Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Determine the general term of a Recurrent Rational Sequence How to determine the general term of the $(u_n)$ sequence with $u_0 \in \mathbb R_+$ and $u_{n+1}=\frac{u_n^2+1}{1+2u_n}$ ?
Source : les dattes à Dattier
| 1/The trick :
Let $t,u,v$ functions with $t\circ v=v\circ u$, then : $t^n\circ v=v\circ u^n$.
Here, we have choosen, $t,u,v$ of the type : $$t(x)=\frac{x^2+1}{2x+1},
u(x)=x^2 \text{ and } v(x)=\frac{ax+b}{cx+d}$$
2/For the solution proposed by Francesco, the form is :
$$h_n=f\circ g^n\circ f^{-1}$$
with $$f(x)=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2114710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
What number should be subtracted from What number should be subtracted from $4x^3+5x+3$
so that the resulting polynomial leaves remainder $-80$ when divided by $2x+5$?.
Let the required number to be subtracted be $K$.
let: $$P(x)=4x^3+5x+3-k$$
$$g(x)=2x+5=2(x+5/2)$$
Comparing $g(x)$ with $x-a $ , $a =\frac {-5}{2}$
... | Let $f (x)=2x+5$ and $p (x)=4x^3+5x-3$. We have that: $$p (-\frac {5}{2}) -k = -80$$ $$\Rightarrow 4 (-\frac {5}{2})^3 +5 (-\frac {5}{2}) +3 - k = -80$$ $$-75+3-k = -80$$ $$\boxed {k = 8} $$ Hope it helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2117512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Find y from given information.
Find $y$ given that $$x^{2}+\frac{1}{x^{2}} = 98;$$ and
$$x^{3}+\frac{1}{x^{3}} = y.$$
I've got $970$, but that's not the full answer, since I got half of the points. What else could I get here?
| We know,
$(x + \frac 1x)^2 = (x^2 + \frac 1{x^2} + 2.x.\frac 1x)$
$(x + \frac 1x)^2 = (98 + 2)$
$(x + \frac 1x)^2 = 100$
$(x + \frac 1x) = \pm10$
Case 1 - $(x + \frac 1x) > 0$
$(x + \frac 1x) = 10$
Now cube of above equation,
$(x + \frac 1x)^3 = 1000$
$x^3 + \frac 1{x^3} + 3.x.\frac 1x(x + \frac 1x) = 1000$
$x^3 + \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2117951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Any hints on how to show that $\frac{2 - \sqrt{3}}{2 \sqrt{3}} < \frac{1}{10} $ I am trying to prove that $$\frac{2 - \sqrt{3}}{2 \sqrt{3}} < \frac{1}{10} $$
My attempt was to suppose that $\frac{2 - \sqrt{3}}{2 \sqrt{3}} \geq \frac{1}{10}$, and show that it was absurd.
Here's what I did:
Suppose that $\frac{2 - \sqrt{... | $$
\frac{2-\sqrt 3}{2 \sqrt3} = \frac{1}{\sqrt 3} - \frac{1}{2}
$$
Now, suppose we assume that the opposite of what is given is true. We get a contradiction:
$$
\frac{1}{\sqrt 3} - \frac 12 \geq \frac 1{10} \implies\frac{1}{\sqrt 3} \geq \frac{3}{5} \implies \frac{1}{3} \geq \frac{9}{25} \implies 25 \geq 27
$$
which i... | {
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"url": "https://math.stackexchange.com/questions/2119085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Recursion - series examples Given:
$$p(n+1) = 2^n \cdot \sqrt{2\cdot {\left( 1 - \sqrt{1 - \left(\frac{p(n)}{2^n}\right)^2}\right)}}$$
And:
$$p(2) = 2\cdot \sqrt{2}$$
Find $p(n)$. I an unable to come up with a generalization for $p(n)$. Please help.
| \begin{align*}
P_{2} &= 2\sqrt{2} \\
P_{3} &= 4\sqrt{2-\sqrt{2}} \\
P_{4} &= 8\sqrt{2-\sqrt{2+\sqrt{2}}} \\
P_{5} &= 16\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}
\end{align*}
$P_{n}$ looks like the perimeter of regular $2^{n+1}$-sided polygon of "diameter" $1$:
$$P_{n}=2^{n} \sin \frac{\pi}{2^{n}}$$
It's not difficu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2119525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is it possible to integrate $\int \frac{\mathrm{d}x}{\sin^2 x+\sin x+1}$ without involving complex numbers? The integral is $$\int \frac{\mathrm{d}x}{\sin^2 x+\sin x+1}$$
Consider
$$\int \frac{\mathrm{d}x}{ax^2+bx+c}$$
There are a total of three cases, depending on the discriminant of $ax^2+bx+c$. Two of which are sho... | Let $\sin x =\frac{1-t^2}{1+t^2}$, or $t= \tan(\frac\pi4-\frac x2)$
\begin{align}
&\int \frac{1}{\sin^2 x+\sin x+1}\ dx
=\int\frac{2+2{t^2}}{3+t^4} \ dt\\
=&\ (1+\frac1{\sqrt3}) \int \frac{1+\frac{\sqrt3}{t^2}}{t^2+\frac3{t^2}}dt +(1-\frac1{\sqrt3}) \int \frac{1-\frac{\sqrt3}{t^2}}{t^2+\frac3{t^2}}dt\\
=&\ \frac{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2120557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Calculate the sum of a function series with indefinite integral We have the series
$$
f(x) = \sum\limits_{n=0}^{\infty}\frac{n}{n+2}(x^2-4)^{2n}
$$
and it's asked to find the result of the series.
I've tried this approach
$$
\frac{n}{n+2} = \left(\frac{n+2}{n+2}-\frac{2}{n+2}\right) = 1 - \frac{2}{n+2}
$$$$
y = (x^2-4)... | $$
\begin{align}
-\sum_{n=0}^{\infty}\frac{y^n}{n+2} &= -\frac{1}{y^2}\,\sum_{n=0}^{\infty}\frac{y^{n+2}}{n+2} = -\frac{1}{y^2}\,\sum_{n=0}^{\infty}\,\int_{\color{red}{0}}^{\color{red}{y}}t^{n+1}\,dt = -\frac{1}{y^2}\,\int_{0}^{y}\,\sum_{n=0}^{\infty}\,t^{n+1}\,dt \\[2mm]
&= -\frac{1}{y^2}\,\int_{0}^{y}\frac{t}{1-t}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2121463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Find the rank of the following matrix depending on $\lambda\in\Bbb R$ Find the rank of the following matrix depending on $\lambda\in\Bbb R$.
$$A=\begin{pmatrix}
1&2&3&4\\
2&\lambda&6&7\\
3&6&8&9\\
4&7&9&10
\end{pmatrix}$$
My attempt:
$$\begin{pmatrix}
1&2&3&4\\
2&\lambda&6&7\\
3&6&8&9\\
4&7&9&10
\end{pmatrix}\sim\begin... | It looks fine to me!! Or, we have $A=\begin{pmatrix}
1&2&3&4\\
2&\lambda&6&7\\
3&6&8&9\\
4&7&9&10
\end{pmatrix}$ with $\lambda \in \Bbb{R}$ and we know: $$r(A) \leq 4 \text{ and } \big[r(A)= 4 \leftrightarrow \det(A)\neq 0\big]$$
We have $\displaystyle\det(A)= 13 - 3 \lambda$ therefore: $$ r(A) =4\leftrightarrow \bigg(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Determine convergence $\sum_{n=0}^{\infty}\frac{n!\cdot 2^n}{1\cdot 3\cdot 5\cdot \cdot \cdot (2n+1)} $ I tried Ratio test for this series and got 1.
$$\sum_{n=0}^{\infty}\frac{n!\cdot 2^n}{1\cdot 3\cdot 5\cdot \cdot \cdot (2n+1)}$$
Any suggestions?
I believe the limit of the sequence goes to $\infty$ , but I don't kno... | Since
\begin{align}\frac{n!2^n}{1\cdot3\cdot5\cdots(2n+1)}=&\frac{2\cdot4\cdot6\cdots2n}{1\cdot3\cdot5\cdots(2n+1)}\\=&2\cdot\frac{4}{3}\cdot\frac{6}{5}\cdots\frac{2n}{2n-1}\cdot\frac{1}{2n+1}\\ \geq&\frac{2}{2n+1}\geq\frac{2}{2n+2}=\frac{1}{n+1}\end{align}
we have that
$$
\sum_{n=0}^\infty\frac{n!2^n}{1\cdot3\cdot5\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2124300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Solve the system ${x}^{2}+ \left( y-1 \right) ^{2}=4,z^{4}+y{z}^{2}+xz+1=0. $ Find all real solutions of the system of equation
\begin{cases}
{x}^{2}+ \left( y-1 \right) ^{2}=4,\\{z}^{4}+y{z}^{2}+xz+1=0.
\end{cases}
| From second equation we get $y=-z^2-\frac{x}{z}-\frac{1}{z^2}$ (it's obvious that $z\neq0$).
Thus, $$x^2+\left(z^2+\frac{1}{z^2}+1+\frac{x}{z}\right)^2=4$$ or
$$\left(1+\frac{1}{z^2}\right)x^2+\frac{2}{z}\left(z^2+\frac{1}{z^2}+1\right)x+\left(z^2+\frac{1}{z^2}+1\right)^2-4=0,$$
which gives
$$\frac{1}{z^2}\left(z^2+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2124568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Fourier Series Proof where $f(x)$ satisfies the symmetry condition $f(L-x)=f(x)$ Let $f(x)$, (where $−L < x < L$), be an odd function that satisfies the symmetry condition $f (L − x) = f (x)$, for all $x$.
Show that $A_n =0$, $\forall n$ while $Bn =0$ for all even $n$.
Some definitions:
Fourier Series: $A_0+\sum_{n=1}... | (Hoping no calculation mistakes will happen)
For all $n$ you have
$$
B_{n} = \frac{2}{L} \int_{0}^{L} f(x) \sin \left( \frac{2 \pi x}{L} \right) dx
$$
Specifically for $2n$ you have
$$
B_{2n} = \frac{2}{L} \int_{0}^{L} f(x) \sin \left( \frac{2 \pi n x}{L} \right) dx
$$
Defining $g(x) = f\left(x - \frac{L}{2} \right)$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2125525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to show this inequality $\sum\sqrt{\frac{x}{x+2y+z}}\le 2$ Let $x,y,z,w>0$ show that
$$\sqrt{\dfrac{x}{x+2y+z}}+\sqrt{\dfrac{y}{y+2z+w}}+\sqrt{\dfrac{z}{z+2w+x}}+\sqrt{\dfrac{w}{w+2x+y}}\le 2$$
I tried C-S, but without success.
| By C-S:
$(LHS)^2\le \sum_{cyc}a(b+2c+d) \sum_{cyc}\frac{1}{(a+2b+c)(b+2c+d)}$
$\sum_{cyc}a(b+2c+d) \sum_{cyc}\frac{1}{(a+2b+c)(b+2c+d)}\le 4 \ \ \iff \ \ (a-c)^2(b-d)^2\ge 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2128211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
How can we show that $4\arctan\left({1\over \sqrt{\phi^3}}\right)-\arctan\left({1\over \sqrt{\phi^6-1}}\right)={\pi\over 2}$
$$4\arctan\left({1\over \sqrt{\phi^3}}\right)-\arctan\left({1\over \sqrt{\phi^6-1}}\right)={\pi\over 2}\tag1$$
$\phi$;Golden ratio
I understand that we can use
$$\arctan{1\over a}+\arctan{1\... | according to "How else can we show that"
Take $f(x)=4\arctan\left({1\over \sqrt{x^3}}\right)-\arctan\left({1\over \sqrt{x^6-1}}\right)$
now you can show $f(1.618)=\dfrac{\pi}{2}$
This question is very interesting to me .I was working on $\phi $ properties ,like $\phi^2=\phi+1 \to \phi^3=\phi^2+\phi$ or $\psi=\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2129898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
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converting $\frac{-1 - \sqrt{3}i}{2^{2/3}}$ to $-(-2)^{1/3}$ I have a question. It might seem silly.
Suppose I have $(x^3-2)$.
I know that $ 2^{1/3}$ is a root. $\frac{-1 \pm \sqrt{3}i}{2^{2/3}}$ are also roots of $(x^3-2)$.
How do you convert $\frac{-1 - \sqrt{3}i}{2^{2/3}}$ to $-(-2)^{1/3}$?
Thanks.
| The principal value of the third root of $-2\;$ is $$(-2)^{1/3}= |-2|^{1/3}\left(\cos(\frac{2}{3}\pi) + i \sin(\frac{2}{3}\pi)\right)
=\frac{2^{1/3}}{2}\left(1 + \sqrt{3}i\right)$$
Multiply with $-2$ to get
$$-(-2)^{1/3}=-\frac{2^{1/3}}{2}\left(1 + \sqrt{3}i\right)$$
Now multiply your expression with $2^{1/3}$ to get
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2131813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Use induction to prove that the inequality, $n! < \left(\frac{n}{2}\right)^n$ continues to hold after $n=6$. I am confused how you would continue with the inductive step. Thanks
| Basis step,
$P_6 : 6!<(\frac{6}{2})^6\rightarrow 720<3^6=729$. True
Inductive step, suppose
$P_n$ is true, prove that $P_{n+1}$ is also true.
Note $P_n:n!<(\frac{n}{2})^n$ and $P_{n+1}:(n+1)!<(\frac{n+1}{2})^{n+1}$
Observe that,
$$\begin{align}(\frac{n+1}{2})^{n+1}&=(\frac{n}{2}+\frac{1}{2})^{n+1} \\&>(\frac{n}{2})^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2134256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Probability to find 'X' number of Balls. A bag contains 5 Red, 4 Blue and x green balls. Two balls are drawn at random from the bag. If the probability of both being green is 1/7, find x.
I am clear with the proceedings but stuck at solving equation of factorials, so please show detailed steps.
| Total balls = x + 9
Green balls = x
Probability of 2 balls both green $$= \frac{\binom x2}{\binom{x + 9}{2}}$$
$$\frac 17= \frac{\frac{x!}{2! × (x - 2)!}}{\frac{(x+9)!}{2! × (x + 7)!}}$$
$\frac 17= \frac{x!}{2! × (x - 2)!} × \frac{2! × (x + 7)!}{(x+9)!}$
$\frac 17= \frac{x!}{(x - 2)!} × \frac{(x+7)!}{(x+9)!}$
$\frac 17... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2134442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given the positive real numbers $0\le a,b,c\le 2$ and $a+b+c=3$. Prove that $a^3+b^3+c^3\le 9$ Given the positive real numbers $$0\le a,b,c\le 2$$ and $$a+b+c=3$$. Prove that $$a^3+b^3+c^3\le 9$$
| Let $a\geq b\geq c$.
Since $f(x)=x^3$ is a convex function on $[0,3]$ and $(2,1,0)\succ(a,b,c)$,
by Karamata we obtain $$9=2^3+1^3+0^2\geq a^3+b^3+c^3$$
and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2134807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all matrices $X$ such that $ABXB^tA^t=I$ Find all matrices $X$ such that:
$$ABXB^tA^t=I$$ if
$A=\begin{pmatrix}
1 &-2 &2\\
3 &-5 &6\\
-1 &2 &-1
\end{pmatrix}$ and $B=\begin{pmatrix}
-3 &-2 &-2\\
2 & 1 &1\\
6 &3 &4
\end{pmatrix}$.
So I managed to get that $AB=\begin{pmatrix}
5 &2 &4\\
17 &7 &13\\
1&1&0
\end{pmat... | From $ABXB^tA^t=I$, we see that $det(A) \ne 0 \ne det(B)$. Hence $A$ and $B$ are invertible.
Again , from $ABXB^tA^t=I$, we see
$X=(AB)^{-1}((AB)^{-1})^t$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2134933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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What is the sum of $100$ terms of the sequence $a_{k} = \frac{1}{k} - \frac{1}{k+1}$? In the sequence $a_{1}, a_{2}, a_{3}, ..., a_{100}$, the $k$th term is defined by $$a_{k} = \frac{1}{k} - \frac{1}{k+1}$$ for all integers $k$ from $1$ through $100$. What is the sum of $100$ terms of this sequence?
The answer given ... | $$a_1+a_2+a_3+...+a_{100}=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{99}-\frac{1}{100}\right)+\left(\frac{1}{100}-\frac{1}{101}\right)=1-\frac{1}{101}=\frac{100}{101}$$
Note that $-\frac{1}{2}$ cancel $\frac{1}{2}$, $-\frac{1}{3}$ ca... | {
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"url": "https://math.stackexchange.com/questions/2135191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\sin\frac{\pi}n·\sin\frac{2\pi}n···\sin\frac{(n-1)\pi}n=\frac{n}{2^{n-1}}$ How to prove that
$$
\sin\dfrac{\pi}n·\sin\dfrac{2\pi}n···\sin\dfrac{(n-1)\pi}n=\dfrac{n}{2^{n-1}}
$$
using the roots of $(z+1)^n-1=0$?
My rough idea is to solve $(z+1)^n-1=0$ and use De Moivre's Theorem to find the product of roo... | Given the equation
$$(z+1)^n = 1 \tag{1}$$
Its roots are, $$z = e^{2\pi i \frac{k}{n}} - 1, \;\; 0 \leq k < n$$
Now,
$$\prod_{k = 1}^{n-1} (e^{2\pi i \frac{k}{n}} - 1)$$
is the product of all roots except $z = 0$. If you open up (1) and cancel one out of from both sides, you can take z common and the remaining polynomi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2138997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Prove that $\cos(\pi/7)$ is root of equation $8x^3-4x^2-4x+1=0$
Prove that $\cos\theta$ is root of equation $8x^3-4x^2-4x+1=0$, given $\theta=\frac{\pi}{7}$.
I put $\cos\theta$ in equation, but couldn't show the left-hand side to be zero.
| $$ \cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=\frac{2\sin\frac{\pi}{7}\cos\frac{2\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{4\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{6\pi}{7}}{2\sin\frac{\pi}{7}}= $$
$$ =\frac{\sin\frac{3\pi}{7}-\sin\frac{\pi}{7}+\sin\frac{5\pi}{7}-\sin\frac{3\pi}{7}+\sin\frac{7\pi}{7}-\sin\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
How could I know that $X^4+1$ is $(X^2+\sqrt 2X+1)(X^2-\sqrt 2X+1)$? I thought that $X^4+1$ was irreducible, but in fact, $$X^4+1=(X^2+\sqrt 2X+1)(X^2-\sqrt 2X+1).$$
In general, how can I have the intuition of such a factorisation if I don't know it ?
| In fact $$(x^2 - x+1)(x^2 + x+1)$$ $$ = (x^2+1 -x)(x^2+1 +x)$$ $$= (x^2+1)^2 -x^2$$ (remember the $\alpha^2-\beta^2 =(\alpha - \beta)(\alpha + \beta)$ formula? here $\alpha = x^2+1$ and $\beta = -x $) $$ = x^4 + x^2 +1$$
Well, if you really want to factor $x^4+1$, see here. Hope it helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 11,
"answer_id": 2
} |
Volume of tetrahedron given $6$ sides. I'm given $6$ side lengths and asked to find the volume. I feel lost
Tetrahedron $ABCD$ has $AB=1.5$, $AC=1$, $BC=2$, $BD=2.5$, $AD=1.5$, and $CD=3$.
I got the area of triangle ABC by Heron's formula to be $\sqrt{2.25(2.25-1)(2.25-2)(2.25-1.5)}$.
How do I find the height of the ... | By the Cayley-Menger determinant
$$ 288 V^2=\det\begin{pmatrix}0 & 1 & 1 & 1 & 1 \\ 1 & 0 & d_{12}^2 & d_{13}^3 & d_{14}^2\\ 1 & d_{21}^2 & 0 & d_{23}^2 & d_{24}^2\\1 & d_{31}^2 & d_{32}^2 & 0 & d_{34}^2 \\1 & d_{41}^2 & d_{42}^2 & d_{43}^2 & 0\end{pmatrix} $$
and in our case $d_{12}=d_{21}=\frac{3}{2}$, $d_{13}=d_{31}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2140214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Various methods to find value of $\sin 18^\circ$ To find value of $\sin 18^\circ$. Now my textbook gives a proof in which it takes $\theta=18^\circ$ and then multiply it out by 5 and write again as sum of $2\theta+3\theta$ and then taking sin on both sides forms a quadratic equation.
Are there any other elegant ways to... | Let $x = \sin(18^{\circ})\implies x^2 = \dfrac{1-y}{2}$, whereas $y = \cos(36^{\circ})$. And square again $y^2 = \dfrac{1+\cos(72^{\circ})}{2}= \dfrac{1+x}{2}\implies 2x^2+y = 1 = 2y^2-x\implies 2(x^2-y^2) + (x+y) = 0\implies (x+y)(2x-2y+1) = 0\implies 2x= 2y-1\implies 4x^2=2-2y = 1-2x\implies 4x^2+2x-1=0$. You can no... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2140356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Splitting the function $ \frac{x}{(1+x^2)^2}$ into partial fractions. I have a function that I am trying to split into partial fractions in order to integrate the function.
The function is:
$$\int \frac{x}{(1+x^2)^2}dx$$
I am trying to split $\dfrac{x}{(1+x^2)^2}$ into partial fractions.
While trying I am splitting... | Since
$1+x^2
=(1+ix)(1-ix)
$,
if we ask Wolfy for
$\dfrac{1}{(1+ax)^2(1+bx)^2}
$
we get
$-\dfrac{2 a^2 b}{(a - b)^3 (a x + 1)}
+ \dfrac{a^2}{(a - b)^2 (a x + 1)^2}
+ \dfrac{2 a b^2}{(a - b)^3 (b x + 1)}
+ \dfrac{b^2}{(a - b)^2 (b x + 1)^2}
$.
Putting $a=i, b=-i$,
we get both
$\dfrac{i}{4 (x + i)}
- \dfrac{1}{4 (x +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2141382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Substitutions for the trigonometric integral $\int \cos \theta \cos^5\left(\sin \theta\right) d\theta$ I evaluated
$$\int \cos \theta \cos^5\left(\sin \theta\right) d\theta$$
and got to a result that is very similar to what I can check with Mathematica but I am not sure if it is equivalent, what I did is the following... | Try this method once -
Put $\sin\theta =t$
$\cos\theta d\theta = dt$
So we have,
$\int \cos^5t$ dt
Now solve it.
Any doubt feel free to ask.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2141785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Let F be a field of characteristic 2. Find the roots of $x^2+bx+c$ if we know $d$ such that $d(d+1)=b^{-2}c$.
Let $F$ be a field of characteristic 2. Find the roots of $x^2+bx+c$ if we know $d$ such that $d(d+1)=b^{-2}c$, where ,b,c,d $\in F$.
I tried using the formula $\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-b\pm ... | If you multiply by $b^{-2}$ you get that $x^2+bx+c=0$ is equivalent to $(b^{-1}x)^2+b^{-1}x+d(d+1)=0$, which is equivalent to $b^{-1}x(b^{-1}x+1)=d(d+1)$. It therefore reduces to solving $y(y+1)=d(d+1)$ with $y=b^{-1}x$. Those solutions are $y=d$ and $y=d+1$, hence the original solutions are $x=bd$ and $x=b(d+1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2141905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to solve for $x$ algebraically, given that $x^2 = 4- \sqrt{12}$ This is a question I have been chewing on for a couple days but haven't quite solved yet. The value of $x^2$ is given as $4-\sqrt{12}$ and then the result given as $±( 1-\sqrt{3})$. How would I solve this problem algebraically without prior knowledge o... | Given an expression like $\sqrt{a\pm\sqrt{b}}$, can we simplify it to an expression of the form $\left|c\pm\sqrt{d}\right|$?
Well, suppose we could. Then we would find that
$$
(c\pm\sqrt{d})^2 = a\pm\sqrt{b}
$$
Carrying out the square, we get
$$
c^2+d\pm2\sqrt{4c^2d} = a\pm\sqrt{b}
$$
We might therefore try to equate ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2142995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove: $\left(3^{2^9}-3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6}\right) > 10^{2009}$
Show that the number $ 3^{{4}^{5}} {+} 4^{{5}^{6}}$ can be expressed as the product of two integers greater than $ 10^{2009}$
By Sophie Germain:
$ 3^{{4}^{5}} {+} 4^{{5}^{6}}=\left(3^{2^9}+3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6}\right... | You have $x^4+4y^4 = (x^2+2xy+2y^2)(x^2-2xy+2y^2)$, but $x^2-2xy+2y^2 = (x-y)^2+y^2\geq y^2$, where $y = 2^{\frac{5^6-1}{2}}$. Therefore, $$y^2 = (2^{10})^{\tfrac{5^6-1}{10}}>10^{\tfrac{3(5^6-1)}{10}} = 10^{2009},$$
because $2^{10} = 1024>1000 = 10^3.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2143979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Can a factor of $x^2+4$ for odd $x$ be $\equiv 3\mod 4$? When $x$ is odd, then $x^2+4 \equiv 5 \mod 8$. Can any factor of $x^2+4$ be $\equiv 3\mod 4$?
Examples:
$$7^2+4=53$$
$$11^2+4= 5^3$$
$$31^2+4= 5 \cdot 193$$
$$47^2+4=2213$$
$$89^2+4=5^2 \cdot 317$$
All of the prime factors above are $\equiv 1 \mod 4$.
Is there no... | $x^2 + 4 \equiv 3 \pmod 4$ is the same as proving $x^2 + 0 \equiv 3 \pmod 4$ since $4 \equiv 0\pmod 4$.
If $x$ is even is easy to see that $x^2 \equiv 0 \pmod 4$ and when $x$ is odd it has the form $2n+1$ then:
$x=(2n+1)^2= 4n^2 + 4n + 1 = 4n(n+1) + 1$ thus $4n(n+1) \equiv 0 \pmod 4$ and $4n(n+1) + 1\equiv 1 \pmod 4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2144662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Find all positive integers $n$ such that $\displaystyle \left({-1 + i\sqrt{3}\over 2}\right)^n + \left({-1 - i\sqrt{3}\over 2}\right)^n = 2$?
Find all positive integers $n$ such that $\displaystyle \left({-1 + i\sqrt{3}\over 2}\right)^n + \left({-1 - i\sqrt{3}\over 2}\right)^n = 2$ ?
I wrote $\displaystyle \left({-1... | First of all, note that $\left| \frac{-1 \pm i\sqrt 3}{2} \right| = 1$, hence this holds for all $n$th powers of these quantities also.
The triangle inequality states that $|a| + |b| \geq |a+b|$, with equality holding if and only if $a$ is a scalar multiple of $b$.
In our case, if we assume that the final condition hol... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $a^2+b^2+c^2=1$ so $\sum\limits_{cyc}\frac{1}{(1-ab)^2}\leq\frac{27}{4}$
Let $a$, $b$ and $c$ be real numbers such that $a^2+b^2+c^2=1$. Prove that:
$$\frac{1}{(1-ab)^2}+\frac{1}{(1-ac)^2}+\frac{1}{(1-bc)^2}\leq\frac{27}{4}$$
This inequality is stronger than $\sum\limits_{cyc}\frac{1}{1-ab}\leq\frac{9}{2}$ with ... | Let $x = \dfrac{1}{1-ab}$, $y = \dfrac{1}{1-bc}$, $z = \dfrac{1}{1-ac}$.
We have proven that
$$x + y + z \leq \dfrac{9}{2}.$$
Therefore, $x + y + z = \dfrac{9}{2} - \varepsilon$ for some $0 \leq \varepsilon \leq \dfrac{9}{2}$. If we assume that $x,y,z \geq 0$, which we can, we can square both sides and obtain
$$(x^2 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2146712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove: $\exists M\in\mathbb{R}$ such that $\forall x\in (1,3)$, $\left|\frac{5x^2 - 2x - 4}{5(x^2 + 1)}\right|\leq M$ I am not sure how to find the correct upper and lower limits for this problem. I did find the breakdown of the inequality to be $\frac{5x^2+|2x|+4}{|5(x^2 + 1)|}$. Thanks for any help you can give!
| $|\frac {5x^2 - 2x -4}{5(x^2 + 1)}|\le M \iff$
$-M \le \frac {5x^2 - 2x -4}{5(x^2 + 1)} \le M \iff$
$-M*5(x^2 + 1) \le 5x^2 - 2x - 4 \le M*5(x^2 + 1)\Leftarrow$
$10M \le 5x^2 - 2x - 4 \le 10M$.
Now $5 < 5x^2 < 45$ for $1 < x < 3$.
And $-10 < -2x - 4 < -6$ for $1 < x < 3$.
So $-5 < 5x^2 - 2x - 4 < 39$.
So if $M = 3.9$ t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2146810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Is the sequence $\left\{\frac{2^n}{n!}\right\}$ convergent? If so, what is the limit? Is the sequence $\left\{\frac{2^n}{n!}\right\}$ convergent? If so, what is the limit?
$$ \frac{2^n}{n!} - 0 = \frac{2^n}{n!} < \frac {2^n}{n} <\; ? < \epsilon$$
I dont know how to simplify $\frac{2^n }{ n}$.
I cannot just do $\frac{... | For $\; n \geq 3 \;$ we have
$$\frac{2^n}{n!} \;\; = \;\; \frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3} \cdot \frac{2}{4} \cdot \frac{2}{5} \cdot \frac{2}{6} \cdot \;\; \cdots \;\; \cdot \frac{2}{n} \;\; \leq \;\; \frac{2}{1} \cdot \frac{2}{2} \cdot \left(\frac{2}{3}\right)^{n-2} \;\; \longrightarrow \;\; 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2147899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
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Prove using induction the inequality.
$\forall$:n∈${N}$
$\binom{2n}{n}$ $\ge \frac{4^n}{2n+1}$
I tried with no any success...
| For $n=1$ we get $2\geq\frac{4}{3}$, which is true.
If $\binom{2n}{n}\geq\frac{4^n}{2n+1}$ we obtain:
$$\binom{2n+2}{n+1}=\binom{2n}{n}\cdot\frac{(2n+2)(2n+1)}{(n+1)^2}>\frac{4^n}{2n+1}\cdot\frac{2(2n+1)}{n+1}$$ and it remains to prove that
$$\frac{4^n}{2n+1}\cdot\frac{2(2n+1)}{n+1}>\frac{4^{n+1}}{2n+3}$$ or
$$2(2n+3)>... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2148627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Evaluating $\lim_{x \to 0} \frac{\sqrt{1- \cos x^2}}{1 - \cos x}$ I'm trying to evaluate the following limit:
$$\lim_{x \to 0} \frac{\sqrt{1- \cos x^2}}{1 - \cos x}$$
I've tried multiplying by the conjugate and variable substitution. I had a look at wolfram alpha and it said that $\lim_{x \to 0} \frac{\sqrt{1- \cos x^2... | Note: I am using the limit $\lim_{\theta \to 0}\frac{\sin \theta}{\theta}=1$ and the identity $1-\cos 2A=2\sin^2 A$.
\begin{align*}
\lim_{x \to 0} \frac{\sqrt{1- \cos x^2}}{1 - \cos x} & = \lim_{x \to 0} \frac{\sqrt{2 \sin^2 \left(x^2/2\right)}}{2 \sin^2 \left(x/2\right)}\\
& = \lim_{x \to 0} \frac{\sin \left(x^2/2\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2150505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to calculate the series $\sum\limits_{n=1}^{\infty} \arctan(\frac{2}{n^{2}})$? I encountered the series
$$
\sum_{n=1}^{\infty} \arctan\frac{2}{n^{2}}.
$$
I know it converges (by ratio test), but if I need to calculate its limit explicitly, how do I do that? Any hint would be helpful..
| \begin{align*}
\sum_{n=1}^\infty\arctan\left ( \frac{2}{n^2} \right ) &=-arg \prod_{n=1}^\infty\left (1-\frac{2i}{n^2} \right ) \\
&=-arg \prod_{n=1}^\infty\left (1-\frac{(\sqrt{2i})^2}{n^2} \right ) \\
&=-arg\left(\frac{\sin(\pi\sqrt{2i})}{\pi\sqrt{2i}} \right ) \\
&=-arg\left(-\frac{(1/2+i/2)\sinh\left(\pi \rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2153365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
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Validating the inequality. $x-\frac{x^2}{2}+\frac{x^3}{3(1+x)}<\log(1+x)<x-\frac{x^2}{2}+\frac{x^3}{3}$, $x>0$
Here I can only see that the right side of second inequality i.e. $x-\frac{x^2}{2}+\frac{x^3}{3}$ comes in the expansion of $\log(1+x)$.
We have done the Lagrange's mean value theorem and intermediate value th... | I expand my comment into an answer.
We have for $t > 0$ $$\frac{1}{1 + t} = 1 - t + \frac{t^{2}}{1 + t}$$ and hence if $0 < t < x$ then $$1 - t + \frac{t^{2}}{1 + x} < \frac{1}{1 + t} < 1 - t + t^{2}$$ and integrating the above with respect to $t$ on interval $[0, x]$ we get $$x - \frac{x^{2}}{2} + \frac{x^{3}}{3(1 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2154246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to integrate this $\int \frac{\sqrt{x}}{1+x^4}dx$? How to integrate this ? $$\int \frac{\sqrt{x}}{1+x^4}dx$$
Any hint or idea on how to proceed?
Edit: Here is the final answer using Wolfram Alpha
| Let $x= \frac1{t^2}$. Then
$$\int \frac{\sqrt{x}}{1+x^4}dx=
-\int \frac{2t^4}{1+t^8}dt
=\frac1{\sqrt2}\int \left(
\frac{t^2}{t^4+\sqrt2t^2+1}
-\frac{t^2}{t^4-\sqrt2t^2+1} \right)dt $$
Integrate
$$\begin{align}
I(a)& =\int\frac{t^2}{t^4+at^2+1}dt\\
&=\frac12\int \frac{t^2+1}{t^4+at^2+1}dt -
\frac12\int \frac{1-t^2}{t^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2158375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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The following equation to solve :$ \tan x+\cot x=\sqrt{2}(\cos x+\sin x)$ The following equation to solve :
$$ \tan x+\cot x=\sqrt{2}(\cos x+\sin x)$$
My try:
$$\frac{2}{\sin 2x}=\sqrt{2}(\cos x+\sin x)$$
$$\left(\frac{2}{\sin 2x}\right)^2=(\sqrt{2}(\cos x+\sin x))^2$$
$$\left(\frac{2}{\sin 2x}\right)^2=2(1+\sin 2x)$$
... | Avoid squaring whenever possible as it immediately introduces extraneous roots.
Method $\#1:$ Put $u=\dfrac{\cos x+\sin x}{\sqrt2}\implies2u^2=1+\sin2x$
$$\implies\dfrac2{2u^2-1}=2u\iff2u^3-u-1=0$$
Clearly, $u=1$ is a root. Please check for the other roots.
$\implies1=\dfrac{\cos x+\sin x}{\sqrt2}=\cos\left(x-\dfrac\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2161337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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How can I show that this complicated expression with square and cube roots reduces to the value 7? How would I go about reducing the complicated-looking expression $$ \sqrt[3]{19\sqrt{5} + 56} + \frac{11}{\sqrt[3]{19\sqrt{5} + 56}} $$ to show that it is equal to 7?
I came across the complicated expression while calcula... | I'll outline the basic method to simplify$$\sqrt[3]{56+19\sqrt{5}}+\dfrac {11}{\sqrt[3]{56+19\sqrt{5}}}\tag{1}$$
To simplify the nested radical, note that we have this general outline:$$\sqrt[n]{A+B\sqrt[m]{C}}=a+b\sqrt[m]{C}\tag{2}$$
So therefore, we have$$\begin{align*}\sqrt[3]{56+19\sqrt{5}} & =a+b\sqrt{5}\\56+19\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2161428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Find a $3\times3$ orthogonal matrix $A$ . How can we find a $3\times3$ orthogonal matrix $A$ such that $$A\,\left[\frac{1}{\sqrt 2},\frac{1}{\sqrt 2},0\right]^T=\left[\frac{1}{\sqrt 3},\frac{1}{\sqrt 3},\frac{1}{\sqrt 3}\right]^T?$$
My try:I don't know how to proceed in order to get that matrix.Thank you
| step 1, Find an otho-normal matrix that maps one of your principle component vectors to $(\frac {1}{\sqrt 3},\frac {1}{\sqrt 3},\frac {1}{\sqrt 3})$
$(\frac {1}{\sqrt 3},\frac {1}{\sqrt 3},\frac {1}{\sqrt 3})$ can be our first vector.
a vector othogonal to that (but not necessarily normal) is (1,-1,0).
and vector to th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2164047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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$|a_{n+1}-a_n| \leq \frac{1}{2} |a_{n}-a_{n-1}|$. Prove $(a_n)$ is Cauchy $|a_{n+1}-a_n| \leq \frac{1}{2} |a_{n}-a_{n-1}|$. I'm trying to show that $(a_n)$ is convergent by showing that $(a_n)$ is Cauchy.
let $|a_2 - a_1 | =x$.
$|a_3-a_2| \leq \frac{1}{2}x$
$|a_4-a_3| \leq \frac{1}{2}|a_3-a_2|\leq \frac{1}{4}x$
In ge... | Your first steps are correct. To finish
$$|a_m - a_n| \leqslant \frac{1}{2^{m-2}} + \frac{1}{2^{m-3}} + \dots + \frac{1}{2^{n-1}} \\ = \frac{1}{2^{n-1}} \sum_{k=0}^{m-n-1}\frac{1}{2^k} \ < \frac{1}{2^{n-1}} \sum_{k=0}^{\infty}\frac{1}{2^k} = \frac{1}{2^{n-1}}\frac{1}{1- 1/2} = \frac{1}{2^{n-2}} .$$
Since $1/2^{n-2} \to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2164254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find $\angle BDC$ Quadrilateral $ABCD$ , $\angle ABD = 17^{\circ}, \angle DBC = 34^{\circ}, \angle ACB = 43^{\circ}, \angle ADB = 13^{\circ}$, Find $\angle BDC$.
| Nice problem!
Let $X$ be a point symmetric to $A$ with respect to $BD$. Let $Y$ be a point symmetric to $A$ with respect to $BX$. Let $Z$ be a point symmetric to $A$ with respect to $DX$.
Then $DA=DX=DZ$, $BA=BX=BY$, and $AX=XY=XZ$. Angle chasing gives
$$\angle ZXY=360^\circ - \angle AXZ - \angle YXA = 360^\circ - 2\a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2165108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Number of triangles with integer sides Total number of right angle triangles whose inradius is $2013$ and sides are integer
Attempt: assuming that $a,b,c>0$ are the sides of a triangle.
so form a right angle triangle with sides $a,b,c$ and right angle at $C$
$\displaystyle r=(s-c)\tan \frac{C}{2}=(s-c)=\frac{a+b-c}{2}$... |
$\displaystyle a^2+b^2 = (a+b)^2+4r^2-4r(a+b) = a^2+b^2+2ab-4r(a+b)$
This should be
$$a^2+b^2=(a+b)^2+4r^2-4r(a+b)=a^2+b^2+2ab\color{red}{+4r^2}-4r(a+b)$$
from which we have
$$(a-2r)(b-2r)=2r^2=2^1\times 3^2\times 11^2\times 61^2\tag1$$
where $r=2013=3\times 11\times 61$.
We may suppose that $$-2r\lt a-2r\le b-2r\tag... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2166524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Find $h(85)$ if $h(x^2+x+3)+2h(x^2-3x+5)=6x^2-10x+17$ let : $h: \mathbb{R}\to \mathbb{R}$ ane for any real number
$$h(x^2+x+3)+2h(x^2-3x+5)=6x^2-10x+17$$
then :
$$h(85)=?$$
My Try: $$x=0:h(3)+2h(5)=17\\x=1:h(5)+2h(3)=13\\+\\3h(3)+3h(5)=30\\h(3)+h(5)=10$$
now ?
| Let $h(x) \equiv Ax+B$, then $$A(x^2+x+3)+B+2A(x^2-3x+5)+2B\equiv 6x^2-10x+17$$
$$
\left \{
\begin{align*}
3A &= 6 \\
-5A &= -10 \\
13A+2B &= 17 \\
\end{align*}
\right.$$
On solving, $(A,B)=\left( 2, -\dfrac{9}{2} \right)$
Alternatively,
*
*$x=\dfrac{-1 \pm \sqrt{329}}{2} \implies x^2+x+3=85 \quad \text{and} ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Checking if two matrices are similar I have two matrices
$$ \begin{pmatrix}
2 & 1 & 0 \\
0 & 2 & 0 \\
0 & 0 & 3
\end{pmatrix} $$
and $$ \begin{pmatrix}
2 & 2 & 0 \\
0 & 2 & 0 \\
0 & 0 & 3
\end{pmatrix} $$
They are not diagonalizable. Share the same characteristic polynomial, the same trace, same determin... | The first of your matrices is in Jordan canonical form, which must also be the Jordan canonical form of the second one (because the eigenvalue $2$ has algebraic multiplicity $2$ but geometric multiplicity $1$). So they are similar.
More concretely, we have
$$ \begin{pmatrix}
2 & 1 & 0 \\
0 & 2 & 0 \\
0 & 0 & 3
\... | {
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"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Determine: $S = \frac{1}{2}{n \choose 0} + \frac{1}{3}{n \choose 1} + \cdots + \frac{1}{n+2}{n \choose n}$ I am studying the book Equations and Inequalities by Herman et al, and am stuck on the following exercise:
Determine: $S = \frac{1}{2}{n \choose 0} + \frac{1}{3}{n \choose 1} + \cdots + \frac{1}{n+2}{n \choose n}$... | Integration Approach
Since
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}\sum_{k=0}^n\frac1{k+2}\binom{n}{k}x^{k+2}
&=\sum_{k=0}^n\binom{n}{k}x^{k+1}\\
&=x(1+x)^n
\end{align}
$$
we have
$$
\begin{align}
\sum_{k=0}^n\frac1{k+2}\binom{n}{k}
&=\int_0^1x(1+x)^n\,\mathrm{d}x\\
&=\int_1^2(x-1)x^n\,\mathrm{d}x\\
&=\frac{2^{n... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Calculate $\int^4_2 xln(x) \:dx$ using integration by parts. Let $f'(x) = x$ and $g(x) = \ln(x) \implies f(x) = \dfrac{x^2}{2}$ and $g'(x) = \dfrac{1}{x}$.
$$\int^4_2 x\ln(x) \:dx = \left[ \dfrac{x^2}{2}\ln(x) \right]^4_2 - \int^4_2 \dfrac{1}{x}\dfrac{x^2}{2} \: dx = 8\ln(4) - 2\ln(2) - 1$$
Apparently, this answer is ... | $$\int _{ 2 }^{ 4 } xln(x)\: dx=\left[ { \frac { x^{ 2 } }{ 2 } }ln(x) \right] ^{ 4 }_{ 2 }-\int _{ 2 }^{ 4 }{ \frac { { x } }{ 2 } } \: dx=\left[ { \frac { x^{ 2 } }{ 2 } }ln(x) \right] ^{ 4 }_{ 2 }-\left[ { \frac { x^{ 2 } }{ 4 } } \right] ^{ 4 }_{ 2 }=\frac { 16 }{ 2 } \ln { 4 } -\frac { { 2 }^{ 2 } }{ 2 } \ln {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2173250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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$\lim_{x \to 0}\frac{(\sqrt{1+x^2}+x)^m-(\sqrt{1+x^2}-x)^m}{x}$ Help with this limit:
$$\lim_{x \to 0}\frac{(\sqrt{1+x^2}+x)^m-(\sqrt{1+x^2}-x)^m}{x}$$
Using L'Hopital rule gives $2m$ but I need it without using L' Hopital.
| In the same spirit as Yves Daoust's answer, start developing $\sqrt{1+x^2}$ using the generalized binomial theorem or Taylor series. This would give
$$\sqrt{1+x^2}=1+\frac{x^2}{2}+O\left(x^3\right)$$ $$\sqrt{1+x^2}+x=1+x+\frac{x^2}{2}+O\left(x^3\right)$$
$$\sqrt{1+x^2}-x=1-x+\frac{x^2}{2}+O\left(x^3\right)$$ Now, t... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} = 9$ Let $$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} = 9,$$
$$\frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} = 32,$$
$$\frac{a^3}{b+c} + \frac{b^3}{c+a} + \frac{c^3}{a+b} = 122.$$
Find the value of $abc$.
Please check if my answer is correct or not.
$a+b+c + \frac... | $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=9$, eqI
$\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}=32$, eqII
$\frac{a^3}{b+c}+\frac{b^3}{c+a}+\frac{c^3}{a+b}=122$ eqIII
Let $a+b+c=k \Rightarrow$
$ a=k-(b+c), b=k-(a+c), c=k-(a+b)$. (eqIV)
Using (eqI): $\frac{k-(b+c)}{b+c}+\frac{k-(a+c)}{a+c}+\frac{k-(a+b)}{a+b}=9 \Rig... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Maximum value of equation $ab+bc+ca$ if $a+2b+c=4$ If $a,b,c$ ${\in}$ $\mathbb{R} $ such that
$a+2b+c=4$, then find the $max$ value of
$ab+bc+ca$.
I always get stuck with max, min questions. We cannot apply AM:GM here, I have not studied much calculus yet.
Can you do this with graphs or by plane algebra?
I don't reall... | $$a+b+c=4-b$$
$$a^2+b^2+c^2+2(ab+ac+bc)=16-8b+b^2$$
$$ab+bc+ca=\frac{16-8b-a^2-c^2}2$$
Since $b=\frac{4-a-c}2$,
$$ab+bc+ca=\frac{16-16+4a+4c-a^2-c^2}2=\frac{a(4-a)+c(4-c)}2\le\frac{4+4}2=4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2174458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Trig Integration $\displaystyle\int\sec^3(x)\tan^2(x)\,\mathrm{d}x$ $$\int \sec^3\left(x\right)\tan^2\left(x\right)\,\mathrm{d}x$$
Hi, for the question above, I think that substituting $\tan(x)$ would be ok but I couldn't figure out the final step.
$$\int\sec^2(x)\sec(x)\tan^2(x)\,\mathrm{d}x$$
$$\int \sec\left(x\right... | The popular approach is to convert it into integrals of odd powers of secant and then use reduction formulas derived from integration by parts, but there is an alternative.
You can make the substitution $u = sec(x) + tan(x)$.
Note that $1/u = sec(x) - tan(x)$,
because $(sec(x)+tan(x))((sec(x)-tan(x)) = sec^2(x)-tan^2(x... | {
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"source": "stackexchange",
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"answer_id": 1
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Factorise the expression Factorise of the expression $$ a^4+b^4-c^4-2a^2b^2+4abc^2~ \\(S:(a + b - c) (a + b + c) (a^2+ b^2 + c^2 - 2 a b))$$
I tried to solve this problem:
$$ a^4+b^4-c^4-2a^2b^2+4abc^2\rightarrow\\\
(a^2-b^2)^2-c^4+4abc^2\rightarrow\\\
(a^2-b^2-c^2)(a^2-b^2+c^2)+4abc^2\\\
((a-b)(a+b)-c^2)((a-b)(a+b)+... | As a quadratic in $c^2\,$, the reduced discriminant of $-(c^4-4ab \cdot c^2-a^4-b^4+2a^2b^2)$ is:
$$
\frac{1}{4}\Delta = 4a^2b^2 +a^4+b^4-2a^2b^2=(a^2+b^2)^2
$$
Therefore the roots are $c^2=2ab \pm (a^2+b^2) = \{(a+b)^2, -(a-b)^2\}\,$ giving the factorization:
$$
\begin{align}
-\left(c^2-(a+b)^2\right)\left(c^2+(a-b)^2... | {
"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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Solve the the given equation: $\sqrt{3x^2+x+5} = x-3$ We have to find the number of solution for the given equation:
$$\sqrt{3x^2+x+5} = x-3.$$
There are two solution one is
By using graph we get one solution
By squaring both sides we get no solution
I want to know which solution is correct
| we have $$\sqrt{3x^2+x+5}=x-3$$ it must be $$x\geq 3$$ and after squaring we get
$$3x^2+x+5=x^2-6x+9$$ this is equivalent to $$2x^2+7x-4=0$$ solving this quadratic equation we obtain
$$x_{1,2}=-\frac{7}{4}\pm\sqrt{\frac{49}{16}+\frac{32}{16}}$$
Can you finish this?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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then minimum number of number of roots of $p(x) = 0$ is let $p(x)=x^6+ax^5+bx^4+x^3+bx^2+ax+1.$ given that $x=1$ is a one rot of $p(x)=0$
and $-1$ is not a root. then minimum number of number of roots of $p(x) = 0$ is
Attempt: $x=0$ in not a root of $p(x)=0.$
So $\displaystyle \left(x^3+\frac{1}{x^3}\right)+a\left(x... | The minimum number of real roots is $2$ counted with multiplicity, or $1$ counted without multiplicity.
Since $p(x)$ is a reciprocal polynomial (that is, $x^6p(\frac1x)=p(x)$), its roots come in pairs $r,\frac1r$. In particular, if $1$ is a root of $p(x)$ then it is a double root. It is possible for these to be the onl... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Solving for $p$ and $k$. For the question,
$$F(x)= x^{2} + 6x + 20 + k(x^{2} -3x -12)$$
I have to find the value of k and the value of $p$ given that the minimum value of $F(x)$ is $p$ and $F(-2) = p$.
What I did is: I expanded the expression of $F(x)$ and reached a term for $p= 12 - 2p$. I tried to complete the square... | You can try completing the square directly. Notice that the minimum of $F(x)$ lies at $x = -2$, and $F(-2) = p$. This means that
$$F(x) = a(x + 2)^2 + p$$
for some constant $a$. We can't assume $a = 1$ because any non-zero value of $a$ can work, so we leave it there as it is.
Now we go ahead and solve as usual by comp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2184116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Show that $I=J={\ln(7+4\sqrt{3})\over 4\sqrt3}?$ Consider the integrals $(1)$ and $(2)$, how does on show that
(1): $I=J$
(2): and $I=J={\ln(7+4\sqrt{3})\over 4\sqrt3}?$
$$\int_{0}^{\pi/2}{\tan x\over \sqrt{(1+\tan^2 x)(1+\tan^6 x)}}\mathrm dx=I\tag1$$
$$\int_{0}^{\pi/2}{\tan^3 x\over \sqrt{(1+\tan^2 x)(1+\tan^6 x)}}\... | On the path of Lab Bhattacharjee,
$J=\displaystyle \int_0^{+\infty} \dfrac{1}{\sqrt{(1+x)(1+x^3)}}dx$
Perform the change of variable $y=\dfrac{1-x}{1+x}$,
$\begin{align}\displaystyle J&=\int_{-1}^{1} \dfrac{1}{\sqrt{1+3x^2}}dx\\
\end{align}$
Perform the change of variable $y=\sqrt{3}x$,
$\begin{align}
J&=\dfrac{1}{\sqr... | {
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"url": "https://math.stackexchange.com/questions/2184647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
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Improper integral $\int_0^\infty\frac{{\rm d}x}{1+x^4\sin^2(x)} $ I know that this integral is convergent, but I don't know how to prove it.
$$\int_0^\infty\frac{{\rm d}x}{1+x^4\sin^2(x)}$$
| Split the integral up into portions where $\sin(x)$ is small (i.e., where the best bound we can hope for is $\frac{1}{1+x^4\sin^2(x)} \le 1$), and portions where $\sin(x)$ is relatively large so that we can use the growth of $x^4$ to our advantage. For the former, we use intervals $I_n =(n\pi - \frac{1}{n^\alpha}, n\pi... | {
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"url": "https://math.stackexchange.com/questions/2185515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $1^2+3^2+...99^2$ given $1^1+2^2+...+100^2$ and $1^1+2^2+...+50^2$ We know that $1^1+2^2+...+100^2=338350$ and $1^1+2^2+...+50^2=42925$. Find $1^2+3^2+...99^2$.
I don't know really where to start. I tried to find a pattern in the sequences, but there was none. Can I substitute values for the equations?
| This should help: $$2^2+4^2+6^2+\dots+100^2=2^2(1^2+2^2+3^2+\dots+50^2)=\dots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2186459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Given $x^9 = e$ and $x^{11} = e$ prove $x = e$. Full Problem: Prove that for any element $x$ in a group $G$ that satisfies
$$x^9 = e \\
x^{11} = e,$$
where $e$ is the identity element, that $x$ itself must be $e$.
Is this as simple as showing that
*
*$x^{11} = x^{9} \cdot x^{2} = e \cdot x^2 \Rightarrow x^2 = e$
*... | Yes, it is that simple. It can be done even shorter, because after showing $x^2=e$, you can go straight to
$$e=x^9=x(x^2)^4=xe^4=x$$and you're done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2192881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
$f(\frac{2\pi}{7})+f(\frac{4\pi}{7})+f(\frac{6\pi}{7})=1$ let $$f(x)=\frac{1}{1+2\cos x}$$
prove that :
$$f(\frac{2\pi}{7})+f(\frac{4\pi}{7})+f(\frac{6\pi}{7})=1$$
My Try :
$$f(\frac{2\pi}{7})=\frac{1}{1+2\cos (\frac{2\pi}{7})}$$
$$f(\frac{2\pi}{7})=\frac{1}{1+2\cos (\frac{4\pi}{7})}$$
$$f(\frac{2\pi}{7})=\frac{1}{1+2\... | Use factor $z^7-1$ into linear and quadratic factors and prove that $ \cos(\pi/7) \cdot\cos(2\pi/7) \cdot\cos(3\pi/7)=1/8$
or if $7x=2m\pi$ where $m$ is any integer
$\sin4x=\sin(2m\pi-3x)=-\sin3x$
$\implies4\sin x\cos x\cos2x=\sin x(4\sin^2x-3)$
$\implies4\sin x\cos x(2\cos^2x-1)=\sin x\{4(1-\cos^2x)-1\}$
So, the ro... | {
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"url": "https://math.stackexchange.com/questions/2193379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Factoring a quartic polynomial over the integers with roots that are not integers The quartic polynomial
$$
1728(x - 3) - x^2(12^2 - x^2)
$$
factors "nicely" as
$$
(x^2 - 12x + 72) (x^2 + 12x - 72) = (x^2 - 12x + 72)(x - 6\sqrt{3} + 6)(x + 6\sqrt{3} + 6) \, .
$$
(Note that $1728 = 3(24^2)$.) How is this factorization ... | This solution was suggested by dxiv.
Solution
The given quartic polynomial factors into a product of two monic, quadratic polynomials:
\begin{equation*}
x^{4} - 144x^{2} + 1728x - 5184 = (x^{2} + ax + b)(x^{2} + cx + d) ;
\end{equation*}
\begin{equation*}
(x^{2} + ax + b)(x^{2} + cx + d) = x^{4} + (a + c)x^{3} + (ac + ... | {
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"url": "https://math.stackexchange.com/questions/2193908",
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"answer_id": 2
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How to factor $r^6 -3r^4 +3r^2 - 1 = 0$ I tried two ways to factor $r^6 -3r^4 +3r^2 - 1 = 0$
When I factor $r^4$ out of $r^6 -3r^4$:
$r^4(r^2-3)+(3r^2-1) = 0$
When I factor $r^2$ out of $r^6 + 3r^2$:
$r^2(r^4+3)-(3r^4 + 1) = 0$
For both methods, I'm stuck at an equation that isn't factorable. But Wolfram Alpha says $r^... | Observe you have
\begin{align}
r^6-3r^4+3r^2-1 = (r^2-1)^3
\end{align}
by the binomial theorem. Then it follows $(r^2-1) = (r+1)(r-1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2194225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Finding $\int\frac{2x^3-1}{x^6-x^4+2x^3-2x^2-x+1}dx$ Finding $\displaystyle \int\frac{2x^3-1}{x^6-x^4+2x^3-2x^2-x+1}dx$
Attempt: Assume $\displaystyle \int\frac{2x^3-1}{x^6-x^4+2x^3-2x^2-x+1}dx =\int \frac{2x-x^{-2}}{\bigg[x^4-x^2+2x-\frac{2}{x}-\frac{1}{x}+\frac{1}{x^2}\bigg]}dx$
could some help me how to solve,thank... | $\dfrac{2x - \frac{1}{x^2}}{x^4 -x^2 +2x -2 -\frac{1}{x} + \frac{1}{x^2}}$
We want to write the denominator in $x^2 + \frac{1}{x}$ since it is derivative in the nominator
Reordering
$$x^4 +2x +\frac{1}{x^2}-2 -x^2 - \frac{1}{x} = \left(x^2 +\frac{1}{x}\right)^2 - 2 -(x^2 + \frac{1}{x}) $$
Let $ u = x^2 + \frac{1}{x^2}... | {
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"source": "stackexchange",
"question_score": "1",
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$3\mid a^3+b^3+c^3$ if only if $3\mid a+b+c $
Prove the following equivalence: $3\mid (a^3 + b^3 + c^3) $ if and only if $3\mid (a + b + c) $.
My try:
I know $a^3 + b^3 + c^3 = (a + b + c) (a^2 + b^2 + c^2 – ab – bc – ca) + 3abc$, but I can't seem to proceed from here.
Thanks all!
| Compare $a^2+b^2+c^2-ab-bc-ca$ with $(a+b+c)^2$
| {
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"url": "https://math.stackexchange.com/questions/2196500",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
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Integrating $\frac{1}{u}$ Say I have $\int_{}{}\frac{\frac{1}{2}}{u + 1}du$
Why is $\frac{1}{2}\ln(u + 1) \neq \frac{1}{2}\ln(2u + 2)$?
Assuming I did this:
$$\int_{}{}\frac{\frac{1}{2}}{u + 1}\ du = \int_{}{}\frac{1}{2u + 2}\ du$$
| You are forgetting the constant for an indefinite integral.
Note that
$$\frac{1}{2} \int \frac{du}{u+1} = \frac{1}{2} \ln (u+1) + C,$$
and
$$\frac{1}{2} \ln(2u+2) = \frac{1}{2} \ln (u+1 ) + \frac{\ln 2}{2}.$$
| {
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The minimum value of $x^8 – 8x^6 + 19x^4 – 12x^3 + 14x^2 – 8x + 9$ is The minimum value of $$f(x)=x^8 – 8x^6 + 19x^4 – 12x^3 + 14x^2 – 8x + 9$$ is
(a)-1
(b)9
(c)6
(d)1
Apart from trying to obtain $1$, which in this case is simple and $f(2)=1$ is there a standard method to approach such problems.
Please keep in mind tha... | $f(x)=x^8 – 8x^6 + 19x^4 – 12x^3 + 14x^2 – 8x + 9$
$= x^4(x^4 - 8x^2 + 16) +19x^4 - 16x^4 - 12 x^3 + 14x^2 - 8x + 9$
$= x^4(x^2 - 4)^2 + 3x^2(x^2 - 4x + 4) + 14x^2 -12x^2 - 8x + 9$
$= x^4(x^2 - 4)^2 + 3x^2(x - 2)^2 + 2(x^2 - 4x + 4) + 1$
$= (x - 2)^2(x^4(x+2) + 3x^2 +2) + 1 \ge 1$ with equality holding iff $x = 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2198967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
If $B= ((X-1)^2, X-1, 1)$ is a base of $\mathbb{R}_2[X]$, how do I the coordinates of a polynomial in this base? Suppose you have a base: $B_2=((X-1)^2, (X-1), 1)$ and a polynomial of the form $a + bX + cX^2$ in base $B_1(X^2, X, 1)$, how do I find the its coordinates in $B_2$?
I figured maybe trying to find the matrix... | \begin{align}
x^2 &= 1\cdot(x-1)^2 + 2\cdot(x-1) + 1\cdot1 \\
x &= 1\cdot(x-1) + 1\cdot1 \\
1 &= 1\cdot1 \\ \\
\Rightarrow ax^2 + bx + c &= a((x-1)^2 + 2(x-1) + 1) + b((x-1) + 1) + c \\
&= a\cdot(x-1)^2 + (2a + b)\cdot(x-1) + (a+b+c)\cdot 1
\end{align}
That is we need something that maps $(a,b,c)$ to $(a, 2a+b, a+b+c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2202558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How many different knight's moves are there on an $n \times n$ chessboard? I have found there to be $48$ total moves on a $4 \times 4$ board... and $96$ on a $5 \times5$... but I can not see the relevance to each other in terms of a "$n \times n$" board.
By "moves" I am referring to from each space the amount of move... | Extending your example ...
For the top row you get 2,3,4,4,....,4,4,3,2
For the second row you get 3,4,6,6,....,6,6,4,3
For the third 4,6,8,8,...,8,8,6,4
For the fourth also 4,6,8,8,...,8,8,6,4
...
[and then the mirror of that for the bottom 4 rows]
So:
With $n\ge 4$ you get:
$2+3+(n-4)*4+3+2=10 +4*(n-4)$ for rows 1 a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2204736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
If $\log\left(\left(\frac{b}{a}\right)^{\frac{b}{3}}\right) + \log\left(\left(\sqrt[3]{\frac{a}{b}}\right)^{9a}\right)=1$, then what is $a^2+b^2$? Let $a$ and $b$ be two positive integers where $b$ is a multiple of $a$. If $\log\left(\left(\frac{b}{a}\right)^{\frac{b}{3}}\right) + \log\left(\left(\sqrt[3]{\frac{a}{b}}\... | $b = na$
$b/3 (\log b - \log a) + 3a (\log a - \log b) = 1$
$(na/3 - 3a)(\log n + \log a - \log a) = 1$
$a (n/3 - 3) \log n = 1$
$n = 10, a = 3, b = 30$ is a solution
you can see that you won't find more solution in integers (e.g. n = 10^k)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2206089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
prove an inequality with conditions Let $a,b, c \in \mathbb{R}$ such that $0\le a\le1,0\le b\le1 , 0\le c\le1.$
If $$a+b\leq c+1,
\\ a+c \leq b+1,
\\b+c\leq a +1$$
can we prove that
$a^2+b^2+c^2\le 1+2abc$ ?
| Let $a+1-b-c=x$, $b+1-a-c=y$ and $c+1-a-b=z$.
Hence, $x$, $y$ and $z$ are non-negatives such that $x+y+z=3-a-b-c\leq3$,
$a=\frac{2-y-z}{2}$, $b=\frac{2-x-z}{2}$, $c=\frac{2-x-y}{2}$
and we need to prove that
$$\frac{1}{4}\sum_{cyc}(2-x-y)^2\leq1+\frac{1}{4}\prod_{cyc}(2-x-y)$$ or
$$\sum_{cyc}(4+2x^2-8x+2xy)\leq4+8-\pr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2206648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find $\lim\limits_{h\to 0}\frac{f(2-h^2)-f(2)}{h^2}$ let $f(x)= (x^3+2x)[\frac{x}{2}]$
[x]:floor function
then :
$$\lim\limits_{h\to 0}\frac{f(2-h^2)-f(2)}{h^2}=?$$
My try :
$$\lim\limits_{h\to 0}\frac{f(2-h)-f(2)}{h}=f'(2)$$
$$\lim\limits_{x\to 2}\frac{(x^3+2x)[\frac{x}{2}]-f(2)}{x-2}=f'(2)$$
$$\lim\limits_{x\to 2}\... | Observe that as $h\rightarrow 0$, $h^2\rightarrow 0^+$ since $h^2$ is always nonnegative. Therefore, we can write:
$$
\lim_{h\rightarrow 0}\frac{f(2-h^2)-f(2)}{h^2}=\lim_{k\rightarrow 0^+}\frac{f(2-k)-f(2)}{k}
$$
by setting $k=h^2$. Then, by substituting the formula for $f$, we get
\begin{align*}
\lim_{k\rightarrow 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2206971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $f:R\to R$ is defined by $f(x)=x^2-3$, find $f^{-1} (-1)$ If $f:R\to R$ is defined by $f(x)=x^2-3$, find $f^{-1} (-1)$
My Attempt :
$$f(x)=x^2-3$$
$$y=x^2-3$$
Then how to proceed further?
| $$
f\left( x \right) = x^{2} - 3
$$
The function has two $x$ values that map to $y = -1$.
$$
\color{blue}{f^{-1}(y) = \sqrt{x+3}} \qquad \Rightarrow \qquad f^{-1}\left( -1 \right) = \sqrt{2}
$$
$$
\color{red}{f^{-1}(y) = -\sqrt{x+3}}\qquad \Rightarrow \qquad f^{-1}\left(-1 \right) = -\sqrt{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2212583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $Col(A)$ and $Col(B)$ are linearly independent, then $AB = 0$ What does $Col(A)$ and $Col(B)$ are linearly independent mean?
It is from an old linear algebra book and I don't see this used very often, does it mean a basis of $Col(A)$ and a basis of $Col(B)$ are linearly independent?
Assume the above definiton, then... | First, an example:
$$
\begin{align}
\mathbf{A} \mathbf{B} &= \mathbf{0} \\[5pt]
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{array}
\right)\,
\left(
\begin{array}{cccc}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{array}
\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2213429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve in $\mathbb Z^3$.
Let $x$, $y$ and $z$ be integer numbers. Solve the following equation.
$$x^2+y^2+z^2=45(xy+xz+yz)$$
My trying.
It's a quadratic equation of $z$ and we need $\Delta=n^2$ for an integer $n$,
but it gives a very ugly expression.
Thank you!
| For Will Jagy, I am sorry!
Let $x-y=a$, $y-z=b$ and $z-x=c$.
Hence, $a^2+b^2+c^2\vdots11$ and $a+b+c=0$.
Thus, $a^2+ab+b^2\vdots11$, which says that $a\vdots11$ and $b\vdots11$ and $x\equiv y\equiv z(\mod11)$,
which gives $x\vdots11$, $y\vdots11$ and $z\vdots11$ (if $x\equiv y\equiv z\equiv r(\mod11)$ then $r^2\vdots... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2215023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Area inside outer curve of limacon $r=1+2\cos\theta$ I need to find the area inside the outer curve of limacon $r=1+2\cos\theta$
Here is what I tried:
$$
0=1+2\cos\theta
$$
$$
\theta=\cos^{-1}(-\frac{1}{2})
$$
$$
\theta=\frac{2}{3}\pi , \theta=\frac{8}{3}\pi
$$
$$
A = 2\int^{\frac{8}{2}\pi}_{\frac{2}{3}\pi}r^2d\theta... | The outer curve of the limacon lies between $\theta=\pm\frac{2\pi}{3}$ but using symmetry it is twice the area between $\theta=0$ and $\theta=\frac{2\pi}{3}$
\begin{eqnarray}
A&=&2\int_0^{2\pi/3}\frac{1}{2}r^2d\theta\\
&=&\int_0^{2\pi/3}(1+2\cos\theta)^2d\theta\\
&=&\int_0^{2\pi/3}3+4\cos\theta+2\cos(2\theta)\,d\theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2216825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Show that each of the following is an isomorphism Verify that each of the following is an isomorphism:
1) $T: \mathbb{R}^3 \to \mathbb{R}^3$ defined by $T(x,y,z)=(x+y,y+z,z+x)$.
2) $T: M_{2,2} \to \mathbb{R}^4$ defined by $T\left(\begin{bmatrix}a & b\\ c & d \end{bmatrix}\right) = (a+b, d, c, a-b)$.
For the 1), shoul... | I will leave you to check that these are actually linear transformations. However, the fact that I am about to express them as matrices implies that they are linear transformations.
1) We can represent $T$ by the matrix $$A = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1\end{pmatrix}_.$$
By this I mean that $T(v... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2218007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluation of two Euler type sums We know that the harmonic number sum (also called Euler type sum) enter link description here
$$\sum\limits_{n = 1}^\infty {\frac{{H_n^{\left( 2 \right)}}}{{{n^2}{2^n}}}} = {\rm{L}}{{\rm{i}}_4}\left( {\frac{1}{2}} \right)\; + \frac{1}{{16}}\zeta (4) + \frac{1}{4}\zeta (3)\log 2 - \f... | By Cauchy product we have
$$\operatorname{Li}_2^2(x)=\sum_{n=1}^\infty x^n\left(\frac{4H_n}{n^3}+\frac{2H_n^{(2)}}{n^2}-\frac{6}{n^4}\right)$$
divide both sides by $x$ then integrate from $x=0$ to $1/2$ and use the fact that $\int_0^{1/2}x^{n-1}=\frac1{n2^n}$ we have
$$\int_0^{1/2}\frac{\operatorname{Li}_2^2(x)}{x}\ dx... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2218866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
If P congruent to $-1 \bmod$ 8 then $2^{(p-1)/2} - 1$ divisible by $p$ I am trying to prove the following statement:
If $p$ congruent to $-1 \bmod 8$, then $$2^{\frac{p - 1}{2}} - 1$$ divisible by $p$.
I can assume $p$ is congruent to $-1 \bmod 8$, then I must prove
$2^{\frac{p - 1}{2}} - 1$ is divisible by $p$.
So ... | Let $p=8k-1$ and let $n=(p-1)/2 = 4k-1$. Then note that
$$1\cdot 2\cdot 3\cdots n \equiv (-(p-1))\cdot 2 (-(p-3)) \cdot 4 (-(p-5)) \cdots (n-1)(-(p-n)) \pmod{p},$$
where we replaced each odd number $x$ by a congruent even number $-(p-x)$.
Now count the minus signs in the last product. Because $n$ is odd, there are $(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2221582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Does $\lim_{m \to \infty}\sum_{n=1}^m (-1)^n (\sum_{k=n^2}^{(n+1)^2-1}\frac{1}{\sqrt{k}}-2) $ exist? This question is based on
an answer and comment
to this question:
convergence of $\sum\limits_{n=1}^\infty \frac{(-1)^{\lfloor \sqrt{n}\rfloor}}{\sqrt{n}}$
Does
$\displaystyle \lim_{m \to \infty} \sum_{n=1}^m (-1)^n
\le... | We can express $2$ as the telescoping sum $$\sum_{k=n^2}^{(n+1)^2-1} (2\sqrt{k+1}-2\sqrt{k}),$$
which lets us rewrite the sum over $n$ as $$\sum_{n=1}^\infty (-1)^n \sum_{k=n^2}^{(n+1)^2-1} \left(\frac1{\sqrt k}-2\sqrt{k+1}+2\sqrt{k}\right).$$
I claim that this series is actually absolutely convergent, for which we ju... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2223199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Singular value decomposition works only for certain orthonormal eigenvectors, not all? I'm trying to find the SVD of the following matrix:
$$A=
\begin{pmatrix}
1 & 1 \\
2 & -2 \\
2 & 2 \\
\end{pmatrix}
$$
I found the eigenvalues and vectors for $A'A$:
$$
\begin{array}{cc}
\lambda_1=10 & \lambda_2=8 \\
e_1'=(1,1) ... | You need to match the left singular vectors to the right ones, or vice versa. E.g. after you have computed $e_1'$ and $e_2'$, you could get the two corresponding left singular vectors as $e_1=Ae_1'/\|Ae_1'\|=\frac1{\sqrt{5}}(1,0,2)^T$ and $e_2=Ae_2'/\|Ae_2'\|=(0,\color{red}{-1},0)^T$ (note: the sign of $e_2$ here is di... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2225393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Rationalise the Denominator Rationalise the denominator and simplify fully:
$$\dfrac{6}{\sqrt{7} + 2}$$
I got the answer $\dfrac{2 \sqrt{7}}{3}$, but didn't get the mark. Is that not fully simplified?
I did $\displaystyle \frac{6}{\sqrt{7} + 2} \times \frac{\sqrt{7}}{\sqrt{7}} = \frac{6 \sqrt{7}}{(7 + 2)} = \frac{6 \sq... | I'll assume your expression was $\frac{6}{\sqrt{7}+2}$, since it's the only way you obtain that denominator.
$$\frac{6}{\sqrt{7}+2}\times\frac{\sqrt{7}-2}{\sqrt{7}-2}=\frac{6\sqrt{7}-12}{\sqrt{7}^2-2^2}=\frac{6\sqrt{7}-12}{3}={2\sqrt{7}-4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2225618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $f(x+1)+f(x-1)=4x^2-2x+10$ then what is $f(x)$
If $$f(x+1) +f(x-1)= 4x^2 -2x +10$$ then what is $f(x)$
What is strategy of solving this kind of problems ?
Thank you for help
| Using Taylor's formula for polynomials:
*
*$f(x+1)=f(x)+f'(x)\cdot 1+f''(x)\cdot\dfrac 12$,
*$f(x-1)=f(x)-f'(x)\cdot 1+f''(x)\cdot\dfrac 12$.
Therefore $\;f(x+1)+f(x-1)=2f(x)+f''(x)=4x^2-2x+10$. If we set $f(x)=ax^2+bx+c$, this equation simplifies to
$$ax^2+bx+c+a=2x^2-x+5,\enspace\text{whence}\quad a=2, b=-1, c=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2226077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Show that $\int_0^{\pi/2}sin^p\,\theta\;cos^q\,\theta\;d\theta = \frac{\sqrt{\frac{p+1}{2}}\sqrt{\frac{q+1}{2}}}{2\sqrt{\frac{p+q+2}{2}}},\; p,q > -1$ Show that: $$\int_0^{\pi/2}sin^p\,\theta\;cos^q\,\theta\;d\theta = \frac{\sqrt{\frac{p+1}{2}}\sqrt{\frac{q+1}{2}}}{2\sqrt{\frac{p+q+2}{2}}},\; p,q > -1$$
Here's the ques... | if $p=q=1$, we get
$$\int_0^\frac {\pi}{2} \frac {\sin (2\theta)d\theta}{2} $$
$$=\frac {1}{4}[-\cos (2\theta)]_0^\frac {\pi}{2} $$
$$=\frac {1}{2} \neq \frac {1}{2\sqrt {2}} $$
Your formula doesn't seem to be correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2227742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Prove that if $(x+y)^2 \equiv 0 \pmod{xy}$, then $x = y$
Let $x,y$ be integers. Prove that if $(x+y)^2 \equiv 0 \pmod{xy}$, then $x = \pm y$.
The given condition is equivalent to $x^2+y^2 \equiv 0 \pmod{xy}$. How do we continue from here to prove that $x = \pm y$?
| Let $p$ be a prime, write $x=p^au, y=p^bv$ where $u,v$ are relatively prime with $p$. $xy=p^{a+b}uv$. Suppose $a<b$ If $xy$ divides $x^2+y^2=p^{2a}u^2+p^{2b}v^2$, we have ${{p^{2a}u^2+p^{2b}v^2}\over{p^{a+b}}}$ is an integer. This implies that $p^{a+b}$ divides $p^{2a}u^2$. This is impossible. We deduce that $a\geq b$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2228505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Find the value of $3+7+12+18+25+\ldots=$ Now, this may be a very easy problem but I came across this in an examination and I could not solve it.
Find the value of
$$3+7+12+18+25+\ldots=$$
Now here is my try
$$3+7+12+18+25+\ldots=\\3+(3+4)+(3+4+5)+(3+4+5+6)+(3+4+5+6+7)+\ldots=\\3n+4(n-1)+5(n-2)+\ldots$$
After that, I co... | If you know the value of $n$ then $$3+7+12+18+25+\ldots=\\3n-3n+3+7+12+18+25+\ldots$$ As $3=1+2$ we can write as $$\\1+2+3+(1+2+3+4)+(1+2+3+4+5)+(1+2+3+4+5+6)+\ldots-3n=\\\left(\sum_{n=3}^n{\frac{n(n+1)}{2}}\right)-3n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2230870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Find a,b and c. Any method? Given that $$\frac{1}{a+\frac{1}{b+\frac{1}{c+1}}}=\frac{16}{38}$$, find $a,b,c$.
I've been figuring for this quite some time? Is it possible to solve?
| Sure it is, but you'll end up with infinite possibilities:
$$\begin{align*}
\frac{1}{a+\frac{1}{b+\frac{1}{c+1}}}&=\frac{16}{38}\\
a+\frac{1}{b+\frac{1}{c+1}}&=\frac{38}{16}\\
\frac{1}{b+\frac{1}{c+1}}&=\frac{38-16a}{16}\\
b+\frac{1}{c+1}&=\frac{16}{38-16a}\\
\frac{1}{c+1}&=\frac{16-(38-16a)b}{38-16a}\\
c&=\frac{38-16a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2232110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Need help to prove the last inequality of this mathematical induction Here is the original question:
Let $$x_n=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdots\frac{2n - 1}{2n}$$
Then show that $$x_n\leq\frac{1}{\sqrt{3n+1}} \text{, for all } n = 1, 2, 3, \ldots \text{ .}$$
It is trivial to show that the statement... | Say this inequality is correct for $x_n$. So we can say that: $x_n \leq \dfrac{1}{\sqrt{3n+1}}$ $x_{n+1} = x_n * \dfrac{2n+1}{2n+2} \leq \dfrac{1}{\sqrt{3n+1}}*\dfrac{2n+1}{2n+2}$ So know if we prove the statement below, the inequality is proved: $\dfrac{1}{\sqrt{3n+1}}*\dfrac{2n+1}{2n+2} \leq \dfrac{1}{\sqrt{3n+4}}$ I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2232895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
inertia of a lamina I am given a lamina and the area that it covers is rectangular. Its density, $d$, varies with position: $d=x^2+y^2$. If I want to find the moment of inertia about the $x$ and $y$ axis, can I do the integrals in polar form? I say this because I am aware that I need to (double) integrate $y^2(x^2+y^2)... | It's easier to use Cartesian coordinates,
\begin{align*}
I_{xx} &= \int_{a}^{b} \int_{c}^{d} \rho y^2 \, dy \, dx \\
&= \int_{a}^{b} \int_{c}^{d} y^2(x^2+y^2) \, dy \, dx \\
&= \int_{a}^{b}
\left[
\frac{x^2(d^3-c^3)}{3}+\frac{d^5-c^5}{5}
\right] \, dx \\
&= \frac{(b^3-a^3)(d^3-c^3)}{9}+\frac{(b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2233272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to solve the partial fraction decomposition $\frac{x^3+5x^2+3x+6}{2x^2+3x}$. I have the following integral:
$$\int\frac{x^3+5x^2+3x+6}{2x^2+3x}dx$$
I'm trying to use partial fraction decomposition but I'm getting stuck at the following formula:
$$\int\frac{(x+6)(1+5x+x^2)}{x(2x+3)}-\frac{x+27}{2x+3}dx$$
I can't n... | I will let you work out the detail, but the solution should be:
$ \cfrac{x^3 + 5x^2 + 3x + 6}{2x^2 + 3x} =\cfrac{7}{4} + \cfrac{2}{x} + \cfrac{x}{2} - \cfrac{25}{4(2x+3)} $
Now, you can evaluate the integral:
\begin{equation} \int \cfrac{x^3 + 5x^2 + 3x + 6}{2x^2 + 3x} dx = \int \cfrac{7}{4} dx + \int \cfrac{2}{x} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2235907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
partial fractions for a function I need help finding the partial fraction decomposition for this function, I am just lost on it, here it is:
$(x^2 + x + 1)/(2x^4+3x^2+1)$. the help is appreciated. thank you.
| As suggested by one user.
$\frac{(x^2 + x + 1)}{(2x^4+3x^2+1)} = \frac{ax+b}{2x^2+1} + \frac{cx+d}{x^2+1}$
$x^2+x+1 = (ax+b)(x^2+1) + (cx+d)(2x^2+1)$
Comparing terms containing $x^3$,
$0 = a + 2c$
Comparing terms containing $x^2$,
$1 = b + 2d$
Comparing terms containing $x$,
$1 = a + c$
Comparing terms containing const... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2237858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find Jordan form of matrix 6x6 Find the Jordan form of matrix $$A = \begin{bmatrix}
2 & 0 & 1 & 0 & -1 & 0\\
0 & 2 & 0 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 1 & 0\\
0 & 0 & 0 & 5 & 0 & -9\\
2 & 0 & 1 & 0 & 3 & 0\\
0 & 0 & 0 & 1 & 0 & -1\\
\end{bmatrix}$$
I found $\lambda$ from $|A-\lambda I|= 0$ and $(\lambda -2)^6=0$
$$B=(... | Since $\operatorname{rank} B = 3$, you have three linearly independent eigenvectors associated to the (only) eigenvalue $\lambda = 2$ so there are three Jordan blocks in the Jordan form of $A$. Since $B^3 = 0$ while $B^2 \neq 0$, the minimal polynomial of $A$ is $(x - 2)^3$ which means that the largest Jordan block in ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2239186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Angle between vectors a and b if a + 3b is perpendicular to.... Determine the angle between vectors $a$ and $b( a \ne 0, b \ne0)$ if $a + 3b$ is perpendicular to $2a - b$ and $a + 7b$ is perpendicular to $2a + b$.
I've done this:
$(a + 3b) \cdot (2a - b) = 0$
$\dots 5a⋅b = 3b⋅b - 2a⋅a$
$(a + 7b) \cdot (2a + b) = 0$
$\c... | You have $$5a\cdot b=3b\cdot b-2a\cdot a$$ $$15a\cdot b=-7b\cdot b-2a\cdot a$$so subtract the 2nd equation from the first to get $$-10a\cdot b=10b\cdot b\Rightarrow b\cdot b=-a\cdot b$$
Also $$35a\cdot b=21b\cdot b-14a\cdot a$$ $$45a\cdot b=-21b\cdot b-6a\cdot a\Rightarrow80a\cdot b=-20a\cdot a$$ so similarly $$a\cdot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2240175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
prove that there are infinitely many primitive pythagorean triples $x,y,z$ such that $y=x+1$ Here is my question: Prove that there are infinitely many primitive pythagorean triples $x,y,z$ such that $y=x+1$.
They gave me a hint that I should consider the triple $3x+2z+1, 3x+2z+2, 4x+3z+2$, but I honestly do not know h... | You may just ignore the hint and prove that $x^2+(x+1)^2=y^2$ has an infinite number of solutions in $\mathbb{N}\times\mathbb{N}$. Such identity is equivalent to
$$ (2x+1)^2-2 y^2 = -1 $$
hence it is enough to show that there are infinite natural solutions to $a^2-2b^2=-1$ with $a$ being odd. But, wait, if $a,b\in\math... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2241186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Factor $ab^3-ac^3+bc^3-ba^3+ca^3-cb^3$
a) Use the remainder theorem to prove that $(a+b+c)$ is a factor of $(a^3+b^3+c^3-3abc)$ . Then find the other factor.
b) Hence factor $(ab^3-ac^3+bc^3-ba^3+ca^3-cb^3)$
So far I have managed to find the other remainder being $(a^2+b^2+c^2-ab-ac-bc)$ but I don't understand how to... | Hint (without using the first part): considering it as a $3^{rd}$ degree polynomial in $a\,$:
$$ab^3-ac^3+bc^3-ba^3+ca^3-cb^3 = (c-b)\,a^3+(b^3-c^3)\,a+bc\,(b^2-c^2)$$
It can be easily verified that it has $a=b$ and $a=c$ as roots. Since the term in $a^2$ is missing, the sum of all three roots is $0$ by Vieta's relatio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2241289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Let a general term $T_n$ be defined as $T_n =\left(\frac{1\cdot 2\cdot 3 \cdot 4 \cdots n}{1 \cdot 3 \cdot 5 \cdot 7 \cdots (2n+1)}\right)^2$ Let a general term $T_n$ be defined as
$$T_n =\left(\frac{1\cdot 2\cdot 3 \cdot 4 \cdots n}{1 \cdot 3 \cdot 5 \cdot 7 \cdots (2n+1)}\right)^2$$
Then prove that
$\lim_{n\to\infty... | I do not think $\sum_{n\geq 1}T_n$ has a closed form, but such inequality can be improved a bit.
We have
$$ T_n = \left(\frac{n!}{(2n+1)!!}\right)^2 = \left(\frac{n!(2n)!!}{(2n+1)!}\right)^2 = \left(\frac{n!^2 2^n}{(2n+1)!}\right)^2 = \frac{4^n}{(2n+1)^2 \binom{2n}{n}^2}$$
and we may borrow a couple of useful lemmas fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2241494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Solve a simple equation: $x+x\sqrt{(2x+2)}=3$
$x+x\sqrt{(2x+2)}=3$
I must solve this, but I always get to a point where I don't know what to do. The answer is 1.
Here is what I did:
$$\begin{align}
3&=x(1+\sqrt{2(x+1)}) \\
\frac{3}{x}&=1+\sqrt{2(x+1)} \\
\frac{3}{x}-1&=\sqrt{2(x+1)} \\
\frac{(3-x)^{2}}{x^{2}}&=2(x+1... | So you got to the cubic equation $f(x)=-2x^{3}-x^{2}-6x+9=0$. When you come across a cubic like this, try evaluating $f(\pm1), f(\pm2)$, etc, to try and figure out some roots so you can factor it (you know $x_0$ a root if $f(x_0)=0$). Here you can see $1$ is a root, so factoring out $(x-1)$ gives $f(x)=(x-1)\underbrace... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2245631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Why is $\frac{a_k}{z^{n-k}},(0\leq k < n)$ less than $\frac{\lvert a_n \rvert}{2n}$ Suppose there is a polynomial:
$$P(z) = a_0 + a_1z +a_2z^2 +\dots +a_nz^n,\quad (a_n \ne 0)$$
Let $$w = \frac{a_0}{z^n} + \frac{a_1}{z^{n-1}} + \frac{a_2}{z^{n-2}} + \dots + \frac{a_{n-1}}{z},$$
Why is it that for a sufficiently large p... | Since, $a_n\not=0$ we have $|a_n|\not=0$. Now, we have to find a positive real number $R_0$ such that for any complex number $z$ with $|z|> R_0$ we have $$\left |\frac{a_0}{z^n}\right|=\frac{|a_0|}{|z^n|}=\frac{|a_0|}{|z|^n}<\frac{|a_n|}{2n},$$$$\left |\frac{a_1}{z^{n-1}}\right|=\frac{|a_1|}{|z^{n-1}|}=\frac{|a_1|}{|z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2246099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Minimum value of a trigonometric function
Prove that the minimum value of $$a \sec \theta + b \csc \theta$$ is $$(a^{2/3} + b^{2/3})^{3/2}$$ where $0 < \theta < \frac{\pi}{2}$ and $a$ and $b$ are positive real numbers.
This is one part of a larger question whose first part I solved. I tried to express $a \sec \theta ... | $f(\theta)=a\sec\theta+b\csc\theta,\quad 0<\theta<\frac{\pi}{2}$
$f'(\theta) = a\cdot(-1)\cdot(-\sin\theta)\cdot\sec^2\theta+b\cdot(-1)\cdot(\cos\theta)\cdot\csc^2\theta$
$$f'(\theta)= \frac{a\sin\theta}{\cos^2\theta} -\frac{b\cos\theta}{\sin^2\theta}$$
Setting $f'(\theta)=0:$
$$f'(\theta) = \frac{a\sin^3\theta-b\cos^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2246447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"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.