Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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If $a^{2}+b^{2} \leq 4$, prove that $a+b \leq 4$ \begin{equation}
\text { If } a^{2}+b^{2} \leq 4 \text { prove that } a+b \leq 4 \text { }
\end{equation} What I have tried:
\begin{equation}
a^{2} \leq a^{2}+b^{2} \leq 4 \text { and } b^{2} \leq a^{2}+b^{2} \leq 4
\end{equation}
\begin{equation}
|a| \leq 2 \text { and }|b| \leq 2
\end{equation}
\begin{equation}\text { So } a+b
\text { can get the maximum value, then } a \geq 0 \text { and } b \geq 0 \text { }
\end{equation}\begin{equation}
\text { } 0 \leq a \leq 2 \text { and } 0 \leq b \leq 2 \text {}
\end{equation}
\begin{equation}
(a-b)^{2}=a^{2}+b^{2}-2 a b \geq 0
\end{equation}\begin{equation}
a^{2}+b^{2} \geq 2 a b \text { So at this step I am not certain what to do next. }
\end{equation}
| Alternatively, $(a+b)^2 = a^2+b^2+2ab \le a^2+b^2+(a^2+b^2) = 2(a^2+b^2) = 2\cdot 4 = 8\implies a+b \le \sqrt{8} < 4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4230740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Why does a boundedness proof of $\exp(x)$ depend on the sign of $x$? To prove $\lim\limits_{n\to\infty} \left(1+\frac{x}{n}\right)^{n}$ exists, we prove that the sequence
$$f_n=\left(1+\frac{x}{n}\right)^n$$
is bounded and monotonically increasing toward that bound.
Proof Attempt:
We begin by showing $f_n=\left(1+\frac{x}{n}\right)^n$ is monotonically increasing by looking at the ratio of consecutive terms:
\begin{align*}
\frac{f_{n+1}}{f_n}
&=\frac{\left(1+\frac{x}{n+1}\right)^{n+1}}{\left(1+\frac{x}{n}\right)^{n}} \tag{Definition of $f_n$} \\
&=\frac{\left(1+\frac{x}{n+1}\right)^{n+1}\left(1+\frac{x}{n}\right)}{\left(1+\frac{x}{n}\right)^{n}\left(1+\frac{x}{n}\right)} \tag{Multiplication by $\frac{\left(1+\frac{x}{n}\right)}{\left(1+\frac{x}{n}\right)}$} \\
&=\frac{\left(1+\frac{x}{n+1}\right)^{n+1}}{\left(1+\frac{x}{n}\right)^{n+1}}\left(1+\frac{x}{n}\right) \tag{Simplify $a^n\cdot a = a^{n+1}$} \\
&=\left(\frac{1+\frac{x}{n+1}}{1+\frac{x}{n}}\right)^{n+1}\left(1+\frac{x}{n}\right) \tag{Simplify $\frac{a^{n+1}}{b^{n+1}}=\left(\frac{a}{b}\right)^{n+1}$} \\
&=\left(\frac{\frac{n+1+x}{n+1}}{\frac{n+x}{n}}\right)^{n+1}\left(1+\frac{x}{n}\right) \tag{Common denominators} \\
&=\left(\frac{n+1+x}{n+1}\cdot \frac{n}{n+x}\right)^{n+1}\left(1+\frac{x}{n}\right) \tag{Simplify $\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a}{b}\cdot \frac{d}{c}$} \\
&=\left(\frac{n^2+n+nx}{(n+1)(n+x)}\right)^{n+1}\left(1+\frac{x}{n}\right) \tag{Distribute $(n+1+x)n$} \\
&=\left(\frac{n^2+n+nx+x-x}{(n+1)(n+x)}\right)^{n+1}\left(1+\frac{x}{n}\right) \tag{Add and subtract $x$} \\
&=\left(\frac{(n+1)(n+x)-x}{(n+1)(n+x)}\right)^{n+1}\left(1+\frac{x}{n}\right) \tag{Factor $n^2+n+nx+x$} \\
&=\left(1+\frac{-x}{(n+1)(n+x)}\right)^{n+1}\left(1+\frac{x}{n}\right) \tag{Simplify $\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}$} \\
&\ge\left(1+\frac{-x}{(n+x)}\right)\left(1+\frac{x}{n}\right) \tag{Bernoulli: $(1+x)^n \ge 1+nx$} \\
&=\left(\frac{n}{n+x}\right)\left(\frac{n+x}{n}\right) \tag{Common denominators} \\
&=1 \tag{Simplify $\frac{a}{b} \cdot \frac{b}{a}=1$}
\end{align*}
Since $\frac{f_{n+1}}{f_n}>1$, then $f_{n+1}>f_n$, which shows the sequence $f_n$ is monotonically increasing for all $n \in \mathbb{N}$.
Next, we show $f_n=\left(1+\frac{x}{n}\right)^n$ is bounded above. Note that
\begin{align*}
f_n
&=\left(1+\frac{x}{n}\right)^n \tag{Definition of $f_n$} \\
&=\sum_{k=0}^n \binom{n}{k} (1)^{n-k} \left(\frac{x}{n}\right)^{k} \tag{Binomial Theorem} \\
&=1+\frac{n}{1!}\left(\frac{x}{n}\right)+\frac{n(n-1)}{2!}\left(\frac{x}{n}\right)^2+\frac{n(n-1)(n-2)}{3!}\left(\frac{x}{n}\right)^3+\cdots+\frac{n!}{n!}\left(\frac{x}{n}\right)^n \\
&=1+\frac{\frac{n}{n}}{1!}x+\frac{\frac{n(n-1)}{n^2}}{2!}x^2+\frac{\frac{n(n-1)(n-2)}{n^3}}{3!}x^3+\cdots+\frac{\frac{n!}{n^n}}{n!}x^n \tag{Simplify}\\
&=1+\frac{1}{1!}x+\frac{\left(1-\frac{1}{n}\right)}{2!}x^2+\frac{\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)}{3!}x^3+\cdots+\frac{\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots \left(1-\frac{n-1}{n}\right)}{n!}x^n \\
& \le 1+\frac{1}{1!}x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+\cdots+\frac{1}{n!}x^n \tag{$1-\frac{k}{n}<1$} \\
& = \sum_{k=0}^n \frac{1}{k!} x^k \tag{Sigma notation}\\
& \to %\underset{n \to \infty}{\to}
\sum_{k=0}^\infty \frac{1}{k!}x^k \tag{as $n \to \infty$} \\
& = \sum_{k=0}^K \frac{1}{k!} x^k + \sum_{k=K+1}^\infty \frac{1}{k!}x^k \tag{$\exists K$, $k>K$ implies $k! \ge (2x)^k$}\\
& \le \sum_{k=0}^K \frac{1}{k!} x^k + \sum_{k=K+1}^\infty \frac{1}{(2x)^k} x^k \tag{$k! \ge (2x)^k$ implies $\frac{1}{k!} \le \frac{1}{(2x)^k}$}\\
& = \sum_{k=0}^K \frac{1}{k!} x^k + \sum_{k=K+1}^\infty \frac{1}{2^k} \tag{$\frac{1}{(2x)^k}x^k=\frac{1}{2^k x^k}x^k = \frac{1}{2^k}$}\\
&= \sum_{k=0}^K \frac{1}{k!} x^k + \frac{1}{2^K} \tag{Geometric series evaluation}
\end{align*}
which is finite. Thus, the sequence $f_n$ is bounded. Since it is both monotonically increasing and bounded, it is convergent by the Monotone convergence theorem.
Is my proof correct? I am suspicious of the step which says "$\rightarrow \sum_{k=0}^n \frac{1}{k!}x^k$", and would like to avoid taking another limit in the middle of the boundedness proof.
I also compared my proof to the following references and saw something worrisome:
*
*Reference 1 <- Assumes $x\ge 0$ (Why?)
*Reference 2 <- Assumes $x \ge -1$ (Why?)
*Reference 3 <- Considers $x=0$, $x>0$, and $x<0$ separately (Why?)
All of the above proofs either assumed $x>0$ or considered cases where $x>0$ and $x<0$ separately, but I do not know why. In fact, the third reference considers $\left(1-\frac{x}{n}\right)^{-n}$ for $x>0$ (I think this is a typo and should read $x<0$), but I am not sure why the negative exponent is needed (we are talking about a negative value of $x$, not negative $n$.)
I could only find one proof that did not consider different cases on the sign of $x$:
*
*Bonus reference 4 <- Uses absolute values, but I am not sure why these are necessary either.
I would like to verify my proof and ask 3 questions:
*
*Why is it necessary to consider cases $x>0$ and $x<0$ separately? Did any step in my proof implicitly assume that $x>0$? If so, which one?
*Is there any way to avoid taking a limit in the middle of the boundedness proof?
*Substituting $n=1$ in my boundedness proof shows $1+x \le \sum_{k=0}^n \frac{1}{k!}x^k$. Does this imply $1+x \le \lim\limits_{n\to \infty} \left(1+\frac{x}{n}\right)^n$, since $f_n$ is an increasing function of $n$? Can this be seen explicitly, or would that require a separate proof?
Thank you.
| Let's observe that $$\left(1+\frac{x}{n}\right) ^n=1+x+\sum_{k=2}^n\frac{x^k}{k!}\left(1-\frac{1}{n}\right)\dots\left(1-\frac{k-1}{n}\right)=\sum_{k=0}^{n}a_k\frac{x^k}{k!}\tag{1}$$ where $$a_0=a_1=1,a_k=\left(1-\frac{1}{n}\right)\dots\left(1-\frac{k-1}{n}\right),k=2,\dots,n$$ Note that the coefficients $a_k$ are positive and do not exceed $1$. If $x>0$ then we can use the implication $$a_k\leq 1\implies a_kx^k\leq x^k\tag{2}$$ to bound the sum in $(1)$ with $\sum_{k=0}^n x^k/k! $. But if $x<0$ the implication $(2)$ does not hold for odd values of $k$ (rather the inequality gets reversed) and hence we can't find an upper bound.
Next I explain the argument used in reference 3 (my blog). Let $$F(x, n) =\left(1+\frac{x}{n}\right)^n,G(x,n)=\left(1-\frac{x}{n}\right)^{-n}\tag{3}$$ then we have $$F(-x, n) G(x, n) =1\tag{4}$$ Let $x>0$ and then we have already established (as in my blog or your question) that $F(x, n) $ is increasing (as function of $n$) and also bounded above and hence the limit $\lim_{n\to\infty} F(x, n) $ exists.
To handle negative values of $x$ I treat the expression $F(-x, n) $ with $x>0$. A better approach would have been to assume $x<0,x=-y,y>0$ and focus on $F(-y, n) $ but I have reused the symbol $x$ instead of inventing another symbol $y$. Thus I keep $x>0$ and handle both $F(x, n), F(-x, n) $.
Now as in reference 3 we have $x>0$ and $n>x$ so that we can apply general binomial theorem to write an infinite power series (in $x$) for $G(x, n) $ as $$G(x, n) =\sum_{k=0}^{\infty}b_k\frac{x^k}{k!}\tag{5}$$ where $$b_0=b_1=1,b_k=\left(1+\frac{1}{n}\right)\dots\left(1+\frac {k-1}{n}\right), k>1$$ Note that $b_k\geq 1$ and if $n$ increases then $b_k$ decreases and thus $G(x, n) $ is a decreasing sequence and bounded below by $\sum_{k=0}^{\infty}x^k/k!$ and hence bounded below by $1+x$. Thus $G(x, n) $ tends to a limit which is not less than $1+x$ (so that the limit is positive). Then $F(-x, n) =1/G(x,n)$ also tends to a positive limit.
Your own approach is smarter because it can show the increasing nature of $F(x, n) $ for all $x$ using Bernoulli inequality whereas I had to invent a $G(x, n) $ to deal with some cases.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4232341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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If $f(x)={{x}^{4}}+a{{x}^{3}}+b{{x}^{2}}+ax+1$ and $f(x) = 0$ has two distinct negative roots and equal positive roots, find least integral value of a If $f(x)={{x}^{4}}+a{{x}^{3}}+b{{x}^{2}}+ax+1$ be a polynomial where a and b are real numbers and $f(x) = 0$ has two distinct negative roots and equal positive roots, find least integral value of a.
Since the function has equal positive roots, I expect $f'(\alpha) = 0$ and $f''(\alpha) > 0$ for the positive root $\alpha$ But I'm not sure how I should proceed in this question. What are the different methods of doing this problem?
| Note that if $r\neq 0$ is a root of $f(x)=0$, then so is $\frac{1}{r}$. This is easily noted by the symmetry of the coefficients and can algebraically shown by the fact that
$$r^4f(\frac{1}{r})=f(r)$$
Hence, if the positive root is $r$ and the negative root of smaller magnitude is $s$, then our four roots are
$$r,\frac{1}{r},s,\frac{1}{s}$$
However, since $r$ and $\frac{1}{r}$ are given as equal positive numbers, we have $$r=1$$
Hence, $f(x)$ is of the form
$$f(x)=(x-1)^2(x-s)(x-\frac{1}{s})$$
$$f(x)=(x^2-2x+1)(x^2-(s+\frac{1}{s})x+1)$$
$$f(x)=x^4-(2+s+\frac{1}{s})x^3+(2+2s+\frac{2}{s})x^2-(2+s+\frac{1}{s})x+1$$
Hence, for all $s\neq 0$ we have that $f(x)$ fits the form $x^4+ax^3+bx^2+ax+1$.
Note that the value of $a$ is
$$a=-2-s-\frac{1}{s}$$
$$a=-2+(-s)+\frac{1}{(-s)}$$
Since $s$ is given as a negative number, we have that $(-s)$ must be a positive number. Hence, by AM-GM we have
$$\frac{(-s)+\frac{1}{(-s)}}{2}\geq \sqrt{(-s)\left(\frac{1}{(-s)}\right)}$$
$$(-s)+\frac{1}{(-s)}\geq 2$$
with equality when $(-s)=\frac{1}{(-s)}$. This occurs when $-s=1$. However, since this results in both negative roots being the same, equality is never attained.
Moreover, we have
$$a=-2+\left((-s)+\frac{1}{(-s)}\right)\geq -2+2$$
$$a\geq 0$$
As we showed before, this equality is unattainable. The next integral value of $a$ is $\boxed{1}$. This is attainable when
$$(-s)+\frac{1}{(-s)}=3$$
$$(-s)^2-3(-s)+1=0$$
$$(-s)=\frac{3\pm\sqrt{5}}{2}$$
Note that the $2$ solutions are positive, and will be the solutions to $(-s)$ and $\frac{1}{(-s)}$. Since we defined $s$ as the root with smaller magnitude, we have $s=-\frac{3-\sqrt{5}}{2}$.
Hence, the equality of $\boxed{a=1}$ is attained when
$$f(x)=(x^2-2x+1)(x^2+3x+1)$$
$$f(x)=x^4+x^3-4x^2+x^3+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4232734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Evaluate : $S=\frac{1}{1\cdot2\cdot3}+\frac{1}{5\cdot6\cdot7}+\frac{1}{9\cdot10\cdot11}+\cdots$ Evaluate:$$S=\frac{1}{1\cdot2\cdot3}+\frac{1}{5\cdot6\cdot7} + \frac{1}{9\cdot10\cdot11}+\cdots$$to infinite terms
My Attempt:
The given series$$S=\sum_{i=0}^\infty \frac{1}{(4i+1)(4i+2)(4i+3)} =\sum_{i=0}^\infty \left(\frac{1}{2(4i+1)}-\frac{1}{4i+2}+\frac{1}{2(4i+3)}\right)=\frac{1}{2}\sum_{i=0}^\infty \int_0^1 \left(x^{4i}-2x^{4i+1}+x^{4i+2}\right) \, dx$$
So,$$S=\frac{1}{2}\int_{0}^{1}\left(\frac{1}{1-x^4}-\frac{2x}{1-x^4} + \frac{x^2}{1-x^4}\right)dx=\frac{1}{2} \int_0^1 \left(\frac{1+x^2}{1-x^4}-\frac{2x}{1-x^4}\right)\,dx = \frac{1}{2} \int_0^1 \left(\frac{1}{1-x^2}-\frac{2x}{1-x^4}\right)\,dx$$
$$=\frac{1}{2}\int_{0}^1\frac{1}{1-x^2}dx-\int_{0}^{1}\frac{2x}{1-x^4}dx=\frac{1}{2}\int_{0}^1\frac{1}{1-x^2}dx-\frac{1}{2}\int_{0}^{1}\frac{1}{1-y^2}dy=0(y=x^2)$$
which is obviously absurd since all terms of $S$ are positive.
But if I do like this then I am able to get the answer, $$S=\frac{1}{2}\int_{0}^{1}\left(\frac{1}{1-x^4}-\frac{2x}{1-x^4}+\frac{x^2}{1-x^4}\right)dx=\frac{1}{2}\int_{0}^{1}\frac{(1-x)^2}{1-x^4}dx=\frac{1}{2}\int_{0}^{1}\left(\frac{1}{1+x}-\frac{x}{1+x^2}\right)dx=\frac{\ln2}{4}$$
What is wrong with the previous approach
| Another similar approach: consider the Beta integral
$$\int_0^1 (1-x)^2 x^n d x = B(3, n+1) = \frac{\Gamma(3)\Gamma(n+1)}{\Gamma(3 + n+1)} = \frac{2! \cdot n!}{(n+3)!} = \frac{2}{(n+1)(n+2)(n+3)}$$
We get
$$\sum_{n\ge 0} \frac{1}{(4n+1)(4n+2)(4n+3)} = \frac{1}{2} \int_{0}^1 (1-x)^2 (\sum_{n\ge 0}x^{4n}) dx=\\ = \frac{1}{2} \int_0^1 \frac{(1-x)^2}{1 - x^4} d x = \frac{\log 2}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4232859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 1
} |
Solving the equation $2x^2-\left(4x-7\right)\sqrt{x-1}-5x+4=0$
Solve the equation $$2x^2-\left(4x-7\right)\sqrt{x-1}-5x+4=0$$
Here is my work:
$$2x^2-5x+4=(4x-7)\sqrt{x-1}$$
$$2(x-1)^2-(x-1)+1=(4(x-1)-3)\sqrt{x-1}$$
Using the substitution $\sqrt{x-1}=t$,
$$2t^4-t^2+1=t(4t^2-3)$$
$$2t^4-4t^3-t^2+3t+1=0$$
Here I plugged in some natural numbers like $1,2,3$ but neither of them worked. So I don't know how to factor the polynomial to find the values of $t$.
|
$2t^4-4t^3-t^2+3t+1=0$
Alt. hint: $\;\;\displaystyle
2t^4-4t^3\color{red}{+2t^2-2t^2}-t^2+3t+1 = \underbrace{2t^2(t-1)^2-3t(t-1)+1}_{\small{\displaystyle 2u^2-3u+1=(u-1)(2u-1)}}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4233420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Infinite product involving triangular numbers $\prod_{n=1}^{\infty} \frac{T_n}{T_{n}+1}$ The question is about the value for the following infinite product involving triangular numbers $T_n=\frac{n(n+1)}2$:
$$\prod_{n=1}^{\infty} \frac{T_n}{T_{n}+1}=\prod_{n=1}^{\infty} \frac{n(n+1)}{n(n+1)+2}=\prod_{n=1}^{\infty} \frac{1}{1+\frac 2{n(n+1)}}$$
Searching on MSE I've found solutions for other similar problems related to Weierstrass' product for the sine function:
$$\frac{\sin(\pi z)}{\pi z}=\prod_{n\geq 1}\left(1-\frac{z^2}{n^2}\right)$$
but nothing specific on that particular case.
I've tried to use Weierstrass' product but without success.
Also expansion in logarithmic sum allow me to find good estimation with a small number of terms but nothing more than that.
According to wolfram the value for the infinite product should be: $\frac{2\pi}{\cosh\left(\frac{\sqrt 7}2\pi\right)} \approx 0.197$.
How can we determine the value for the partial and the infinite product in a closed form?
| Start from the infinite product expansion of normalized sinc function:
$${\rm sinc}(z) =\frac{\sin \pi z}{\pi z} = \prod_{n=1}^\infty \left(1 - \frac{z^2}{n^2}\right)
$$
Substitute $z$ by $2z$, grouping factors in RHS in pairs and then divide it by expansion of ${\rm sinc}(z)$, we obtain an infinite product expansion of $\cos \pi z$:
$$\cos\pi z = \frac{{\rm sinc}(2z)}{{\rm sinc}(z)}
= \prod_{k=1}^\infty\left(1 - \frac{z^2}{(k-\frac12)^2}\right)
$$
Move first factor on RHS to LHS and let $n = k - 1$, this becomes
$$\frac{\cos\pi z}{1 - 4z^2} = \prod_{n=1}^\infty\left(1 - \frac{z^2}{(n+\frac12)^2}\right)\tag{*1}$$
Taking limit at $z = \frac12$ and apply L'Hopital's rule, LHS becomes
$$\lim_{z\to\frac12} \frac{\cos\pi z}{1 - 4z^2} = \lim_{z\to\frac12}
\frac{-\pi \sin\pi z}{-8z} = \frac{\pi}{4}$$
This leads to
$$\frac{\pi}{4} = \prod_{n=1}^\infty\left(1 - \frac{\frac14}{(n+\frac12)^2}\right) = \prod_{n=1}^\infty\frac{n(n+1)}{(n+\frac12)^2}\tag{*2a}$$
In $(*1)$, substitute $z$ by $i\frac{\sqrt{7}}{2}$, we obtain
$$\frac{\cosh(\frac{\pi\sqrt{7}}{2})}{8} = \prod_{n=1}^\infty\left(1 + \frac{\frac74}{(n+\frac12)^2}\right)
= \prod_{n=1}^\infty \frac{n(n+1)+2}{(n+\frac12)^2}\tag{*2b}$$
Divide $(*2a)$ by $(*2b)$, we obtain:
$$\frac{2\pi}{\cosh(\frac{\pi\sqrt{7}}{2})} = \prod_{n=1}^\infty\frac{n(n+1)}{n(n+1)+2} = \prod_{n=1}^\infty \frac{T_n}{T_n + 1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4234613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Determine the minimum value of: $P=\frac{\sqrt{x^2+xy+y^2}}{(x+y)^2+1}+\frac{\sqrt{y^2+yz+z^2}}{(y+z)^2+1}+\frac{\sqrt{z^2+zx+x^2}}{(z+x)^2+1}.$ Given $x,y,z>0$ satisfy $x+y+z=\dfrac{3}{2}$.Determine the minimum value of:
$$P=\frac{\sqrt{x^2+xy+y^2}}{(x+y)^2+1}+\frac{\sqrt{y^2+yz+z^2}}{(y+z)^2+1}+\frac{\sqrt{z^2+zx+x^2}}{(z+x)^2+1}.$$
I have tried:
$\bullet$ The minimum value is $\dfrac{3\sqrt{3}}{4}$ occur when $x=y=z=\dfrac{1}{2}$
$ \bullet P\ge \dfrac{\dfrac{\sqrt{3}}{2}(x+y)}{(x+y)^2+1}+\dfrac{\dfrac{\sqrt{3}}{2}(y+z)}{(y+z)^2+1}+\dfrac{\dfrac{\sqrt{3}}{2}(z+x)}{(z+x)^2+1}$
$\bullet$ So we need to prove $\dfrac{\dfrac{\sqrt{3}}{2}(x+y)}{(x+y)^2+1}\ge \dfrac{\sqrt{3}}{4}$ as well as prove $\dfrac{\dfrac{\sqrt{3}}{2}(y+z)}{(y+z)^2+1} \ge \dfrac{\sqrt{3}}{4}$ and $\dfrac{\dfrac{\sqrt{3}}{2}(z+x)}{(z+x)^2+1}\ge \dfrac{\sqrt{3}}{4}$
$\bullet$ Let $a=x+y$, the problem is $\dfrac{\dfrac{\sqrt{3}}{2}a}{a^2+1}\ge \dfrac{\sqrt{3}}{4}$ or $\dfrac{-(a-1)^2}{a^2+1}\ge0$ (which isn't true)
Pls help me with this problem and it would be nice if you could explain why my work is wrong.
| The minimum does not exist.
Let $x\rightarrow\frac{3}{2}^-$ and $y=z\rightarrow0^+$.
Thus, we get a value $\frac{12}{13}.$
We'll prove that it's an infimum.
If it's given that $x\geq0$, $y\geq0$ and $z\geq0$ so for $(x,y,z)=\left(\frac{3}{2},0,0\right)$ we obtain a value $\frac{12}{13}$ again
and we'll prove that it's a minimal value.
Indeed, let $x=\max\{x,y,z\}$.
Thus, by C-S $$\sum_{cyc}\frac{\sqrt{x^2+xy+y^2}}{(x+y)^2+1}\geq\frac{x+\frac{y}{2}}{(x+y)^2+1}+\frac{x+\frac{z}{2}}{(x+z)^2+1}+\frac{\frac{5}{6}(y+z)}{(y+z)^2+1}\geq$$
$$\geq\frac{\left(x+\frac{y}{2}+x+\frac{z}{2}+\frac{5}{6}(y+z)\right)^2}{(x+y)^2\left(x+\frac{y}{2}\right)+(x+z)^2\left(x+\frac{z}{2}\right)+\frac{5}{6}(y+z)^3+2x+\frac{4}{3}(y+z)}=$$
$$=\tfrac{\frac{2}{3}(x+y+z)\left(2x+\frac{4}{3}(y+z)\right)^2}{(x+y)^2\left(x+\frac{y}{2}\right)+(x+z)^2\left(x+\frac{z}{2}\right)+\frac{5}{6}(y+z)^3+\frac{8}{9}\left(x+\frac{2}{3}(y+z)\right)(x+y+z)^2}$$ and it's enough to prove that:
$$\tfrac{\frac{2}{3}(x+y+z)\left(2x+\frac{4}{3}(y+z)\right)^2}{(x+y)^2\left(x+\frac{y}{2}\right)+(x+z)^2\left(x+\frac{z}{2}\right)+\frac{5}{6}(y+z)^3+\frac{8}{9}\left(x+\frac{2}{3}(y+z)\right)(x+y+z)^2}\geq\frac{12}{13}$$ or
$$303(y+z)x^2+4(43y^2+248yz+43z^2)x\geq(y+z)(104y^2-35yz+104z^2),$$
which is obvious
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Derivative of $\sin^{-1}\frac{x+\sqrt{1-x^2}}{\sqrt 2}$ I'm a calculus beginner. I was asked to find the derivative of the function: $$\sin^{-1}\frac{x+\sqrt{1-x^2}}{\sqrt 2}.$$
I'm able to solve it in the following way:
I first calculate the derivative of $\frac{x+\sqrt{1-x^2}}{\sqrt 2}$ and get $\frac{1}{\sqrt 2}(1-\frac{x}{\sqrt{1-x^2}})$. Then the derivative of the given function is $\frac{1}{\sqrt{1-(\frac{x+\sqrt{1-x^2}}{\sqrt 2}})^2}\cdot \frac{1}{\sqrt 2}(1-\frac{x}{\sqrt{1-x^2}})$. Simplifying this gives the final answer $\frac{1}{\sqrt{1-x^2}}$. But the simplication process is quite lengthy and involves some bizarre calculations. Is there tricks/ways to solve these kinds of derivatives that do not involve too much calculations like above?
| Substitution is a good way to solve these kinds of problems. In the original problem, the term $\sqrt{1-x^2}$ suggests the substitution $x=\sin\theta\ \text{or}\cos\theta$. Also, notice that $\sin\frac\pi 4=\cos\frac\pi 4=\frac{1}{\sqrt2}$. Now Substituting $x=\sin\theta$ where $\theta\in[-\frac\pi 2,\frac\pi 2]$, we get
$$\begin{align}\sin^{-1}\left(\frac{x+\sqrt{1-x^2}}{\sqrt 2}\right) &= \sin^{-1}\left(\frac{1}{\sqrt2}(\sin\theta+\cos\theta)\right) \\ &= \sin^{-1}\left(\frac{1}{\sqrt2}\sin\theta+\frac{1}{\sqrt2}\cos\theta\right)\\ &= \sin^{-1}\left(\cos\frac\pi 4\sin\theta+\sin\frac\pi 4\cos\theta\right) \\ &= \sin^{-1}\left(\sin\left(\frac\pi 4+\theta\right)\right).
\end{align}$$
Now we have $\sin^{-1}\left(\sin\left(\frac\pi 4+\theta\right)\right)=\frac\pi 4+\theta=\frac\pi 4+\sin^{-1}x$ when $\theta\in[-\frac\pi 2,\frac\pi 4]$ and $\sin^{-1}\left(\sin\left(\frac\pi 4+\theta\right)\right)=\pi-\left(\frac\pi 4+\theta\right)=\frac{3\pi} 4-\theta=\frac{3\pi} 4- \sin^{-1}x\ \text{when}\ \theta\in[\frac\pi 4,\frac\pi 2]$.
You just have to find the derivatives in both cases now.
| {
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If $ab+bc+ca=1$ what is the minimum value of $10a^2+10b^2+c^2$?
$a,b,c$ are positive real numbers such that $ab+bc+ca=1$. What is the minimum value of $10a^2+10b^2+c^2$?
I want to solve this problem without using Lagrange multipliers or calculus. I tried the following with some basic inequalities:
From AM-GM inequality $$5a^2+5b^2\geq10ab\\ 5b^2+\frac 12c^2\geq\sqrt{10}bc\\ \frac 12c^2+5a^2\geq\sqrt{10}ca$$
Summing them gives $$10a^2+10b^2+c^2\geq\sqrt{10}(\sqrt{10}ab+bc+ca)$$
But this doesn't help.
| Let $k$ be the minimal value.
Thus, $k>0$ and the inequality $$10a^2+10b^2+c^2\geq k(ab+ac+bc)$$ or $$c^2-k(a+b)c+10a^2+10b^2-kab\geq0$$ is true for any reals $a$,$b$ and $c$ (because after homogenization the condition $ab+ac+bc=1$ is not relevant already), which says $$k^2(a+b)^2-4(10a^2+10b^2-kab)\leq0,$$ which gives
$$(40-k^2)a^2-2(2k+k^2)ab+(40-k^2)b^2\geq0,$$ for which we need $40-k^2>0$ and $$(2k+k^2)^2-(40-k^2)^2\leq0$$ or $$(k+20)(k+5)(k-4)\leq0$$ or
$$0<k\leq4,$$ which gives that a maximal value of $k$, for which the inequality $$10a^2+10b^2+c^2\geq k(ab+ac+bc)$$ is true it's $4$.
The equality occurs for $a=b$ and $c=\frac{k(a+b)}{2}=4a$.
Id est, $$\min_{ab+ac+bc=1}(10a^2+10b^2+c^2)=4.$$
| {
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Prove that for every positive integer $n$, $1/3 + 1/9 + \cdots + 1/{3^n} < 1/2$ Base case is $n=1$: $\frac {1}{3} < \frac{1}{2}$. So the base case holds.
Inductive hypothesis for $n = k$: $\frac{1}{3} + \frac{1}{9} + \cdots + \frac {1}{3^k} < \frac{1}{2}$
Inductive Step for $n = k + 1$:
$$ \left( \frac{1}{3} + \frac{1}{9} + \cdots + \frac {1}{3^k} \right) + \frac {1}{3^{k+1}}< \frac{1}{2}.$$
Multiplying the $n = k + 1$ step by $3$:
$$ \left( 1 + \frac{1}{3} + \cdots + \frac {1}{3^{k-1}} \right) + \frac {1}{3^{k}}< \frac{3}{2}$$
$$\implies \frac{1}{3} + \cdots + \frac {1}{3^{k-1}} + \frac {1}{3^{k}} < \frac{3}{2} - 1 = \frac {1}{2}.$$
We know this to be true from our inductive hypoethsis. Hence, $\frac{1}{3} + \frac{1}{9} + \cdots + \frac {1}{3^n} < \frac{1}{2}.$
Is this proof correct?
| Consider, $$S=\frac{1}{3} + \frac{1}{9} + \frac{1}{27}+\cdots \tag{1}$$ Now multiply $S$ by 3, $$3S=1+\frac{1}{3} + \frac{1}{9} + \frac{1}{27}+\cdots \tag{2}$$ $(2)-(1)$, $$2S=1$$ $$S=\frac12$$
Thus, for finite terms, $$\frac{1}{3} + \frac{1}{9} + \frac{1}{27}+\cdots+\frac{1}{3^n}\lt\frac12$$
| {
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Computing functions with discontinuity what is the best way to compute the value of the following function in a computer program, e.g., C/Fortran, in double precision?
$$
g(a,b)={{e^a/a - e^b/b}\over{a-b}} + {1\over{ab}}.
$$
The above function has removable discontinuities at a=0, b=0, and a=b. I also need to code a similar three-parameter function that has removable discontinuities at a=0, b=0, c=0, a=b, b=c, c=a, and a=b=c. I am trying to avoid too many if-else in the program, and wondering if there are any mathematical tricks that can help.
As for the one-parameter function $g(x)=(e^x-1)/x$, it is currently coded as follows:
If |x| > $\epsilon$, return g(x) directly; else return $e^{x/2}{{sinh(x/2)}\over{(x/2)}}$, where $sinh(t)\over{t}$ is computed from a truncated Tayler series.
This method can easily be applied to function $f(a,b)=(e^a-e^b)/(a-b)$, as it is equal to $e^{(a+b)/2}{{sinh((a-b)/2)}\over{(a-b)/2}}$, but the best solution for $g(a,b)$ is still not obvious to me.
Thanks.
| The answer is the uppermost rightmost element of the matrix exponential
$$
\exp\begin{pmatrix} 0 & 1 & 0 \\ 0 & a & 1 \\ 0 & 0 & b \end{pmatrix}
$$
So you can use the Taylor series of $e^x$ and apply it to
$$
A= \begin{pmatrix} 0 & 1 & 0 \\ 0 & a & 1 \\ 0 & 0 & b \end{pmatrix}
$$
Proof:
We have
$$ A= SDS^{-1} = \\ \phantom{x} \\
\frac{1}{ab(a-b)}
\begin{pmatrix} 1 & 1 & 1 \\ 0 & a & b \\ 0 & 0 & -b(a-b) \end{pmatrix}
\begin{pmatrix} 0 & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & b \end{pmatrix}
\begin{pmatrix} ab(a-b) & -b(a-b) & a-b \\ 0 & b(a-b) & b \\ 0 & 0 & -a \end{pmatrix}
$$
Therefore
$$
\exp(A) = S
\begin{pmatrix} e^0 & 0 & 0 \\ 0 & e^a & 0 \\ 0 & 0 & e^b \end{pmatrix}
S^{-1}
$$
and the uppermost rightmost element of this is
$$
\frac{1}{ab(a-b)}
\begin{pmatrix} 1 & 1 & 1 \end{pmatrix}
\begin{pmatrix} e^0 & 0 & 0 \\ 0 & e^a & 0 \\ 0 & 0 & e^b \end{pmatrix}
\begin{pmatrix} a-b \\ b \\ -a \end{pmatrix}
$$
which is
$$
\frac{(a-b)e^0 +be^a-ae^b}{ab(a-b)}
= \frac{e^a/a-e^b/b}{a-b}
+\frac{1}{ab}
$$
Note that the evaluation of the Taylor series can be simplified due to the zeroes in $A$.
We can show that the same Taylor series can also be used in the special cases $a=0$, $b=0$ and $a=b$. So we do not have any distinction of cases at all.
Edit
While the approach described above avoids all distinctions of cases, it might not be the best one for all cases. If your main goal is numeric stability, you should at least distinguish $5$ cases, as described below.
In the following, I use $\mathrm{sinhc}(x)=\frac{\sinh x}{x}$
*
*$|a|>\epsilon$, $|b|>\epsilon$ and $|a-b|>\epsilon$. In this case, simply use the original definition of $g(a,b).$
*$|a-b|\leq \epsilon$, $|a|>\epsilon$ and $|b|>\epsilon$. Rewrite $g$ as follows
\begin{align}
g(a,b) & = \frac{be^a-ae^b}{ab(a-b)}+\frac{1}{ab} \\
& = \frac{be^a-ae^b + be^b - be^b}{ab(a-b)}+\frac{1}{ab} \\
& = \frac{b(e^a-e^b) + e^b(b-a)}{ab(a-b)}+\frac{1}{ab} \\
& = \frac{1}{a} \cdot \left(\frac{e^a-e^b}{a-b} - \frac{e^b-1}{b}\right) \\
& = \frac{1}{a} \cdot \left(\exp\left(\frac{a+b}{2}\right)\mathrm{sinhc}\left(\frac{a-b}{2}\right) - \frac{e^b-1}{b}\right) \\
& = \frac{1}{b} \cdot \left(\exp\left(\frac{a+b}{2}\right)\mathrm{sinhc}\left(\frac{a-b}{2}\right) - \frac{e^a-1}{a}\right)
\end{align}
*$|a-b|> \epsilon$, $|a|\leq\epsilon$ and $|b|>\epsilon$. Rewrite $g$ as follows:
\begin{align}
g(a,b) & = \frac{be^a-ae^b}{ab(a-b)}+\frac{1}{ab} \\
& = \frac{be^a-ae^b+a-b}{ab(a-b)} \\
& = \frac{b(e^a-1) - a(e^b-1)}{ab(a-b)} \\
& = \frac{1}{a-b}\cdot\left(\frac{e^a-1}{a} - \frac{e^b-1}{b} \right) \\
& = \frac{1}{a-b}\cdot\left(\exp\left(\frac{a}{2}\right)\mathrm{sinhc}\left(\frac{a}{2}\right) - \frac{e^b-1}{b} \right)
\end{align}
*$|a-b|> \epsilon$, $|a| > \epsilon$ and $|b| \leq \epsilon$. Rewrite $g$ as follows:
\begin{align}
g(a,b) & = \frac{1}{a-b}\cdot\left(\frac{e^a-1}{a} - \frac{e^b-1}{b} \right) \\
& = \frac{1}{b-a}\cdot\left(\exp\left(\frac{b}{2}\right)\mathrm{sinhc}\left(\frac{b}{2}\right) - \frac{e^a-1}{a} \right)
\end{align}
*None of the above, i.e. at least two of the values $|a|$, $|b|$ and $|a-b|$ are smaller than or equal to $\epsilon$ (which means that the third one is smaller than or equal to $2\epsilon$). In this case, you can use the matrix method proposed above. It will converge quickly due to the small values of $a$ and $b$. However, I do not know how many terms in the truncated Taylor series should be used.
| {
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Prove $\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}\ge 4(\frac{z}{x+y}+\frac{x}{y+z}+\frac{y}{z+x})$ with Popoviciu's inequality. I want to prove the inequality $$\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}\ge 4(\frac{z}{x+y}+\frac{x}{y+z}+\frac{y}{z+x})$$, for positive reals. I have a hint that it is solved with Popoviciu's inequality applied to $f(x)=x+\frac{1}{x}$. I can show the the function $f$ is convex. Moreover, I see that the LHS of the inequality can be expressed as $f(\frac{x}{y})+f(\frac{x}{z})+f(\frac{z}{y})$. There are other representations as we can choose $2^3=8$ different ways to express the LHS as the sum of $f$'s. However, none of these seem to lead to the desired inequality. I considered rewriting the Popoviciu's inequality as $$f(a)+f(b)+f(c)\ge 2(f(\frac{a+b}{2})+f(\frac{a+c}{2})+f(\frac{b+c}{2}) - \frac{3}{2}f(\frac{a+b+c}{3}))$$, where $a,b,c$ equal $x/y, x/z, z/y$ respectively.
However, it doesn't seem to be the solution. After simplifying the RHS becomes: $$\frac{4}{a+b}+\frac{4}{a+c}+\frac{4}{b+c}-\frac{9}{a+b+c}+a+b+c$$.
Trying slightly different $a, b, c$, namely $x/z, y/z, y/x$ still gives the same result that I can't simplify further. Note, however, that in this case $\frac{1}{a+b}=\frac{z}{x+y}$.
I thought that maybe the RHS of my Popoviciu's inequality can be shown to be greater than or equal than the desired inequality, but I couldn't come up with anything.
| We dont really need Popoviciu's Inequality here,Basically use the following lemma.
If $x,y,z>0\in \mathbb{R}$ $$\dfrac{x}{y}+\dfrac{x}{z}\geq \dfrac{4x}{y+z}$$
Proof : Note that this is equivalent to $x\left(\dfrac{1}{y}+\dfrac{1}{z}-\dfrac{4}{y+z}\right)\Longleftrightarrow\dfrac{x(y-z)^2}{yz(y+z)}\geq 0$ which is clearly true.
Back to the original problem using our lemma we have $$\dfrac{y+z}{x}+\dfrac{z+x}{y}+\dfrac{x+y}{z}=\left(\dfrac{x}{y}+\dfrac{x}{z}\right)+\left(\dfrac{y}{z}+\dfrac{y}{x}\right)+\left(\dfrac{z}{x}+\dfrac{z}{y}\right)\geq \dfrac{4x}{y+z}+\dfrac{4y}{z+x}+\dfrac{4z}{x+y}$$
| {
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Doubt when something is some fraction less than some quantity. Why is $\frac{1}{4}$ less than X equals $\left(\text{X}-\frac{\text{X}}{4}\right)$ and not equals $\left(\text{X}-\frac{{1}}{4}\right)$ ?
For example: I was given the following problem to solve:
Q) One fourth less than 50% of 120 equals to ?
My solution:
50% of 120 = 60
then I did $60-\frac{1}{4}= \frac{239}{4}$
| \begin{align*}
\dfrac{x}{4} &\lt x-\dfrac{x}{4}\Large{?}\\ \\
x-\dfrac{x}{4}
&=\dfrac{4x-x}{4}=\frac{3x}{4}\\ \\
\dfrac{x}{4} &\lt \frac{3x}{4}\\ \\
\therefore\space \dfrac{x}{4} &\lt x-\dfrac{x}{4}
\end{align*}
| {
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Find $\int\frac{x+1}{x^2+x+1}dx$ Find
$\int\frac{x+1}{x^2+x+1}dx$
$\int \frac{x+1dx}{x^2+x+1}=\int \frac{x+1}{(x+\frac{1}{2})^2+\frac{3}{4}}dx$
From here I don't know what to do.Write $(x+1)$ = $t$?
This does not work.Use partial integration?I don't think it will work here.
And I should complete square then find.
| \begin{align}
\int \frac{x+1}{x^2+x+1}\, dx &= \int \frac{x+\frac12 + \frac12}{x^2+x+1} \, dx \\
&= \int \frac{x+\frac12}{x^2+x+1} \, dx + \frac12 \int \frac1{x^2+x+1}\, dx \\
&=\frac12\ln |x^2+x+1| + \frac12 \int\frac1{(x+\frac12)^2 + \frac34} \, dx
\end{align}
I leave the rest as an exercise.
| {
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Why does $-1<\frac{7}{x+4}$ imply $x<-11$? I keep getting $x>-11$ I'm trying to solve for the $|\frac{7}{x+4}|<1$. Which becomes $-1<\frac{7}{x+4}<1$.
Evaluating the positive side is fine, $3<x,$ but for the negative side:
$-1<\frac{7}{x+4}$ should evaluate to $-11 > x$. But I keep getting $-11 < x.$
My working:
$$-1<\frac{7}{x+4}\\
-1(x+4)<7\\-x < 11\\
x>-11.$$
This results in the answer $x>-11, x>3,$ which doesn't make sense.
| Just to add to the noise, this is how I would personally attack this problem. This is a matter of personal style and may not work for you.
$|\frac 7{x+4}| < 1$.
Either $\frac 7{x+4}$ is non-negative (positive or zero) or it is negative.
It is positive if $x+4$ is positive. It is negative if $x+4$ is negative. (And it is never $0$ as $\frac ab \ne 0; a=0$ and we can't have $x+4 = 0$ as we can't have $0$ in a denominator).
Case 1:
If $\frac 7{x+4}$ is positive then $x+4$ is positive and then 1) $x > -4$ and 2) $|\frac 7{x+4}| =\frac 7{x+4}$ and $x + 4 > 0$.
So.... $\frac 7{x+4} > 1\implies 7 < x+4 \implies 3<x $ and combining that with $x > -4$ we get.... $x > 3$.
Case 2:
If $\frac 7{x+4}$ is negative then $x+4$ is negative and 1) $x < -4$ and 2) $|\frac 7{x+4}| =\frac 7{-(x+4)}$ and $-(x + 4) > 0$.
So.... $\frac 7{-(x+4)} < 1\implies 7 < -(x+4) = -x -4\implies x< -11$ and combining that with $x< -4$ we get.... $x<-11$.
....
So Case 1: $x > 3$ and Case 2: $x < -11$.
So $x < -11$ or $x > 3$ and the solution set is:
$(-\infty, -11) \cup (3, \infty) =$
$\{$ everything that isn't between $-11$ and $3\}=$
$\mathbb R \setminus [-11,3]$.
| {
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Find $s,t$ such that $\gcd(509,1177) = 509s+1177t$? How to find $s,t\in\mathbb{Z}$ such that $\gcd(509,1177) = 509s+1177t$?
I know that $509$ is prime and that $1177=107\cdot 11$ so we have $\gcd(509,1177)=1$. How can I proceed from here? Kind of system of equations in two variables but only one equation.
Actually I can do:
$$\begin{align*}
1177&=509 \cdot 2 + 159\\
509&=159 \cdot 3 + 32\\
159&=32 \cdot 4 + 31\\
32&=31 \cdot 1 + 1\\
31&=1 \cdot 31 + 0\\
\end{align*}$$
But not sure how to get from here that
$$1 = 37\cdot 509-16\cdot 1177$$
| $$ \gcd( 1177, 509 ) = ??? $$
I prefer, for the "extended" part, writing it as a continued fraction, rather than "back substitution." The reason this is possible is that the cross product of consecutive convergents is always $\pm 1$
$$ \frac{ 1177 }{ 509 } = 2 + \frac{ 159 }{ 509 } $$
$$ \frac{ 509 }{ 159 } = 3 + \frac{ 32 }{ 159 } $$
$$ \frac{ 159 }{ 32 } = 4 + \frac{ 31 }{ 32 } $$
$$ \frac{ 32 }{ 31 } = 1 + \frac{ 1 }{ 31 } $$
$$ \frac{ 31 }{ 1 } = 31 + \frac{ 0 }{ 1 } $$
Simple continued fraction tableau:
$$
\begin{array}{cccccccccccc}
& & 2 & & 3 & & 4 & & 1 & & 31 & \\
\frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 2 }{ 1 } & & \frac{ 7 }{ 3 } & & \frac{ 30 }{ 13 } & & \frac{ 37 }{ 16 } & & \frac{ 1177 }{ 509 }
\end{array}
$$
$$ $$
$$ 1177 \cdot 16 - 509 \cdot 37 = -1 $$
| {
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For all diagonals $a$ of Pascal's triangle (figurate numbers), $\sum_{k=a}^\infty {k\choose a}\frac{1}{2^k}=2$? I am looking for the derivation of the closed form along any given diagonal $a$ of Pascal's triangle,
$$\sum_{k=a}^n {k\choose a}\frac{1}{2^k}=?$$
Numbered observations follow. As for the limit proposed in the title given by:
Observation 1
$$\sum_{k=a}^\infty {k\choose a}\frac{1}{2^k}=2,$$
when I calculate the sums numerically using MS Excel for any $a$ within the domain ($0\le a \le100$) the sum approaches 2.000000 in all cases within total steps $n\le285$. The first series with $a=0$ is a familiar geometric series, and perhaps others look familiar to you as well:
$$\sum_{k=0}^\infty {k\choose 0}\frac{1}{2^k}=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+... =2,$$
$$\sum_{k=1}^\infty {k\choose 1}\frac{1}{2^k}=\frac{1}{2}+\frac{1}{2}+\frac{3}{8}+\frac{1}{4}+\frac{5}{32}... =2,$$
$$\sum_{k=2}^\infty {k\choose 2}\frac{1}{2^k}=\frac{1}{4}+\frac{3}{8}+\frac{3}{8}+\frac{5}{16}+\frac{15}{64}... =2,$$
but it is both surprising and elegantly beautiful that these sums across all diagonals appear to approach the same value. Some additional observations from the numerically determined sums:
Observation 2
The maximum value of any term ${k\choose a}\frac{1}{2^k}$ within a diagonal $a$ for the domain $(a>0)$ is attained at $k=2a-1$ and repeated for the term immediately following ($k=2a$).
Observation 3
$$\sum_{k=a}^{2a} {k\choose a}\frac{1}{2^k}=1$$
Observation 4
$$\sum_{k=a}^{n} {k\choose a}\frac{1}{2^k} + \sum_{k=n-a}^{n} {k\choose n-a}\frac{1}{2^k}=2$$
It's very likely that the general closed form has been derived before, but searching for the past several days has produced no results. It appears that setting up the appropriate generating function may play a role, but I am at a loss as to how to proceed. Looking forward to the responses.
| Using algebra
$$f_n(x)=\sum_{k=a}^n {k\choose a}x^k=\frac 1x \,\left(\frac{x}{1-x}\right)^{a+1}-x^{n+1} \binom{n+1}{a} \, _2F_1(1,n+2;n+2-a;x)$$ where appears the gaussian hypergeometric function.
$$g(x)=\sum_{k=a}^\infty {k\choose a}x^k=\frac 1x \,\left(\frac{x}{1-x}\right)^{a+1}$$
So
$$f_n\left(\frac{1}{2}\right)=2-\frac{ \Gamma (n+2) }{2^{n-a}\Gamma (a+1) \Gamma (n+2-a)}\,
_2F_1\left(-a,n+1-a;n+2-a;\frac{1}{2}\right)$$
$$g\left(\frac{1}{2}\right)=2$$
Now
$$f_{2a}(x)=\frac 1x \,\left(\frac{x}{1-x}\right)^{a+1}-\binom{2 a+1}{a} x^{2 a+1} \, _2F_1(1,2 a+2;a+2;x)$$
$$f_{2a}\left(\frac{1}{2}\right)=1$$
| {
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Find $\int \frac{1}{(1+m\cos(\theta))^2} \mathrm d\theta$ where $0I tried Weistrass substitution, integration by parts and partial fractions for this integral, but it made the integral even worse. Actually I got this integral when I was working with the Kepler's second law.
When I did Weistrass substitution I got some results as follow:
$$∫ \frac{2(1+z^2)}{(1+z^2+m(1-z^2))^2} \mathrm dz$$
If you can solve this please solve it.
| This can be done via e.g. Mathematica. Try the substitution $w = \tan(\theta/2)$, so $$\textrm{d}w = \frac{1}{2} \sec^2\left(\frac{\theta}{2}\right)\mathrm{d}\theta$$ and also, $$\cos(\theta) = \frac{1-w^2}{1+w^2}$$ Then, using $\sec^{2}(\theta) = 1 + \tan^{2}(\theta)$, we have $\sec^{2}(\theta/2) = 1 + w^2$, and therefore
$$\int \frac{1}{(1+m \cos(\theta))^2} \textrm{d}\theta = \int \frac{2}{(1+w^2)(1+m\left(\frac{1-w^2}{1+w^2}\right))^2} \textrm{d}w$$
we then apparently have, via Mathematica,
$$
\int \frac{2}{(1+w^2)(1+m\left(\frac{1-w^2}{1+w^2}\right))^2} \textrm{d}w \\ = \frac{2}{1+m}\left(\frac{m w}{-m^2+(m-1)^2 w^2+1}+\frac{\tan ^{-1}\left(\sqrt{\frac{2}{m+1}-1} w\right)}{(1-m)^{3/2} \sqrt{m+1}}\right) + C
$$
and so we have the closed form
$$\int \frac{1}{(1+m \cos(\theta))^2} \textrm{d}\theta = \frac{m \sin (\theta )}{\left(m^2-1\right) (m \cos (\theta )+1)}-\frac{2 \tanh ^{-1}\left(\frac{(m-1) \tan \left(\frac{\theta }{2}\right)}{\sqrt{m^2-1}}\right)}{\left(m^2-1\right)^{3/2}}+C$$
| {
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Given $a + b = c + d$ and $a^2 + b^2 = c^2 + d^2$ prove $a=c, b=d$ or $a=d, b=c$
Given $$a + b = c + d\space \text{and}\space a^2 + b^2 = c^2 + d^2\quad\forall\space a,b,c,d \in \mathbb{R}$$ Prove: $$a=c, b=d\space \text{or} \space a=d, b=c $$
*
*I managed to get $ab=cd$. Don't know how to proceed further.
| Introduce $x$, $y$, $\alpha$ and $\delta$, so
$$x=\frac {a+b} 2$$
$$y=\frac {c+d} 2$$
$$\alpha=\frac {a-b} 2$$
$$\delta=\frac {c-d} 2$$
Now
$$a=x-\alpha$$
$$b=x+\alpha$$
$$c=y-\delta$$
$$d=y+\delta$$
Rewrite $a+b=c+d$ using $x$, $y$, $\alpha$ and $\delta$:
$$x-\alpha+x+\alpha=y-\delta+y+\delta$$
$$2x=2y$$
$$x=y$$
Similarly rewrite $a^2+b^2=c^2+d^2$:
$$(x-\alpha)^2+(x+\alpha)^2=(y-\delta)^2+(y+\delta)^2$$
$$x^2-2x\alpha+\alpha^2+x^2-2x\alpha+\alpha^2=y^2-2y\delta+\delta^2+y^2+2y\delta+\delta^2$$
$$2x^2+2\alpha^2=2y^2+2\delta^2$$
We already know $x=y$, therefore
$$\alpha^2=\delta^2$$
This means $\alpha=\delta$ or $\alpha=-\delta$. The former case leads to $a=c,b=d$; the latter case leads to $a=d,b=c$.
| {
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How to find the probability of $P(Y=6)$ Anna writes down a random sequence created by the following process. She repeatedly rolls a fair
6-sided die. If the number she rolls is larger than all of the numbers she has previously rolled (if
any), then she writes the new number down and then continues rolling. Otherwise, she does not
write the new number down and the process ends.
Let X be the length of Anna’s sequence, and Y be the last number in her sequence.
For example, if Anna rolled 1 then 4 then 5 then 4, her sequence would be 1,4,5 and the random
variables X, Y would take values X = 3 and Y = 5.
Question:
what is $P(Y=6)$
My method
I know I can write down all the possible ways that the last number of the sequence could be 6, and then I add up all the probabilities.
*
*$1,2,3,4,5,6 = (\frac{1}{6})(\frac{1}{5})(\frac{1}{4})(\frac{1}{3})(\frac{1}{2})(1)$
*$2,3,4,5,6 = (\frac{1}{6})(\frac{1}{5})(\frac{1}{4})(\frac{1}{3})(\frac{1}{2})$
*$3,4,5,6 = (\frac{1}{6})(\frac{1}{5})(\frac{1}{4})(\frac{1}{3})$
*$1,2,4,6 = (\frac{1}{6})(\frac{1}{5})(\frac{1}{4})(\frac{1}{3})$
etc.
But this method is way too time-consuming as there are too many possible sequences.
I also tried to do it a different way by using $1 - $(The possible sequences that the last number is not 6), but there are even more sequences to take into account this way.
Can anyone show me a much more efficient way of finding the solution to this problem?
|
Question:
what is P(Y=6)
observe that the sequence's length can only be $X \in\{1,2,3,4,5,6\}$
$X=1$ when you roll immediately 6
$X=5$ when you roll 4 numbers in increasing order, and then 6. The only favourable sequences are, in this case
$1234$
$1235$
$1245$
$1345$
$2345$
...and 6 on the fifth roll. Observe that these cases are exactly $\binom{5}{4}$
thus here the probability to get $Y=6$ is $\frac{5}{6^4}\cdot\frac{1}{6}$
Using this procedure, you easy get
$$\mathbb{P}[Y=6]=\frac{1}{6}+\frac{5}{6}\cdot\frac{1}{6}+\frac{10}{6^2}\cdot\frac{1}{6}+\frac{10}{6^3}\cdot\frac{1}{6}+\frac{5}{6^4}\cdot\frac{1}{6}+\frac{1}{6^6}=\approx 0.36$$
| {
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Minimum of $abc$ when $a^3+b^3+c^3\geq a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}.$
Let $a,b,c$ be positive real numbers with $abc=k$ such that the inequality $$a^3+b^3+c^3\geq a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}$$
holds for all $a,b,c$. Find the minimum value of $k$.
I found that $abc=2$ works. Here is my proof for $abc=2$:
By Cauchy-Schwarz and Muirhead inequality
$$\begin{align} a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b} &\leq \sqrt{2(a^2+b^2+c^2)(a+b+c)}\\ &= \sqrt{abc(a^2+b^2+c^2)(a+b+c)}\\ &= \sqrt{\sum_{cyc}a^4bc+\sum_{cyc}(a^3b^2c+a^2b^3c)}\\ &\leq \sqrt{\sum_{cyc}a^6+2\sum_{cyc}a^3b^3}\\ &= a^3+b^3+c^3
\end{align}$$
as desired. Equality holds when $a=b=c=\sqrt[3]2$.
The inequality seems quite tight also. But I am not sure if it is the minimum of $abc$. And if it is, how do I prove that?
| Let $abc=k$.
Thus, after homogenization we need to prove that
$$\sum_{cyc}\left(a^3-a\sqrt{\frac{abc(b+c)}{k}}\right)\geq0,$$ which for $a=b=c$ gives $k\geq2$ and it's enough to prove that:
$$\sum_{cyc}\left(a^3-a\sqrt{\frac{abc(b+c)}{2}}\right)\geq0,$$ which you proved already by C-S.
Also, it's true by AM-GM and Muirhead:
$$\sum_{cyc}\left(a^3-a\sqrt{\frac{abc(b+c)}{2}}\right)=\sum_{cyc}\left(a^3-a\sqrt{\frac{2bc(ab+ac)}{4}}\right)\geq$$
$$\geq\sum_{cyc}\left(a^3-\frac{a(2bc+ab+ac)}{4}\right)=\frac{1}{4}\sum_{cyc}(4a^3-a^2b-a^2c-2abc)\geq0.$$
| {
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Showing LI without determinant or Reduced row echelon form. Consider the $\mathbb{R}$-vector space $ P^3(\mathbb{R})$ of polynomials over $\mathbb{R}$ with degree at most 3. Define $$W= \{f \in P^3(\mathbb{R}): f^{'}(2)=f(1)=0\}$$
EDIT: Calculate $\dim(W)$. by finding a basis for W.
We need to find a basis for W the easiest way to do this is to look for as many linearly independent vectors as we can, we start by writing down the equations we get $ ax^3 + bx^2 + cx +d =0$ and $ 3ax^2 + 2bx +c =0$ where $ x=1$ in the first equation and $x=2$ in the second this is from $f(1)=f^{'}(2)=0$ we get
$\begin{bmatrix}
a & b &c &d & &0 \\
12a & 4b & c & 0& &0
\end{bmatrix}=\begin{bmatrix}
a & b &c &d& &0 \\
0 & -8b & -11c & -12d& &0
\end{bmatrix} $
$\begin{bmatrix}
a & 0 &c -\frac{11c}{8} &d -\frac{3d}{2}& &0 \\
0 & -b & -\frac{11c}{8} & -\frac{3d}{2}& &0
\end{bmatrix}=\begin{bmatrix}
a & 0 & -\frac{3c}{8} & -\frac{d}{2}& &0 \\
0 & b & \frac{11c}{8} & \frac{3d}{2}& &0
\end{bmatrix} $
Now here i am a bit lost we can solve for a and b in terms of c and d but we cant actually find a value for anything. I do have a theorem that tells me that this spans my subspace W but it doesn't tell me that my two vectors are linearly independent, in fact I don't even really have 2 vectors I got infinitely many of them in some sense, I tried writing out $$c_1(ax^3 -\frac{3c}{8} x -\frac{d}{2})+c_2 (bx^2 +\frac{11c}{8}x+\frac{3d}{2})=0$$ but i just have too many variables to say that $c_1=c_2=0$ is the only solution. Normally i would just guess some values but my prof said we apparently are not able to do that and assume what we found was spanning but I don't see anywhere to go from here. Can someone give me a hint how to proceed?
We are going to need a and b to be nonzero for this to make any sense so lets pick $a=b=1$ we know by theorem that what we have is spanning but must show linearly independent.
we have that $1= \frac{3c}{8} +\frac{d}{2}$ and $1= -\frac{11c}{8} -\frac{3d}{2}$ this tells us that $1+3 = -\frac{11c}{8}+\frac{9c}{8} -\frac{3d}{2} + \frac{3d}{2}$ or $4=-\frac{2c}{8}$ or $c=-16$ and $d = (1+6)2=14$.\ We now wish to show that
$c_1 ( x^3 +6x -7) +c_2 (x^2 -22x +21) =0 $ if and only if $c_1=c_2=0$. $ c_1 x^3 +c_2 x^2 + x(6c_1-22c_2) -7c_1 +21c_2 =0$ for this to be true for all x this tells us that $ -7c_1 +21c_2 =0$ or $ c_1 =3c_2$ Now we have that $ 3c_2 x^3 +c_2 x^2 + x(18c_2-22c_2) -21c_2 +21c_2 =0$ but this tells us that $-4c_2 =0$ so we have that $c_2=0 $ but from this we know that $c_1=0$. This tells us that the vector $( x^3 +6x -7)$ and the vector $(x^2 -22x +21) $ are linearly independent. However, we know that they span our subspace W as well so they form a basis for W.
| For a polynomial $f = ax^3 + bx^2 + cx +d$ in $P^3(\mathbb R)$ one has that $f(1) = a+b+c+d$ and $f'(2) = 12a+4b+c$, so
$$
f \in W \implies
\begin{cases} 12a+4b+c = 0 \\[1mm] \qquad \quad \& \\[0.5mm]
a+b+c+d = 0
\end{cases} \implies
\begin{cases} c = -12a-4b \\[1mm] \qquad \quad \& \\[0.5mm]
d = -a-b-c = 11a+3b \end{cases}
$$
that is,
\begin{align}
f &= ax^3+bx^2+(-12a-4b)x+(11a+3b) \\ &= a(x^3-12x+11) + b(x^2-4x+3).
\end{align}
In other words, every polynomial in $W$ can be written as a linear combination of the polynomials $f_1 = x^3-12x+11$ and $f_2 = x^2-4x+3$. Finally, since $f_1$ and $f_2$ are not scalar multiples, they are linearly independent; hence they form a basis for $W$.
| {
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Finding the maximum value of the following product.
Let $x,y,z$ be three reals which satisfies $x^2+y^2+z^2=1$. Find the maximum value of $$P = (x^2-yz)(y^2-zx)(z^2-xy).$$
I think expanding would be a bad idea. I tried to apply the Weirstrass product inequality but it may give the minimum not maximum value.
| Lagrange multiplier is the correct tool for solving this type of constrained optimization problem. I don't know your backgroud, but the following is a completely elementary approach (which I only found out with the help of Lagrange).
First it's not hard to show that the maximum (which exists as the sphere is compact) is positive. If the maximum is achieved when two of the three factors are negative, say $x^2-yz<0$ and $y^2-xz<0$, then $yz>x^2\ge 0, xz>y^2\ge 0$, therefore $x,y,z$ all have the same sign. But then we may flip the sign of $z$, so that $(x^2+yz)(y^2+xz)(z^2-xy)\ge (x^2-yz)(y^2-xz)(z^2-xy)$. Therefore the max can only be achieved when all three factors are positive.
Use the geometric mean $\le$ the arithmetic mean, we have $f(x,y,z)^{1/3}\le\frac{x^2+y^2+z^2-xy-yz-xz}{3}=\frac{1-xy-yz-xz}{3}$. And equality holds only when $x^2-yz=y^2-xz=z^2-xy \Leftrightarrow (x-y)(x+y+z)=0, (y-z)(x+y+z)=0$. As $x,y,z$ cannot be all equal, we must have $x+y+z=0$. If we can show further that $\frac{1-xy-yz-xz}{3}$ achieves a maximum when $x+y+z=0$, then we can conclude maximum can be achieved for $f(x,y,z)$ for the same triple.
We are left to find when $\frac{xy+yz+xz}{3}$ achieves minimum on the sphere. This is equivalent to find the minimum of $(x+y+z)^2=(x^2+y^2+z^2) + 2(xy+yz+xz)=1+2(xy+yz+xz)$ on the sphere, and clearly the minimum of $(x+y+z)^2$ is achieved when $x+y+z=0$.
Put all the work together and reverse the reasoning, we conclude that $f(x)$ achieves a maximum at $(x,y,z)\in \mathbb S^2$ if and only if $x^2-yz, y^2-xz, z^2-xy$ are all positive and $x+y+z=0$. (In fact, e.g. $x^2-yz = y^2+z^2+yz\ge 0$ on $x+y+z=0$, hence the condition of the three factors being positive is redundant. And the intersection of the sphere with the plane are exactly all the solutions to the optimization problem.)
Now we may use any point on the intersection of the plane and the sphere to calculate the maximum. E.g. $x=\frac{1}{\sqrt 2}, y = -\frac{1}{\sqrt 2}, z = 0$, then the maximum is just $\frac{1}{8}$ as pointed out in the comment.
| {
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Evaluate the partial sum $\sum_{n=1}^k \frac{1}{n(n+1)(n+2)}$ Evaluate the partial sum $$\sum_{n=1}^k \frac{1}{n(n+1)(n+2)}$$
What I have tried:
Calculate the partial fractions which are (for sake of brevity) :
$$\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{n(n+2)}$$
So we get:
$$\sum_{n=1}^k \frac{1}{n(n+1)(n+2)} = \sum_{n=1}^k \left(\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{n(n+2)}\right)$$
Then calculating a few numbers for $n$ we get:
$$\left(\frac{1}{2}-\frac{1}{2}+\frac{1}{6} \right) + \left(\frac{1}{4} - \frac{1}{3} + \frac{1}{8} \right) + \left(\frac{1}{6} - \frac{1}{4} + \frac{1}{10}\right) . . . \left(\frac{1}{2n} - \frac{1}{n+1} + \frac{1}{n+2}\right)$$
The first two fractions cancel out in the first bracket and we're left with $\frac{1}{6}$, as for the second bracket the first fraction is cancelled out by the second fraction in the third bracket.
I have noticed that the first fractrion so $\frac{1}{2n}$ cancel out by every even term in the denominator for $-\frac{1}{n+1}$ so the equation becomes:
$$\left(-\frac{1}{2n+1}+\frac{1}{n+2}\right) = \left(\frac{n-1}{(2n+1)(n+2)} \right)$$
Have I approached this correctly? I would greatly appreciate some assistance on any improvements!
| $$\sum_{n=1}^{k}\frac{1}{(n)(n+1)(n+2)}$$
By partial fraction decomposition,
$$\frac{1}{(n)(n+1)(n+2)}=\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2(n+2)}$$
$$\sum_{n=1}^{k}\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2(n+2)}$$
By splitting the terms and taking the constants outside,
$$\frac{H_{k}}{2}-(H_{k+1}-1)+\frac{H_{k+2}-\frac{3}{2}}{2}$$
Where $H_{k}$ denotes the k-th harmonic number.
Using the fact that, $$H_{k+a}=H_{k}+\sum_{u=a+1}^{k+a}\frac{1}{u}$$
$$H_{k+1}=H_{k}+\frac{1}{k+1}$$ $$H_{k+2}=H_{k}+\frac{1}{k+1}+\frac{1}{k+2}$$
Placing them in summation and after some simplification (which I leave to reader) we get as follows,
$$\frac{1}{2(k+2)}-\frac{1}{2(k+1)}-\frac{3}{4}$$
| {
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Is $g(x,y)= \frac{f(x^{2y+1},y)}{f(x,y)}$ always an integer? This question is similar to this other question:
Let $$ f(x,y):= \frac{x^y -1}{x+(-1)^y}$$
and $$ g(x,y):= \frac{f(x^{2y+1},y)}{f(x,y)}.$$
Let $y\ge1$ be an integer. Show that $g(x,y)$ is a polynomial function of $x$ with integer coefficients.
Moreover, it seems that that $g$ is irreducible in $\mathbb{Z}[x]$ if and only if both $y$ and $2y+1$ are prime numbers (in other words if and only if $y$ is a Sophie Germain prime).
Here are the first $g(x,y)$ for the integers $y$ such that $1\le y\le 4$:
$$g(x,1)=1 $$
$$g(x,2)=1+x+x^2+x^3+x^4=\Phi_5(x)$$
$$g(x,3)=1-x+x^3-x^4+x^6-x^8+x^9-x^{11}+x^{12}=\Phi_{21}(x)$$
$$g(x,4)=\Big(1+x+x^2\Big) \Big(1-x^2+x^4\Big) \Big(1+x^3+x^6\Big) \Big(1-x^6+x^{12}\Big)=\Phi_{3}(x)\Phi_{9}(x)\Phi_{12}(x)\Phi_{36}(x)$$
It seems to me that $g(x,y)$ should be a product of cyclotomic polynomials $\Phi_k(x)$ where $k$ are divisors of $y(2y+1)$, but I could not derive a general formula.
| If $y$ is even, then
$$g(x,y)=\frac{\frac{x^{y(2y+1)}-1}{x^{2y+1}+1}}{\frac{x^y-1}{x+1}}=\frac{(x^{y(2y+1)}-1)(x+1)}{(x^{2y+1}+1)(x^y-1)}.$$
If $y$ is odd, then
$$g(x,y)=\frac{\frac{x^{y(2y+1)}-1}{x^{2y+1}-1}}{\frac{x^y-1}{x-1}}=\frac{(x^{y(2y+1)}-1)(x-1)}{(x^{2y+1}-1)(x^y-1)}.$$
One has
$$x^t-1=\prod_{d\mid t}\Phi_d(x),\qquad x^t+1=\prod_{\substack{d\mid 2t \\ d\nmid t}}\Phi_d(x).$$
As a result, for example, using that $\gcd(y,2y+1)=1$ you can show that for $y$ odd
$$g(x,y)=\prod_{\substack{d\mid y(2y+1)\\ d\nmid y\text{ or }2y+1}}\Phi_d(x).$$
Can you do something similar when $y$ is even, and use these results to deduce exactly when $g(x,y)$ is irreducible?
| {
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Prove that $\frac{a}{1+b^2c}+\frac{b}{1+c^2d}+\frac{c}{1+d^2a}+\frac{d}{1+a^2b}\ge 2$
Given $a+b+c+d=4, $ and $a,b,c,d$ are positive real numbers. Prove that $$\frac{a}{1+b^2c}+\frac{b}{1+c^2d}+\frac{c}{1+d^2a}+\frac{d}{1+a^2b}\ge 2$$
Can someone give me hints for this?
To show
$$\frac{a^2}{a+b^2ca}+\frac{b^2}{b+c^2db}+\frac{c^2}{c+d^2ac}+\frac{d^2}{d+a^2bd}\ge 2\implies \frac{(a+b+c+d)^2}{a+b^2ca+b+c^2db+c+d^2ac+d+a^2bd}$$
$$\implies \frac{16}{4+ b^2ca+ c^2db+ d^2ac+ a^2bd}\ge 2.$$
It is enough to show that $$ \frac{8}{4+ b^2ca+ c^2db+ d^2ac+ a^2bd}\ge 1.$$
| It's enough to prove that:
$$a^2bd+b^2ca+c^2bd+d^2ac\leq4,$$
Now, use AM-GM twice.
| {
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Isn't my book using logarithm wrongly to differentiate? Problem:
Differentiate with respect to $x$: $\frac{e^{-3x}(3x+5)}{7x-1}$
My book's attempt:
Let,
$$y=\frac{e^{-3x}(3x+5)}{7x-1}$$
$$\ln(y)=\ln(\frac{e^{-3x}(3x+5)}{7x-1})\tag{1}$$
$$\text{rest of the math...}$$
Question:
*
*Isn't taking $\ln()$ unfounded in $(1)$: $\frac{e^{-3x}(3x+5)}{7x-1}$ could a negative number? Don't we have to just use the product and quotient rule to find the derivative here?
| This is called "logarithmic differentiation". It can be used to make taking derivatives of expressions involving lots of products, quotients, and powers, easier by invoking the properties of the logarithm.
For example, say we want the derivative of
$$y = \frac{(x^2+1)^3\sqrt{x^3+2x}}{e^x(x+1)}.$$
Taking logarithms on both sides, then using the properties of the logarithm to rewrite the side with lots of products/quotients/powers/etc, and finally differentiating implicitly, we get:
$$\begin{align*}
y &= \frac{(x^2+1)^3\sqrt{x^3+2x}}{e^x(x+1)}\\
\ln|y| &= \ln\left|\frac{(x^2+1)^3\sqrt{x^3+2x}}{e^x(x+1)}\right|\\
\ln|y| &= 3\ln|x^2+1| + \frac{1}{2}\ln|x^3+2x| - \ln|e^x| - \ln|x+1|\\
\frac{d}{dx}\ln|y| &=\frac{d}{dx}\left( 3\ln|x^2+1| + \frac{1}{2}\ln|x^3+2x|-x-\ln|x+1|\right)\\
\frac{y'}{y} &= 3\left(\frac{2x}{x^2+1}\right) + \frac{1}{2}\left(\frac{3x^2+2}{x^3+2x}\right) - 1 - \frac{1}{x+1}\\
y' &= y\left(3\left(\frac{2x}{x^2+1}\right) + \frac{1}
{2}\left(\frac{3x^2+2}{x^3+2x}\right) - 1 - \frac{1}
{x+1}\right)\\
y'&= \left(\frac{(x^2+1)^3\sqrt{x^3+2x}}{e^x(x+1)}\right)\left(3\left(\frac{2x}{x^2+1}\right) + \frac{1}{2}\left(\frac{3x^2+2}{x^3+2x}\right) - 1 - \frac{1}{x+1}\right).
\end{align*}$$
This is usually simpler than the compounded quotient and power rules needed to deal with $y$ directly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4275161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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In triangle. Prove that: $2(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9+\sum_{cyc}{\sqrt{17+\frac{4(b+c)((b-c)^2-a^2)}{abc}}}$ Problem: Given a,b,c are length of triangle. Prove that: $$2(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9+\sum_{cyc}{\sqrt{17+\frac{4(b+c)((b-c)^2-a^2)}{abc}}}$$
Happy Vietnamese Women's Day
Phan Ngoc Chau, Oct 20th 2021
My approach: Note that: $(b-c)^2-a^2=(b-c-a)(b-c+a)<0$ so $\sqrt{17+\frac{4(b+c)((b-c)^2-a^2)}{abc}}<\sqrt{17}$
And as well- known result: $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9$
But it is not true to show: $3\sqrt{17}\le9$. It is quite ugly for me to get a proof. Anyone help me ? Thanks!
| Let $a=y+z$, $b=x+a$ and $c=x+y$.
Thus,$x$, $y$ and $z$ are positives and we need to prove that:
$$4(x+y+z)\sum_{cyc}\frac{1}{x+y}-9\geq\sum_{cyc}\sqrt{17-\frac{16yz(2x+y+z)}{\prod\limits_{cyc}(x+y)}}$$ or
$$\sum_{cyc}(4x^3+7x^2y+7x^2z+6xyz)\geq\sum_{cyc}\sqrt{\prod_{cyc}(x+y)(17\prod_{cyc}(x+y)-16yz(2x+y+z))}.$$
Now, by AM-GM
$$\sum_{cyc}\sqrt{\prod_{cyc}(x+y)(17\prod_{cyc}(x+y)-16yz(2x+y+z))}=$$
$$=\frac{1}{3}\sum_{cyc}\sqrt{9\prod_{cyc}(x+y)\cdot(17\prod_{cyc}(x+y)-16yz(2x+y+z))}\leq$$
$$\leq\frac{1}{6}\sum_{cyc}\left(9\prod_{cyc}(x+y)+17\prod_{cyc}(x+y)-16yz(2x+y+z)\right)$$ and it's enough to prove that:
$$6\sum_{cyc}(4x^3+7x^2y+7x^2z+6xyz)\geq\sum_{cyc}\left(9\prod_{cyc}(x+y)+17\prod_{cyc}(x+y)-16yz(2x+y+z)\right)$$ or
$$\sum_{cyc}(6x^3-5x^2y-5x^2z+4xyz)\geq0$$ or
$$5\sum_{cyc}(x^3-x^2y-x^2z+xyz)+\sum_{cyc}(x^3-xyz)\geq0,$$ which is true by Schur and Muirhead.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4276061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Winning strategy for a game with cubic equation The following problem is from USSR $1990$:
The following equation with erased coefficients is written on a blackboard: $$x^3+\dots x^2+\dots x+\dots =0$$
Two players are playing a game. In one move the first player chooses a number and the second player puts it instead of dots into one of the vacant places. After three moves the game is over. Is it possible for the first player to choose three numbers that will secure three distinct integer roots for the equation, no matter how the second player plays?
The solution to this problem which I found from a book is as follows:
Solution from book: Yes, it is possible. One strategy is as follows. At the beginning the first player chooses $0$.
$\color\green{\text{Case}\ 1}$: If the second player makes it the constant term of the equation, the equation will be $$x^3+\dots x^2+\dots x=0$$ and the first player chooses successively $2$ and $-3$ to obtain $$x(x-1)(x+3)=0\ \ \text{or}\ \ x(x-1)(x-2)=0$$
$\color\red{\text{Case}\ 2}$: If the second player puts $0$ as the coefficient of $x^2$, then the equation becomes $x^3+bx+c=0$ with $b$ and $c$ not fixed yet. The first player chooses the number $\color\red{-(3\cdot4\cdot5)^2}$ and then depending on the move of the second player, either $c=0$ or $b=3^2\cdot4^2-4^2\cdot5^2-3^2\cdot5^2$. This will result in the equations $$x(x-3\cdot4\cdot5)(x+3\cdot4\cdot5)=0$$ or $$(x+3^2)(x+4^2)(x-5^2)=0$$
$\color\red{\text{Case}\ 3}$: If after the move of the second player the equation is $x^3+ax^2+c=0$, the first player chooses $\color\red{6^2\cdot7^3}$ and then either $c=-49$ or $c=-6^8\cdot7^6$ to get the equations $$(x+2\cdot7)(x-3\cdot7)(x-6\cdot7)=0$$ or $$(x-2\cdot6^2\cdot7^2)(x+3\cdot6^2\cdot7^2)(x+6\cdot6^2\cdot7^2)=0$$
In all cases, we got equations with three distinct integer roots and we are done.$\ \ \blacksquare$
I understand the case 1 of the solution. But I don't understand the case 2 and case 3 (which is in red) of the solution, especially how the numbers in the second move (numbers in red) are chosen in these two cases. I know I have to use Vieta's formula. I have seen that choosing the numbers in the second move arbitrarily doesn't work.
| In case 2, the idea is:
*
*If $N$ is a perfect square, then $x^3 - Nx$ factors as $x(x+\sqrt N)(x-\sqrt N)$. This means we can win by playing $-N$, then $0$ if $-N$ is chosen as the coefficient of $x$.
*If $N = pq(p+q)$, then the expansion of $(x+p)(x+q)(x-p-q)$ has an $x^2$ coefficient of $0$ and constant term of $-N$. This means that we can win by playing $-N$ if it is chosen as the constant term: our next play is whatever the coefficient of $x$ in $(x+p)(x+q)(x-p-q)$ is.
The easiest way to make sure $pq(p+q)$ is a perfect square is to make $p$, $q$, and $p+q$ all perfect squares. (That's not the only way: $2 \cdot 16 \cdot (2+16)$ is a perfect square, for example. But such solutions are harder to find.) And the easiest way to make $p$, $q$, and $p+q$ all perfect squares is to set $p=3^2$ and $q=4^2$ so that $p+q = 5^2$.
Case 3 is the hardest, because our three roots $p,q,r$ must satisfy $pq+pr+qr=0$, which is not a terribly nice condition. In both subcases, the idea we are inspired by is that $(x+2)(x-3)(x-6)$ has this property (which we can check) and therefore $(x+2k)(x-3k)(x-6k)$ also has this property (because this scales all three terms of $pq+pr+qr$ by $k^2$).
The expansion of $(x+2k)(x-3k)(x-6k)$ is $x^3 - 7kx^2 + 36k^3$. So to force a polynomial of this form we need a value of $N$ that is
*
*of the form $-7k_1$ for some $k_1$, and
*of the form $36k_2^3$ for some $k_2$.
The most straightforward way to satisfy both conditions is to set $N = 36 \cdot 7^3$.
*
*If $N$ is chosen as the coefficient of $x^2$, then we can get the polynomial $(x+2k)(x-3k)(x-6k)$ with $k = -36 \cdot 7^2$.
*If $N$ is chosen as the constant term, then we can get the polynomial $(x+2k)(x-3k)(x-6k)$ with $k = 7$.
| {
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"url": "https://math.stackexchange.com/questions/4277439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Simplifying a derivative of a function involving roots. I am working on a question where I need to differentiate $$y=\{x+\sqrt{1+x²}\}^\frac{3}{2}\tag1$$
Using the chain rule I have found the first derivative $$\frac{dy}{dx} =\frac{3}{2}\{x+\sqrt{1+x²}\}^\frac{1}{2}\{1+\frac{x}{\sqrt{1+x²}}\}\tag2$$
Which can be written as
$$\frac{dy}{dx} = \frac{3y}{2\{x+\sqrt{1+x²}\}}\{1+\frac{x}{\sqrt{1+x²}}\}\tag3$$
This is as far as I have got in terms of simplifying. The solution says that $(2)$ is equal to $(4)$ $$\frac{\frac{3}{2}\{x+\sqrt{1+x²}\}^\frac{3}{2}}{\sqrt{1+x²}}=\frac{3y}{2\sqrt{1+x^2}}\tag4$$
How do you get from $(2)$ to $(4)$? I have included $(3)$ in order to show my workings.
| You can also consider taking the $\ln$ from both sides:
\begin{align*}
y = (x + \sqrt{1 + x^{2}})^{3/2} & \Rightarrow \ln(y) = \frac{3\ln(x + \sqrt{1 + x^{2}})}{2}\\\\
& \Rightarrow \frac{y'}{y} = \frac{3}{2(x + \sqrt{1 + x^{2}})}\times\left(1 + \frac{x}{\sqrt{1 + x^{2}}}\right)\\\\
& \Rightarrow y' = \frac{3y}{\sqrt{1 + x^{2}}}
\end{align*}
Hopefully this helps!
| {
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"url": "https://math.stackexchange.com/questions/4277518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding $\int_0^{\frac{\pi}{2}} \frac{x}{\sin x} dx $ Is there a way to show $$\int_0^{\frac{\pi}{2}} \frac{x}{\sin x} dx = 2C$$ where $C=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^2} $ is Catalan’s constant, preferably without using complex analysis?
The following is an attempt to expand it as as a series:
\begin{align*}
\int_0^{\frac{\pi}{2}} \frac{x}{\sin x} dx
&= \int_0^{\frac{\pi}{2}} \frac{x}{1-\cos^2 x}\sin x\ dx \\
&= \sum_{n=0}^{\infty} \int_0^{\frac{\pi}{2}} x\sin x \ \cos^{2n}x \ dx \\
&= \sum_{n=0}^{\infty}\frac{1}{2n+1} \int_0^{\frac{\pi}{2}} \cos^{2n+1}x \ dx \\
&= \sum_{n=0}^{\infty} \frac{4^n}{\binom{2n}{n}(2n+1)^2}
\end{align*}
which is close but not quite there.
| As you posted:
$$ \int_0^{\frac{\pi}{2}} \frac{x}{\sin x} dx \\ \sum_0^{\infty} \frac{4^n}{\binom{2n}{n}(2n+1)^2}$$
You can rewrite it as the following using the definition of the central binomial coefficient:
$$\sum_0^{\infty} \frac{4^n}{\binom{2n}{n}(2n+1)^2} = \sum_0^{\infty} \frac{4^n}{\frac{(2n)!}{n!^2}(2n+1)^2} = \sum_0^{\infty} \frac{4^n}{\frac{4^nΓ\left(n+\frac12\right)}{\sqrt{\pi}Γ(n+1)}(2n+1)^2}=\frac{\sqrt \pi}2\sum_0^\infty \frac{n!}{\left(n+\frac12\right)!(2n+1)} $$
Now let’s apply the main definition of the $\,_3\text F_2$ Hypergeometric function using Pochhammer Symbol:
$$\,_3\text F_2(a_1,a_2,a_3;b_1,b_2;z)=\sum_{n=0}^\infty \frac{(a_1)_n(a_2)_n(a_3)_n x^n}{(b_1)_n(b_2)_nn!}$$
The constant is then as seen in formula $(4)$ of this arxiv article:
$$\,_3\text F_2\left(\frac12,1,1;\frac32,\frac32;\frac{4x}{(x+1)^2}\right)=(x+1)\sum_{n=0}^\infty \frac{(-x)^n}{(2n+1)^2}$$
Therefore
$$ \int_0^{\frac{\pi}{2}} x\csc(x)= \,_3\text F_2\left(\frac12,1,1;\frac32,\frac32;1\right)=2\text C$$
Please correct me and give me feedback!
| {
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"timestamp": "2023-03-29T00:00:00",
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What's the measure of height relative to $BC$ in the triangle below?. For reference:In the isosceles triangle $ABC$ ($AB = BC$) is marked
the interior point O whose distances to sides $AB, BC and AC$ are $2, 3$ and $5$ respectively,
If the sum of the height measurements of triangle $ABC$ is $30$, calculate the
measure of height relative to $BC$.
My progress.. (drawing without scale)
I try :
$BF = h_1, CL = HK = h_2$
$HM \parallel FB \implies\\
\angle IMO \sim IBF:\\
\triangle AMH \sim ABF\\
\triangle OIM \sim \triangle AFB \implies\\
\frac{IM}{FB}=\frac{IO}{AH}=\frac{OM}{AB} \rightarrow \frac{IM}{h_1}=\frac{2}{AF}=\frac{OM}{AB}\\
\triangle ION \sim \triangle KCA \sim \triangle LAC$
but I can't finish...
| If altitude from $A$ and $C$ to the opposite sides is $h_1$ and from $B$, it is $h_2$, then
$2 h_1 + h_2 = 30 \tag1$
If $A$ is the area of $\triangle ABC$,
$ \displaystyle a = c = \frac{2A}{h_1}, b = \frac{2A}{h_2}$
Now note that $A = Area ~ \triangle OAB + Area ~ \triangle OBC + Area ~ \triangle OAC$
$ \displaystyle A = \frac{1}{2} \left[ 3 a + 2c + 5 b \right] = 5A \left(\frac{1}{h_1} + \frac{1}{h_2}\right) $
$ \implies \displaystyle \frac{1}{h_1} + \frac{1}{h_2} = \frac{1}{5} \tag2$
Using $(1)$ and $(2)$, can you solve for $h_1$?
| {
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"url": "https://math.stackexchange.com/questions/4285007",
"timestamp": "2023-03-29T00:00:00",
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Calculate $E[|X^{k}|]$ when k is odd, $X$ is a random variable with standard normal distribution I want to calculate $E[|X^{k}|]$ when k is odd, $X$ is a random variable with standard normal distribution. My approach is as follows.
Since $E[X^{k}]=0$ when $k$ is odd. We have $$E[X^{k}] = 0 = \int_{-\infty}^{\infty}x^{k}(\frac{1}{\sqrt{2\pi}}e^{\frac{-x^{2}}{2}})dx = \int_{-\infty}^{\infty}(x^{k})^{+}(\frac{1}{\sqrt{2\pi}}e^{\frac{-x^{2}}{2}})dx - \int_{-\infty}^{\infty}(x^{k})^{-}(\frac{1}{\sqrt{2\pi}}e^{\frac{-x^{2}}{2}})dx$$, where $(x^{k})^{+} = x^{k}$ when $x^{k}\geq 0$ and $0$ otherwise. $(x^{k})^{-} = -x^{k}$ when $x^{k}\leq 0$ and $0$ otherwise.
From the above identity, we have $$\int_{-\infty}^{\infty}(x^{k})^{+}(\frac{1}{\sqrt{2\pi}}e^{\frac{-x^{2}}{2}})dx = \int_{-\infty}^{\infty}(x^{k})^{-}(\frac{1}{\sqrt{2\pi}}e^{\frac{-x^{2}}{2}})dx$$
Now $$E[|X^{k}|] = \int_{-\infty}^{\infty} ((x^{k})^{+} + (x^{k})^{-})(\frac{1}{\sqrt{2\pi}}e^{\frac{-x^{2}}{2}})d = 2\int_{-\infty}^{\infty}(x^{k})^{+}(\frac{1}{\sqrt{2\pi}}e^{\frac{-x^{2}}{2}})dx$$
Since $(x^{k})^{+}=0$ when $x\in[-\infty,0)$, we can write
$$E[|X^{k}|] = 2\int_{0}^{\infty}x^{k}(\frac{1}{\sqrt{2\pi}}e^{\frac{-x^{2}}{2}})dx$$
Let $A_{k} = \int_{0}^{\infty}x^{k}(\frac{1}{\sqrt{2\pi}} e^{\frac{-x^{2}}{2}}) dx = \frac{1}{\sqrt{2\pi}} \int_{0}^{\infty}x^{k} (e^{\frac{-x^{2}}{2}}) dx$.
We use integration by parts. Take $u=x^{k-1}, dv = xe^{\frac{-x^{2}}{2}}$, then $du=(k-1)x^{k-2}, v = -e^{\frac{-x^{2}}{2}}$.
Thus
$$A_{k} = \frac{1}{\sqrt{2\pi}}(-x^{k-1} e^{\frac{-x^{2}}{2}}\Big|_{0}^{\infty} + (k-1)\int_{0}^{\infty}x^{k-2} (e^{\frac{-x^{2}}{2}}) dx) = (k-1)A_{k-2}$$
Since $A_{1} = 1$, we have $E[|X^{k}|] = 2A_{k} = 2((k-1)(k-3)\cdots 1)$.
I think my approach is wrong since using my solution, $E[|X^{5}|]=16$ and $E[|X^{6}|]=15$. I want to ask which step is wrong? Any help is appreciated, thank you.
| Let $k$ be a positive integer and $$I(k) = \int_{x=0}^\infty x^{2k-1} e^{-x^2/2} \, dx.$$ Then the substitution
$$u = x^2/2, \quad du = x \, dx$$ gives
$$I(k) = \int_{u=0}^\infty (2u)^{k-1} e^{-u} \, du = 2^{k-1} \int_{u=0}^\infty u^{k-1} e^{-u} \, du = 2^{k-1} \Gamma(k).$$ So we have
$$\operatorname{E}[|X^{2k-1}|] = \frac{2}{\sqrt{2\pi}} I(k) = \frac{2^k \Gamma(k)}{\sqrt{2\pi}}.$$ A table for the first few values of $k$ are as follows:
$$\begin{array}{c|c|c}
k & 2k-1 & \operatorname{E}[|X^{2k-1}|] \\
\hline
1 & 1 & \sqrt{\frac{2}{\pi}} \\
2 & 3 & 2 \sqrt{\frac{2}{\pi}} \\
3 & 5 & 8 \sqrt{\frac{2}{\pi}} \\
4 & 7 & 48 \sqrt{\frac{2}{\pi}} \\
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4285731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Substitution to evaluate $\int{\sqrt{\sin x}\cos^3x\ dx}$? I'm already solving more daunting exercises, but for some reason I can't tackle this one:
$$
\int{\sqrt{\sin x}\cos^3x\ dx}
$$
None of the "obvious" substitutions ($\sin x$, $\sqrt{\sin x}$, $\cos x$, $\cos^3x$) seem to make sense. I'm probably missing something obvious. Can someone give a nudge please?
Solution based on the hints below:
*
*$\int\sqrt{\sin x }\cos^3x\ dx = \int{\sqrt{\sin x}(1-\sin^2{x}})\cos x\ dx$
*Now if we substitute $$u=\sin x \\ du=\cos x\ dx$$ We'll get: $$\int{u^{1/2}(1-u^2)\ du} = \int{u^{1/2}}\ du - \int{u^{5/2}}\ du = \frac{2}{3}u^{3/2} - \frac{2}{7}u^{7/2} + c$$
*This results in $$\frac{2}{3}\sin^{3/2}x - \frac{2}{7}\sin^{7/2}x + c$$
| Another trivial substitution is $u^2=\sin x$.
So we have
$$\cos x\,\mathrm{d}x=2u\,\mathrm{d}u$$
As a result,
\begin{align}
\int \sqrt{\sin x}\cos^3 \,\mathrm{d}x
&= \int \sqrt{\sin x}\, \cos^2 x \,\cos x \, \mathrm{d}x \\
&= \int \sqrt{u^2}\, (1-u^4) \,2u\, \mathrm{d}u \\
&= 2\int u^2(1-u^4)\,\mathrm{d}u\\
&= \frac{2}{3}u^3 -\frac{2}{7} u^7 + c\\
&= \frac{2}{3}\sin^{3/2}x -\frac{2}{7} \sin^{7/2}x + c
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4286737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Demonstrate that $3ba^2 = a^3\sqrt3 +b^3$ in the triangle below For reference: In triangle ABC: $AC=BC=a, AB=b$ and $\angle C= 40^\circ$.
Demonstrating identity: $3ba^2 = a^3\sqrt3+b^3$
My progress:
Draw $CD \perp AB, D\in AB$ and $AF \perp CD,(F \in CD)$
$\triangle ABF \sim \triangle CBD \implies:
\frac{AF}{CD}=\frac{BF}{BD}=\frac{AB}{BC}\rightarrow \frac{BF}{\frac{b}{2}} = \frac{b}{a}\\
\therefore BF = \frac{b^2}{2a}\\
\triangle AGD \sim ABF \implies\\
\frac{DG}{BF}=\frac{AG}{b}=\frac{\frac{b}{2}}{AF}\rightarrow \frac{DG}{BF}=\frac{AG}{b}=\frac{b}{2AF}\implies \frac{DG}{AG} = \frac{b}{2a}(I)\\
\triangle AGD \sim \triangle CAD \implies\\
\frac{AD}{DC}=\frac{AG}{AC}=\frac{DG}{AD}\rightarrow \frac{b}{2DC}=\frac{AG}{a}=\frac{2DG}{b} \implies \frac{DG}{AG} = \frac{b}{2a}\\
\therefore (I): \frac{b}{2a} = \frac{BF}{b} \implies BF = \frac{b^2}{2a}\\
\triangle ADC: DC^2 = a^2-(\frac{b}{2})^2\implies DC = \frac{\sqrt{4a^2-b^2}}{2} $
But I am not able to equate...
| Draw line segment $\small AD$ as in the figure.
$\small \triangle ACB\sim\triangle BAD\implies\dfrac{BD}b=\dfrac ba\implies BD=\dfrac{b^2}a$
Therefore, $\small CD=a-\dfrac{b^2}a$.
Now applying cosine rule to $\small \triangle ACD$ with the angle $\small 30^\circ$ gives the desired formula.
$\small \left(a-\dfrac{b^2}a\right)^2=a^2+b^2-2ab\cos30^\circ$
If you are not interested in using cosine rule directly,
Drop perp from $\small D$ to $\small AC$ and mark the foot $\small E$. Now using the fact that $\small \triangle ADE$ is a half of an equilateral triangle, we see $\small DE=b/2$ and $\small AE=\sqrt3b/2$. Now find $\small EC$ and already found $\small CD$, apply Pythagorean theorem to $\small \triangle CED$, from where you can derive the same expression.
| {
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"timestamp": "2023-03-29T00:00:00",
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Non-linear 1st order differential equation of $~x\frac{dy}{dx}+y=x\sqrt{y}~~~$where$~\sqrt{y}~$exists I think this is the first time when I handle of non-linear differential equation.
$$ \underbrace{x \frac{dy}{dx} + y = x\sqrt{ y } }_{x > 0} \tag{1} $$
$$ \frac{dy}{dx} + \frac{1}{ x } y = \sqrt{ y } $$
$$ \underbrace{\frac{d}{dx}\left(yx\right)}_{\text{LHS of eqn1} } = x \sqrt{ y } ~~ \leftarrow~~ \text{What can I do for next?} $$
Which website is suitable for this problem?
ADD
I got the following after I have read the post of @Gerd
$$ \frac{d}{dx}\left(yx\right) = x \sqrt{ y } \tag{2} $$
$$ \frac{d}{dx}\left(xy\right) = x^{\frac{1}{2} } \cdot x^{\frac{1}{2} } \sqrt{ y } $$
$$ = \sqrt{ x } \sqrt{ xy } $$
$$ z:=xy \tag{3} $$
$$ \frac{ dz }{ dx } = \sqrt{ x } \sqrt{ z } \tag{4} $$
$$ \frac{ 1 }{ \sqrt{ z } } \frac{dz}{dx} = \sqrt{ x } $$
$$ \int_{ }^{ } \frac{1}{ \sqrt{ z } } \frac{dz}{dx} \,dx = \int_{ }^{ } \sqrt{ x } \,dx $$
$$ \int_{ }^{ } \frac{1}{ \sqrt{ z } } \,dz = \int_{ }^{ } x^{\frac{1}{ 2 } } \,dx $$
$$ \int_{ }^{ } \frac{ 1 }{ z^{\frac{1}{2} } } \,dz = \frac{ x^{\frac{ 3 }{ 2 } } }{ \frac{ 3 }{ 2 } } + \text{const}_{1} $$
$$ \int_{ }^{ } z^{-1/2} \,dz = \frac{ 2 }{ 3 } x\sqrt{ x } + \text{const}_{1} $$
$$ \frac{ z^{\frac{1}{2}} }{ \frac{1}{2} } = \frac{ 2 }{ 3 } x\sqrt{ x } + \text{const}_{1} $$
$$ 2z^{\frac{1}{2}} = \frac{ 2 }{ 3 } x\sqrt{ x } + \text{const}_{1} $$
$$ 2 \sqrt{ z } = \frac{ 2 }{ 3 } x\sqrt{ x } + \text{const}_{1} $$
$$ 2 \sqrt{ yx } = \frac{ 2 }{ 3 } x\sqrt{ x } + \text{const}_{1} $$
$$ 2 \sqrt{ yx } - \frac{ 2 }{ 3 } x\sqrt{ x } = \text{const}_{1} $$
$$ \underbrace{\sqrt{ x } \left( 2 \sqrt{ y } - \frac{ 2 }{ 3 } x \right) = \text{const}_{1} }_{\text{general solution} } ~~ \leftarrow~~ \text{Is this also correct?} $$
| This is a special case of a Bernoulli differential equation, wich has a well know method of solution by transforming in a linear equation.
| {
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"timestamp": "2023-03-29T00:00:00",
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In a room of $n \geq 7$ people, what is the probability that, for each day of the week, there is at least one person whose birthday falls on that day? In a room of $n \geq 7$ people, what is the probability that, for each day of the week, there is at least one person whose birthday falls on that day? Assume each day of the week has equal probability.
I am thinking about allocating each day one person and then stars and bars the rest? But cant think of the sample space.
I also consider using complement:
$$1- \binom{7}{1}\left(\frac{6}{7}\right)^n - \binom{7}{2}\left(\frac{5}{7}\right)^n \ldots$$
but then I will double count. I tried inclusion-exclusion, but I can't figure it out.
| Since there are seven possible days of the week on which each person could have been born, there are $7^n$ possible birthday assignments.
If at least one person is born on each day of the week, we must exclude those birthday assignments in which at least one day is missing. There are $\binom{7}{k}$ ways to exclude $k$ days of the week and $(7 - k)^n$ ways to distribute the $n$ birthdays to the remaining $7 - k$ days of the week. Hence, by the Inclusion-Exclusion Principle, the number of favorable cases is
$$\sum_{k = 0}^{7} (-1)^k\binom{7}{k}(7 - k)^n = 7^n - \binom{7}{1}6^n + \binom{7}{2}5^n - \binom{7}{3}4^n + \binom{7}{4}3^n - \binom{7}{5}2^n + \binom{7}{6}1^n - \binom{7}{7}0^n$$
Hence, the probability that at least one person is born on each day of the week is
$$\frac{7^n - \dbinom{7}{1}6^n + \dbinom{7}{2}5^n - \dbinom{7}{3}4^n + \dbinom{7}{4}3^n - \dbinom{7}{5}2^n + \dbinom{7}{6}1^n - \dbinom{7}{7}0^n}{7^n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4301055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that: $\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\geq\frac{21}{2}$ Given 3 positive real numbers $x, y, z$ satisfies $xy+yz+xz=1$. Prove that: $$\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\geq\frac{21}{2}$$
(Only use AM-GM, Cauchy-Schwarz inequalities)
My progress:
Till now, I have not made much of a progress besides finding out that $x^2+y^2+z^2\geq1$ and $xyz\leq\frac{\sqrt{3}}{9}$ and turn the LHS into $$\frac{x}{\sqrt{(x+y)(x+z)}}+\frac{y}{\sqrt{(x+y)(y+z)}}+\frac{z}{\sqrt{(x+z)(y+z)}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}$$
| The desired inequality is written as
$$\frac{2x}{\sqrt{1 + x^2}}
+ \frac{2y}{\sqrt{1 + y^2}} +
\frac{2z}{\sqrt{1 + z^2}}
+ 8 \cdot \frac{1 + x^2}{4x^2}
+ 8 \cdot \frac{1 + y^2}{4y^2}
+ 8 \cdot \frac{1 + z^2}{4z^2} \ge 27.$$
Using AM-GM, we have
\begin{align*}
\mathrm{LHS}
&\ge 27\,\sqrt[27]{\frac{2x}{\sqrt{1 + x^2}}
\frac{2y}{\sqrt{1 + y^2}}
\frac{2z}{\sqrt{1 + z^2}}
\left(\frac{1 + x^2}{4x^2}\right)^8
\left(\frac{1 + y^2}{4y^2}\right)^8
\left(\frac{1 + z^2}{4z^2}\right)^8}\\
&= 27\,\sqrt[27]{
\left(\frac{\sqrt{(1 + x^2)(1 + y^2)(1 + z^2)}}{8xyz}\right)^{15}}\\
&= 27\,\sqrt[27]{
\left(\frac{(x + y)(y + z)(z + x)}{8xyz}\right)^{15}}\\
&\ge 27\,\sqrt[27]{
\left(\frac{2\sqrt{xy} \cdot 2\sqrt{yz} \cdot 2\sqrt{zx}}{8xyz}\right)^{15}}\\
&= 27
\end{align*}
where we have used
\begin{align*}
&(1 + x^2)(1 + y^2)(1 + z^2)\\
=\, & (xy + yz + zx + x^2)(xy + yz + zx + y^2)(xy + yz + zx + z^2)\\
=\, & (x + y)^2(y + z)^2(z + x)^2.
\end{align*}
We are done.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $R_{ab} = \frac{1}{2} S g_{ab}$ (2 dimensional Ricci curvatures) In the title $R_{ab}$ is the usual Ricci curvature tensor and $S$ is the scalar curvature. As we are in a two dimensional Riemanian manifold with metric $g$, we can write $S= g^{11} R_{11} + g^{12} R_{12} + g^{21} R_{21} + g^{22} R_{22}$, where $g^{ij} = (g_{ij})^{-1}$.
I have read here that $g^{ik} g_{kj} = \delta_i^j$ where $\delta_i^j$ is the Kronecker delta function.
Again, because we are in two dimensions we really just need to compute $R_{12}=R_{21}, R_{11}$ and $R_{22}$. The first one is straightforward. If we see what we obtain on the right hand side of the equation in the title ($R_{ab} = \frac{1}{2} S g_{ab}$):
\begin{align*}
\frac{1}{2} S g_{12} &= \frac{1}{2} (g^{11} R_{11} + g^{12} R_{12} + g^{21} R_{21} + g^{22} R_{22}) g_{12} \\
&= \frac{1}{2} (g^{11} R_{11} + 2g^{12} R_{12} + g^{22} R_{22}) g_{12} \\
&= \frac{1}{2} (g^{11} g_{12} R_{11} + 2g^{12} g_{12} R_{12} + g^{22} g_{21} R_{22}) \\
&= \frac{1}{2} ( 2g^{12} R_{12} g_{12}) = R_{12}
\end{align*}
When dealing however with the remaining 2 I am not sure on how to deal with expressions like $g^{11} R_{11} g_{22}$. Any hints are indeed appreciated.
| You have to use that in dimension $2$, the curvature tensor has a very specific formula. Namely, it holds that $$R(X,Y)Z=K(g(Y,Z)X-g(X,Z)Y),$$where $K$ is the Gaussian curvature of the surface. Trace in $X$ to obtain $${\rm Ric}(Y,Z) = Kg(Y,Z).$$Now take the $g$-trace to obtain $S = 2K$. So $K=S/2$ and so ${\rm Ric} = (S/2)g$.
| {
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"url": "https://math.stackexchange.com/questions/4303515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to turn into a product of $2$ factors the expression : $ x^4 -3x^2+1$? Source : Lebossé & Hémery, Algèbre et Analyse ( Classe de seconde , 1965).
The exercice requires : turn into a product of $2$ factors the expression $ x^4 -3 x^2 +1$.
The only way I see is to use the property : $P(x)$ being a polynomial , if $P(a) =0$ then $P(x)$ is divisible by $(x-a)$.
Using the substitution $X = x^2$, the expression can be turned into : $X^2 -3X +1$
Setting $X^2 -3X +1= 0$ yields $ X = \frac {3+\sqrt5} {2}$ or $ X = \frac {3-\sqrt5} {2}$.
This implies that : $x= \pm\sqrt{ \frac {3+\sqrt5} {2}}$ or $x= \pm\sqrt{ \frac {3-\sqrt5} {2}}$.
Finally, using the property referred to above, one gets :
$ x^4 -3 x^2 +1 = ( x- \sqrt{\frac {3+\sqrt5} {2}}) ( x+ \sqrt{\frac {3+\sqrt5} {2} })( x - \sqrt{\frac {3-\sqrt5} {2}})(x+ \sqrt{\frac {3-\sqrt5} {2}})$.
But this method yields 4 factors instead of 2.
Is there a better way to meet the requirement of the exercise?
Is there a classical identity hidden below the original expression?
| It is quite easy question. You said two factors.
$$x^4-3x^2+1=(x^4-2x^2+1)-x^2=(x^2-1)^2-x^2=(x^2+x-1)(x^2-x-1).$$
Is it your answer?
| {
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Olympiad Math Question - If the sum of the positive inverses of 4 positive integers equals $1.1$, what’s the lowest possible sum of the integers? I was going through some Olympiad pass papers and came across this question:
Given four different positive integers $A, B, C, D$ so that $\frac{1}{A}+\frac{1}{B}+ \frac{1}{C}+\frac{1}{D}=1.1$. Find the smallest possible value of $A+B+C+D$.
Does the value $1.1$ have to do with anything? Also, what trick can I use to solve this question? Is there an inverse equation formula of some type I can use?
I tried doing this:
$$\frac{1}{A}+\frac{1}{B}+ \frac{1}{C}+\frac{1}{D}=1.1$$
$$ABC+BCD+ACD+ABD=1.1ABCD$$
But now I don’t know where to continue. Also, this is an Olympiad math question, which means I probably need an answer that can solve the question in 2 minutes or less.
| Without loss of generality, suppose $A < B < C < D$.
Notice that if $A \ge 3$, then $B \ge 4$, $C \ge 5$, $D \ge 6$, and then $\tfrac{1}{A}+\tfrac{1}{B}+\tfrac{1}{C}+\tfrac{1}{D} \le \tfrac{1}{3}+\tfrac{1}{4}+\tfrac{1}{5}+\tfrac{1}{6} = \tfrac{19}{20} < 1.1$. So we need $A = 1$ or $A = 2$.
Case 1: $A = 2$. Then we need $\tfrac{1}{B}+\tfrac{1}{C}+\tfrac{1}{D} = \tfrac{3}{5}$ with $2 < B < C < D$.
Since $\tfrac{3}{B} > \tfrac{1}{B}+\tfrac{1}{C}+\tfrac{1}{D} = \tfrac{3}{5}$, we must have $B < 5$, i.e. $B = 3$ or $B = 4$.
If $B = 4$, we need $\tfrac{1}{C}+\tfrac{1}{D} = \tfrac{7}{20}$ with $4 < C < D$. Since $\tfrac{2}{C} > \tfrac{1}{C}+\tfrac{1}{D} = \tfrac{7}{20}$, we must have $C < \tfrac{40}{7}$, i.e. $C \le 5$. Since $4 < C \le 5$, we must have $C = 5$, but then $D = \tfrac{20}{3}$, which is not an integer. So there are no solutions with $A = 2$ and $B = 4$.
If $B = 3$, we need $\tfrac{1}{C}+\tfrac{1}{D} = \tfrac{4}{15}$ with $3 < C < D$. Since $\tfrac{2}{C} > \tfrac{1}{C}+\tfrac{1}{D} = \tfrac{4}{15}$, we must have $C < \tfrac{15}{2}$, i.e. $C \le 7$. Testing $C = 4, 5, 6, 7$ yields $D = 60, 15, 10, \tfrac{105}{13}$ respectively. In this case, the smallest sum where $C$ and $D$ are integers is $21$ which occurs for $(A,B,C,D) = (2,3,6,10)$.
Case 2: $A = 1$. Then, we need $\tfrac{1}{B}+\tfrac{1}{C}+\tfrac{1}{D} = \tfrac{1}{10}$. But since $\tfrac{3}{D} < \tfrac{1}{B}+\tfrac{1}{C}+\tfrac{1}{D} = \tfrac{1}{10}$, any solution in this case will have $D > 30$, and thus, $A+B+C+D > 30 > 21$. So we will not find a smaller sum in this case.
Therefore, the minimum sum is $21$.
Note that if you just need to get an answer quickly without a rigorous proof, then you can probably just guess and check until you find something reasonably small. In problems with Egyptian fractions (fractions with numerator $1$), the sum $\tfrac{1}{2}+\tfrac{1}{3}+\tfrac{1}{6} = 1$ comes up a lot, namely it is the smallest set of distinct Egyptian fractions that add up to $1$. So it's not too hard to build off of that to get $\tfrac{1}{2}+\tfrac{1}{3}+\tfrac{1}{6}+\tfrac{1}{10} = \tfrac{11}{10}$. I'm not sure if there is an easy way to convince yourself that's the smallest sum though.
| {
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"timestamp": "2023-03-29T00:00:00",
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Gaussian integral $\int_{0}^{\infty} dx \, x \, e^{-a^2 x^2} \left( \sqrt{(x - c)^2 + b^2} - \sqrt{(x + c)^2 + b^2} \right)$ I've been having trouble evaluating the Gaussian integral of the form
$$\int_{0}^{\infty} dx \, x \, e^{-a^2 x^2} \left( \sqrt{(x - c)^2 + b^2} - \sqrt{(x + c)^2 + b^2} \right) \, ,$$
which can be rearranged into integrals of the form
$$\int_{0}^{\infty} dx \, x^k \sqrt{x^2 + b^2} \left( e^{-a^2 (x + c)^2} \pm e^{-a^2 (x - c)^2} \right) \, ,$$
for $a, \, b, \, c > 0$ and $k \in \mathbb{Z}$. Does anyone know how to evaluate these expressions analytically? Or know any good integral tables on Gaussians I should refer to?
Edit: While browsing Gradshteyn and Ryzhik, I came across notation for my difference of square roots in the section on Bessel functions, see $\S 6.52$ and e.g. notation $6.522$ with $\ell_1 = \frac{1}{2} \left( \sqrt{(b + c)^2 + a^2} - \sqrt{(b - c)^2 + a^2} \right)$. I didn't realise functions of $\ell_1$ showed up in other contexts, although here $\ell_1$ appears in the integrand and as a function of $x$. Based on this connection though, I wonder if the integral above may evaluate to a Bessel function or combination thereof. Of course, this integral is still very non-trivial, so any help would be much appreciated.
| As @rtem Alexandrov commented, a formal expansion around $x=0$ would perfectly work and would probably converge fast since
$$\sqrt{(x - c)^2 + b^2} - \sqrt{(x + c)^2 + b^2}=-2 c+\frac{b^2 c}{x^2}+\frac{b^2 c \left(4 c^2-3 b^2\right)}{4x^4}+O\left(\frac{1}{x^6}\right)$$
Expanded as series around $x=0$, we have $$\sqrt{(x - c)^2 + b^2}=\sum_{n=0}^\infty \alpha_n x^n$$
$$\alpha_0=\sqrt{b^2+c^2}\qquad \alpha_1=-\frac{c}{\sqrt{b^2+c^2}} \qquad\alpha_n=\frac{c (2 n-3) \alpha_{n-1}-(n-3) \alpha_{n-2}}{n \left(b^2+c^2\right)} $$
$$\sqrt{(x + c)^2 + b^2}=\sum_{n=0}^\infty \beta_n x^n$$
$$\beta_0=\sqrt{b^2+c^2}\qquad \beta_1=\frac{c}{\sqrt{b^2+c^2}} \qquad\beta_n=-\frac{c (2 n-3) \beta_{n-1}+(n-3) \beta_{n-2}}{n \left(b^2+c^2\right)} $$
So, making the problem more general
$$I_k=\int_{0}^{\infty} x^k \, e^{-a^2 x^2} \left( \sqrt{(x - c)^2 + b^2} - \sqrt{(x + c)^2 + b^2} \right) \,dx$$ Since $\alpha_{2n}=\beta_{2n}$, then
$$\large\color{red}{I_k=\frac 1{2a^{k+2}}\sum_{n=0}^\infty \frac{\alpha_{2n+1} -\beta_{2n+1}}{a^{2n}}\Gamma \left(n+1+\frac{k}{2} \right)}$$
For illustration purposes, trying with only ten terms and $a=1$,$b=2$, $c=3$, $k=1$ we get
$$-\frac{21175391684262887968287 }{14394292089850546320512}\sqrt{\frac{\pi }{13}}=\color{red}{-0.723176}66$$ while numerical integration gives $\color{red}{-0.72317639}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Manipulating a polar equation I started with this:
$\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$
And via substitution, got this far:
$r=\left(\frac{144}{9-25\sin^{2}\theta}\right)^{.5}$
For the fact that Desmos plots these the same, I assume I'm right so far.
The goal in this section is to end with the form
$r=\frac{ep}{1-e\sin\theta}$
and I am at a loss what the next manipulation might be.
| The right focus is $(5, 0)$. Shifting the hyperbola to the left by $5$ units results in the new equation:
$ \dfrac{(x + 5)^2}{16} - \dfrac{y^2}{9} = 1 $
Now using polar coordinates, but measuring the angle $\theta $ from the positive $y$ axis direction, then $x = -r \sin \theta , y = r \cos \theta $
Hence,
$ \dfrac{ (-r \sin \theta + 5 )^2 }{16} - \dfrac{ (r \cos \theta)^2 }{9} = 1 $
Multiplying through by $144$,
$ 9 ( r^2 \sin^2 \theta - 10 r \sin \theta + 25 ) - 16 r^2 \cos^2 \theta = 144 $
Simplifiying,
$ r^2 ( 25 \sin^2 \theta - 16 ) - 90 r \sin \theta + 81 = 0 $
Using the quadratic formula,
$ r = \dfrac{1}{ 2(25 \sin^2 \theta - 16) } \left( 90 \sin \theta - \sqrt{ 8100 \sin^2 \theta - (8100 \sin^2 \theta - 5184 ) } \right)$
And this simplifies to,
$ r = \dfrac{ 90 \sin \theta -72 }{2(5 \sin \theta - 4)(5 \sin \theta + 4) } $
which simplifies further to
$ r = 9 \dfrac{ 5 \sin \theta - 4 }{ (5 \sin \theta - 4)(5 \sin \theta + 4 ) } = \dfrac{ 9 }{ 5 \sin \theta + 4 } $
$e = \sqrt{1 + \dfrac{3}{4}^2 } = \dfrac{5}{4}$
Therefore,
$ r = \dfrac{ 9/4 }{ 1 + e \sin \theta } $
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving a problem in Coordinate Geometry Suppose, $2x \cos \alpha - 3y \sin \alpha = 6$ is equation of a variable straight line. From two point $A(\sqrt{5},0$) and $B(-\sqrt{5},0)$ foot of the altitude on that straight line is $P$ and $Q$ respectively. Show that the product of the length of two line segment $AP$ and $BQ$ is free of $\alpha$.
This question appeared in my exam today. The way I did it is first constructed the equation of two perpendicular line to $2x \cos \alpha - 3y \sin \alpha = 6$ which goes through the points $A(\sqrt{5},0$) and $B(-\sqrt{5},0)$. In this way, for $A$ and $B$, I got the following equations respectively-
$$3x \sin \alpha + 2y \cos \alpha -3 \sqrt{5}\sin \alpha=0 \qquad(1)$$$$3x \sin \alpha + 2y \cos \alpha +3 \sqrt{5}\sin \alpha=0 \qquad(2)$$
Then I found that the line $(1)$ intersects the line $2x \cos \alpha - 3y \sin \alpha = 6$ at point $$P\left ( \frac{9\sqrt5 \sin^2 \alpha+12 \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha}, \frac{-18 \sin \alpha+6\sqrt{5} \sin \alpha \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha} \right )$$
and the line $(2)$ intersects the line $2x \cos \alpha - 3y \sin \alpha = 6$ at point $$Q\left ( \frac{-9\sqrt5 \sin^2 \alpha+12 \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha}, \frac{-18 \sin \alpha-6\sqrt{5} \sin \alpha \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha} \right )$$
Now using distance formula $$AP = \sqrt{ \left ( \frac{9\sqrt5 \sin^2 \alpha+12 \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha}- \sqrt5 \right )^2 + \left ( \frac{-18 \sin \alpha+6\sqrt{5} \sin \alpha \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha} \right )^2}$$ and $$ BQ= \sqrt{ \left ( \frac{-9\sqrt5 \sin^2 \alpha+12 \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha} +\sqrt5 \right )^2 + \left ( \frac{-18 \sin \alpha-6\sqrt{5} \sin \alpha \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha} \right )^2}$$
Now the multiplication product $AP \cdot BQ$ indeed gives a constant value of $4$ which is free of the arbitrary variable $\alpha$ as you can see here is the simplified version of product of those two quantity. But this is tedious and I do not think this is the only way to do it and an appropriate way to follow in exam with limited time. So, I am looking for an alternative, time saving proof of it. I was wondering if using parametric form would help me, but I think it would get as difficult equally.
I created a visualisation for you on desmos to help my problem understand better.
| $ \displaystyle 2x \cos \alpha - 3y \sin \alpha = 6 \implies y = \frac{2 \cot \alpha}{3} x - 2 \csc \alpha = mx + c$ where $m = \tan \theta = \dfrac{2 \cot \alpha}{3}$
Consider $0 \leq \alpha \leq 2\pi$ that gives all possible lines for the given equation. If $\alpha = \frac{\pi}{2}$ or $\frac{3 \pi}{2}$, the line is $y = -2$ or $y = 2$ respectively and the perp distance from both points on x-axis would be $2$. So the product is $4$. For other values of $\alpha$
The line intersects x-axis at $~ \displaystyle x = \frac{3}{\cos\alpha}$
So distance of the given points to the intersection point is,
$ \left|\displaystyle \sqrt5 \pm \frac{3}{\cos\alpha}\right|$
The product of perpendicular lengths to the line is then given by,
$ \displaystyle \left| 5 - \frac{9}{\cos^2\alpha}\right| \cdot \sin^2\theta = \left| \frac{5 \cos^2\alpha - 9}{\cos^2\alpha} \cdot \frac{\tan^2\theta}{1 + \tan^2\theta} \right|$
Now, $ \displaystyle \frac{\tan^2\theta}{1 + \tan^2\theta} = \frac{4 \cot^2\alpha}{9 + 4 \cot^2\alpha} = \frac{4 \cos^2\alpha}{9 \sin^2\alpha + 4 \cos^2\alpha}$
$ \displaystyle = \frac{4 \cos^2\alpha}{9 - 5 \cos^2\alpha}$
That leads to product being $4$.
| {
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Showing that $b \leq \frac{b}{\frac{1}{4}+b^2} \Leftrightarrow b^2-2b+\frac{1}{4} \leq 0$ For an ODE where the solution should be on the interval $[-\frac{1}{2},\frac{1}{2}]$ it has to be $b \leq \frac{b}{\frac{1}{4}+b^2}$.
The task is:
Let $f: [-\frac{1}{2}, \frac{1}{2}]\times[-b,b]\to \mathbb{R}$ with $f(x,y):=x^2+y^2$. Find $b>0$ with the theorem of Picard-Lindelöf such that the IVP $y'=f(x,y),\ y(0)=0$ has an unique solution for all $|x|\leq\frac{1}{2}$.
The solution says
$b \leq \frac{b}{\frac{1}{4}+b^2} \Leftrightarrow b^2-2b+\frac{1}{4} \leq 0$.
So $b \in [1-\frac{\sqrt{3}}{2}, 1+\frac{\sqrt3}{2}]$.
I get $b^3-\frac{3}{4}b \leq 0$, so $b\leq \frac{-\sqrt{3}}{2}$ or $b \in [0, \frac{\sqrt3}{2}]$.
How to get $b \leq \frac{b}{\frac{1}{4}+b^2} \Leftrightarrow b^2-2b+\frac{1}{4} \leq 0$?
| The "butterfly" (double cone) condition that you have to check is that
$$
Ma\le b,
$$
with $a,b$ the sizes of the box under consideration, $a=\frac12$, and $M=a^2+b^2=\frac14+b^2$ the maximum of the function value inside the box. Inserted this gives
$$
\frac12(\frac14+b^2)\le b\iff \frac14-2b+b^2\le 0.
$$
To compute a non-constant upper bound, one can check that for the differential inequality
$$
|y'|\le \frac14+y^2
$$
one gets
$$
|y(x)|\le\frac12\tan(\frac{|x|}2),
$$
so that in principle the height $b=\frac12\tan(\frac14)$ is sufficient to contain the solution inside the box.
| {
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"url": "https://math.stackexchange.com/questions/4308015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate $\lim_{x\rightarrow \infty} \frac{1}{\ln(x+1)-\ln(x)}-x$ I'm supposed to compute
$$\lim_{x\rightarrow \infty}\left( \frac{1}{\ln(x+1)-\ln(x)}-x\right).$$
However, I keep getting the wrong answer, so I'll present my solution for you, and I hope you can give me any tips on how to solve it.
Rewriting using logarithm laws, we have
$$\lim_{x\rightarrow \infty} \left(\dfrac{1}{\ln\frac{x+1}{x}}-x\right).$$
Simplyfing further, we have:
$$\lim_{x\rightarrow \infty} \dfrac{1-x\ln\frac{x+1}{x}}{\ln\dfrac{x+1}{x}}= \lim_{x\rightarrow \infty} \frac{1-\ln(1+\frac{1}{x})^x}{\ln(1+\frac{1}{x})} = \frac{1-e}{0} \rightarrow -\infty.$$
However, the answer sheet tells me that it's $1/2$, and I don't really see where I did something wrong in the solution. Thanks.
| If we are allowed to use series expansions, then the problem is fairly trivial. We want to evaluate $$\lim_{x\to \infty} \frac{1 - x \log \left(1+ \frac1x\right)}{\log \left(1+ \frac1x\right)}$$
The series expansion of $\log \left(1+ \frac1x\right)$ is given by $$\log \left(1+ \frac1x\right) = \frac{1}{x} - \frac{1}{2x^2} + \frac{1}{3x^3} - \frac{1}{4x^4} + \ldots$$
Plugging this in, we find that $$\frac{1 - x \log \left(1+ \frac1x\right)}{\log \left(1+ \frac1x\right)} = \frac{1 - x\left(\frac{1}{x} - \frac{1}{2x^2} + \frac{1}{3x^3} + \ldots \right)}{\left(\frac{1}{x} - \frac{1}{2x^2} + \frac{1}{3x^3} + \ldots\right) } = \frac{\left(\frac{1}{2} - \frac{1}{3x} + \frac{1}{4x^2} \ldots \right)}{\left(1 - \frac{1}{2x} + \frac{1}{3x^2} + \ldots \right)} \xrightarrow{x\to\infty} \frac12$$
Thus, $$\boxed{\lim_{x\rightarrow \infty} \frac{1}{\ln(x+1)-\ln(x)}-x = \frac12}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is $\lim_{n\to\infty} n (\sum_{k = 1}^{n} \frac{1}{\sqrt {n^2 + k}} - 1)$? I am trying to find this limit: $\lim_{n \to \infty} a_n^{b_n}$, where $a_n = \sum_{k = 1}^{n} \frac{1}{\sqrt{n^2+ k}}$ and $b_n = n$. I have managed to prove that $\lim_{n \to \infty} a_n = 1$, by using the squeeze lemma (bounding $a_n$ between $1$ and $\frac{n}{\sqrt {n^2 + n}}$) and I want to use this theorem: if $\lim_{n \to \infty} = 0$, then $\lim_{n\to\infty} (1 + a_n) ^ {\frac{1}{a_n}} = e$. To apply this theorem, I did:
$$\lim_{n\to\infty}a_n^{b_n} = \lim_{n\to\infty}((1 + a_n - 1) ^ \frac{1}{a_n - 1}) ^ {b_n(a_n - 1)} = e^{\lim_{n\to\infty}b_n(a_n - 1)}$$. Now I am having problems finding the limit in the exponent. Could you help?
| We can rewrite the limit as
$$\lim_{n\to\infty}\sum_{k=1}^n \frac{n}{\sqrt{n^2+k}}-n = \lim_{n\to\infty}\sum_{k=1}^n \frac{n-\sqrt{n^2+k}}{\sqrt{n^2+k}} = \lim_{n\to\infty}\sum_{k=1}^n \frac{1-\sqrt{1+\frac{k}{n^2}}}{\sqrt{1+\frac{k}{n^2}}}$$
Since the numerator approaches zero we will rationalize and continue to simplify
$$\lim_{n\to\infty}\sum_{k=1}^n \frac{-\frac{k}{n^2}}{\sqrt{1+\frac{k}{n^2}}+1+\frac{k}{n^2}} = \lim_{n\to\infty}\sum_{k=1}^n \frac{-\frac{k}{n}}{\sqrt{1+\frac{k}{n^2}}+1+\frac{k}{n^2}}\cdot\frac{1}{n}$$
We can sandwich this complicated expression with two other limits
$$\lim_{n\to\infty}-\frac{1}{2}\sum_{k=1}^n \frac{k}{n}\cdot\frac{1}{n} < \lim_{n\to\infty}\sum_{k=1}^n \frac{-\frac{k}{n}}{\sqrt{1+\frac{k}{n^2}}+1+\frac{k}{n^2}}\cdot\frac{1}{n} < \lim_{n\to\infty}\frac{-1}{\sqrt{1+\frac{1}{n}}+1+\frac{1}{n}}\sum_{k=1}^n \frac{k}{n}\cdot\frac{1}{n}$$
Both of which approach the Riemann sum
$$\longrightarrow -\frac{1}{2}\int_0^1 x\:dx = -\frac{1}{4}$$
Thus the limit is $\boxed{-\frac{1}{4}}$ by squeeze theorem.
| {
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"timestamp": "2023-03-29T00:00:00",
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Optimization Triangle I can't seem to finish this task. I found the derivative and the critical numbers, except when the derivative $= 0$. I don't understand how I can find it manually.
Task:
$2$ points A and C lie on a line $l$. a point B is $4$ away from line $l$. The sum of AB + BC is always equal to $10$. Find the shortest distance for AC. (A and C can not be on the same side of the perpendicular, however, A or C can be on the perpendicular).
What I got:
I have these equations
let AB $= x$, and BC $= y$.
$$\text{AC} = \sqrt{x^2 - 4^2} + \sqrt{y^2 - 4^2}$$
$$\text{AC} = \sqrt{x^2 - 16} + \sqrt{y^2 - 16}$$
$$\text{AC} = \sqrt{x^2 - 16} + \sqrt{x^2 - 20x + 84}$$
$$\text{AC}' = \frac{x}{\sqrt{x^2 - 16}} + \frac{(x - 10)}{\sqrt{x^2 - 20x + 84}}$$
$$\text{AC}' = \frac{x}{\sqrt{(x+4)(x-4)}} + \frac{x - 10}{\sqrt{(14-x)(6-x)}}$$
I can now find that $x = 4, x = -4, x = 14, x = 6$ are critical numbers, and only $x = 4$ and $x = 6$ apply. But how do I now find AC' $= 0?$
| From $$0 = \frac{x}{\sqrt{(x+4)(x-4)}} + \frac{x-10}{\sqrt{(14-x)(6-x)}},$$ simply multiply both sides by $\sqrt{(x+4)(x-4)} \cdot \sqrt{(14-x)(6-x)}$ to get
$$0 = x\sqrt{(14-x)(6-x)} + (x-10){\sqrt{(x+4)(x-4)}}.$$
Move one of the two terms to the left hand side, and then square both sides to get
$$x^2(14-x)(6-x) = (x-10)^2(x+4)(x-4).$$
Expanding, we get
$$x^4 - 20x^3 + 84x^2 = x^4 - 20x^3 + 84x^2 + 320x - 1600.$$
This boils down to $x = 5$.
The most tedious part is the expansion step, but it is still doable by hand.
| {
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"timestamp": "2023-03-29T00:00:00",
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What is $ \lim_{x \rightarrow 0}\left(\frac{1}{\ln\cos (x)}+\frac{2}{\sin ^{2}(x)}\right) $? The answer of the following limit:
$$
\lim_{x \rightarrow 0}\left(\frac{1}{\ln \cos (x)}+\frac{2}{\sin ^{2}(x)}\right)
$$
is 1 by Wolfram Alpha.
But I tried to find it and I got $2/3$ :
My approach :
$1)$
$
\ln(\cos x)=\ln\left(1-\frac{x^{2}}{2}+o\left(x^{3}\right)\right)=-\frac{x^{2}}{2}+o\left(x^{3}\right)
$
$2)$
$
\sin ^{2}(x)=\left(x-\frac{x^{3}}{3!}+o\left(x^{3}\right)\right)^{2}=x^{2}-\frac{x^{4}}{3}+o\left(x^{4}\right)
$
$3)$
$\begin{aligned} \frac{1}{-\frac{x^{2}}{2}+o\left(x^{3}\right)}+\frac{2}{x^{2}-\frac{x^{4}}{3}+o\left(x^{4}\right)}=\frac{-x^{2}+x^{2}-\frac{x^{4}}{3}+o\left(x^{3}\right)}{-\frac{x^{4}}{2}+o\left(x^{5}\right)}=\frac{-\frac{1}{3}+o\left(x^{3}\right)}{-\frac{1}{2}+o\left(x^{5}\right)} \end{aligned}$
$4)$
$\lim _{x \rightarrow 0} \frac{-\frac{1}{3}+o\left(x^{3}\right)}{-\frac{1}{2}+o\left(x^{5}\right)}=\frac{-\frac{1}{3}}{-\frac{1}{2}}=\frac{2}{3}$
So where is the mistake in my approach?
Note: $o$ denotes the little-o notation
Edit : I've understood where's my mistake is, but another question popped up reading the answers which is : does $o(1/x)$ tends to zero as x tends to zero?
| We need more terms to obtain
$$\ln(\cos x)=\ln\left(1-\frac{x^{2}}{2}+\frac{x^{4}}{24}+o\left(x^{4}\right)\right)=-\frac{x^{2}}{2}-\frac{x^{4}}{12}+o\left(x^{4}\right)$$
and then by binomial expansion
$$\frac{1}{-\frac{x^{2}}{2}-\frac{x^{4}}{12}+o\left(x^{4}\right)}+\frac{2}{x^{2}-\frac{x^{4}}{3}+o\left(x^{4}\right)}=\frac2{x^2}\left(-\frac{1}{1+\frac{x^2}6+o\left(x^{2}\right)}+\frac{1}{1-\frac{x^{2}}{3}+o\left(x^{2}\right)}\right)=\frac2{x^2}\left(-1+\frac{x^2}6+1+\frac{x^{2}}{3}+o\left(x^{2}\right)\right)= \frac2{x^2}\left(\frac{x^2}2+o(x^2)\right)=1+o(1) \to 1$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Are there any two real nonzero numbers $a$ and $b$ such that $a^2+ab+b^2 = 0$ Why or why not? Are there any two real nonzero numbers $a$ and $b$ such that $a^2+ab+b^2 = 0$ Why or why not?
$$a^2+ab+b^2=(a+b)^2-ab=0$$ iff $$(a+b)^2=ab \tag{1}$$
but $(a+b)^2 = a^2+2ab+b^2 $ so equation 1 couldn't possibly be true.
Also, when $a=b\ne 0$, $(a^2+ab+b^2)(a-b) = a^3-b^3 =0$.
| Hint :
$$a^2 + ab + b^2 = \left( a + \frac{b}{2}\right)^2 + \frac{3b^2}{4}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that there is a unit $u \in R$ such that $ub = bu = a$
Let $R$ be a ring with identity containing elements $a$ and $b$ with $ab = b$ and $b^2 = a$. Prove that there is a unit $u \in R$ such that $ub = bu = a$.
Source: Problem $15.3.6$, Algebra in Action: A Course in Groups, Rings, and Fields by Shahriar Shahriari.
My work:
From $ab = b$ and $b^2 = a$, we have $ab^2 = a^2 = b^2$. So, $a^2 = a$, i.e. $a$ is idempotent. $u = b$ satisfies $ub = bu = a$, but we don't know if $b$ is a unit. I believe the main reason I'm stuck is that $a$ and $b$ may not be units, but from the given information, we likely want to find $u$ in the form $u = a^\alpha b^\beta$ for some $\alpha,\beta\in \mathbb N$. How do I proceed?
Thank you!
| We know that $b^3 = b$ and so $b^4 = b^2$. We may then compute that
$$
(b^2+b-1)^2 = b^4 + 2 b^3 - b^2 - 2 b + 1 = b^2 + 2b - b^2 - 2b + 1 = 1.
$$
Hence $b^2+b-1$ is a unit. Furthermore, we have
$$
(b^2+b-1)b = b(b^2+b-1) = b^3 + b^2 - b = b + b^2 - b = b^2,
$$
so it is the unit we are after.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve $(y+1)dx+(x+1)dy=0$ Solve $(y+1)dx+(x+1)dy=0$
$$\frac{dy}{y+1}=-\frac{dx}{x+1}$$
then we get $\ln|y+1|=-\ln|x+1|+c$
$$\ln(|(y+1)(x+1)|)=c$$
$$|(y+1)(x+1)|=e^c=c_1$$
but answer is $y+1=\frac{c}{x+1}$ can you help to find where is my mistake?
| The equation can be properly written as $y(x)+1+(x+1)y'(x)=0,$ which suggests the substitution $z(x)=(x+1)[y(x)+1].$ Hence $z'(x)=y(x)+1+(x+1)y'(x),$ implying $z'(x)=0.$ Since $y$ need not be differentiable at $-1$ to satisfy the equation everywhere else, we account for this, so we have that $(x+1)[y(x)+1]=A$ for every $x\lt-1,$ and $(x+1)[y(x)+1]=B$ for every $x\gt-1.$ $A$ and $B$ need not be equal. As such, $y(x)=\frac{A}{x+1}-1$ for every $x\lt-1$ and $y(x)=\frac{B}{x+1}-1$ for every $x\gt-1.$
In your case, you did this instead by writing the equation as $\frac{y'(x)}{y(x)+1}=-\frac{1}{x+1},$ which ignores the solution $y(x)=-1.$ Once you antidifferentiated, you wrote $\ln|y(x)+1|=-\ln|x+1|+c,$ but a more careful way to antidifferentiate is to use different constants of antidifferentiation in different intervals. In other words: $\ln(-(y(x)+1))=-\ln(-(x+1))+A$ for $y(x)\lt-1,x\lt-1,$ $\ln(-(y(x)+1))=-\ln(x+1)+B$ for $y(x)\lt-1,x\gt-1,$ $\ln(y(x)+1)=-\ln(-(x+1))+C$ for $y(x)\gt-1,x\lt-1,$ $\ln(y(x)+1)=-\ln(x+1)+D$ for $y(x)\gt-1,x\gt-1.$ Respectively, this simplifies to $$y(x)+1=\frac{\exp(A)}{x+1},$$ $$y(x)+1=-\frac{\exp(B)}{x+1},$$ $$y(x)+1=-\frac{\exp(C)}{x+1},$$ $$y(x)+1=\frac{\exp(D)}{x+1}.$$ This can be simplified to the answer I gave, provided that you include $y(x)=-1$ by allowing the constants to be $0.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Ways to find $\frac{1}{2\cdot4}+\frac{1\cdot3}{2\cdot4\cdot6}+\frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8}+\cdots$
$$\frac{1}{2\cdot4}+\frac{1\cdot3}{2\cdot4\cdot6}+\frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8}+\frac{1\cdot3\cdot5\cdot7}{2\cdot4\cdot6\cdot8\cdot10}+\cdots$$
is equal to?
My approach:
We can see that the $n^{th}$ term is \begin{align}a_n&=\frac{1\cdot3\cdot5\cdot\space\dots\space\cdot(2n-3)\cdot(2n-1)}{2\cdot4\cdot6\cdot\space\dots\space\cdot(2n)\cdot(2n+2)}\\&=\frac{1\cdot3\cdot5\cdot\space\dots\space\cdot(2n-3)\cdot(2n-1)}{2\cdot4\cdot6\cdot\space\dots\space\cdot(2n)\cdot(2n+2)}\color{red}{[(2n+2)-(2n+1)}]\\&=\frac{1\cdot3\cdot5\cdot\space\dots\space\cdot(2n-3)\cdot(2n-1)}{2\cdot4\cdot6\cdot\space\dots\space\cdot(2n)}-\frac{1\cdot3\cdot5\cdot\space\dots\space\cdot(2n-3)\cdot(2n-1)\cdot(2n+1)}{2\cdot4\cdot6\cdot\space\dots\space\cdot(2n)\cdot(2n+2)}\\
\end{align}
From here I just have a telescopic series to solve, which gave me $$\sum_{n=1}^{\infty}a_n=0.5$$
Another approach : note : $$\frac{(2n)!}{2^nn!}=(2n-1)!!$$
Which gives $$a_n=\frac{1}{2}\left(\frac{(2n)!\left(\frac{1}{4}\right)^n}{n!(n+1)!}\right)$$
So basically I need to compute
$$\frac{1}{2}\sum_{n=1}^{\infty}\left(\frac{(2n)!\left(\frac{1}{4}\right)^n}{n!(n+1)!}\right) \tag{*}$$
I'm not able to determine the binomial expression of $(*)$ (if it exists) or else you can just provide me the value of the sum
Any hints will be appreciated, and you can provide different approaches to the problem too
| I tried to rewrite the sum:
$$\sum _{n=1}^k \frac{\left(\frac{1}{4}\right)^n (2 n)!}{n! (n+1)!}$$
computationally using the Mathematica Code:
Sum[Factorial[2*n]*(1/4)^n/(Factorial[n]*Factorial[n + 1]), {n, 1, k}]
As a result I got:
$$\frac{2^{-2 k-1} \left(-k (2 (k+1))!-2 (2 (k+1))!+2^{2 k+1} (k+1)! (k+2)!\right)}{(k+1)! (k+2)!}\\=1-\frac{2^{-2 k-1} (k+2) (2 (k+1))!}{(k+1)! (k+2)!}=1-\frac{2 \Gamma \left(k+\frac{3}{2}\right)}{\sqrt{\pi } \Gamma (k+2)}$$
where $\Gamma$ is Euler Gamma Function. By setting $k=\infty$, the complete sum becomes 1.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Find the area of region ABC. For reference: In figure $T$ and $K$ are points of tangency,
$MT = a$ and $KN = b$; calculate area of region $ABC$.
(Answer:$2\sqrt{ab}(\sqrt a+\sqrt b)^2$)
My progress:
$$S_{ABC} = p \cdot r = \frac{r \cdot (AB+BC+AC)}{2}\\
AC +2R = AB+BC\\
S_{ABC} = AG \cdot GC \qquad \text{(property)} \\
S_{ABC} = (AC+R)R \qquad \text{(property)} \\
OTBQ:~\text{square} \implies TK = R\sqrt2 \\
\ldots ?$$
I'm not able to use segments a and b in the resolution
| Here is a geometrical solution without much algebra.
For a right triangle, if you draw a line through the points of tangency of the incircle with the perpendicular sides, it does bisect the arcs of the circumcircle on both sides. In other words, $M$ and $N$ are midpoints of minor arcs $AB$ and $BC$ respectively. At the end of the answer, I have shown a proof.
With that, note that $\triangle BTM \sim \triangle NKB$. That leads to,
$\frac{r}{a} = \frac{b}{r} \implies r = \sqrt{ab}$
As $FM$ is perpendicular bisector of $AB$ and $FN$ is perpendicular bisector of $BC$,
$\frac{AB}{2} = \sqrt{ab} + \frac{a}{\sqrt2}$ and $\frac{BC}{2} = \sqrt{ab} + \frac{b}{\sqrt2}$
As we know $AB$ and $BC$ in terms of $a$ and $b$, we are done, for $S_{\triangle ABC} = \frac 12 \cdot AB \cdot BC$.
Proof of the property that I used in the above answer -
Say $M$ and $N$ are midpoints of the arcs $AB$ and $BC$ and segment $MN$ intersects $AB$ and $BC$ at $T$ and $K$ respectively.
$\angle BIN = 45^\circ + \angle A/2$
So, $\angle KPN = 90^\circ + \angle A/2$
Also, $\angle PNK = \angle C/2$
That leads to $\angle BKM = \angle PKN = 45^\circ$
Also note that $\angle INK = \angle ICK = \angle C / 2$ so $ICNK$ is cyclic and therefore $\angle KIN = \angle KCN = \angle A / 2$
That leads to $\angle IKM = \angle A / 2 + \angle C / 2 = 45^\circ$
$\angle BKI = \angle BKM + \angle IKM = 90^\circ$.
So $K$ must be point of tangency of incircle with side $BC$. Finally since $KI \parallel BT$ and $BT = BK = KI$, $T$ is the point of tangency of incircle with side $AB$.
| {
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Solving $\sqrt3\;\cos(x)= \sin\;(\frac {x}{2})$ I'm currently stuck on this trigonometric equation:
$$\sqrt3 \;\cos(x)= \sin\;(\frac {x}{2})$$
Here's what I've tried so far (2 methods):
Method n.1
$$\sqrt3 \;\cos(x)= \sin\;(\frac {x}{2})$$
$$\sqrt3 \;(1-\sin^2x)= \sin\;(\frac {x}{2})$$
$$\sqrt3-\sqrt3\;\sin^2x= \sin\;(\frac {x}{2})$$
$$2\sqrt3-2\sqrt3\;\sin^2x- \sin\;x=0$$
$$// \; let\; \sin \;(x)=y \;//$$
$$2\sqrt3-2\sqrt3\;y^2- y=0$$
$$y=\frac {1\pm\sqrt{1^2-4(-2\sqrt3)(2\sqrt3)}}{2(-2\sqrt3)}$$
$$y=\frac {1\pm\sqrt{1^2(8\sqrt3)(2\sqrt3)}}{-4\sqrt3}$$
$$y=\frac {1\pm\sqrt{1^2+16\times3}}{-4\sqrt3}$$
$$y=\frac {1\pm\sqrt{49}}{-4\sqrt3}$$
$$y=\frac {1\pm7}{-4\sqrt3}$$
$$y_1=\frac {1+7}{-4\sqrt3}=\frac{-2}{\sqrt3}=\frac{-2\sqrt3}{3}$$
$$y_2=\frac {1-7}{-4\sqrt3}=\frac{-3}{-2\sqrt3}=\frac{\sqrt3}{2}$$
$$\arcsin(\frac{-2\sqrt3}{3})=\nexists$$
$$\arcsin(\frac{\sqrt3}{2})=60°$$
$$x_1= 60°+360°k$$
$$x_1= 120°+360°k$$
Method n.2
$$\sqrt3 \;\cos(x)= \sin\;(\frac {x}{2})$$
$$\sqrt3 \; \cos(x)=\pm \frac{\sqrt{ 1-cos (x)}}{2}$$
Any help is appreciated, thank you in advance! :D
| We begin with $\sqrt{3}\cos(x)=\sin(\frac{x}{2})$. We can start by using our half-angle formula to replace $\sin(\frac{x}{2})$ with $\sqrt{\frac{1-\cos(x)}{2}}$. Making the substitution and squaring both sides, we now see that $3\cos^2 (x)=\frac{1-\cos(x)}{2}$. Multiplying both sides by $2$, we get $6\cos^2(x)=1-\cos(x)$, and rearranging the equation, we get $6\cos^2(x)+\cos(x)-1=0$.
Now we can say that $y=\cos(x)$. This turns the equation into $6y^2+y-1=0$. This is an elementary quadratic equation, the solution of which is $y=\frac{-1\pm\sqrt{1^2-4(6)(-1)}}{2(6)}=\frac{-1\pm5}{12}$, or $y=\frac{1}{3}$ and $y=\frac{-1}{2}$.
If $y=\cos(x)$, then of course, $x=\arccos(y)$. And so, $x=\arccos(\frac{1}{3})\approx1.231$ and $x=\arccos(\frac{-1}{2})=\frac{-2\pi}{3}$.
One thing briefly to note is that there's a small amount of strangeness here due to the fact that $\cos(x)=\cos(-x)$. Specifically, a calculator will tell you that $\arccos(\frac{-1}{2})=\frac{2\pi}{3}$, which is true, but does not satisfy our given equation: $\sqrt{3}\cos(\frac{2\pi}{3})=\frac{-\sqrt{3}}{2}$ and $\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}$. In order to remedy this, we switch the sign from positive to negative, keeping the value of the cosine expression negative while switching the sine expression's value to also be negative.
| {
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"answer_id": 0
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$(x^2+2mx+7m-12)(4x^2-4mx+5m-6)=0$ have two distinct real roots Find the range of $m$ so that the equation $(x^2+2mx+7m-12)(4x^2-4mx+5m-6)=0$ have two distinct real roots.
Either $(x^2+2mx+7m-12)=0$ or $(4x^2-4mx+5m-6)=0$
The discriminant of $(x^2+2mx+7m-12)=0$ is $4(m-3)(m-4)$
and the discriminant of $(4x^2-4mx+5m-6)=0$ is $16(m-2)(m-3)$
I don't know how to do it further.
| We can also look at the "vertex form" of each quadratic factor of $ \ (x^2+2mx+7m-12)·(4x^2-4mx+5m-6) \ $ to see what they contribute to their product function.
The first factor can be written as
$$ f(x) \ \ = \ \ x^2 \ + \ 2mx \ + \ (7m - 12) \ \ = \ \ (x \ + \ m)^2 \ + \ (7m - 12) \ - \ m^2 $$ $$ = \ \ (x \ + \ m)^2 \ - \ (m^2 \ - \ 7m \ + \ 12) \ \ = \ \ (x \ + \ m)^2 \ - \ [ \ (m \ - \ 3)·(m \ - \ 4) \ ] \ \ $$
[as expected, the factors of the discriminant you found appear in the bracketed term of the "completed square"]. The vertex of the "upward-opening" parabola corresponding to this factor is then "above" the $ \ x-$axis when the term in brackets is negative, which occurs for $ \ 3 < m < 4 \ \ . $ We can then say that this factor has
• no real zeroes for $ \ 3 \ < \ m \ < \ 4 \ \ , $
• one real zero at $ \ x \ = \ -m \ $ when $ \ m \ = \ 3 \ $ or $ \ m \ = \ 4 \ \ , \ \ $ and
• two real zeroes for $ \ m \ < \ 3 \ $ or $ \ m \ > \ 4 \ \ . $
Considering the second factor in this way, we find
$$ g(x) \ \ = \ \ 4x^2 \ - \ 4mx \ + \ (5m - 6) \ \ = \ \ 4·\left(x \ - \ \frac{m}{2} \right)^2 \ + \ (5m - 6) \ - \ 4·\left(\frac{m}{2} \right)^2 $$ $$ = \ \ 4·\left(x \ - \ \frac{m}{2} \right)^2 \ - \ (m^2 \ - \ 5m \ + \ 6) \ \ = \ \ 4·\left(x \ - \ \frac{m}{2} \right)^2 \ - \ [ \ (m \ - \ 2)·(m \ - \ 3) \ ] \ \ . $$
The vertex for the "upward-opening" parabola here is again "above" the $ \ x-$axis when the term in brackets is negative for $ \ 2 < m < 3 \ \ . $ So we conclude that this factor has
• no real zeroes for $ \ 2 \ < \ m \ < \ 3 \ \ , $
• one real zero at $ \ x \ = \ \frac{m}{2} \ $ when $ \ m \ = \ 2 \ $ or $ \ m \ = \ 3 \ \ , \ \ $ and
• two real zeroes for $ \ m \ < \ 2 \ $ or $ \ m \ > \ 3 \ \ . $
The quartic polynomial that is the product of these two factors has the number of real zeroes that is the sum of the number of real zeroes in the two factors. Therefore, $ \ f(x) \ · \ g(x) \ $ has, for varying $ \ m \ \ : $
$ \mathbf{m \ < \ 2 \ \ : } \ \ \ 2 + 2 \ = \ 4 \ $ real zeroes;
$ \ \ \mathbf{m \ = \ 2 \ \ : } \ \ \ 2 + 1 \ = \ 3 \ $ real zeroes (one of them at $ \ x \ = \ \frac{m}{2} \ = \ \frac{2}{2} \ = \ +1 \ \ ) \ ; $
$ \mathbf{2 \ < \ m \ < \ 3 \ \ : } \ \ \ 2 + 0 \ = \ 2 \ $ real zeroes;
$ \ \ \mathbf{m \ = \ 3 \ \ : } \ \ \ 1 + 1 \ = \ 2 \ $ real zeroes (one of them at $ \ x \ = \ -m \ = \ -3 \ \ , $ the other at $ \ x \ = \ \frac{m}{2} \ = \ +\frac{3}{2} \ \ ) \ ; $
$ \mathbf{3 \ < \ m \ < \ 4 \ \ : } \ \ \ 0 + 2 \ = \ 2 \ $ real zeroes;
$ \ \ \mathbf{m \ = \ 4 \ \ : } \ \ \ 1 + 2 \ = \ 3 \ $ real zeroes (one of them at $ \ x \ = \ -m \ = \ = \ -4 \ \ ) \ ; $
and $ \ \ \mathbf{ m \ > \ 4 \ \ : } \ \ \ 2 + 2 \ = \ 4 \ $ real zeroes.
Hence, the quartic polynomial has two distinct real zeroes for $ \ \mathbf{2 \ < \ m \ < \ 4} \ \ ; \ \ $ these are the zeroes of either factor alone or from both vertices at once $ \ ( \ m \ = \ 3 \ ) \ \ . $
The polynomial never has less than two real zeroes for any value of $ \ m \ \ . \ $ The third zero for $ \ m \ = \ 2 \ $ and $ \ m \ = \ 4 \ $ must be a "double zero"; the two zeros for $ \ m \ = \ 3 \ $ are both "double zeroes". In fact, this polynomial never has complex zeroes.
[There are places where one could be deceived by using a low-resolution graph or not examining a graph carefully enough. The function $ \ f(x) \ · \ g(x) \ $ has a turning point extremely close to the $ \ x-$axis for
$ m \ \approx \ -2.25 \ $ near $ \ x \ \approx \ -3.482 \ \ $ (slightly below); $ \ m \ \approx \ +1.12 \ $ near $ \ x \ \approx \ +1.205 \ \ $ (slightly below); and $ \ m \ \approx \ +1.53 \ $ near $ \ x \ \approx \ +0.362 \ \ $ (slightly above). ]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4340466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does $\ \sum_{n=1}^{\infty} \frac{\sin(2^n x)}{n}\ $ converge for all $x\ ?$ Does $\ \displaystyle\sum_{n=1}^{\infty} \frac{\sin(2^n x)}{n}\ $ converge for all $x\ ?$
It obviously converges for any $x\ $ of the form $\ 2^mk \pi\ $ where $\ m,k\in\mathbb{Z},\ $ but for any other values of $\ x\ $ the question is interesting because I don't see how to answer it.
I tried using Cauchy's condensation test with $\ 2^nf(2^n) = \frac{\sin(2^n x)}{n},\ $which implies that $\ f(n) = \frac{\sin(nx)}{n\log_2(n)},\ $ and so $\ \displaystyle\sum_{n=1}^{\infty} \frac{\sin(2^n x)}{n}\ $ converges if and only if $\ \displaystyle\sum_{n=1}^{\infty} \frac{\sin(nx)}{n\log_2(n)}\ $ converges.
I then found out from this video that we can prove using some basic complex analysis that $\ \displaystyle\sum_{k=1}^{\infty} \frac{\sin(n x)}{n}\ $ converges for all $\ x,\ $ and then I thought we were done by the limit comparison test. However, this attempt is wrong because $$\ \frac{\frac{\sin(nx)}{n}}{\frac{\sin(nx)}{n\log_2(n)}}\ \not\to c>0. $$
We also cannot use the integral test for convergence because $\ \displaystyle\sum_{n=1}^{\infty} \frac{\sin(2^n x)}{n}\ $ is not monotone.
Finally, I do not think we can use Abel's test or Dirichlet's test because $\ \displaystyle\sum_{n=1}^{\infty} \sin(2^n x)\ $ probably diverges. Edit: Maybe we can use Dirichlet's test here. I just realised we only need to show the series $\ \displaystyle\sum_{n=1}^{\infty} \sin(2^n x)\ $ is bounded, not that it converges! But I don't know how to do this... Apparently $\ \displaystyle\sum_{n=1}^{\infty} \sin(kx)\ $ is bounded but $\ \displaystyle\sum_{n=1}^{\infty} \sin(k^2)\ $ isn't, although proving this is difficult! So whether or not $\ \displaystyle\sum_{n=1}^{\infty} \sin(2^n x)\ $ is bounded is probably difficult too. So we should look to use a different test...
Maybe some version of the Alternating Series test or Absolute convergence test ? Although I think that determining convergence of $\ \displaystyle\sum_{n=1}^{\infty} \frac{\lvert \sin(2^n x) \rvert }{n}\ $ is more difficult than the original question here. But Alternating series test might be promising, not sure...
Or maybe there is some other test from complex analysis that is applicable here?
| As it has been over a year and no answer/edit has been written to encompass the answer in the comments by @achille_hui, I'll present their answer here for posterity:
At $x=\frac{\pi}{7}$ and $n\geq 1$ we have
$$\sin\left(2^n\frac{\pi}{7}\right)=\sin\left(2\pi\frac{2^{n-1}}{7}\right)=\begin{cases}
\sin\left(2\pi\frac{1}{7}\right) & n\equiv 1\ (\text{mod }3) \\
\sin\left(2\pi\frac{2}{7}\right) & n\equiv 2\ (\text{mod }3) \\
\sin\left(2\pi\frac{4}{7}\right) & n\equiv 3\ (\text{mod }3)
\end{cases}$$
These are bounded by
$$=\begin{cases}
\sin\left(2\pi\frac{1}{7}\right)>\frac{3}{4} & n\equiv 1\ (\text{mod }3) \\
\sin\left(2\pi\frac{2}{7}\right)>\frac{3}{4} & n\equiv 2\ (\text{mod }3) \\
\sin\left(2\pi\frac{4}{7}\right)>-1 & n\equiv 3\ (\text{mod }3)
\end{cases}$$
Taken together, the sum is bounded from below:
$$\sum_{n=1}^\infty \frac{\sin(2^n \pi/7)}{n}=\sum_{k=0}^\infty \left[\sin\left(2\pi \frac{1}{7}\right)\frac{1}{3k+1}+\sin\left(2\pi \frac{2}{7}\right)\frac{1}{3k+2}+\sin\left(2\pi \frac{4}{7}\right)\frac{1}{3k+3}\right]$$
$$>\sum_{k=0}^\infty \left[\frac{3}{4}\cdot\frac{1}{3k+1}+\frac{3}{4}\cdot \frac{1}{3k+2}-\frac{1}{3k+3}\right]=\sum_{k=1}^\infty \frac{18k^2+45k+19}{108 k^3+216k^2+132k+24}$$
$$>\sum_{k=1}^\infty \frac{18k^2+36k+18}{108 k^3+324k^2+324k+108}=\frac{18}{108}\sum_{k=1}^\infty \frac{(k+1)^2}{(k+1)^3}=\frac{18}{108}\sum_{k=1}^\infty \frac{1}{k+1}=\infty$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4340640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
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Power series solution of $y'-y=x^2$ $y'-y=x^2$
I try to get a power series solution.
$y(x)=\sum_{n=0}^{\infty}a_{n}X^{n},y'(x)=\sum_{n=1}^{\infty}na_{n}X^{n-1}=\sum_{n=0}^{\infty}(n+1)a_{n+1}X^{n}$.
$\sum_{n=0}^{\infty}(n+1)a_{n+1}X^{n}-\sum_{n=0}^{\infty}a_{n}X^{n}=X^{2}$
Then $ \space a_{n+1}=\frac{a_n}{n+1} \space \forall n\neq 2$
$n=2 \implies 3a_3-a_2=1$
Then $a_1=a_0, a_2=\frac{a_0}{2}, a_3=\frac{2+a_0}{6}, a_4=\frac{2+a_0}{4!}, a_5=\frac{2+a_0}{24\cdot 4}$
Then the solution is $y=a_0+a_0x^1+\frac{a_0}{2}x^2+\frac{2+a_0}{6}x^3+\frac{2+a_0}{4!}x^4+\frac{2+a_0}{24\cdot 4}x^5 \cdots$
Is it correct? Am I supposed to find $a_0$ ?
Thanks!
| Your solution amounts to
$$
y - a_0 \left( 1 + x + \dfrac{x^2}{2} \right) = (2 + a_0) \left( \mathrm{e}^x - 1 - x - \dfrac{x^2}{2} \right) .
$$
Notice you've indeed made a mistake. You have $a_{n + 1} = \dfrac{a_n}{n + 1}$, so $a_5 = \dfrac{a_4}{5} = \dfrac{2 + a_0}{5!}$, $a_6 = \dfrac{a_5}{6} = \dfrac{2 + a_0}{6!}$, and so forth.
Back to the topic,
$$
y = \left( 2 + a_0 \right) \mathrm{e}^x - 2 \left(1 + x + \dfrac{x^2}{2} \right) .
$$
It seems that any $a_0$ will fit $y^\prime - y = x ^ 2$. So just discard the foremost $2$ and the solution is $y = a_0 \mathrm{e}^x - 2-2 x - x^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4354247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How does the theorem 2 of algebra of limits contradict in this Q? Q: $\lim _{x \rightarrow 1} \frac{x^{15}-1}{x^{10}-1}$
Solution in my textbook using theorem 2 of algebra of limits I.e $\lim _{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}=n a^{n-1}$ .
$\begin{aligned} \lim _{x \rightarrow 1} \frac{x^{15}-1}{x^{10}-1} &=\lim _{x \rightarrow 1}\left[\frac{x^{15}-1}{x-1} \div \frac{x^{10}-1}{x-1}\right] \\ &=\lim _{x \rightarrow 1}\left[\frac{x^{15}-1}{x-1}\right] \div \lim _{x \rightarrow 1}\left[\frac{x^{10}-1}{x-1}\right] \\ &=15(1)^{14} \div 10(1)^{9} \text {} \\ &=15 \div 10=\frac{3}{2} \end{aligned}$
I agree with it using the formula but I also thought of thinking how would it be solved from the long way [Inserting x = 1]
We get :
$\frac{(1)^{15}-1}{1-1} \div\left[\frac{(1)^{10}-1}{1-1}\right]$
$\rightarrow \frac{0}{0} \div \frac{0}{0}=0$
My answer is completely different from the solution of my textbook. How is that possible ?
If anyone wants to argue that we cannot put value of x = 1 just like that. There is another example in my textbook according to which , we can.
In all of them , the Q simply says: Find the limit.
1)$\begin{aligned} \lim _{x \rightarrow 2} \frac{x^{2}-4}{x^{3}-4 x^{2}+4 x} &=\lim _{x \rightarrow 2} \frac{(x+2)(x-2)}{x(x-2)^{2}} \\ &=\lim _{x \rightarrow 2} \frac{(x+2)}{x(x-2)}=\frac{2+2}{2(2-2)}=\frac{4}{0} \end{aligned}$
Here , x is tending to 2. This is a rational function.
EDIT: I will share a few more examples.
*We have $\lim _{x \rightarrow 1} \frac{x^{2}+1}{x+100}=\frac{1^{2}+1}{1+100}=\frac{2}{101}$
*(iii) $\lim _{x \rightarrow-1}\left[1+x+x^{2}+\ldots+x^{10}\right]=1+(-1)+(-1)^{2}+\ldots+(-1)^{10}$
| $x^{15}-1=(x-1)(1+x+x^2+...+x^{13}+x^{14}).$
$x^{10}-1=(x-1)(1+x+x^2+...+x^{8}+x^{9}).$
So if $x\ne 1$ then $$\frac {x^{15}-1}{x^{10}-1}=\frac {1+x+x^2+...+x^{13}+x^{14}}{1+x+x^2+...+x^{8}+x^{9}}$$ which obviously converges to $\frac {15}{10}$ as $x\to 1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4358139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Laurent series of $f(z) = \frac{z - 4}{(z+1)^2(z - 2)}$ I want to expand function $$f(z) = \frac{z - 4}{(z+1)^2(z - 2)}$$ for $1 < |z| < 2$.
My work so far
By partial fraction decomposition we can get that:
$$\frac{z - 4}{(z + 1)^2(z - 2)} = \frac{-3(z - 4)}{(z + 1)^2} + \frac{\frac 1 9 (z - 4)}{z - 2}$$
Now deriving Laurent expansion for second term is not so hard:
$$\frac{1}{z - 2} = -\frac{1}{2} \frac{1}{1 - \frac z 2} = - \frac 1 2 \sum_{n = 0}^\infty (\frac z 2)^n = - \sum_{n = 0}^\infty \frac{z^2}{2^{n + 1}}$$
So we have that:
$$ \frac{\frac 1 9 (z - 4)}{z - 2} = \sum_{n = 0}^\infty - \frac 1 9 \cdot \frac{z^n}{2^{n + 1}}(z - 4)$$
But I'm quite struggling with expanding $\frac{1}{(z + 1)^2}$. I tried several possibilities but those all led me nowhere. Can I ask you for a hand in expanding form $\frac{1}{(z + 1)^2}$?
EDIT
We know that $\frac{1}{1 + z} = \sum_{n = 0}^\infty \frac{(-1)^n}{z^{n + 1}}$
As I understood the idea with derivative is to:
$$\frac{d(\frac{1}{1 + z})}{dz} = \sum_{n = 0}^\infty \frac{d(\frac{(-1)^n}{z^{n + 1}})}{dz}$$
$$- \frac{1}{(1 + z)^2} = \sum_{n = 0}^\infty - (n + 1)z^{-n-2}(-1)^n$$
$$\frac{1}{(1 + z)^2} = \sum_{n = 0}^\infty (n + 1)z^{-(n + 2)}(-1)^n$$
Does it make any sense to you?
| You want a series for $\frac1{(1+z)^2}$ that’s good outside the unit circle, don’t you? Thus a series in $1/z$. So write
$$
\frac1{(1+z)^2}=\frac1{z^2}\frac1{(1+1/z)^2}\,.
$$
If you find that confusing, expand $t^2\frac1{(1+t)^2}$ in powers of $t$, and substitute $1/z$ for $t$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4358469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A curious family of Chebyshev-like polynomials Consider the family $f_n(x)$ of functions of $x$ for $0\leq x\leq1$, each indexed by a variable $n \in \mathbb{N}$, described by the following equation:
$$f_n(x) = \sin^2\left(n \arcsin\left(\sqrt{x}\right)\right)$$
Evaluation in numerical algebra programs yields the following polynomial forms of each function $f_n(x)$ for the first few $n$:
*
*$f_1(x) = x$
*$f_2(x) = 4x - 4x^2$
*$f_3(x) = 16 x^3 - 24 x^2 + 9 x$
*$f_4(x) = -64 x^4 + 128 x^3 - 80 x^2 + 16 x$
and so on.
Is there any way to analytically derive the polynomial form of this family of functions in terms of $n$ and $x$ (i.e., $f_n(x) = g(n,x)$ where $g$ is a polynomial in $x$)? Or perhaps this family of polynomials is related to another "named" family of polynomials under an appropriate transformation?
COMMENT: The "Chebyshev-like" qualifier in the title comes from the fact that the Chebyshev polynomials $T_n(x)$ can be defined by a similar trigonometric identity: $T_n(x) = \cos(n\arccos(x))$.
| Chebyshev polynomials of the second kind satisfy $\,U_{n-1}(\cos \theta)\,\sin \theta =\sin(n\theta)\,$.
With $\,\theta=\arcsin\left(\sqrt{x}\right)\,$ it follows that:
$$
\sqrt{x} \; U_{n-1}\left(\sqrt{1-x}\right) = \sin\left(n \arcsin(\sqrt{x})\right) \;\;\iff\;\; f_n(x) = x \cdot U_{n-1}^2\left(\sqrt{1-x}\right)
$$
Since $\,U_n(x)\,$ contains only powers of the same parity, the square root in $\,\sqrt{1-x}\,$ will either vanish or factor out then get squared again, so the end result is in fact a polynomial.
For example:
*
*$n=3:$
$$
\begin{align}
U_2(x) &= 4x^2-1
\\ f_3(x) &= x \cdot U_2^2\left(\sqrt{1-x}\right)
\\ &= x \cdot \left(4\,(1-x)-1\right)^2
\\ &= x\,\left(-4x+3\right)^2
\\ &= 16 x^3 - 24 x^2 + 9 x
\end{align}
$$
*
*$n=4:$
$$
\begin{align}
U_3(x) &= 4x\,(2x^2-1)
\\ f_4(x) &= x \cdot U_3^2\left(\sqrt{1-x}\right)
\\ &= x \cdot 16(1-x) \, \left(2\,(1-x)-1\right)^2
\\ &= 16x(1-x)\,\left(-2x+1\right)^2
\\ &= -64 x^4 + 128 x^3 - 80 x^2 + 16 x
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4362134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Doubt regarding a differential equation. There's a problem in a book regarding exact differential equation :
Solve :
$$x\,dx+y\,dy = \frac{a^2(x\,dy-y\,dx)}{x^2+y^2}$$
The author further proceeds to rearrange above in the form $M\,dx+N\,dy=0$
where
$$M=x+\frac{a^2y}{x^2+y^2} ; N= y-\frac{a^2x}{x^2+y^2}$$
Which further implies
$$M_y=N_x=\frac{a^2(x^2-y^2)}{(x^2+y^2)^2}$$
This is the required condition for the given equation to be exact and the solution is obtained using standard formula.
But, what I did is as follows :
$$x\,dx+y\,dy = \frac{a^2(x\,dy-y\,dx)}{x^2+y^2}$$
$$\to (x^2+y^2)x\,dx + (x^2+y^2)y\,dy=a^2(x\,dy -y\,dx)$$
Therefore
$$(x^3+xy^2+a^2y)\,dx + (y^3+yx^2-a^2x)\,dy =0$$
Comparing it with the equation $M\,dx+N\,dy=0$ we get,
$$M=x^3+xy^2+a^2y; N=y^3+yx^2-a^2x$$
But
$$M_y=2xy+a^2; N_x=2xy-a^2$$
Evidently, $M_y\neq N_x$.
So what am I doing wrong here?
At first I thought I shouldn't/couldn't just simply multiply both sides of the given equation with the denominator ($x^2+y^2$) of right hand side if it's zero, but since as it's already in denominator isn't it understood that it isn't/can't be zero ??
Any help is really appreciated.
| $$x\,dx+y\,dy = \frac{a^2(x\,dy-y\,dx)}{x^2+y^2}\tag 1$$
As you correctly found : Equation $(1)$ is an exact ODE.
You transformed Eq.$(1)$ into Eq.$(2)$:
$$(x^3+xy^2+a^2y)\,dx + (y^3+yx^2-a^2x)\,dy =0\tag 2$$
Again you correctly found : Equation $(2)$ is not an exact ODE.
This is not contradictory because Eqs.$(1)$ and $(2)$ are not the same ODE.
In order to transform Eq.$(2)$ into an exact ODE one have to multiply it with an "integration factor", say $\mu(x,y)$.
One find that $\mu(x,y)=\frac{1}{x^2+y^2}$.
$$\mu(x,y)\Big( (x^3+xy^2+a^2y)\,dx + (y^3+yx^2-a^2x)\,dy\Big) =0\tag 3$$
$$ \frac{x^3+xy^2+a^2y}{x^2+y^2}\,dx + \frac{y^3+yx^2-a^2x}{x^2+y^2}\,dy =0\quad\text{is exact.}$$
In fact Eq.$(3)$ is equivalent to Eq.$(1)$ thanks to the correct $\mu(x,y)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Support for marginals of joint PDF I have two joint PDF-s and three questions below.
I have a random vector' $(U,V)'$ and it's PDF. I need to find the marginal PDF-s for both $U$ and $V$. I tried deploying the idea from this anwer, but I still have some questions.
This is the joint PDF:
$$
f_{U,V}(u,v)=\begin{cases}
\frac{12}{4\pi-3\sqrt{3}}, \hspace{2mm} |2u|\leq \sqrt{3}, \hspace{2mm} 1\leq 2v\leq 2\sqrt{1-u^2}\\
0 \hspace{12mm} \text{elsewhere}
\end{cases}
$$
The biggest problem for me is finding the support i.e the bounds for integration.
I think I have correctly found the marginal $f_U(u):$
$$f_{U}(u)=
\int_{v=\frac{1}{2}}^{v=\sqrt{1-u^2}}f(u,v)dv=
\frac{12}{4\pi-3\sqrt{3}} \int_{v=\frac{1}{2}}^{v=\sqrt{1-u^2}} 1dv=
\frac{12\sqrt{1-u^2}-6}{4\pi-3\sqrt{3}}
$$
The problem appears with $f_V(v)$. From the PDF plot below we can see that the area is symmetric around $0$. Therefore we can find integral with respect to $u$ from $0$ until the curved line above it, and the whole area would be twice the value?
*
*Is it correct that I can only look at the upper bound: $2v\leq 2\sqrt{1-u^2} \implies v^2\leq 1-u^2 \implies u^2 \leq 1-v^2 \implies u\leq \sqrt{1-v^2} $
and now
$$
f_V(v) = 2\int_{u=0}^{u=\sqrt{1-v^2}}f(u,v)du=
2\frac{12}{4\pi-3\sqrt{3}} \int_{u=0}^{u=\sqrt{1-v^2}} 1du=
\frac{24\sqrt{1-u^2}}{4\pi-3\sqrt{3}}
$$
*Is it correct that for the expected values the integration areas are now $EX=\int_{-\frac{\sqrt3}{2}}^{\frac{\sqrt3}{2}} xf_X(x)$ as well as $EY=\int_{\frac{1}{2}}^1 yf_Y(Y)$?
*I have another example where the PDF is as $f_{X,Y}(x,y)=\begin{cases}
\frac{3}{10}xy, \hspace{2mm} 0<x<2, \hspace{2mm} 0<y <\sqrt{x(4-x)}\\
0 \hspace{12mm} \text{elsewhere}
\end{cases}$
and for the $f_Y(y)$ I can find integration area as looking at the upper bound again, because lower is $0$. So $y = \sqrt{x(4-x)} \implies y^2 = 4x-x^2 \implies y^2 + 4 = 4x-x^2 +4 \implies (x^2-4x+4) + y^2 = 4 \implies (x-2)^2=4-y^2 \\
\implies x-2 =\sqrt{4-y^2} \implies x = \sqrt{4-y^2}+2$.
Now, I would integrate $f_Y(y)=\int_0^{\sqrt{4-y^2}+2}f_{X,Y}(x,y)dx$, but I already can see that $x$ should be maximum $2$ so the answer is incorrect. My textbook says is should be integrated as $f_Y(y)=\int_{2-\sqrt{4-y^2}}^2f_{X,Y}(x,y)dx$.
Where am I going wrong with this one?
Thanks a bunch!
| Your work for $1$ and $2$ is mostly correct except some typos.
$ \displaystyle f_U(u) = \frac{12\sqrt{1-u^2}-6}{4\pi-3\sqrt{3}}, ~2 |u| \leq \sqrt3$
Also, $ \displaystyle f_V(v) = \frac{24\sqrt{1-v^2}}{4\pi-3\sqrt{3}}, 1/2 \leq v \leq 1$
So, $ \displaystyle E[U] = \int_{-\sqrt3/2}^{\sqrt3/2} u \cdot f_U(u) ~ du~,$ and similarly for $E[V]$
For the third question, the below diagram may help understand the limits of integration to obtain marginal of $Y$.
Note that $~(x-2)^2 \leq 4-y^2$
$$\implies 2 - \sqrt{4-y^2} \leq x \leq 2 + \sqrt{4-y^2}$$
But the support of the given pdf has $x \leq 2$. So for a given $y$,
$2 - \sqrt{4-y^2} \leq x \leq 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4365768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Are there any formula for the integral $\int_{0}^{\infty} \frac{d x}{\left(x^{n}+a\right)^{m}}$, where $m$ and $n$ are natural numbers and $a>0$? As mentioned in my post, I started to investigate the integral $$
I(m,n,a)=\int_{0}^{\infty} \frac{d x}{\left(x^{n}+a\right)^{m}},
$$
where $m$ and $n$ are natural number and $a$ is positive.
First of all, let’s start with the ‘simple’ case, $$
I(1, n, 1)=\int_{0}^{\infty} \frac{d x}{x^{n}+1}.
$$
However, $\displaystyle \int_{0}^{\infty} \frac{d x}{x^{n}+1}$ is itself not simple. I was forced to use a ready made formula
$$
\int_{0}^{\infty} \frac{d x}{x^{n}+1}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right).
$$
which I can’t prove by elementary method yet. Please tell me if you have any.
Then $$I(m,n,a) =\int_{0}^{\infty} \frac{\sqrt[n]{a} d\left(\frac{x}{\sqrt[n]{a}}\right)}{a\left[\left(\frac{x}{\sqrt[n]{a}}\right)^{n}+1\right]}
=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right) a^{-\frac{n-1}{n}}$$
Now differentiating $I(1, n, a)$ w.r.t. $a$ by $(n-1)$ times yields
$$
\begin{aligned}
\int_{0}^{\infty} \frac{(-1)^{m-1}(m-1) ! d x}{\left(x^{n}+a\right)^{m}}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right)\left(-\frac{n-1}{n}\right)\left(-\frac{2 n-1}{n}\right)\left(-\frac{3 n-1}{n}\right)
\cdots\left(-\frac{m n-n-1}{n}\right) a^{-\frac{m n-1}{n}}
\end{aligned}
$$
Rearranging and simplifying yields
$$
\boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{n}+a\right)^{m}}=\frac{\pi \csc \left(\frac{\pi}{n}\right)}{(m-1) ! n^{m} a^{\frac{m n-1}{n}} }\prod_{k=1}^{m-1}(k n-1)}
$$
Putting $a=1$ gives our formula
$$
\boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{n}+1\right)^{m}}=\frac{\pi \csc \left(\frac{\pi}{n}\right)}{(m-1) ! n^{m} }\prod_{k=1}^{m-1}(k n-1)}
$$
For verification, $$
\begin{aligned}
\int_{0}^{\infty} \frac{d x}{\left(x^{3}+1\right)^{10}} =\frac{\pi \csc \left(\frac{\pi}{3}\right)}{9 ! 3^{10}} \cdot 2 \cdot 5 \cdot 8 \cdot 11 \cdot 14 \cdot 17 \cdot 20 \cdot 23 \cdot 26
=\frac{1118260 \pi}{4782969 \sqrt{3}},
\end{aligned}
$$
which is checked by Wolframalpha
$$\begin{aligned}
\int_{0}^{\infty} \frac{d x}{\left(x^{12}+1\right)^{5}}&=\frac{\pi \csc \left(\frac{\pi}{12}\right)}{4 ! 12^{5}} \cdot 11 \cdot 23 \cdot 35 \cdot 47\\
&=\frac{416185 \pi(\sqrt{6}+\sqrt{2})}{5971968} \\
&\doteq 0.845906950943631,
\end{aligned}
$$
which is checked by Wolframalpha
My question is whether we can prove the formula without using $\int_{0}^{\infty} \frac{d x}{x^{n}+1}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right).$
| Proving $\displaystyle\int_{0}^{\infty} \frac{d x}{x^{n}+1}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right)$:
Set $a=z$ and $b=1-z$ in the Beta function:
$$\operatorname{B}(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}=\int_0^\infty \frac{x^{a-1}}{(1+x)^{a+b}}\mathrm{d}x,$$
we obtain
\begin{gather}
\Gamma(z)\Gamma(1-z)=\int_0^\infty\frac{x^{z-1}}{1+x}\mathrm{d}x\\
=\left(\int_0^1+\int_1^\infty\right) \frac{x^{z-1}}{1+x}\mathrm{d}x=\int_0^1 \frac{x^{z-1}}{1+x}\mathrm{d}x+\underbrace{\int_1^\infty \frac{x^{z-1}}{1+x}\mathrm{d}x}_{x\to 1/x}\\
=\int_0^1 \frac{x^{z-1}}{1+x}\mathrm{d}x+\int_0^1 \frac{x^{-z}}{1+x}\mathrm{d}x\\
\left\{\text{expand $\frac1{1+x}$ in series in both integrals}\right\}\\
=\int_0^1 x^{z-1}\left(\sum_{n=0}^\infty(-1)^nx^n\right)\mathrm{d}x+\int_0^1 x^{-z}\left(\sum_{n=0}^\infty(-1)^nx^n\right)\mathrm{d}x\\
\{\text{interchange integration and summation}\}\\
=\sum_{n=0}^\infty(-1)^n\int_0^1 x^{n+z-1}\mathrm{d}x+\sum_{n=0}^\infty(-1)^n\int_0^1 x^{n-z}\mathrm{d}x\\
=\sum_{n=0}^\infty\frac{(-1)^n}{n+z}+\sum_{n=0}^\infty\frac{(-1)^n}{n-z+1}\\
\{\text{seperate the first term of the first sum and shift the index of the second}\}\\
=\frac1z+\sum_{n=1}^\infty\frac{(-1)^n}{n+z}-\sum_{n=1}^\infty\frac{(-1)^n}{n-z}\\
=\frac1z-\sum_{n=1}^\infty\frac{2z(-1)^n}{n^2-z^2}\\
=\frac{\pi}{\sin(\pi z)}.
\end{gather}
To prove the last equality, set $x=0$ in the Fourier series of $\cos(zx)$:
\begin{equation}
\cos(zx)=\frac{2z\sin(\pi z)}{\pi}\left[\frac1{2z^2}-\sum_{n=1}^\infty\frac{(-1)^n\cos(nx)}{n^2-z^2}\right],\quad z\notin\mathbb{Z}.\label{cos(zx)}
\end{equation}
Thus,
$$\Gamma(z)\Gamma(1-z)=\int_0^\infty\frac{x^{z-1}}{1+x}\mathrm{d}x=\frac{\pi}{\sin(\pi z)}.$$
To complete the proof, replace $z$ by $1/n$ then let $x=y^n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4366270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Find all positive integers s.t. $\frac{1}{a^2} - \frac{1}{b^2} = \frac{1}{c^2} - \frac{1}{d^2}$ Find all positive integers $a, b, c, d$ such that :
$$\frac{1}{a^2} - \frac{1}{b^2} = \frac{1}{c^2} - \frac{1}{d^2}$$
The original problem came from atomic electron transitions :
I would like to find out non-trivial positive integer solutions; since it is trivial if $a = b$ and $c = d$, or $a = c$ and $b = d$.
I found some of the solutions, and they look like :
There is some pattern in these integers, but it seems difficult to obtain a gerneralized form of the solution.
| we take, $(a,b,c,d)=(pq,rq,st,wt)$
where, $r^2=p^2+q^2$ & $w^2=s^2+t^2$
And, $(r,p,q)=((u^2+v^2),(u^2-v^2),(2uv))$
$(w,s,t)=((m^2+n^2),(m^2-n^2),(2mn))$
We have Identity,
$(1/rp)^2=(1/pq)^2-(1/rq)^2$ ---(1)
$(1/ws)^2=(1/st)^2-(1/wt)^2$ ----(2)
we want, eqn (1)=eqn (2)
Hence, $pr=sw$
or
$(u^4-v^4)=(m^4-n^4)$ ----(3)
eqn (3) has numerical solution:
$(m,n,u,v)=(59,134,133,158)$
hence,
$(p,q,r)=(7275,42028,42653)$
$(s,t,w)=(14475,15812,21437)$
$a=305753700$
$b=1792620284$
$c=228878700$
$d=338961844$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4370392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
} |
Continuity of this function as $(x,y)$ tends to $(0,0)$
Here's a function in $x$ and $y$ defined piecewise as
$$f(x,y)=
\left\{\begin{array}{ll}
0 & (x,y)=(0,0)\\
\frac{x^2y}{x^4+y^2} & (x,y) \neq (0,0) \\
\end{array}\right.$$
Examine its continuity as the ordered pair tends to $(0,0)$.
Okay, so I first tried this by assuming $x=\frac{1}{n}$ and $y=\frac{1}{n^2}$, where $n\rightarrow \infty$ . The limit of the function as $(x,y) \rightarrow (0,0)$ came out to be $\frac{1}{2}$, and since this is not equal to the value of the function at the said point, its discontinuous at the origin.
But when I assumed $x=\frac{1}{n}$ and $y=\frac{1}{n}$ , I got the limit zero:$$f(\frac{1}{n},\frac{1}{n})= \frac{\frac{1}{n^3}}{\frac{1}{n^4}+ \frac{1}{n^2}} = \frac{\frac{1}{n}}{\frac{1}{n^2}+ 1} $$ Since $x,y \rightarrow 0$,$f(\frac{1}{n},\frac{1}{n}) \rightarrow \frac{0}{0+1}=0$
Where am I going wrong?
| Notice that :
$$f(t, t^2) = \dfrac{t^2 t^2}{t^4 + t^2} = \dfrac{t^4}{2 t^4} = \dfrac{1}{2} \underset{t \to 0}{\to} \dfrac{1}{2} \neq f(0, 0)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4373662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the limit of $(1+\frac{1}{n^2})(1+\frac{2}{n^2})...(1+\frac{n}{n^2})$. To find the limit of $(1+\frac{1}{n^2})(1+\frac{2}{n^2})...(1+\frac{n}{n^2})$.
First notice that $(1+\frac{1}{n^2})$ is the smallest term in the product and $(1+\frac{n}{n^2})$ is the greatest. So: $ (1+\frac{1}{n^2})^n \leq (1+\frac{1}{n^2})(1+\frac{2}{n^2})...(1+\frac{n}{n^2}) \leq (1+\frac{n}{n^2})^n$.
First this seemed like a good idea but then I found $ 1 \leq (1+\frac{1}{n^2})(1+\frac{2}{n^2})...(1+\frac{n}{n^2}) \leq e$. Which is not really helpful.
How can this problem be done?
P.S: In previous questions of the problem I had to prove that $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$ and $x-\frac{x^2}{2} \leq \ln(x+1) \leq x$.
| You have proved that $x-\frac{x^2}{2}\leq\ln(x+1)\leq x$
Now as @Gary said $$\prod_{i=1}^n(1+\frac{i}{n^2})=e^{\sum_{i=1}^n\ln(1+\frac{k}{n^2})}$$
From $(1)$ we get , $$\sum_{i=1}^n(\frac{i}{n^2}-\frac{i^2}{2n^4})\leq\sum_{i=1}^n\ln(1+\frac{i}{n^2})\leq\sum_{i=1}^n(\frac{i}{n^2})$$
$$\frac{n(n+1)}{2n^2}-\frac{n(n+1)(2n+1)}{12n^4}\leq S\leq\frac{n(n+1)}{2n^2}$$
where $\displaystyle S=\sum_{i=1}^n\ln(1+\frac{i}{n^2})$
Taking $\lim_{n\to\infty}$ on each side and using squeeze theorem, we get
$$\frac{1}{2}\leq\lim_{n\to\infty}S\leq\frac{1}{2}$$
Therefore $$\lim_{n\to\infty}\prod_{i=1}^n(1+\frac{i}{n^2})=e^{\lim_{n\to\infty}S}=\sqrt e$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4375377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Given $\varphi$ is golden ratio, how do I prove this $\sum \limits_{j=1}^{\infty}\frac{(1-\varphi)^j}{j^2}\cos{\frac{3j\pi}{5}}=\frac{\pi^2}{100}$? Given $ \varphi$ is golden ratio, how do I prove this:
$ \displaystyle \tag*{}\sum \limits_{j=1}^{\infty}\frac{(1-\varphi)^j}{j^2}\cos{\frac{3j\pi}{5}}=\frac{\pi^2}{100}$
My approach:
We can reduce the sum into simpler parts by using the property of golden ratio, we have:
$\displaystyle \tag*{} \phi^2 - \phi - 1 = 0 \Leftrightarrow 1- \phi = - \dfrac{1}{\phi}$
And I also found the values of $\cos \dfrac{3j\pi}{5}$ in terms of $\varphi$:
$\displaystyle \tag*{} \begin{align} \cos \dfrac{3\pi}{5} &= \dfrac{-1}{2 \varphi} \\\\ \cos \dfrac{6\pi}{5} &= \dfrac{-\varphi}{2} \\\\ \cos \dfrac{9\pi}{5} &= \dfrac{\varphi}{2} \\\\ \cos \dfrac{12\pi}{5} &= \dfrac{1}{2 \varphi} \\\\ \cos \dfrac{15\pi}{5} &= {-1} \end{align}$
And this repeats, periodically with alternate opposite signs. I don't know how to connect these information I found to prove the question. Maybe my approach is wrong. Any help would be appreciated. Thanks.
| Hint
For $x \in \mathbb R$
$$\cos x =\frac{e^{ix}+e^{-ix}}{2}$$
Now let $$f(x)= \sum_{n=1}^\infty \frac{x^n}{n^2}.$$
We have $f(0)=0$ and $$f^\prime(x)=\sum_{n=1}^\infty \frac{x^{n-1}}{n}=-\frac{1}{x}\ln(1-x)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4379575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int {\frac{{dx}}{{{{(\sin \frac{x}{2} + \cos \frac{x}{2})}^2}}}} $ My first solution was:
$$\begin{array}{l}
\int {\dfrac{{dx}}{{{{(\sin \dfrac{x}{2} + \cos \dfrac{x}{2})}^2}}}} \\
= \int {\dfrac{{dx}}{{{{\sin }^2}\dfrac{x}{2} + 2\sin \dfrac{x}{2}\cos \dfrac{x}{2} + {{\cos }^2}\dfrac{x}{2}}}} \\
= \int {\dfrac{{dx}}{{1 + \sin x}}} \\
= \int {\dfrac{{1 - \sin x}}{{1 - {{\sin }^2}x}}dx} \\
= \int {\dfrac{1}{{1 - {{\sin }^2}x}}dx - \int {\dfrac{{\sin x}}{{1 - {{\sin }^2}x}}dx} } \\
= \int {\dfrac{{dx}}{{{{\cos }^2}x}}} - \int {\dfrac{{d(\cos x)}}{{{{\cos }^2}x}}} \\
= \tan x - \dfrac{1}{{\cos x}} + C
\end{array}$$
However, the answer from Wolfram Alpha is: $\dfrac{{ - 2}}{{\tan \dfrac{x}{2} + 1}} + C$, it tells me that the step:
$$\int {\dfrac{{1 - \sin x}}{{1 - {{\sin }^2}x}}dx} \\
= \int {\dfrac{1}{{1 - {{\sin }^2}x}}dx - \int {\dfrac{{\sin x}}{{1 - {{\sin }^2}x}}dx} }$$ is false.
What is the problem with that step? If it is correct then where is the error in my solution?
| If $f(x) = \tan x - \sec x + C $
Then $f'(x) = \sec^2 x - \sec \tan x = \sec x ( \sec x - \tan x ) $
Multiplying top and bottom by $(\sec x + \tan x ) $
$f'(x) = \dfrac{\sec x}{\sec x + \tan x } $
Multiplying top and bottom by $\cos x $
$f'(x) = \dfrac{1}{1 + \sin x }$
So, your answer is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4380188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Is (f, g, h) an orthonormal string for $\langle f,g \rangle = \frac{1}{2\pi}\int_{0}^{2\pi}f(x)g(x)dx? I'm working on this exercise about orthogonality related to Fourier series. Basically, the exercise is as follows:
Study the inner product space of the space $C([0,2\pi])$:
$$
\langle f,g\rangle=\frac{1}{2\pi}\int_{0}^{2\pi}f(x)g(x)dx.
$$
Is the string $(f,g,h)$ orthogonal, when
$f(x)=1,\, g(x)=\sin x,\, h(x)=\cos x$?
Here's what I tried:
The conditions of orthonormality:
a) $\langle v_i,v_j\rangle = 0$ for all $j \neq i$
b) $||v_j|| = 1$ for all $j\in\{1,\cdots,n\}$.
Studying all the function pairs:
\begin{align*}\langle f,g\rangle &= \frac{1}{2\pi}\int_{0}^{2\pi}f(x)g(x)dx\\
&= \frac{1}{2\pi}\int_{0}^{2\pi}1sin(x)dx&\text{| }\int sin(x)dx = -cos(x)\\
&= \frac{1}{2\pi}(-cos(2\pi)+cos(0)) = \frac{1}{2\pi}(-1+1) = 0
\end{align*}
\begin{align*}\langle f,h\rangle &= \frac{1}{2\pi}\int_{0}^{2\pi}f(x)h(x)dx\\
&= \frac{1}{2\pi}\int_{0}^{2\pi}1cos(x)dx&\text{| }\int cos(x)dx = sin(x)\\
&= \frac{1}{2\pi}(sin(2\pi)-sin(0)) = \frac{1}{2\pi}(0-0) = 0
\end{align*}
\begin{align*}\langle g,h\rangle &= \frac{1}{2\pi}\int_{0}^{2\pi}g(x)h(x)dx&\\
&= \frac{1}{2\pi}\int_{0}^{2\pi}sin(x)cos(x)dx&\text{| }sin(x)cos(x) = \frac{1}{2}sin(2x)\\
&= \frac{1}{2\pi}\int_{0}^{2\pi}\frac{1}{2}sin(2x)dx&\text{| }2x = z, 2dx = dz\\
&= \frac{1}{2\pi}\cdot\frac{1}{4}\int_{0}^{2\pi}sin(u)du&\text{| }\int sin(x)dx = -cos(x)\\
&= \frac{1}{2\pi}(-\frac{1}{4}cos(4\pi)-(-\frac{1}{4})cos(2\cdot0))&\text{| }u = 2x \iff cos(u) = cos(2x)\\
&= \frac{1}{2\pi}(-\frac{1}{4}cos(4\pi)+(\frac{1}{4})cos(2\cdot0))&\\
&= \frac{1}{2\pi}(-\frac{1}{4}\cdot1+(\frac{1}{4})1) = 0&
\end{align*}
So the first condition is filled. Then:
\begin{align*}||f|| = \sqrt{\langle f,f\rangle} &= sqrt{\frac{1}{2\pi}\int_{0}^{2\pi}f(x)f(x)dx}\\
&= \sqrt{\frac{1}{2\pi}\int_{0}^{2\pi}1dx}&\text{| }\int 1 = x\\
&= \sqrt{\frac{1}{2\pi}(2\pi-0)} = sqrt{\frac{1}{2\pi}(2\pi)} = 1
\end{align*}
Then, to count $\langle g,g\rangle$ I need the formula for integrating $sin^2(x)$:
\begin{align*}1 - 2sin^2(x) &= cos(2x)&\\
2sin^2(x) &= cos(2x) + 1&\\
sin^2(x) &= \frac{1}{2}(cos(2x) + 1)&\\
\int sin^2(x)dx &= \frac{1}{2}\int cos(2x) + 1dx&\\
\int sin^2(x)dx &= \frac{1}{2}(-\frac{1}{2}sin(2x) + x) + C&\\
\end{align*}
\begin{align*}
||g|| = \sqrt{\langle g,g\rangle} &= sqrt{\frac{1}{2\pi}\int_{0}^{2\pi}g(x)g(x)dx}\\
&= \sqrt{\frac{1}{2\pi}\int_{0}^{2\pi}sin^2(x)dx}\\
&= \sqrt{\frac{1}{2\pi}(\frac{1}{2}((-\frac{1}{2}sin(4\pi)+2\pi)-(-\frac{1}{2}sin(2\cdot0)+0)}\\
&= \sqrt{\frac{1}{2\pi}(\frac{1}{2}(0 + 2\pi)+0)} = \frac{1}{\sqrt{2}}
\end{align*}
Therefore, because $||g|| \neq 1$, we can conclude that the series $(f, g, h)$ is not orthogonal.
This is where I'm uncertain. From the setup of the exercise, it feels natural to me that the exercise is constructed so that the series should be orthogonal. But I can't find a mistake in what I've done. What did I mess up or is my assumption wrong?
| It seems like you are using the terms orthogonal and orthonormal interchangeably in your question. This is not correct. In fact, orthonormality is a stronger condition than orthogonality.
A set of vectors $\{v_i\}_{i\in I}$ in an inner product space is orthogonal if $\langle v_i,v_j\rangle=0$ for all $i\neq j$. That set is orthonormal if it is orthogonal and if $\|v_i\|=1$ for all $i\in I$.
In conclusion, the set $\{f,g,h\}\subset C([0,2\pi])$ is orthogonal (as you have shown) but not orthonormal (since $\|g\|\neq 1$).
| {
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"url": "https://math.stackexchange.com/questions/4382157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Decomposing $\sum_{i = 0}^{2n} x^i$ as a simple sum of squares As we have $\sum_{i = 0}^{2n} x^i = (x^{2n + 1} - 1) / (x - 1)$, the polynomial is positive.
So we know that there is a decomposition as a sum of squares. Is there a closed simple form for such a decomposition? For small value of $n$ we have
$$ x^2 + x + 1 = 1/4 ((2x+1)^2+ 3)$$
$$ x^4 + x^3 + x^2 + x + 1 = 1/16((4 x^2 + 2 x + 1)^2 + (2 x + 3)^2 + 6)$$
$$ x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 = 1/144 ((12x^3+6x^2+4x+3)^2+ 3(2x^2+2x+5)^2 + 5(2x+3)^2 +15)$$
| $$
P(x) = \prod_{d \mid 2n+1, d \neq 1} \Phi_d (x)\\
$$
So focus on getting a good decomposition for $\Phi_d (x)$.
$\Phi_d (x)$ is a product of $\phi(d)$ terms. $d$ is not even because it divides $2n+1$ and it is greater than $1$ by assumption. For the factor corresponding to $k$ there is also the factor corresponding to $d-k$ and those two are conjugate and distinct so $\Phi_d (x) = Q(x) \overline{Q}(x)$ for some $Q$ as before but with the caveat that now we are taking only other primitive roots of unity rather than all powers of $\omega$. But that doesn't really help because the decomposition is still using polynomials with irrational and hard to compute coefficients.
Another property $\Phi_n (x) = \Phi_q (x^{n/q})$ where $q$ is the radical of $n$ by only keeping the same prime factors but only with multiplicity $1$.
So we focus on providing good sum of squares decomposition for $\Phi_q (x)$ where $q$ is a square free number dividing $2n+1$ and greater than $1$.
If $q=3$, then
$$
\Phi_3 (x) = x^2 + x + 1\\
= (x+\frac{1}{2})^2 + (\frac{1}{2})^2 + (\frac{1}{2})^2 + (\frac{1}{2})^2\\
$$
Otherwise Gauss' formula applies
$$
\Phi_q (x) = (\frac{1}{2} A_n (x))^2 - (-1)^{(q-1)/2} q (\frac{1}{2} x B_n(x))^2\\
$$
When $q \equiv 3 (4)$ this gives a decomposition into a sum of $q+1$ squares of polynomials. All the coefficients involved are half-integers in this case.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4382334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
how to prove $A+A'B+A'B'C+A'B'C'D=A+B+C+D$ Prove the above relationship by using the Boolean definition. I tried $A+A'B=A+B$, but end up with $A+B+A'B'(C+D)$, how can I go next?
|
I tried $A+A'B=A+B$
Apply this to $\,A+A'B'C=A+B'C\,$ then again to $\,B+B'C=B+C\,$:
$$
\begin{align}
A+A'B+A'B'C &= \color{red}{A+B}+A'B'C
\\ &= \color{red}{A}+B+\color{red}{B'C}
\\ &= A+\color{red}{B+C}
\end{align}
$$
Repeat the steps to prove the equality with all four terms.
[ EDIT ] $\;$ Another way also using OP's attempt, only working from the end backwards.
$$
\begin{align}
A+A'B+A'B'C+A'B'C'D &= A+A'B+A'B'(\color{blue}{C+C'D})
\\ &= A+A'B+A'B'(\color{blue}{C+D})
\\ &= A+A'\big(\color{red}{B+B'(C+D)}\big)
\\ &= A+A'\big(\color{red}{B+(C+D)}\big)
\\ &= \color{green}{A+A'(B+C+D)}
\\ &= \color{green}{A+(B+C+D)}
\\ &= A+B+C+D
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4382451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Area of shaded region in a square
My approach
The area of shaded region = 8×the area of
Given that the points of the shaded region are closer to the center than the boundary of the square.
Let's talk about the boundary of the shaded region
The boundary of the shaded region therefore must be the locus of all the points whose distance from the center of the square = distance from the boundary.
Let's find the locus of the boundary of the shaded region
From the second figure
${\sqrt {h^2+k^2} } = {\sqrt {(h-h)^2 + (k-{a\over 2})^2 }}$
This simplifies to be:
$k = {a^2 - 4h^2\over 4a}$
$y = {a^2 - 4x^2\over 4a}$
Also the curve intersects the line( the hypotenuse of the triangle) y= x
For point of intersection :
$ {a^2 - 4x^2\over 4a} = x $
$4x^2 +4ax - a^2 = 0$
$x = {-4a ± \sqrt{16a^2 - 4×4×(-a^2)}\over 8}$
$x = a{(\sqrt{2} ± 1)\over 2}$
Solution 1: $x = a{(\sqrt{2} + 1)\over 2}$
Solution 2: $x = a{(\sqrt{2} - 1)\over 2}$
Solution 1 can be discarded as $ x = a{(\sqrt{2} + 1)\over 2} ≈ 1.207106 a > {a\over 2} $
Solution 2: $ a{(\sqrt{2} - 1)\over 2} ≈ 0.2071 a < {a\over 2} $
from 0 to $a{(\sqrt{2} - 1)\over 2}$ :
The curve (boundary of shaded region) lies above the line $ y=x$
so $dA = ({a^2\over 4a} - {4x^2\over 4a} - x)dx $
$dA = ({a\over 4} - {x^2\over a} - x)dx $
$\int_{0}^{A} dA = \int_{0}^{a{(\sqrt{2} - 1)\over 2}} ({a^2\over 4a} - {4x^2\over 4a} - x)dx $
A = $({a\over 4}x - {x^3\over 3a} - {x^2\over 2})]_{0}^{{a{(\sqrt{2} - 1)\over 2}}}$
$ A = {a^2\over 8}(\sqrt{2}-1) - {a^2\over 24}(\sqrt{2}-1)^3 - {a^2\over 8}
(\sqrt{2}-1)^2 $
$ = {a^2\over 8}(\sqrt{2}-1) \Biggl( 1 - {(\sqrt{2}-1)^2\over 3} - (\sqrt{2} -1)\Biggr)$
This simplifies to be equal to
${a^2\over 8}(\sqrt{2}-1)({3+5\sqrt{2}\over 3})$
$= {(7- 2\sqrt{2})\over 8×3}a^2$
Area of shaded figure = 8A
$A_{total}$ = ${(7- 2\sqrt{2})\over 3}a^2$
But the answer is :
${4\sqrt{2}-5\over 3}a^2 $
I don't know where I got it wrong, and also I have re calculated this and the result is same.
Did I miss something important or calculated wrongly
Any help of hint or suggestion or worked out solution would be appreciated.
| Using polar coordinates
$(x, y) = (r \cos \theta, r \sin \theta)$
In the range of $ 0 \le \theta \le \dfrac{\pi}{4} $ we want
$ r \le \left( \dfrac{a}{2} - r \cos \theta \right ) $
Hence,
$ r \le \dfrac{ a } { 2 (1 + \cos \theta ) } $
The area enclosed by this polar curve
$ r(\theta) = \dfrac{ a } { 2 (1 + \cos \theta ) } =\dfrac{a}{4 \cos^2(\dfrac{\theta}{2}) } $ is given by
$A = \dfrac{1}{2} \displaystyle \int_0^\dfrac{\pi}{4} r^2(\theta) \text{d} \theta = \dfrac{1}{2} \int_0^\dfrac{\pi}{4} \dfrac{a^2}{16 \cos^4 (\dfrac{\theta}{2})} \text{d} \theta = \dfrac{a^2}{32} \int_0^{\dfrac{\pi}{4}} \sec^4(\dfrac{\theta}{2}) \text{d} \theta $
Now, let $ u = \dfrac{\theta}{2} $ then
$A = \dfrac{a^2}{16} \displaystyle \int_0^\dfrac{\pi}{8} \sec^4(u) du $
And since $\sec^2(u) = \tan^2(u) + 1 $, the above becomes
$A = \dfrac{a^2}{16} \displaystyle \int_0^\dfrac{\pi}{8} (\tan^2(u) + 1) \sec^2(u) du $
And this reduces to
$A = \dfrac{a^2}{16} ( \dfrac{t^3}{3} + t ) $
where $t = \tan(\dfrac{\pi}{8} ) = \dfrac{\sin(\frac{\pi}{4})}{\cos(\frac{\pi}{4}) + 1} = \dfrac{ 1 }{ \sqrt{2} + 1 } =
\sqrt{2} - 1 $
Therefore,
$A = \dfrac{a^2}{48} ( 2 \sqrt{2} - 3(2) + 3 \sqrt{2} - 1 + 3 \sqrt{2} - 3 ) = \dfrac{a^2}{24} (4 \sqrt{2} - 5) $
And since there is $8$ of these areas in the square, then the answer is
$ \text{Area} = 8 A = \dfrac{a^2}{3} ( 4 \sqrt{2} - 5) $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4385102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Finding $\int_0^\pi\frac{f\left(\alpha+e^{ix}\right)+f\left(\alpha+e^{-ix}\right)}{1+2p\cos(x)+p^2}\mathrm dx$ Let $f$ be a analytic function in the closed unit circle with its center
at the point $\alpha\in\mathbb{R}$, then:
\begin{equation*}
\int_0^\pi\frac{f\left(\alpha+e^{ix}\right)+f\left(\alpha+e^{-ix}\right)}{1+2p\cos(x)+p^2}\mathrm dx=\frac{2\pi}{1-p^2}f(\alpha+p) ,
\end{equation*}
for $|p|<1$.
My attempt: By
\begin{align*}
\therefore\quad \sum_{n=1}^{\infty}p^{n}\sin(nx)=\frac{p\sin (x)}{1-2p\cos (x)+p^2},\qquad|p|<1
\end{align*}
adjusting $p\to -p$ and highlighting $\displaystyle \frac1{1+2p\cos(x)+p^2}$:
\begin{align*}
\frac1{1+2p\cos(x)+p^2}=-\frac{1}{\sin(x)}\sum_{n=1}^\infty(-p)^{n}\sin(nx).
\end{align*}
Thus:
\begin{align*}
\int_0^\pi\frac{f\left(\alpha+e^{ix}\right)+f\left(\alpha+e^{-ix}\right)}{1+2p\cos(x)+p^2}\mathrm dx=-\sum_{n=1}^\infty(-p)^{n}\int_0^\pi\left\{f\left(\alpha+e^{ix}\right)+f\left(\alpha+e^{-ix}\right)\right\}\frac{\sin(nx)}{\sin(x)}\mathrm dx\tag{1}
\end{align*}
by the \textit{Dirichlet Kernel}: $\displaystyle \sum_{k=0}^{N-1}e^{2ikx}=e^{(N-1)x}\frac{\sin(Nx)}{\sin(x)}$ setting $N\to n$, $n\in\mathbb{N}$ and then taking $(1)$, follows that:
\begin{align*}
&=-\sum_{n=1}^\infty(-p)^{n}\sum_{k=0}^{n-1}\int_0^\pi\left\{f\left(\alpha+e^{ix}\right)+f\left(\alpha+e^{-ix}\right)\right\}e^{-(n-1)x}e^{2ikx}\mathrm dx,\quad\left(e^{ix}\to z\right)\\
&=...
\end{align*}
At this point I'm out of ideas. I would like some light on my last step, or another approach that is similar to this one.
| If we convert the integral into a contour integral, we get
$$ \begin{align} \int_{0}^{\pi} \frac{f(\alpha+e^{ix})+f(\alpha+e^{-ix})}{1+2p \cos x +p^{2}} \, \mathrm dx &= \frac{1}{2}\int_{-\pi}^{\pi} \frac{f(\alpha+e^{ix})+f(\alpha+e^{-ix})}{1+2p \cos x +p^{2}} \, \mathrm dx \\ &= \int_{-\pi}^{\pi}\frac{f(\alpha+e^{ix})}{1+2p \cos x +p^{2}} \, \mathrm dx \\&= \int_{-\pi}^{\pi} \frac{f(\alpha +e^{ix})}{(e^{ix}+p)(e^{-ix}+p)} \, \mathrm dx \\ &= \int_{|z|=1} \frac{f(\alpha + z)}{(z+p)(\frac{1}{z}+ p)} \frac{dz}{iz} \\ &= \int_{|z|=1} \frac{f(\alpha +z)}{(z+p)(1+pz)} \, \frac{\mathrm dz}{i}. \end{align}$$
Since $|p| <1$, the only singularity inside the unit circle is a simple pole at $z=-p$.
Therefore, $$ \begin{align} \int_{0}^{\pi}\frac{f(\alpha+e^{ix})+f(\alpha+e^{-ix})}{1+2p \cos x +p^{2}} \, \mathrm dx &= 2 \pi i \operatorname{Res} \left[ \frac{f(\alpha +z)}{i(z+p)(1+pz)}, -p \right] \\ &= \frac{2\pi f(\alpha \color{red}{-}p)}{1-p^{2}}. \end{align}$$
If $|p| <1$ and not equal to zero, we also have $$ \begin{align} \int_{0}^{\pi} \frac{f(\alpha+e^{ix})+ f(\alpha+e^{-ix})}{1+2p \cos (x) +p^{2}} \, \cos (x) \, \mathrm dx &= \frac{1}{2}\int_{-\pi}^{\pi} \frac{f(\alpha+e^{ix})+f(\alpha+e^{-ix})}{1+2p \cos (x) +p^{2}} \, \cos (x) \, \mathrm dx \\ &=\int_{-\pi}^{\pi} \frac{f(\alpha+e^{ix})}{1+2p \cos (x) +p^{2}} \, \cos (x) \, \mathrm dx \\ &= \int_{-\pi}^{\pi} \frac{f(\alpha+e^{ix})}{(e^{ix}+p)(e^{-ix}+p)} \, \frac{\left(e^{ix}+e^{-ix} \right)}{2} \, \mathrm dx \\ &= \int_{|z|=1} \frac{f(\alpha+z)}{(z+p)(\frac{1}{z}+p)} \, \frac{1}{2} \left(z+\frac{1}{z} \right) \, \frac{\mathrm dz}{iz} \\ &= \int_{|z|=1} \frac{f(\alpha+z)}{(z+p)(1+pz)} \, \frac{1+z^{2}}{2iz} \, \mathrm dz \\ &= 2 \pi i \left(\frac{f(\alpha - p)}{1-p^{2}} \frac{1+p^{2}}{-2ip}+ \frac{f(\alpha)}{2ip} \right) \\ &= \frac{\pi}{p} \left(f(\alpha)- \frac{1+p^{2}}{1-p^{2}} \, f(\alpha -p) \right). \end{align}$$
At $p=0$, we get $$ \int_{0}^{\pi} \left(f(\alpha+e^{ix})+f(\alpha+e^{-ix}) \right) \cos(x) \, \mathrm dx = \lim_{p \to 0} \frac{\pi}{p} \left(f(\alpha)- \frac{1+p^{2}}{1-p^{2}} \, f(\alpha -p) \right) = \pi f^{\prime}(\alpha). $$
Finally, if $|p| <1$ and not equal to zero, $$ \begin{align} \int_{0}^{\pi} \frac{f(\alpha+e^{ix})\color{red}{-} f(\alpha +e^{-ix})}{1+2p \cos(x) + p^{2}} \, \sin(x) \, \mathrm dx &= \int_{|z|=1} \frac{f(\alpha+z)}{(z+p)(1+pz)} \, \frac{1-z^{2}}{2z} \, \mathrm dz \\ &= 2 \pi i \left(\frac{f(\alpha-p)}{1-p^{2}} \frac{1-p^{2}}{-2p} + \frac{f(\alpha)}{2p}\right) \\ &=\frac{\pi i}{p} \left(f(\alpha) - f(\alpha-p) \right). \end{align}$$
And at $p=0$, we get $$ \int_{0}^{\pi}\left(f(\alpha+e^{ix})\color{red}{-} f(\alpha +e^{-ix})\right)\, \sin(x) \mathrm dx = \pi i f^{\prime}(\alpha)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4388974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Distribution density of the sum of two independent, uniform ditributioned random variables Random variable $ξ_1$ and $ξ_2$ are independent and have uniform distribution on interval [-a;a]. What is the disribution density of random variable $\eta = 2ξ_1 + ξ_2^2$? Obviously, we should use
convolution $f_{\eta}(y) = \int\limits_{-\infty}^{\infty} f_{2ξ_1}(x) \cdot f_{ξ_2^2}(y - x)dx$. I found $f_{2ξ_1}(x) = \frac{1}{2c - (-2c)} = \frac{1}{4c}, -2c \leq x \leq 2c$. $f_{ξ_2^2}(x) = \frac{1}{c^2 - 0^2} = \frac{1}{c^2}, 0 \leq x \leq c^2$. And as I know, now we need to set intervals of $y$ and then we will find intervals of $x$ and solve system of equations at different intervals of $y$. But I can't understand how to set these intervals to solve system of equations and how to find disribution density of $\eta = 2ξ_1 + ξ_2^2$ at the finish. Can someone explain me it, please?
| I will refer to $~ξ_1$ as $X$, $ξ_2$ as $Y$ and $\eta$ as $Z$.
$X, Y$ are independent and we have $X, Y \sim U(-a, a)$ and we need to find density of $Z = 2X + Y^2$. We can first find $F_z(z)$ and then differentiate to find the density.
We first note that $-2a \leq z \leq 2a + a^2$.
$i$) For $-2a \leq z \lt a^2 - 2a$, $-a \leq x \leq \frac{z}{2}, |y| \leq \sqrt {z - 2x}~$
$ii$) For $~a^2-2a \leq z \lt 2a$, $-a \leq x \leq \frac{z-a^2}{2}, |y| \leq a$ and for $\frac{z-a^2}{2} \leq x \leq \frac{z}{2}, |y| \leq \sqrt{z-2x}$
$iii$) For $~2a \leq z \leq a^2 + 2a$, $-a \leq x \leq \frac{z-a^2}{2}, |y| \leq a$ and for $\frac{z-a^2}{2} \leq x \leq a, |y| \leq \sqrt{z-2x}$
The above becomes easier to understand from the below diagram showing curve $z = 2x + y^2$ for three different values of $z$ -
The blue curve is for a value of $z$ in $(i)$
The red curve is for a value of $z$ in $(ii)$
The black curve for a value of $z$ in $(iii)$
For $(i)$,
$ \displaystyle F_Z(z) = 2 \int_{-a}^{z/2} \int_0^{\sqrt{z-2x}} \frac{1}{4a^2} ~dy ~ dx $
For $(ii)$,
$ \displaystyle F_Z(z) = \frac{1}{4a^2} \cdot \left(\frac{z-a^2}{2} + a\right) \cdot 2a + $
$$ \displaystyle 2 \int_{(z-a^2)/2}^{z/2} \int_0^{\sqrt{z-2x}} \frac{1}{4a^2} ~dy ~ dx $$
For $(iii)$,
$ \displaystyle F_Z(z) = \frac{1}{4a^2} \cdot \left(\frac{z-a^2}{2} + a\right) \cdot 2a + $
$$ \displaystyle 2 \int_{(z-a^2)/2}^{a} \int_0^{\sqrt{z-2x}} \frac{1}{4a^2} ~dy ~ dx $$
You could also write the integral in the order $dx ~ dy$, which may be easier to evaluate.
$(i) ~ F_Z(z) = \displaystyle 2 \int_0^{\sqrt{z+2a}} \int_{-a}^{(z - y^2)/2} \frac{1}{4a^2} ~dx~dy$
$(ii)~F_Z(z) = \displaystyle 2 \int_0^a \int_{-a}^{(z - y^2)/2} \frac{1}{4a^2} ~dx~dy$
$(iii) ~ F_Z(z) = \displaystyle 1 - 2 \int_{\sqrt{z-2a}}^a \int_{(z - y^2)/2}^a \frac{1}{4a^2} ~dx~dy$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4392245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Weird Problem on Polynomial Roots The polynomial $f(x)=x^3-3x^2-4x+4$ has three real roots $r_1$, $r_2$, and $r_3$. Let $g(x)=x^3+ax^2+bx+c$ be the polynomial which has roots $s_1$, $s_2$, and $s_3$, where
\begin{align*}
s_1 &= r_1+r_2z+r_3z^2, \\
s_2 &= r_1z+r_2z^2+r_3, \\
s_3 &= r_1z^2+r_2+r_3z,
\end{align*}and $z=\dfrac{-1+i\sqrt3}2$. Find the real part of the sum of the coefficients of $g(x)$.
I know the sum of the coefficients is $g(1)$, $g(x)=(x-s_1)(x-s_2)(x-s_3)$, and $z^3=1$. This means $s_1z=s_2$, and $s_2z=s_3$. Since $s_1^3=s_2^3=s_3^3$, I have $g(x)=x^3-s_1^3$. Since the answer is $g(1)$, I need to calculate $$1-s_1^3.$$ I expanded $s_1^3$ to get $$s_1^3=r_1^3+r_1^2r_2z+3r_1^2r_3z+3r_1r_2^2z^2+6r_1r_2r_3+3r_1r_3^2z+r_2^3+3r_2^2r_3z+3r_2^2r_3z+3r_2r_3^2z^2+r_3^3.$$ I'm pretty sure using Vieta's can finish this, but I'm not sure where else to apply Vieta's other than $r_1r_2r_3$. I also tried substituting $z^2=-z-1$, but it didn't do much. I also tried using $(r_1+r_2+r_3)^2$, but this also failed. Could someone give me some guidance?
Thanks in advance!
|
Since the answer is g(1), I need to calculate
$1−s_1^3$.
Well, the real part of $1−s_1^3$ is to be calculated, thus we need to calculate $\frac{ 1−s_1^3 + \overline{1−s_1^3}}{2}$ where $\overline{z}$ is the complex conjugate of z.
$\frac{ 1−s_1^3 + \overline{1−s_1^3}}{2}$ $\implies$$\frac{ 2−(s_1^3 + \overline{s_1^3})}{2}$
$\implies$ $\frac{ 2−(s_1+ \overline{s_1})(s_1^2+\overline{s_1}^2-s_1\overline{s_1})}{2}$
Also,
$s_1= r_1+r_2z+r_3z^2$
$\overline{s_1}=r_1+r_3z+r_2z^2$ (as $z^3=1$)
On simplifying $ (s_1+ \overline{s_1})(s_1^2+\overline{s_1}^2-s_1\overline{s_1})$,we obain an expression $2\displaystyle\sum_{i=1}^{3} r_i^3-3\displaystyle\sum_{1\leq i , j\leq 3,(i≠j) } r_i r_j^2+12\displaystyle\prod_{i=1}^{3} r_i$ (where $r_i$ are roots of f(x)).
Can you proceed further from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4392446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Tracing a circle by a sliding triangle An isosceles triangle with a unit length base is sliding on two lines which make an angle of $60^\circ$ between them. The third vertex traces a circle centered at the intersection of the two lines. What is the altitude of the triangle, and what is the radius of the circle ?
What I have tried:
I found the coordinates of points $X$ and $Y$ as shown above on the two lines in terms of the angle $\theta$, then found the coordinates of the tip of the triangle (the third vertex) as a function of $\theta$, and finally found the altitude $h$ that results in the distance of this tip from the origin being constant. And that constant is the radius of the circle.
|
Let one line be horizontal and the other making an angle of $\phi$ with it.
I'll take the origin of the coordinate system at their intersection. Placing the triangle at a general orientation with its base making an angle of $\theta$ with the horizontal as shown above, we can write the following equation from the law of sines
$ \dfrac{y}{\sin \theta } = \dfrac{1}{\sin \phi} $
Hence, $ y = \dfrac{\sin \theta}{\sin \phi} $
Since the unit vector pointing from the origin to $Y$ is $(-\cos \phi, -\sin \phi) $, then point $Y$ has the following coordinates
$Y = (- \sin \theta \cot \phi, - \sin \theta ) $
Now, the unit vector along $\vec{YX}$ is $(-cos \theta, \sin \theta)$, therefore,
$X = Y + (1) (-\cos \theta, \sin \theta) = (- \sin \theta \cot \phi - \cos \theta, 0 ) $
The midpoint of the base has coordinates
$ M = \frac{1}{2} (X + Y) = Y + \frac{1}{2} ( - \cos \theta, \sin \theta ) $
Hence,
$M = (- \sin \theta \cot \phi - \frac{1}{2}\cos \theta, - \frac{1}{2} \sin \theta ) $
Finally the tip of the triangle $T$ is a distance $h$ in the direction $(\sin \theta, \cos \theta) $, thus its coordinates are
$T = ( \sin \theta (h - \cot \phi) - \frac{1}{2} \cos \theta, -\frac{1}{2} \sin \theta + h \cos \theta ) $
If $T$ traces a circle centered at the origin, then the sum of squares of its $x$ and $y$ coordinates must be constant, hence
$ T_x^2 + T_y^2 = \sin^2 \theta ( (h - \cot \phi)^2 + \frac{1}{4} ) + \cos^2 \theta ( \frac{1}{4} + h^2 ) - \sin \theta \cos \theta ( 2 h - \cot \phi ) $
For this to be constant, we must have
$ (h - \cot \phi)^2 = h^2 + \frac{1}{4} $
and
$ 2 h - \cot \phi = 0 $
We can see that the two equations are identical and have the solution
$ h = \frac{1}{2} \cot \phi $
And the radius of the circle is $\sqrt{ \frac{1}{4} + h^2 } = \dfrac{1}{2 \sin \phi } $
For $\phi = \dfrac{\pi}{3} $, we get
$ h = \dfrac{1}{2 \sqrt{3}} $ and $R = \dfrac{1}{\sqrt{3} } $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4392844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Differentiability of $f(x,y)= \frac{xy^3}{x^2+y^2}$ for $(x,y)\neq (0,0)$ and $0$ for $(x,y)=(0,0)$.
Let's consider $f:\mathbb{R}^2\to\mathbb{R}$ with\begin{align*} f:=\begin{cases} \frac{xy^3}{x^2+y^2},&(x,y)\neq (0,0)\\0, & (x,y)=(0,0). \end{cases} \end{align*}
Show that $f$ is differentiable at $(0,0)$.
My approach:
We prove that ${0\choose 0}$ is the (total) derivative at point $(0,0)$.
If $|x|\geq|y|>0$ then
$$
\Big|\frac{xy^3}{(x^2+y^2)\sqrt{x^2+y^2}}\Big|\leq\frac{|x||y|y^2}{(2y^2)|y|\sqrt{2}}=\frac{|x|}{4}.
$$
If $|y|\geq|x|>0$ then
$$
\Big|\frac{xy^3}{(x^2+y^2)\sqrt{x^2+y^2}}\Big|=\frac{|y|}{(\frac{x^2}{y^2}+1)\sqrt{1+\frac{y^2}{x^2}}}\leq |y|.
$$
Keeping those bounds in mind we look at the limit:
\begin{align*} \lim\limits_{{x\choose y}\to{0\choose 0}}\frac{f(x,y)-f(0,0)-{0\choose 0}\left({x\choose y}-{0\choose 0}\right)}{\Vert{x\choose y}-{0\choose 0}\Vert_2 }=\lim\limits_{{x\choose y}\to{0\choose 0}}\frac{\frac{xy^3}{x^2+y^2}}{\Vert{x\choose y}\Vert_2 }=\lim\limits_{{x\choose y}\to{0\choose 0}}\frac{xy^3}{(x^2+y^2)\sqrt{x^2+y^2}},\text{ where } (x,y)\neq (0,0).
\end{align*}
If $x=0$ or $y=0$ then the above limit is $0$. If $|y|\geq|x|>0$ or $|x|\geq|y|>0$ then we use the above upper bounds and see that the limit is again $0$. Hence, $f$ is (total) differentiable at $(0,0)$.
Is this correct? Our tutor told us that when it comes to total differentiability we should rather prove that the partials are continuous. Maybe in this case it was just luck that one could easily see the matrix/derivative.
| It looks correct, but it's easier to see that, if $x=\rho\cos(\theta)$ and $y=\rho\sin(\theta)$, then$$\left|\frac{xy^3}{(x^2+y^2)^{3/2}}\right|=\rho|\cos(\theta)\sin^3(\theta)|\leqslant\rho=\sqrt{x^2+y^2}.$$Therefore,$$\lim_{(x,y)\to(0,0)}\frac{xy^3}{(x^2+y^2)^{3/2}}=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4393649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
$f(x)$ and $g(x)$ are monic cubic polynomials, with $f(x)-g(x)=r$. If $f$ has roots $r+1$ and $r+7$, and $g$ has roots $r+3$ and $r+9$, then find $r$.
Let $f(x)$ and $g(x)$ be two monic cubic polynomials, and let $r$ be a real number. Two of the roots of $f(x)$ are $r+1$ and $r+7$. Two of the roots of $g(x)$ are $r + 3$ and $r + 9,$ and$$f(x) - g(x) = r$$for all real numbers $x.$ Find $r.$
So far, I have $$f(x)=(x-r-1)(x-r-7)(x-p)$$ and $$g(x)=(x-r-3)(x-r-9)(x-q).$$ From $f(x)-g(x)=r$, I know that their constant terms differ by $r$. I expanded the two functions but it was too complicated. I also plugged in $x=r+1,r+7,r+3,r+9$ into $f(x)-g(x)=r$, but it didn't do much.
Thanks in advance!!!!!
| Some properties of these two polynomials may be worth remarking upon first:
• because $ \ f(x) \ = \ g(x) + r \ \ , \ \ r \ $ real, the curves of the two functions never intersect;
• for the same reason, the quadratic $ \ (b) \ $ and linear $ \ (c) \ $ coefficients are identical ;
• from the given information about their respective zeroes, we determine that $ g(r + 1) \ = \ -r \ \ , $ $ g(r + 3) \ = \ 0 \ \ , \ \ g(r + 7) \ = \ -r \ \ , \ \ g(r + 9) \ = \ 0 \ \ . $
We will also use your notation to write $ \ f(p) \ = \ 0 \ \ , \ \ g(q) \ = \ 0 \ \ . $ Since the polynomials have three real zeroes, their curves have two relative extrema (the cubic "S - curve"). This tells us something significant about the zeroes: "shifting" $ \ g(x) \ $ vertically by $ \ r \ $ moves the zeroes of $ \ g(x) \ $ at $ \ (r + 3) \ $ and $ \ (r + 9) \ $ "leftward" to the zeroes of $ \ f(x) \ $ at $ \ (r + 1) \ $ and $ \ (r + 7) \ \ , $ implying that $ \ r \ > \ 0 \ \ . $ (The relative maximum moves away from the $ \ x-$axis and the relative minimum, towards it.) The third zero of these polynomials must therefore be between the other two $ \ ( \ r + 1 \ \le \ p \ \le \ r + 7 \ \ , \ \ r + 3 \ \le \ q \ \le \ r + 9 \ ) \ \ $ and $ \ q \ $ should "move rightward" to $ \ p \ \ . $ This last statement is confirmed from the Viete relations: the quadratic coefficient is $$ b \ \ = \ \ -[ \ (r + 1) \ + \ p \ + \ (r + 7) \ ] \ \ = \ \ -[ \ (r + 3) \ + \ q \ + \ (r + 9) \ ] $$ $$ \Rightarrow \ \ 2r \ + \ p \ + \ 8 \ \ = \ \ 2r \ + \ q \ + \ 12 \ \ \Rightarrow \ \ p \ = \ q + 4 \ \ . $$
For what follows, we will re-label the given zeroes in terms of $ \ \rho \ = \ (r + 3) \ \ . \ $ The linear coefficient of these polynomials is then
$$ c \ \ = \ \ (\rho - 2)·(q + 4) \ + \ (\rho + 4)·(q + 4) \ + \ (\rho - 2)·(\rho + 4) $$ $$ = \ \ \rho · q \ + \ (\rho + 6) · q \ + \ \rho · (\rho + 6) $$
$$ \Rightarrow \ \ \rho^2 \ + \ 2·q·\rho \ + \ 10·\rho \ + \ 2·q \ \ = \ \ \rho^2 \ + \ 2·q·\rho \ + \ 6·\rho \ + \ 6·q \ \ \Rightarrow \ \ \rho \ \ = \ \ q \ \ . $$
So we discover that $ \ q \ = \ (r + 3) \ $ is in fact a double zero of $ \ g(x) \ \ . $ In turn, we find that $ \ p \ = \ q + 4 \ = \ (r + 3) + 4 \ = \ (r + 7) \ $ is a double zero of $ \ f(x) \ \ . $
If we designate $ \ d \ $ as the "constant term" of $ \ g(x) \ \ , \ $ we find
$$ d \ = \ -(r + 3)^2·(r + 9) \ \ \Rightarrow \ \ d \ + \ r \ = \ r \ - \ (r + 3)^2·(r + 9) \ \ = \ \ -(r + 1)·(r + 7)^2 $$
$$ \Rightarrow \ \ -r^3 \ - \ 15r^2 \ - \ 62r \ - \ 81 \ \ = \ \ -r^3 \ - \ 15r^2 \ - \ 63r \ - \ 49 \ \ \Rightarrow \ \ r \ = \ 32 \ \ . $$
Our polynomials are therefore
$$ g(x) \ \ = \ \ (x - 35)^2 \ · \ (x - 41) \ \ = \ \ x^3 \ - \ 111x^2 \ + \ 4095x \ - \ 50225 \ \ \ \text{and} $$
$$ f(x) \ \ = \ \ (x - 33) \ · \ (x - 39)^2 \ \ = \ \ x^3 \ - \ 111x^2 \ + \ 4095x \ - \ 50193 \ \ = \ \ g(x) \ + \ 32 \ \ . $$
[The relative extrema of $ \ g(x) \ $ are located at $ \ (35 \ , \ 0) \ $ and $ \ (39 \ , \ -32) \ \ , $ while those of $ \ f(x) \ $ are $ \ (35 \ , \ 32) \ $ and $ \ (39 \ , \ 0) \ \ . \ ] $
Incidentally, there is a complementary pair of polynomials for $ \ r \ = \ -32 \ \ : $
$$ g(x) \ \ = \ \ (x + 25)^2 \ · \ (x + 31) \ \ = \ \ x^3 \ + \ 81x^2 \ + \ 2175x \ + \ 19375 \ \ \ \text{and} $$
$$ f(x) \ \ = \ \ (x + 23) \ · \ (x + 29)^2 \ \ = \ \ x^3 \ + \ 81x^2 \ + \ 2175x \ + \ 19343 \ \ = \ \ g(x) \ - \ 32 \ \ . $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4395949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Is there an elementary method to evaluate the indefinite integral $\int \frac{1}{1+\sin ^{6} x} d x?$ Inspired by the post, I want to increase the power of $\sin x$ by $2$ to $6$,
$$
I=\int\frac{1}{1+\sin ^{6} x} d x.
$$
As usual, we multiply both the numerator and denominator by $\sec^6 x$ and get
$$
\begin{aligned}
I &=\int \frac{\sec ^{6} x}{\sec ^{6} x+\tan ^{6} x} d x \\
& \stackrel{t=\tan x}{=} \int\frac{\left(1+t^{2}\right)^{2} d t}{\left(1+t^{2}\right)^{3}+t^{6}}
\end{aligned}
$$
Factorizing the denominator yields $$
I=\int \frac{1+2 t^{2}+t^{4}}{\left(1+t^{2}+t^{4}\right)\left(2 t^{2}+1\right)} d t
$$
Resolving the rational function $$
\frac{1+2 x+x^{2}}{\left(1+x+x^{2}\right)(2 x+1)}=\frac{1}{3(2 x+1)}+\frac{x+2}{3\left(x^{2}+x+1\right)},
$$
into partial fractions yields $$
\int\frac{1+2 t^{2}+t^{4}}{(1 +t^{2}+t^{4})(2t^2+1)} d t=\frac{1}{3} \left(\underbrace{\int\frac{d t}{2 t^{2}+1}}_{K} +\underbrace{\int\frac{t^{2}+2}{t^{4}+t^{2}+1} d t}_{J}\right)
$$
$$
K=\int \frac{d t}{2 t^{2}+1}=\frac{1}{\sqrt{2}} \tan ^{-1}(\sqrt{2} t) +C_1 =\frac{1}{\sqrt{2}} \tan ^{-1}(\sqrt{2} \tan x)+C_1
$$
We now focus on evaluating the last integral.
\begin{aligned}
J &=\int \frac{t^{2}+2}{t^{4}+t^{2}+1} d t \\
&=\int \frac{1+\frac{2}{t^{2}}}{t^{2}+\frac{1}{t^{2}}+1} d t \\
&=\int \frac{\frac{3}{2}\left(1+\frac{1}{t^{2}}\right)-\frac{1}{2}\left(1-\frac{1}{t^{2}}\right)}{t^{2}+\frac{1}{t^{2}}+1} d t \\
&=\frac{3}{2} \int \frac{d\left(t-\frac{1}{t}\right)}{\left(t-\frac{1}{t}\right)^{2}+3}-\frac{1}{2} \int \frac{d\left(t+\frac{1}{t}\right)}{\left(t+\frac{1}{t}\right)^{2}-1} \\
&=\frac{\sqrt{3}}{2} \tan ^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{3}}\right)+\frac{1}{4 \sqrt{2}} \ln \left|\frac{t+\frac{1}{t}+1}{t+\frac{1}{t}-1}\right|+C \\
&=\frac{\sqrt{3}}{2} \tan ^{-1}\left(\frac{\tan ^{2} x-1}{\sqrt{3} \tan x}\right)-\frac{1}{4 \sqrt{2}} \ln \left|\frac{\tan ^{3} x+\tan x+1}{\tan ^{2} x-\tan x+1}\right|+C_2
\end{aligned}
Now we can conclude that
$$
\begin{aligned}
I=& \frac{1}{3}\left[\frac{1}{\sqrt{2}} \tan ^{-1}(\sqrt{2} \tan x)+\frac{\sqrt{3}}{2} \tan ^{-1}\left(\frac{\tan ^{2} x-1}{\sqrt{3} \tan x}\right)\right.\left.-\frac{1}{4 \sqrt{2}} \ln \left|\frac{\tan ^{2} x+\tan x+1}{\tan ^{2} x-\tan x+1}\right|\right]+C
\end{aligned}
$$
My Question
Can we go further with $n\geq 8$,
$$
I_n= \int\frac{1}{1+\sin ^{n} x} d x?
$$
| Thanks to @Claude Leibovici for his shortest solution. I want to add an elementary but a bit long one. $$
\begin{aligned}
\int \frac{d x}{1+\sin ^{4} x} &=\int \frac{\sec ^{4} x}{\sec ^{4} x + \tan ^{2} x} d x \\
\\
&=\int \frac{1+t^{2}}{\left(1+t^{2}\right)^{2}+t^{4}} d t, \quad \textrm{ where } t =\tan x \\
&=\int \frac{1+t^{2}}{2 t^{4}+2 t^{2}+1} d t \\
&=\frac{1}{2} \int \frac{\frac{1}{t^{2}}+1}{t^{2}+1+\frac{1}{2 t^{2}}} d t \\
&=\frac{1}{2} \int \frac{A\left( 1+\frac{1}{\sqrt{2} t^{2}}\right)+B\left(1-\frac{1}{\sqrt{2} t^{2}}\right)}{t^{2}+\frac{1}{2 t^{2}}+1} d t ,\\
&\quad\textrm{ where } \int \frac{\sqrt{2}+1}{2} \text { and } B=-\frac{\sqrt{2}-1}{2}.
\end{aligned}
$$
Let’s play a little trick now.
$$
\begin{aligned}
I_4 &=\frac{1}{2}\left[A\int \frac{d\left(t-\frac{1}{\sqrt{2} t}\right)}{\left(t-\frac{1}{\sqrt{2} t}\right)^{2}+(\sqrt{2}+1)}+B \int \frac{d\left(t+\frac{1}{\sqrt{2} t}\right)}{\left(t+\frac{1}{\sqrt{2} t}\right)^{2}-(\sqrt{2}-1)}\right]\\
&=\frac{\sqrt{2}+1}{4 \sqrt{\sqrt{2} +1}} \tan \left(\frac{t-\frac{1}{\sqrt{2} t}}{\sqrt{\sqrt{2}+1}}\right)+\frac{\sqrt{2}-1}{8 \sqrt{\sqrt{2}-1}} \ln \left|\frac{t+\frac{1}{\sqrt{2} t}+\sqrt{\sqrt{2}-1}}{t+\frac{1}{\sqrt{2} t}-\sqrt{\sqrt{2}-1}}\right|+C \\
&=\frac{1}{8}\left[2\sqrt{\sqrt{2}+1} \tan ^{-1}\left(\frac{\sqrt{2} \tan ^{2} x-1}{\sqrt{\sqrt{2}+1}}\right)\right.+\sqrt{\sqrt{2}-1} \ln \left|\frac{\sqrt{2} \tan ^{2} x+\sqrt{2 \sqrt{2}-2} \tan x+1}{\sqrt{2} \tan ^{2} x-\sqrt{2 \sqrt{2}-2} \tan x+1}\right|+C
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4397004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the minimum value of $\frac{2a}{a+2b+2c} + \frac{2b}{b+2c+2a} +\frac{2c}{c+2a+2b}$ given $ a+b+c=2$? What is the minimum value of $\frac{2a}{a+2b+2c} + \frac{2b}{b+2c+2a} +\frac{2c}{c+2a+2b}$ given $ a+b+c=2$?
I know the answer is going to be $\frac{6}{5}$ because it will occur when there is equality between the three pronumerals.
The issue is this question was given to a group of students who have only learned the AM/GM inequality for two pronumerals.
I can do the question by using the AM/GM inequality with three pronumerals (along with a bit of not immediately obvious algebraic manipulation) however I'm struggling to see how I can solve this using the AM/GM inequality with only two values.
I've tried making 3 separate equations and adding them together, however so far I haven't been able to find how to do it.
| Here's a "Only 2-variable AM-GM" approach.
(Per the comments: Since the expression is homogenous, I'm working with $a+b+c = 3 $ to simplify the working. There isn't a huge difference with working with $a+b+c = 2$, but the constants get uglier.
Also I'm restricting to positive values, since otherwise there is no minimum like when $a \rightarrow 6^+$.)
Some of the takeaways of using AM-GM (I have a much longer list from my class.)
*
*We can apply it termwise to a sum (likewise product).
*We can cancel a denominator with a numerator, so we might want to force that term in.
*Equality holds when the terms of the AM-GM are equal. If not, try to tweak with constants.
Given those specific ideas, what ways can we bound $ \frac{ 2a} {6-a}$ from below via AM-GM, with equality at $ a = 1$, in the domain $ a \in [ 0, 3 ] $?
Notice that $ \frac{2a}{6-a} = \frac{12}{6-a} - 2. $
Let's apply AM-GM to $ \frac{12}{6-a} + ( 6-a) \times k \geq 2\sqrt{12\times k } $. We want equality at $ a = 1$, which means $ k = \frac{12}{25}$. Thus:
$$ \frac{12 }{6-a} + (6-a) \times \frac{12}{25} \geq \frac{24}{5}$$
As such, this gives us
$$ \frac{2a}{6-a} \geq \frac{24}{5} - (6-a)\times \frac{12}{25} - 2 = \frac{12a}{25} - \frac{2}{25}. $$
Finally, sum it up to get
$$ \sum \frac{2a}{6-a} \geq \sum \left( \frac{12a}{25} - \frac{2}{25} \right) = \frac{ 12\times 3 }{ 25} - 3 \times \frac{2}{25} = \frac{6}{5}.$$
Notes:
*
*A more direct approach is to realize that the inequality $ \frac{ 2a}{6-a} \geq \frac{ 12a}{25} - \frac{2}{25} $ is known as the "Tangent Line Method". It can easily be verified if we knew the inequality beforehand (EG polynomial inequality), but deriving it takes some work (EG Calculus, or the above).
*It's hard to AM-GM $\frac{2a}{6-a}$ directly, since we likewise can't easily manipulate $\frac{6-a}{2a}$, and using $ k\times (6-a)$ leaves us with $ \sqrt{a}$ that we can't manipulate. Hence, the switch to a term where the numerator is constant, which allows us to do$ \frac{12}{6-a} + k \times ( 6 -a ) \geq 2 \sqrt{12 \times k } $.
*For a further challenge, find the minimum subject to $ a^2 + b^2 + c^2 = 3$ using the above ideas.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4400668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\sum_{n=0}^\infty \big( \frac{1}{3n+4}-\frac{3}{3n+2}+\frac{2}{3n+1}\big)$ without power series or integration? In a calculus, I have to find a direct way to compute the sum
$$\sum_{n=0}^\infty \left( \frac{1}{3n+4}-\frac{3}{3n+2}+\frac{2}{3n+1}\right).$$
I tried to reform a partial sum:
$$\sum_{n=0}^N a_n = \frac12+3\sum_{n=1}^N \left(\frac{1}{3n+1}-\frac{1}{3n+2}\right).$$
A classic way then is to introduce the missing terms to use Euler's formula. Problem is, I don't know how to deal with the alternating signs.
What I can't do: use power series (the sum comes from one), neither integration (same problem). I need to find a direct way to sum this, that is, a way which only involves classic techniques with series.
Anyone to help me? Thanks.
| Update: This uses integration, which is something OP sought to avoid.
$$\begin{split}
\sum_{n=0}^\infty \left( \frac{1}{3n+4}-\frac{3}{3n+2}+\frac{2}{3n+1}\right)
&=\sum_{n=0}^\infty \int_0^1x^{3n}(x^3-3x+2)dx\\
&= \int_0^1\frac{x^3-3x+2}{1-x^3}dx\\
&= \int_0^1 \left( \frac 3 {x+x^2+1}-1\right)dx\\
&= \left[ 2{\sqrt 3}\arctan\left(\frac{2x+1}{\sqrt 3}\right) - x\right]_0^1 \\
&= 2{\sqrt 3} \left( \frac \pi 3 - \frac \pi 6\right)-1\\
&= \frac{\pi}{\sqrt 3}-1
\end{split}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4401993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving integral identity via Substitution my task is to prove that $\int_{0}^{\infty} \frac{t^a}{sinh(t)} \frac{x}{x^2+t^2} dt= x^a \int_{0}^{\infty} \frac{t^a}{sinh(xt)} \frac{1}{1+t^2} dt$
so starting with the integral on the right hand side and by substituting u=xt one gets
$x^a \int_0^{\infty} \frac{\frac{u^a}{x^a}}{sinh(u)} \frac{1}{1+ (\frac{u}{x})^2} \frac{1}{x} du$ which in the end is equal to
$\int_{0}^{\infty} \frac{u^a}{sinh(u)} \frac{x}{x^2+u^2} du$.
So my question is: Is this a legitimate proof since we can rename u=t?
| I'm going to assume that $x>0$, so
\begin{align}
x^a \int_{0}^{\infty} \frac{t^a}{\sinh(xt)} \frac{1}{1+t^2} dt&=x^a \lim_{b\to+\infty}\int_{0}^{b} \frac{t^a}{\sinh(xt)} \frac{1}{1+t^2} dt\qquad(\text{doing $u=xt$})\\
&=x^a \lim_{b\to+\infty}\int_{0}^{bx} \frac{\frac{u^a}{x^a}}{\sinh(u)} \frac{1}{1+ (\frac{u}{x})^2} \frac{1}{x} du\\
&=\lim_{b\to+\infty}\int_{0}^{bx} \frac{u^a}{\sinh(u)} \frac{x}{x^2+ u^2}du\\
&=\lim_{b\to+\infty}\int_{0}^{bx} \frac{t^a}{\sinh(t)} \frac{x}{x^2+ t^2}dt\\
&=\int_{0}^{+\infty} \frac{t^a}{\sinh(t)} \frac{x}{x^2+ t^2}dt
\end{align}
In case $x<0$, then we arrive at the following:
\begin{align}
x^a \int_{0}^{\infty} \frac{t^a}{\sinh(xt)} \frac{1}{1+t^2} dt&=\lim_{b\to+\infty}\int_{0}^{bx} \frac{u^a}{\sinh(u)} \frac{x}{x^2+ u^2}du\\
&=\int_{0}^{-\infty} \frac{u^a}{\sinh(u)} \frac{x}{x^2+ u^2}du\qquad(\text{doing $t=-u$})\\
&=\int_{0}^{+\infty} \frac{(-1)^a t^a}{-\sinh(t)} \frac{x}{x^2+ t^2}(-dt)\\
&=(-1)^a\int_{0}^{+\infty} \frac{t^a}{\sinh(t)} \frac{x}{x^2+ t^2}dt
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What technique can be used to prove this inequality? For $a,b\in(0,1)$, prove
$$\sqrt{a^2+b^2}+\sqrt{\frac{1}{4}+\left(1-a\right)^2}+\sqrt{\frac{4}{9}+\left(1-b\right)^2}\ge \frac{\sqrt{181}}{6}$$
where the equality holds for $a=\frac49$ and $b=\frac25$.
Attempted for a while, it seems no standard inequality tricks can be utilized at all in stead of solving the complicated derivative equation group.
$$\frac{d}{da}\left(\sqrt{a^2+b^2}+\sqrt{\frac{1}{4}+\left(1-a\right)^2}+\sqrt{\frac{4}{9}+\left(1-b\right)^2}\right)=\frac{a}{\sqrt{a^2 + b^2}}-\frac{2(1 - a)}{\sqrt{4 a^2 - 8 a + 5}}$$
and
$$\frac{d}{db}\left(\sqrt{a^2+b^2}+\sqrt{\frac{1}{4}+\left(1-a\right)^2}+\sqrt{\frac{4}{9}+\left(1-b\right)^2}\right)=\frac{b}{\sqrt{a^2 + b^2}}-\frac{1 - b}{\sqrt{(1 - b)^2 + \frac49}}.$$
| If you're looking for 'standard inequalities', then check Minkowski's inequality,
$$\sqrt{\color{red}{a^2}+\color{blue}{b^2}} + \sqrt{\color{red}{(1-a)^2}+\color{blue}{\tfrac14}} + \sqrt{\color{red}{\tfrac49}+\color{blue}{(1-b)^2}} \geqslant \sqrt{\color{red}{(a+1-a+\tfrac23)^2} + \color{blue}{\left(b+\tfrac12+1-b \right)^2}} \\= \sqrt{\color{red}{\frac{25}9}+\color{blue}{\frac94}}$$
P.S. Maybe simpler to rephrase this in terms of the triangle inequality, if you can see an appropriate choice of points.
| {
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"url": "https://math.stackexchange.com/questions/4405386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integral with elliptical coordinates Good evening everyone. I'm doing an integral of which I know the result but it comes out different to me.
Anyone able to tell me where am I wrong?
The result is $\frac{64}{27} \sqrt{3} \pi$.
The starting integral was a triple integral. Passing in elliptical coordinates I found:
$x = \frac 23 + \frac 43\rho cos\theta; y=\frac {2\sqrt{3}}{3}\rho sen \theta $
with $\rho \in [0,1), \theta \in [0, 2\pi).$
$\int \int (-3x^2+4x+4-4y^2) dx dy = \int_0^{2\pi} \int_0^1 [-3(\frac 49+ \frac {16}{9} \rho^2cos^2\theta+\frac {16}{9} \rho\cos\theta)+\frac 83 + \frac {16}{3}\rho\cos\theta+4-\frac {16}{3}\rho^2sen^2\theta] d\rho d\theta= \int_0^{2\pi} \int_0^1 [- \frac 43 - \frac {16}{3}\rho^2cos^2\theta -\frac {16}{3}\rho cos\theta + \frac 83 + \frac {16}{3}\rho cos \theta +4 - \frac {16}{3}\rho^2sen^2\theta] d\rho d\theta =\int_0^{2\pi} \int_0^1 [\frac {16}{3}-\frac {16}{3}\rho^2] d\rho d\theta = \frac {64\pi}{9}$
The starting integral is:
$\int_\Omega (2z) dxdydx$
$\Omega = (x,y,z) \in R^3 | 2 \sqrt {x^2+y^2}<z<x+2$
| In short, you are missing the Jacobian of transformation in your integral. If you plug that in, rest of your work is correct.
If the projection of $z = x + 2$, intersected by the paraboloid surface, in xy-plane is $E$,
$x + 2 \geq 2 \sqrt{ (x^2 + y^2)} \implies 3x^2 - 4x + 4 y^2 \le 4$
$ \displaystyle E: ~\left(x - \frac 23\right)^2 + \frac 43 y^2 \leq \frac {16}9$
We use substitution $ \displaystyle x = \frac 23 + \frac 43 \rho \cos\theta~; ~y=\frac {2\sqrt{3}}{3}\rho \sin \theta$
$$0 \leq \rho \leq 1, 0 \leq \theta \leq 2\pi$$
$ \displaystyle |J| = \frac{ 8 \rho}{3 \sqrt3}$
Also note that,
$\begin {aligned}
-3x^2+4x+4-4y^2 &= -3 \left(x^2 - \frac{4x}{3} + \frac{4y^2}{3} - \frac 43\right) \\
&= \frac{16}3 - 3 \left[\left(x - \frac{2}{3}\right)^2 + \frac{4y^2}{3}\right] \\
&= \frac{16}3 (1 - \rho^2) \\
\end {aligned}$
$ \begin {aligned}
\iiint_{\Omega} 2 z ~dV &= \iint_E [(x + 2)^2 - 4 (x^2 + y^2)] ~dx ~ dy \\
&= \iint_E (-3x^2+4x+4-4y^2) ~dx ~ dy \\
&= \int_0^{2\pi} \int_0^1 \frac {16}{3} (1 -\rho^2) ~|J| ~ d\rho ~ d\theta \\
&= \frac{128}{9 \sqrt3} \int_0^{2\pi} \int_0^1 (\rho - \rho^3) ~d\rho ~d\theta = \frac{64 \sqrt 3 \pi}{27} \\
\end {aligned}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Prove that: $ab\sqrt{ab}+bc\sqrt{bc}+ca\sqrt{ca}\le abc+\frac{1}{2}\sqrt[3]{\frac{(a^{2}+bc)^{2}(b^{2}+ca)^{2}(c^{2}+ab)^{2}}{abc}}$
Let $a,b,c>0$. Prove that: $$a b \sqrt{a b}+b c \sqrt{b c}+c a \sqrt{c a} \leqslant a b c+\frac{1}{2} \sqrt[3]{\frac{\left(a^{2}+b c\right)^{2}\left(b^{2}+c a\right)^{2}\left(c^{2}+a b\right)^{2}}{a b c}}$$
I really don't have many ideas in this problem. First I thought of using AM-GM: $a^2+bc\ge 2a\sqrt{bc}$ and so on, or using Holder $(\dfrac{a^3}{bc}+2a+\dfrac{bc}{a})(\dfrac{b^3}{ac}+2b+\dfrac{ca}{b})(\dfrac{c^3}{ab}+2c+\dfrac{ab}{c })$ but all make the inequality into the form $x^3+y^3+z^3\le3xyz$, which is of course incorrect. I thought about using derivatives but they don't work either
Can anyone give me a different way of thinking?
| Using AM-GM, it suffices to prove that
$$ab\frac{a + b}{2} + bc\frac{b + c}{2} + ca\frac{c + a}{2}
\le abc + \frac12\sqrt[3]{\frac{(a^2 + bc)^2(b^2 + ca)^2(c^2 + ab)^2}{abc}}$$
or
$$a^2b + b^2c + c^2a + ab^2 + bc^2 + ca^2 - 2abc
\le \sqrt[3]{\frac{(a^2 + bc)^2(b^2 + ca)^2(c^2 + ab)^2}{abc}}.$$
Since the inequality is homogeneous, WLOG, assume that $abc = 1$.
Let $p = a + b + c, q = ab + bc + ca, r = abc = 1$.
We have $p, q \ge 3$.
The inequality is written as
$$pq - 5r \le \sqrt[3]{\frac{(p^3r + q^3 - 6pqr + 8r^2)^2}{r}}$$
or
$$pq - 5 \le \sqrt[3]{(p^3 + q^3 - 6pq + 8)^2}.$$
Using AM-GM, we have
$$p^3 + q^3 = (p + q)(p^2 - pq + q^2)
\ge (p + q)pq \ge 2\sqrt{pq}\, pq.$$
Using $p, q \ge 3$, we have
$$2\sqrt{pq}\, pq \ge 6pq.$$
It suffices to prove that
$$pq - 5 \le \sqrt[3]{(2\sqrt{pq}\,pq - 6pq + 8)^2}.$$
Letting $x = \sqrt{pq} \ge 3$, the inequality becomes
$$x^2 - 5 \le \sqrt[3]{(2x^3 - 6x^2 + 8)^2}$$
or
$$(x^2 - 5)^3 \le (2x^3 - 6x^2 + 8)^2$$
or
$$(3x^4 - 6x^3 - 12x^2 + 14x + 21)(x - 3)^2 \ge 0$$
which is true.
We are done.
| {
"language": "en",
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"source": "stackexchange",
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How to evaluate $\int_{0}^{1} \sqrt{\frac{x^{m}}{1-x^{n}}} d x$? Stuck by the integral
$$\int_{0}^{1} \sqrt{\frac{x}{1-x^{3}}} d x,$$
I finally solve the the problem using Beta Function.
Then I generalize the result to
$$
I(m,n)=\int_{0}^{1} \sqrt{\frac{x^{m}}{1-x^{n}}} d x
$$
can be tackled by the Beta function by letting $$
y=x^{n} \text {, then } x=y^{\frac{1}{n}} \textrm{ followed by} \quad d x=\frac{1}{n} y^{\frac{1}{n}-1} d y
$$
$$
\begin{aligned}
I &=\int_{0}^{1} y^{\frac{m}{2 n}}(1-y)^{-\frac{1}{2}} \cdot\frac{1}{n} y^{\frac{1}{n}-1} d y \\
&=\frac{1}{n} \int_{0}^{1} y^{\frac{m+2}{2 n}-1}{(1-y)^{\frac{1}{2}-1}} d y \\
&=\frac{1}{n} B\left(\frac{m+2}{2 n} , \frac{1}{2}\right) \\
&=\frac{\Gamma\left(\frac{m+2}{2 n}\right) \Gamma\left(\frac{1}{2}\right)}{n\left(\frac{m+2+n}{2 n}\right)}\\&=\frac{\sqrt{\pi} \Gamma\left(\frac{m+2}{2 n}\right)}{n \Gamma\left(\frac{m+2}{2 n}+\frac{1}{2} \right)}
\end{aligned}
$$
For example, $$
\begin{aligned}
I(1,3) &=\frac{1}{3} B\left(\frac{1}{2}, \frac{1}{2}\right) \\
&=\frac{1}{3} \Gamma^{2}\left(\frac{1}{2}\right) \\
&=\frac{\pi}{3}
\end{aligned}
$$
My Question:
Can we simplify the last answer? Your suggestion or help are highly appreciated.
| $$\int \sqrt{\frac{x^{m}}{1-x^{n}}}\, dx=\frac 2{2+m}x^{\frac{m}{2}+1} \, _2F_1\left(1,\frac{m+n+2}{2 n};\frac{m+2n+2}{2
n};x^n\right)$$
$$\int_0^1 \sqrt{\frac{x^{m}}{1-x^{n}}}\, dx=\frac{\sqrt{\pi }\, \Gamma \left(\frac{m+2}{2 n}\right)}{n\, \Gamma
\left(\frac{m+n+2}{2 n}\right)} \quad \text{if} \quad \Re(n)>0\land \Re(m)>-2$$
Making the problem more general
$$\int\left(\frac{x^m}{1-x^n}\right)^p\,dx=\frac 1{1+mp} x^{mp+1}\, _2F_1\left(p,\frac{m p+1}{n};\frac{m p+n+1}{n};x^n\right)$$
$$\int_0^1 \left(\frac{x^m}{1-x^n}\right)^p\,dx=\frac{\Gamma (1-p)\, \Gamma \left(\frac{m p+1}{n}\right)}{n\,\Gamma
\left(\frac{(m-n)p+n+1}{n}\right)}\quad \text{if} \quad \Re(p)<1\land \Re(m p)>-1\land \Re(n)>0$$
| {
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Application of squeeze theorem to $\displaystyle\lim_{x \rightarrow 0}x^4\cos(\frac{\pi}{x^2})+1$ Use squeeze theorem to evaluate
$$\lim_{x \rightarrow 0}x^4\cos(\frac{\pi}{x^2})+1$$
Can I sandwich it between $-x^4\cos(\frac{\pi}{x^2})+1$ and $x^4\cos(\frac{\pi}{x^2})+1$?
| You can squeeze using $1-x^4$ and $1+x^4$. First, show $1-x^4 \leq 1+x^4cos\left(\frac{\pi}{x^2}\right) \leq 1+x^4$. Then we can use the squeeze theorem to show $\displaystyle\lim_{x\rightarrow 0} 1+x^4cos\left(\frac{\pi}{x^2}\right) = 1$ as $\displaystyle\lim_{x\rightarrow 0} 1 + x^4 = 1 = \displaystyle\lim_{x\rightarrow 0} 1 - x^4$.
| {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Examples where $(a+b+\cdots)^2 = (a^2+b^2+\cdots)$ Consider the two infinite series
$$
\frac{\pi}{\sqrt{8}} = 1 + \frac{1}{3} - \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - \cdots
$$
and
$$
\frac{\pi^2}{8} = 1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \frac{1}{9^2} + \frac{1}{11^2} + \cdots
$$
(Notice that the first series has signs that go two-by-two rather than every-other.)
Squaring the first equality also gives $\pi^2/8$ and so these two, when put together, satisfy the 'highschooler's dream' for squaring a sum: just square each term and sum,
$$
(a + b + c + \cdots)^2 = (a^2 + b^2 + c^2 + \cdots)
$$
with nothing like $2ab + 2ac + 2bc + \cdots$ needed.
A trivial example of this would be
$$
(a + 0)^2 = a^2 + 2a0 + 0^2 = a^2 + 0^2
$$
but it only succeeds because one addend is zero.
My questions are
*
*Are there any other simple nontrivial examples? I believe any other nontrivial example must be an infinite sum. edit: John Omielan provides the simple finite example $(1+1-\frac{1}{2})^2 = 1^2 + 1^2 + \frac{1}{2^2}$.
*Is there an "obvious" demonstration that the above sum (other than the direct evaluation) satisfies the highschooler's dream? Put another way, is there a simple demonstration that the infinite sum of "cross terms" vanishes?
| There are arbitrary length examples with real numbers. Moreover, you can take almost arbitrary first $k-1$ numbers $a_1$, $a_2$, ... , $a_{k-1}$ and always exists $a_k$ such that $$(a_1+a_2+...+a_{k-1}+a_k)^2=a_1^2+a_2^2+...+a_{k-1}^2+a_k^2$$
Let's prove it: Mark partial sums as $a_1 +...+a_{k-1}=A \neq 0$ and $a_1^2+a_2^2+...+a_{k-1}^2=B$. Then $a_k$ must satisfy $$(A+a_k)^2=B+a_k^2$$
$$A^2+2Aa_k+a_k^2=B+a_k^2 \Rightarrow 2Aa_k=B-A^2 \Rightarrow a_k=\frac{B-A^2}{2A}$$
The only limitation put on $a_1$, ..., $a_{k-1}$ for existence of $a_k$ is that $A\neq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4411628",
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"source": "stackexchange",
"question_score": "27",
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} |
$\frac{2-a}{a+\sqrt{bc+abc}}+\frac{2-b}{b+\sqrt{ca+abc}}+\frac{2-c}{c+\sqrt{ab+abc}}\ge1$
For $a,b,c\ge0: ab+bc+ca+abc=4$ then: $$\frac{2-a}{a+\sqrt{bc+abc}}+\frac{2-b}{b+\sqrt{ca+abc}}+\frac{2-c}{c+\sqrt{ab+abc}}\ge1$$
I used the condition and get: $a+\sqrt{bc+abc}=a+\sqrt{4-a(b+c)}\le a+2$
So we need to prove that: $$\frac{2-a}{a+2}+\frac{2-b}{b+2}+\frac{2-c}{c+2}\ge1$$
I tried to full expand but the rest seems complicated for me.
Can anyone help me full my idea? Every thinking is welcomed, thanks!
| Let $x = a + 2, \; y = b + 2, \; z = c + 2$
Hence, we need to prove that
$$\frac{4-x}{x} + \frac{4-y}{y} + \frac{4-z}{z} \geq 1$$
$$\implies \frac{4}{x} - \frac{x}{x} + \frac{4}{y} - \frac{y}{y} + \frac{4}{z} - \frac{z}{z} \geq 1$$
$$\implies \frac{4}{x} + \frac{4}{y} + \frac{4}{z} - 3 \geq 1$$
$$\implies \frac{4}{x} + \frac{4}{y} + \frac{4}{z} \geq 4$$
$$\implies \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq 1$$
$$\implies \frac{1}{a+2} + \frac{1}{b+2} + \frac{1}{c+2} \geq 1$$
$$\implies \frac{4(a + b + c) + ab + bc + ca + 12}{4(a + b + c) + 2(ab + bc + ca) + abc + 8} \geq 1$$
$$\implies 4(a + b + c) + ab + bc + ca + 12 \geq 4(a + b + c) + 2(ab + bc + ca) + abc + 8$$
$$\implies ab + bc + ca +abc -4 \leq 0$$
We know that $ab + bc + ca + abc = 4 \implies ab + bc + ca +abc - 4
= 4 - 4 = 0$.
| {
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} |
$\text{Prove}\left( 1-\frac{1}{\sqrt{2}} \right)...\left( 1-\frac{1}{\sqrt{n}} \right)\lt \frac{2}{n^{2}}\text{ for }n\ge 2$ I'm completely lost on this question:
$$\text{Prove}\left( 1-\frac{1}{\sqrt{2}} \right)\left( 1-\frac{1}{\sqrt{3}} \right)...\left( 1-\frac{1}{\sqrt{n}} \right)\lt \frac{2}{n^{2}} \text{ for }n\ge 2$$
alternatively,
$$\prod_{a=2}^{n} {\left(1-\frac{1}{\sqrt{a}}\right)} \lt \frac{2}{n^2} \text{ for } n\geq2$$
I tried to do it by induction but ended up with $\frac{2}{k^2}(1-\frac{1}{\sqrt{k+1}})$ in the RHS which I have no idea what to do with. I also tried to do it with AM/GM inequality but also had little luck. The question did not specifically specify what type of proof to use so I'm assuming you can use any type of proof. The first that came to mind is induction obviously, but I'm really lost on this one. Any help or hints would be appreciated!
With induction, the thing that I'll need to prove is:
$$\frac{2}{k^2}(1-\frac{1}{\sqrt{k+1}}) \lt \frac{2}{(1+k)^{2}}$$
But I'm not sure how to approach it.
SOLVED
$$\text{Prove}\left( 1-\frac{1}{\sqrt{2}} \right)\left( 1-\frac{1}{\sqrt{3}} \right)...\left( 1-\frac{1}{\sqrt{n}} \right)\lt \frac{2}{n^{2}} \text{ for }n\ge 2$$
Prove by induction:
A. When $n=2$,
$$\begin{aligned}LHS &= 1-\frac{1}{\sqrt{2}}\newline
&\approx 0.293\newline
RHS&=\frac{2}{2^2}\newline
&=0.5\newline
&\gt LHS\newline
\therefore \text{true for }n&=2\end{aligned}$$
B. Assume the statement is true for the positive integer $n=k$, where $k\in\mathbb{N}$.
That is, assume that
$$\left( 1-\frac{1}{\sqrt{2}} \right)\left( 1-\frac{1}{\sqrt{3}} \right)...\left( 1-\frac{1}{\sqrt{k}} \right)\lt \frac{2}{k^{2}}$$
Now prove the statement for $n=k+1$. That is, prove that
$$\left( 1-\frac{1}{\sqrt{2}} \right)\left( 1-\frac{1}{\sqrt{3}} \right)...\left( 1-\frac{1}{\sqrt{k+1}} \right)\lt \frac{2}{(k+1)^{2}}$$
By the induction hypothesis:
$$\left( 1-\frac{1}{\sqrt{2}} \right)\left( 1-\frac{1}{\sqrt{3}} \right)...\left( 1-\frac{1}{\sqrt{k}} \right)\left( 1-\frac{1}{\sqrt{k+1}} \right)\lt \frac{2}{k^{2}}\left( 1-\frac{1}{\sqrt{k+1}} \right)$$
Hence, we are required to prove for $k\geq2$ that $$\frac{2}{k^2}\Big(1-\frac{1}{\sqrt{k+1}}\Big)\leq\frac{2}{(k+1)^2}.$$
For $k\geq2,$ \begin{aligned}&\text{LHS-RHS}\\
=&\frac{2}{k^2}\Big(1-\frac{1}{\sqrt{k+1}}\Big)-\frac{2}{(k+1)^2}\\
=&\frac{2\sqrt{k+1}}{1+\sqrt{k+1}}\left(\frac{1}{k^2}\left(1-\frac{1}{\sqrt{k+1}}\right)\left(1+\frac{1}{\sqrt{k+1}}\right)-\frac{1}{(k+1)^2}\left(1+\frac{1}{\sqrt{k+1}}\right)\right)\\
=&\cdots\\ =&\frac{2\sqrt{k+1}}{1+\sqrt{k+1}}\left(\frac{1}{k}-\frac{1}{\sqrt{k+1}}\right)\\
=&\frac{2(\sqrt{k+1}-k)}{k(1+\sqrt{k+1})} \\ \lt&0,\end{aligned} as required.
$\therefore$ It follows from parts A and B by mathematical induction that the result is true for all integers $n\gt2$
| I've wanted to give U a little help with the thinking here.
we have the base case: $1-\frac{1}{\sqrt{2}}<\frac{2}{2^{2}} \longrightarrow -\frac{1}{\sqrt{2}}\lt -\frac{1}{2} \\\\\text{and that's true because}\longrightarrow \sqrt{2}\lt 2$
let there be n such that: $\left( 1-\frac{1}{\sqrt{2}} \right)...\left( 1-\frac{1}{\sqrt{n}} \right)\lt \frac{2}{n^{2}}\text{ for }n\ge 2$
now we just need to prove the following: $\left( 1-\frac{1}{\sqrt{2}} \right)...\left( 1-\frac{1}{\sqrt{n+1}} \right)\lt \frac{2}{(n+1)^{2}}$
we know that: $\left( 1-\frac{1}{\sqrt{2}} \right)...\left( 1-\frac{1}{\sqrt{n}} \right)\lt \frac{2}{n^{2}}$
so here when we think about the question we end up in a "problem", how to continue.
We have: $\frac{2}{n^2}(1-\frac{1}{\sqrt{n+1}})\leq\frac{2}{(n+1)^2}$
from here what we need to do is the following thing: move $\frac{1}{\sqrt{n+1}}$ to the other side and then group by similarities. from here you'll get the answer. I hope I helped :)
| {
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"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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A problem related to coefficient of $\int{(t-a)^n}dt$ Let's make some antiderivative of $(t-a)^n$.
When $n=1$, we get $\frac{1}{2}t^2-at$ for the antiderivative (ignoring some constant). Now when we put $t=a$, we get $-\frac{1}{2}a^2$ and $-\frac{1}{2}$ is the coefficient.
When $n=3$, we get $\frac{1}{4}t^4-at^3+\frac{3}{2}a^2t^2-a^4t$ for the antiderivative and we get $-\frac{1}{4}$ for the coefficient by putting $t=a$.
When $n=5$, in the same manner, we get $-\frac{1}{6}$ for the coefficient.
The case when exponent $n$ is even, it is straightforward to know that the coefficient always comes out as $\frac{1}{n+1}$. However I'm stuck with finding a proof for when $n$ is odd.
Any help would be grateful, thanks.
Edit: I've already gotten a great answer however my original purpose of posing this question was to approach this question in regard of the pattern of binomial coefficients. If there is someone who knows some proofs that looks somewhat more 'discrete', I would be grateful to know any of them. Thanks.
Edit: My question can be substituted as follows.
Let's make an array of coefficients of $(a-b)^n$ as Pascal's Triangle.
Then we can see the following.
\begin{array}{c}
+1 \\
+1 \quad -1 \\
+1 \quad -2 \quad +1 \\
+1 \quad -3 \quad +3 \quad -1 \\
+1 \quad -4 \quad +6 \quad -4 \quad +1 \\
+1 \quad -5 \quad +10 \quad -10 \quad +5 \quad -1 \\
+1 \quad -6 \quad +15 \quad -20 \quad +15 \quad -6 \quad +1 \\
+1 \quad -7 \quad +21 \quad -35 \quad +35 \quad -21 \quad +7 \quad -1 \\
\end{array}
Now, let's multiply each coefficient by $\frac{1}{n+1}$, $\frac{1}{n}$, $\frac{1}{n-1}$, $\frac{1}{n-2}$ $\cdots$, $\frac{1}{1}$ as the following.
\begin{array}{c}
\frac{+1}{1} \\
\frac{+1}{2} \quad \frac{-1}{1} \\
\frac{+1}{3} \quad \frac{-2}{2} \quad \frac{1}{1} \\
\frac{+1}{4} \quad \frac{-3}{3} \quad \frac{+3}{2} \quad \frac{-1}{1} \\
\frac{+1}{5} \quad \frac{-4}{4} \quad \frac{+6}{3} \quad \frac{-4}{2} \quad \frac{1}{1} \\
\frac{+1}{6} \quad \frac{-5}{5} \quad \frac{+10}{4} \quad \frac{-10}{3} \quad \frac{+5}{2} \quad \frac{-1}{1} \\
\frac{+1}{7} \quad \frac{-6}{6} \quad \frac{+15}{5} \quad \frac{-20}{4} \quad \frac{+15}{3} \quad \frac{-6}{2} \quad \frac{1}{1} \\
\frac{+1}{8} \quad \frac{-7}{7} \quad \frac{+21}{6} \quad \frac{-35}{5} \quad \frac{+35}{4} \quad \frac{-21}{3} \quad \frac{+7}{2} \quad \frac{-1}{1} \\
\end{array}
When $n$ is even, then it is straightforward that the sum of each floor equals to $\frac{1}{n+1}$.
However, when $n$ is odd, the numbers get quite messed up. So I'm stuck with deducing some 'discrete' proof that it equals to $-\frac{1}{n+1}$.
If someone knows some other approach than using the fundamental theorem of calculus it would be grateful if I could know them. Thanks.
| Setting the constant equal to $0$ and taking $t=a$ means you are computing :
$$I_n = \int_0^a (t-a)^n\text dt$$
A change of variable $x = t-a$ tells us that :
$$I_n = \int_0^a(-x)^n\text dx$$
And we can compute :
$$I_n = \frac{(-1)^n a^{n+1}}{n+1}$$
In particular when $n$ is odd, you have :
$$I_n = -\frac{a^{n+1}}{n+1}$$
Another proof
\begin{align}
I_n &= \int_0^a (t-a)^n \text dt \\
&= \sum_{k=0}^n {n \choose k}\int_0^a t^{k} (-a)^{n-k}\text dt \\
&= a^{n+1}\sum_{k=0}^n {n\choose k} \frac{(-1)^{n-k}}{k+1} \\
\end{align}
Then, we use :
$$\frac{1}{k+1}{n\choose k} = \frac{1}{n+1}{n+1\choose k+1}$$
Therefore :
\begin{align}
I_n &= \frac{a^{n+1}}{n+1} \sum_{k=0}^{n} {n+1 \choose k+1} (-1)^{n-k} \\
&=\frac{a^{n+1}}{n+1} \sum_{k=1}^{n+1} {n+1 \choose k} (-1)^{n+1-k}\\
&= \frac{a^{n+1}}{n+1}\left( \sum_{k=0}^{n+1} {n+1 \choose k} (-1)^{n+1-k} - (-1)^{n+1} \right) \\
&= \frac{a^{n+1}}{n+1}\left(- (-1)^{n+1} \right) \\
&= \frac{(-1)^n a^{n+1}}{n+1}
\end{align}
| {
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Why can't a recurring pattern be found by using the second term in sine graph? In the question
$\sin(x) = \sin (4x), 0 < x < \pi$
I can use the equation
$4x = \pi - x$ ,
$4x = 2\pi + x$ ,
$4x = 3\pi - x$ ,
to work out the solutions, which are $\pi/5, 2\pi/3, 3\pi/5$.
But $\sin(x) = \sin(\pi - x)$.
I am thinking that multiple of the answer of second quadrant should also work.
So, if I try
$4(\pi - x) = 2\pi + x$ , it gives $2\pi/5$, and
$\sin\left(\dfrac{2\pi}{5}\right) = 0.951$
$\sin\left(4 \cdot \dfrac{2\pi}{5}\right) = -0.951$
No good.
If I try
$4(\pi - x) = 3\pi - x$ , it gives $\pi/4$, and
$\sin\left(\dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt{2}}$
$\sin\left(4 \cdot \frac{\pi}{4}\right) = 0$
Also, no good.
Can someone explain why does it not work please?
| The sine of an angle in standard position (vertex at the origin, initial side along the positive $x$-axis) is the $y$-coordinate of the point where the terminal side of the angle intersects the unit circle.
Clearly, $\sin\theta = \sin\varphi$ if $\theta = \varphi$. By symmetry, $\sin\theta = \sin\varphi$ if $\theta = \pi - \varphi$. Also, $\sin\theta = \sin\varphi$ if $\theta$ is coterminal with $\varphi$ or $\pi - \varphi$. Hence, $\sin\theta = \sin\varphi$ if
$$\theta = \varphi + 2k\pi, k \in \mathbb{Z}$$
or
$$\theta = \varphi + 2m\pi, m \in \mathbb{Z}$$
At the risk of obscuring the symmetry argument above, $\sin\theta = \sin\varphi$ if
$$\theta = (-1)^n\varphi + n\pi, n \in \mathbb{Z}$$
You wish to solve the equation $\sin x = \sin(4x)$ in the interval $(0, \pi)$. Let's apply the above formulas.
\begin{align*}
4x & = x + 2k\pi, k \in \mathbb{Z} & 4x & = \pi - x + 2m\pi, m \in \mathbb{Z}\\
3x & = 2k\pi, k \in \mathbb{Z} & 5x & = \pi + 2m\pi, m \in \mathbb{Z}\\
x & = \frac{2k\pi}{3}, k \in \mathbb{Z} & x & = \frac{\pi}{5} + \frac{2m\pi}{5}, m \in \mathbb{Z}
\end{align*}
which is the general solution of the equation $\sin(4x) = \sin x$.
In order to obtain values in the interval $(0, \pi)$, we must take $k = 1$ and can take $m = 0, 1$. That yields the solutions
\begin{align*}
x & = \frac{2\pi}{3} & x & = \frac{\pi}{5}, \frac{3\pi}{5}
\end{align*}
which you found by applying the formula
$$\theta = (-1)^n\varphi + n\pi, n \in \mathbb{Z}$$
for particular values of $n$.
Let's apply the above formulas again, with the roles of $x$ and $4x$ reversed.
\begin{align*}
x & = 4x + 2j\pi, j \in \mathbb{Z} & x & = \pi - 4x + 2\ell \pi, \ell \in \mathbb{Z}\\
-3x & = 2j\pi, j \in \mathbb{Z} & 5x & = \pi + 2\ell \pi, \ell \in \mathbb{Z}\\
x & = -\frac{2j}{3}, j \in \mathbb{Z} & x & = \frac{\pi}{5} + \frac{2\ell\pi}{3}, \ell \in \mathbb{Z}
\end{align*}
If we set $k = -j$ and $m = \ell$, we obtain
\begin{align*}
x & = \frac{2k}{3}, k \in \mathbb{Z} & x & = \frac{\pi}{5} + \frac{2m\pi}{5}, m \in \mathbb{Z}
\end{align*}
which is the same general solution as we obtained above. To find the particular solutions in the interval $(0, \pi)$, we proceed as above.
Notice I obtained the same solutions by reversing the roles of $x$ and $4x$, not $x$ and $\pi - x$. While $\sin x = \sin(\pi - x)$, it does not follow that $\sin(4x) = \sin[4(\pi - x)] = \sin(4\pi - 4x)$, as your examples illustrate. Observe that
$$\sin(4\pi - 4x) = \sin(4\pi)\cos(4x) - \cos(4\pi)\sin(4x) = -\sin(4x)$$
Therefore, by replacing $\sin(4x)$ with $\sin[4(\pi - x)]$, you replaced $\sin(4x)$ by $-\sin(4x)$, which is why you obtained
$$\sin\left(4 \cdot \frac{2\pi}{5}\right) = -\sin\left(\frac{2\pi}{5}\right)$$
You made an error when you solved the equation
$$4(\pi - x) = 3\pi - x$$
You should have obtained $x = \dfrac{\pi}{3}$. Notice that
$$\sin\left(4 \cdot \frac{\pi}{3}\right) = -\sin\left(\frac{\pi}{3}\right)$$
which is also attributable to replacing $\sin(4x)$ by $\sin[4(\pi - x)] = -\sin(4x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4419956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is $d(\mathbb{Q}(\sqrt{2},\sqrt{3}))$? $K=\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2}+\sqrt{3})$ and $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}]=[\mathbb{Q}(\sqrt{2}+\sqrt{3}):\mathbb{Q}]=4$. We also know the conjugates of $\sqrt{2}+\sqrt{3}$ are,
$$x_1=\sqrt{2}+\sqrt{3}$$
$$x_2=\sqrt{2}-\sqrt{3}$$
$$x_3=-\sqrt{2}+\sqrt{3}$$
$$x_4=-\sqrt{2}-\sqrt{3}$$
$\mathbb{Q}(\sqrt{2}+\sqrt{3})=\{a+b(\sqrt{2}+\sqrt{3})+c(\sqrt{2}+\sqrt{3})^2+d(\sqrt{2}+\sqrt{3})^3|a,b,c,d \in \mathbb{Q}\}$. Thus an integral basis of $K$ is,
$$S=\{1,\sqrt{2}+\sqrt{3},(\sqrt{2}+\sqrt{3})^2,(\sqrt{2}+\sqrt{3})^3\}$$
$$S=\{1,\sqrt{2}+\sqrt{3},5+2\sqrt{6},(\sqrt{2}+\sqrt{3})^3\}$$
and hence,
$$d(K)=d(\mathbb{Q}(\sqrt{2}+\sqrt{3}))=D(S)$$
$$D(S)=\begin{vmatrix}
1 & \sqrt{2}+\sqrt{3} & (\sqrt{2}+\sqrt{3})^2 & (\sqrt{2}+\sqrt{3})^3\\
1 & \sqrt{2}-\sqrt{3} & (\sqrt{2}-\sqrt{3})^2 & (\sqrt{2}-\sqrt{3})^3\\
1 & -\sqrt{2}+\sqrt{3} & (-\sqrt{2}+\sqrt{3})^2 & (-\sqrt{2}+\sqrt{3})^3\\
1 & -\sqrt{2}-\sqrt{3} & (-\sqrt{2}-\sqrt{3})^2 & (-\sqrt{2}-\sqrt{3})^3
\end{vmatrix}^2$$
$$D(S)=\begin{vmatrix}
1 & \sqrt{2}+\sqrt{3} & 5+2\sqrt{6} & 11\sqrt{2}+9\sqrt{3}\\
1 & \sqrt{2}-\sqrt{3} & 5-2\sqrt{6} & 11\sqrt{2}-9\sqrt{3}\\
1 & -\sqrt{2}+\sqrt{3} & 5-2\sqrt{6} & -11\sqrt{2}+9\sqrt{3}\\
1 & -\sqrt{2}-\sqrt{3} & 5+2\sqrt{6} & -11\sqrt{2}-9\sqrt{3}
\end{vmatrix}^2$$
Using Mathematica,
$$D(S)=2677248 + 1078272\sqrt{6}$$
Why am I not getting an integer answer?
Appararently I don't know how to type, $D(S)= 147456$
| There are two issues here: First, you seem to have made a data entry error, which resulted in a non-integer answer. Second, $\mathbb{Z}[\sqrt{2} + \sqrt{3}]$ isn't integrally closed, so you're computing the discriminant of some non-maximal order of $\mathbb{Q}(\sqrt{2} + \sqrt{3})$.
The following SageMath code verifies that $\mathbb{Z}[\sqrt{2} + \sqrt{3}]$ in fact has index $8$ in the maximal order:
sage: K.<b> = NumberField((sqrt(2) + sqrt(3)).minpoly())
sage: R = K.order(b)
sage: OK = K.maximal_order()
sage: R.index_in(OK)
8
Thus, the discriminant of this order is $8^2$ times the discriminant of the number field. (For reference, this is number field 4.4.2304.1 in the LMFDB.)
| {
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"url": "https://math.stackexchange.com/questions/4420178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Number of ordered triples $(a,b,c)$ such that $a+b+c=2019$ **How many ordered triples $(a, b, c)$ are there satisfy the following criteria **
*
*$a,b,c$ are positive numbers
*$a+b+c=2019$
*a, b, c form increasing arithmetic progression
*a , b and c are suitable to be sides of triangle
My idea includes application of stars and bars theorem to solve
$ a+b+c=2019$ where a, b, c are positive numbers
and the result will be
$1+a'+1+b'+1+c' = 2019$
$a'+b'+c'=2016$
By stars and bars
Number of triples will be
$2018 C 2$
but i didn't know how can i involve the other conditions to exclude the unappropriate cases
| $\color{blue}{1-)}$ $a ,b,c$ form increasing arithmetic progression , so $a<b<c$
Let say $b=a+x$ , and $c=b+x$ where $x$ is positive integer
*
*$c=a+x+x$
*$b=a+x$
$a+b+c =3a+3x$ where $a,x$ all are positive integers
$\color{blue}{2-)}$ $a , b$ and $c$ are suitable to be sides of triangle , then
*
*$a+b > c \rightarrow 2a+x > a+2x \rightarrow a>x$
*$a+c >b \rightarrow 2a+2x >a+x \rightarrow a >-x$
*$b+c >a \rightarrow 2a+3x > a \rightarrow a>-3x$
Lets say $a=x+k$ where $k$ is positive integer
$a+b+c =6x+3k=2019$ where $k,x$ all are positive integers. Then i lets use generating functions such that
*
*$3k=3,6,9,12,15,...=\frac{x^3}{1-x^3}$
*$6x=6,12,18,24,30,...=\frac{x^6}{1-x^6}$
Now , find $$[x^{2019}]\bigg(\frac{x^3}{1-x^3}\bigg)\bigg(\frac{x^6}{1-x^6}\bigg)$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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If $-1 \le x \le 1$, what is the maximum value of $x+\sqrt{1-x^2}$? (Cannot use calculus method) If $-1 \leq x \leq 1$, what is the maximum value of $x+\sqrt{1-x^2}$? (Cannot use calculus method)
As stated in the problem, I can't use calculus. Therefore, I'm using things I've learnt so far instead:
One of the things I have tried most successfully is using trig substitutions...
For example, if I substitute $x = \sin \phi$, this yields $\sin \phi + \cos \phi$
But what should I do next? Or there are any methods else to solve the problem?
| We have
$$(x + \sqrt{1 - x^2})^2
+ (x - \sqrt{1 - x^2})^2
= 2x^2 + 2(1 - x^2) = 2 \tag{1}$$
which results in
$$x + \sqrt{1 - x^2} \le \sqrt 2.$$
(Note: Alternatively, we my use $(a + b)^2 \le 2(a^2 + b^2)$ to get
$(x + \sqrt{1 - x^2})^2 \le 2[x^2 + (1 - x^2)] = 2$.)
Also, when $x = 1/\sqrt 2$, we have $x + \sqrt{1 - x^2} = \sqrt 2$.
(Note: From (1), letting $(x - \sqrt{1 - x^2})^2 = 0$, we get $x = \sqrt{1 - x^2}$ or $x = 1/\sqrt 2$.)
Thus, the maximum of $x + \sqrt{1 - x^2}$ on $[-1,1]$ is $\sqrt 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4428309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to tackle $\int_{0}^{\frac{\pi}{2}}y \ln (1+\cos y)\,d y$? I recently encounter an integral problem consisting the integral
$$
I:=\int_{0}^{\frac{\pi}{2}} y \ln (1+\cos y) d y,
$$
I tried to tackle $I$ using the double angle formula and the result of $$
\int_{0}^{\frac{\pi}{4}} y\ln (\cos y) d y
$$
\begin{aligned}
I &=\int_{0}^{\frac{\pi}{2}} y \ln \left(2 \cos ^{2} \frac{y}{2}\right) d y \\
&=\ln 2 \int_{0}^{\frac{\pi}{2}} y d y+2 \int_{0}^{\frac{\pi}{2}} y \ln \left(\cos \frac{y}{2}\right) d y \\
&=\frac{\pi^{2}}{8} \ln 2+8 \int_{0}^{\frac{\pi}{4}} y \ln (\cos y) d y
\end{aligned}
By my post,$$\int_{0}^{\frac{\pi}{4}} y\ln (\cos y) d y = \frac{\pi G}{8}-\frac{\pi^{2}}{32} \ln 2-\frac{21}{128} \zeta(3) $$
Now we can conclude that $$
\boxed{\int_{0}^{\frac{\pi}{2}} y \ln (1+\cos y) d y = \pi G-\frac{21}{16} \zeta(3)-\frac{\pi^{2}}{8} \ln 2}
$$
Suggestions for improvement and alternative methods are warmly welcome!
| Another solution, obtained by one integration by parts is
$$\int y \log (1+\cos (y))\,dy=$$ $$\frac{1}{2} y^2 \log (1+\cos (y))+2 i y \text{Li}_2\left(-e^{i y}\right)-2 \text{Li}_3\left(-e^{i y}\right)+\frac{i
y^3}{6}-y^2 \log \left(1+e^{i y}\right)$$ giving for the definte integral
$$\int_0^{\frac \pi 2} y \log (1+\cos (y))\,dy=\pi C-2 \text{Li}_3(-i)-\frac{1}{4} \pi ^2 \log (1+i)-\frac{3 }{2} \zeta (3)$$
$$\int_0^{\frac \pi 2} y \log (1+\cos (y))\,dy=\pi C-\frac{21 }{16}\zeta (3)-\frac{1}{8} \pi ^2 \log (2)$$
Just for the fun
Using the $1,400$ years old approximation
$$\cos(y) \simeq\frac{\pi ^2-4y^2}{\pi ^2+y^2}\qquad \text{for} \qquad -\frac \pi 2 \leq y\leq\frac \pi 2$$ we should have
$$\int_0^{\frac \pi 2} y \log (1+\cos (y))\,dy\sim \pi ^2 \log \left(\frac{4}{5^{5/6}}\right) $$ (relative error equal to $0.07$%).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4428744",
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"source": "stackexchange",
"question_score": "2",
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Relation between roots and coefficients of equation If the roots of the equation $x^4 - x^3 +2x^2+x+1 = 0 $ are given by $a,b,c,d$ then find the value of $(1+a^3)(1+b^3)(1+c^3)(1+d^3)$
I found out that:
$$ (1+a^3)(1+b^3)(1+c^3)(1+d^3) = (abcd)^3+\sum(abc)^3 +\sum(ab)^3+\sum(a)^3 +1$$
But how do I calculate $\sum(abc)^3$ and $\sum(ab)^3$?
| The expression is the product of the roots of the quartic equation satisfied by $\,z = x^3 + 1\,$. Using polynomial resultants, the equation in $\,z\,$ is $\,\text{res}(x^4 - x^3 + 2 x^2 + x + 1, z - x^3 - 1, x)=0\,$ which results in $\,z^4 + 4 z^3 - 4 z^2 - 16 z + 16 = 0\,$, so the expression evaluates to $\,16\,$.
If not familiar with polynomial resultants, the same result can be derived as follows.
*
*Let $\,y = x^3\,$, then $\,x^4 = xy\,$ and $\,x^4 - x^3 + 2 x^2 + x + 1 = 0\,$ can be written as:
$$
2 x^2 + (y+1) x - z + 1= 0 \tag{1}
$$
Multiplying $\,(1)\,$ by $\,x\,$ and replacing $\,x^3 = y, \,x^4 = xy\,$ again:
$$
(y+1) x^2 - (y -1) x + 2 y = 0 \tag{2}
$$
Repeating one more time:
$$
(y-1) x^2 - 2 xy - y^2 - y = 0 \tag{3}
$$
Considering $\,(1), (2), (3)\,$ as a linear homogeneous system in $\,x^2, x, 1\,$, non-trivial solutions exist, so the determinant must be zero.
$$
0 =
\begin{vmatrix}
2 & y+1 & -y+1
\\ y+1 & -y+1 & 2 y
\\ y-1 & -2y & -y^2-y
\end{vmatrix}
= y^4 + 8 y^3 + 14 y^2 - 8 y + 1 \tag{4}
$$
*
*Let $\,z = y+1\,$, then substituting $\,y = z-1\,$ in $\,(4)\,$:
$$
\begin{align}
0 &= (z-1)^4 + 8 (z-1)^3 + 14 (z-1)^2 - 8 (z-1) + 1
\\ &= z^4 + 4 z^3 - 4 z^2 - 16 z + 16 \tag{5}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4432146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Generating Function Approach giving wrong combination count while normal brute force casework approach giving correct answer I am solving a problem: How many words are less than four letters long and contain only the letters A, B, C, D, and E? Here, 'word' refers to any string of letters.
My Solution 1 uses Generating functions gives wrong answer
$(1+x+x^2+x^3)(1+x+x^2+x^3)(1+x+x^2+x^3)(1+x+x^2+x^3)(1+x+x^2+x^3)$
where $(1+x+x^2+x^3)$ is Generating function for each letter. This approach gave a wrong answer $(35+15+5+1)$ which is the sum of coefficients of $(x^3, x^2, x, 1)$ in $$x^{15} + 5 x^{14} + 15 x^{13} + 35 x^{12} + 65 x^{11} + 101 x^{10} + 135 x^9 + 155 x^8 + 155 x^7 + 135 x^6$$ $$+ 101 x^5 + 65 x^4 + 35 x^3 + 15 x^2 + 5 x + 1$$
Correct Approach using Case Work gives the correct answer
Case 1: The word is one letter long. Clearly, there are $5$ of these words.
Case 2: The word is two letters long. Constructing the set of these words, there are $5$ options for the first letter and $5$ options for the second letter, so there are $5^2 = 25$ of these words.
Case 3: The word is three letters long. By similar logic as above, we have $5$ options for the first letter, $5$ options for the second, and $5$ options for the third. Then there are $5^3 = 125$ of these letters.
Adding all our cases up, there are $5 + 25 + 125 = 155$ words that are less than four letters long and contain only the letters A, B, C, D, and E.
Could Someone help me in explaining why the Generating function approach failed here?
| You will have to consider whether the e.g.f. (exponential generating function) approach is worth while here for such a short problem. However, the approach would stand you in good stead for more complex problems.
You will have to extract from the expansion of $(1+x+x^2/2!+x^3/3!)^5$,
(coefficient of $x$) + $2!$(coefficient of $x^2$) + $3!$(coefficient of $x^3$)
The factorials in the denominator of the exponential generating function effectively compensate the permutations for a letter occurring multiple times
| {
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"url": "https://math.stackexchange.com/questions/4434822",
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"question_score": "2",
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$A^2 - B^2$ is an invertible matrix and $A^5 = B^5$ and $A^3B^2 = A^2B^3$ . Then what is the determinant of the matrix $A^3 + B^3$? $A,B$ are two $3 \times 3$ real-matrices with following three properties.
*
*$A^2 - B^2$ is an invertible matrix
*$A^5 = B^5$
*$A^3B^2 = A^2B^3$
Then what is the determinant of the matrix $A^3 + B^3$ ?
| $A$ and $B$ are real $3\times3$ matrices, so their eigenvalue polynomials are cubic and hence have at least one real zero. Let $\lambda\in\Bbb{R}$ be an eigenvalue of $B$, and $x\in\Bbb{R}^3$ be the corresponding eigenvector. Then
$$
A^5x=B^5x=\lambda^5x.\tag{1}
$$
We also have
$$
\lambda^3A^2x=(A^2B^3)x=A^3B^2x=\lambda^2A^3x.\tag{2}
$$
Apply $A^3$ to both sides of $(2)$, and use $(1)$ to arrive at
$$
\lambda^8x=A^3(\lambda^3A^2x)=A^3(\lambda^2A^3x)=\lambda^2A^6x=\lambda^7Ax.\tag{3}
$$
If $\lambda\neq0$ then $(3)$ implies that $Ax=\lambda x$ also. This violates the first assumption as then
$$(A^2-B^2)x=\lambda^2x-\lambda^2x=0.$$
We are left with the case $\lambda=0$. Again, by $(1)$ we see that $A^5x=0$. Because $A$ is a $3\times3$ matrix this implies (think Jordan canonical form) that already $A^3x=0$.
We have shown that $(A^3+B^3)x=0$ for a non-zero vector $x$, and can conclude that $\det(A^3+B^3)=0$.
| {
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.