - a^2 - b^2}.$$
Thus, we have
$$\mathrm{Re}\left(1- \frac{2}{\pi} \arctan(r^{\rho}e^{i\rho\theta})\right)
= 1 - \frac{1}{\pi}\arctan \frac{2r^\rho \cos \rho\theta}{1 - r^{2\rho}}.$$
(2) If $r > 1$ and $a \ge 0$, we have
$$\mathrm{Re} [\arctan (a + b\mathrm{i})]
= \mathrm{Re} \left(\frac{1}{2\mathrm{i}}\ln \frac{1 - a^2 - b^2 + 2a \mathrm{i}}{a^2 + (1 + b)^2} \right) = \frac{\pi}{2} - \frac12 \arctan \frac{2a}{a^2 + b^2 - 1}.$$
Thus, we have
$$\mathrm{Re}\left(1- \frac{2}{\pi} \arctan(r^{\rho}e^{i\rho\theta})\right)
= \frac{1}{\pi}\arctan \frac{2r^\rho \cos \rho\theta}{r^{2\rho} - 1}.$$
(3) If $r > 1$ and $a < 0$, we have
$$\mathrm{Re} [\arctan (a + b\mathrm{i})]
= \mathrm{Re} \left(\frac{1}{2\mathrm{i}}\ln \frac{1 - a^2 - b^2 + 2a \mathrm{i}}{a^2 + (1 + b)^2} \right) = \frac12 \arctan \frac{2a}{1 - a^2 - b^2} - \frac{\pi}{2}.$$
Thus, we have
$$\mathrm{Re}\left(1- \frac{2}{\pi} \arctan(r^{\rho}e^{i\rho\theta})\right)
= 2 - \frac{1}{\pi}\arctan \frac{ - 2 r^\rho \cos \rho\theta}{r^{2\rho} - 1}.$$
(4) If $r = 1$ and $a > 0$, we have
$$\mathrm{Re} [\arctan (a + b\mathrm{i})]
= \mathrm{Re} \left(\frac{1}{2\mathrm{i}}\ln \frac{2a \mathrm{i}}{a^2 + (1 + b)^2}\right) = \frac{\pi}{4}.$$
Thus, we have
$$\mathrm{Re}\left(1- \frac{2}{\pi} \arctan(r^{\rho}e^{i\rho\theta})\right)
= \frac12.$$
(5) If $r = 1$ and $a < 0$, we have
$$\mathrm{Re} [\arctan (a + b\mathrm{i})]
= \mathrm{Re} \left(\frac{1}{2\mathrm{i}}\ln \frac{2a \mathrm{i}}{a^2 + (1 + b)^2}\right) = -\frac{\pi}{4}.$$
Thus, we have
$$\mathrm{Re}\left(1- \frac{2}{\pi} \arctan(r^{\rho}e^{i\rho\theta})\right)
= \frac32.$$
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4481000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\int_{0}^{1} \lim_{n \rightarrow \infty} \sum_{k=1}^{4n-2}(-1)^\frac{k^2+k+2}{2} x^{2k-1} dx$ for $n \in \mathbb{N}$ For $n \in \mathbb{N}$, evaluate
$$\int_{0}^{1} \lim_{n \rightarrow \infty} \sum_{k=1}^{4n-2}(-1)^\frac{k^2+k+2}{2} x^{2k-1} dx$$
I could not use wolframalpha, I do not know the reason.
For $n = 1$, the integrand $=x+x^3$
For $n = 2$, the integrand $=x+x^3-x^5-x^7+x^9+x^{11}$
For $n = 3$, the integrand $=x+x^3-x^5-x^7+x^9+x^{11}-x^{13}-x^{15}+x^{17}+x^{19}$
and so on.
For $n = 1000$, the integrand $=x+x^3-x^5-x^7+x^9+x^{11}-x^{13}-x^{15}+x^{17}+x^{19}-\dots+x^{7993}+x^{7995}$
Using MS-EXCEL with $n=1000$, I found that the value is approximately $0.56...$.
I do not know if $n \rightarrow \infty$ , will the required expression have a closed form or no.
Your help would be appreciated. THANKS!
| Hint: Notice that the sum is of consecutive odd powers of $x$ changing sign every two terms. What if you break it up into two sums, one of which is of the powers $4k+1$ and one of which is the powers of $4k+3$, both of alternating sign, and compare them to the Taylor series for $\frac{1}{1+y}$ for some suitable $y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4481299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Assistance with solving the integral Can you give me an idea how to handle this integral?
$\int_{0}^{2\pi} (1+\cos{x})\sqrt{3+\cos{x}}\,dx$
| Since $1 - \cos(x) = 2 \sin^2\left(\frac{x}{2} \right) $ then
\begin{align}
\int_{0}^{2\pi} (1+\cos{x})\sqrt{3+\cos{x}}\, \mathrm{d}x & \overset{u= x -\pi}{=} 2\int_{0}^{\pi}2 \sin^2\left(\frac{u}{2} \right) \sqrt{2+2\sin^2\left(\frac{u}{2} \right)}\, \mathrm{d}u \\
& \overset{t = \frac{u}{2}}{=} 8 \sqrt{2} \int_0^{\frac{\pi}{2}} \sin^2(t) \sqrt{1 + \sin^2(t)}\, \mathrm{d}t\\
\end{align}
And from this answer we know
$$
\int_0^{\frac{\pi}{2}} \sin^2\theta \sqrt{1-k^2\sin^2 \theta}\, \mathrm{d}\theta = \frac{(1-k^2)K(k)-(1-2k^2)E(k)}{3k^2}
$$
So with $k=i$ you get
$$
\int_{0}^{2\pi} (1+\cos{x})\sqrt{3+\cos{x}}\, \mathrm{d}x =8 \sqrt{2} E(i) - \frac{16 \sqrt{2}}{3} K(i)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4483052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the extrema of $f(x,y)=max(x,y)$ constrained to $\mathscr A=\{(x,y) ∈ \mathbb R^2 ∣ x^2+y^2=1\}$ I know the maxima occurs at $f(x,y) = 1$ at the points $(1,0)$ and $(0,1)$... but I can't seem to find all the minima.
I've come to the conclusion that the minima occurs at a point $(a,a)$ where $a ∈ \mathbb R_<0$, such that $2a^2=1\rightarrow a=\frac{\sqrt 2}{2} \lor a=-\frac{\sqrt 2}{2}$.
Does the minima occur at ($-\frac{\sqrt 2}{2}, -\frac{\sqrt 2}{2}$)?
| The original curve can be expressed by the following equation in polar coordinates $(\rho,\theta)$:
$$
\rho=1.
$$
$(1)$ When $y\ge x$, there is $\theta\in[\frac{\pi}{4},\frac{3\pi}{4}]$, we have $f(x,y)=y=\text{sin}(\theta)$, so $\min f(x,y)=-\frac{\sqrt{2}}{2}$ when $\theta=\frac{3\pi}{4}$ or $x=y=-\frac{\sqrt{2}}{2}$.
$(2)$ When $y\le x$, there is $\theta\in[-\frac{3\pi}{4},\frac{\pi}{4}]$, we have $f(x,y)=x=\text{cos}(\theta)$, so $\min f(x,y)=-\frac{\sqrt{2}}{2}$ when $\theta=\frac{-3\pi}{4}$ or $x=y=-\frac{\sqrt{2}}{2}$.
To sum up, the minima occurs at $\left(-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)$ under the rectangular coordinate system $(x,y)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4486697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Irrational equation $\left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0$ I saw the problem from one math competition:
$$\left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0$$.
I tried to solve it this way:
\begin{align*}
& \left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0\\
\Leftrightarrow\,\, & \left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x\left(x^{3}-3x+1\right)+1=0\\
\Leftrightarrow\,\, & \left(x^{3}-3x+1\right)\left(\sqrt{x^{2}-1}+x\right)=-1
\end{align*}
Now I tried to use substitutions like $a=x^3-3x$ and $b=\sqrt{x^{2}-1}$ but I cannot get anything. I know that solutions are $x=1$ and $x=\pm\sqrt{2}$.
| HINT
Based on the suggestion of @JohnOmielan in the comments, assuming that $|x| > 1$, it yields that
\begin{align*}
(x^{3} - 3x + 1)(\sqrt{x^{2} - 1} + x) = -1 & \Longleftrightarrow x^{3} - 3x + 1 = \sqrt{x^{2} - 1} - x\\\\
& \Longleftrightarrow x^{3} - 2x + 1 = \sqrt{x^{2} - 1}\\\\
& \Longleftrightarrow (x^{3} - 2x + 1)^{2} = x^{2} - 1\\\\
& \Longleftrightarrow x^{6} - 4x^{4} + 2x^{3} + 4x^{2} - 4x + 1 = x^{2} - 1\\\\
& \Longleftrightarrow x^{6} - 4x^{4} + 2x^{3} + 3x^{2} - 4x + 2 = 0
\end{align*}
Since the sum of the coefficients equals zero, the value $x = 1$ is a root.
Taking advantage of such fact, can you proceed from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4490883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
} |
Prove, using a combinatorial argument, that $\binom{m}{n}\cdot\binom{n}{r}= \binom{m}{r}\cdot \binom{m-r}{n-r}$
Prove, using a combinatorial argument, that
$$\binom{m}{n}\cdot\binom{n}{r}= \binom{m}{r}\cdot \binom{m-r}{n-r}$$
attempt: In $m$, suppose that $n$ went for a ride, and out of those $n$, $r$ were awarded. This can be done in $\binom{m}{n}\cdot\binom{n}{r}$ ways. Now suppose we choose $r$ people to be rewarded from these $m$, and then we choose the $n-r$ people who only went on the tour but were not rewarded: $\binom{m}{r}\cdot \binom{m-r}{n-r}$ . So $$\binom{m}{n}\cdot\binom{n}{r}= \binom{m}{r}\cdot \binom{m-r}{n-r}$$
I'm right?
| Solution:
For the left part, we have
$$
\left( \begin{array}{c}
m\\
n\\
\end{array} \right) \cdot \left( \begin{array}{c}
n\\
r\\
\end{array} \right) =\frac{m!}{n!\left( m-n \right) !}\cdot \frac{n!}{r!\left( n-r \right) !}=\frac{m!}{\left( m-n \right) !\left( n-r \right) !r!},
$$
and for the right part, we have
$$
\left( \begin{array}{c}
m\\
r\\
\end{array} \right) \cdot \left( \begin{array}{c}
m-r\\
n-r\\
\end{array} \right) =\frac{m!}{r!\left( m-r \right) !}\cdot \frac{\left( m-r \right) !}{\left( n-r \right) !\left( m-n \right) !}=\frac{m!}{\left( m-n \right) !\left( n-r \right) !r!}.
$$
Therefore, we have
$$
\left( \begin{array}{c}
m\\
n\\
\end{array} \right) \cdot \left( \begin{array}{c}
n\\
r\\
\end{array} \right) =\left( \begin{array}{c}
m\\
r\\
\end{array} \right) \cdot \left( \begin{array}{c}
m-r\\
n-r\\
\end{array} \right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4491006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $A+B+C+D=\pi$ then find $\sum\cos A\cos B-\sum\sin A\sin B$ Question:
If $A+B+C+D=\pi$ then find $\sum\cos A\cos B-\sum\sin A\sin B$
My Attempt:
$$\cos((A+B)+(C+D))=-1\\\implies\cos(A+B)\cos(C+D)-\sin(A+B)\sin(C+D)=-1\\\implies(\cos A\cos B-\sin A\sin B)(\cos C\cos D-\sin C\sin D)-(\sin A\cos B+\cos A\sin B)(\sin C\cos D+\sin D\cos C)=-1\\$$
By opening the brackets, I am not getting the desired expression. How to approach this?
| Using $\cos(x+y)=\cos x\cos y-\sin x\sin y$, we have
$$\cos A\cos B-\sin A\sin B=\cos(A+B),$$
$$\cos B\cos C-\sin B\sin C=\cos(B+C),$$
$$\cos C\cos D-\sin C\sin D=\cos(C+D),$$
$$\cos D\cos A-\sin D\sin A=\cos(D+A),$$
hence
$$\sum\cos A\cos B-\sum\sin A\sin B=\cos(A+B)+\cos(C+D)+\cos(B+C)+\cos(D+A).$$
Now using $\cos(\pi-x)+\cos x=0$, we have
$$\cos(A+B)+\cos(C+D)=0\qquad\text{and}\qquad\cos(B+C)+\cos(D+A)=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4491129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to solve this function $f(x)$ for degree $3$ or $4$
If $f(x)=0$ is a polynomial whose coefficients are either $1$ or $-1$ and whose roots are all real, then the degree of $f(x)$ can be equal to$:$
$A$. $1$
$B$. $2$
$C$. $3$
$D$. $4$
My work$:$
For linear only four polynomials are possible which are $x+1$ and $-x+1$ and $x-1$ and $-x-1$. All of which have real roots. So the answer is $A$.
For quadratic we have $8$ polynomials which are $x^2+x+1$ and $x^2+x-1$ and $x^2-x+1$ and $x^2-x-1$ and $-x^2+x+1$ and $-x^2-x+1$ and $-x^2+x-1$ and $-x^2-x-1$ Here four of the polynomials have real roots which are $x^2+x-1$ and $-x^2-x+1$ and $-x^2+x+1$ and $x^2-x-1$. So we can say that $f(x)$ can be of degree two. So $B$ is also the answer.
For cubic and quartic I don't know how to proceed. Any hints or solutions are appreciated.
| The cubic case is handled by noting that
$(x-1)^2(x+1)=x^3-x^2-x+1=0$
has all of its roots real.
For the quartic case, we do not have such a simply soluble case. But we can use Descartes' Rule of Signs, if we are careful with it.
For instance, when we have all coefficients positive, as in $x^4+x^3+x^2+x+1=0$, four negative roots are allowed. As we will see, that particular equation does not have such roots, but $x^4+10x^3+35x^2+50x+24=0$ with the same sign pattern does (the polynomial is $(x+1)(x+2)(x+3)(x+4)$). The question, then, is how to refine the Rule of Signs so as to distinguish $x^4+x^3+x^2+x+1=0$ from $x^4+10x^3+35x^2+50x+24=0$.
We can improve the bounds from the Rule of Signs by multiplying in factors that will cause the product to have zero coefficients, which is likely to cut down on the sign changes and thus lead to fewer allowed real roots. In the problem here, the coefficients of the polynomial are $\pm1$, which suggests the factor $x+1$ or $x-1$ would likely work.
Going back to $x^4+x^3+x^2+x+1=0$, then we see that choosing $x-1$ as the auxiliary factor gives
$x^5-1=0$
with sign pattern $+0000-$. This allows only one positive root, and reversing alternating signs reveals that there are no negative ones. Thus $x^5-1=0$ has only one real root, which is the root $x=1$ from the multiplier, leaving none for the original quartic equation $x^4+x^3+x^2+x+1=0$.
Now try $x^4+x^3+x^2+x-1=0$. With $x-1$ as a multiplier we get
$x^5-2x+1=0$
$+000-+$
This gives two possible positive roots and one possible (sure) negative root; discounting the introduced root $x=1$ this means $x^4+x^3+x^2+x-1=0$ could have up to two real roots.
By itself that does not mean the equation does not have all real roots; the roots could correspond to repeated factors. But if there is a root with repeated factors the Rule of Signs will actually count the root with that multiplicity. For instance, $x^3-x^2+x+1=(x+1)(x-1)^2=0$ technically has only one positive root, but the corresponding factor appears twice (squared) and the Rule of Signs will count it as two. With this qualification, the bound of two real roots for $x^4+x^3+x^2+x-1=0$ already includes whatever multiplicity of factors there may be, and we are forced to accept that this case actually does not give all real roots.
We go through all the possible cases and determine whether four real roots are possible after trying out the multiplier $x-1$ as above, and if the equation survives that test we then try it also with the auxiliary factor $x+1$ (which the four-root bound must also survive).
Did I say all the possible cases? Actually if we are smart about our trial choices, we need to test only half of them because if any function $P(x)$ fails with $x-1$ as the multiplier, $P(-x)$ will fail with $x+1$ and vice versa.
Then, if any of our eight trials results in a quartic equation that could have four real roots including appropriate multiple factor-counts, we must examine that case more precisely to see if all those real roots actually exist.
All of the possible cases fail with the Rule of Signs augmented by a factor of $x+1$ or $x-1$. So no fourth-degree equation with all coefficients $\pm1$ can have all roots real. Further testing shows that Descartes' Rule of Signs with the auxiliary factors $x\pm1$ also rules out all possibilities fir degree 5, and it seems likely for higher degrees as well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4492077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How to find $\int_0^1 x^4(1-x)^5dx$ quickly? This question came in the Rajshahi University admission exam 2018-19
Q) $\int_0^1 x^4(1-x)^5dx$=?
(a) $\frac{1}{1260}$
(b) $\frac{1}{280}$
(c)$\frac{1}{315}$
(d) None
This is a big integral (click on show steps):
$$\left[-\dfrac{\left(x-1\right)^6\left(126x^4+56x^3+21x^2+6x+1\right)}{1260}\right]_0^1=\frac{1}{1260}$$
It takes a lot of time to compute. How can I compute this quickly (30 seconds) using a shortcut?
| If you don't have the beta function formula memorized, you can use integration by parts repeatedly
\begin{align*}
\int_{0}^{1}x^4(1-x)^5\,dx &= \left[x^4 \cdot -\dfrac{1}{6}(1-x)^6\right]_{0}^{1} - \int_{0}^{1}4x^3 \cdot -\dfrac{1}{6}(1-x)^6\,dx
\\
&= \dfrac{4}{6}\int_{0}^{1}x^3(1-x)^6\,dx
\\
&= \dfrac{4}{6}\left[x^3 \cdot -\dfrac{1}{7}(1-x)^7\right]_{0}^{1} - \dfrac{4}{6}\int_{0}^{1}3x^2 \cdot -\dfrac{1}{7}(1-x)^7\,dx
\\
&= \dfrac{4 \cdot 3}{6 \cdot 7}\int_{0}^{1}x^2(1-x)^7\,dx
\\
&= \dfrac{4 \cdot 3}{6 \cdot 7}\left[x^2 \cdot -\dfrac{1}{8}(1-x)^8\right]_{0}^{1} - \dfrac{4 \cdot 3}{6 \cdot 7}\int_{0}^{1}2x \cdot -\dfrac{1}{8}(1-x)^8\,dx
\\
&= \dfrac{4 \cdot 3 \cdot 2}{6 \cdot 7 \cdot 8}\int_{0}^{1}x(1-x)^8\,dx
\\
&= \dfrac{4 \cdot 3 \cdot 2}{6 \cdot 7 \cdot 8}\left[x \cdot -\dfrac{1}{9}(1-x)^9\right]_{0}^{1} - \dfrac{4 \cdot 3 \cdot 2}{6 \cdot 7 \cdot 8}\int_{0}^{1}1 \cdot -\dfrac{1}{9}(1-x)^9\,dx
\\
&= \dfrac{4 \cdot 3 \cdot 2}{6 \cdot 7 \cdot 8 \cdot 9}\int_{0}^{1}(1-x)^9\,dx
\\
&= \dfrac{4 \cdot 3 \cdot 2}{6 \cdot 7 \cdot 8 \cdot 9}\left[-\dfrac{1}{10}(1-x)^{10}\right]_{0}^{1}
\\
&= \dfrac{4 \cdot 3 \cdot 2}{6 \cdot 7 \cdot 8 \cdot 9 \cdot 10}
\\
&= \dfrac{1}{1260}
\end{align*}
This looks fairly long since I wrote out all the steps in detail. However, if you notice the pattern, you can probably arrive at the answer faster. Also, if are familiar with the Tabular Integration Method, then you only need to do a fairly short amount of scratchwork.
| {
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"url": "https://math.stackexchange.com/questions/4493842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A golden question $\sqrt{2+\sqrt{2-x}}=\sqrt{x-\frac 1x} + \sqrt{1-\frac 1x}$ How would you solve this problem for real $x$?
$$\sqrt{2+\sqrt{2-x}}=\sqrt{x-\frac 1x} + \sqrt{1-\frac 1x}$$
It can be easily shown that both equations
$$x=\sqrt{2+\sqrt{2-x}}\tag{1}$$
and $$x=\sqrt{x-\frac 1x} + \sqrt{1-\frac 1x}\tag{2}$$
have the same real solution, i.e. the golden ratio, $$x = \frac {\sqrt 5 + 1}2=\phi.$$
The main question is derived by combining (1) and (2) into one equation. The final solution is still the same, but the approach might not be as straightforward.
| With some difficulty.
You can turn the equation into a polynomial through lots of squaring and rearranging, starting with:
$\begin{eqnarray} \sqrt{2+\sqrt{2-x}} & = & \sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}} \\
& = & \sqrt{\frac{x - 1}{x}}\left(\sqrt{x + 1} + 1\right) \\
2 + \sqrt{2 - x} & = & \frac{x - 1}{x} \left( x + 2 + 2 \sqrt{x + 1} \right)
\\
\frac{(x-1)(x+2)}{x} - 2 & = & \sqrt{2 - x} - 2 \frac{x - 1}{x}\sqrt{x + 1} \\
x^2 - x - 2 & = & x\sqrt{2 - x} - 2(x-1)\sqrt{x+1} \end{eqnarray}$
After a few more rounds, you should get to something vaguely resembling $x^8-10x^7+39x^6-22x^5-63x^4+32x^3+32x^2 = 0$. You can divide out $x^2$ (which is clearly not a solution to the original equation), and then you can use the knowledge that $\varphi$ is one of the roots to divide out the quadratic $x^2 - x - 1$ (since the polynomial has integer coefficients, you can deduce that $-\varphi^{-1}$ is also a root, which is an erroneous solution introduced in one of the squaring steps).
At this point, you're left with $x^4-9x^3+31x^2-32 = 0$. This quartic has two complex roots and two real roots, but actually figuring them out analytically is what mathematicians call "damn hard", and is commonly achieved via the method known as "chuck it into Mathematica".
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating $\lim _{x \rightarrow 0} \frac{12-6 x^{2}-12 \cos x}{x^{4}}$ $$
\begin{align*}
&\text { Let } \mathrm{x}=2 \mathrm{y} \quad \because x \rightarrow 0 \quad \therefore \mathrm{y} \rightarrow 0\\
&\therefore \lim _{x \rightarrow 0} \frac{12-6 x^{2}-12 \cos x}{x^{4}}\\
&=\lim _{y \rightarrow 0} \frac{12-6(2 y)^{2}-12 \cos 2 y}{(2 y)^{4}}\\
&=\lim _{y \rightarrow 0} \frac{12-24 y^{3}-12 \cos 2 y}{16 y^{4}}\\
&=\lim _{y \rightarrow 0} \frac{3(1-\cos 2 y)-6 y^{2}}{4 y^{4}}\\
&=\lim _{y \rightarrow 0} \frac{3.2 \sin ^{2} y-6 y^{2}}{4 y^{4}}\\
&=\lim _{y \rightarrow 0} \frac{ 3\left\{y-\frac{y^{3}}{3 !}+\frac{y^{5}}{5 !}-\cdots \infty\right\}^{2}-3 y^{2}}{2 y^{4}}\\
&=\lim _{y \rightarrow 0} \frac{3\left[y^{2}-\frac{2 y^{4}}{3 !}+\left(\frac{1}{(3 !)^{2}}+\frac{2}{3 !}\right) y^{4}+\cdots \infty\right)^{2}-3 y^{2}}{2 y^{4}}\\
&=\lim _{y \rightarrow 0} \frac{3\left\{y^{2}-\frac{2 y^{4}}{3 !}+\left\{\frac{1}{(3 !)^{2}}+\frac{2}{5 !}+y^{4}+\cdots \infty\right)-3 y^{2}\right.}{2 y^{4}}\\
&=\lim _{y \rightarrow 0}\left[\frac{-\frac{6}{3 !}+3\left\{\frac{1}{(3 !)^{2}}+\frac{2}{5 !}\right\} y^{2}+\cdots \infty}{2}\right]\\
&=\lim _{y \rightarrow 0}\left[\frac{-1+3\left\{\frac{1}{(3 !)^{2}}+\frac{2}{5 !}\right\} y^{2}+\cdots \infty}{2}\right]\\
&=-\frac{1}{2} \text { (Ans.) }
\end{align*}
$$
Doubt
Can anyone please explain the 5,6,7 equation line? Thank you
| While I'm not entirely sure what's going on with the answer you have provided (exponents seem to appear and disappear, multiplication turns into addition etc.), I shall offer you two methods to evaluate your given limit.
Method 1 : As both the numerator and denominator both tend to $0$ as $x\to0 $, we can use L'Hopital's Rule
$$\displaystyle{\lim_{x \to 0}} \frac{12-6x^2-12\cos(x)}{x^4}=\displaystyle{\lim_{x \to 0}}\frac{-12x+12\sin(x)}{4x^3}$$
As both numerator and denominator still tend to $0$, we can use the rule again,
$$\displaystyle{\lim_{x \to 0}}\frac{-12x+12\sin(x)}{4x^3}=\displaystyle{\lim_{x \to 0}} \frac{-12+12\cos(x)}{12x^2}=\displaystyle{\lim_{x \to 0}}\frac{\cos(x)-1}{x^2}$$
and again,
$$\displaystyle{\lim_{x \to 0}}\frac{\cos(x)-1}{x^2}=\displaystyle{\lim_{x \to 0}}\frac{-\sin(x)}{2x}=-\frac{1}{2}$$
to get the answer.
Method 2 : We could also use the Taylor series expansion for $\cos(x)$ directly, without the cumbersome $y=2x$ substitution or the conversion of $\cos(2y)$ into $\sin^2(y)$.
$$\displaystyle{\lim_{x \to 0}} \frac{12-6x^2-12\cos(x)}{x^4}=\displaystyle{\lim_{x \to 0}} \frac{12-6x^2-12(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}\dotsb \infty)}{x^4}=\displaystyle{\lim_{x \to 0}}\frac{-12(\frac{x^4}{4!}-\frac{x^6}{6!}\dotsb \infty)}{x^4}$$
$$\implies{\lim_{x \to 0}} \frac{12-6x^2-12\cos(x)}{x^4}= $$$$\displaystyle{\lim_{x \to 0}}\frac{-12(\frac{x^4}{4!}-\frac{x^6}{6!}\dotsb \infty)}{x^4}=-\frac{12}{4!}=-\frac{1}{2}$$
Links that may be helpful:
L'Hopital's Rule
Taylor Series of cos(x)
| {
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"url": "https://math.stackexchange.com/questions/4496409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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On Réalis’s solution of the “cubic Markov equation” I am interested in the Diophantine equation
$$X^3+Y^3+Z^3=3XYZ$$
and its solutions. In Nouv. Corr. Math. 1879 (page 7), Réalis claims that, other than the trivial solution $x=y=z$, the complete solution in integers is given by
$$
x=(a-b)^3+(a-c)^3, \quad y=(b-c)^3+(b-a)^3, \quad z=(c-a)^3+(c-b)^3.
$$
It’s easy to show that these values suffice — by simply substituting and doing the algebra — but the result seems just a little too convenient for my tastes…
Does anyone know how to prove [or disprove] the claim? Are there any known solutions that don’t fit Réalis’s parameterization?
| This is quite wrong. We have
$$x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x + y \omega + z \omega^2)(x + y \omega^2 + z \omega)$$
where $\omega = e^{ \frac{2 \pi i}{3} }$ is a primitive third root of unity. This identity comes from recognizing the LHS as the determinant of the circulant matrix $\left[ \begin{array}{ccc} x & y & z \\ z & x & y \\ y & z & x \end{array} \right]$ and using a standard fact about the eigenvalues of circulant matrices. Setting this product equal to zero, we need to solve
$$x + y + z = 0 \text{ or } x + y \omega + z \omega^2 = 0 \text{ or } x + y \omega^2 + z \omega = 0.$$
The first equation is very easy and we get a parameterized family of solutions $\boxed{ x = t, y = s, z = - t - s }$, which amounts to the identity
$$t^3 + s^3 - (t + s)^3 = - 3 ts(t + s)$$
which is not hard to check. The given parametric solution only allows values of $x, y, z$ which are a sum of two cubes so it doesn't include all of these solutions.
For the second two equations, we have $1 + \omega + \omega^2 = 0$ so $\omega^2 = -1 - \omega$. This means the first equation is equivalent to
$$x + y \omega - z(1 + \omega) = (x - z) + (y - z) \omega = 0$$
so $\boxed{ x = y = z }$, and the second equation also gives this. We can also argue without using complex numbers by instead writing
$$\begin{eqnarray*} (x + y \omega + z \omega^2)(x + y \omega^2 + z \omega) &=& x^2 + y^2 + z^2 - xy - yz - zx \\
&=& \frac{(x - y)^2 + (y - z)^2 + (z - x)^2}{2} \end{eqnarray*}$$
which is clearly $\ge 0$ with equality case $x = y = z$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\int_0^1 \frac{\arcsin x\arccos x}{x}dx$ Someone on Youtube posted a video solving this integral.
I can't find on Math.stack.exchange this integral using search engine https://approach0.xyz
It is related to $\displaystyle \int_0^\infty \frac{\arctan x\ln x}{1+x^2}dx$
Following is a solution that is not requiring the use of series:
\begin{align}J&=\int_0^1 \frac{\arcsin x\arccos x}{x}dx\\
&\overset{\text{IBP}}=\underbrace{\Big[\arcsin x\arccos x\ln x\Big]_0^1}_{=0}-\underbrace{\int_0^1 \frac{\arccos x\ln x}{\sqrt{1-x^2}}dx}_{x=\cos t }+\underbrace{\int_0^1 \frac{\arcsin x\ln x}{\sqrt{1-x^2}}dx}_{x=\sin t}\\
&=\int_0^{\frac{\pi}{2}} t\ln(\tan t)dt\\
&\overset{u=\tan t}=\int_0^\infty \frac{\arctan u\ln u}{1+u^2}du\\
&\overset{\text{IBP}}=\underbrace{\left[\arctan u\left(\int_0^u \frac{\ln t}{1+t^2}dt\right)\right]_0^\infty}_{=0}-\int_0^\infty \frac{1}{1+u^2}\left(\underbrace{\int_0^u \frac{\ln t}{1+t^2}dt}_{y(t)=\frac{t}{u}}\right)du\\
&=-\int_0^\infty \left(\int_0^1 \frac{u\ln(uy)}{(1+u^2)(1+u^2y^2)}dy\right)du\\
&=-\int_0^\infty \left(\int_0^1 \frac{u\ln u}{(1+u^2)(1+u^2y^2)}dy\right)du-\int_0^1 \left(\int_0^\infty \frac{u\ln y}{(1+u^2)(1+u^2y^2)}du\right)dy\\
&=-\int_0^\infty \left[\frac{\arctan(uy)}{1+u^2}\right]_{y=0}^{y=1}\ln udu-\frac{1}{2}\int_0^1 \left[\frac{\ln\left(\frac{1+u^2}{1+u^2y^2}\right)}{1-y^2}\right]_{u=0}^{u=\infty}\ln ydy\\
&=-J+\int_0^1 \frac{\ln^2 y}{1-y^2}dy\\
&=\frac{1}{2}\int_0^1 \frac{\ln^2 y}{1-y}dy-\frac{1}{2}\underbrace{\int_0^1 \frac{y\ln^2 y}{1-y^2}dy}_{z=y^2}\\
&=\frac{7}{16}\int_0^1 \frac{\ln^2 y}{1-y}dy\\
&=\frac{7}{16}\times 2\zeta(3)=\boxed{\frac{7}{8}\zeta(3)}
\end{align}
NB:I assume $\displaystyle \int_0^1 \frac{\ln^2 y}{1-y}dy=2\zeta(3)$
Feel free to post your solution.
| Letting $y=\arcsin x$ transforms our integral
$$
\begin{aligned}
I &=\int_{0}^{\frac{\pi}{2}} \frac{y\left(\frac{\pi}{2}-y\right) \cos y d y}{\sin y} \\
&=\frac{\pi}{2} \underbrace{\int_{0}^{\frac{\pi}{2}} \frac{y \cos y}{\sin y} d y}_{J}-\underbrace{\int_{0}^{\frac{\pi}{2}} \frac{y^{2} \cos y}{\sin y} d y}_{K}
\end{aligned}
$$
$$
\begin{aligned}
J &=\int_{0}^{\frac{\pi}{2}} y d(\ln \sin y)=-\int_{0}^{\frac{\pi}{2}} \ln (\sin y) d y =\frac{\pi}{2} \ln 2
\end{aligned}
$$
where the last result comes from my post 1.
$$
\begin{aligned}
K=\int_{0}^{\frac{\pi}{2}} y^{2} d(\ln (\sin y)) \stackrel{IBP}{=} 2 \int_{0}^{\frac{\pi}{2}} y \ln (\sin y) d y
\end{aligned}
$$
By the result that $\int_{0}^{\frac{\pi}{2}} y \ln (\sin y) d y=\frac{\pi^2}{8}\ln 2 -\frac{7}{16}\zeta(3)$ shown below as Footnote, we can conclude that
$$
\boxed{I=\frac{\pi}{2}\left(\frac{\pi}{2} \ln 2\right)-\left(\frac{\pi^{2}}{4} \ln 2-\frac{7}{8} \zeta(3)\right)=\frac{7}{8} \zeta(3)}
$$
Footnote:
By the Fourier series of $\ln(\sin x)$,
$$\ln (\sin x)+\ln 2=-\sum_{k=1}^{\infty} \frac{\cos (2 k x)}{k}$$
Multiplying the equation by $x$ followed by integration from $0$ to $\infty$, we have
$$
\begin{aligned}
\int_{0}^{\frac{\pi}{2}} x \ln (\sin x) d x+\int_{0}^{\frac{\pi}{2}} x\ln 2 d x&=-\sum_{k=1}^{\infty} \int_{0}^{\frac{\pi}{2}} \frac{x \cos (2 k x)}{k} d x\\
\int_{0}^{\frac{\pi}{2}} x \ln (\sin x) d x+\left[\frac{x^{2}}{2} \ln 2\right]_{0}^{\frac{\pi}{2}}&=-\sum_{k=1}^{\infty} \frac{1}{2 k^{2}} \int_{0}^{\frac{\pi}{2}} x d(\sin 2 k x)\\
\int_{0}^{\frac{\pi}{2}} x \ln (\sin x) d x&=-\frac{\pi^{2}}{8} \ln 2-\frac{1}{2} \sum_{k=1}^{\infty} \frac{1}{k^{2}}\left[\frac{\cos 2(x)}{2 k}\right]_{0}^{\frac{\pi}{2}} \\
&=-\frac{\pi^{2}}{8} \ln 2-\frac{1}{4} \sum_{k=1}^{\infty} \frac{(-1)^{k}-1}{k^{3}}\\
&=\frac{\pi^{2}}{8} \ln 2-\frac{1}{4}\left(\sum_{k=1}^{\infty} \frac{2}{(2 k+1)^{3}}\right)\\
&=\frac{\pi^{2}}{8} \ln 2-\frac{1}{2}\left[\sum_{k=1}^{\infty} \frac{1}{k^{3}}-\sum_{k=1}^{\infty} \frac{1}{(2 k)^{3}}\right]\\
&=\frac{\pi^{2}}{8} \ln 2-\frac{7}{16}\zeta(3) \blacksquare
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4500073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Use the definition of limit to prove that $\lim_{x \to 0} \frac{x}{x+1}=0$ Use the definition of limit to prove that $\lim_{x \to 0} \dfrac{x}{x+1}=0$
My attempt:
Let $\epsilon >0$ then I need to find $\delta>0$ such that if $0<|x|<\delta$ then $\left| \dfrac{x}{x+1} \right| < \epsilon$
What I was thinking is that, to get a $\delta$ that satisfies the definition, I should see what $\epsilon$ satisfies $|x|<\epsilon |x+1|$ and then I could define $\delta$ in terms of such an $\epsilon$.
If $\epsilon =1$ then: $0\leq|x|<|x+1|$ implies that $x^2 < x^2+2x+1$ and then $x>-\dfrac{1}{2}$. In particular, if $-\dfrac{1}{2}<x<\dfrac{1}{2} \Rightarrow |x|<\dfrac{1}{2} \Rightarrow \dfrac{1}{2} < x+1<\dfrac{3}{2}$ so that $|x+1|=x+1$ and then $\dfrac{1}{2}<|x+1| \Rightarrow \dfrac{1}{2|x+1|}<1$. Since $|x|<\dfrac{1}{2}$, then $\left| \dfrac{x}{x+1}\right| = \dfrac{|x|}{|x+1|} <\dfrac{1}{2|x+1|}<1$.
So $\delta=\dfrac{\epsilon}{2}$ stisfies the definition.
If $\epsilon \neq 1$ then:
$0 \leq |x|<\epsilon |x+1|$ Implies that: $x^2<{\epsilon}^2(x^2+2x+1) \Rightarrow (\epsilon ^2-1)x^2+2 \epsilon ^2x + \epsilon ^2 > 0$.
If $0< \epsilon <1 \Rightarrow \epsilon^2-1<0 \Rightarrow x^2 + \dfrac{2 \epsilon^2}{\epsilon^2-1}x+\dfrac{\epsilon^2}{\epsilon^2-1}<0 \Rightarrow \left(x+ \dfrac{\epsilon}{\epsilon-1} \right) \left(x+ \dfrac{\epsilon}{\epsilon+1} \right) < 0$
Let $ x_1 = -\dfrac{\epsilon}{\epsilon-1}$, $x_2=-\dfrac{\epsilon}{\epsilon+1}$. We see that $x_1 >0>x_2$. Then solving the inequality for $x$ we would have $x_2=-\dfrac{\epsilon}{\epsilon+1} < x < -\dfrac{\epsilon}{\epsilon-1} = x_1$.
I am expecting to get something like: $-\phi(\epsilon) < x < \phi(\epsilon) \Rightarrow |x|< \phi(\epsilon)$ on each case so that $\delta = \phi(\epsilon)$ satisfies the definition, but I don't know how to continue here. Could someone give me a hint? Maybe there is a simpler way to approach the exercise.
| Suppose $\delta < \min\{\frac12, \frac{\epsilon}{2}\}$. Then $|x + 1| \ge \min\{\frac12, 1 - \frac{\epsilon}{2}\} \ge \frac12$ so that $\frac{|x|}{|x+1|} \le 2\delta < \epsilon$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
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Show that $\frac{\sin^3 \beta}{\sin \alpha} + \frac{\cos^3 \beta}{\cos \alpha} = 1$ with certain given $\alpha, \beta$ Let $$\frac{\sin (\alpha)}{\sin (\beta)} + \frac{\cos (\alpha)}{\cos (\beta)} = -1 \tag{$1$}$$ where $\alpha, \beta$ are not multiples of $\pi / 2$. Show that
$$\frac{\sin^3 (\beta)}{\sin (\alpha)} + \frac{\cos^3 (\beta)}{\cos (\alpha)} = 1\tag{$2$}$$
I've tried to rewrite $(1)$ and insert into $(2)$ to get
$$ - 1 - \frac{\sin^4 \beta \cos^2 \alpha + \cos^4 \beta \sin^2 \alpha}{\sin \alpha \sin \beta \cos \alpha \cos \beta} = 1 \\
\iff \frac{\sin^4 \beta \cos^2 \alpha + \cos^4 \beta \sin^2 \alpha}{\sin \alpha \sin \beta \cos \alpha \cos \beta} = -2$$
But I can't simplify any further, maybe there are some trig. identities I'm missing?
| Multiplying by a common denominator gives:
$$\sin \alpha \cos \beta + \cos \alpha \sin \beta = -\cos \beta \sin \beta$$
which is the same as
$$\sin(\alpha + \beta) = -\cos \beta \sin \beta. \tag{1}$$
Now:
$$(\sin \alpha \cos \beta + \cos \alpha \sin \beta)(\sin^2 \beta + \cos^2 \beta) = -\cos \beta \sin \beta$$
$$\sin \alpha \cos \beta \sin^2 \beta + \cos \alpha \sin^3 \beta + \sin \alpha \cos^3 \beta + \cos \alpha \sin \beta \cos^2 \beta = -\cos \beta \sin \beta$$
$$\cos \alpha \sin^3 \beta + \sin \alpha \cos^3 \beta = -(\cos \beta \sin \beta + \sin \alpha \cos \beta \sin^2 \beta + \cos \alpha \sin \beta \cos^2 \beta)$$
We want the RHS to equal $\sin \alpha \cos \alpha$. Repeatedly using $(1)$ we get:
$$-(-\sin(\alpha + \beta) + \sin \alpha \cos \beta \sin^2 \beta + \cos \alpha \sin \beta \cos^2 \beta)$$
$$-(-\sin(\alpha + \beta) + \sin \beta \cos \beta (\sin \beta \sin \alpha + \cos \alpha \cos \beta))$$
$$-(-\sin(\alpha + \beta) + (-\sin(\alpha + \beta))(\sin \beta \sin \alpha + \cos \alpha \cos \beta))$$
$$-(-\sin(\alpha + \beta) - \sin(\alpha + \beta) \cos(\alpha - \beta))$$
$$= \sin(\alpha + \beta) + \sin(\alpha + \beta) \cos(\alpha - \beta)$$
and the rest follows from MathFail's answer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How many method are there to handle the integral $\int \frac{\sin x}{1-\sin x \cos x} d x$ $$
\begin{aligned}
\int \frac{\sin x}{1-\sin x \cos x} d x =& \frac{1}{2} \int \frac{(\sin x+\cos x)+(\sin x-\cos x)}{1-\sin x \cos x} d x \\
=& \int \frac{d(\sin x-\cos x)}{2-2 \sin x \cos x}-\int \frac{d(\sin x+\cos x)}{2-2 \sin x \cos x} \\
=& \int \frac{d(\sin x-\cos x)}{1+(\sin x-\cos x)^{2}}-\int \frac{d(\sin x+\cos x)}{3-(\sin x+\cos x)^{2}} \\
=& \tan ^{-1}(\sin x-\cos x)+\frac{1}{2 \sqrt{3}} \ln \left| \frac{\sin x+\cos x-\sqrt{3}}{\sin x+\cos x+\sqrt{3}} \right|+C
\end{aligned}
$$
Is there any other method?
| Letting $y=\frac{\pi}{2}-x$ converts the integral into
$$
I=\int \frac{\cos y}{1-\sin y \cos y} d y
$$
Using the Weierstrass substitution, $t =\tan \frac{y}{2} $, we simplify and get
$$
I=-2 \int \frac{t^{2}-1}{t^{4}-2 t^{3}+2 t^{2}-2 t+1} d t
$$
Playing a little trick on the $I$ gives
$$
I =-2 \int \frac{1-\frac{1}{t^{2}}}{t^{2}+\frac{1}{t^{2}}-2\left(t+\frac{1}{t}\right)+2} d t =-2 \int \frac{d\left(t+\frac{1}{t}\right)}{t^{2}+\frac{1}{t^{2}}-2\left(t+\frac{1}{t}\right)+2}
$$
Letting $T=t+\frac{1}{t}$ yields
$$
\begin{aligned}
I &=-2 \int \frac{d T}{T^{2}-2 T} \\
&=-2 \int\left(\frac{1}{T-2}-\frac{1}{T}\right) d T \\
&=2 \ln \left|\frac{T}{T-2}\right|+C \\
&=2 \ln \left|\frac{t+\frac{1}{t}}{t+\frac{1}{t}-2}\right|+C\\&= 2 \ln \left|\frac{ \tan^2 \frac{y}{2}+1}{\tan ^{2} \frac{y}{2}-2 \tan \frac{y}{2}+1}\right|+C\\&= 2 \ln \left|\frac{\tan ^{2}\left(\frac{\pi}{4}+\frac{x}{2}\right)+1}{\tan ^{2}\left(\frac{\pi}{4}+\frac{x}{2}\right)-2 \tan \left(\frac{\pi}{4}+\frac{x}{2}\right)+1}\right|+C
\end{aligned}
$$
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"answer_id": 1
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Determination of order of cosets in a factor group of a finite abelian group. Consider the group $G=\frac{\mathbb{Z}_{3^{10}}\times \mathbb{Z}_{3^7} }{\langle (3^2, 3^3)\rangle}$. Let $a=(1,0)+\langle (3^2, 3^3)\rangle$ and $b=(0,1)+\langle (3^2, 3^3)\rangle$. Since the order of the group $G$ is $3^9$, so order of $a$ and $b$ must be a divisor of $3^9$.
I am willing to find $|a|$ and $|b|$. Assume $\alpha=|a|$. Then
$\alpha(1,0)=m(3^2, 3^3)$ for some positive integer $m$ and so $\alpha\equiv 3^2 m[3^{10}], 0\equiv 3^3 m[3^7]$.
The second congruence yields $m=3^4 t$ where $t$ is chosen to be non-negative integers. Hence $\alpha\equiv 3^6 t[3^{10}]$. Do not understand what to do next to derive $\alpha$.
For the determination of order of $(0,1)+\langle (3^2, 3^3)\rangle$, let $\beta$ be its order. Then $\beta(0,1)=m'(3^2, 3^3)$ for some positive integer $m'$. So $0\equiv 3^2 m'[3^{10}], \beta\equiv 3^3 m'[3^7]$. The first case gives $m'=3^8 t, t\geqslant 1$ and so the second equation becomes $\beta\equiv 3^3(3^8 t')\equiv 0[3^7]$. How come this arrived ? I am really confused here.
Please help what to be done here.
| Let $G_1=\Bbb{Z}_{3^{10}} $,$G_2=\Bbb{Z}_{3^{7}}$ , $H=\langle (3^2, 3^3)\rangle$
Then $|3^2|_{G_{1}}=3^8$ , $|3^3|_{G_{2}}=3^4$
Let $|a|=n$ where $a=(1,0)+H$
Then $n$ is the least positive integer such that $na=n(1, 0) +H=H$
$(n, 0) +H=H$ implies $(n, 0) \in H$
Hence $n$ is the least positive integer such that $(n, 0) \in H$
$|3^3|_{G_{2}}=3^4 \implies 3^7=0$.
Hence $3^4(3^2, 3^3) =(3^6, 0) \in H$
$3^6$ is the least positive integer in the first component such that $(3^6, 0) \in H$ implies $n=3^6$
$|(1, 0) +H|=3^6$
Edit:
Let $|(0, 1) +H|=m$
Then $m$ is the least positive integer such that $(0, n) \in H$
$3^7$ is the least positive integer such that $(0, 3^7) =(0,0)\in H$.
$|(0, 1) +H|=3^7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4510985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Coefficients of the polynomial? I have the following function
$$ p_n(x,y) = \sum_{i=0}^n c_{n}(i)~x^{n-i}y^i . $$
I know that first couple of $n$'s have a form
$$
n=1, ~~~ p_1(x,y) = 3 x + y\, , \\
n=2, ~~~ p_2(x,y) = 15 x^2 + 10 x y + 3 y^2\, , \\
n=3, ~~~ p_3(x,y) = 105 x^3 + 105 x^2 y + 63 x y^2 + 15 y^3\, , \\
n=4, ~~~ p_4(x,y) = 945 x^4 + 1260 x^3 y + 1134 x^2 y^2 + 540 x y^3 + 105 y^4\, ,\\
n=5, ~~~ p_5(x,y) = 10395 x^5 + 17325 x^4 y + 20790 x^3 y^2 + 14850 x^2 y^3 +
5775 x y^4 + 945 y^5\, .
$$
How do I figure out the general expression for $c_{n}(i)$ for arbitrary $n$ and $i$?
I managed to figure out edge cases, when $i=0$ and $i=n$, where
$$ c_n(0) = (2 n + 1)!!,~~{\rm while}~~ c_n(n) = (2 n - 1)!! ~.$$
| Thanks to the information provided in OPs comment we can show
\begin{align*}
\color{blue}{c_n(i)=\frac{2n+1}{2i+1}\,\frac{(2n)!}{2^nn!}\binom{n}{i}\qquad\qquad 0\leq i\leq n}\tag{1}
\end{align*}
We are looking according to OPs comment for the coefficients $c_i(n)$ of the bivariate polynomial
\begin{align*}
p_n(x,y)=\sum_{i=0}^nc_n(i)x^{n-i}y^i:=(2n+1)!!\lim_{p\to 0}\left(x-y\frac{d^2}{dp^2}\right)^n\left(\frac{\sin p}{p}\right)\tag{2}
\end{align*}
We obtain from (2)
\begin{align*}
\color{blue}{\lim_{p\to 0}}&\color{blue}{\left(x-y\frac{d^2}{dp^2}\right)^n\left(\frac{\sin p}{p}\right)}\\
&=\lim_{p\to 0}\sum_{j=0}^n\binom{n}{j}x^{n-j}(-1)^j\left(y\frac{d^2}{dp^2}\right)^{j}\frac{\sin p}{p}\tag{3.1}\\
&=\lim_{p\to 0}\sum_{j=0}^n\binom{n}{j}x^{n-j}y^j(-1)^j\frac{d^{2j}}{dp^{2j}}\frac{\sin p}{p}\tag{3.2}\\
&=\lim_{p\to 0}\sum_{j=0}^n\binom{n}{j}x^{n-j}y^j(-1)^j
\frac{d^{2j}}{dp^{2j}}\sum_{k=0}^{\infty}\frac{p^{2k}}{(2k+1)!}(-1)^k\tag{3.3}\\
&=\lim_{p\to 0}\sum_{j=0}^n\binom{n}{j}x^{n-j}y^j(-1)^j
\sum_{k=0}^{2j}\frac{(2k)^{\underline{2j}}p^{2k-2j}}{(2k+1)!}(-1)^k\tag{3.4}\\
&=\sum_{j=0}^n\binom{n}{j}x^{n-j}y^j(-1)^j\frac{(2j)^{\underline{2j}}}{(2j+1)!}(-1)^j\tag{3.5}\\
&\,\,\color{blue}{=\sum_{j=0}^n\binom{n}{j}\frac{1}{2j+1}}x^{n-j}y^j\tag{3.6}
\end{align*}
Comment:
*
*In (3.1) we apply the binomial theorem.
*in (3.2) we can separate the factor $y^j$ since it does not affect the derivation of the variable $p$.
*In (3.3) we use the series expansion of $\sin x$.
*In (3.4) we differentiate $2j$ times and use the falling factorial notation $q^{\underline{j}}=q(q-1)\cdots(q-j+1)$. We also set the upper limit of the sum to $2j$ since other terms cancel.
*In (3.5) we do the limit operation which leaves us with the term with index $k=j$ only and we make some simplifications in the last step noting that $q^{\underline{q}}=q!$.
Denoting with $[y^i]$ the coefficient of $y^i$ in a series we finally obtain from (2) and (3.6)
\begin{align*}
\color{blue}{c_n(i)}&=[x^{n-i}y^i]p_n(x,y)\\
&=[x^{n-i}y^i](2n+1)!!\sum_{j=0}^n\binom{n}{j}\frac{1}{2j+1}x^{n-j}y^j\\
&=(2n+1)\frac{(2n)!}{(2n)!!}\binom{n}{i}\frac{1}{2i+1}\\
&\,\,\color{blue}{=\frac{2n+1}{2i+1}\,\frac{(2n)!}{2^nn!}\binom{n}{i}}
\end{align*}
and the claim (1) follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4511437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the sum of radicals without squaring, Is that impossible?
Find the summation:
$$\sqrt {3-\sqrt 5}+\sqrt {3+\sqrt 5}$$
My attempts:
\begin{align*}
&A = \sqrt{3-\sqrt{5}}+ \sqrt{3+\sqrt{5}}\\
\implies &A^2 = 3-\sqrt{5} + 3 + \sqrt{5} + 2\sqrt{9-5}\\
\implies &A^2 = 6+4 = 10\\
\implies &A = \sqrt{10}
\end{align*}
So I was wondering about a way to find this sum without squaring? It seems impossible, but I still want to ask.
| I want to solve the problem with a geometric approach. Consider a right triangle $ABC$ as below:
If we suppose that $BH = \sqrt{3 - \sqrt{5}}$ and $HC = \sqrt{3 + \sqrt{5}}$, then we have $AH^2 = BH.HC = \sqrt{9 - 5} = 2$. So $AH = \sqrt{2}$. Now use Pythagorean theorem:
\begin{align*}
\overset{\triangle}{ABH}&: AB^2 = AH^2 + BH^2 = 2 + 3 - \sqrt{5} = 5 - \sqrt{5} \implies AB = \sqrt{5 - \sqrt{5}}\\
\overset{\triangle}{ACH}&: AC^2 = AH^2 + HC^2 = 2 + 3 + \sqrt{5} = 5 + \sqrt{5} \implies AC = \sqrt{5 + \sqrt{5}}\\
\overset{\triangle}{ABC}&: BC^2 = AB^2 + AC^2 = 5 - \sqrt{5} + 5 + \sqrt{5} = 10 \implies BC = \sqrt{10}
\end{align*}
Therefore:
$$\bbox[5px, border: 2px solid magenta]{\sqrt{10} = BC = BH + HC = \sqrt{3 - \sqrt{5}} + \sqrt{3 + \sqrt{5}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4514668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 5
} |
If a and b are the distinct roots of the equation $x^2+3^{1/4}x + 3^{1/2} =0$ then find the value of $a^{96}(a^{12}-1)+b^{96}(b^{12}-1)$.
If $a$ and $b$ are the distinct roots of the equation $x^2+3^{1/4}x +
3^{1/2} =0$ then find the value of $a^{96}(a^{12}-1)+b^{96}(b^{12}-1)$
My attempt:
LHS = $(a^{108}+b^{108})-(a^{96}+b^{96})$
$a^{n} + b^{n} = -[3^{1/4}(a^{n-1}+b^{n-1})+3^{1/2}(a^{n-2}+b^{n-2})]$
$a^{108} + b^{108} = -[(3^{1/4}(a^{107}+b^{107})+3^{1/2}(a^{106}+b^{106})]$
I suppose i could repeat this process until every term in is terms of $a^{96}$ and $b^{96}$ but that would be very tedious
| As suggested in Eric Snyder's comment, let $c = \sqrt[4]{3}$. Then rearrange your equation to get
$$x^2 = -cx - c^2 \tag{1}\label{eq1A}$$
Multiply both sides by $x$ and use \eqref{eq1A} to get
$$\begin{equation}\begin{aligned}
x^3 & = -cx^2 - c^{2}x \\
& = -c(-cx - c^2) - c^{2}x \\
& = c^{2}x + c^3 - c^{2}x \\
& = c^3
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Thus, with $x = a$ and $x = b$ being the $2$ distinct roots of the equation in \eqref{eq1A}, we get
$$a^3 = b^3 = c^3 = (\sqrt[4]{3})^3 \; \; \to \; \; a^{12} = b^{12} = 3^3 = 27 \tag{3}\label{eq3A}$$
This results in
$$a^{96}(a^{12}-1)+b^{96}(b^{12}-1) = 2(27^{8})(27 - 1) = 52(3^{24}) \tag{4}\label{eq4A}$$
Alternatively, note $(x-c)(x^2+cx+c^2)=x^3-c^3=0 \; \to \; a^3=b^3=c^3$. Also, we could've used the quadratic formula with \eqref{eq1A} to get $a,b = c\left(\frac{-1\pm\sqrt{3}i}{2}\right)$. Cubing this (instead of doing the explicit calculations, it's simpler to use that $\frac{-1\pm\sqrt{3}i}{2}$ are third roots of $1$) then also gives $a^3 = b^3 = c^3$, as previously shown in \eqref{eq3A}.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4516013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solving floor function system of equations given $x \lfloor y \rfloor + y \lfloor x \rfloor =66$ and $x \lfloor x \rfloor + y \lfloor y \rfloor=144$
$x$ and $y$ are real numbers satisfying $x \lfloor y \rfloor + y \lfloor x \rfloor =66$ and $x \lfloor x \rfloor + y \lfloor y \rfloor=144.$ Find $x$ and $y.$
I first assumed $x>y.$ Then I approximated $x$ and $y$ respectively to be in ranges $11<x<12$ and $2<y<3.$ Then this means $\lfloor x \rfloor =11$ and $\lfloor y \rfloor =2.$ So I wrote the system of equations as $2x+11y=66$ and $11x+2y=144,$ but solving the system gave me solutions of $x$ and $y$ that didn't satisfy the $\lfloor x \rfloor =11$ and $\lfloor y \rfloor =2$ constraint. I'm not really sure how to approach this problem from here other than trying values. May I have some help?
| Here is an idea: you already showed that $(x+y)(\lfloor x \rfloor + \lfloor y \rfloor)=210$. If you denote $k=\lfloor x \rfloor + \lfloor y \rfloor$ then $k\leq x+y<k+2$. So we have $k^2\leq k(x+y)<k(k+2)$. Since $k(x+y)=210$, we now know that $k^2\leq 210<k(k+2)$. Solving this inequality for integer $k$ yields $k=14$.
Now, use the substruction part: $(x-y)(\lfloor x \rfloor - \lfloor y \rfloor)=78$. Denote $m=\lfloor x \rfloor - \lfloor y \rfloor$ and assume WLOG that $x \geq y$. Now $m-1<x-y<m+1$, so, similarly we will get that $m(m-1)<78<m(m+1)$. Solving this for integer $m$ yields $m=9$.
This seems impossible, as $\lfloor x \rfloor$ and $\lfloor y \rfloor$ are integers, so $m,k$ should have the same parity - so I guess there is a mistake somewhere, but the idea seems to have potential - that's why I decided to write it up.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4516294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Sum of squares of ratios of diagonal of a regular heptagon A problem from the 1998 Lower Michigan Math Competition...
A regular heptagon has diagonals of two different lengths. Let $a$ be the length of a side, $b$ the length of a shorter diagonal, and $c$ the length of a longer diagonal. Prove that
$$
\frac{a^2}{b^2} + \frac{b^2}{c^2} + \frac{c^2}{a^2} = 6 \quad \text{ and } \quad \frac{b^2}{a^2} + \frac{a^2}{c^2} + \frac{c^2}{b^2} = 5.
$$
What I have so far:
*
*Introduce coordinates so that the heptagon consists of the seventh roots of unity in the complex plane.
*Let $\omega = e^{2 \pi i/7}$ so that the vertices are $1, \omega, \omega^2, \dots, \omega^6$. Then $a = |1-\omega|$ and $b = |1-\omega^2| = |1-\omega||1+\omega|$ and $c = |1-\omega^3| = |1-\omega||1+\omega+\omega^2|$.
*The first requested expression then becomes
$$
\frac{1}{|1+\omega|^2} + \frac{|1+\omega|^2}{|1+\omega+\omega^2|^2} + |1+\omega+\omega^2|^2.
$$
*All those modulus-squared expressions can be written as (thingy) times (conjugate of thingy), and the conjugate of a power of $\omega$ is another power of $\omega$. For example,
$$
|1+\omega|^2 = (1+ \omega)(1 + \overline{\omega}) = (1 + \omega)(1 + \omega^6).
$$
Using this, the first requested expression becomes
$$
\frac{1}{(1+\omega)(1+\omega^6)} + \frac{(1+\omega)(1+\omega^6)}{(1+\omega+\omega^2)(1+\omega^5+\omega^6)} + (1 + \omega + \omega^2)(1 + \omega^5 + \omega^6).
$$
Annnnd...that's about where I ran out of steam. Not sure how much of this is progress, but I really want the complex-roots-of-unity approach to pan out!
| Adding details to David's answer---too much algebra to fit in a comment. ;-)
Some milestones along the way:
*
*Taking the modulus-squared on both sides of David's expression for $\frac{1}{1+\omega}$ yields
$$
\frac{1}{|1+\omega|^2} = |1+\omega^2 + \omega^4|^2
= (1+\omega^2+\omega^4) \overline{(1+\omega^2+\omega^4)}
= (1+\omega^2+\omega^4)(1+\omega^5+\omega^3)
$$
*Expanding the polynomial and reducing powers of $\omega$ modulo 7 yields
$$
\frac{1}{|1+\omega|^2} = 1 + \omega^5 + \omega^3 + \omega^2 + \omega^7 + \omega^5 + \omega^4 + \omega^9 + \omega^7 = 3 + 2 \omega^2 + \omega^3 + \omega^4 + 2 \omega^5
$$
*Likewise,
$$
|1+\omega+\omega^2| = (1+\omega+\omega^2)(1+\omega^6+\omega^5) = \cdots = 3 + 2 \omega + \omega^2 + \omega^5 + 2 \omega^6
$$
*Multiplying both sides of David's expression for $\frac{1}{1+\omega+\omega^2}$ by $1+\omega$ yields
$$
\frac{1+\omega}{1+\omega+\omega^2} = - \omega(1+\omega)(1+\omega^3)
$$
and taking modulus-squared on both sides,
$$
\left|\frac{1+\omega}{1+\omega+\omega^2}\right|^2 = (1 + \omega)(1 + \omega^6)(1+\omega^3)(1+\omega^4) = (2 + \omega + \omega^6)(2 + \omega^3 + \omega^4) = \cdots = 4 + 2 \omega + \omega^2 + 3 \omega^3 + 3\omega^4 + \omega^5 + 2 \omega^6.
$$
-- Adding the three equations together,
$$
\frac{1}{|1+\omega|^2} + |1+\omega+\omega^2|^2 + \left|\frac{1+\omega}{1+\omega+\omega^2}\right|^2 = 10 + 4 \omega + 4 \omega^2 + 4 \omega^3 + 4 \omega^4 + 4 \omega^5 + 4 \omega^6.
$$
*But that latter expression is
$$
6 + 4 (1 + \omega + \omega^2 + \cdots + \omega^6) = 6 + 4(0) = 6.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Show that for any two positive real numbers $a, b: \frac a{a + 2b} + \frac b{b + 2a} ≥ \frac12$ Question:
Show that for any two positive real numbers $a, b: \frac a{a + 2b} + \frac b{b + 2a} ≥ \frac12$.
So for this question, I began by expanding all terms and moving them all to one side. However, I do not know how to definitively say that the statement is proved.
This is my "work" so far:
$a, b: \frac a{a + 2b} + \frac b{b + 2a} ≥ \frac12$
$\frac {2a(b + 2a) + 2b(a + 2b) - 1(b + 2a)(a + 2b)}{2(a + 2b)(b+2a)} ≥ 0$
$\frac {4ab + 4a^2 + 4b^2 - 2b^2 - 5ab - 2a^2}{4b^2 + 10ab + 4a^2} ≥ 0$
$\frac {2a^2 - ab + 2b^2}{4b^2 + 10ab + 4a^2} ≥ 0$
| first let's do some cleaning:
$$\frac{a}{a+2b}+\frac{b}{b+2a} = \frac{2a^2 + 2ab + 2b^2}{2a^2 + 4ab + 2a^2} = \frac{2(a+b)^2 - 2ab}{2(a+b)^2} = 1-\frac{ab}{(a+b)^2}$$
and we have :
$$A = \frac{a}{a+2b}+\frac{b}{b+2a} = \frac{1}{1+\frac{2b}{a}} + \frac{1}{1+\frac{2a}{b}}$$
let $x = \frac{a}{b} $ then we have :
$$f(x) = \frac{1}{1+\frac{2}{x}} + \frac{1}{1+2x} = \frac{x}{x+2} + \frac{1}{1+2x}$$
therefore:
$$f^\prime(x) = \frac{2}{(x+2)^2} - \frac{2}{(1+2x)^2} = 0 \Rightarrow x^2 + 4x + 4 = 4x^2 + 4x + 1\Rightarrow 3x^2-3 = 0 \Rightarrow x = \pm 1 \Rightarrow f(x) = \frac{2}{3}, -2$$
but since both a and b are positive x = -2 is not an option and therefore we have that with x>0 f(x) is always larger than 2/3 which is a subset of the desired result.$\blacksquare$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Is there any method to compute $\frac{d^{n}}{d x^{n}}\left(\cos ^{k} x\right)$ other than complex numbers? In this couple of days, I need to know the high derivatives of $cos^kx$ whose power $k$ make the differentiation much harder. Then I attempt to
use the identity $$
\cos x=\frac{1}{2}\left(e^{x i}+e^{-x i}\right),
$$
to expand it with Binomial expansion
$$
\begin{aligned}
\cos ^{k} x &=\frac{1}{2 ^k} \sum_{j=0}^{k}\binom{k}{j}e^{x(k-j) i} e^{-x i j}
=\frac{1}{2^{k}} \sum_{j=0}^{k}\binom{k}{j} e^{x(k-2 j) i}
\end{aligned}
$$
Differentiating it by $n$ times yields
$$
\begin{aligned}
\frac{d^{n}}{d x^{n}}\left(\cos ^{k} x\right) &=\frac{1}{2^k} \sum_{j=0}^{k}\binom{k}{j}[(k-2 j) i]^{n} e^{x(k-2 j) i} \\
&=\frac{i^{n}}{2 ^k} \sum_{j=0}^{k}\binom{k}{j}(k-2 j)^{n} e^{x(k-2 j) i} \\
&=\frac{i^{n}}{2^{k}} \sum_{j=0}^{k}\binom{k}{j}(k-2 j)^{n}[\cos ((k-2 j) x)+i \sin ((k-2 j) x)]
\end{aligned}
$$
If $x$ is real, then comparing the imaginary and real parts on both sides yields \begin{equation}
\displaystyle \frac{d^{n}}{d x^{n}}\left(\cos ^{k} x\right)=\left\{\begin{array}{ll}
\displaystyle \frac{(-1)^{\frac{n}{2}}}{2^{k-1}} \sum_{j=0}^{\left\lfloor\frac{k}{2}\right\rfloor}\binom{k}{j}(k-2 j)^{n}\cos((k-2j)x) \quad \textrm{ if n is even.}\\\displaystyle
\frac{(-1)^{\frac{n+1}{2}}}{2^{k-1}} \sum_{j=0}^{\left\lfloor\frac{k}{2}\right\rfloor}\binom{k}{j}(k-2 j)^{n} \sin((k-2j)x) \quad \textrm{ if n is odd.}
\end{array}\right.
\end{equation}
For examples,
*
*When $n$ is even,
$$
\left.\frac{d^{n}}{d x^{n}}\left(\cos ^{k} x\right)\right|_{x=0}=\frac{(-1)^{\frac{n}{2}}}{2^{k-1}} \sum_{j=0}^{\left\lfloor\frac{k}{2}\right\rfloor}\binom{k}{j}(k-2 j)^{n}
$$
In particular,
$$
\begin{aligned}
\left.\frac{d^{6}}{d x^{6}}\left(\cos ^{5} x\right)\right|_{x=0}&=\frac{(-1)^{3}}{2^{4}} \sum_{j=0}^{2}\binom{5}{j}(5-2 j)^{6} \\
&=\frac{1}{16}\left[5^{6}+5 \cdot 3^{6}+10 \cdot 1^{6}\right] \\
&=1205
\end{aligned}
$$
*When $n$ is odd,$$
\left.\frac{d^{n}}{d x^{n}}\left(\cos ^{k} x\right)\right|_{x=0}=0;
$$
$$\left.\frac{d^{n}}{d x^{n}}\left(\cos ^{k} x\right) \right|_{x=\frac{\pi}{4} }=
\frac{(-1)^{\frac{n+1}{2}}}{2^{k-1}} \sum_{j=0}^{\left\lfloor\frac{k}{2}\right\rfloor}\binom{k}{j}(k-2 j)^{n} \sin \frac{(k-2 j) \pi}{4} $$
Can we find its closed form without using the complex numbers?
| Although essentially equivalent, you can say
$$\cos^k(x)=\frac1{2^k}\sum_{j=0}^k{k\choose j}\cos(k-2j)x,$$
which can be proved by induction on $k$, together with the sum formula. Now, differentiating is fairly straightforward, based on casework on $n\pmod 4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4519048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Combinatorial Identity involving Bernoulli numbers For all $n\geq 1$ and $m\geq0$, I'm trying to prove that
$\sum_{k=0}^n\sum_{l=0}^m\binom{n}{k}\binom{m}{l}\frac{(n-k)!(m-l)!}{(n+m-k-l+1)!}(-1)^l B_{k+l}=0$
where $B_n$ are the Bernoulli numbers with $B_{1}=-\frac{1}{2}$.
I made a couple of attempts using the recursive relationship of Bernoulli numbers and induction, but unsuccessful. Now I'm wondering if this is a viable proof strategy. Any comments are appreciated!
Edit: This is a conjecture, I evaluated it numerically for $n,m\leq 20$.
| With the work by @RenéGy and @epi163sqrt we are trying to prove the
statement
$$(-1)^n \sum_{g=0}^m \frac{B_{n+g+1}}{n+g+1} {m\choose g}
+ (-1)^m \sum_{g=0}^n \frac{B_{m+g+1}}{m+g+1} {n\choose g}
= - \frac{1}{n+m+1} {n+m\choose m}^{-1}$$
We prove this for $n\ge m$, it then follows by symmetry for $m\ge n.$
Using
$$B_n = (-1)^n \sum_{k=0}^n {n\choose k} B_k$$
we get for the first piece
$$\sum_{g=0}^m \frac{1}{n+g+1} {m\choose g}
(-1)^{g+1} \sum_{k=0}^{n+g+1} {n+g+1\choose k} B_k.$$
Extracting $k=0$ we get
$$\sum_{g=0}^m \frac{1}{n+g+1} {m\choose g} (-1)^{g+1}
= \sum_{g=0}^m \frac{1}{n+m-g+1} {m\choose g} (-1)^{m-g+1}
\\ = [z^{n+m+1}] \log\frac{1}{1-z}
\sum_{g=0}^m z^g {m\choose g} (-1)^{m-g+1}
\\ = (-1)^{m+1} [z^{n+m+1}] \log\frac{1}{1-z}
(1-z)^m
= - [z^{n+m+1}] \log\frac{1}{1-z} (z-1)^m.$$
Recall from MSE
4316307
that with $1\le k\le n$
$${n\choose k}^{-1}
= k [z^n] \log\frac{1}{1-z}
(z-1)^{n-k}.$$
We put $n := n+m+1$ and $k := n+1$ to get
$$-\frac{1}{n+1} {n+m+1\choose n+1}^{-1}
= -\frac{1}{n+m+1} {n+m\choose n}^{-1}.$$
This also could have been obtained by summing residues of a suitable
function. Good, we have the RHS. Now we get for the remainder
$$\sum_{g=0}^m \frac{1}{n+g+1} {m\choose g}
(-1)^{g+1} \sum_{k=1}^{n+g+1} {n+g+1\choose k} B_k
\\ = \sum_{g=0}^m {m\choose g}
(-1)^{g+1} \sum_{k=1}^{n+g+1} {n+g\choose k-1} \frac{B_k}{k}
\\ = \sum_{g=0}^m {m\choose g}
(-1)^{g+1} \sum_{k=0}^{n+g} {n+g\choose k} \frac{B_{k+1}}{k+1}.$$
We get two components, the first is,
$$\sum_{g=0}^m {m\choose g}
(-1)^{g+1} \sum_{k=0}^{m-1} {n+g\choose k} \frac{B_{k+1}}{k+1}
\\ = \sum_{k=0}^{m-1} \frac{B_{k+1}}{k+1}
\sum_{g=0}^m {m\choose g} (-1)^{g+1} {n+g\choose k}.$$
The inner sum is
$$[z^k] (1+z)^n \sum_{g=0}^m {m\choose g} (-1)^{g+1} (1+z)^g
= - [z^k] (1+z)^n (-1)^m z^m = 0$$
since $k\lt m$. That leaves
$$\sum_{g=0}^m {m\choose g}
(-1)^{g+1} \sum_{k=m}^{n+g} {n+g\choose k} \frac{B_{k+1}}{k+1}
\\ = \sum_{g=0}^m {m\choose g}
(-1)^{g+1} \sum_{k=0}^{n-m+g} {n+g\choose m+k}
\frac{B_{m+k+1}}{m+k+1}
\\ = \sum_{k=0}^n \frac{B_{m+k+1}}{m+k+1}
\sum_{g=k+m-n}^m {m\choose g} (-1)^{g+1} {n+g\choose m+k}.$$
Now when $n+g\lt m+k$ or $g\lt m-n+k$ the second binomial coefficient
is zero, so we may lower $g$ to zero (observe that the first binomial
coefficient is zero when $g\lt 0$ so we also may raise to zero when $k+m-n \lt 0$):
$$\sum_{k=0}^n \frac{B_{m+k+1}}{m+k+1}
\sum_{g=0}^m {m\choose g} (-1)^{g+1} {n+g\choose m+k}.$$
The inner sum is
$$[z^{m+k}] (1+z)^n
\sum_{g=0}^m {m\choose g} (-1)^{g+1} (1+z)^g
= - [z^{m+k}] (1+z)^n (-1)^m z^m
\\ = - (-1)^m [z^k] (1+z)^n = - (-1)^m {n\choose k}.$$
We have obtained
$$- (-1)^m \sum_{k=0}^n \frac{B_{m+k+1}}{m+k+1} {n\choose k},$$
which is minus the second piece and concludes the argument.
Addendum. Obviously what we have proved here is with $b_n=
(-1)^n \sum_{k=0}^n {n\choose k} a_k$ then
$$(-1)^n \sum_{g=0}^m \frac{b_{n+g+1}}{n+g+1} {m\choose g}
+ (-1)^m \sum_{g=0}^n \frac{a_{m+g+1}}{m+g+1} {n\choose g}
= - \frac{a_0}{n+m+1} {n+m\choose m}^{-1}.$$
We get for an ordinary binomial transform $b_n = \sum_{k=0}^n {n\choose
k} a_k$ the relation
$$\sum_{g=0}^m (-1)^{g+1} \frac{b_{n+g+1}}{n+g+1} {m\choose g}
+ (-1)^m \sum_{g=0}^n \frac{a_{m+g+1}}{m+g+1} {n\choose g}
= - \frac{a_0}{n+m+1} {n+m\choose m}^{-1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4520057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
"answer_id": 2
} |
Proving $(b + c) \cos \frac{B + C}{2} = a \cos \frac{B - C}{2}$ for any $\triangle ABC$ I found this problem on Mathematics, Textbook for Class XI by NCERT, ed. January 2021 that uses the sine and cosine formulas along with standard trigonometric identities for proving an identity.
By triangle ABC, the question assumes that the angles of the triangle are A, B and C, and the sides opposite the angles A, B and C, are a, b and c respectively.
For any triangle $ABC$, prove that:
$$(b + c) \cos \frac{B + C}{2} = a \cos \frac{B - C}{2}$$
I've tried to approach this problem by simplifying the
$$\cos \frac{B+C}{2}$$
part into
$$\sin\frac{A}{2}$$
and assuming that
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$$
keeping in mind the sine law. However, I haven't been able to solve it yet.
A guidance would be helpful even if the question is not solved in its entirety.
| $$\sin(B)+\sin(C)=2\sin\left(\frac{B+C}{2}\right)\cos\left(\frac{B-C}{2}\right)$$
Multiply $\cos\left(\frac{B+C}{2}\right)$ on both sides, and use double angle formula for sine
$$(\sin(B)+\sin(C))\cos\left(\frac{B+C}{2}\right)=\sin(B+C)\cos\left(\frac{B-C}{2}\right)$$
$\sin(B+C)=\sin(\pi-A)=\sin(A)$
$$(\sin(B)+\sin(C))\cos\left(\frac{B+C}{2}\right)=\sin(A)\cos\left(\frac{B-C}{2}\right)$$
Use sine law: $\sin(A)=\frac{a}k, \sin(B)=\frac{b}k, \sin(C)=\frac{c}k$
$$(b+c)\cos\left(\frac{B+C}{2}\right)=a\cos\left(\frac{B-C}{2}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4521409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
why exactly doesn't logarithmic differentiation work here? As part of a larger problem, I have to differentiate $y^x$
so if we let $y=y^x$
we get $\ln(y)=x\ln(y)$
which means the differential is zero
alternatively, let $y=e^{\ln(x^y)}$
so $y=e^{x\ln(y)}$
which can be differentiated as follows
$y'=x^y\left(\ln(y)+y'\ \frac xy\right)$
which is $y'\left(1-{(x^y)}\ \frac xy\right)=x^{y}\ln(y)$
which means $y'=x^y\ln\left(\frac {y}{\left(1- {(x^y)}\frac{x}{y}\right)}\right)$
I apologise for not using MathJaX. I can't seem to get it to work for me
| Differentiating the function $f(x) = y^x$ (for some $y$ which is a function of $x$) is not the same thing as using implicit differentiation on $y=y^x$.
For example, if $y=e^x$, then $f(x)=y^x$ is the function $f(x)=(e^x)^x = e^{x^2}$; but $y=y^x$ gives $e^x=e^{x^2}$, which gives $e^{x^2-x}=1$, so the only values of $x$ that satisfy the equation are $x=0$ and $x=1$ and the graph of $y=y^x$ consists of exactly two points, $(0,1)$ and $(1,e)$.
The graph of $y=y^x$ consists of all points of the form $(x,0)$ with $x\gt 0$; the line $y=1$; and the line $x=1$. Indeed, all of these satisfy the equation: if $y=0$, then we need $x\gt 0$ for $y^x$ to make sense, and we get $0=0^x$; if $y=1$ then we get $1=1^x$, which holds for all $x$; and if $x=1$ we get $y=y^1$ which holds for all $y$. And these are the only solutions. If $y=0$, then we must have $x\gt 0$ (the positive $x$-axis); if $y\neq 0$, then $y^{x-1}=1$, so either $y=1$ and $x$ is arbitrary (the line $y=1$); or else $x-1=0$ (the line $x=1$ with $y\neq 0$).
Here, the derivative exists only on the positive $x$ axis and the line $y=1$. Of those, we can only use logarithmic differentiation for $y=1$, where we get
$\ln |y| = \ln|y^x|$, or $\frac{y'}{y} = \frac{xy'}{y}$, or $y'(x-1)=0$. This means $y'=0$, as expected.
The main problem with what you are doing is that you are trying to differentiate $f(x) = y^x$, but you are actually finding the implicit derivative for $y=y^x$, and there only for values with $y\neq 0$... where you just get $y'=0$ because of what that graph is.
You can use logarithmic differentiation for $y^x$ (leaving $y'$ indicated) if you set it up correctly. Setting $f(x)=y^x$, we have:
$$\begin{align*}
\ln |f(x)| &= \ln |y^x|\\
\ln |f(x)| &= x\ln|y|\\
\frac{d}{dx} \ln|f(x)| &= \frac{d}{dx}\left(x\ln|y|\right)\\
\frac{f'(x)}{f(x)} &= \ln|y| + x\left(\frac{y'}{y}\right)\\
f'(x) &= f(x)\left(\ln|y| + x\frac{y'}{y}\right)\\
f'(x) &= y^x\left(\ln|y| + x\frac{y'}{y}\right)\\
f'(x) &= (\ln|y|)y^x + xy^{x-1}y'\\
f'(x) &= (\ln y)y^x + xy^{x-1}y',
\end{align*}$$
with last equality because $y^x$ is only defined for $y\gt 0$.
If you do it directly using $a^b=e^{b\ln a}$, we have:
$$\begin{align*}
\frac{d}{dx}y^x &= \frac{d}{dx} e^{x\ln y}\\
&= (x\ln y)'e^{x\ln y}\\
&= \left(\ln y + x\frac{y'}{y}\right) e^{x\ln y}\\
&= \left( \ln y + x\frac{y'}{y}\right)y^x\\
&= (\ln y)y^x + xy^{x-1}y',
\end{align*}$$
exactly the same result as before.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 1,
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} |
Power series expansion of $f(x)=\frac{x}{2x^2+1}$ I was requested to find a series representation of $$f(x)=\frac{x}{2x^2+1}$$
and its convergence radius. I attempted a solution but it is wrong according to online calculators, and I can't find my mistake.
My approach was the standard procedure of rewriting $f(x)$ as $\frac{1}{1-u}$ where $u$ is a function of $x$, and using the fact that $\frac{1}{1-u}=\sum_{n=0}^\infty u^n $. I noticed that
$$f(x)=\frac{x}{2x^2+1} = x \cdot \frac{1}{1-(-2x^2)} = x \sum_{n=0}^\infty (-2x^2)^n = x\sum_{n=0}^\infty (-2)^n (x^2)^n = \sum_{n=0}^\infty (-2)^n (x^2)^{n+1}$$
The expansion I get is therefore $f(x) = \frac{x}{2x^2+1}=x^2 - 2x^4 + 4x^6-8x^8+...$ with a convergence radius of $\frac{1}{\sqrt{2}}$.
As stated earlier, this does not match with the results I get from automated computations, and yet I can't find an error in my logic. I only started learning power series a couple of hours ago, so I must be clearly missing something. Any help would be much appreciated. Thanks in advance.
| Another way to do it
$$f(x)=\frac{x}{2x^2+1}\quad \implies \int f(x)\,dx=\frac{1}{4} \log \left(1+2 x^2\right)$$ Let $t=2x^2$
$$\log(1+t)=\sum_{n=1}^\infty (-1)^{n+1} \frac {t^n}n$$
$$\frac{1}{4} \log \left(1+2 x^2\right)=\sum_{n=1}^\infty (-1)^{n+1} \frac {2^{n-2}}n x^{2n}$$
Now, differentiate
$$f(x)=\frac{x}{2x^2+1}=\sum_{n=1}^\infty (-1)^{n+1} 2^{n-1} x^{2 n-1}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solve the equation $x^6-2x^5+3x^4-3x^2+2x-1=0$ Solve the equation $$x^6-2x^5+3x^4-3x^2+2x-1=0$$ Let's divide both sides of the equation by $x^3\ne0$ (as $x=0$ is obviously not a solution, we can consider $x\ne0$). Then we have $$x^3-2x^2+3x-3\cdot\dfrac{1}{x}+2\cdot\dfrac{1}{x^2}-\dfrac{1}{x^3}=0\\\left(x^3-\dfrac{1}{x^3}\right)-2\left(x^2-\dfrac{1}{x^2}\right)+3\left(x-\dfrac{1}{x}\right)=0$$ What do we do now? If we say $y=x-\dfrac{1}{x}$, we won't be able to express $\left(x^2-\dfrac{1}{x^2}\right)$ in terms of $y$ because of the minus sign as $(a-b)^2=a^2-2ab\color{red}{+b^2}$. On the other side, $$y^3=x^3-\dfrac{1}{x^3}-3\left(x-\dfrac{1}{x}\right)\\x^3-\dfrac{1}{x^3}=y^3+3y$$ I think this is a traditional issue when solving reciprocal equations, but I can't figure out how to deal with it.
|
$$\left(x^3-\dfrac{1}{x^3}\right)-2\left(x^2-\dfrac{1}{x^2}\right)+3\left(x-\dfrac{1}{x}\right)=0$$
There are two roots $\pm1$ associated with the factor of $x-1/x$. To find the other roots, divide by $x-1/x$ to get
$$
x^2+1+{1\over x^2}-2\left(x+{1\over x}\right)+3=0
$$
Noticing that
$(x+1/x)^2=x^2+2+1/x^2$ we have
$$
(x+1/x)^2-2(x+1/x)+2=0,
$$
This quadratic has solutions $x+1/x=1\pm i$.
Then you need to solve $x^2-(1\pm i)x+1=0$ to get the final roots.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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If $g(x)=ax^2+bx+c$ and $f(x)= \begin{cases} {g(x)} & {x\ge k} \\ {g'(x)} & {x< k} \end{cases} $ , $\max (k)=?$ if $f$ is a differentiable function
If $g(x)=ax^2+bx+c$ and $f(x)= \begin{cases} {g(x)} & {x\ge k} \\
{g'(x)} & {x< k} \end{cases} $. If $f(x)$ is a differentiable
function, what is the maximum value of $k$, provided that $b+c=a$?
$1)\frac34\qquad\qquad2)1\qquad\qquad3)3\qquad\qquad4)4$
In order to $f(x)$ be differentiable function, we should have $g(k)=g'(k)=g''(k)$,
$$ak^2+bk+c=2ak+b=2a$$ $$(b+c)k^2+bk+c=(2b+2c)k+b=2b+2c$$
Here for each equation I tried to equate the coefficients of $k^2 , k^1 , k^0$ but I get $a=b=c=0$ which doesn't make sense at all. I don't know how to continue form here.
| Assume $a\neq 0$:
Statement 1:
$$ak^2+bk+c=2ak+b$$
$$ak^2+(b-2a)k+(c-b)=0$$
$$k=\frac{2a-b\pm\sqrt{b^2-4ab+4a^2+4ab-4ac}}{2a}$$
$$k=\frac{2a-b\pm\sqrt{b^2+4a^2-4ac}}{2a}$$
Statement 2:
$$2ak+b=2a$$
$$k=\frac{2a-b}{2a}$$
Statement 3, combining 1 and 2:
$$\frac{2a-b\pm\sqrt{b^2+4a^2-4ac}}{2a}=\frac{2a-b}{2a}$$
$$2a-b\pm\sqrt{b^2+4a^2-4ac}=2a-b$$
$$\sqrt{b^2+4a^2-4ac}=0$$
$$b^2+4a^2-4ac=0$$
Substituting $b=a-c$:
$$a^2-2ac+c^2+4a^2-4ac=0$$
$$5a^2-6ac+c^2=0$$
$$a=\frac{6c\pm\sqrt{36c^2-20c^2}}{10}$$
$$a=\frac{6c\pm4c}{10}$$
This gives two solutions, $a_+=c, a_-=\frac c5$
Then, $b_+=0, b_-=-\frac{4c}5$
We then solve using each of these values:
For $+$, we can substitute into statement $2$ and get that $k=1$, and the solution $g(x)=ax^2+a$ results in $f(x)$ being continuous and differentiable everywhere and satisfies $b+c=a$
For $-$, we substitute into statement $2$ and get that $k=1+\frac45c=1+4a$. This gives us $g(x)=ax^2-4ax+5a, g'(x)=2ax-4a, g''(x)=2a$. As such, $g(k)=16a^3-8a^2+2a$, $g'(k)=8a^2-2a$, $g''(k)=2a$. Solving $g'(k)=g''(x)$ gives $-a^2=\frac a2$, or possible solutions of $a=0$ and $a=\frac12$. As we assumed $a\neq 0$, we have that $a=\frac 12$, which can easily be verified is also a solution for $g(k)=g'(k)$. We then have $k=3$ and $g(x)=\frac 12 x^2 -2x+\frac 52$, which results in $f(x)$ being continuous and differentiable everywhere and satisfies $b+c=a$
Lastly, consider what if $a=0$. In this case, $g(x)=bx+c$ and $g''(x)=0$, which means that in order for $f$ to be differentiable, $g'(k)=0$. This means that $b=0$, and therefore $g(x)=c$. In order for $f$ to be continuous, this means that $c=0$ as well.
In answer to the question, if we allow $g(x)$ to be $0$ everywhere, then there is no maximum value of $k$ that makes $f$ differentiable everywhere. If we restrict that there must be some value of $x$ for which $g(x)\neq0$, we
found two solutions for $g(x)$ and $k$ for which $f$ was continuous everywhere, and the greater value for $k$ was $3$
| {
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"url": "https://math.stackexchange.com/questions/4523509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Finding the value of $\mathop {\lim }\limits_{n \to \infty } f\left( n \right)$ of a series having the value "n" in both numerator and denominator If $f\left( n \right) = n + \frac{{16 + 5n - 3{n^2}}}{{4n + 3{n^2}}} + \frac{{32 + n - 3{n^2}}}{{8n + 3{n^2}}} + \frac{{48 - 3n - 3{n^2}}}{{12n + 3{n^2}}} + .. + \frac{{25 - 7{n^2}}}{{7{n^2}}}$. Then find the value of $\mathop {\lim }\limits_{n \to \infty } f\left( n \right) = \_\_\_$
My approach is as follow
$f\left( n \right) = n + \frac{{16 + 5n - 3{n^2}}}{{4n + 3{n^2}}} + \frac{{32 + n - 3{n^2}}}{{8n + 3{n^2}}} + \frac{{48 - 3n - 3{n^2}}}{{12n + 3{n^2}}} + .. + \frac{{25 - 7{n^2}}}{{7{n^2}}}$
$16 + \left( {n' - 2} \right)16 = 16n' - 16::\left( {5 + \left( {n' - 2} \right)\left( { - 4} \right)} \right) = 13 - 4n'::4 + \left( {n - 2} \right)4 = 4n' - 4$
${T_{n'}} = \frac{{16n' - 16 + 13n - 4n'n - 3{n^2}}}{{4n'n - 4n + 3{n^2}}}$
How will I proceed form here
| $f(n) = \sum_{r =1}^n (1 + \frac{16r + (9-4r)n -3n^2}{4rn + 3n^2}) $
$ = \sum_{r =1}^n ( \frac{16r +9n}{4rn +3n^2} ) \\ $
$ \lim_{n\to\infty} f(n) = \lim_{n\to\infty} \frac{1}{n} ( \sum_{r =1}^n ( \frac{ 16(\frac{r}{n}) +9}{4(\frac{r}{n}) +3}) )$
$ = \int_{0}^1 \frac{16x +9}{4x+3} dx$
$ = 4 + \frac{3}{4} ln \frac{3}{7} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4524114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve inequality $\sinh (2x) - 3 \sinh x \ > 0$
Solve inequality $\sinh (2x) - 3 \sinh x \ > 0$
Using hyperbolic trigo indenties:
$ 2 \sinh x \cosh x - 3 \sinh x \ > 0$
$\sinh x (2 \cosh x -3) \ > 0$
To solve this, we need to take
*
*$\sinh x (2 \cosh x -3) \ > 0$
*$\sinh x (2 \cosh x -3) \ < 0$
why is this the case ? I thought we are only solving for the inequality of $ > 0$
| $\sinh x = \frac {e^x - e^{-x}}{2}$
$\sinh 2x - 3\sinh x = \frac {e^{2x} - 3e^{x} + 3e^{-x} - e^{-2x}}{2}$
Let $u = e^x$
$\frac {u^2 - 3u + u^{-1} + u{-2}}{2}$
Multiply numerator and denominator by $u^2$
$\frac {u^4 - 3u^3 + 3u - 1}{2u^2}$
Factor the numerator
$\frac {(u-1)(u+1)(u - \frac {3 - \sqrt {5}}{2})(u - \frac {3 + \sqrt {5}}{2})}{2u^2}$
$e^x > 0$ for all $x$ so $u> 0$ for all $x$
This gives us the intervals to consider $(0,\frac {3-\sqrt {5}}{2}),(\frac {3-\sqrt {5}}{2},1),(1,\frac {3+\sqrt {5}}{2},1),(\frac {3+\sqrt {5}}{2},\infty)$
The LHS is greater than $0$ when $u\in (\frac {3-\sqrt {5}}{2},1)\cup(\frac {3+\sqrt {5}}{2},\infty)$
or $x\in (\ln(\frac {3-\sqrt {5}}{2}),0)\cup(\ln(\frac {3+\sqrt {5}}{2}),\infty)$
Okay, for the work you have done....
$\sinh x(\cosh x - 3) > 0$
either $\sinh x > 0$ and $\cosh x > 3$
or $\sinh x < 0$ and $\cosh x < 3$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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How did people come up with the formula $a^3+b^3=(a+b)(a^2-ab+b^2)$? Every resource that I've read proves the formula
$$ a^3 + b^3 = (a+b)(a^2-ab+b^2) \tag1$$
by just multiplying $(a+b)$ and $(a^2 - ab + b^2)$.
But how did people come up with that formula? Did they think like, "Oh, let's just multiply these polynomials, I don't know why, let's just do it." I don't think that people just pointed a finger at the sky and came up with that formula.
So, how to prove $(1)$?
| In this answer I wanted to suggest a method where I get results by applying substitutions (based on the idea that it is the most basic algebraic technique).
Let, $a=m+n$ and $b=m-n$ then:
$$
\begin{aligned}a^3+b^3=m^3+3m^2n+3mn^2+n^3+m^3-3m^2n+3mn^2-n^3=2m^3+6mn^2=2m(m^2+3n^2)\end{aligned}
$$
and note that,
$$
\begin{cases} m+n=a\\m-n=b\end{cases}\implies \begin{cases}m=\frac {a+b}{2}\\n=\frac {a-b}{2}\end{cases}
$$
Putting $m=\frac {a+b}{2}$ and $n=\frac {a-b}{2}$, we have:
$$
\begin{aligned}a^3+b^3&=2m(m^2+3n^2)\\
&=2\times \frac {a+b}{2}\left(\frac {(a+b)^2}{4}+3\times \frac {(a-b)^2}{4}\right)\\
&=(a+b)\left(\frac {(a+b)^2+3(a-b)^2}{4}\right)\end{aligned}
$$
Now, let's open the parentheses and do some basic simplifications:
$$
\begin{aligned}
(a+b)^2+3(a-b)^2&=a^2+2ab+b^2+3(a^2-2ab+b^2)\\
&=a^2+2ab+b^2+3a^2-6ab+3b^2\\
&=4a^2-4ab+4b^2\\
&=4(a^2-ab+b^2)\end{aligned}
$$
Thus, we get the following result:
$$
\begin{aligned}a^3+b^3&=(a+b)\times \frac {4(a^2-ab+b^2)}{4}\\
&=(a+b)(a^2-ab+b^2).\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4525498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Can we show algebraically that this is true? This problem is part of a larger proof I've been working on. Long story short, I'm trying to prove a particular explicit formula for the recursive function
$$
t(p, 0) = 0 \\
t(p, k+1) = \left\lfloor \dfrac{t(p, k)}{3} \right\rfloor + p
$$
where $p=2n+1, n \in \mathbb{N}$. The inductive step of proving my particular explicit formula includes the following question:
Can we show algebraically that
$$ \left\lfloor \dfrac{1}{3} - \dfrac{1}{3} \left\lfloor \dfrac{2n-1}{2(3^{k-1})} + \dfrac{1}{2} \right\rfloor \right\rfloor \ \ = \ \ (-1) \left\lfloor \dfrac{2n-1}{2(3^k)} + \dfrac{1}{2} \right\rfloor$$
is true for all $k,n = 1, 2, 3, ...$ ? I used Python to test it and it seems to be true, at least for all $k≤1000, \ n≤1000$. Can anyone prove this rigorously? I tried using induction on $k$ but was not successful. I did not try induction on $n$.
Any help is appreciated.
|
Claim. Let $m,n$ be integers with $n>0$ and $x$ a real number. Then
$$\Bigl\lfloor \frac{-\lfloor x \rfloor+m}{n} \Bigr\rfloor = -\Bigl\lfloor \frac{x-m+n-1}{n} \Bigr\rfloor.$$
Proof.
We have
\begin{align}
\Bigl\lfloor \frac{-\lfloor x \rfloor+m}{n} \Bigr\rfloor
&=\Bigl\lceil \frac{-\lfloor x \rfloor+m-n+1}{n} \Bigr\rceil\tag{1}\\
&=-\Bigl\lfloor \frac{\lfloor x \rfloor-m+n-1}{n} \Bigr\rfloor\tag{2}\\
&=-\Bigl\lfloor \frac{\lfloor x-m+n-1 \rfloor}{n} \Bigr\rfloor\tag{3}\\
&=-\Bigl\lfloor \frac{ x-m+n-1 }{n} \Bigr\rfloor\tag{4}
\end{align}
where we used well-known identities:
*
*$\left \lfloor \frac{n}{m} \right \rfloor = \left \lceil \frac{n-m+1}{m} \right \rceil$
*$\lfloor -x \rfloor=-\lceil x \rceil$ (see Proving that the $\lfloor-x\rfloor= -\lceil x\rceil$)
*$\lfloor x+m \rfloor=\lfloor x \rfloor +m$ (see Find the value of $\lfloor x+y \rfloor$ where $x \in \mathbb{R}$, $y \in \mathbb{Z}$)
*$\lfloor \frac{\lfloor x \rfloor}{n} \rfloor = \lfloor \frac{x}{n}\rfloor$ (see How to prove or disprove $\forall x\in\Bbb{R}, \forall n\in\Bbb{N},n\gt 0\implies \lfloor\frac{\lfloor x\rfloor}{n}\rfloor=\lfloor\frac{x}{n}\rfloor$.)
$\square$
Set $m=1,n=3,x=\dfrac{2n-1}{2(3^{k-1})} + \dfrac{1}{2}$ to get your result.
Note. Adding the ceiling version of the claim as an exercise and to make search engines happy (same $m,n,x$ restrictions):
$$
\Bigl\lceil \frac{-\lceil x \rceil+m}{n} \Bigr\rceil = -\Bigl\lceil \frac{x-m-n+1}{n} \Bigr\rceil.
$$
| {
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Inequality $xy+yz+zx-xyz \leq \frac{9}{4}.$ Currently I try to tackle some olympiad questions:
Let $x, y, z \geq 0$ with $x+y+z=3$. Show that
$$
x y+y z+z x-x y z \leq \frac{9}{4}.
$$
and also find out when the equality holds.
I started by plugging in $z=3-x-y$ on the LHS and got
$$
3y-y^2+3x-x^2-4xy+x^2y+xy^2 = 3y-(y^2+x^2)+3x-4xy+x^2y+xy^2\leq 3y-((y+x)^2)+3x-4xy+x^2y+xy^2
$$
But this got me nowhere.
Then I started again with the left hand side
$$
x y+y z+z x-x y z \Leftrightarrow yz(1-x)+xy+zx
$$
and $x+y+z=3 \Leftrightarrow y+z-2=1-x$ so
$$
yz(y+z-2)+x(y+z)
$$
But this also leaves no idea. Do I have to use a known inequality?
| This is equivalent to $$(1-x)(1-y)(1-z)\leq \frac14.\tag 1$$
This is true if any $x,y,z$ is exactly one, since the the left side of $(1)$ is zero.
It is not possible for all $x,y,z$ to be greater than $1,$ nor for all $x,y,z$ to be less than $1,$ since $x+y+z=3.$
If only one of the values is $>1,$ then the left side of $(1)$ is negative, so it is true, again.
So we are left with $x=1+a,y=1+b,z=1-(a+b)$ where $a,b>0$ and $a+b\leq 1.$
Then the left side of $(1)$ is $ab(a+b).$ If $a+b<1,$ we can always increase this value by using $a'=\frac{a}{a+b},b'=\frac{b}{a+b}.$ So the maximum value is when $a+b=1.$
That means, substituting $b=1-a,$ we want to maximize $a(1-a)$ for $0<a<1.$ That is easy to do with calculus, with a maximum at $a=\frac12$ for a value of $\frac14.$
So the maximum of your expression occurs when $(x,y,z)=\left(\frac32,\frac 32,0\right).$
| {
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Integral of infinite G.P. : $ \int 1+ 2x+ 3x^2 + 4x^3 ...... \, dx $ In this $$\int 1+ 2x+ 3x^2 + 4x^3 ...... \, dx $$
textbook solution goes like let $$ S = 1+ 2x+ 3x^2 + 4x^3 ...... $$ (call this equation 1, then multiple both sides by x , we get $$ Sx= x + 2x^2 + 3x^3 + 4x^4 ...... $$ (call this equation 2), now subtract equation 2 from 1, we get $$ S-Sx= 1 + x^2 + x^3 + x^4 .... $$ now in this right hand side is sum of an infinite G.P. with a=1 and r=x, sum of infinite G.P. is $ \frac{a}{(1-r)} $ so we finally get $$ S(1-x) = \frac{1}{(1-x)} $$ thus $$ S = \frac{1}{(1-x)^2} $$ Now integrating $$ \int \frac{1}{(1-x)^2} \ dx = \frac{1}{(1-x)} + c $$ (where c is an arbitrary constant).
however I can also integrate series as it is, by separating the integration over addition, $$\int 1 \ dx + \int 2x \ dx + \int 3x^2 \ dx + ..... $$ I will get $$ x + x^2 + x^3 + x^4.... + c $$ this is correct as derivative of series $ x + x^2 + x^3 + x^4+.... $ is $ 1+ 2x+ 3x^2 + 4x^3 ....... $, Now $ x + x^2 + x^3 + x^4+... $ is also a sum of infinite G.P. with a=x and r=x so I finally get $\frac{x}{(1-x)} + c$, why I am getting a different answer, I don't understand.
| They are actually the same, up to the famous constant:
$$\frac{x}{1-x}=\frac{1}{1-x}-1$$
| {
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Find the all possible values of $a$, such that $4x^2-2ax+a^2-5a+4>0$ holds $\forall x\in (0,2)$
Problem: Find the all possible values of $a$, such that
$$4x^2-2ax+a^2-5a+4>0$$
holds $\forall x\in (0,2)$.
My work:
First, I rewrote the given inequality as follows:
$$
\begin{aligned}f(x)&=\left(2x-\frac a2\right)^2+\frac {3a^2}{4}-5a+4>0\end{aligned}
$$
Then, we have
$$
\begin{aligned}
0<x<2\\
\implies -\frac a2<2x-\frac a2<4-\frac a2\end{aligned}
$$
Case $-1:\,\,\,a≥0 \wedge 4-\frac a2≤0$.
This leads,
$$
\begin{aligned}\frac {a^2}{4}>\left(2x-\frac a2\right)^2>\left(4-\frac a2\right)^2\\
\implies \frac {a^2}{4}+\frac {3a^2}{4}-5a+4>f(x)>\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4\end{aligned}
$$
For $f(x)>0$, it is enough to take $\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4>0$ with the restriction $a≥0\wedge 4-\frac a2≤0$.
Case $-2:\,\,\,a≤0 \wedge 4-\frac a2≥0$.
We have:
$$
\begin{aligned}\frac {a^2}{4}<\left(2x-\frac a2\right)^2<\left(4-\frac a2\right)^2\\
\implies \frac {a^2}{4}+\frac {3a^2}{4}-5a+4<f(x)<\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4\end{aligned}
$$
Similarly, for $f(x)>0$, it is enough to take $\frac{a^2}{4}+\frac {3a^2}{4}-5a+4>0$ with the restriction $a≤0\wedge 4-\frac a2≥0$.
Case $-3:\,\,\,a≥0 \wedge 4-\frac a2≥0$.
This case implies, $\left(2x-\frac a2\right)^2≥0$. This means, $f(x)≥\frac {3a^2}{4}-5a+4$
Thus, for $f(x)>0$, it is enough to take $\frac {3a^2}{4}-5a+4>0$ with the restriction $a≥0\wedge 4-\frac a2≥0$.
Finally, we have to combine all the solution sets we get.
I haven't done the calculation, because I want to make sure that the method I use is correct. Do you see any flaws in the method?
| I found the solution sets in your cases:
Case-1) $((-\infty,4)\cup(5,\infty))\cap[8,\infty)=[8,\infty)$
Case-2) $((-\infty,1)\cup(4,\infty))\cap[-\infty,0)=(-\infty,0]$
Case-3) $((-\infty,\frac{10-2\sqrt{13}}{3})\cup(\frac{10+2\sqrt{13}}{3},\infty))\cap[0,8]=[0,\frac{10-2\sqrt{13}}{3})\cup(\frac{10+2\sqrt{13}}{3},8]$
The union of sets we found in all cases is $S=(-\infty,\frac{10-2\sqrt{13}}{3})\cup(\frac{10+2\sqrt{13}}{3},\infty)$. S is the solution of the problem.
My solution:
$f'(x)=0$ gives the minumum of the quadratic function $f(x)=4x^2-2ax+a^2-5a+4$ at $x=\frac{a}{4}$ and the solution of the inequality $f(\frac{a}{4})=\frac{3}{4}a^2-5a+4>0$ is the set $S$ above.
If $a$ were in $[\frac{10-2\sqrt{13}}{3},\frac{10+2\sqrt{13}}{3}]$ then $x=\frac{a}{4}$ would be in $(0,2)$ and $f(\frac{a}{4})$ would be non-positive. So, we can conclude that $S$ is the solution set of the problem.
| {
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Is this proof of $\left\lfloor \frac{n}{m} \right\rfloor = \left\lceil \frac{n-m+1}{m} \right\rceil$ correct? I've been practicing proving things about floor and ceiling functions, so I thought I'd try to prove this well-known identity:
$$\left\lfloor \frac{n}{m} \right\rfloor = \left\lceil \frac{n-m+1}{m} \right\rceil$$
for all $n,m \in \mathbb{Z}$, $m>0$.
This is what I came up with. Is my proof correct?
Proof:
[see edit below]
Case 1: $m=1$ $$\left\lfloor \frac{n}{1} \right\rfloor = \lfloor n \rfloor = n$$ $$\left\lceil \frac{n-1+1}{1} \right\rceil = \lceil n \rceil = n$$
Case 2: $m>1$
If $\frac{n}{m}$ is an integer, then
$$\left\lceil \frac{n-m+1}{m} \right\rceil = \left\lceil \frac{n}{m} -1 + \frac{1}{m}\right\rceil = \frac{n}{m} -1 + \left\lceil \frac{1}{m} \right\rceil = \frac{n}{m} -1 + 1 = \frac{n}{m} = \left\lfloor \frac{n}{m} \right\rfloor$$
If $\frac{n}{m}$ is NOT an integer, then
$$\left\lfloor \frac{n}{m} \right\rfloor = \left\lceil \frac{n}{m} \right\rceil - 1 = \left\lceil \frac{n}{m} + \frac{1}{m} \right\rceil - 1 = \left\lceil \frac{n}{m} + \frac{1}{m} -1 \right\rceil = \left\lceil \frac{n-m+1}{m} \right\rceil$$
$\blacksquare$
If it's correct but you know a simpler/better way to prove it, please include that in your answer. Thank you.
EDIT: As pointed out by user peterwhy, "Case 1: $m=1$" is simply a special case of "$\frac{n}{m}$ is an integer" and therefore is not needed; hence I have grayed it out, and we don't need to separate the $m=1$ and $m>1$ cases anymore.
| Another succinct proof is
$$\left\lfloor\frac nm\right\rfloor = \left\lfloor\frac{n + \frac12}m\right\rfloor = \left\lceil\frac{n - m + \frac12}m\right\rceil = \left\lceil\frac{n - m + 1}m\right\rceil,$$
since $\frac{n + \frac12}m$ and $\frac{n - m + \frac12}m$ are always non-integers with a difference of $1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The locus of the intersection point of two perpendicular tangents to the curve $xy^2=1$. Find the locus of the intersection point of two perpendicular tangents to the curve $xy^2=1$.
I find out that the tangents to this curve with slope $m$ has this general form:
$y = mx+(-2m)^{\frac{1}{3}}-m \left( -\frac{1}{2m} \right)^{\frac{2}{3}}$
The perpendicular tangent would have this form then:
$y = -\frac{1}{m}x+ \left(\frac{2}{m} \right)^{\frac{1}{3}}+\frac{1}{m} \left( \frac{m}{2} \right)^{\frac{2}{3}}$
I solved geometrically that the locus is the circle: $2x^2+2y^2-3\sqrt[3]{2}x=0$ (you can check it here: desmos.com/calculator/xt2xquh72l), but I couldn't solve it algebraically. I wonder if you could help me in eliminating $m$ in this system:
$\begin{cases}
y &= mx+(-2m)^{\frac{1}{3}}-m \left( -\frac{1}{2m} \right)^{\frac{2}{3}} \\
y &= -\frac{1}{m}x+ \left(\frac{2}{m} \right)^{\frac{1}{3}}+\frac{1}{m} \left( \frac{m}{2} \right)^{\frac{2}{3}} \\
\end{cases}
$
| The curve is $x=1/t^2,y=t$ with dual curve (The dual curve lives in the dual plane of lines in the plane and is the set of tangents to the original curve. I use the parametric form $X=\frac{-y'}{xy'-yx'},Y=\frac{x'}{xy'-yx'}$) $X=-t^2/3,Y=-2/(3t)$ so two independent tangents are
$(-t^2/3)x -2y/(3t)+1=0 \\
(-s^2/3)x -2y/(3s)+1=0 $
(this is two independent parameters $s,t$ in the universal line $xX+yY+1=0$ that links the plane and the dual plane.)
using that the slope of the perpendicular to slope $m$ is $-1/m$ two perpendicular tangents are (determine $s$ i.t.o. $t$)
$-\frac12 t^3 x +\frac32 t =y\\
\frac{2}{t^3}x - \frac32\frac{\sqrt[3]{4}}{t} =y$
which intersect in
$x = (3 t^2)/(t^4 - 2^\frac23 t^2 + 2 \cdot 2^\frac13)\\y = (3 (2^\frac23 t^3 - 2 \cdot2^\frac13 t))/(2 (-t^4 + 2^\frac23 t^2 - 2 \cdot2^\frac13))$
which implicitizes to your circle.
Edit (or your solution w/o dual curves also leads to the same conclusion)
For your specific question solve the system for $x,y$ (I used wolframalpha Solve[{y == 2^(1/3) (-m)^(1/3) - ((-m^(-1))^(2/3) m)/2^(2/3) + m x, y == 2^(1/3) (m^(-1))^(1/3) + 1/(2^(2/3) m^(1/3)) - x/m}, {x, y}]), then substitute $m=n^3$ to get
$x=(3\cdot 2^\frac13 n^2)/(2 n^4-2 n^2+2) \\y=(3\cdot 2^\frac13 n^3-3\cdot 2^\frac13 n)/(2 n^4-2 n^2+2)$
set $c=2^\frac13$ and implicitize, in say M2,
R=QQ[c]
S=R[n,x,y,MonomialOrder=>Eliminate 1]
I=ideal(c^3-2,(2*n^4-2*n^2+2)*x-3*c*n^2,(2*n^4-2*n^2+2)*y-3*c*(n^3-n)) -- matrix {{c^3-2, 2*x^2+2*y^2-3*c*x, 2*n^2*y+2*n*x-3*c*n-2*y, n^2*x-n*y-x}}
that is, the locus is
$$2x^2+2y^2-3cx=0$$
| {
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Representing $\frac{1}{x^2}$ in powers of $(x+2)$ I was asked to represent $f(x)=\frac{1}{x^2}$ in powers of $(x+2)$ using the fact that $\frac{1}{1 − x} = 1 + x + x^2 + x^3 + ...$. I am able to represent $\frac{1}{x^2}$ as a power series, but I am struggling withdoing it in powers of $(x+2)$. This is what I attempted.
$I.$ Firstly, I used the simple expansion
$$\ln(x+1) = \sum_{n=1}^\infty (-1)^n\frac{x^n}{n}$$
which comes from the fact that $\frac{1}{1 − x} = 1 + x + x^2 + x^3 + ...$. I then noticed that
$$\ln(x) = \ln(1+(x-1))=\sum_{n=1}^\infty (-1)^n\frac{(x-1)^n}{n}$$
$II$. $-f(x)=-\frac{1}{x^2}$ is the second derivative of $\ln(x)$, so that
$$\frac{d}{dx}\sum_{n=1}^\infty (-1)^n\frac{(x-1)^n}{n}= \sum_{n=1}^\infty(-1)^n \frac{(x-1)^{n-1}}{n^2}$$
and
$$\frac{d}{dx} \sum_{n=1}^\infty(-1)^n \frac{(x-1)^{n-1}}{n^2} = \sum_{n=1}^\infty(-1)^n \frac{(x-1)^{n-2}}{n^2(n-1)}=-\frac{1}{x^2}$$
from what plainly follows that $$\frac{1}{x^2} = -\sum_{n=1}^\infty(-1)^n \frac{(x-1)^{n-2}}{n^2(n-1)}$$
$III.$ Of course, this expansion is not in powers of $(x+2$). What I pressumed would be the logical thing to do is to repeat these steps but, instead of the using $\ln(1 + (x-1))$, using the equivalent
$$\ln(-2 + (x+2))=\ln((-2)(1+\frac{x+2}{(-2)})) = \ln(-2)+\ln(1+\frac{x+2}{(-2)})$$
I expected to be able to use this because the second term is of the form $\ln(1+ u)$ where $u$ is some function of $x$, and we know the expansion of such expression. However, $\ln(-2)$ is nonsense, since $\ln(x)$ is defined only for $\mathbb{R}^+$.
Is there an alternative, better way to do this or am I missing something? Thanks.
| There is a better way.
$\frac 1 {x^{2}}=\frac 1 {((x+2)-2)^{2}}=\frac 1 4 \frac 1 {(1-\frac {x+2} 2)^{2} } =\frac 1 2\frac d {dx} \frac 1{1-\frac {x+2} 2}=\frac 1 2\frac d {dx} [\sum\limits_{n=0}^{\infty} (\frac {x+2} 2 )^{n}]$. Can you continue?
| {
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Find $f(7)$ given $f(x)f(y)=f(x+y)+f(x-y)$ and $f(1)=3$
Let $f:\Bbb R\to\Bbb R$ such that $f(1)=3$ and $f$ satisfies the functional equation
$$f(x)f(y) = f(x+y) + f(x-y)$$
Find the value of $f(7)$.
Attempt:
If $x=1$ and $y=0$, we find
$$f(1) f(0) = 2f(1) \implies f(0) = 2$$
If we fix $y=1$, we get the recurrence relation
$$\begin{cases} f(0) = 2 \\ f(1) = 3 \\ f(x + 1) - 3f(x) + f(x - 1) = 0 & \text{for } x\ge1 \end{cases}$$
From here I can solve for $f(x)$ or sequentially compute $f(2),f(3),f(4),\ldots$ to arrive at $f(7) = 843$.
Question:
Is there a more elegant way of finding $f(7)$ directly from the functional equation?
| Well, is not too fancy, but you could do
$f(7)=f(4+3)=f(4)f(3)-f(1),$
and using the fact that
$f(4)=f(3)f(1)-f(2),$
you could compute $f(7)$ by knowing only $f(1),\;f(2),$ and $f(3)$.
| {
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Roll two balanced dice until the sum of the faces equals 7 appears for the first time. Determine the expected value of tosses in this experiment. Roll two balanced dice until the sum of the faces equals 7 appears for the first time. After that, roll the same two dice until some face 3 appears for the first time. Determine the expected value of tosses in this experiment.
All the possibilities are bellow:
(1,1)(2,1)(3,1)(4,1)(5,1)(6,1)
(1,2)(2,2)(3,2)(4,2)(5,2)(6,2)
(1,3)(2,3)(3,3)(4,3)(5,3)(6,3)
(1,4)(2,4)(3,4)(4,4)(5,4)(6,4)
(1,5)(2,5)(3,5)(4,5)(5,5)(6,5)
(1,6)(2,6)(3,6)(4,6)(5,6)(6,6)\
So, there are 6 possibilities for the sum to be 7. So the probability is $6\over36$ $=$ $1 \over 6$.\
And the probability of face three is going to be $11 \over 36$.
But how do I determine the number of expected tosses?
| Let us represent
*
*$X$: number of tosses until dice sum is 7, $X\sim$ Geometric$(p=1/6)$, $E(X)=1/p=6$.
*$Y$: number of tosses until some face is 3, $Y\sim$ Geometric$(q=11/36)$, $E(Y)=1/q=36/11$.
*$X+Y$: total number of tosses
It is true that $E(X+Y)=E(X)+E(Y)=6+36/11=102/11.$
Therefore the expected number of tosses is $E(X+Y)=102/11$.
| {
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Why am I getting $\beta=90^{\circ}$ Consider the geometry below, where the small circle is touching both semi circles of radius $5$ and the side of the square. Find the radius of the small circle.
My try: $M$ and $N$ are centers of semicircles and $G$ is center of small circle indicated below.Let $r$ be the radius of small circle.
We have
$$\beta=\frac{3\pi}{4}-\alpha$$
So we have
$$\sin \beta=\frac{\sin \alpha+\cos \alpha}{\sqrt{2}}$$
Also $$\sin \beta=\frac{r}{5-r}=\frac{\sin \alpha+\cos \alpha}{\sqrt{2}}\tag{1}$$
By cosine law
$$\cos \alpha=\frac{(5-r)^2+(5 \sqrt{2})^2-(5+r)^2}{2(5 \sqrt{2})(5-r)}=\frac{5-2 r}{\sqrt{2}(5-r)}\:\Rightarrow \sin \alpha=\sqrt{\frac{2(5-r)^2-(5-2 r)^2}{2(5-r)^2}}=\frac{\sqrt{25-2r^2}}{\sqrt{2}(5-r)}$$
Hence from $(1)$ we get
$$\frac{(5-2 r)+\sqrt{-2 r^2+25}}{\sqrt{2}(5-r)}=\frac{r}{5-r}$$
Solving the above equation we get $r=\frac{5}{2}$, but in that case $\beta=90^{\circ}$. Where I went wrong?
| The strategy is fine, but you have made an algebraic mistake: $\frac{(5-2 r)+\sqrt{-2 r^2+25}}{\sqrt{2}(5-r)}$ is only $\cos \alpha + \sin \alpha$, not $\frac{\cos \alpha + \sin \alpha}{\sqrt 2}$. Solving $$\frac{(5-2 r)+\sqrt{-2 r^2+25}}{2(5-r)} = \frac{r}{5-r}$$ instead gives $r = \frac{20}{9}$.
A simpler approach is to use the two right triangles below:
The larger right triangle (with height $h$) has $h^2 + (5-r)^2 = (5+r)^2$, from which $h = \sqrt{20r}$. Then, the smaller right triangle has $(h-5)^2 + r^2 = (5-r)^2$, from which $r = \frac{20}{9}$.
| {
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Deriving exact value of $\sin \pi/12$ using double angle identity
Deriving the exact value of $\sin \pi/12$ using double angle identity
Double angle identity - $\sin 2A = 2\sin A \cos A$
So, $\sin (2 \frac{\pi}{12}) = 2 \sin \frac{\pi}{12} \cos \frac{\pi}{12} $
$\sin^2 A + \cos^2 A = 1 $ so, $\cos \frac{\pi}{12} = \sqrt{1- \sin^2 \frac{\pi}{12}}$
$\sin (2 \frac{\pi}{12}) =2 \sin \frac{\pi}{12}\sqrt{1- \sin^2 \frac{\pi}{12}} $
I square both sides of the equation to get:
$(1/2)^2 = 4 \sin^2 \frac{\pi}{12} (1- \sin^2 \frac{\pi}{12})$
With this, I am moving away from finding the exact value of $\sin \pi/12$ which is $\frac{\sqrt{3}-1}{2\sqrt{2}}$
| Let $x := \sin^2\frac{\pi}{12}$. Then your last equality is
$$\frac{1}{4} = 4x(1-x) = 4x - 4x^2.$$
The solutions of this quadratic equation are $x_{1,2} = \frac{1}{2} \pm \frac{\sqrt{3}}{4}$. Clearly, $x$ is smaller than $\frac{1}{2}$, so we must have $x = \frac{1}{2} - \frac{\sqrt{3}}{4}$. Taking the square root gives
$$\sin\frac{\pi}{12} = \frac{\sqrt{2 - \sqrt{3}}}{2}.$$
Now since $(\sqrt{3}-1)^2 = 4 - 2\sqrt{3} = 2(2 - \sqrt{3})$, we get
$$\sin\frac{\pi}{12} = \frac{\sqrt{3}-1}{2\sqrt{2}}$$
as expected.
| {
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show that $f(x)=x$ where $\,\forall x,y\in\mathbb{R}^+, f(x+f(y+xy)) = (y+1)f(x+1)-1$
Let $f:\mathbb{R}^+\to\mathbb{R}^+$ be a function such that $\,\forall x,y\in\mathbb{R}^+, f(x+f(y+xy)) = (y+1)f(x+1)-1$. Prove that $f(x)=x$ for all real numbers $x>0$.
I think it might be useful to prove that f is injective and then make a substitution. It should be possible to prove $f$ is surjective too. It might also help to prove additional properties of f such as multiplicativity, that f has infinitely many fixed points, that f satisfies the functional equation $f(x+y) = f(x)+f(y)$ for all positive real numbers x,y, etc. First substituting $x=y$ gives $f(x+f(x+x^2)) = (x+1)f(x+1)-1.$ Swapping $x$ and $y$ gives $f(y+f(x+xy)) = (x+1)f(y+1)-1.$ Substituting $x=1$ into the latter equation gives $f(1+f(2)) = 2 f(2) - 1.$ In general for any positive integer $n, f(n+f(n+1)) = 2f(n+1)-1$. Suppose $f(x)=f(y).$ Then It might be useful to make the substitution $u=y+1, v= x+1$. Then we have $f(u-1 + f(u(v-1))) = uf(v) - 1$ for all $u,v > 1.$
If $f(\dfrac{x+y}2) = \dfrac{f(x)+f(y)}2$ for $x,y \in \mathbb{R}_{\ge 0}$, then letting $g(x) = f(x) - f(0)$ we get $g((x+y)/2) = (g(x)+g(y))/2$ for all $x\in \mathbb{R}^+$ so we may assume $f(0) = 0$. Then $f(x) = f(2x)/2$ for all nonnegative x and so $f(x+y) = (f(2x)+f(2y))/2 = f(x)+f(y)$, implying that $f(nx) = nf(x)$ for all rationals $n > 0$ and $x \ge 0$.
| Let $P(x,y)$ be the assertion, as usual.
$P(1,x/2)\implies f(1+f(x))=(1+x/2)f(2)-1$. Thus if $f(a)=f(b)$, $a=b$ i.e. $f$ is injective.
Now we put the fairly obvious $P(x,\frac{1}{f(x+1)})$ to get $f(x+f\left(\frac{x+1}{f(x+1)}\right))=f(x+1)$ which, by injectivity, implies $f\left(\frac{x+1}{f(x+1)}\right)=1$. Again by the injectivity $\frac{x+1}{f(x+1)}$ is fixed for all $x$, so $f(x)=cx$ for some constant $c$. A quick plug back into the original problem yields $c=1$, as wanted.
| {
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Being given a cyclic summation on 3 letters equal to $1$, deduce the value of another cyclic summation If
$$
\dfrac{a}{b+c} + \dfrac{b}{a+c} + \dfrac{c}{a+b} = 1
$$
then
$$
\dfrac{a^2}{b+c} + \dfrac{b^2}{a+c} + \dfrac{c^2}{a+b} = \;?
$$
I tried to manipulated the equation above using some properties as
$(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+ac+bc)$ or
$$
\dfrac{a(a+c)(a+b)+b(b+c)(a+b)+c(b+c)(a+c)}{(b+c)(a+c)(a+b)} = 1.
$$
But I don't have some success ahead.
| By assumption we have
$$
\frac{a^3 + abc + b^3 + c^3}{(a + b)(a + c)(b + c)}=0.
$$
The second term equals
$$
\dfrac{a^2}{b+c} + \dfrac{b^2}{a+c} + \dfrac{c^2}{a+b} =\frac{(a^3 + abc + b^3 + c^3)(a+b+c)}{(a + b)(a + c)(b + c)}
$$
So it is equal to zero.
| {
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Solve the equation $(x-1)^5+(x+3)^5=242(x+1)$ Solve the equation $$(x-1)^5+(x+3)^5=242(x+1)$$ My idea was to let $x+1=t$ and use the formula $$a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4),$$ but I have troubles to implement it. The equation becomes $$(t-2)^5+(t+2)^5=242t\\(t-2+t+2)\left[(t-2)^4-(t-2)^3(t+2)+(t-2)^2(t+2)^2-\\-(t-2)(t+2)^3+(t+2)^4\right]=242t$$ Let $A=(t-2)^4-(t-2)^3(t+2)+(t-2)^2(t+2)^2-(t-2)(t+2)^3+(t+2)^4.$
Then $$A=(t-2)^4-(t-2)^2(t^2-4+t^2+4t+4)-(t+2)^3(2-t+t+2)\\=(t-2)^4-2t(t+2)(t-2)^2-4(t+2)^3.$$
| You could also solve it in a much simpler way, like this:
$$
\begin{align*}
(x - 1)^{5} + (x + 3)^{5} &= 242 \cdot (x + 1) \quad\mid\quad -(242 \cdot (x + 1))\\
(x - 1)^{5} + (x + 3)^{5} - 242 \cdot (x + 1) &= 0\\
2 \cdot x^{5} + 10 \cdot x^{4} + 100 \cdot x^{3} + 260 \cdot x^{2} + 168 \cdot x &= 0 \quad\mid\quad \div 2\\
x^{5} + 5 \cdot x^{4} + 50 \cdot x^{3} + 130 \cdot x^{2} + 84 \cdot x &= 0\\
x \cdot (x^{4} + 5 \cdot x^{3} + 50 \cdot x^{2} + 130 \cdot x + 84) &= 0 \quad\mid\quad \text{factor into four terms}\\
x \cdot (x + 1) \cdot (x + 2) \cdot (x^{2} + 2 \cdot x + 42) &= 0 \quad\mid\quad \text{use the zero product theorem}\\
\\
x &= 0 \tag{1.}\\
x + 1 &= 0 \quad\mid\quad -1 \tag{2.}\\
x + 2 &= 0 \quad\mid\quad -2 \tag{3.}\\
x^{2} + 2 \cdot x + 42 &= 0 \quad\mid\quad -41 \tag{4.}\\
\\
x &= 0 \tag{1.}\\
x &= -1 \tag{2.}\\
x &= -2 \tag{3.}\\
x^{2} + 2 \cdot x + 1 &= -41 \tag{4.}\\
\\
x &= 0 \tag{1.}\\
x &= -1 \tag{2.}\\
x &= -2 \tag{3.}\\
(x + 1)^{2} &= -41 \quad\mid\quad \sqrt{~~~} \tag{4.}\\
x + 1 &= \pm\sqrt{41} \cdot \mathrm{i} \quad\mid\quad -1\\
\\
x &= 0 \tag{1.}\\
x &= -1 \tag{2.}\\
x &= -2 \tag{3.}\\
x &= -1 \pm \sqrt{41} \cdot \mathrm{i} \tag{4.}\\
\end{align*}
$$
| {
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Probability that 2 out of 3 balls will be of the same colour if the first one drawn is red Three balls are drawn from a box containing 3 red, 4 black and 5 white balls. Find the probability that 2 out of 3 balls will be of the same colour (event A) if the first one drawn is red (event B).
I tried to solve this question using combinations and conditional probability. First, I calculated $P(A \cap B) = \frac{{3\choose1}{4\choose2}+{3\choose1}{5\choose2}+{3\choose2}{4\choose1}+{3\choose2}{5\choose1}} {12\choose3} = 15/44 $ and then I calculated $P(B) = \frac{3\choose1} {12\choose1} = 1/4$. Then I used the two results to find $P(A|B) = \frac{15/44}{1/4} = 1.36$, which is wrong (correct result is $34/55$). I know there are other ways of solving this exercise, but if possible, I would like to solve it using combinations.
Thanks in advance!
|
First, I calculated $P(A \cap B) = \frac{{3\choose1}{4\choose2}+{3\choose1}{5\choose2}+{3\choose2}{4\choose1}+{3\choose2}{5\choose1}} {12\choose3} = 15/44 $
This is the probability that exactly two among the three are the same colour and at least one among the three is red. That is not the required event.
Event $B$ specifies the first position's colour. So therefore, you must count ways to fill it separately from the later two in both the numerator and denominator.
The probability that the first one is red and exactly two among the first three are the same colour
$\qquad\begin{align}\mathsf P(A\cap B)&=\dfrac{\dbinom{3}{1}\left[\dbinom{2}{1}\dbinom{4+5}{1}+\dbinom{4}{2}+\dbinom{5}{2}\right]}{\dbinom{12}{1}\dbinom{11}{2}}\end{align}$
And so:
$\qquad\begin{align}\mathsf P(A\mid B)&=\dfrac{\dbinom{2}{1}\dbinom{4+5}{1}+\dbinom{4}{2}+\dbinom{5}{2}}{\dbinom{11}{2}}\end{align}$
| {
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Is it possible for two non real complex numbers a and b that are squares of each other? ($a^2=b$ and $b^2=a$)? Is it possible for two non real complex numbers a and b that are squares of each other? ($a^2=b$ and $b^2=a$)?
My answer is not possible because for $a^2$ to be equal to $b$ means that the argument of $b$ is twice of arg(a) and for $b^2$ to be equal to $a$ means that arg(a) = 2.arg(b) but the answer is it is possible.
How is it possible when arg(b) = 2.arg(a) and arg(a) = 2.arg(b) contradict each other?
| It is entirely possible. Consider the complex conjugate cube roots of one. $\omega = -\frac 12 + \frac{\sqrt 3}{2}i$ and $\overline \omega = -\frac 12 - \frac{\sqrt 3}{2}i$
Note that $\omega = (\overline \omega)^2$, but also $\overline \omega = \omega^2$.
There is no contradiction here, because from the two equations, you get $\omega = \omega^4 = \omega^3\omega = 1\cdot \omega$.
Going by arguments, $\mathrm{arg}(\omega) = \frac{2\pi}{3}$ while $\mathrm{arg}(\overline \omega) = \frac{4\pi}{3}$. Clearly, $2\cdot \frac{2\pi}{3} = \frac{4\pi}{3}$ but also $2\cdot \frac{4\pi}{3} = \frac{8\pi}{3} \pmod{2\pi} = \frac{2\pi}{3}$. Essentially, you're forgetting that arguments work modulo $2\pi$.
Note that that pair is the only possible solution. If you write $x = y^2$ and $y = x^2$, then $x = (x^2)^2 = x^4 \implies x^4 - x = 0 \implies x(x^3-1)=0 \implies x = 0$ or $x^3 - 1 = 0$. The latter implies $(x-1)(x^2 + x + 1)=0 \implies x=1$ or $x^2 + x + 1=0$. Solving the quadratic gives you the complex conjugate roots of one. Therefore, $x = 0, 1, \omega, \overline \omega$ are the only solutions. If you impose the additional constraint that $x \neq y$, you're left with the only solutions being $\omega$ and $\overline \omega$, as already stated.
| {
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Is this proof (scratch work) of $\lim_{x\rightarrow c}x^3=c^3$ correct? Is this scratch work for the proof of $\lim_{x\rightarrow c}x^3=c^3$ correct?
We have that $|x^3-c^3|=|x-c||x^2+xc+c^2|$.
If $|x-c|<|c|$ then $|x|=|x-c+c|\leq |x-c|+|c|<2|c|$, it follows that $|x^2+xc+c^2|\leq |x|^2+|x||c|+|c|^2< 4|c|^2+2|c|^2+|c|^2=7|c|^2$.
So if $|x-c|<|c|$ and $|x-c|<\frac{\epsilon}{7|c|^2}$ we have that $|x^3-c^3|=|x-c||x^2+xc+c^2|<\frac{\epsilon 7|c|^2}{7|c|^2}=\epsilon$.
So if we let $\epsilon>0$ be given and $\delta=\min(|c|,\frac{\epsilon}{7|c|^2})$ then $|x^3-c^3|<\epsilon$ if $0<|x-c|<\delta$.
| (After reading the comments.) With a little more care, you can include the case $c=0$:
If $|x-c|\le\delta\le1$ then $|x|=|x-c+c|\leq |x-c|+|c|\le1+|c|,$ so
$|x^3-c^3|\le\delta\left(3(1+|c|)^2\right),$
which is $\le\epsilon$ if $\delta$ is not only $\le1$ but also $\le\frac\epsilon{3(1+|c|)^2}.$
| {
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Why is $ab+\frac{1}{2} - \frac{a+b}{2} >0$ for $0I am trying to show that the function $$ab+\frac{1}{2}-\frac{a+b}{2}$$ is positive when $0<a,b<1.$
Here is what I have done.
*
*The arithmetic-geometric mean inequality doesn't seem like the way to go because it implies $$ab < \sqrt{ab} \le \frac{a+b}{2}.$$
This implies $$ab+\frac{1}{2}-\frac{a+b}{2}\le \frac{1}{2}.$$
*When I fix values for $a$ (or $b$) and sketch traces of the function $ab+\frac{1}{2}-\frac{a+b}{2}$, I obtain positive values on the graph.
| As a rule, when you are dealing with numbers in $[0,1]$, you can write them as $a = \frac{u}{u+v}$ and $b= \frac{p}{p+q}$, where $u$, $v$, $p$, $q$ are positive. Substituting we get
$$a b + \frac{1}{2} - \frac{a+b}{2} = \frac{p u + q v}{2 (p + q) (u + v)}$$
clearly positive. We can substitute back if we wish, $u+v= p+q=1$, $u = a$, $p=b$ and get the expression
$$\frac{ a b + (1-a)(1-b)}{2}$$
| {
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Find $\lim_{n \to \infty} \prod_{r=1}^n (n^2+r^2)^{\frac{1}{n}}/n^2$
Find $\lim_{n \to \infty} \prod_{r=1}^n \frac{(n^2+r^2)^{\frac{1}{n}}}{n^2}$
Let $L = \lim_{n \to \infty} \prod_{r=1}^n \frac{(n^2+r^2)^{\frac{1}{n}}}{n^2}$
Then $\log(L) = \lim_{n \to \infty} \log(\prod_{r=1}^n \frac{(n^2+r^2)^{\frac{1}{n}}}{n^2})$
$\log(L) = \lim_{n \to \infty} \sum_{r=1}^n \log( \frac{(n^2+r^2)^{\frac{1}{n}}}{n^2})$
$\log(L) = \lim_{n \to \infty} \sum_{r=1}^n(\log( (n^2+r^2)^{\frac{1}{n}})-\log(n^2))$
$\log(L) = \lim_{n \to \infty} \sum_{r=1}^n( \frac{1}{n} \log( n^2(1+(\frac{r}{n})^2)^{})-\log(n^2))$
$\log(L) = \lim_{n \to \infty} \sum_{r=1}^n( \frac{1}{n}( 2\log(n)+\log(1+(\frac{r}{n})^2)^{})-\log(n^2))$
We see that 1 part becomes $\int_{0}^1 \log(1+x^2)dx$
How do I manipulate the remaining terms $\frac{1}{n}2log(n) - log(n^2)$?
| Notice that:
$$\prod\nolimits_{r = 1}^{n} \frac {(n^2+r^2)^{\frac{1}{n}}}{n^2}\le \prod\nolimits_{r = 1}^{n}\frac {(n^2+n^2)^{\frac{1}{n}}}{n^2}\le \frac {2n^2}{n^{2n}};$$
hence $L=0$.
| {
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The least value of $f(x)$=$\sqrt{x^2-2x+2}+\sqrt{x^2-4x+29}$ If $m$ is the least value of $f(x)$=$\sqrt{x^2-2x+2}+\sqrt{x^2-4x+29}$, occur at $x = α$ , then $[m]+[α]$ is equal to (where [.] denotes greatest integer function)
(A)6
(B)7
(C)5
(D)4
My approach is as follow
$f(x)=\sqrt{(x-1)^2+1}+\sqrt{(x-2)^2+5^2}$
$\alpha$ lies between 1&2 and $m$ lies between 6 and 7 hence the answer is 7. I checked it and ur us correct. But solution is required
|
In the figure, let $X(x,0)$ be any point on the $x$-axis.
Note that $AX=\sqrt{(x-1)^2+1}$ and $BX=\sqrt{(x-2)^2+5^2}$.
Therefore our job is to minimize $AX+BX$ where $X$ is an arbitrary point on the $x$-axis
Let $A'(1,-1)$ be the mirror image of $A(1,1)$ with respect to the $x$-axis.
From perpendicular bisector theorem,
$AX+XB=A'X+XB \ge A'B=AC+CB$
Thus the least value of $AX+XB$ is $A'B$. It is attained when $X=C$.
Since $A'B=\sqrt{(2-1)^2+(5+1)^2}=\sqrt{37}$
The least value of $f(x)$ is $\sqrt{37}$
Finally let $C=(\alpha, 0)$, then $\frac{0+1}{\alpha-1}=\frac{5+1}{2-1} \implies \alpha = \frac{7}{6}$
| {
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For $5b^2+4=n^2$, prove that $b=\frac{1}{5}(\lambda_2-\lambda_1)(\lambda_2^k-\lambda_1^k)$ where $\lambda_1, \lambda_2$ are roots of $x^2-3x+1=0$. If $b\in \mathbb{Z}_{>0}$ and $5b^2+4$ is a perfect square, then $b$ can be written in the form $b=\frac{1}{5}(\lambda_2-\lambda_1)(\lambda_2^k-\lambda_1^k)$, where $\lambda_1, \lambda_2$ are the two distinct solutions of the equation $x^2-3x+1=0$.
I have computed that $b=1,3,8, 21$ satisfies the condition and their $k$s are respectively $1,2,3,4$
| COMMENT.-Your equation $5b^2+4=k^2$ is equivalent to $$\left(\dfrac k2\right)^2-5\left(\dfrac b2\right)^2=1$$ and the Pell-Fermat equation $$x^2-5y^2=1$$ has the infinite set of integer solutions where $n$ is natural integer.
$$x=\frac{(9+4\sqrt5)^n+(9-4\sqrt5)^n}{2}\\y=\frac{(9+4\sqrt5)^n-(9-4\sqrt5)^n}{2\sqrt5}$$ so you have integer solution for $k,b$
$$k=(9+4\sqrt5)^n+(9-4\sqrt5)^n\\b=\frac{(9+4\sqrt5)^n-(9-4\sqrt5)^n}{\sqrt5}$$ With these solutions you can verify if it is true what you say.
| {
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Does $\sum_{j=0}^n\sum_{k=0}^n\binom{j}{a}\binom{k}{b}\binom{n-j-k}{c}=\binom{n+1}{a+b+c+2}?$ I'm working on a problem,
(a) Let $a,b,n\geq1$ with $a+b\leq n$. By considering choosing $a+b+1$ numbers from the set $\{0,1,...,n\}$, and the possibilities for the number in position $a+1$ when the chosen numbers are listed in increasing order, show that $$\binom{n+1}{a+b+1}=\sum_{k=0}^n\binom{k}{a}\binom{n-k}{b}.$$
(b) Hence, or otherwise, express $$\sum_{j=0}^n\sum_{k=0}^n\binom{j}{a}\binom{k}{b}\binom{n-j-k}{c},$$ where $a+b+c\leq n$, as a single binomial coefficient.
Does the following argument make sense for (b)? (I leave a certain amount implicit.)
Let \begin{align}
& j && \text{ be the } (a+1)^\text{th} && \text{number chosen}, \\
& j+1+k && \text{ be the } (a+1)+(b+1)=(a+b+2)^\text{th} && \text{number chosen;}
\end{align}
Then there are \begin{align}
& \binom{j}{a} && \text{ways of choosing the first $a$ numbers from } [0,j-1], \\
& \binom{k}{b} && \text{ways of choosing the next $b$ numbers from } [j+1,j+k], \\
& \binom{n-j-k}{c} && \text{ways of choosing the next $c$ numbers from } [j+1+k,n];
\end{align}
therefore the given expression is equal to $$\binom{n+1}{a+b+c+2}.$$
| We can also derive a closed formula of the double sum by applying the identity
\begin{align*}
\sum_{k=0}^n\binom{k}{a}\binom{n-k}{b}=\binom{n+1}{a+b+1}\tag{1}
\end{align*}
twice.
We obtain
\begin{align*}
\color{blue}{\sum_{j=0}^n}&\color{blue}{\sum_{k=0}^{n-j}\binom{j}{a}\binom{k}{b}\binom{n-j-k}{c}}\tag{2}\\
&=\sum_{j=0}^n\binom{j}{a}\sum_{k=0}^{n-j}\binom{k}{b}\binom{(n-j)-k}{c}\tag{3}\\
&=\sum_{j=0}^n\binom{j}{a}\binom{n-j+1}{b+c+1}\tag{$\to$ (1)}\\
&=\sum_{j=0}^{n+1}\binom{j}{a}\binom{(n+1)-j}{b+c+1}\tag{4}\\
&\,\,\color{blue}{=\binom{n+2}{a+b+c+2}}\tag{$\to$ (1)}\\
\end{align*}
and observe that lower and upper index of the result have a summand $2$.
Comment:
*
*In (2) we set the upper limit of the inner sum to $n-j$ since the upper index of $\binom{n-j-k}{c}$ is non-negative.
*In (3) we make a rearrangement as preparation to apply (1).
*In (4) we set the upper limit of the sum to $n+1$ which does not change the sum, since $\binom{0}{b+c+1}=0$. We observe we can apply (1) again.
| {
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How can we guess one solution of the equation $y''+\frac{2}{x}y'+y=0$? I want to solve the equation $$\frac{1}{x^2} \, \frac{d}{dx}\left(x^2\frac{dy}{dx}\right)=-y.$$ I converted it to this equation $y''+\frac{2}{x}y'+y=0$. How can we guess one solution of the equation $y''+\frac{2}{x}y'+y=0$? Is there any way to solve it generally?
| One thing to try for non-constant coefficients second order linear ode's is the Liouville transformation. This involves no guessing and is one of the methods to try for such ode's if other methods fail.
Solve
\begin{align*}
y''+\frac{2 y'}{x}+y&=0\\ \tag{1}
A y''+ B y' + C y&=0
\end{align*}
Hence $A=1,B=\frac{1}{2},C=1$. Applying the Liouville transformation on the dependent variable gives
\begin{align*}
z(x) &= y e^{\int \frac{B}{2 A} \,dx} \tag{2}
\end{align*}
Then (1) becomes
\begin{align*}
z''(x) = r z(x)\tag{3}
\end{align*}
Where $r$ is given by
\begin{align*}
r &= \frac{2 A B' - 2 B A' + B^2 - 4 A C}{4 A^2} \\
&= -1
\end{align*}
Hence (3) becomes
\begin{align*}
z''(x) + z(x) = 0
\end{align*}
This is constant coefficient ode now. Its solution is easily found to be $z=\cos x$. Applying (2) again in reverse to find the first basis solution for the $y$ ode gives
\begin{align*}
y_1 &= z e^{-\int \frac{B}{2 A} \,dx}\\
&= z e^{ -\int \frac{1}{2} \frac{\frac{2}{x}}{1} \,dx}\\
&= z e^{-\ln \left(x \right)}\\
&= z \left(\frac{1}{x}\right)\\
&= \frac{\cos \left(x \right)}{x}
\end{align*}
The second basis solution $y_2$ to the original ode is found using reduction of order
\begin{align*}
y_2 &= y_1 \int \frac{ e^{\int -\frac{B}{A} \,dx}}{y_1^2} \,dx \\
y_2 &= y_1 \int \frac{ e^{\int -\frac{\frac{2}{x}}{1} \,dx}}{\left(y_1\right)^2} \,dx\\
& = y_1 \int \frac{ e^{-2 \ln \left(x \right)}}{\left(y_1\right)^2} \,dx\\
&= y_1 \left(\tan \left(x \right)\right)
\end{align*}
Therefore the solution is
\begin{align*}
y &= c_1 y_1 + c_2 y_2\\
&= c_1 \frac{\cos \left(x \right)}{x} + c_2 \frac{\cos \left(x \right)}{x} \tan \left(x \right)\\
&= c_1 \frac{\cos \left(x \right)}{x} + c_2 \frac{\sin \left(x \right)}{x} \\
\end{align*}
Another thing to also always try with non-constant coefficient is solving as Bessel ODE using Bowman method. This works if one is able to solve for the unknown using linear algebra solver
Start by writing the ode as
\begin{align*}
x^{2} y^{\prime \prime}\left(x \right)+2 x y^{\prime}\left(x \right)+x^{2} y \left(x \right) = 0\tag{1}
\end{align*}
But Bessel ode has the form
\begin{align*}
x^{2} y^{\prime \prime}\left(x \right)+x y^{\prime}\left(x \right)+\left(-n^{2}+x^{2}\right) y \left(x \right) = 0\tag{2}
\end{align*}
The generalized form of Bessel ode is given by Bowman (1958) is the following
\begin{align*}
x^{2} y^{\prime \prime}\left(x \right)+\left(1-2 \alpha \right) x y^{\prime}\left(x \right)+\left(\beta^{2} \gamma^{2} x^{2 \gamma}-n^{2} \gamma^{2}+\alpha^{2}\right) y \left(x \right) = 0\tag{3}
\end{align*}
With is standard solution being
\begin{align*}
y \left(x \right)&=x^{\alpha} \left(c_{1} \operatorname{BesselJ} \left(n , \beta \,x^{\gamma}\right)+c_{2} \operatorname{BesselY} \left(n , \beta \,x^{\gamma}\right)\right)\tag{4}
\end{align*}
Comparing (3) to (1) and solving for $\alpha,\beta,n,\gamma$ gives (little algebra)
\begin{align*}
\alpha &= -{\frac{1}{2}}\\
\beta &= 1\\
n &= {\frac{1}{2}}\\
\gamma&= 1
\end{align*}
Substituting all the above into (4) gives
\begin{align*}
\frac{1}{\sqrt x}\left(c_1 \operatorname{BesselJ}\left(\frac{1}{2},x\right)
+ \operatorname{BesselY}\left(\frac{1}{2},x\right)\right) \tag{5}
\end{align*}
But
\begin{align*}
\operatorname{BesselJ}\left(\frac{1}{2},x\right)&=\frac{\sqrt{2}\, \sin \left(x \right)}{\sqrt{\pi}\, \sqrt{x}}\\
\operatorname{BesselY}\left(\frac{1}{2},x\right) &= -\frac{\sqrt{2}\, \cos \left(x \right)}{\sqrt{\pi}\, \sqrt{x}}
\end{align*}
Hence (5) becomes
\begin{align*}
y \left(x \right) = \frac{c_{1} \sqrt{2}\, \sin \left(x \right)}{x \sqrt{\pi}}-\frac{c_{2} \sqrt{2}\, \cos \left(x \right)}{x \sqrt{\pi}}
\end{align*}
The constants can be rewritten and the above becomes
\begin{align*}
y \left(x \right) &= C_{1} \frac{\sin \left(x \right)}{x }+ C_{2} \frac{\cos \left(x \right)}{x}
\end{align*}
| {
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} |
Show that $\ln(1/1-x) = \sum_{n=1}^{\infty} \frac{x^n}{n}$ for $-1
Show that $\ln(1/1-x) = \sum_{n=1}^{\infty} \frac{x^n}{n}$ for $-1<x<1$ using the power series $\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$
I did this using term by term integration
$\int \frac{1}{1-x} dx = (-1) \ln (1-x) + C = \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1}$
LHS: $\ln (1) - \ln (1-x) = \ln \frac{1}{1-x} + C$
What do I have to do with the right hand side of the equation? And afterwards, what is the reason for substituting $x=0$ to find the value of $C$ ? Where did the $x=0$ came from?
| You are on the right track. For $ |x|<1$ let
$$f(x)= \sum_{n=1}^{\infty}\frac{x^n}{n}.$$
Then
$$f'(x)= \sum_{n=1}^{\infty}x^{n-1}= \sum_{n=0}^{\infty}x^{n}= \frac{1}{1-x}.$$
Hence
$$f(x)= - \ln (1-x)+c.$$
With $x=0$ we see that $c=0.$
Therefore
$$f(x)= - \ln (1-x)= \ln (1)-\ln (1-x)= \ln ( \frac{1}{1-x}).$$
| {
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} |
Laurent Series of f On Given Annulus I am attempting to solve the following question and I am encountering some difficulties:
Expand the function:
$$
f(z) = \frac{z}{z^2 + 2z -3}
$$
in powers of z to find a series that is valid for an annulus containing z=2. For what values of z does this series converge?
My attempt at a solution was to write out the function using partial fraction decomposition as:
$$
f(z) = \frac{3/4}{z+3} - \frac{1/4}{1-z}
$$
I then tried using the formula for a power series to write the two terms as:
$$
f(z) = \frac{3}{4}\frac{1}{1-(-\frac{z}{3})} - \frac{1}{4}\frac{1}{1-z} = \frac{3}{4}\sum_{k=0}^{\infty}(-1)^k \cdot (z/3)^k + \frac{1}{4}\sum_{k=0}^{\infty}(z)^k
$$
Though, it is clear that the second series is only valid for $|z| < 1$, so I am stuck. How should I move forward with this? Thank you for your help!
| We want to expand a series valid for an annulus containing $z=2$. When looking at
\begin{align*}
\color{blue}{f(z)=\frac{3}{4}\left(\frac{1}{z+3}\right)-\frac{1}{4}\left(\frac{1}{1-z}\right)}\tag{1}
\end{align*}
*
*the left summand of (1) can be expanded as usual. We derive using the geometrics series expansion
\begin{align*}
\color{blue}{\frac{1}{z+3}}=\frac{1}{3}\left(\frac{1}{1+\frac{z}{3}}\right)\color{blue}{=\frac{1}{3}\sum_{k=0}^{\infty}(-1)^k\left(\frac{z}{3}\right)^k\qquad\qquad |z|<3}\tag{2}
\end{align*}
We derive the region of convergence from $\left|\frac{z}{3}\right|<1$ which is a disc with center $0$ and radius $3$.
*Doing the same with the right summand of (1) is not helpful, since
\begin{align*}
\frac{1}{1-z}=\sum_{k=0}^{\infty}z^k\qquad\qquad |z|<1
\end{align*}
gives as region of convergence the unit disc with center $0$ which is not helpful since it doesn't contain $z=2$. But, if we expand $\frac{1}{1-z}$ in terms of $\frac{1}{z}$ instead of $z$ we get the complement of the unit disc as region of convergence (besides the boundary of the unit disc). We obtain
\begin{align*}
\color{blue}{\frac{1}{1-z}}&=\left(-\frac{1}{z}\right)\frac{1}{1-\frac{1}{z}}=\left(-\frac{1}{z}\right)\sum_{k=0}^{\infty}\left(\frac{1}{z}\right)^k\\
&=-\sum_{k=0}^{\infty}\frac{1}{z^{k+1}}\color{blue}{=-\sum_{k=1}^{\infty}\frac{1}{z^k}\qquad\qquad\qquad|z|>1}\tag{3}
\end{align*}
We derive the region of convergence from $\left|\frac{1}{z}\right|<1$ which is the complement of the unit disc with center $0$.
*The region of convergence of $f(z)$ is consequently the annulus
\begin{align*}
\color{blue}{1<|z|<3}
\end{align*}
and the series expansion is combining (1) to (3)
\begin{align*}
\color{blue}{f(z)=\frac{1}{4}\sum_{k=0}^{\infty}(-1)^k\left(\frac{z}{3}\right)^k+\frac{1}{4}\sum_{k=1}^{\infty}\frac{1}{z^k}}
\end{align*}
| {
"language": "en",
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$\lim_{n \to \infty} \frac{1}{\sqrt{n^k}} (1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}})^k$ by Stolz–Cesàro $$\lim_{n \to \infty} \frac{1}{\sqrt{n^k}} (1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}})^k, k \in \mathbb{N}$$
Putting $\sqrt{n^k}$ in denominator, we get:
$$\lim_{n \to \infty} \frac{(1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}})^k}{\sqrt{n^k}}$$
Apllying Stolz–Cesàro theorem:
$$\lim_{n \to \infty} \frac{(1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}})^k - (1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n-1}})^k }{\sqrt{n^k} - \sqrt{(n-1)^k}}$$
I don't know how to use properly multinomial formula, but the biggest number in numerator among denominators in first brackets with $k$ degree is $\sqrt{n^k}$, in second brackets $\sqrt{(n-1)^k}$, their coefficients are $1$
$$\lim_{n \to \infty} \frac{ \frac{1}{\sqrt{n^k}} - \frac{1}{\sqrt{(n-1)^k}} + \frac{1}{\sqrt{(n-1)^k}} - \cdots + 1 - 1 + \cdots}{\sqrt{n^k} - \sqrt{(n-1)^k}}$$
Multiplying numbers in first brackets by $\frac{1}{\sqrt{n}}$ gives us $\frac{1}{\sqrt{2^{k-1} \times n}} + \frac{1}{\sqrt{3^{k-1} \times n}} + \cdots + \frac{1}{\sqrt{(n-1)^{k-1} \times n}}$ etc., but I suspect we can't get same numbers in second brackets.
My guess, after evaluating sum (or difference) in numerator, we need to divide each number of our expression by $\sqrt{n^k}$.
In denominator we get 0, because $$\sqrt{\frac{n^k}{n^k}} - \sqrt{\frac{(n-1)^k}{n^k}} = 1 - \sqrt{(\frac{n-1}{n})^k} = 1 - 1 = 0, n \to \infty$$
Multiplication and division of denominator by conjugate ($\sqrt{n^k} + \sqrt{(n-1)^k}$) gives "slightly" different picture (unless I've done several big mistakes in a row):
$$\frac{ n^k - (n-1)^k }{ \sqrt{n^k} + \sqrt{(n-1)^k} } = \frac{ n^k - n^k + kn^{k-1} - \cdots }{ \sqrt{n^k} + \sqrt{(n-1)^k} } = \frac{kn^{k-1} - {k\choose 2}n^{k-2} +\cdots }{ \sqrt{n^k} + \sqrt{(n-1)^k} }, n \to \infty$$
Dividing by $n^{k-1}$:
$$\frac{k - 0 + 0}{ \sqrt{\frac{n^k}{n^{2(k-1)} } } + \sqrt{ \frac{(n-1)^k}{n^{2(k-1)}} }} = \frac{k - 0 + 0 - \cdots}{0 + 0}$$
So my guess was a bad guess.
Unfortunately, by this point I'm pretty much lost. Hence, I need some help... Thank you!
| There is a nice proof already provided by iskander using Stolz–Cesàro.
Also, using definite integrals it can be shown the limit is $2^k$
$\left( \lim_{n \to \infty} \frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{1}{\sqrt{i}} \right)^k=\left( \lim_{n \to \infty} \frac{1}{n}\sum_{i=1}^{n}\sqrt{\frac{n}{i}} \right)^k=\left( \int_{0}^{1}\frac{dx}{\sqrt
{x}} \right)^k=2^k$
| {
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Proof verification: $\frac{a}{(bc-a^2)^2}+\frac{b}{(ca-b^2)^2}+\frac{c}{(ab-c^2)^2}=0$ The question states:
Let $a, b, c$ be real numbers such that $$\frac{1}{(bc-a^2)}+\frac{1}{(ca-b^2)}+\frac{1}{(ab-c^2)}=0$$ Prove that $$\frac{a}{(bc-a^2)^2}+\frac{b}{(ca-b^2)^2}+\frac{c}{(ab-c^2)^2}=0$$
Now this can be solved with algebraic manipulation, but I want to check if my solution is valid too:
By Cauchy-Schwarz,
$$\left (\frac{a^2}{(bc-a^2)^3}+\frac{b^2}{(ca-b^2)^3}+\frac{c^2}{(ab-c^2)^3}\right )\left (\frac{1}{(bc-a^2)}+\frac{1}{(ca-b^2)}+\frac{1}{(ab-c^2)}\right ) \ge \left (\frac{a}{(bc-a^2)^2}+\frac{b}{(ca-b^2)^2}+\frac{c}{(ab-c^2)^2}\right )^2 = P^2$$
Hence, $0 \ge P^2 \implies P = 0$.
| Your approach is not valid, and by coincidence, you don't arrive at an obvious contradiction because the final answer is zero too.
Let's examine a numerical example.
For $a=1, b=1, c=\frac{-1}{2}$, we get $\frac {1}{bc-a^2}+\frac {1}{ca-b^2}+\frac {1}{ab-c^2}=0$. If your approach was supposed to be correct, in a similar way, we would get:
$$(a^2(bc-a^2)+b^2(ac-b^2)+c^2(ab-c^2))(\frac {1}{bc-a^2}+\frac {1}{ca-b^2}+\frac {1}{ab-c^2})\ge (a+b+c)^2,$$
therefore we would have $a+b+c=0$, which is impossible according to the set of $a=1, b=1, c=\frac{-1}{2}$.
Indeed, the correct inequality we are allowed to use is:
$$(|(a^2(bc-a^2)|+|b^2(ac-b^2)|+|c^2(ab-c^2)|)(|\frac {1}{bc-a^2}|+|\frac {1}{ca-b^2}|+|\frac {1}{ab-c^2}|)\ge (a+b+c)^2,$$
and the same goes for the inequality you used (considering the absolute values).
Similar to Michael Rozenberg's solution, you can see this link.
| {
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Integration of Root What will be the integral
$$\int_0^1\sqrt{1-y^2}dy$$
I initially substituted $y =\sin t$ to solve (but couldn't since I'm bad at trigonometry).
| If you are not able to work around trig substitutions, you can use other elementary functions to substitute as well.
$$\sqrt{1 - y^2} \to y\sqrt{\frac{1}{y^2} - 1}$$
Let $u = y^2, du = 2y\cdot dy$
$u(0) = 0^2 = 0$
$u(1) = 1^2 = 1$
Hence, we have
$$y\sqrt{\frac{1}{y^2} - 1} \cdot dy \overset{u = y^2}{\to} \frac{1}{2}\sqrt{\frac{1}{u} - 1} \cdot du \to \frac{1}{2}\sqrt{\frac{1 - u}{u}} \cdot du \to \frac{1}{2}\frac{1 - u}{\sqrt{u}\sqrt{1 - u}} \cdot du \to \frac{1}{2}\frac{1 - u}{\sqrt{u - u^2}} \cdot du$$.
We substitute again, $w = u - u^2 = \frac{1}{4} - \big(u - \frac{1}{2}\big)^2, dw = 1- 2u \cdot du$.
$w(0) = 0$
$w(1) = 0$
$$\frac{1}{2}\frac{1 - u}{\sqrt{u - u^2}} \cdot du \to \frac{1}{2} \big[\frac{1}{2}\frac{1-2u + 1}{\sqrt{\frac{1}{4} - \big(u - \frac{1}{2}\big)^2}}\cdot du\big] \overset{w = \frac{1}{4} - \big(u - \frac{1}{2}\big)^2}{\to} \frac{1}{4}\big[\frac{dw}{\sqrt{w}} + \frac{du}{\sqrt{\frac{1}{4} - \big(u - \frac{1}{2}\big)^2}}\big]$$
Hence,
$$\frac{1}{4}\big[\int_{0}^{0}{\frac{dw}{\sqrt{w}}} + \int_{0}^{1}{\frac{du}{\sqrt{\frac{1}{4} - \big(u - \frac{1}{2}\big)^2}}}\big]$$
$$\to \frac{1}{4} \big{[} \sin^{-1}{(2u - 1)} \big{]}_{0}^{1}$$
$$= \frac{\pi}{4}$$
| {
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"answer_count": 4,
"answer_id": 3
} |
Find the value of: $\sqrt[3]{a+b}+\sqrt[3]{b+c}+\sqrt[3]{a+c}$
Let $a,b,c$ be roots of the cubic
$$x^3-x^2-2x+1=0$$
Then, find the value of:
$$\sqrt[3]{a+b}+\sqrt[3]{b+c}+\sqrt[3]{a+c}$$
My attempt.
I used the substitutions $$a+b=x^3, b+c=y^3, a+c=z^3$$
$$x^3+y^3+z^3=2(a+b+c)=2$$
Then I used the identity $$x^3+y^3+z^3=(x+y+z)^3-3(x+y+z)(xy+yz+xz)+3xyz$$
But I stuck here. I can not simplify $$xy+yz+xz$$
| From the given cubic, we know that $a+b+c=1$, therefore
$$\sqrt[3]{a+b}+\sqrt[3]{b+c}+\sqrt[3]{a+c}=\sqrt[3]{1-c}+\sqrt[3]{1-a}+\sqrt[3]{1-b}=\sqrt[3]{A}+\sqrt[3]{B}+\sqrt[3]{C}$$
where $A,B,C$ are the roots of
$(1-x)^3-(1-x)^2-2(1-x)+1=0$, that is $x^3-2 x^2-x+1=0$.
It can be shown (see for example this link) that
$$x^3+(\alpha^3−3\alpha\beta +3\gamma)x^2+
(\beta^3−3\alpha\beta \gamma+3\gamma^2)x+\gamma^3=0$$
is the cubic equation whose roots are the cubes of the roots of $x^3+\alpha x^2+\beta x+\gamma=0$. So it remains to find $-\alpha$ in the system
$$\begin{cases}
\alpha^3−3\alpha\beta +3\gamma=-2\\
\beta^3−3\alpha\beta \gamma+3\gamma^2=-1\\
\gamma^3=1
\end{cases}$$
which boils down to $\gamma=1$, $\beta=\frac{\alpha^3+5}{3\alpha}$ and
$$(\alpha^3+5)^3-27\alpha^3(\alpha^3+5)+4\cdot 27\alpha^3=0$$
or
$$\alpha^9-12\alpha^6+48\alpha^3+125=(\alpha^3-4)^3+189=0$$
which has just one real root at $\alpha=-\sqrt[3]{3\sqrt[3]{7}-4}$.
Hence
$$\sqrt[3]{a+b}+\sqrt[3]{b+c}+\sqrt[3]{a+c}=-\alpha=\sqrt[3]{3\sqrt[3]{7}-4}\approx 1.20249.$$
P.S. Notice that the roots $a,b,c$ of the cubic polynomial $f(x)=x^3-x^2-2x+1$ are three distinct real numbers (one is negative and two are positive) because $f(−2)<0$, $f(0)>0$, $f(1)<0$, $f(2)>0$.
| {
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How should I go about proving this identity? Found this on a math worksheet but wasn't able to prove it.
$$4(\cos^6x+\sin^6x)=1+3\cos^2(2x)$$
| Looking at the LHS of this equation, we can lower the exponent as follow:
$4(cos^6( x) +sin^6( x ) )= 4(cos^{2\cdot 3}(x) + sin^{2\cdot 3}(x) )= 4[(\dfrac{1+cos(2x)}{2})^3+(\dfrac{1-cos(2x)}{2})^3] = 4 \cdot \dfrac{2 + 2 cos^2(2x) + 4 cos^2(2x)}{2^3}= 4 \cdot \dfrac{2 + 6cos^2(2x)}{8} = \dfrac{2 + 6 cos^2(2x)}{2} = 1 + 3cos^2(2x)$
So we get $4(cos^6( x) +sin^6( x ) )= 1 + 3cos^2(2x)$ as wanted.
| {
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Solve the radical equation for all reals: $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1+x^2}$ Question:
Solve the radical equation for all reals: $$x\left(1+\sqrt{1-x^2}\right)=\sqrt{1+x^2}$$
My approach:
$$1+\sqrt {1-x^2}=\frac {\sqrt {1+x^2}}{x}\\1+2\sqrt {1-x^2}+1-x^2=\frac{1+x^2}{x^2}\\4(1-x^2)=\left(\frac{1+x^2}{x^2}+x^2-2\right)^2\\4x^4(1-x^2)=(x^4-x^2+1)^2$$
I don't know how can I proceed from here.
Is there a way so that not using the complicated expansion of polynomials?
I'm looking for methods that doesn't use $4$ or higher degree polynomial expansions.
| Starting from: $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1+x^2}$ then
$$ x^2 \, (2 - x^2 + 2 \, \sqrt{1-x^2}) = 1 + x^2 \\
2 \, x^2 \, \sqrt{1-x^2} = 1 - x^2 + x^4 \\
4 \, x^4 \, (1-x^2) = 1 - 2 x^2 + 3 x^4 - 2 x^6 + x^8 \\
x^8 + 2 x^6 - x^4 - 2 x^2 + 1 = 0 \\
(x^4 + x^2 - 1)^2 = 0 \\
x^4 + x^2 - 1 = 0 \\
(x^2 + \alpha)(x^2 + \beta) = 0 \\
(x + i \sqrt{\alpha})(x - i \sqrt{\alpha})(x + i \sqrt{\beta})(x - i \sqrt{\beta}) = 0.$$
This gives the solutions as
$$ x \in \{ i \sqrt{\alpha}, - i \sqrt{\alpha}, i \sqrt{\beta}, - i \sqrt{\beta} \}, $$
or
$$ x \in \left\{ \frac{1}{\sqrt{\beta}}, - \frac{1}{\sqrt{\beta}}, \frac{1}{\sqrt{\alpha}}, - \frac{1}{\sqrt{\alpha}} \right\} $$
where $2 \, \alpha = 1 + \sqrt{5}$ (the golden ratio), and $2 \, \beta = 1 - \sqrt{5}$ (inverse golden ratio).
Now that there are values to consider it can be shown that $x \in \{ i \sqrt{\alpha}, - i \sqrt{\alpha} \}$ do not satisfy the original equation. Of the remaining values only one satisfies the equation, namely,
$$ x = \frac{1}{\sqrt{\alpha}} = i \, \sqrt{\beta} = \sqrt{\frac{\sqrt{5}-1}{2}} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4594841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Basic question about floor function and limit ( $\lim \limits_{x \to 0} \lfloor{x-2}\rfloor \cdot \lfloor{x+3}\rfloor$)
$\lim \limits_{x \to 0} \lfloor{x-2}\rfloor \cdot \lfloor{x+3}\rfloor$
calculate the limit if it exists if not then prove it does not exist
I tried approaching by squeeze theorem and floor function property and got
$(x-2) \cdot (x+3)-1<\lfloor{x-2}\rfloor \cdot \lfloor{x+3}\rfloor \leq (x-2) \cdot (x+3)$
but then if I calculate the limits as $x$ approaches zero I get
$-7<\lfloor{x-2}\rfloor \cdot \lfloor{x+3}\rfloor \leq-6$
which did not give me an answer according to squeeze theorem so I tried a different approach by side limits
$\lim \limits_{x \to 0^+} \lfloor{x-2}\rfloor \cdot \lfloor{x+3}\rfloor = \lfloor{0-2}\rfloor \cdot \lfloor{0+3}\rfloor = -6$
and $\lim \limits_{x \to 0^-} \lfloor{x-2}\rfloor \cdot \lfloor{x+3}\rfloor = \lfloor{-1-2}\rfloor \cdot \lfloor{-1+3}\rfloor = -6$
so the limit exists and $L=-6$
is this correct? is there a different way?
thank you !
| As $x$ approaches $0$ from above (i.e. the right side limit), you have that
$\lfloor x-2\rfloor~$ stays at $~-2~$ and
$\lfloor x+3\rfloor~$ stays at $~3.~$
Therefore, the product stays at $~-6.~$
As $x$ approaches $0$ from below (i.e. the left side limit), you have that
$\lfloor x-2\rfloor~$ stays at $~-3~$ and
$\lfloor x+3\rfloor~$ stays at $~2.~$
Therefore, the product stays at $~-6.~$
So, the limit, as $x$ approaches $0$ from above does in fact equal the limit as $x$ approaches $0$ from below, and this limit is $-6.$
What makes this problem unusual is that you have the limit of the product of two functions, $~\lfloor x-2\rfloor~$ and $~\lfloor x+3\rfloor,~$ where for each function, as $x$ approaches $0$, the left side limit of the function is not equal to the right side limit of the function.
Despite that, when examining the product of the two functions, as $x$ approaches $0$, the left side limit of the product does equal the right side limit of the product.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4595678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How do I find the equation of an ellipse formed by a projection of a tilted rotating circle? Suppose I have a circle with radius $R$ on $xy$ plane with it's center at $(0,0,0)$. Then I rotate this circle about the y-axis by $0 \le \varphi \le {\pi \over 2}$, and then start rotating it about z-axis by $0 \le \alpha \le 2\pi$. What is the expression for the projection of this circle onto a $yz$ plane?
I know the projection is an ellipse with a semi-major axis always staying equal to $R$, and semi-minor axis ranging from $R\sin \varphi$ at $\alpha=\pi n$ to $0$ at $\alpha= {\pi\over2}n $. When semi-minor axis equals $0$ the projection becomes a segment of a line $y=\pm x\tan\varphi$ with the length $2R$. It's similar to the most basic Lissajoux curve, but not quite the same.
I need a non-parametric expression for this projection that is a function of $R, \alpha, \varphi$. Would greatly appreciate the expression itself as well as it's derivation. Thank you in advance.
| The circle is initially (before any rotation) spanned by two vectors
$ u_1 = (R, 0, 0) $ and $u_2 = (0, R, 0) $
So that points on the circle are parametrically given by
$ P(t) = \cos t u_1 + \sin t u_2 $
Apply the rotation about the $y$ axis gives
$ Q(t) = \cos t R_y(\varphi) u_1 + \sin t R_y(\varphi) u_2 $
where
$ R_y(\varphi) = \begin{bmatrix} \cos \varphi && 0 && \sin \varphi \\
0 && 1 && 0 \\ -\sin \varphi && 0 && \cos \varphi \end{bmatrix} $
Thus
$Q(t) = \cos t v_1 + \sin t v_2 $
where
$ v_1 = \begin{bmatrix} R \cos \varphi \\ 0 \\ - R \sin \varphi \end{bmatrix} , \hspace{50pt} v_2 = \begin{bmatrix} 0 \\ R \\ 0 \end{bmatrix} $
Next, apply the rotation about the $z$-axis by an angle $\alpha$, this we give a new parametric curve,
$ S(t) = \cos t R_z(\alpha) v_1 + \sin t R_z(\alpha) v_2 $
where
$ R_z(\alpha) = \begin{bmatrix} \cos \alpha && - \sin \alpha && 0 \\ \sin \alpha && \cos \alpha && 0 \\ 0 && 0 && 1 \end{bmatrix} $
Hence
$ S(t) = \cos t \ w_1 + \sin t \ w_2 $
where
$ w_1 = R \begin{bmatrix} \cos \alpha \cos \varphi \\ \sin \alpha \cos \varphi \\ - \sin \varphi \end{bmatrix} $
$ w_2 = R \begin{bmatrix} - \sin \alpha \\ \cos \alpha \\ 0 \end{bmatrix} $
Now projecting $S(t)$ onto the $xy$ plane gives
$ r(t) = \cos t \ V_1 + \sin t \ V_2 $
where $V_1 = R \begin{bmatrix} \cos \varphi \cos \alpha \\ \cos \varphi \sin \alpha \end{bmatrix} , V_2 = R \begin{bmatrix} -\sin \alpha \\ \cos \alpha \end{bmatrix} $
Clearly, $V_1 \perp V_2 $, therefore these two vectors are the semi-minor and semi-major axes of the ellipse of projection, respectively.
Writing r(t) in matrix-vector form , we get
$ r(t) = [x, y]^T = [V_1, V_2] u(t) = V u(t)$
where $ u(t) = [\cos t , \sin t ]^T $
From the above equation, we get
$u(t) = V^{-1} r(t) $
Now,
$V^{-1} = \dfrac{1}{R \cos \varphi} \begin{bmatrix} \cos \alpha && \sin \alpha \\ - \cos \varphi \sin \alpha && \cos \varphi \cos \alpha \end{bmatrix} = \dfrac{1}{R} \begin{bmatrix} \sec \varphi \cos \alpha && \sec \varphi \sin \alpha \\ - \sin \alpha && \cos \alpha \end{bmatrix}$
Now since $u^T(t) u(t) = 1 $ then
$ r^T V^{-T} V^{-1} r = 1 $
what remains is to find $V^{-T} V^{-1}$ and this is given by
$ \dfrac{1}{R^2} \begin{bmatrix} 1 + \tan^2 \varphi \cos^2 \alpha && \sin \alpha \cos \alpha \tan^2 \phi \\ \sin \alpha \cos \alpha \tan^2 \varphi && 1 + \tan^2 \varphi \sin^2 \alpha \end{bmatrix} $
Thus the final equation of the projected ellipse is
$ \bigg(1 + \tan^2 (\varphi) \cos^2 (\alpha) \bigg) x^2 + \bigg(\sin(2 \alpha) \tan^2( \varphi) \bigg) x y + \bigg(1 + \tan^2(\varphi) \sin^2(\alpha) \bigg) y^2 = R^2 $
If you're looking the projection onto the $yz$ plane then
$ r(t) = [y, z]^T = \cos t V_1 + \sin t V_2 $
where $V_1 = R \begin{bmatrix} \sin \alpha \cos \varphi \\ - \sin \varphi \end{bmatrix} , \hspace{50pt} V_2 = R \begin{bmatrix} \cos \alpha \\ 0 \end{bmatrix} $
Writing r(t) in matrix-vector form , we get
$ r(t) = [y, z]^T = [V_1, V_2] u(t) = V u(t)$
where $ u(t) = [\cos t , \sin t ]^T $
From the above equation, we get
$u(t) = V^{-1} r(t) $
Now,
$V^{-1} = \dfrac{1}{R \cos \alpha \sin \varphi} \begin{bmatrix} 0 && - \cos \alpha \\ \sin \varphi && \sin \alpha \cos \varphi \end{bmatrix} = \dfrac{1}{R} \begin{bmatrix} 0 && -\csc \varphi \\ \sec \alpha && \tan \alpha \cot \varphi \end{bmatrix}$
Now since $u^T(t) u(t) = 1 $ then
$ r^T V^{-T} V^{-1} r = 1 $
what remains is to find $V^{-T} V^{-1}$ and this is given by
$ \dfrac{1}{R^2} \begin{bmatrix} \sec^2 \alpha && \sec \alpha \tan \alpha \cot \varphi \\ \sec \alpha \tan \alpha \cot \varphi &&
1 + \cot^2 \varphi \sec^2 \alpha \end{bmatrix} $
Thus the final equation of the projected ellipse on the $yz$ plane is
$ \bigg(\sec^2 \alpha \bigg) y^2 + \bigg(2\sec \alpha \tan \alpha \cot \varphi \bigg) yz + \bigg(1 + \cot^2 \varphi \sec^2 \alpha \bigg) z^2 = R^2 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4597523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Representation of number as a sums and differences of natural numbers Lets consider all the combinations of:
$$1+2+3+4=10,\ \ 1+2+3-4=2,\ \ 1+2-3+4=4,\ \ 1+2-3-4=-4, $$
$$1-2+3+4=6,\ \ 1-2+3-4=-2,\ \ 1-2-3+4=0,\ \ 1-2-3-4=-8,$$
$$-1+2+3+4=8,\ \ -1+2+3-4=0,\ \ -1+2-3+4=2,\ \ -1+2-3-4=-6,$$
$$-1-2+3+4=4,\ \ -1-2+3-4=-4,\ \ -1-2-3+4=-2,\ \ -1-2-3-4=-10$$
or
we have one 10, two 2, etc.
Let now create a list of tuples with integer as a first element and sum of the oddity of minus signs in a representation as a second one.
So we will have
$${(10,1),(8,-1),(6,-1),(4,-1+1),(2,-1+1),(0,1+1),(-2,1-1),(-4,1-1),(-6,-1),(-8,-1),(-10,1)} $$
or
$${(10,1),(8,-1),(6,-1),(4,0),(2,0),(0,2),(-2,0),(-4,0),(-6,-1),(-8,-1),(-10,1)} $$
or after removing the tuples with 0 second element:
$${(10,1),(8,-1),(6,-1),(0,2),(-6,-1),(-8,-1),(-10,1)} $$
Trying to create a generating function for the general case of the last sequence for even counts (for example 6, 8, etc instead of 4 elements). For the odd ones obviously all the second elements will be equal to $0$ due to oddity.
Actually I am familiar with creation of generating functions for restricted partitions, etc. But for this particular case not sure how to deal with negative numbers?
| I believe you just need
$$ \prod_{k=1}^n (q^k - q^{-k}).$$
For example, $$(q-q^{-1})(q^2-q^{-2})(q^3-q^{-3})(q^4-q^{-4}) = q^{10} - q^8 - q^6 + 2 - q^{-6} - q^{-8} + q^{-10}$$
which corresponds to your final list of ordered pairs.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4598476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Algebraic calculation with polynomial and complex root.
Let $f=X^{3}-7 X+7$ be in $\mathbb{Q}[X]$.
Let $\alpha \in \mathbb{C}$ be a root of $f$ and hence $1, \alpha, \alpha^{2}$ be a basis of the $\mathbb{Q}$ vector space $\mathbb{Q}(\alpha)$. Let $\beta=3 \alpha^{2}+4 \alpha-14$. Write $\beta^{2}$ and $\beta^{3}$ as linear combinations of $1, \alpha, \alpha^{2}$ over $\mathbb{Q}$ and conclude that $\beta$ is a zero of $f$. Is $\beta=\alpha$?
I found my mistake with your help.
The solution: We have
$$
\beta^2=(3\alpha^{2}+4 \alpha-14)^2=9 \alpha^4 + 24 \alpha^3 - 68 \alpha^2 - 112 \alpha + 196
$$
To remove the cubic and quartic powers, we use the fact that $\alpha^3-7\alpha+7=0$ holds and gradually eliminate partial polynomials.
We get
$$
\beta^2=9 \alpha^4- 68 \alpha^2 + 24 \alpha^3 + 16\cdot (-7 \alpha) + 28\cdot 7
=9 \alpha^4- 68 \alpha^2 -8\cdot (-7 \alpha) \
+ 4\cdot 7\alpha(9 \alpha^3+10\cdot(-7) \alpha +8\cdot 7+2\alpha)+4\cdot 7=\alpha(-5\alpha -7) + 4\cdot 7=-5\alpha^2-7\alpha+28.
$$
Now we try $\beta^3$ with the same method
$$
\beta^3=\beta^2\beta=(3\alpha^{2}+4 \alpha-14)^2(3\alpha^{2}+4 \alpha-14)=(-5\alpha^2-7\alpha+28)(3\alpha^{2}+4 \alpha-14)\
=-15\alpha^4+ 126 \alpha^2 -(41 \alpha^3 +30\cdot(-7\alpha) + 56\cdot 7)=-15\alpha^4+ 126 \alpha^2 -( -11\cdot(-7\alpha) + 15\cdot 7)\
=-\alpha(15\alpha^3+18\cdot(-7\alpha) +7\cdot 11) - 15\cdot 7=-\alpha(3\cdot(-7\alpha) -4\cdot7) - 15\cdot 7=21\alpha^2 +28\alpha-105.
$$
And with this i get indeed
$$
f(\beta)=0.
$$
Thanks for your help! Maybe there are different reasonings for why $\beta\neq \alpha$?
| $\beta$ cannot equal $\alpha$. If so, then $\alpha$ would be a root of the polynomial $3x^2+3x-14$. But your starting polynomial $x^3-7x+7$ is irreducible over $\mathbb{Q}[x]$ (use Eisenstein criterion for instance), so $x^3-7x+7$ would divide $3x^2+3x-14$, which is impossible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4600336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Is this quadratic polynomial monotone at solutions of this cubic? Let $0<s < \frac{4}{27}$. The equation $x(1-x)^{2}=s$ admits exactly two solutions in $(0,1)$: Denote by $a,b$ be these solutions, and suppose that $a<b$.
Does
$$
(1-a)^2+2a^2<(1-b)^2+2b^2
$$
hold?
The limitation on the range of $s$ is due to $$\max_{a \in (0,1)}a(1-a)^{2}=\frac{4}{27},$$
which is attained at $a=\frac{1}{3}$.
I checked this numerically. I hope for an analytic argument which does not require solving explicitly the cubic.
Here is some partial information:
Define $F(x)= (1-x)^2+2x^2$. Then
$$
F'(x)=2(3x-1)>0 \iff x > \frac{1}{3}.
$$
Thus $F|_{(0,\frac{1}{3})}$ is decreasing, while $F|_{(\frac{1}{3},1)}$ is increasing.
Unfortunately, this does not seem to help.
The motivation for this problem come from this question.
| The given inequality
$$(1-a)^2 + 2a^2 < (1-b)^2 + 2b^2$$
is equivalent to
$$0 < (b-a)\cdot(3(b+a) - 2)$$
i.e.
$$ b+a > \frac{2}{3}.$$
If $c$ is the third root of the equation $x(1-x)^2 = s$, then Vieta's formula says $a+b+c = 2$, so the inequality is equivalent to
$$ c < \frac{4}{3}.$$
But if $c \ge \frac{4}{3}$, then
$$c(c-1)^2 \ge \frac{4}{3}\left(\frac{4}{3}-1\right)^2 = \frac{4}{27} > s,$$
so no value of $c\ge \frac{4}{3}$ is a root of $x(x-1)^2 = s$.
Therefore the given inequality is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4600892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show that $(x+1)^{p}x^{1-p}-(x+1)^{1-p}x^{p}$ is strictly increasing Let $1/2<p< 1$. I am asked to show that
$$f(x)=(x+1)^{p}x^{1-p}-(x+1)^{1-p}x^{p}$$
is strictly increasing for $x\geq 0$ and to compute $\lim_{x\to\infty} f(x)$.
I first computed the derivative, but I don't see why it must be positive:
$$\frac{d f(x)}{d x}=p(x+1)^{p-1}x^{1-p}+(1-p)(x+1)^px^{-p}-(1-p)(x+1)^{-p}x^p -p(x+1)^{1-p}x^{p-1}$$
Any ideas?
Thanks a lot for your help.
Case $p=3/4$. Then
$$f(x)=(x+1)^{1/4}x^{1/4}(\sqrt{x+1}-\sqrt{x})=\frac{(x+1)^{1/4}x^{1/4}}{\sqrt{x+1}+\sqrt{x}}$$
It suffices to show that $\ln f(x)$ is strictly increasing. We have
$$\ln f(x) =\frac{1}{4} \ln(x+1)+\frac{1}{4} \ln(x)-\ln(\sqrt{x+1}+\sqrt{x})$$
Taking derivative w.r.t. $x$ we get
$$\frac{d \ln f(x) }{d x}=\frac{1}{4}\frac{1}{x+1}+\frac{1}{4}\frac{1}{x}-\frac{1}{\sqrt{x+1}+\sqrt{x}}(\frac{1}{2}\frac{1}{\sqrt{x+1}}+\frac{1}{2}\frac{1}{\sqrt{x}})$$
Hence $\frac{d \ln f(x) }{d x}>0$ is equivalent to
$$(\frac{1}{4}\frac{1}{x+1}+\frac{1}{4}\frac{1}{x})(\sqrt{x+1}+\sqrt{x})>\frac{1}{2}\frac{1}{\sqrt{x+1}}+\frac{1}{2}\frac{1}{\sqrt{x}}$$
which is equivalent to
$$\frac{1}{4}\frac{\sqrt{x}}{x+1}+\frac{1}{4}\frac{\sqrt{x+1}}{x}>\frac{1}{4}\frac{1}{\sqrt{x+1}}+\frac{1}{4}\frac{1}{\sqrt{x}}$$
or
$$\frac{\sqrt{x+1}-\sqrt{x}}{x}>\frac{\sqrt{x+1}-\sqrt{x}}{x+1}$$
which holds. Hence $f(x)$ is strictly increasing.
| Rather a long comment than a solution:
My idea was to set $t=p-\frac 12\in(0,\frac 12)$ to introduce symmetry in the formulas, and then rewrite the expression as hyperbolic trig.
$\begin{align}f(x)
&=(x+1)^{\frac 12+t}x^{\frac 12-t}-x^{\frac 12+t}(x+1)^{\frac 12-t}\\\\
&=\sqrt{x}\sqrt{x+1}\,\Big(\tfrac{(x+1)^t}{x^t}-\tfrac{x^t}{(x+1)^t}\Big)\\\\
&=2\sqrt{x}\,\sqrt{x+1}\,\sinh\Big(t\ln(1+\tfrac 1x)\Big)
\end{align}$
We can calculate the derivative and factor out the positive quantity which doesn't change the sign:
$\underbrace{\sqrt{x}\,\sqrt{x+1}\,\cosh\Big(t\ln(1+\tfrac 1x)\Big)}_{\ge 0}\times f'(x)=(2x+1)\tanh\Big(t\ln(1+\tfrac 1x)\Big)-2t$
RHS seems decreasing and positive, but so far I'm stuck at proving it...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4604904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Is there another simpler method to evaluate the integral $\int_0^{2 \pi} \frac{1}{1+\cos \theta \cos x} d x , \textrm{ where } \theta \in (0, \pi)?$ Using ‘rationalization’, we can split the integral into two manageable integrals as:
$\displaystyle \begin{aligned}\int_0^{2 \pi} \frac{1}{1+\cos \theta \cos x} d x = & \int_0^{2 \pi} \frac{1-\cos \theta \cos x}{1-\cos ^2 \theta \cos ^2 x} d x \\= & \int_0^{2 \pi} \frac{d x}{1-\cos ^2 \theta \cos ^2 x}-\cos \theta \int_0^{2 \pi} \frac{\cos x}{1-\cos ^2 \theta \cos ^2 x} d x \\= & 4 \int_0^{\frac{\pi}{2}} \frac{\sec ^2 x}{\sec ^2 x-\cos ^2 \theta} d x+\int_0^{2 \pi} \frac{d(\cos \theta \sin x)}{\sin ^2 \theta+\cos ^2 \theta \sin ^2x} \\= & 4 \int_0^{\frac{\pi}{2}} \frac{d\left(\tan x\right)}{\sin ^2 \theta+\tan ^2 x}+\frac{1}{\sin \theta}\left[\tan ^{-1}\left(\frac{\cos ^2 \theta \sin x}{\sin \theta}\right)\right]_0^{2 \pi} \\= & \frac{4}{\sin \theta}\left[\tan ^{-1}\left(\frac{\tan x}{\sin \theta}\right)\right]_0^{\frac{\pi}{2}} \\= & \frac{4}{\sin \theta} \cdot \frac{\pi}{2} \\= & \frac{2 \pi}{\sin \theta}\end{aligned}\tag*{} $
Is there another simpler method to evaluate the integral?
Your comments and alternative methods are highly appreciated.
| Utilize the Fourier series below\begin{equation}
\frac{\sin\theta}{1+\cos\theta\cos x}=1+2\sum_{k=1}^\infty \left(\frac{\sin\theta-1}{\cos\theta}\right)^k\cos kx
\end{equation}
and recognize that only the constant term survives upon integration, i.e.
$$\int_0^{2 \pi} \frac{\sin\theta}{1+\cos \theta \cos x} d x=2\pi
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4605188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
Find the greatest integer in the expression below as a function of the given conditions Find the greatest integer less than $\sqrt{2^{100}+10^{10}}.$
Answer: $2^{50}$
I tried, but I can't finish:
$\sqrt{2^{100}+(2\cdot5)^{10}}=\sqrt{2^{10}(2^{90}+5^{10})}=2^5\sqrt{2^{90}+5^{10}}$
| A little GM-AM can also be helpful:
\begin{eqnarray*} \sqrt{2^{100}+10^{10}}
& = & 2^{50} \sqrt{ 1+\frac{10^{10}}{2^{100}}} \\
& \stackrel{GM-AM}{<} & 2^{50}\frac{1+1+\frac{10^{10}}{2^{100}}}2 \\
& = & 2^{50} + \frac{10^{10}}{2^{51}} \\
& < & 2^{50} + \frac{10^{10}}{2^{50}} \\
& = & 2^{50} + \left( \frac{10}{2^5}\right)^{10} \\
& < & 2^{50} + 1
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4607743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How To Prove $a^3b＋b^3c＋c^3a>a^2b^2＋b^2c^2＋a^2c^2 $ if $ a > b > c > 0\,$? How to prove this :
$$a^3b＋b^3c＋c^3a>a^2b^2＋b^2c^2＋a^2c^2 $$
if we know: $$ a > b > c > 0 $$
My attempt:
$$\frac {a^3b＋b^3a}{2}>a^2b^2
...(1)$$
$$\frac {b^3c＋c^3b}{2}>c^2b^2...(2)$$
$$\frac {a^3c＋c^3a}{2}>a^2c^2...(3)$$
(1) + (2)+ (3) :
$$\frac {a^3b＋b^3a＋a^3c＋c^3a＋b^3c＋c^3b}{2}>\;a^2b^2＋b^2c^2＋a^2c^2$$
$$a^3b＋b^3c＋c^3a>ab^3＋bc^3＋ca^3\\~\\ab(a^2－b^2)＋bc(b^2－c^2)>ca(a^2-c^2)\\~\\ab(a－b)(a＋b)＋bc(b－c)(b＋c)>ca(a－c)(a＋c)$$
I appreciate your help
| Here is a no-brainer approach suggested by Calvin Lin, compared to Michaels Rozenberg's magical answer.
We will use a positive variable to stand for the difference between two given variables and then eliminate the larger one of the two given variables.
For example since $b>c$, we let $b=c+x$ where $x>0$. We will substitute $c+x$ for $b$, eliminating $b$. We introduce a new "free" variable, $x$, which is very easy to use since the only condition on $x$ is $x>0$.
Similarly, let $a=b+y$ where $y>0$. So, we have $a=c+x+y$. We will also substitute $c+x+y$ for $a$, eliminating $a$ and welcoming the "free" variable $y$.
With some straightforward labor, we can find $\text{LHS}-\text{RHS}$ is
$$\begin{aligned}
&\quad a^2b(a-b) + b^2c(b-c) + c^2a(c-a)\\
&=(c+x+y)^2(c+x)y + (c+x)^2cx + c^2(c+x+y)(-x-y)\\
&=c^2 x^2 + c^2 x y + c^2 y^2 + c x^3 + 3 c x^2 y + 4 c x y^2 + c y^3 + x^3 y + 2 x^2 y^2 + x y^3
\end{aligned}$$
which is obvious $>0$ since all variables are $>0$.
Here are two exercises.
*
*Suppose $a>b>c>0$. Prove $a^3b+b^3c+c^3a > ab^3+bc^3+ca^3.$
*Suppose $a>b>c>d>0$. Prove $$a^3b+b^3c+c^3d+d^3a > a^2b^2+b^2c^2+c^2d^2+d^2a^2$$
(The computation might be long and boring. The point is no magic trick is needed)
| {
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"url": "https://math.stackexchange.com/questions/4608532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Indefinite integral $\int \frac{x^2}{\sqrt{x^2 + x^4}} \, dx $ involving $\mathrm{sgn}()$ function How to solve integrals like
$$\int \frac{x^2}{\sqrt{x^2 + x^4}} \, dx $$
If we factor out $x^2$ from denominator it become
$=\int \frac{x}{\left| x \right|} \frac{x}{\sqrt{1 + x^2}} \, dx $
$=\int \mathrm{sgn}(x) \frac{x}{\sqrt{1 + x^2}} \, dx $
What to do next ?
can't we do this like
$=\int \frac{x}{x} \frac{x}{\sqrt{1 + x^2}} \, dx $
$=\int \frac{x}{\sqrt{1 + x^2}} \, dx $ it becomes an easy integral with substitution now.
Is it wrong ?
So my quetion is that can we write $\sqrt{x^2}$ as x in this case ?
| Too long for a comment.
The integrand is well defined on $\mathbb{R}\backslash\{0\}$. This is the crucial point then. So how do you work with such cases?
You observe that indeed
$$\int \dfrac{x^2}{\sqrt{x^2+x^4}}\ \text{d}x = \int \dfrac{\vert x\vert }{x} \dfrac{x}{\sqrt{x^2+1}}\ \text{d}x = \int \dfrac{\vert x\vert }{\sqrt{x^2+1}}\ \text{d}x$$
The integrand is well defined in $0$ (indeed for example you can compute the integral from $-1$ to $+1$). Split into positive and negative $x$ and observe.
Consider also the integrand is an even function.
Eventually,
$$\int \dfrac{x^2}{\sqrt{x^2+x^4}}\ \text{d}x = \frac{\sqrt{x^4+x^2}}{x} + C \equiv \dfrac{\vert x\vert \sqrt{x^2+1}}{x} + C = \dfrac{\sqrt{x^2+1}}{\text{sgn(x)}} + C = \sqrt{x^2+1}\ \text{sgn}(x) + C$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculating double integral using variables substitution $\displaystyle
D = \left\lbrace \left. \rule{0pt}{12pt} (x,y) \; \right| \; 3 x^2 + 6 y^2 \leq 1 \right\rbrace$
Calculate $\displaystyle
\iint_D \frac{ x^2 }{ ( 3 x^2 + 6 y^2 )^{ 3/2 } } \; dx dy{}$.
Attempt:
$x=\frac{r}{\sqrt3}cost,y=\frac{r}{\sqrt6}sint \implies |J|=\sqrt{\frac{2}{3}}r$
$3 x^2 + 6 y^2 \leq 1 \implies 0\leq r \leq 1$
$\iint_D \frac{ x^2 }{ ( 3 x^2 + 6 y^2 )^{ 3/2 } } \; dx dy{} =\int _0^1\:\int _0^{2\pi }\:\frac{cos^2t\sqrt{2}}{3\sqrt{3}}dtdr = \frac{\sqrt{2}\pi }{3\sqrt{3}}$
My answer isn't corect , can't find out what is wrong.
Appreciate any help.
| $$J = \begin{bmatrix} x_r & x_t \\ y_r & y_t \end{bmatrix} = \begin{bmatrix} \frac1{\sqrt3}\cos(t) & -\frac r{\sqrt3} \sin(t) \\ \frac1{\sqrt6}\sin(t) & \frac r{\sqrt 6}\cos(t) \end{bmatrix}$$
$$\implies \det(J) = x_r y_t - x_t y_r = \frac r{\sqrt{18}} \cos^2(t) + \frac r{\sqrt{18}} \sin^2(t) = \frac r{\sqrt{18}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4610204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
If $A$ is any matrix and $n$ is an integer, what is $A^n$? If $A$ is any matrix and $n$ is an integer, what is $A^n$? What I know is $A_n$ is a matrix of order $n$. It seems that the superscript symbol (integer) is not common. What does it imply?
| The superscript symbol typically means the product of $A$ with itself $n$-many times. So $$A^3 = A AA$$ for instance, in the same way that $$5^3 = 5 \cdot 5 \cdot 5$$ If negative and $A$ is invertible, then it is the inverse being raised to $|n|$; for instance, if $A$ is invertible, then $A^{-4}$ means $$A^{-4} = \left( A^{-1} \right)\left( A^{-1} \right)\left( A^{-1} \right)\left( A^{-1} \right)$$
in the same way that
$$5^{-4} = \left( 5^{-1} \right)\left( 5^{-1} \right)\left( 5^{-1} \right)\left( 5^{-1} \right)$$
(though you might be more accustomed to just writing $1/5$ instead of $5^{-1}$).
To demonstrate with some explicit matrices, we have
$$\begin{bmatrix}
2 & 3 \\ 0 & 4 \end{bmatrix}^3
= \begin{bmatrix}
2 & 3 \\ 0 & 4 \end{bmatrix}\begin{bmatrix}
2 & 3 \\ 0 & 4 \end{bmatrix}\begin{bmatrix}
2 & 3 \\ 0 & 4 \end{bmatrix}$$
and
$$\begin{bmatrix}
2 & 3 \\ 0 & 4 \end{bmatrix}^{-4}
= \left( \begin{bmatrix}
2 & 3 \\ 0 & 4 \end{bmatrix}^{-1} \right) \left( \begin{bmatrix}
2 & 3 \\ 0 & 4 \end{bmatrix}^{-1} \right) \left( \begin{bmatrix}
2 & 3 \\ 0 & 4 \end{bmatrix}^{-1} \right) \left( \begin{bmatrix}
2 & 3 \\ 0 & 4 \end{bmatrix}^{-1} \right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4611189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Without calculator prove that $9^{\sqrt{2}} < \sqrt{2}^9$ Without calculator prove that $9^{\sqrt{2}} < \sqrt{2}^9$.
My effort: I tried using the fact $9^{\sqrt{2}}<9^{1.5}=27.$
Also We have $512 <729 \Rightarrow 2^9<27^2 \Rightarrow 2^{\frac{9}{2}}<27 \Rightarrow \sqrt{2}^9=2^{4.5}<27$. But both are below $27$.
| What you want is a suitable rational upper bound to $\sqrt{2} \approx 1.4142$ such as $\frac{17}{12} \approx 1.4167$ which you can find for example with continued fractions. As a check, $\left(\frac{17}{12}\right)^2=\frac{289}{144}>2$.
Then you can say $$9^{\sqrt{2}} \lt 9^{17/12}=3^{17/6} = 129140163^{1/6} < 134217728^{1/6} =2^{27/6} =2^{9/2} = \sqrt{2}^9$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4611390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 4
} |
Find $a>0$ if the floor of $(n^{2}-n)(\sqrt[n]{a}-1)$ is equal to $n-1$. Find $a>0$ knowing that for every n non-zero natural number, the floor of $(n^{2}-n)(\sqrt[n]{a}-1)$ is equal to $n-1$.
I know a is e, because taking the limit of the expresion we find that a is e. But i dont know how to prove that this works for every n. I am thniking at induction, but i dont know how i can use it. Can someone help me solve this problem in a logic form.
| We have that $$ \lfloor (n^2 - n)(a^{1/n} - 1)\rfloor = n-1$$
Since $x-1< \lfloor x\rfloor \le x,$ we can conclude that for every $n$, $$ n -1 \le (n^2-n)(a^{1/n} - 1) \le n \\ \iff \left(1 + \frac{1}{n}\right)^n \le a \le \left(1 + \frac{1}{n-1}\right)^n.$$ Now taking the limit shows that if $a$ exists then $a = e$.
So, the only possibility for $a$ is $e$, but we need to verify that this works. For $n = 1,$ this is obvious. For $n \ge 2,$ the following fact can be exploited: $$1 + \frac{1}{n} < e^{1/n} < 1 + \frac{1}{n-1}. $$ Note the strict inequalities. This does the job, since the relations imply
$$ (n^2 - n)(e^{1/n} - 1)< n \implies \lfloor (n^2 - n) (e^{1/n} - 1)\rfloor \le n-1, \\ (n^2 - n)(e^{1/n} - 1) > n-1 \implies \lfloor (n^2 - n) (e^{1/n} - 1)\rfloor \ge n-1,$$ forcing the floor in question to be $n-1$.
It remains to show the bounds I posited. The lower bound is trivial (truncate the Taylor series). For the upper bound, it suffices to show that for $n \ge 2,$ $$ \sum_{k \ge 2} \frac{1}{k!n^k} < \frac{1}{n(n-1)}$$
But notice that for $k \ge 4,$ $$ \frac{1}{k! n^k} \le \frac{1}{(2n)^k}.$$ Summing the geometric series, and using $n \ge 2,$ we have that $$ \sum \frac{1}{k! n^k} \le \frac{1}{2n^2} + \frac{1}{6n^3} + \frac{1}{2^4n^4} \cdot \frac{1}{1 - 1/2n}\\\le \frac{1}{2n^2} + \frac{1}{6n^2} + \frac{1}{2^4n^4} \cdot \frac{4}{3} \\ \le \frac{1}{2n^2} + \frac{1}{12n^2} + \frac{1}{48n^2 } <\frac{29}{48n^2} < \frac{5}{8n(n-1)},$$
and we're done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4613067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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