Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Simple-looking bound on root of unity I am trying to prove some bound and stuck with the following:
If $|n|\leq 3N/4$, then $\left|e^{2\pi in/N}-1\right|\geq\dfrac{n}{N}$
($n,N$ are integers)
How can I prove it?
| $\sqrt{2-2\cos \frac{2\pi n}{N}}\geq\frac{n}{N}$
Let $\alpha=n/N$. We want to prove $\sqrt{2(1-\cos (2\pi \alpha))}\geq\alpha$ for $-\frac{3}{4}\leq\alpha\leq\frac{3}{4}$ This is plainly true for $\alpha<0$, so we will only conider positive $\alpha$.
Note $\sqrt{2(1-\cos (2\pi\alpha))}\geq\alpha\iff 1\geq \frac{1}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/587313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Matrix raised to 14th power Calculate $\left(\begin{matrix} 6&1&0\\0&6&1\\0&0&6\end{matrix}\right)^{14}$
Whould I do it one by one, and then find a pattern? I sense $6^{14}$ on the diagonal, and zeroes in the "lower triangle", but the "upper triangle" I'm not sure. Was thinking $14 \cdot 6^{13} $ but that's not correct... | I didn't notice that this was suggested by Harald Hanche-Olsen until just now. Consider this an expansion on his answer.
Since the identity matrix commutes with any matrix, we can use the binomial theorem with
$$
\left(6\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}+\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/588314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 2
} |
Integer $a$ , If $x ^ 2 + (a-6) x + a = 0 (a ≠ 0)$ has two integer roots If the equation $x ^ 2 + (a-6) x + a = 0 (a ≠ 0)$ has two integer roots. Then the integer value of $a$ is
$\bf{My\; Try}::$ Let $\alpha,\beta\in \mathbb{Z}$ be the roots of the equation . Then $\alpha+\beta = (6-a)$ and $\alpha\cdot \beta = a$
Now... | You are on the right track. We need
$$(a+k-8)(a-k-8) = 28$$
Now note that $(a+k-8)$ and $(a-k-8)$ are of the same parity (Why?) and the product is even. Hence, both have to be even. Hence, the possible cases are:
$1$. $(a+k-8)(a-k-8) = 2 \cdot 14$
$2$. $(a+k-8)(a-k-8) = (-2) \cdot (-14)$
$3$. $(a+k-8)(a-k-8) = 14 \cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/590463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Minimum value of $\left|z^2-z+1\right|+\left|z^2+z+1\right|$ for $z\in \mathbb{C}$ (1) If $\left|z\right| = 1$. Then find minimum value of $\left|z^2+z+4\right|$
(2) If $z\in \mathbb{C}.$ Then minimum value of $\left|z^2-z+1\right|+\left|z^2+z+1\right|.$
$\bf{My\; Try}::$ (1) Given $\left|z\right| = 1\Rightarrow z \bar... | For part 2:
Let $a = z^2 + 1$ and $b = z$. We have $a - 1 = b^2$.
We have
\begin{align*}
&(|z^2 - z + 1| + |z^2 + z + 1|)^2\\
=\,& |a - b|^2 + |a + b|^2 + 2|a - b|\cdot |a + b| \\
=\,& 2(|a|^2 + |b|^2) + 2|a^2 - b^2|\\
=\,& 2 |a|^2 + 2|a - 1| + 2|a^2 - a + 1|\\
\ge\,& 2|a|^2 + 2\mathrm{Re}(1 - a) + 2\mathrm{Re}(a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/591264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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What's the algebraic property where you can flip the fractions in an equation? Earlier in algebra, we spent over 20 minutes trying to figure out
$$ \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R_e} \,\,\,\, \text{solve for }R_2 $$
when the teacher said "What you start out with is the same as what you learned in pre-algebra... | You can't just flip all willy-nilly.
If what you said was true, then the resistance of electrical circuits in series and parallel connections would be the same! That just isn't true.
$$\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots + \frac{1}{R_n}\mathbf{\neq} \frac{1}{R_1 + R_2 + R_3 + \dots + R_n}$$
Fastest wa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/591674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 2
} |
Solve $x^2+(-7-4i)x+9+15i=0.$ Solve $$x^2+(-7-4i)x+9+15i=0.$$
Using the quadratic formula, I get $$\frac12 (7+4i \pm \sqrt{-4i})$$ but that's not correct. How do you solve this? I get no help from looking at wolfram alpha.
| Given:
$$x^2+(-7-4i)x+(9+15i) = 0$$
We use the well known formula to find
$$\begin{align}
x & = \frac{-b \pm \sqrt{b^2 - 4 \cdot a \cdot c}}{2 \cdot a} \\
& = \frac{-(-7-4i)\pm \sqrt{(-7-4i)^2 - 4 \cdot 1 \cdot (9+15i)}}{2 \cdot 1} \\
& = \frac{7+4i \pm\sqrt{(49 + 56i -16) - (36+60i)}}{2} \\
& = \frac{7+4i \pm\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/592295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Equal integral but only one of them converges absolutely . Consider the following integral
$$\int_0 ^\infty \frac{\sin x}{1+x} \, dx.$$
By integration by parts we get $$\int_0^\infty \frac{\cos x}{(1+x)^2}\,dx.$$
But according to Rudin , one of them is absolutely convergent and the other isn't. How do i prove it.
$$\i... | The integral $\int_0^\infty\left|\frac{\sin x}{1+x}\right|\,dx$ diverges. In fact, we have
\begin{eqnarray}
\int_0^\infty\left|\frac{\sin x}{1+x}\right|\,dx&=&\sum_{k=0}^\infty\int_{k\pi}^{(k+1)\pi}\left|\frac{\sin x}{1+x}\right|\,dx=\sum_{k=0}^\infty\int_0^\pi\left|\frac{\sin(x+k\pi)}{1+k\pi+x}\right|\,dx\\
&=&\sum_{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/592533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Prove that Q($\sqrt{2}$, $\sqrt{3}$) is a field Prove that $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \{a+b\sqrt{2} +c\sqrt{3} +d\sqrt{6}\ |\ a,b,c,d \in \mathbb{Q}\}$ is a field.
I am doing the subfield test, but having trouble in showing how to express the inverse in such a form. Anyone can help?
| Consider doing this step-wise. Since $\mathbb{Q}(\sqrt{2})$ is obviously a ring, to show it is a field, we just need to find an inverse for $a + b \sqrt{2}$. Let $a, b \in \mathbb{Q}$:
$$(a + b \sqrt{2})(a - b \sqrt{2}) = a^2 - 2b^2 \implies (a + b \sqrt{2})(\frac{a}{a^2 - 2b^2} - \frac{b}{a^2 - 2b^2} \sqrt{2}) = 1$$
N... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/592773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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If $3^n+81$ is a perfect square, then positive integer value $n$ is If $3^n+81$ is a perfect square, Then calculation of a positive integer value of $n$.
$\bf{My\; Try}::$ When $n≤4,$ then easy to know that $3^n+81$ is not a perfect square.
Now let $n=k+4(k∈Z^{+}),$ then $3^{n}+81=81(3^{k}+1).$
So $3^{n}+81$ is a pe... | $(x-1)$ and $(x+1)$ must be power of 3. Then $x-1=3^a$ and $x+1=3^b$
$\Rightarrow 3^b-3^a = 2 \Rightarrow 3^a(3^{b-a}-1)=2$.
So $a=0$ and $b-a=1 \Rightarrow b=1$. Thus $x-1 = 3^a = 1$ and $x+1 = 3^b = 3$.
$\Rightarrow 3^k=1\cdot 3=3 \Rightarrow k = 1 \Rightarrow n=k+1=5$.
Therefore, the unique solution is $n=5$ with $3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/593572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to find the minimum of $a+b+\sqrt{a^2+b^2}$ let $a,b>0$, and such
$$\dfrac{2}{a}+\dfrac{1}{b}=1$$
Find this minimum
$$a+b+\sqrt{a^2+b^2}$$
My try: since
$$2b+a=ab$$
so
$$a+b+\sqrt{a^2+b^2}=\sqrt{a^2+b^2+2ab}+\sqrt{a^2+b^2}=\sqrt{a^2+b^2+4b+2a}+\sqrt{a^2+b^2}$$
then I can't
maybe this problem can use AM-GM or Cauchy... | As you stated, $2b + a = ab$. We can get this even nicer: $a = \frac{2b}{b - 1}$.
Substitute into the square root expression:
$$
\begin{align*}
a + b + \sqrt{a^2 + b^2} &= \frac{2b}{b - 1} + b + \sqrt{\left( \frac{2b}{b - 1} \right)^2 + b^2} \\
&= \frac{b^2 + b}{b - 1} + \sqrt{ \frac{b^4 - 2b^3 + 5b^2}{(b - 1)^2} } \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/593703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
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finding minimum of function Can you please give me some hints finding minimum of this function:
$ (r-1)^2 + (\frac{s}{r} -1)^2 + (\frac{t}{s}-1)^2 + (\frac{4}{t}-1)^2$
where $ 1 \le r \le s \le t \le 4 $,
$r,s,t \in \Bbb R $
| Another way... As Daniel Fischer has done, define $a=r, b=\frac{s}r, c=\frac{t}s, d=\frac4t$. Then we have $a, b, c, d \ge 1$ and $abcd = 4$.
Now using the QM- AM-GM inequalities, we have
$$\sqrt{\frac{(a-1)^2 + (b-1)^2 + (c-1)^2+(d-1)^2}{4}}\ge \frac{a+b+c+d - 4}4 \ge \sqrt[4]{abcd}-1 = \sqrt{2}-1$$
with equality ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/595465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Let ${a_n}$ be the sequence given by $a_1 = 3$ and $a_{n+1} = 2a_n + 5$. Use induction to prove that $a_n > 2^n$ for all $n \in N$ Let ${a_n}$ be the sequence given by $a_1 = 3$ and $a_{n+1} = 2a_n + 5$. Use induction to prove that $a_n > 2^n$ for all $n \in \mathbb{N}$
Attempt:
For $n = 1,$ we have $3>2$ so the base ... | You could infact obtain $a_n$ with reasonable ease; Note that
$$a_{n+1} + 5 = 2(a_n+5)$$
Hence, we have
$$a_{n+1} + 5 = 2(a_n+5) = 2(2(a_{n-1}+5)) = 2^k (a_{n-k+1}+5) = 2^n(a_1 + 5)$$
Hence, we get
$$a_{n+1} +5 = 2^n(3+5) = 2^{n+3} \implies a_n = 2^{n+2} - 5$$
Now show that for $n \geq 1$, we have
$$2^{n+2} - 5 > 2^n$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/597484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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How would I show this problem through mathematical induction? I am trying to learn a lot of this on my own but I have never tried proving something through mathematical induction. Here is the problem below.
$$1+3+3^2 + \cdots + 3^n = \frac{3^{(n+1)}-1}{2}$$
for all $n\in\mathbb{N}_0$, using mathematical induction.
N... | For the inductive step, assume your claim holds for $n$ and then try to prove it for $n+1.$ That is,
Assume: $1+3+3^2+...+3^n=\frac{3^{n+1}-1}{2}$.
Then see if you can prove that $1+3+3^2+...+3^n+3^{n+1}=\frac{3^{(n+1)+1}-1}{2}$.
How to do this? Start by transforming the left hand side using what you already know:
$1+3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/598306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Stewart's Calculus (6 edn 2007) : p. 1097. §16.8, Exercise #4. Source: Stewart. Calculus: Early Transcendentals (6 edn 2007). p. 1097. §16.8, Exercise #4.
Use Stokes' Theorem to evaluate $\int \int_S curl \ \vec F \cdot d \vec S $
$4. \; F(x,y,z) = x^2 \ y^3\ z \ \vec i + \sin(x\ y\ z)\ \vec j + x\ y\ z\ \vec k$,
$S... | As Fantini points out, the boundary is a circle - we have a winner - of radius $3$.
Then, we fix $y=3$ and parameterise $(x,z) = (\sqrt{9-z^2},z)$ and then integrate from $z=3$ to $z=-3$ (going counterclockwise). One must also note that by the Implicit Function Theorem, there no unique function $x(z)$ in an open neighb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/600706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Induction Proof: Sum of Products of $3$ Consecutive Numbers I'm still trying to learn induction. But stuck on the following question. It will be highly appreciated if someone can show me how to do it.
I need to prove the following using math induction.
$$1\cdot2\cdot3 + 2\cdot3\cdot4 + 3\cdot4\cdot5 + \cdots + n(n+1)(n... | The formula is true for the case $n=1$, since $1\times 2\times 3 = 6 = \frac{1\times 2 \times 3 \times 4}{4}$.
Now, suppose that the formula is true for $n$. We have that
$1\times 2 \times 3 + \cdots + n(n+1)(n+2) + (n+1)(n+2)(n+3) \\= \frac{n(n+1)(n+2)(n+3)}{4} + (n+1)(n+2)(n+3) \\
= \frac{n(n+1)(n+2)(n+3)}{4} + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/600818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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finding the value of $f(\frac{1}{7})$ $f$ is a function mapping positive reals between $0$ and $1$ to reals. Let $f$ be given by, $f( \frac{x+y}{2} ) = (1-a)f(x)+af(y)$ where $y > x$ and $a$ being a constant. Also,$f(0) = 0$ and $f(1) = 1$. Find $f( \frac{1}{7} )$.
I have tried some things but it is not working out. Ho... | Here are the calculations:
$$f(\frac{1}{7})=f(\frac{0+\frac{2}{7}}{2})=(1-a)f(0)+af(\frac{2}{7})=af(\frac{2}{7})$$
$$f(\frac{2}{7})=f(\frac{0+\frac{4}{7}}{2})=(1-a)f(0)+af(\frac{4}{7})=af(\frac{4}{7})$$
$$f(\frac{4}{7})=f(\frac{\frac{1}{7}+1}{2})=(1-a)f(\frac{1}{7})+af(1)$$
So we have
$$f(\frac{1}{7})=a^2f(\frac{4}{7})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/601410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that if $m^2 + n^2$ is divisible by $4$, then both $m$ and $n$ are even numbers. Let $m$ and $n$ be two integers. Prove that if $m^2 + n^2$ is divisible by $4$, then both $m$ and $n$ are even numbers.
I think I have to use the contrapositive to solve this. So I assume $\neg P\implies Q$ and I have to derive $\ne... | I choose to use a and b instead of m and n because I already answered this somewhere else using a and b. Let $a=2\cdot x+1$, $b=2\cdot y+1$, where x and y are integers. This way a and b are odd numbers by definition. For example if x is 0, $a=2\cdot 0+1=1$. If x is 1, $a=2\cdot 1+1=3$. Repeating the pattern x and y can... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/604717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 3
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Finding a Correlation between Bernoulli Variables? Let X and Y be Bernoulli random variables. We don't assume independence or identical distribution, but we do assume that all 4 of the following probabilities are nonzero.
Let a := P[X = 1, Y = 1], b := P[X = 1, Y = 0], c := P[X = 0, Y = 1],
and d := P[X = 0, Y = 0].
H... | Stefan Hansen's hint is a good one. Here is the complete derivation:
$${\rm E}[X]=a+b=p$$
$${\rm E}[Y]=a+c=q$$
\begin{align}
\mathrm{Var}(X) & ={\rm E}[(X - {\rm E}[X])^2] \\
& = {\rm E}[(X - p)^2] \\
&= p(1-p)^2 + (1-p)(-p)^2 \\
& = p (1-2p+p^2) + p^2 - p^3 \\
& = p - 2p^2 + p^3+p^2-p^3 \\
& = p - p^2 \\
& = p(1-p)
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/610443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
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Proving that a number is composite I have proved that the number $10^{5}2^{17}+1$ is composite by showing that it is divisible by 3 , using remainders. I want an alternative proof.I am looking for a very elementary proof that does not mention remainders.
| $10^5 \pmod 3 = (1)^5 \pmod 3 = 1 \pmod 3$.
$2^{17} \pmod 3 = (-1)^{17} \pmod 3 = -1 \pmod 3$.
Thus, $10^5 \times 2^{17} \pmod 3 = 1 \times (-1) \pmod 3 = -1 \pmod 3$.
Now, $10^5 \times 2^{17} + 1 \pmod 3 = -1 + 1 \pmod 3 = 0 \pmod 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/610665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
$\inf$ and $\sup$ of a set given by $\sum\limits_{k=1}^{n}\frac{a_{k}}{a_{k}+a_{k+1}+a_{k+2}}$ Let $n\geq3$ be an arbitrarily fixed integer. Take all the possible finite sequences $(a_{1},...,a_{n})$ of positive numbers. Find the supremum and the infimum of the set of numbers $$\sum_{k=1}^{n}\frac{a_{k}}{a_{k}+a_{k+1}... | Your bounds are sharp (to my surprise).
Example: $a_k = x^k$. Then
\begin{align*}
\sum_{k=1}^n \frac{a_k}{a_k + a_{k+1} + a_{k+2}}
&= \sum_{k=1}^{n-2} \frac{x^k}{x^k + x^{k+1} + x^{k+2}}
+ \frac{x^{n-1}}{x^{n-1} + x^n + x}
+ \frac{x^n}{x^n + x + x^2} \\
&= \frac{n-2}{1 + x + x^2} + \frac{1}{1 + x + x^{-n+1}} + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/611874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Find $\lim_{x\to 0} \left(\frac{x}{\arcsin x}\right)^{\frac{1}{x^2}}$. Find $\lim_{x\to 0} \left(\dfrac{x}{\arcsin x}\right)^{\frac{1}{x^2}}$.
$\lim_{x\to 0}f(x)=e^a$ with $a=\lim_{X\to 0}\dfrac{1}{x^2}\left(\dfrac{x}{\arcsin x}-1\right)=\lim_{X\to 0}\dfrac{x-\arcsin x}{x^3}$
Continue i used formular L'hospotal .
| My hint:
Put $$L=\lim_{x\to 0}\left(\frac{x}{\arcsin x}\right)^{\frac{1}{x^2}}=\exp\left\{\lim_{x\to 0}\frac{1}{x^2}\ln\frac{x}{\arcsin x}\right\}=\exp\left\{-\lim_{x\to 0}\frac{1}{x^2}\ln\frac{\arcsin x}{x}\right\}$$
We use Taylor' expansion, we get:
$$\arcsin x=x+\frac{x^3}{6}+o(x^3)\to \frac{\arcsin x}{x}=1+\frac{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/614879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Real roots of the equation $1+\sum_{r=1}^{7}\frac{x^{r}}{r} = 0$ The number of real roots of the equation $\displaystyle 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}+\frac{x^6}{6}+\frac{x^7}{7} = 0$
$\bf{My\; Try}::$ Let $\displaystyle f(x) = 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}... | Law of signs is covered in pre-calc/algebra. Using the standard law of signs, one has to show that there is exactly one real root between -1 and -2. Now I am rusty on how this is done (I am 60+ and I learnt it 40 years ago!) but goes something like this.
$$
\begin{align}
f(x) &= x^7/7 + x^6/x + x^5/5 + x^4/4 + x^3/3 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/617030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Mclaurin on $\arccos(\frac{n^2-1}{n^2+1})$ I have expanded $\lim_{n\to \infty} \arccos(\frac{n^2-1}{n^2+1})$ to $\arccos(1-\frac{2}{n^2})$ and now i dont know what to do. I wrote the function on walfram alpha and he told me that the result is $\frac{2}{n}+\frac{1}{3n^3}+O((\frac{1}{n})^6)$ you can explain to me how tha... | Actually, for $x\in [0,+\infty[$,
$$\arccos \frac{1-x^2}{1+x^2}=2\arctan x$$
To see why, write $x=\tan \theta$ for $\theta \in[0,\pi/2[$, then $\theta=\arctan x$ and
$$\frac{1-x^2}{1+x^2}=\frac{1-\tan^2 \theta}{1+\tan^2 \theta}=\frac{\cos^2\theta -\sin^2\theta}{\cos^2\theta +\sin^2\theta}=\cos 2\theta$$
And since $2\th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/617264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluate the limit $\lim\limits_{n \to \infty} \frac{1}{1+n^2} +\frac{2}{2+n^2}+ \ldots +\frac{n}{n+n^2}$ Evaluate the limit
$$\lim_{n \to \infty} \dfrac{1}{1+n^2} +\dfrac{2}{2+n^2}+ \ldots+\dfrac{n}{n+n^2}$$
My approach :
If I divide numerator and denominator by $n^2$ I get :
$$\lim_{ n \to \infty} \dfrac{\frac{1}... | For each $1\leq i\leq n$, $\frac{1}{n^2+i}\leq \frac{1}{n^2}$ and $\frac{1}{n^2+i}\geq \frac{1}{n^2+n}$, and so we may bound the sum from above and below by $$\sum_{i=1}^{n}\frac{i}{n+n^{2}}\leq\sum_{i=1}^{n}\frac{i}{i+n^{2}}\leq\sum_{i=1}^{n}\frac{i}{n^{2}}.$$ Since $\sum_{i=1}^{n}i=\frac{n(n+1)}{2},$ this becomes $$\... | {
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"url": "https://math.stackexchange.com/questions/617407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 5,
"answer_id": 1
} |
How to get the simplest form of this radical expression: $3\sqrt[3]{2a} - 6\sqrt[3]{2a}$. How to get the simplest form of this radical expression:
$$3\sqrt[3]{2a} - 6\sqrt[3]{2a}$$
Here is my work:
$$3\sqrt[3]{2a} - 6\sqrt[3]{2a}$$
Since the radicands are the same, we just add the coefficients.
$$-3\sqrt[3]{2a} \sqrt... | $$3\sqrt[3]{2a}-6\sqrt[3]{2a}=\sqrt[3]{2a}(3-6)=\sqrt[3]{2a}(-3)= -3\sqrt[3]{2a}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/618297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$a,b$ are roots of $x^2-3cx-8d = 0$ and $c,d$ are roots of $x^2-3ax-8b = 0$. Then $a+b+c+d =$ (1) If $a,b$ are the roots of the equation $x^2-10cx-11d=0$ and $c,d$ are the roots of the equation
$x^2-10ax-11b=0$. Then the value of $\displaystyle \sqrt{\frac{a+b+c+d}{10}}=,$ where $a,b,c,d$ are distinct real numbers.
(2... | Thanks mathlove.
My Solution for (2) one::
Given $a,b$ are the roots of $x^2-3cx-8d=0.$ So $a+b=3c.........(1)$ and $ab=-8d........(2)$
and $c,d$ are the roots of $x^2-3ax-8b=0.$ So $c+d=3a..........(3)$ and $cd=-8b..........(4)$
So $a+b+c+d = 3(a+c)......................(5)$
Now $\displaystyle \frac{a+b}{c+d} = \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/619294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How prove this $\frac{a-b}{a+2b+c}+\frac{b-c}{b+2c+d}+\frac{c-d}{c+2d+e}+\frac{d-e}{d+2e+a}+\frac{e-a}{e+2a+b}\ge 0$ Let $a,b,c,d,e$ are postive real numbers,show that
$$\dfrac{a-b}{a+2b+c}+\dfrac{b-c}{b+2c+d}+\dfrac{c-d}{c+2d+e}+\dfrac{d-e}{d+2e+a}+\dfrac{e-a}{e+2a+b}\ge 0$$
My try: since
$$\Longleftrightarrow\sum_{sy... | You need to try a different fraction to add, so that you get better terms on applying Cauchy-Schwarz. For e.g. here we have the equivalent inequality:
$$\sum_{cyc} \left(\frac{a-b}{a+2b+c}+\frac15\right)\ge 1$$
$$\iff \sum_{cyc} \frac{6a-3b+c}{a+2b+c} \ge 5$$
By Cauchy-Schwarz, we have:
$$ \sum_{cyc} \frac{6a-3b+c... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove $ \cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) + \cos\left(\frac{6\pi}{5}\right)+ \cos\left(\frac{8\pi}{5}\right) = -1 $ $$
\cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) + \cos\left(\frac{6\pi}{5}\right)+ \cos\left(\frac{8\pi}{5}\right) = -1
$$
Wolframalpha shows that it i... | If $z^{5}-1=0$, then $$(z^{4}+z^{3}+z^{2}+z+1)(z-1)=0,$$ because $z^{5}-1=(z^{4}+z^{3}+z^{2}+z+1)(z-1)$. This implies that $$
z^{4}+z^{3}+z^{2}+z+1=0$$ or $z-1=0$. Since
\begin{equation*}
z=\cos \frac{2\pi }{5}+i\sin \frac{2\pi }{5}=e^{i2\pi /5}\ne 1
\end{equation*}
is a root of $z^5-1=0 $ and
\begin{equation*}
z^{k}=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/625101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 0
} |
Sum with Exponent I've got the following sum which I'm trying to figure out:
$$\sum_{x=0}^n 2^{-x}$$
Wolframalpha tells me that it's equal to $2 - 2^{-n}$ but I am interested in figuring out why, and how to get that result by hand.
Any help is greatly appreciated. Thanks.
Also, as you can tell, I could really use some ... | This is a geometric series with first term $1$ and common ratio $1/2$.
There is a well-known formula for calculating such sums.
Your sum, $S$ is given by:
$$S = \frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2} + \cdots + \frac{1}{2^n}$$
This is a geometric series because to get from one term to the next we multiply the the co... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Integration of $\int \frac{dx}{a+f^2(x)}$ I want to solve a integral of the form:
$$
\int \frac{dx}{a+f^2(x)}
$$
in my particular case I got
$$
\int \frac{dx}{5+\cos^2(x)}
$$
in my case I followed this process:
$$
\int \frac{dx}{5+\cos^2(x)} \\
let \ t = tg(\frac{x}{2}) =>
dx = \frac{2dt}{1+t^2}\\
\int \frac{dx}{5+\... | I you want a solution of $\displaystyle \int\frac{1}{5+\cos^2 x}dx$
$\bf{Solution::}$ Divide both $\bf{N_{r}}$ abd $\bf{D_{r}}$ by $\cos^2 x$, we get $\displaystyle \int\frac{\sec^2 x}{5\sec^2 x+1}dx = \int\frac{\sec^2 x}{6+5\tan^2 x}dx$
Now Let $\sqrt{5}\tan x= t$, Then $\displaystyle \sec^2 xdx = \frac{1}{\sqrt{5}}dt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/625913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How find this sum $I_n=\sum_{k=0}^{n}\frac{H_{k+1}H_{n-k+1}}{k+2}$ $$I_n=\sum_{k=0}^{n}\dfrac{H_{k+1}H_{n-k+1}}{k+2}$$
where $$H_{n}=1+\dfrac{1}{2}+\cdots+\dfrac{1}{n}$$
my try:since
$$I_n=\dfrac{1+\dfrac{1}{2}+\cdots+\dfrac{1}{n+1}}{2}+\dfrac{\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}+\cdots+\dfrac{1}{n}\right)}{... | Continuing using both Greg Martin and Jack D'Aurizio's answers, I start with:
$$
\begin{align}
\sum_{k=1}^{m+1}\frac{(H_{k-1})^2}{k} &= (H_m)^2 H_{m+1} - \sum_{k=1}^m{ \frac{(H_k+H_{k-1} )H_k}{k} } \\
&= (H_m)^2 H_{m+1} - \sum_{k=1}^m{ \frac{ \left( (H_{k-1}+\frac{1}{k}) + H_{k-1} \right)(H_{k-1}+\frac{1}{k})}{k} } \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/627936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 3,
"answer_id": 2
} |
How to construct change of basis matrix How do I construct a change of basis matrix?
For example in $\mathbb R^3$, how to construct matrix changing basis from $A$ to $B$?
$A=\begin{pmatrix} 1 \\ 0 \\5 \end{pmatrix}\begin{pmatrix} 4 \\ 5 \\5 \end{pmatrix}\begin{pmatrix} 1 \\ 1 \\4 \end{pmatrix}$
$B=\begin{pmatrix} 1 ... | Let $M_{u,v}$ be the base change matrix from base $u$ to base $v$. Then the following identity holds for bases $u$,$v$, and $w$: $$ M_{u,v}=M_{w,v}M_{u,w}$$ In this case $A$ can be the role of $u$ and $B$ is $v$. Use the standard basis $e$ as the intermediate basis $w$. Then the expression is: $$M_{A,B}=M_{e,B}M_{A,e} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/628061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
Prove the series identity
Prove an identity:
$$\sum_{n=2}^{ \infty } \frac{2}{(n^3-n)3^n}=- \frac{1}{2}+ \frac{4}{3} \cdot \sum_{n=1}^{ \infty } \frac{1}{n \cdot 3^n}$$
I've checked that the left-hand-side of this identity is convergent absolutely, hence I can write it as:
$$\sum_{n=2}^{ \infty } \frac{2}{(n-1)(n+... | Using Partial Fraction Decomposition,
$$\frac1{n^3-n}=\frac An+\frac B{n-1}+\frac C{n+1}$$
multiply either sides $n^3-n=n(n-1)(n+1)$ and compare the coefficients of the different powers of $n$ to find $A,B,C$
Alternatively,
$$\frac2{n(n-1)(n+1)}=\frac{n+1-(n-1)}{n(n-1)(n+1)}=\frac1{n(n-1)}-\frac1{(n+1)n}$$
Again, $\di... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/628123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Joint To Marginal Density : Can't figure it out. Here goes the problem:
Problem:
Suppose $X$ and $Y$ have the joint density function:
$f(X,Y) = c \sqrt{1 - x^2 - y^2}, \,\,\,\,\, x^2 + y^2 \leq 1$
*
*Find $c$.
*Find the marginal densities of $X$ and $Y$
My Approach For Solution:
Part 1:
To obtain $c$, we need to... | defint $y=\sqrt{1-x^2}\sin(t)$, then $dy=\sqrt{1-x^2}\cos(t)dt$.
$$
f_X(x)=\int_{-\sqrt{1-x^2}}^\sqrt{1-x^2}c\sqrt{1-x^2-y^2}dy = c(1-x^2)\int_{-\pi/2}^{\pi/2}\cos^2(t)dt=c\frac{\pi}{2} (1-x^2)=\frac{3}{4}(1-x^2)
$$
p.s. your integral limit for $y$ is wrong since $y$ must be smaller than $\sqrt{1-x^2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/628763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Trigonometry or inequality problem Today, I saw this question:
If $x,y,z \in [0,\frac\pi 2]$, $x+y+z=\frac{3\pi}{4}$ and $\sec^2(x)\sec^2(y)\sec^2(z)=8$, calculate $E=\tan x\tan y+\tan y\tan z+\tan z\tan x$
My first thought was getting everything in term of $\tan$, and everything went as expected. Let $a=\tan x, b=\t... | Note that each integer has a unique prime factorisation..
$8=2*2*2$
$$(1+a^2)(1+b^2)(1+c^2)=8=2*2*2=(1+1)*(1+1)*(1+1) \ \ \ \ \ \ (*)$$
Since all terms on the the $R.H.S$ are prime numbers ...
By Euclid's Lemma
$2$ must divide either $(1+a^2)$ or $(1+b^2)$ or $(1+c^2)$ or all of them
Let us assume $2$ divides $(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/629751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
determinant of the linear transformation $T(X) =\frac{1}{2} (AX+XA)$ Let $V$ vector space of all matrices $3\times3$, and let $A$ be the diagonal matrix :
$$
\begin{pmatrix}
1 & 0 & 0\\
0 & 2& 0 \\
0 & 0& 1\end{pmatrix}
$$
Compute thee determinant of the linear transformation $T(X) =\frac{1}{2} (AX+XA)$.
Any hi... | Note that your transformation is realized by a $9\times 9 $ matrix, as $X$ has $9$ elements.
So
$$
\left(
\begin{matrix}
x_{11} \\
x_{12} \\
x_{13} \\
x_{21} \\
x_{22} \\
x_{23} \\
x_{31} \\
x_{32} \\
x_{33}
\end{matrix}
\right) \quad\longmapsto\quad
\left(
\begin{matrix}
x_{11} \\
\tfrac{3}{2}x_{12} \\
x_{13} \\
\tf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/631682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Prove that $ \frac12 < 4\sin^2\left(\frac{\pi}{14}\right) + \frac{1}{4\cos^2\left(\frac{\pi}{7}\right)} < 2 - \sqrt{2} $ I can't figure out how to prove the following inequality:
$$
1/2 < 4\sin^2\left(\frac{\pi}{14}\right) + \frac{1}{4\cos^2\left(\frac{\pi}{7}\right)} < 2 - \sqrt{2}
$$
Thanks
| As suggested by Lucian, use Taylor series for each piece and expand around zero. Then
$$
4\sin^2\left(x\right) + \frac{1}{4\cos^2\left(2x\right)}
$$
gives 1/4 + 5 x^2 + (4 x^4)/3 + (56 x^6)/9 + (3964 x^8)/315 + ...
Replacing x by Pi/7 and taking into account that Pi^2 is almost 10, you notice that terms above x^4 ar... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Help with limit calculation Can anyone help me with this limit please:
I have been trying to solve this for 2 hours with no success:
$$\lim_{n\to \infty } \frac {1^3+4^3+7^3+...+(3n-2)^3}{[1+4+7+...+(3n-2)]^2}$$
| You can simplify the fraction using
$$\sum_{k=1}^{n}k=\frac{n(n+1)}{2},\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6},\sum_{k=1}^{n}k^3=\left(\frac{n(n+1)}{2}\right)^2.$$
The numerator is
$$\sum_{k=1}^{n}(3k-2)^3=\cdots=\frac 14 (27 n^4-18 n^3-9 n^2+4n).$$
The dinominator is
$$\left(\sum_{k=1}^{n}(3k-2)\right)^2=\cdots=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/633770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Generating function and combinatorics I am studying now the concept of generating function, and I have a solved question in my book which I don't understand, completely. There it is:
What is the number of options to roll 10 different dice, so the sum
of the results will be 25?
Now for the solution:
f(x) = (x+x^2+x^3+x... | Since you have 10 dice, and each die can be any integer from 1 to 6, your generating function is
$f(x)=(x+x^2+x^3+x^4+x^5+x^6)^{10}=(x(1+x+x^2+x^3+x^4+x^5))^{10}=x^{10}(1+x+x^2+x^3+x^4+x^5)^{10}$
$=\displaystyle x^{10}\big(\frac{1-x^6}{1-x}\big)^{10}=x^{10}(1-x^6)^{10}(1-x)^{-10}.$
Now we have to find the coefficient ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/635041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Find a limit in an efficent way I'm trying to calculate the following limit:
$$\mathop {\lim }\limits_{x \to {0^ + }} {\left( {\frac{{\sin x}}{x}} \right)^{\frac{1}{x}}}$$
What I did is writing it as:
$${e^{\frac{1}{x}\ln \left( {\frac{{\sin x}}{x}} \right)}}$$
Therefore, we need to calculate:
$$\mathop {\lim }\l... | Elementary proof using well known limits and inequalities:
$$\frac{{\ln \left( {\frac{{\sin x}}{x}} \right)}}{x} = \frac{{\ln \left( \left( \frac{{\sin x}}{x}-1 \right) + 1 \right)}}{\frac{{\sin x}}{x}-1 } \cdot \frac{\frac{{\sin x}}{x}-1 }{x} =\frac{{\ln \left( \left( \frac{{\sin x}}{x}-1 \right) + 1 \right)}}{\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/635195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
What is the limit of $\mathrm{e}^{At}$ when $t\to \infty$, for $A$ a square matrix? I was doing a matrix calculation and need to find
$$\lim_{t\to \infty} \mathrm{e}^{At}=?$$
What is the limit of $\mathrm{e}^{At}$ when $t\to \infty$, for $A$ a matrix?
| Recall the definition of exponential function of one variable in terms of power series:
$$
e^x := \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots
$$
Similarly, we can define the matrix exponent of $n\times n$ matrix $A$:
$$
e^A : = I + A + \frac{1}{2!}A^2 + \frac{1}{3!}A^3 + \dot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/641207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding $ \lim_{x \to 0^+} (\frac{\arctan (x)}{x})^{\frac{1}{x^2}} $ without power series. I have to find: $\lim_{x \to 0^+} (\frac{\arctan (x)}{x})^{\frac{1}{x^2}}$
"Minimized" the problem to finding: $ \lim_{x \to 0^+} \frac {\ln(\frac{\arctan x}{x})} {x^2}
$
L'Hôpital's rule once is not enough, second L'Hôpital's ru... | \begin{align*}
\lim_{x \to 0^+} (\frac{\arctan (x)}{x})^{\frac{1}{x^2}}&=\lim_{x \to 0^+} (\frac{x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} +... }{x})^{\frac{1}{x^2}}\\
&=\lim_{x \to 0^+}(1-\frac {x^2}3+ \frac{x^4}{5} - \frac{x^6}{7} +...)^{1/x^2}\\
&=e^{-1/3}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/646297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Prove that the series is bounded Good afternoon everybody! I have the following problem:
" Prove that the series $s_n=1+\frac{1}{2}+...+\frac{1}{n}-\ln n$ is decreasing and bounded, therefore convergent".
In order to prove it is decreasing, I must prove that $s_n-s_{n+1} >0$. But $s_n- s_{n+1}=[1+\frac{1}{2}+...+\frac... | Note that ${1 \over n+1} \le { 1 \over t } \le {1 \over n}$ for $t \in [n,n+1]$, and $\ln({n+1 \over n}) = \int_n^{n+1} {dt \over t}$, hence ${1 \over n+1} \le \ln({n+1 \over n}) \le {1 \over n}$.
This gives $s_n -s_{n+1} \ge 0$ as you showed above.
Using the other side of the bound and the fact that $\ln$ is non-decre... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/646383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find all roots of $z^2=3-4i$. Find all roots of $z^2=3-4i$.
$z^2=3-4i$
$z^2+4i-3=0$
But how do I go on from here?
| Solve: $z^2 = 3 - 4i$
Let
$$z= x + yi$$
Rewrite as
$$(x + yi)^2 = 3 - 4i$$
Expand
$$x^2 +2xyi + y^2i^2 = 3 - 4i$$
Simplify
$$x^2 - y^2 + 2xyi = 3 - 4i$$
real/imaginary parts
$$x^2 - y^2 = 3$$
and
$$2xy = -4$$
$$xy = -2$$
Therefore $y = -\frac{2}{x}$.
Substitute $y = -\frac 2x$ into $x^2 - y^2 = 3$, giving
$$x^2 - \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/646547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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Find the all the pairs $(n,m) \in \mathbb{N} \times \mathbb{N}$ with the property that $ 2^n+3^m $ is divisible by $23$. Find the all the pairs $(n,m) \in \mathbb{N} \times \mathbb{N}$ with the property that $ 2^n+3^m $ is divisible by $23$.
I'm not really sure how to start this one, but since I found it in a book on g... | As $\displaystyle 2^3\cdot3=24\equiv1\pmod{23},$
$\displaystyle 2^n+3^m\equiv0\pmod{23}\iff 2^{n+3m}\equiv-3^m\cdot2^{3m}\equiv-(3\cdot2^3)^m$
$\displaystyle\implies2^{n+3m}\equiv-1\pmod{23}$
Now as $23$ is prime, $\displaystyle a^2\equiv1,a\equiv\pm1\pmod{23}$ for $(a,23)=1$
and $\displaystyle2^5=32\equiv9,2^{10}\equi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/649841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
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Prove : $\left | a\sqrt{2}+b\sqrt{3} \right |> \frac{1}{350}$ Problem :
Let $a,b\in \mathbb{Z}$ such that $a\neq 0,b\neq 0$ ; $\left | a \right |\leq 100,\left | b \right |\leq 100$.
Prove that:
$$\left | a\sqrt{2}+b\sqrt{3} \right |> \frac{1}{350}$$
Thanks :)
P/s : I have no ideas about this problem ! :(
| Let $\lambda = a\sqrt{2} + b\sqrt{3}$ where $a$, $b$ not both zero and $|a|,|b| \le 100$. We have
$$\begin{array}{rrcl}
& ( \lambda - a\sqrt{2})^2 - 3b^2 &=& 0\\
\implies & \lambda^2 + 2a^2 - 3b^2 &=& \sqrt{8} a \lambda\\
\implies & \lambda^4 - (4a^2+6b^2)\lambda^2 + (2a^2-3b^2)^2 &=& 0\\
\iff & \lambda^2 (4a^2 + 6b^2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/650915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
The factors of $5^n-3^n-2^n$ I have been assigned the following question.
Let $f(n):= 5^n-3^n-2^n$. Prove that
(a) $p$ divides $f(p)$ for each prime $p$;
(b) $p^{k+1}$ divides $f(n)$ for $n=p^k$, with $p=2,3,5$ and $k\geq 0$;
(c) $p^{k+2}$ divides $f(n)$ for $n=p^k$, with $p=19$ and $k\geq 0$.
Progress
I have been ab... | The cases $k=0$ are trivial as $f(p^k)=f(1)=0$.
Lemma. $p^{q^k}\equiv p\mod q$.
By simple induction on $k$, as for $k=0$ is trivially true, suppose $p^{q^k}\equiv p \mod q$. We have
$$
p^{q^{k+1}}=p^{q·q^k}=(p^{q^k})^q
$$
and, by induction, we can complete as
$$
p^{q^{k+1}}=p^{q·q^k}=(p^{q^k})^q \equiv p^q \equiv p
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/652205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Given $x+y$ and $x\cdot y$, what is $x^3+ y^3$ ? I have been looking at an assortment of high school number sense tests and I noticed a reoccurring problem that states what x+y is and what $x\cdot y$ is then asks for $x^3+ y^3$. I want to know how to work these problems. I have a couple of examples.
$x+y=5$ and $x\cdot... | If the problem cannot be reduced to the symmetric case then Groebner basis can be used to solve the problem. But the calculation by hand is very cumbersome so I use Maxima and got the following output. poly_reduced_grobnercalculates a Groebner base and poly_pseudo_dividedoes the division. The first element of the resul... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/652252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 5
} |
Double Integral Confusion A buddy was asking me for help with one of his MV Calc problems, and I ended up getting the same answer as him so I figured I'd ask it here...
Question Find $$\iint_{R} (x-1) \, dA$$ where $R$ is the region enclosed by $y=x$ and $y=x^3$ in the first quadrant. So naturally I told him to comp... | As mentioned in the comments, your answer is correct. I'll provide a solution here for future readers.
First, draw a picture. I'm going to use Wolfram Alpha to show one online--it should be simple enough to graph by hand. We can see the intersections are at $(0, 0)$ and a $(1, 1)$. Thus, we have our limits of integ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/652798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Parabola $\sqrt {x}+\sqrt {y}=1 $ How do I prove that the equation $\sqrt {x}+\sqrt {y}=1 $ is part of parabola.
My attempt:rotation in 45 degrees brings the equation to $ -2a^2=1-2\sqrt {2}b $ when $ x= \frac {a-b} {\sqrt {2} } $ and $ y= \frac {a+b} {\sqrt {2} } $. It is a parabola, why is it only part of it? (also ... | Squaring, we get $x + y + 2\sqrt{xy} = 1$; moving the $2\sqrt{xy}$ term to one side, and squaring again, we obtain $$
x^2 -2xy + y^2 -x -y + 1 = 0$$
The value of the determinant $$\begin{vmatrix}A & B \\ B & C\\\end{vmatrix}
$$ will tell us which type of curve is represented by a quadratic relation $$Ax^2 + 2Bxy + Cy^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/655104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How to solver this equation $\sum_{i=1}^{6}x_{i}x_j=-3,j=1,2,3,4,5,6,j\neq i$ Let $x_{i}\in R,i=1,2,3,4,5,6$ such that
$$\begin{cases}
x_{1}x_{2}+x_{1}x_{3}+x_{1}x_{4}+x_{1}x_{5}+x_{1}x_{6}=-3\\
x_{2}x_{1}+x_{2}x_{3}+x_{2}x_{4}+x_{2}x_{5}+x_{2}x_{6}=-3\\
x_{3}x_{1}+x_{3}x_{2}+x_{3}x_{4}+x_{3}x_{5}+x_{3}x_{6}=-3\\
x_{... | Following on David Peterson's hint: This shows there are at most two different values among the $x_i$ (solve the quadratic). Call them $a,b$. We go through the cases:
1)all the same. Each equation becomes $5a^2=-3$, which gives $x_i=\pm\sqrt{\frac 35}i$, which is acceptable over $\Bbb C$ but not $\Bbb R$
2)five... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/655402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solution for $4^{2x+1}-3^{3x+1}=4^{2x+3}-3^{3x+2}$ I'm trying to get a solution for:
$4^{2x+1}-3^{3x+1}=4^{2x+3}-3^{3x+2}$
My main problem is that I don't know how to combine this potencys!
Ive also thought about another function that would bring me same difficulties:
$6^x=36*9.75^{x-2}$
What am I supposed to do?
| For the first one, we have
$$
4^{2x+1}-3^{3x+1}=4^{2x+3}-3^{3x+2} \implies\\
3^{3x+2}-3^{3x+1} = 4^{2x+3}-4^{2x+1} \implies\\
(3-1)3^{3x+1} = (4^2 - 1)4^{2x+1} \implies\\
2\cdot 3^{3x+1} = 5\cdot 3\cdot4^{2x+1} \implies\\
3^{3x} = 5 \cdot 2^{4x+1}
$$
I think that's the simplest we can get it. From there, I suppose we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/656560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$n$th derivative of $e^x \sin x$ Can someone check this for me, please? The exercise is just to find a expression to the nth derivative of $f(x) = e^x \cdot \sin x$. I have done the following:
Write $\sin x = \dfrac{e^{ix} - e^{-ix}}{2i}$, then we have $f(x) = \dfrac{1}{2i} \cdot (e^{(1+i)x} - e^{(1-i)x})$.
Taking the... | Yes, your work is correct. But there is an easier way. Put $g(x)= e^{(1+i)x}$ and note that $f(x)$ is the imaginary part of $g(x)$ for real $x$. Now $g^{(n)}(x) =(1+i)^n e^{(1+i)x}$. Since $1+i=2^{1/2} e^{\pi i/4}$, we have $g^{(n)}(x) =2^{n/2}e^{n\pi i/4}e^{(1+i)x}=2^{n/2}e^{x + i(x +n\pi /4)}$. Thus $f^{(n)}(x)$ is t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/659104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
find the limits find the limit
$$\mathop {\lim }\limits_{n \to \infty } \frac{{\sum\limits_{i = 0}^{n + 1} {{{\left( {\begin{array}{*{20}{c}}
{n + 1}\\
i
\end{array}} \right)}^3}} }}{{\sum\limits_{i = 0}^n {{{\left( {\begin{array}{*{20}{c}}
n\\
i
\end{array}} \right)}^3}} }}$$
I use Maple to get the
$$\sum\limits_{i =... | Let $\;\displaystyle\text{Fr}_n = \sum\limits_{k=0}^n \binom{n}{k}^3\;$ be the $n^{th}$
Franel number (OEIS
A000172 ).
Method 1 (Asymptotic expansion)
By my answer to a related question, $\text{Fr}_n$ has following asymptotic expansion:
$$\text{Fr}_n = 8^n \frac{2}{\pi\sqrt{3}n}\left[
1
- \frac{1}{3n}
+ \frac{1}{27... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/660502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Identity for this recursive relation with multiple terms I have a recursive relation algorithm which is defined as follows:
$$F_n = 3(F_{n-1} - F_{n-2}) + F_{n-3}$$
$$F_0 = 0$$
$$F_1 = 1$$
$$F_2 = 4$$
From calculating the first few values, I know this is equivalent to $F_n = n^2$, however I don't quite know how to deri... | Proof
We want to show that $F_n = n^2$ for all $n \geq 0$. This is clearly true for $n=0,1,2$ since $F_0=0, F_1=1, F_2=4$. For $F_3$, we get that $F_3 = 3(F_2-F_1)+F_0 = 3(4-1)+0 = 9 = 3^2$. So it is true for $n=3$.
Assume for some $k\geq 3$ that $F_i = i^2$ for all $i \leq k$. We want to show that $F_{k+1} = (k+1)^2$.... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Finding a real number c for polynomial (proof) The question is to find a real number c for which $ x\ge c%+$ implies $$x^4-4x^3+7x-9 \ge1000$$.
I was given the hint that $x>10$, then $4x^3<0.4x^4$, so $x^4-4x^3>0.6x^4$.
Problem is, I'm not understanding this line of reasoning, particularly how x>10 goes to the next ste... | If $x > 10$ then $0.4x > 4$ and so $4x^3 < (0.4x)x^3 = 0.4x^4$. Therefore $x^4 - 4x^3 > x^4 - 0.4x^4 = 0.6x^4$.
Furthermore, if $x > 10$, then $7x - 9 \geq 7(10) -9 = 61$. Therefore, if $x > 10$, $x^4 - 4x^3 + 7x - 9 > 0.6x^4 + 61$. So how big should $x$ be so that $0.6x^4 \geq 1000?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/664068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Area of a triangle using vectors I have to find the area of a triangle whose vertices have coordinates
O$(0,0,0)$, A$(1,-5,-7)$ and B$(10,10,5)$
I thought that perhaps I should use the dot product to find the angle between the lines $\vec{OA}$ and $\vec{OB}$ and use this angle in the formula:
area $= \frac{1}{2}ab\sin... | Since your vectors are in $\mathbb{R}^3$, you can find the area of the parallelogram generated by the vectors by computing the magnitude of the cross product. The area of the triangle is half that value:
$Area=(1/2) | a \times b |$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/667560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
How do I prove $\lim_{n\to\infty} \frac{n^z n!}{z(z+1)\cdots (z+n)}=\frac{1}{z}\prod_{n=1}^\infty \frac{(1+\frac{1}{n})^z}{1+\frac{z}{n}}$? How do I prove $$\lim_{n\to\infty} \frac{n^z n!}{z(z+1)\cdots (z+n)}=\frac{1}{z}\prod_{n=1}^\infty \frac{(1+\frac{1}{n})^z}{1+\frac{z}{n}}$$?
This could be proven if $$\lim_{n\to\i... | Hint:
\begin{align}
\lim_{n\to\infty} \frac{n^z n!}{z(z+1)\cdots (z+n)} &= \frac{1}{z} \lim_{n \to \infty} \frac{ \left( \frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdots \frac{n}{n-1} \right)^z n!}{(1+\frac{z}{1})(1 + \frac{z}{2}) (1 + \frac{z}{3}) \cdots (1 + \frac{z}{n}) \cdot n! } \\
&= \frac{1}{z} \lim_{n \to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/668180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find a formula for $\sum\limits_{k=1}^n \lfloor \sqrt{k} \rfloor$ I need to find a clear formula (without summation) for the following sum:
$$\sum\limits_{k=1}^n \lfloor \sqrt{k} \rfloor$$
Well, the first few elements look like this:
$1,1,1,2,2,2,2,2,3,3,3,...$
In general, we have $(2^2-1^2)$ $1$'s, $(3^2-2^2)$ $2$'s e... | Hint in the form of an example: If you start the sum at $0$, you can write
$$\begin{align}
\lfloor\sqrt0 \rfloor+\lfloor\sqrt1 \rfloor+\lfloor\sqrt2 \rfloor+\cdots+\lfloor\sqrt{11} \rfloor&=0+1+1+1+2+2+2+2+2+3+3+3\\
&=3+3+3+3+3+3+3+3+3+3+3+3\\
&\quad-1-1-1-1-1-1-1-1-1\\
&\quad-1-1-1-1\\
&\quad-1\\
&=12\cdot3-(1+4+9)\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/669460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 6,
"answer_id": 3
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Integration by parts. How can I integrate $ \int_{3}^8 \ln \sqrt{x+1}\ dx$ by parts ?
Is this step right ?
$ \int_{3}^8 \frac{1}{2}\ln(x+1)\ dx $ = $ \frac{1}{2} \int_{3}^8\ln(x+1)\ dx$
$f^{'}(x) = 1 , f(x) = x , g(x)= ln(x+1) $
$ \left[ x\ln(x+1) \right]_{x=3}^{x=8} - \int_{3}^8 x\frac{1}{x+1}\ dx $
| Everything you wrote is right.
Now take the derivative of $g(x)=\ln (x+1)$. So $g'(x)=\frac{1}{x+1}$. Then by the integration by parts formula you have
$$\frac{1}{2}\int_3^8 \ln(x+1)dx= \frac{1}{2}\left( x\ln(x+1) \bigg|_{3}^8 - \int_3^8 \frac{x}{x+1}dx\right).$$
Altogether
$$\frac{1}{2}\int_3^8 \ln(x+1)dx= 4\ln (9)-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/671386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find the least nonnegative residue of a number Find the least nonnegative residue of $4^{47} \bmod 12$. I began by having $4^2$ congruent to $4 \bmod 12$ but I am not sure where to go from there.
| Since $4^2\equiv 4$ we\ have
$$
4^4\equiv 4 \mod 12 \\
4^8 \equiv 4 \mod 12 \\
4^{16}\equiv 4 \mod 12 \\
4^{32}\equiv 4 \mod 12 \\
4^{40}= 4^{32}\cdot 4^8 \equiv 4^2 \equiv 4 \mod 12 \\
4^{44}= 4^{40}\cdot 4^4 \equiv 4^2 \equiv 4 \mod 12 \\
4^{46}= 4^{44}\cdot 4^2 \equiv 4^2 \equiv 4 \mod 12 \\
4^{47}= 4^{46}\cdot 4^1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/671875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the surface area obtained by rotating $y=1+3x^2$ from $x=0$ to $x=2$ about the y-axis. Find the surface area obtained by rotating $y= 1+3 x^2$ from $x=0$ to $x = 2$ about the $y$-axis.
Having trouble evaluating the integral:
Solved for $x$:
*
*$x=0, y=1$
*$x=2, y=13$
$$\int_1^{13} 2\pi\sqrt\frac{y-1}3 \cdot... | This can be done using Pappus's Theorem and integrating in $x$:
$$
\begin{align}
\int_0^22\pi x\,\overbrace{\sqrt{y'^2+1}\,\mathrm{d}x}^{\mathrm{d}s}
&=\int_0^22\pi x\sqrt{36x^2+1}\,\mathrm{d}x\\
&=\frac\pi{36}\int_0^2\sqrt{36x^2+1}\,\mathrm{d}(36x^2+1)\\
&=\frac\pi{36}\int_1^{145}\sqrt{u}\,\mathrm{d}u\\
&=\frac\pi{36}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/672744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Answer Says $\lim_{(x,y)\rightarrow(0,0)}\frac{x^3y^2}{x^4+y^6} = 0$. I say DNE. What did I do wrong? I was asked to find
$$\lim_{(x,y)\rightarrow(0,0)}\frac{x^3y^2}{x^4+y^6}$$
Observe that setting y=mx results in $$\lim_{(x,mx)\rightarrow(0,0)}\frac{x^3(mx)^2}{x^4+(mx)^6} = 0$$
The textbook solution then proved that... | Another approach:
$$\frac{x^3y^2}{x^4+y^6}\stackrel{\text{Polar Coord.}}=\frac{r^5\cos^3\theta\sin^2\theta}{r^4(\cos^4\theta+r^2\sin^6\theta)}=r\frac{\cos^3\theta\sin^2\theta}{\cos^4\theta+r^2\sin^6\theta}\xrightarrow[r\to0]{}0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/675196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
If $a+b+c = 7\;\;,a^2+b^2+c^2 = 23$, then $a^3+b^3+c^3=$ If $a,b,c\in \mathbb{R}$ and $a+b+c = 7\;\;,a^2+b^2+c^2 = 23$ and $\displaystyle \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1} = 31$. Then $a^3+b^3+c^3 = $
$\bf{My\; Trial\; Solution::}$ Given $a^2+b^2+c^2 = 23$ and
$a+b+c = 7\Rightarrow (a+b+c)^2 = 49\Rightarrow (a... | HINT:
Where you have left of,
we can derive $$a^3+b^3+c^3-3abc=(a+b+c)[a^2+b^2+c^2-(ab+b+ca)]$$
Now, $(a+b+c)^2-(a^2+b^2+c^2)=2(ab+bc+ca)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/677184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
How can this trig equation be simplified? We have $9+40\sin^2x=-42\sin x\cos x$.
I know this simplifies to $7\sin x+3\cos x=0$, but how?
| Along with Trafalgar Law's hint, we can use the fact that $\sin^2(x) + \cos^2(x) = 1$. Multiply both sides by $9$ to get $9 = 9\sin^2(x) + 9\cos^2(x)$. We then have
$$\begin{aligned}
9\sin^2(x) + 9\cos^2(x) + 40\sin^2(x) &= -42\sin(x)\cos(x)\\
49\sin^2(x) + 42\sin(x)\cos(x) + 9\cos^2(x) &= 0\\
(7\sin(x) + 3\cos(x))^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/677752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Calculate $\iint_D 3dxdy$ where $D = \{(x, y) : (x+y)^2 + (2x - y)^2 \le 4 \}$ Calculate $\displaystyle\iint_D 3dxdy$ where $D = \{(x, y) : (x+y)^2 + (2x - y)^2 \le 4 \}$.
I tried to solve this can I failed.
Can you please give me some hints?
| Assume that $D:5x^2+2y^2-2xy=4$. Since $4-10<0$, $D$ is an ellipse and it is sufficient that we calculate area of it. Now, consider $A=\left[ \begin{array}{cc} 5 & -1 \\ -1 & 2 \end{array} \right]$. Let us call $\lambda_1$ and $\lambda_2$ the eigenvalues of $A$. One can see, $D$ in new coordinate is $\lambda_1X^2+\lam... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/679204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proof of one inequality $a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}$ How to prove this for positive real numbers?
$$a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}$$
I tried AM-GM, CS inequality but all failed.
| I have come up with an answer with myself. Using CS inequality
$$(a^4+b^4+c^4)(1+1+1)\geq(a^2+b^2+c^2)^2$$
$$(a^2+b^2+c^2)(1+1+1)\geq(a+b+c)^2$$
Hence we have
$$a^4+b^4+c^4\geq\frac{(a+b+c)^4}{27}=(a+b+c)\left(\frac{a+b+c}{3}\right)^3\geq abc(a+b+c)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/679544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
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Find the minimum of : $P=(1+\frac{1}{a^3})(1+\frac{1}{b^3})(1+\frac{1}{c^3})$ $a;b;c\in \mathbb{R}^+$ such that $a+b+c=6$.
Find the minimum of : $P=(1+\frac{1}{a^3})(1+\frac{1}{b^3})(1+\frac{1}{c^3})$
Thanks :)
I have no ideas about this problem ! :(
| Use Holder inequality and we have:
$$\left(1 + \frac 1{a^3}\right)\left(1 + \frac 1{b^3}\right)\left(1 + \frac 1{c^3}\right) \ge \left(1 + \frac 1{abc}\right)^3$$
From $AM-GM$ we have:
$$\frac{a+b+c}{3} \ge \sqrt[3]{abc}$$
$$\frac 63 \ge \sqrt[3]{abc}$$
$$2^3 \ge {abc}$$
$$8 \ge abc$$
So using this we minimize the righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/683397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Infinite Sum with Combination I am trying to figure out what the following sum converges to:
$$\sum_{n=0}^\infty {6+n\choose n}x^n(6+n),\qquad\qquad0<x<1$$
An answer would be great, but if you have an explanation, that'd be better!
| Related techniques: (I), (II). Follow the steps:
1) simplify $(n+6){ n+6\choose n} $ as
$$ (n+6){ n+6\choose n} = \frac{1}{6!}(n+1)(n+2)(n+3)(n+4)(n+5)(n+6)^2. $$
2) use the series identity
$$ \sum_{n=0}^{\infty} x^{n+1}=\frac{x}{1-x} \longrightarrow (*) $$
3) Applying the operators $D(xD)(x^2D)^5 $ to both sides of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/685634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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An equation, where the solution does not exist, but on solving the equation we got a solution. why this is happening? The solution of the equation
$\sqrt{(x+1)} -\sqrt{(x-1)}= \sqrt{(4x-1)}$
is $\frac{5}{4}$,but when we put $x=\frac{5}{4}$ in the given equation, then it does not satisfy the equation.
Actually, if we ta... | $-\frac{x}{x}= \frac{x}{x}$ has no solutions at all, since $-\frac{x}{x}\neq \frac{x}{x}$ no matter what $x$ is.
But we can square both sides, and then what happens?
$\frac{x^2}{x^2}= \frac{x^2}{x^2}$ is an equation that is true for all nonzero numbers.
By applying a non-invertible operation to both sides, we can turn... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/685687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 4
} |
Find the equation of a circle containing three given points Problem:
Determine the equation of the circle that passes through three points, $J(-3, 2)$, $K(4, 1)$, and $L(6, 5)$.
I thought of using systems like so:
$$\left\{ \begin{array}{rcl} (x+3)^2 + (y-2)^2 = r^2 \\ (x-4)^2 + (y-1)^2 = r^2 \\ (x-6)^2 + (y-5)^2 = r... | $\begin{vmatrix}
x^2+y^2&x&y&1\\
(-3)^2+2^2&-3&2&1\\
4^2+1^2&4&1&1\\
6^2+5^2&6&5&1\\
\end{vmatrix}=0$
$30(x^2+y^2)-60x-300y+30=0$
$x^2+y^2-2x-10y+1=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/685777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Solve $2^x+7=y^2$ for integer $(x,y)$ How many ordered solutions $(x,y)$ are there to the equation $2^x+7=y^2$ , where $x$ and $y$ are integers?
I tried taking cases for $x$ and $y$ like they are even or odd but I couldn't solve further.
| Given: $2^x+7=y^2$
Since $L.H.S.$ is odd, $R.H.S$. must be odd.
Putting $y=2m+1$,
$2^x+6=4m^2+4m$,
or $2^{x-1}+3=2m^2+2m$
this forces $x=1$
putting it in original equation, we get $y=3$ and $y=-3$
Thus we have solutions $(1,3)$ and $(1,-3)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/686147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
A definite integral involving logarithmic functions: $\int_{0}^{1} \frac{\ln(x) \ln (1+x)}{1-x} \, \mathrm{d} x$ I'm trying to show $$
\int_0^1 \frac{\ln x \cdot \ln(1+x)}{1-x}dx=-\frac{1}{4}\pi^2 \ln(2)+\zeta(3).
$$
I am unsure how to approach this integral as I do not know how to use a power series representation for... | \begin{align}J&=\int_{0}^{1} \frac{\ln(x) \ln (1+x)}{1-x} \, \mathrm{d} x\\&\overset{\text{IBP}}=\left[\left(\int_0^x \frac{\ln t}{1-t}dt\right)\ln(1+x)\right]_0^1-\int_0^1 \frac{1}{1+x}\left(\int_0^x \frac{\ln t}{1-t}dt\right)dx\\&=-\zeta(2)\ln 2-\int_0^1\int_0^1 \frac{x\ln(tx)}{(1+x)(1-tx)}dtdx\\
&=-\zeta(2)\ln 2-\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/689581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Prove that $\frac{d^n}{dx^n}\left(\frac{\sin x}{x}\right)=\frac{1}{x^{n+1}}\int_0^x y^n\cos\left(y+\frac{n\pi}{2}\right) \, dy,\: n\in \mathbb{N}$ Prove that $$\frac{d^n}{dx^n}\left(\frac{\sin x}{x}\right)=\frac{1}{x^{n+1}}\int_0^x y^n\cos\left(y+\frac{n\pi}{2}\right) \, dy,\: n\in \mathbb{N}$$
My try
*
*$n=0$ then ... | Let $f_n(x)=\int_0^x y^n \cos(y+n\pi/2) dy$. Then, using integration by parts,
$$\int_0^z \frac{1}{x^{n+1}} f_n(x) dx =
-\frac{1}{nz^n} f_n(z)
+ \frac{1}{n} \sin(z+n\pi/2).$$
Now, again using integration by parts, this time on $f_n(z)$, it follows that the right hand side equals
$$\frac{1}{z^n} \int_0^z y^{n-1} \sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/690697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
If $\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}=0$ with unequal $a,b,c$, Prove that $\dfrac{a}{(b-c)^2}+\dfrac{b}{(c-a)^2}+\dfrac{c}{(a-b)^2}=0$ If $\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}=0$ with unequal $a,b,c$, Prove that $\dfrac{a}{(b-c)^2}+\dfrac{b}{(c-a)^2}+\dfrac{c}{(a-b)^2}=0$
I could not approach the p... | You only need to prove that
$$
\begin{multline}
\left(\frac{a}{b - c} + \frac{b}{c - a} + \frac{c}{a - b}\right)\left(\frac{1}{b - c} + \frac{1}{c - a} + \frac{1}{a - b}\right) \\
= \frac{a}{(b - c)^2} + \frac{b}{(c - a)^2} + \frac{c}{(a - b)^2}
\end{multline}$$
Where did that come from?
Well, consider this simple equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/693087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
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Integral $\int_0^\infty \frac{1}{(1+x^m)(1+x^2)}\,dx$ I saw somewhere that the above integral is equal to $\pi/4$ for all real number $m$.
This seems to be surprising. Does anyone have a nice proof?
| Let
$$\begin{align}
I(m) &= \int_0^\infty \frac{dx}{(1+x^m)(1+x^2)}\tag{$x = t^{-1}$}\\
&= \int_0^\infty \frac{t^{-2}\,dt}{(1+t^{-m})(1+t^{-2})}\\
&= \int_0^\infty \frac{dt}{(1+t^{-m})(1+t^2)}\\
&= I(-m).
\end{align}$$
But
$$\begin{align}
I(m) + I(-m) &= \int_0^\infty \left(\frac{1}{1+x^m} + \frac{1}{1+x^{-m}}\right)\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/696269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Functional Equation : If $(x-y)f(x+y) -(x+y)f(x-y) =4xy(x^2-y^2)$ for all x,y find f(x). Problem :
If $(x-y)f(x+y) -(x+y)f(x-y) =4xy(x^2-y^2)$ for all x,y find f(x).
My approach :
The given equation can be written as $$(x-y)f(x+y) -(x+y)f(x-y) =4xy(x-y)(x+y)$$
$$\Rightarrow \frac{(x-y)f(x+y)}{(x-y)(x+y)} -\frac{(x+y... | The additional term comes from the relation $(x-y)k(x+y)-(x+y)k(x-y)=0$ for any $k$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/702695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Polynomials with Integer Coefficients and irrational roots Is there a polynomial with integer coefficients which has √2 +√7  as a root?
| $$
x = \sqrt{2} + \sqrt{7} \\ (x-\sqrt{2})^2=7 \\ x^2-2\sqrt{2}x+2=7 \\ x^2-5=2\sqrt{2}x \\ (x^2-5)^2=8x^2 \\ x^4-18x^2+25=0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/708168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Proving that $\frac{\sec\theta\cdot\sin\theta}{\tan\theta+\cot\theta}=\sin^2\theta$ The question is:
Prove that: $$\dfrac{\sec\theta\cdot\sin\theta}{\tan\theta+\cot\theta}=\sin^2\theta$$
My proof is shown below. If anyone has an alternate proof please, please post it. Thanks!
| "I know! Let's stand him on his head!"
$$ \frac{\tan \theta \ + \ \cot \theta}{\sec \theta \ \cdot \ \sin \theta} \ = \ \frac{\tan \theta \ + \ \cot \theta}{\tan \theta } \ = \ 1 \ + \ \frac{\cos^2 \theta }{\sin^2 \theta } \ = \ \frac{\sin^2 \theta \ + \ \cos^2 \theta }{\sin^2 \theta } \ = \ \frac{1}{\sin^2 \theta } $$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/709134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Express y in terms of x Question:
$$
\text{It is given that } y= \frac{3a+2}{2a-4} \text{and }x= \frac{a+3}{a+8} \\
$$
$$
\text{Express } y \text{ in terms of } x.
$$
From using $x$ to solve for $a$, I discovered that
$$
a = \frac{8x-3}{1-x}
$$
Then I proceeded to substitute $a$ into $y$. I did this twice to ensure n... | You got the correct answer. Just multiply both numerator and denominator by -1
$$\frac{22x-7}{20x-10} = \frac{-1}{-1} \times \frac{7-22x}{10-20x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/711856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
How find the minimum of $ab+\frac{1}{a^2}+\frac{1}{b^2}$ Let $a,b>0$ such that $$a+b\le 1$$ Find the minimum of
$$ab+\dfrac{1}{a^2}+\dfrac{1}{b^2}$$
My try:
I can find this minimum，use Holder inequality
$$(\dfrac{1}{a^2}+\dfrac{1}{b^2})(a+b)^2\ge (1+1)^3=8$$
But
$$ab\le \dfrac{(a+b)^2}{4}\le\dfrac{1}{4}$$
so for
$$a... | By AM-GM
$$16ab+16ab+a^{-2}+b^{-2}\ge16$$
But
$$2\sqrt{ab}\le a+b\le 1\implies ab\le\frac14\implies -31ab\ge-\frac{31}4$$
Summing we get, with equality iff $a=b=\frac12$
$$S\ge\frac{33}4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/713950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Find the minimum value of $\frac{1}{2}(x^{2}+2bxy+9y^{2})-y$ Find the minimum value of $f=\frac{1}{2}(x^{2}+2bxy+9y^{2})-y$
I know that there's a theorem which states $P(x)=\frac{1}{2}x^{T}Ax-x^{T}b$ reaches min at the point $Ax=b$ at $P_{min}=-\frac{1}{2}b^{T}A^{-1}b$.
I first tried to write the function $f$ in the fo... | Hint: Your matrix A is not right. It should be:
1 b
b 9.
In other words, a(11) = 1, a(12) = b = a(21), and a(22) = 9.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/714613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Safe prime mod 24 Given a safe-prime $p = 2q + 1$ where $q$ is also a prime and $p \gt 7$, I've read in a crypto.se answer that either $p \equiv 11 \pmod {24}$ or $p \equiv 23 \pmod {24}$.
I understand the proofs of why $p^2 \equiv 1 \pmod {24}$, and $p \equiv 1 \pmod 6$ or $p \equiv 5 \pmod 6$ for any prime $p$, and I... | I suggest you change to "otherwise $q$ would have to be even: $q \equiv 2$ or $14 \pmod{24}$, or $q \equiv 8$ or $20\pmod{24}$ respectively."
I would approach it from the other direction, checking candidates for $q$, because multiplication is safer than division in modular atithmetic. There are $12$ odd congruency clas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/718097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Simplifying a radical with fraction and sum What are the steps to simplify
$(1+(\frac{1}{2}(x^3-\frac{1}{x^3})^2))^ \frac{1}{2}$
to
$\frac{1}{2}(x^3+\frac{1}{x^3})$ ?
| If what you have is $$ \sqrt{ 1 + \frac{1}{2}(x^3 - \frac{1}{x^3})^2} $$
$$ \sqrt{ 1 + \frac{(x^6 - 1)^2}{2x^6}} = \sqrt{ \frac{x^{12 } + 1}{2x^6}} = \sqrt{ \frac{1}{2}(x^6 + \frac{1}{x^6})}$$
So they are not equal.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/720465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Average Value Half-Disk Find the average value of the x-coordinate of a point in the half-disk $R = {(x,y) : x^2 + y^2 \leq 25, x\geq 0}$ Also, out of curiosity what would be the value of the y coordinate?
I'm not really sure how to approach this problem, since we're not given a function f over R.
| $$x_c = \dfrac{\int_{x=0}^{5} \int_{y=-\sqrt{25-x^2}}^{\sqrt{25-x^2}} x dy dx}{\int_{x=0}^{5} \int_{y=-\sqrt{25-x^2}}^{\sqrt{25-x^2}} dydx}$$
First the denominator:
$$\int_{x=0}^{5} \int_{y=-\sqrt{25-x^2}}^{\sqrt{25-x^2}} dydx=\int_{x=0}^{5} 2\sqrt{25-x^2}dx$$
Using $x=5\sin(\theta)$ then $dx=5\cos{\theta}d\theta$
$$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/721148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Compute the square root of a complex number This is a follow up to a previous question. I solved the equation $z^4 - 6z^2 + 25 = 0$ and I found four answer to be $z = \pm\sqrt{3 \pm 4i}$.
However someone in the comment said that the answer is going to be $2+i$, $2-i$, $-2+i$, $-2-i$. I cannot understand how we can fin... | Hint:
Let $x + yi = \sqrt{a + bi}$. Then $(x+ yi)^2 = a + bi$. Then solve for $x$ and $y$ and you will generally have two sets of values for the square root $ \sqrt{a + bi}$
Example:
Say you want to compute $\sqrt{3 + 4i}$. Then assume the square root is $a + bi$. That is $a + bi = \sqrt{3 + 4i} \implies (a + bi)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/722852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Is there some way to simplify $\sum_{i=1}^n \sum_{j\neq i}(\frac{j-1}{2})(\frac{i-1}{2}) $ To obtain a closed form. Is there some way to simplify $\sum_{i=1}^n \sum_{j\neq i}(\frac{j-1}{2})(\frac{i-1}{2}) $?
Does it have a closed form? It's the last piece of a puzzle I need to solve a similar question Differentiate $P... | To answer my own question,
$\sum_{i=1}^n \sum_{j\neq i}(\frac{j-1}{2})(\frac{i-1}{2}) = \sum_{i=1}^n \sum_{j=1}^n(\frac{j-1}{2})(\frac{i-1}{2}) - \sum_{i=1}^n\frac{(i-1)^2}{4} $
Which simplifies to
$\frac{n(n-1)}{4}\frac{n(n-1)}{4} - \frac{1}{4}(\frac{(n-1)n(2(n-1)+1)}{6})$
or just $\frac{(n - 2) (n - 1) n (3 n - 1)}{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/723245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Inequality: $x^2+y^2+z^2 \geq \sqrt{2}x(z+y)$ How can I prove the following inequality:
$$x^2+y^2+z^2 \geq \sqrt{2}x(z+y)?$$
Thanks!
| Obviously, is enough to prove for $x\ge0$, $y\ge0$, $z\ge0$. By homogeneity, can be supposed wlog that $x^2+y^2+z^2=1$ Using Lagrange multipliers with the problem
$$f(x,y,z)=\sqrt{2}x(z+y),\qquad x^2+y^2+z^2=1,$$
we have:
$$\sqrt{2}(z+y)=2\lambda x,$$
$$\sqrt{2}x=2\lambda y,$$
$$\sqrt{2}x=2\lambda z,$$
with nontrivial ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/724617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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To Find $A^{50}$ $$A=\begin{bmatrix}1 & 0&0\\1 & 0&1\\0&1&0\end{bmatrix}$$
Find $A^{50}$ ?
Now from Cayley–Hamilton theorem, I get $A^3-A^2-A+I=0$ and $A^{50}=(A^4)^{12}A^2$ so I found $A^4$ which is $-2A-I$, then we have $A^{50}=B^{12}A^2$ where $B =A^4$ was calculated, now should I again use Cayley–Hamilton theorem ... | Let $f(x) = x^3-x^2-x-1 = (x^2-1)(x-1) = (x-1)^2(x+1)$.
Since $f(A) = 0$, we want to find polynomial $q(x), r(x)$ such that
$$x^{50} = q(x)f(x) + r(x)\tag{*1}$$
with $\deg r(x) \le \deg f(x) - 1 = 2$. If we can figure out what is $r(x)$, then
$$A^{50} = q(A)f(A) + r(A) = r(A)$$
Write $r(x)$ as $a x^2 + b x + c$. To fi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/731571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\overline{x} \times \overline{a} = \overline{b}$ has a solution when $ \langle\overline{a},\overline{b} \rangle =0$ I'm trying to solve this exercise:
Let $\overline{a} \neq \overline{0}$, $\overline{b}$ be two vectors of the Euclidean vector space $V_{3}$. Prove the equation $\overline{x} \times \overline{a} = \over... | $$
0 \cdot x_1+ a_3\cdot x_2 -a_2\cdot x_3=b_1\\
-a_3\cdot x_1+0\cdot x_2+a_1\cdot x_3=b_2\\
a_2 \cdot x_1 -a_1 \cdot x_2 + 0 \cdot x_3=b_3\\
$$
This is a system of linear equation. We are given that the vector a is nonzero. Without loss of generosity, let us assume that a_3 is nonzero.
Then by using Gaussian eliminat... | {
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"url": "https://math.stackexchange.com/questions/732753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to integrate $\int_{1}^{3} {\frac{x^2 - 1}{x^4 + 1}}\, dx$ $$\int_{1}^{3} {\frac{x^2 - 1}{x^4 + 1}}\, dx$$
Well, I can simplify the numerator:
$$\int_{1}^{3} {\frac{(x - 1)(x+1)}{x^4 + 1}}\, dx$$
But I have no idea how to simplify the denumenator:
$$\ x^4+1=? \ $$
How to solve this integral?
| We will first compute the integral without bounds. Start off by rewriting the integral using partial fractions:
$$\int \frac{-2\sqrt{2}x - 2}{4(x^2 + \sqrt{2}x + 1)} + \frac{2 - 2\sqrt{2}x}{4(-x^2 + \sqrt{2}x - 1)} \ \mathrm{d}x.$$
Note that we can split this into two integrals to get
$$\int \frac{-2\sqrt{2}x - 2}{4(x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/736897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to evaluate Ahmed's integral? How to show that:
$$\int_{0}^{1}\frac{\tan^{-1}\sqrt{x^{2}+2}}{(x^{2}+1)\sqrt{x^{2}+2}}\mathop{\mathrm{d}x}=\frac{5\pi ^{2}}{96}$$
I saw this on Wolfram.
| Define $f\left(t\right):=\int_{0}^{1}\frac{\arctan\left(t\sqrt{x^{2}+2}\right)}{\left(x^{2}+1\right)\sqrt{x^{2}+2}}dx$ so$$\begin{align}f\left(\infty\right)&=\frac{\pi}{2}\int_{0}^{1}\frac{dx}{\left(x^{2}+1\right)\sqrt{x^{2}+2}}\\&=\frac{\pi}{2}\left[\arctan\frac{x}{\sqrt{x^{2}+2}}\right]_{0}^{1}\\&=\frac{\pi^{2}}{12},... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/741933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 1
} |
$f$ is an even function defined on $(-5,5)$
If $f$ is an even function defined on $(-5,5)$ on the interval, then find four real values of x satisfying $f(x)=f(\frac{x+1}{x+2})$.
My book gives the answer as $\frac{\pm3 \pm\sqrt{5}}{2}$
And the solution is given as:
Since, f is an even function,
then, $f(x... | You are correct; we can't conclude that $x=\frac{1-x}{2-x}$ just from $f(x) = f\left(\frac{1-x}{2-x}\right)$. But we don't need to; we need to make the opposite inference: If $x = \frac{1-x}{2-x}$, then $f(x) = f\left(\frac{1-x}{2-x}\right)$, which is what we are looking for.
There might of course be many other $x$ fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/742588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Exponential and Logarithmic Functions - $\log_2(x^2-4x-28)= 2$ I got $x= \frac{45}{x-4} $
I know its wrong. Can someone show me how to solve.
Thanks
| Problem: We need to solve for $x$ in the given equation
$$\log_2(x^2-4x-28)=2$$
Remember that if $\log_a(b)=y$, then $a^y=b$. In our equation, $a=2$, $b=x^2-4x-28$, and $y=2$. Therefore $x^2-4x-28=2^2=4$. This is easy to solve for.
$$x^2-4x-28=4$$
$$x^2-4x-32=0$$
Factor it
$$(x-8)(x+4)=0$$
$$\color{green}{\boxed{x=8, \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/742928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$19 \mid 2^{2^{n}} + 3^{2^{n}} + 5^{2^{n}}$ I tried to demonstrate the next equation is divisible by 19:
$$ 2^{2^{n}} + 3^{2^{n}} + 5^{2^{n}} $$
When $n$ is $1$:
$$ 2^{2^1} + 3^{2^1} + 5^{2^1} $$
$$ 4 + 9 + 25 = 38 $$
When $n$ is $k$:
$$ 2^{2^k} + 3^{2^k} + 5^{2^k} $$
Finally, when $n$ is $k+1$:
$$ 2^{2^{k+1}} + 3^{2^{... | Not with induction:
The residues of each term cycle since $a^{2^{n+1}} = (a^{2^n})^2$
$2^{2^n} \pmod {19}$ with $4, 16, 9, 5, 6, 17, 4, \dots$
$3^{2^n} \pmod {19}$ with $9, 5, 6, 17, 4, 16, 9, \dots$
$5^{2^n} \pmod {19}$ with $6, 17, 4, 16, 9, 5, 6, \dots$
By adding terms pairwise, you can see that the statement is tru... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/744328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Improper integrals in curve length I am supposed to find the length of curve of the following:
*
*$ y = \sqrt{2-x^2}$ ; $0\le x\le 1$
*$y =\ln(\cos x) $; $0\le x\le \frac{\pi}{3}$
I followed the directions found from this question : Length of a curve y = 1 - √x to solve till the integral. So currently i have this... | In comments you've said that you've already managed to calculate the first integral, so let's talk about the second.
We have $$ \sqrt{1+\left(\frac{-\sin x}{\cos x}\right)^2} = \sqrt{1+\frac{\sin^2 x}{\cos^2 x}} =\sqrt{\frac{\cos^2 x + \sin^2 x}{\cos^2 x}} = \frac{1}{|\cos x|}$$
For $x\in(0,\frac\pi 3)$ we have $\cos x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/745696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluate the integral $\int\frac{1}{2+3\sin x}\,\text{d}x$. Please evaluate the integral,
$$\int \frac{1}{2+3\sin x}\,\text{d}x.$$
What I have tried is to substitute $\sin x = \sqrt{1-x^x}$ but I was stuck in a maze. Also, I did look a the wolfram solution. Can anyone propose a different solution from Wolfram, perhaps ... | use Weierstrass substitution: $\sin\left(x\right)=\frac{2t}{1+t^2}$, where $t=\tan\left(\frac{x}{2}\right)$
$⇔2\arctan\left(t\right)=x$,$dx=\frac{2dt}{1+t^2}$
$$\int_{ }^{ }\frac{dx}{2+3\sin\left(x\right)}
$$$$=\int_{ }^{ }\frac{\frac{2dt}{1+t^2}}{2+3\left(\frac{2t}{1+t^2}\right)}
$$$$=\int_{ }^{ }\frac{2dt}{2t^2+6t+2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/750810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
LogSine Integral $I=-\int_0^{\pi/3} \ln^2\big(2\cos \frac{\theta}{2}\big) d\theta$ These are known as LogSine integrals at $2\pi/3$, so I will call the integral Ls as this is common in the literature. I am trying to prove
$$
Ls=-\int_0^{\pi/3} \ln^2\big(2\cos \frac{\theta}{2}\big) d\theta=-\frac{13\pi^3}{162}-2Gl_{2,1... | Using the principal brach of $\log z$,
$$\log(1+e^{2ix}) = \log(e^{i x}(e^{-ix}+ e^{i x})) = \log(e^{ix})+ \log(2 \cos x) = ix + \log(2 \cos x) .$$
Squaring both sides,
$$ \int_{0}^{\pi /6} \log^{2}(1+e^{2ix}) \ dx = \int_{0}^{\pi /6} \Big( ix + \log(2 \cos x) \Big)^2 \ dx .$$
Then equating the real parts on both sides... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/751227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
If $a_{n+1}=1+\frac{1}{a_{1}+a_{2}+\cdots+a_{n}-1}$ then $0define the sequence $\{a_{n}\}$ and such
$$a_{1}=a,a_{n+1}=1+\dfrac{1}{a_{1}+a_{2}+\cdots+a_{n}-1},n\ge 1$$
Find the all real number $a>0$,such
$$0<a_{n}<1,n\ge 2$$
My try: since $$a_{2}=1+\dfrac{1}{a_{1}-1}=1+\dfrac{1}{a-1}$$
then
$$0<1+\dfrac{1}{a-1}<1$$
so
... | First, we try to simplify the given sequence:
$$a_{n+1} - 1 = \frac{1}{a_1 + \dots + a_n - 1}$$
$$a_1 + \dots + a_n - 1 = \frac{1}{a_{n+1} - 1}$$
$$a_1 + \dots + a_n = 1 + \frac{1}{a_{n+1} - 1}$$
Now, we do some subtraction for $n\ge2$:
$$(a_1 + \dots + a_{n-1} + a_n) - (a_1 + \dots + a_{n-1}) = \left(1 + \frac{1}{a_{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/753010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}}\le\frac{3}{2}$ For positive real numbers with $a+b+c=abc$ prove that
$$\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}}\le\frac{3}{2}$$
I made the substitution $a=\tan(\alpha), b = \tan(\beta), c= \tan(\gamma)$ with th... | By AM-GM
$$\sum_{cyc}\frac{1}{\sqrt{1+a^2}}=\sum_{cyc}\frac{1}{\sqrt{\frac{abc}{a+b+c}+a^2}}=\sum_{cyc}\sqrt{\frac{a+b+c}{a(a+b)(a+c)}}=$$
$$=\sum_{cyc}\sqrt{\frac{bc}{(a+b)(a+c)}}\leq\frac{1}{2}\sum_{cyc}\left(\frac{b}{a+b}+\frac{c}{a+c}\right)=\frac{1}{2}\sum_{cyc}\left(\frac{b}{a+b}+\frac{a}{a+b}\right)=\frac{3}{2}.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/759592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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