Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Prove for any positive real numbers $a,b,c$ $\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2} \geq \frac{a+b+c}{3}$ Since the problem sheets says I should use Cauchy-Schwarz inequality, I used
$\frac{{a_1}^2}{x_1}+\frac{{a_2}^2}{x_2}+\frac{{a_3}^2}{x_3}$
$\geq \frac{(a_1+a_2+a_3)^2}{x_1+x_2+x_3}$
I... | $
\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}
=\frac{a^4}{a(a^2+ab+b^2)}+\frac{b^4}{b(b^2+bc+c^2)}+\frac{c^4}{c(c^2+ca+a^2)}
\geq \frac{(a^2+b^2+c^2)^2}{a^3+b^3+c^3+ab(a+b)+bc(b+c)+ca(c+a)} \tag{1}
$
Remember
$\frac{{a_1}^2}{x_1}+\frac{{a_2}^2}{x_2}+\frac{{a_3}^2}{x_3} \geq \frac{(a_1+a_2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/122741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
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Evaluate and prove by induction: $\sum k{n\choose k},\sum \frac{1}{k(k+1)}$
*
*$\displaystyle
0\cdot \binom{n}{0} + 1\cdot \binom{n}{1} + 2\binom{n}{2}+\cdots+(n-1)\cdot \binom{n}{n-1}+n\cdot \binom{n}{n}$
*$\displaystyle\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3}+\frac{1}{3\cdot 4} +\cdots+\frac{1}{(n-1)\cdot n}$
Ho... | 1) Take $\displaystyle f(x)= \sum_{i=0}^n\binom{n}{i} x^i=(x+1)^n$. Now consider $f'(1)$.
2) Use that $\frac{1}{(k-1)\cdot k} = \frac{1}{k-1} -\frac{1}{k}$.
| {
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"url": "https://math.stackexchange.com/questions/123655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to compute $\int{\frac{5x^3+8x^2+x+2}{x^2(2x^2+1)}} dx$? $$\int{\frac{5x^3+8x^2+x+2}{x^2(2x^2+1)}} dx$$
So ... how do I start? Numerator cant be factorized it seems, and this looks like a complicated expression ...
I tried expanding the denominator to see if integration by substitution will work, but it didn't giv... | Hint :
Use partial fraction decomposition :
$$\frac{5x^3+8x^2+x+2}{x^2(2x^2+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{2x^2+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/129305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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No Integer Solutions and Congruences Using congruences, I seek to prove two things:
1) $x^2 - 4y^2 = 3$ has no solutions in integers $x,y,z$.
I think this can be done using modulo 4? How so?
2) $3x^3 - 7y^3 + 21 z^3 = 2$ has no solutions in integers $x,y,z$.
Not sure how to attack this one...
| (1) If $x$ is even, is $x^2-4y^2$ odd or even? If $x$ is odd, can $x^2-3$ be divisible by $4$ if $x$ is an integer?
(2) $3x^3-7y^3+21z^3=7(3z^3-y^3)+3x^3$; can $3x^3-2$ be divisible by $7$ if $x$ is an integer?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/131186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Differentiation of $y = \tan^{-1} \Bigl\{ \frac{\sqrt{1+x^{2}} - \sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+ \sqrt{1-x^{2}}}\Bigr\}$ How do i differentiate the following: $$y = \tan^{-1} \biggl\{ \frac{\sqrt{1+x^{2}} - \sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+ \sqrt{1-x^{2}}}\biggr\}$$
I know that $\text{derivative}$ of $\tan^{-1}{x}$ is $\... | Here are all the steps in excruciating detail.
Using the tangent of a sum formula gives
$$
\frac{1-\tan{\phi}}{1+\tan{\phi}}=\tan\left(\frac\pi4-\phi\right)\tag{1}
$$
and letting $\tan(\phi)=\sqrt{\frac{1-x^2}{1+x^2}}$ yields
$$
\begin{align}
\tan^{-1}\left(\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/131679",
"timestamp": "2023-03-29T00:00:00",
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How to prove $n^5 - n$ is divisible by $30$ without reduction How can I prove that prove $n^5 - n$ is divisible by $30$?
I took $n^5 - n$ and got $n(n-1)(n+1)(n^2+1)$
Now, $n(n-1)(n+1)$ is divisible by $6$.
Next I need to show that $n(n-1)(n+1)(n^2+1)$ is divisible by $5$.
My guess is using Fermat's little theorem but ... | \begin{align*}
\dfrac{n(n^4-1)}{30}&=\dfrac{n(n^2-1)\color{#c00}{(n^2+1)}}{30}\\&=\dfrac{(n-1)n(n+1)\color{#c00}{(5+(n+2)(n-2))}}{30}\\&=\dfrac{(n-1)n(n+1)}{6}+\dfrac{\color{#c00}{(n-2)}(n-1)n(n+1)\color{#c00}{(n+2)}}{30}\\&=\binom{n+1}{3}+4\binom{n+2}{5}\in\Bbb Z
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/132210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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How do I solve $(x-1)(x-2)(x-3)(x-4)=3$ How to solve $$(x-1) \cdot (x-2) \cdot (x-3) \cdot (x-4) = 3$$
Any hints?
| I assume the hints would already have given you the answer. If not, here is the full answer:
Let y = x-2.5
(y+1.5)(y-1.5)(y+0.5)(y-0.5) = 3.
so $(y^2-2.25)(y^2-0.25) = 3$.
Let $z = y^2-1.25$.
(z-1)(z+1) = 3.
So $z^2-1 = 3$.
Hence $z^2 = 4$.
z = -2 gives $y^2 = z + 1.25 = -0.75$.
So $y = \pm \sqrt{0.75}i$. Clearly, this... | {
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"timestamp": "2023-03-29T00:00:00",
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Prime number in a polynomial expression Will be glad for a little hint: let x and n be positive integer such that $1+x+x^2+\dots+x^{n-1}$ is a prime number then show that n is prime
| Assuming that you meant that $1+x+x^2+\ldots+x^{n-1}$ is prime, note that this sum is $\frac{x^n-1}{x-1}$. If $n=ab$, we have $$\frac{x^n-1}{x-1}=\frac{(x^a)^b-1}{x-1}=\frac{(x^a-1)(1+x^a+x^{2a}+\ldots+x^{(b-1)a})}{x-1}\;,$$ and $\dfrac{x^a-1}{x-1}=\;$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/132633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Taking the derivative of $y = \dfrac{x}{2} + \dfrac {1}{4} \sin(2x)$ Again a simple problem that I can't seem to get the derivative of
I have $\frac{x}{2} + \frac{1}{4}\sin(2x)$
I am getting $\frac{x^2}{4} + \frac{4\sin(2x)}{16}$
This is all very wrong, and I do not know why.
| $y = \dfrac{1}{2} x + \dfrac {1}{4} \sin(2x)$
$y' = \dfrac{1}{2} + \dfrac{1}{4}\cos(2x)*2$
$y' = \dfrac{1}{2} + \dfrac{2}{4}\cos(2x)$
$y' = \dfrac{1}{2} + \dfrac{1}{2}\cos(2x)$
$y' = \dfrac{1}{2} \big(1 + \cos (2x)\big) $
Which is equivalent to $\cos^2x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/134855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is awry with this proof? Let $x=5$, $y=7$, $z=6$
$x+y = 2z$
Rearranging, $x-2z = -y$
and $x = -y+2z$
Multiply both sides respectively. $x^2-2xz = y^2-2yz$
$$x^2-2xz+z^2 = y^2-2yz+z^2$$
$$(x-z)^2 = (y-z)^2$$
$$x-z = y-z$$
Hence $x=y$, or $5 = 7$
Well, the conclusion is clearly false, but what went wrong? I think it... | If $(x-z)^2 = (y-z)^2$ then $x-y = \pm(y-z)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/137859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What is this series called? $\frac{x^1}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!} \dots \pm \frac{x^n}{n!}$ I remember learning about this series in Precalculus the other day but I neglected to get the name of it. It looks something like this:
$
\begin{align*}
\frac{x^1}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x... | This
$$
\frac{x^1}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!} \dots \pm \frac{x^n}{n!}
$$
Is called a Maclaurin polynomial for the sine function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/138470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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More highschool math $\frac{3^n(n^3+3n^2+3n+1)}{3^{n+1}\cdot n^3} = \frac{n^3+3n^2+3n+1}{3n^3} \to \frac{1}{3}$ So the question I am trying to work through is:
Test the series
$$\frac{1}{3}+\frac{2^3}{3^2}+\frac{3^3}{3^3}+\frac{4^3}{3^4}+\frac{5^3}{3^5}+\cdot\cdot\cdot$$
for convergence.
The solution (using D'Alember... | Here is one way to simplify the limit and arrive at the answer. Hopefully, it will let you see how terms cancel out.
\begin{align*}
\left|\frac{u_{n+1}}{u_{n}}\right| &= \frac{(n+1)^3}{3^{n+1}}\cdot \frac{3^n}{n^3} \\
&= \frac{(n+1)^3}{n^3}\cdot \frac{3^n}{3^{n+1}} \\
&= \left(\frac{n+1}{n}\right)^3 \cdot \frac{1}{3} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/138600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Determine convergence of $\sum_{n=1}^{\infty} (\cos{\frac{2}{n}}-\cos{\frac{4}{n}})$ Determine convergence of $$\sum_{n=1}^{\infty} \left(\cos{\frac{2}{n}}-\cos{\frac{4}{n}}\right)$$
In the answer, it says
$$\cos{\frac{2}{n}}-\cos{\frac{4}{n}} = 2\sin{\frac{3}{n}}\sin{\frac{1}{n}} \le 2\cdot \frac{3}{n} \cdot \frac{1}... | The result follows from the Addition Law for Cosines.
We have $\cos(x+y)=\cos x\cos y-\sin x\sin y$ and $\cos(x-y)=\cos x\cos y+\sin x\sin y$. Subtract. We get
$$\cos(x-y)-\cos(x+y)=2\sin x\sin y.$$
Let $x-y=\frac{2}{n}$ and $x+y=\frac{4}{n}$. Solve for $x$ and $y$. We get $x=\frac{3}{n}$ and $y=\frac{1}{n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/139272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluating $\int y^4(1-y)^3 dy$ using integration by parts Here is the function which could easily be solved using expansion method but how could I solve it using integration by parts
$$\int y^4(1-y)^3 dy$$
The problem is, when I apply integration by parts to solve it, it is never ending solution and I am not able to g... | $$I(4,3) = \int y^4 (1-y)^3 \mathrm{d}y$$
You were heading in the right direction, i.e. $\mathrm{d}v=y^4 \Rightarrow v = \frac{y^5}{5}$
$$
\begin{align*}
I(4,3) &= \frac{y^5(1-y)^3}{5} + \frac{3}{5} \int y^5 (1-y)^2 \mathrm{d}y\\
&= \frac{y^5(1-y)^3}{5} + \frac{3}{5} I(5,2)\\
\end{align*}
$$
Similarly use $\mathrm{d}v... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\lim\limits_{n\to\infty} \left(\frac{1^p+2^p+3^p + \cdots + n^p}{n^p} - \frac{n}{p+1}\right)$ Evaluate $$\lim_{n\to\infty} \left(\frac{1^p+2^p+3^p + \cdots + n^p}{n^p} - \frac{n}{p+1}\right)$$
| If we draw the graph of $x^p$ from $x=1$ to $x=n,$ divide it into unit length intervals and approximate each segment of area by a trapezium (this is known as the trapezoidal rule) then we see that $$\int^n_1 x^p dx \approx \sum_{k=1}^n k^p - \frac{n^p+1}{2}.$$ The integral on the left is precisely $\displaystyle \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/149142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "39",
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"answer_id": 5
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Complex number: calculate $(1 + i)^n$. I have to solve the following complex number exercise: calculate $(1 + i)^n\forall n\in\mathbb{N}$ giving the result in $a + ib$ notation.
Basically what I have done is calculate $(1 + i)^n$ for some $n$ values.
$$(1 + i)^1 = 1 + i$$
$$(1 + i)^2 = 2i$$
$$(1 + i)^3 = - 2 + 2i$$
$$\... | Perhaps, proceed via the trigonometric form using De Moivre's formula:
$$\left(1+i\right)^{n}=2^{\frac{n}{2}}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)^{n}=2^{\frac{n}{2}}\cos\frac{\pi n}{4}+i2^{\frac{n}{2}}\sin\frac{\pi n}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/149803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Integral of $\int \frac{du}{u \sqrt{5-u^2}}$ I am trying to find this integral and I can get the answer on wolfram of course but I do not know what is wrong with my method, having gone through it twice.
$$\int \frac{du}{u \sqrt{5-u^2}}$$
$u = \sqrt{5} \sin\theta$ and $du= \sqrt{5} \cos \theta$
$$\int \frac{\sqrt{5}... | You made several mistakes throughout your answer. Once corrected the answer should be
$$\int \frac{du}{u \sqrt{5-u^2}}$$
$u = \sqrt{5} \sin\theta$ and $du= \sqrt{5} \cos \theta d\theta$
\begin{eqnarray}
\int \frac{du}{u \sqrt{5-u^2}} &=& \int\dfrac{ \sqrt{5} \cos \theta d\theta}{\sqrt{5} \sin\theta \sqrt{5 - (\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/151966",
"timestamp": "2023-03-29T00:00:00",
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Factorize the polynomial $P(z) = z^4 - 2z^3-z^2+2z+10$, into linear and/or quadratic factors with real coefficients $2+i$ is given to be one of the roots of the polynomial.
I am doing this as a practice for exam prep.
Since $2+i$, is a root, then $(z-2-i)$ is a factor?
So I have:
$(z-2-i)(z^3-Az^2-Bz+C) = z^4-2z^3-z^2+... | As Thomas has said, if $2+i$ is a root then $2-i$ is also root. Hence you have two factors: $z-(2-i)$ and $z-(2+i)$ whose product is $z^2-4z+5$. So we can can assume that there is another quadratic factor $z^2+bz+c$ where $b$ and $c$ are constants to be determined. Obviously their product should yield the given polynom... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Numbers arranged in square arrays - observe patterns, make a conjecture, and prove or disprove it Please can anyone tell me what the below question means and how to solve it:
Observe patterns. Make a conjecture. Prove or disprove.
$$\begin{matrix}1\end{matrix}\qquad\begin{matrix}1 & 1 \\ 1 & 2\end{matrix}\qquad\begin... | Assuming the pattern is $A_{ij}=\min(i,j)+1$
Each number $m$ in the $n \times n$ matrix is written $2(n-m)+1$ times. So we have
$$\sum_{m=1}^n m(2n-2m+1)$$
$$(2n+1)\sum_{m=1}^n m - 2\sum_{m=1}^n m^2$$
If you know the formula for sums of values and sums of squares, then you're golden.
$$=(2n+1)\cdot n(n+1)/2-n(n+1)(2n+1... | {
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"url": "https://math.stackexchange.com/questions/152720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Minimal polynomial of the root of algebraic number I have obtained the minimal polynomial of $9-4\sqrt{2}$ over $\mathbb{Q}$ by algebraic operations:
$$ (x-9)^2-32 = x^2-18x+49.$$
I wonder how to calculate the minimal polynomial of $\sqrt{9-4\sqrt{2}}$ with the help of this sub-result? Or is there a smarter way to do t... | Since $x = 9 - 4\sqrt{2}$ satisfies $x^2 - 18x + 49 = 0$, your number $y = \sqrt{x} = \sqrt{9 - 4\sqrt{2}}$ satisfies $y^4 - 18y^2 + 49 = 0$. This could be your minimal polynomial, but the polynomial factorizes as $$y^4 - 18y^2 + 49 = (y^2 + 2y - 7)(y^2 - 2y - 7).$$ Since the product is zero if and only if at least one... | {
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"timestamp": "2023-03-29T00:00:00",
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Finding $\int \frac {dx}{\sqrt {x^2 + 16}}$ I can not get the correct answer.
$$\int \frac {dx}{\sqrt {x^2 + 16}}$$
$x = 4 \tan \theta$, $dx = 4\sec^2 \theta$
$$\int \frac {dx}{\sqrt {16 \sec^2 \theta}}$$
$$\int \frac {4 \sec^ 2 \theta}{\sqrt {16 \sec^2 \theta}}$$
$$\int \frac {4 \sec^ 2 \theta}{4 \sec \theta}$$
$$\in... | $$\int \frac {dx}{\sqrt {x^2 + 16}}$$
Let $$t=\sqrt {x^2+16}$$ this will change the integral into $$ \int \frac {dt}{\sqrt {t^2 - 16}}$$
$$t=4\sec (\theta)$$ changes the integral into $$\int \sec \theta d \theta =ln| \sec \theta + \tan \theta|+c$$
$$=\ln (t/4 + \frac {\sqrt {t^2-16}}{4} )+c$$
$$= \ln ( \sqrt \frac {x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/153787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Integral of $\int \sqrt{1-4x^2}$ I know I am messing up something with the substitutions but I am not sure what.
$$\int \sqrt{1-4x^2}$$
$$u = 4x, du = 4 \,dx$$
$$\frac{1}{4}\int \sqrt{1-u^2}$$
$u = \sin \theta$
$$\frac{1}{4}\int \sqrt{1-\sin^2 \theta} = \frac{1}{4}\int \sqrt{ \cos^2 \theta} = \frac{1}{4}\int \cos \the... | $$\int \sqrt{1-4x^2}\ dx$$
Substitute
$$\begin{align*}x &= \frac{\sin (\theta)}{2}\\ dx &= \frac{\cos(\theta)}{2}d \theta\end{align*}$$
So the intregral become
$$\begin{align*}
\int \sqrt{1-4x^2}\ dx &=\frac{1}{2}\int \cos^2(\theta)d\theta\\
&=\frac{1}{2}\int \frac{\cos(2\theta) + 1}{2}d\theta\\
&=\frac{\sin(2\theta)}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/153838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Integral of $\int \frac{5x+1}{(2x+1)(x-1)}$ I am suppose to use partial fractions
$$\int \frac{5x+1}{(2x+1)(x-1)}$$
So I think I am suppose to split the top and the bottom. (x-1)
$$\int \frac{A}{(2x+1)}+ \frac{B}{x-1}$$
Now I am not sure what to do.
| Bring the expression $\frac{A}{2x+1}+\frac{B}{x-1}$ to the common denominator $(2x+1)(x-1)$. We get
$$\frac{A(x-1)+B(2x+1)}{(2x+1)(x-1)}.$$
The top is $(A+2B)x -A+B$. We want this to be identically equal to $5x+1$.
The coefficients of $x$ must match, and the constant terms must match.
So we want $A+2B=5$, and $-A+B=1... | {
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"timestamp": "2023-03-29T00:00:00",
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Need help with the integral $\int \frac{2\tan(x)+3}{5\sin^2(x)+4}\,dx$ I'm having a problem resolving the following integral, spent almost all day trying.
Any help would be appreciated.
$$\int \frac{2\tan(x)+3}{5\sin^2(x)+4}\,dx$$
| $$
\begin{aligned}
& \int \frac{2 \tan x+3}{5 \sin ^{2} x+4} d x \\
=& \int \frac{2 \sec ^{2} x \tan x+3 \sec ^{2} x}{5 \tan ^{2} x+4 \sec ^{2} x} d x \\
=& 2 \int \frac{\sec ^{2} x \tan x}{5 \tan ^{2} x+4 \sec ^{2} x} d x+3 \int \frac{\sec ^{2} x}{5 \tan ^{2} x+4 \sec ^{2} x} d x \\
=& \int \frac{2sd s}{9 s^{2}-5}+\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/155022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
What is wrong with my solution? $\int \cos^2 x \tan^3x dx$ I am trying to do this problem completely on my own but I can not get a proper answer for some reason
$$\begin{align}
\int \cos^2 x \tan^3x dx
&=\int \cos^2 x \frac{ \sin^3 x}{ \cos^3 x}dx\\
&=\int \frac{ \cos^2 x\sin^3 x}{ \cos^3 x}dx\\
&=\int \frac{ \sin^3 ... | You have been simplifying things up until line 6 and then kind of turned back into complications. It would be natural to notice that
$$\sin x dx = -d\left(\cos x\right)$$
Then the integral looks as follows:
$$I=-\int\frac{1-\cos^2x}{\cos x}d\left(\cos x\right)$$
So it appears that $\cos x$ plays the role of a variable ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/155829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 7,
"answer_id": 1
} |
Formula to estimate sum to nearly correct : $\sum_{n=1}^\infty\frac{(-1)^n}{n^3}$ Estimate the sum correct to three decimal places :
$$\sum_{n=1}^\infty\frac{(-1)^n}{n^3}$$
This problem is in my homework. I find that n = 22 when use Maple to solve this. (with some programming) But, in my homework, teacher said find th... | Averaging the 9th and 10th partial sums will do it, as in robjohn's answer and my comment there. Just for fun, as an alternative to alternating series methods, you could group consecutive terms to give a positive series and then use integral approximation.
$$\begin{align}
S:=\sum_{n=1}^\infty\frac{(-1)^n}{n^3}
&=-1+\su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/156518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Spivak Calculus Prologue I'm completely blown away by the difficulty of Spivak. I've managed to work through the first 3 problems, but I feel I'm missing something important to solve these basic inequalities in his 4th problem:
$$x^2 + x + 1 > 2$$
&
$$x^2 + x + 1 > 0$$
any suggestions?
| Complete the square:
$$
x^2 + x + 1 = (x^2 + x + \tfrac 1 4) + \frac 3 4 = \left(x + \frac 1 2 \right)^2 + \frac 3 4.
$$
That gets you the second one.
For the first one, put everything on one side of the inequality and $0$ on the other side, and procede similarly.
Later addendum in response to vitno's question in the c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/156577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Another Congruence Proof I've been asked to attempt a proof of the following congruence. It is found in a section of my textbook with Wilson's theorem and Fermat's Little theorem. I've pondered the problem for a while and nothing interesting has occurred to me.
$1^23^2\cdot\cdot\cdot(p-4)^2(p-2)^2\equiv (-1)^{(p+1)/2... | To Prove : $1^2.2^2.3^2....(p-1)^2 \equiv (-1)^{\frac{p+1}{2}} \pmod p$
We know that
$$k \equiv -(p-k) \pmod p$$
Applying this repeatedly we will get
$$2.4.6.......(p-1) \equiv (-1)^{\frac{p-1}{2}} 1.3.5.......(p-2) \pmod p$$
$$\implies 1.3.5.......(p-2) \equiv (-1)^{\frac{p-1}{2}} 2.4.6.......(p-1) \pmod p$$
$$\implie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/156823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Limit involving $(\sin x) /x -\cos x $ and $(e^{2x}-1)/(2x)$, without l'Hôpital Find:
$$\lim_{x\to 0}\ \frac{\dfrac{\sin x}{x} - \cos x}{2x \left(\dfrac{e^{2x} - 1}{2x} - 1 \right)}$$
I have factorized it in this manner in an attempt to use the formulae.
I have tried to use that for $x$ tending to $0$, $\dfrac{\sin x}{... | $$\dfrac{\sin(x)}{x} = 1 - \dfrac{x^2}{3!} + \mathcal{O}(x^4)$$
$$\cos(x) = 1 - \dfrac{x^2}{2!} + \mathcal{O}(x^4)$$
$$\exp(2x) = 1 + 2x + \dfrac{(2x)^2}{2!} + \mathcal{O}(x^3)$$
Hence, $$\dfrac{\dfrac{\sin(x)}{x} - \cos(x)}{\exp(2x) - 1 -2x} = \dfrac{-\dfrac{x^2}{3!} + \dfrac{x^2}{2!} + \mathcal{O}(x^4)}{2x^2 + \mathc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/157100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Complex numbers lead a trigonometric equality Times ago, I used to think about some trigonometric equalities. Now, I have faced a new one with different one:
Show that if $z^7+1=0$ then cos$(\frac{\pi}{7})$+cos$(\frac{3\pi}{7})$+cos$(\frac{5\pi}{7})=\frac{1}{2}$ wherein $z\in\mathbb C$.
Thanks
| if $z^7+1=0$ then $$(z+1)(z^6-z^5+z^4-z^3+z^2-z+1)=0$$
Putting $z=e^{\frac{i \pi}{7}}$ we have $z + 1 \neq 0$ and $z^7 +1 =0$, then :
$$z^6-z^5+z^4-z^3+z^2-z+1=0$$
This gives :
$$\mathcal Re(z^6-z^5+z^4-z^3+z^2-z+1)=0$$
Since : $\cos \frac{\pi}{7} = - \cos \frac{6\pi}{7}$ and $\cos \frac{2\pi}{7} = - \cos \frac{5\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/160251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Simplifying $\frac{2^{n + 4} + 2^{n + 2} + 2^{n - 1}}{2^{n - 2} + 2^{n - 1}}$ I'm stuck in the follow equation:
$$\dfrac{2^{n + 4} + 2^{n + 2} + 2^{n - 1}}{2^{n - 2} + 2^{n - 1}}$$
As all the bases are equal, I got $\dfrac{3n + 5}{2n - 3}$
Where I've to go now ?
Thanks
EDIT:
Then, my initial idea was totally wrong, sta... | You’ve nothing to solve. My guess is that you’re supposed to simplify the fraction:
$$\begin{align*}
\dfrac{2^{n + 4} + 2^{n + 2} + 2^{n - 1}}{2^{n - 2} + 2^{n - 1}}&=\frac{2^{n-1}(2^5+2^3+1)}{2^{n-2}(1+2)}\\
&=\frac{2^{n-1}}{2^{n-2}}\cdot\frac{32+8+1}{3}\;,
\end{align*}$$
and you should have no trouble finishing it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/161843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Smallest number in a set
A is the set of seven consequtive two digit numbers, none of these being multiple of 10. Reversing the numbers in set A forms numbers in set B. The difference between the sum of elements in set A and those in set B is 63. The smallest number in set A can be :
I tried to write some sets and re... | Let $A = \{10a+b,10a+b+1,10a+b+2,10a+b+3,10a+b+4,10a+b+5,10a+b+6\}$, where $a \in \{1,2,\ldots,9\}$ and since none of them is divisible by $10$, we have that $b\in\{1,2,3\}$. Then $$B =\{10b+a,10b+a+10,10b+a+20,10b+a+30,10b+a+40,10b+a+50,10b+a+60\}$$
Sum of elements in $A$ is $70a+7b+21$ and sum of elements in $B$ is $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/163982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
$|2^x-3^y|=1$ has only three natural pairs as solutions Consider the equation $$|2^x-3^y|=1$$ in the unknowns $x \in \mathbb{N}$ and $y \in \mathbb{N}$. Is it possible to prove that the only solutions are $(1,1)$, $(2,1)$ and $(3,2)$?
| Assume that $x>3$ and $y>2$.
$3^2=1\pmod{8}$. Since $3\not=-1\pmod{8}$ we know that $3^y\not=-1\pmod{8}$. Thus, if $|2^x-3^y|=1$, we must have $2^x-3^y=-1$ and $y$ must be even. Thus, we get that
$$
2^x=(3^{y/2}-1)(3^{y/2}+1)\tag{1}
$$
The only factors of a power of $2$ are other powers of $2$, and the only powers of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/164874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 0
} |
Two proofs I'm having difficulty with I've been given an assignment. Almost done except the last two are tripping me up. They are as follows:
1) if $2x^2-x=2y^2-y$ then $x=y$
2) if $x^3+x=y^3+y$ then $x=y$
I imagine they use a similar tactic as they both involve powers, but I've tried factoring,completing the square, d... | Question $1$: The equation is equivalent to $2x^2-2y^2=x-y$. The left-hand side factors as $2(x+y)(x-y)$. So our equation can be rewritten as
$$2(x+y)(x-y)=x-y.$$
Thus any pair $(x,y)$ such that $x\ne y$ and $2(x+y)=1$ is a counterexample to the assertion that $x$ must be equal to $y$. This was pointed out by ncmaths... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/169732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Some method to solve $\int \frac{1}{\left(1+x^2\right)^{2}} dx$ and some doubts. First approach.
$\int \frac{1}{1+x^2} dx=\frac{x}{1+x^2}+2\int \frac{x^2}{\left(1+x^2\right)^2} dx=\frac{x}{1+x^2}+2\int \frac{1}{1+x^2}dx-2\int \frac{1}{\left(1+x^2\right)^2}dx$
From this relationship, I get:
$2\int \frac{1}{\left(1+x^2\... | One way to make the trigonometric solution substitution solution end nicely, without worries about signs, is to note that
$$\sin t \cos t =\frac{\sin t}{\cos t} \cos^2 t.$$
Since $\cos^2 t=\frac{1}{\sec^2 t}=\frac{1}{1+\tan^2 t}$, we get that
$$\sin t \cos t =\tan t\frac{1}{1+\tan^2 t}=\frac{x}{1+x^2}.$$
Another way: ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/170207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Prove $(-a+b+c)(a-b+c)(a+b-c) \leq abc$, where $a, b$ and $c$ are positive real numbers I have tried the arithmetic-geometric inequality on $(-a+b+c)(a-b+c)(a+b-c)$ which gives
$$(-a+b+c)(a-b+c)(a+b-c) \leq \left(\frac{a+b+c}{3}\right)^3$$
and on $abc$ which gives
$$abc \leq \left(\frac{a+b+c}{3}\right)^3.$$
Since ... | Norbert's answer explains the case when $a,b,c$ are not the sides of a triangle.
Let $x,y,z$ be as in this picture, where $AB=c, BC=a, CA=b$.
Then $(a,b,c)=(y+z,x+z,x+y)$.$\,$ Inequality becomes $$8xyz\le (x+y)(y+z)(z+x),$$
which as Norbert says is true by $2\sqrt{xy}\le x+y$ (proof: $\Leftrightarrow (\sqrt{x}-\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/170813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 7,
"answer_id": 3
} |
Trigonometric Identities To Prove
*
*$\tan\theta+\cot\theta=\dfrac{2}{\sin2\theta}$
Left Side:
$$\begin{align*}
\tan\theta+\cot\theta={\sin\theta\over\cos\theta}+{\cos\theta\over\sin\theta}={\sin^2\theta+\cos^2\theta\over\cos\theta\sin\theta}
= \dfrac{1}{1\sin\theta\cos\theta}
\end{align*}$$
Right Side:
$$\begin{a... | I'll just put together what you wrote...
$$\begin{align*}
\tan\theta+\cot\theta=\dfrac{\sin\theta}{\cos\theta}+\dfrac{\cos\theta}{\sin\theta}=\dfrac{\sin^2\theta+\cos^2\theta}{\cos\theta \cdot\sin\theta} = \dfrac{1}{\cos \theta \cdot \sin \theta}
\end{align*}$$
Where the penultimate inequality is what you should have... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/170951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to evaluate $\int_{-\infty}^\infty \frac{\cos x}{\cosh x}\,\mathrm dx$ by hand How can I evaluate$$\int_{-\infty}^\infty \frac{\cos x}{\cosh x}\,\mathrm dx\text{ and }\int_0^\infty\frac{\sin x}{e^x-1}\,\mathrm dx.$$
Thanks in advance.
| For the second one,
$$ \begin{align*}
\int_{0}^{\infty} \frac{\sin x}{e^x - 1} \; dx
&= \int_{0}^{\infty} \frac{\sin x \, e^{-x}}{1 - e^{-x}} \; dx \\
&= \int_{0}^{\infty} \left( \sum_{n=1}^{\infty} \sin x \, e^{-nx} \right) \; dx \\
&\stackrel{\ast}{=} \sum_{n=1}^{\infty} \int_{0}^{\infty} \sin x \, e^{-nx} \; dx \\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/171073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 4,
"answer_id": 1
} |
How to prove that $a^2(1+b^2)+b^2(1+c^2)+c^2(1+a^2)\geq6abc$ Help me prove $a^2(1+b^2)+b^2(1+c^2)+c^2(1+a^2)\geq6abc$
| LHS=$\sum a^2\ +\ \sum(ab)^2$
Applying A.M. ≥ G.M.,
$\sum a^2 ≥ 3(abc)^{\frac{2}{3}} $
$\sum (ab)^2 ≥ 3(abc)^{\frac{4}{3}} $
Taking summation,$ LHS ≥ 3((abc)^{\frac{2}{3}} + (abc)^{\frac{4}{3}}) $
But $(abc)^{\frac{2}{3}} + (abc)^{\frac{4}{3}}$ ≥ 2 abc (applying A.M. ≥ G.M.,)
LHS ≥ 3(2abc)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/171136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Sum of the series : $1 + 2+ 4 + 7 + 11 +\cdots$ I got a question which says
$$ 1 + \frac {2}{7} + \frac{4}{7^2} + \frac{7}{7^3} + \frac{11}{7^4} + \cdots$$
I got the solution by dividing by $7$ and subtracting it from original sum. Repeated for two times.(Suggest me if any other better way of doing this).
However now ... | Note that starting with your first element, $a_0=1$; to get to $a_1=2$ we sum $1$, to get to $a_2=4$, we sum $2$, to get to $a_3=7$, we sum $3$. So in general, we can say that
$$a_{n}=a_{n-1}+n$$
This is
$$
\begin{align}
2&=1+1 \\[8pt]
4& =2+2\\[8pt]
7&=4+3\\[8pt]
11& =7+4\\[8pt]
\cdots&=\cdots\\[8pt]
a_n&=a_{n-1}+n
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/171754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 2
} |
Evaluating $\int (x^6+x^3)\sqrt[3]{x^3+2}dx$ I am trying to evaluate:
$$\int (x^6+x^3)\sqrt[3]{x^3+2} \ \ dx$$
My solution:
$$\int (x^5+x^2)\sqrt[3]{x^6+2x^3} \ \ dx$$
Let $$(x^6+2x^3) = t^3 \ \ \text{and} \ \ (x^5+x^2) \ \ dx = \frac{1}{2}t^2 \ \ dt$$
$$\frac{1}{2}\int t^2\cdot t \ \ dt = \frac{1}{2}.\frac{t^4}{4}+C ... | The $t$ stuff is not necessary. You can directly let $u=x^6+2x^3$. Then $(x^5+x^2)\,dx=\frac{1}{6}\,du$. But the initial step was the key one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/171799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
A question about arithmetic progressions, a.k.a. arithmetic sequences What is the relation between the general formula of the sum of $n$ terms of an arithmetic progression, $a n^2+b n+c$, and the first term $a+b+c$ and the common difference?
| If the sum of $n$ terms of an arithmetic sequence is
$$
s(n)=\color{#C00000}{an^2+bn+c}\tag{1}
$$
then $c$ is the sum of $0$ terms; that is,
$$
c=s(0)=0\tag{2}
$$
The general term of the arithmetic sequence would be
$$
\begin{align}
a(n)
&=s(n)-s(n-1)\\
&=(b-a)+(2a)n\tag{3}
\end{align}
$$
The first term of the arithmet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/172001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
General solution of $C_{n+2}(x)=xC_n(x)+nC_{n-1}(x)$ Airy differential equation.
$y''(x)=xy(x)$
$y'''(x)=y(x)+x y'(x)$
$y'^v(x)=x^2y(x)+2 y'(x)$
$y^v(x)=4xy(x)+x^2 y'(x)$
$y^{(6)}(x)=(x^3+4)y(x)+6x y'(x)$
.
.
$y^{(n)}(x)=A_n(x)y(x)+B_n(x) y'(x)$
Where $A_n(x)$ and $B_n(x)$ are polynomials
($y^{(n)}(x)$ means $n$-th ... | Let $G(z)$ be the exponential generating function defining $G(z) = \sum_{n=0}^\infty C_n \frac{z^n}{n!}$. The recurrence equation $C_{n+2} = x C_n + n C_{n-1}$ translates into a differential equation:
$$
G^{\prime\prime}(z)- (x+z) G(z)= C_2 - x C_0
$$
This equation admits a closed form solution:
$$
G(z) = \kappa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/172130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Solving $E=\frac{1}{\sin10^\circ}-\frac{\sqrt3}{\cos10^\circ}$ $$E=\frac{1}{\sin10^\circ}-\frac{\sqrt3}{\cos10^\circ}$$
I got no idea how to find the solution to this. Can someone put me on the right track?
Thank you very much.
| We have
\begin{eqnarray*}
E&=&\frac{\cos 10^\circ-\sqrt{3}\sin 10^\circ}{\sin 10^\circ \cos 10^\circ}\\
&=&4\frac{(1/2)\cos 10^\circ-(\sqrt{3}/2)\sin 10^\circ}{2\sin 10^\circ \cos 10^\circ}\\
&=&4\frac{\cos 60^\circ\cos 10^\circ-\sin 60^\circ\sin 10^\circ}{\sin 20^\circ}\\
&=&4\frac{\cos(60^\circ+10^\circ)}{\sin 20^\ci... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/172471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
What can be the possible value of $a+b+c$ in the following case? What can be the possible value of $a+b+c$ in the following case?
$$a^{2}-bc=3$$
$$b^{2}-ca=4$$
$$c^{2}-ab=5$$
$0, 1, -1$ or $1/2$?
After doing $II-I$, $III-I$ and $III-II$, I got,
$$(a+b+c)(b-a)=1$$
$$(a+b+c)(c-a)=2$$
$$(a+b+c)(c-b)=1$$
I'm unable to s... | $(a-b)^2+(b-c)^2+(c-a)^2=2(3+4+5)=24$
Now $a-b=\dfrac1{a+b+c}$ etc.
Putting the values of $a-b, b-c, c-a;$
$$\frac1{(a+b+c)^2}(1^2+2^2+1^2)=24$$
$$\implies(a+b+c)^2=\frac14$$
$$\implies a+b+c=\pm\frac12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/173133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to solve this System of Polynomial Equations? I have to complete a summer packet of 90 Algebra 2 questions. I have completed 89 of them, the only one I could not get was this. I know the answer is $y = \frac {47}2$, $\frac 17$ according to WolframAlpha, but I have no idea how to reach that answer, my Algebra 2 Hono... | How about this approach: It's not hard to see that $2 \cdot \left(\frac{3}{x-1}\right) = \frac{6}{x-1}$. Therefore, we can do a linear combination approach:
Add $-2$ times the first equation to the second equation. We get:
$$ \left(\frac{6}{x-1} - \frac{7}{y+2} \right) - 2 \left( \frac{3}{x-1} + \frac{4}{y+2} \right) =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/174783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Trigonometric eigenvalue equation In solving an eigenvalue problem, I've come to following equation ($\lambda=1$):
$$\begin{pmatrix}
\cos(\theta) & \sin(\theta) \\ \sin(\theta) & -\cos(\theta)
\end{pmatrix}\begin{pmatrix}
a \\ b
\end{pmatrix}=\begin{pmatrix}
a \\ b
\end{pmatrix}$$
Now, the solution says, "This... | $$\begin{pmatrix}
\cos(\theta) & \sin(\theta) \\ \sin\theta & -\cos\theta
\end{pmatrix}\begin{pmatrix}
a \\ b
\end{pmatrix}\\=\begin{pmatrix}
a\cos\theta+b\sin\theta \\ a\sin\theta-b\cos\theta
\end{pmatrix}=\begin{pmatrix}
a \\ b
\end{pmatrix}$$
Which gives us the following system of equations:
$$a\cos\thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/176445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
For which angles we know the $\sin$ value algebraically (exact)? For example:
*
*$\sin(15^\circ) = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}$
*$\sin(18^\circ) = \frac{\sqrt{5}}{4} - \frac{1}{4}$
*$\sin(30^\circ) = \frac{1}{2}$
*$\sin(45^\circ) = \frac{1}{\sqrt{2}}$
*$\sin(67 \frac{1}{2}^\circ) = \sqrt{ \frac{\sqrt... | We use radian notation. Every rational multiple of $\pi$ has trigonometric functions that can be expressed using the ordinary arithmetic operations, plus $n$-th roots for suitable $n$.
This is almost immediate if we use complex numbers, since $(\cos(2\pi/n)+i\sin(2\pi/n)^n=1$.
But it is known, for example, that ther... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/176889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 3
} |
Prove that $(n+\sqrt{n^2 -1})^k$ will always be of the form$ (t+\sqrt{t^2 -1})$ where $n$, $k$, $t$ are natural numbers Show that $(n+\sqrt{n^2 -1})^k$ will always be of the form$ (t+\sqrt{t^2 -1})$ where $n$, $k$, $t$ are natural numbers
| I enjoyed working this out-cool question man. Here's a straightforward inductive proof:
I'm writing $c$ instead of $n$.
Inductively assume $(c + \sqrt{c^2 - 1})^k = a + b \sqrt{c^2 - 1}$, where $b^2(c^2 - 1) = a^2 - 1$ (the base case is clear). That is, not only can you write it as $t + \sqrt{t^2 - 1}$, but also $t^2 -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/178318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 1
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How to show x and y are equal? I'm working on a proof to show that f: $\mathbb{R} \to \mathbb{R}$ for an $f$ defined as $f(x) = x^3 - 6x^2 + 12x - 7$ is injective. Here is the general outline of the proof as I have it right now:
Proof: For a function to be injective, whenever $x,y \in A$ and $x\neq y$, then $f(x) \neq ... | Note that $w^3-6w^2+12w-8=(w-2)^3$. So $w^3-6w^2+12w-7=(w-2)^3+1$.
So we want to show that if $(x-2)^3+1=(y-2)^3+1$ then $x=y$.
Equivalently, we want to show that if $(x-2)^3=(y-2)^3$ then $x=y$. This is easy, the cube function is increasing.
Remark: We can use the basic algebra of ordered fields to show that if $s^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/179224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
How do I proceed with these quadratic equations? The question is
$$ax^2 + bx + c=0 $$ and $$cx^2+bx+a=0$$ have a common root, if $b≠ a+c$, then what is $$a^3+b^3+c^3$$
| The value of $a^3+b^3+c^3$ is not determined. Just choose $a=c$. But leaving out the condition $a\ne c$ is probably an oversight, so assume from now on that $a\ne c$.
If $q$ is a common root, then $aq^2+bq+c=cq^2+bq+a=0$. Subtracting, we find that $(a-c)q^2-(a-c)=0$. Since $a\ne c$, we get $q^2=1$.
We cannot have $q=-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/180479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Proving $\sin A + \sin B + \sin C = 4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$
Possible Duplicate:
Prove that $\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)$ when $A,B,C$ are angles of a triangle
Prove trigonometry identity?
If $A$, $B$, and $C$ are to be taken as the angles of a triangle, then I beg ... | How about a proof with a geometric flavor?
Let $a$, $b$, $c$ be the sides that oppose angles $A$, $B$, $C$, respectively. By the Law of Sines,
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = d$$
where $d$ is the circumdiameter of the triangle. If we conveniently scale the triangle so that $d=1$, then we can... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/180860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
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Finding the conjugate hyperbola Assume we are given a general proper hyperbola $ a_{11}x^2 + 2 a_{12} x y + a_{22} y^2 + 2 a_{13}x +2 a_{23}y + a_{33} = 0$ with
$D =\det \begin{pmatrix}
a_{11} & a_{12} \\
a_{12} & a_{22}
\end{pmatrix} < 0$
and
$
A = \det \begin{pmatrix}
a_{11} & a_{12} & a_{13} \\
a_{12} & a_... | Using the right translation and rotation one can obtain for a general hyperbola, as given in the OP, a canonical congruent hyperbola of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = \pm 1$. The canonical conjugate is then given by $\frac{x^2}{a^2} - \frac{y^2}{b^2} = \mp 1$, and can be transformed back (using inverse t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/182429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the value of $x$ in the displayed figure Find $x$ in the following figure.
$AB,AC,AD,BC,BE,CD$ are straight lines.
$AE=x$, $BE=CD=x-3$, $BC=10$, $AD=x+4$
$\angle BEC=90^{\circ}$, $\angle ADC=90^{\circ}$
NOTE: figure not to scale.
| By the Pythagorean theorem we have
$$\begin{equation*}
CE=\sqrt{10^{2}-\left( x-3\right) ^{2}}=\sqrt{91-x^{2}+6x}
\end{equation*}$$
and
$$\begin{equation*}
CD^{2}+AD^{2}=AC^{2}=\left( CE+AE\right) ^{2}
\end{equation*}.$$
So we have to solve the following irrational equation
$$\begin{equation*}
\left( x-3\right) ^{2}+\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/183088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
How to find $\lim\limits_{x\to0}\frac{e^x-1-x}{x^2}$ without using l'Hopital's rule nor any series expansion? Is it possible to determine the limit
$$\lim_{x\to0}\frac{e^x-1-x}{x^2}$$
without using l'Hopital's rule nor any series expansion?
For example, suppose you are a student that has not studied derivative yet (and... | Consider fundamental limit: $e = \lim\limits_{n\to \infty}(1+\frac{1}{n})^n$ and $e^x = \lim\limits_{n\to\infty}(1+\frac{x}{n})^n$
Proof
$e^x = [\lim\limits_{k\to\infty}(1+1/k)^k]^x = \lim\limits_{k\to \infty}((1+1/k)^{kx})\Rightarrow kx = n
\Rightarrow e^x = \lim\limits_{n\to\infty}(1+\frac{x}{n})^n$.
Understand the f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/184053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
"answer_id": 5
} |
$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\geq \frac{3}{\sqrt{2}}$ Let $a, b, c$ be positive real numbers such that $a\geq b\geq c$ and $abc=1$
prove that $$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\geq \frac{3}{\sqrt{2}}$$
| Let $a=b$. Then $c=\frac{1}{a^2}$ and the formula is :
$$f(a)=\sqrt{\frac{a}{2}}+\sqrt{a+\frac{1}{a^2}}$$
$$f'(a)=\frac{1}{2\sqrt{2a}}+(\frac{1}{2}-\frac{1}{a^3}).\frac{1}{\sqrt{a+\frac{1}{a^2}}}$$
The only root in $[0,\infty)$ of f'(a) is 1, hence $f(1)=\frac{3}{\sqrt{2}}$ is a minimum.
Now, what happens if $a\neq b$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/185825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
"answer_id": 1
} |
Finding all $x$ for $\frac{2x - 13}{2x + 3} \lt \frac{15}{x}$ I'm trying to find all $x$ for the inequality $\frac{2x - 13}{2x + 3} \lt \frac{15}{x}$. In order to do this, I want to factor one side so that I can find all values where $x$ determines the term to equal $0$.
$$\frac{2x - 13}{2x + 3} \lt \frac{15}{x} \iff \... | One approach is to find the (potential) "switchover" points. Note that one rational function can only become bigger than another after either intersecting or after a hole or vertical asymptote (of one or both). (Why?) There our no holes, and our asymptotes occur at $x=0$ and $x=-\frac32$. Intersection occurs when $x\ne... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/186770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Numerically evaluating the limit of $\frac{x^4-1}{x^3-1}$ as $x\rightarrow 1$ What is the limit as $x \to 1$ of the function
$$ f(x) = \frac{x^4-1}{x^3-1} . $$
| Applying L'Hôpital's rule makes it quite straightforward.
$$ \lim_{x \rightarrow 1}\; \frac{x^4-1}{x^3-1} = \lim_{x \rightarrow 1} \;\frac{4x^3}{3x^2} = \lim_{x \rightarrow 1} \;\frac{4}{3}x = \frac{4}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/188607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Why is $a^n - b^n$ divisible by $a-b$? I did some mathematical induction problems on divisibility
*
*$9^n$ $-$ $2^n$ is divisible by 7.
*$4^n$ $-$ $1$ is divisible by 3.
*$9^n$ $-$ $4^n$ is divisible by 5.
Can these be generalized as
$a^n$ $-$ $b^n$$ = (a-b)N$, where N is an integer?
But why is $a^n$ $-$ $b^n$$ ... | It’s a standard identity:
$$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\ldots+a^2b^{n-3}+ab^{n-2}+b^{n-1})\;.$$
It’s most neatly verified using summation notation, but you can also see what’s going on when you write everything out in extended form, as I did above. First,
$$\begin{align*}
a(a^{n-1}&+a^{n-2}b+a^{n-3}b^2+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/188657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "45",
"answer_count": 8,
"answer_id": 6
} |
Inequality $(a+\frac{1}{b})^2+(b+\frac{1}{c})^2+(c+\frac{1}{a})^2\ge 16$ For every real positive number $a,b,c$ such that $ab+bc+ca=1$, how to prove that:
$$\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2\ge 16$$
| Expanding the left hand side, we find that because for all non-zero real numbers $x$, $x^2 + x^{-2} \geq 2$ that
$$\left(a + \frac{1}{a}\right)^2 + \ldots \left(c + \frac{1}{c}\right)^2 \geq \frac{2a}{b} + \frac{2b}{c} + \frac{2c}{a} + 6.$$
Can you complete the argument?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/188790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Easy polynomials question? Please do this without using the quadratic formula.
If $\alpha$ and $\beta$ are zeroes of the polynomial $x^2 -6x + a$ then find the value of "$a$" if $3\times \alpha + 2\times \beta = 20$
Thank you for the help
There is also a second question of this sort, i dont get that either. Would help ... | First of all don't confuse $a$ with $\alpha$ !(It is better to substitute A or m for $a$)
We know that the sum of the roots of a quadratic equation $ax^2+bx+c$ is $x_1+x_2=-\frac{b}{a}$
So: $\alpha +\beta =6$
On the other hand:$3\times \alpha + 2\times \beta =20$
Solving this two equations two variables yields:$\beta =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/189593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Perimeter of an ellipse How can I calculate the perimeter of an ellipse? What is the general method of finding out the perimeter of any closed curve?
| If the semi-major axis has length $a$ and the eccentricity is $e$, then the perimeter is
$$
2πa \left(\frac{1}{1} \left(\frac{(-1)!!}{0!!} e^0\right)^2 + \frac{1}{-1} \left(\frac{1!!}{2!!} e^1\right)^2 + \frac{1}{-3} \left(\frac{3!!}{4!!} e^2\right)^2 + \frac{1}{-5} \left(\frac{5!!}{6!!} e^3\right)^2 + ⋯\right)
$$
wher... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/191307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Showing that $ \frac{a}{a^2+3}+\frac{b}{b^2+3}\leq\frac{1}{2}$ for $a,b > 0$ and $ab = 1$ using rearrangement inequalities Please help to solve the following inequality using rearrangement inequalities.
Let $a \gt 0$, $b \gt0$ and $ab=1$. Prove that
\begin{equation}\frac{a}{a^2+3}+\frac{b}{b^2+3}\leq\frac{1}{2}.\e... | We can assume $a \leq 1 \leq b$. Applying rearrangement inequalities to
$$
\begin{align}
a &\leq 1 \\
1 &\leq b
\end{align}
$$
we get
$$
a + b \geq 1 + ab = 2
$$
and
$$
b + 3a \geq 2a + 2 \\
a + 3b \geq 2b + 2
$$
Therefore
$$
\begin{align}
\frac{a}{a^2+3}+\frac{b}{b^2+3} &= \frac{1}{a + 3b} + \frac{1}{b + 3a} \leq\\ &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/191431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Evaluation of $ \sum_{k=0}^n \cos k\theta $ I just wanted to evaluate
$$ \sum_{k=0}^n \cos k\theta $$
and I know that it should give
$$ \cos\left(\frac{n\theta}{2}\right)\frac{\sin\left(\frac{(n+1)\theta}{2}\right)}{\sin(\theta / 2)} $$
I tried to start by writing the sum as
$$ 1 + \cos\theta + \cos 2\theta + \cdots... | Use:
$$
\sin \frac{\theta}{2} \cdot \cos(k \theta) = \underbrace{\frac{1}{2} \sin\left( \left(k+\frac{1}{2}\right)\theta\right)}_{f_{k+1}} - \underbrace{\frac{1}{2} \sin\left( \left(k-\frac{1}{2}\right)\theta\right)}_{f_{k}}
$$
Thus
$$ \begin{eqnarray}
\sin \frac{\theta}{2} \cdot \sum_{k=1}^n \cos(k \theta) &=&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/192065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
An intriguing definite integral: $\int_{\pi/4}^{\pi/2} \frac{x^2+8}{(x^2-16)\sin (x) + 8 x \cos(x)} \ dx$ I need some hints, suggestions for the following integral
$$\int_{\pi/4}^{\pi/2} \frac{x^2+8}{(x^2-16)\sin (x) + 8 x \cos(x)} \ dx$$
Since it's a high school problem, I thought of some variable change, integration ... | Following up on Blue's comment:
In general
$$
a\sin x+b\cos x = \sqrt{a^2+b^2}\Big( \frac{a}{\sqrt{a^2+b^2}}\sin x + \frac{b}{\sqrt{a^2+b^2}}\cos x \Big) = \sqrt{a^2+b^2} \left(\cos\varphi\sin x+\sin\varphi\cos x\right) = \sqrt{a^2+b^2}\sin(x+\varphi),
$$
so $\tan\varphi = \dfrac b a$.
Now putting in $a=x^2-16$ and $b=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/192710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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how to calculate the exact value of $\tan \frac{\pi}{10}$ I have an extra homework: to calculate the exact value of $ \tan \frac{\pi}{10}$.
From WolframAlpha calculator I know that it's $\sqrt{1-\frac{2}{\sqrt{5}}} $, but i have no idea how to calculate that.
Thank you in advance,
Greg
| $$\tan\frac{3\pi}{10}=\tan(\frac{\pi}{2}-\frac{2\pi}{10})=\cot\frac{2\pi}{10}$$
$$\frac{3\tan\frac{\pi}{10}-\tan^3\frac{\pi}{10}}{1-3\tan^2\frac{\pi}{10}}=\frac{\cot^2\frac{\pi}{10}-1}{2\cot\frac{\pi}{10}}$$
$$(3\tan\frac{\pi}{10}-\tan^3\frac{\pi}{10})(2\cot\frac{\pi}{10})=(\cot^2\frac{\pi}{10}-1)(1-3\tan^2\frac{\pi}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/196067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 4
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Could there be a formula of this? Say you have a function: f (m,n)
f (m,n) = m if n = 1
f (m,n) = n if m = 1
otherwise f (m,n) = f (m - 1, n) + f (m, n - 1)
Pre-calculated value:
1 2 3 4 5 6
2 4 7 11 16
3 7 14 25
4 11 25
5 16
6
Just wondering if there could be formula of f(m,n), instead of doing a dynami... | There is indeed. Let $g(m,n)=f(m-n+1,n)$, so that for instance $g(5,3)=f(3,3)=14$. The corresponding table for $g$ is:
$$\begin{array}{}
1\\
2&2\\
3&4&3\\
4&7&7&4\\
5&11&14&11&5\\
6&16&25&25&16&6
\end{array}$$
Now
$$\begin{align*}
g(m,n)&=f(m-n,n)\\
&=f(m-n-1,n)+f(m-n,n-1)\\
&=f\big((m-1)-n,n\big)+f\big((m-1)-(n-1),n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/200161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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finding $\frac{\sin 2x}{\sin 2y}+\frac{\cos 2x}{\cos 2y}$ If:
$$\frac{\cos x}{\cos y}=\frac{1}{2}$$ and $$\frac{\sin x}{\sin y}=3$$
How to find
$$\frac{\sin 2x}{\sin 2y}+\frac{\cos 2x}{\cos 2y}$$
| $$\frac{\sin{2x}}{\sin{2y}}=\frac{2\sin{x}\cos{x}}{2\sin{y}\cos{y}}=\frac{\sin{x}}{\sin{y}}\cdot \frac{\cos{x}}{\cos{y}}=3\cdot \frac{1}{2}=\frac{3}{2}. \tag{1}$$
$$\frac{\cos{2x}}{\cos{2y}}=\frac{\cos^{2}{x}-\sin^{2}{x}}{\cos^{2}{y}-\sin^{2}{y}}. \tag{2}$$
$$\frac{\cos{x}}{\cos{y}}=\frac{1}{2} \Leftrightarrow 4 \cdot\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/200621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Help me prove $\sqrt{1+i\sqrt 3}+\sqrt{1-i\sqrt 3}=\sqrt 6$ Please help me prove this Leibniz equation: $\sqrt{1+i\sqrt 3}+\sqrt{1-i\sqrt 3}=\sqrt 6$. Thanks!
| I'm pretty new to proofs, but it seems like you could just evaluate $\sqrt{1+i\sqrt{3}}+\sqrt{1-i\sqrt{3}}$ and if it equals $\sqrt{6}$, then it's proof enough (though this is admittedly a convoluted step-by-step that I derived from WolframAlpha):
\begin{aligned}
\sqrt{1+i\sqrt{3}}+\sqrt{1-i\sqrt{3}} &= \sqrt{\frac{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/203462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 6,
"answer_id": 5
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Problem using Jensen's Inequality For some convex function $f$, and elements $x_{1},x_{2},x_{3}$ of its domain, show that:
$$f(x_{1})+f(x_{2})+f(x_{3})+f\left( \frac{1}{3}(x_{1}+x_{2}+x_{3}\right)\ge \frac{4}{3}\left( f\left(\frac{x_{1}+x_{2}}{2} \right)+f\left(\frac{x_{2}+x_{3}}{2} \right)+f\left(\frac{x_{3}+x_{1}}{2}... | Use these:
$f\left(\displaystyle\frac{x_i+x_j}2\right) \le \displaystyle\frac{f(x_i)+f(x_j)}{2}$, and
$f\left(\displaystyle\frac{y_1+y_2+y_3}3\right) \le \displaystyle\frac{f(y_1)+f(y_2)+f(y_3)}{3}$.
This applied to $y_1=\displaystyle\frac{x_2+x_3}2$,.. we also have
$f\left(\displaystyle\frac{x_1+x_2+x_3}3\right) \le \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/204430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove $\sqrt[3]{60}>2+\sqrt[3]{7}$ Prove $$\sqrt[3]{60}>2+\sqrt[3]{7}$$
I try to both sides of the cubic equation, but it is quite complicated
| Well, I don't see any alternative way..
$$60\overset{?}> 8+6\sqrt[3]{7^2}+12\sqrt[3]7+7 $$
$$45\overset{?}> 6\sqrt[3]{7^2}+12\sqrt[3]7 $$
$$15\overset{?}> 2\sqrt[3]{7^2}+4\sqrt[3]7 $$
Well, we could raise it to cubic, but that's really not nice.
What about considering the roots of $2x^2+4x-15 =2(x+1)^2-17$, and finally... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/207036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Evaluate $x+4/x$ when $x$ is a solution of $x-6/\sqrt{x}=11$ I have a question that I see in a math text.
I asked to my teacher but she couldn't find the solution, too.
If
$$x - \frac{6}{\sqrt{x}}=11,$$
then to what is equal
$$x + \frac{4}{x}\text{ ?}$$
| AS $x-\frac{6}{\sqrt{x}}-11=(\sqrt{x}+3)(\sqrt{x}-\frac{2}{\sqrt{x}}-3)$, so from $x-\frac{6}{\sqrt{x}}=11$ and $\sqrt{x}\geq0$, we can obtain that $\sqrt{x}-\frac{2}{\sqrt{x}}=3$, squaring at both sides, i.e. $(\sqrt{x}-\frac{2}{\sqrt{x}})^{2}=x+\frac{4}{x}-4=9$, so $x+\frac{4}{x}=13$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/207183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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How to prove an Inequality I'm beginner with proofs and I got the follow exercise:
Prove the inequality $$(a + b)\Bigl(\frac{1}{a} + \frac{4}{b}\Bigr) \ge 9$$ when $a > 0$ and $b > 0$. Determine when the equality occurs.
I'm lost, could you guys give me a tip from where to start, or maybe show a good resource for beg... | The first thing to do is simplify the expression on the lefthand side of the inequality.
$$(a + b)\left(\frac{1}{a} + \frac{4}{b}\right)=\frac{(a+b)(4a+b)}{ab}=\frac{4a^2+5ab+b^2}{ab}=\frac{4a}b+5+\frac{b}a\;.$$
Now notice that the resulting expression contains both $\frac{a}b$ and $\frac{b}a$; this is an indication th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/207521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 8,
"answer_id": 7
} |
Two Diophantine equations What is known about solutions in integers of the equations; $x^4 - y^2 = z^6$
I got $x=4st(s^4 - t^4)$ , $z=4st(s^2 - t^2)$ , $y=(4st(s^2 - t^2))^2 (s^4 + t^4 - 6 (st)^2) $
and, the equation $x^2 - y^4 = z^6 $
I got $y=4st(s^4 - t^4) , z=4st(s^2 + t^2) , x=(4st(s^2 + t^2))^2 (s^4 + t^4 + 6 (s... | Here is another solution of $x^4 - y^2 = z^6$. Suppose we have a Pythagorean triple $a^2 + b^2 = c^2$, and c is itself a square, say $d^2$. There are many such solutions, a sufficient condition being that c can be factorised into primes of the form $4n+1$, with an even exponent of each prime. Where this is the case e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/207668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to solve the cubic equation $x^3-12x+16=0$ Please help me for solving this equation $x^3-12x+16=0$
| This is a cubic equation. The forme of cubic equation is $x^3+px+q=0$ where
$1)$ $p$, $q$ $\in$ R
$2)$ $D=(\frac{q}{2})^2+(\frac{p}{3})^3$
$3)$ $x=u+v=\sqrt[3]{-\frac{q}{2}+\sqrt{(\frac{q}{2})^2 +(\frac{p}{3})^3}}+\sqrt[3]{-\frac{q}{2}-\sqrt{(\frac{q}{2})^2 +(\frac{p}{3})^3}}$
For the given equation we have:
$p=-12$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/208183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Check my answer for the derivative of $y=\tan^{-1}(40/h) -\tan^{-1}(32/h)$
Differentiate
$$y=\tan^{-1}{\frac {40}{h}}-\tan^{-1}{\frac {32}{h}}$$
My answer
Using the identity $\frac{d}{dy}\tan^{-1}(x)=\frac{1}{1+x^2}$, can I conclude that
$$\frac{dy}{dh}=\frac{1}{1+(\frac{40}{h})^2}(-\frac {40}{h^2})-\frac{1}{1+(\frac... | Why don't we use
$$\tan^{-1}\frac a h= \frac \pi 2-\cot^{-1}\frac a h=\frac \pi 2-\tan^{-1}\frac h a.$$
So, $$\frac{d(\tan^{-1}\frac a h)}{dh}=\frac{d(\frac \pi 2-\tan^{-1}\frac h a)}{dh}=-\frac{d(\tan^{-1}\frac h a)}{dh}=-\frac{1}{1+(\frac h a)^2}\frac 1 a=-\frac{a}{h^2+a^2}$$
So, $$\frac{d(\tan^{-1}\frac {40} h)}{dh}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/209175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find the distance between the line and plane Find the distance between the line $x-z=3, x+2y+4z=6$ and the plane $3x+2y+2z=5$.
What I have done so far,
Found the vector by crossing $1,0-1$ and $1,2,4$ from the first line
The line is parallel to plane $\langle 2,-5,2\rangle \cdot \langle 3,2,2 \rangle = 0$
I think my ne... | Put $(P_1): x - z -3 = 0$, $(P_2): x+2y+4z=6$, $(P_3): 3x + 2y +2z -5 = 0$ and $\Delta$ is intersection of two plane $(P_1)$ and $(P_2)$. We have, a normal vector of the plane $(P_1)$ is $a = (1,0,-1)$, a normal vector of the plane $(P_2)$ is $b = (1,2,4)$. A direction vector $v$ of $\Delta$ is cross product of $a$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/209910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Simplification of Sum I can't see how to simplify from step 1 to step 2 in the following example:
*
*$$
\frac{1}{3}n(n+1)(n+2)+(n+2)(n+1)
$$
*$$
(\frac{1}{3}n+1)(n+1)(n+2)
$$
Thanks to the answers this is how I got from 1 to 2:
1.1
$$
\frac{1}{3}n(n+1)(n+2)+1(n+2)(n+1)
$$
1.2
$$
(n+2)\left(\frac{1}{3}n(n+1)+1(n+... | Factor out $(n+1)(n+2)$. What's left in each term? What's the sum of those two expressions?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/210102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve $5a^2 - 4ab - b^2 + 9 = 0$, $ - 21a^2 - 10ab + 40a - b^2 + 8b - 12 = 0$
Solve $\left\{\begin{matrix} 5a^2 - 4ab - b^2 + 9 = 0\\ - 21a^2 - 10ab + 40a - b^2 + 8b - 12 = 0. \end{matrix}\right.$
I know that we can use quadratic equation twice, but then we'll get some very complicated steps. Are there any elegant wa... | By arrange the terms, you can note that
$$\begin{cases}
5a^{2}-4ab-b^2+9=0\\
-21a^{2}-10ab+40a-b^{2}+8b-12=0
\end{cases}
\Leftrightarrow
\begin{cases}
9(a^{2}+1)=(2a+b)^{2}\\
4(a^{2}+1)=(5a+b-4)^{2}
\end{cases}$$
So we get
$$\begin{cases}
9(a^{2}+1)=(2a+b)^{2}\\
4(2a+b)^{2}=9(5a+b-4)^{2}
\end{cases}$$
According to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/210454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
} |
number of integral solutions for $x^2+y^2=5^k$ Prove that the equation $x^2+y^2=5^k$ has $4k+4$ integral solution.
Any ideas would be appreciated.
Thanks
| Lets use Strong form of mathematical induction.
Base Case: $k=0$ and $k=1$ are obvious.
Assume $x^2+y^2=5^k$ has exactly $8$ solutions $(x,y)$ such that $x$ and $y$ are not divisible by $5$ along with $4k-4$ solutions of the form $(5a,5b)$ where $(a,b)$ are solution of $a^2+b^2=5^{k-2}$. We need to show similar for $k+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/211270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to construct a bijection from $(0, 1)$ to $[0, 1]$?
Possible Duplicate:
Bijection between an open and a closed interval
How do I define a bijection between $(0,1)$ and $(0,1]$?
I wonder if I can cut the interval $(0,1)$ into three pieces:
$(0, \frac{1}{3})\cup(\frac{1}{3},\frac{2}{3})\cup(\frac{2}{3},1)$, in whi... | The idea you mention, with mild modification, will work. Please note that what is below is a minor variant of the solution given by Patrick da Silva.
Your decomposition of $(0,1)$ is not quite complete. We want to write
$$(0,1)=\left(0,\frac{1}{3}\right) \cup\left\{\frac{1}{3}\right\} \cup \left(\frac{1}{3},\frac{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/213391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Get the equation of a circle when given 3 points Get the equation of a circle through the points $(1,1), (2,4), (5,3) $.
I can solve this by simply drawing it, but is there a way of solving it (easily) without having to draw?
| Based on what I've learnt from this page by Stephen R. Schmitt (thanks to @Dan_Uznanski for pointing the concept out), there's a faster way to do it using matrices:
$$\text{let }〈x_1, y_1〉, 〈x_2, y_2〉, 〈x_3, y_3〉\text{ be your 3 points, and let }〈x_0, y_0〉\text{ represent the center of the circle.}\\\
\\
\text{let }A =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/213658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 16,
"answer_id": 4
} |
Proving trigonometric Identity: $\frac{1+\sin x}{\cos x} = \frac{1+\sin x+\cos x}{1-\sin x+\cos x}$ I would like to try and prove
$$\frac{1+\sin x}{\cos x} = \frac{1+\sin x+\cos x}{1-\sin x+\cos x}$$
using $LHS=RHS$ methods, i.e. pick a side and rewrite it to make it identical to the other side.
I found a quick way by ... | We'll work on the RHS. We'll multiply $\frac{1+\sin x+\cos x}{1-\sin x+\cos x}$ with the conjugate of the denominator. We'll do this from right to left, i.e. we pick the RHS and walk through to LHS.
Solution 1:
$$\require{cancel}\begin{aligned}\frac{1+\sin x+\cos x}{1-\sin x+\cos x}&=\frac{1+\sin x+\cos x}{1-\sin x+\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/213788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Algebraic manipulation help I'm self-teaching geometric progressions and was asked to find the sum to $n$ terms of $$x + 1 + \frac{1}{x} + \cdots$$ so I used the formula $$\frac{a(r^n - 1)}{r - 1}$$ with $a = x$ and $r = \frac{1}{x}$, and I arrived at $$\frac{x^{1-n} - x}{x^{-1} - 1}$$ The answer in the book is $$\frac... | Here’s a computational verification.
$$\begin{align*}
\frac{x^{1-n}-x}{x^{-1}-1}&=\frac{x^{1-n}-x}{x^{-1}-1}\cdot\frac{x^{n-1}}{x^{n-1}}\\
&=\frac{1-x^n}{x^{n-2}-x^{n-1}}\\
&=\frac{1-x^n}{x^{n-2}\left(1-x\right)}\\
&=\frac{1-x^n}{x^{n-2}\left(1-x\right)}\cdot\frac{-1}{-1}\\
&=\frac{x^n-1}{x^{n-2}(x-1)}
\end{align*}$$
H... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/214150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Mechanism Behind Dot Product and Least Square Sorry for my ignorance, but I want to know how the mechanism of finding the least square solutions or the closest points in Euclidean space works.
For example:
Find the closest point or points to $b =(−1,2)^T$ that lie on the line $x + y = 0$.
I know the answer is
$$\f... | Clearly your first equation must be incorrect -- the dot product returns a scalar, so there is no way to obtain a vector on the right hand side, unless you are overloading your nomenclature.
The dot product returns the sum of the element-wise products of vectors. When you take the dot product of a vector with itself, t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/214577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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If $\dfrac{4x^2-1}{4x^2-y^2}$ is an integer, then it is $1$ The problem is the following:
If $x$ and $y$ are integers such that $\dfrac{4x^2-1}{4x^2-y^2}=k$ is
also an integer, does it implies that $k=1$?
This equation is equivalent to $ky^2+(1-k)4x^2=1$ or to $(k-1)4x^2-ky^2=-1$. The first equation is a pell equat... | This is a fun problem! Where did you find it?
There are no solutions except for $k=1$. Assume from now on that $k \neq 1$. Since $k$ is clearly odd, it is also not $0$ and we deduce that $k(k-1)>0$. It is convenient to set $M = k(k-1)$. Also, we may assume WLOG that $x$ and $y\geq0$.
As you did, rewrite the equation to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/215372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 0
} |
Find the value of : $\lim_{n\to\infty}\prod_{k=1}^n\cos\left(\frac{ka}{n\sqrt{n}}\right)$ Find the limit (where a is a constant)
$\lim_{n\to\infty}\prod_{k=1}^n\cos\left(\frac{ka}{n\sqrt{n}}\right)$
I think the answer is $1-a^2/6$
| Let $$f(n) = \prod_{k=1}^n \cos \left(\dfrac{ka}{n \sqrt{n}}\right)$$
$$g(n) = \log (f(n)) = \sum_{k=1}^{n} \log \left(\cos \left(\dfrac{ka}{n \sqrt{n}}\right) \right) = \sum_{k=1}^{n} \log \left(1 - \dfrac{\left(\dfrac{ka}{n \sqrt{n}}\right)^2}2 + \mathcal{O} \left( \dfrac{k^4}{n^6}\right)\right)$$
$$\log \left(1 - \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/216617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
} |
$x^2 \equiv 2x \pmod m$ Toward counting the solutions for the congruence $x^2 \equiv 2x \pmod n$, if we write $m$
as $m = p_1^{a_1}p_2^{a_2}...p_r^{a_r}$
we have the following equivalent system of congruence equations:
$p_1^{a_1}\mid x(x-2)$
$p_2^{a_2}\mid x(x-2)$
.
.
.
$p_r^{a_r}\mid x(x-2)$
I am thinking of the two... | Using your approach, let for prime $p,$ $p\mid x$ and $p\mid (x-2)$
$\implies p$ divides $x-(x-2)=2\implies p=2$ for $p^n\mid x(x-2)$
So, if prime $p>3,$ either $p^n\mid x$ or $p^n\mid (x-2)$
So, there are $2$ in-congruent solutions, namely, $0,2\pmod {p^n}$ of $x^2\equiv2x \pmod {p^n}$.
If case $p=2,x$ must be even $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/216784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Solving the recurrence relation $T(n)=2T(n/4)+\sqrt{n}$ I've solved $T(n)=2T(n/4)+\sqrt{n}$ to equal $2^{\log_{4}n}(\log_{4}n+1)$, but I'm not sure how to solve it directly.
I have:
$2(2T(\frac{n}{16})+\sqrt{\frac{n}{4}})+\sqrt{n} = 4T(\frac{n}{16})+2\sqrt{n}$
$2(4T(\frac{n}{64})+2\sqrt{\frac{n}{16}})+\sqrt{n} = 8T(\fr... | Let's turn the equation $T(n) = 2 T(n/4) + \sqrt{n}$ into a recurrence equation. To this end, let $f(m) = T(4^m p)$ for some $p>0$. Then
$$
f(m) = 2 f(m-1) + \sqrt{p} 2^m
$$
which can be systematically solved. First rewrite it as
$$
2^{-m} f(m) - 2^{-(m-1)} f(m-1) = \sqrt{p}
$$
Then sum equations from $m=1$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/217211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
General term of the series Given the series:
$$\sqrt c + \sqrt{c\sqrt c} + \sqrt{c\sqrt{c\sqrt c}} + \ldots$$
where $0 < c < 1$
What is the general term of this series?
| $a_1 = \sqrt{c} = c^{1/2}$, $a_2 = \sqrt{c\sqrt{c}} = \sqrt{c^{3/2}} = c^{3/4}$, $a_3 = \sqrt{c\sqrt{c\sqrt{c}}} = \sqrt{c \times c^{3/4}} = c^{7/4}$. In general, $$a_{n+1} = \sqrt{c a_{n-1}}$$ with $a_0 = 1$.
This gives us
$a_n = c^{1-1/2^n}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/218202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
integral of exponential divided by polynomial I would like to solve the integral $$A\int_{-\infty}^\infty\frac{e^{-ipx/h}}{x^2+a^2}dx$$
where h and a are positive constants.
Mathematica gives the solution as $\frac\pi{a}e^{-|p|a/h}$, but I have been trying to reduce my reliance on mathematica. I have no idea what metho... | $\mathbf{Method\;1: }$ Integral Fourier Transform
Consider the function $f(t)=e^{-a|t|}$, then the Fourier transform of $f(t)$ is given by
$$
\begin{align}
F(\omega)=\mathcal{F}[f(t)]&=\int_{-\infty}^{\infty}f(t)e^{-i\omega t}\,dt\\
&=\int_{-\infty}^{\infty}e^{-a|t|}e^{-i\omega t}\,dt\\
&=\int_{-\infty}^{0}e^{at}e^{-i\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/219098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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Inequality $\frac{x^3+1}{\sqrt{x^4+y+z}}+\frac{y^3+1}{\sqrt{y^4+z+x}}+\frac{z^3+1}{\sqrt{z^4+x+y}}\geq 2\sqrt{3}$ If we Let $x,y,z>0$ such that $x+y+z=3$. how to prove that
$$\frac{x^3+1}{\sqrt{x^4+y+z}}+\frac{y^3+1}{\sqrt{y^4+z+x}}+\frac{z^3+1}{\sqrt{z^4+x+y}}\geq 2\sqrt{3}$$
| We shall prove the inequality with the weaker condition $x, y, z \geq 0$.
$$\sum_{cyc}{\frac{x^3+1}{\sqrt{x^4+y+z}}}=\sum_{cyc}{\frac{x^3+1}{\sqrt{x^4-x+3}}}$$
Let $f(x)={\frac{x^3+1}{\sqrt{x^4-x+3}}}$. Differentiate twice, and note that $f''(x)$ is positive for $0 \leq x \leq 1$, and that $f''(x)=0$ for exactly 1 valu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/220942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How to take the gradient of the quadratic form? It's stated that the gradient of:
$$\frac{1}{2}x^TAx - b^Tx +c$$
is
$$\frac{1}{2}A^Tx + \frac{1}{2}Ax - b$$
How do you grind out this equation? Or specifically, how do you get from $x^TAx$ to $A^Tx + Ax$?
| Yet another approach.
We will utilize the following the identities
*
*Trace and Frobenius product relation $$\left\langle A, B \right\rangle={\rm tr}(A^TB) = A:B$$ or $$\left\langle A^T, B \right\rangle ={\rm tr}(AB) = A^T:B$$
*Cyclic property of Trace/Frobenius product
\begin{align}
\left\langle A, B C \right\ra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/222894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "71",
"answer_count": 5,
"answer_id": 2
} |
Inequality. $\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a} \geq \frac{ab+bc+ca}{2}$ prove the following inequality:
$$\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a} \geq \frac{ab+bc+ca}{2},$$ for $a,b,c$ real positive numbers.
Thanks :)
| $$\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a} =\frac{a^4}{a^2+ab}+\frac{b^4}{b^2+bc}+\frac{c^4}{c^2+ac}\geq \frac{(a^2+b^2+c^2)^2}{a^2+b^2+c^2+bc+ca+ab}.\tag{1}$$ But $$a^2+b^2+c^2 \geq ab+bc+ca $$ so
$$a^2+b^2+c^2+ab+bc+ca \leq 2(a^2+b^2+c^2)$$
$$(1) \geq \frac{(a^2+b^2+c^2)^2}{2(a^2+b^2+c^2)} \geq \frac{ab+bc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/222934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
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Solving 4 Quadratic Simultaneous Equations
Consider the equations:
$$\begin{cases}
ab+c+d=3\\
bc+d+a=5\\
cd+a+b=2\\
da+c+b=6
\end{cases}$$
Where $a,b,c,d \in \mathbb{R}$. How can we find $a,b,c,d$?
The furthest I've got is by adding the first two equations and the last two equations together to get:
$ab+bc+c+d+d+a=... | Take $\#1 - \#2 + \#3 - \#4$ to get $(b-d)(a-c)=-6$
$\#1 + \#2 - \#3 - \#4$ gives $(b-d)(a+c-2) = 0$
$\#1 - \#2 - \#3 + \#4$ gives $(b+d-2)(a-c)=2$.
So now $a+c-2 = 0$, and $$\frac{1}{a-c} = \frac{b-d}{-6} = \frac{b+d-2}{2}$$
That gives us $d=3-2b$ and $a-c = \dfrac{-2}{b-1}$ (in particular, $b \ne 1$).
Together with ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/227100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Triple integral for Volume of a region $Q$ is the solid bounded by the plane $x+2y+2z=2$ and above paraboloid $x=z^2+y^2$. Setup the triple integral for the volume of $Q$.
I tried to find vertices and go from there bu t got confused. I need help with this problem.
Thanks in advance.
| When I approach a problem like this I consider a point $(x,y,z)$ and work towards discovering a set of 3 inequalities which describe an arbitrary point in the given region. The paraboloid $x=z^2+y^2$ opens into the positive $x$-half volume. On the other hand $x = 2-2y-2z$ intersects the paraboloid when $z^2+y^2=2-2y-2z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/229329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do I find a value for $ 2^{156221} - 1\pmod 9$? Again the problem is: Calculate the value of:
$$\left(2^{156221} - 1\right) \bmod 9$$
I have no idea how to find a solution to this and need help urgently!!
Thank you in advance.
| The problem has already been solved several times. Perhaps a marginally different point of view will be helpful. The hard part of the problem is to calculate
$$2^{156221}\pmod{9}.$$
Let us calculate $2^n\pmod{9}$, starting at $n=0$.
We have $2^0\pmod{9}=1$, $2^1\pmod{9}=2$, $2^2\pmod{9}=4$, $2^3\pmod{9}=8$, and $2^{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/229972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
A functional equation: $f\left(x^2-1\right)+2f\left(\frac{2x-1}{(x-1)^2}\right)=2-\frac{4}{x}+\frac{3}{x^2}$
If for all $ x > 1 $
$$
f\left(x^2-1\right)+2f\left(\frac{2x-1}{(x-1)^2}\right)=2-\frac{4}{x}+\frac{3}{x^2} \text,
$$
then $f(x)=?$
I don't know how to solve such equations. Help me please. Thank you.
| The solution is based on finding a change of variable $x\to\phi(x)$ such that $\phi(\phi(x))=x$ leading to a system of 2 equations in 2 variables.
Let's rewrite
$$\frac{2x-1}{\left(x-1\right)^{2}}=\frac{-x^{2}+2x-1+x^{2}}{\left(x-1\right)^{2}}=\frac{x^{2}-\left(x-1\right)^{2}}{\left(x-1\right)^{2}}=\frac{x^{2}}{\left(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/237506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Riemann-integrable function The following multiple choice question was asked in my exam but I don't know how to proceed:
Define $f:[0,1]\to [0,1]$ by $\displaystyle f(x)=\frac{2^{k}-1}{2^{k}}$ for $\displaystyle x\in [\frac{2^{k-1}-1}{2^{k-1}},\frac{2^{k}-1}{2^{k}}],k\geq 1$. Then $f$ is a Riemann-integrable function s... | $$\int_0 ^1f=\int_0^{\frac{1}{2}} \frac{1}{2}+\int_{\frac{1}{2}}^{\frac{3}{4}}\frac{3}{4}+\int_{\frac{3}{4}} ^{\frac{7}{8}}\frac{7}{8}+... =\sum_{k=1}^\infty \int_{\frac{2^{k-1}-1}{2^{k-1}}} ^{\frac{2^k-1}{2^k}} \frac{2^k-1}{2^k}=\sum_{k=1}^\infty (\frac{2^k-1}{2^k})(\frac{2^k-1}{2^k}-\frac{2^{k-1}-1}{2^{k-1}})=\sum_{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/238540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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If $(a, b, c)$ is a Pythagorean triple, then $(a+1, b+1, c+1)$ is not a Pythagorean triple. Pythagoras stated that there exist positive natural numbers, $a$, $b$ and $c$ such that $a^2+b^2=c^2$. These three numbers, $a$, $b$ and $c$ are collectively known as a
Pythagorean triple. For example, $(8, 15, 17)$ is one of th... | We must suppose that $(a+1,b+1,c+1)$ is in fact a Pythagorean triple, and that $(a,b,c)$ is, too. Then we have $$(a+1)^2+(b+1)^2=(c+1)^2\tag{1}$$ and $$a^2+b^2=c^2.\tag{2}$$
Expand $(1)$--using for example that $(a+1)^2=a^2+2a+1$--and then use $(2)$ to eliminate all the squared terms from the resulting equation. You sh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/239312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.