Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Convergence of $\sum_{n = 1}^{\infty} \frac{(-1)^n}{n}$ I was messing around in Mathematica with infinite sums until I tried taking the sum of the following:
$$\sum_{n = 1}^{\infty} \frac{(-1)^n}{n}$$
Mathematica spat out -Log[2]. Can somebody give me proof explaining this answer?
| A formal argument (ignoring convergence questions)
Start with the Geometric series $$\frac 1{1-x}=1+x+x^2+\cdots$$
Integrate to obtain $$-\ln(1-x)=x+\frac {x^2}2+\frac {x^3}3+\cdots$$
Now evaluate at $x=-1$ to get your result.
To be more rigorous, note that (inductively) it is easy to prove
$$\frac{1}{1}-\frac{1}{2}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1892018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve the equations for x.
a. $10e^{2x}=7\cdot 2^{ax}$
b. $\frac{2(a-x)}{3\sqrt[3]{x}}-x^{2/3}=0$
I did both, but I'm not sure if I did them correctly. Here is my work for both questions:
a. $10e^{2x}=7\cdot 2^{ax}$
$e^{2x}=\frac{7\cdot 2^{ax}}{10}$
$In(\frac{7\cdot 2^{ax}}{10})=2x$
$In(\frac{7}{10})+In(2^{ax})=2x$... | For (a), the real solution (correct):
$$10e^{2x}=7\cdot2^{\text{a}x}\Longleftrightarrow$$
$$\ln(10e^{2x})=\ln(7\cdot2^{\text{a}x})\Longleftrightarrow$$
$$\ln(10)+\ln(e^{2x})=\ln(7)+\ln(2^{\text{a}x})\Longleftrightarrow$$
$$2x+\ln(10)=\ln(7)+\text{a}x\ln(2)\Longleftrightarrow$$
$$2x-\text{a}x\ln(2)=\ln(7)-\ln(10)\Longle... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1892192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Rational Solutions to Trigonometric Equation Consider the equation
$$
\cos(\pi a) + \cos(\pi b)
=
\cos(\pi c) + \cos(\pi d),
$$
with
$$
a,b,c,d \in \mathbb{Q} \cap \left[0,\frac{1}{2}\right].
$$
Clearly, this equation admits some trivial solutions, namely $(a,b) = (c,d)$ or $(a,b) = (d,c)$. Are there any rational solut... | For nontrivial solutions, $a,b,c,d$ are distinct. WLOG let $a = \max(a,b,c,d)$.
Taking a common denominator $N$ and writing $a=A/N$, ..., $d = D/N$,
we can write this as
$$ \omega^A + \omega^{-A} + \omega^B + \omega^{-B} - \omega^C - \omega^{-C} - \omega^D - \omega^{-D} = 0$$
or equivalently
$$ P(\omega) = \omega^{2... | {
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"url": "https://math.stackexchange.com/questions/1892263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Evaluation of $\int_{\pi/6}^{\pi/3} \frac{1}{\sqrt{1+\tan(x)}}dx$ Evaluate the given integral:
$$\int_{\pi/6}^{\pi/3} \frac{1}{\sqrt{1+\tan(x)}}dx$$
After using $\int_{a}^{b}f(x)dx= \int_{a}^{b} f(a+b-x)dx$, we get $\int_{\pi/6}^{\pi/3} \sqrt{\sin(x)+\cos(x)} dx$ but I am not able to solve this integral. Could someon... | Hint: Let $u=\sqrt{1+\tan x}$, so $du=\frac{\sec^2x}{2\sqrt{1+\tan x}}dx\;$. Then
$\displaystyle\int\frac{1}{\sqrt{1+\tan x}}dx=2\int\frac{1}{1+\tan^2x}\frac{\sec^2 x}{2\sqrt{1+\tan x}}dx=2\int\frac{1}{1+(u^2-1)^2}du=2\int\frac{1}{u^4-2u^2+2}du$.
Next use partial fractions:
Since $u^4-2u^2+2=[(u^2)^2+2\sqrt{2}u^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1892353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find The Last Two Digits Of $9^{8^7}$ Find the last two digits of $9^{8^7}$.
I tried finding a secure pattern for the last two digits of powers of $9$ but that didn't work. Any answers?
| Since \begin{align}(10-1)^{10}&=\sum_{i=0}^{10} \begin{pmatrix} 10 \\i \end{pmatrix}10^i(-1)^{10-i}\\&=(-1)^{10}+10(10)(-1)^9+\sum_{i=2}^{10} \begin{pmatrix} 10 \\i \end{pmatrix}10^i(-1)^{10-i} \\&\equiv1\mod 100\end{align}
We have
$$9^{10} \equiv 1 \mod 100$$
Since $8^7 \equiv (-2)^7 \equiv -128 \equiv 2 \mod 10,$
$9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1896566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Solve this $6$-th degree polynomial equation: $x^6-x^5-8x^4+2x^3+21x^2-9x-54=0$ The question is as follows:
Solve the equation: $x^6-x^5-8x^4+2x^3+21x^2-9x-54=0$, one root being $\sqrt{2}+i$
It's trivial that another root is $\sqrt{2}-i$. But, I can go no further. Can anyone please help me how to solve this?
| As @avs answered, after dividing the polynomial by $$(x - \sqrt{2} + i)(x - \sqrt{2} - i)$$ we obtain $$x^4+\left(2 \sqrt{2}-1\right) x^3-\left(3+2 \sqrt{2}\right) x^2-\left(3+12
\sqrt{2}\right) x-18=0$$ By inspection $x=-2$ and $x=3$ are roots. Continuing the division, we are left with $$x^2+2 \sqrt{2} x+3=0$$ the... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the limit $\lim_{n\rightarrow\infty}(1-\frac{1}{3})^2(1-\frac{1}{6})^2(1-\frac{1}{10})^2...(1-\frac{1}{n(n+1)/2})^2$ Any hints on how to find the following limit ?.
I haven't been able to figure it out still.
$$
\lim_{n \to \infty}\left(\, 1 - \frac{1}{3}\, \right)^{2}
\left(\, 1 - \frac{1}{6}\, \right)^{2}\lef... | We have
$$ \prod_{n=2}^{+\infty}\left(1-\frac{2}{n(n+1)}\right)=\prod_{n=2}^{+\infty}\left(\frac{n+2}{n+1}\cdot\frac{n-1}{n}\right)=\frac{1}{3} $$
(it is a telescopic product) hence you limit equals $\color{red}{\large\frac{1}{9}}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to compute the Pythagorean triple by one of the numbers that belonged to it? I have a positive number $n>2$. How to compute the Pythagorean triple containing $n$? $n$ may be the hypotenuse and leg.
| Using Euclid's formula, we can find triples matching any hypotenuse by solving the C-function for k and seeing which of a range of m-values yield integers. The limitation is that, for primitive triples, C must take the form $C-4n+1$ and not all of these are valid. Some valid ones are $5,13,17,25,29,37,41 ... 65$
\begin... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Mathematical induction problem: $\frac12\cdot \frac34\cdots\frac{2n-1}{2n}<\frac1{\sqrt{2n}}$ This is a problem that I tried to solve and didn't come up with any ideas
.?$$\frac{1}{2}\cdot \frac{3}{4}\cdots\frac{2n-1}{2n}<\frac{1}{\sqrt{2n}}.$$
All I get is $\frac{1}{\sqrt{2n}}\cdot\frac{2n+1}{2n+2}<\frac{1}{\sqrt{2n+2... | Induction is a really efficient way for proving that
$$\frac{(2n-1)!!}{(2n)!!} = \frac{(2n)!}{4^n n!^2} = \frac{1}{4^n}\binom{2n}{n}<\frac{1}{\sqrt{2n}}\tag{1}$$
but it isn't the only option. For instance, we may notice that the LHS of $(1)$ can be written as $$\prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)=\sqrt{\prod_{k=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1899857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find the value of $\tan A + \tan B$, given values of $\frac{\sin (A)}{\sin (B)}$ and $\frac{\cos (A)}{\cos (B)}$ Given
$$\frac{\sin (A)}{\sin (B)} = \frac{\sqrt{3}}{2}$$
$$\frac{\cos (A)}{\cos (B)} = \frac{\sqrt{5}}{3}$$
Find $\tan A + \tan B$.
Approach
Dividing the equations, we get the relation between $\tan A$ and $... | HINT:
$$\tan A=\frac{\sin A}{\cos A}=\frac{\frac{\sqrt{3}}2\sin B}{\frac{\sqrt{5}}3\cos B}=\frac{3\sqrt{3}}{2\sqrt{5}}\tan B\tag{1}$$
And
$$1=\sin^2 A+\cos^2 A=\frac{3}{4}\sin^2 B+\frac{5}{9}\cos^2 B=\frac59\left(\sin^2 B+\cos^2B\right)+\frac7{36}\sin^2 B$$
Last equation implies
$$\sin^2B=\frac{16}7$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1901177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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How to determine that the solution to the following problem is unique? The question is:
[BMO2 2000 Q3] Find positive integers a and b such that $$(\sqrt[3]{a} + \sqrt[3]{b} - 1)^2 = 49 + 20\sqrt[3]{6}$$
It suffices to find one solution to gain full marks, which I did by expanding and assuming that neither $a$ nor $b... | We do not need to account for your second case, since it is trivially impossible due to the fact that for positive integers $ a, b, c, d $ the left hand side is positive and the right hand side is negative. Your first case is settled by the following theorem:
Theorem. Let $ a, b \in \mathbf Q^{\times} $ be non-perfect ... | {
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Show that $(1,1,1)$, $(a,b,c)$, $(a^2, b^2, c^2)$ are linearly indepdenent for distinct $a,b,c$ Show that $(1,1,1)$, $(a,b,c)$, $(a^2, b^2, c^2)$ are linearly indepdenent, where $a,b,$ and $c$ are distinct real numbers.
I will show my attempt and then state where I get stuck.
Suppose $c_1(1,1,1) + c_2(a,b,c) + c_3(a^2,... | You have a linear system of three equations in three unknowns, which you can write in matrix form as
$$
A\mathbf{c} = \mathbf{0},
$$
where
$$
A = \begin{pmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{pmatrix} \in \mathbb{R}^{3 \times 3}
$$
is your coefficient matrix and
$$
\mathbf{c} = \begin{pmatrix}c_1\\c_2\\c_3\end{pmatri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1903886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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Let a, b, c, x and y be integers. Why are there no solutions for $\sqrt {a^2 + b^2 + c^2} = x$, where $x=2^y$ or $x=5 \times 2^y$ Edit - An addition condition: $a$, $b$, and $c$ do not equal $0$.
I'm really digging into 3D vectors and their properties. I've decided to look and see which combinations of three integers w... | There is no solution with $y= 0.$
Let $a,b,c,y$ be positive. If there exists a solution to $a^2+b^2+c^2\in \{4^y, 25\cdot 4^y\}$ for some $(a,b,c,y)$ then there is a solution with a least $y$. Such a solution cannot have $a,b,c$ all even, else $(a/2)^2+(b/2)^2+(c/2)^2\in \{4^{y-1},4^{y-1}\cdot 25\},$ which contra... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the area under the curve I apologize, I have asked this question previously and I expect I will be down voted but thats okay because I really need help on this problem not just the answer.
Find the area of the figure rotated about the x-axis $0 \leq t \leq 1$
$x=9t-3t^3$ and $y=9t^2$
Setting up the integral:
$\int... | I would say:
$\displaystyle S=18\pi\int_0^1 t^2 \sqrt{(9-9t^2)^2+(18t)^2}\,dt=18\pi\int_0^1 t^2 \sqrt{81(1-t^2)^2+81\cdot 4t^2}\,dt=$
$\displaystyle=162\pi\int_0^1 t^2 \sqrt{(1-t^2)^2+4t^2}\,dt=162\pi\int_0^1 t^2 \sqrt{1-2t^2+t^4+4t^2}\,dt=$
$\displaystyle =162\pi\int_0^1 t^2 \sqrt{(1+t^2)^2}\,dt=162\pi\int_0^1 t^2
(1... | {
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"url": "https://math.stackexchange.com/questions/1909967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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simplify $\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}$ simplify $$\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}$$
1.$\frac{3}{2}$
2.$\frac{\sqrt[3]{65}}{4}$
3.$\sqrt[3]{2}$
4.$1$
I equal it to $\sqrt[3]{a}+\sqrt[3]{b}$ but I cant find $a$ and $b$
| Hint: Rewrite $x_{1,2}=5\pm 2\sqrt{13}$ in polar coorinates $x=|x|\exp(i\phi_{1,2})$. Determine $|x_{1,2}|=\sqrt{5^2+(2\sqrt{13})^2}$ and $\tan(\phi_{1,2})=\frac{\pm 2\sqrt{13}}{4}$
Use the periodicity of the exponential function in the complex domain
$\sqrt[3]{x_{1,2}}=|x_{1,2}|^{1/3}\exp(i\phi_{1,2}/3+2\pi k/3)$, in ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1910728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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How can I prove this inequality:$(x+y)^4\le8(x^4+y^4)$? I have to proce that $(x+y)^4\le8(x^4+y^4)$
I have seen that the equality is when $x=y$...
I have tried to develop $(x+y)^4$ but it leads me nowhere... I think that I have to pass by the intermediate of the averages, but I don't see witchone... Could someone plea... | $$(x+y)^4= x^4+4x^3y+6x^2y^2+4xy^3+y^4$$
Now, by AM-GM you have
$$
x^3y \leq \frac{x^4+x^4+x^4+y^4}{4} \\
x^2y^2 \leq \frac{x^4+y^4}{2} \\
xy^3\leq \frac{x^4+y^4+y^4+y^4}{4}$$
Thus
$$(x+y)^4 \leq x^4+4 \cdot \frac{x^4+x^4+x^4+y^4}{4} +6 \frac{x^4+y^4}{2} +4 \frac{x^4+y^4+y^4+y^4}{4} +y^4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1910948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Secondary school level mathematical induction
*
*It is given that
$$1^3+2^3+3^3+\cdots+n^3=\frac{n^2(n+1)^2}{4}$$
Then, how to find the value of
$2^3+4^3+\cdots+30^3$?
Which direction should I aim at?
*Prove by mathematical induction, that $5^n-4^n$ is divisible by 9 for all positive even numbers $n$.
$$5^n-... | Hints
1) $2^3+4^3+...+30^3=2^3(1^3+2^3+...+15^3)$ now apply the formula you got.
2) Assume that $9k=5^n-4^n$ then $5^{n+2}-4^{n+2}=25.5^n-16.4^n=25(9k+4^n)-16.4^n=9k'+9.4^n$
3) Change the questions to show that $(a+b)|(a^{2k+1}+b^{2k+1})$ for all $k\in\mathbb{N}$.Let $(a+b)x=a^{2k+1}+b^{2k+1}$ now $a^{2k+3}+b^{2k+3}=a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1911131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Solving exponential equation of form $(a+b)^x + (a-b)^x = c$ I'm given equation:
$$\left(2-\sqrt3\right)^\frac{x}{2} + \left(2+\sqrt3\right)^\frac{x}{2} = 4,$$
and I'm stuck with it. Which way shall I dig into to solve it?
| Let $u = (2+\sqrt{3})^{x/2}$. Using the observation that $2-\sqrt{3} = \frac{1}{2+\sqrt{3}}$, we have that $(2-\sqrt{3})^{x/2} + (2+\sqrt{3})^{x/2} = u+\frac{1}{u}$. We thus aim to solve
$$ u + \frac{1}{u} = 4. $$
Since $u\ne 0$, this is equivalent to $u^2 - 4u + 1 = 0$. Using the quadratic formula, we obtain $u = 2\pm... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Want help for prove formula by combinatoric argument $1\cdot 2+2\cdot 3+3\cdot 4+\dots +n(n+1)=\frac{n(n+1)(n+2)}{3}$ I have a problem.I just read first combinatorics textbook and I found the identity
$1\cdot 2+2\cdot 3+3\cdot 4+\dots +n(n+1)=\frac{n(n+1)(n+2)}{3}$.
I can prove this by induction,but the problem is from... | On the right side, you have the number of ways of selecting three items without replacement from a set of $n+2$ distinct items, multiplied by $2$. For simplicity, I'll assume there are $n$ items first, then I'll up the index later.
Now, suppose you certainly select the first item. The number of ways of choosing the res... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find all positive natural $x$ and $y$ so $(x+y)^7-x^7-y^7$ is divisble by $7^7$, but so that $xy(x+y)$ is not divisble by $7$ I thought of looking modulo $7$ but it lead me nowhere... Does anyone have a better idea?
| Note that
$$\frac{(x+y)^7-x^7-y^7}{7xy(x+y)}= x^4 + 2yx^3 + 3y^2x^2 + 2y^3x + y^4 $$
We immediately check that $x\equiv y\pmod 7$ cannot lead to a solution because $9x^4$ cnnot be a multiple of $7^6$. Hence we are allowed to rewrite this as
$$x^4 + 2yx^3 + 3y^2x^2 + 2y^3x + y^4 =\frac{x^6-2x^3y^3+y^6}{(x-y)^2}=\frac{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1916833",
"timestamp": "2023-03-29T00:00:00",
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What is the maximal value of $2 \sin x - 7 \cos x$? What is the maximal value of $2 \sin x - 7 \cos x$?
How do I calculate this? Do I have to write out the $\sin$ and the $\cos$?
| Let us first develope a general methodology by considering the general identity:
$$\ \tag{1} f(x)=A \cos(x) + B \sin(x)=C\left(\dfrac{A}{C}\cos(x) + \dfrac{B}{C} \sin(x)\right)$$
with $C:=\sqrt{A^2+B^2}$.
As $\left(\dfrac{A}{C}\right)^2 + \left(\dfrac{B}{C}\right)^2=1$, point $\left(\dfrac{A}{C},\dfrac{B}{C}\right)$ ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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I can't find 'the easiest way' to simplify this expression? $\large\dfrac{3^{2008} (10^{2013} + 5^{2012} + 2^{2011})}{5^{2012} (6^{2010} + 3^{2009} + 2^{2008})}$
$\large=\dfrac{3^{2008} ((2\cdot5)^{2013} + 5^{2012} + 2^{2011})}{5^{2012} ((2\cdot3)^{2010} + 3^{2009} + 2^{2008})}$
$\large=\dfrac{3^{2008} (2^{2013}\cd... | The expression is extremely close to $\dfrac{40}{9}$. This is because in the numerator, $2^{2013}\cdot 5^{2013}$ is much larger than the remaining terms $5^{2012}$ and $2^{2011}$, so the remaining terms are almost negligible. Similarly for the denominator. The expression then simplifies to$\dfrac{2^{2013}\cdot 3^{2008}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1919027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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integrate $\int \frac{1}{\sqrt{x^2 - 1}} \, dx$ I am solving an ODE
$$y^2 \, dx + \left(x\sqrt{y^2 - x^2} - xy\right)dy=0$$
I let $y=xu$ and did some algebra, and I ended up with this
$$\ln(x)+C=\int \frac{1}{\sqrt{u^2 - 1}} \, du - \int \frac{1}{u} \, du$$
I don't know how to solve the first right hand side integral. ... | If the hyperbolic trigonometry is not known, we can see that
$\int\frac{dt}{\sqrt{t}\sqrt{t+1}}=\int\frac{\sqrt{t+1}}{\sqrt{t}}-\frac{\sqrt t}{\sqrt{t+1}}dt.$
Then $z=1+\frac{1}{t}$ gives $t=\frac{1}{z-1}$, $dt=\frac{-dz}{(1-z)^2}$ and
$\frac{1}{2}\int\frac{\sqrt{t+1}}{\sqrt{t}}-\frac{\sqrt t}{\sqrt{t+1}}dt=\frac{1}{2}... | {
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"url": "https://math.stackexchange.com/questions/1919767",
"timestamp": "2023-03-29T00:00:00",
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Prove: $\cos^3{A} + \cos^3{(120°+A)} + \cos^3{(240°+A)}=\frac {3}{4} \cos{3A}$ Prove that:
$$\cos^3{A} + \cos^3{(120°+A)} + \cos^3{(240°+A)}=\frac {3}{4} \cos{3A}$$
My Approach:
$$\mathrm{R.H.S.}=\frac {3}{4} \cos{3A}$$
$$=\frac {3}{4} (4 \cos^3{A}-3\cos{A})$$
$$=\frac {12\cos^3{A} - 9\cos{A}}{4}$$
Now, please help m... | You can use this particular formula: $$\cos{x}\cos{y}=\frac{1}{2}\left(\cos(x+y)+\cos(x-y)\right)$$
twice and simplify $\cos^3{A}$, $\cos^3{(A+2\pi/3)}$, and $\cos^3{(A-2\pi/3)}$ in this way:
$$\begin{aligned} \cos^3{A}&=\cos{A}\left(\cos{A}\cos{A}\right) \\
&=\frac{1}{2}\cos{A}\left(1+\cos{2A} \right) \\
&=\frac{1}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1921191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 1
} |
Calculate $(1+i)^{11}$ I'm studying complex numbers and I wanted to know if this solution is correct.
The problem is to calculate $(1+i)^{11}$, here's my attempt:
I can express $(1+i)^{11}$ using the argument notation: $z=\rho(\cos \phi +i\sin \phi)$
$$z=\left(\cos \frac{\pi}{2}+i\sin\frac{\pi}{2}\right)$$
This is very... | There's another approach:
\begin{align}
(1+i)^{11}&=\sum_{k=0}^{11}\binom{11}{k}i^k
\\\\&=\sum_{\substack{0\le k\le 11 \\k\equiv 0\!\pmod 4}}\binom{11}{k} + \sum_{\substack{0\le k\le 11 \\k\equiv 1\!\pmod 4}}\binom{11}{k}\cdot i + \sum_{\substack{0\le k\le 11 \\k\equiv 2\!\pmod 4}}\binom{11}{k} \cdot (-1)+ \sum_{\subst... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1921525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Epsilon-delta limit definition I got $\lim_{x \rightarrow 3} \frac{x^2-2x+1}{x-2} = 4$
I need to prove that by delta epsilon. I came to delta ={1/2epsilon,1/2}
Is that right? If not can you explain me how?
| Let $\forall \epsilon >0$ and we have to show that ${ \delta }>0$ such that
if $0<\left| x-3 \right| <{ \delta }$ then $\left| \frac { { x }^{ 2 }-2x+1 }{ x-2 } -4 \right| <\epsilon $
$$\left| \frac { { x }^{ 2 }-2x+1 }{ x-2 } -4 \right| =\left| \frac { { x }^{ 2 }-2x+1-4x+8 }{ x-2 } \right| =\left| \frac { { x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1921647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proving the identity $\frac{\cos^2\theta+\tan^2\theta-1}{\sin^2\theta}=\tan^2\theta$ I am stuck with this trigonometric identity. It appeared in a question paper of mine, and I am wondering whether there is a print error or something, because I absolutely have no idea how to solve this.
$$\frac{\cos^2\theta+\tan^2\th... | $$\frac{\cos^2\theta + \frac{\sin^2 \theta}{\cos^2\theta}-1}{\sin^2\theta}=$$
$$= \frac{\cos^2 \theta +\frac{\sin^2\theta}{\cos^2 \theta}-\sin^2 \theta-\cos^2\theta}{\sin^2 \theta}$$
$$= \frac{\frac{\sin^2\theta-\cos^2\theta \cdot \sin^2\theta}{\cos^2\theta}}{\sin^2\theta}$$
$$=\frac{\sin^2\theta \cdot (1-\cos^2\theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1922605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
How to integrate $\int\frac{du}{u\sqrt{c-2u}} $ I was looking through these notes, and at the top of the second page it says you can integrate
$$\mathrm{d}\zeta = \frac{\mathrm{d}u}{u\sqrt{c-2u}} $$
to get
$$u(\zeta) = \frac{c}{2}\mathrm{sech}^2(\frac{1}{2}\sqrt{c}(\zeta - \zeta_0)).$$
I can't see where that comes from... | Take the integral:
$$\int \frac{1}{u \sqrt{c-2 u}} \ du$$
For the $$\int \frac{1}{u \sqrt{c-2 u}}$$, substitute $u = c-2 u$ and $du = -2 du$. Then we have :
$$\int \frac{1}{\sqrt{u} (u-c)} \ du$$
For the integrand $\frac{1}{\sqrt{u} (u-c)}$, substitute $s = \sqrt{u}$ and $ds = \frac{1}{2 \sqrt{u}} du$, then we have :... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1922791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Symmetric polynomials with Vieta's and Newton's theorems
Let $ x_{1}, x_{2}, x_{3}$ be the solutions of the equation $ x^3 -3x^2 + x - 1 = 0.$
Determine the values of $$\frac{1}{{x_{1}x_{2}}} + \frac{1}{{x_{2}x_{3}}} + \frac{1}{{x_{3}x_{1}}}$$
and also
$$ x_{1}^{3}+x_{2}^{3}+x_{3}^{3}$$
| We have
$$x^3-3x^2+x-1=(x-x_1)(x-x_2)(x-x_3)$$
so by developing the RHS we see that $\sum x_i x_j=1:$ the coefficient of $x$ and similarly $x_1x_2x_3=1$ and $x_1+x_2+x_3=3$.
Denote $V_1$ the first desired value than $V_1\times(x_1x_2x_3)=x_1+x_2+x_3$. And denote $V_2$ the second value then
$$(x_1+x_2+x_3)^3=V_2+3(x_1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1923980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove $(a^2+b^2)(c^2+d^2)\ge (ac+bd)^2$ for all $a,b,c,d\in\mathbb{R}$. Prove $(a^2+b^2)(c^2+d^2)\ge (ac+bd)^2$ for all $a,b,c,d\in\mathbb{R}$.
So $(a^2+b^2)(c^2+d^2) = a^2c^2+a^2d^2+b^2c^2+b^2d^2$
and $(ac+bd)^2 = a^2c^2+2acbd+b^2d^2$
So the problem is reduced to proving that $a^2d^2+b^2c^2\ge2acbd$ but I am not sure ... | By Lagrange's identity
$$(a^2+b^2)(c^2+d^2)=(ac+bd)^2\color{red}{+(ad-bc)^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1925766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Finding number of roots of $1+2x+3x^2+...+(n+1)x^n=0$ Find the number of roots of the equation
$1+2x+3x^2+4x^3+....+(n+1)x^n=0$
where $n$ is even.
My attempt:
Let $f(x)=x+x^2+x^3+....+x^n$
Clearly,$x=0$ is a root and $f(x)=0$
cannot have a positive root.
$f(x)=x\left(1+x+x^2+...+x^n\right)$
How to determine whether the... | We have $$ 1+2x+3x^2+...+(n+1)x^n = \sum_{k=0}^n (k+1)x^k = \sum_{k=0}^n \sum_{m=k}^n x^m = \sum_{k=0}^n \sum_{m=0}^{n-k} x^{m+k} =$$
$$= \sum_{k=0}^n x^k \sum_{m=0}^{n-k} x^m = \sum_{k=0}^n x^k \cdot \frac{x^{n-k+1}-1}{x-1} = \frac{1}{x-1} \sum_{k=0}^n {(x^{n+1}-x^k)} = $$
$$= \frac{(n+1)x^{n+1}}{x-1} - \frac{1}{x-1}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1926080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove the inequality $\sqrt{a^2+b^2}+\sqrt{c^2+b^2}+\sqrt{a^2+c^2}<1+\frac{\sqrt2}{2}$ Let $a,b,c -$ triangle side and $a+b+c=1$. Prove the inequality
$$\sqrt{a^2+b^2}+\sqrt{c^2+b^2}+\sqrt{a^2+c^2}<1+\frac{\sqrt2}{2}$$
My work so far:
1) $a^2+b^2=c^2-2ab\cos \gamma \ge c^2-2ab$
2) $$\sqrt{a^2+b^2}+\sqrt{c^2+b^2}+\sqrt{... | Just a sketch of a possible proof (a variational, mixing variables approach). Assume that $a\leq b\leq c$ and show that if you replace $a$ with $a-\varepsilon$ and $b$ with $b+\varepsilon$, the convexity of $f(x)=\sqrt{1+x^2}$ grants that the LHS increases (after such substitution, the order of $b$ and $c$ may change. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1929245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Polar to cartesian form of r=sin(4θ)? The Polar to cartesian form of $ r = \sin(2\theta)$ is fairly simple.
What is the Cartesian form of the polar equation r=sin(4θ)?
[edit]
$$r=4sin(θ)cos(θ)(cos(θ)^2-sin(θ)^2)$$, so $$r^5=4rsin(θ)rcos(θ)(r^2cos(θ)^2-r^2sin(θ)^2)$$,so $$r^5=4xy(x^2-y^2)$$, so $$(x^2+y^2)^{5/2}=4xy(x^... | $$\rho=\sin 4\theta=2\sin2\theta\cos2\theta=4\sin\theta\cos\theta(\cos^2\theta-\sin^2\theta)=\frac{4xy(x^2-y^2)}{\rho^4},$$ hence
$$(x^2+y^2)^5=(4xy(x^2-y^2))^2,$$
$$x^{10}+10x^4y^6+32x^4y^4+5x^2y^8-16x^2y^6+y^{10}+10y^4x^6+5y^2x^8-16y^2x^6=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1930325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find sum of first $n$ terms of the series : $1+\frac{1^3+2^3}{1+2}+\frac{1^3+2^3+3^3}{1+2+3}+\dots$ The main question is:
Find sum of first $n$ terms of the series : $1+\frac{1^3+2^3}{1+2}+\frac{1^3+2^3+3^3}{1+2+3}+\dots$
My approach:
Initially, nothing clicked, so I went forward with simplifying the series.
So, we g... | Recall that
$$x_n=\sum_{k=1}^n k=\frac{n(n+1)}{2}$$
and
$$y_n=\sum_{k=1}^n k^3=\frac{n^2(n+1)^2}{4}$$
so
$$\frac{y_n}{x_n}=x_n$$
so you need to recall the sum
$$\sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6}$$
to compute the desired sum $\sum_{k=1}^n \frac{y_k}{x_k}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1931771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Recurrence relation of type $a_{n+1} = a^2_{n}-2a_{n}+2$
A sequence $\{a_{n}\}$ is defined by $a_{n+1} = a^2_{n}-2a_{n}+2\forall n\geq 0$ and $a_{0} =4$
And another sequence $\{b_{n}\}$ defined by the formula $\displaystyle b_{n} = \frac{2b_{0}b_{1}b_{2}..........b_{n-1}}{a_{n}}\forall n\geq 1$ and $\displaystyle b_{... | Define $c_{n}=a_{n}-1$. Note that $c_{n+1}=c_{n}^{2}$. Then,
$c_{4} = c_{3}^{2} = c_{2}^{4} = c_{1}^{8} = c_{0}^{16} = 3^{16}$, so $a_{4}=3^{16}+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1932519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
For how many positive integers $a$ is $a^4−3a^2+9$ a prime number? I understand that there are many posts on the problems similar to mine. I have tried my best, but still get different answers from the answer sheet. Can anyone help me? Also is there a simple way to find $a$?
For how many positive integers $a$ is $a^4-... | You already have the product $p = \left(a^2-3 a+3\right) \left(a^2+3 a+3\right)$, and since for both factors you have $0 < a^2-3 a+3 < a^2+3 a+3$ for positive $a$ (check), and since a prime number is only divisible by $1$ and itself, you have $a^2-3 a+3 = 1$ and $a^2+3 a+3 = p$. Solving first equation gives you $a=1$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1933220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Easier way to calculate the derivative of $\ln(\frac{x}{\sqrt{x^2+1}})$? For the function $f$ given by
$$
\large \mathbb{R^+} \to \mathbb{R} \quad x \mapsto \ln \left (\frac{x}{\sqrt{x^2+1}} \right)
$$
I had to find $f'$ and $f''$.
Below, I have calculated them.
But, isn't there a better and more convenient way to do t... | The expression $$\frac{x}{\sqrt{x^2 + 1}}$$
is the sine of the angle adjacent to $1$, in a right triangle with legs of $1$ and $x$ and hypotenuse $\sqrt{x^2 + 1}$.
Therefore, it is equal to
$$
\sin \arctan x
$$
and we have
$$
\bbox[5px,border:2px solid blue]{f(x) = \ln (\sin (\arctan x)).}
$$
Computing $f'(x)$ for thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1933460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Prove the inequality $\frac a{a+b^2}+\frac b{b+c^2}+\frac c{c+a^2}\le\frac14\left(\frac1a+\frac1b+\frac1c\right)$ Let $a,b,c>0; a+b+c=1$. Prove the inequality
$$\frac a{a+b^2}+\frac b{b+c^2}+\frac c{c+a^2}\le\frac14\left(\frac1a+\frac1b+\frac1c\right)$$
My work so far:
I tried AM-GM and used fact $a+b+c=1$.
| $$\frac{a}{a+b^2}= \frac{a}{a^2+ab+ac+b^2} = \frac{a}{(a^2+ac)+(ab+b^2)} \le \frac{a}{4} \cdot \left( \frac{1}{a^2+ac}+\frac{1}{ab+b^2} \right)=$$
$$= \frac{1}{4}\cdot \left( \frac{1}{a+c}+\frac{1}{b}-\frac{1}{a+b} \right)$$
$$\frac{a}{a+b^2}+\frac{b}{b+c^2}+\frac{c}{c+a^2} \le \frac{1}{4}\cdot \left( \frac{1}{a}+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1934520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Determine $ax^4 + by^4$ for system of equations I found the following recreational problem without further specification for $a,b$.
Let $x,y$ be real numbers s.t.
$a + b = 6$,
$ax + by = 10$,
$ax^2 + by^2 = 24$,
$ax^3 + by^3 = 62$.
Determine $ax^4 + by^4$.
I am new to problem solving exercises like this and therefo... | This is not a nice solution.
Considering the equations $$a+b-6=0\tag 1$$ $$ax+by-10=0\tag 2$$ $$ax^2+by^2-24=0\tag 3$$ $$ax^3+by^3-62=0\tag 4$$ Let us eliminate the variables one at the time and express them as a function of $a$.
From $(1)$, $b=6-a$. Replacing in $(2)$ and assuming $a\neq6$, we have $$a x+(6-a) y-10=0 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1936350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
$a_1=3;\ a_{n+1}=3^{a_n}$; find $a_{2004}\bmod100$ A detailed solution will be helpful. Given that $a_1=3$ and $a_{n+1}=3^{a_n}$, find the remainder when $a_{2004}$ is divided by 100.
| By the CRT, in order to find $a_{2004}\pmod{100}$ it is enough to find $a_{2004}\pmod{8}$ and $a_{2004}\pmod{25}$. Now $a_n=3^{a_{n-1}}\pmod{8}$ just depends on $a_{n-1}\pmod{4}$, and $a_{n-1}=3^{a_{n-2}}\pmod{4}$ just depends on $a_{n-2}$ being even or odd. Since $a_{2002}$ is clearly odd, $a_{2003}\equiv 3\pmod{4}$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1938353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\sum_{k=0}^nk{m+k \choose m}=n{m+n+1\choose m+1}-{m+n+1 \choose m+2}$ Can someone please see the work I have so far for the following proof and provide guidance on my inductive step?
Prove that if $m,n\in\mathbb{N}$, then $\sum_{k=0}^nk{m+k \choose m}=n{m+n+1\choose m+1}-{m+n+1 \choose m+2}$
Base Case. Let ... | Here is a somewhat different approach. We use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series. This way we can write e.g.
\begin{align*}
[x^k](1+x)^n=\binom{n}{k}
\end{align*}
We obtain for $m,n\geq 0$
\begin{align*}
\sum_{k=0}^n&k\binom{m+k}{m}\\
&=\sum_{k=1}^nk[x^m](1+x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1939505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 2
} |
Calculating $\int \frac{x^2 - 2x + 3}{x^4 -x^2 +1}dx$ I was calculating
$$\int \frac{x^7 + 3x^6 - x^4 + 6x^3 -1}{x^4 - x^2 + 1}dx$$
I performed Euclidean division and the integral reduced to
$$\frac{x^4}4 + x^3 + \frac{x^2}2 + 2x + \frac32 \ln(x^4 - x^2 + 1) - J(x)$$
with
$$J(x) = \int \frac{x^2 - 2x +3}{x^4 - x^2 + 1}... | The denominator readily factors into $$x^4 - x^2 + 1 = (x^2 + \sqrt{3}x + 1)(x^2 - \sqrt{3}x + 1),$$ where upon the usual partial fraction decomposition procedure requires us to solve $$x^2-2x+3 = (Ax + B)(x^2 - 3x+1) + (Cx + D)(x^2 + 3x + 1)$$ for suitable coefficients $A, B, C, D$. I trust that this is something you... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1942393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the points on the ellipsoid $x^2 + 2y^2 + 3z^2 = 1$ where the tangent plane is parallel to the plane $3x - y + 3z = 1$. Find the points on the ellipsoid $x^2 + 2y^2 + 3z^2 = 1$, where the tangent plane is parallel to the plane $3x - y + 3z = 1$.
I'm not sure how to go about solving this. I'd appreciate some help.
| Let the contact point be $(X,Y,Z)$, now the tangent plane is
$$Xx+2Yy+3Zz=1$$
Comparing coefficients,
$$X:2Y:3Z=3:-1:3$$
That is $$\frac{X}{3}=\frac{2Y}{-1}=\frac{3Z}{3}=k$$
Now substitute $(X,Y,Z)=\left( 3k,-\dfrac{k}{2},k \right)$ into $x^{2}+2y^{2}+3z^{2}=1$
We have
\begin{align*}
(3k)^{2}+2\left( -\frac{k}{2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1943008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How to prove that $k^n=\frac{d^n}{dx^n} (1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....+\frac{x^n}{n!})^k|_{x=0}$ $$k^n=\frac{d^n}{dx^n} (1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....+\frac{x^n}{n!})^k|_{x=0}$$
It is easy to show $k=1$ and $k=2$
$k=1$
$$\frac{d^n}{dx^n} (1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....+\frac{x^n}{n!})|_{x=0}=1... | Let
$$ s_n = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+…+\frac{x^n}{n!} $$
We can use induction on $n$. $n=1$ is easy to prove. Assume the hypothesis is true for $1 .. (n-1)$.
We have
$$ \frac{d}{dx}s_n = s_{n-1}$$
Now:
$$ \frac{d^n}{dx^n}s_n^k = \frac{d^{n-1}}{dx^{n-1}}(\frac{d}{dx}s_n^k) $$
And, using the chain rule:
$$ \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1943301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Elegant way to prove that this vector set is linear independent. I'm trying to find an elegant way to prove that this vector set $$\begin{bmatrix}1\\ 1\\1\\0\end{bmatrix}, ~~\begin{bmatrix}1\\ 1\\0\\1\end{bmatrix},~~\begin{bmatrix}1\\ 0\\1\\1\end{bmatrix},~~\begin{bmatrix}0\\ 1\\1\\1\end{bmatrix}$$ is linearly independ... | You can observe that, if
$$
M=\begin{bmatrix}
1 & 1 & 1 & 0 \\
1 & 1 & 0 & 1 \\
1 & 0 & 1 & 1 \\
0 & 1 & 1 & 1
\end{bmatrix}
$$
then
$$
M\begin{bmatrix}1\\1\\1\\1\end{bmatrix}=
3\begin{bmatrix}1\\1\\1\\1\end{bmatrix},
\qquad
M\begin{bmatrix}1\\-1\\-1\\1\end{bmatrix}=
-\begin{bmatrix}1\\-1\\-1\\1\end{bmatrix},
\qquad
M\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1946491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Integration by completing the square I need to complete the square on the following integral. Once this is done apparently I will be able to use on of the integration tables in the back of my book.
$\int x \sqrt{x^2 + 6x +3} dx $
This is what I have come up with so far:
$\int x \sqrt{(x+3)^2 -6} $
I really am at a l... | Substitute $u=x+3 \implies x=u-3, dx=du$:$$\int x\sqrt{(x+3)^2-6}dx=\int (u-3)\sqrt{u^2-6} du$$Integrating the integral gives:
$$\int u\sqrt{u^2-6}du-\int 3\sqrt{u^2-6} du=\frac{(u^2-6)^{\frac{3}{2}}}{3}-3\left(\frac{u\sqrt{u^2-6}}{2}-3\ln\left(u+\sqrt{u^2-6}\right)\right)$$In terms of $x$:$$\int x\sqrt{(x+3)^2-6}dx=$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1947649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to prove that $|z^2| = |z|^2$ where $z = a+bi$? I just started my topic on complex numbers and I'm stuck on this question.
What I have managed to get (I go wrong here, I don't know where though):
$z^2 = (a+bi)^2 = a^2 + b^2$, so $|z^2| = \sqrt{(a^2 + b^2)^2 + 0^2} = \sqrt{a^4+2a^2b^2+b^4}$
and
$|z| = a^2+b^2$ so $|... | $$z^2 = (a+bi)^2 = a^2 - b^2 +2iab $$
$$|z^2| = \sqrt{(a^2 - b^2)^2 + (2ab)^2} = \sqrt{a^4+2a^2b^2+b^4}=\sqrt{(a^2+b^2)^2}$$
and
$$|z| = \sqrt {a^2+b^2}$$
$$\implies |z|^2 = (\sqrt{a^2+b^2})^2 = a^2+b^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1949316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
How do I prove that $f(n) = 0$ for all $n$ where $n$ is a positive integer? In particular, $f$ is defined such that $f(mn)=f(m)+f(n)$, where $m$ and $n$ are positive integers. Also $f(n)\ge 0$ for all $n$, $f(10)=0$, and $f(n)=0$ if $n$ ends in a $3$. Thank you.
| *
*If $n\equiv 1\pmod{10}$ then $3n\equiv 3\pmod{10}$ and so $f(n)=f(3n)-f(3)=0$.
*If $n\equiv 7\pmod{10}$ then $3n\equiv 1\pmod{10}$ and so $f(n)=f(3n)-f(3)=0$.
*If $n\equiv 9\pmod{10}$ then $3n\equiv 7\pmod{10}$ and so $f(n)=f(3n)-f(3)=0$.
From $f(2)\ge 0$ and $f(5)\ge 0$ and $f(2)+f(5)=f(10)=0$, we obtain $f(2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1950831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to formally prove that we cannot find a polynomial in $\textbf Z[x]$ with degree $2$ with such a root? I am trying to find the kernel of the map from $\textbf Z[x]$ to $\textbf C$. The map is evaluating at $\sqrt 2 + \sqrt 3$.
A solution says that we cannot find polynomials of degree $2$ or $3$ that has such a root... | Proceed by contradiction.
Assume that there exist integer $a,b,c$ such that
$$ax^2+bx+c=0$$
with $x=\sqrt{2}+\sqrt{3}$.
$$a(5+2\sqrt{6})+b(\sqrt{2}+\sqrt{3})+c=0$$
Note that $1,\sqrt{2},\sqrt{3},$ and $\sqrt{6}$ are linearly independent, which means that there exist no integers $n_1,n_2,n_3,n_4$, not all $0$, such tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1952831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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Prove that $\binom{2n}{n}>\frac{4n}{n+1}\forall \; n\geq 2, n\in \mathbb{N}$
Prove that $$\binom{2n}{n}>\frac{4n}{n+1}\forall \; n\geq 2, n\in \mathbb{N}$$
$\bf{My\; Try::}$ Using $$(1+x)^n = \binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+........+\binom{n}{n}x^n$$
and $$(x+1)^n = \binom{n}{0}x^n+\binom{n}{1}x^{n-1}+\bin... | $$\binom{2n}n=\frac{n+1}{1}\frac{n+2}{2}\cdots\frac{2n-1}{n-1}\frac{2n}{n}\ge2^n>\frac{4n}{n+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1954608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
If $a$, $b$ and $c$ are three positive real numbers, prove that $\frac{a^2+1}{b+c}+\frac{b^2+1}{c+a}+\frac{c^2+1}{a+b} ≥ 3$ If $a$, $b$ and $c$ are three positive real numbers, prove that $$\frac{a^2+1}{b+c}+\frac{b^2+1}{c+a}+\frac{c^2+1}{a+b} ≥ 3$$
I think we are supposed to use the AM-GM inequality here; or maybe thi... | After using AM-GM we need to prove that
$$(a^2+1)(b^2+1)(c^2+1)\geq(a+b)(a+c)(b+c)$$
which follows from C-S: $(a^2+1)(1+b^2)\geq(a+b)^2$.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1955141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Forming basis in Linear Algebra Looking for some help with the following question. For which value or values of $k$ do the vectors below form a basis of $ \Bbb R^4$.
$\begin{bmatrix}
1&0&0&2
\\0&1&0&3
\\0&0&1&4
\\2&3&4&k
\end{bmatrix}$
My thinking is that the columns of this matrix are linearly independent if and... | Hint:
$$ \begin{vmatrix}
1&0&0&2
\\0&1&0&3
\\0&0&1&4
\\2&3&4&k
\end{vmatrix} = \begin{vmatrix}
1&0&0&2
\\0&1&0&3
\\0&0&1&4
\\0&3&4&k-2(2)
\end{vmatrix} = \begin{vmatrix}
1&0&0&2
\\0&1&0&3
\\0&0&1&4
\\0&0&4&k-2(2)-3(3)
\end{vmatrix}$$
Are you able to finish the working?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1957040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Integrate $\int \frac{\sqrt{x^2-8x+5}}{x}dx$ I've been trying this one for days and I can't seem to get it.
I tried:
$$\int \frac{\sqrt{x^2-8x+5}}{x}dx = \int \frac{\sqrt{\left(x-4\right)^2-11}}{x}dx$$
and set
$$x-4 = \sqrt{11}\sec\theta $$
so
$$dx=\sqrt{11}\sec\theta \tan\theta d\theta $$
and I had
$$\int \frac{\sqrt{... | Another possibility is to do an Euler substitution.
Let $$y=\sqrt{x^2-8x+5}-x$$
Then $$(y+x)^2=x^2-8x+5$$ and one solves to get
$$x=\frac{5-y^2}{2(y+4)}$$
That implies that
$$\sqrt{x^2-8x+5}=y+x=y+\frac{5-y^2}{2(y+4)}$$
Thus the integral reduces to a rational function. The denominator is not that bad, I get $(y+4)^2(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1958447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solve $x=\sqrt{x-x^{-1}}+\sqrt{1-x^{-1}}$ Find all real numbers of $x$ such that $$x=\left(x-\frac 1x\right)^{\frac 12}+\left(1-\frac 1x\right)^{\frac 12}$$
My Attempt: Square both sides to get$$x^2=x-\frac 1x+1-\frac 1x+2\sqrt{\left(x-\frac 1x\right)\left(1-\frac 1x\right)}$$
Then move all the terms to the left side e... | Multiply the original equation
$$
(x-x^{-1})^{1/2}+(1-x^{-1})^{1/2}=x\tag{1}
$$
by $(x-x^{-1})^{1/2}-(1-x^{-1})^{1/2}$ to get
$$
x-1=x((x-x^{-1})^{1/2}-(1-x^{-1})^{1/2})
$$
That is
$$
(x-x^{-1})^{1/2}-(1-x^{-1})^{1/2}=1-x^{-1}\tag{2}
$$
Denote $a=(x-x^{-1})^{1/2}$ and $b=(1-x^{-1})^{1/2}$. Then $(1)$ and $(2)$ can be ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1959363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Divisibility Proof $8\mid (x^2 - y^2)$ for $x$ and $y$ odd $x,y \in\Bbb Z$. Prove that if $x$ and $y$ are both odd, then $8\mid (x^2 - y^2)$.
My Proof Starts:
Assume $x$ and $y$ are both odd. So, $x = 2k + 1$ and $y = 2l +1$ for some integers $k$ and $l$. Thus,
\begin{align}
x^2 - y^2 &= (2k + 1)^2 - (2l + 1)^2 \\
&= 4... | if $K$and $l$ are even, then $K=2K_1$ and $l=2K_2$
then $4K^2+4K-4l^2-4l=16K_1^2+8K_1-16K_2^2-8K_2$ which is clearly divisble by 8
now if $K$and $l$ are odd, then $K=2K_1+1$ and $l=2K_2+1$
then $4K^2+4K-4l^2-4l=4(4K_1^2+4K_1+1)+8K_1+4 -4(4K_2^2+4K_2+1)-8K_2+4= 16K_1^2+16K_1+8K_1+8-16K_2^2-16K_2-8K_2-8$ which is clearly... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1959713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Systematic way to find the answers of such question Q.What is the largest perfect square that divides:
$2014^3-2013^3+2012^3-2011^3+\ldots+2^3-1^3$.
My efforts:
$2014^3-2013^3=(2014-2013)(2014^2+2013^2+2014 \cdot2013)=2014^2+2013^2+2014*2013$
lly,
$2014^3-2013^3+2012^3-2011^3+\ldots+2^3-1^3=(2014^2+2013^2+2012^2+2011^... | $$ \sum_{k=1}^n \left((2k)^3 - (2k-1)^3\right) = \sum_{k=1}^n \left(12 k^2 - 6 k + 1\right) = (4n+3) n^2$$
In this case $n=1007$ and $4n+3 = 29 \times 139$ which is coprime to $n$ and squarefree. Thus the answer is $n^2 = 1007^2 = 1014049$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1960359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\arccos(\cos\frac{15\pi }{4})$ I have a problem with understanding of this exercise:
$\arccos(\cos\frac{15\pi }{4})= ?$
$\cos(\frac{15\pi}{4}-2\pi)=\cos(\frac{7\pi}{4})$
$\cos(\frac{7\pi}{4}+\pi)= -\cos(\frac{3\pi}{4})$
then $\arccos(-\cos\frac{3\pi}{4})$
All above I understand pretty clearly. We did it at ... | Here's one way to do it
$$\arccos\left(\cos\left(\frac{15}{4}\pi\right)\right)$$
$$=\arccos\left(\cos\left(\pi+2\pi+\frac{3}{4}\pi\right)\right)$$
$$=\arccos\left(\cos\left(\pi+\frac{3}{4}\pi\right)\right)$$
$$=\arccos\left(-\cos\left(\frac{3}{4}\pi\right)\right)$$
$$=\pi-\arccos\left(\cos\left(\frac{3}{4}\pi\right)\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1961155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find the limit of the following expression: $$\lim_{n\to\infty}\frac {1-\frac {1}{2} + \frac {1}{3} -\frac {1}{4}+ ... + \frac {1}{2n-1}-\frac{1}{2n}}{\frac {1}{n+1} + \frac {1}{n+2} + \frac {1}{n+3} + ... + \frac {1}{2n}}$$
I can express the value of the geometric sum of ${\frac {1}{2} + \frac {1}{4}+...+\frac {1}{2n}... | In the denominator we have:
$$\sum_{i=1}^{n} \frac{1}{n+i}$$
And as $n \to \infty$ this is:
$$\lim_{n \to \infty} \sum_{i=1}^{n} \frac{1}{n+i}=\lim_{n \to \infty} \sum_{i=1}^{n} \frac{1}{1+(0+i \frac{1-0}{n})}\frac{1-0}{n}=\int_{0}^{1} \frac{1}{1+x} dx=\ln (1+1)-\ln(1+0)=\ln (2)$$
For the numerator as $n \to \infty$, c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1961724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
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Solving $x+x^3=5$ without using the cubic equation. In lessons, I get quite bored and recently throughout these lessons I have been trying to solve for x in:
$$x+x^3=5$$
I've figured out how to do it for squares using the quadratic equation, but the cubic equation looks so dauntingly massive it actually makes my bladde... | A trick for solving this equation goes back to Viète (but is not much different from Tartaglia's and Cardano's method).
Set $x=y-\dfrac{1}{3y}$. Then
$$
x^3+x=y^3-y+\frac{1}{3y}-\frac{1}{27y^3}+y-\frac{1}{3y}=
y^3-\frac{1}{27y^3}
$$
so your equation becomes, after setting $z=y^3$,
$$
z-\frac{1}{27z}=5
$$
This becomes a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1964176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
How do I find an analytical solution of $x x' + \frac 3 2 (x')^2 + A \frac {x'} x = W$? I am trying to solve the Rayleigh-Plesset equation of a bubble made of super-critical fluid, and its radius varies as a function of time. A generalised form of the equation I got is:
$$x x' + \frac 3 2 (x')^2 + A \frac {x'} x = W .$... | $$x x' + \frac 3 2 (x')^2 + A \frac {x'} x = W .$$
We observe that $x(t)$ is an even function. This draw to the change of function :
$$X=x^2 \quad\to\quad dX=2xdx \quad\to\quad x'=\frac{dx}{dt}=\frac{dx}{dX}\frac{dX}{dt}=\frac{1}{2x}\frac{dX}{dt}$$
$$x \left(\frac{1}{2x}\frac{dX}{dt} \right) + \frac 3 2 \left(\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1965098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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if $abc=1$,show that $a^3+b^3+c^3+\frac{256}{(a+1)(b+1)(c+1)}\ge 35$
Let $a,b,c>0,abc=1$ show that
$$a^3+b^3+c^3+\dfrac{256}{(a+1)(b+1)(c+1)}\ge 35\tag{1}$$
iff $a=b=c=1$
I know use AM-GM inequality
$$a^3+b^3+c^3\ge 3abc=3$$
and
$$(a+1)(b+1)(c+1)\ge 2\sqrt{a}\cdot 2\sqrt{b}\cdot 2\sqrt{c}=8$$
In this way, will ... | Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that $f(u)\geq0$, where
$$f(u)=27u^3-27uv^2+\frac{256w^5}{2w^2+3uw+3v^2}-32w^3$$
But by AM-GM $f'(u)=81u^2-27v^2-\frac{768w^6}{(2w^2+3uw+3v^2)^2}\geq81u^2-27v^2-12w^2>0$,
which says that $f$ is an increasing function.
Thus, it's enough to prove our... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1966306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Find minimal value $ \sqrt {{x}^{2}-5\,x+25}+\sqrt {{x}^{2}-12\,\sqrt {3}x+144}$ without derivatives. Find minimal value of $ \sqrt {{x}^{2}-5\,x+25}+\sqrt {{x}^{2}-12\,\sqrt {3}x+144}$ without using the derivatives and without the formula for the distance between two points.
By using the derivatives I have found ... | Note that the expression in question, $f(x) = \sqrt {{x}^{2}-5\,x+25}+\sqrt {{x}^{2}-12\,\sqrt {3}x+144}$, can be written as follows:
$$
f(x) = \sqrt {(a(x))^2 + (b(x))^2}+\sqrt {(a(x))^2 + (13 - b(x))^2}
$$
where
$$a(x) =\frac{x (- 5\sqrt3 - 12) + 120}{26} $$
$$b(x) =\frac{x (12\sqrt3 -5) + 50}{26} $$
Now from $(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1967129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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A better way to evaluate a certain determinant
Question Statement:-
Evaluate the determinant:
$$\begin{vmatrix}
1^2 & 2^2 & 3^2 \\
2^2 & 3^2 & 4^2 \\
3^2 & 4^2 & 5^2 \\
\end{vmatrix}$$
My Solution:-
$$
\begin{align}
\begin{vmatrix}
1^2 & 2^2 & 3^2 \\
2^2 & 3^2 & 4^2 \\
3^2 & 4^2 & 5^2 \\
\end{vmatrix} &=
(1^2\t... | If we consider
$$ f(x)=\det\begin{pmatrix}x^2 & (x+1)^2 & (x+2)^2 \\ (x+1)^2 & (x+2)^2 & (x+3)^2 \\ (x+2)^2 & (x+3)^2 & (x+4)^2 \end{pmatrix}$$
we have:
$$\scriptstyle f'(x) = \det\begin{pmatrix}2x & (x+1)^2 & (x+2)^2 \\ 2x+2 & (x+2)^2 & (x+3)^2 \\ 2x+4 & (x+3)^2 & (x+4)^2 \end{pmatrix}+\det\begin{pmatrix}x^2 & 2x+2 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1969290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 1
} |
Solving for the roots of a sextic I was solving the equation $3(1+x^2+x^3)^2=(2+x)^4$ for $x$, and after expanding it out, I got $$3x^6+6x^5+2x^4-2x^3-18x^2-32x-13=0\tag{1}$$
which should be solvable because it has a Galois group of order $72$. But since it's a degree six, I have no method for solving this.
I have atte... | From $3(1+x^2+x^3)^2=(2+x)^4$, we can move all terms to one side and use $a^2-b^2=(a+b)(a-b)$ to get\begin{align*} & (\sqrt3(1+x^2+x^3)-(2+x)^2)(\sqrt3(1+x^2+x^3)+(2+x)^2)=0\tag1\\ & (g\sqrt3+(\sqrt3-1)g^2-4g+\sqrt3-4)(g\sqrt3+(\sqrt3+1)g^2+4g+\sqrt3+4)=0\tag2\end{align*}
And solving for the roots of the cubic gives al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1971255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Basis for the annihilator Let the field $F$ be given by $F=\mathbb{Z}/5\mathbb{Z}$ and let $V=F^3$. Let $S$ be a subspace of $V$ spanned by $v_1=(1,2,3),v_2=(2,1,2)$. I'm now asked to find a basis for $S^0.$
I've tried solving this and I get $(-1/3,-4/3,1)$. Is it right that the $S^0$ has dimension $1$?
| Since $(v_1,v_2)$ are linearly independent vectors, we have $\dim S = 2$ and indeed
$$ \dim S^{0} = \dim V - \dim S = 3 - 2 = 1. $$
To check your answer, note that we indeed have
$$ \left( -\frac{1}{3}, -\frac{4}{3}, 1 \right) \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = -\frac{1}{3} - \frac{8}{3} + 3 = 0, \\
\left( -\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1971525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solve the differential equation $\frac{dy}{dx} + 3yx = 0$ for the values $x = 0$ when $y = 1$ - Solution Review
Solve the differential equation $\frac{dy}{dx} + 3yx = 0$; $x = 0$ when $y = 1$.
I solved this DE using the integration factor method. However, online calculators are giving me a different answer, where the... | For the Initial Value Problem (IVP) $y′+3xy=0, ~y(0)=1$ (which you are most probably trying to solve) you are correct in all your steps above, except the last part: substitute $x=0, ~y=1$ and get $C=−1$, so the solution to the IVP is $y(x)=e^{−3x^2/2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1971730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Proof by induction that $4^3+4^4+4^5+⋅⋅⋅+4^n = \frac{4(4^n-16)}{3}$ I am stuck on this problem for my discrete math class.
Prove the equation by induction for all integers greater than or equal to $3$:
$$4^3+4^4+4^5+⋅⋅⋅+4^n = \frac{4(4^n-16)}{3}.$$
I know that base case $n=3$:
$4^3=64$ as well as $4(4^3-16)/3 = 64$
M... | Hint:
$$\color{red}{4^3+\cdots 4^n}+4^{n+1}=\color{red}{\frac{4(4^n-16)}{3}}+4\cdot4^{n}=\left(\frac43+4\right)4^n-4\cdot\frac {16}3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1972847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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What is the sum of imaginary roots of this equation? What is the sum of imaginary roots of equation:
$$x^3+3x^2+3x+3$$
Here is my attempt:
$$x^3+3x^2+3x+3=(x^3+3x^2+3x+1)+2=(x+1)^3+2$$
Since $f(x')=0$ if x' is a root of $f(x)$
$$(x'+1)^3+2=0$$
$$(x'+1)^3=-2$$
$$x'=-1+\sqrt[3]{-2}$$
$$x'=-1+\sqrt[3]{2}\sqrt[3]{-1}$$....... | So, the values of $x+1$ are $$\sqrt[3]2, \sqrt[3]2w,\sqrt[3]2w^2$$ where $w$ is a complex cube root of unity
So, the only real root of the given equation is $$-1+\sqrt[3]2$$
Now the sum of all roots is $$-\dfrac31$$
| {
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Computing image under coordinate map How do I compute the image?
Compute the image of (1,2,3,4) under the coordinate map
$$ \begin{bmatrix}t^3&t^3 + t^2&t^3+t^2+t&t^3+t^2+t+1\end{bmatrix}: \Bbb F^4\to P_3.$$
| I assume they expect you to do
\begin{align}
\begin{bmatrix}t^3&t^3 + t^2&t^3+t^2+t&t^3+t^2+t+1\end{bmatrix}\begin{bmatrix}1\\2\\3\\4\end{bmatrix}
&=t^3+2(t^3+t^2)+3(t^3+t^2+t)+4(t^3+t^2+t+1)
\end{align}
| {
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Why doesn't the derivative of integration by trig substitution match the original function? For one of my assignments in Calc II, I had to solve ${\displaystyle\int} \frac{\sqrt{x^2-4}}{x} \text{ d}x$. By using Trig substitution where $x = 2 \text{ sec } \theta$, my solution was $\sqrt{x^2-4} - 2\text{ arcsec}\left(\fr... | The trouble is when getting rid of the square root. In general we have $\sqrt{x^2} = |x|$, so
$$\sqrt{\sec^2\theta-1} = \sqrt{\tan^2\theta} = \lvert\tan\theta\rvert.$$
So the integrand is really
$$\lvert\tan\theta\rvert \tan \theta = \begin{cases}
\sec^2\theta - 1 & 0\le\theta<\pi/2,\\
-(\sec^2\theta - 1) & \pi/2<\thet... | {
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Prove that: $\int_{0}^{1}\frac{dx}{\sqrt{1-x^4}}=\sum_{0}^{+\infty}\frac{C_{2n}^{n}}{4^n(4n+1)}$ Prove that: $\int_{0}^{1}\frac{dx}{\sqrt{1-x^4}}=\sum_{0}^{+\infty}\frac{C_{2n}^{n}}{4^n(4n+1)}$
Could you give me some hint? Thank for helping.
| Using
$$\frac{1}{\sqrt{1-x^2}} = \sum_{n=0}^{\infty} \binom{2n}{n} \, \frac{x^{2n}}{4^{n}}$$
then
\begin{align}
\int \frac{dx}{\sqrt{1-x^4}} &= \int \left( \sum_{n=0}^{\infty} \binom{2n}{n} \, \frac{x^{4n}}{4^{n}} \right) \, dx \\
&= \sum_{n=0}^{\infty} \binom{2n}{n} \, \frac{x^{4n+1}}{4^{n} \, (4n+1)}
\end{align}
Wit... | {
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value of $a$ for which $25^x+(a+2)5^x-(a+3)<0$for at least one real $x$
Find the values of $a$ for which the inequality is satisfied for $25^x+(a+2)5^x-(a+3)<0$
for at least one real value of $x$
$\bf{My\; Try::}$ We can write it as $a(5^x-1)<-\left[25^x+2\cdot 5^x-3\right]$
So $$a < -\left(\frac{25^x-5^x+3\cdot 5^x-... | Let $$(5^x)^2+(a+2)5^x-(a+3)=-y$$
As $5^x$ is real, we need the discriminant $$(a+2)^2-4(y-a-3)\ge0$$
$$4y\le a^2+8a+16=(a+4)^2$$
Now we need $y>0$ which will be true if $(a+4)^2>0$ which is true for $a+4\ne0\iff a\ne-4$
| {
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Find the sum $\sum_{n=2}^{\infty} \frac{\binom n2}{4^n} ~~=~~ \frac{\binom 22}{16}+\frac{\binom 32}{64}+\frac{\binom 42}{256}+\cdots$. The sum
$$\sum_{n=2}^{\infty} \frac{\binom n2}{4^n} ~~=~~ \frac{\binom 22}{16}+\frac{\binom 32}{64}+\frac{\binom 42}{256}+\cdots$$
has a finite value. Use what you know about generating... | Another approach is to use an arithmetico-geometric sequence, although this isn't using generating functions. To use generating functions, see the accepted answer.
Call the sum $S$, so $$S = \frac{1}{16}+\frac{3}{64}+\frac{6}{256}+\frac{10}{1024}+\dots$$
Divide by $4$, so
$$\frac{1}{4}S = \quad \frac{1}{64}+\frac{3}{2... | {
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Polynomial of degree 5 such that $P(x)-1$ is divisible by $(x-1)^3$ and $P(x)$ is divisible by $x^3$
$P(x)$ is a polynomial of degree 5 such that $P(x)-1$ is divisible by $(x-1)^3$ and $P(x)$ is divisible by $x^3$. Find $P(x)$.
No idea where to start, would $P(x)$ be of the form $x^3(Ax^2+Bx+C)$?
| Expanding $(x-1)^3$ we find that
$$P(x)=(Ax^2+Bx+C)(x^3-3x^2+3x-1)+1$$
$$=Ax^5+(B-3A)x^4+(C-3B+3A)x^3+(-3C+3B-A)x^2+(3C-B)x-C+1$$
Yet we also know that $P(x)$ is divisible by $x^3$ and is of the form
$$Dx^5+Ex^4+Fx^3$$
Comparing constant, linear and quadratic coefficients we have
$$1-C=0$$
$$3C-B=0$$
$$-3C+3B-A=0$$
and... | {
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Sequence of fractions that converges to $\sqrt{n}$ Begin with any two positive integers $a,b$. Let
$$s_1 = \frac{a}{b}$$
and recursively we define $s_{i+1}$ from $s_i$ as follows: If ${s_i}$ is the fraction $\frac{a'}{b'}$, then set
$$s_{i+1} = \frac{a'+2b'}{a'+b'}$$
(Notice that the actual value of the $s_{i+1}$ does ... | Let $t_i$ denote the top part of the fraction and $\ell_i$ the bottom part. Then $t_0 = a$, $\ell_0 = b$ and
$$
\begin{pmatrix} t_{i+1} \\ \ell_{i+1} \end{pmatrix} = \begin{pmatrix} 1 & n \\ 1 & 1 \end{pmatrix}\begin{pmatrix} t_{i} \\ \ell_{i} \end{pmatrix}.
$$
You can then diagonalize the middle matrix
$$
\begin{pmatr... | {
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Prove: $ \frac{1}{\sin 2x} + \frac{1}{\sin 4x } + \cdots + \frac{1 }{\sin 2^n x} = \cot x - \cot 2^n x $
Prove
$$ \frac{1}{\sin 2x} + \frac{1}{\sin 4x } + \cdots + \frac{1 }{\sin 2^n x} = \cot x - \cot 2^n x $$
where $n \in \mathbb{N}$ and $x$ not a multiple of $\frac{ \pi }{2^k} $ for any $k \in \mathbb{N}$.
My... | Skipping over the base case ($n=1$) for the moment, let's do the inductive step: If we take the equation
$${1\over\sin2x}+{1\over\sin4x}+\cdots+{1\over\sin2^nx}=\cot x-\cot2^nx$$
and replace $x$ with $2x$, we get
$${1\over\sin4x}+{1\over\sin8x}+\cdots+{1\over\sin2^{n+1}x}=\cot2x-\cot2^{n+1}x$$
Adding $1/\sin2x$ to bot... | {
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Finding limit involving a recursively defined sequence I have the following recursive sequence:
$$
S(n) = 1 + S(n-1) + \frac{2S(n-2)}n\\S(1) = 1\\S(0) = 0
$$
I've written a program that computes elements of $S(n)$ and seems to imply that:
$$
\lim_{n\to\infty} \frac{S(n) - S(n-1)}n = \frac{1 - e^{-2}}2
$$
I don't believ... | Hint: You could try the following. Multiply the equation with $x^n$ and sum over $n=2$ to $n= \infty$.
$$\sum_{n=2}^{\infty}S(n)x^n=\sum_{n=2}^{\infty}x^n+\sum_{n=2}^{\infty}S(n-1)x^n+2\sum_{n=2}^{\infty}\frac{S(n-2)}{n}x^n$$
Now, switch the index for the second and third sum on the right hand side to get:
$$\sum_{n=2}... | {
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Determine using the comparison test whether the series $\sum_{n=1}^\infty \frac{1}{\sqrt[3]{n^2 + 1}}$ diverges How can I determine whether or not the following series converges using the comparison test?
$$\sum_{n=1}^\infty \frac{1}{(n^2 + 1)^\frac{1}{3}}$$
As $n$ goes to infinity, the sum is roughly equal to $\sum_{... | By the limit comparison test, we have
$$(\frac{1}{n^2+1})^\frac{1}{3} \sim \frac{1}{n^\frac{2}{3}}\;\; (n\to +\infty)$$
$\frac{2}{3}<1$ thus
$\;\;\;\;\;\sum (\frac{1}{n^2+1})^\frac{1}{3}$ diverges.
| {
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Prove equality $\sqrt{1+x^2}\cdot\ln (x+\sqrt{1+x^2}) = x+\frac{x^3}{3} - \frac{2}{3}\frac{x^5}{5}+ \frac{2}{3}\frac{4}{5}\frac{x^7}{7}-...$ $\sqrt{1+x^2}\cdot\ln (x+\sqrt{1+x^2}) = x+\frac{x^3}{3} - \frac{2}{3}\frac{x^5}{5}+ \frac{2}{3}\frac{4}{5}\frac{x^7}{7}-...$
I stuck with performing right part. Where are these c... | We are looking for the Taylor series of
$$f(x)=\sqrt{1+x^2}\int\frac{dx}{\sqrt{1+x^2}}\tag{1}$$
that is clearly related with the Taylor series of
$$ g(x) = \sqrt{1-x^2}\int\frac{dx}{\sqrt{1-x^2}} = \sqrt{1-x^2}\arcsin(x).\tag{2}$$
We may now invoke some heavy artillery. For instance, the following identity:
$$ \arcsin... | {
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Basic inequality problem For $a,b,m,n\in \mathbb{R}$, such that $m^2n^2>a^2m^2+b^2n^2$, what is the relationship between $M=\sqrt{m^2+n^2}$ and $N=a+b$?
I deduced the following: $$m^2n^2>a^2m^2+b^2n^2\geq 2abmn$$, $m,n$ is nonzero from the given inequality, so $mn\geq 2ab$. However, I fail to draw any conclusion betwee... | Geometrically, $(a,b)$ lies on the $(x,y)$-plane inside the ellipse
$$
\frac{x^2}{n^2} + \frac{y^2}{m^2}=1.\tag1
$$
Since $a+b$ is constant on lines with slope $-1$, the value of $a+b$ is no greater than $A+B$, where $(A,B)$ is the point in the first quadrant on the ellipse where the slope of the tangent line is $-1$. ... | {
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"timestamp": "2023-03-29T00:00:00",
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10-permutations with exactly 4 inversions Let $p = p_1p_2\dots p_n$ A pair of elements $(p_i,p_j)$ is called an inversion in p if i > j and $p_i < p_j $.
Now, I know that the generating function for 10-permutations with k inversions is $I_{10}=(1+x)(1+x+x^2)\dots (1+x+\dots +x^9) $ where the coefficient of $x^4$ is the... | As pointed out by @Jack D'Aurizio, it's not that hard to compute the coefficient of $x^4$ in the generating function.
We have
$$\begin{aligned}I_{10} &= (1+x)(1+x+x^2)(1+x+x^2+x^3) \cdots (1+x+\dots +x^9) \\
&= \frac{1-x^2}{1-x} \cdot \frac{1-x^3}{1-x} \cdot \frac{1-x^4}{1-x} \cdots \frac{1-x^{10}}{1-x} \\
&= (1-x^2)(1... | {
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Coefficient of a term using binomial theorem I was just wondering how would I find the coefficient of any term let's say $x^3$ in the expansion of $(x^2+2x+2)^{10}$ using binomial expansion or any other technique. Please let me know if this can be found directly using a shortcut if any.
| $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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\newcommand{\ic}{\... | {
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Show that $5\cdot10^n+10^{n-1}+3$ is divisible by 9 Prove by induction the following:
$5*10^n+10^{n-1}+3$ is divisible by 9
Base case: $n=1$ $5*10+10^{1-1}+3=5*10+10^0+3=50+1+3=54$
$9|54=6$
Inductive Hypothesis: If $k$ is a natural number such that $9|5*10^k+10^{k-1} +3$
Inductive step:
Show that $S_k$ is true $\Rig... | Hint:$$5\cdot10^{n+1}+10^{n} +3=10\cdot5\cdot10^{n}+10\cdot10^{n-1}+30-27=$$
$$=10(5\cdot10^{n}+10^{n-1} +3)-3\cdot9$$
| {
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Deriving Taylor series without applying Taylor's theorem. First, a neat little 'proof' of the Taylor series of $e^x$.
Start by proving with L'Hospital's rule or similar that
$$e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n$$
and then binomial expand into
$$e^x=\lim_{n\to\infty}1+x+\frac{n-1}n\frac{x^2}2+\dots+\frac{(n-... | The series for $\sin$ and $\cos$ can be derived from the expansion of $e^x$.
$$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+\frac{x^6}{720}+\frac{x^7}{5040}+\cdots$$
Sub in $ix$ to get:
$$e^{ix}=1+ix+\frac{(ix)^2}{2}+\frac{(ix)^3}{6}+\frac{(ix)^4}{24}+\frac{(ix)^5}{120}+\frac{(ix)^6}{720}+\frac{(i... | {
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binomial congruence $\sum_{i=1}^{\frac{p-1}{2}}\binom{2i}{i}\equiv 0~or (-2)\pmod p$
Let $p\ge 5$ be a prime number. Show that $$\sum_{i=1}^{\frac{p-1}{2}}\binom{2i}{i}\equiv 0 \text{ or } -2\pmod p .$$
Examples:
If $p=5$, then
$$f=\sum_{i=1}^{\frac{p-1}{2}}\binom{2i}{i}=\binom{2}{1}+\binom{4}{2}=8\equiv -2\pmod 5 .$... | It's more convenient to start the sum at $i=0$ instead of $i=1$,
adding the term ${0 \choose 0} = 1$ and aiming for
$$
\sum_{i=0}^{(p-1)/2} {2i \choose i} \equiv \pm 1 \bmod p.
$$
To prove this, note that ${2i \choose i} \equiv 0 \bmod p$ for
$(p-1)/2 < i \leq p-1$. Therefore
$$
\left(
\sum_{i=0}^{(p-1)/2} {2i \cho... | {
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Does the following series converges or diverges? $\sum_{n=1}^{+\infty}\left[n \log\left ( \frac{2n+1}{2n-1} \right )-1\right]$ ,converges or not
My attempt
$a_n=n \log\left ( \frac{2n+1}{2n-1} \right )-1 \\ \\
\log \left ( \frac{2n+1}{2n-1} \right )=\log \left ( 1+\frac{2}{2n-1} \right ) \sim \frac{2}{2n-1} \ (n\righta... | $$\log\left(\frac{2n+1}{2n-1}\right) = 2\,\text{arctanh}\left(\frac{1}{2n}\right)=\frac{1}{n}+\frac{1}{12n^3}+O(n^{-5})$$
leads to
$$ n\log\left(\frac{2n+1}{2n-1}\right)-1 = \frac{1}{12n^2}+O(n^{-4}) $$
so the given series is convergent. Your initial approximation was simply not accurate enough.
The exact value of the... | {
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Simultaneous equations I have the question
Solve the simultaneous equations
$$\begin{cases}
3^{x-1} = 9^{2y} \\
8^{x-2} = 4^{1+y}
\end{cases}$$
source image
I know that $x-1=4y$ and $3X-6=2+2y$ .
However when I checked the solutions this should become $6X-16=4Y$ .
How is this?
| From $$\begin{align*} & 3^{x-1}=9^{2y}\\ & 8^{x-2}=4^{y+1}\end{align*}\tag1$$
We can rewrite $9$ as $3^2$, $8$ as $2^3$ and $4$ as $2^2$. Doing so and setting the powers equivalent, we get$$\begin{align*} & x-1=4y\\ & 3x-6=2+2y\end{align*}\tag2$$
To answer your question, look at the second equation. Moving the $2$ to t... | {
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Find a limit without l'Hospital: $\lim_{x\to -\infty}\left(\frac{\ln(2^x + 1)}{\ln(3^x+ 1)}\right)$ Find $\displaystyle \lim_{x\to -\infty}\left(\frac{\ln(2^x + 1)}{\ln(3^x+ 1)}\right)$.
$\displaystyle \lim_{x\to -\infty}\left(\frac{\ln(2^x + 1)}{\ln(3^x+ 1)}\right)$ =
$\displaystyle \lim_{x\to -\infty}\left(\frac{\l... | If we use the well known limit
$$\lim_{X\to 0}\frac{\ln(1+X)}{X}=1$$
with $X=2^x\;$ and $\;X=3^x,\;\;$we find
$$\lim_{x\to-\infty}\left( \frac{2}{3} \right)^x=+\infty$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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locus of orthocentre of given triangle if $\tan B+\tan C = c$
Vertex $A$ of $\triangle ABC$ moves in such a way that $\tan B+\tan C = c(\bf{constant}),$
Then locus of orthocentre of $\triangle ABC$ is ( side $BC$ is fixed)
$\bf{My\; Try::}$ We can write $$\tan B+\tan C = c\Rightarrow \frac{\sin B}{\cos C}+\frac{\sin ... | Suppose $B(-a,0)$ and $C(a,0)$, suppose $A(x_A,y_A)$ (in the first quadrant for easyness), then, the relation $tan(b) + tan(c) =C$ translates into $\frac{y_A}{x_A+a}+\frac{y_A}{a-x_A}=C$. Performing calculations you will arrive to $y_A=\frac{-C}{2a}{x_{A}}^2+\frac{aC}{2}$.
Now observe that $H$ is the intersection of a... | {
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Find the integer $m$ such that $(\frac{1}{5})^m \cdot (\frac{1}{4})^ {24} = \frac{1}{2(10)^{47}}$ I have this problem
$$(\frac{1}{5})^m \cdot (\frac{1}{4})^ {24} = \frac{1}{2(10)^{47}}$$
and I need to find the integer $m$.
I know that I can write $5^m \cdot 4^{24} = 2(10)^{47} \Leftrightarrow 5^m = \frac{2(10)^{47}}{(1... | $$\frac{1}{2(10)^{47}}=\frac{1}{2}\frac{1}{2^{47}}\frac{1}{5^{47}}=\frac{1}{2^{48}}\frac{1}{5^{47}}=\frac{1}{4^{24}}\frac{1}{5^{47}}=\left(\frac{1}{4}\right)^{24}\left(\frac{1}{5}\right)^{47}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Showing $\frac{x+y+z}{\sqrt 2}\le\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{x^2+z^2}$
Showing $\frac{x+y+z}{\sqrt 2}\le\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{x^2+z^2}$ $(x,y,z>0)$
Can I apply Jensen here, sum of the squareroots is greater then squareroot of the sum, so I make the RHS smaller then show also then the inequality ... | Hint: Note that $ \sqrt{2(a^2+b^2)} \geq a+b, ~\forall a,b\in \mathbb{R}$ and use it to derive the desired inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2015163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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A congruence for the product of quadratic residues + the product of quadratic non-residues Let $p \gt 3$ be an odd prime,
$$p \equiv 3 \pmod 4 \iff \prod_{1\le k \le p-1\\ \ \
\left(\frac{k}{p}\right)=1}k \ \ \ \ + \prod_{1\le k \le p-1\\ \
\left(\frac{k}{p}\right)=-1}k \ \ \ \equiv 0 \pmod {p^2}$$
... | Please check the following proposed proof (for $\Rightarrow$) .
We have:
$$ \prod_{1\le k \le p-1\\ \ \
\left(\frac{k}{p}\right)=1 }k \ \ \ \ \equiv \prod_{1\le a \le \frac{p-1}{2}\\ \ \
}a^2 \pmod {p}$$ because it is well known that all the squares $\pmod p$ are obtained by squaring all the $a$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2015830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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probability last draw I have a bag initially containing $r$ red fruit pastilles (my favourites) and $b$ fruit pastilles of other colours. From time to time I shake the bag thoroughly and remove a pastille at random. (It may be assumed that all pastilles have an equal chance of being selected.) If the pastille is red I ... | I believe that an easier method will be by induction.
For n=1 the conditional probability must be 1 which it is if you substitute.
Assume that the formula holds true for n=k and prove that it is also true for n=k+1
Consider k pickings:
The tree will be k long and k high as follows:
rbbbbb........b (1st line) $P_{d1} =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2016648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Solve the equation $x^2+\frac{9x^2}{(x+3)^2}=27$
Problem Statement:-
Solve the equation $$x^2+\dfrac{9x^2}{(x+3)^2}=27$$
I have tried to turn it into a quadratic equation so as to be saved from solving a quartic equation, but have not been able to come up with anything of much value.
These are the things that I have... | Simplifying and reducing we get
$$x^4+6x^3-9x^2-162x-243=0\,\text{and}\,x\neq 3$$
Let's look for a factorisation in the form
$$x^4+6x^3-9x^2-162x-243=(x^2+ax+b)(x^2+cx+d)$$
Developing and comparing the coefficients we get
$$\begin{align}a+c=&6\\b+d+ac=&-9\\ad+bc=&-162\\bd=&-243\end{align}$$
We obviously look first for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2020139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Showing $\frac12 3^{2-p}(x+y+z)^{p-1}\le\frac{x^p}{y+z}+\frac{y^p}{x+z}+\frac{z^p}{y+x}$
Showing $\frac12 3^{2-p}(x+y+z)^{p-1}\le\frac{x^p}{y+z}+\frac{y^p}{x+z}+\frac{z^p}{y+x}$ with $p>1$, and $x,y,z$ positive
By Jensen I got taking $p_1=p_2=p_3=\frac13$, (say $x=x_1,y=x_2,z=x_3$)
$\left(\sum p_kx_k\right)^p\le\sum... | WLOG, $x+y+z=1\tag1$ then by power mean inequality;
$\sum\limits_{cyc}\left(\frac13(x+y)^{-1}\right)^{-1}\le(x+y)^{\frac 13}(y+z)^{\frac13}(x+z)^{\frac1 3}$
and AM-GM
$(x+y)^{\frac 13}(y+z)^{\frac13}(x+z)^{\frac13}\le\frac13(x+y)+\frac13(y+z)+\frac13(x+z)\overset{(1)}=\frac23$
$\displaystyle\implies \frac{1}{x+y}+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2021324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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The point on $y = \sqrt{\ln x}$ closest to $(4,0)$ Find the point of the graph of $y = \sqrt{\ln x}$ that is closest to the point $(4,0)$.
I started by using the distance formula and plugging in the values where:
$$d = \sqrt{(x-4)^2 + (\sqrt{\ln x} - 0)^2}$$
$$d = \sqrt{(x^2-8x+16+\ln x)}$$
I know we should set this eq... | You don't want to set the distance itself to $0$, but rather the rate of change of the distance (as a function of $x$). So using the formula you correctly derived: $$\begin{array}{rcl} d(x) & = & \sqrt{(x-4)^2+\ln x} \\ & = & \left[(x-4)^2 + \ln x \right]^{\frac{1}{2}} \end{array} $$ we now differentiate it (using the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2021611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find the poles and residue of the function $\frac{1}{z-\sin z }$ at $z=0$? $$\frac{1}{z-\sin z}$$
$$\frac{1}{z(1-\frac{\sin z}{z})}$$
the laurent's transform is also used to simplify the function.
the residue is found a $\frac{2*36*2}{5}$
but the actual answer is given as $\frac{3}{10}$
the residue is given as
The solu... | Use the Taylor Series for the $\sin z$ function:
$$\sin z \approx z - \frac{z^3}{6} + \frac{z^5}{5!} + \ldots$$
In this way your denominator becomes
$$\frac{1}{\frac{z^3}{6} - \frac{z^5}{5!} + \ldots}$$
Collect a $z^3/6$ term:
$$\frac{1}{\frac{z^3}{6}\left(1 - \frac{z^2}{20}\right)} = \frac{6}{z^3} \cdot \frac{1}{1 - \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2022571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How can we show $\cos^6x+\sin^6x=1-3\sin^2x \cos^2x$? How can we simplify $\cos^6x+\sin^6x$ to $1−3\sin^2x\cos^2x$?
One reasonable approach seems to be using $\left(\cos^2x+\sin^2x\right)^3=1$, since it contains the terms $\cos^6x$ and $\sin^6x$. Another possibility would be replacing all occurrences of $\sin^2x$ by $1... | Here's a slick way:
$(\cos^2(\theta)+\sin^2(\theta))^3=1$
Use Pascal's Triangle to get
$\cos^6(\theta)+3 \cos^4(\theta) \sin^2(\theta)+3 \cos^2(\theta) \sin^4(\theta)+\sin^6(\theta)=1$
$\cos^6(\theta)+\sin^6(\theta)=1-3 \cos^2(\theta) \sin^4(\theta)-3 \cos^4(\theta) \sin^2(\theta)$
$\cos^6(\theta)+\sin^6(\theta)=1-3\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2023931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 11,
"answer_id": 5
} |
Show that $1^3 + 2^3 + ... + n^3 = (n(n+1)/2)^2$ by induction I have some with proving by induction. I cannot find a solution for the inductive step: $1^3 + 2^3 + ... + n^3 = (n(n+1)/2)^2$
I already did the induction steps:
Basis: P(1) = $1^3 = (1(1+1)/2)^2$ (This is true)
Inductive step: Assume $P(k) = ((k)(k+1)/2)^2$... | Show that $S_n-S_{n-1}=n^3$ (or if you prefer, $S_n=S_{n-1}+n^3$).
Indeed,
$$\left(\frac{n(n+1)}2\right)^2-\left(\frac{(n-1)n}2\right)^2=n^2\left(\frac{n+1}2+\frac{n-1}2\right)\left(\frac{n+1}2-\frac{n-1}2\right).$$
Note that induction is unnecessary. As both members are quartic polynomials, it suffices to prove equal... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2024028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Solving an inequality involving floor. We have this inequality :
$$\lfloor x+1 \rfloor < \sqrt x +1. $$
I want to find algebraic way to solve this inequality.(not geometric solution)
Note : It is obvious when $x \in \Bbb Z$ we have $0<x<1$ but I can't solve it when $x \notin \Bbb Z$
| To show another possible way , we have that
$$
\begin{array}{l}
\left\lfloor {x + 1} \right\rfloor < \sqrt x + 1 \\
\left\lfloor x \right\rfloor < \sqrt x \\
x - \left\{ x \right\} < \sqrt x \quad \left| {\;0} \right. \le \left\{ x \right\} < 1 \\
x - 1 < x - \left\{ x \right\} < \sqrt x \\
\end{array}
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2028673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
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.