Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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whether $x^4+x^3+x^2+x+1$ is irreducible over $\mathbb Z$ How to check whether $x^4+x^3+x^2+x+1$ is irreducible over $\mathbb Z?$
My guess: If $x^4+x^3+x^2+x+1$ is reducible then $$x^4+x^3+x^2+x+1=f(x)g(x)$$ where $1\le\deg f(x),\deg g(x)\le3$ and $\deg f(x)+\deg g(x)=4$
Checking all the cases for $\deg f(x),\deg g(x)$ is a lengthy exercise. Is there any other way?
| Substitute $\;x\to x+1\;$ , which in fact is an automorphism of the ring of pol's:
$$x^4+x^3+x^2+x+1\to x^4+5x^3+10x^2+10x+5$$
Now just apply Eisenstein with $\;p=5\;$ to the above right polynomial...
| {
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"url": "https://math.stackexchange.com/questions/609021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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sum of 4 squares Is there any natural number $A$ which cannot be written as:
$$A=W^2+X^2+Y^2+Z^2$$
where $W,X,Y,Z \in \mathbb N \cup 0$
I was considering the fact that $a^2+b^2 \not = 1 \mod 4$ and was attempting to determine simular results for $3$ and $4$ squares when I realised that I could not find a number which did not work for $4$. I did a quick search and found that no numbers less than $1000$ contradict this. I have been trying to find a proof but have been, so far, unsuccessful. This is too general a concept to be new but I have not been exposed to it previously.
| Just for completion, here is an elementary proof using modular arithmetic, which I wrote up previously for a number theory course.
Claim For any $n \in \mathbf{N}$ there exists $w,x,y,z \in \mathbf{Z}$ such that $n=w^2+x^2+y^2+z^2$.
Proof: First, we observe that sums of four squares are closed under multiplication. That is given $a,b,c,d,e,f,g,h$ integers, there are $w,x,y,z$ such that
$$(a^2+b^2+c^2+d^2)(e^2+f^2+g^2+h^2)=w^2+x^2+y^2+z^2.$$ I will not prove this, it follows from expanding the terms and finding the right expressions for $w,x,y,z$. Alternatively, there is a proof considering certain matrices over $\mathbf{C}$.
So, let $n \in \mathbf{N}$. By prime decomposition, there are primes $p_1,\dots_,p_k$ such that $n=p_1^{e_1}\cdots p_k^{e_k}$. If one of the $p_i$ is $2$ then notice that $2=1^2+1^2+0^2+0^2$. So that any power of $2$ is a sum of four squares by the previous result. Now, let $p \vert n$ such that $p \equiv 1 \mod 4$ or $p \equiv 3 \mod 4$.
Consider the sets $S_1=\{u^2 : u \in \{0,1,\dots,p-1\}\}$ and $S_2 = \{-1-v^2 : v \in \{0,1,\dots,p-1\}\}$. Since $p$ is a prime, we have there are $\frac{p-1}{2}$ nozero squares modulo $p$ and $\frac{p+1}{2}$ including $0$. $S_1$ and $S_2$ both contain precisely these numbers so that the cardinality of both is $\frac{p+1}{2}$. Thus, since there are only $p$ elements of the set $\{0,1,\dots,p-1\}$, we see $S_1 \cap S_2 \neq \emptyset$. Thus, there is $u,v\in \{0,1,\dots,p-1\}$ such that $u^2\equiv-1-v^2 \mod p$. Thus, we have $u^2+v^2+1\equiv 0 \mod p$. So that there is an integer $k$ such that $$u^2+v^2+1=kp.$$
Now, if $k=1$ we are done, since $p$ is the sum of three squares, we can add $0^2$ and the result holds. Suppose $k\neq 1$. We can see immediately that $k >1$, since $u^2+v^2+1 \geq 1$. So, we have $k \in \mathbf{N}$. We will then produce a strictly decreasing sequence of integers $k_1,k_2,\dots$ which are bounded below by $1$. As a result, there is $m$ such that $k_m=1$. In which case we have that $p$ is the sum of four squares.
Now, we examine our $k$. We have, by renaming $u$ and $v$ four integers $a,b,c,d$ such that $a^2+b^2+c^2+d^2=kp$. Now $k$ is either even or odd. If $k$ is even, then $kp$ must be even. Since the square of an even number is even and an odd number is odd, we then can only have 0,2 or 4 of $a,b,c,d$ even. Without loss of generality, take $a\equiv b\mod 2$ and $c \equiv d \mod 2$. Then, let $a_1=\frac{a+b}{2}, b_1=\frac{a-b}{2},c_1=\frac{c+d}{2}, d_1=\frac{c-d}{2}$ and observe that
$${a_1}^2+{b_1}^2+{c_1}^2+{d_1}^2= \frac{a^2+b^2+c^2+d^2}{2}=\frac{kp}{2}.$$ Thus, there is $k_1 = k/2$ such that $k_1p$ is the sum of four squares.
Now, by repeating the above argument, we can assume we have found $k_l$ such that $k_l$ is odd. Thus, $k_l p={a_l}^2+{b_l}^2+{c_l}^2+{d_l}^2$. So, we can find $e,f,g,h \in \{\frac{-k+1}{2},\dots,\frac{k-1}{2}\}$ such that $a_l\equiv e,b_l\equiv f, c_l \equiv g, d_l \equiv h \mod k_l$. Thus, $e^2+f^2+g^2+h^2 < 4\left( \frac{k_l}{2}\right)^2=k_l^2$.
Thus $$({a_l}^2+{b_l}^2+{c_l}^2+{d_l}^2)(e^2+f^2+g^2+h^2) < k_l^2 k_l p.$$
So we can divide through by $k_l^2$ and find integers $a_{l+1},b_{l+1},c_{l+1},d_{l+1}$ such that ${a_{l+1}}^2+{b_{l+1}}^2+{c_{l+1}}^2+{d_{l+1}}^2=k_{l+1}p$ where $k_{l+1} < k_{l}$.
Thus, by repeating the arguments above, we have constructed a decreasing sequence of natural numbers $(k_i)$. So, we must have some $m$ such that $k_m=1$. In which case, we see that there are integers $w_0,x_0,y_0,z_0$ such that $$w_0^2+x_0^2+y_0^2+z_0^2 = k_m p=p.$$
As we've seen before, we have seen that sum of four squares are closed under multiplication, so that there are integers $w,x,y,z$ such that $n=w^2+x^2+y^2+z^2.$ as desired. Thus any integer is the sum of four squares.
| {
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"timestamp": "2023-03-29T00:00:00",
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what is the value of $\tan\theta$ and $\csc\theta$? Please help me to solve this problem.I can not understand what to do.
If $(a^2-b^2)\sin\theta + 2ab\cos\theta = a^2+b^2$ and $\theta$ is acute and positive angle then what is the value of $\tan\theta$ and $\csc\theta$ ?
| Your equation can be written
$$\frac{1-t^2}{1+t^2} \sin \theta + \frac{2t}{1+t^2}\cos \theta = 1$$
Where $t = \frac b a$.
Now, if $\frac b a = t=\tan \frac{\alpha}{2}$ (with $\alpha \in ]0, \pi[$ if both $a$ and $b$ are positive), then $\frac{1-t^2}{1+t^2}=\cos \alpha$ and $\frac{2t}{1+t^2}=\sin \alpha$, and the equation becomes:
$$\cos \alpha \sin \theta + \sin \alpha \cos\theta = 1$$
Or
$$\sin (\alpha + \theta) = 1$$
You get $\theta = \frac{\pi}2 - \alpha + 2k\pi$, thus
$$\tan \theta = \frac{1}{\tan \alpha} = \frac {\cos \alpha}{\sin \alpha}=\frac{a^2-b^2}{2ab}$$
And
$$\csc \theta = \frac{1}{\sin \theta} = \frac{1}{\cos \alpha} = \frac{a^2+b^2}{a^2-b^2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to demonstrate the equality of these integral representations of $\pi$? Each of the three following definite integrals are well known to have the same value of $\pi$: $$\int_{-1}^1\frac{1}{\sqrt{1-x^2}}\,\mathrm{d}x=2\int_{-1}^1\sqrt{1-x^2}\,\mathrm{d}x=\int_{-\infty}^{\infty}\frac{1}{1+x^2}\,\mathrm{d}x=\pi.$$
I like taking the first definite integral as the definition of $\pi$, since it represents half the circumference of the unit circle. The second integral obviously represents the area of the unit circle, but as an exercise I wanted to prove its equality with integral #1 using just elementary integration rules. I was successful once I tried integrating by parts:
$$\begin{align}
\int\sqrt{1-x^2}\,\mathrm{d}x
&=x\sqrt{1-x^2}-\int\frac{-x^2}{\sqrt{1-x^2}}\,\mathrm{d}x\\
&=x\sqrt{1-x^2}+\int\frac{1}{\sqrt{1-x^2}}\,\mathrm{d}x-\int\frac{1-x^2}{\sqrt{1-x^2}}\,\mathrm{d}x\\
&=x\sqrt{1-x^2}+\int\frac{1}{\sqrt{1-x^2}}\,\mathrm{d}x-\int\sqrt{1-x^2}\,\mathrm{d}x\\
\implies2\int\sqrt{1-x^2}\,\mathrm{d}x&=x\sqrt{1-x^2}+\int\frac{1}{\sqrt{1-x^2}}\,\mathrm{d}x\\
\implies2\int_{-1}^1\sqrt{1-x^2}\,\mathrm{d}x&=\int_{-1}^1\frac{1}{\sqrt{1-x^2}}\,\mathrm{d}x.
\end{align}$$
Having accomplished this much, I decided I'd like to demonstrate the equality of these two integrals to $\int_{-\infty}^{\infty}\frac{1}{1+x^2}\mathrm{d}x$ as well, in a similarly elementary manner, but I'm stumped as to what to try. Can anyone suggest a substitution or transformation that demonstrates their equality?
| These are both immediate using two subsitutions. Everything is even, so split all of them at $0$.
$$\begin{aligned} t=\sqrt{1-x^2}:\quad\int_0^1 \frac{dx}{\sqrt{1-x^2}}= \int_0^1 2\sqrt{1-t^2}\,dt\end{aligned}$$
And:
$$\begin{aligned} t=\frac{x}{\sqrt{1-x^2}}:\quad\int_0^{1} \frac{dx}{\sqrt{1-x^2}}= \int_0^{\infty}\frac{dt}{1+t^2}\end{aligned}$$
| {
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Proof by induction that $3^{4n + 1} + 5^{2n + 1}$ is divisble by $8$ This is a homework problem:
Prove that: $$ 3^{4n+1} + 5^{2n+1}$$ is divisible by $8$ for every natural number $n$.
Base case: $$n = 0$$
$$ 3^{0 + 1} + 5^{0 + 1} = 8$$
$$8\bmod8 = 0 $$ Base case verified.
Inductive Hypothesis: Assume for some $n$, $$ 3^{4n+1} + 5^{2n+1} =8k$$ where $k$ is any non-negative integer.
Prove that: $$ 3^{4(n+1)+1} + 5^{2(n+1)+1} =8l$$
Indeed: $$ 3^{4n+4+1} + 5^{2n+2+1} = 3^4\cdot3^{4n+1} + 5^2\cdot5^{2n+1}$$
$$ = 81\cdot3^{4n+1} + 25\cdot5^{2n+1} $$
$$ = (80+1)\cdot3^{4n+1} + (24+1)\cdot5^{2n+1} $$
$$ = 80\cdot3^{4n+1} + 24\cdot5^{2n+1} + 3^{4n+1} + 5^{2n+1} $$
$$ = 8(10\cdot3^{4n+1} + 3\cdot5^{2n+1}) + 8k $$
$$ = 8(10\cdot3^{4n+1} + 3\cdot5^{2n+1} + k) $$
Is this wrong in any particular way? My answer is different from the textbook, but I feel like I sufficiently proved that $ 3^{4(n+1)+1} + 5^{2(n+1)+1} =8l$.
Edit: What the book did:
$$ = 81\cdot3^{4n+1} + 25\cdot5^{2n+1} $$
$$ = (56+25)\cdot3^{4n+1} + 25\cdot5^{2n+1} $$
$$ = 56\cdot3^{4n+1}+25\cdot3^{4n+1} + 25\cdot5^{2n+1} $$
$$ = 56\cdot3^{4n+1}+25(3^{4n+1} + 5^{2n+1}) $$
$$ = 56\cdot3^{4n+1}+25(8k) $$
$$ = 8(7\cdot3^{4n+1}+25k) $$
| Your method looks fine. Another method that is slightly cleaner: Let $S_n=3^{4n+1}+5^{2n+1}=3\cdot 81^n+5\cdot25^n$. Note that $S_n$ satisfies $$ S_n=106S_{n-1}-2025S_{n-2} $$
(This can be found by that fact that $u_n:=c_1A^n+c_2B^n$ satisfies the recursion $u_n=(A+B)u_{n-1}-ABu_{n-2}$)
This recursion implies that if $k|S_{n}$ and $k|S_{n+1}$ then $k|S_{n+2}$. Since the initial values of $S_n$ are $S_0=8$ and $S_1=368$, we can see inductively that $8|S_n$ for all $n\in\Bbb N_0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the remainder when $1.1!+2.2!+3.3!+ ... +10.10! +2$ is divided by $11!$ Find the remainder when $1.1!+2.2!+3.3!+ ... +10.10! +2$ is divided by $11!$
An attempt: Rearranging:
$$\frac{1}{11!}+\frac{2.2!}{11}+\frac{3.3!}{11} \cdots +\frac{10.10!}{11}+\frac{2}{11!}$$
$$\frac{1}{11!}+\frac{2}{11!}+\frac{2.2!}{11}+\frac{3.3!}{11} \cdots +\frac{10.10!}{11}$$
$$\frac{1}{11!}+\frac{(2+1)2!}{11!}+\frac{3.3!}{11} \cdots +\frac{10.10!}{11}$$
$$\frac{1}{11!}+\frac{3.2!}{11!}+\frac{3.3!}{11} \cdots +\frac{10.10!}{11}$$
$$\frac{1}{11!}+\frac{(3+1)3!}{11!}+\frac{3.3!}{11} \cdots +\frac{10.10!}{11}$$
$$.$$
$$.$$
$$.$$
$$\frac{1}{11!}+\frac{(10+1)10!}{11!}$$
$$\frac{11!+1}{11!}$$
What is the remainder?
| HINT:
Observe that $$r\cdot r!=(r+1-1)\cdot r!=(r+1)!- r!$$
Do you recognize the Telescoping series? Just set $\displaystyle r=1,2,\cdots,9,10$ and add
| {
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Understanding $n \left(\frac{2n \choose n}{4^n}\right)^2$ for large $n$ My answer over at cstheory.stackexchange.com involved the expression $$\lim_{n\to \infty} n \left(\frac{2n \choose n}{4^n}\right)^2$$ According to Wolfram Alpha, this expression is at most $\frac{1}{\pi}$.
I would like to know why this is the case, but I don't know enough about these kinds of expressions to figure this out myself.
| Use Stirling's formula:
$$ n! \sim \sqrt{2\pi n} \left(\frac{n}{e}\right)^n. $$
Using this formula, we get that
$$ \binom{2n}{n} \sim \frac{\sqrt{4\pi n} (2n/e)^{2n}}{2\pi n (n/e)^{2n}} = \frac{4^n}{\sqrt{\pi n}}, $$
and so
$$ n \left( \frac{\binom{2n}{n}}{4^n} \right)^2 \sim \frac{1}{\pi}. $$
There are variants of Stirling's formula with bounds on the error terms, for example
$$ n! = \sqrt{2\pi n} \left(\frac{n}{e}\right)^n e^{\Theta(1/n)} $$
(Here $\Theta(1/n)$ is some quantity in the range $[-C/n,C/n]$ for some absolute constant $C > 0$.)
Using this, we get
$$ \binom{2n}{n} = \frac{\sqrt{4\pi n} (2n/e)^{2n} e^{\Theta(1/n)}}{2\pi n (n/e)^{2n} e^{\Theta(1/n)}} = \frac{4^n}{\sqrt{\pi n}} e^{\Theta(1/n)}, $$
and so
$$ n \left( \frac{\binom{2n}{n}}{4^n} \right)^2 = \frac{e^{\Theta(1/n)}}{\pi} = \frac{1}{\pi} + \Theta\left(\frac{1}{n}\right). $$
For even more terms, check the Wikipedia article.
Your experiments seem to imply that the sequence $f(n) = n (\binom{2n}{n}/4^n)^2$ is increasing. To check this, let's compute the ratio of two adjacent terms:
$$
\begin{align*}
\frac{f(n+1)}{f(n)} &= \frac{n+1}{n} \frac{(2n+2)!^2 n!^4}{(2n)!^2 (n+1)!^4} \frac{4^{2n}}{4^{2n+2}} \\ &= \frac{n+1}{n} \frac{(2n+2)^2(2n+1)^2}{(n+1)^4} \frac{1}{16} \\ &=
\frac{(2n+1)^2}{4n(n+1)} = \frac{4n^2 + 4n + 1}{4n^2 + 4n} > 1.
\end{align*}
$$
This shows that $f(n) < 1/\pi$ and $f(n) \to 1/\pi$. Since $f(1) = 1/4$, we get the following explicit formula:
$$ f(n) = \frac{1}{4} \prod_{k=1}^{n-1} \frac{4k^2+4k+1}{4k^2+4k}. $$
This also shows that
$$ \pi = 4 \prod_{k=1}^\infty \frac{4k^2+4k}{4k^2+4k+1}. $$
Therefore
$$
\begin{align*}
f(n) &= \frac{1}{\pi} \prod_{k=n}^\infty \frac{4k^2+4k}{4k^2+4k+1} \\ &=
\frac{1}{\pi} \prod_{k=n}^\infty \left(1 - \Theta\left(\frac{1}{k^2}\right)\right) \\ &=
\frac{1}{\pi} \left(1 - \sum_{k=n}^\infty \Theta\left(\frac{1}{k^2}\right)\right) \\ &=
\frac{1}{\pi} \left(1 - \Theta\left(\frac{1}{n}\right)\right) = \frac{1}{\pi} - \Theta\left(\frac{1}{n}\right),
\end{align*}
$$
as we have obtained before. (Here $\Theta(1/n)$ is some quantity in the range $[C_0/n,C_1/n]$ for some positive constants $0 < C_0 < C_1$.) If we sweat more, we can get better estimates on the error from this formula for $f(n)$; this is left to the reader.
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Closed form of this arithmetic series? I am computing the partial sums $s_{nn}$ of a series given by
$a_{ij} = {1 \over 2}^{j-i}$ if $j>i$, $a_{ij} = -1$ if $j=i$ and $a_{ij} = 0$ if $j < i$
The $s_{nn}$ is the partial sum $\sum_{i,j =1}^n a_{ij}$. I computed that $$s_{nn}=\sum_{k=1}^{n-1} (n-k) {1 \over 2^k}$$
and now I am trying to find a closed formula for $s_{nn}$ but I'm stuck. Please can someone help me how to do this?
| Assuming your expression $s_{nn} = \sum_{k=1}^{n-1} (n-k) {1 \over 2^k}$ is correct, I work from there.
$$s_{nn} = \sum_{k=1}^{n-1} (n-k) {1 \over 2^k} = n \sum_{k=1}^{n-1}{1 \over 2^k} - \sum_{k=1}^{n-1}{k \over 2^k}$$
As the first term is clearly a geometric series, I focus on the latter term.
$$ \sum_{k=1}^{n-1}{k \over 2^k} = \frac{1}{2} + (\frac{1}{4} + \frac{1}{4}) + (\frac{1}{8} + \frac{1}{8} + \frac{1}{8}) + \cdot \cdot \cdot + \frac{n-1}{2^{n-1}} $$
$$ \sum_{k=1}^{n-1}{k \over 2^k} = (\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdot \cdot \cdot + \frac{1}{2^{n-1}}) + (\frac{1}{4} + \frac{1}{8} + \cdot \cdot \cdot + \frac{1}{2^{n-1}}) + (\frac{1}{8} + \cdot \cdot \cdot + \frac{1}{2^{n-1}}) + \cdot \cdot \cdot + \frac{1}{2^{n-1}} $$
Which is a sum of geometric series. Plugging in the closed form expression for each individual series will result in another geometric series, finally providing the partial sum.
| {
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"timestamp": "2023-03-29T00:00:00",
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Cubic equation $ax^3+3bx^2+3cx+d = 0$ has $2$ equal roots. How can I find their value in terms of $a,b,c$? If the equation $ax^3+3bx^2+3cx+d = 0$ has $2$ equal roots, then equal root must
be equal to $\displaystyle \frac{bc-ad}{2(ac-b)^2}.$
My Try:: Let $x=\alpha,\alpha,\beta$ be the roots of given equation. Then using Vieta's formula
$$ \alpha+\alpha+\beta = -\frac{3b}{a}\Rightarrow 2\alpha +\beta = -\frac{3b}{a}$$
$$ \alpha \cdot \alpha +\alpha \cdot \beta +\alpha \cdot \beta = \frac{3c}{a}\Rightarrow \alpha^2+2\alpha \cdot \beta = \frac{3c}{a}$$
$$\alpha \cdot \alpha \cdot \beta = -\frac{d}{a}\Rightarrow \alpha^2 \cdot \beta = -\frac{d}{a}.$$
Now I did not understand how can I find the value of $\alpha$ in terms of $a,b$ and $c$.
Help is required.
Thanks
| I made a formula to solve a cubic equation with all equal roots.
$$ax^3+bx^2+cx+d=0$$
To find such an equation you will have to check the condition $b=\sqrt{3ac}$ if it is so then we will use this formula $\frac{-b+\sqrt[3]{b^3 - 27a^2d}}{3a}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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how do i write $y = 2\sin2(x + \frac{\pi}{4}) - \cos2(x + \frac{\pi}{4} )$ in format $y = A\sin(kx) + B\cos(kx)$ The problem is the double angles. I tried to simplify them and change them around but no luck,
$$\begin{align} y&=A\sin(kx)+B\cos(ky)\\
y&=2\sin2(x+\pi/4)-\cos2(x+\pi/4)\\
&=2\left(2\sin(x+\pi/4)\cos(x+\pi/4)\right)-\cos^2(x+\pi/4)+\sin^2(x+\pi/4)\\
&=4\left(\sin x\cos\frac\pi 4+\cos x\sin\frac\pi4\right)\left(\cos x\cos\frac\pi 4-\sin x\sin\frac\pi4\right)-\cos^2(x+\pi/4)+\sin^2(x+\pi/4)\\
&=4\left(\sin x\frac1{\sqrt2}+\cos x\frac1{\sqrt2}\right)\left(\cos x\frac1{\sqrt2}-\sin x\frac1{\sqrt2}\right)-\cos^2(x+\pi/4)+\sin^2(x+\pi/4)\\
&=\frac 4{\sqrt 2}\left(\sin x+\cos x\right)\left(\cos x-\sin x\right)-\cos^2(x+\pi/4)+\sin^2(x+\pi/4)\\
&=2\sqrt 2(\cos^2x-\sin^2x)+\left(\sin x\frac1{\sqrt2}+\cos x\frac1{\sqrt2}\right)^2-\left(\cos x\frac1{\sqrt2}-\sin x\frac1{\sqrt2}\right)^2\\
&=2\sqrt 2(\cos^2x-\sin^2x)+4\left(\frac1{\sqrt2}\sin x\frac1{\sqrt2}\cos x\right)\\
&=2\sqrt 2(\cos^2x-\sin^2x+\sin x\cos x) \end{align}$$
| $$\begin{align} 2\sin\left(2\left(x+\frac \pi4\right)\right)-\cos \left(2\left(x+\frac \pi4\right)\right)&=2\sin\left(2x+\frac \pi2\right)-\cos\left(2x+\frac \pi2\right)\\
&=2\cos(2x)+\sin(2x) \end{align}$$
Shifting sine by $\pi/2$ obtains cosine, and cosine shifted by $\pi/2$ is negative sine, so the result follows.
To be slightly more concrete: \begin{align}\sin(2x+\pi/2)&=\sin(2x)\cos(\pi/2)+\cos(2x)\sin(\pi/2)=\cos(2x)\\
\cos(2x+\pi/2)&=\cos(2x)\cos(\pi/2)-\sin(2x)\sin(\pi/2)=-\sin(2x)\end{align}
The problem with your approach is that you consistently fail to square $\frac1{\sqrt 2}$ after factoring it out of two terms. e.g. $$[(\sin x)/\sqrt 2+(\cos x)/\sqrt 2][(\cos x)/\sqrt 2-(\sin x)/\sqrt 2]=\frac1{\sqrt 2}(\sin x+\cos x)\frac 1{\sqrt 2}(\cos x-\sin x)=\frac12(\sin x+\cos x)(\cos x-\sin x)$$
Once you adjust for this, your final expression is $$ 2(\cos^2x-\sin^2x+\sin x\cos x)=2\cos(2x)+\sin(2x) $$ So you weren't too far from the finish line.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/619167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
On the roots of a polynomial Let $f(x)= x^3 - 3x + 1$. Show that if a complex number $a$ is root of $f(x)$, then $a^2-2$ is too.
| HINT : Since we know $f(a)=a^3-3a+1=0\iff a^3=3a-1,$ use this to prove $$f(a^2-2)=0.$$
$$f(a^2-2)=(a^2-2)^3-3(a^2-2)+1=a^6-6a^4+9a^2-1$$$$=(a^3-3a+1)^2-2(a^3-3a+1)=0^2-2\cdot 0=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/619512",
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"source": "stackexchange",
"question_score": "3",
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} |
$\omega^2+\omega+1$divides a polynomial The question is
Show that $f(n)=n^5+n^4+1$ is not prime for $n>4$.
The solution is given as
Let $\omega$ be the third root of unity. Then $\omega^2+\omega+1=0$. Since $\omega^5+\omega^4+1=\omega^2+\omega+1$, we see that $\omega^2+\omega+1$ is a *factor of the polynomial. So *$n^2+n+1|n^5+n^4+1$.
Which polynomial are we referring to in the bold typeface above? And how is $n^2+n+1|n^5+n^4+1$ true due to $\omega^5+\omega^4+1=\omega^2+\omega+1$?
| See that $\omega$ is a root of both $f(x)=x^5+x^4+1$ and $g(x)=x^2+x+1$. Now $g(x)$ has all coefficients real, and it is a polynomial of degree 2. Hence it has two roots and the second root must be the complex conjugate of $\omega$, that is $\omega^2$. Hence $g(x)=(x-\omega)(x-\omega^2)$. Also note that $f(\omega^2)=0$. Hence $f(x)=(x-\omega)(x-\omega^2)h(x)=g(x)h(x)$ for some real polynomial $h(x)$. Now you can easily show by equating coefficients that $h(x)$ has integer coefficients. Now put $x=n$ on both sides. Then since $h(n)$ turns out to be an integer $n^2+n+1|n^5+n^4+1$.
| {
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"source": "stackexchange",
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Decomposition as a product of factors For several days, I try to do this excercise. Without success.
Prove that for any pair of positive integers $k$ and $n$, there exist $k$ positive integers $m_1, m_2, \dots, m_k$ (not necessarily different) such that:
$$1+ \frac{2^k - 1}n = \left(1+\frac 1{m_1}\right)\cdot\left(1+\frac 1{m_2}\right)\cdots\left(1+\frac 1{m_k}\right)$$
| $k=1$ is clear. We now suppose that the statement is true for some fixed $k$ (and all $n$), and proceed to demonstrate it for $k+1$.
If $n$ is even, then we can write $1+\frac{2^k-1}{\frac{n}{2}}$ as a product of the given form with $k$ terms. And so, the following can be written as a product with $k+1$ terms:
$$\left(1+\frac{1}{n+2^{k+1}-2}\right)\left(1+\frac{2^k-1}{\frac{n}{2}}\right) = \left(\frac{n+2^{k+1}-1}{n+2^{k+1}-2}\right)\left(\frac{n+2^{k+1}-2}{n}\right)$$
$$= \frac{n+2^{k+1}-1}{n} = 1 + \frac{2^{k+1}-1}{n}$$
This verifies the claim when $n$ is even. If $n$ is odd, then we can write $1+\frac{2^k-1}{\frac{n+1}{2}}$ as a product with $k$ terms. As above, the following is a product with $k+1$ terms:
$$\left(1+\frac{1}{n}\right)\left(1+\frac{2^k-1}{\frac{n+1}{2}}\right) = \left( \frac{n+1}{n}\right)\left( \frac{n+2^{k+1}-1}{n+1}\right)$$
$$= \frac{n+2^{k+1}-1}{n} = 1 + \frac{2^{k+1}-1}{n}$$
That about does it.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
how can i evaluate this indefinite-integral? $$\int \frac{x^2}{(x\cos{x}-\sin{x})^2} dx$$
I tried to turn it to $\tan$ and $\sec$ but it didn't work out very well. How can I evaluate this indefinite integral using regular methods?
| $\bf{My\; Solution::}$ Given $\displaystyle \int\frac{x^2}{(x\cos x-\sin x)^2}dx$
we can write $\displaystyle (x\cos x-\sin x) = \sqrt{1+x^2}\left(\frac{x}{\sqrt{1+x^2}}\cdot \cos x - \frac{1}{\sqrt{x}}\cdot \sin x\right) = \sqrt{1+x^2}\cdot \cos (x+\phi)$
where $\displaystyle \cos \phi = \frac{x}{\sqrt{1+x^2}}$ and $\displaystyle \sin \phi = \frac{1}{\sqrt{1+x^2}}$. So we get $\;\; \cot \phi = x\Rightarrow \phi = \cot^{-1}(x)$
So Integral convert into $\displaystyle \int \frac{x^2}{(1+x^2)\cdot \cos^2 \left(x+\phi\right)}dx = \int \sec^2(x+\phi)\cdot \frac{x^2}{1+x^2}dx$
$\displaystyle = \int\sec^2\left(x+\cot^{-1}(x)\right)\cdot \frac{x^2}{1+x^2}dx$
Now Let $(x+\cot^{-1}(x)) = u$, Then $\displaystyle \left(1-\frac{1}{1+x^2}\right)dx = du\Rightarrow \frac{x^2}{1+x^2}dx = du$
So Integral convert into $\displaystyle \int \sec^2(u)du = \tan (u)+\mathbb{C}$
So we get $\displaystyle = \tan \left(x+\cot^{-1}(x)\right)+\mathbb{C} = \tan \left(\frac{\pi}{2}+x-\tan^{-1}(x)\right)+\mathbb{C}$
$\displaystyle = -\cot(x-\tan^{-1}(x)) = -\frac{1}{\tan(x-\tan^{-1}(x))}+\mathbb{C} = -\left(\frac{1+\tan (x)\cdot x }{\tan (x)-x}\right)+\mathbb{C}$
So $\displaystyle \int \frac{x^2}{(x\cos x-\sin x)^2}dx = \left(\frac{\cos x +x\cdot \sin x}{x\cdot \cos x-\sin x}\right)+\mathbb{C}$
| {
"language": "en",
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"source": "stackexchange",
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Computing an integral Find what is equal $I$ to:
$$I=\int_{0}^{\pi/2} \sqrt[3]{\tan x}dx$$
I found this simple looking integral, but I don't know if my logic is right.
Let $J$ be the next integral with $x\in(0, \frac{\pi}{2})\Rightarrow$
$$J=\int \sqrt[3]{\cot x}dx$$
Then
$$\begin{align}
I\pm J & =\int (\sqrt[3]{\tan x}\pm \sqrt[3]{\cot x})dx \\
& =\int \frac{(\sqrt[3]{\tan x})^2 \pm 1}{\sqrt[3]{\tan x}}dx \\
& =\int \frac{\tan x \pm \sqrt[3]{\tan x}}{\sqrt[3]{\tan x}\sqrt[3]{\tan x}}dx
\end{align}$$
And so forth using integration by parts (haven't done yet). But if I compute $I+J$ and $I-J$, how can I get $I$? Is my approach correct until now?
| $$ \int^{\pi / 2}_{0} \tan^{a} x \ dx \ \ (|a| <1 ) $$
$$= \int^{\pi / 2}_{0} \frac{ \sin^{a} x}{\cos^{a} x} dx = \int^{\pi / 2}_{0} \sin^{2(\frac{a}{2} + \frac{1}{2})-1} (x) \cos ^{2(-\frac{a}{2} + \frac{1}{2}) -1} (x) \ dx $$
$$ = \frac{1}{2} \ B(\frac{a}{2}+ \frac{1}{2}, -\frac{a}{2}+ \frac{1}{2}) = \frac{1}{2} \ \frac{\Gamma(\frac{a+1}{2}) \Gamma(1 - \frac{a+1}{2})}{\Gamma(1)}$$
$$ = \frac{1}{2} \ \frac{\pi}{\sin \ (\frac{\pi (a+1)}{2})} = \frac{1}{2} \frac{\pi}{\cos (\frac{\pi a}{2})}$$
Therefore,
$$\int^{\pi / 2}_{0} \tan^{\frac{1}{3}} x \ dx = \frac{1}{2} \frac{\pi}{\cos \left(\frac{\pi}{6} \right)} = \frac{\pi}{\sqrt{3}}$$
| {
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"question_score": "3",
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How prove this $|P_{n}(x)|\le 1$ let $0<r<1$,and
$$\dfrac{1}{\sqrt{1-2tr+r^2}}=\sum_{n=0}^{\infty}P_{n}(t)r^n$$
show that
$$|P_{n}(t)|\le 1,-1\le t\le 1$$
My try:
I know this coefficients $P_{n}$ are called Legendre polynomials,
$$P_{n}(t)=\sum_{k=0}^{[n/2]}\dfrac{n!}{2^{2k}(k!)^2(n-2k)!}t^{n-2k}(t^2-1)^k$$maybe prove $|P_{n}(t)|\le 1$ have some methods? Thank you for post you solution,Thank you
| For $|t|\le 1$ we can put $t=\cos\theta$ so that we can write $$1-2tr+r^2=1-2r\cos\theta+r^2=\left(1-r\operatorname{e}^{+i\theta}\right)\left(1-r\operatorname{e}^{-i\theta}\right)$$
and using the expansion $(1+y)^m=\sum_{n=0}^{\infty}\binom{m}{n}x^n$ we have
$$
\begin{align}
\frac{1}{\sqrt{1-2r\cos\theta+r^2}}
&=\left(1-r\operatorname{e}^{+i\theta}\right)^{-1/2}\left(1-r\operatorname{e}^{-i\theta}\right)^{-1/2}\\
&=\sum_{m=0}^\infty a_mr^m\operatorname{e}^{+im\theta}\sum_{m=0}^\infty a_mr^m\operatorname{e}^{-im\theta}
\end{align}
$$
where $a_0=1$ and $a_m=\tfrac{1\cdot3\cdots(2m-1)}{2\cdot 4\cdots(2m)}$ for $m=1,2,\ldots$, so that
$$
\begin{align}
P_n(\cos\theta)&=\sum_{m=0}^n a_m\operatorname{e}^{+im\theta}a_{n-m}\operatorname{e}^{-i(n-m)\theta}\\
&=\sum_{m=0}^n a_m a_{n-m}\operatorname{e}^{-i(n-2m)\theta}\\
&=\sum_{m=0}^n a_m a_{n-m} \cos(n-2m)\theta\\
&=2a_0a_n\cos n\theta+2a_1a_{n-1}\cos (n-2)\theta+\cdots+
\begin{cases}2 a_{(n-1)/2}a_{(n+1)/2}\cos \theta& \text{for $n$ odd }\\
a^2_n/2 & \text{for $n$ even}
\end{cases}\\
&=\tfrac{1\cdot3\cdots(2n-1)}{2\cdot 4\cdots(2n)}\left(2\cos n\theta+\tfrac{1\cdot (2n)}{2\cdot (2n-1)}2\cos(n-2)\theta+\tfrac{1\cdot 3\cdot (2n)\cdot (2n-2)}{2\cdot 4\cdot(2n-1)\cdot(2n-3)}2\cos(n-4)\theta+\cdots\right)
\end{align}
$$
Consequently $P_n(\cos\theta)$ is a trigonometric cosine polynomial with non-negative coefficients.
If $\theta$ is a real angle
$$
|P_n(\cos\theta)|\le \tfrac{1\cdot3\cdots(2n-1)}{2\cdot 4\cdots(2n)}\left(2+\tfrac{1\cdot (2n)}{2\cdot (2n-1)}2+\tfrac{1\cdot 3\cdot (2n)\cdot (2n-2)}{2\cdot 4\cdot(2n-1)\cdot(2n-3)}2+\cdots\right)=P_n(1)
$$
so that $|P_n(\cos\theta)|\le 1$.
| {
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"url": "https://math.stackexchange.com/questions/622412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the exact value of $\tan\left ( \sin^{-1} \left ( \sqrt 2/2 \right )\right )$
Find the exact value of $\tan\left ( \sin^{-1} \left ( \dfrac{\sqrt{2}}{2} \right )\right )$ without using a calculator.
I started by finding $\sin^{-1} \left ( \dfrac{\sqrt{2}}{2} \right )=\dfrac{\pi}{4}$
So, $\tan\left ( \sin^{-1} \left ( \dfrac{\sqrt{2}}{2} \right )\right )=\tan\left( \dfrac{\pi}{4}\right)$.
The answer is $1$. Can you show how to solve $\tan\left( \dfrac{\pi}{4}\right)$ to get $1$? Thank you.
| We have to find the exact value of $\;\tan\left(\sin^{-1}\dfrac{1}{\sqrt{2}}\right)$.
Since $\;\sin(45)=\dfrac{1}{\sqrt{2}}\;,\;x=45\;$ so $\;\tan(45)=1\;,\;$ hence we are done. Now $\;\sin(45)=\cos(45)\;,\;$ thus $\;\tan(45)=1\;.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $1 + \cos\alpha + \cos\beta + \cos\gamma = 0$ If $\alpha + \beta + \gamma = \pi $ and $\tan(\frac{-\alpha + \beta + \gamma}4)\tan(\frac{\alpha - \beta + \gamma}4)\tan(\frac{\alpha + \beta - \gamma}4) = 1$
Then prove that:
$1 + \cos\alpha + \cos\beta + \cos\gamma = 0$.
I have no idea how to go about this.
Please help.
| A little Generalization :
Let $\alpha+\beta+\gamma=4C$
and $\displaystyle\tan\left(\frac{-\alpha + \beta + \gamma}4\right)\tan\left(\frac{\alpha - \beta + \gamma}4\right)\tan\left(\frac{\alpha + \beta - \gamma}4\right) = \cot C$
As $\displaystyle \frac{-\alpha+\beta+\gamma}4=\frac{-\alpha+4C-\alpha}4=C-\frac\alpha2,$
$\displaystyle \tan\frac{-\alpha+\beta+\gamma}4=\tan\left(C-\frac\alpha2\right)=\frac{\sin\left(C-\frac\alpha2\right)}{\cos\left(C-\frac\alpha2\right)}$
So, the problem reduces to $\displaystyle\frac{\sin\left(C-\frac\alpha2\right)\sin\left(C-\frac\beta2\right)\sin\left(C-\frac\gamma2\right)}{\cos\left(C-\frac\alpha2\right)\cos\left(C-\frac\beta2\right)\cos\left(C-\frac\gamma2\right)}=\cot C$
$\displaystyle\implies\frac{\sin\left(C-\frac\alpha2\right)\sin\left(C-\frac\beta2\right)}{\cos\left(C-\frac\alpha2\right)\cos\left(C-\frac\beta2\right)}
=\frac{\cos\left(C-\frac\gamma2\right)\cos C}{\sin\left(C-\frac\gamma2\right)\sin C}$
Applying $\displaystyle2\sin A\sin B,2\cos A\cos B$ formula,
$\displaystyle\implies\frac{\cos\frac{\beta-\alpha}2-\cos\frac{4C-\beta-\alpha}2}{\cos\frac{\beta-\alpha}2+\cos\frac{4C-\beta-\alpha}2}
=\frac{\cos\left(-\frac\gamma2\right)+\cos\frac{4C-\gamma}2}{\cos\left(-\frac\gamma2\right)-\cos\frac{4C-\gamma}2}$
Now as $\displaystyle\alpha+\beta+\gamma=4C,$ this becomes
$\displaystyle\implies\frac{\cos\frac{\beta-\alpha}2-\cos\frac{\gamma}2}{\cos\frac{\beta-\alpha}2+\cos\frac{\gamma}2}
=\frac{\cos\frac\gamma2+\cos\frac{\alpha+\beta}2}{\cos\frac\gamma2-\cos\frac{\alpha+\beta}2}$
Applying Componendo and dividendo,
$\displaystyle\implies\frac{\cos\frac{\beta-\alpha}2}{-\cos\frac{\gamma}2}
=\frac{\cos\frac{\alpha+\beta}2}{\cos\frac{\alpha+\beta}2}$
$\displaystyle\implies\cos\frac{\beta-\alpha}2\cos\frac{\beta+\alpha}2=-\cos^2\frac{\gamma}2 $
Applying $2\cos A\cos B,\cos2x=2\cos^2x-1,$
$\displaystyle\frac{\cos\alpha+\cos\beta}2=-\frac{1+\cos\gamma}2$
Here $\displaystyle C=\frac\pi4\implies \alpha+\beta+\gamma=4C=\pi$
If $\displaystyle C=-\frac\pi4\implies \alpha+\beta+\gamma=4C=-\pi\equiv\pi\pmod{2\pi}$
So, $\displaystyle\tan\left(\frac{-\alpha + \beta + \gamma}4\right)\tan\left(\frac{\alpha - \beta + \gamma}4\right)\tan\left(\frac{\alpha + \beta - \gamma}4\right)=-1$ and $\alpha+\beta+\gamma=-\pi$ or $\pi$ will satisfy the required identity
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/624189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Integral Of $\sqrt{9-y^2}dy$ I would like to integrate the following:
$$\int\sqrt{9-y^2}dy$$
What I did:
*
*$y=3\sin t$
*$dy=3\cos tdt$
$$\int\sqrt{9-9\sin ^2 t}\cdot 3\cos t dt=\int\sqrt{9}\cdot\sqrt{1-\sin^2 t}\cdot 3\cos tdt=\int 9\cos ^2tdt$$
$$9\cdot \int(\frac{1}{2}+\frac{\cos 2t}{2}) dt=9\cdot\left [ t+\frac{\sin 2t}{2}\right]$$
I'm pretty stuck now, I would like to get some advice
How its suppose to be now?
$$\dots=9\left[ \arcsin(\frac{y}{3})+\frac{\sin( 2 \cdot \arcsin(\frac{y}{3}))}{2}\right]?$$
| Note that $$\sin 2t = 2 \sin t \cos t$$
Now use the fact that $\sin^2 t + \cos^2 t = 1$ to rewrite $\cos t$ in terms of $y$. This leads to
\begin{align*}
\frac{\sin 2t}{2} &= \sin t \cos t \\
&= \sin t \sqrt{1 - \sin^2 t} \\
&= \frac{y}{3} \sqrt{1 - \left(\frac{y}{3}\right)^2} \\
&= \frac{1}{9} y \sqrt{9 - y^2}
\end{align*}
You've got a very small mistake, though, in your application of the double angle formula: You should have $$\frac 1 2 + \frac{\cos 2t}{2}$$
Fixing this and applying the above leads to the simplified solution from W|A.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How find this limit $\lim_{x\to 1}\frac{f(2001)-f(2002)}{f(2002)-f(2003)}$ Let $$f(m)=\dfrac{m+1}{\dfrac{m+m+1}{\dfrac{m}{1-x^{m}}-\dfrac{m+1}{1-x^{m+1}}+\dfrac{1}{2}}+\dfrac{m+1+m+2}{\dfrac{m+1}{1-x^{m+1}}-\dfrac{m+2}{1-x^{m+2}}+\dfrac{1}{2}}}$$
Find this limit
$$I=\lim_{x\to 1}\dfrac{f(2001)-f(2002)}{f(2002)-f(2003)}$$
My try: since Use (L'Hôpital's rule） we have
\begin{align*}&\lim_{x\to 1}\left(\dfrac{n}{1-x^n}-\dfrac{n+1}{1-x^{n+1}}\right)\\
&=\lim_{x\to1}\dfrac{n(1-x^{n+1})-(n+1)(1-x^n)}{(1-x^n)(1-x^{n+1})}\\
&=\lim_{x\to 1}\dfrac{n-nx^{n+1}-n-1+(n+1)x^n}{1-x^n-x^{n+1}+x^{2n+1}}\\
&=\lim_{x\to 1}\dfrac{-n(n+1)x^n+n(n+1)x^{n-1}}{-(n+1)x^n-nx^n+（2n+1)x^{2n}}\\
&=\lim_{x\to 1}\dfrac{-n^2(n+1)x^{n-1}+n(n-1)(n+1)x^{n-2}}{-n(n+1)x^{n-1}-n^2x^{n-1}+2n(2n+1)x^{2n-1}}\\
&=\dfrac{n(n+1)[n-1-n]}{-n^2-n-n^2+4n^2+2n}=-\dfrac{1}{2}
\end{align*}
This problem is creat a teacher of China, zhejiang university,it is said that teacher want Floored the arrogance of the students.
then I can't,Thank you
| Just for fun, I'm adding this long computation which doesn't involve the Hospital's rule and explains your (...).
$$
\lim_{x \to 1}\left(\frac{n}{1-x^n}-\frac{n+1}{1-x^{n+1}}\right) \\
= \lim_{x \to 1} \frac {\frac{n}{1 + x + \cdots + x^{n-1}} - \frac{n+1}{1+x+\cdots+x^n} }{1-x} \\
= \lim_{x \to 1} \frac 1{1+x+\cdots+x^{n-1}} \frac 1{1+x+\cdots +x^n} \frac{n(1 + x + \cdots + x^n) - (n+1)(1+x+\cdots+x^{n-1}) }{1-x} \\
= \frac 1{n(n+1)} \lim_{x \to 1} \frac {nx^n - (1+x+\cdots+x^{n-1})}{1-x} \\
= \frac{-1}{n(n+1)} \lim_{x \to 1} \frac{(x^n - 1) + (x^n - x) + \cdots + (x^n - x^{n-1})}{x-1} \\
= \frac{-1}{n(n+1)} \lim_{x \to 1} \frac{(x^n - 1) + x(x^{n-1} - 1) + \cdots + x^{n-1}(x - 1)}{x-1} \\
= \frac{-1}{n(n+1)} \lim_{x \to 1} \left( (1+x+\cdots+x^{n-1}) + x(1+x+\cdots + x^{n-2}) + \cdots + x^{n-2}(1+x) + x^{n-1}(1) \right) \\
= \frac{-1}{n(n+1)} \lim_{x \to 1} \sum_{i=0}^{n-1} \sum_{j=0}^{n-i} x^{i+j}\\
= \frac{-1}{n(n+1)} \left( n + (n-1) + \cdots + 1 \right) \\
= \frac {-1}2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/624649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How prove this $H_{2n}-H_{n}+\frac{1}{4n}>\ln{2}$
Show that, for every positive integer $n$,
$$\dfrac{1}{n+1}+\dfrac{1}{n+2}+\cdots+\dfrac{1}{2n}+\dfrac{1}{4n}>\ln{2}$$
I know this
$$\lim_{n\to\infty}\dfrac{1}{n+1}+\dfrac{1}{n+2}+\cdots+\dfrac{1}{2n}=\ln2$$
and use this
$$\ln{(1+\dfrac{1}{n})}<\dfrac{1}{n}$$ is not useful
But for this inequality I can't. Thank you
| A not at all tedious computation shows that the expression - which I rearrange a tiny bit -
$$\begin{align}
\frac{1}{2n} + \frac{1}{n+1} + \dotsc + \frac{1}{2n-1} + \frac{1}{4n} &= \sum_{k=0}^{n-1} \frac{1}{2}\left(\frac{1}{n+k} + \frac{1}{n+k+1}\right)\\
&= \frac{1}{n} \sum_{k=0}^{n-1} \frac{1}{2}\left(\frac{1}{1 + \frac{k}{n}} + \frac{1}{1+\frac{k+1}{n}}\right)
\end{align}$$
is a trapezium sum for the integral
$$\int_0^1 \frac{dt}{1+t} = \log 2,$$
and since $\frac{1}{1+t}$ is convex, the trapeziums have greater area than the corresponding part of the integral.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/624984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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If $a=\sqrt{4+\sqrt{5+a}}\;\;,b=\sqrt{4-\sqrt{5+b}}\;\;,c=\sqrt{4+\sqrt{5-c}}\;\;d=\sqrt{4-\sqrt{5-d}}$. Then $abcd $ If $a,b,c,d$ are $4$ distinct positive real no. such that
$a=\sqrt{4+\sqrt{5+a}}\;\;,b=\sqrt{4-\sqrt{5+b}}\;\;,c=\sqrt{4+\sqrt{5-c}}\;\;d=\sqrt{4-\sqrt{5-d}}$.
Then $abcd = $
$\bf{My\; Try}::$ Let here we form an equation $\displaystyle x=\sqrt{4\pm \sqrt{5\pm x}},$ where $x=a,b,c,d$
$\Longrightarrow \displaystyle x^2=4\pm \sqrt{5\pm x}$
$\Longrightarrow \displaystyle \left(x^2-4\right)^2=\left(\sqrt{5\pm x}\right)^2 = (5+x)\;\;,\;\;(5-x)$
$\Longrightarrow \displaystyle x^4+16-8x^2=5+x\;\; ,\;\; 5-x$
$\Longrightarrow \displaystyle x^4-8x^2-x+11=0\;\;,x^4-8x^2+x+11=0$
Now i did not understand how can i solve it.
Help me
Thanks
| Hint $a,b,-c,-d$ are the roots of the same equation.
| {
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Find all $z\in\Bbb C$ such that $|z+1|+ |z-1|=4$ I'd like to find all points of the complex plane which satisfy
$$|z+1| + |z-1| = 4. $$
I know this is an ellipsis with foci $1$ and $-1$, and I know that the answer is
$$3 x^2+4 y^2 \leq 12,$$
but I can't find a correct way of getting there.
First, I write $z$ as $x + i y$ and square both sides of the equation, then divide by 2 and get
$$x^2+y^2+1+\sqrt{(x-1)^2+y^2} \sqrt{(x+1)^2+y^2} =8.$$
Pass $x^2+y^2+1$ to the RHS (right hand side), then
$$\sqrt{(x-1)^2+y^2} \sqrt{(x+1)^2+y^2} =7-x^2-y^2.\tag{1} $$
Now, I would have to square both sides of the equation like this,
$$((x-1)^2+y^2)((x+1)^2+y^2) = (7-x^2-y^2)^2, \tag{2}$$
but the problem is that I cannot assure that the RHS is not negative, so there could be a value for $z$ such that (2) is satisfied but not (1), i.e it could exist $z=x + i y$ which satisfies
$$\sqrt{(x-1)^2+y^2} \sqrt{(x+1)^2+y^2} =-(7-x^2-y^2)\tag{3} $$
in which case also satisfy (2) but not (1)!
So I would get an incorrect solution.
| Note that for $z=x+iy$, we have the following three equalities:
$$|z|^2 = x^2+y^2$$
$$|z+1|^2 = |z|^2+1 + 2x $$
$$|z-1|^2 = |z|^2+1 - 2x$$
Now, from the given equation we have:
$$|z+1| = 4-|z-1|$$
By squaring once, we get:
$$|z+1|^2 = 16+|z-1|^2 - 8|z-1|.$$
From the second and third equalities above, we get:
$$4x = 16-8|z-1|$$
Hence:
$$2|z-1| = 4-x$$
Squaring AGAIN:
$$4|z-1|^2 = (4-x)^2$$
Using the third equality:
$$4|z|^2+4-8x = (4-x)^2$$
Finally, using the first equality gives:
$$4(x^2+y^2)+4-8x = (4-x)^2,$$
from where we have:
$$3x^2+4y^2 = 12.$$
| {
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Prove Inequality without induction. I showed this inequality by induction. I want other methods to prove it.
$(\frac{2n}{3}+\frac{1}{3})\sqrt{n}\leq \sum_{k=1}^{n}\sqrt{k}\leq (\frac{2n}{3}+\frac{1}{2})\sqrt{n}$
Thank
| ($\frac{2n}{3}$+$\frac{1}{3})$$\sqrt{n}$$\leq $$\sum_{k=1}^{n}$$\sqrt{k}$$\leq $$(\frac{2n}{3}$+$\frac{1}{2})$$\sqrt{n}$ ?
By these inequalities :
$\forall$ k $\geq$ 1 : $\frac{2}{3}$[$k^\frac{3}{2}$-$(k-1)^\frac{3}{2}$]+$\frac{1}{3}$[$k^\frac{1}{2}$-$(k-1)^\frac{1}{2}$]$\leq$ $k^{\frac{1}{2}}$$\leq$ $\frac{2}{3}$[$k^{\frac{3}{2}}$-$(k-1)^{\frac{3}{2}}$]+$\frac{1}{2}$[$k^{\frac{1}{2}}$-$(k-1)^{\frac{1}{2}}$] $(*)$
and we sum by telescoping we get the result.
show (*)
3$a_{k}$=2[$k^{\frac{3}{2}}$-$(k-1)^{\frac{3}{2}}$]+[$k^{\frac{1}{2}}$-$(k-1)^{\frac{1}{2}}$] - 3$k^{\frac{1}{2}}$= 2$k^{\frac{1}{2}}$(k-1)-$(k-1)^{\frac{1}{2}}(2(k-1)+1)$ $\\$
3$a_{k}$=2$k^{\frac{1}{2}}$(k-1)-$(k-1)^{\frac{1}{2}}$(2k-1)$\leq $0 because 4k$(k-1)^{2}$$\leq$ (k-1)$(2k-1)^{2}$ indeed 4k(k-1)-(2k-1)²= -1$\leq $0
6$b_{k}$=6$k^{\frac{1}{2}}$-4[$k^{\frac{3}{2}}$-$(k-1)^{\frac{3}{2}}$]-3[$k^{\frac{1}{2}}$-$(k-1)^{\frac{1}{2}}$]
6$b_{k}$=$(k-1)^{\frac{1}{2}}$(4k-1)-$k^{\frac{1}{2}}$(-3+4k)$\leq$ 0 because (k-1)$(4k-1)^{2}$$\leq$ k$(4k-3)^{2}$
indeed (k-1)$(4k-1)^{2}$ - k$(4k-3)^{2}$= -1$\leq$0
| {
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"timestamp": "2023-03-29T00:00:00",
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finding sides of a triangle when circumradius and inradius are given
The radius of the circumscribed circle of a right triangle is $15 cm$ and the radius of its inscribed circle is $6 cm$. Find sides of triangle.
From another site I got, $c=30$, $a+b=2(15+6)=42$. $a+b+c=72$. $ab=6\times 72=432$. So, sides are $18$, $24$.
I didn't get how we wrote $a+b$ and $ab$ equations. What's the relation of sum and product of sides with circumscribed and inscribed radii.
| Because radius of circumcircle is $15$, it means yes that $c$ or hypotenuse is $30$.
Also,
$r=(a+b-30)/2$
Putting values we get,
$a+b=42$
Now we have
$a^2+b^2=900$
$a+b=42$
From which, we have $b=42-a$.
We get,
$a^2+(42-a)^2=900$
$a^2+1764-84\cdot a+a^2=900$
$2\cdot a^2-84\cdot a+864=0$
Or, $a^2-42\cdot a+432-0$
$D=1764-1728=36$
Could you continue please? Also reject negative values.
EDITED:
$a_1=(42+6)/2=24$
and $a_2=(42-6)/2=18$
Therefore,
$b_1=18$ and $b_2=24$
Now you know what is the relationship between small radius and sides, and also hypotenuse and big radius. Sure you can find a general formula for relationship between sides combination and radius, but it would be a little tricky, you should express from Pythagorean theorem sides and put in radius calculation formulas, or at least use angles formula, which you can find easily on the internet.
| {
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find the second derivative $(x^3+x-1)(x^3+1)$ $$(x^3+x-1)(x^3+1)$$
$$=x^6+x^4-x^3+x^3+x-1$$
$$f'(x)= 6x^5+4x^3-3x^2+3x^2+1$$
Am I suppose to cancel out $-3x^2+3x^2$?
$$f''(x)= 30x^4+12x^2-6x+6x$$
Can someone check my work please? Thank you so much!
| You could also work it out this way:
The Product Rule gives us $ \ [ fg ]' \ = \ f' \cdot g \ + \ f \cdot g' $ . Applying it again gives us
$$ [fg]'' \ = \ f'' \cdot g \ + \ 2 \cdot f' \cdot g' \ + \ f \cdot g'' \ . $$
(The "higher-derivative" Product Rule has a resemblance to the Binomial Theorem. [It's more like a relation to the Theorem if you use derivative operators.])
For the functions $ \ f(x) \ = \ x^3 \ + \ x \ - \ 1 \ \ $ and $ \ g(x) \ = \ x^3 \ + \ 1 \ \ , $ we have
$$ f'(x) \ = \ 3x^2 \ + \ 1 \ \ , \ \ f''(x) \ = \ 6x \ \ , \ \ g'(x) \ = \ 3x^2 \ \ , \ \ g''(x) \ = \ 6x \ \ ,$$
making the second derivative of the product $ \ fg \ $
$$ 6x \ \cdot \ (x^3 + 1 ) \ + \ 2 \ \cdot \ (3x^2 + 1) \ \cdot \ 3x^2 \ + \ (x^3 \ + \ x \ - \ 1 ) \ \cdot \ 6x $$
$$ = \ 6x^4 \ + \ 6x \ + \ 18x^4 \ + \ 6x^2 \ + \ 6x^4 \ + \ 6x^2 \ - \ 6x $$
$$ = \ 30x^4 \ + \ 12x^2 \ \ . $$
Granted, it's not much of a saving in calculation effort here, but it can be handy for more complicated functions...
| {
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Evaluate $\sum_{n=1}^{\infty} \left( \arctan \frac{4n - 1}{2} - \arctan \frac{4n - 3}{2} \right)$ I got stuck on the following series:
$$ \sum_{n=1}^{\infty} \left( \arctan \frac{4n - 1}{2} - \arctan \frac{4n - 3}{2} \right). $$
I can't seem to make an approach because there's $-3$ not $+3$. Please help!
| Let $a_n$ be the $n^{th}$ term of the series at hand. We have
$$\begin{align}
a_n = & \tan^{-1}\frac{4n-1}{2} - \tan^{-1}\frac{4n-3}{2}\\
= & \tan^{-1}\left(\frac{\frac{4n-1}{2}-\frac{4n-3}{2}}{1 + \frac{4n-1}{2}\frac{4n-3}{2}}\right)
= \tan^{-1}\left(\frac{1}{(2n-1)^2 + \frac34}\right)\\
\end{align}$$
Notice $\;\tan^{-1}(x) = \Im\log(1 + ix)\;$ for real $x$, we can rewrite $a_n$ as
$$a_n = \Im\left\{\log\left( 1 + \frac{i}{(2n-1)^2 + \frac34}\right)\right\}
= \Im\left\{\log\left( 1 + \frac{\frac34 + i}{(2n-1)^2}\right)\right\}
$$
Compare this with the factors in the
infinite product expansion
of $\cosh x$:
$$\cosh x = \prod_{n=1}^{\infty}\left(1 + \frac{4x^2}{(2n-1)^2\pi^2}\right)$$
We find$\color{blue}{^{[1]}}$
$$\begin{align}
\sum_{n=1}^\infty a_n
= &\Im\left\{\log\cosh\left(\frac{\pi}{2}\sqrt{\frac34+i}\right)\right\}
= \Im\left\{\log\cosh\left[\frac{\pi}{2}\left(1 + \frac{i}{2}\right)\right]\right\}\\
= & \Im\left\{\log\left[\cosh\frac{\pi}{2} \cos\frac{\pi}{4} + \sinh\frac{\pi}{2}\sin\frac{\pi}{4}i\right]\right\}
= \Im\left\{\log\left[ 1 + \tanh\frac{\pi}{2} i\right]\right\}\\
= & \tan^{-1}\left[\tanh\left(\frac{\pi}{2}\right)\right]
\end{align}$$
Notes
*
*$\color{blue}{[1]}$ Given any two complex numbers $u$ and $v$, $\log(uv)$ need not equal to $\log u + \log u$ in general. Instead, we have
$$\log(uv) = \log u + \log v + i2\pi N$$
for some integer $N$. So in principle,
$$\begin{align}
a_n &= \Im\left\{\log\left( 1 + \frac{\frac34 + i}{(2n-1)^2}\right)\right\}\\
\implies \sum_{n=1}^\infty a_n
&= \Im\left\{\log\prod_{n=1}^\infty\left( 1 + \frac{\frac34 + i}{(2n-1)^2}\right)\right\}
+ 2\pi N
\end{align}$$
for some integer $N$ only. However, $a_n$ is small enough and the sum falls within the range $(-\frac{\pi}{2},\frac{\pi}{2})$, the $N$ here is actually zero. The naive looking replacement:
$$\sum_{n=1}^\infty a_n \quad\longrightarrow\quad
\Im\left\{\log\cosh\left(\frac{\pi}{2}\sqrt{\frac34+i}\right)\right\}$$
does work.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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How to find what the sum of this infinite series is. I have the following series:
$\sum_{n\geq3} \frac{4n-3}{n^3-4n}$
which I've simplified into the following:
$\frac{3}{4}\sum_{n\geq3}{ \frac{1}{n}} + \frac{5}{8}\sum_{n\geq3} \frac{1}{n-2} - \frac{11}{8}\sum_{n\geq3} \frac{1}{n+2}$
And this is where I'm stuck... How do I calculate the total sum?
Thanks in advance!
| Hint
Do a change of index to express all the sums on the form $\sum_n \frac 1 n$ and simplify.
Edit First notice as user127001 said you are not allowed to split the sum on three divergent sum so the key is to use a partial sum and finaly you pass to the limit.
Now from your work
$$\frac{3}{4}\sum_{n=3}^N{ \frac{1}{n}} + \frac{5}{8}\sum_{n=3}^N \frac{1}{n-2} - \frac{11}{8}\sum_{n=3}^N \frac{1}{n+2}=\frac{3}{4}\sum_{n=3}^N{ \frac{1}{n}}+\frac{5}{8}\sum_{n=1}^{N-2} \frac{1}{n}- \frac{11}{8}\sum_{n=5}^{N+2} \frac{1}{n}$$
and we cancel all the terms from $n=5$ to $n=N-2$ and we pass to the limit $N\to\infty$ to conclude.
| {
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Explicit closed form $\sum \frac{1}{an^2+bn+c}$ I know that $\sum_{n\geq 1} \frac{1}{n^2}=\pi^2/6$, is there a simple way to get an explicit closed form $\sum \frac{1}{an^2+bn+c}$, where $a,b,c$ are integers, $a\neq 0$?
| For all $z\in\mathbb{C}$,
$$
\sum_{n=1}^\infty\left(\frac1n-\frac1{n+z}\right)=\psi(z+1)+\gamma\tag{1}
$$
where $\psi(z)=\frac{\mathrm{d}}{\mathrm{d}z}\log(\Gamma(z))$ is the digamma function.
Partial fractions says
$$
\begin{align}
\frac1{an^2+bn+c}
&=\frac1{\sqrt{b^2-4ac}}\left(\frac1{n+\frac{b-\sqrt{b^2-4ac}}{2a}}-\frac1{n+\frac{b+\sqrt{b^2-4ac}}{2a}}\right)\\
&=\frac1{\sqrt{b^2-4ac}}\left(\frac1n-\frac1{n+\frac{b+\sqrt{b^2-4ac}}{2a}}\right)\\
&-\frac1{\sqrt{b^2-4ac}}\left(\frac1n-\frac1{n+\frac{b-\sqrt{b^2-4ac}}{2a}}\right)\tag{2}
\end{align}
$$
If $b^2\ne4ac$, applying $(1)$ to $(2)$ yields
$$
\begin{align}
\sum_{n=1}^\infty\frac1{an^2+bn+c}
&=\frac1{\sqrt{b^2-4ac}}\psi\left(\frac{b+\sqrt{b^2-4ac}}{2a}+1\right)\\
&-\frac1{\sqrt{b^2-4ac}}\psi\left(\frac{b-\sqrt{b^2-4ac}}{2a}+1\right)\tag{3}
\end{align}
$$
Taking the derivative of $(1)$ gives
$$
\sum_{n=1}^\infty\frac1{(n+z)^2}=\psi^{\,\prime}(z+1)\tag{4}
$$
In the case of $b^2=4ac$, we have
$$
\frac1{an^2+bn+c}=\frac1a\frac1{\left(n+\frac{b}{2a}\right)^2}\tag{5}
$$
Applying $(4)$ to $(5)$ yields
$$
\sum_{n=1}^\infty\frac1{an^2+bn+c}
=\frac1a\psi^{\,\prime}\left(\frac{b}{2a}+1\right)\tag{6}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Two trigonometric eliminations I have a couple of trigonometric elimination questions for you...
$1) a \sin \theta = b \sin (2 \theta)$; $c \cos \theta = d \cos (2 \theta)$
*NOTE: the first equation was incorrect...I had $\cos 2 \theta$ and it should be $\sin 2 \theta$.
I was able to get the answer to $2(abc - a^2 d + 2b^2 d)=0$, but the actual answer is $(c^2 - d^2)(abc - a^2 d + 2b^2 d)=0$. Where does the $(c^2 -d^2)$ come from?
UPDATE: I think I know $(c^2-d^2)$ comes from. If $\theta$ =0, then cos $\theta$ = 1 and in turn cos 2 $\theta$ = 1; plugging these values into the first equation gives us $c = d$. (I already did the second condition where $\frac {a} {2b}$)) If we square the first condition and subtract, that gives us $(c^2-d^2)$; since this also equals $2(abc - a^2 d + 2b^2 d)=0$, we can multiply both sides by $(c^2-d^2)$ to get the final result.
UPDATE #2: Actually, the more I thought about it, this is correct.
$2) \displaystyle x \cos \theta + y \sin \theta = \cos (3 \theta)$; $x \sin \theta - y \cos \theta = 3 \sin (3 \theta)$
I'm not sure where to start...I've tried squaring both equations, multiplying, etc. (The answer for this is $\displaystyle(x^2 + y^2)(x^2+y^2+18)+8x(x^2-3y^2) = 27$.)
UPDATE: I was able to get the following for x and y (thanks to the suggestion!)
$\displaystyle x= 2\cos 2 \theta - \cos 4 \theta$
$\displaystyle y = 2\sin 2 \theta + \sin 4 \theta$
UPDATE 2/16/14: BREAKTHROUGH! Never mind...that answer bombed out.
UPDATE 2/17/14: Not giving up and using the advice below, I finally got $\displaystyle c = \frac {(x-1)^2 + (y^2-4)}{12}$...I'm going to plug this in to the equation for x and see how it goes. If all works well, I should finally get the answer, save all the cleanup work. Never mind.
UPDATE 4/8/17: I reposted question #2 to get a fresh perspective...see Trigonometric elimination (reprise from 2014). If it's a duplicate, my apologies!
Thanks for your help!
| Here's a brute force approach to #2. For simplicity, I'll write $s$ and $c$ for $\sin\theta$ and $\cos\theta$.
*
*Expand $\cos 3\theta$ and $\sin 3\theta$ $$\begin{align}
x c + y s &= c \; ( 4 c^2 - 3 ) \\
x s - y c &= s \; ( 4 c^2 - 1 )
\end{align}$$
*Isolate $s$ terms, and square, so that we can re-write in terms of $c$. $$\begin{align}
(y s)^2 &= c^2 ( 4 c^2 - 3 - x )^2 \quad \to \quad y^2 ( 1 - c^2 ) - c^2 ( 4 c^2 - 3 - x )^2 = 0\\
(y c)^2 &= s^2 ( 4 c^2 - 1 - x )^2 \quad \to \quad y^2 c^2 - ( 1 - c^2 )( 4 c^2 - 1 - x )^2 = 0
\end{align}$$
*Invoke one of my favorite tools ---the "method of resultants" (which I describe a bit in this answer)--- to eliminate $c$. Mathematica's (and/or WolframAlpha's) Resultant[] function makes it easy, yielding this polynomial equation: $$\begin{align}(\;x^4 - 4 y^4 + 3 x^2 y^2 + 8 x^3 - 18 x y^2 + 18 x^2 + 27 y^2 - 27 \;) & \\ \cdot\;(\;x^4 + y^4 + 2 x^2 y^2 + 8 x^3 - 24 x y^2 + 18 x^2 + 18 y^2 - 27 \;) &= 0\end{align}$$
*The second factor corresponds to the answer you expect. Presumably, the first factor is extraneous in the current context (as is often the case with resultant results), but it's not obvious to me why that is.
The same process works for #1.
| {
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$2 - 4 + 6 - 8 + \cdots = 1 - 1 + 1 -1 + \cdots$? I want to talk about the weirdness of $2\sum\limits_{n=1}^\infty n(-1)^{n-1}$ ,
$$\sum\limits_{n=1}^\infty n(-1)^{n-1} = 1 - 2 + 3 - 4 + 5 - 6 + \dots$$
$$\times 2 \implies 2\sum\limits_{n=1}^\infty n(-1)^{n-1} = 2 - 4 + 6 - 8 + 10 - 12 + \dots\quad\longleftarrow (1)$$
But
$2\sum\limits_{n=1}^\infty n(-1)^{n-1}$
$=\sum\limits_{n=1}^\infty n(-1)^{n-1} + \sum\limits_{n=1}^\infty n(-1)^{n-1}$
$= 1 - 2 + 3 - 4 + 5 - 6 + \dots$
$\quad\quad+ 1 - 2 + 3 - 4 + 5 - 6 +\dots$
$= 1 - 1 + 1 - 1 + 1 - 1 + 1 - \dots \quad\longleftarrow (2)$
Is it true that these series are equal? The manipulation that I've made to get $(2)$ seems very unnatural but in my defense, the way I add them shouldn't matter because the result should be the same. For example, using the same manipulation on a finite series, $2( 1 + 2 + 3) = 1 + 3 + 5 +3$. It yields the same sum but I'm not sure if this is applicable to infinite series a well.
Using $S_\infty = \large \frac{a}{1-r}$ on $(2)$, it can be seen that it's sum is $\large \frac{1}{2}$. Can I say that the sum of $(1)$ is the same?
I'm a high school student and I have no real background in sequences, so a simple and understandable answer is greatly appreciated. I may not understand many complicated terms but I am willing to learn.
| Sangchul Lee commented above that the manipulations are correct for Abel sums.
In fact, the series $1-2+3-4+\cdots$ corresponds to the function
$$
f_1(x)=\frac{x}{(1+x)^2}=x-2x^2+3x^3-4x^4+\cdots~,
$$
the series $0+1-2+3-4+\cdots$ corresponds to the function
$$
f_2(x)=\frac{x^2}{(1+x)^2}=x^2-2x^3+3x^4-4x^5+\cdots
$$
and the series $1-1+1-1+\cdots$ corresponds to the function
$$
f_3(x)=\frac{x}{1+x}=x-x^2+x^3-x^4+\cdots~,
$$
which are convergent for $|x|<1$.
The manipulations correspond to the identity
$$
f_1(x)+f_2(x)=f_3(x)~,
$$
that is,
$$
\frac{x}{(1+x)^2}+\frac{x^2}{(1+x)^2}=\frac{x}{1+x}~,
$$
which is also valid in the limit as $x\to 1^-$.
The Abel sum of $1-2+3-4+\cdots$ is defined by
$\lim_{x\to 1^-} f_1(x)=1/4$,
the Abel sum of $0+1-2+3-4+\cdots$ is defined by
$\lim_{x\to 1^-} f_2(x)=1/4$,
and the Abel sum of $1-1+1-1+\cdots$ is defined by
$\lim_{x\to 1^-} f_3(x)=1/2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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The value of $\cos^4\frac{\pi}{8} + \cos^4\frac{3\pi}{8}+\cos^4\frac{5\pi}{8}+\cos^4\frac{7\pi}{8}$ Problem :
The value of $\cos^4\frac{\pi}{8} + \cos^4\frac{3\pi}{8}+\cos^4\frac{5\pi}{8}+\cos^4\frac{7\pi}{8}$
If this could have been like this $\cos\frac{\pi}{8} + \cos\frac{3\pi}{8}+\cos\frac{5\pi}{8}+\cos\frac{7\pi}{8}$ then we can take the terms like
$(\cos\frac{3\pi}{8}+\cos\frac{5\pi}{8} ) + (\cos\frac{\pi}{8} +\cos\frac{7\pi}{8})$
and solve further , but in this case due to power of 4 I am unable to proceed please suggest... thanks..
| HINT
From $\cos(2x) = 2 \cos^2(x) - 1$, replace each term already squared by something which looks that the cosine of the double angle. Since your base angle is $\pi/8$, the double angle is $\pi/4$ and you know the value of the cosine of this angle.
I am sure you can take from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/645874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
An upper bound on certain finite trigonometric series given a lower bound
Let $f$ be the function $f(x)=1+a\sin{x}+b\cos x+c\sin{(2x)}+d\cos{(2x)}$, where $a,b,c,d$ are arbitrary real numbers. Prove that if $f(x)>0$ for all $x\in \mathbb R$, then $f(x)<3$ for all $x\in \mathbb R$.
My try:
\begin{align*}f(x)&=1+a\dfrac{e^{ix}-e^{-ix}}{2i}+b\dfrac{e^{ix}+e^{-ix}}{2}+c\dfrac{e^{2ix}-e^{-2ix}}{2i}+d\dfrac{e^{2ix}+e^{-2ix}}{2}\\
&=1+e^{ix}\left(\dfrac{a}{2i}+\dfrac{b}{2}\right)+e^{-ix}\left(-\dfrac{a}{2i}+\dfrac{b}{2}\right)+e^{2ix}\left(\dfrac{c}{2i}+\dfrac{d}{2}\right)+e^{-2ix}\left(\dfrac{d}{2}-\dfrac{c}{2i}\right).
\end{align*}
I don't see a way to proceed from there. Thank you very much.
| Today,I have solve this problem.
we only note this problem have follow form
$$f\left(x-\dfrac{2\pi}{3}\right)+f(x)+f\left(x+\dfrac{2\pi}{3}\right)=3$$
because use
$$\sin{(x+y)}+\sin{(x-y)}=2\sin{x}\cos{x}$$
$$\cos{(x+y)}+\cos{(x-y)}=2\cos{x}\cos{y}$$
$$\Longrightarrow \sin{(x-\dfrac{2\pi}{3})}+\sin{(x+\dfrac{2\pi}{3})}=-sin{x}$$
$$\cdots\cdots$$
$$\cos{(2x+\dfrac{4\pi}{3})}+\cos{(2x-\dfrac{4\pi}{3})}=-\cos{(2x)}$$
so
$$\sin{(x-\dfrac{2\pi}{3})}+\sin{x}+\sin{(x+\dfrac{2\pi}{3})}=0$$
$$\cos{(x-\dfrac{2\pi}{3})}+\cos{x}+\cos{(x+\dfrac{2\pi}{3})}=0$$
$$\sin{(2x-\dfrac{4\pi}{3})}+\sin{2x}+\sin{(2x+\dfrac{4\pi}{3})}=0$$
$$\cos{(2x-\dfrac{4\pi}{3})}+\cos{2x}+\cos{(2x+\dfrac{4\pi}{3})}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/646272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 2,
"answer_id": 0
} |
Solving the differential equation: $ x = 11 (y')^{10} \cdot y'' $ $$ x = 11 (y')^{10} \cdot y'' $$
I proceed as follows:
$$
\begin{align}
\text{Assume } v \equiv v(y) &= \dfrac{dy}{dx} \\
\implies \dfrac{d^2y}{dx^2} &= \dfrac{dv}{dy} \cdot \dfrac{dy}{dx} \\
&= v \cdot \dot{v} \tag{ $ \frac{dv}{dy} = \dot{v} $ } \\
\therefore \qquad x &= 11 v^{10} \dot{v}
\end{align}
$$
Solving which, I get: $ v^{11} = xy + c $. $ c $ being the constant of integration. What to do now?
How do I solve:
$$ \dfrac{dy}{dx} = \left( xy + c \right)^{\frac{1}{11}} $$
| You made a mistake in your derivations. Note that
$$ 11 v^{10}v' = x \implies v^{11} = \frac{x^2}{2}+c_1. $$
Now, one possible solution of the last equation is
$$ v = \left(\frac{x^2}{2}+c_1 \right)^{\frac{1}{11}},$$
which implies
$$ \frac{dy}{dx} = \left(\frac{x^2}{2}+c_1 \right)^{\frac{1}{11}}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/647129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How prove this $xy+yz+xz\le 2xyz+\frac{1}{2}$ let $x,y,z>0$ and such
$$x^2+y^2+z^2+2xyz=1$$
show that
$$xy+yz+xz\le 2xyz+\dfrac{1}{2}$$
My try: since
$$1=x^2+y^2+z^2+2xyz\ge xy+yz+xz+2xyz$$
then
$$xy+yz+xz\le 1-2xyz$$
so
we only prove follow this
$$1-2xyz\le 2xyz+\dfrac{1}{2}$$
$$\Longleftrightarrow xyz\ge\dfrac{1}{4}$$
But this is not true,so How prove my inequality? Thank you
| We first we first notice that there are always two of the three numbers, both greater than $\dfrac{1}{2}$,because of symmetry,we may assume that
$x,y\le\dfrac{1}{2},$ or $x,y\ge \dfrac{1}{2}$ and then
$$(2x-1)(2y-1)\ge 0\Longleftrightarrow x+y-2xy\le\dfrac{1}{2}$$
on the other hand,
$$1=x^2+y^2+z^2+2xyz\ge 2xy+z^2+2xyz$$
then
$$2xy(1+z)\le 1-z^2\Longrightarrow 2xy\le 1-z$$
we only have to multiply side by side the inequality from above
$$x+y-2xy\le\dfrac{1}{2},z\le 1-2xy$$
then
$$xz+yz-2xyz\le\dfrac{1}{2}-xy\Longleftrightarrow xy+xz+yz\le\dfrac{1}{2}+2xyz$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/650614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Convergence Proof: $\lim_{x\rightarrow\infty} \sqrt{4x+x^2}- \sqrt{x^2+x}$ I have to check whether the following expression converges; if yes I have to give the limit.
$$\lim_{x\rightarrow\infty} \sqrt{4x+x^2}- \sqrt{x^2+x}$$
Now I did the following:
$$\lim_{x\rightarrow\infty} \sqrt{4x+x^2}- \sqrt{x^2+x}$$
$$\lim_{x\rightarrow\infty} x\sqrt{\frac{4}{x}+1}- x\sqrt{1+\frac{1}{x}}$$
$$\lim_{x\rightarrow\infty} x \lim_{x\rightarrow\infty}(\sqrt{\frac{4}{x}+1}- \sqrt{1+\frac{1}{x}})$$
That confuses me. The left limit approaches $\infty$ while the right approaches $0$. What is wrong here or what can I conclude from that?
Thank you very muvh for your help in advance!
FunkyPeanut
| Hint: Multiply by $\frac{\sqrt{4x+x^2}+\sqrt{x^2+x}}{\sqrt{4x+x^2}+\sqrt{x^2+x}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/652245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Ramanujan's partial fraction decomposition of $\frac{1}{(x^2+a^2)\cdots(x^2+(a+n)^2)}$. \begin{align*}
\frac{1}{(x^2+a^2)\cdots(x^2+(a+n)^2)} &= \frac{2\Gamma(2a)}{\Gamma(n)\Gamma(2a+n)}\left(\frac{a}{x^2+a^2}-\frac{2a}{1!}\frac{n-1}{n+2a}\frac{a+1}{x^2+(a+1)^2}\right. \\
& \qquad + \left.\frac{2a(2a+1)}{2!}\frac{(n-1)(n-2)}{(n+2a)(n+2a+1)}\frac{a+2}{x^2+(a+2)^2}-\cdots\right).
\end{align*}
The preceding was by Ramanujan, appearing in one of his notebooks. How does one prove this?
Especially interesting is motiving the proof: given only the complete fraction on the left, is there a method that makes the right side almost immediately obvious? (Basically, it would be nice if the answers imagined the RHS didn't exist in the above equation).
| First of all there is a error in the expression on the left hand side. The correct right hand side is:
\begin{equation}
lhs=\frac{1}{(x^2+a^2)\dot \cdots \dot (x^2+(a+n-1)^2)}
\end{equation}
The left hand side is correct.
Now from partial fraction decomposition the right hand side clearly equals:
\begin{equation}
rhs = \sum\limits_{k=0}^{n-1} \frac{1}{x^2+(a+k)^2} \cdot C_k
\end{equation}
where:
\begin{eqnarray}
C_k&=&\frac{1}{\prod\limits_{j=0,j\neq k}^{n-1} (-(a+k)^2 + (a+j)^2)}=\frac{1}{\prod\limits_{j=0,j\neq k}^{n-1} (2 a+k+j)(j-k)} \\
&=&\frac{(-1)^{k}}{(2a+k)^{(k)} (k)! (2a+2 k+1)^{(n-1-k)}(n-1-k)!}\\
&=& \frac{1}{\Gamma(n)} \cdot \binom{n-1}{k} \cdot (-1)^k \cdot \frac{\Gamma(2a+k)}{\Gamma(2 a+n+k)} (2a+2 k) \\
&=& \frac{2 \Gamma(2a)}{\Gamma(n) \Gamma(2 a+n)} \cdot \binom{n-1}{k} (-1)^k \cdot \frac{(2 a)^{(k)}}{(2a+n)^{(k)}} \cdot (a+k)
\end{eqnarray}
which is exactly what we have on the right hand side. Here of course $k=0,\cdots,n-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/655968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Sum of radii of exspheres I am interested in finding some results for tetrahedron that would be analoguous to known results for triangle. In triangle with circumradius R and inradius r, if we consider the excircles $r_i$, then for their sum the formula
$$\sum r_i = 4R + r$$ holds.
I wonder if there is similar formula for the sum of the
exradii of the exspheres of the tetrahedron in terms
of the radii of the circumradius and inradius of the circumscribed and inscribed spheres?
$$\sum r_i = ? \quad i = 1,..,4.$$
Maybe you have seen similar problems or you can give me
references where to search further. Thank you.
| Such a relation does not and cannot exist.
Let us first write down something we know about in-sphere and ex-spheres for higher dimensions.
For any $d \ge 2$, consider $d+1$ points $\vec{v}_1, \vec{v}_2,\ldots,\vec{v}_{d+1} \in \mathbb{R}^{d}$ in general position.
Let $\Delta$ be the $d$-simplex formed by these $d+1$ points. Let $A_k$ be the area of the face on $\partial\Delta$ opposite to $\vec{v}_k$.
Let $A = \sum\limits_{k=1}^{d+1} A_k$ be the total "area" and $V$ be the "volume" of the simplex $\Delta$.
Let $r$ and $\vec{c}$ be the radius and center of the in-sphere.
Let $r_k$ and $\vec{c}_k$ be the radius of center of the ex-sphere on the face opposite to vertex $\vec{r}_k$.
In terms of $A_k$ and $V$, they are given by
$$\begin{cases}
r =& \frac{dV}{A}\\
\\
\vec{c} =& \frac{1}{A}\sum\limits_{k=1}^{d+1} A_k \vec{v}_k\\
\end{cases}
\quad\text{ and }\quad
\begin{cases}
r_k =& r\frac{A}{A-2A_k} = \frac{dV}{A - 2A_k}\\
\\
\vec{c}_k =& \frac{r_k}{r} \vec{c} + \left(1 - \frac{r_k}{r}\right) \vec{v}_k\\
\end{cases}$$
In particular, this means in any dimension, we have a relation between $r_k$ and $r$.
$$\sum_{k=1}^{d+1} \frac{1}{r_k} = \frac{d-1}{r}$$
Apply $AM \ge HM$ to the list of exradii $r_k$, we obtain a lower bound for their sum:
$$\frac{1}{d+1}\sum_{k=1}^{d+1}r_k \ge \frac{1}{\frac{1}{d+1}\sum_{k=1}^{d+1}\frac{1}{r_k}}
\quad\implies\quad
\sum_{k=1}^{d+1}r_k \ge \frac{(d+1)^2}{d-1} r$$
Let us back to our original problem at $d = 3$. We will like to show
For any $R > 3$ and $K \gtrsim \frac{(d+1)^2}{d-1} = 8$, there is a
tetrahedron whose circumradius is $R$, inradius $1$ and
$\sum\limits_{k=1}^{4} r_k = K$ whenever $K$ is sufficiently close to $8$.
For any $0 < s < p < R$, let $q = \sqrt{R^2-p^2}$ and $t = \sqrt{R^2 - s^2}$. Consider
the tetrahedron formed by following 4 points on a sphere of radius $R$:
$$\vec{v}_1 = (p, -q, 0),\; \vec{v}_2 = (p, q, 0),\;\vec{v}_3 = (-s,0,t)\;\text{ and }\;
\vec{v}_4 = (-s,0,-t)$$
It is easy to check
$$V = \frac23 (p+s)qt \quad\text{ and }\quad
\begin{cases}
A_1 = A_2 = t\sqrt{(p+s)^2 + q^2}\\
A_3 = A_4 = q\sqrt{(p+s)^2 + t^2}
\end{cases}
$$
This implies
$$r = \frac{3V}{A} = \frac{(p+s)qt}{t\sqrt{(p+s)^2 + q^2} + q\sqrt{(p+s)^2 + t^2}}\\
\sum_{k=1}^4 r_k = \sum_{k=1}^4\frac{3V}{A-2A_k} = 2(p+s)qt \left(\frac{1}{t\sqrt{(p+s)^2 + q^2}} + \frac{1}{q\sqrt{(p+s)^2+t^2}}\right)$$
Introduce a small parameter $\sigma$ such that $K = \frac{8}{1-\sigma^2}$. The condition that $r = 1$ and $\sum_{k=1}^4 r_k = K$ can be rewritten as
$$
\begin{cases}
t\sqrt{(p+s)^2 + q^2} = \frac{1+\sigma}{2} (p+s)qt\\
q\sqrt{(p+s)^2 + t^2} = \frac{1-\sigma}{2} (p+s)qt
\end{cases}
\quad\iff\quad
\begin{cases}
\frac{1}{q^2} + \frac{1}{(p+s)^2} = \left(\frac{1+\sigma}{2}\right)^2\\
\frac{1}{t^2} + \frac{1}{(p+s)^2} = \left(\frac{1-\sigma}{2}\right)^2\\
\end{cases}
$$
Let $p+s = 2\mu$, we can rewrite the condition on RHS as
$$
\begin{cases}
R^2 - p^2 = q^2 = 4\mu^2\Lambda_{+}(\mu)\\
R^2 - s^2 = t^2 = 4\mu^2\Lambda_{-}(\mu)
\end{cases}
\quad\text{ where }\quad
\begin{cases}
\Lambda_{+}(\mu) = \frac{1}{(1+\sigma)^2\mu^2 - 1}\\
\Lambda_{-}(\mu) = \frac{1}{(1-\sigma)^2\mu^2 - 1}
\end{cases}
$$
This will lead to a solution for $(p,s)$ of the from
$$\begin{cases}
p = \mu \left( 1 + \Lambda_{-}(\mu) - \Lambda_{+}(\mu) \right)\\
s = \mu \left( 1 - \Lambda_{-}(\mu) + \Lambda_{+}(\mu) \right)
\end{cases}
$$
provided one can find a root $\mu$ for following consistency condition:
$$R^2 = p^2 + 4\mu^2\Lambda_{+}(\mu) = \mu^2 \left( 1 + 2( \Lambda_{+}(\mu) + \Lambda_{-}(\mu)) + (\Lambda_{+}(\mu) - \Lambda_{-}(\mu))^2\right)$$
Let $\Omega(\mu)$ be the horrible expression at RHS. Notice in the limit of small $\sigma$,
we have
$$\Omega(\sqrt{3}) \sim 3\left( 1 + 2( \frac{1}{3-1} + \frac{1}{3-1} )\right) = 3^2 < R^2$$
Since $\Omega(R) > R^2$, there is a $\mu \in (\sqrt{3}, R)$ which satisfy $R^2 = \Omega(\mu)$. The corresponding $(p,s)$ will then give us
a tetrahedron with fixed circumradius and inradius and yet the sum of exradii is sort of arbitrary.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/656332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Infinite Series $1+\frac12-\frac23+\frac14+\frac15-\frac26+\cdots$ Was given the following infinite sum in class as a question, while we were talking about Taylor series expansions of $\ln(1+x)$ and $\arctan(x)$:
$$1+\frac12-\frac23+\frac14+\frac15-\frac26+\cdots$$
The question isn't homework or anything, just a thought tease.
I tried for a long while but couldn't find anything remotely close.
Thanks in advance for the help.
| This is a direct proof using $$\int_0^1 x^n dx= \frac{1}{n+1},$$ the sum of a geometric progression and integral evaluation.
$$\begin{align}
S&=\sum_{k=0}^\infty \left(\frac{1}{3k+1}+\frac{1}{3k+2}-\frac{2}{3k+3}\right)\\
&=\sum_{k=0}^\infty \int_0^1 \left(x^{3k}+x^{3k+1}-2x^{3k+2}\right)dx\\
&=\int_0^1 \sum_{k=0}^\infty \left(x^{3k}+x^{3k+1}-2x^{3k+2}\right)dx\\
&=\int_0^1 \frac{1+x-2x^2}{1-x^3}dx\\
&=\int_0^1 \frac{1+2x}{1+x+x^2}dx\\
&=\log(1+x+x^2)|_{0}^1\\
&=\log(3)
\end{align}$$
Thus, similarly to
$$\log(1+x)=1-\frac{1}{2}x+\frac{1}{3}x^2-\frac{1}{4}x^3...$$
we have
$$\log(1+x+x^2)=1+\frac{1}{2}x-\frac{2}{3}x^2+\frac{1}{4}x^3+\frac{1}{5}x^4-\frac{2}{6}x^5+...$$
Moreover, the pattern generalizes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/657241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
Number theory problem assignment $x^2 + x - 3 \equiv 0 \pmod {p^2}$. $p$ is a prime number and satisfies $13^{(p-1)/2} \equiv 1 \pmod p$. I need to find positive value of $x$. Please help me.
| If $p=2m+1$ then $x^2+x-3\equiv x^2 -2m-3\equiv (x-m)^2-m^2-3\pmod p$.
So your first task is to verify that $m^2+3$ is a square $\pmod p$.
With a root $a^2\equiv m^2+3\pmod p$ (so $a^2=m^2+3+kp $ for some $k$), let $x=(m\pm a)+rp$ and compute
$$\begin{align} x^2+x-3&=\left((m\pm a)^2+2(m\pm a)rp+r^2p^2\right)+(m\pm a)+rp-3\\
&={(m^2+a^2-3)}\pm (2m+1)a+m+(2m+1\pm a)pr+r^2p^2\\
&=2m^2+kp\pm pa+m+p^2r\pm apr+r^2p^2\\
&=pm+kp\pm pa \pm apr+p^2(r+r^2)\\
&\equiv p(m+k\pm a\pm ar)\end{align}$$
to obtain the condition $r\equiv \mp a^{-1}(m+k)-1$ and hence
$$x\equiv m-p\pm (a-a^{-1}(m+k))\pmod{p^2}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/663422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that the inequality $(1+ \frac{1}{n})^n < n$ holds for all $n \geq 3$ First we need to prove the basis. If we let $n=3$, then $(1+ \frac{1}{3})^3 < 3$
$(\frac{3}{3}+ \frac{1}{3})^3 < 3$
$(\frac{4}{3})^3 < 3$
$(\frac{64}{27}) < 3$
The inequality statement is true
For $P(n), (1+ \frac{1}{n})^n < n$
We assume that $(1+ \frac{1}{n})^n < n$ is true for $P(n+1)$
$(1+ \frac{1}{n+1})^{n+1} < n+1$
$(1+ \frac{1}{n+1})^{n})(1+ \frac{1}{n+1})^{1}) < n+1$
And then I'm stuck afterwards. I know that there are a variety of problems that use induction and they have different methods, but I only know the ones that are similar to $1+2+3+...+n = n+2$ or $7^n-8^n$ is divisible by $8$. Is there any technique to tackle this type of problem?
| As canaaerus points out in the comments, we can in fact prove
$$\left(1+\frac{1}{n}\right)^n<3$$
In fact there's a proof of this that doesn't even use induction! I think it's worth writing down here since it's so simple (and it won't be giving away the answer to the homework problem, since the teacher presumably expects induction). We will binomially expand and use the following facts:
$${{n}\choose{r}}=\frac{n\cdot(n-1)\cdot\ldots\cdot(n-r+1)}{r!}\leq\frac{n\cdot n\cdot\ldots\cdot n}{r!}=\frac{n^r}{r!}$$
and (for $r\geq1$)
$$\frac{1}{r!}=\frac{1}{1\cdot2\cdot3\cdot4\cdot\ldots\cdot r}\leq\frac{1}{1\cdot2\cdot2\cdot2\cdot\ldots\cdot 2}=\frac{1}{2^{r-1}}$$
Here's the proof:
$$\begin{align*}
\left(1+\frac{1}{n}\right)^n&=\sum^{n}_{r=0}{{n}\choose{r}}\frac{1}{n^r}\\
&\leq\sum^{n}_{r=0}\frac{n^r}{r!}\frac{1}{n^r}\\
&=\sum^{n}_{r=0}\frac{1}{r!}\\
&=1+\sum^{n}_{r=1}\frac{1}{r!}\\
&\leq1+\sum^{n}_{r=1}\frac{1}{2^{r-1}}\\
&<1+\sum^{\infty}_{r=1}\frac{1}{2^{r-1}}\\
&=1+2=3
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/669380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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rearranging a trig equation this is a problem that im not 100% can be done as i derived the equation myself but please help if you can and if it cant be done let me know:
i need the end product to be $\tan(\theta) = \ldots$
and for there to be no $\theta$ on the RHS.
equation is:
$$\frac{96.04 \cdot D^2}{\left(U \cdot \cos(\theta)\right)^2} = 2\left(U \cdot \sin(\theta)\right)^2 + (19.6 \cdot W)$$
where D, U and W are constants.
thanks so much in advance!!
------edit-------
i believe i owe you both an apology, it turns out i made a mistake earlier on in my equation so the one i posted was not possible. i only came to this conclusion after subbing in the data values i had into both answers and getting imaginary numbers back. so if i could be a pain, here is the updated (and verified) equation:
\begin{equation*}
\frac{D}{U\cos (\theta )}=\frac{U\sin (\theta )}{C}+( \frac{A+U^{2}\sin
^{2}(\theta )}{B}) ^{0.5}
\end{equation*}
once again im after $\tan(\theta)=$ etc with no ($\theta$) on the RHS
sorry for the mess around and thanks for helping :)
| UPDATE (answer to the revised question). If we multiply the edited equation by $C\sqrt{B}U\cos \theta \neq 0$, we get
the equivalent equation
\begin{equation*}
\sqrt{B}CD=\sqrt{B}U^{2}\sin \theta \cos \theta +CU\left( A+U^{2}\sin
^{2}\theta \right) ^{1/2}\cos \theta .
\end{equation*}
Using the definition of $\tan \theta =\frac{\sin \theta }{\cos \theta }$ and
the fundamental trigonometric identity $\sin ^{2}\theta +\cos ^{2}\theta =1$
, we are able to express $\sin \theta $ and $\cos \theta $, as well as $\sin
\theta \cos \theta $, in terms of $t=\tan \theta $. A possible derivation is
as follows. From
\begin{equation*}
\tan ^{2}\theta =\frac{\sin ^{2}\theta }{\cos ^{2}\theta }=\frac{\sin
^{2}\theta }{1-\sin ^{2}\theta }=\frac{1-\cos ^{2}\theta }{\cos ^{2}\theta }=
\frac{1}{\cos ^{2}\theta }-1
\end{equation*}
we obtain successively
\begin{eqnarray*}
\cos ^{2}\theta &=&\frac{1}{1+\tan ^{2}\theta } \\
&\Rightarrow &\cos \theta =\pm \frac{1}{\sqrt{1+\tan ^{2}\theta }}=\pm \frac{
1}{\sqrt{1+t^{2}}} \\
&& \\
\sin ^{2}\theta &=&\left( 1-\sin ^{2}\theta \right) \tan ^{2}\theta
\Leftrightarrow \sin ^{2}\theta =\frac{\tan ^{2}\theta }{1+\tan ^{2}\theta }=
\frac{t^{2}}{1+t^{2}} \\
&\Rightarrow &\sin \theta =\pm \frac{\tan \theta }{\sqrt{1+\tan ^{2}\theta }}
=\pm \frac{t}{\sqrt{1+t^{2}}}
\end{eqnarray*}
and
\begin{eqnarray*}
\sin \theta &=&\cos \theta \tan \theta \\
&\Leftrightarrow &\sin \theta \cos \theta =\cos ^{2}\theta \tan \theta =
\frac{\tan \theta }{1+\tan ^{2}\theta }=\frac{t}{1+t^{2}}.
\end{eqnarray*}
Substituting these expressions in the first equation yields
\begin{eqnarray*}
\sqrt{B}CD &=&\frac{\sqrt{B}U^{2}t}{1+t^{2}}+CU\left( A+\frac{U^{2}t^{2}}{
1+t^{2}}\right) ^{1/2}\left( \pm \frac{1}{\sqrt{1+t^{2}}}\right) \\
\sqrt{B}CD\left( 1+t^{2}\right) -\sqrt{B}U^{2}t &=&\pm CU\left( A+\left(
A+U^{2}\right) t^{2}\right) ^{1/2}
\end{eqnarray*}
Squaring now both sides and rearranging the terms we obtain the following
quartic equation in $t$
\begin{equation*}
at^{4}+bt^{3}+ct^{2}+dt+e=0
\end{equation*}
whose coefficients are
\begin{eqnarray*}
a &=&BC^{2}D^{2} \\
b &=&-2BCDU^{2} \\
c &=&2BC^{2}D^{2}-C^{2}U^{4}+BU^{4}-AC^{2}U^{2} \\
d &=&-2BCDU^{2} \\
e &=&-2BCDU^{2}-AC^{2}U^{2}+BC^{2}D^{2}.
\end{eqnarray*}
unfortunately this is the simplest form i can get it to. is it still possible to do or am i wasting my time?
This equation (after depressed) although solvable algebraically by an auxiliary cubic equation has huge solutions (see Wikipedia entry Ferrari's solution).
Multiplying your equation by $U^{2}\cos ^{2}\theta\ne 0 $ we get the equivalent
equation
\begin{equation*}
2U^{4}\sin ^{2}\theta \cos ^{2}\theta +19.6WU^{2}\cos ^{2}\theta
-96.04D^{2}=0.
\end{equation*}
Using the identity $\sin ^{2}\theta =1-\cos ^{2}\theta $ and rearranging the
terms we obtain the following quadratic equation in $y=\cos ^{2}\theta $:
\begin{equation*}
ay^{2}+by+c=0\Leftrightarrow y=\frac{1}{2a}\left( -b\pm \sqrt{b^{2}-4ac}
\right) ,
\end{equation*}
whose coefficients are
\begin{equation*}
a=-2U^{4},\quad b=2U^{4}+19.6WU^{2},\quad c=-96.04D^{2}.
\end{equation*}
To get the result we just need to express $\tan \theta $ in terms of $y$
\begin{equation*}
\sin ^{2}\theta +\cos ^{2}\theta =1\Leftrightarrow \tan ^{2}\theta +1=\frac{1
}{\cos ^{2}\theta }=\frac{1}{y}\Leftrightarrow \tan \theta =\pm \sqrt{\frac{1}{y}-1}.
\end{equation*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/669615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
A closed form for the antiderivative of $\frac 1{\sin^5 x +\cos^5 x} $ Does there exist a closed form expression for $$\int \dfrac {dx}{\sin^5 x +\cos^5 x}? $$
| Utilize the identity
$$\sin^5 (t+\frac\pi4) +\cos^5(t+\frac\pi4)
=\sqrt2 \cos t\left(\frac{5}4-\cos^4 t\right)
$$
to integrate
\begin{align}
&\int \dfrac {1}{\sin^5 (t+\frac\pi4) +\cos^5(t+\frac\pi4)}dt\\
=& \frac{2\sqrt2}5 \int\bigg(\frac1{\cos^2 t}-\frac{1}{2\cos^2 t+\sqrt5}
-\frac{1}{2\cos^2 t-\sqrt5} \bigg)\cos t \>dt \\
=& \frac25\bigg(\sqrt2\tanh^{-1}\sin t
-\frac{\tanh^{-1}\frac{\sqrt2\sin t}{\sqrt{\sqrt5+2}}}{\sqrt{\sqrt5+2}}
+\frac{\tan^{-1}\frac{\sqrt2 \sin t}{\sqrt{\sqrt5-2}}}{\sqrt{\sqrt5-2}}\bigg)+C
\end{align}
Then, replace $t=x-\frac\pi4$ to arrive at the original integral.
In particular
\begin{align}
&\int_0^{\frac\pi2}\frac {dx}{\sin^5 x +\cos^5 x}
=\int_{-\frac\pi4}^{\frac\pi4} \dfrac {1}{\sin^5 (t+\frac\pi4) +\cos^5(t+\frac\pi4)}dt
\\
=&\frac45\bigg( \sqrt2\coth^{-1}\sqrt2-\frac{\coth^{-1} \sqrt{\sqrt5+2}}{\sqrt{\sqrt5+2}}
+ \frac{\cot^{-1} \sqrt{\sqrt5-2}}{\sqrt{\sqrt5-2}}\bigg)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/670916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
A simple series $\sum_{i=1}^\infty \frac{i}{2^i} = 2$ I don't do math a long time, so I completely don't remember how to prove that:
$$
\sum_{i=1}^\infty \frac{i}{2^i} = 2
$$
Can anybody help me?
| $S = x + 2x^2 + 3x^3 + \ldots $
It can be written as
$ \Rightarrow S = (x + x^2 + x^3 + \ldots)+(x^2 + x^3 + \ldots)+(x^3 + \ldots)+\ldots $
$\Rightarrow S = (x + x^2 + x^3 + ...)+x(x + x^2 + ...)+x^2(x + ...) + \ldots $
$\Rightarrow S = ( 1+x+x^2+ .. )\times( x+x^2+.. )$
$\Rightarrow S = \frac{1}{1-x}\times\frac{x}{1-x}$
$\Rightarrow S = \frac{x}{(1-x)^2} $
Put $x =0.5$ you will get the answer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/674220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
If $\gcd(a, b) = 1$ and if $ab = x^2$, prove that $a, b$ must also be perfect squares; where $a,b,x$ are in the set of natural numbers Problem:
If $\gcd(a, b) = 1$ and If $ab = x^2$ ,prove that $a$, $b$ must also be perfect squares; where $a$,$b$,$x$ are in the set of natural numbers
I've come to the conclusion that $a \ne b$ and $a \ne x$ and $b \ne x$ but I guess that won't really help me.. I understand that if the $\gcd$ between two numbers if $1$ then they obviously have no common divisors but where do I go from this point?
Any tips at tackling this would be great. It looks quite easy though I'm still trying to get my hand around these proofs! Any pointers in the right direction would be great.
Thank you in advance,
| Here is a proof using Bezout's Identity.
Let $x^2=ab$ and $\gcd(a,b)=1$, where $a,b\gt0$.
There are $u,v$ so that
$$
au+bv=1\tag{1}
$$
Let $s_a=\gcd(x,a)$. Rewriting $(1)$, we have
$$
\begin{align}
s_a\left(\dfrac{a}{s_a}\right)u+bv=1
&\implies s_a^2\left(\dfrac{a}{s_a}\right)^2u^2=1-b(2v-bv^2)\tag{2}\\
&\implies s_a^2\left(\dfrac{a}{s_a}\right)^2u^2a+ab(2v-bv^2)=a\tag{3}\\
&\implies s_a^2\left(\dfrac{a}{s_a}\right)^2u^2a+s_a^2\left(\dfrac{x}{s_a}\right)^2(2v-bv^2)=a\tag{4}\\[8pt]
&\implies s_a^2\mid a\tag{5}
\end{align}
$$
Justification:
$(2)$: Move $bv$ to the right side and square
$(3)$: Move $b(2v-bv^2)$ to the left side and multiply by $a$
$(4)$: $x^2=ab$
$(5)$: $s_a^2$ divides each term on the left side
There are $u_a,v_a$ so that
$$
\begin{align}
xu_a+av_a=s_a
&\implies\frac{x}{s_a}u_a+\frac{a}{s_a}v_a=1\tag{6}\\
&\implies\frac{x^2}{s_a^2}u_a^2=1-\frac{a}{s_a}\left(2v_a-\frac{a}{s_a}v_a^2\right)\tag{7}\\
&\implies\frac{ab}{s_a^2}u_a^2+\frac{a}{s_a}\left(2v_a-\frac{a}{s_a}v_a^2\right)=1\tag{8}\\
&\implies abu_a^2+as_a\left(2v_a-\frac{a}{s_a}v_a^2\right)=s_a^2\tag{9}\\[7pt]
&\implies a\mid s_a^2\tag{10}
\end{align}
$$
Justification:
$\ \:(6)$: Divide by $s_a$
$\ \:(7)$: Move $\frac{a}{s_a}v_a$ to the right side and square
$\ \:(8)$: Move $\frac{a}{s_a}\left(2v_a-\frac{a}{s_a}v_a^2\right)$ to the left side, $x^2=ab$
$\ \:(9)$: Multiply by $s_a^2$
$(10)$: $a$ divides each term on the left side
Combining $(5)$ and $(10)$ yields $a=\gcd(x,a)^2$. Symmetry yields, $b=\gcd(x,b)^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/675917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Writing $x^2+y^2+z^2$ as a polynomial combination of $xyz$, $x+y+z$, and $\frac1x+\frac1y+\frac1z$ Can we write $x^2+y^2+z^2$ as a polynomial combination of $xyz$, $x+y+z$, and $\dfrac1x+\dfrac1y+\dfrac1z$?
What about $x^3+y^3+z^3$?
| $$x^2+y^2+z^2 = (x+y+z)^2 - 2(xy+yz+zx) = (x+y+z)^2 - 2(xyz)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$$
$$x^3+y^3+z^3 = (x+y+z)^3 - (3x^2y + 3x^2z + 3xy^2 + 3y^2z + 3xz^2 + 3yz^2 + 6xyz) = (x+y+z)^3 - (3xy(x+y+z) + 3yz(x+y+z) + 3zx(x+y+z) - 3xyz)$$
$$= (x+y+z)^3 - 3(x+y+z)(xy+yz+zx) + 3xyz$$
$$= (x+y+z)^3 - 3(x+y+z)(xyz)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) + 3xyz$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/676935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
What went wrong? [One-dimensional-inverse-square-law] Intrigued by this question,
one-dimensional inverse square laws,
I started to try to find an answer and came up with what follows. However, I calculated the derivatives to double check myself, and this does not work. However, it seems straightforward enough that I do not see my mistake.
Question: Is there a closed form solution to all
$$ \ddot x = \frac{k}{x^2} \quad (x(0)=x_0), \ \dot{x}(0) = v_0, \ k \in \mathbb{R}.$$
Here is my very naive two cents. Multiplying both sides by $\dot{x}$ and obtain,
$$ \dot{x} \ddot{x} = \frac{k \dot{x}}{x^2}.$$
Integrating both sides, we obtain
\begin{equation} \tag{1} \frac{ (\dot{x})^2}{2} = -\frac{k}{x} + C. \end{equation}
Here it seems natural to re-write this as,
\begin{equation} \tag{2} \dot{x} = \sqrt{2} \sqrt{C - \frac{k}{x}}. \end{equation}
I do not know if that is fully acceptable. However, if it is we then have via integration one more time, or rather letting WolframAlpha integrate one side,
$$x(t) = \sqrt{2} x \sqrt{C - \frac{k}{x}} - \frac{ k \log \left( 2 \sqrt{C} x \sqrt{C - \frac{k}{x}} + 2Cx - k \right)}{\sqrt{2C}} + \tilde{C}. $$
Note, that both $C$ and $\tilde{C}$ are determined by initial conditions. Using the given initial conditions and (1), we can solve for $C$, to find that $C = \frac{v_0^2}{2} + \frac{k}{x_0}.$ Similarly we could solve for $\tilde{C}$, but before I did that I discovered my solution was wrong.
If you substitute my "solution" back into the differential equation, you find (much to my dismay)
$$ \ddot{x} = \frac{k}{x^2 \sqrt{2c - \frac{k}{x}}}. $$
Insight on my mistake and/or a proper way to solve this is greatly appreciated.
| Lets write
$$
\dot{x} = \sqrt{C-\frac{k}{x}} = \sqrt{C}\sqrt{1 - \frac{k}{Cx}}.
$$
where i have absorbed the factor 2 into the constants k and C.
We can solve as follows:
$$
\int \frac{1}{\sqrt{1 - \frac{k}{Cx}}}dx = \sqrt{C}t + \lambda_{1}
$$
If we use the transformation $u = x - \frac{k}{C}$, we transform the r.h.s as,
$$
\int \sqrt{1 + \frac{k}{Cu}}du = \sqrt{C}t + \lambda_{1}.
$$
Integrating by parts with,
$$
d\bar{u} = 1,\\
v = \sqrt{1 + \frac{k}{Cu}}.
$$
This leads to:
$$
\int \sqrt{1 + \frac{k}{Cu}}du = u\sqrt{1 + \frac{k}{Cu}} - \int u\frac{-\frac{k}{2Cu^{2}}}{\sqrt{1 + \frac{k}{Cu}}}du,\\
=u\sqrt{1 + \frac{k}{Cu}} + \frac{1}{2}\int \frac{1}{\sqrt{1 + \frac{k}{Cu}}}\frac{k}{Cu}du
$$
We can sort out the first term in terms of x already as
$$
u\sqrt{1 + \frac{k}{Cu}} = \sqrt{u}\sqrt{u + \frac{k}{C}} = \sqrt{x -\frac{k}{C}}\sqrt{x},\\
=x\sqrt{1 - \frac{k}{Cx}}
$$
Now the integral:
$$
\frac{1}{2}\int \frac{1}{\sqrt{1 + \frac{k}{Cu}}}\frac{k}{Cu}du
$$
using another transformation $w^{2} = 1 + \frac{k}{Cu} $ which has an inverse of $u = \frac{k/C}{w^{2}-1}$, we can transform the integral to,
$$
\frac{1}{2}\int \frac{w^{2} - 1}{w}\left(\frac{-2(k/C) w}{\left(w^{2}-1\right)^{2}}\right)dw = -\frac{k}{C}\int \frac{1}{w^{2}-1}dw
$$
which integrates to:
$$
-\frac{k}{2C} \mathrm{log}\left(\frac{w-1}{w+1}\right) = \frac{k}{2C} \mathrm{log}\left(\frac{w+1}{w-1}\right) = \frac{k}{2C} \mathrm{log}\left(\frac{(w+1)^{2}}{w^{2} - 1}\right).
$$
The log argument can be manipulated to yield:
$$
\frac{(w+1)^{2}}{w^{2} - 1} = \frac{\frac{k}{Cu} + 2 + 2w}{\frac{k}{Cu}} = 1 + \frac{2Cu}{k} - 2w\frac{Cu}{k},\\
=\frac{2Cx - k + 2wCu}{k} = \frac{2Cx - k - 2\sqrt{1 + \frac{k}{Cu}}Cu}{k},\\
=\frac{2Cx - k + 2C\sqrt{u + \frac{k}{C}}\sqrt{u}}{k} = \frac{2Cx - k + 2Cx\sqrt{1 - \frac{k}{Cx}}}{k}
$$
Tying everything together we find:
$$
x\sqrt{1 - \frac{k}{Cx}} +\frac{k}{2C} \mathrm{log}\left(\frac{2Cx - k + 2Cx\sqrt{1 - \frac{k}{Cx}}}{k}\right)\\
= x\sqrt{1 - \frac{k}{Cx}} +\frac{k}{2C} \mathrm{log}\left(2Cx - k + 2Cx\sqrt{1 - \frac{k}{Cx}}\right) +\frac{k}{2C}\mathrm{log}\left(\frac{1}{k}\right)\\
=\sqrt{C}t + \lambda_{1}
$$
The third term is a constant and can be absorbed by $\lambda_{1}$
$$
\frac{1}{\sqrt{C}}x\sqrt{C - \frac{k}{x}} +\frac{k}{2C} \mathrm{log}\left(2Cx - k + 2\sqrt{C}x\sqrt{C - \frac{k}{x}}\right) = \sqrt{C}t + \lambda_{1}
$$
multiplying the $\sqrt{C}$ we finally reach
$$
\sqrt{2}x\sqrt{C - \frac{k}{x}} +\frac{k}{\sqrt{2C}} \mathrm{log}\left(2Cx - k + 2\sqrt{C}x\sqrt{C - \frac{k}{x}}\right) = \sqrt{2}Ct + \lambda_{2}
$$
I final not worth respect to getting back to the original equation, you have to do implicit differentiation and re-arrange for $\frac{dx}{dt}$, thats why you had a issues when you neglect the r.h.s of the equation i.e. the t term.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/681838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 1
} |
Using Lagrange multipliers to solve for minimum I am having troubles with one part of this homework problem. Hopefully somebody can help me out:
Find the minimum and maximum values of the function $f(x,y)=x^2+y^2$ subject to the given constraint $x^4+y^4=18$.
Using Lagrange multipliers, I can easily solve for the maximum:
$f_x(x,y)=2x$ and $f_y(x,y)=2y$.
If we call the second equation $g$, then:
$g_x=4x^3$ and $g_y=4y^3$.
Then we apply the Lagrange multiplier:
$2x=\lambda 4x^3$ and $2y = \lambda 4y^3$
By solving for $x$ and $y$ and plugging in to $g$, we get $\lambda=1/\sqrt{36}=\pm1/6$.
To find the maximum, I will use the positive $1/6$, and solve for $x^2=3$ and $y^2=3$, resulting in a maximum of $6$, which the system spits out as correct.
For a minimum, I originally thought $0$ because $x^2$ and $y^2$ must be positive numbers, but that is not correct. Then I noticed that since $x^2=1/(2\lambda)$, and $y$ also, then when $\lambda$ is negative that would be the minimum, resulting in $-6$, which is also not correct.
So long story short, how do I find the minimum value in this case?
Thanks!
| $$\mathcal{L}(x,y,\lambda)=x^2+y^2-\lambda(x^4+y^4-18)$$
We have
$$\nabla \mathcal{L}(x,y,\lambda) = \left(\begin{array}{c}2x-4\lambda x^3\\2y-4\lambda y^3 \\ x^4+y^4-18 \end{array}\right).$$ Setting this equal to zero gives us $$\frac{1}{2\lambda}=x^2 \;\text{ -or- }\; x=0$$ and $$\frac{1}{2\lambda}=y^2 \;\text{ -or- }\; y=0$$
Clearly both $x$ and $y$ cannot be zero at the same time. So we have three cases:
Case 1) $$x^2=y^2=\frac{1}{2\lambda} \;\Longrightarrow\; \frac{1}{4\lambda^2}+\frac{1}{4\lambda^2}=18 \;\Longrightarrow\; \lambda = \pm \frac{1}{6}.$$ Since the function is of $x^2$ and $y^2$ we need only consider one of the two cases; let's pick $\lambda=\frac{1}{6}$ which results in $$f(x,y)=x^2+y^2=\frac{1}{2\lambda}+\frac{1}{2\lambda}=6.$$
Case 2) $$x=0, \qquad y^2=\frac{1}{2\lambda}$$
Putting this into the constraint gives us
$$0+y^4=18\;\Longrightarrow\; \frac{1}{4\lambda^2}=18 \;\Longrightarrow\; \frac{1}{72}=\lambda^2 \;\Longrightarrow\; \lambda = \pm \frac{1}{6\sqrt{2}}.$$
This results in $$f(x,y)=x^2+y^2=0+\frac{1}{2\lambda}=\frac{6\sqrt{2}}{2}=3\sqrt{2}.$$
Case 3) is symmetric with case (2) since the function and constraint is symmetric w.r.t. $x$ and $y$.
We have enumerated all possible values of $f$, so you can easily see which are maxima and minima. You will have to use some care if you need to enumerate all minimizers and maximizers, as there are a few cases and some symmetry to consider.
P.S., the accepted capitalization of Joseph-Louis Lagrange's surname is with lower-case `g's. This is different from some other similar words, e.g., LaGrange County, LaGrange College, etc. I cannot recommend strongly enough sticking with "Lagrange" for capitalization.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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How do I solve $yy'+x=\sqrt{x^2+y^2}$? I tried this:
$yy'+x=\sqrt{x^2+y^2}$
$y'=-\frac{x}{y}+\frac{1}{y}\sqrt{x^2+y^2}$
$y'=-\frac{x}{y}+\sqrt{(\frac{x}{y})^2+1}$
Substitution: $v=\frac{y}{x}$
$v'x+v=-\frac{1}{v}+\sqrt{(\frac{1}{v})^2+1}$
$v'x=-\frac{1-v^2}{v}+\sqrt{(\frac{1}{v})^2+1}$
$v'x=-\frac{1-v^2+ v\sqrt{(\frac{1}{v})^2+1}}{v}$
$v'x=-\frac{1-v^2+\sqrt{1+v^2}}{v}$
If I have to separate at this point, it's going to be pretty awkward to work with.
Is there a better way?
| One should have $2yy'=(y^2)'$ burned into one's memory. Now notice
$$\color{Blue}{x}+(\color{Red}{y^2/2})'=\sqrt{\color{Blue}{x^2}+\color{Red}{y^2}}$$
The red observation tells us we could use the substitution $v=y^2$. The blue observation tells us we can go further, since $2x=(x^2)'$ (which is just our original observation a second time). Now:
$$\frac{d}{dx}\left(\frac{x^2+y^2}{2}\right)=\sqrt{x^2+y^2}\iff \frac{dr^2}{dx}=2r.$$
Simplifyig, $dr/dx=1\implies r=x+c\iff y^2=(x+c)^2-x^2=c(2x+c)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/695566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How do I find the number of sides of a regular polygon given only its side length and area? How do I find the number of sides of a regular polygon given only its side length and area?
Absolutely no IDEA where to start. Anyone?
| The approach I was describing in my comment (which was all I had time to write then) is similar to the one given by Michael Hardy, but does not require finding the altitude of a triangle. The regular polygon with $ \ n \ $ sides ("regular $ \ n-$ gon") is divided up into $ \ n \ $ isosceles triangles arranged around the centroid of the figure, giving them an "apex angle" of $ \ \frac{2 \pi}{n} \ . $ The base of each triangle is a side $ \ s \ $ of the polygon, and the other (congruent) legs of the triangle will be said to have length $ \ L \ . $
The Law of Cosines gives us
$$ s^2 \ = L^2 \ + \ L^2 \ - \ 2 \cdot L \cdot L \cdot \cos \frac{2 \pi}{n} \ = \ 2 L^2 \ (1 \ - \ \cos \frac{2 \pi}{n} ) $$
and the "included angle" formula for the area of a triangle yields
$$ A_{tri} \ = \ \frac{1}{2} \cdot L \cdot L \cdot \sin \frac{2 \pi}{n} \ . $$
From these results, we can write the area of the triangle in terms of $ \ n \ $ and $ \ s \ $ as
$$ A_{tri} \ = \ \frac{1}{2} \ \left[ \frac{s^2}{2 \ (1 \ - \ \cos \frac{2 \pi}{n} )} \right] \cdot \sin \frac{2 \pi}{n} \ = \ \left[ \frac{\sin \frac{2 \pi}{n}}{4 \ (1 \ - \ \cos \frac{2 \pi}{n} )} \right] \cdot s^2 \ . $$
The polygon comprises $ \ n \ $ of these triangles, so its area is
$$ A(n) \ = \ \left[ \frac{n \ \sin \frac{2 \pi}{n}}{4 \ (1 \ - \ \cos \frac{2 \pi}{n} )} \right] \cdot s^2 \ . \ \ \ \mathbf{ [1] }$$
Using the "small-angle approximations" for the trigonometric functions of the apex angle as $ \ n \ \rightarrow \ \infty \ , $ we find that
$$ A(n) \ \rightarrow \ \left[ \frac{n \ \cdot \frac{2 \pi}{n}}{4 \ (1 \ - \ [ \ 1 \ - \ \frac{1}{2}\left( \frac{2 \pi}{n} \right)^2 \ ] \ )} \right] \cdot s^2 \ = \ \frac{n^2 \cdot s^2}{4 \pi} \ , $$
producing the relation described by Henry (with the appropriate dimension -- I believe he is using unit side lengths).
Equation 1 above will give us the usual area formulas for equilateral triangles, squares, etc., but the enclosed area also tends to infinity as $ \ n \ $ does, since we are using a fixed side length. If we instead consider the perimeter $ \ p \ $ for the polygon, and write the share of that perimeter represented by each side as $ \ s = \frac{p}{n} \ , $ then we may also express our result as
$$ A(n) \ = \ \left[ \frac{n \ \sin \frac{2 \pi}{n}}{4 \ (1 \ - \ \cos \frac{2 \pi}{n} )} \right] \cdot \left( \frac{p}{n} \right)^2 \ = \ \left[ \frac{ \sin \frac{2 \pi}{n}}{4n \ (1 \ - \ \cos \frac{2 \pi}{n} )} \right] \cdot p^2 \ \rightarrow \ \frac{p^2}{4 \pi} \ . $$
The limit does indeed give the relation between the area and circumference of a circle.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the product of the series? Can anybody please help me to evaluate the following product?
$$\prod_{n=2}^\infty\dfrac{n^3 - 1}{n^3 + 1} $$
| Note that
$$
\frac{n^3-1}{n^3+1}=\frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}=\frac{(n-1)}{(n+1)}\cdot\frac{\big((n+1)^2-(n+1)+1\big)}{(n^2-n+1)}.
$$
We have
$$
\prod_{n=2}^N\frac{n-1}{n+1}=\frac{1\cdot 2\cdots(N-2)(N-1)}{3\cdot 4\cdots N(N+1)}=\frac{1\cdot 2}{N(N+1)},
$$
while
\begin{align}
\prod_{n=2}^N\frac{n^2+n+1}{n^2-n+1}&=\prod_{n=2}^N\frac{\big((n+1)^2-(n+1)+1\big)}{(n^2-n+1)}\\&=\frac{\big((N+1)^2-(N+1)+1\big)}{2^2-2+1}=\frac{N^2+N+1}{3},
\end{align}
and hence
$$
\prod_{n=2}^N\frac{n^3-1}{n^3+1}=\frac{1\cdot 2\cdot(N^2+N+1)}{3\cdot N(N+1)}\to \frac{2}{3},
$$
as $N\to \infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/699003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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} |
Using the Chain Rule Find $dy/dx$. $y=x/\sqrt{x^2+1}$. I started with the quotient rule and got $y'=(x^2+1)^{1/2}-x^2(x^2+1)^{1/2 }$all over $((x^2+1)^{1/2})^2$. I can not find my mistake.
| Another way:
Product rule
\begin{align*}
y' = [x(x^2+1)^\frac{-1}{2}]' &= x\frac{-1}{2}(x^2+1)^\frac{-3}{2}(2x) + 1(x^2+1)^\frac{-1}{2}\\
&= -x^2(x^2+1)^\frac{-3}{2} + (x^2+1)^\frac{-1}{2}\\
&= \frac{-x^2}{(x^2+1)^{\frac{3}{2}}}+ \frac{1}{(x^2+1)^\frac{1}{2}}\\
&= \frac{-x^2}{(x^2+1)^{\frac{3}{2}}}+ \frac{1}{(x^2+1)^\frac{1}{2}}\frac{x^2+1}{x^2+1}\\
&= \frac{1}{(x^2+1)^\frac{3}{2}}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/699863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integration do not know how to proceed: $\int_0^{\infty}\frac{\ln(x+1)}{x\cdot(x+1)^\frac14} dx$ Integration 0 to infinity
$$\int_0^\infty\frac{\ln(x+1)}{x\cdot(x+1)^\frac14}dx$$
How to proceed ?
Answer : $8C+\pi^2$
Procedure wanted
| First, assume $\displaystyle \sum_{n=1}^{+\infty} \dfrac{1}{n^2}=\dfrac{\pi^2}{6}$
Then:
$\displaystyle \dfrac{\pi^2}{6}=\sum_{n=1}^{+\infty} \dfrac{1}{n^2}=\sum_{n=1}^{+\infty} \dfrac{1}{(2n)^2}+\sum_{n=0}^{+\infty} \dfrac{1}{(2n+1)^2}=\dfrac{\pi^2}{24}+\sum_{n=0}^{+\infty} \dfrac{1}{(2n+1)^2}$
Therefore:
$\displaystyle \sum_{n=0}^{+\infty} \dfrac{1}{(2n+1)^2}=\dfrac{\pi^2}{8}$ (1)
Let $\displaystyle I=\int_0^{+\infty}\dfrac{\ln(x+1)}{x (x+1)^\frac14}dx$
Perform change of variable $u=(x+1)^\frac14$:
$\displaystyle I=\int_1^{+\infty} \dfrac{16x^2\log(x)}{x^4-1}dx$
Therefore:
$\displaystyle I=\int_1^{+\infty} \dfrac{8\log(x)}{x^2-1}dx+\int_1^{+\infty} \dfrac{8\log(x)}{x^2+1}dx$
The last integral equals $8G$ where $G$ is the Catalan's constant.
Let $\displaystyle J=\int_1^{+\infty} \dfrac{8\log(x)}{x^2-1}dx$
Perform change of variable $u=\dfrac{x-1}{x+1}$:
$\displaystyle J=\int_0^1 4\log\Big(\dfrac{1+x}{1-x}\Big)\dfrac{dx}{x}=4\Big(\int_0^1 \dfrac{\log(1+x)dx}{x}-\int_0^1 \dfrac{\log(1-x)dx}{x}\Big)$
$\displaystyle \int_0^1 \dfrac{\log(1+x)dx}{x}=\int_0^1 \Big(\sum_{n=0}^{+\infty} \dfrac{(-1)^n x^n}{n+1}\Big) dx=\sum_{n=0}^{+\infty} \Big( \int_0^1\dfrac{(-1)^nx^n}{n+1}dx \Big)$
$\displaystyle \int_0^1 \dfrac{\log(1+x)dx}{x}=\sum_{n=0}^{+\infty} \dfrac{(-1)^n}{(n+1)^2}$
$\displaystyle \int_0^1 \dfrac{\log(1+x)dx}{x}=\sum_{n=1}^{+\infty} \dfrac{(-1)^{2n+1}}{(2n)^2}+\sum_{n=0}^{+\infty} \dfrac{(-1)^{2n+2}}{(2n+1)^2}=-\dfrac{\pi^2}{24}+\sum_{n=0}^{+\infty} \dfrac{1}{(2n+1)^2}$
Apply (1) :
$\displaystyle \int_0^1 \dfrac{\log(1+x)dx}{x}=-\dfrac{\pi^2}{24}+\dfrac{\pi^2}{8}=\dfrac{\pi^2}{12}$
$\displaystyle -\int_0^1 \dfrac{\log(1-x)dx}{x}=\int_0^1 \Big(\sum_{n=0}^{+\infty} \dfrac{x^n}{n+1}\Big) dx=\sum_{n=0}^{+\infty} \Big(\int_0^1\dfrac{x^n}{n+1}dx \Big)=\sum_{n=0}^{+\infty} \dfrac{1}{(n+1)^2}=\sum_{n=1}^{+\infty} \dfrac{1}{n^2}= \dfrac{\pi}{6}$
Therefore:
$J=4 \times \dfrac{\pi^2}{12}+4 \times \dfrac{\pi^2}{6}=\pi^2$
$I=8G+\pi^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/701528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Probability of an airplane crash Some time ago I read a problem about probability of an airplane crash. I can't recall it correctly but I ll try to write it here. Please tell me if it doesn't make sense and correct me.
Any engine of an airplane has 50% chance of failure during travel and if half of the engines survive during travel then airplane will reach its destination.
Which plane voyage would be safer, four engine one or two engine one?
| The four-engine plane will crash if more than half of its engines fail during travel. This happens when $3$ of the engines fail or all $4$ fail.
The probability of $3$ engines failing is $$\dbinom{4}{3} \left(\frac{1}{2}\right)^4 = \frac{1}{4}$$
and the probability of $4$ engines failing is $$\dbinom{4}{4} \left(\frac{1}{2}\right)^4 = \frac{1}{16}$$
so the probability of a four-engine plane crashing is $\displaystyle \frac{1}{4} + \frac{1}{16} = \frac{5}{16}$.
The two-engine plane will crash if more than half of its engines fail during travel. This happens only when both fail.
The probability of $2$ engines failing is $$\dbinom{2}{2} \left(\frac{1}{2}\right)^2 = \frac{1}{4}$$
so the probability of a two-engine plane crashing is $\displaystyle \frac{1}{4}$.
So we get the ironic result that the two-engine plane is safer than the four-engine one.
For the general case, if we're worried about $n+1$ to $2n$ engines failing, the probability would be $$\sum_{i=n+1}^{2n} \dbinom{2n}{i} \left(\frac{1}{2}\right)^{2n}$$
Let's call this sum $S$. As $n$ approaches $\infty$, we note that $$\sum_{i=0}^{2n} \dbinom{2n}{i} \left(\frac{1}{2}\right)^{2n} = 1 = \sum_{i=0}^{n-1} \dbinom{2n}{i} \left(\frac{1}{2}\right)^{2n} + \sum_{i=n+1}^{2n} \dbinom{2n}{i} \left(\frac{1}{2}\right)^{2n} + \dbinom{2n}{n} \left(\frac{1}{2}\right)^{2n}$$
$$ = 2S + \dbinom{2n}{n} \left(\frac{1}{2}\right)^{2n} \approx 2S$$
and we can see that $S$ approaches $\displaystyle \frac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/702278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$ \dfrac a{\sqrt[4] {16+b-d}} +\dfrac b{\sqrt[4] {16+c-a}}+ \dfrac c{\sqrt[4] {16+d-b}}+\dfrac d{\sqrt[4] {16+a-c}} \ge 8$ Let $a,b,c,d>0 ; a+b+c+d=16$ , then how to prove that
$ \dfrac a{\sqrt[4] {16+b-d}} +\dfrac b{\sqrt[4] {16+c-a}}+ \dfrac c{\sqrt[4] {16+d-b}}+\dfrac d{\sqrt[4] {16+a-c}} \ge 8$ ?
| Using Holder's Inequality, we have:
$$\left(\sum_{cyc} \frac{a}{\sqrt[4]{16+b-d}}\right)^4 \left(\sum_{cyc} a(16+b-d) \right) \ge \left(\sum_{cyc} a \right)^5 = 16^5$$
$$\implies (LHS)^4 \times 16 \sum_{cyc} a \ge 16^5 \implies LHS \ge 8$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/704051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Integration of $1/(1+\sin x)$ I solved it using $t=\tan(\frac{x}{2})$ substitution and got $-2/(1+\tan(x/2))+C$, but in my math book solution is $\tan(x/2-\pi/4)+C$. Are those the same expressions and if they are, how do I transform from one to another, or are one(or both) solutions incorrect ?
| To some extent:
$$I=\int \frac{d x}{1+\varepsilon \sin x}=\int \frac{\sin^2 \frac{x}{2}+\cos^2 \frac{x}{2}}{\sin^2 \frac{x}{2}+\cos^2 \frac{x}{2}+2\varepsilon\sin(\frac{x}{2})\cos(\frac{x}{2})}dx\\=2 \int \frac{d t}{(1+t^2)+2\varepsilon t} \quad(t=\tan\frac{x}{2})\\=2\int\frac{dt}{(t+\varepsilon)^2-\varepsilon^2+1}=2\int\frac{d(t+\varepsilon)}{(t+\varepsilon)^2-\varepsilon^2+1}$$
Let $\varepsilon=\pm1$, then
$$I=2\int\frac{dm}{m^2}=-\frac{2}{m}+C=-\frac{2}{t\pm1}+C=-\frac{2}{\pm1+\tan\frac{x}{2}}+C$$
Moreover, for $\varepsilon\in(-1,1)$:
$$I=\int \frac{d x}{1+\varepsilon \sin x}=\frac{2}{\sqrt{1-\varepsilon^2}}\int\frac{dm}{1+m^2}\\=\frac{2}{\sqrt{1-\varepsilon^2}}\arctan m+C=\frac{2}{\sqrt{1-\varepsilon^2}}\arctan(\frac{\tan\frac{x}{2}+\varepsilon}{\sqrt{1-\varepsilon^2}})+C$$
for $\varepsilon>1$ or $\varepsilon<-1$:
$$I=\int \frac{d x}{1+\varepsilon \sin x}=-2\int\frac{d(t+\varepsilon)}{-(t+\varepsilon)^2+\varepsilon^2-1}\\=\frac{1}{\sqrt{\varepsilon^2-1}}\ln |\frac{\tan\frac{x}{2}+\varepsilon-\sqrt{\varepsilon^2-1}}{\tan\frac{x}{2}+\varepsilon+\sqrt{\varepsilon^2-1}}|+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/704163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Finding $\lim\limits_{n\rightarrow \infty}\sum\limits_{r=1}^{n}\frac{1}{T_r}$ given $\sum\limits_{r=1}^{n}T_r=\frac{n(n+1)(n+2)(n+3)}{8}$ If $\displaystyle\sum_{r=1}^{n}T_r=\frac{n(n+1)(n+2)(n+3)}{8}$, then how can we find $\displaystyle\lim_{n\rightarrow \infty}\sum_{r=1}^{n}\frac{1}{T_r}$?
| $$T_{n+1}=\sum_{r=1}^{n+1}T_r - \sum_{r=1}^{n}T_r = \frac{1}{2}(n+1)(n+2)(n+3)$$
So, $$\begin{align}
\sum_{n=1}^{m}\frac{1}{T_r} = \sum_{n=1}^{m}\frac{2}{n(n+1)(n+2)}
&= \sum_{n=1}^{m}\frac{2(n+1)^2-2n(n+2)}{n(n+1)(n+2)} \\
&= \sum_{n=1}^{m}\frac{2(n+1)^2}{n(n+1)(n+2)} -\sum_{n=1}^{m}\frac{2n(n+2)}{n(n+1)(n+2)} \\
&= \sum_{n=1}^{m}\frac{1}{n}+\sum_{n=1}^{m}\frac{1}{n+2}-\sum_{n=1}^{m}\frac{2}{n+1}\end{align}$$
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Values of the limit $\lim\limits_{x\to+\infty}\left(\sqrt{(x+a)(x+b)}-x\right)$ Hi I have a question regarding finding the values of limit for the following question.
Let $a, b \in \mathbb R$. Find the limit
$$\lim_{x\to+\infty}\left(\sqrt{(x+a)(x+b)}-x\right)$$
| There is another way to do it and even to get a bit more information.
Let
$$y=\sqrt{(x+a) (x+b)}-x=x \left(\sqrt{1+\frac{a}{x}}\times\sqrt{1+\frac{b}{x}} -1\right)$$ and now, remembering the binomial expansion or Taylor expansion
$$\sqrt{1+\epsilon}=1+\frac{\epsilon }{2}-\frac{\epsilon ^2}{8}+O\left(\epsilon ^3\right)$$ replace $\epsilon$ successively by $\frac{a}{x}$ and $\frac{b}{x}$ to get
$$y=x \left(\left(1+\frac{a}{2 x}-\frac{a^2}{8 x^2}+O\left(\frac{1}{x^3}\right)\right)\times\left(1+\frac{b}{2 x}-\frac{b^2}{8 x^2}+O\left(\frac{1}{x^3}\right)\right) -1\right)$$ Expand the whole stuff to get
$$y=x \left(1+\frac{a+b}{2 x}-\frac{(a-b)^2}{8 x^2}+O\left(\frac{1}{x^3}\right)-1 \right)=\frac{a+b}{2}-\frac{(a-b)^2}{8 x}+O\left(\frac{1}{x^2}\right)$$ which shows the limit and also how it is approached.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show $\lim\limits_{(x,y)\to(0,0)} \frac{xy}{\sqrt{x^2+y^2}}=0$ using Delta Epsilon Proof I tried following this answer (https://math.stackexchange.com/a/542303/86425), but I cannot see how this shows the limit is 0. Can someone please provide guidance
| Note that $\sqrt{x^2 + y^2} \geq \sqrt{y^2} = |y|$. Then
$$
\left|\frac{xy}{\sqrt{x^2 + y^2}}\right| = |x|\frac{|y|}{\sqrt{x^2 + y^2}} \leq |x|
$$
so
$$
\lim_{(x,y) \to (0,0)}\left|\frac{xy}{\sqrt{x^2 + y^2}}\right| \leq \lim_{(x,y) \to (0,0)} |x| = 0 \, .
$$
Personally, I prefer thinking of these problems in polar coordinates. So let $x = r \cos\theta$, $y = r \sin \theta$ where $r = \sqrt{x^2 + y^2}$. Then
\begin{align*}
\lim_{(x,y) \to (0,0)} \frac{xy}{\sqrt{x^2 + y^2}} &= \lim_{r \to 0} \frac{r^2 \cos \theta \sin \theta}{r} = \lim_{r \to 0} r \cos \theta \sin \theta = 0 \, .
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove this trigonometric identity? $$\sum_{k=1}^n\frac{1}{\displaystyle \sin^4\left(\frac{k\pi}{2n+1}\right)}=\frac8{45}n(n+1)(n^2+n+3)$$
I think it maybe use
$\cos(nx)+i\sin(nx) = \displaystyle\sum_{k=0}^n\binom nki^k(\sin^kx)(\cos^{n-k}x)$.
But I can't .
| From this, $$\sin(2n+1)x$$
$$=(2n+1)s-\frac{(2n+1)\{(2n+1)^2-1^2\}}{3!}s^3+\frac{(2n+1)\{(2n+1)^2-1^2\}\{(2n+1)^2-3^2\}}{5!}s^5+\cdots+(-1)^n2^{2n}s^{2n+1}$$
If $\displaystyle\sin(2n+1)x=0,(2n+1)x=k\pi$ where $k$ is any integer
$\displaystyle\implies x=\frac{k\pi}{2n+1}$ where $-n\le k\le n$
So, $\displaystyle\sin\frac{k\pi}{2n+1},-n\le k\le n$ are the roots of $$(2n+1)s-\frac{(2n+1)\{(2n+1)^2-1^2\}}{3!}s^3+\frac{(2n+1)\{(2n+1)^2-1^2\}\{(2n+1)^2-3^2\}}{5!}s^5+\cdots+(-1)^n2^{2n}s^{2n+1}=0$$
So, $\displaystyle\sin\frac{k\pi}{2n+1},-n\le k\le n,k\ne0$ are the roots of $$(2n+1)-\frac{(2n+1)\{(2n+1)^2-1^2\}}{3!}s^2+\frac{(2n+1)\{(2n+1)^2-1^2\}\{(2n+1)^2-3^2\}}{5!}s^4+\cdots+(-1)^n2^{2n}s^{2n}=0$$
If we set $\displaystyle\sin^2\frac{k\pi}{2n+1}=\frac1{y_k},1\le k\le n$
$y_k$ will be roots of $$(2n+1)-\frac{(2n+1)\{(2n+1)^2-1^2\}}{3!}\frac1t+\frac{(2n+1)\{(2n+1)^2-1^2\}\{(2n+1)^2-3^2\}}{5!}\frac1{t^2}+\cdots+(-1)^n2^{2n}\frac1{t^n}=0$$
$$\iff t^n(2n+1)-\frac{(2n+1)\{(2n+1)^2-1^2\}}{3!}t^{n-1}+\frac{(2n+1)\{(2n+1)^2-1^2\}\{(2n+1)^2-3^2\}}{5!}t^{n-2}+\cdots+(-1)^n2^{2n}=0$$
Now, we need $\displaystyle\sum_{k=1}^ny_k^2=\left(\sum_{k=1}^n y_k\right)^2-2\sum_{i,j=1,i\ne j}^ny_iy_j$ which can be safely managed by Vieta's Formula
| {
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"timestamp": "2023-03-29T00:00:00",
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Quadratic equation cancellation error. $$(x-1)^2 + 2x - 2 = (x+1)(x+2)$$
$$(x-1)^2 + 2(x-1) = (x+1) (x+2)$$
$$(x-1) (x-1+2) = (x+1) (x+2)$$
$$(x-1)\require{cancel}\cancel{(x+1)} = \require{cancel}\cancel{(x+1)}(x+2)$$
$$(x-1) = (x+2)$$
$$x-x = 2+1$$
$$0 = 3$$
But when i do it like this
$$(x-1)^2 + 2x - 2 = (x+1) (x+2)$$
$$x^2 - 2x + 1 + 2x -2 = x^2 + 2x + x +2$$
$$x^2 - 1 = x^2 + 3x +2$$
$$-1 -2 = 3x$$
$$-3 = 3x$$
$$-1 = x$$
So in the first solution is my simplifying is wrong?
| In the first set of equations, where you reached a contradiction, you divided both sides of the equation by $x+1$, but you can't divide by zero.
So you should have said at that point, either $x=-1$ or I can divide ... and since dividing gives a contradiction, I must have $x=-1$.
Then the two methods become consistent.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the distance of the point $( 1,2,3 )$ to ,,,,,,,,,,,,,, Find the distance of the point $( 1,2,3 )$ of the plane
$3x-2y+5z+17=0$ .
| Hint: from analytic formulas you know that the distance beteween a plane and a point is $$\frac{ax_0 + by_o+cz_o+d}{\sqrt{a^2+b^2+c^2}}$$ given a plane of equation $ax+by+cz+d=0$, like in this problem...
So substituting we get $$ d = \frac{3\cdot 1 -2\cdot 2 +5\cdot 3 +17}{\sqrt{3^2+(-2)^2+5^2}} = \frac{3-4+15+17}{\sqrt{9+4+25}}=\frac{31}{\sqrt{38}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/717213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Resolving a Limit I'm studying for college exams and I don't know how to solve this type of limit:
$$\begin{align}
\lim_{x\to -3} \sqrt{\frac{x^2-9}{2x^2+7x+3}}
\end{align}$$
Any help?
Update: I know that the solution is:
$$\begin{align}
\frac{1}{5} \sqrt{30}
\end{align}$$
| If a polynomial has $-3$ as a root, then $x+3$ can always be factored out:
$$\lim_{x\to-3}\sqrt{\frac{x^2-9}{2x^2+7x+3}}=\lim_{x\to-3}\sqrt{\frac{(x+3)(x-3)}{(x+3)(2 x+1)}}=\lim_{x\to-3}\sqrt{\frac{x-3}{2x+1}}=\sqrt\frac{-6}{-5}=\sqrt\frac65$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/723209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to apply LU factorization to find a polynomial of degree 3 that passes through given points I am familiar with LU factorization when given a matrix to begin with, but I don't know how to apply it in order to solve a problem of this sort.
Find the polynomial (of degree 3) passes through the following points:
(1,-5),(2,-8),(3,9),(4,31)
| Let us write
$$f(x)=ax^3+bx^2+cx+d$$
Then
$$-5=a+b+c+d\\
-8=8a+4b+2c+d\\
9=27a+9b+3c+d\\
31=64a+16b+4c+d\\
\implies\begin{pmatrix}1&1&1&1\\8&4&2&1\\27&9&3&1\\64&16&4&1\end{pmatrix}\begin{pmatrix}a\\b\\c\\d\end{pmatrix}=\begin{pmatrix}-5\\-8\\9\\31\end{pmatrix}\\
\implies \begin{pmatrix}a\\b\\c\\d\end{pmatrix}=\begin{pmatrix}-\dfrac{1}{6}&\dfrac{1}{2}&-\dfrac{1}{2}&\dfrac{1}{6}\\\dfrac{3}{2}&-4&\dfrac{7}{2}&-1\\-\dfrac{13}{3}&\dfrac{19}{2}&-7&\dfrac{11}{6}\\4&-6&4&-1\end{pmatrix}\begin{pmatrix}-5\\-8\\9\\31\end{pmatrix}$$
You should be able to take it from here.
| {
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Proof that the function $\cot(\pi z)$ is uniformly bounded on the sides of the square with vertices $\pm(N+1/2)\pm i(N+1/2),n∈ℕ$.
Proof that the function $\cot(\pi z)$ is uniformly bounded on the
sides of the square with vertices $\pm(N+1/2)\pm i(N+1/2),n∈ℕ$.
My idea was that since those squares are compact and this function is continuous on those squares the image must be compact and therefore bounded.
But I'm not sure what to do with the "uniform" notion.
| Uniformly bounded just means that the bound doesn't depend on $N$.
First note that
$$ |\cot \pi z|^{2} = \Big|\frac{\cos \pi z}{\sin \pi z} \Big|^{2}$$
$$ = \Big| \frac{\cos \pi z \cosh \pi y -i \sin \pi x \sinh \pi y}{\sin \pi x \cosh \pi y + i \cos \pi x \sinh \pi y }\Big|^{2}$$
$$ =\frac{\cos^{2} \pi x \cosh^{2} \pi y +\sin^{2} \pi x \sinh^{2} \pi y}{\sin^{2} \pi x \cosh^{2} \pi y + \cos^{2} \pi x \sinh^{2} \pi y }$$
$$ = \frac{\cos^{2} \pi(1+ \sinh^{2} \pi y) + \sin^{2} \pi x \sinh^{2} \pi y}{\sin^{2} \pi x(1+\sinh^{2} \pi y) + \cos^{2} \pi x \sinh^{2} \pi y} $$
$$ = \frac{\cos^{2} \pi x + \sinh^{2} \pi y}{\sin^{2} \pi x +\sinh^{2} \pi y}$$
Next note that $\cos^{2}\Big( \pi (N+\frac{1}{2}) \Big) = 0$ and $\sin^{2} \Big( \pi (N+ \frac{1}{2}) \Big) =1$.
So on the vertical sides of the square (where $x = N + \frac{1}{2}$ or $x = - N -\frac{1}{2}$),
$$ |\cot \pi z|^{2} =\frac{\sinh^{2} \pi y}{1+ \sinh^{2} \pi y} = \frac{\sinh^{2} \pi y}{\cosh^{2} \pi y} = \tanh^{2} \pi y $$
$$ \implies |\cot \pi z| \le |\tanh \pi y| \le 1 $$
And on the horizontal sides of the square (where $y = N + \frac{1}{2}$ or $y = -N - \frac{1}{2}$),
$$ |\cot \pi z|^{2} \le \frac{1 + \sinh^{2} \pi y}{\sin^{2} \pi x + \sinh^{2}\pi y} $$
$$ = \frac{\cosh^{2} \pi y}{\sin^{2} \pi x + \sinh^{2} \pi y} \le \frac{\cosh^{2}\pi y}{\sinh^{2} \pi y}= \coth^{2} \pi y$$
$$ \implies |\cot \pi z| \le |\coth \pi y| \le \coth \frac{\pi}{2} \approx 1.09$$
Thus $|\cot \pi z|$ is bounded by $\coth \frac{\pi}{2}$ on the sides the square, which implies that $\cot \pi z$ is uniformly bounded on the sides of the square.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/726336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Compute the limit of an integral This is a nice calculus problem, which seems not to be easy.
Compute the limit: $$\lim_{n\rightarrow\infty}n\left(\frac{1}{2}-(n-1)\int_{0}^{1}\frac{x^n}{x^2+1} dx\right)$$. A solution would really be appreciated!!
| A solution without assuming the knowledge of limiting behaviour of special functions:
The catch of this limit is that $x^n$ becomes very flat and zero-ish everywhere except for a steep part at the end. Only there, the $1+x^2$ has any effect when $n$ is large. We can't expand $x$ around $0$ because the series diverges at $x=1$. But we can expand around $1$. Say $u=1-x$. The idea is that $u$ is small where $x^n$ matters at all (essentially we are doing asymptotic expansion).
$$\frac{1}{1+x^2}=\frac{1}{2-2u+u^2}=\frac{1}{2}\frac{1}{1-(u-u^2/2)}=\frac{1}{2}(1+((1-x)-(1-x)^2/2)+((1-x)-(1-x)^2/2)^2+\cdots)$$
We used the geometric expansion, $\frac{1}{1-q}=1+q+q^2+\cdots$ for $q=u-u^2/2$ that represents the lowest order effects of $\frac{1}{1+x^2}$ around $x=1$, where the contribution of $x^n$ is strongest.
We still need to integrate by $x$ (remember, expansion didn't affect the $x^n$ term, just the denominator). So we get
$$\int_0^1 \frac{x^n}{1+x^2}dx=\frac12\int_0^1 x^n+x^n(1-x)-x^n(1-x)^2/2+x^n(1-x)^2+\cdots dx$$
$$\frac{1}{2}\left(\frac{1}{n+1}+\frac{1}{(n+2)(n+1)}+\cdots\right)$$
Other terms don't matter because the power of $n$ terms in the denominator rises with power of $(1-x)$, which can be verified by noticing $\int_0^1 x^n (1-x)^m dx=B(n+1,m+1)=\frac{n!m!}{(n+m+1)!}$, where $B$ is the Beta function.
Now, plug this back in the limit:
$$\lim_{n\to \infty} n\left(\frac{1}{2}-\frac{1}{2}\left(\frac{n-1}{n+1}+\frac{n-1}{(n+2)(n+1)}+\cdots\right)\right)$$
Now we can see that the quadratic term in $n$ is the last one that matters: the next one would limit to $0$ by default. This limit is easily evaluated to
$$\frac{1}{2}\lim_{n\to \infty} n\left(\frac{n+1}{n+1}-\frac{n-1}{n+1}-\frac{n-1}{(n+2)(n+1)}+\cdots\right)$$
$$\frac{1}{2}\lim_{n\to \infty} \left(\frac{2n}{n+1}-\frac{n(n-1)}{(n+2)(n+1)}+\cdots\right)$$
$$\frac{1}{2}\lim_{n\to \infty} \left(2-1\right)=\frac12$$
It's not pretty, but it uses the idea inspired by the behaviour of the functions in the limit, and expansions under the integral sign are pretty standard, you just have to be careful to stay within the convergence radius.
| {
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"source": "stackexchange",
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Concept Clarification: Sum of two subspaces Suppose that $U$ and $W$ are subspaces of $V$. Then the sum $U+W$ is the subspace:
$U+W = \{u+w: u \in U, w \in W\}$
I'm having trouble understand what this means. So if I have the following subspaces $U$ and $W$ of $\mathbb{R}^3$ say:
$U = \{\begin{pmatrix}
x\\
y\\
z
\end{pmatrix}: x+y+z=0\}$
$W = \{\begin{pmatrix}
-5x\\
3x\\
4x
\end{pmatrix}:$ $x$ is an arbitrary number$\}$
So, if I list some elements of each subspace,
$U = \{\begin{pmatrix}
1\\
1\\
-2
\end{pmatrix}$, $\begin{pmatrix}
2\\
3\\
-5
\end{pmatrix},...\}
$
$W = \{\begin{pmatrix}
-5\\
3\\
4
\end{pmatrix}$, $\begin{pmatrix}
-10\\
6\\
8
\end{pmatrix},...\}
$
Does this mean $U+W = \{\begin{pmatrix}
1\\
1\\
-2
\end{pmatrix} + \begin{pmatrix}
-5\\
3\\
4
\end{pmatrix}$, $\begin{pmatrix}
1\\
1\\
-2
\end{pmatrix} + \begin{pmatrix}
-10\\
6\\
8
\end{pmatrix}$, $\begin{pmatrix}
2\\
3\\
-5
\end{pmatrix} + \begin{pmatrix}
-5\\
3\\
4
\end{pmatrix}$, $\begin{pmatrix}
2\\
3\\
-5
\end{pmatrix} + \begin{pmatrix}
-10\\
6\\
8
\end{pmatrix}, ....
\}
$
| What you're writing is exactly correct, although I can see why it would be a bit confusing from this perspective. It can be tedious to list out infinitely many elements...
A better way to view the space $U+W$ is to write it in formal set notation to get a sense of what an arbitrary vector looks like.
$$U+W=\left\{\begin{pmatrix}a-5d\\b+3d\\c+4d\end{pmatrix}\middle|a,b,c,d\in\mathbb{R},\;a+b+c=0\right\}$$
So now we can see the form that all vectors have in $U+W$. Now, can you show that any vector in $\mathbb{R}^3$ has this form, so that in fact $U+W=\mathbb{R}^3$?
| {
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} |
Description of the boundary of a ball Solving an optimization problem in multiple variables, I had to examine a function
$$
f(x,y,z)=x^2+2yz
$$
defined on a ball
$$
\{ (x,y,z)\enspace|\enspace x^2+y^2+z^2\leq1 \}.
$$
The boundary is then described by the sphere $x^2+y^2+z^2=1$. According to my textbook, one can search for maxima on the boundary by rewriting $x^2+y^2+z^2=1$ as $x^2=1-y^2-z^2$, then substituting this into $f$ to obtain
$$
f(x,y,z)=1-y^2-z^2+2yz,\quad y^2+z^2\leq1
$$
Now what I don't understand is, how come we need to specify that $y^2+z^2\leq1$? The unit sphere is described by $x^2+y^2+z^2=1$ with no need to specify any restrictions on $y$ or $z$, but once we rewrite this as $x^2=1-y^2-z^2$ and plug it into the function, suddenly it becomes necessary to specify that $y^2+z^2\leq1$. Why?
| Because $0 \leq x^2 = 1-y^2-z^2$.
| {
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} |
Knowing the length of two sides of a triangle and the angle bisector in between , find the length of one of the altitude. In $\triangle ABC$, $AB = 6, AC = 8$ and internal angle bisector $AD = 6$ such that $D$ lies on
segment $ BC$. Compute the length of altitude $CF$ where $F$ is a point on line $AB$.
For calculating $CF$ , we will need area of the triangle.
For calculating area , we will need $BC$ .( Then we can use heron's formula)
How can I calculate $BC$?
Also it is given that , the angle bisector , $AD$ is $6$.
How can I utilize this information?
| Summing up the area of two triangles we get, $3\sin (\frac{A}{2})+4\sin (\frac{A}{2})=7\sin (\frac{A}{2})=4\sin A$. Thus, $\cos (\frac{A}{2})=\frac78$. Or $\cos A=2(\frac78)^2-1$.
So: $$CF=8\sin A=8\sqrt{1-(2(\frac78)^2-1)^2}=8\sqrt{1-(\frac{17}{32})^2}=8\sqrt{1-\frac{17}{32}}\sqrt{1+\frac{17}{32}}=\frac74\sqrt{15}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Limit of: $\lim_{x\to0}\frac{1-\sqrt[3]{\cos{x}}}{1-\cos{\sqrt[3]{x}}}$ I calculated this limit with L'Hospital's rule, but at the end it got rather complicated.
$\lim_{x\to0}\frac{1-\sqrt[3]{\cos{x}}}{1-\cos{\sqrt[3]{x}}}$
Is there some other more effective way for this limit?
| Writing $x=u^3$ for convenience, we have
$$\begin{align}
{1-\sqrt[3]{\cos x}\over1-\cos\sqrt[3]x}
&={1-\sqrt[3]{\cos u^3}\over1-\cos u}\\
&={1-\cos u^3\over1-\cos u}\cdot{1\over1+\sqrt[3]{\cos u^3}+\sqrt[3]{\cos^2u^3}}\\
&={1-\cos^2u^3\over1-\cos^2u}\cdot{1+\cos u\over(1+\cos u^3)(1+\sqrt[3]{\cos u^3}+\sqrt[3]{\cos^2u^3})}\\
&=u^4\left(\sin u^3\over u^3\right)^2\left(u\over\sin u\right)^2{1+\cos u\over(1+\cos u^3)(1+\sqrt[3]{\cos u^3}+\sqrt[3]{\cos^2u^3})}\\
&\to0\cdot1\cdot1\cdot{1+1\over(1+1)(1+1+1)}\\
&=0
\end{align}$$
| {
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"url": "https://math.stackexchange.com/questions/739919",
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"source": "stackexchange",
"question_score": "3",
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"answer_id": 4
} |
coefficient of $x^2$, in $(1+x+x^2)^{10}$
How to find coefficient of $x^2$, in $(1+x+x^2)^{10}$, without actually expanding it?
I think the fact $\dfrac{1-x^3}{1-x}=1+x+x^2$ may help.
But can't use it!
| As you noted:
$$
[x^2]\left(1+x+x^2\right)^{10} = [x^2] \frac{(1-x^3)^{10}}{(1-x)^{10}}
$$
Because $x^3$ will only contribute to terms $x^3$ and higher order
$$
[x^2] \frac{(1-x^3)^{10}}{(1-x)^{10}} = [x^2] \frac{1}{(1-x)^{10}} = [x^2] \frac{1}{(1-x)^{10}}
$$
Now using $n \cdot [x^n] g(x) = [x^{n-1}] g^\prime(x)$ with $n=2$ and $g(x) = \frac{1}{9}(1-x)^{-9}$:
$$
[x^2] \frac{1}{(1-x)^{10}} = 3 [x^{3}] \frac{1}{9 (1-x)^9} = 3 \cdot 4 [x^4] \frac{1}{9 \cdot 8 (1-x)^8} = \frac{(11)!}{2} \cdot [x^{11}] \frac{1}{9! \cdot (1-x)}
$$
Using $[x^n] \left(1-x\right)^{-1} = 1$ for $n\geq 1$ we arrive at:
$$
[x^2] \frac{1}{(1-x)^{10}} = \frac{(11)!}{2 \cdot 9!} = \frac{11 \cdot 10}{2} = 55
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $
How to find infinite sum How to find infinite sum $$1+\dfrac13+\dfrac{1\cdot3}{3\cdot6}+\dfrac{1\cdot3\cdot5}{3\cdot6\cdot9}+\dfrac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $$
I can see that 3 cancels out after 1/3, but what next? I can't go further.
| Consider denominator and numerator separately at first,
$$G_n = 2^n \prod_{m=1}^n m-1/2, \qquad F_n = \frac{1}{3^n n!}$$
Thus we have
$$T_n = \prod_{m=1}^n \frac{m-1/2}{n!} \left(\frac{2}{3}\right)^n \qquad \text{or} \qquad T = \sum_{n=0}^\infty \prod_{m=1}^n \frac{m-1/2}{n!}\left(\frac{2}{3}\right)^n$$
Looking these series elements up we arrive at $T=\sqrt{3}$.
EDIT
The final form of the series is
$$\sum_{n=0}^\infty \frac{\Gamma (n+1/2) }{\sqrt{\pi} n!} \left( \frac{2}{3}\right)^n =\sqrt{3}$$
where $\Gamma(n)$ is the well-known Gamma function.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 7,
"answer_id": 5
} |
Prove that $f(x)=(e^x-1)(x^2+3x-2)+x$ has exactly one positive root, exactly one negative root and one root at $x=0$. Prove that the function
$f(x)=(e^x-1)(x^2+3x-2)+x$
has exactly one positive root, exactly one negative root and one root at $x=0$.
My work so far:
$f(0)=0$
Thus, $x=0$ is a root.
For the positive root,
$f(\frac{1}{4})=(e^{\frac{1}{4}}-1)((\frac{1}{4})^2+3(\frac{1}{4})-2)+(\frac{1}{4})
=(e^{\frac{1}{4}}-1)(\frac{1}{16}+\frac{3}{4}-2)+\frac{1}{4}
=-\frac{19}{16}e^{\frac{1}{4}}+\frac{23}{16}
<0$
$f(1)=(e^1-1)((1)^2+3(1)-2)+(1)
=(e^1-1)(1+3-2)+1=2e^1-1
>0$
Thus, there is at least one positive root, by the intermediate value theorem. I have also done the same thing for the negative root, as $f(-4)<0$ and $f(-2)>0$.
But how do I show that there is only one positive root and only one negative root?
| I bet, you have to find any two positive numbers a and bsuch that f(a)<0 and f(b)>0. Thus, it will prove that there is one positive root that is between a and b.
Same logic applies for the negative root.
| {
"language": "en",
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} |
How prove this irrational $x\in(0,1)$,then have $0
show that:
for any irrational $x\in(0,1)$,and positive integer $n$,there exsit positive integer $p_{1},p_{2},\cdots,p_{n}$ where
$$p_{1}<p_{2}<\cdots<p_{n}$$
such
$$0<x-\sum_{i=1}^{n}\dfrac{1}{p_{i}}<\dfrac{1}{n!(n!+1)}$$
This problem is from Mathematical contest in jiangxi province at last problem.
I think this reslut maybe involves
Irrational Approximation?
Note $$\dfrac{1}{n!(n!+1)}=\dfrac{1}{n!}-\dfrac{1}{n!+1}$$
Thank you for you help
| The greedy algorithm works. We let $x_0 = x$, and for all $n$ define
$$p_{n+1} = \left\lfloor \frac{1}{x_n}\right\rfloor + 1; \quad x_{n+1} = x_n - \frac{1}{p_{n+1}}.$$
It is easily verified that
$$0 < x_1 = x - \frac{1}{p_1} < \frac{1}{2} = \frac{1}{1!(1!+1)},$$
and since all $x_n$ are irrational, we have $p_{n+1}-1 < \frac{1}{x_n} < p_{n+1}$, whence
$$0 < x_n - \frac{1}{p_{n+1}} < \frac{1}{p_{n+1}-1} - \frac{1}{p_{n+1}},$$
and by induction $p_{n+1} > n!(n!+1)$, which yields
$$\begin{align}
\frac{1}{p_{n+1}-1} - \frac{1}{p_{n+1}} &< \frac{1}{n!(n!+1)} - \frac{1}{n!(n!+1)+1}\\
&= \frac{1}{n!(n!+1)(n!(n!+1)+1)}\\
&\leqslant \frac{1}{(n+1)!((n+1)!+1)}.
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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number of terms The following problem maybe tedious if done by hand and requires patience. After factorizing the following variables find the number of terms and the sum of the number of terms.
$(a^0),(a+b)^0,(a+b+c)^0,(a+b+c+d)^0,(a+b+c+d+e)^0,(a+b+c+d+e+f)^0$
$(a^1),(a+b)^1,(a+b+c)^1,(a+b+c+d)^1,(a+b+c+d+e)^1,(a+b+c+d+e+f)^1$
$(a^2),(a+b)^2,(a+b+c)^2,(a+b+c+d)^2,(a+b+c+d+e)^2,(a+b+c+d+e+f)^2$
$(a^3),(a+b)^3,(a+b+c)^3,(a+b+c+d)^3,(a+b+c+d+e)^3,(a+b+c+d+e+f)^3$
$(a^4),(a+b)^4,(a+b+c)^4,(a+b+c+d)^4,(a+b+c+d+e)^4,(a+b+c+d+e+f)^4$
$(a^5),(a+b)^5,(a+b+c)^5,(a+b+c+d)^5,(a+b+c+d+e)^5,(a+b+c+d+e+f)^5$
$(a^6),(a+b)^6,(a+b+c)^6,(a+b+c+d)^6,(a+b+c+d+e)^6,(a+b+c+d+e+f)^6$
| What you are searching for is the multinomial theorem.
The formula is
$$ (x_1+x_2+...+x_m)^n = \sum_{k_1+...+k_m=n} {n \choose k_1,...,k_m} x_1^{k_1}...x_m^{k_n} $$
where
$$ {n \choose k_1,...,k_m} = \frac{n!}{k_1!k_2!...k_m!}.$$
For a proof take a look at the Wikipedia post: http://en.wikipedia.org/wiki/Multinomial_theorem
| {
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Prove that $\sum_{k=0}^n k^2{n \choose k} = {(n+n^2)2^{n-2}}$ Prove that: $$\sum_{k=0}^n k^2{n \choose k} = {(n+n^2)2^{n-2}}$$ i know that: $$\sum_{k=0}^n {n \choose k} = {2^n}$$ how to get the (n + n^2)?
| Its a bit easiler to evaluate up to $n+1$.
I will use the identity that $\dbinom{a}k = \frac{a}k \dbinom{a-1}{k-1}$.
$\sum_{k = 0}^{n+1} k^2 \dbinom{n+1}k$
$ = 0+1 \dbinom{n+1}1 + 4 \dbinom{n+1}2 + 9 \dbinom{n+1}3 + \cdots + (n+1)^2 \dbinom{n+1}{n+1}$
$ = 0+(n+1) \dbinom{n}0 + 2(n+1) \dbinom{n}1 + 3 \dbinom{n}2 + \cdots + (n+1) \dbinom{n}n$
$ = (n+1) \left( \sum_{k = 0}^{n} (k+1) \dbinom{n}k \right)$.
This can be easily evaluated from the derivative of $x(1+x)^n$.
| {
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Finding circumcentre Tangents are draw from $P(2,3)$ to $x^2+y^2=4$ meeting at $Q,R$ on circle. Parallelogram $PQSR$ is completed. Find the circumcentre of triangle $QSR$.
My attempt: Clearly, the parallelogram is a rhombus. $S$ will be mirror image of $P$ in chord of contact. But, I want to know a good way of finding the circumcentre other than it lies on $PS$ and dropping perpendiculars and other boring stuff. Maybe a nice geometrical solution? Or combining it with coordinate?
| The equation of the circle with center $(0, 0)$ and radius $r$ is
$$x^2 + y^2 = r^2.$$
The equation of tangent to this circle at the point $(x_p, y_p)$ on the circle is
$$x_px + y_py = r^2,$$
then the equation of tangent at the point $(2,3)$ is
$$2x + 3y = 4\quad\Rightarrow\quad y=\frac{4-2x}3.$$
Now, we determine the points of tangency (coordinates of $R$ and $Q$).
$$
\begin{align}
x^2+y^2&=4\\
x^2+\left(\frac{4-2x}3\right)^2&=4\\
x_1=2\quad&;\quad x_2=-\frac{10}{13}\\
&\text{and}\\
y_2=0\quad&;\quad y_2=\frac{24}{13}.\\
\end{align}
$$
We obtain $Q(2,0)$ and $R\left(-\dfrac{10}{13},\dfrac{24}{13}\right).$ Since $PQRS$ is a parallelogram, then $PQ\parallel RS$ and $PR\parallel QS$. It's easy to notice that $PQ$ is perpendicular with $x$-axis, therefore $RS$ is also perpendicular with $x$-axis and line $RS: x=-\dfrac{10}{13}$.
Line $PQ:\ x=2$
Line $PR:$
$$
\begin{align}
\frac{y-y_P}{y_R-y_P}&=\frac{x-x_P}{x_R-x_P}\\
\frac{y-3}{\frac{24}{13}-3}&=\frac{x-2}{-\frac{10}{13}-2}\\
y&=\frac{5}{12}x+\frac{13}{6}
\end{align}
$$
Line $RQ:y=-\dfrac{2}{3}x+\dfrac{4}{3}$
Since $PR\parallel QS$, then gradient line $PR$ is equal to gradient line $QS$, $m=\dfrac{5}{12}$. Therefore, line $QS:$
$$
\begin{align}
y-y_Q&=m(x-x_Q)\\
y-0&=\dfrac{5}{12}(x-2)\\
y&=\dfrac{5}{12}x-\dfrac{5}{6}\\
\end{align}
$$
The coordinate of $S$ is the intersection point line $RS$ and $QS$. Hence $x_S=-\dfrac{10}{13}$ and
$$
y_S=\dfrac{5}{12}x_S-\dfrac{5}{6}=-\dfrac{15}{13}.
$$
Now, we determine line that is perpendicular to line $RQ$ and passes through point $S$, let's call this line as $SS'$. The gradient of $SS'$ is $-\dfrac1{m_{RQ}}=\dfrac32$, then the equation is
$$
\begin{align}
y-y_S&=m(x-x_S)\\
y+\dfrac{15}{13}&=\dfrac32\left(x+\dfrac{10}{13}\right)\\
y&=\frac32 x.
\end{align}
$$
The midle point of line $RS$ is
$$
y_M=\frac12(y_R+y_S)=\frac12\left(\dfrac{24}{13}-\dfrac{15}{13}\right)=\frac9{26}.
$$
It is clearly that the circumcentre $\Delta QSR$ on the intersection point of line $SS'$ and line $QM$. I think you can handle it the rest.
P.S. : Unfortunately, I cannot find the fastest way than this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/753517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How Find this maximum $P=\frac{4}{\sqrt{a^2+b^2+c^2+4}}-\frac{9}{(a+b)\sqrt{(a+2c)(b+2c)}}$
let $a,b,c>0$, find the maximum
$$P=\dfrac{4}{\sqrt{a^2+b^2+c^2+4}}-\dfrac{9}{(a+b)\sqrt{(a+2c)(b+2c)}}$$
I think this inequality we can use AM-GM inequality to solve it,and Now first we must sure this equality when $a,b,c$ hold maximum
Thank you
| First, we have by AM-GM:
$$(a+b)\sqrt{(a+2c)(b+2c)} \le (a+b)\frac{a+b+4c}2 = \frac{a^2}2 + \frac{b^2}2+ab+2ac+2bc \le 2(a^2+b^2+c^2)$$
where there is equality iff $a=b=c$. So let $s = a^2+b^2+c^2$. Then we have:
$$P \le \frac4{\sqrt{s+4}}- \frac9{2s}$$
Using one variable calculus, we can find the derivative condition for maximising the RHS, which gives $16s^4=81(s+4)^3$. This is equivalent to $(s-12)(16s^3+111s^2+360s+432)=0$, where the cubic factor cannot have any positive roots by the rule of signs.
Hence $P$ has a maximum for positive $s$ only when $s=12$. Thus $a=b=c=2$ at the maximum, and $P\le \frac58$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/754813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
William Lowell Putnam Integral Problem Prove That
$$
\frac{22}{7}-\pi
=
\int_{0}^{1}\frac{x^{4}\left(1 - x\right)^{4}}{1 + x^{2}}\,{\rm d}x
$$
| Literally all you have to do is expand it out, divide, and integrate.
$$
x^4(1-x)^4 = x^8 - 4x^7 + 6x^6 - 4x^5 + x^4
$$
Now use long division:
$$
x^4(1-x)^4 = x^6(x^2+1)- 4x^5(x^2+1) + 5x^4(x^2 + 1) - 4x^2(x^2+1)+4(x^2+1) -4 \\
\frac{x^4(1-x)^4}{1+x^2} = (x^6 -4x^5 + 5x^4 - 4x^2 + 4) - \frac{4}{x^2+1} \\
$$
We integrate this:
$$
\int_0^1 \! (x^6 -4x^5 + 5x^4 - 4x^2 + 4) \,\mathrm{d}x - \int_0^1 \! \frac{4}{x^2+1} \, \mathrm{d}x \\
= \frac{22}{7} - 4(\arctan(1) - \arctan(0)) \\
= \frac{22}{7} - \pi
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/755865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Problem solving a simple integral equation Could someone help me finding q(t) so that
\begin{equation}
\int_x^{x+2\pi/3}q(t)\,\mathrm{d}t = a\cdot\sin{(x)}+b
\end{equation}
Thanks in advance!
| My try:
\begin{equation}
\int_0^{\frac{2\pi}{3}}q\left(v+x\right)\,\mathrm{d}v = a \sin{x}+b
\end{equation}
In the other hand:
$$\int_0^{\frac{2\pi}{3}} \frac{3xa}{2\pi}\cos\left(\frac{xv}{2\pi/3}\right) dv=a\sin\left(x\right)$$
$$\int_0^{\frac{2\pi}{3}} \frac{b}{2\pi/3} dv=b$$
Then:
$$\int_0^{\frac{2\pi}{3}}q\left(v+x\right)\,\mathrm{d}v= \int_0^{\frac{2\pi}{3}} \frac{3xa}{2\pi}\cos\left(\frac{xv}{2\pi/3}\right) dv+\int_0^{\frac{2\pi}{3}} \frac{b}{2\pi/3}dv$$
$$\int_0^{\frac{2\pi}{3}} \left( q\left(v+x\right)-\frac{3xa}{2\pi}\cos\left(\frac{xv}{2\pi/3}\right)-\frac{b}{2\pi/3} \right) \,\mathrm{d}v=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/756396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding (sin(A+B))^2 given roots of a quadratic equation. If tan A and tan B are the roots of the equation x^2 -ax + b = 0, then the value of sin(A+B)^2 is?
Options are:
((a^2)/((a^2)+(1-b)^2),
(a^2)/(a^2+b^2),
a^2/(b+a)^2,
a^2/(b^2*(1-a)^2)
The value is needed in terms of a and b.
The farthest i have gotten is the sum of roots giving me:
(sinAcosB + CosAsinB)/CosAcosB = a
which simplifies to sin^2(A+B) = (a*cosA*cosB)^2
after which i substituted the value obtained from product of roots which was ((sinAsinB)/b)^2 = (CosACosB)^2 which is substituted again to obtain:
(a*SinA*sinB)^2/(b^2)
Please find the value with steps as i am very confused about this question.
| So, $\displaystyle\tan A+\tan B=\frac a1$ and $\displaystyle \tan A\tan B=\frac b1$
$\displaystyle\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$
$\displaystyle\sin^2(A+B)=\frac1{\csc^2(A+B)}=\frac1{1+\cot^2(A+B)}=\frac{\tan^2(A+B)}{\tan^2(A+B)+1}=\cdots$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/756512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\int \frac{\sqrt{x^2-1}}{x} \mathrm{d}x$ My try, using $x = \sec(u)$ substitution:
$$
\begin{eqnarray}
\int \frac{\sqrt{x^2-1}}{x} \mathrm{d}x &=& \int \frac{\sqrt{\sec^2(u) - 1}}{\sec(u)}\tan(u)\sec(u) \mathrm{d}u \\
&=& \int \tan^2(u) \mathrm{d}u \\
&=& \tan(u) - u + C \\
&=& \tan(arcsec(x)) - arcsec(x) + C
\end{eqnarray}
$$
However, according to Wolfram Alpha, the answer should be:
$$
\int \frac{\sqrt{x^2-1}}{x} \mathrm{d}x = \sqrt{x^2-1}+\arctan \left( \frac{1}{\sqrt{x^2-1}} \right)+C
$$
When I derive this last answer I don't get back the integrand, but rather:
$$
\frac{\mathrm d}{\mathrm d x}\left(\sqrt{x^2-1}+\arctan \left( \frac{1}{\sqrt{x^2-1}} \right)+C\right) = \frac{x}{\sqrt{x^2-1}}- \frac{x}{(x^2-1)^{3/2}\left(1+\frac{1}{x^2-1}\right)}
$$
I don't know how to simplify this expression more. Also, I am unable to check whether my answer is correct because I don't know how to find the derivative of $arcsec(x)$.
Can someone check my calculations and tell me where I've done something wrong and how one can simplify the last expression to get back the integrand?
| Calculating this integral can be done in this way:
$$\int { \sqrt{x^2-a^2} \over x } dx =\int { x^2-a^2 \over x\sqrt{x^2-a^2} } dx= \int { x \over \sqrt{x^2-a^2} } dx-\int { a^2 \over x\sqrt{x^2-a^2} } dx =\sqrt{x^2-a^2}(+-)a\cdot\int {(\frac{x}{a})'\over \sqrt{1-(\frac{x}{a})^2)} } dx=\sqrt{x^2-a^2}(+-)a\cdot\arcsin(\frac{x}{a})+C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/757059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Find $4\cos\theta-3\sin\theta$, given that $4\sin \theta +3\cos \theta = 5$ Another problem that I already wasted hours on.
Given
$$4\sinθ +3\cosθ = 5$$
Find
$$4\cosθ -3\sinθ$$
Help me guys (PS:I'm not that good in maths)
| If $a\sin\theta+b\cos\theta=c$
Squaring we get $$a^2\sin^2\theta+b^2\cos^2\theta+2ab\cos\theta\sin\theta=c^2$$
$$\iff a^2(1-\cos^2\theta)+b^2(1-\sin^2\theta)+2ab\cos\theta\sin\theta=c^2$$
$$\iff c^2-a^2-b^2=(a\cos\theta-b\sin\theta)^2$$
Here $\displaystyle a=4,b=3,c=5\implies c^2-a^2-b^2=?$
Alternatively,
Let $a=r\cos\phi, b=r\sin\phi\ \ \ \ (1)$ where $r\ge0$
So, $\displaystyle a\sin\theta+b\cos\theta=c\implies r\sin(\theta+\phi)=c$
$\displaystyle a\cos\theta-b\sin\theta=r\cos(\theta+\phi)=r\cdot\pm\sqrt{1-\frac{c^2}{r^2}}=\pm\sqrt{r^2-c^2}$
Now, squaring & adding $(1)$ we get $\displaystyle r^2=a^2+b^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/757497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 14,
"answer_id": 4
} |
what method should I use? The power formula? or The logarithmic formula? (indefinite integral) Should I use the power formula or logarithmic to evaluate this integral:
$$\int \frac {x+1} {\sqrt[3]{x^2 + 2x + 1}}~dx$$
| I assume the integral is
\begin{align}
\int\frac{x+1}{\sqrt[3]{x^2+2x+1}}\,dx
\end{align}
Then
\begin{align}
\require{cancel}
\int\frac{x+1}{\sqrt[3]{x^2+2x+1}}\,dx&=\int\frac{x+1}{\sqrt[3]{(x+1)^2}}\,dx\\
&=\int\frac{x+1}{(x+1)^{\frac{2}{3}}}\,dx\\
&=\int\frac{(x+1)^{\frac{2}{3}+\frac{1}{3}}}{(x+1)^{\frac{2}{3}}}\,dx\\
&=\int\frac{\cancel{(x+1)^{\frac{2}{3}}}(x+1)^{\frac{1}{3}}}{\cancel{(x+1)^{\frac{2}{3}}}}\,dx\\
&=\int(x+1)^{\frac{1}{3}}\,dx\qquad;\;\text{let}\;u=x+1\;\text{then}\;dx=du\\
&=\int u^{\frac{1}{3}}\,du\\
&=\frac{1}{\frac{1}{3}+1}u^{\frac{1}{3}+1}+C\\
&=\frac{3}{4}u^{\frac{4}{3}}+C\\
&=\frac{3}{4}(x+1)^{\frac{4}{3}}+C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/761484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
show that inequality holds for real numbers There are given real numbers $a,b,c,d \in [0,1]$ show that:
$a+b+c+d \le 1+a(b+c+d)+b(c+d)+cd$
I tried to transform it to
$b(1-a)+c(1-a)+d(1-a)-(1-a)\le b(c+d)+cd$
$(1-a)(b+c+d-1) \le b(c+d)+cd$
| Expanding $0 \le (1-a)(1-b)(1-c)(1-d)$ gives
$$
0 \le 1-a-b-c-d+ab+ac+ad+bc+bd+bd - abc-abd-acd-bcd+abcd
$$
Each of the terms $abc$, $abd$, $acd$, $bcd$ is $\ge abcd$, so it follows that
$$
0 \le 1-a-b-c-d+ab+ac+ad+bc+bd+bd - 3 abcd
$$
which gives the slightly stronger statement
$$
a+b+c+d+3abcd \le 1+a(b+c+d)+b(c+d)+cd
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/764257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Maximize $\sqrt{2x + 13} + \sqrt[3]{3y+5} + \sqrt[4]{8z+12}$ Given three non-negative (as pointed out by Calvin Lin) real numbers $x+y+z = 3$, find the maximum value of $\sqrt{2x + 13} + \sqrt[3]{3y+5} + \sqrt[4]{8z+12}$.
(Source : Singapore Math Olympiad 2012, Senior section, Round 1, question 29).
I tried using the fact that $2x +13, 8z + 12\ge 0$ to deduce that $y \le 11$, but I couldn't continue from there on. The answer should be an integer, since only integer answers were allowed.
The competition was designed for 15/16 year olds. A simple yet elegant solution would be nice.
For reference of the original problem,
| \begin{align*}&\sqrt{\dfrac{2x+13}{4}}\cdot\sqrt{4}+\sqrt[3]{\dfrac{3y+5}{4}}\cdot\sqrt[3]{2}\cdot\sqrt[3]{2}+\sqrt[4]{\dfrac{8z+12}{8}}\cdot\sqrt[4]{2}\cdot\sqrt[4]{2}\cdot\sqrt[4]{2}\\
&\le\dfrac{\dfrac{2x+13}{4}+4}{2}+\dfrac{\dfrac{3y+5}{4}+2+2}{3}+\dfrac{\dfrac{8z+12}{8}+2+2+2}{4}\\
&=\dfrac{1}{4}(x+y+z)+\dfrac{29}{4}\\
&=8
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/766869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
If $(1 + 2i)$ and $(3 - 2i)$ are two roots of $x^5 + ax^4 + bx^3 + cx^2 + dx + 4$, then $a$ =? Consider the polynomial $x^5 + ax^4 + bx^3 + cx^2 + dx + 4$ where $a, b, c, d$ are real
numbers. If $(1 + 2i)$ and $(3 - 2i)$ are two roots of this polynomial then what is the
value of a?
Well, I know only the 4 roots, which are obvious from what are given (the conjugates serve as the other two), but what next?
| i think the solution will come from vieta's theorem .
after this multiple of all the roots will give a 4 . let the fifth root be r then ,
Acc to vieta theorem: product of roots = (-1)^n (a0/an) = -4
-4 = (1+ 2i)(1-2i)(3-2i) (3+2i) r
find the value of r from this ..
r= -65/4
now after this again from vietas theorem
a = (-1)((1+2i) + (1-2i) + (3+2i) +(3-2i) + r ( r is the value u got above )
a= 33/4
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/767147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Why is $ \int \frac{\sin x (b-a\cos x)}{(b^2+a^2-2ab \cos x)^{3/2}}\,dx = \frac{a-b\cos x}{b^2 \sqrt{a^2-2ab\cos x + b^2}}$? Why is $$ \int \frac{\sin x (b-a\cos x)}{(b^2+a^2-2ab \cos x)^{3/2}}\,dx = \frac{a-b\cos x}{b^2 \sqrt{a^2-2ab\cos x + b^2}}\text{ ?}$$
Constant of integration omitted.
| Hint: $~t=\cos x\iff dt=-\sin x,~$ or $~u=a^2+b^2-2ab\cos x\iff du=2ab\sin x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/767497",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Elementary problems Find $n$ such that $n$ is a positive integer satisfying the following equation.
$$2(2^2)+3(2^3)+4(2^4)+\ldots+n(2^n)=2^{n+10}$$
Can anybody help ? I can't believe this is an elementary school problem...
| $\begin{eqnarray}
{\bf Hint} \
&& \color{#0a0}0\cdot 2^2\! &+& \color{#c00}2\cdot 2^2\! &+& 3\cdot 2^3\! &+& 4\cdot 2^4\! &+&\, \cdots &\!+& n\cdot 2^n \\
&& &=& \color{#0a0}1\cdot 2^3\! &+& \color{#c00}3\cdot 2^3\! &+& 4\cdot 2^4\! &+&\, \cdots &\!+& n\cdot 2^n \\
&& && &=& \color{#0a0}2\cdot 2^4\! &+& \color{#c00}4\cdot 2^4\! &+&\, \cdots &\!+& n\cdot 2^n \\
&& && && && \ddots &&\ \ \, \vdots &&\\
&& && && && &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! = \color{#0a0}{(n\!-\!2)}\, 2^n\!\! &+& \color{#c00} n\cdot 2^n\\
&& && && && && &&\!\!\!\!\!\!\!\!\!\!\!\!\!\! =\color{#0a0}{(n\!-\!1)}\, 2^{n+1}
\end{eqnarray}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/770021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\tan70^\circ - \tan50^\circ + \tan10^\circ = \sqrt{3}$ The question is:
Prove $$\tan70^\circ - \tan50^\circ + \tan10^\circ = \sqrt{3}.$$
I had no idea how to do it & proceeded by making RHS = $\tan60^\circ$ but this doesn't make any help. Please help me solving this.
| If you know following cotangent identity,
$$n \cot(n\theta) = \sum_{k=0}^{n-1}\cot\left(\theta+\frac{k\pi}{n}\right)
\quad\forall n \in \mathbb{Z}_{+}\tag{*1}$$
What you need to show is pretty simple,
$$\tan10^\circ + \tan 70^\circ - \tan 50^\circ = \cot 80^\circ + \cot 20^\circ + \cot( -40^\circ)\\
= \sum_{k=0}^{2}\cot\left(-40^\circ + \frac{k}{3} \times 180^\circ\right)
= 3\cot(3 \times -40^\circ) = 3 \cot 60^\circ = \frac{3}{\sqrt{3}} = \sqrt{3}.
$$
Since I didn't find a proof of the cotangent identity online. I will give a proof here.
For any $\theta$ and integer $n > 0$, notice
$$\begin{align}
\prod_{k=0}^{n-1}\sin\left(\theta + \frac{k\pi}{n}\right)
&= \prod_{k=0}^{n-1}\left[\frac{
e^{i\left(\theta + \frac{k\pi}{n}\right)} - e^{-i\left(\theta + \frac{k\pi}{n}\right)}
}{2i}\right]
= \frac{e^{-i\left(n\theta - \sum_{k=0}^n \frac{k\pi}{n}\right)}}{(2i)^n}
\prod_{k=0}^{n-1}\left(e^{2i\theta}-e^{-i\frac{2k\pi}{n}}\right)\\
&= \frac{e^{-in\theta}}{2^ni}\left(e^{2in\theta} - 1\right)
= 2^{1-n} \sin(n\theta)
\end{align}
$$
Taking logarithm and differentiate on both sides immediately give you identity $(*1)$.
| {
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"url": "https://math.stackexchange.com/questions/770391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
If $f$ is uniformly continuous on $\mathbb{R}$, $f(x) \ge a >0$ and $g(x) = 1/f(x)^2$, then $g(x)$ is uniformly continuous I get to the this point
$$
|g(x)-g(y)| =
\left|\frac{1}{f(x)^2} - \frac{1}{f(y)^2}\right| =
\left|\frac{f(x)^2 - f(y)^2}{f(x)^2f(y)^2}\right| \leq \frac{1}{a^4} |f(x)-f(y)|\,|f(x)+f(y)|.
$$
I want to use my assumption $|x-y| < \delta$, but I don't know how to do that given $|f(x)-f(y)|$.
| An upper bound that is not as resourceful but also serves is as follows:
\begin{align*}
|g(x)-g(y)|
=& \left|\frac{1}{f(x)^2} - \frac{1}{f(y)^2}\right|
\\
=&
\left|\frac{f(x)^2 - f(y)^2}{f(x)^2f(y)^2}\right|
\\
=&
\left|\frac{1}{f(x)^2} \right|\cdot \left|\frac{1}{f(y)^2}
\right|\cdot \left|\color{red}{f(x)\cdot f(x)} - f(y)\cdot f(y)\right|
\\
=
&
\left|\frac{1}{f(x)^2} \right|\cdot \left|\frac{1}{f(y)^2} \right|
\cdot
\left|\Big[ \color{red}{\big(f(x)-f(y)\big)}+ \color{blue}{f(y)} \Big]
\cdot \Big[ \color{red}{\big(f(x)-f(y)\big)}+ \color{blue}{f(y) }\Big] - f(y)\cdot f(y)\right|
\\
=
&
\left|\frac{1}{f(x)^2} \right|\cdot \left|\frac{1}{f(y)^2} \right|
\cdot
\bigg|
\color{red}{\Big(f(x)-f(y)\Big)\cdot \Big(f(x)-f(y)\Big)}
+2\cdot\color{blue}{f(y)}\cdot \color{red}{\Big(f(x)-f(y)\Big)}
\bigg|
\end{align*}
Due to do the triangle inequality, $\cfrac{1}{f(x)}\leq \cfrac{1}{a}$ and$ \cfrac{1}{f(y)}\leq \cfrac{1}{a}$:
\begin{align*}
|g(x)-g(y)|
\leq
&
\left|\frac{1}{a^4} \right|
\cdot
\Big|\color{red}{\big(f(x)-f(y)\big)}\Big|\cdot
\Big|\color{red}{\big(f(x)-f(y)\big)}\Big|
\\
&
\qquad\qquad +
\left|\frac{1}{a^3} \right|
\cdot
\left|\frac{1}{f(y)} \right|
\cdot2\cdot\big|\color{blue}{f(y)}\big|\cdot \Big|\color{red}{\big(f(x)-f(y)\big)}\Big|
\\
\leq
&
\left|\frac{1}{a^4} \right|
\cdot
\Big|\color{red}{\big(f(x)-f(y)\big)}\Big|\cdot \Big|\color{red}{\big(f(x)-f(y)\big)}\Big|
+\left|\frac{2}{a^3} \right|
\cdot
\Big|\color{red}{\big(f(x)-f(y)\big)}\Big|
\\
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/771955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Help with integral I seem to be stuck trying to prove the following integral
$$
\int\frac{\cos^mx}{\sin^nx}dx = -\frac{\cos^{m+1}x}{(n-1)\sin^{n-1}x}-\frac{m-n+2}{n-1}\int\frac{\cos^mx}{\sin^{n-2}x} dx + C\,\,(n \neq 1)
$$
My thinking so far has been that if I take
$$
I = \int\frac{\cos^mx}{\sin^nx}dx
$$
I have been able to prove that
$$
I = -\frac{\cos^{m-1}x}{(n-1)\sin^{n-1}x} - \frac{m-1}{n-1}\int\frac{\cos^{m-2}x}{\sin^{n-2}x}\,dx+C\,\,\,\,\,(1)
$$
and
$$
I = \frac{\cos^{m-1}x}{(m-n)\sin^{n-1}x} + \frac{m-1}{m-n}\int\frac{\cos^{m-2}x}{\sin^nx}\,dx+C\,\,\,\,\,(2)
$$
but showing that
$$
I = -\frac{\cos^{m+1}x}{(n-1)\sin^{n-1}x}-\frac{m-n+2}{n-1}\int\frac{\cos^mx}{\sin^{n-2}x} dx + C
$$
seems to be eluding me. I attempted to apply a similar technique what I used on $(1)$ to get $(2)$ to try to obtain this integral, but it didn't seem to work.
I can also show that
$$
I = -\frac{\cos^{m+1}x}{(m+1)\sin^{n+1}x} - \frac{n+1}{m+1}\int\frac{\cos^{m+2}x}{\sin^{n+2}x}\, dx + C\,\,\,\,\,(3)
$$
but there's obviously more to it.
Any broad hints would be more than welcome.
| After a little more thought (and a lot more sleep) I finally figured out the answer:
Let
$$
I_1 = \int\frac{\cos^mx}{\sin^{n-2}x}\, dx = \int\frac{\cos^mx}{\sin^{n-1}x}. \sin x\, dx
$$
Integrate $I_1$ by parts, choose $u=\frac{\cos^mx}{\sin^{n-1}x}$, $dv = \sin x\, dx$ so $v = -\cos x$
$$
du = \frac{\sin^{n-1}x(-m\cos^{m-1}x\sin x)-\cos^mx((n-1)\sin^{n-2}x\cos x)}{(\sin^{n-1}x)^2}
$$
$$
=\frac{sin^{n-1}x\cos^{m-1}x(-m\sin x)-\sin^{n-1}x\cos^{m-1}x((n-1)\sin^{-1}x\cos^2x)}{(\sin^{n-1}x)^2}
$$
$$
= \frac{\cos^{m-1}x}{\sin^{n-1}x}(-m\sin x - (n-1)\frac{\cos^2 x}{\sin x})
$$
So
$$
I_1 = -\frac{\cos^{m+1}x}{\sin^{n-1}x} - \int\frac{\cos^{m-1}x}{\sin^{n-1}x}(-m\sin x - (n-1)\frac{\cos^2 x}{\sin x})(-\cos x)\, dx
$$
$$
= -\frac{\cos^{m+1}x}{\sin^{n-1}x} - \int\frac{\cos^{m-1}x}{\sin^{n-1}x}(m\sin x\cos x + (n-1)\frac{\cos^3x}{\sin x})\, dx
$$
$$
= -\frac{\cos^{m+1}x}{\sin^{n-1}x} - m\int\frac{\cos^mx}{\sin^{n-2}x}\,dx - (n-1)\int\frac{\cos^{m+2}x}{\sin^nx}\, dx
$$
which implies
$$
(1+m)I_1 = -\frac{\cos^{m+1}x}{\sin^{n-1}x} - (n-1)\int\frac{\cos^{m+2}x}{\sin^nx}\, dx
$$
$$
= -\frac{\cos^{m+1}x}{\sin^{n-1}x} - (n-1)\int\frac{\cos^mx. \cos^2 x}{\sin^nx}\, dx
$$
$$
= -\frac{\cos^{m+1}x}{\sin^{n-1}x} - (n-1)\int\frac{\cos^mx.(1-\sin^2 x)}{\sin^nx}\, dx
$$
$$
= -\frac{\cos^{m+1}x}{\sin^{n-1}x} - (n-1)(\int\frac{\cos^mx}{\sin^nx}\,dx - \int\frac{\cos^mx}{\sin^{n-2}x}\,dx )
$$
Let
$$
I = \int\frac{\cos^mx}{\sin^nx}\,dx
$$
so we now have
$$
I_1 = \frac{-\cos^{m+1}x}{(1+m)\sin^{n-1}x} - \frac{n-1}{m+1}I + \frac{n-1}{m+1}\int\frac{\cos^mx}{\sin^{n-2}x}\,dx
$$
$\implies$
$$
(1-\frac{n-1}{m+1})I_1 = \frac{-\cos^{m+1}x}{(1+m)\sin^{n-1}x} - \frac{n-1}{m+1}I
$$
$\implies$
$$
\frac{m-n+2}{m+1}I_1 = \frac{-\cos^{m+1}x}{(1+m)\sin^{n-1}x} - \frac{n-1}{m+1}I
$$
$\implies$
$$
\frac{n-1}{m+1}I = \frac{-\cos^{m+1}x}{(1+m)\sin^{n-1}x} - \frac{m-n+2}{m+1}I_1
$$
$\implies$
$$
I = \frac{m+1}{n-1}.\frac{-\cos^{m+1}x}{(1+m)\sin^{n-1}x} - \frac{m+1}{n-1}.\frac{m-n+2}{m+1}I_1
$$
$$
= \frac{-\cos^{m+1}x}{(n-1)\sin^{n-1}x} - \frac{m-n+2}{n-1}I_1
$$
All comments greatly appreciated!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/772470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Property of quadratic residues A simple question. If $p$ is an odd prime - just to avoid trivial cases, assume $p>3$ - and $z\in\mathbb{N}$. What can we say about $\left(p-z\mid p\right)$, where the notation $\left(a\mid b\right)$ denotes the Legendre symbol of $a$ and $b$?
Thanks
| Note that since $x^{p-1}=1$ for all $x\not\equiv0\pmod{p}$, we have
$$
\left(x^{\frac{p-1}2}\right)^2\equiv1\pmod{p}\tag{1}
$$
Since there are at most $\frac{p-1}2$ roots of
$$
x^{\frac{p-1}2}\equiv1\pmod{p}\tag{2}
$$
and at most $\frac{p-1}2$ roots of
$$
x^{\frac{p-1}2}\equiv-1\pmod{p}\tag{3}
$$
and there are exactly $p-1$ roots of $(1)$ there must be exactly $\frac{p-1}2$ roots of each $(2)$ and $(3)$. Since $x^2\equiv y^2\pmod{p}$ is true if and only if $x\equiv y\pmod{p}$ or $x\equiv-y\pmod{p}$, and $p$ is odd, exactly half of the non-zero residue classes mod $p$ are quadratic residues. Furthermore, if $x=y^2$, then
$$
x^{\frac{p-1}2}=y^{p-1}\equiv1\pmod{p}\tag{4}
$$
Therefore, $\left(x\over p\right)=1\implies(2)$ and $\left(x\over p\right)=-1\implies(3)$; that is,
$$
\left(x\over p\right)\equiv x^{\frac{p-1}2}\pmod{p}\tag{5}
$$
Therefore, using $(5)$, we can say
$$
\begin{align}
\left(p-x\over p\right)
&=\left(-x\over p\right)\\
&=\left(-1\over p\right)\left(x\over p\right)\\
&=\left\{\begin{array}{rl}
\left(x\over p\right)&\text{if }p\equiv1\pmod{4}\\
-\left(x\over p\right)&\text{if }p\equiv3\pmod{4}
\end{array}\right.\tag{6}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/773280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Integral $\int_0^{\pi/2} \ln(1+\alpha\sin^2 x)\, dx=\pi \ln \frac{1+\sqrt{1+\alpha}}{2}$ $$
I_1:=\int_0^{\pi/2} \ln(1+\alpha\sin^2 x)\, dx=\pi \ln \frac{1+\sqrt{1+\alpha}}{2}, \qquad \alpha \geq -1.
$$
I am trying to prove this integral $I_1$. We can write
$$
\int_0^{\pi/2} \ln(\alpha(1/\alpha+\sin^2 x))dx=\int_0^{\pi/2} \left(\ln \alpha+\ln (\frac{1}{\alpha}+\sin^2 x)\right)dx=\frac{\pi}{2} \ln \alpha+I_2
$$
where
$$
I_2=\int_0^{\pi/2}\ln (\frac{1}{\alpha}+\sin^2 x) \,dx
$$
however I am not sure what that will do for us.... I also tried differentiating wrt $\alpha$ but didn't get placed. How can we prove $I_1$ result? Thanks
| Using Feynman’s Technique Integration, we first differentiate the integral w.r.t. $\alpha$ and obtain
$$
\begin{aligned}
I^{\prime}(\alpha) &=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} x}{1+\alpha \sin ^{2} x} d x \\
&=\frac{1}{\alpha} \int_{0}^{\frac{\pi}{2}} \frac{1+\alpha \sin ^{2} x-1}{1+\alpha \sin ^{2} x} d x \\
&=\frac{\pi}{2 \alpha}-\frac{1}{\alpha} \int_{0}^{\frac{\pi}{2}} \frac{d x}{1+\alpha \sin ^{2} x} \\
&=\frac{\pi}{2 \alpha}-\frac{1}{\alpha} \int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} x}{\sec ^{2} x+\alpha \tan ^{2} x} d x \\
&=\frac{\pi}{2 \alpha}-\frac{1}{\alpha} \int_{0}^{\infty} \frac{d t}{1+(1+\alpha) t^{2}}, \text { where } t=\tan x \\
&=\frac{\pi}{2 \alpha}-\frac{1}{\alpha \sqrt{1+\alpha}}\left[\tan ^{-1}(\sqrt{1+\alpha }t)\right]_{0}^{\infty} \\
&=\frac{\pi}{2 \alpha}-\frac{\pi}{2 \alpha \sqrt{1+\alpha}}
\end{aligned}
$$
Integrating give back the integal
$$
\begin{aligned}
I(\alpha) &=\frac{\pi}{2} \ln |\alpha|-\frac{\pi}{2} \int \frac{d \alpha}{\alpha \sqrt{1+\alpha}} \\
&=\frac{\pi}{2} \ln |\alpha|-\pi\int \frac{d \sqrt{1+\alpha}}{(\sqrt{1+\alpha})^{2}-1}\\
&=\frac{\pi}{2}\left[\ln |\alpha|+\ln \left(\frac{\sqrt{1+\alpha}+1}{\sqrt{1+\alpha}-1}\right) \right ]+C\\
&=\frac{\pi}{2}\left[\ln |\alpha|+\ln \left(\frac{\sqrt{1+\alpha}+1)^{2}}{\mid \alpha \mid}\right) \right]+C\\
&=\pi \ln (\sqrt{1+\alpha}+1)+C
\end{aligned}
$$As $I(0)=0$ gives the value $C=-\pi\ln2 $ and hence we can conclude that $$ I(\alpha) =\pi \ln (1+\sqrt{1+\alpha})-\pi\ln 2 =\pi \ln \left(\frac{1+\sqrt{1+\alpha}}{2}\right) .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/775206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 5,
"answer_id": 4
} |
Differential equation with initial value problem I am trying to solve the differential equation $$y'= \frac{1}{ty}$$ where $y(1)=2$.
I have no idea how to solve this, and any help would be great!
| \begin{align}
\frac{dy}{dt}&=\frac{1}{ty} \\
y \, dy &= \frac{dt}{t} & \text{separate the variables}\\
\int y \, dy &= \int \frac{dt}{t} \\
\frac{y^2}{2}&=\ln|t|+C &\text{integrate both sides} \\
y(t)&=\pm \sqrt{2\ln|t|+C} & \text{solve for $y=y(t)$}
\end{align}
Applying $y(1)=2$, we get
\begin{align}2&=\pm\sqrt{2\ln|1|+C} \\
4&=2\ln|1|+C & \text{square both sides} \\
4&=2\cdot0+C & \text{because } \ln|1|=0 \\
4&=C
\end{align}
Thus, $\boxed{y= \sqrt{2\ln|t|+4}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/776563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Inverse matrix for a matrix with sinus and cosinus functions I have this matrix A:
$$\left(\begin{array}{cc} \cos x & -\sin x \\ \sin x & \cos x \end{array}\right)$$
and I need to create an inverse matrix for this matrix A. The sinus and cosinus functions in there makes me confused, I don't know how to start and proceed.
To count a determinant from this matrix is kinda easy, but how to count an inverse matrix to this?
Thank you
| If $M$ is a matrix where
$$
M=\left(\begin{array}{cc} a & b \\ c & d \end{array}\right)
$$
then inverse of $M$ is
$$
M^{-1}=\frac{1}{ad-bc}\left(\begin{array}{cc} d & -b \\ -c & a \end{array}\right)
$$
Similarly
$$
A=\left(\begin{array}{cc} \cos x & -\sin x \\ \sin x & \cos x \end{array}\right)
$$
then inverse of $A$ is
\begin{align}
A^{-1}&=\frac{1}{(\cos x)(\cos x)-(-\sin x)(\sin x)}\left(\begin{array}{cc} \cos x & \sin x \\ -\sin x & \cos x \end{array}\right)\\
&=\frac{1}{\cos^2 x+\sin^2 x}\left(\begin{array}{cc} \cos x & \sin x \\ -\sin x & \cos x \end{array}\right)\\
&=\left(\begin{array}{cc} \cos x & \sin x \\ -\sin x & \cos x \end{array}\right)
\end{align}
where $\cos^2 x+\sin^2 x=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/777178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Working out the value of $a^4+b^4$ If $ab = 2$ and $a+b = 5$ then calculate the value of $a^4+b^4$
My approach:
$$a^4+b^4 = (a+b)^4-4a^3b-6a^2b^2-4ab^3$$
$$=(5)^4-6(ab)^2-4ab.a^2-4ab.b^2$$
$$=(5)^4-6(24)-4ab(a^2-b^2)$$
$$=(5)^4-6(24)-8(a+b)(a-b)$$
$$=(5)^4-6(24)-8(5)(a-b)$$
I am a little stuck now and any help will be appreciated.
| There is a general recursive identity for such questions: let $f(n) = a^n + b^n$. Then $$f(n+1) = f(1)f(n) - ab f(n-1).$$ Note that $f(0) = 2$ provided $ab \ne 0$. Then in your particular case, we wish to find $f(4)$, where $$f(n+1) = 5f(n) - 2f(n-1).$$ With starting values $f(0) = 2$ and $f(1) = 5$, we easily compute $f(2) = 5(5)-2(2) = 21$, $f(3) = 5(21) - 2(5) = 95$, $f(4) = 5(95) - 2(21) = 433$. The advantage of this approach is that it can be used to compute sums of higher powers quite easily. It also leads to a general solution via the solution of the associated linear recurrence; e.g., with generating functions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/778184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Cube root of two $\sqrt[3]2$ continued fraction I know there is a nice way of getting the continued fraction expansion of quadratic irrationals mainly because they recur after a point, and if they recur after a point they are quadratic irrationals. When constructing the expansion you can multiply by conjugates (kind of), e.g.
$\sqrt 3 =1+\sqrt 3 -1 = 1+\frac {1}{\frac {\sqrt 3 +1}{2}} $
Where you use $(\sqrt 3 - 1)(\sqrt 3 +1)=2$.
Are there identities that would help with the construction for $ \sqrt[3]{2} $?
One I thought was useful in the first step to get [1; 3,...] was
$ (\sqrt[3]{2}-1)( \sqrt[3]{4} + \sqrt[3]{2}+1 )=1$,
So you get:
$ \sqrt[3]{2}=1+( \sqrt[3]{2}-1 )=1+\frac {1}{ \sqrt[3]{4} + \sqrt[3]{2}+1 }= 1+\frac {1}{3+ (\sqrt[3]{4} + \sqrt[3]{2}-2)} $
Thanks for the help.
| Starting from the column vector $(1,0,0,-2)$, consider the following steps:
Step a) Repeat multiplication by the matrix $A$
$$A=\begin{bmatrix}
1&0&0&0\\
3&1&0&0\\
3&2&1&0\\
1&1&1&1
\end{bmatrix}$$
while the coefficients of the resulting vector have different signs.
Step b) Reverse the coefficients of the vector, or equivalently multiply by
$$B=\begin{bmatrix}
0&0&0&1\\
0&0&1&0\\
0&1&0&0\\
1&0&0&0
\end{bmatrix}$$
Then the number of times you multiply by $A$ in step a gives the partial quotients of continued fraction of $\sqrt[3]{2}$.
For, starting from $(1,0,0,-2)$, successive multiplication by $A$ gives:
\begin{align}
(1,0,0,-2)
&\xrightarrow A\color{red}{(1,3,3,-1)}\\
&\xrightarrow A(1,6,12,6)
\end{align}
hence in step a we multiply by $A$ one time only, because $(1,6,12,6)$ have positive coefficients only, hence the first partial quotient is $1$:
$$\sqrt[3]{2}=1+\cdots$$
Apply step b to $(1,3,3,-1)$ we get $(-1,3,3,1)$.
Then applying step a to $(-1,3,3,1)$, successive multiplication by $A$ gives:
\begin{align}
(-1,3,3,1)
&\xrightarrow A(-1,0,6,6)\\
&\xrightarrow A(-1,-3,3,11)\\
&\xrightarrow A\color{red}{(-1,-6,-6,10)}\\
&\xrightarrow A(-1,-9,-21,-3)\\
\end{align}
hence the second partial quotient is $3$:
$$\sqrt[3]{2}=1+\frac 1{3+}\cdots$$
and so on...
This algorithm holds for every algebraic number of third degree which is the only positive root of it minimal polynomial.
For higher degree the matrix $A$ is enlarged as in the Tartaglia-Pascal triangle; for example for fourth degree:
$$A=\begin{bmatrix}
1&0&0&0&0\\
4&1&0&0&0\\
6&3&1&0&0\\
4&3&2&1&0\\
1&1&1&1&1
\end{bmatrix}$$
For the intuition behind this algorithm.
Then vector $(1,0,0,-2)$ corresponds to the polynomial $p(x)=x^3-2$.
Multyplication by $A$ corresponds to $p(x)\mapsto p(x+1)$, while revesing in step b corresponds to $p(x)\mapsto x^3p(1/x)$.
Finally Descartes's signs rule provide the stopping criterion in step a.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/779509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
} |
Logarithm base transformation I am trying to solve a problem which, I think, revolves around base transformation of logarithms. It goes like this:
$\log_5\,{\log_6\,{\frac{6x-1}{x+1}}} < \log_\frac{1}{5}\,{\log_\frac{1}{6}\,{\frac{x+1}{6x-1}}}$
I tried transforming "first" logarithms to base 5, yielding
$\log_6\,{\frac{6x-1}{x+1}}< \frac{1}{\log_\frac{1}{6}\,{\frac{x+1}{6x-1}}}$ (if I am right, of course..) Further transformation to base 6 leaves me helpless with :
$\log_6\,{\frac{6x-1}{x+1}}<- \frac{1}{\log_6\,{\frac{x+1}{6x-1}}}$
Thanks.
| First of all, the existence conditions: the argument of a logarithm must always be $>0$. So we must impose
$$
\log_{6}\frac{6x-1}{x+1}>0\;\;,\;\;\log_{\frac16}\frac{x+1}{6x-1}>0
$$
But observe that $\log_{\frac16}\frac{x+1}{6x-1}=\log_{6}\frac{6x-1}{x+1}$ (from the changing base formula... see later!); hence we can work only on the first.
The same holds for this one: the arguments needs to be $>0$... but we want also this logarithm $>0$ so the conditions on the argument is simply
$$
\frac{6x-1}{x+1}>1\;,\;\;\mbox{i.e.}\;\;x>5/2
$$
Moreover $x$ must be different from $-1$ (the denominator has to be different from zero), but this is included in $x>5/2$.
Let's now go to the computations: you approached to
$$
\log_{6}\frac{6x-1}{x+1}<-\frac{1}{\log_{6}\frac{x+1}{6x-1}}
$$
using the change base formula for logarthims: $\log_ax=\frac{\log_bx}{\log_ba}$. And you're right.
Now note that
$$
-\frac{1}{\log_{6}\frac{x+1}{6x-1}}=\frac{1}{-\log_{6}\frac{x+1}{6x-1}}=\frac{1}{\log_{6}\frac{6x-1}{x+1}}$$
where the last equality follows from the basic properties of logarithms, i.e. $-\log_ax=\log_a{\frac1x}$.
So our inequality is now turned in the following one:
$$
\log_{6}\frac{6x-1}{x+1}<\frac1{\log_{6}\frac{6x-1}{x+1}}
$$
Now simply multiply every side for $\log_{6}\frac{6x-1}{x+1}$, so we have:
$$
\left(\log_{6}\frac{6x-1}{x+1}\right)^2<1
$$
i.e.
$$
-1<\log_{6}\frac{6x-1}{x+1}<1
$$
Then taking the power of $6$ of all sides we came to
$$
\frac16<\frac{6x-1}{x+1}<6
$$
that leads to $x>1/5$. But the existence conditions impose that $x>2/5$.
Hence the solution is $x>2/5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/781120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Gaussian elimination-complex numbers How a row of zero can be reached with this matrix....$\begin{pmatrix}1-i&0&-1\\1&-i&0\\1&-1&-i\end{pmatrix}$I have used conjugate of the complex number, but I can't reach the right answer....may you please show me the way?
| $$
\begin{align}
\begin{bmatrix}
1-i & 0 & -1 \\
1 & -i & 0 \\
1 & -1 & -i
\end{bmatrix}
&\sim
\begin{bmatrix}
1-i & 0 & -1 \\
1 & -i & 0 \\
0 & -1+i & -i
\end{bmatrix} && {\color{blue}{(3):=(3)-(2)}}\\
&\sim
\begin{bmatrix}
1-i & 0 & -1 \\
1-i & -1-i & 0 \\
0 & -1+i & -i
\end{bmatrix} && \color{blue}{(2):=(1-i)\times(2)}\\
&\sim
\begin{bmatrix}
1-i & 0 & -1 \\
0 & -1-i & 1 \\
0 & -1+i & -i
\end{bmatrix} && \color{blue}{(2):=(2)-(1)}\\
&\sim
\begin{bmatrix}
1-i & 0 & -1 \\
0 & -1-i & 1 \\
0 & -1-i & 1
\end{bmatrix} && \color{blue}{(3):=i\times(3)}\\
&\sim
\begin{bmatrix}
1-i & 0 & -1 \\
0 & -1-i & 1 \\
0 & 0 & 0
\end{bmatrix} && \color{blue}{(3):=(3)-(2)}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/783478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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.