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$\triangle ABC$ is an equilateral triangle with circumcentre $O(1,2)$ and vertex $A$ lying on the line $5x+2y=4$.Find area of quadrilateral $BCDE$. $\triangle ABC$ is an equilateral triangle with circumcentre $O(1,2)$ and vertex $A$ lying on the line $5x+2y=4$. A circle with
centre $I(2,0)$ passes through the vertices $B$ and $C$ and intersects the sides $AC$ and $AB$ at $D$ and $E$ respectively. Find area of quadrilateral $BCDE$.
My Attempt
All I could do here was finding the perpendicular distance of $O$ from the given line. How do I use $I(2,0)$
|
As can be seen in figure OI is coincident on altitude of AH. The equation of AH is:
$y-2=\frac{-2}{2-1}(x-1)\Rightarrow y=-2x+4$
This with line $2y+6x=4$ gives $A(-4, 12)$ and we hav:
$AO=\sqrt {(1+5)^2+(12-2)^2}\approx 11.2$
$OH=\frac{AO}2=5.6$
$AH=\frac{3\times AO}2=16.8$
$AC=\frac{AH}{\sin 60}\approx 19.4$
$A_{ABC}=\frac{19.4\times 16.8}2=163$
$OI=\sqrt{1^2+5^2}=\sqrt 5\approx 2.24$
Now :
$IH=OH-OI=5.6-2.24=3.36$
$R=IC=\sqrt {3.36^2+(\frac{19.4}2)^2}\approx 10$
Equation of circle is:
$(x-2)^2+y^2=10^2$
This with line AH , $y=-2x+4$ gives:
$p(1-\sqrt 5, 2+2\sqrt 5$ and $Q(1+\sqrt 5, 2-2\sqrt 5)$
$AP=\sqrt{(-4-1+\sqrt 5)^2+(12-2-2\sqrt 5}\approx 6.2$
$ED=\frac{6.2}{16.8}\times 19.4\approx 7\Rightarrow A_{AED}=\frac 12(7\times 6.2)\approx 22$
$A_{BCDE}\approx 163-22=141$
.
| {
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Interesting Inequality using AM-GM and other identities. Let $a, b, c > 0$. Prove that $$\sqrt{a^2-ab+b^2} + \sqrt{b^2 - bc + c^2} + \sqrt{c^2 - ca + a^2} \le \frac{ab}{c} + \frac{bc}{a} + \frac{ca}{b}.$$ This should be solvable with AM-GM and a few other inequalities, but I am a little stuck on this problem.
My idea was to remove the radical. $\sqrt{a^2-ab+b^2} \le \frac{a^2-ab+b^2}{a+b} + \frac{a+b}{4}$ by AM-GM. Adding this up cyclically, it suffices to show the inequality
$$\frac{5}{2}\left(a+b+c\right)-3\left(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ac}{a+c}\right) \le \frac{ab}{c} + \frac{bc}{a} + \frac{ca}{b},$$
which I'm pretty sure is true, but I have no clue how to prove.
This inequality resembles https://artofproblemsolving.com/community/c6h1288310p6804993 and https://artofproblemsolving.com/community/q2h1817483p12130020, the latter of which is a weaker version of this inequality.
| WLOG, assume $c = \max(a, b, c)$.
By AM-GM inequality, we have
\begin{align*}
\left(\frac{b^2}{c} - b + c\right) + c
&\ge 2\sqrt{b^2 - bc + c^2}, \\
\left(\frac{a^2}{c} - a + c\right) + c
&\ge 2\sqrt{a^2 - ac + c^2}, \\
\left(\frac{b^2}{a} - b + a\right) + a
&\ge 2\sqrt{b^2 - ab + a^2}.
\end{align*}
It suffices to prove that
$$\frac{b^2}{a} + \frac{a^2 + b^2}{c}
+ a - 2b + 4c \le 2\left(\frac{ab}{c} + \frac{bc}{a} + \frac{ca}{b}\right)$$
that is
$$\frac{(2c^2 - ab - bc)(a - b)^2}{abc} \ge 0$$
which is true.
We are done.
| {
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Number of integer solutions of $a^2+b^2=10c^2$ Find the number of integer solutions of the equation $a^2+b^2=10c^2$.
I can only get by inspection that $a=3m, b=m,c=m$ satisfies for any $m \in Z$.
Is there a formal logic to find all possible solutions? Any hint?
Also i tried taking $a=p^2-q^2$, $b=2pq$ and $10c^2=(p^2+q^2)^2$
which gives $$\frac{p^2+q^2}{c}=\sqrt{10}$$ which is invalid, since a rational can never be an irrational.
| Solving the $C$-function of Euclid's formula
$\quad A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2\quad$ for $(k), \space $ we can find Pythagorean triples for any given $C$-values, if they exist, that are primitive, doubles, or square multiples of primitives. This will not find, for example $(9,12,15)\space$ or $(15,20,25),$ but it will find $(3,4,5),\space (6,8,10),\space (12,16,20),\space (27,36,45), \space$ etc. We begin with the following formula. Any
$m$-value that yields an integer $k$-value indicates a valid $(m,k)$ pair for generating a Pythagorean triple.
\begin{equation}
C=m^2+k^2\implies k=\sqrt{C-m^2}\\
\text{for}\qquad \bigg\lfloor\frac{ 1+\sqrt{2C-1}}{2}\bigg\rfloor \le m \le \lfloor\sqrt{C-1}\rfloor
\end{equation}
The lower limit ensures $m>k$ and the upper limit ensures $k\in\mathbb{N}.$
Here is an example for $C=40\implies 10c=4$ where $c$ is the one shown in the OP equation.
$$C=40\implies \bigg\lfloor\frac{ 1+\sqrt{80-1}}{2}\bigg\rfloor=4 \le m \le \lfloor\sqrt{40-1}\rfloor=6\\
\land \quad m\in\{6\}\Rightarrow k\in\{2\}\\$$
$$F(6,2)=(32,24,40)\implies (32,24,10\times 4)$$
This method will not find all Pythagorean triples that match the criteria but it will find an infinite number of triples that do such as:
$$c=1\longrightarrow (8,6,10\times 1)\\
c=2\longrightarrow (12,16,10\times 2)\\
c=4\longrightarrow (32,24,10\times 4)\\
c=9\longrightarrow (72,54,10\times 9)\\
$$
Note that any multiple of a triple found also yields a valid triple so
$3\times (8,6,10)\longrightarrow (24,18,10\times3)$ and provides the "missing" $c=3$ triple in the list above. The combination of the two will find all Pythagorean triples where the $c$ in $10c$ is an integer except for the most unusual case like $(3,1,10\times 1)$ mentioned in another post.
| {
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Computing $\int_{-2}^{2}\frac{1+x^2}{1+2^x} dx$ I am trying to compute the following integral by different methods, but I have not been able to come up with the result analytically.
$$\int_{-2}^{2}\frac{1+x^2}{1+2^x}dx$$
First I tried something like: $2^{x}=e^{x\ln{2}}\Rightarrow u=x\ln{2} \iff x=\frac{u}{\ln{2}}$ $\Rightarrow$ $\frac{du}{\ln{2}}=dx$.
On the other hand, $1+x^{2}=(x-1)^{2}-2x$
Replacing
$$\int_{-2}^{2}\frac{1+x^2}{1+2^x}dx=\int_{-2}^{2}\frac{(x-1)^{2}-2x}{1+e^{x\ln{2}}}dx=\int_{-2}^{2}\frac{(x-1)^{2}-2x}{1-(-e^{x\ln{2}})}dx$$
$$\int_{-2}^{2}\frac{1+x^2}{1+2^x}dx=\int_{-2}^{2}((x-1)^{2}-2x)\sum_{n=0}^{\infty}(-e^{x\ln{2}})^{n}dx$$
$$\int_{-2}^{2}\frac{1+x^2}{1+2^x}dx=\int_{-2}^{2}((x-1)^{2}-2x)\sum_{n=0}^{\infty}((-1)^{n}e^{nx\ln{2}})dx=\int_{-2}^{2}((x-1)^{2}-2x)\sum_{n=0}^{\infty}\frac{((-1)^{n}n^{n}x^{n}\ln^{n}{2})}{n!}dx$$
$$\int_{-2}^{2}\frac{1+x^2}{1+2^x}dx=\sum_{n=0}^{\infty}\frac{((-1)^{n}n^{n}\ln^{n}{2})}{n!}\int_{-2}^{2}(1+x^2)x^{n}dx$$
I do not know if the reasoning is correct. I hope someone can help me.
Note: By symmetry the integral can be reduced to $f(-x)=2^{x}f(x)$ so $2I=I+I=\int_{-2}^{2}(1+2^{x})\frac{1+x^2}{1+2^x}dx=\frac{14}{3}$
| My approach:
$$\int_{-2}^{2}\frac{1+x^2}{1+2^x} dx = \int_{-2}^{0}\frac{1+x^2}{1+2^x}dx + \int_{0}^{2}\frac{1+x^2}{1+2^x} dx $$$$ \overset{t = -x}= \int_{0}^{2}\frac{1+t^2}{1+2^{-t}} dt+\int_{0}^{2}\frac{1+x^2}{1+2^x}dx = \int_{0}^{2}x^2 + 1 dx =\frac{14}{3}. $$
| {
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How do I calculate the sum of sum of triangular numbers? As we know, triangular numbers are a sequence defined by $\frac{n(n+1)}{2}$. And it's first few terms are $1,3,6,10,15...$. Now I want to calculate the sum of the sum of triangular numbers. Let's define
$$a_n=\frac{n(n+1)}{2}$$
$$b_n=\sum_{x=1}^na_x$$
$$c_n=\sum_{x=1}^nb_x$$
And I want an explicit formula for $c_n$. After some research, I found the explicit formula for $b_n=\frac{n(n+1)(n+2)}{6}$. Seeing the patterns from $a_n$ and $b_n$, I figured the explicit formula for $c_n$ would be $\frac{n(n+1)(n+2)(n+3)}{24}$ or $\frac{n(n+1)(n+2)(n+3)}{12}$.
Then I tried to plug in those two potential equations,
If $n=1$, $c_n=1$, $\frac{n(n+1)(n+2)(n+3)}{24}=1$, $\frac{n(n+1)(n+2)(n+3)}{12}=2$. Thus we can know for sure that the second equation is wrong.
If $n=2$, $c_n=1+4=5$, $\frac{n(n+1)(n+2)(n+3)}{24}=5$. Seems correct so far.
If $n=3$, $c_n=1+4+10=15$, $\frac{n(n+1)(n+2)(n+3)}{24}=\frac{360}{24}=15$.
Overall, from the terms that I tried, the formula above seems to have worked. However, I cannot prove, or explain, why that is. Can someone prove (or disprove) my result above?
| One approach is to calculate $5$ terms of $c_n$, recognize that it's going to be a degree-4 formula, and then solve for the coefficients. Thus:
$$c_1 = T_1=1 \\ c_2 = c_1 + (T_1+T_2) = 5 \\ c_3 = c_2+(T_1+T_2+T_3) = 15 \\ c_4 = c_3 + (T_1+T_2+T_3+T_4) = 35 \\ c_5 = c_4 + (T_1+T_2+T_3+T_4+T_5) = 70$$ Now we can find coefficients $A,B,C,D,E$ so that $An^4+Bn^3+Cn^2+Dn+E$ gives us those results when $n=1,2,3,4,5$. This leads to a linear system in 5 unknowns, which we can solve and obtain $A=\frac1{24},B=\frac14,C=\frac{11}{24},D=\frac14,E=0$. Thus taking a common denominator, we have $$c_n=\frac{n^4+6n^3+11n^2+6n}{24}=\frac{n(n+1)(n+2)(n+3)}{24}$$
So that agrees with your result.
Another way is to use the famous formulas for sums of powers. Thus, we find $b_n$ first: $$b_n = \sum_{i=1}^n \frac{i(i+1)}{2} = \frac12\left(\sum i^2 + \sum i\right) = \frac12\left(\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}\right)\\ =\frac{n^3+3n^2+2n}{6}$$
Now, we find $c_n$: $$c_n = \sum_{i=1}^n \frac{i^3+3i^2+2i}{6}=\frac16\sum i^3 + \frac12\sum i^2 + \frac13\sum i \\ = \frac16\frac{n^2(n+1)^2}{4} + \frac12\frac{n(n+1)(2n+1)}{6} + \frac13\frac{n(n+1)}{2} \\ = \frac{n^4+6n^3+11n^2+6n}{24}=\frac{n(n+1)(n+2)(n+3)}{24}$$
So we have confirmed the answer 2 different ways. As is clear from the other solutions given here, there are other ways as well.
| {
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Prove by induction that $3 \mid n^4-n^2 \forall n \in \mathbb{Z^+}, n \ge 2$. Proposition: $3 \mid n^4-n^2$ for all $n \in \mathbb{Z^+}, n \ge 2$
My attempt
Lemma: $3 \mid (m-1)m(m+1)$
Proof.
Suppose $m \in \mathbb{Z}$. By the QRT, we have $m=3q+r$ $\,\ni\, r \in \{0,1,2\}$, and $q=\lfloor{\frac{m}{3}}\rfloor$.
We want to show that $(m-1)m(m+1)$ is divisible by 3.
Clearly, if any one of the three factors in $(m-1)m(m+1)$ is divisible by 3, then the product must also be divisible by 3.
$\cdot$ Case 1 ($r=0$): If $r=0$, then $m=3q$ is divisible by 3.
$\cdot$ Case 2 ($r=1$): If $r=1$, then $m-1=(3q+1)-1=3q$ is divisible by 3.
$\cdot$ Case 3 ($r=2$): If $r=2$, then $m+1=(3q+2)+1=(3q+3)=3(q+1)$ is divisible by 3.
In either case, one of the three factors in $(m-1)m(m+1)$ is divisible by 3. Thus $3 \mid (m-1)m(m+1)$. Therefore, the product of any 3 consecutive integers is divisible by 3.
Proof.
Base case: Let $n=2$. Then $\exists k \in \mathbb{Z}$ such that $12=3k$, namely, $k=4$. Thus $3 \mid 12$ and hence the claim holds for the base case.
Assume for positive integer $m \ge 2$ that $m^4-m^2=3k$ for some $k \in \mathbb{Z}$. We need only to show that this implies that $3 \mid (m+1)^4-(m+1)^2$.
$(m+1)^4-(m+1)^2= \sum_{i=0}^{4} {4\choose i} m^i \cdot 1^{4-i}-(m+1)^2=(m^4-m^2)+4m^3+6m^2+2m=(m^4-m^2)+3(m^3+2m^2+m)+(m^3-m)$.
By the inductive hypothesis, $3 \mid(m^4-m^2)$. And since $(m^3+2m^2+m) \in \mathbb{Z}$, it follows that $3 \mid 3(m^3+2m^2+m)$. Note that $(m^3-m)=m(m^2-1)=(m-1)m(m+1)$. By the Lemma, $3 \mid (m^3-m)$.
Thus we have $3 \mid (m^4-m^2)$, $3 \mid 3(m^3+2m^2+m)$, and $3 \mid (m^3-m)$.
To show that their sum is divisible by 3, let $k_1,k_2,k_3 \in \mathbb{Z} \,\ni\, m^4-m^2=3k_1, 3(m^3+2m^2+m)=3k_2$, and $(m^3-m)=3k_3$. Thus $3k_1+3k_2+3k_2=3(k_1+k_2+k_3)=m^4-m^2+3(m^3+2m^2+3m)+(m^3-m)$, where $(k_1+k_2+k_3) \in \mathbb{Z}$.
Hence $3 \mid (m+1)^4-(m+1)^2$.
Therefore, by induction, $3 \mid n^4-n^2, \forall n \in \mathbb{Z^+}, n \ge 2$.
| You don't need induction, write $$n^4-n^2=n^2(n-1)(n+1)$$ a one line proof.
| {
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Find the largest number that $ n(n^2-1)(5n+2) $ is always divisible by? My Solution:
$$ n(n^2-1)(5n+2) = (n-1)n(n+1)(5n+2) $$
*
*This number is divisible by 6 (as at least one of 2 consecutive integers is divisible by 2 and one of 3 consecutive integers is divisible by 3.
*$ 5n+2 \equiv 5n \equiv n \mod 2 $ then $n$ and $5n+2$ have the same pairness and at least one of $n+1$ and $5n+2$ is divisible by 2.
*$ n \equiv 5n \equiv 5n+4 \mod 4 \to $
if $ 2\ | \ n+1 \to n - 1 $ or $ n + 1 $ is divisible by 4
if $ 2\ | \ 5n+2 \to n $ or $ 5n + 2 $ is divisible by 4
The expression is divisible by 6 and has 2 even integers and one of them is divisible by 4 $\to$ is divible by 24.
| Your proof seems correct to all of us, as it appears in the comments.
I would consider applying a method like this:
$$\begin{align}f(n)&=n(n^2-1)(5n+2)
\\&=n(n^2-1)(4n+n+2)\\
&=\underbrace{4n^2(n-1)(n+1)}_{\equiv ~0~(\text{mod}~~ 48)}
\\
&+\underbrace{(n-1)n(n+1)(n+2)}_{\equiv ~0~(\text{mod}~ 24)}\end{align}$$
If $n=3$, then $5n+2$ is prime and if the largest number to which the function is always divided was greater than $24$, the next factor must be $17$. But, $f(2)$ is not divisible by $17.$ Therefore, the largest number should only be $24$.
Explanations:
*
*$24|(n-1)n(n+1)(n+2)$
Because, the product of $4$ consecutive positive integers are always divisible by $24$.
Applying $$n=8k±m, ~0≤m≤4, m\in\mathbb Z$$ shows that, $8|(n-1)n(n+1)(n+2)$ and we already know that, $6|(n-1)n(n+1)(n+2)$. This means $24|(n-1)n(n+1)(n+2)$.
*
*$48|4n^2(n-1)(n+1)$
Because, $48|4n^2(n-1)(n+1)=12|(n-1)n^2(n+1)$
Observing at the cases where $n$ is odd or even completes the proof.
| {
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What is the minimum value of $8 \cos^2 x + 18 \sec^2 x$? As per me the answer should be $26$.
But when we apply AM-GM inequality it gives $24$ as the least value but as per the graph 24 can never come.
What I think is that in AM-GM, it gives $8 \cos^2 x = 18 \sec^2 x$ which gives $\cos x > 1$ which is not possible and because of this, AM-GM is giving a wrong minimum value.
If we had $18 \cos^2 x + 8 \sec^2 x$, then AM-GM would have worked and $24$ would be a right answer since $18 \cos^2 x = 8 \sec^2 x$, which gives $\cos x < 1$ which is true.
Is this reason correct?
| No need for AM-GM.
Differentiate wrt $x$ and set $f'(x)=0$ as follows:
$$\implies -16\sin x\cdot \cos x + 36\sec^2 x\cdot \tan x=0$$
$$4\cos x\sin x=\frac{9\sin x}{\cos^3 x}$$
If $\sin x$ is non-zero, then:
$$\cos^4 x=\frac{9}{4}\implies \cos x=±\sqrt{\frac{3}{2}}>1\implies \text{no solution}$$
Hence $\sin x=0\implies \cos x=±1$ and hence $\cos^2 x=1$.
Substitute this into your original equation and you get $8+18=26.$
NOTE: You can confirm that this is a minima by evaluating $f''(x).$
| {
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Let $f\in\mathbb{Q}[X]$ such that $f(1)=-2, f(2)=1$ and $f(-1)=0$. Find the remainder of the division of $f$ by $X^3-2X^2-X+2$. Let $f\in\mathbb{Q}[X]$ such that $f(1)=-2, f(2)=1$ and $f(-1)=0$. Find the remainder of the division of $f$ by $X^3-2X^2-X+2$.
So, I figured: $f=(X+1)q$. Assumming that $f$ has degree 3, I solve $\begin{cases} (2+1)(2-a)(2-b)=1 \\ (1+1)(1-a)(1-b)=-2\end{cases}$ to find that $\begin{cases} a=\frac{5}{6}-\frac{\sqrt{37}}{6}, b=\frac{5}{6}+\frac{\sqrt{37}}{6} \\ a=\frac{5}{6}+\frac{\sqrt{37}}{6}, b=\frac{5}{6}-\frac{\sqrt{37}}{6} \end{cases}$. I divide $X^3-2X^2-X+2$ by $(X+1)(X-\frac{5}{6}-\frac{\sqrt{37}}{6})(X-\frac{5}{6}+\frac{\sqrt{37}}{6})$. The remainder is $\frac{4}{3}(X+1)(X-\frac{7}{4})$. Is it correct to assume that the remainder is always this, no matter the degree of $f$? Since that's what the problem asks for, I'm lead to assume this.
| You can not assume that $f$ has degree $3$. Also the remainder that you have calculated is not correct, the actual remainder is $\frac{1}{3} \, {\left(4 \, x - 7\right)} {\left(x + 1\right)}$.
For example take $f(x) =
\frac{3}{4} \, x^{4} + \frac{1}{6} \, x^{3} - \frac{11}{4} \, x^{2} - \frac{7}{6} \, x + 1$.Observe that $f$ satisfies all the requirements given in the question. So when you divide (perhaps using long division or some other methods) $f$ by $x^3 - 2x^2 - x + 2$
you get the same remainder.
A big hint:
Let $g(x) = x^3 - 2x^2 - x + 2$ and let $r(x)$ be the remainder when you divide $f$ by $g$. Then there exist a $q(x)$ such that $f(x) = g(x)q(x) + r(x)$. Observe that $r(x)$ is atmost quadratic polynomial. Also see that $g(1) = g(2) = g(-1) = 0$. So $f(1) = r(1) = -2$, $f(2) = r(2) = 1$ and $f(-1) = r(-1) = 0$. Can you find $r(x)$ satisfieng these conditions?
| {
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Given a general region, find the double integral bounded between $y = x$ and $y=3x-x^2 $ $$ J = \iint_R (x^2-xy)\,dx \,dy, $$
Suppose region R is bounded between $y = x$ and $y=3x-x^2 $
My attempt using vertical integration:
$$ \int^{x=2}_{x=0} \int^{y=3x-x^2}_{y=x} \left({x^2-xy}\right)dy\ dx$$
$$\int^2_0 \left[x^2y-x\frac{y^2}{2}\right]^{3x-x^2}_{x}\, dx$$
$$\int^2_0 \frac{-x^5+4x^4-4x^3}{2} \,dx $$
$$\boxed{J = -\frac{8}{15}}$$
My attempt using horizontal integration :
$$ \int^{y=2}_{y=0} \int^{x=y}_{x=3\,\pm \sqrt{9-y}} \left({x^2-xy}\right)dx\ dy$$
For $ x = 3+\sqrt{9-y}$
$$ \int^2_0 \left[\frac{x^3}{3}-\frac{x^2}{2}y\right]^y_{3+\sqrt{9-y} }\,dy$$
For $ x = 3-\sqrt{9-y}$
$$ \int^2_0 \left[\frac{x^3}{3}-\frac{x^2}{2}y\right]^y_{3-\sqrt{9-y} }\,dy$$
My doubts :
1.) How do I set my limit of integration for horizontal integration, if there is $\pm$ to be considered ?
2.) the answer as negative what does that imply in questions related to double integrals?
Could you guys please help
| Your first approach is correct, but your second approach is not. The possible values of $y$ lie in the interval $\left[0,\frac94\right]$, because $\frac94$ is the maximum of $3x-x^2$ when $x\in[0,2]$. When $y\in\left[0,\frac94\right]$, then (see the picture below):
*
*if $y\in[0,2]$, then the possible values of $x$ lie in $\left[\frac{3-\sqrt{9-4y}}2,y\right]$;
*if $y\in\left[2,\frac94\right]$, then the possible values of $x$ lie in $\left[\frac{3-\sqrt{9-4y}}2,\frac{3+\sqrt{9-4y}}2\right]$.
So, the answer is\begin{multline}\int_0^2\int_{\frac{3-\sqrt{9-4y}}2}^yx^2-xy\,\mathrm dy\,\mathrm dx+\int_2^{9/4}\int_{\frac{3-\sqrt{9-4y}}2}^{\frac{3+\sqrt{9-4y}}2}x^2-xy\,\mathrm dy\,\mathrm dx=\\=\int_0^2\frac{1}{12} \left(-2 y^3-6 y^2+\left(45-11 \sqrt{9-4 y}\right) y+18 \left(\sqrt{9-4 y}-3\right)\right)\,\mathrm dy+\\+\int_2^{9/4}\frac{1}{6}(18-11 y)\sqrt{9-4 y}\,\mathrm dy=-\frac{47}{120}-\frac{17}{120}=-\frac8{15}.\end{multline}
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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How to evaluate $\lim\limits_{x\to \infty}\frac {2x^4-3x^2+1}{6x^4+x^3-3x}$?
Evaluate: $\lim\limits_{x\to \infty}\frac {2x^4-3x^2+1}{6x^4+x^3-3x}$.
I've just started learning limits and calculus and this is an exercise problem from my textbook.
To solve the problem, I tried factorizing the numerator and denominator of the fraction. The numerator can be factorized as $(x-1)(x+1)(2x^2-1)$ and the numerator can be factorized as $x(6x^3+x^2-3x)$. So, we can rewrite the problem as follows: $$\lim\limits_{x\to\infty}\frac {2x^4-3x^2+1}{6x^4+x^3-3x}=\lim\limits_{x\to\infty}\frac {(x-1)(x+1)(2x^2-1)}{x(6x^3+x^2-3)}$$ But this doesn't help as there is no common factor in the numerator and denominator. I've also tried the following: $$\lim\limits_{x\to\infty}\frac {(x-1)(x+1)(2x^2-1)}{x(6x^3+x^2-3)}=\lim\limits_{x\to\infty}\frac{x-1}{x}\cdot \lim\limits_{x\to\infty}\frac {(x+1)(2x^2-1)}{6x^3+x^2-3}=1\cdot \lim\limits_{x\to\infty}\frac {(x+1)(2x^2-1)}{6x^3+x^2-3}$$ Here I used $\frac{x-1} x=1$ as $x$ approaches infinity. Yet this does not help.
The answer is $\frac 1 3$ in the book but the book does not include the solution.
So, how to solve the problem?
| Divide the numerator and denominator of the expression by $x^4$:$$\lim\limits_{x\to \infty}\frac {2-3/x^2+1/x^4}{6+1/x-3/x^3}$$and use the fact that $\lim\limits_{x\to\infty}\frac{1}{x^n} = 0$ for any positive $n$
| {
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For what primes $p$ and positive integers $k$ is this algebraic expression divisible by $3$? My initial question is as is in the title:
For what prime $p$ and positive integers $k$ is this algebraic expression divisible by $3$?
$$A(p,k):=\dfrac{p^{2k+2} - 4p^{2k+1} + 6p^{2k} + 2p^{k+1} - 8p^k + 3}{2(p - 1)^2}$$
I would like to qualify that I am specifically interested in those values for $p$ and $k$ satisfying the congruence $p \equiv k \equiv 1 \pmod 4$.
MY ATTEMPT
Here, I will evaluate my expression for
$$p_1 = 5, k_1 = 1,$$
which gives $A(p_1,k_1)=9$ (which is divisible by $3$),
$$p_2 = 13, k_2 = 1,$$
which gives $A(p_2,k_2)=73$ (which is not divisible by $3$), and
$$p_3 = 17, k_3 = 1,$$
which gives $A(p_3,k_3)=129$ (which is divisible by $3$).
Here is my final question:
Will it be possible to consider the cases $p \equiv 1 \pmod 3$ and $p \equiv 2 \pmod 3$ separately, then use the Chinese Remainder Theorem afterwards? (I know the concept, but I would have forgotten how to do that.)
CONJECTURE: If $p \equiv 2 \pmod 3$, then $3 \mid A(p,k)$.
Alas, this is where I get stuck.
| Consider the point that the sum of coefficiant of numerator is 0, so for $p\equiv 1\bmod3$ the numerator is divisible by 3 for any k. For the case $p\equiv 2\bmod 3$ we have:
$p\equiv 2 \bmod 3\Rightarrow p^2\equiv 1 \bmod 3\Rightarrow p^{2(k+1)}\equiv 1 \bmod 3$
$p^{2k+1}=p^{2k}\cdot p\equiv(1\bmod 3)(2\bmod 3)\equiv 2\bmod 3\Rightarrow 4p^{2k+1}\equiv 8 \bmod 3\equiv 2 \bmod 3$
$6p^{2k}\equiv 0 \bmod 3$
$p^{k+1}p^k\cdot p\equiv (2^k \bmod 3)(2\bmod 3)\Rightarrow 2p^{k+1}\equiv 2^{k+2} \bmod 3$
$8p^k\equiv 2^{k+3} \bmod 3$
Summing these we get for numerator:
$r=1-2+0+2^{k+2}-2^{k+3}-3\equiv 2^{k+2}(1-2)-1\equiv -1-2^{k+2}\bmod 3$
$2^{k+1}=(3-1)^{k+1}$
So if $k=2m+1$ we have:
$r=-1 -3t+1=-3t\equiv 0 \bmod 3$
| {
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How to make use of angle sum and difference identities to find the value of sine and cosine?
Calculate: $\cos\left({5\pi\over12}\right)$ and $\cos\left({\pi\over12}\right)$
What is the easiest way to find $\cos\left({5\pi\over12}\right)$ and $\cos\left({\pi\over12}\right)$ (without a calculator)? If I know that $\frac{5\pi }{12}=\frac{\pi }{4}+\frac{\pi }{6}$ and $\frac{\pi }{12}=\frac{\pi }{3}-\frac{\pi }{4}$, then I can apply angle sum and difference identities. But how do I know $\frac{5\pi }{12}= \frac{\pi }{4}+\frac{\pi }{6}$ and $\frac{\pi }{12}= \frac{\pi }{3}-\frac{\pi }{4}$ in the first place. I know $ \frac{\pi }{4}+\frac{\pi }{6} = \frac{5\pi }{12}$ and $ \frac{\pi }{3}-\frac{\pi }{4}=\frac{\pi }{12}$ but I can't go the other way round.
I gave $\frac{5\pi}{12}$ and $\frac{\pi}{12}$ as an example, I want the general solution for any value in pi rational form $\frac{\pi p}{q}$.
| We know that
$$\begin{align}
\cos{\pi\over 4}&=\sin{\pi\over 4}={\sqrt{2}\over 2}\\
\cos{\pi\over 6}&=\sin{\pi\over 3}={\sqrt{3}\over 2}\\
\cos{\pi\over 3}&=\sin{\pi\over 6}={1\over 2}\\
\cos(x+y)&=\cos{x}\cos{y}-\sin{x}\sin{y}\\
\cos(x-y)&=\cos{x}\cos{y}+\sin{x}\sin{y}\\
\sin(x+y)&=\sin{x}\cos{y}+\cos{x}\sin{y}\\
\cos(x-y)&=\sin{x}\cos{y}-\cos{x}\sin{y}\\
\end{align}$$
With the above you should be done.
| {
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Find all primes $(p,q)$ satisfying the condition that $pq$ divides $2^p+2^q.$
Find all primes $(p,q)$ satisfying the condition that $pq$ divides $2^p+2^q.$
I get answers for $p =2$ or $q = 2$ but I want to know the general approach for this question.
| Assume both $p$ and $q$ are primes at least 3. Let $a_p$ denote the smallest integer satisfying $2^{a_p} \equiv_p -1$ and let $a_q$ denote the smallest integer satisfying $2^{a_q} \equiv_q -1$. [There indeed exists such an $a_p,a_q$ if there exists a $p,q$ satisfying the equation $pq|(2^p+2^q)$; indeed [assuming $q \ge p$]
$$pq|(2^p+2^q) \quad \Rightarrow \quad pq|(1+2^{q-p})$$ $$\Rightarrow \quad 2^{q-p} \equiv_p -1; \ 2^{q-p} \equiv_q -1.$$ Thus given $2^{q-p} \equiv_p -1$ and $2^{q-p} \equiv_q -1$ there indeed exists such an $a_p,a_q$ as claimed.]
Also note that $2^{p} \equiv_p 2$ and $2^{q} \equiv_q 2$. Then $$pq|(2^p+2^q); \ 2^p \equiv_p 2; \ 2^q \equiv_q 2 \quad \Rightarrow \quad 2^{q} \equiv_p -2; \ 2^p \equiv_q -2$$
$$\Rightarrow \quad 2^{q-1} \equiv_p -1; \ 2^{p-1} \equiv_q -1.$$ As both $$2^{q-1} \equiv_p -1$$ and $$2^{q-1} \equiv_q 1$$ hold then, it follows that $q-1 = \ell a_q$ for some even integer $\ell$ and $q-1 = ka_p$ for some odd integer $k$. Thus what follows is the strict inequality $$\nu_2(a_p) > \nu_2(a_q),$$ where $\nu_2(M)$ for each integer $M$ is defined as the largest power of $2$ dividing $M$.
Likewise, $$2^{p-1} \equiv_q -1.$$ And, $$2^{p-1} \equiv_p 1.$$ So it follows that $p-1 = b a_p$ for some even integer $b$ and $p-1 = ca_q$ for some odd integer $c$. Thus what also follows is the strict inequality: $$\nu_2(a_p) < \nu_2(a_q).$$
However, the strict inequalities $\nu_2(a_p) > \nu_2(a_q)$ and $\nu_2(a_p)<\nu_2(a_q)$ contradict each other, so indeed, there are no two odd primes $p,q \ge 3$ satisfying $pq|(2^p+2^q)$. Thus the result follows at least for any two primes $p,q \ge 3.$
If $p=2$ then $$2^p+2^q \equiv_q 4(1+2^{q-2}) \equiv_q 4(1+2^{-1}),$$ where $2^{-1}$ is defined to be the multiplicative inverse of $2$ in $\mathbb{Z}/q\mathbb{Z}$. [Indeed as $2^{q-1} \equiv_q 1$ it follows that $2^{q-2} \equiv_q 2^{q-1}2^{-1} \equiv_q 1$]. So for $q$ to divide $2^{p}+2^{q}$ for $p=2$ it follows that either the equation $4\equiv_q 0$ or the equation
$1+2^{-1} \equiv_q 0$ holds. The former equation gives $q=2$. The latter equation gives $q=3$. [Indeed, $2^{-1} \equiv_q -1$ which gives $2 \times 2^{-1} \equiv_q 2 \times -1$ which gives $1 \equiv_q -2$ which indeed gives $q=3$.] So if $p$ is $2$ then $q$ must be $2$ or $3$. However, note that $p=2,q=2$ and $p=2,q=3$ are both solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving an inequality given $\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}\le 1$ Given that $a,b,c > 0$ are real numbers such that $$\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}\le 1,$$ prove that $$\frac{1}{b+c+1}+\frac{1}{c+a+1}+\frac{1}{a+b+1}\ge 1.$$
I first rewrote $$\frac{1}{a+b+1} = 1 - \frac{a+b}{a+b+1},$$ so the second inequality can be rewritten as $$\frac{b+c}{b+c+1} + \frac{c+a}{c+a+1} + \frac{a+b}{a+b+1} \le 2.$$ Cauchy-Schwarz gives us $$\sum \frac{a+b}{a+b+1} \geq \frac{(\sum \sqrt{a+b})^2}{\sum a+ b+ 1}.$$ That can be rewritten as $$\frac{2(a+b+c) + 2\sum \sqrt{(a+b)(a+c)}}{2(a+b+c) + 3},$$ which is greater than or equal to $$\frac{2(a+b+c) + 2 \sum(a + \sqrt{bc})}{2(a+b+c) + 3} = \frac{4(a+b+c) + 2 \sum \sqrt{bc}}{2(a+b+c) + 3} \geq 2,$$ which is the opposite of what I want. Additionally, I'm unsure of how to proceed from here.
| Applying Jensens to $ f(x) = \frac{ x} { (a+b+c+1) - x } $, we have $$ 1\geq \sum \frac{a}{b+c+1} = f(a) + f(b) + f(c) \geq 3 f ( \frac{a+b+c } { 3} ) = 3 \times \frac{ a + b + c } { 2a + 2b + 2c + 3 } \Rightarrow a + b + c \leq 3.$$
Applying Jensens to $ g(x) = \frac{ 1 } { (a+b+c+1) - x }$, we have
$$ \sum \frac{1}{ b+c+1} = g(a) + g(b) + g(c) \geq 3 g( \frac{ a+b+c} { 3 } ) = 3 \times \frac{ 3 } { 2a+2b+2c + 3 } \geq 1. $$
| {
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Laplace transform of $t^2 \cos \omega t$ I have to find the Laplace transform of: $$f(t) = t^2 \cos \omega t$$
We know that,
$$\mathcal{L} (t^n f(t)) = (-1)^n \frac{d^n}{ds^n}{F(s)}$$
so , $$\mathcal{L}(\cos \omega t) = \frac{s}{s^2 + \omega^2}$$
$$\therefore \mathcal{L} (t^2 \cos \omega t) = \frac{d^2}{ds^2} \frac{s}{s^2 + \omega^2}$$
let $$\frac{d^2}{ds^2} = F^{\prime\prime}(s)$$
so,
Using quotient rule of differentiation
$$F^\prime(s) = \frac{\omega^2 -s^2}{(s^2 + \omega^2)^2}$$
however, when I try to differentiate $F^\prime(s)$ again to find $F^{\prime\prime}(s)$ it is an entire mess. Can someone help me to differentiate it or is there any other way to find its Laplace Transformation.
The answer is:
$$\frac{2s(s^2 - 3 \omega^2)}{(s^2 + \omega^2)^3}$$
| OP, your process is correct and the only thing left is to find $F^{\prime\prime}(s)$.
$$F^\prime(s) = \frac{\omega^2 -s^2}{(s^2 + \omega^2)^2}$$
$$F^{\prime\prime}(s) = \frac{d}{ds} \frac{\omega^2 -s^2}{(s^2 + \omega^2)^2}$$
Quotient Rule:
$$\frac{d}{ds} \frac{f(s)}{g(s)} = \frac{f'(s)g(s)-g'(s)f(s)}{g(s)^2}$$
So,$$\frac{d}{ds} \left(\frac{\omega^2 -s^2}{(s^2 + \omega^2)^2}\right) = \frac{(\frac{d}{ds}(\omega^2 -s^2))\ (s^2 + \omega^2)^2\ \ -\ \ (\frac{d}{ds}(s^2 + \omega^2)^2)\ (\omega^2 -s^2)}{((s^2 + \omega^2)^2)^2}$$
$$=\ \frac{(-2s)(s^2+\omega^2)^2\ -\ (2(s^2+\omega^2)\ (2s))\ (\omega^2 -s^2)}{(s^2 + \omega^2)^4}$$
$$=\ \frac{(-2s)(s^2+\omega^2)\ -\ (4s)\ (\omega^2 -s^2)}{(s^2 + \omega^2)^3}$$
$$=\ \frac{(-2s)(s^2+\omega^2)\ +\ (4s)\ (s^2-\omega^2)}{(s^2 + \omega^2)^3}$$
$$=\ \frac{(2s)\left(2(s^2-\omega^2)-(s^2+\omega^2)\right)}{(s^2 + \omega^2)^3}$$
$$=\ \frac{2s(s^2 - 3 \omega^2)}{(s^2 + \omega^2)^3}$$
$$\therefore\ F^{\prime\prime}(s) = \frac{d}{ds} \left(\frac{\omega^2 -s^2}{(s^2 + \omega^2)^2}\right)\ =\ \frac{2s(s^2 - 3 \omega^2)}{(s^2 + \omega^2)^3}$$
Thus, the Laplace transform of $f(t) = t^2 \cos \omega t$ is
$$\frac{2s(s^2 - 3 \omega^2)}{(s^2 + \omega^2)^3}$$
| {
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How many solutions the equation $(x-2)(x+1)(x+6)(x+9)+108=0$ has in the interval $(-10,-1)$? How many solutions the equation $(x-2)(x+1)(x+6)(x+9)+108=0$ has in the interval $(-10,-1)$ ?
Here is my work:
By expanding the expression we get, $$(x^2-x-2)(x^2+15x+54)+108=x^4+14x^3+37x^2-84x$$
So I got $x(x^3+14x^2+37x-84)=0$. One root is zero which doesn't lie in the interval $(-10,-1)$. But I don't know how many roots the cubic equation has in that interval.
| I just solved the equation with the following method:
$$\color{red}{(x-2)}\color{green}{(x+1)(x+6)}\color{red}{(x+9)}+108=0$$
$$(x^2+7x-18)(x^2+7x+6)+108=0$$
By using the substitution $t=x^2+7x$ we get,
$$t^2-12t=0\Rightarrow t_1=0\quad,t_2=12$$So we have,
$x^2+7x=0\Rightarrow \quad x_1=0 ,\quad x_2=-7$
$x^2+7x-12=0\Rightarrow\quad x_{3}=\dfrac{-7+\sqrt{97}}{2},\quad x_4=\dfrac{-7-\sqrt{97}}{2}$
Hence $x_2,x_4\in(-1,-10)$
| {
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Help evaluating $\lim_{n\to \infty} \sum_{k=1}^n \frac{e^{\frac{-k}{n}}}{n}$ $$ \lim_{n\to \infty} \sum_{k=1}^n \frac{e^{\frac{-k}{n}}}{n} $$
is what I am asked to evaluate. My working:
\begin{align}
&= \lim_{n\to \infty} \frac{1}{n}\sum_{k=1}^n e^{\frac{-k}{n}} \\
&= \lim_{n\to \infty} \frac{1}{n} \left( \frac{1}{e^\frac{1}{n}} + \frac{1}{e^\frac{2}{n}} + \frac{1}{e^{\frac{3}{n}}} + \cdots + \frac{1}{e^\frac{n}{n}} \right) \\
&= \lim_{n\to \infty} \frac{1}{n}\frac{1}{e^\frac{1}{n} -1} = \lim_{\frac{1}{n} \to 0} \frac{\frac{1}{n}}{e^\frac{1}{n} -1} \\ &= 1.
\end{align}
Wrong. The answer is $1-\frac{1}{e}$. Where did I go wrong here?
| The equality
$$\left( \frac{1}{e^\frac{1}{n}} + \frac{1}{e^\frac{2}{n}} + \frac{1}{e^{\frac{3}{n}}} + \cdots + \frac{1}{e^\frac{n}{n}} \right)=\frac{1}{e^\frac{1}{n} -1}$$
is false. I think what you want to see is a geometric sum
$$\sum_{k=1}^n e^{-k/n} = \sum_{k=1}^n \left(e^{-1/n}\right)^k =\frac{e^{-1/n-1}-1}{e^{-1/n}-1} $$
where the numerator takes the form
$$\lim_{n\to\infty}e^{-1/n-1}-1=\frac{1}{e}-1$$
and by L'Hopital's rule the denominator becomes
$$\lim_{n\to\infty} n \left (e^{-1/n}-1 \right ) = \lim_{n\to\infty}\frac{e^{-1/n}-1}{1/n}=-1$$
| {
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Proving $x^2 + x + 1$ is a factor of $P_n (x)=(x+1)^{2n+1} + x^{n+2}$. Further says to consider $P_n(\omega)$ and $P_n(\omega^2)$ wherein $\omega$ is a cube root of unity. $\omega \not=1$.
Found this from a examination paper with no solutions. I understand it relates to roots of unity, but I'm unsure how I can bring that into play. Any explanation on how to approach these types of questions is appreciated thank you.
| Induction is really helpful here, did you try it?
lets check the case $n=0$:
$$(x+1)^{2\cdot0+1}+x^{0+2} = x+1+x^2$$ which is of course multiplier of $x^2+x+1$.
Assume, that your claim is true for every $k\leq n$, and lets prove the next step ($n+1$):
we need to prove that $x^2+x+1$ is factor of $(x+1)^{2(n+1)+1}+x^{n+1+2}$
and this can be proved easily:
$$\begin{align}(x+1)^{2n+3}+x^{n+3}&=(x+1)^2\cdot (x+1)^{2n+1}+x\cdot x^{n+2}\\
&=(x^2+2x+1)\cdot(x+1)^{2n+1}+x\cdot x^{n+2}\\
&=(x^2+x+1)\cdot(x+1)^{2n+1}+x\cdot(x+1)^{2n+1}+x\cdot x^{n+2}\\
&=(x^2+x+1)\cdot(x+1)^{2n+1}+x\cdot((x+1)^{2n+1}+x^{n+2})
\end{align}$$
where both terms are multipliers of $x^2+x+1$
| {
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If $x^3 + \frac{1}{x^3} = 2\sqrt{5}$, then find $x^2 + \frac{1}{x^2}$. Question: If $x$ is a real number satisfying $x^3 + \frac{1}{x^3} = 2\sqrt{5}$, determine the exact value of $x^2 + \frac{1}{x^2}$.
My partial solution: $(x + \frac{1}{x})^3 = x^3 + 3x + \frac{3}{x} + \frac{1}{x^3}.$
We know that $x^3 + \frac{1}{x^3} = 2\sqrt5 \Longrightarrow (x + \frac{1}{x})^3 = 2\sqrt5 + 3x + \frac{3}{x}.$
So, $(x + \frac{1}{x})^3 = 2\sqrt5 + 3(x + \frac{1}{x})$. Let $y = x + \frac{1}{x}$.
Then, $y^3 = 2\sqrt5 + 3y$. We want to solve for $x^2 + \frac{1}{x^2}$. $(x + \frac{1}{x})^2 = x^2 + \frac{1}{x^2} + 2$.
So, we want to solve for $y^2 - 2$. But, I'm confused about how to solve the equation $y^3 = 2\sqrt5 + 3y$, should I check random values?
Any help is appreciated, thanks in advance! (Thanks for the suggestions in the comments :)
| Reading the comments, I now realize that from the equation $x^3 + \frac{1}{x^3} = 2\sqrt5$, if we let $z = x^3$, then we get $z + \frac{1}{z} = 2\sqrt5 \Longrightarrow z^2 + 1 - 2\sqrt5z = 0 \Longrightarrow$ using the quadratic formula, we get $z = \sqrt5 \pm 2$. Hence, from here, we can get that the answer is $5 - 2 = 3$.
I could have also tried different values of $y$ to get $\sqrt5$ as the answer.
| {
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Using Rearrangement Inequality .
Let $a,b,c\in\mathbf R^+$, such that $a+b+c=3$. Prove that $$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{a^2+b^2+c^2}{2}$$
$Hint$ : Use Rearrangement Inequality
My Work :-$\\$
Without Loss of Generality let's assume that $0\le a\le b\le c$ then we can infer that $c^2\ge b^2\ge a^2$ also $b+c\ge c+a \ge a+b$. Thus $\frac{1}{b+c}\le \frac{1}{c+a}\le \frac{1}{a+b}$. Hence by Rearrangement Inequality $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}$ is the greatest permutation however I am not able to express $\frac{a^2+b^2+c^2}{2}$ as another permutation of the same, hence I require assistance.
Thank You
| I apply rearrangement inequality 3 times to obtain
$$
3\left(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right) \geqq (a^2+b^2+c^2)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)
$$
Next we can show that
$$
\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\geqq \frac{3}{2}
$$
by using Cauchy-Schwarz inequality and the fact that $a+b+c=3$.
Combining both inequalities give the desired result.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $a^2 + b^2 + c^2 + ab + bc +ca \ge 6$ given $a+b+c = 3$ for $a,b,c$ non-negative real. I want to solve this problem using only the AM-GM inequality. Can someone give me the softest possible hint? Thanks.
Useless fact: from equality we can conclude $abc \le 1$.
Attempt 1:
Adding $(ab + bc + ca)$ to both sides of inequality and using the equality leaves me to prove: $ab + bc + ca \le 3$.
Final edit: I found a easy way to prove above.
$18 = 2(a+b+c)^2 = (a^2 + b^2) + (b^2 + c^2) + (c^2 + a^2) + 4ab + 4bc + 4ca \ge 6(ab + bc + ca) \implies ab + bc + ca \le 3$
(please let me know if there is a mistake in above).
Attempt 2: multiplying both sides of inequality by $2$, we get:
$(a+b)^2 + (b+c)^2 + (c+a)^2 \ge 12$. By substituting $x = a+b, y = b+c, z = c+a$ and using $x+y+z = 6$ we will need to show:
$x^2 + y^2 + z^2 \ge 12$. This doesnt seem trivial either based on am-gm.
Edit: This becomes trivial by C-S.
$(a+b).1 + (b+c).1 + (c+a).1 = 6 \Rightarrow \sqrt{((a+b)^2 + (b+c)^2 + (c+a)^2)(1 + 1 + 1)} \ge 6 \implies (a+b)^2 + (b+c)^2 + (c+a)^2 \ge 12$
Attempt 3:
$x = 1-t-u$, $y = 1+t-u$, $z = 1 + 2u$
$(1-u-t)^2 + (1-u+t)^2 + (1+2u)^2 + (1-u-t)(1-u+t) + (1+t-u)(1+2u) + (1-t-u)(1+2u)$
$ = 2(1-u)^2 + 2t^2 + (1 + 2u)^2 + (1-u)^2 - t^2 + 2(1+2u)$
expanding we get:
$ = 3(1 + u^2 -2u) + t^2 + 1 + 4u^2 + 4u + 2 + 4u = 6 + 7u^2 + t^2 + 2u\ge 6$.
Yes, this works.. (not using am-gm or any such thing).
| For Attempt 1, you can use the rearrangement inequality: $ab+bc+ca\le a^2+b^2+c^2$ to get:
$$ab+bc+ca\le \frac{(a+b+c)^2}{3}=3$$
For Attempt 2, you can use $a+b=3-c$:
$$(3-a)^2+(3-b)^2+(3-c)^2\ge 12 \Rightarrow \\
27+a^2+b^2+c^2-6(a+b+c)\ge 12\Rightarrow \\
a^2+b^2+c^2\ge 3,$$
which is true by QM-AM.
| {
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Solving the equation $3\cos x+\frac{1}{2}=\cos ^{2}x\left(1+2\cos x\left(1+4\sin ^2x\right)\right)$ Solve the equation:$$3\cos x+\frac{1}{2}=\cos ^{2}x\left(1+2\cos x\left(1+4\sin ^2x\right)\right)$$
To solve it, I tried writing the equation in term of $\cos x$. ( I denote $\cos x$ by $c$):
$$3c+\frac12=c^2(1+2c(5-4c^2))$$
$$3c+\frac12=c^2+10c^3-8c^5$$
$$16c^5-20c^3-2c^2+6c+1=0 \qquad\text{Where $c\in[-1,1]$}$$
I tried $c=\pm1,\pm\frac12,0$, but neither of them satisfied the equation. So I don't know how to find $\cos x$.
| $3\cos x+\frac{1}{2}=\cos ^{2}x\left(1+2\cos x\left(1+4\sin ^2x\right)\right)$
$ \displaystyle 6\cos x + 1 = (1 + \cos 2x) \left(1 + 2 \cos x (3 - 2 \cos2x)\right)$
$ \displaystyle 6\cos x + 1 = (1 + \cos 2x) (1 + 6 \cos x - 4 \cos x \cos 2x)$
$ \displaystyle 6\cos x + 1 = (1 + \cos 2x) (1 + 4 \cos x - 2 \cos 3x)$
$6\cos x + 1 = 5 \cos x + \cos 2x - \cos 5x + 1$
$\cos 2x = \cos x + \cos 5x$
$\cos 2x = 2 \cos 2x \cos 3x$
$\cos 2x = 0$ is a solution and if $\cos 2x \ne 0$,
$\cos 3x = \frac{1}{2}$ is a solution.
| {
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Denoting $2\le\frac{6-4x}5\le3$ by $|mx-n|\le5$. What is the value of $|n-m|$?
If we denote solution set of the inequality $2\le\frac{6-4x}5\le3$ by
$|mx-n|\le5$, what is the value of $|n-m|$?
$1)7\qquad\qquad2)5\qquad\qquad3)21\qquad\qquad4)23$
I solved this problem with following approach:
$$2\le\frac{6-4x}5\le3\quad\Rightarrow-\frac94\le x\le-1$$
And $|mx-n|\le5$ is equivalent to $\frac{-5+n}m\le x\le\frac{5+n}{m}$. Hence we have $\frac{-5+n}m=-\frac94$ and $\frac{5+n}m=-1$ and by solving system of equations I got $m=8$ and $n=-13$. So the final answer is $21$.
I wonder, can we solve this problem with other approahces?
| ....or... from the other direction
$|mx- n| \le 5$
$-5 < mx - n < 5$
Stretch those so the end points are only one apart
$\frac {-5}{10} < \frac {mx-n}{10} < \frac {5}{10}$.
Then shift so that the endpoints line up where we want.
$\frac {-5}{10} + 2\frac 12 < \frac {mx-n}{10}+2\frac 12 < \frac 5{10} +2\frac 12$.
We need $2<\frac m{10}x + (\frac 52 -\frac n{10}) < 3 \iff 2< \frac{6-4x}5 < 3$
.... I suppose we can give a handwavy argument for why that would require $\frac m{10} = -\frac 45$ and for $\frac 52-\frac n{10} = \frac 65$
that is to say why $a < kx + j < b \iff a< wx + v < b$ must imply $k=w$ and (if $k\ne 0$) $j=v$. I think your method of solving directly does that. I could argue that if $j\ne v$ or $k\ne w$ there will always be wiggle room to slide an $x$ so that one inequality is true and the other not. Handwavy.... but true.
=====
An astute observer might notice that once I chose the assymetrce shift of adding $2\frac 12$ to every term I committed $m$ to equaling $-8$ and $n$ to equalling $13$ (the exact opposite of your values).
That is because as long as we have $-M < something < M$ we could just as well have $-M < -something < M$. But once we commit to one and we "shift the origin" ... well, the die is cast.
| {
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For which value of $t \in \mathbb R$ the equation has exactly one solution : $x^2 + \frac{1}{\sqrt{\cos t}}2x + \frac{1}{\sin t} = 2\sqrt{2}$ For which value of $t \in R $ the equation has exactly one solution : $x^2 + \frac{1}{\sqrt{\cos t}}2x + \frac{1}{\sin t} = 2\sqrt{2}$
Here $t \neq n\pi, t \neq (2n+1)\frac{\pi}{2}$
Therefore , for the given equation to have exactly one solution we should have :
$(\frac{2}{\sqrt{\cos t}})^2 -4.(\frac{1}{\sin t} - 2\sqrt{2}) = 0 $
$\Rightarrow \frac{4}{\cos t} - 4 (\frac{1}{\sin t} - 2\sqrt{2}) = 0 $
$\Rightarrow \sin t -\cos t +2 \sqrt{2}\sin t\cos t = 0 $
$\Rightarrow \sqrt{2}( \frac{1}{\sqrt{2}}\sin t - \frac{1}{\sqrt{2}}\cos t) = -2\sqrt{2} \sin t\cos t$
$\Rightarrow \sqrt{2}(\cos(\pi/4)\sin t -\sin(\pi/4)cos t = -\sqrt{2}\sin2t $ [Using $\sin x\cos y -\cos x\sin y = \sin(x-y)$]
$\Rightarrow \sqrt{2}\sin(\frac{\pi}{4}-t) =-\sqrt{2}\sin2t$
$\Rightarrow \sin(\frac{\pi}{4}-t) =-\sin2t $ [ Using $-\sin x = \sin(-x)$ and comparing R.H.S. with L.H.S. ]
$\Rightarrow \frac{\pi}{4}-t = -2t $
$\Rightarrow t = - \frac{\pi}{4}$
Is it correct answer, please suggest.. thanks
| Note:
$$\sin t\cos \frac{\pi}{4}-\cos t \sin \frac{\pi}{4}=\sin\left(t-\frac{\pi}4\right)$$
So, you must have
$$\sin\left(t-\frac{\pi}{4}\right) =-\sin2t$$
Then you can not simply equate the arguments, because you must remember the period.
Here is how you can continue:
$$\sin\left(t-\frac{\pi}{4}\right)+\sin 2t=0 \stackrel{{\sin A + \sin B=2\sin \frac{A+B}{2}}\cos \frac{A-B}{2}}{\Rightarrow} \\
2\sin{\left(\frac{3t}2-\frac{\pi}{8}\right)}\cos\left(-\frac{t}{2}-\frac{\pi}{8}\right)=0 \Rightarrow \\
\sin\left(\frac{3t}2-\frac{\pi}{8}\right)=0 \quad \text{or} \quad \cos\left(-\frac{t}{2}-\frac{\pi}{8}\right)=0 \Rightarrow \\
\frac{3t}2-\frac{\pi}{8}=\pi n \quad \text{or} \quad -\frac{t}{2}-\frac{\pi}{8}=-\frac{\pi}{2}+\pi n \Rightarrow \\
t=\frac{\pi}{12}+\frac{2\pi n}{3} \quad \text{or} \quad t=\frac{3\pi}4-2\pi n,n\in Z.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to solve this complicated trigonometric equation for the solution of this triangle.
In $\triangle ABC$, if $AB=AC$ and internal bisector of angle $B$ meets $AC$ at $D$ such that $BD+AD=BC=4$, then what is $R$ (circumradius)?
My Approach:- I first drew the diagram and considered $\angle ABD=\angle DBC=\theta$ and as $AB=AC$, $\angle C=2\theta$. Therefore $\angle A=180-4\theta$. Also as $AE$ is the angular bisector and $AB=AC$, then $BE=EC=2.$ Now applying the sine theorem to $\triangle ADB$ and $ \triangle BDC$ gives $$\frac{BD}{\sin {(180-4\theta)}}=\frac{AD}{\sin \theta}$$
$$\frac{BC}{\sin(180-3\theta)}=\frac{BD}{\sin 2\theta}$$
Now we know that $BC=4$ and then solving both the equations by substituting in $BD+AD=4$, we get $$\sin 2\theta .\sin4\theta+\sin2\theta.\sin\theta=\sin3\theta.\sin4\theta$$
$$\sin4\theta+\sin\theta=\frac{\sin3\theta.\sin4\theta}{\sin2\theta}$$
Now I have no clue on how to proceed further from here. Though I tried solving the whole equation into one variable ($\sin\theta$), but it's getting very troublesome as power of $4$ occurs. Can anyone please help further or else if there is any alternative method to solving this problem more efficiently or quickly?
Thank You
| The values are such that equations do not simplify. Nonetheless, here is an alternate approach.
Say $AB = AC = x$ and we know $BC = 4$,
By angle bisector theorem,
$\cfrac{4}{x} = \cfrac{x-AD}{AD} \implies AD = \cfrac{x^2}{4+x}$
Now by angle bisector length formula,
$BD^2 = \cfrac{4x}{(4+x)^2} [(4+x)^2 - x^2] = \cfrac{32x(x+2)}{(4+x)^2}$
Now, $BD + AD = 4 = \cfrac{x^2}{4+x} + \cfrac{\sqrt{32x(x+2)}}{4+x}$
$16+4x-x^2 = \sqrt{32x(x+2)}$
Solving using WolframAlpha, the only valid solution is $x \approx 2.61$
Now to find circumradius, use $R = \cfrac{abc}{4 \triangle} = \cfrac{4 x^2}{8 \sqrt{x^2 - 4}}$
| {
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Given $d = \min(\gcd(a, b+1),\gcd(a+1,b))$, prove $d \leqslant \frac{\sqrt{4a+4b+5}-1}{2}$ The question is stated above:
Given $d = \min(\gcd(a, b+1),\gcd(a+1,b))$, prove $d \leqslant \frac{\sqrt{4a+4b+5}-1}{2}$.
Here's what I've done so far:
$$d \leqslant \frac{\sqrt{4a+4b+5}-1}{2}$$
then
$$2d + 1 \leqslant \sqrt{4a+4b+5}$$
Squaring both sides, we have
$$4d^2+4d+1 \leqslant 4a+4b+5$$
$$4d^2+4d \leqslant 4a+4b+4$$
$$d^2+d \leqslant a+b+1$$
From here I'm stuck and unable to continue. I've also noticed that the sum $a+b+1$ can have some relations with the two GCDs, since $a+b+1=(a+1)+b=a+(b+1)$, but I'm unable to work out how.
| Let $d_1=\gcd(a,b+1)$ and $d_2=\gcd(a+1,b)$.
Then, as you've mentioned already, $a+b+1=a+(b+1)=(a+1)+b$, so $d_1$ and $d_2$ are divisors of the $a+b+1$. Since $d_1\mid a$ and $d_2\mid a+1$ we also have $\gcd(d_1,d_2)=1$. Thus, $d_1d_2\mid a+b+1$. In particular, $d_1d_2\le a+b+1$.
Finally, if $d_1=d_2$, then $d_1=d_2=1$ and the statement is trivial and otherwise due to $d=\min\{d_1,d_2\}$ we have
$$
a+b+1\ge d_1d_2\ge d(d+1),
$$
as desired.
| {
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$f(n) =$ the sum of digits of $n$ and number of digits in decimal notation. How many solutions does $f(x)=M$ have? For any integer $n$ let us denote $f(n)$ as the sum of digits in decimal notation of $n$ and number of digits in decimal notation. For example, $f(12) = (1+2)+2=5$. Now, consider the equation $f(x) = M$. Can I estimate the number of solutions?
| I have a proof that it is $2^{M-2}$ for $M>1$ if digits were not capped at 9
First, $M=1$ does not follow this rule, but obviously there is $1$ solution. Now for $M>2$
Each digit except for the leftmost ranges from $0$ to $\infty$ and contributes a value ranging from $1$ to $\infty$, while the leftmost digit contributes a value ranging from $2$ to $\infty$.
Using generating functions, we want the coefficient of $x^{M}$ in the expansion of
$$\left(x^2+x^3+x^4+\ldots\right)\left( 1+\left(x+x^2+x^3+\ldots\right)+\left(x+x^2+x^3+\ldots\right)^2+\left(x+x^2+x^3+\ldots\right)^3+\ldots\right)$$
This simplifies to
$$\left(\frac{x^2}{1-x}\right)\left(1+\left(\frac{x}{1-x}\right)+\left(\frac{x}{1-x}\right)^2+\left(\frac{x}{1-x}\right)^3+\ldots\right)$$
$$\left(\frac{x^2}{1-x}\right)\left(\frac{1}{1-\frac{x}{1-x}}\right)$$
$$\left(\frac{x^2}{1-x}\right)\left(\frac{1}{\frac{1-2x}{1-x}}\right)$$
$$\frac{x^2}{1-2x}$$
$$x^2\left(1+2x+4x^2+8x^3+\ldots\right)$$
$$\sum_{k=2}^\infty 2^{k-2}x^k$$
Hence, the coefficient of $x^M$ is $2^{M-2}$.
When our digits are capped at $9$, we have to also cap most of our infinite geometric series to finite geometric series and this makes the generating function math a lot messier. However, the actual answer should still be somewhat close to $2^{M-2}$ for relatively small values of $M>10$.
The actual generating function would be the coefficient of $x^M$ in
$$\frac{x^2-x^{11}}{1-2x+x^{11}}$$
$$\left(x^2-x^{11}\right)\left(1+(2x-x^{11})+(2x-x^{11})^2+(2x-x^{11})^3+\ldots\right)$$
We can probably use this to estimate some smaller values, but an exact solution is rather complicated.
| {
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Evaluating $\lim\limits_{x\rightarrow 0}\frac{\sin(x)}{\exp(x)-\exp(\sin(x))}$?
Calculate this limit
$$\lim\limits_{x\rightarrow 0}\frac{\sin(x)}{\exp(x)-\exp(\sin(x))}$$
My attempt: using the limit development : we find
$$\exp(\sin(x)-x)=\exp(x-\frac{x^{3}}{3!}+o (x^{3})-x)=\exp(-\frac{x^3}{3!}+o(x^3))=1-\frac{x^3}{3!}+o(x^3)$$
So:
\begin{align}
\lim\limits_{x\rightarrow 0}\frac{\sin(x)}{\exp(x)-\exp(\sin(x))}
&=\lim\limits_{x\rightarrow 0}\frac{\sin(x)}{\exp(x)}\left(\frac{1}{\frac{x^3}{3!}+o(x^3)}\right)\\\
& \sim \lim\limits_{x\rightarrow 0}\left(\frac{\sin(x)}{x}\right)\left(\frac{3!}{x^{2}\exp(x)}\right)=\pm\infty.
\end{align}
Is this correct?
| Your calculations are correct except for your final result which should obviously be only $+\infty$.
Here another way which uses the standard limits $\lim_{t\to 0}\frac{e^t-1}{t}= 1$ and $\lim_{t\to 0}\frac{\sin t}{t}=1$.
First of all note that for $x>0$ you have $\sin x < x$ and for $x<0$ you have $x < \sin x$. Hence
$$\frac{\sin x}{e^x - e^{\sin x}} >0 \text{ for } x \in \left[-\frac{\pi}2 , \frac{\pi}2\right]\setminus\{0\}$$
Now, just consider the reciprocal
\begin{eqnarray*} \frac{e^x - e^{\sin x}}{\sin x}
& = & \frac{e^x -1 - \left(e^{\sin x}-1\right)}{\sin x} \\
& = & \frac{e^x -1}{x}\cdot \frac{x}{\sin x} - \frac{e^{\sin x}-1}{\sin x} \\
& \stackrel{x \to 0}{\longrightarrow} & 1\cdot 1 - 1 = 0
\end{eqnarray*}
Hence,
$$\lim_{x\to 0} \frac{\sin x}{e^x - e^{\sin x}} = +\infty$$
| {
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Maximum value of a expression Problem: If $\alpha+\beta+\gamma=20$, then what is $\max(\sqrt{3\alpha+5}+\sqrt{3\beta+5}+\sqrt{3\gamma+5})$?
My attempt: Assume $\alpha \geq \beta \geq \gamma$. Then $\alpha+\beta+\gamma \leq 3\alpha$ so $25 \leq 3\alpha+5$.
Also $\sqrt{3\alpha+5}+\sqrt{3\beta+5}+\sqrt{3\gamma+5}\leq 3\sqrt{3\alpha+5}$
At this point, I don't have idea what to do next. What should I do next?
Am I doing it incorrectly?
| Use Cauchy-Bunyakovsky-Schwarz inequality:
$$\left(\sum_{i=1}^3u_iv_i\right)^2\le \left(\sum_{i=1}^3u_i^2\right)\left(\sum_{i=1}^3v_i^2\right)$$
We choose $u_1=\sqrt{3\alpha+5}$ and so on, and $v_i=1$. Then $$\sqrt{3\alpha+5}+\sqrt{3\beta+5}+\sqrt{3\gamma+5}\le\sqrt{(3\alpha+5+3\beta+5+3\gamma+5)3}=15$$
The equality is achieved when all the terms are equal, for $\alpha=\beta=\gamma=20/3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4223021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving an integral equivalence involving floor and ceiling functions During the course of looking at the Euler–Mascheroni constant I have run across the following result:
$$\gamma
= \int \limits_1^\infty \Bigg( \frac{1}{\lfloor x \rfloor} - \frac{1}{x} \Bigg) \ dx
= \int \limits_1^\infty \Bigg( \frac{\lceil x \rceil}{x^2} - \frac{1}{x} \Bigg) \ dx.$$
Representation of this constant using the first integral is a well-known result. What is the simplest way to prove the equivalence of the two integrals?
| This is actually easier than it seems, since for any positive integer $k$,
\begin{align*}
\int_k^{k+1} \bigg( \frac{\lceil x \rceil}{x^2} - \frac{1}{x} \bigg) \, dx
&= \int_k^{k+1} \bigg( \frac{k+1}{x^2} - \frac{1}{x} \bigg) \, dx = \frac1k - \log\bigg( 1+\frac1k \bigg) \\
\int_k^{k+1} \bigg( \frac{1}{\lfloor x \rfloor} - \frac{1}{x} \bigg) \, dx &= \int_k^{k+1} \bigg( \frac{1}{k} - \frac{1}{x} \bigg) \, dx = \frac1k - \log\bigg( 1+\frac1k \bigg).
\end{align*}
Then one can sum both sides over all positive integers $k$ (yielding the Euler–Mascheroni constant as it turns out).
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrate via substitution: $x^2\sqrt{x^2+1}\;dx$ I need to integrate the following using substitution:
$$
\int x^2\sqrt{x^2+1}\;dx
$$
My textbook has a similar example:
$$
\int \sqrt{x^2+1}\;x^5\;dx
$$
They integrate by splitting $x^5$ into $x^4\cdot x$ and then substituting with $u=x^2+1$:
$$
\int \sqrt{x^2+1}\;x^4\cdot x\;dx\\
=\frac{1}{2}\int \sqrt{u}\;(u-1)^2\;du\\
=\frac{1}{2}\int u^{\frac{1}{2}}(u^2-2u+1)\;du\\
=\frac{1}{2}\int u^{\frac{5}{2}}-2u^{\frac{3}{2}}+u^{\frac{1}{2}}\;du\\
=\frac{1}{7}u^{\frac{7}{2}}-\frac{2}{5}u^{\frac{5}{2}}+\frac{1}{3}u^{\frac{3}{2}}+C
$$
So far so good. But when I try this method on the given integral, I get the following:
$$
\int x^2\sqrt{x^2+1}\;dx\\
=\frac{1}{2}\int \sqrt{x^2+1}\;x\cdot x\;dx\\
=\frac{1}{2}\int \sqrt{u}\;\sqrt{u-1}\;du\;(u=x^2+1)\\
=\frac{1}{2}\int u^{\frac{1}{2}}(u-1)^\frac{1}{2}\;du
$$
Here is where it falls down. I can't expand the $(u-1)^\frac{1}{2}$ factor like the $(u-1)^2$ factor above was, because it results in an infinite series. I couldn't prove, but I think any even exponent for the $x$ factor outside the square root will cause an infinite series to result. Odd exponents for $x$ will work, since it will cause the $(u-1)$ term to have a positive integer exponent.
How should I proceed? I don't necessarily want an answer. I just want to know if I'm missing something obvious or if it is indeed above first year calculus level and probably a typo on the question.
| I have used the trigonometric substitution $x=tan\theta$ and proceeded to get the same answer as @Aman Kushwaha got via completing the square and @user773458 got via hyperbolic trig identies. Below is my steps for reference:
$$
\int (x^2-\sqrt{x^2 +1})dx \\
=\int tan^2\theta \sqrt{tan^2 \theta + 1}sec^2 \theta \;d\theta \\
=\int tan^2 \theta \sqrt{sec^2 \theta}sec^2 \theta \;d\theta \\
=\int tan^2 \theta sec^3 \theta \;d\theta \\
=\int sec^3 \theta(sec^2 \theta - 1) \; d\theta \\
=\int sec^5 \theta - sec^3 \theta \; d\theta \\
=\int sec^5 \theta \; d\theta - \int sec^3 \theta \; d\theta \\
=sec^3 \theta \; tan \theta \; - \; 3 \int sec^3 \theta \; tan^2 \theta \; d\theta - \Big(sec \theta \; tan \theta \; - \; \int sec \theta \; tan^2 \theta \; d\theta\Big) \\
\therefore 4 \int sec^3 \theta \; tan^2 \theta \; d\theta = sec^3 \theta \; tan \theta \; - \Big(sec \theta \; tan \theta \; - \; \int sec \theta \; tan^2 \theta \; d\theta\Big) \\
\therefore \int sec^3 \theta \; tan^2 \theta \; d\theta = \frac{1}{4} \; \bigg(sec^3 \theta \; tan \theta \; - \Big(sec \theta \; tan \theta \; - \; \int sec \theta(sec^2 \theta - 1)\; d\theta\Big)\bigg) \\
= \frac{1}{4} \; \bigg(sec^3 \theta \; tan \theta \; - \Big(sec \theta \; tan \theta \; - \; \int sec^3 \theta \; d\theta \; - \int sec \theta \; d\theta\Big)\bigg) \\
= \frac{1}{4}\big(sec^3 \theta \; tan \theta \; - \; \frac{1}{2}(\sec \theta \; tan \theta - \ln|sec \theta + tan \theta|)\big) + C\\
=\frac{1}{4}\big(\sqrt{tan^2 \theta + 1}(tan^2 \theta + 1)tan \theta \; - \; \frac{1}{2}(\sqrt{tan^2 \theta + 1}(tan \theta) - \ln | \sqrt{tan^2 \theta + 1} + tan \theta|)\big) + C\\
=\frac{1}{4}\big(\sqrt{x^2 + 1}(x^2 + 1)x \; - \; \frac{1}{2}(\sqrt{x^2 + 1}(x) - \ln | \sqrt{x^2 + 1} + x|)\big) + C\\
=\frac{(2x^3+2x)\sqrt{x^2 +1}-x\sqrt{x^2 + 1} - \ln|\sqrt{x^2 + 1} + x|}{8} + C \\
=\frac{(2x^3 + x)\sqrt{x^2 + 1} - \ln|\sqrt{x^2 + 1} + x|}{8} + C \\
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4223596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
} |
Having trouble with finding a limit for this sequence $a_n=\frac{n+1}{n}\ln(\sqrt{n})-\frac{1}{n^2}\sum^n_{k=1}\ln(k+n)^k$ We got this question as a bonus for our homework assignment and
I'm having trouble figuring out how to start to solve this limit. We need to find $\lim_{n\rightarrow\infty}(a_n)$ for this sequence:
$$a_n=\frac{n+1}{n}\ln(\sqrt{n})-\frac{1}{n^2}\sum^n_{k=1}\ln(k+n)^k$$
Can anyone give suggestions on how to solve this? (any help would be appreciated)
p.s. This was in the 'Riemann sums' section of our homework assignment
| We define
\begin{align}
g(n) = \frac{1}{n^2}\sum^n_{k=1}k\ln\left( k+n \right).
\end{align}
Since $f(x) = x \ln(x + n)$ is an increasing function, we find
\begin{align}
n^{-2}\int_{0}^{n}\mathrm{d} x~f(x) < ~& g(n) < n^{-2}\int_{1}^{n+1}\mathrm{d} x~f(x) \\
\frac{1+2\ln n}{4}<~&g(n) < \frac{n^2 - 1}{2n^{2}}\ln(n+1) + \frac{n-2}{4n} + \frac{2n+1}{2n}\frac{\ln(2n+1)}{n}.
\end{align}
Therefore, we obtain
\begin{align}
\frac{\ln n}{2n} - \frac{1}{4}< a_{n} < \frac{n+1}{2n}\left(\ln n - \frac{n - 1}{n}\ln(n+1)\right) - \frac{n-2}{4n} + \frac{2n+1}{2n}\frac{\ln(2n+1)}{n}.
\end{align}
Since both LHS and RHS approach $-1/4$ in the limit of $n \to \infty$, we can conclude
$$
\lim_{n \to \infty} a_{n} = -\frac{1}{4}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4224363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $A$ and $B$ are solutions to $7\cos\theta+4\sin\theta+5=0$, then $\cot\frac{A}{2}+\cot\frac{B}{2}=-\frac{2}{3}$ If $A$ and $B$ are the solutions to ${\displaystyle 7\cos\theta+4\sin\theta+5=0\mbox{ where }A>0,0<B<2\pi;}$
Without finding the solutions to the trig equation, show that;
$$\cot\left(\frac{A}{2}\right)+\cot\left(\frac{B}{2}\right)=-\frac{2}{3}$$
This is my effort so far:
Since A and B are solutions then:
\begin{align*}
7\cos A+4\sin A+5 & =0\\
7\left( 2\cos^{2}\frac{A}{2}-1\right) +4\times2\sin\frac{A}{2}\cos\frac{A}{2}+5 & =0\\
14\cos^{2}\frac{A}{2}+8\sin\frac{A}{2}\cos\frac{A}{2}-2 & =0\\
\mbox{Now divide by }\sin\frac{A}{2}\cos\frac{A}{2} & \mbox{ gives:}\\
14\cot\frac{A}{2}+8-\frac{2}{\sin\frac{A}{2}\cos\frac{A}{2}} & =0\\
\mbox{similarly for B;}\\
14\cot\frac{B}{2}+8-\frac{2}{\sin\frac{B}{2}\cos\frac{B}{2}} & =0
\end{align*}
When I add these two equations I have $\cot\left(\frac{A}{2}\right)+\cot\left(\frac{B}{2}\right)$,
but am unsure how to progress.
| We use following relations:
$sin(\theta)=\frac {2 tan\frac{(\theta)}2}{1+tan^2(\frac{\theta}2}$
$cos(\theta)=\frac {1-tan^2(\frac{\theta}2}{1+tan^2(\frac{\theta}2}$
plugging these in equation we get:
$6cot^2(\frac{(\theta)}2)+4cot(\frac{(\theta)}2)-1=0$
Comparing with $ax^2+bx+c=0$ we have:
$x_1+x_2=-\frac ba=-\frac 23$
which gives what is required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4226367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find $m$ if $x^3 - mx^2 - 4 = 0$ and $x^3 + mx + 2 =0$ have a common root. Good Day
Today, I was trying to solve this problem below:
Find the values real number $m$ can take if $x^3 - mx^2 - 4 = 0$ and $x^3 + mx + 2 =0$ have a common root.
Here's what I did:
Assume the common root to be $a$. Substituting:
$$(1)\ a^3 - ma^2 - 4 = 0$$
$$(2)\ a^3 + ma + 2 = 0$$
Subtracting $(1)$ from $(2)$ gives:
$$ma^2 + ma + 6 = 0$$
Using this as a quadratic in $a$, we use the fact that $D \geq 0$ because $a$ is real.
Thus,
$$-4(m)(6) + (m)^2 \geq 0$$
So, $$m(m - 24) \geq 0$$
which is true precisely when $m \in (-\infty, 0]$ $\cup$ $[24, \infty)$
However, the answer given to me is only $m = -3$. Where am I going wrong?
Thanks!
| The intersections equation you found, $ \ mx^2 + mx + 6 \ = \ 0 \ , $ is useful, but does not by itself tell us about the zeroes of the polynomial; rather, it only locates the $ \ x-$coordinates of the two intersection points between the curves for $ \ x^3 - mx^2 - 4 \ \ $ and $ \ x^3 + mx + 2 \ \ . $ These points lie at
$$ x \ \ = \ \ \frac{-m \ \pm \ \sqrt{m^2 \ - \ 24m}}{2m} \ \ = \ \ -\frac12 \ \pm \ \frac{\sqrt{1 \ - \ \frac{24}{m}}}{2} \ \ . $$
As you found, this means that the two function curves intersect only for $ \ m \ \ge \ 24 \ \ $ or $ \ m \ < \ 0 \ \ . $ (We can discard $ \ m = 0 \ $ as this produces the two "parallel" cubic curves $ \ y \ = \ x^3 + 2 \ \ $ and $ \ y \ = \ x^3 - 4 \ \ . ) $
A differing approach we may take is to consider the properties of the functions and their curves and what this quadratic relation tell us. Something we can say immediately (although it is not clear how it is helpful) is that the common zero $ \ r \ $ must be a common factor of $ \ 4 \ $ and $ \ -2 \ \ . $ If we "complete-the-cube" for the function $ \ x^3 + bx^2 + cx + d \ , \ \ $ we obtain $$ \ \left(x + \frac{b}{3} \right)^3 \ + \ \left(c - \frac{b^2}{3} \right)x \ + \ \left( d - \frac{b^3}{27} \right) \ \ , $$
which indicates a symmetry axis (neglecting "vertical displacement") at $ \ x \ = \ -\frac{b}{3} \ $ and that the curve will have local extrema ("turning-points") if $ \ c < \frac{b^2}{3} \ \ . $ Consequently, $ \ y \ = \ x^3 - mx^2 - 4 \ $ has a symmetry axis about $ \ x \ = \ \frac{m}{3} \ \ $ and always has turning-points since $ \ c = 0 \ \ ; $ the curve for $ \ y \ = \ x^3 + mx + 2 \ $ has a symmetry about the $ \ x-$axis $ \ ( b = 0 ) \ $ and turning-points when $ \ m < 0 \ \ . $ [Calculus would enable us to say rather more, but we are limiting ourselves to algebra of polynomials and analytic geometry.] We will be able to narrow down possibilities with this information.
It might also be noted that the Rule of Signs indicates that:
• for $ \ m > 0 \ , \ \ x^3 - mx^2 - 4 \ $ has one positive real zero and no negative real zeroes; $ \ x^3 + mx + 2 \ $ has no positive real zeroes and one negative real zero
• for $ \ m < 0 \ , \ \ x^3 - mx^2 - 4 \ $ has one positive real zero and two or no negative real zeroes; $ \ x^3 + mx + 2 \ $ has two or no positive real zeroes and one negative real zero
So the Rule of Signs analysis permits us to reject the case for $ \ m \ \ge \ 24 \ \ . $ This is confirmed by investigating the function values. For $ \ m = 24 \ \ , $ the single intersection at $ \ x = -\frac12 \ $ gives us $ \ y \ = \ -\frac{81}{8} \ \ . $ As $ \ m \ $ increases, the intersection bifurcates into two points with $ \ x \rightarrow -1^{+} \ $ and $ \ x \rightarrow 0^{-} \ \ . $ For the former intersection, $ \ y \rightarrow -\infty \ \ , $ while the latter approaches the $ \ y-$intercept at $ \ ( 0 \ , -4 ) \ \ . $ Hence, no common zero is possible for this case.
In a similar fashion, we can show that $ \ m \ $ cannot have a large negative value if there is to be a common zero for the two polynomials. (The Rule of Signs leaves the possibility open for $ \ m < 0 \ \ . ) $ If we choose a convenient value of $ \ m = -8 \ \ , $ the intersections are found at $ \ x = -\frac32 \ $ and $ \ x = +\frac12 \ \ , $ with the corresponding function values $ \ y = \frac{85}{8} \ $ and $ \ -\frac{15}{8} \ \ ; $ we find that the S-turns in the two cubic curves are intertwining. As $ \ m \ $ decreases from this value, the $ \ x-$coordinates of the intersections behave as $ \ x \rightarrow -1^{-} \ $ and $ \ x \rightarrow 0^{+} \ \ , $ with the former intersection having $ \ y \rightarrow +\infty \ $ and the other intersection again approaching $ \ ( 0 \ , -4 ) \ \ . $ So there can be no common zero for this situation either.
Thus, if a common zero between the polynomials exists, we must have $ \ -8 < m < 0 \ \ . $ We could continue to analyze the behavior of the intersections, but we see that they must lie not far from the origin. So we could just try common factors of $ \ 4 \ $ and $ \ -2 \ $ at this point to see what happens.
It might be remarked here that there is something suspicious about these real zeroes. The Rule of Signs shows that there is one positive and one negative real zero among the two polynomials. But a cubic polynomial with real coefficients must have either one or three real zeroes: could it be that there is a double zero?
On testing possible integer zeroes, we observe for the intersections that
$ \mathbf{x = +1 \ \ } \Rightarrow \ \ 1 \ - \ \frac{24}{m} \ = \ 9 \ \ \Rightarrow \ \ m \ = \ -3 \ \ ; $
$ \mathbf{x = -1 \ \ } $ [not admissible: asymptotic value] ;
$ \mathbf{x = +2 \ \ } \Rightarrow \ \ 1 \ - \ \frac{24}{m} \ = \ 25 \ \ \Rightarrow \ \ m \ = \ -1 \ \ ; $
$ \mathbf{x = -2 \ \ } \Rightarrow \ \ 1 \ - \ \frac{24}{m} \ = \ 9 \ \ \Rightarrow \ \ m \ = \ -3 \ \ . $
For the $ \ m = -1 \ $ case, the function curves intersect at $ \ (2 \ , 8) \ \ , $ so $ \ x = 2 \ $ is not a zero of either function. On the other hand, for $ \ m = -3 \ $ , we find two intersections at $ \ x = -2 \ $ and $ \ x = +1 \ \ ; $ we find the function values to be
$ \mathbf{f(x) \ = \ x^3 \ + \ 3x^2 \ - \ 4 \ \ : } \quad f(-2) \ = \ -8 + 12 - 4 \ = \ 0 \ \ \ , \ \ \ f(1) \ = \ 1 + 3 - 4 \ = \ 0 \ \ ; $
$ \mathbf{g(x) \ = \ x^3 \ - \ 3x \ + \ 2 \ \ : } \quad g(-2) \ = \ -8 + 6 + 2 \ = \ 0 \ \ \ , \ \ \ f(1) \ = \ 1 - 3 + 2 \ = \ 0 \ \ . $
[More thorough analysis would show that the intersections move away from the $ \ x-$axis as $ \ m \ $ is "shifted away" from $ \ -3 \ \ . ] $
Polynomial or synthetic division shows that
$$ x^3 \ + \ 3x^2 \ - \ 4 \ \ = \ \ (x - 1) · (x + 2)^2 \ \ \ \text{and} \ \ \ x^3 \ - \ 3x \ + \ 2 \ \ = \ \ (x - 1)^2 · (x + 2) \ \ , $$
which means that the singleton zero of each of these polynomials makes an intersection with the double zero of the other (an initially unanticipated property!). The only solution for the problem then is $ \ \mathbf{m \ = \ -3 \ \ } . $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4227185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
Compute upper limit and lower limit Consider the sequence, $0<a_{1}<a_{2}$ and
\begin{eqnarray*}
a_{n}= \frac{a_{n-1}+a_{n-2}}{2}
\end{eqnarray*}
for $n\geq 3$. Show that:
\begin{eqnarray*}
\overline{\lim a_{n}}=\underline{\lim a_{n}}
\end{eqnarray*}
For this, the bounds are $a_{1}$ and $a_{2}$ so I can describe the sequence:
\begin{align*}
\overline{a_{k}} &= \begin{cases}
a_{k} & \text{if } k \text{ is even} \\
a_{k+1} & \text{if } k \text{ is odd}
\end{cases} \\
\underline{a_{k}} &= \begin{cases}
a_{k} & \text{if } k \text{ is odd} \\
a_{k+1} & \text{if } k \text{ is even}
\end{cases}
\end{align*}
So $\{\overline{a_{k}}\}=\{a_{2},a_{4},a_{6},...\}$ and $\{\underline{a_{k}}\}=\{a_{1},a_{3},a_{5},...\}$, the sequence $\{\overline{a_{k}}\}$ is decreasing and the sequence $\{\underline{a_{k}}\}$ is increasing, so, by Weirstrass's theorem are convergent but when I try to compute $\overline{\lim a_{n}}$ or $\underline{\lim a_{n}}$ I try to do this, Since:
\begin{eqnarray*}
a_{2k}=\frac{a_{2k-1}+a_{2k-2}}{2}
\end{eqnarray*}
for $n\geq 3$ and $\lim a_{2k}= x = \lim a_{2k-1}= \lim a_{2k-2}$, but this don't say a great information, Can you give some hint or advice to start? Thank you
| At each step, you take the center of the two previous terms. That means that $a_3$ is the center of $[a_1,a_2]$, samely $a_4$ is the center of $[a_2,a_3]$, etc... This suggests that $a_{n+1}-a_n$ behaves like $\frac{1}{2^n}$, indeed
$$ a_{n+2}-a_{n+1}=\frac{a_{n+1}+a_n}{2}-a_{n+1}=-\frac{1}{2}(a_{n+1}-a_n) $$
Therefore $a_{n+1}-a_n=\lambda\left(-\frac{1}{2}\right)^n$ where $\lambda$ is a constant. Summing this gives that $a_n=\alpha+\beta\left(-\frac{1}{2}\right)^n$ where $\alpha,\beta$ are constants. Therefore $(a_n)$ converges and thus $\liminf\limits_{n\rightarrow +\infty} a_n=\limsup\limits_{n\rightarrow +\infty} a_n=\lim\limits_{n\rightarrow +\infty} a_n$.
Note : You can directly show that $a_n=\alpha+\beta\left(-\frac{1}{2}\right)^n$ by searching the roots of $X^2-\frac{1}{2}X-\frac{1}{2}$, which are $1$ and $-\frac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4227839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Optimising the given function I am required to find minimum value of the following function without calculus.
$$f(a,b)=\sqrt{a^2+b^2}+2\sqrt{a^2+b^2-2a+1}+\sqrt{a^2+b^2-6a-8b+25}$$
My attempt:
I realised that I can write the function as $$f(a,b)=\sqrt{(a-0)^2+(b-0)^2}+2\sqrt{(a-1)^2+(b-0)^2}+\sqrt{(a-3)^2+(b-4)^2}$$ which when $a$ and $b$ replaced with coordinates in Cartesian plane represents distances from points $(1,0),(0,0)$ and $(3,4).$ But I'm not sure how I can minimise this.
| The function to be minimized is composed of three addends, one of which has a double coefficient. If we eliminate it, imposing that
$\sqrt{a^{2}+b^{2}-2a+1}=0$
give zero contribution, we get:
$a=1$ and $b=0$.
Therefore the minimum is:
$f(1,0)=\sqrt{1^2+0^2}+2\sqrt{1^2+0^2-2+1}+\sqrt{1^2+0^2-6-0+25}=1+\sqrt{20}=1+2\sqrt{5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4228156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 3
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Factoring $\frac{n(n+1)}2x^2-x-2$ for $n\in\mathbb Z$ I was factoring quadratic polynomials for high-school practice and I noticed a pattern: $$\begin{align} x^2-x-2 &=(x+1)(x-2) \\ 3x^2-x-2 &=(x-1)(3x-2) \\ 6x^2-x-2 &= (2x+1)(3x-2) \\ 10x^2-x-2 &= (2x-1)(5x+2) \\ 15x^2-x-2 &= (3x+1)(5x-2) \\ 21x^2-x-2 &= (3x-1)(7x+2) \\ &\vdots\end{align}$$ So it seemed that, for any integer $n$, we have:
$$\frac{n(n+1)}2x^2-x-2=\Big(\left\lceil{\frac n2}\right\rceil x+(-1)^{n+1}\Big)\Big(\left\lceil{n+\frac{(-1)^n}2}\right\rceil x + 2(-1)^n\Big).$$
where the pattern I showed above begins with $n=1$ and ends with $n=6$.
I am not sure how to prove this (assuming it is true). I know that $-2=(-1)^{n+1}\cdot 2(-1)^n$ but I don't know how to prove that: $$\left\lceil{\frac n2}\right\rceil\cdot \left\lceil{n+\frac{(-1)^n}2}\right\rceil=\frac{n(n+1)}2\tag{$\star$}$$ and $$2(-1)^n\left\lceil{\frac n2}\right\rceil+(-1)^{n+1}\left\lceil{n+\frac{(-1)^n}2}\right\rceil=-1$$ which I believe is necessary to prove this conjecture. Since $n$ is an integer, I was thinking of letting $n=\left\lceil\frac k2\right\rceil$ for any real $k$ or something, and I'd assume that $$\left\lceil{\frac n2}\right\rceil=\left\lceil{\frac{\left\lceil{\frac k2}\right\rceil}2}\right\rceil=\left\lceil{\frac k4}\right\rceil$$ but I'm not sure how "round-off arithmetic" works (informally speaking). Any help is appreciated.
Edit:
Thanks to @lone_student's comment, I have shown $(\star)$ to be true for all $n\in\mathbb Z$ by considering $n$ even and odd.
Lemma:$$\left\lceil {\frac nm}\right\rceil=\left\lfloor{\frac{n-1}m+1}\right\rfloor\tag1$$
Here, $m=2$.
Also: $$n-\left\lfloor\frac n2\right\rfloor=\left\lceil\frac n2\right\rceil\tag2$$
Using these, we can show that:
$$\frac{n(n+1)}2=\left\lceil\frac n2\right\rceil\cdot\left\lceil{n+\frac{(-1)^n}2}\right\rceil=\underbrace{\Big(n-\left\lfloor\frac n2\right\rfloor\Big)}_{\text{By } (2)}\cdot\underbrace{\left\lfloor\frac{2n+(-1)^n-1}2+1\right\rfloor}_{\text{By (1)}}$$
When $n=2k\in\mathbb Z$ we have $$k(2k+1)=\big(2k-\left\lfloor k\right\rfloor\big)\lfloor2k+1\rfloor=k(2k+1)$$ since $k$ and $2k+1$ are integers, and, by definition, $\lceil \alpha\rceil =\lfloor \alpha\rfloor = \alpha$ iff $\alpha\in\mathbb Z$.
Similarly, when $n=2k-1$, we have: $$k(2k-1)=\Big(2k-1-\left\lfloor k-\frac 12\right\rfloor\Big)\lfloor 2k-1\rfloor = k(2k-1)$$ since obviously $\left\lfloor k-\frac 12\right\rfloor = k-1$ for $k\in\mathbb Z$.
I believe there was some confusion towards my question: did I mean to factorise the quadratic in terms of $n$ or did I mean to prove specifically the ceiling-function product identity? I did intend to ask a question on the latter subject, but I had falsely assumed that the case-by-case polynomial pattern I showed above could only be represented through the ceiling functions. This was wrong.
| Alternative factoring method:
When $B^2 - 4AC \geq 0,$ with $A > 0$, then $Ax^2 + Bx + C$ factors into
$A\left(x^2 + \frac{B}{A}x + \frac{C}{A}\right)$
$= A \left[\left(x + \frac{B}{2A}\right)^2 - \frac{B^2}{4A^2} + \frac{C}{A}\right]$.
$= A \left[\left(x + \frac{B}{2A}\right)^2 - \frac{B^2 - 4AC}{4A^2}\right]$
$= A \left[\left(x + \frac{B}{2A}\right)^2 - \left(\frac{\sqrt{B^2 - 4AC}}{2A}\right)^2\right]$
Under the assumptions, this factors into the difference of two squares as
$$= A\left[\left(x + \frac{B}{2A} + \frac{\sqrt{B^2 - 4AC}}{2A}\right) \times \left(x + \frac{B}{2A} - \frac{\sqrt{B^2 - 4AC}}{2A}\right)\right].\tag1$$
With the posted problem, you have that
$A = \frac{n(n+1)}{2}$
$B = -1$
$C = -2$.
From this you can immediately conclude (since $A>0, C<0$) that $B^2 - 4AC > 0$. Therefore, the formula in (1) above applies.
$\frac{B}{2A} = \frac{-1}{n(n+1)}.$
$\frac{\sqrt{B^2 - 4AC}}{2A} = \frac{\sqrt{1 + 4n(n+1)}}{n(n+1)} = \frac{(2n+1)}{n(n+1)}.$
Therefore, $Ax^2 + Bx + C$ factors into
$\frac{n(n+1)}{2}\left[\left(x + \frac{-1}{n(n+1)} + \frac{(2n+1)}{n(n+1)}\right) \times
\left(x + \frac{-1}{n(n+1)} - \frac{(2n+1)}{n(n+1)}\right)\right].$
This simplifies into
$$\frac{1}{2n(n+1)} ~\left\{~
\left[~n(n+1)x + 2n\right] ~\times
~\left[~n(n+1)x - 2(n+1)~\right]~
\right\}.$$
This further simplifies to
$$\frac{1}{2} \times \left[(n+1)x + 2\right]
\times \left[nx - 2\right].\tag2$$
In (2) above, either $(n+1)$ or $n$ will be even, thus allowing the factor of $\frac{1}{2}$ to be cleared.
Therefore, you can forgo any consideration of the floor or ceiling functions, and simply divide the formula mentioned in (2) above into two cases: either $n$ is odd or $n$ is even.
Finally, while I showed the derivation of the formula in (2) above, my answer could have been significantly shorter, if I had instead provided a (sanity-checking) verification of the formula in (2) above.
That is, if you manually multiply the factors in (2) above, the product will be $Ax^2 + Bx + C$, where $A,B,C$ are as specified in your original question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4232401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Calculate maximum and minimum when second partial derivative test fail
Calculate the maximum and minimum of function $z = f(x,y) = x^2 - y^2 +2$ subjected to the inequality constraint $D=\{(x,y)| x^2 + \frac{y^2}{4} \leq 1\}$.
My solution:
First form the function
$$
g(x,y) = x^2 + \frac{y^2}{4} - c, ~0 \leq c \leq 1.
$$
Then form the Lagrangian function
$$
L(x,y,\lambda) = x^2 - y^2 + 2 + \lambda\left(x^2 + \frac{y^2}{4} - c\right).
$$
Therefore we have
$$
\left \{
\begin{array}{ll}
L_x' = 2x +2\lambda x = 0\\
L_y' = -2y + \frac{\lambda}{2}y = 0 \\
L_\lambda' = x^2 + \frac{y^2}{4} - c
\end{array} \right.
$$
After solving above equations, so we can got saddle point like $(\varphi(c),\psi(c))$.
*
*when $c = 0$, we have $x = y = 0$.
*when $c \neq 0$, there are two kind of solutions: (1 $x = 0$, we have $y = \pm 2\sqrt{c}$; (2 $y=0$, we have $ x = \pm \sqrt{c}$.
My problem is that i cant use second partial derivative test for judging $(0,\pm 2\sqrt{c})$ and $(\pm \sqrt{c},0)$ are maximum or minimum, obviously $AC-B^2 = 0$. How can i do next? Thanks in advance!
| Since the the maximum and the minimum can be attained only at the boundary we can proceed by direct substitution of the constraint (ellipse) that is
$$h(y) = 1-\frac{y^2}4 - y^2 +2=3-\frac 54 y^2$$
therefore we have
*
*minimum $f(x,y)=3$ at $y=0$, $x=\pm 1$
*maximum $f(x,y)=-2$ at $y=\pm 2$, $x=0$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find all analytic functions in the disk $|z-2|<2$ such that $f \left(2+\frac{1}{n} \right) = \frac{1-n}{5n+3} + \sin{\left(\frac{n \pi}{2} \right)}$ Find all analytic functions (or prove that no such exist) inside the disk $|z-2|<2$ that satisfy the following condition:
$$f \left(2+\frac{1}{n} \right) = \frac{1-n}{5n+3} + \sin{\left(\frac{n \pi}{2} \right)}, \ n \in \mathbb{N}$$
For $n=2k$ the expression simplifies quite a bit since $\sin{\left(\frac{n \pi}{2} \right)}= \sin{\left(k \pi \right)} = 0$ so we're left with $$f \left(\frac{4k+1}{2k}\right) = \frac{1-2k}{10k+3}$$
Similarly for $n=4k-3$ and $n=4n-1$ we can simplify the sinus expression to $1$ and $-1$. So we're getting $3$ different expressions for $f\left(2+\frac{1}{n}\right)$ based on whether $n=2k$ , $n=4k-3$ or $n=4k-1$. My idea is to use the identity theorem for any of these two expressions and prove that no such function exists but I can't quite tie it all together.
| Your idea is correct, there is also a simpler one: assume that such a function exists. Then using continuity of $f$ we get
$$f(2) = \lim_{n \to \infty} f \left( 2+\frac{1}{n} \right) = \lim_{n \to \infty} \left( \frac{1-n}{5n+3} + \sin \frac{n \pi}{2} \right).$$
But the limit on the right clearly does not exist, which is a contradiction.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\left(\frac{u}{a}\right)^a.\left(\frac{v}{b}\right)^b.\left(\frac{w}{c}\right)^c \le \left(\frac{u+v+w}{a+b+c}\right)^{(a+b+c)} $ Let $u,v,w>0$ and $a,b,c$ are positive constant. Prove that $\left(\frac{u}{a}\right)^a.\left(\frac{v}{b}\right)^b.\left(\frac{w}{c}\right)^c \le \left(\frac{u+v+w}{a+b+c}\right)^{(a+b+c)} $
First, I prove with $x+y+z=1$ so $x^ay^bz^c\le\left(\frac{a}{a+b+c}\right)^a\left(\frac{b}{a+b+c}\right)^b\left(\frac{c}{a+b+c}\right)^c$ by Largrange theorem
And it become $\left(\frac{x}{a}\right)^a\left(\frac{y}{b}\right)^b\left(\frac{z}{c}\right)^c\le\left(\frac{1}{a+b+c}\right)^{a+b+c}=\left(\frac{x+y+z}{a+b+c}\right)^{a+b+c}$
So it true with $x+y+z=1$ but i can't prove it true with $x,y,z>0$. Please help me! Thank you.
| Since we have positive numbers involved therefore we can use weighted $A.M-G.M$ on $\frac{u}{a},\frac{v}{b},\frac{w}{c}$ with weights respectively $a,b,c$
$$\displaystyle\bigg({\frac{\sum_ia_ix_i}{\sum_i a_i}}\bigg)^{\sum_i a_i}\geq\bigg(\Pi (x_i)^{(a_i)}\bigg) $$
Here $x_1=\frac{u}{a} $ and $a_1=a$ , $x_2=\frac{v}{b} $ and $a_2=b$ and $x_3=\frac{w}{c} $ and $a_3=c$
Equality holds when $\frac{u}{a}=\frac{v}{b}=\frac{w}{c}$
| {
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Proving divisibility by $9$ for $10^n + 3 \times 4^{n+2} + 5$. I am trying to prove that for all $n$, $9$ divides $10^n + 3 \times 4^{n+2} + 5$. I first tried induction on $n$, but couldn't get anywhere with the induction step. I then tried to use modular arithmetic. The furthest I could get was:
As $10 \equiv 1 \mod 9$, $10^n \equiv 1^n = 1$ for all $n$, so modulo $9$, we have
\begin{align*}
10^n + 3 \cdot 4^{n+2} + 5 & = 3 \cdot 4^{n+2} + 6 \\
& = 3\left(4^{n+2} + 2\right) \\
& = 3\left(\left(2^2\right)^{n+2} + 2 \right) \\
& = 3\left(2^{2(n+2)} + 2 \right) \\
& = 3\left(2\left(2^{2(n+2) - 1} + 1\right) \right)
\end{align*}
I need to something get a factor of $3$ to show that the entire expression is divisible by $9$ and hence equal to $0$, mod $9$. But, with a sum of even terms, this does not appear possible.
Any hints on how to proceed would be appreciated. Is induction the standard way to prove something like this?
| Here is another way if you know the binomial theorem.
Observe that $$10^n = (1+9)^n = 1 + {n \choose1}9 + \ldots + 9^n = 1 + 9A$$ where $A= {n \choose1}+ \ldots + 9^{n-1}$ and similarly $$4^{n+2} = (1+3)^{n+2} = 1 + (n+2) 3 + {{n+2} \choose 2}3^2 + \ldots + 3^{n+2} = 1 + 3(n+2) + 3^2B$$
So $$10^n + 3 \cdot 4^{n+2} + 5 = 1 + 9A + 3( 1 + 3(n+2) + 3^2B ) + 5 = \color{red}{9} {\left(1 + A + (n+2) + 3B \right) }$$
| {
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"timestamp": "2023-03-29T00:00:00",
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inequality for positive real numbers Given $$a,b,c,x,y,z \in R^+$$ how to show the following inequality:
$$\frac{a^3}{x} + \frac{b^3}{y} + \frac{c^3}{z} \geq \frac{(a+b+c)^3}{3(x + y + z)}$$
I rearranged the inequality since all are positive, the inequality would be true iff
$$\frac{a^3}{x(a+b+c)^3}+\frac{b^3}{y(a+b+c)^3}+\frac{c^3}{z(a+b+c)^3}\geq\frac{1}{3(x+y+z)}$$
which further simplifies to
$$\frac{a^3yz}{(a+b+c)^3}+\frac{b^3xz}{(a+b+c)^3}+\frac{c^3xy}{(a+b+c)^3}\geq\frac{xyz}{3(x+y+z)}$$
and now I am struggling how to rearrange further to apply Holding's inequality
| Apply Sedakryan's inequality https://en.m.wikipedia.org/wiki/Sedrakyan%27s_inequality
$\\$ For $n=3$ , $\small{\frac{a_{1}^2 }{b_1 } +\frac{a_{2}^2 }{b_2 } +\frac{a_{3}^2 }{b_3 } \ge \frac{( a_1 +a_2 +a_3 )^2 }{b_1 + b_2 +b_3 }}$.
$\\$ Set $a_1 =a^{3/2} ,a_2 =b^{3/2} , a_3 =c^{3/2} , b_1=x ,b_2 =y , b_3 =z$.
$\\$ The inequality must be simplified like $a^3 /x +b^3 /y +c^3 /z \ge \frac{(a^{3/2} +b^{3/2} +c^{3/2 } )^2 }{(x+y+z)}$.
$\\$ Apply $ (a^{3/2} + b^{3/2}+c^{3/2} )^2 \ge \frac{1}{3} (a+b+c)^3 $ to complete the proof.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove by contradiction that if $n^3$ is a multiple of $3$, then $n$ is a multiple of $3$ Problem statement:
Using proof by contradiction, prove that if $n^3$ is a multiple of $3$ , then $n$ is a multiple of $3.$
Attempt 1:
Assume that there is exist $n$ which is a multiple of $3$ such that $n^3$ is not a multiple of $3.$
Then $n = 3k $ , $n^3 = 27 k^3 $ which a multiple of $3$, which contradicts the assumption that $n^3$ is a multiple of $3.$
Attempt 2:
Assume that there exist $n^3$ which is a multiple of $3$ such that $n$ is not a multiple of $3.$
Then $n = k+ 1 $ or $n= k+2.$
Then $n^3 = k^3 + 3k^2 + 3k + 1$
or $n^3 = k^3 + 6 k^2 + 12 k + 8.$
Both cases contradict the assumption that $n^3 $ is a multiple of $3.$
Which solution is correct ? Or do both work ?
| If $n$ is not a multiple of $3$, then $n$ can be $3k+1$ or $3k+2$.
If $n = 3 k + 1$ then $n^3 = 27 k^3 + 27 k^2 + 9 k + 1 = 3 (9 k^3 + 9 k^2 + 3k) + 1$
; so $n^3$ is not divisible by $3$. If $n = 3 k + 2$ then $n^3 = 27 k^3 + 54 k^2 + 36 k + 8 = 3 (9 k^3 + 18 k^2 + 12 k + 2) + 2 $. So, again, $n^3$ is not divisible by $3$. Thus we have be contradiction that if $n^3$ is a multiple of $3$ then so is $n$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Showing $\int_0^\infty\frac{Li_2\frac{(1-x^4)^2}{(1+x^4)^2}}{1+x^4}dx=\sqrt2\,\Re\left(\int_0^1\frac{Li_2\frac{(1+x^4)^2}{(1-x^4)^2}}{1+x^2}dx\right)$ A beautiful equality:
$$\int_{0}^{\infty }\frac{Li_{2}\left(\frac{(1-x^{4})^{2}}{(1+x^{4})^{2}}\right)}{1+x^{4}}dx=\sqrt{2} \Re \left(\int_{0}^{1 }\frac{Li_{2}\left(\frac{(1+x^{4})^{2}}{(1-x^{4})^{2}}\right)}{1+x^{2}}dx\right)$$
This question was proposed by Sujeethan Balendran in RMM(Romanian Mathematical Magazine)
I did by using complex number but i didn't by
Real method by using
$$Li _{2}(a)=-a\int_{0}^{1}\frac{\ln x}{1-ax}dx$$
I believe some of you know some nice proofs of this, can you please share it with us?
| $$\int_{0}^{\infty }\frac{Li_{2}(\frac{(1-x^{4})^{2}}{(1+x^{4})^{2}})}{1+x^{4}}dx $$
$$=\frac{1}{2}\int_{0}^{\infty }\frac{Li_{2}(\frac{(1-x^{2})^{2}}{(1+x^{2})^{2}})}{\sqrt{x}(1+x^{2})}dx$$
$$substitute\quad x=tan(\frac{x}{2})$$
$$=\frac{1}{4}\int_{0}^{\pi}Li_{2}(cos^{2}(x))(\sqrt{tan(\frac{x}{2})})dx
=\frac{1}{8}\int_{0}^{\pi}Li_{2}(cos^{2}(x))(\sqrt{tan(\frac{x}{2})}+\sqrt{cot(\frac{x}{2})})dx$$
$$=\frac{-2\sqrt{2}}{8}\Re (\int_{0}^{\pi}\frac{iLi_2(cos^{2}(x))e^{ix}}{\sqrt{1-e^{2ix}}})dx$$
$$=\frac{-2\sqrt{2}}{8}\Re (\oint_ {\left | z \right |=1}\frac{Li_2(\frac{(z+\frac{1}{z})^{2}}{4})}{\sqrt{1-z^{2}}})dz\\=\frac{1}{2\sqrt{2}}\Re \int_{-1}^{1}\frac{Li_2((\frac{x^{2}+1}{2x})^{2})}{\sqrt{1-x^{2}}}dx$$
$$=\frac{1}{2\sqrt{2}}\Re \int_{-1}^{1}\frac{Li_2((\frac{x^{2}+1}{2x})^{2})}{\sqrt{1-x^{2}}}dx$$=$$\frac{1}{\sqrt{2}}\Re \int_{0}^{1}\frac{Li_2((\frac{x^{2}+1}{2x})^{2})}{\sqrt{1-x^{2}}}dx$$
$$substitute \quad x=\frac{1-x}{1+x}\\=\frac{1}{\sqrt{2}}\Re \int_{0}^{1}\frac{Li_2((\frac{x^{2}+1}{1-x^{2}})^{2})}{(1+x)\sqrt{x}}dx$$$$
so \quad =\sqrt{2}\Re (\int_{0}^{1}\frac{Li_2((\frac{x^{4}+1}{1-x^{4}})^{2})}{(1+x^{2})}dx)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Ivan has 3 red blocks, 4 blue blocks and 2 green blocks. He builds the tower with randomly selected blocks but only stops when the tower consists of all three colours.
*
*What is the probability that the tower is 4 blocks tall?
My approach is the following but I am not sure at all:
to make it 3 colours it builds a tower that is either 3 blocks or 8 blocks.
8 blocks would happen in the case where the first 7 blocks are 4Blue and 3Red, so the 8th must be the green block.
so $E(8) = 4B,3R,1G = \binom{4}{4}*\binom{3}{3}*\binom{2}{1}= 2$
$E(7) = 4B,2G,1R = \binom{4}{4}*\binom{2}{2}*\binom{3}{1}= 3$
E(6) = let's reason the other way around. First $\binom{9}{6} = 84$. then the combinations of 6 blocks with only 2 colours is 4B,2R or 4B,2G or 3B,3R.
This makes:
$ \binom{4}{4}*\binom{3}{2} + \binom{4}{4}*\binom{2}{2}+ \binom{4}{3}*\binom{3}{3}= 8$.
So $E(6) = 84 - 8 = 76$
E(5) = 3B,1R,1G or 3R,1B,1G or 2B,2R,1G or 2B,2G,1R or 2R,2G,1B =
$E(5) = \binom{4}{3}*\binom{3}{1}*\binom{2}{1}+ \binom{3}{3}*\binom{4}{1}*\binom{2}{1}+ \binom{4}{2}*\binom{3}{2}*\binom{2}{1}+ \binom{4}{2}*\binom{2}{2}*\binom{3}{1}+ \binom{3}{2}*\binom{2}{2}*\binom{4}{1}= 98$
E(4) = 2B,1R,1G or 2R,1B,1G or 2G,1B,1R =
$\binom{4}{2}*\binom{3}{1}*\binom{2}{1}+ \binom{3}{2}*\binom{4}{1}*\binom{2}{1}+ \binom{2}{2}*\binom{4}{1}*\binom{3}{1}= 72$
$E(3) = \binom{4}{1}*\binom{3}{1}*\binom{2}{1}= 24$
total combinations $(T) = 24 + 72 + 98 + 76 + 3 + 2 = 275$.
The probability of stopping at 4 blocks is $E(4)/(T) = 72/275$
but this is a very long solution. Do you think that is correct? if so, is there any other method to make it shorter?
| There are three cases to consider.
Case 1: The fourth block is green.
Of the first three blocks, none are green $\binom{7}{3}$, and they are not all blue $\binom{4}{3}$ or all red $\binom{3}{3}$. Then the probability of the fourth block being green is $\frac{2}{6}$
$$\frac{\binom{7}{3}-\binom{4}{3}-\binom{3}{3}}{\binom{9}{3}}\times \frac{2}{6}=\frac{10}{84}$$
Case 2: The fourth block is red.
Of the first three blocks, none are red $\binom{6}{3}$, and they are not all blue $\binom{4}{3}$. Then the probability of the fourth block being red is $\frac{3}{6}$
$$\frac{\binom{6}{3}-\binom{4}{3}}{\binom{9}{3}}\times \frac{3}{6}=\frac{8}{84}$$
Case 3: The fourth block is blue.
Of the first three blocks, none are blue $\binom{5}{3}$, and they are not all red $\binom{3}{3}$. Then the probability of the fourth block being blue is $\frac{4}{6}$
$$\frac{\binom{5}{3}-\binom{3}{3}}{\binom{9}{3}}\times \frac{4}{6}=\frac{6}{84}$$
The total probability is $$\frac{10+8+6}{84}=\frac{24}{84}=\frac{2}{7}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Using calculus to prove an algebraic inequality Let $$f(x,y):=4\pi \left(\frac{y}{\tan(\frac{\pi}{2}x)}+\frac{x}{\tan(\frac{\pi}{2}y)}\right)+16(1+x+y)-\frac{16}{3}\frac{(x+y+xy)^2}{xy}. $$
Prove that $f(x,y) \ge 0$, for $0\le x <1, 0\le y<1$.
The figure of the function from WolframAlpha indicates the above claim is true. I'm wondering how to show this by hand?
| Using $\frac{2}{\pi}u \le \sin u \le u$
for all $u\in [0, \pi/2]$, we have
$$\frac{1}{\tan(\pi x/2)}
= \frac{\sin(\pi/2 - \pi x/2)}{\sin(\pi x/2)} \ge \frac{\frac{2}{\pi}\cdot (\pi/2 - \pi x/2) }{\pi x/2} = \frac{2}{\pi}\cdot \frac{1 - x}{x}.$$
Then, it suffices to prove that
$$4\pi \cdot \left(y\cdot \frac{2}{\pi}\cdot \frac{1 - x}{x} + x\cdot \frac{2}{\pi}\cdot \frac{1 - y}{y}\right) + 16(1 + x + y) - \frac{16}{3}(x + y + xy)^2/(xy) \ge 0$$
or
$$\frac{8(x + y + xy)(x + y - 2xy)}{3xy} \ge 0$$
which is clearly true.
We are done.
| {
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Trigonometric Substitution Absolute value issue Evaluate $ \, \displaystyle \int _{0}^{4} \frac{1}{(2x+8)\, \sqrt{x(x+8)}}\, dx. $
$My\ work:-$
by completing the square and substitution i.e. $\displaystyle \left(\begin{array}{rl}x+4 & = 4\sec (\theta )\\ dx & = 4\sec \theta \tan \theta \, d\theta \end{array}\right) \qquad$ $\Rightarrow \displaystyle \int \frac{4\sec \theta \tan \theta \, d\theta }{2(4\sec (\theta ))( |4\tan (\theta )|) }$ $\Rightarrow \displaystyle \frac{1}{8} \int \frac{\tan \theta \, d\theta }{|\tan (\theta )|}$ now because my limits are positive so $sec\ \theta \geq 0\ $ and $sec\ \theta\ $ is positive in $\ Ist\ $ and $\ IVth\ $ Quadrant. At this stage i have 2 options either i consider $\ Ist\ $ Quadrant and take postive $|tan\ \theta|\ =\ tan\ \theta\ $ or i consider $\ IVth\ $ Quadrant where $\ |tan\ \theta|\ = -tan\ \theta$ So when in 1st quadrant i.e. $\ |tan\ \theta|\ = tan\ \theta ,$ $\ 0\ \geq\ \theta\ \geq\ \pi/2\ $ i get
$\Rightarrow \displaystyle \frac{1}{8} \theta +C$ $\Rightarrow \displaystyle \frac{1}{8} \mathrm{arcsec}\left(\frac{x+4}{4}\right)+C$ $\Rightarrow \displaystyle \left. \frac{1}{8} \mathrm{arcsec}\left(\frac{x+4}{4}\right)\, \right|_{x=0}^{x=4}$ $\Rightarrow \displaystyle \frac{1}{8} \left(\mathrm{arcsec}(2)-\mathrm{arcsec}(1)\right)$
Now if i consider 4th quadrant i.e. $\ |tan\ \theta|\ = -tan\ \theta ,$ $\ 3 \pi /2 \ \geq\ \theta\ \geq\ 2\pi\ $ i get $\Rightarrow \displaystyle -\frac{1}{8} \theta +C$ $\Rightarrow \displaystyle -\frac{1}{8} \mathrm{arcsec}\left(\frac{x+4}{4}\right)+C$ $\Rightarrow \displaystyle \left. -\frac{1}{8} \mathrm{arcsec}\left(\frac{x+4}{4}\right)\, \right|_{x=0}^{x=4}$ $\Rightarrow \displaystyle -\frac{1}{8} \left(\mathrm{arcsec}(2)-\mathrm{arcsec}(1)\right)$ So i am getting positive value in 1st case whereas in 4th quadrant my answer is Negative. Why is so ? am i making some mistake when considering 4th quadrant ? or both are acceptable answer ? also in MIT lecture they said whatever acceptable quadrant you choose you'll get the same answer. so why i m getting Negative answer ?
| HINT
I propose another way to tackle this integral so that you can compare both methods.
Notice that $x(x+8) = x^{2} + 8x = (x + 4)^{2} - 16$.
Hence, if we make the substitution $4\cosh(z) = x + 4$, we arrive at
\begin{align*}
\int\frac{\mathrm{d}x}{(2x+8)\sqrt{x(x+8)}} & = \int\frac{4\sinh(z)}{8\cosh(z)\sqrt{16\cosh^{2}(z) - 16}}\mathrm{d}z\\\\
& = \frac{1}{8}\int\frac{\mathrm{d}z}{\cosh(z)}\\\\
& = \frac{1}{8}\int\frac{\cosh(z)}{\cosh^{2}(z)}\mathrm{d}z\\\\
& = \frac{1}{8}\int\frac{\cosh(z)}{\sinh^{2}(z) + 1}\mathrm{d}z
\end{align*}
where the absolute value was omitted because the function $\sinh(z)$ is positive whenever $z\geq 0$.
In the last expression, we can make the change of variable $w = \sinh(z)$, whence it results that
\begin{align*}
\int\frac{\mathrm{d}x}{(2x+8)\sqrt{x(x+8)}} = \frac{\arctan(w)}{8} + c
\end{align*}
Now it remains to apply the integration limits.
Can you take it from here?
| {
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How can I solve $\frac{(1-x)(x+3)}{(x+1)(2-x)}\leq0$ and express solution in interval notation? The inequality I need to solve is
$$\frac{(1-x)(x+3)}{(x+1)(2-x)}\leq0$$
My attempt:
Case I: $x<2$
Then we have $(1-x)(x+3)\leq0$
$x\leq-3, x\geq1$
We reject $x\leq-3$ (not sure why exactly, I just looked at the graph on desmos and it shows that $x\geq1$
Case II: $x<-1$
Then we have $(1-x)(x+3)\geq0$
$x\geq-3, x\leq-1$
Reject $x\leq-1$ (again, not sure exactly why aside from the graph I saw on desmos)
So the final solution is $x\in[-3,-1) \cup [1,2)$
I know that my final solution is correct, however I am a little confused on why I reject certain values.
| Case I: $\ (1-x)(x+3) \le 0$ and $(x+1)(2-x) > 0$
\begin{align}
&(1-x)(x+3) \le 0 &\text{AND} \qquad&(x+1)(2-x) > 0\\
&\Rightarrow\begin{cases}
x \ge 1\ \&\ x \ge -3 \text{ OR}\\
x \le 1\ \&\ x \le -3
\end{cases}&\text{AND} \qquad
&\Rightarrow\begin{cases}
x > -1\ \&\ x < 2 \text{ OR}\\
x < -1\ \&\ x > 2 \text{ (Impossible)}\\
\end{cases} \\
&\Rightarrow\begin{cases}
x \ge 1 \text{ OR}\\
x \le -3
\end{cases}&\text{AND} \qquad
&\Rightarrow
-1 < x < 2
\end{align}
$\therefore\ 1 \le x < 2$
Case II: $\ (1-x)(x+3) \ge 0$ and $(x+1)(2-x) < 0$
\begin{align}
&(1-x)(x+3) \ge 0 &\text{AND} \qquad&(x+1)(2-x) < 0\\
&\Rightarrow\begin{cases}
x \le 1 \ \&\ x \ge -3 \text{ OR}\\
x \ge 1 \ \&\ x \le -3 \text{ (Impossible)}
\end{cases}&\text{AND} \qquad
&\Rightarrow\begin{cases}
x < -1 \ \&\ x < 2 \text{ OR}\\
x > -1 \ \&\ x > 2 \\
\end{cases} \\
&\Rightarrow -3 \le x \le 1
&\text{AND} \qquad
&\Rightarrow\begin{cases}
x < -1 \text{ OR}\\
x > 2
\end{cases} \\
\end{align}
$\therefore\ -3 \le x < -1$
Hence, $x\in[-3,-1)\ \cup\ [1, 2)$.
| {
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Finding $x^8+y^8+z^8$, given $x+y+z=0$, $\;xy +xz+yz=-1$, $\;xyz=-1$
The system says
$$x+y+z=0$$
$$xy +xz+yz=-1$$
$$xyz=-1$$
Find
$$x^8+y^8+z^8$$
With the first equation I squared and I found that $$x^2+y^2+z^2 =2$$
trying with
$$(x + y + z)^3 =
x^3 + y^3 + z^3 + 3 x^2 y + 3 x y^2 + 3 x^2 z++ 3 y^2 z + 3 x z^2 +
3 y z^2 + 6 x y z$$
taking advantage of the fact that there is an $xyz=-1$ in the equation, but I'm not getting anywhere, someone less myopic than me.how to solve it?
Thanks
Edit : Will there be any university way to solve this problem , they posed it to a high school friend and told him it was just manipulations of remarkable products.
His answers I understand to some extent but I don't think my friend understands all of it.
| Note that $x$, $y$, $z$ are the roots of the polynomial $\lambda^3 - \lambda + 1$, so they are the eigenvalues of the companion matrix
$$A=\left(\begin{matrix}
0 & 0 & -1 \\
1 & 0 & 1\\
0 & 1 & 0
\end{matrix} \right)$$
Conclude that the eigenvalues $A^{8}$ are $x^8$, $y^8$, $y^8$. Calculate $A^8$ by squaring three times ( @Theo Bendit:'s idea in the comments). We get
$$A^8=\left(\begin{matrix}
2 & -2 & 3 \\
-3 & 4 & -5\\
2 & -3 & 4
\end{matrix} \right)$$
so the sum $x^8+y^8+z^8$ equals the trace of $A^8$, that is $10$.
| {
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"answer_count": 8,
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} |
Q: Epsilon-delta proof for $\lim_{x\to1} \frac{x}{x^2+1}=\frac{1}{2}$ In proving $\lim_{x\to1} \frac{x}{x^2+1}=\frac{1}{2}$, I have done this working:
Let $ϵ>0$.
$0<|x-1|< \Rightarrow|\frac{x}{(x^2+1)}-\frac{1}{2}|<$
$ |\frac{x}{(x^2+1)}-\frac{1}{2}|< = |\frac{2x-x^2+1}{2(x^2+1)}| = |\frac{-2(x-1)^2}{x(x-1)^2+4x}| = |\frac{-(x-1)^2}{(x-1)^2+2(x-1)+2}| = \frac{(x-1)^2}{|(x-1)^2+2(x-1)+2|}|$
after which I get stuck at trying to implement triangle inequality to push in the modulo in the denominator to get $|x-1|$ terms to substitute .
| Alternative approach:
Let
$$0 < |x - 1| < \delta \leq \frac{1}{10}.\tag1$$
Such a constraint on $\delta$ is viable by developing a relationship between $\epsilon$ and $\delta$ that looks like
$\displaystyle \delta = \min\left[f(\epsilon), \frac{1}{10}\right]~~$
rather than
$~~\displaystyle \delta = f(\epsilon).$
Using (1) above, you have that
$~~~\displaystyle 1 - \delta < x < 1 + \delta$.
Then, since $\displaystyle \delta \leq \frac{1}{10} \implies \delta^2 < \delta$
you have that $~~~\displaystyle 1 - 2\delta < x^2 < 1 + 2\delta + \delta^2 < 1 + 3\delta.$
This implies that $~~~\displaystyle 2 - 2\delta < x^2 + 1 < 2 + 3\delta.$
Then, using (1) above, and considering the minimum and maximum possible values of the numerator $(x)$, and denominator $(x^2 + 1)$, you have that
$$\frac{1 - \delta}{2 + 3\delta} < \frac{x}{x^2 + 1} < \frac{1 + \delta}{2 - 2\delta}. \tag2$$
$\displaystyle \frac{1 - \delta}{2 + 3\delta}
= \frac{1 + (3/2) \delta}{2 + 3\delta} - \frac{(5/2) \delta}{2 + 3\delta}
= \frac{1}{2} - \frac{5\delta}{4 + 6\delta} > \frac{1}{2} - 2\delta ~: ~\delta \leq \frac{1}{10}.
$
Similarly,
$~~\displaystyle \frac{1 + \delta}{2 - 2\delta}
= \frac{1 - \delta}{2 - 2\delta} + \frac{2 \delta}{2 - 2\delta}
= \frac{1}{2} + \frac{\delta}{1 - \delta} < \frac{1}{2} + 2\delta ~: ~\delta \leq \frac{1}{10}.
$
Therefore, using (2) above,
$$\frac{1}{2} - 2\delta < \frac{x}{x^2 + 1} < \frac{1}{2} + 2\delta \implies
- 2\delta < \frac{x}{x^2 + 1} - \frac{1}{2} < 2\delta.$$
This implies that
$$\left|\frac{x}{x^2 + 1} - \frac{1}{2}\right| < 2\delta.\tag3$$
Consequently, set $\displaystyle \delta = \min\left(\frac{\epsilon}{2}, \frac{1}{10}\right).$
Then, based on (1), (2), and (3) above,
$~~\displaystyle0 < |x-1| < \delta~~$ will imply that
$~~\displaystyle \left|\frac{x}{x^2 + 1} - \frac{1}{2}\right| < \epsilon.$
| {
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} |
Caculate $\iint_D \frac{x^2+y^2}{\sqrt{4-(x^2+y^2)^2}}dxdy$, with D:$\frac{x^2}{2}+y^2\leq1$ I found some difficulty with this exercise:
Calculate
$$\iint_D \frac{x^2+y^2}{\sqrt{4-(x^2+y^2)^2}}dxdy$$
with $D := \left\{(x,y)\in\mathbb{R}^{2}\mid \dfrac{x^2}{2}+y^2\leq1\right\}$
I use change of Variables in Polar Coordinates, but the integral become so hard to calculate.
I think maybe we change variables $u = x^2 + y^2$, the integral will be easier, but I can't find $v(x,y$) to have the Jacobi easy to calculate.
| Here is a possible way to evaluate explicitly.
We use the change of variables $x=r\sqrt 2\; \cos t$, $y=r\sin t$. Then we have formally
$$
\begin{aligned}
dx &= \sqrt 2\; \cos t\; dr - r\sqrt 2\; \sin t\; dt\ ,\\
dy &= \sin t\; dr + r\; \cos t\; dt\ ,\\
dx\wedge dy &=
dr\wedge dt\cdot
\sqrt 2\; \cos t\cdot r\; \cos t
- dt\wedge dr\cdot r\sqrt 2\; \sin t
\cdot
\sin t
\\
&=
dr\wedge dt\cdot r\sqrt 2\cdot (
\cos^2 t +\sin^2 t)
\\
&=
dr\wedge dt\cdot r\sqrt 2\ .\\[2mm]
x^2 + y^2
&=2r^2\cos ^2t + r^2\sin^2 t\\
&=r^2(1+\cos^2 t)\ .
\end{aligned}
$$
So the given integral $I$ can be computed as follows:
$$
\begin{aligned}
I
&=\iint_D \frac{x^2+y^2}{\sqrt {4-(x^2+y^2)^2}}\; dx\; dy
\\
&=4
\iint_{substack{(x,y)\in D\\x,y\ge 0}}
\frac{x^2+y^2}{\sqrt {4-(x^2+y^2)^2}}\; dx\; dy
\\
&=
4
\iint_{\substack{0\le r\le 1\\0\le t\le \pi/2}}
\frac{r^2(1+\cos^2 t)}{\sqrt{4-r^4(1+\cos^2 t)^2}}
\;r\sqrt 2\; dr\; dt
\\
&\qquad\text{ ... now use $s=r^4$, $ds=4r^3\; dr$}
\\
&=
\sqrt 2
\iint_{\substack{0\le s\le 1\\0\le t\le \pi/2}}
\frac
{\color{blue}{1+\cos^2 t}}
{\sqrt{4-s\color{red}{(1+\cos^2 t)^2}}}
\;ds\; dt
\\
&\qquad\text{ ... now use
$\int_0^1\frac{ds}{\sqrt{4-s\color{red}{a}}}
=\frac 1{\color{red}{a}}(4-2\sqrt {4-\color{red}{a}})$}
\\
&=
\sqrt 2
\int_0^{\pi/2}
\frac{\color{blue}{1+\cos^2 t}}{\color{red}{(1+\cos^2 t)^2}}
\left(4-2\sqrt{4-\color{red}{(1+\cos^2 t)^2}}\right)
\; dt
\\
&=
2\sqrt 2
\int_0^{\pi/2}
\frac1{1+\cos^2 t}
\left(2-\sqrt{4-(1+\cos^2 t)^2}\right)
\; dt
\\
&=
2\pi
-
2\sqrt 2
\int_0^{\pi/2}
\frac1{1+\cos^2 t}
\sqrt{2^2-(1+\cos^2 t)^2}
\; dt
\\
&=
2\pi
-
2\sqrt 2
\int_0^{\pi/2}
\frac1{1+\cos^2 t}
\sqrt{(3+\cos^2 t)(1-\cos^2 t)}
\; dt
\\
&=
2\pi
-
2\sqrt 2
\int_0^{\pi/2}
\frac1{1+\cos^2 t}\cdot
\sqrt{3+\cos^2 t}
\; \sin t\; dt
\\
&\qquad\text{ ... now use $u=\cos t$}
\\
&=
2\pi
-
2\sqrt 2
\int_0^1
\frac1{1+u^2}\cdot
\sqrt{3+u^2}
\; du
\\
&=
\color{forestgreen}{
2\pi
-
\sqrt 2\log 3
-
4\arctan\frac 1{\sqrt 2}}\ .
\end{aligned}
$$
$\square$
Numerical check:
sage: 4 * numerical_integral( lambda x:
....: numerical_integral( lambda y :
....: (x^2 + y^2) / sqrt(4 - (x^2 + y^2)^2),
....: (0, sqrt(1-x^2/2)) )[0],
....: (0, sqrt(2)) )[0]
2.2675940740738505
sage: ( 2*pi - sqrt(2)*log(3) - 4*atan(1/sqrt(2)) ).n()
2.26759407407385
| {
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"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find all values of $x$ such that $\frac{4x-5}{3x+5}\geq3$ Find all values of $x$ such that $\frac{4x-5}{3x+5}\geq3$.
If we consider two cases: $3x+5 > 0$ for the first and $3x + 5 < 0$ for the second.
When $3x+5 > 0$:
$\frac{4x-5}{3x+5}\geq3$.
Sloving for $x$ we find $x\leq -4$.
When $3x+5 < 0$:
$\frac{4x-5}{3x+5}\geq3$.
Sloving for $x$ we find $x\geq -4$
Additionaly through trail and error I discovered $x=-2$, $x=-3$ and $x=-4$... satisfy the original inequality.
$-2$, $-3$, $-4$ satisfy the inequality chain $-4 \leq x < -5/3$. All other values the satisify the inequality chain are also solutions.
In the case that $3x+5 > 0$, the inequality chain $-4 \geq x > -5/3$ is illogical.
Is this soultion correct? How can I improve it?
| Your way to proceed also works, indeed we have
*
*for $3x+5>0 \iff x>-\frac 5 3$
$$\frac{4x-5}{3x+5}\geq3 \iff 4x-5\ge 9x+15 \iff 5x \le -20 \iff x \le -4$$
that is no solution for this first case, then
*
*for $3x+5<0 \iff x<-\frac 5 3$
$$\frac{4x-5}{3x+5}\geq3 \iff 4x-5\le 9x+15 \iff 5x \ge -20 \iff x \ge -4$$
that is $x \in \left[-4,-\frac 5 3\right)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\operatorname{Mat}_2(\mathbb{R})$ as a field I'm trying to solve this question:
Prove that the set of $2 \times 2$-matrices of the form
$$
\left(\begin{array}{ll}
a & b \\
c & d
\end{array}\right), \quad a, b, c, d \in \mathbb{R}
$$
with the usual matrix addition and multiplication is a non-commutative ring (that is, the multiplication is not commutative). Can you find conditions on $a, b, c, d$ to make it into a field?
I already did the first part and showed that this is a non-commutative ring. Now, I'm trying to find the conditions to make it be a field. For this, we just have to find the multiplicative inverse, which from linear algebra we know that a $2\times 2$ matrix is invertible if $\operatorname{det} (A) \neq 0$, which means $ad-bc \neq 0$ or equivalently, one row isn't a multiple of the other. Now, I just want to know that if this is enough!?
| This is how I finally did:
if $a=d$ and $b=-c$, the set $SM_2(\mathbb{R})=\bigg\{\begin{pmatrix} a & -b \\ b & a\\ \end{pmatrix} \bigg| a, b \in \mathbb{R}\bigg\}$ turns into a field because:
$SM_2(\mathbb{R})$ is commutative, because:
\begin{equation*}
\begin{aligned}
\begin{pmatrix} a & -b \\ b & a\\ \end{pmatrix} \begin{pmatrix} c & -d \\ d & c\\ \end{pmatrix} &= \begin{pmatrix} ac-bd & -ad-bc \\ bc+ad & -bd+ac\\ \end{pmatrix}
\\ &= \begin{pmatrix} c & -d \\ d & c\\ \end{pmatrix} \begin{pmatrix} a & -b \\ b & a\\ \end{pmatrix}
\end{aligned}
\end{equation*}
The multiplicative inverse exists and is the inverse of matrix $A$, because $\det(A)=a^2-(-b)b=a^2+b^2\neq 0$, so $A$ is invertible and $A^{-1}=\frac{1}{a^2+b^2}\begin{pmatrix} a & b \\ -b & a\\ \end{pmatrix}$ and
\begin{equation*}
\begin{aligned}
\begin{pmatrix} a & -b \\ b & a\\ \end{pmatrix} \begin{pmatrix} \frac{a}{a^2+b^2} & \frac{b}{a^2+b^2} \\ \frac{-b}{a^2+b^2} & \frac{a}{a^2+b^2}\\ \end{pmatrix}
&= \begin{pmatrix} \frac{a^2+b^2}{a^2+b^2} & \frac{ab-ba}{a^2+b^2} \\ \frac{ba-ab}{a^2+b^2} & \frac{b^2+a^2}{a^2+b^2}\\ \end{pmatrix}
\\ &= \begin{pmatrix} 1 & 0 \\ 0 & 1\\ \end{pmatrix}
\end{aligned}
\end{equation*}
Thus, $SM_2(\mathbb{R})$ is a field. Plus, it cannot be an ordered field. For this, consider $\varphi: SM_2(\mathbb{R}) \longrightarrow \mathbb{C}$ such that $\varphi\left( \begin{pmatrix}
a & -b \\ b & a\\
\end{pmatrix}\right)=a+bi$. We have:
For any $A, B \in SM_2(\mathbb{R})$, we have:
\begin{equation*}
\begin{aligned}
\varphi(AB)&= \varphi\left( \begin{pmatrix}
a & -b \\ b & a\\
\end{pmatrix}\begin{pmatrix}
c & -d \\ d & c\\
\end{pmatrix}\right)= \varphi\left( \begin{pmatrix} ac-bd & -ad-bc \\ bc+ad & -bd+ac\\ \end{pmatrix} \right) \\
&= (ac-bd)+(ad+bc)i= ac+adi+bci+bdi^2\\
&= a(c+di)+bi(c+di)= (a+bi)(c+di)\\
&= \varphi\left( \begin{pmatrix}
a & -b \\ b & a\\
\end{pmatrix}\right) \varphi\left(\begin{pmatrix}
c & -d \\ d & c\\
\end{pmatrix}\right)=\varphi(A)\varphi(B)
\end{aligned}
\end{equation*}
So, $\varphi$ is a homomorphism.
For any $A, B \in SM_2(\mathbb{R})$, assume that $\varphi(A)=\varphi(B)$, then $a+bi=c+di$, so $(a-c)+(b-d)i=0$, and hence, $a=c$ and $b=d$ which means that $A=\begin{pmatrix}
a & -b \\ b & a\\
\end{pmatrix}= \begin{pmatrix}
c & -d \\ d & c\\
\end{pmatrix} =B$. So, $\varphi$ is injective.
Now, let $a+bi\in \mathbb{C}$. Then, since we defined matrices in $SM_2(\mathbb{R})$ to be all the $2\times 2$ matrices with $a=d$, and $b=-c$, there exists $A=\begin{pmatrix}
a & -b \\ b & a\\
\end{pmatrix} \in SM_2(\mathbb{R})$ such that $\varphi\left( \begin{pmatrix}
a & -b \\ b & a\\
\end{pmatrix}\right)=a+bi$.
So $\varphi$ is an isomorphism. Hence, $SM_2(\mathbb{R})$ is isomorphic to $\mathbb{C}$ and since no order can be defined on $\mathbb{C}$, the aforementioned field cannot take any orders as well.
| {
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"answer_count": 1,
"answer_id": 0
} |
Evaluate $\tan(\frac{1}{2}\sin^{-1}\frac{3}{4})$ Evaluate $\tan(\frac{1}{2}\sin^{-1}\frac{3}{4})$
My attempt:
Method:- 1
Let $\sin^{-1}\frac{3}{4} = \theta \implies \sin\theta = \frac{3}{4}$ and $\theta \in [0,\frac{π}{2}]$
Therefore $$\cos\theta = \sqrt{1-\sin^2\theta} = \frac{\sqrt 7}{4}$$
Now $\tan(\frac{1}{2}\sin^{-1}\frac{3}{4}) = \tan\frac{\theta}{2} = \frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}} = \sqrt{ \frac{2\sin^2\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}}} = \sqrt{\frac{1-\cos\theta}{1+\cos\theta}} = \frac{4-\sqrt7}{3}$
Method:-2
Let $\sin^{-1}\frac{3}{4} = \theta \implies \sin\theta = \frac{3}{4} \implies 2\sin\frac{\theta}{2} \cos\frac{\theta}{2} = \frac{3}{4} \implies
\sin\frac{\theta}{2} \cos\frac{\theta}{2} = \frac{3}{8}$
Now $\tan\frac{\theta}{2} = \frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}$
Please help me in 2nd method. Thanks in advance.
| Your Method 2:
Since $$\sin\theta=2\sin\frac\theta2\cos\frac\theta2\\=2\tan\frac\theta2\cos^2\frac\theta2\\=\frac{2\tan\frac\theta2}{\sec^2\frac\theta2}\\=\frac{2\tan\frac\theta2}{1+\tan^2\frac\theta2}$$ and $$\sin\theta=\frac34,$$ therefore $$\frac{2\tan\frac\theta2}{1+\tan^2\frac\theta2}=\frac34\\3\tan^2\frac\theta2-8\tan\frac\theta2+3=0\\\tan\frac\theta2=\frac{4-\sqrt7}3.$$
| {
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Is this function can be differentiable at (0, 0)? Let we have $$f(x, y) = \left\{\begin{matrix}
\log(1 + x\sin\sqrt[3]{\frac{y^4}{x}}) & x \neq 0\\
0 & x = 0
\end{matrix}\right.$$
I wanna know if this fuction can be differentiated in $(0, 0)$.
Here's what I have:
$f'_x = f'_y = 0$ cuz $f(x, 0) = f(0, y) = 0$
In this case we only need to prove that $f(x, y) = o(||h||)$ for $h \to 0$
$$\lim_{(x, y) \to (0, 0)} \frac{\log(1 + x\sin\sqrt[3]{\frac{y^4}{x}})}{\sqrt{x^2 + y^2}} = 0$$
But how can I find this limit?
| We have that by squeeze theorem
$$\left|x\sin\sqrt[3]{\frac{y^4}{x}}\right| \le |x| \to 0$$
therefore
$$ \frac{\log\left(1 + x\sin\sqrt[3]{\frac{y^4}{x}}\right)}{\sqrt{x^2 + y^2}}=\frac{\log\left(1 + x\sin\sqrt[3]{\frac{y^4}{x}}\right)}{ x\sin\sqrt[3]{\frac{y^4}{x}}}\frac{x\sin\sqrt[3]{\frac{y^4}{x}}}{\sqrt{x^2 + y^2}}$$
with $\frac{\log\left(1 + x\sin\sqrt[3]{\frac{y^4}{x}}\right)}{ x\sin\sqrt[3]{\frac{y^4}{x}}}\to 1$ therefore all boils down in the following
$$\lim_{(x, y) \to (0, 0)} \frac{x\sin\sqrt[3]{\frac{y^4}{x}}}{\sqrt{x^2 + y^2}}=0$$
indeed assuming wlog $x>0$ we have
$$\frac{x\sin\sqrt[3]{\frac{y^4}{x}}}{\sqrt{x^2 + y^2}} \le \frac{x\sqrt[3]{\frac{y^4}{x}}}{\sqrt{x^2 + y^2}}=\frac{\sqrt[3]{y^4x^2}}{\sqrt{x^2 + y^2}}\to 0$$
indeed by polar coordinates
$$\frac{\sqrt[3]{y^4x^2}}{\sqrt{x^2 + y^2}}=\rho \sqrt[3]{\sin^4 \theta \cos^2 \theta} \to 0$$
| {
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How to find the lower bound of $f(x, y) = -2xy+x^2y^2+x^2$? $f(x, y) = -2xy+x^2y^2+x^2, x\in \mathbb{R}, y\in \mathbb{R}$, how to find its lower bound?
Here are my thoughts, I don't know if it is rigorous.
$f(x, y) = -2xy+x^2y^2+x^2=(xy-1)^2+x^2-1$, as $(xy-1)^2 \geq 0$, $x^2\geq0$, and they can not get to $0$ at the same time, therefore $f(x, y) > -1$.
Therefore, the lower bound of $f(x, y)$ is $-1$.
| As you stated the function is $f(x, y)=(xy-1)^2+x^2-1$, and $x^2\ge0$, and $(xy-1)^2\ge0$. so it has a minimum when both of the inequalities are at their smallest
now $x^2$ is minimised when $x=0$, and at $x=0$, $(xy-1)^2=(-1)^2=1$, therefore $f(0,y)=0+1-1=0$,
But looking at $(x,y)=(\frac{1}{2},1)$, therefore $f(\frac{1}{2},1)=(\frac{1}{2}-1)^2+\frac{1}{2}^2-1=\frac{1}{4}+\frac{1}{4}-1=-\frac{1}{2}$ so minimising along $x=0$ won't give us the minima of the function
So maybe tying along the line where $(xy-1)^2$ is minimised (ie when $xy-1=0$) and seeing if it also allows you to minimises $x^2$
hopefully that helps
| {
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Proving an identity relating to binomial coefficients This question is from the book Discrete mathematics and its applications, by K. Rosen, 6th chapter and 27th question.
Show that if n is a positive
integers $2C(2n, n+1) + 2C(2n, n) =
C(2n+2,n+1)$
I know how to prove this algebraically, applying the formula and this is how it is also given in hints. Since I want to understand counting arguments I am thinking to give a proof of this using combinatorial arguments (double counting).
The right side is the number of ways to select n+1 elements from 2n+2 elements. But I am not able to get an argument for the left side. I'm not getting how can I argue that I need to choose n or n+1 elements from 2n elements twice.
Any other (counting) approaches are also fine. Thank you.
| Here we show the validity of the binomial identity
\begin{align*}
\color{blue}{2\binom{2n}{n+1}+2\binom{2n}{n}=\binom{2n+2}{n+1}}\tag{1}
\end{align*}
by counting lattice paths. We consider lattice paths with horizontal steps $(1,0)$ and vertical steps $(0,1)$. The binomial coefficient
\begin{align*}
\binom{n}{k}\tag{2}
\end{align*}
gives the number of lattice paths of length $n$ from $(0,0)$ to $(k,n-k)$ consisting of $k$ horizontal $(1,0)$ steps and $n-k$ vertical $(0,1)$ steps.
Let's have a look at the graphic below. We see an $(n+1)\times(n+1)$ rectangle where we study lattice paths from the origin $(0,0)$ to $(n+1,n+1)$.
We observe
*
*The number of lattice paths from $(0,0)$ to $(n+1,n+1)$ is according to (2) equal to
\begin{align*}
\color{blue}{\binom{2n+2}{n+1}}\tag{3.1}
\end{align*}
*A lattice path from $(0,0)$ to $(n+1,n+1)$ has to cross either the blue point $(n,n)$ or one of the orange points $(n-1,n+1)$ or $(n+1,n-1)$ respectively. This way we can partition the lattice paths into three sets.
*
*Blue point $(n,n)$: The number of lattice path from $(0,0)$ to the blue point $(n,n)$ is $\binom{2n}{n}$. From the blue point $(n,n)$ there are two ways to go to $(n+1,n+1)$. Either by a horizontal step $(n,n)\to(n+1,n)\to (n+1,n+1)$ or by a vertical step $(n,n)\to(n,n+1)\to(n+1,n+1)$. So we have two possibilities giving a total of
\begin{align*}
\color{blue}{2\binom{2n}{n}}\tag{3.2}
\end{align*}
lattice paths.
*Orange point $(n+1,n-1)$: The number of lattice paths from $(0,0)$ to the orange point $(n+1,n-1)$ is
\begin{align*}
\binom{2n}{n-1}=\color{blue}{\binom{2n}{n+1}}\tag{3.3}
\end{align*}
From $(n+1,n-1)$ there is just one possibility to go to $(n+1,n+1)$: namely $(n+1,n-1)\to(n+1,n)\to(n+1,n+1)$. The total number of lattice paths from $(0,0)$ to $(n+1,n+1)$ crossing $(n+1,n-1)$ is therefore $\binom{2n}{n-1}=\binom{2n}{n+1}$.
*Orange point $(n-1,n+1)$: A symmetrical situation. The number of lattice paths from $(0,0)$ to the orange point $(n-1,n+1)$ is
\begin{align*}
\color{blue}{\binom{2n}{n+1}}\tag{3.4}
\end{align*}
Since there is just one possibility to go from $(n-1,n+1)$ to $(n+1,n+1)$: namely $(n-1,n+1)\to(n,n+1)\to(n+1,n+1)$, the total number of lattice paths from $(0,0)$ to $(n+1,n+1)$ crossing $(n-1,n+1)$ is $\binom{2n}{n+1}$.
Putting (3.1) to (3.4) together we see
\begin{align*}
\color{blue}{2\binom{2n}{n+1}+2\binom{2n}{n}=\binom{2n+2}{n+1}}
\end{align*}
and the claim (1) follows.
| {
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"url": "https://math.stackexchange.com/questions/4256337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Find value of $x^{5} + y^{5} + z^{5}$ given the values of $x + y + z$, $x^{2} + y^{2} + z^{2}$ and $x^{3} + y^{3} + z^{3}$ If$$x+y+z=1$$
$$x^2+y^2+z^2=2$$
$$x^3+y^3+z^3=3$$
Then find the value of $$x^5+y^5+z^5$$
Is there any simple way to solve this problem ? I have tried all my tricks tried to multiply two equations , substitute $z=1-x-y$ , but things got messy nothing seems to work out .
| Here is yet another way, in case you are interested:
$$(x+y+z)^2=1\implies\sum x^2+2\sum xy=1\implies \sum xy=-\frac12$$
Therefore the polynomial equation whose roots are $x,y,z$ takes the form:
$$t^3-t^2-\frac12t+c=0$$
Summing this equation for each of the roots,
$$\sum x^3-\sum x^2-\frac12\sum x +3c=0$$
$$\implies3-2-\frac12(1)+3c=0\implies c =-\frac16$$
Multiplying the polynomial by $t$ and summing again,
$$\sum x^4-\sum x^3-\frac12\sum x^2-\frac16\sum x=0$$
$$\implies\sum x^4=3+\frac12(2)+\frac16(1)$$
$$\implies \sum x^4=\frac{25}{6}$$
Multiplying the polynomial by $t^2$ and summing again,
$$\sum x^5-\sum x^4-\frac12\sum x^3-\frac16\sum x^2=0$$
$$\implies\sum x^5=\frac{25}{6}+\frac12(3)+\frac16(2)=6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4256912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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$\int_0^\infty x^5e^{-x^3}\ln(1+x)dx$ Can anyone help me to cope with this integral? I have tried solving it but I had no breakthrough whatsoever ...
$$\int_0^\infty x^5e^{-x^3}\ln(1+x)dx$$
| \begin{align}
\int_0^\infty x^5 e^{-x^3} \ln(1+x)\,dx
&= -\frac{1}{3} e^{-x^3}\left(x^3+1\right)\ln(1+x)\Big|_0^\infty + \frac{1}{3} \int_0^\infty \frac{e^{-x^3}\left(x^3+1\right)}{1+x}\,dx \\
&= \frac{1}{3} \int_0^\infty \frac{e^{-x^3}\left(x^3+1\right)}{1+x}\,dx \\
&= \frac{1}{3} \int_0^\infty e^{-x^3}(x^2 - x + 1)\,dx \\
&= \frac{1}{3} \left[\int_0^\infty x^2e^{-x^3}\,dx - \int_0^\infty xe^{-x^3}\,dx + \int_0^\infty e^{-x^3}\,dx\right] \\
&= \frac{1}{3} \left[\frac{1}{3}\int_0^\infty e^{-u}\,du - \frac{1}{3}\int_0^\infty u^{-\frac{1}{3}}e^{-u}\,du + \int_0^\infty u^{-\frac{2}{3}}e^{-u}\,du\right] \\
&= \frac{1}{9} \left[1-\Gamma\left(\frac{2}{3}\right) + \Gamma\left(\frac{1}{3}\right)\right]
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4260086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A combinatorial question about odd and even numbers Consider an $n$ tuple. Each index of the tuple is filled with a symbol chosen from the following set comprising of four symbols:
\begin{equation}
S = \{a, b, c, d\}.
\end{equation}
Note that there are $4^{n}$ possible fillings.
I want to count the number of possible tuples such that $b$-s, $c$-s, and $d$-s all occur in it an even number of times (not necessarily the same even number for each.)
For example, if $n = 4$, then $bbba$ is not a valid tuple --- as it has $3$ $b$-s, which is an odd number. However $bb cc$ is a valid tuple — as both $b$ and $c$ occur an even number of times.
For $n = 2$, the valid choices are $aa, bb, cc, dd$.
Hence, the number of such tuples is $4$.
For $n = 3$, the valid choices are $aaa, abb, acc, add, bba, cca, dda, bab, cac, dad$.
So, $10$ choices.
Is there a general pattern?
| Since this is tagged combinatorial-proofs, here’s a purely bijective solution. Let the “signature” of a tuple be the numbers of $b$s, $c$s, and $d$s mod 2, and let $x_s$ be the number of tuples with signature $s$ (so we seek $x_{000}$).
By the symmetry $b → c → d → b$, we have $x_{100} = x_{010} = x_{001}$ and $x_{110} = x_{101} = x_{011}$. We can count all $4^n$ tuples as follows:
$$x_{000} + 3x_{100} + 3x_{011} + x_{111} = 4^n. \tag{1}$$
Consider the transformation that flips the first letter that’s $a$ or $b$ to $b$ or $a$, respectively. This is a self-inverse transformation that exchanges signatures $000 ↔ 100$, $010 ↔ 110$, $001 ↔ 101$, $011 ↔ 111$, except that it fails on the $2^n$ tuples made entirely from $c$ and $d$. If $n$ is even, the signatures of these failing tuples are half $000$ and half $011$; if $n$ is odd, they’re instead half $010$ and half $001$. So:
\begin{gather*}
x_{000} - x_{100} = 2^{n - 2} + (-2)^{n - 2}, \tag{2} \\
x_{011} - x_{111} = 2^{n - 2} + (-2)^{n - 2}, \tag{3} \\
x_{010} - x_{110} = 2^{n - 2} - (-2)^{n - 2}, \tag{4} \\
x_{001} - x_{101} = 2^{n - 2} - (-2)^{n - 2}. \tag{5} \\
\end{gather*}
By $\text{(1)} + 7·\text{(2)} + \text{(3)} + 4·\text{(4)}$, we have
$$8x_{000} = 4^n + 8(2^{n - 2} + (-2)^{n - 2}) + 4(2^{n - 2} - (-2)^{n - 2}), \\
x_{000} = \frac{4^n + 3·2^n + (-2)^n}{8}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4266460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Non-probabilistic/non-combinatoric proof of $\sum_{k=n}^{2n} \frac{{k \choose n}}{2^k}=1$ The summation $$S_n=\sum_{k=n}^{2n} \frac{{k \choose n}}{2^k}=1~~~~(1)$$
has been proved using probabilistic/combinatoric method in MSE earlier:
Combinatorial or probabilistic proof of a binomial identity
Here we give an analytic proof for (1):
$$S_n=\sum_{k=n}^{2n} \frac{{k-1 \choose n-1}+{k-1 \choose n}}{2^k}~~~~~(2)$$
Let $k-1=p$, then
$$S_n=\sum_{p=n-1}^{2n-1} \frac{{p \choose n-1}+{p \choose n}}{2^{p+1}}~~~~(3)$$
Take the first term
$$\sum_{p=n-1}^{2n-1} \frac{{p \choose n-1}}{2^{p+1}}=\sum_{p=n-1}^{2n-2} \frac{{p \choose n-1}}{2^{p+1}}+\frac{{2n-1 \choose n-1}}{2^{2n-1}}=\frac{S_{n-1}}{2}+\frac{{2n-1 \choose n-1}}{2^{2n}}~~~~~(4)$$
Now take the second term
$$\sum_{p=n-1}^{2n-1} \frac{{p \choose n}}{2^{p+1}}=\sum_{p=n}^{2n} \frac{{p \choose n}}{2^{p+1}}+\frac{{n-1 \choose n}}{2^{n}}-\frac{{2n \choose n}}{2^{2n+1}}~~~~(5)$$
The second term in RHS vanishes and the third term is equal and opposite to the second term of the RHS of (4). Adding (4) and (5), we get
$$S_n=\frac{S_{n-1}}{2}+\frac{S_{n}}{2} \implies S_n=S_{n-1}$$So $S_n$ is a constant (independent of $n$) we can have $S_n=S_{n-1}=S_1=1$
The question is: What could be other analytic proofs of (1) without using the ideas of probability/combinatorics?
| We seek to show that
$$\sum_{k=n}^{2n} 2^{-k} {k\choose n} = 1$$
or alternatively
$$S_n = \sum_{k=0}^n 2^{-k} {k+n\choose n} = 2^n.$$
Using an Iverson bracket we find
$$\sum_{k\ge 0} 2^{-k} {k+n\choose n}
[z^n] \frac{z^k}{1-z}
= [z^n] \frac{1}{1-z}
\sum_{k\ge 0} 2^{-k} {k+n\choose n} z^k
\\ = [z^n] \frac{1}{1-z}
\frac{1}{(1-z/2)^{n+1}}.$$
This is
$$\mathrm{Res}_{z=0} \frac{1}{z^{n+1}} \frac{1}{1-z}
\frac{1}{(1-z/2)^{n+1}}
\\ = (-1)^n 2^{n+1} \mathrm{Res}_{z=0}
\frac{1}{z^{n+1}} \frac{1}{z-1}
\frac{1}{(z-2)^{n+1}}.$$
Residues sum to zero and the residue at infinity is zero by inspection.
Hence we take minus the sum of the residues at $z=1$ and $z=2.$ We
obtain for $z=1$
$$-(-1)^n 2^{n+1} (-1)^{n+1} = 2^{n+1}.$$
We write for $z=2$
$$- (-1)^n 2^{n+1} \mathrm{Res}_{z=2}
\frac{1}{((z-2)+2)^{n+1}} \frac{1}{(z-2)+1}
\frac{1}{(z-2)^{n+1}}
\\ = (-1)^{n+1} \mathrm{Res}_{z=2}
\frac{1}{(1+(z-2)/2)^{n+1}} \frac{1}{(z-2)+1}
\frac{1}{(z-2)^{n+1}}.$$
This yields
$$(-1)^{n+1} \sum_{q=0}^n (-1)^q 2^{-q} {n+q\choose n} (-1)^{n-q}
= - \sum_{q=0}^n 2^{-q} {n+q\choose n} = - S_n.$$
We have shown that $S_n = 2^{n+1} - S_n$ or $S_n = 2^n$ as claimed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4266830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Proving that expression can be expressed as a sum of $3n-1$ natural square numbers that are bigger than one $n$ is a natural number. Prove that
$(2n+1)^3 - 2$ can be expressed as sum of $3n - 1$ natural square numbers, which are bigger than one.
Is this a valid proof?
Showing that $((4+1)^3 - 2) - ((2+1)^3 - 2)=98= 3^2 + 5^2 + 8^2$ can be expressed as a sum of $(6-1)-(3-1) = 3$ natural square numbers that are bigger than one. Then showing for $n=k+1$ and $n=k$, that it can be expressed as $(3(k+1) - 1) - (3k-1) = 3$ natural square numbers?
| You have made a good start on using induction to prove the hypothesis that
$$p(n) = (2n + 1)^3 - 2 \tag{1}\label{eq1A}$$
can be expressed as the sum of $3n - 1$ perfect squares, each greater than $1$. First, the base case is $n = 1$, where $p(1) = 3^3 - 2 = 25$, which can be expressed as the sum of $3n - 1 = 2$ perfect squares as $25 = 9 + 16 = 3^2 + 4^2$.
Next, assume the hypothesis is true for $n = k$ for some $k \ge 1$. Then, as you indicated, showing that $p(k+1) - p(k)$ can be expressed as the sum of $3$ squares means these squares can be added to the $3k - 1$ squares already used for $p(k)$, thus showing that $p(k + 1)$ is a sum of $(3k - 1) + 3 = 3(k + 1) - 1$ perfect squares. Note that
$$\begin{equation}\begin{aligned}
p(k + 1) - p(k) & = ((2(k + 1) + 1)^3 - 2) - ((2k + 1)^3 - 2) \\
& = (2k + 3)^3 - (2k + 1)^3 \\
& = 8k^3 + 3(4)(3)k^2 + 3(2)(9)k + 27 - (8k^3 + 3(4)k^2 + 3(2)k + 1) \\
& = 8k^3 + 36k^2 + 54k + 27 - (8k^3 + 12k^2 + 6k + 1) \\
& = 24k^2 + 48k + 26 \\
& = (4k^2 + 4k + 1) + (4k^2 + 12k + 9) + (16k^2 + 32k + 16) \\
& = (2k + 1)^2 + (2k + 3)^2 + (4k + 4)^2
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Since $k \ge 1$, each perfect square above is $\gt 1$. As explained earlier, this means the hypothesis is also true for $n = k + 1$ so, by induction, it's true for all $n \ge 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why is $\frac{1}{m+1} = \sum_{n=0}^\infty \frac{2^{-2n-m-1} (2n+m)!}{(n+m+1)!n!}$? I have the strong assumption that for $m \in \mathbb N_0$ we have
$$\frac{1}{m+1} = \sum_{n=0}^\infty \frac{2^{-2n-m-1} (2n+m)!}{(n+m+1)!n!}=\sum_{n=0}^\infty \frac{2^{-2n-m-1}}{2n+m+1}\binom{2n+m+1}{n}.$$
But I didn't find an explicit way to show it.
In case you wanna know where this series is from: I'm still trying to solve How to prove an equality involving an infinity series and the modified Bessel functions?. If you plug in $s=\alpha=\beta=1$ you eventually come up with the series equation from above. At least the series equation would show the special case (in fact the other direction is not true).
| I guess I post my approch which is based on the idea of Claude Leibovici for clarity.
Using Legendre duplication we have
$$(2n+m)!=\Gamma(2n+m+1) = \Gamma(n+ \tfrac{m+1}{2})\Gamma(n+ \tfrac{m+2}{2})2^{2n+m}/\sqrt{\pi}.$$
So we have
$$\sum_{n=0}^\infty \frac{2^{-2n-m-1} (2n+m)!}{(n+m+1)!n!}=\frac{1}{2\sqrt{\pi}} \sum_{n=0}^\infty \frac{ \Gamma(n+ \tfrac{m+1}{2})\Gamma(n+ \tfrac{m+2}{2})}{\Gamma(n+m+2)n!}\\
= \frac{1}{2\sqrt{\pi}} \frac{ \Gamma(\tfrac{m+1}{2})\Gamma(\tfrac{m+2}{2})}{\Gamma(m+2)} F(\tfrac{m+1}{2},\tfrac{m+2}{2},m+2,1)\\
\frac{1}{2\sqrt{\pi}} \frac{ \Gamma(\tfrac{m+1}{2})\Gamma(\tfrac{m+2}{2})}{\Gamma(m+2)} \frac{\Gamma(m+2)\Gamma(m+2-\tfrac{m+1}{2}-\tfrac{m+2}{2})}{\Gamma(m+2-\tfrac{m+1}{2})\Gamma(m+2-\tfrac{m+2}{2})}\\
= \frac{1}{2\sqrt{\pi}} \frac{ \Gamma(\tfrac{m+1}{2})\Gamma(\tfrac{m+2}{2})}{\Gamma(m+2)} F(\tfrac{m+1}{2},\tfrac{m+2}{2},m+2,1)\\
\frac{1}{2\sqrt{\pi}} \frac{ \Gamma(\tfrac{m+1}{2})\Gamma(\tfrac{m+2}{2})}{\Gamma(m+2)} \frac{\Gamma(m+2)\Gamma(\tfrac{1}{2})}{\Gamma(\tfrac{m}{2}+\tfrac{3}{2})\Gamma(\tfrac{m}{2}+\tfrac{2}{2})}
=\frac{1}{m-1}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\lim_{n \to \infty}\left[\,\sqrt{\,2\,}\,\frac{\Gamma\left(n/2 + 1/2\right)}{\Gamma\left(n/2\right)} - \,\sqrt{\,n\,}\right]$
I am trying to find this limit:
$$\lim_{n \to \infty}\left[\,\sqrt{\,2\,}\,\frac{\Gamma\left(n/2 + 1/2\right)}{\Gamma\left(n/2\right)} - \,\sqrt{\,n\,}\right]$$
I know that
$$\frac{\sqrt{2}\Gamma(\frac{n+1}{2})}{\Gamma(\frac{n}{2})}=\begin{cases}\sqrt{\frac{\pi}{2}} \frac{(n-1)!!}{(n-2)!!},\ n\ is\ even\sim \sqrt{n-1}\\ \sqrt{\frac{2}{\pi}} \frac{(n-1)!!}{(n-2)!!},\ n\ is\ odd \sim\sqrt{n+1}\end{cases}$$
However, I can not find the limit above.
Based on the numeric study, it seems like the limit is 0:
| $$f(n)=\sqrt{2}\,\frac{\Gamma(\frac{n+1}{2})}{\Gamma(\frac{n}{2})} - \sqrt{n}=\sqrt{2}\,y- \sqrt{n}$$
Take logarithms
$$\log(y)=\log \left(\frac{\Gamma \left(\frac{n+1}{2}\right)}{\Gamma
\left(\frac{n}{2}\right)}\right)=\log \left(\Gamma \left(\frac{n+1}{2}\right)\right)-\log \left(\Gamma \left(\frac{n}{2}\right)\right)$$ Use Stirling approximation twice
$$\log(y)=\frac{1}{2} \log \left(\frac{n}{2}\right)-\frac{1}{4 n}+\frac{1}{24
n^3}+O\left(\frac{1}{n^5}\right)$$ Continue with Taylor
$$y=e^{\log(y)}=\sqrt{\frac n 2}\Bigg[1-\frac{1}{4 n}+\frac{1}{32 n^2}+O\left(\frac{1}{n^3}\right)\Bigg]$$
$$f(n)=-\frac 1{4\sqrt {n}}\Bigg[1-\frac{1}{8 n}-\frac{5}{32 n^2}+O\left(\frac{1}{n^3}\right)\Bigg]$$
Use it for $n=9$; rhe above truncated series gives
$$f(9) \sim -\frac{2551}{31104}=-0.0820152\cdots$$ while the exact value is
$$f(9)=-3+\frac {128}{35} \sqrt{\frac 2 \pi}=-0.0820222\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4270745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Calculate $\lfloor \log _{2} \pi + \log _{\pi} 5 +\log_5 8 \rfloor=?$ Calculate $$\lfloor \log _{2} \pi + \log _{\pi} 5 +\log_5 8 \rfloor=?$$
I suppose that:
$$\log_2 3 < \log _{2} \pi <\log_2 4 $$
But how can I approximate the $\log_2 3$?
Can somebody give an idea?
| $\newcommand{\fl}[1]{\left\lfloor #1 \right\rfloor}$Let $a = \log_{2}(\pi)$, $b = \log_{\pi}(5)$, and $c = \log_{5}(8)$.
Then, we have $abc = \log_{2}(8) = 3$.
Thus, by the AM-GM inequality, we have $$a + b + c \geqslant 3 \sqrt[3]{3} \geqslant 4.$$
(The second inequality follows by showing that $\sqrt[3]{3} \geqslant \frac{4}{3}$. To check that, simply note that $3^4 > 4^3$ and then take cube roots and rearrange.)
Thus, $\fl{a + b + c} \geqslant 4$. To show that it is equal to $4$, it suffices to show that $$a + b + c < 5.$$
Claim 1. $a < 2$.
Proof. $\log_{2}(\pi) < \log_{2}(4) = 2$. $\square$
Claim 2. $b < \frac{5}{3}$.
Proof. Note that $$\log_{\pi}(5) < \frac{5}{3} \iff 5^3 < \pi^5.$$
But the second statement is clearly true since
$$5^3 = 125 < 243 = 3^5 < \pi^5. \qquad \square$$
Claim 3. $c < \frac{4}{3}$.
Proof. As before, this reduces to showing that $$8^3 < 5^4,$$ which is easily checked. $\square$
Thus, we have $$a + b + c < 2 + \frac{5}{3} + \frac{4}{3} = 5,$$
as desired.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving that $\rho =3\cos\varphi$ is a circle Practising some exercises in Physics. In one of the questions they said that $\rho =3\cos\varphi$ is a circle and asked to prove that it is a circle.
I'm trying to figure out how to do it. On the internet I saw that the general form of a circle in central $r_0,\varphi$ and radius $a$ is:
$$
r^2 - 2 r r_0 \cos(\theta - \varphi) + r_0^2 = a^2
$$
Do I need to compare the two?
| You are given
$\rho = 3 \cos \varphi $
From which it follows that
$x = \rho \cos \varphi = 3 \cos^2 \varphi $
and
$y = \rho \sin \varphi = 3 \cos \varphi \sin \varphi $
Simplifying the expressions, we get
$x = \frac{3}{2} (1 + \cos 2 \varphi) $
and
$y = \frac{3}{2} \sin 2 \varphi $
Thus
$\cos 2 \varphi = -1 + \dfrac{2}{3} x $
and
$\sin 2 \varphi = \frac{2}{3} y $
And since $\cos^2 \theta + \sin^2 \theta = 1$ for any $\theta$ , then
$\left(-1 + \dfrac{2}{3} x \right)^2 + \left(\dfrac{2}{3} y \right)^2 = 1$
Multiplying through by $\left( \dfrac{3}{2} \right)^2$ , we get,
$ \left(x - \dfrac{3}{2} \right)^2 + y^2 = \left( \dfrac{3}{2} \right)^2 $
Which is an equation of a circle centered at $ \left(\dfrac{3}{2}, 0 \right) $ and having a radius of $\left( \dfrac{3}{2} \right) $.
| {
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"timestamp": "2023-03-29T00:00:00",
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Asymptotic integration of $\int_0^\infty\frac{x^{-\frac{1}{2}+a}J_{-\frac{1}{2}+a}(x\alpha)}{e^x-1}{\rm d}x$ when $\alpha \gg 1$ What is the asymptotic integration of
$$\int_0^\infty\frac{x^{-\frac{1}{2}+a}J_{-\frac{1}{2}+a}(x\alpha)}{e^x-1}{\rm d}x$$
when $\alpha\rightarrow\infty$. How to compute that using standard identities? Please give a general expression and then consider a special case where $a=\frac{5}{2}$.
| We assume that $a>\frac{1}{2}$ is fixed. Note that
$$
\frac{{x^{a - 1/2} }}{{e^x - 1}} \sim \sum\limits_{n = 0}^\infty {\frac{{B_n }}{{n!}}x^{n + a - 3/2} }
$$
as $x\to 0+$, where $B_n$ denotes the Bernoulli numbers. Then, by Theorem 2 in https://doi.org/10.1137/0507061, we find
\begin{align*}
\int_0^{ + \infty } {\frac{{x^{a - 1/2} }}{{e^x - 1}}J_{a - 1/2} (\alpha x)dx} & \sim \frac{1}{{2}}\sum\limits_{n = 0}^\infty {\frac{{B_n }}{{n!}}\frac{\Gamma\! \left( {\frac{n}{2} + a - \frac{1}{2}} \right)}{\Gamma\! \left( {1-\frac{n}{2}} \right)}\!\left( {\frac{2}{\alpha }} \right)^{n + a - 1/2} }
\\ &
= \frac{1}{2}\Gamma \!\left( {a - \frac{1}{2}} \right)\!\left( {\frac{2}{\alpha }} \right)^{a - 1/2} - \frac{{\Gamma (a)}}{{4\sqrt \pi }}\left( {\frac{2}{\alpha }} \right)^{a + 1/2}
\end{align*}
as $\alpha \to +\infty$. Note that the Bernoulli numbers of odd index $n\geq 3$ or the reciprocal gamma function eliminate all terms in the asymptotic expansion except the first two. The absolute error of this two-term approximation decays to zero faster in $\alpha$ than any negative power of $\alpha$. Thus, this asymptotics is rather accurate for large $\alpha$.
In the special case $a=\frac{5}{2}$, the aproximation is
$$
\int_0^{ + \infty } {\frac{{x^2 }}{{e^x - 1}}J_2 (\alpha x)dx} \sim \frac{2}{{\alpha ^2 }} - \frac{3}{{2\alpha ^3 }}.
$$
The right-hand side apporaches $0$ from above for large positive values of $\alpha$.
A different type of expansion may be obtained as follows. We can expand the denominator of the integrand and integrate term-by-term using http://dlmf.nist.gov/10.22.E49, to deduce
\begin{align*}
\int_0^{ + \infty } {\frac{{x^{a - 1/2} }}{{e^x - 1}}J_{a - 1/2} (\alpha x)dx} & = \int_0^{ + \infty } {x^{a - 1/2} \left( {\sum\limits_{n = 1}^\infty {e^{ - nx} } } \right)J_{a - 1/2} (\alpha x)dx}
\\ &
= \sum\limits_{n = 1}^\infty {\int_0^{ + \infty } {x^{a - 1/2} e^{ - nx} J_{a - 1/2} (\alpha x)dx} }
\\ &
= \left( {\frac{\alpha }{2}} \right)^{a - 1/2} \Gamma (2a)\sum\limits_{n = 1}^\infty {\frac{1}{{n^{2a} }}{\bf F}\!\left( {a,a + \frac{1}{2};a + \frac{1}{2}; - \left( {\frac{\alpha }{n}} \right)^2 } \right)}
\\ & = \left( {\frac{\alpha }{2}} \right)^{a - 1/2} \frac{{\Gamma (2a)}}{{\Gamma \!\left( {a + \frac{1}{2}} \right)}}\sum\limits_{n = 1}^\infty {\frac{1}{{(n^2 + \alpha ^2 )^a }}}
\\ &
= (2\alpha )^{a - 1/2} \frac{{\Gamma (a)}}{{\sqrt \pi }}\sum\limits_{n = 1}^\infty {\frac{1}{{(n^2 + \alpha ^2 )^a }}}
\end{align*}
for $a>\frac{1}{2}$ and $\alpha>0$. Here $\mathbf F$ stands for the regularised hypergeometric function. The last series is a generalised Mathieu series. Thus applying Theorem 1 in https://arxiv.org/abs/1601.07751v1, we obtain
\begin{align*}
\int_0^{ + \infty } {\frac{{x^{a - 1/2} }}{{e^x - 1}}J_{a - 1/2} (\alpha x)dx} = & \;\frac{1}{2}\Gamma\! \left( {a - \frac{1}{2}} \right)\!\left( {\frac{2}{\alpha }} \right)^{a - 1/2} - \frac{{\Gamma (a)}}{{4\sqrt \pi }}\left( {\frac{2}{\alpha }} \right)^{a + 1/2} \\ & +(2\pi )^{a - 1/2} \frac{{e^{ - 2\pi \alpha } }}{{\sqrt \alpha }}\left( {1 + \mathcal{O}\!\left( {\frac{1}{\alpha }} \right)} \right)
\end{align*}
as $\alpha \to +\infty$, with $a>\frac{1}{2}$ being fixed. This shows the exponential accuracy of our original two-term asymptotics.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
"answer_count": 1,
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} |
Induction inequality/ n in exponential I need to find out through induction for which $n \geq 0$ the following inequality holds
$$3n+2^{n} \leq 3^n$$
clearly it does not work for n=1 and n=2, so my induction hypothesis is that it works for $n \geq 3$,
however when performing the induction step, I am kind of stuck
because
$ n \rightarrow n+1$
$3 (n+1) + 2^{n+1} \leq 3^{n+1}$
$3(n+1) + 2^{n} \cdot 2 \leq 3^n \cdot 3$
I recreated the starting inequality within (overline)
$3n + 3 + 2^n + 2^n \leq (2+1) \cdot 3^n$
$2^n + 3 + \overline{3n + 2^n \leq 3^n} + 2\cdot 3^n$
I tried to show that what was added on the LHS is always smaller than on the RHS, therefor
$2n+3 \leq 2\cdot 3^n$
$2n + 3 \leq 3^n + 3^n$
so it is clear that this holds,but would the proof work that way? if yes, are there other, more elegant approaches?
| Another way.
For $n\geq3$ by AM-GM we obtain:
$$3^n-2^n=(3-2)\left(3^{n-1}+3^{n-2}\cdot2+...+3\cdot2^{n-2}+2^{n-1}\right)\geq$$
$$\geq n\sqrt[n]{3^{n-1+n-2+...+1}2^{1+...+n-1+n-2}}=n\cdot6^{\frac{n-1}{2}}\geq n\cdot6^{\frac{3-1}{2}}=6n>3n.$$
| {
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"answer_count": 2,
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} |
Dividing numerator and denominator of integrand in definite integral I'm trying to solve the following definite integral $$\int_0^{\frac{\pi}{4}} \sqrt{\tan x}\,dx$$
First, I made the substitution $u = \sqrt{\tan x}$
And arrived with: $$\int_0^1 \frac{2u^2}{u^4+1}\,du$$
When divided both numerator and denominator of the integrand by $u^2$ and with some manipulation I got
$$\int_0^1 \frac{1+\frac{1}{u^2}}{(u-\frac{1}{u})^2+2}+\frac{1-\frac{1}{u^2}}{(u+\frac{1}{u})^2-2}\, du$$
This integral could be solved easily with u sub, giving the result:
$$\left.\frac{1}{2\sqrt{2}} \ln\left|\frac{u^2-\sqrt{2}u+1}{u^2+\sqrt{2}u+1}\right|+\frac{1}{\sqrt{2}}\arctan\left(\frac{1}{\sqrt{2}}\left(u-\frac{1}{u}\right)\right)\ \right|_0^1$$
The integral however is not defined at $u=0$, my question is: Are we allowed to do this kind of manipulation to definite integrals with rational function, and will proceeding with improper integral yield the correct result?
| I don't think the best thing to do is divide by $u^2$. I think this can solve your problem:
\begin{align*}
\int_0^1 \frac{2u^2}{u^4+1}\ du&=\frac{1}{\sqrt2}\int_0^1\frac{u}{u^2-\sqrt2u+1}\ du-\frac{1}{\sqrt2}\int_0^1\frac{u}{u^2+\sqrt2u+2}\ du\\[2mm]
&=\frac{1}{\sqrt2}\int_0^1\frac{u}{\left(u-\frac{\sqrt2}{2}\right)^2+\frac{1}{2}}\ du-\frac{1}{\sqrt2}\int_0^1\frac{u}{\left(u+\frac{\sqrt2}{2}\right)^2+\frac{1}{2}}\ du\\[2mm]
&=\frac{1}{2\sqrt 2}\ln\left|\frac{2x^2-2\sqrt2 x+2}{2x^2+2\sqrt2x+2}\right|-\frac{1}{\sqrt2}\arctan(\sqrt2x-1)+\frac{1}{\sqrt2}\arctan(\sqrt2x+1)\Big\vert_0^1\\[2mm]
&=\frac{\pi +\ln(3-2\sqrt 2)}{2\sqrt2}\\[2mm]
&\approx 0.487495
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4285216",
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} |
$\sum\limits_{cyc} \sqrt{\frac{a^3}{1+bc}} \geq 2$ for $a, b, c > 0$ which satisfies $abc=1$. $\displaystyle \sum_{cyc} \sqrt{\dfrac{a^3}{1+bc}} \geq 2$ for $a, b, c > 0$ which satisfies $abc=1$.
My attempt:
\begin{align}
&\text{let } a=\frac{y}{x}, b=\frac{x}{z}, c=\frac{z}{y}. \\
&\text{Substituting for the original F.E.: }\displaystyle \sum_{cyc}\sqrt{\frac{(\frac{y}{x})^3}{1+\frac{x}{y}}} \geq 2 \text{ for }x, y, z\in \mathbb{R}^+. \\
&\therefore \text{ETS) }\displaystyle \sum_{cyc} \sqrt{\frac{y^4}{x^3y+x^4}} \geq 2. \\
\ \\
& \text{Two ways to think: }\\
\ \\
& (1) \\
&\therefore \text{Using Cauchy-Schwarz inequality, } \displaystyle \Bigg(\sum_{cyc} \sqrt{\frac{y^4}{x^3y+x^4}}\Bigg) \Bigg(\sum_{cyc} \frac {x}{y}\Bigg) \geq \sum_{cyc} \frac{y^2}{x^2+xy} \\
&\text{ETS) }\displaystyle \sum_{cyc} \frac{y^2}{x^2+xy} \geq 2\Bigg( \sum_{cyc} \frac {x}{y} \Bigg).
\ \\
&(2)\\
&\therefore \text{Using Cauchy-Schwarz inequality, } \displaystyle \Bigg(\sum_{cyc} \sqrt{\frac{y^4}{x^3y+x^4}}\Bigg) \Bigg(\sum_{cyc} \sqrt{\frac {x+y}{y}}\Bigg) \geq \sum_{cyc} \frac{y^3}{x^3} \\
&\text{ETS) }\displaystyle \sum_{cyc} \frac{y^3}{x^3} \geq 2\Bigg( \sum_{cyc} \sqrt{\frac{x+y}{y}} \Bigg).
\end{align}
p.s. I think we can't use the AM-GM one, but I'll try.
| Using Titu's Lemma,
$$\sum\sqrt{\frac{a^3}{1+bc}}=\sum \frac{a^2}{\sqrt{a+1}}\ge\frac{(a+b+c)^2}{\sum \sqrt{a+1}} \tag{1}$$
It remains to show,
$$(a+b+c)^2\ge 2(\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1})$$
Using A.M-G.M inequality,
$$\sum \frac{(a+1)+1}{2}\ge \sum\sqrt{(a+1)\cdot 1}\implies a+b+c+6\ge2(\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}) $$
It remains to show,
$$(a+b+c)^2 \ge (a+b+c)+6\tag{2}$$
which is true, since $a+b+c\ge 3\sqrt[3]{abc}=3.$
Proof of $(2)$:$$(a+b+c)^2\ge 3(a+b+c)\ge (a+b+c)+6$$
Given below is a proof of a stronger result.
Claim:
$$\sum \sqrt{\frac{a^3}{1+bc}}\ge \frac{3}{\sqrt{2}}$$
Proceeding from $(1)$, it remains to show,
$$\sqrt{2}(a+b+c)^2\ge3\sum{\sqrt{a+1}}\implies 4(a+b+c)^2\ge 3\sum2\sqrt{2(a+1)}$$
Using A.M-G.M inequality,
$$\frac{\sum(a+1)+2}{2}\ge \sum \sqrt{2(a+1)}\implies a+b+c+9\ge \sum2\sqrt{2(a+1)} $$
It remains to show,
$$4(a+b+c)^2\ge 3(a+b+c)+27$$
which is true, since $a+b+c\ge 3.$
Also, equality can be achieved, unlike the original problem, at $a=b=c=1$.
| {
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} |
Given triangle ABC with $\alpha=2\beta$ and $b, c, a$ is forming an arithmetic sequence. Find $\alpha, \beta, \gamma$. Given $\triangle ABC$ with $\alpha=2\beta$ and $b, c, a$ is forming an arithmetic sequence ($b$ is the $1$'st term, and $a$ is the last term), find $\alpha, \beta, \gamma$.
My attempts so far:
Let $c=b+k$ and $a=b+2k$. Then, from $\alpha=2\beta$, I got $\gamma = 180^\circ - 3\beta$.
With the Law of Sines , I can write
$$\dfrac{b}{\sin \beta}=\dfrac{b+k}{\sin 3\beta}=\frac{b+2k}{\text{sin } 2\beta}$$
From $\dfrac{b}{\sin\beta}=\dfrac{b+k}{\sin 3\beta}$, it gives $\cos 2\beta = \dfrac{k}{2b}$.
From $\dfrac{b}{\sin \beta}=\dfrac{b+2k}{\sin 2\beta}$, it gives $\cos \beta = \dfrac{b+2k}{2b}$.
After this, I don't know what should I do. Is there any other theorem that can be used to solve this problem?
| You can continue from where you left off. Using the fact that $\cos 2 \beta = 2 \cos^2 \beta - 1$:
$$2 \left( \frac{b+2k}{2b} \right)^2 - 1 = \frac{k}{2b}$$
$$2 (b + 2k)^2 - 4b^2 = 2bk$$
$$2b^2 + 8bk + 8k^2 - 4b^2 = 2bk$$
$$-2b^2 + 6bk + 8k^2 = 0$$
$$b^2 - 3bk - 4k^2 = 0$$
$$(b+k)(b-4k) = 0$$
but since $c = b + k \ne 0$, $b = 4k$ which gives side lengths $(4k, 5k, 6k)$ and has the same angles as $(4,5,6)$. Now the cosine rule $\cos C = \frac{a^2+b^2-c^2}{2ab}$ and so on can be used to find angles $\alpha,\beta,\gamma$.
| {
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asymptotic approximation of Fresnel integrals with complex argument It turns out that SciPy's Fresnel values are wrong for complex arguments and large enough absolute value. I'm trying to fix that.
The implementation is based on Zhang/Jin, Computation of special functions, which in turn is based on Abramowitz/Stegun, Handbook of Mathematical Functions. There we find for the Fresnel S integral (7.3.10)
$$
S(z) = \frac{1}{2} - f(z) \cos\left(\frac{\pi}{2} z^2\right) - g(z) \sin\left(\frac{\pi}{2} z^2\right)
$$
for all $z$ with the auxiliary functions (7.3.5), (7.3.6)
$$
\begin{split}
f(z) &= \left[\frac{1}{2} - S(z) \right] \cos\left(\frac{\pi}{2} z^2\right) - \left[\frac{1}{2} - C(z) \right] \sin\left(\frac{\pi}{2} z^2\right),\\
g(z) &= \left[\frac{1}{2} - C(z) \right] \cos\left(\frac{\pi}{2} z^2\right) + \left[\frac{1}{2} - S(z) \right] \sin\left(\frac{\pi}{2} z^2\right).
\end{split}
$$
Computation of $S$ for large values is done via the asymptotic expansion of $f$ (7.3.27)
$$
\DeclareMathOperator\arg{arg}
\pi z f(z)\sim 1 + \sum_{m=1}^\infty (-1)^m \frac{1\cdot 3\cdot\cdots \cdot (4m-1)}{(\pi z^2)^{2m}}, \quad z\to\infty, |\arg(z)|<\frac{\pi}{2}.
$$
This is where I have problems understanding the approximation.
Consider $f(iz)$; for the asymptotic $f_a$, we have $f_a(iz) = -if_a(z)$. However no such thing is true for $f$ itself. Numerical computation via the representations
$$
\DeclareMathOperator\erf{erf}
\begin{split}
S(z) &= \frac{1 + i}{4} \left[\erf\left(\frac{1 + i}{2} \sqrt{\pi} z\right) - i \erf\left(\frac{1 - i}{2} \sqrt{\pi} z\right)\right]\\
C(z) &= \frac{1 - i}{4} \left[\erf\left(\frac{1 + i}{2} \sqrt{\pi} z\right) + i \erf\left(\frac{1 - i}{2} \sqrt{\pi} z\right)\right]
\end{split}
$$
shows that the approximation incorrect for everything off of the real axis, but the issue here could be numerical instability in the $\erf$ represenation too.
My current guess is that the above infinite sum is valid only in the area $|\arg{z}|<\pi/4$, which would already change how we compute the Fresnel integral values significantly.
To finish things off, here's a cplot of $f$ (for smaller $|z|$):
| I will consider the function $\operatorname{f}(z)$, the treatment of $\operatorname{g}(z)$ is analogous. By http://dlmf.nist.gov/7.12.ii, we have
$$
\operatorname{f}(z) = \frac{1}{{\pi z}}\sum\limits_{m = 0}^{N - 1} {( - 1)^m \left( {\frac{1}{2}} \right)_{2m} \frac{1}{{(\pi z^2 /2)^{2m} }}} + R_N^{(\operatorname{f})} (z)
$$
where
\begin{align*}
R_N^{(\operatorname{f})} (z) & = \frac{{( - 1)^N }}{{\pi \sqrt 2 }}\int_0^{ + \infty } {\frac{{e^{ - \pi z^2 t/2} t^{2N - 1/2} }}{{1 + t^2 }}dt} \\ & = \frac{1}{{\pi z}}( - 1)^N \left( {\frac{1}{2}} \right)_{2N} \frac{1}{{(\pi z^2 /2)^{2N} }}\frac{1}{{\Gamma \left( {2N + \frac{1}{2}} \right)}}\int_0^{ + \infty } {\frac{{e^{ - s} s^{2N - 1/2} }}{{1 + s^2 /(\pi z^2 /2)^2 }}ds}
\\ &
= \frac{1}{{\pi z}}( - 1)^N \left( {\frac{1}{2}} \right)_{2N} \frac{1}{{(\pi z^2 /2)^{2N} }}\Pi _{2N + 1/2} (\pi z^2 /2),
\end{align*}
provided that $|\arg z|<\frac{\pi}{4}$ and $N\geq 0$. Here $\Pi_p(w)$ denotes one of Dingle's basic terminants:
$$
\Pi _p (w) = \frac{1}{{\Gamma (p)}}\int_0^{ + \infty } {\frac{{e^{ - s} s^{p - 1} }}{{1 + (s/w)^2 }}ds}
$$
for $|\arg w|<\frac{\pi}{2}$ and by analytic continuation elswhere. Using the expression for $R_N^{(\operatorname{f})} (z)$ in terms of this terminant, we can extend $R_N^{(\operatorname{f})} (z)$ to the universal covering of $\mathbb C \setminus \left\{ 0\right\}$. Now employing the estimates for the basic terminant established in https://doi.org/10.1007/s10440-017-0099-0, we obtain the bound
\begin{align*}
\left| {R_N^{(\operatorname{f})} (z)} \right| \le &\; \left| {\frac{1}{{\pi z}}( - 1)^N \left( {\frac{1}{2}} \right)_{2N} \frac{1}{{(\pi z^2 /2)^{2N} }}} \right| \\ & \times \begin{cases} 1 & \text{ if } \; \left|\arg z\right| \leq \frac{\pi}{8}, \\ \min\!\Big(\left|\csc ( 4\arg z)\right|,1 + \cfrac{1}{2}\chi(2N+1/2)\Big) & \text{ if } \; \frac{\pi}{8} < \left|\arg z\right| \leq \frac{\pi}{4}, \\ \cfrac{\sqrt {2\pi (2N + 1/2)} }{2\left| {\sin (2\arg z)} \right|^{2N+1/2} } + 1 + \cfrac{1}{2}\chi (2N +1/2) & \text{ if } \; \frac{\pi}{4} < \left|\arg z\right| < \frac{\pi}{2}. \end{cases}
\end{align*}
Here $\chi(p) =\sqrt{\pi}\Gamma(p/2+1)/\Gamma(p/2+1/2)$ for $p>0$. It is seen that the asymptotic expansion of $\operatorname{f}(z)$ is valid in every closed sub-sector of $|\arg z|<\frac{\pi}{2}$ in the sense of Poincaré. The Stokes lines are $\arg z =\pm \frac{\pi}{4}$ where terms are swiched on that remain exponentially small compared to any negative power of $z$ as long as we stay away from the rays $\arg z =\pm \frac{\pi}{2}$ (the anti-Stokes lines).
To obtain a better result and reveal the exponentially small terms, we can use the functional equation
$$
\Pi _p (w) = \pm \pi i\frac{{e^{ \mp \frac{\pi }{2}ip} }}{{\Gamma (p)}}w^p e^{ \pm iw} + \Pi _p (we^{ \mp \pi i} )
$$
where $p>0$ and $w$ is any element of the universal covering of $\mathbb C \setminus \left\{ 0\right\}$ (the Riemann surface of the logarithm). With this functional equation, we find, after some algebra,
$$
R_N^{(\operatorname{f})} (ze^{ \mp \frac{\pi }{2}i} ) = \pm iR_N^{(\operatorname{f})} (z) + \frac{{1 \mp i}}{2}e^{ \pm \frac{\pi }{2}iz^2 } .
$$
This result is valid for all $N\geq 0$ and $z$ on the universal covering of $\mathbb C \setminus \left\{ 0\right\}$. You can see that if we omit the second term (which is exponentially small when $
0 < \pm \arg z < \frac{\pi}{2}$) then, with $N=0$, we get the false result
$$
\operatorname{f}(ze^{ \mp \frac{\pi }{2}i} ) = \pm i\operatorname{f}(z).
$$
See also http://dlmf.nist.gov/7.4.
In summary, use
$$
\operatorname{f}(z) \sim \frac{1}{{\pi z}}\sum\limits_{m = 0}^\infty {( - 1)^m \left( {\frac{1}{2}} \right)_{2m} \frac{1}{{(\pi z^2 /2)^{2m} }}}
$$
when $\left| {\arg z} \right| \le \frac{\pi }{4}$, and use
$$
\operatorname{f}(z) \sim \frac{{1 \pm i}}{2}e^{ \pm \frac{\pi }{2}iz^2 } + \frac{1}{{\pi z}}\sum\limits_{m = 0}^\infty {( - 1)^m \left( {\frac{1}{2}} \right)_{2m} \frac{1}{{(\pi z^2 /2)^{2m} }}}
$$
when $\frac{\pi }{4} < \pm \arg z \le \frac{{3\pi }}{4}$. Of couse by relaying on the symmetry relation http://dlmf.nist.gov/7.4.E8, we can assume $|\arg z|\leq \frac{\pi}{2}$.
| {
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Find $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying $f(x+y)^2=f(x)^2+yf(x)+xf(y)+y^2$ - Solved.
Find $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying $f(x+y)^2=f(x)^2+yf(x)+xf(y)+y^2$
It's solved, but I'm posting it to share it.
\begin{align} &P(x, y): f(x+y)^2=f(x)^2+yf(x)+xf(y)+y^2 \\ &P(y, x):f(x+y)^2=x^2+yf(x)+xf(y)+f(y)^2. \\ &\therefore f(x)^2-x^2=f(y)^2-y^2=c. \\ &\therefore f(x)= \pm \sqrt{x^2+c}. \\ &\text{Substituting to the original F.E.: } x^2+2xy+y^2+c = x^2+y^2+c \pm \Big(y\sqrt{x^2+c}+x\sqrt{y^2+c}\Big). \\ &\therefore 2xy= \pm y\sqrt{x^2+c} \pm x\sqrt{y^2+c}. \\ &\Rightarrow 4x^2y^2=x^2y^2+y^2c + x^2y^2+x^2c + 2xy\sqrt{(x^2+c)(y^2+c)}. \\ &\Rightarrow 2x^2y^2-(x^2+y^2)c=2xy\sqrt{(x^2+c)(y^2+c)}. \\ & \therefore 4x^4y^4 + (x^2+y^2)^2c^2 - 4x^2y^2(x^2+y^2)c = 4x^2y^2(x^2+c)(y^2+c). \\ &\Rightarrow 4x^4y^4+(x^2+y^2)c^2-4x^2y^2(x^2+y^2)c=4x^4y^4+4x^4y^2c + 4x^2y^4c + 4x^2y^2c^2 \\ &\Rightarrow (x^2+y^2)c^2-4x^2y^2(x^2+y^2)c=4x^2y^2(x^2+y^2+c)c \\ & \text{if } c=0: \text{Solution.} \\ &\text{if } c \neq 0: (x^2+y^2)c-4x^2y^2(x^2+y^2)=4x^2y^2(x^2+y^2+c). \\ &\Rightarrow (x^2+y^2-4x^2y^2)c=8x^2y^2(x^2+y^2). \\ & \therefore c = \frac {8x^2y^2(x^2+y^2)}{(x^2+y^2-4x^2y^2)}, \text{ which isn't constant, Contradiction.} \\ & \therefore c = 0. \\ & \text{Contributing this to the original F.E.: } x^2+2xy+y^2=x^2+y^2 \pm 2xy \\ & \Rightarrow f(x)=x.\end{align}
If you have another solution, please post it to the answer with a spoiler code. How to put math equations in a "spoiler" block?
|
Substitute $x = 1$ and $y = 0$ to get $f(0) = 0$.
Substitute $x = 0$ to get $f(y)^2 = y^2$.
The equation then becomes $(x + y)^2 = x^2 + x f(y) + y f(x) + y^2$ which simplifies to $2xy = x f(y) + y f(x)$. Plug in $y = 1$ to get $2x = x f(1) + f(x)$; then $f(x) = (2 - f(1)) x$.
Then in particular $f(1) = 2 - f(1)$. So $f(1) = 1$. Then $f(x) = x$.
We easily verify that $f(x) = x$ is a solution. So the only solution here is $f(x) = x$.
| {
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$\frac{3}{4} \lim_{n \to \infty}\left(\frac{\sum_{r=1}^{n}\frac{1}{\sqrt{r}} \sum_{r=1}^{n}\sqrt{r} }{\sum_{r=1}^{n}r }\right)$ $$\frac{3}{4} \lim_{n \to \infty}\left(\frac{\sum_{r=1}^{n}\frac{1}{\sqrt{r}} \sum_{r=1}^{n}\sqrt{r} }{\sum_{r=1}^{n}r }\right)$$
Apparently, the answer is 2.
My try:
$$\frac{3}{4} \lim_{n \to \infty}\left(\frac{n^2}{\frac{n(n+1)}{2} }\right)$$
$$\frac{3}{2} \lim_{n\to \infty}\left(1-\frac{1}{n+1}\right)$$
Which is $\frac{3}{2}$. Surely I did something stupid or illegal here. What was it?
Also on the forum where I found this, It was said it can be done by converting into integrals, A push in the right direction on that too would be pretty rad.
Hope I am not asking too much.
| Here is a solution using the suggested integral approach. Notice that
\begin{align*}
\frac{{\sum\nolimits_{r = 1}^n {\frac{1}{{\sqrt r }}} \sum\nolimits_{r = 1}^n {\sqrt r } }}{{\sum\nolimits_{r = 1}^n r }} = \frac{{\sum\nolimits_{r = 1}^n {\sqrt {\frac{n}{r}} } \sum\nolimits_{r = 1}^n {\sqrt {\frac{r}{n}} } }}{{\sum\nolimits_{r = 1}^n r }} & = \frac{{\frac{1}{{n^2 }}\sum\nolimits_{r = 1}^n {\sqrt {\frac{n}{r}} } \sum\nolimits_{r = 1}^n {\sqrt {\frac{r}{n}} } }}{{\frac{1}{{n^2 }}\sum\nolimits_{r = 1}^n r }} \\ &= \frac{{\left( {\frac{1}{n}\sum\nolimits_{r = 1}^n {\sqrt {\frac{n}{r}} } } \right)\left( {\frac{1}{n}\sum\nolimits_{r = 1}^n {\sqrt {\frac{r}{n}} } } \right)}}{{\frac{1}{n}\sum\nolimits_{r = 1}^n {\frac{r}{n}} }}.
\end{align*}
Thus, noticing the Riemann sums, we find
\begin{align*}
\mathop {\lim }\limits_{n \to + \infty } \frac{3}{4}\frac{{\sum\nolimits_{r = 1}^n {\frac{1}{{\sqrt r }}} \sum\nolimits_{r = 1}^n {\sqrt r } }}{{\sum\nolimits_{r = 1}^n r }} & = \frac{3}{4}\frac{{\mathop {\lim }\limits_{n \to + \infty } \left( {\frac{1}{n}\sum\nolimits_{r = 1}^n {\sqrt {\frac{n}{r}} } } \right)\mathop {\lim }\limits_{n \to + \infty } \left( {\frac{1}{n}\sum\nolimits_{r = 1}^n {\sqrt {\frac{r}{n}} } } \right)}}{{\mathop {\lim }\limits_{n \to + \infty } \frac{1}{n}\sum\nolimits_{r = 1}^n {\frac{r}{n}} }} \\ & = \frac{3}{4}\frac{{\int_0^1 {\frac{{dx}}{{\sqrt x }}} \int_0^1 {\sqrt x dx} }}{{\int_0^1 {xdx} }} = 2.
\end{align*}
| {
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"question_score": "6",
"answer_count": 1,
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Factorize $x^8+2$ over $\mathbb{F}_3$ using cyclotomic cosets
Determine the splitting field of $x^8 + 2$ over $\mathbb{F}_3$ and factor it into irreducible polynomials over $\mathbb{F}_3$ using cyclotomic cosets.
I'm having issues in factorizing this polynomial. can anyone see if I am worng with the computations or something else please?
My attempt: By a Lemma we had in the lecture:
Lemma: The splitting field $\mathbb{F}_{q^s}$ $\text{split}(f,\mathbb{F}_q)$, of $f=x^n-1$, is characterized through the smallest integer $s$ such that $n|q^s-1$.
Notice that over $\mathbb{F}_3$ we have $x^8+2\equiv x^8-1$.
For $n=8$ and $q=3$, $n|3^2-1=8\Rightarrow \text{split}(x^8+2,\mathbb{F}_3)=\mathbb{F}_{3^2}=\mathbb{F}_9$.
Consider that 2 is a root of $x^8-1$, i.e. $2^8-1=255\equiv 0\mod3$
Then I find the cyclotomic cosets $C_0=\{0\}$, $C_1=\{1,3\}$, $C_2=\{2,6\}$, $C_3=\{1,3\}=C_1$, $C_4=\{4\}$, $C_5=\{5,7\}$
So the factorization gives: $x^8-1=(x-2^0)\cdot(x-2^1)(x-2^3)\cdot(x-2^2)(x-2^6)\cdot(x-2^4)\cdot(x-2^5)(x-2^7)\equiv(x+2)(x^2+2x+1)(x^2+x+2)(x+2)(x^2+2x+1)$
But checking the L.H.S., this doesn't give $x^8-1$...
| Here $x^8+2 = x^8-1 = (x^4-1)(x^4+1)$ with
$x^4+1 = (x^2-1)(x^2+1)$ and $x^2-1 = (x-1)(x+1)$.
Now $x^2+1 = (x+i)(x-i)$ where $i$ is a primitive 4th root of unity in an extension field of $F_3$.
Here $2^2 = 4 = 1$ in $F_3$ and so 2 is a primitive 2nd root of unity. Thus $x^2+1$ does not factor over $F-3$.
Similarly, $x^4+1$ does not factor over $F_3$ as $F_3$ does not contain a primitive 8th root of unity.
This gives the overall decomposition.
| {
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Find all positive integers $n$, such that $(\left\lfloor \sqrt{n} \right\rfloor^{2} +2) | (n^2 + 1) $ I tried to look at the cases when $n$ is a perfect square. Then $\left\lfloor \sqrt{n} \right\rfloor^{2} +2= n+2$, $ n^2 + 1 =(n-2)(n+2) + 5$. Then we must have $(n+2)|5$.
But only $1$ and $5$ divide $5$. Thus, $n=3$, but that is not a solution since we assumed $n$ to be a perfect square. The problem therefore has no perfect-square solutions.
I'm not sure how relevant this is to the general case, but I did not manage to get any further.
Thank you for your help in advance.
| Let $m^2=n-k$ be the largest square less than or equal to $n$.
$\implies0\le k\le 2m$
We have $n-k+2|n^2+1$
$$\implies n-k+2|1+n(k-2)$$
$$\implies n-k+2|k^2-4k+5$$
$$\implies m^2+2|k^2-4k+5$$
$\dfrac{k^2-4k+5}{m^2+2}=1,2\text{ or }3$ otherwise the inequality $0\le k\le 2m$ is violated.
If $\dfrac{k^2-4k+5}{m^2+2}=1$ we then have $(k-m-2)(k+m-2)=1$.
The only solutions for the above equality are $(m,k)=(0,1),(0,3)$, but $k>2m$, so no solution exists.
$\dfrac{k^2-4k+5}{m^2+2}=2$ we then have $(k-2)^2+1=2m^2+4$. $k$ should be odd. The LHS is of the form $8k+2$ while RHS is of the form $8k+4$ or $8k+6$. So no solution exists.
If $\dfrac{k^2-4k+5}{m^2+2}=3$ we then have $(k-2)^2+1=3m^2+6$. $3$ never divides the LHS so there are no solutions.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Strategies to solve a limit $\lim_{x\to \pi/4} \frac{\cos(2x)}{2\cos(x)-\sqrt 2}$ This morning a colleague asked me how I would solve quickly this limit,
$$\lim_{x\to \frac{\pi}{4}} \frac{\cos(2x)}{2\cos(x)-\sqrt 2}$$
Probably because I am a member of this community but I have suggest to do:
$$\frac{\cos(2x)}{2\cos(x)-\sqrt 2}=\frac{\cos(2x)}{(2\cos(x)-\sqrt 2)}\frac{(2\cos(x)+\sqrt 2)}{(2\cos(x)+\sqrt 2)}$$
Being $\cos(2x)=2\cos^2 x-1$,
$$\frac{\cos(2x)(2\cos(x)+\sqrt 2)}{2(2\cos^2 x-1)}=\frac{2\cos(x)+\sqrt 2}{2}$$
and we can found the limit for $x\to \frac{\pi}4$.
But I have now look the denominator $2\cos(x)-\sqrt 2=2\left(\cos x-\cos \frac{\pi}{4}\right)$. Using the prostapheresis formulas I will have:
$$2\left(\cos x-\cos \frac{\pi}{4}\right)=-2\left(\sin\frac{(x+\pi/4)}{2}\cdot\sin\frac{(x-\pi/4)}{2}\right)$$
If $x\to \frac{\pi}4$ then $$-2\left(\sin\frac{(x+\pi/4)}{2}\cdot\sin\frac{(x-\pi/4)}{2}\right)\to 0$$
Hence the strategy with the prostapheresis formulas is it not good or is there a way to use the notable limits?
| Consider rewriting the numerator as $\cos(2x) - \cos(\frac{\pi}2).$ Then we can use the same manipulation to get
$$\cos(2x) - \cos(\frac{\pi}2) = -2\sin\left(\frac{2x + \frac{\pi}{2}}{2}\right)\sin\left(\frac{2x - \frac{\pi}{2}}{2}\right) = -2\sin\left(x + \frac{\pi}{4}\right)\sin\left(x - \frac{\pi}{4}\right)$$
Substituting this in gives us
$$\lim_{x \to \frac{\pi}4} \frac{-2\sin\left(x + \frac{\pi}{4}\right)\sin\left(x - \frac{\pi}{4}\right)}{2\left(-2\sin\left(\frac{x + \frac{\pi}{4}}2\right)\sin\left(\frac{x - \frac{\pi}{4}}2\right)\right)} = \frac12\lim_{x \to \frac{\pi}4}\frac{\sin\left(x + \frac{\pi}{4}\right)}{\sin\left(\frac{x + \frac{\pi}{4}}2\right)} \cdot \lim_{x \to \frac{\pi}4}\frac{\sin\left(x - \frac{\pi}{4}\right)}{\sin\left(\frac{x - \frac{\pi}{4}}2\right)}$$
supposing for the moment that those two limits exist. (also note the additional factor of $2$ in the denominator which seems to have been dropped in your manipulation)
For the first limit we can simply evaluate at $x = \frac{\pi}4$ to get $\frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2}.$
For the second limit, consider rewriting $\sin(x - \frac{\pi}4)$ in the numerator as $\sin\left(2\frac{x - \frac{\pi}{4}}{2}\right) = 2\sin\left(\frac{x - \frac{\pi}{4}}{2}\right)\cos\left(\frac{x - \frac{\pi}{4}}{2}\right).$ This gives us that
$$\lim_{x \to \frac{\pi}4}\frac{\sin\left(x - \frac{\pi}{4}\right)}{\sin\left(\frac{x - \frac{\pi}{4}}2\right)} = \lim_{x \to \frac{\pi}4} 2\cos\left(\frac{x - \frac{\pi}{4}}{2}\right) = 2$$
So our total limit is $\frac12 \cdot \sqrt{2} \cdot 2 = \sqrt{2},$ which agrees with your initial solution.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the minimum value of the constant term.
Let $f(x)$ be a polynomial function with non-negative coefficients such that $f(1)=f’(1)=f’’(1)=f’’’(1)=1$. Find the minimum value of $f(0)$.
By Taylor’s formula, we can obtain
$$f(x)=1+(x-1)+\frac{1}{2!}(x-1)^2+\frac{1}{3!}(x-1)^3+\cdots+\frac{f^{(n)}(1)}{n!}(x-1)^n.$$
Hence $$f(0)=\frac{1}{2}-\frac{1}{6}+\sum_{k=4}^n\frac{f^{(k)}(1)}{k!}(-1)^k.$$
This will help?
| It is perhaps simpler to use Taylor's formula with fixed degree three and a Lagrange remainder:
$$
f(x)=1+(x-1)+\frac{1}{2!}(x-1)^2+\frac{1}{3!}(x-1)^3+\frac{f^{(4)}(\xi)}{4!}(x-1)^4.
$$
where $\xi$ is between $1$ and $x$. For $x=0$ is $\xi \ge 0$ and the last term is non-negative, since $f^{(4)}$ has non-negative coefficients as well. This gives
$$
f(0) \ge 1 - 1 + \frac 12 - \frac 16 = \frac 13 \, .
$$
The bound is sharp, equality holds for the function
$$
f(x)=1+(x-1)+\frac{1}{2!}(x-1)^2+\frac{1}{3!}(x-1)^3 = \frac 13 + \frac 12 x + \frac 16 x^3 \, .
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Extracting coefficients I stuck at the following problem:
Let
\begin{equation}
f(z) = \frac{1 - \sqrt{1 - 4z}}{2}.
\end{equation}
Find $[z^n]$ of $f(z)^r$ for $r \in \mathbb{N}$. Where $[z^n]f(z)$ is the $n$-th coefficient of the power series $f(z) = \sum_{n \geq 0}{a_n z^n}$ therefore $[z^n]f(z) = a_n$.
So far I got
\begin{equation}
\sqrt{1 - 4z} = \sum_{n \geq 0}\binom{1/2}{n}(-4 x)^n.
\end{equation}
and therefore
\begin{align*}
f(z) = \frac{1 - \sqrt{1 - 4z}}{2} &= \frac{\sum_{n \geq 1}\binom{1/2}{n}(-4)^n z^n}{2} \\
&= \sum_{n \geq 1}\binom{1/2}{n}(-1)^n 2^{n} z^n \\
&= \sum_{n \geq 1}\binom{1/2}{n}(-2)^n z^n \\
&= \sum_{n \geq 0}a_n z^n
\end{align*}
with coefficients $a_0 = 0$ and $a_n = \binom{1/2}{n}(-2)^n$.
I wanted to use cauchys integral formula for
$g(z) = f(z)^r$ to extract the $n$ coefficient
but then I get
\begin{align}
\left( \frac{1 - \sqrt{1 - 4z}}{2}\right)^r &= \frac{1}{2^r} \left(1 - \sqrt{ 1 - 4z} \right)^r \\
&= \frac{1}{2^r} \sum^{r}_{m=0}\binom{r}{m}(-1)^m \sqrt{1 - 4z}^m \\
&= \frac{1}{2^r} \sum^{r}_{m=0}\binom{r}{m}(-1)^m (1 - 4z)^{m/2} \\
&= \frac{1}{2^r} \sum^{r}_{m=0}\binom{r}{m}(-1)^m \sum_{k \geq 0}\binom{m/2}{k}(-1)^k 4^k z^k \\
&= \frac{1}{2^r}\sum^{r}_{m=0} \sum_{k \geq 0}\binom{r}{m} \binom{m/2}{k} (-1)^{m+k} 4^k z^k.
\end{align}
Therefore I should have
\begin{align*}
[z^n] (f(z))^r &= [z^n] \sum^{r}_{m=0} \sum_{k \geq 0}\frac{1}{2^r} \binom{r}{m} \binom{m/2}{k} (-1)^{m+k} 4^k z^k \\
&= \sum^{r}_{m=0}[z^n] \sum_{k \geq 0}{\frac{1}{2^r} \binom{r}{m} \binom{m/2}{k} (-1)^{m+k} 4^k z^k} \\
&= \sum^{r}_{m=0}\frac{1}{2^r}\binom{r}{m} \binom{m/2}{n} (-1)^{m + n} 4^n \\
&= \sum^{r}_{m=0}\frac{1}{2^r}\binom{r}{m} \binom{m/2}{n} (-1)^m (-4)^n \\
&= \frac{1}{2^r} \sum^{r}_{m=0} \binom{r}{m} \binom{m/2}{n} (-1)^m (-4)^n \\
&= \frac{1}{2^r} \sum^{r}_{m=2 n} \binom{r}{m} \binom{m/2}{n} (-1)^m (-4)^n
\end{align*}
For $r \geq 2n$ otherwise the coefficient vanishes.
I would be thankful if anyone can give me a hint.
| We seek
$$[z^k] \left(\frac{1-\sqrt{1-4z}}{2}\right)^r.$$
where $k\ge r$ and we get zero otherwise since $\frac{1-\sqrt{1-4z}}{2}
= z + \cdots.$
Using the residue operator this is
$$\; \underset{z}{\mathrm{res}} \;
\frac{1}{z^{k+1}} \left(\frac{1-\sqrt{1-4z}}{2}\right)^r.$$
Now put $\frac{1-\sqrt{1-4z}}{2} = w$ so that $z = w(1-w)$ and $dz\; =
(1-2w) \; dw$ to get
$$\; \underset{w}{\mathrm{res}} \;
\frac{1}{w^{k+1} (1-w)^{k+1}} w^r (1-2w)
= \; \underset{w}{\mathrm{res}} \;
\frac{1}{w^{k-r+1} (1-w)^{k+1}} (1-2w)
\\ = {2k-r\choose k} - 2 {2k-r-1\choose k}
= {2k-r\choose k} - 2 \frac{k-r}{2k-r} {2k-r\choose k}
\\ = \frac{r}{2k-r} {2k-r\choose k}.$$
Following the other post we may write
$$\sum_{k\ge r} \frac{r}{2k-r} {2k-r\choose k} z^k
= z^r \sum_{k\ge 0} \frac{r}{2k+r} {2k+r\choose k} z^k.$$
The residue operator returns zero when $k\lt r.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the maximum of $\sin(A)+\sin(B)+\sin(C)$ for $\triangle ABC$. (Without Jensen Inequality)
Find the maximum of $\sin(A)+\sin(B)+\sin(C)$ for $\triangle ABC$. (Without Jensen Inequality)
Proof of Jensen Inequality:
\begin{align}
&\text{let } f(x)=\sin x. \\
\ \\
\Rightarrow & f(A)+f(B)+f(C) \\
& =3 \bigg( \frac 1 3 f(A) + \frac 1 3 f(B)+ \frac 1 3 f(C) \bigg) \\
& \leq 3 \Bigg( f \bigg( \frac 1 3 A + \frac 1 3 B + \frac 1 3 C \bigg) \Bigg) \\
& = 3\Bigg(f\bigg( \frac {A+B+C} 3 \bigg) \Bigg) \\
& = 3\big(f(60)\big) & (\because A+B+C=180) \\
&=3\sin 60 = 3 \cdot \frac {\sqrt{3}} 2 = \frac {3\sqrt{3}}{2}.
\end{align}
I just wondered if there is another precalculus solution to this. Is there another solution to this?
| Your solution is good; if you want another solution, use $C=\pi-A-B$ and then you look for the maximum value of
$$F=\sin (A)+\sin (B)+\sin (A+B)$$ Compute the partial derivatives
$$\frac{\partial F}{\partial A}=\cos (A)+\cos (A+B)=0\qquad \qquad \frac{\partial F}{\partial B}=\cos (B)+\cos (A+B)=0$$ Then, subtracting, $\cos(A)=\cos(B)$ and $A=B$ and then
$$\frac{\partial F}{\partial A}=\cos (A)+\cos (2A)=0$$
Use the double angle formula $$2\cos^2 (A)+\cos (A)-1=0$$ Solve the quadratic in $\cos(A)$ which gives $\cos(A)=\frac 12$ and then $A=B=C=\frac \pi 3$
| {
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"question_score": "1",
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Ellipse in the triangle Find the equation of an ellipse if its center is S(2,1) and the edges of a triangle PQR are tangent lines to this ellipse. P(0,0), Q(5,0), R(0,4).
My attempt: Let take a point on the line PQ. For example (m,0). Then we have an equation of a tangent line for this point: $(a_{11}m+a_1)x+(a_{12}m+a_2)y+(a_1m+a)=0$, where $a_{11}$ etc are coefficients of our ellipse: $a_{11}x^2+2a_{12}xy+a_{22}y^2+2a_1x+2a_2y+a=0$. Now if PQ: y=0, then $(a_{11}m+a_1)=0$, $a_{12}m+a_2=1$, $a_1m+a=0$.I've tried this method for other 2 lines PR and RQ and I got 11 equations (including equations of a center)! Is there a better solution to this problem?
| Translate the center to the origin.. the triangle translates to
The translated points become $(-2,-1), (3,-1), (-2,3)$
I considered this on the projective plane, I don't know if that makes it any easier, but that is what I did... if you don't know projective geometry, just set $z = 1$ in everything that follows.
$\begin {bmatrix} x&y&z\end{bmatrix}
\begin {bmatrix} A&B&0\\B&C&0\\0&0&D\end{bmatrix}
\begin{bmatrix} x\\y\\z\end{bmatrix} = 0$
Describes a cone, which when it intersects the plane $z = 1$ forms the ellipse
i.e. $Ax^2 + 2Bxy + Cy^2 + D = 0$
The planes $x + 2z = 0, y+z = 0$ and $4x + 5y - 7z = 0$ are tangent to the cone.
Consider the triplet $(B,-A, 1)$
$\begin {bmatrix} A&B&0\\B&C&0\\0&0&D\end{bmatrix}
\begin{bmatrix} B\\-A\\1\end{bmatrix} = \begin{bmatrix} 0 \\ B^2 - AC \\ D \end{bmatrix}$
If $A = 1$ this point lies in the plane $y + z = 0.$
And if $D = B^2 - AC$
$\begin{bmatrix} B&-1&1\end{bmatrix}\begin{bmatrix} 0 \\ B^2 - AC \\ B^2 - AC \end{bmatrix}= 0$
The point is on our cone.
Similarly $(-\frac {C}{2},\frac {B}{2}, 1)$
$\begin {bmatrix} A&B&0\\B&C&0\\0&0&B^2 - AC\end{bmatrix}
\begin{bmatrix} -\frac {C}{2}\\ \frac {B}{2}\\1\end{bmatrix} =
\begin{bmatrix} \frac 12 (B^2 - AC) \\ 0\\ B^2 - AC\end{bmatrix}$
If $C = 4$ the point is on the plane $x + 2z = 0$ and on the cone. Our matrix thus far.
$\begin {bmatrix} 1&B&0\\B&4&0\\0&0&B^2-4\end{bmatrix}$
We just need an equation for the 3rd point of tangency.
$(\frac {4C - 5B}{7}, \frac {5A-4B}{7}, 1)$
If it is in the plane $4x + 5y - 7z = 0$
$4\frac {4C - 5B}{7} + 5\frac {5A - 4B}{7} - 7 = 0\\
4\frac {16 - 5B}{7} + 5\frac {5 - 4B}{7} - 7 = 0\\
64 - 40 B + 25 - 49 = 0\\
40-40B = 0\\
B = 1$
And finally we need to check to see if this point is on the cone.
$\begin {bmatrix} 1&1&0\\1&4&0\\0&0&-3\end{bmatrix}
\begin{bmatrix} \frac {11}{7}\\ \frac {1}{7}\\1\end{bmatrix} =
\begin{bmatrix} \frac {12}7 \\ \frac {15}{7} \\-3\end{bmatrix}$
$\begin{bmatrix} \frac {11}{7} & \frac {1}{7} & 1\end{bmatrix} =
\begin{bmatrix} \frac {12}7 \\ \frac {15}{7} \\-3\end{bmatrix} = \frac {132 + 15}{49} - 3 = 0$
$x^2 + 2xy + 4y^2 - 3 =0$
And finally translate back to the original coordinates
$(x-2)^2 + 2(x-2)(y-1) + 4(y-1)^2 - 3 = 0$
| {
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Recurrence relations in the generating function of binary partitions Let $b(n)$ denote the number of binary partitions of $n$, that is, the number of partitions of $n$ as the sum of powers of $2$. Define
\begin{equation*}
F(x) = \sum_{n=0}^\infty b(n)x^n
= \prod_{n=0}^\infty \frac{1}{1-x^{2^{n}}}.
\end{equation*}
Clearly, $F(x)$ satisfies the functional equation
\begin{equation*}
(1-x)F(x) = F(x^2).
\end{equation*}
Then, we have recurrence relation
$b(2n+1)=b(2n)$ and $b(2n) = b(2n-2) + b(n), n\ge1$.
How to derive these recurrence relation from the given informations?
What I tried:
We have
\begin{equation*}
\sum_{n=0}^\infty b(2n)x^n
= \prod_{n=0}^\infty \frac{1}{1-x^{2^{2n}}}
\end{equation*}
and
\begin{equation*}
\sum_{n=0}^\infty b(2n+1)x^n
= \prod_{n=0}^\infty \frac{1}{1-x^{2^{2n+1}}}.
\end{equation*}
But, I didn't get those recurrence relation.
| The sequence $\,b(n)\,$ is OEIS sequence A018819 "Binary partition function: number of partitions of n into powers of 2.". The FORMULA section
states:
$a(2m+1) = a(2m), a(2m) = a(2m-1) + a(m)$. Proof: If $n$ is odd there is a part of size $1$; removing it gives a partition of $n - 1$. If $n$ is even either there is a part of size $1$, whose removal gives a partition of $n - 1$, or else all parts have even sizes and dividing each part by $2$ gives a partition of $n/2$.
You asked
Then, we have recurrence relation
$b(2n+1)=b(2n)$ and $b(2n) = b(2n-2) + b(n), n\ge1$.
How to derive these recurrence relation from the given informations?
but you did not explicitly state that you wanted to use the ordinary
generating function $\,F(x).\,$ The first recurrence relation for
$\,b(2n+1)\,$ is already given in the FORMULA entry. For $\,b(2n)\,$
we also get from the entry $\,b(2n) = b(2n-1)+b(n)\,$ but we already
know that if $\,n\ge 1\,$ then $\,b(2n-1)=b(2n-2)\,$ and thus
$\,b(2n)=b(2n-2)+b(n)\,$ as the second recurrence relation.
Your attempt using the g.f. $\,F(x)\,$ was flawed, but can be fixed up.
$$ F(x) = \frac1{1-x}\prod_{n=1}^\infty \frac{1}{1-x^{2^{n}}} =
\sum_{n=0}^\infty b(n)x^n. \tag{1}$$
Substitute $\,-x\,$ for $\,x\,$ to get
$$ F(-x) = \frac1{1+x}\prod_{n=1}^\infty \frac{1}{1-x^{2^{n}}} =
\sum_{n=0}^\infty b(n)(-x)^n. \tag{2}$$
Notice that
$$ \frac12\Big(\frac1{1-x} + \frac1{1+x}\Big) = \frac1{1-x^2}, \quad
\frac12\Big(\frac1{1-x} - \frac1{1+x}\Big) = \frac{x}{1-x^2}. \tag{3} $$
Add equation $(2)$ to $(1)$ and use equation $(3)$ to get the even part of $\,F(x)\,$
$$ \frac12(F(x)+F(-x)) =
\frac1{1-x^2}\prod_{n=1}^\infty \frac{1}{1-x^{2^{n}}} =
\sum_{n=0}^\infty \,b(2n)\,x^{2n}.\tag{4}$$
Subtract equation $(2)$ from $(1)$ and use equation $(3)$ to get the odd part
$$ \frac12(F(x)-F(-x)) =
\frac{x}{1-x^2}\prod_{n=1}^\infty \frac{1}{1-x^{2^{n}}} =
\sum_{n=0}^\infty \,b(2n+1)\,x^{2n+1}.\tag{5}$$
Divide both sides of equation $(5)$ by $\,x\,$ to get
$$ \frac1{1-x^2}\prod_{n=1}^\infty \frac{1}{1-x^{2^{n}}} =
\sum_{n=0}^\infty \,b(2n+1)\,x^{2n}.\tag{6}$$
Equate coefficients in equations $(4)$ and $(6)$ to get $\,b(2n+1)=b(2n).\,$
From equation $(1)$ we get
$$ (1-x)F(x) = \prod_{n=1}^\infty \frac{1}{1-x^{2^{n}}} = F(x^2) \tag{7}$$
and use equation $(1)$ again with $\,x\,$ replaced with $\,x^2\,$ to get
$$ F(x^2) = \sum_{n=0}^\infty\, b(n)\,x^{2n} \tag{8}$$
and also
$$ (1-x)F(x) = 1+\sum_{n=1}^\infty \,(b(n)-b(n-1))\, x^n. \tag{9} $$
Using equations $(7)$,$(8)$, and $(9)$ together to get
$$ \sum_{k=0}^\infty\, b(k)\,x^{2k} =
1+\sum_{n=1}^\infty \,(b(n)-b(n-1))\, x^n. \tag{10} $$
Equating coefficients of $\,x^n\,$ when $\,n\,$ is
even and odd gives the recurrence relations in the OEIS entry.
By the way, in equation $(4)$ replace $\,x^2\,$ with $\,x\,$ to get
$$ \frac1{1-x}F(x) = \sum_{n=0}^\infty \,b(2n)\,x^n \tag{11}$$
and since $\,b(2n+1)=b(2n)\,$ the same left side is the generating
function of $\,b(2n+1).\,$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4317636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Rolling a die - Conditional Probability A die is thrown repeatedly.
Let $X$ ~ First 5 is thrown and $Y$ ~ First 6 is thrown
Calculate $\mathbb{E}(X|Y=3)$
You may use the identity: $\sum_{n=k}^\infty nz^{n-k} = \frac{1}{(1-z)^2}+\frac{k-1}{1-z}$
I know from the definition of expectation, we have:
$\mathbb{E}(X|Y=3) = (1*\frac{1}{5})+(2*\frac{4}{5} * \frac{1}{5}) + (3* \frac{4}{5}* \frac{4}{5} * 0) + (5* \frac{4}{5} * \frac{4}{5} * 1 * \frac{5}{6} * \frac{1}{6}) + (6* \frac{4}{5} * \frac{4}{5} * 1 * \frac{5}{6} * \frac {5}{6} * \frac{1}{6}) + ...$, where every following term, has an extra '$*\frac{5}{6}$' term and constant increases by 1.
However I am unsure of how to apply this to the identity given to find the value of the infinite sum?
| $\mathbb{P}(X=1|Y=3)=\frac{1}{5},\mathbb{P}(X=2|Y=3)=\frac{4}{25},\mathbb{P}(X>3|Y=3)=1−(\frac{1}{5}+\frac{4}{25})=\frac{16}{25}.$
Then we multiply these by the expected results, i.e. $1,2$ and $9$, giving $\mathbb{E}(X|Y=3)=(1∗\frac{1}{5})+(2∗\frac{4}{25})+(9∗\frac{16}{25})=6.28$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4320165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
Find inverse of element in a binary field The question states:
Let us consider the field $GF(2^4)$ with multiplication modulo $x^4+ x^3+1$
Find all y such that $1010(y + 0011) = 1111$, in other words find y that satisfies $(x^3+x)(y +x+1) = x^3+x^2+x+1$
I tried to find the inverse for element $1010$ to multiply both sides of the equation and continue from there, I tried by solving the equation:
$(x^3+x)(ax^3+bx^2+cx+d) = 1$
But the solution I get is that $a=0,b=1,c=1,d=0$ which doesn't produce the inverse element.
What is the inverse element of $(1010)$ in this field and how can I arrive at that solution?
Any help is greatly appreciated.
| We know $$x^4+x^3+1=0 .\tag1$$
It will be useful to know
$$
x^4 = x^3+1\tag2$$
$$
x^5 = x^4+x=(x^3+1)+x = x^3+x+1\tag3$$
$$
x^6 = x^5+x^2 = (x^3+x+1)+x^2 = x^3+x^2+x+1\tag4
$$
These will be useful for doing a calculation like
$(x^3+x)(ax^3+bx^2+cx+d) = 1$ that the OP proposes.
Compute
\begin{align}
1 &= (x^3+x)(ax^3+bx^2+cx+d)
\\
&=(a)x^6+(b)x^5+(c+a)x^4+(d+b)x^3+(c)x^2+(d)x+0
\\
&=(a+b+d+c+a+b)x^3+(a+c)x^2+(a+b+d)x+(a+b+c+a)
\\
&=(d+c)x^3+(a+c)x^2+(a+b+d)x+(b+c)
\end{align}
Thus
\begin{align}
d+c&=0\\a+c&=0\\a+b+d&=0\\b+c&=1
\end{align}
So $d=c$, then $a=c$, then $b=a+d=c+c=0$, then $c=d=a=1$. So
$$
(x^3+x)^{-1} = x^3+x+1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4324627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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} |
Find a lower bound of $-\frac{1}{4}\sqrt{(1-2a+x)^2-8(-3-4a-3x-2ax)}-\frac{1}{4}(1-2a+x)$ I have to find a lower bound (that does not depend on $x$) of the following quantity:
$$-\frac{1}{4}\sqrt{(1-2a+x)^2-8(-3-4a-3x-2ax)}-\frac{1}{4}(1-2a+x)$$
where $x\geq 0,\, a>0$.
Really I have tried to observe that:
$-\frac{1}{4}\sqrt{(1-2a+x)^2-8(-3-4a-3x-2ax)}-\frac{1}{4}(1-2a+x)>-\frac{1}{2}\sqrt{(1-2a+x)^2-8(-3-4a-3x-2ax)}=A$
but then I can't find a lower bound of A...can you help me?
| Let $a$ be a fixed positive real. Let $x \geqslant 0$ be allowed to vary.
Let $A=2,\quad B=1-2a+x,\quad C=-3-4a-3x-2ax.$
Consider the quadratic $Ay^2+By+C=0.$
Then your expression is the left-hand root of this quadratic:
$$-\frac{1}{4}\sqrt{(1-2a+x)^2-4A(-3-4a-3x-2ax)}-\frac{1}{4}(1-2a+x).$$
(For we see that $A>0,\; C<-3,\; B^2-4AC>0$ for any $x$, and thus the parabola always has two roots.)
So I think you cannot have such a bound. For we can, by increasing $x$ more and more, move the axis of symmetry of the parabola further and further to the left. This must in turn push the left-hand root towards $-\infty$.
The axis of symmetry: $\;y= \frac{-B}{2A}.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Some hypergeometric transformation In Concrete Mathematics, the identity
$$\sum_{k\geq 0}\binom{m+r}{m-n-k}\binom{n+k}{n}x^{m-n-k}y^k\\=\sum_{k\geq 0}\binom{-r}{m-n-k}\binom{n+k}{n}(-x)^{m-n-k}(x+y)^{k}$$
is proven, where $n\geq 0$ and $m$ are integers, and $x,y,r$ are complex numbers.
From this identity, the hypergeometric equality
$$F(a,-n;c;z)=\frac{(a-c)^{\underline n}}{(-c)^\underline n}F(a,-n;1-n+a-c;1-z)$$
can then "clearly" be derived, where $n\geq 0$ is an integer, and $a,b,c$ are complex numbers.
I don't understand how this hypergeometric equality can be proven using the first identity. Any suggestions?
| We represent the binomial identity
\begin{align*}
\sum_{k\geq 0}&\binom{m+r}{m-n-k}\binom{n+k}{n}x^{m-n-k}y^k\\
&=\sum_{k\geq 0}\binom{-r}{m-n-k}\binom{n+k}{n}(-x)^{m-n-k}(x+y)^k\tag{1}
\end{align*}
using hypergeometric series by following the presentation in Concrete Mathematics.
LHS
We start with the LHS of (1) and calculate the quotient $\frac{t_{k+1}}{t_{k}}$ of consecutive terms of the sum. We obtain
\begin{align*}
t_k&=\binom{m+r}{m-n-k}\binom{n+k}{n}x^{m-n-k}y^k\\
&=\frac{(m+r)!(n+k)!}{(m-n-k)!(r+n+k)!n!k!}x^{m-n-k}y^k\\
\\
\frac{t_{k+1}}{t_k}&=\frac{(n+k+1)(m-n-k)}{(r+n+k+1)(k+1)}\,\frac{y}{x}\\
&=\frac{(k+\color{blue}{n+1})(k+\color{blue}{n-m})}{(k+\color{blue}{n+r+1})(k+1)}\left(\color{blue}{-\frac{y}{x}}\right)
\end{align*}
The first summand $k=0$ of the LHS of (1) is $\binom{m+r}{m-n}x^{m-n}$ and we conclude
\begin{align*}
\sum_{k\geq 0}&\binom{m+r}{m-n-k}\binom{n+k}{n}x^{m-n-k}y^k\\
&=\binom{m+r}{m-n}x^{m-n}\color{blue}{F\left(n+1,n-m;n+r+1;-\frac{y}{x}\right)}\tag{2.1}
\end{align*}
RHS
We proceed in the same way with the RHS and obtain
\begin{align*}
u_k&=\binom{-r}{m-n-k}\binom{n+k}{n}(-x)^{m-n-k}(x+y)^k\\
&=\frac{(-r)!(n+k)!}{(m-n-k)!(-r-m+n+k)!n!k!}\left(-x\right)^{m-n-k}(x+y)^k\\
\\
\frac{u_{k+1}}{u_k}&=\frac{(n+k+1)(m-n-k)(-1)}{(-r-m+n+k+1)(k+1)}\,\frac{x+y}{x}\\
&=\frac{(k+\color{blue}{n+1})(k+\color{blue}{n-m})}{(k+\color{blue}{n-m-r+1})(k+1)}\,\color{blue}{\frac{x+y}{x}}
\end{align*}
The first summand $k=0$ of the RHS of (1) is $\binom{-r}{m-n}(-x)^{m-n}$ and we conclude
\begin{align*}
\sum_{k\geq 0}&\binom{-r}{m-n-k}\binom{n+k}{n}(-x)^{m-n-k}(x+y)^{k}\\
&=\binom{-r}{m-n}(-x)^{m-n}\color{blue}{F\left(n+1,n-m;n-m-r+1;\frac{x+y}{x}\right)}\tag{2.2}
\end{align*}
Replacing $n$ with $q$ in (2.1) and (2.2) we obtain from (1)
\begin{align*}
&\color{blue}{F\left(q+1,q-m;q+r+1;-\frac{y}{x}\right)=\binom{-r}{m-q}\binom{m+r}{m-q}^{-1}(-1)^{m-q}}\\
&\qquad \color{blue}{\cdot F\left(q+1,q-m;q-m-r+1;\frac{x+y}{x}\right)}\tag{2.3}
\end{align*}
Substitutions
Finally we show the identity (2.3) can be written as
\begin{align*}
\color{blue}{F(a,-n;c;z)=\frac{(a-c)^{\underline n}}{(-c)^{\underline n}}F(a,-n;1-n+a-c;1-z)}\tag{3.1}
\end{align*}
Comparing the LHS of (2.3) and (3.1) we use the substitutions
\begin{align*}
a&:=q+1\\
n&:=-q+m\\
c&:=q+r+1\tag{3.2}\\
z&:=-\frac{y}{x}
\end{align*}
With these substitutions the LHS of (2.3) and (3.1) coincide. The upper parameters $q+1$ and $q-m$ are the same at both sides of (2.3). We now use the substitutions to derive the argument, the lower parameter and the factor in front of the hypergeometric series.
*
*From $z=-\frac{y}{x}$ we see
\begin{align*}
1-z=1+\frac{y}{x}=\frac{x+y}{x}
\end{align*}
and the arguments coincide.
*Looking at the lower parameter $n-m-r+1$ in (2.3) we obtain with (3.2)
\begin{align*}
q-m-r+1&=(-n)-r+1\\
&=(-n)-(c-q-1)+1\\
&=(-n)-(c-a)+1\\
&=1-n+a-c
\end{align*}
in accordance with the lower parameter of the RHS in (3.1).
*We obtain
\begin{align*}
\binom{-r}{m-q}&=\binom{a-c}{n}=(a-c)^{\underline{n}}\\
\binom{m+r}{m-q}(-1)^{m-q}&=\binom{-q-r-1}{m-q}(-1)^{m-q}=\binom{-c}{n}\\
&=(-c)^{\underline{n}}
\end{align*}
and the claim (3.1) follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4326420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the area of the shaded region in the triangle below? For reference: In figure $G$ is the centroid of the triangle $ABC$; if the area of the $FGC$ triangle is $9m^2$, the area of the FGB triangle is $16m^2$
Calculate the area of the shaded region. (Answer:$7m^2$)
If possible by geometry
My progress:
$S_{FGC} = \frac{b.h_1}{2} = \frac{FG.h_1}{2}\implies FG = \frac{18}{h_1}\\
S_{FGB}=\frac{b.h_2}{2} = \frac{FG.h_2}{2} \implies FG = \frac{32}{h_2}\\
\therefore \frac{18}{h_1} = \frac{32}{h_2}\implies \frac{h_1}{h_2} = \frac{32}{18}=\frac{16}{9}\\
S_{ABG} = S_{BCG} = S_{ACG}$
...??? I'm not able to develop this
| With the usual conventions:
\begin{aligned}
\overrightarrow{GF}\times\overrightarrow{GB} &= (0,0,32)\\
\overrightarrow{GF}\times\overrightarrow{GC} &= (0,0,-18)\\
\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC} &= \overrightarrow{0}
\end{aligned}
The area you search:
\begin{multline*}
\frac{1}{2}\big\lvert\overrightarrow{GF}\times\overrightarrow{GA}\big\rvert=\frac{1}{2}\Big\lvert\overrightarrow{GF}\times\big(-\overrightarrow{GB}-\overrightarrow{GC}\big)\Big\rvert
=\frac{1}{2}\big\lvert-\overrightarrow{GF}\times\overrightarrow{GB}-\overrightarrow{GF}\times\overrightarrow{GC}\big\rvert=\frac{1}{2}\big\lvert(0,0,-14)\big\rvert = 7
\end{multline*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4326621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Prove $5^n > n^5 $ for $n≥6$ How can I prove
$5^n > n^5 $ for $n≥ 6 $
using mathematical induction ?
So my solution is
Step 1: For $n = 6$
$5^6 = 15625$, $6^5 = 7776$
then $5^n > n^5 $ is true for $n = 6 $
Step 2: Assume it's true for $n = k$, $5^k > k^5 $
Step 3: Prove it's true for $n = k+1$
$5^{k+1} > (k+1)^5 $
Here when things get messy I don't know how to prove it for $n = k+1$
| $$P(n): 5^n > n^5 \text{ : } \forall n \ge6$$
*
*Base Case: $5^6 >6^5$ is True
*Inductive Case:$P(k)\implies P(k+1)$
The induction step, proves that if the statement holds for any given case $n = k$, then it must also hold for the next case $n = k + 1$. These two steps establish that the statement holds for every natural number $n$. The base case does not necessarily begin with $n = 0$, but often with $n = 1$, and possibly with any fixed natural number $n = N$(here, $N = 6$), establishing the truth of the statement for all natural numbers $n\ge N.$
$$\color{blue}{P(k): 5^k > k^5 \implies (5^k5 = 5^{k +1})>5k^5} $$
To prove $P(k+1) $ is true we need to prove $(5^{k+1} = \color{blue}{5^k.5 > 5k^5} )> (k+1)^5$
Now,
$\color{blue}{1< 5k < 10k^2 < 10k^3< 5k^4}$ is true for $k\ge3 \implies \forall k\ge6$ You can prove this by considering that $P(1<n<5)$ is not true but we know this is obvious.
$k^4 = k^4 \implies \color{red}{5k^4 = 5k^4} \text{ for } (k\ge 1 \implies \text{ also true for } k\ge 6)$
Consider $\color{red}{5k^4 = 5k^4}$ for $k> 5$ We have $\color{blue}{5k^4 \le \color{green}k\times k^4} \text{As } k> 5 $
Combining the above results we have
$\color{blue}{1< 5k < 10k^2 < 10k^3< 5k^4 < k^5} \text{ is true for} k > 5 \implies \text{ also true for } k > 6$
$$\begin{align*}
(k+1)^5
& = k^5 + 5k^4 + 10k^3 + 10k^2 + 5k + 1\\
& < k^5 + k^5 + k^5 + k^5 + k^5 = 5k^5 < 5^{k+1}
\text{thus } P(k + 1) \text{is True}
\end{align*}$$
Now, if True of $P(k)$ $\implies $ True of $P(k+ 1) $ for $k> 6$ then $P(N>6) $ is True.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find a point $P_2$ on an ellipse, whose chord with $P_1$ is a max distance $d$ from its nearest side I'm not sure if this solution is available in closed form, but after drawing it out I do think there will be two unique solutions always. I unfortunately have no clue where to start.
Given:
*
*An ellipse with x radius $a$, y radius $b$ and centre point $(0, 0)$
*A point on that ellipse $P_1$
*A distance $d$
Find all points $P_{2j}$ (i.e. find their coordinates) on the ellipse which satisfy the following condition (where $C$ is the chord joining $P_1$ and $P_{2j}$):
*
*The distance between $C$ and a line which is tangent to the ellipse and parallel to $C$ (on the side of the smaller segment formed by $C$ i.e. the "nearer" side) is equal to $d$
In other words, the maximum distance between $C$, and the nearer side of the ellipse from $C$, must equal $d$.
The diagram I've drawn below is an example of this. $P_{21}$ and $P_{22}$ are the desired points, whereas $P_{23}$ is an example of an invalid point. (The distances are not fully accurate)
The reason I believe the solution is closed, is because as you sweep the point $P_2j$ around the ellipse, the distance (required to be $d$) increases, reaches a turning point, and then decreases.
| The parametric equation of the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 $ is
$ r = (x, y) = ( a \cos t , b \sin t ) \hspace{20pt} t \in \mathbb{R} $
Now $P_1 = (x_1, y_1) = (a \cos t_1, b \sin t_1 ) $
Suppose $Q = (a \cos s, b \sin s)$ is the point where the tangent is to be drawn.
The unit normal vector at $Q$ is
$n = \dfrac{(b \cos s , a \sin s )}{\sqrt{b^2 \cos^2 s + a^2 \sin^2 s}} $
The distance between the tangent line and the line $P_1 P_2 $ that is parallel to it is
$(P_1 Q) \cdot n = \dfrac{ ab( (\cos s - \cos t_1 ) \cos s + (\sin s - \sin t_1 ) \sin s ) }{\sqrt{b^2 \cos^2 s + a^2 \sin^2 s}} = d $
Simplifying,
$(P_1 Q) \cdot n = \dfrac{ ab( 1 - ( \cos t_1 \cos s + \sin t_1 \sin s )) }{\sqrt{b^2 \cos^2 s + a^2 \sin^2 s}} = d $
Therefore,
$ d {\sqrt{b^2 \cos^2 s + a^2 \sin^2 s}} = a b( 1- ( \cos t_1 \cos s + \sin t_1 \sin s ) ) $
Squaring,
$ d^2 (b^2 \cos^2 s + a^2 \sin^2 s ) = a^2 b^2 ( 1 + \cos^2 t_1 \cos^2 s + \sin^2 t_1 \sin^2 s + \dfrac{1}{2} \sin 2 t_1 \sin 2 s - 2 ( \cos t_1 \cos s + \sin t_1 \sin s ) ) $
After using the identities $\cos^2 s = \frac{1}{2} (1 + \cos 2 s ) $ and $\sin^2 s = \frac{1}{2} (1 - \cos 2 s)$, the last equation becomes of the form
$A \cos s + B \sin s + C \cos 2 s + D \sin 2 s + E = 0 $
where
$A = 2 a^2 b^2 \cos t_1 $
$B = 2 a^2 b^2 \sin t_1$
$C = -\dfrac{1}{2} \left((a b) ^ 2 \cos(2 t_1) - d ^ 2 (b ^ 2 - a ^ 2) \right) $
$D = -\dfrac{1}{2} a^2 b^2 \sin(2 t_1) $
$E = \dfrac{1}{2} d ^ 2 (a ^ 2 + b ^ 2) - \dfrac{3}{2} a ^ 2 b ^ 2$
Which can be solved for $s$, using the substitution $z = \tan \dfrac{s}{2} $ that results in a quartic polynomial equation in $z$.
Once $s$ is found (there will be two solutions), the equation of the line $P_1 P_2 $ is
$ n \cdot ((x, y) - P_1) = 0 $
where $n = ( b \cos s, a \sin s ) $
This line we need to intersect with the ellipse to find $P_2$. This can be done as follows. Since $(x, y)$ is on the ellipse then $P_2 =(x_2, y_2) = (a \cos t_2 , b \sin t_2)$. Substitute this into the equation of the line, you get,
$ (b \cos s , a \sin s ) \cdot ( a (\cos t_2 - \cos t_1) , b (\sin t_2 - \sin t_1) ) = 0 $
Dividing through by $ ab $,
$ \cos s (\cos t_2 - \cos t_1) + \sin s (\sin t_2 - \sin t_1 ) = 0 $
In which $t_1$ and $s$ are known, and the unknown is $t_2$. This equation is of the form
$ \cos(t_2 - s) = \cos(t_1 - s) $
Since $t_2 \ne t_1$ then
$ t_2 - s = - (t_1 - s) $
from which,
$ t_2 = 2 s - t_1 $
and $P_2 = (a \cos t_2, b \sin t_2 ) $
As a numerical example, I've taken the ellipse with $a = 10, b = 5, t_1 = \dfrac{\pi}{3}$ and the distance $d = 3 $. The figure below shows the two solutions.
| {
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"question_score": "2",
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Let $x+y+z = 5$ where $x,y,z \in \mathbb{R}$, prove that $x^2+y^2+z^2\ge \frac{5}{3}$ My thinking:
Since $x+y+z = 5$, we can say that $x+y+z \ge 5$.
By basic fact: $x^2,y^2,z^2\ge 0$
If $x+y+z \ge 5$, then $\frac{\left(x+y+z\right)}{3}\ge \frac{5}{3}$
If $\frac{\left(x+y+z\right)}{3}\ge \frac{5}{3}$ and $x^2,y^2,z^2\ge 0$, hence $\frac{x^2+y^2+z^2}{3}\ge \frac{5}{3}$...???
I'm not sure about this, can someone help?
| $(x-1)^2+(y-1)^2+(z-1)^2\ge 0$. Hence $x^2+y^2+z^2-2(x+y+z)+3\ge 0$, so $x^2+y^2+z^2\ge 10-3=7>5/3$. Moreover $(x^2+y^2+z^2)/3\ge 7/3>5/3$
Your proof is not justified: the last "hence" statement does not follow from the previous statements.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4327738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Appropriate test for convergence of a series when numerator and denominator have same degree? Say we have a series like this:
$$
\sum^\infty_{n=1} \frac{16n^5+2n^2+4}{6n^5+10n+1}
$$
Somehow this confused me when I saw the same power in the numerator and denominator.
My thinking: for large enough $n$, we can assume that the series behaves like $\sum \frac{16n^5}{6n^5}=\sum \frac{8}{3}$, which is divergent. So I used the Limit Comparison Test with
$$
a_n = \frac{16n^5+2n^2+4}{6n^5+10n+1}
$$
And
$$
b_n = \frac{8}{3}
$$
To find
$$
\frac{a_n}{b_n}= \frac{16n^5+2n^2+4}{6n^5+10n+1} \times \frac{3}{8} = \frac{48n^5+6n^2+12}{48n^5+80n+8} = \frac{48+6 \frac{1}{n^3} + 12 \frac{1}{n^5}}{48+ 80 \frac{1}{n^4}+8 \frac{1}{n^5}} \to 1
$$, as $n \to \infty$.
So this means the initial series is divergent.
Is that the right approach in this case when numerator and denominator have the same degree or is there something more obvious/easier that I overlooked?
| We know that if a series is convergent then the sequence in the infinite sum must converge to zero. So when the sequence does not converge to zero, the series is divergent. Here, our sequence obviously converges to $\frac{16}{6}\neq 0$. Hence, the series is divergent.
| {
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"source": "stackexchange",
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} |
Solve $\sqrt{x-5}-\sqrt{9-x}\gt1,x\in\mathbb Z$
Solve $\sqrt{x-5}-\sqrt{9-x}\gt1,x\in\mathbb Z$
The statement tells us that $x\in[5,9]$. Also,
$$\sqrt{x-5}\gt1+\sqrt{9-x}$$
Since both sides are positive, we can square
$$x-5>1+9-x+2\sqrt{9-x}\\2x-15\gt2\sqrt{9-x}$$
$\implies 2x-15\gt0\implies x\gt7.5$
Since $x\in\mathbb Z\implies x=8,9$
But on back substitution, $x=8$ doesn't satisfy. Is there a way we could get the final answer without back sustitution?
| Starting from where you left off, since both sides of the inequality $$2x - 15 > 2\sqrt{9 - x}$$ are positive, the direction of the inequality is preserved if we square both sides, which yields
\begin{align*}
4x^2 - 60x + 225 & > 4(9 - x)\\
4x^2 - 60x + 225 & > 36 - 4x\\
4x^2 - 56x & > -189
\end{align*}
Since $(2a + b)^2 = 4a^2 + 4ab + b^2$, we can complete the square with $a = x$ and $b = 14$ to obtain
\begin{align*}
4x^2 - 56x + 196 & > 7\\
4(x^2 - 14x + 49) & > 7\\
(x - 7)^2 & > \frac{7}{4}
\end{align*}
Since $(8 - 7)^2 = 1 < \dfrac{7}{4}$, this eliminates $8$. Thus, the only integer solution is $x = 9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4331630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
find the value of the product of roots of this quadratic equation It is given that one of the roots of the quadratic equation : $x^2 + (p + 3)x - p^2 = 0$,
where $p$ is a constant, is negative of the other. The question is : find the value of the product of roots.
| Without applying the factorization of a "difference of two squares" or Viete's relations, we can still use the information stated in the problem. If we call the two roots of the quadratic equation $ \ r \ $ and $ \ -r \ \ , $ then we have
$$ r^2 \ + \ (p + 3)·r \ - \ p^2 \ \ = \ \ 0 $$
and $$ [-r]^2 \ + \ (p + 3)·[-r] \ - \ p^2 \ \ = \ \ r^2 \ - \ (p + 3)·r \ - \ p^2 \ \ = \ \ 0 \ \ . $$
This means that $ \ r^2 \ = \ -(p + 3)·r \ + \ p^2 \ = \ (p + 3)·r \ + \ p^2 \ \Rightarrow \ 2·(p + 3)·r \ = \ 0 \ \ . $ So either $ \ r \ = \ 0 \ $ or $ \ p \ = \ -3 \ \ . $
But if $ \ r \ = \ 0 \ = \ -r \ \ , \ $ then $ \ 0^2 \ + \ (p + 3)·0 \ - \ p^2 \ \ = \ \ 0 \ \ $ would require $ \ p \ = \ 0 \ \ , $ which would then make the quadratic equation $ \ x^2 \ + \ 3·x \ = \ 0 \ \ . $ But that polynomial factors as $ \ x · (x + 3) \ = \ 0 \ \ , $ so we couldn't have both roots equal to zero.
Instead, it must be that $ \ p \ = \ -3 \ \ , $ making the equation $ \ x^2 \ + \ 0·x \ - \ (-3)^2 \ = \ x^2 \ - \ 9 \ = \ 0 \ \ , $ for which the roots are given by $ \ r^2 \ = \ 9 \ \Rightarrow \ r \ = \ +3 \ , \ -3 \ \ ; \ $ the product of the roots is thus $ \ -9 \ \ . $
Another way to arrive at this conclusion is that $ \ y \ = \ x^2 \ + \ (p + 3)·x \ - \ p^2 \ \ $ is the equation of an "upward-opening" parabola, for which we want the $ \ x-$intercepts to be $ \ x \ = \ -r \ $ and $ \ x \ = \ r \ \ . $ Its axis of symmetry is located midway between these intercepts, so we have $ \ h \ = \ 0 \ $ in the "vertex form" of the parabola's equation, $ \ y \ = \ (x - 0)^2 \ - \ p^2 \ \ . $ (The vertex is definitely "below" the $ \ x-$axis at $ \ (0 \ , \ -p^2) \ \ , $ so we know these $ \ x-$intercepts exist.) The equation of the parabola is therefore $ \ y \ = \ x^2 \ - \ p^2 \ \ , \ $ making $ \ p + 3 \ = \ 0 \ \ $ and the rest of the argument above follows.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Compute explicitly $\sum_{x=1}^\infty\sum_{y=1}^\infty \frac{x^2+y^2}{x+y}(\frac{1}{2})^{x+y}$ It is possible to compute explicitly the following series?
$$\sum_{x=1}^\infty\sum_{y=1}^\infty \frac{x^2+y^2}{x+y}\Big(\frac{1}{2}\Big)^{x+y}$$
@Edit I tried to sum and subtract $2xy$ in the numerator.
In this way I get the following
\begin{align}
\sum_{x=1}^\infty\sum_{y=1}^\infty (x+y)\Big(\frac{1}{2}\Big)^{x+y}-2\sum_x x\Big(\frac{1}{2}\Big)^x\sum_y\frac{y}{x+y}\Big(\frac{1}{2}\Big)^y.
\end{align}
The first series should be easy to be computed. It remain the second one, in particular the following
\begin{align}
\sum_y\frac{y}{x+y}\Big(\frac{1}{2}\Big)^y=\sum_y\Big(\frac{1}{2}\Big)^y-x\sum_y\frac{1}{x+y}\Big(\frac{1}{2}\Big)^y.
\end{align}
Therefore the problem is reduced to computing:
\begin{align}
\sum_{y=1}^\infty \frac{1}{x+y}\Big(\frac{1}{2}\Big)^y.
\end{align}
But I don't know well how to do it.
| $$ \sum_{y=1}^\infty \frac{1}{x+y} \left(\frac{1}{2}\right)^y = \Phi(1/2,1,x) - 1/x$$
where $\Phi$ is the Lerch Phi function.
| {
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"answer_count": 2,
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} |
Solving a nonhomogenous system of eqns with one eigenvalue I have the system:
$\left[\begin{array}{@{}c@{}}
x' \\
y'
\end{array} \right]= \left[\begin{array}{@{}c@{}}
3&2 \\
-2 & -1
\end{array} \right]\left[\begin{array}{@{}c@{}}
x' \\
y'
\end{array} \right]+\left[\begin{array}{@{}c@{}}
2e^{-t} \\
e^{-t}
\end{array} \right]$
Which I should solve using the fundamental matrix.
So I start with obtaining the homogenous solution:
I find the eigenvalues;
\begin{pmatrix}
3-\lambda&2 \\
-2 & -1-\lambda
\end{pmatrix}
which gives the determinant: $\lambda^2-2\lambda+1=0$. Thus $\lambda_1=1$. Plugging that in the matrix in the original equation, I get that x=y. So a solution to the homogenous system would be: $y_h=e^{t}\left[\begin{array}{@{}c@{}}
1 \\
1
\end{array} \right]$
Since there is no second solution to the determinant, I would ideally form the fundamental matrix:
\begin{pmatrix}
e^{t} & e^0 \\
e^{t} & e^0
\end{pmatrix}
but this is to no avail. So how do I find the solution of this nonhomogenous system using the fundamental matrix with one eigenvalue?
Thanks
UPDATE:
I set up the generalized eigenvector formula
\begin{equation}
v_2(A-\lambda I)=v_2
\begin{pmatrix}
3-\lambda&2 \\
-2 & -1-\lambda
\end{pmatrix}=v_1
\end{equation}
\begin{equation}
v_2(A-\lambda I)=v_1=
\begin{vmatrix}
3-\lambda&2 & | 1 \\
-2 & -1-\lambda & |-1
\end{vmatrix}
\end{equation}
I now get as given by Moo, with Gaussian elimination, the matrix:
\begin{equation}
\begin{vmatrix}
1 &1 & | 1/2 \\
0 & 0 & |0
\end{vmatrix}
\end{equation}
and have the second eigenvector: $e_2=e^{t}\left[\begin{array}{@{}c@{}}
\frac{1}{2} \\
0
\end{array} \right]$
.
So the homogeneous solution is:
\begin{equation}
y_h=e^{\lambda_1 t}e_1+e^{\lambda_2t}e_2=e^{t}\left[\begin{array}{@{}c@{}}
1 \\
-1
\end{array} \right]+e^{t}\left[\begin{array}{@{}c@{}}
\frac{1}{2} \\
0
\end{array} \right]
\end{equation}
At this stage, it remains to find the particular solution. We know that it must be in the form of:
\begin{equation}
y_p=Ce^{-t}
\end{equation}
and thus the general solution is:
\begin{equation}
y_p=y_h+Ce^{-t}=e^{t}\left[\begin{array}{@{}c@{}}
1 \\
-1
\end{array} \right]+e^{t}\left[\begin{array}{@{}c@{}}
\frac{1}{2} \\
0
\end{array} \right]+Ce^{-t}
\end{equation}
But can this be said?
| You have what is called a deficient matrix, so you need to find a generalized eigenvector.
We have the system
$$\left[\begin{array}{@{}c@{}}
x' \\
y'
\end{array} \right]= \left[\begin{array}{@{}c@{}}
3&2 \\
-2 & -1
\end{array} \right]\left[\begin{array}{@{}c@{}}
x \\
y
\end{array} \right]+\left[\begin{array}{@{}c@{}}
2e^{-t} \\
e^{-t}
\end{array} \right]$$
We find a repeated eigenvalue of $\lambda_{1,2} = 1$ and we can find a single eigenvector of
$$v_1 = \begin{bmatrix} -1 \\ 1 \end{bmatrix}$$
Finding generalized eigenvectors is not a simple topic and requires work to learn the ins and outs, but in this case, we will use this example.
Solve for $v_2$ using the row-reduced-echelon-form (RREF) of $[A-\lambda I]v_2 = [A -I]v_2 = v_1 $
We get the augmented matrix
$$
\left[\begin{array}{rr|r}
1 & 1 & -\dfrac{1}{2} \\ 0 & 0 & 0
\end{array}\right]
$$
We can choose
$$v_2 = \begin{bmatrix} -\dfrac{1}{2} \\ 0 \end{bmatrix}$$
Update For the eigenvalues, we find
$$|A - \lambda I| = \begin{vmatrix}
-\lambda +3 & 2 \\
-2 & -\lambda -1 \\
\end{vmatrix} = (-\lambda+3)(-\lambda - 1) -2(-2) = \lambda ^2-2 \lambda +1 = 0$$
This results in
$$\lambda_{1, 2} = 1$$
To find the generalized eigenvector, we solve (you are actually using the eigenvalue $\lambda = 1$ below)
$$[A - \lambda I]v_2 = [A - 1 I]v_2 = \begin{bmatrix}
2 & 2 \\
-2 & -2 \\
\end{bmatrix}v_2 = v_1 = \begin{bmatrix}
-1 \\
1 \\
\end{bmatrix}$$
That is
$$\begin{bmatrix}
2 & 2 \\
-2 & -2 \\
\end{bmatrix}v_2 = \begin{bmatrix}
-1 \\
1 \\
\end{bmatrix}$$
As an augmented matrix, this is
$$
\left[\begin{array}{rr|r}
2 & 2 & -1 \\ -2 & -2 & 1
\end{array}\right]
$$
The RREF (Gaussian Elimination) is
$$
\left[\begin{array}{rr|r}
1 & 1 & -\dfrac{1}{2} \\ 0 & 0 & 0
\end{array}\right]
$$
From this, we can choose
$$v_2 = \begin{bmatrix} -\dfrac{1}{2} \\ 0 \end{bmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4335192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solve the equation $(2x^2-3x+1)(2x^2+5x+1)=9x^2$ Solve the equation $$(2x^2-3x+1)(2x^2+5x+1)=9x^2$$
The given equation is equivalent to $$4x^4+4x^3-11x^2+2x+1=9x^2\\4x^4+4x^3-20x^2+2x+1=0$$ which, unfortunately, has no rational roots. What else can I try?
| Recall that your last equation can be written as the product of two polynomials of degree two, $$(2x^2-4x+1)(2x^2+6x+1)=0$$ It is then easy to find the roots of both polynomials.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Closed-form for $\sum_{n=1}^{\infty} \left(m n \, \text{arccoth} \, (m n) - 1\right)$ I'm looking for a closed-form for the following sum:
$$\sum_{n=1}^{\infty} \left(m n \, \text{arccoth} \, (m n) - 1\right)$$ for $|m|>1.$
In a previous question of mine, the following similar sum was determined:
$$\sum_{n=1}^{\infty} (-1)^n \left(m n \, \text{arccoth} \, (m n) - 1\right)=\frac{1}{2}+\frac{1}{2} \log \left|\cot\left(\frac{\pi}{2m}\right)\right|+\frac{m}{2\pi} \left(\frac{1}{2}\text{Cl}_2\left(\frac{2\pi}{m}\right)-2\text{Cl}_2\left(\frac{\pi}{m}\right)\right)$$
for $|m|>1,$ where $\text{Cl}_2$ is the Clausen function of order 2.
I determined the following closed-forms for example as mentioned in that question:
$$\sum_{n=1}^{\infty} \left( 4n \, \text{arccoth} \, (4n)-1\right) = \frac{1}{2} - \frac{G}{\pi}- \frac{1}{4} \ln (2)$$
$$\sum_{n=1}^{\infty} \left(6n \, \text{arccoth} \, (6n) - 1\right) = \frac{1}{2} - \frac{3}{2\pi} \, \text{Cl}_2 \left( \frac{\pi}{3}\right)$$
$$\sum_{n=1}^{\infty} \left( 8n \, \text{arccoth} \, (8n) - 1\right) = \frac{1}{2} - \frac{2}{\pi} \, \text{Cl}_2 \, \left(\frac{\pi}{4}\right) - \frac{1}{4} \ln (2-\sqrt{2})$$
I suspect a similar method to that used by @skbmoore involving the Barnes G function might be applicable. Any help would be much appreciated.
Here is my attempt:
$$S(m) := \sum_{n=1}^{\infty} \left(m n \, \text{arccoth} \, (m n) - 1\right) = \frac{m}{2} \log \left( \prod_{n=1}^{\infty} \frac{1}{e} \left(\frac{1+1/(m n)}{1-1/(m n)}\right)^{n}\right)\\ = \frac{1}{2} + \frac{m}{2} \zeta' \left(-1,1-\frac{1}{m}\right)-\frac{m}{2}\zeta' \left(-1,1+\frac{1}{m}\right)+\frac{1}{2} \zeta' \left(0,1-\frac{1}{m}\right)+\frac{1}{2} \zeta' \left(0,1+\frac{1}{m}\right) \\ =\frac{1}{2} + \frac{m}{2} \left(\zeta' \left(-1,1-\frac{1}{m}\right)-\zeta' \left(-1,1+\frac{1}{m}\right)\right) \\+ \frac{1}{2} \ln\left( \Gamma \left(1-\frac{1}{m}\right)\Gamma \left(1+\frac{1}{m}\right)\right) - \frac{1}{2}\ln (2\pi) \\=\frac{1}{2} + \frac{m}{2} \left(\zeta' \left(-1,1-\frac{1}{m}\right)-\zeta' \left(-1,1+\frac{1}{m}\right)\right) + \frac{1}{2} \ln\left( \frac{\pi}{m} \csc \left(\frac{\pi}{m}\right)\right) - \frac{1}{2}\ln (2\pi)$$
However, I would like to write the solution in terms of the Clausen function if possible, like in the alternating sum, but I cannot see how to do that.
| To get your result in terms of the Clausen function of order $2$, we can use the identity $$ \begin{align} \zeta'(-1,x) &= -\log G(x+1) + x \log \Gamma (x) + \zeta'(-1) \\ &= - \log \Gamma(x) - \log G(x) + x \log \Gamma(x) + \zeta'(-1) \\ &= - \log G(x) + (x-1) \log \Gamma(x) + \zeta'(-1), \end{align}$$
which holds for $x>0$,
and the Barnes G reflection formula $$\log \left( \frac{G(1-x)}{G(1+x)} \right)= x\log \left(\frac{\sin \pi x}{\pi} \right) + \frac{\operatorname{Cl}_{2}(2 \pi x)}{2 \pi}, \quad 0 < x< 1, $$ which was used in skbmoore's answer to your previous question.
Then $$ \begin{align} \small\zeta' \left(-1, 1- \frac{1}{m}\right)- \zeta' \left(-1, 1+ \frac{1}{m}\right) &= \small- \log \left(\frac{G \left(1-\frac{1}{m}\right)}{G \left(1+ \frac{1}{m}\right)} \right) - \frac{1}{m} \log \left(\Gamma \left(1+ \frac{1}{m} \right) \Gamma \left(1- \frac{1}{m} \right) \right) \\ &= \small - \log \left(\frac{G \left(1-\frac{1}{m}\right)}{G \left(1+ \frac{1}{m}\right)} \right) - \frac{1}{m} \log \left(\frac{1}{m}\Gamma \left(\frac{1}{m}\right) \Gamma \left(1- \frac{1}{m} \right) \right) \\ &= \small- \frac{1}{m}\log \left(\frac{\sin \left(\frac{\pi}{m} \right)}{\pi} \right) - \frac{\operatorname{Cl}_{2} \left(\frac{2 \pi}{m} \right)}{2 \pi}- \frac{\log\left(\frac{\pi}{m} \csc\left(\frac{\pi}{m}\right)\right)}{m}. \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4343353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How many ways to deal with the integral $\int \frac{d x}{\sqrt{1+x}-\sqrt{1-x}}$? I tackle the integral by rationalization on the integrand first.
$$
\frac{1}{\sqrt{1+x}-\sqrt{1-x}}=\frac{\sqrt{1+x}+\sqrt{1-x}}{2 x}
$$
Then splitting into two simpler integrals yields $$
\int \frac{d x}{\sqrt{1+x}-\sqrt{1-x}}=\frac{1}{2}\left [\underbrace{\int\frac{\sqrt{1+x}}{x}}_{J} d x+\underbrace{\int\frac{\sqrt{1-x}}{x} d x}_{K}\right]
$$
To deal with $J$, we use rationalization instead of substitution. $$
\begin{aligned}
J &=\int \frac{\sqrt{1+x}}{x} d x \\
&=\int \frac{1+x}{x \sqrt{1+x}} d x \\
&=2 \int\left(\frac{1}{x}+1\right) d(\sqrt{1+x}) \\
&=2 \int \frac{d(\sqrt{1+x})}{x}+2 \sqrt{1+x} \\
&=2 \int \frac{d(\sqrt{1+x})}{(\sqrt{1+x})^{2}-1}+2 \sqrt{1+x} \\
&=\ln \left|\frac{\sqrt{1+x}-1}{\sqrt{1+x}+1} \right| +2 \sqrt{1+x}+C_{1}
\end{aligned}
$$
$\text {Replacing } x \text { by } -x \text { yields }$
$$
\begin{array}{l} \\
\displaystyle K=\int \frac{\sqrt{1-x}}{-x}(-d x)=\ln \left|\frac{\sqrt{1-x}-1}{\sqrt{1-x}+1}\right|+2 \sqrt{1-x}+C_{2}
\end{array}
$$
Now we can conclude that $$
I=\sqrt{1+x}+\sqrt{1-x}+\frac{1}{2}\left(\ln \left|\frac{\sqrt{1+x}-1}{\sqrt{1+x}+1}\right|+\ln \left|\frac{\sqrt{1-x}-1}{\sqrt{1-x}+1}\right|\right)+C
$$
My question is whether there are any simpler methods such as integration by parts , trigonometric substitution, etc…
Please help if you have. Thank you for your attention.
| Here is another approach, $$\int\dfrac{\sqrt{1+x}+\sqrt{1-x}}{2x}dx=\int\dfrac{\sqrt{2+2\sqrt{1-x^2}}}{2x}dx=\frac1{\sqrt2}\int\dfrac{\sqrt{1+\sqrt{1-x^2}}}xdx$$
Let, $x=\sin2\theta$,
$$\int\dfrac{\cos\theta}{\sin2\theta}\times2\cos2\theta \;d\theta=\int\dfrac{\cos2\theta}{\sin\theta}d\theta=\int\csc\theta-2\sin\theta \;d\theta$$$$=\ln|\csc\theta-\cot\theta|+2\cos\theta+C$$
Now we need to denote the result in term of $x$,$\cos2\theta=\sqrt{1-x^2}=2\cos^2\theta-1\;\Rightarrow\;2\cos\theta=\sqrt{2+2\sqrt{1-x^2}}$
$\csc\theta-\cot\theta=\dfrac{1-\cos\theta}{\sin\theta}=\dfrac{2\cos\theta-\cos2\theta-1}{\sin2\theta}=\dfrac{\sqrt{2+2\sqrt{1-x^2}}-\sqrt{1-x^2}-1}{x}$
Hence, $$\int\dfrac{\sqrt{1+x}+\sqrt{1-x}}{2x}dx=\sqrt{2+2\sqrt{1-x^2}}+\ln\left|\dfrac{\sqrt{2+2\sqrt{1-x^2}}-\sqrt{1-x^2}-1}{x}\right|+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4344261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
find the number of irrational roots of $\frac{4x}{x^2 + x + 3} + \frac{5x}{x^2 - 5x + 3} = -\frac{3}{2}$
Find the number of irrational roots of the equation $$\dfrac{4x}{x^2 + x + 3} + \dfrac{5x}{x^2 - 5x + 3} = -\dfrac{3}{2}.$$
I got a solution :
divide both denominator and numerator by $x$.
Let $x+\dfrac{3}{x} = y$. Then the equation becomes $$\dfrac{4}{y + 1} + \dfrac{5}{y - 5} = \dfrac{3}{2}.$$
Now simplifying this we get $y = -5, 3$.
Finally, $x+\dfrac{3}{x} = -5$ has 2 irrational roots, and $x+\dfrac{3}{x} = 3$ has 2 imaginary roots.
But my question is, since this question is for a competitive exam, is there any other quick approach to solve this question?
| Here is another way to solve this problem. It is up to you and @dxiv to decide whether this method is longer or shorter than yours.
We start by shifting the $2^{\text{nd}}$ term on LHS of the given identity to its RHS.
$$\dfrac{4x}{x^2 + x + 3} = -\dfrac{3}{2}-\dfrac{5x}{x^2 - 5x + 3}$$
When RHS is simplified, we have,
$$\dfrac{4x}{x^2 + x + 3} = -\dfrac{3x^2 - 5x + 9}{2x^2 - 10x + 6}.$$
Now, we add the denominators to the respective numerators to get,
$$\dfrac{x^2 + 5x + 3}{x^2 + x + 3} = \dfrac{-x^2 - 5x - 3}{2x^2 - 10x + 6}.$$
This implies,
$$ x^2 + 5x + 3 = 0\qquad\text{and}\qquad x^2 + x + 3= -2x^2 + 10x - 6\quad\Longrightarrow\quad x^2 -3 x + 3=0.$$
The first quadratic equation gives us two irrational roots, while the second two imaginary roots
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"language": "en",
"url": "https://math.stackexchange.com/questions/4345694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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