Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Calculate root interval of $x^5+x^4+x^3+x^2+1$ So I have to find an interval (in the real numbers) such that it contains all roots of the following function:
$$f(x)=x^5+x^4+x^3+x^2+1$$
I've tried to work with the derivatives of the function but it doesn't give any information about the interval, only how many possible ... | EDITED for correct function.
$f(x) = 1+x^2+x^3+x^4+x^5$
is positive for x>0.
So consider $x < 0$:
One can write $f(x) = (1+x^2)(1+x^3) + x^4$ which is positive at least for $(1+x^3) >0$, i.e. $x > -1$.
$f'(x) = 2x+3x^2+4x^3+5x^4 = x(2+3x)+x^3(4+5x)$
so this is positive at least for $x < -4/5 $ since then all terms are ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2375023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 1
} |
Convert a equation with fractions into whole numbers So I have this equation:
$$\frac{2}{3}a^2-\frac{4}{9}a^2 = 8a$$
So this is a really easy problem, I could just multiply
$$\frac{2}{3}*\frac{3}{3} = \frac{6}{9}$$
Then subtract
$$\frac{6}{9}a^2 - \frac{4}{9}a^2 = \frac{2}{9}a^2=8a$$
$$36a=a^2$$
$$36=a$$
However, I wa... | $$9\left(\frac{2}{3}a^2-\frac{4}{9}a^2\right)=8a\cdot9$$ it's
$$6a^2-4a^2=72a$$ and it's not $6a^2-9a^2=8a$, which you wrote.
The continue is
$$a^2-36a=0$$ or
$$a(a-18)=0,$$
which gives $a=0$ or $a=18$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2376720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
A problem from the chapter 'theory of equations' Question : Form a biquadratic equation with rational coefficients two of whose roots are √3 + 2 and √3 - 2.
So f(x) = (x- √3-2)(x-√3+2)q(x) where q(x) must be a quadratic equation. This is how far I have done. But I cannot solve it. I am a Bsc student with maths as an a... | Write a quadratic equation which has roots $x_1=\left(\sqrt{3}+2\right)^2=7+4 \sqrt{3};\;x_2=\left(\sqrt{3}-2\right)^2=7-4 \sqrt{3}$
We need the sum and the product of the roots
$x_1+x_2=14;\;x_1x_2=\left(7+4 \sqrt{3}\right)\left(7-4 \sqrt{3}\right)=1$
$$(x-x_1)(x-x_2)=x^2-(x_1+x_2)x+x_1x_2=x^2-14x+1$$
And then substit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2377365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Maximum value of $a^3 + b^3 + c^3$ If $a+b+c=5$ and $ \frac{1}{a} +\frac{1}{b} +\frac{1}{c} =\frac{1}{5}$ then find the maximum value of $a^3 + b^3 +c^3$ where $a,b,c $ is real numbers.
My Attempt
writing $a+b=5-c$ and $ \frac{1}{a} +\frac{1}{b} =\frac{1}{5} -\frac{1}{c}$ and after algebraic manipulation I obtain $(a+... | Here is a lesser known approach to these things, which sometimes helps
Let $xy+yz+zx=p$ so that $x^2+y^2+z^2=(x+y+z)^2-2p=25-2p$.
Set $S_3=a^3+b^3+c^3$
Then clearing fractions in the second equation gives $abc=5p$.
We have also that $a,b,c$ are roots of the polynomial equation $$q(x)=x^3-5x^2+px-5p=0$$
Now observe that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2377466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How can one prove $-\arctan\left(\frac{x}{y}\right)+\pi H(y)=2\arctan\left(\frac{y}{\sqrt{x^2+y^2}+x}\right)+\frac{\pi}{2}$? How can one prove the following?
$$
-\arctan\left(\frac{x}{y}\right)+\pi H(y)=2\arctan\left(\frac{y}{\sqrt{x^2+y^2}+x}\right)+\frac{\pi}{2}$$
with $H(x)$ the Heaviside function.
I have the impres... | Using calculus and derivatives is fast but doesn't show the simplicity of this kind of angle relationships.
Start with the case $x>0$. Then consider the following figure
If $y>0$, let the right-angled triangle $\triangle ABC$ have sides $\overline{AB} = y$ and $\overline{BC} = x$, so that
$$\alpha = \arctan \left(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2378266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\sqrt{m-1}$ is an integer
Let $k$ be a fixed odd positive integer. Let $m,n$ be positive integers such that $(2+\sqrt{3})^k=m+n\sqrt{3}$. Prove that $\sqrt{m-1}$ is an integer.
Let $(2+\sqrt{3})^n = a_n+b_n\sqrt{3}$. From $$a_{n+1}+b_{n+1}\sqrt{3} = (a_n+b_n\sqrt{3})(2+\sqrt{3}) = (2a_n+3b_n)+(a_n+2b_n)\s... | A first step:
Define integer sequences $x_n$ and $y_n$ by
$$ (2+\sqrt{3})^{2n+1} = x_n+y_n\sqrt{3}. $$
Then $x_0=2$ and $y_0=1$, and furthermore you get the recursion
\begin{eqnarray*}
x_{n+1} &=& 7x_n+12y_n\\
y_{n+1} &=& 4x_n+7y_n
\end{eqnarray*}.
The first equation gives $12y_n=x_{n+1}-7x_n$, and plugging this into t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2378922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How to simplify $\left({{\frac{2207+(2207^2-4)^{1/2}}2}}\right)^{1/8}$ into quadratic form? The 1995 Putnam B4 asks to evaluate some infinite expression and I got this value after some computations:$$\left({{\frac{2207+(2207^2-4)^{1/2}}2}}\right)^{1/8}$$
However I am supposed to write it in form:$$\frac{a+b\sqrt{c}}{d}... | Let $\,x \gt 1\,$ be the expression in question, then:
$$\require{cancel}
2x^8=2207 + \sqrt{2207^2-4} \\
(2x^8-2207)^2=2207^2-4 \\
4x^{16}-4\cdot 2207 x^8+\cancel{2207^2}=\cancel{2207^2}-4 \\
x^{16}-2207 x^8 + 1 = 0 \\
x^8 + \frac{1}{x^8}=2207 \\
\left(x^4+\frac{1}{x^4}\right)^2 = 2209 \\
x^4+\frac{1}{x^4} = 47 \\
\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2380627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Infinite square root muliplication $(x=3\sqrt{y\sqrt{3\sqrt{y}....}})$ I have this problem which says
$$
x=3\sqrt{y\sqrt{3\sqrt{y\cdots}}}\\
y=3\sqrt{x\sqrt{3\sqrt{x\cdots}}}
$$
What is $x+y$
I tried
$$
x^2=9y\sqrt{x}\\
x^3=81y^2\\
y^3=81x^2\\
x^3+y^3=81(x^2+y^2)\\
(x+y)(x^2-xy+y^2)=81(x^2+y^2)
$$
I don't know what to... | Suppose $x \neq 0 \neq y$. From $x^3=81y^2$ and $y^3=81x^2$, it follows that $\frac{x^3}{y^3}=\frac{y^2}{x^2}$, hence $x^5=y^5$, so $x=y$. I think you can take it from there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2380992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Maximize linear function over disk of radius $2$
Maximize $f(x, y) = x + 2y$ with constraint $x^2 + y^2 \le 4$.
$f(x, y)$ has no CP's so thats something gone.
I considered the boundary/edge of $g(x, y) = x^2 + y^2$. And I got the point $(0, 2), (2, 0)$, which gave $f(0, 2) = f(2, 0) = 4$. However, the point $(2/\sqrt... | So you have a problem to compute the critical points of $f\mid_{S}$ where
$S=g^{-1}(4)$ with $g(x,y)=x^2+y^2$.
You need to solve the equation $\nabla f(x,y)=\lambda \nabla g(x,y)$ for some $\lambda\in\mathbb R$. You get
$$
\nabla f(x,y)=\lambda\nabla g(x,y)\Leftrightarrow \begin{pmatrix}1\\2\end{pmatrix}=\lambda\begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2382224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Please prove $\frac{1 + \sin\theta - \cos\theta}{1 + \sin\theta + \cos\theta} = \tan \left(\frac \theta 2\right)$
Prove that $\dfrac{1 + \sin\theta - \cos\theta}{1 + \sin\theta + \cos\theta} = \tan\left(\dfrac{\theta}{2}\right)$
Also it is a question of S.L. Loney's Plane Trignonometry
What I've tried by now:
\be... | hint just use
$$1+\cos (X)=2\cos^2 (\frac {X}{2}) $$
$$1-\cos (X)=2\sin^2 (\frac {X}{2}) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2384719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Induction on Fermat Numbers: $F_n = \prod_{j=0}^{n-1}F_j+2$ Is the Following Proof Correct?
Theorem. Given that $\forall n\in\mathbf{N}(F_n = 2^{2^n}+1)$ show that the following is true
$$\forall n\in{1,2,3...}\left(F_n = \prod_{j=0}^{n-1}F_j+2\right)$$
Proof. We construct the proof by recourse to Mathematical-Inductio... | Your proof is fine, but I think it's a little easier if you define the numbers $F_n$ by the recurrence
$$F_n=F_0F_1\cdots F_{n-1}+2$$
and then prove the identity $F_n=2^{2^n}+1$ by induction.
Basis step:
$$F_0=3=2^{2^0}+1$$
Inductive step:
$$F_{n+1}=F_0\cdots F_{n-1}F_n+2$$
$$F_{n+1}-2=(F_0\cdots F_{n-1})F_n=(F_n-2)F_n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2387336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Inequality : $\sqrt[3]{\frac{x^3+y^3+z^3}{xyz}} + \sqrt{\frac{xy+yz+zx}{x^2+y^2+z^2}} \geq 1+\sqrt[3]{3}$
Let $x>0$, $y>0$ and $z > 0$. Prove that $$\sqrt[3]{\frac{x^3+y^3+z^3}{xyz}} + \sqrt{\frac{xy+yz+zx}{x^2+y^2+z^2}} \geq
1+\sqrt[3]{3}.$$
From Micheal Rozenberg's answer :
$(x+2)\sqrt{x^2+2}\left(\sqrt{x^2+2}+\sqr... | A proof by using the Sergic Primazon's idea. Dedicated to dear Hans.
Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, we need to prove that
$$\sqrt[3]{\frac{27u^3-27uv^2}{w^3}+3}+\sqrt{\frac{v^2}{3u^2-2v^2}}\geq\sqrt[3]3+1.$$
Now, since $27u^3-27uv^2\geq0$ and $v^4\geq uw^3$, it's enough to prove that
$$\sqrt[3]{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2387423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Why is the reciprocal of the power series $1-\frac{z^2}{3!}+\frac{z^4}{5!}+\cdots$ equal to $1+\frac{z^2}{6}+\frac{7z^4}{360}+\cdots$? \begin{align}
\frac1{\sin(z)}
&=\frac1z\frac{z}{\sin(z)}\\
&=\frac1z\left(1-\frac{z^2}{3!}+\frac{z^4}{5!}-\frac{z^6}{7!}+\cdots\right)^{-1}\\
&=\frac1z\left(1+\frac{z^2}{6}+\frac{7z^4}{... | The Taylor series of $\frac{\sin z}{z}$ at the origin is straightforward to compute, the Taylor series of the reciprocal function $\frac{z}{\sin z}$ a bit less. However, we may notice that $\frac{z}{\sin z}$ is an even meromorphic function with simple poles at the elements of $\pi\mathbb{Z}\setminus\{0\}$. Additionally... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2388351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Sum of first $1+3+9+\cdots+3^n$ natural numbers. How to prove that the sum of the first $1+3+9+\cdots+3^n$ natural numbers is equal to $1^2+3^2+9^2+\cdots+(3^n)^2$?
I've tried induction, but I can't get through the induction step. The base is simple, but in the step I can only use the induction hypothesis in a way tha... | You don't need induction; you can actually show equality with a relatively well-known (though perhaps at first "tricky") summation technique:
Well first, what is $1+3+9+\cdots+3^n$? Define that quantity to be $N$, and note that $3N = 3+9+\cdots+3^n+3^{n+1} = N + 3^{n+1}-1$, so $2N = 3^{n+1}-1$, and $N = \frac12(3^{n+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2388476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Let $a_{1}>0,a_{2}>0$ and $a_{n}=\frac{2a_{n-1}a_{n-2}}{a_{n-1}+a_{n-2}}, n>2$, then $\{ a_{n}\}$ converges to $\frac{3a_{1}a_{2}}{a_{1}+a_{2}}$. Let $a_{1}>0,a_{2}>0$ and $a_{n}=\dfrac{2a_{n-1}a_{n-2}}{a_{n-1}+a_{n-2}}, n>2$, then $\{ a_{n}\}$ converges to $\dfrac{3a_{1}a_{2}}{a_{1}+a_{2}}$.
My attempt:
\begin{alig... | Hint: let $b_n=1/a_n$ then $b_{n+1}=(b_n+b_{n-1})/2 \iff (b_{n+1}-b_n)=-(b_n-b_{n-1})/2$. Then:
$$
b_{n+1}-b_n=\frac{-1}{2}(b_n-b_{n-1})=\left(\frac{-1}{2}\right)^2(b_{n-1}-b_{n-2}) = \cdots = \left(\frac{-1}{2}\right)^{n-1}(b_{2}-b_{1})
$$
Next, telescope:
$$
b_{n+1} = b_n + \left(\frac{-1}{2}\right)^{n-1}(b_{2}-b_{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2388689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
$\ \log _x\left(\log _{36}\left(2\cdot 9^{2x}-3\cdot 4^{2x}\right)\right)<\:1 $ I have to find x in the inequation below:
$$\ \log _x\left(\log _{36}\left(2\cdot 9^{2x}-3\cdot 4^{2x}\right)\right)<\:1 $$
So I did the following:
$$\ \log _{36}\left(2\cdot \:9^{2x}-3\cdot \:4^{2x}\right)<\:x $$
$$\ 2\cdot \:9^{2x}-3\cdot... | You have to consider two different cases separately: $0<x<1$ and $x>1$. The solution you presented here is only valid for the latter case when $x>1$. But in the case $0<x<1$, logarithm base $x$ is a decreasing function, and thus eliminating it would flip the sign of the inequality — so that your second line will be $\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2389246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Can't solve this limit using $\frac{\sin x}x$ I've been told to solve the following limit using only $\lim_{x \to 0}\frac{\sin x}{x}=1$:
$$\lim_{x \to 0} \frac{1-\cos(1-\cos x)}{x^4}$$
I don't know how to do it except using L'Hospital's Rule but it's just insane amount of math and not the writer's intention(Symbolab do... | $$\begin{array}{rcl}
\displaystyle \lim_{x \to 0} \frac{1-\cos(1-\cos x)}{x^4}
&=&
\displaystyle \lim_{x \to 0} \frac{1-\cos^2(1-\cos x)}{x^4[1+\cos(1-\cos x)]} \\
&=& \displaystyle \lim_{x \to 0} \frac{\sin^2(1-\cos x)}{x^4[1+\cos(1-\cos x)]} \\
&=& \displaystyle \lim_{x \to 0} \frac{(1-\cos x)^2\sin^2(1-\cos x)}{x^4(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2389392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Evaluate $\lim_{x \to 0}\frac{x}{\sqrt{1-\sqrt{1-x^2}}}$ Evaluate $$L=\lim_{x \to 0}\frac{x}{\sqrt{1-\sqrt{1-x^2}}} \tag{1}$$
I have used L Hopital's Rule we get
$$L=\lim_{x \to 0} 2\frac{ \sqrt{1-\sqrt{1-x^2}} \sqrt{1-x^2}}{x}$$
$\implies$
$$L=2 \lim_{x \to 0}\frac{ \sqrt{1-\sqrt{1-x^2}}}{x}$$ and from $(1)$ we get
... | There is a right limit and a left limit and they are different, which means that the limit doesn't exist because when it exists it is unique.
Let substitute $1-x^2=w^2$
Suppose $x>0$ we have $x=\sqrt{1-w^2}$ and the limit becomes
$$\lim_{w\to 1} \, \dfrac{\sqrt{1-w^2}}{\sqrt{1-w}}=\lim_{w\to 1} \, \frac{\sqrt{1-w} \sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2390192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Using Lagrange's multiplier method, find the shortest distance between the line y=10-2x and the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$
Using Lagrange's multiplier method, find the shortest distance between the line $y=10-2x$ and the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$.
My work:
Let the point on ellipse be $(2\c... | $\frac{x^2}{4} + \frac{y^2}{9}$ can be made into $ 1 = 9x^2 + 4y^2 = 36 $ which can be made into $9cos^2(\theta) + 4sin^2(\theta) = 36$. This should help simplify your calculations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2392883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Elliptic Integrals & Gamma Functions of Rational values. I recently saw this question ... https://math.stackexchange.com/questions/2393668/double-factorial-sum-k-0-infty-left-frac-2k-1-2k-right#2393668 ... & I am unable to show it.
The result
$$1-\left( \frac{1}{2} \right)^{3}+\left( \frac{1\cdot 3}{2\cdot 4} \righ... | The sum evaluates to $(1-k^{2})(2K(k)/\pi)^{2}$ for $k=\sqrt{2}-1$. The value of $K$ is obtained from here. Using the value of $K$ we can get the desired closed form sum of the given series.
Using hypergeometric transformations we can prove the formula $$\left(\frac{2K(k)}{\pi}\right)^{2}=\sum_{n=0}^{\infty}\left(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2393810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Matrix similarity is an equivalence relation Recall that for $ A,B \in \mathsf{M}_n$, we say that $A$ is similar to $B$, denoted $A \sim B$, if there is an invertible matrix $S \in \mathsf{M}_n$ such that $A = S B S^{-1}$. Prove that similarity is an equivalence relation on $\mathsf{M}_n$.
I have no idea on what to do ... | I'm not sure how you get any of the equalities in your work, except for $B=BI_n$. Since $S$ is just some invertible matrix, how do you get $A=SAS^{-1}=AI_n$ and $BI_n=SBS^{-1}$? I suspect you are commuting $SAS^{-1}$ into $ASS^{-1}$ and similarly $SBS^{-1}$ into $BSS^{-1}$, but remember that matrix multiplication is no... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2395050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$\sin^3 a\sin(b-c)+\sin^3b\sin(c-a)+\sin^3c\sin(a-b)+\sin(a+b+c)\sin(b-c)\sin(c-a)\sin(a-b)=0$
Verify that$$\sin^3 a\sin(b-c)+\sin^3b\sin(c-a)+\sin^3c\sin(a-b)$$
$$+\sin(a+b+c)\sin(b-c)\sin(c-a)\sin(a-b)=0.$$
I tried to break $\sin^3a$ into $\sin^2\cdot\sin a$ and use it to make sum with $\sin(b-c)$, but eventually... | $$\sum_{cyc}\sin^3a\sin(b-c)=\frac{1}{4}\sum_{cyc}(3\sin{a}-\sin3a)\sin(b-c)=$$
$$=\frac{1}{8}\sum_{cyc}\left(3\cos(a-b+c)-3\cos(a+b-c)-\cos(3a-b+c)+\cos(3a+b-c)\right).$$
$$\sin(a+b+c)\prod_{cyc}\sin(a-b)=\frac{1}{2}\sin(a+b+c)\sin(b-c)(\cos(2a-b-c)-\cos(b-c))=$$
$$=\frac{1}{4}\sin(a+b+c)(-\sin(2c-2a)-\sin(2a-2b)-\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2395234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to evaluate $\binom{2}{2}\binom{10}{3} + \binom{3}{2}\binom{9}{3} + \binom{4}{2}\binom{8}{3} + \ldots + \binom{9}{2}\binom{3}{3}$ I really can't understand how to approach exercise 41.
I tried by comparing the 5th term of binomial expansion and something like that, but getting 12C5, I think the answer should be 13... | We wish to show that
$$\sum_{k = 2}^{9} \binom{k}{2}\binom{10 - k}{3} = \binom{13}{6}$$
The right hand side counts six-element subsets of the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13\}$.
The left-hand side counts six-elements subsets of the set whose third-largest element is $k + 1$. To see this, observe that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2397944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Show that: $\sin ^2 (a+b)+\sin ^2 (a-b)=4(1-\cos ^2 a \cos ^2 b)$ Show that: $$\sin ^2 (a+b)+\sin ^2 (a-b)=4(1-\cos ^2 a \cos ^2 b)$$
I tried two approaches.
Approach 1: $$\sin ^2 (a \pm b)=\sin a\cos b \pm \sin b\cos a$$
This reduced to: $$2(\cos ^2 a +\cos ^2 b -2\cos ^2 a\cos ^2b)$$
I can't see where to go from here... | hint:$$\sin ^2 (a+b)+\sin ^2 (a-b)=4(1-\cos ^2 a \cos ^2 b)=\\1-\cos(2a+2b)+1-\cos(2a-2b)=\\2-(\cos(2a+2b)+\cos(2a-2b))= $$use sum to product formula
$$=2-(\cos(2a+2b)+\cos(2a-2b))=\\2 -2\cos (\frac{2a+2b+2a-2b}{2}).\cos (\frac{2a+2b-(2a-2b)}{2})=\\2-2\cos (2a).\cos (2b)=\\$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2398042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Simplify $\sqrt{6-\sqrt{20}}$ My first try was to set the whole expression equal to $a$ and square both sides. $$\sqrt{6-\sqrt{20}}=a \Longleftrightarrow a^2=6-\sqrt{20}=6-\sqrt{4\cdot5}=6-2\sqrt{5}.$$
Multiplying by conjugate I get $$a^2=\frac{(6-2\sqrt{5})(6+2\sqrt{5})}{6+2\sqrt{5}}=\frac{16}{2+\sqrt{5}}.$$
But I sti... | Here's a useful formula for this kind of problems:
$$\sqrt{a \pm \sqrt{b}} = \sqrt{\frac{a + \sqrt{a^2 - b}}{2}} \pm \sqrt{\frac{a - \sqrt{a^2 - b}}{2}}$$
where we have $a,b \ge 0$ and $a^2 > b$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2400336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
Show that for any natural number n between $n^2$ and$(n+1)^2$ there exist 3 distinct natural numbers a, b, c, so that $a^2+b^2$ is divisible by c Show that for any natural number n ,one can find 3 distinct natural numbers a, b, c, between $n^2$ and$(n+1)^2$, so that $a^2+b^2$ is divisible by c.
It's easy to prove that... | Surprisingly there seems to be another answer to the stricter reading of the problem.
Let $a=(n^2+n)$, $b=(n^2+n+2)$ and $c=(n^2+1)$
and as with @ARoberts Solution if $n\ge 2$ then $n^2 < c < a < b < (n+1)^2$
we have
$a^2+b^2 = (n^2+n)^2 + (n^2 + n + 2)^2 = 2(n^2+1)(n^2+2n+2)$.
So again $c | a^2 + b^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2401167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Show that if $a+b+c=0$, $2(a^4 + b^4+ c^4)$ is a perfect square
Show that for $\{a,b,c\}\subset\Bbb Z$ if $a+b+c=0$ then $2(a^4 + b^4+ c^4)$ is a perfect square.
This question is from a math olympiad contest.
I started developing the expression $(a^2+b^2+c^2)^2=a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2$ but was not able ... | Since $$2(a^2b^2+a^2c^2+b^2c^2)-a^4-b^4-c^4=(a+b+c)(a+b-c)(a+c-b)(b+c-a)=0,$$
we obtain
$$2(a^4+b^4+c^4)=a^4+b^4+c^4+2(a^2b^2+a^2c^2+b^2c^2)=(a^2+b^2+c^2)^2.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2401281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 5
} |
find a and b such that the limit will exist and find the limit I am given this question $$\lim_{x\rightarrow 0} \frac{e^{\sqrt{1+x^2}}-a-bx^2}{x^4}$$
And asked to find the $a$ and $b$ such that the limit exists and also to compute the limit
My solution:
Let $t$=$\sqrt{1+x}$.
Then the Maclaurin polynomial is : $$\sqrt{... | Ignoring the squares, we have
$$\lim_{x\to0}\frac{e^{\sqrt{1+x}}-a-bx}{x^2}.$$
By adjusting $a$ and $b$, we can certainly get rid of the linear terms and the limit will exist.
So let's compute the Taylor development of $f(x):=e^{\sqrt{1+x}}$.
$$f(0)=e=a,$$
$$f'(x)=\frac{e^{\sqrt{1+x}}}{2\sqrt{1+x}},f'(0)=\frac e2=b,$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2402304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Wrong result for a $1^\infty$ limit I wanna know the $$\lim_{x\rightarrow\infty}(\sqrt{x^2+2x+4}-x)^x .$$
I applied $1^\infty$ algorithm and I have in last sentence $$\lim_{x\rightarrow\infty} \left(\frac{2x+4-\sqrt{x^2+2x+4}}{\sqrt{x^2+2x+4}+x}\right)^x.$$ After that I used $x=e^{\ln x}$ and got this: $$\lim_{x\righta... | Sometimes WA gives absolutely wrong results, but in our case he gave a right answer.
Since $f(x)=e^x$ is a continuous function, we obtain:
$$\lim_{x\rightarrow+\infty}(\sqrt{x^2+2x+4}-x)^x =\lim_{x\rightarrow+\infty}\left(\frac{2x+4}{\sqrt{x^2+2x+4}+x}\right)^x =$$
$$=\lim_{x\rightarrow+\infty}\left(1+\frac{2x+4}{\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2403985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solve $\lfloor \frac{2x-1}{3} \rfloor + \lfloor \frac{4x+1}{6} \rfloor=5x-4$ I came across this problem and it doesn't seem so tricky, but I didn't do it.
Solve the equation
$$\lfloor {\frac{2x-1}{3}}\rfloor + \lfloor {\frac{4x+1}{6}}\rfloor=5x-4.$$
My thoughts so far:
Trying to use the inequality $k-1 < \lfloor k\... | Let $\lfloor{\frac{2x-1}{3}}\rfloor + \lfloor{\frac{4x+1}{6}}\rfloor=n$, where $n\in\mathbb Z$.
Hence, $5x-4=n$, which gives $x=\frac{n+4}{5}$.
Thus,
$$\lfloor{\frac{2n+13}{15}}\rfloor + \lfloor{\frac{4x+21}{30}}\rfloor=n,\tag{1}$$
which gives
$$n\leq\frac{2n+13}{15}+ \frac{4x+21}{30}<n+2,\tag{2}$$ which gives
$0\leq n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2405495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Solve $\frac{1}{x}<\frac{x}{2}<\frac{2}{x}$. First I consider the case $\frac{1}{x}<\frac{x}{2}$, manipulating to $\frac{(x-\sqrt{2})(x+\sqrt{2})}{2x}<0\Leftrightarrow x < -\sqrt{2}.$ Did this by subtracting $\frac{x}{2}$ from both sides and writing everything with a common denominator.
Second case to consider is $\fra... | There are some good answers up already but let me address specifically what's wrong with your attempted solution.
1) $1/x < x/2$ is not equivalent to $\frac{(x-\sqrt{2})(x+\sqrt{2})}{2x} < 0$. The inequality is going the wrong way. Subtracting $x/2$ on the left side gives
$$\frac{1}{x} - \frac{x}{2} = \frac{2 - x^2}{2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2407630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
If $a$ is a root of $x^2+ x + 1$, simplify $1 + a + a^2 +\dots+ a^{2017}.$
If $a$ is a root of $x^2 + x + 1$ simplify $$1 + a + a^2 + a^3 + \cdots + a^{2017}.$$
my solution initially starts with the idea that $1 + a + a^{2} = 0$ since $a$ is one of the root then using the idea i grouped $$1 + a + a^2 + a^3 + \cdots +... | You are so close to figuring this out on your own!
Here's a "meta"-question and some ideas for you.
Meta-question: What do think the purpose of this exercise is? Other than busy work, of course. What do you think will be learned?
ideas: 1) If $a$ is a root to $1 + x + x^2$ then as $1+x+x^2$ is a quadratic there are... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2408233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Conversion to base $9$ Where am I going wrong in converting $397$ into a number with base $9$?
$397$ with base $10$ $$= 3 \cdot 10^2 + 9 \cdot 10^1 + 7 \cdot 10^0$$ To convert it into a number with base $9$
$$3 \cdot 9^2 + 9 \cdot 9^1 + 7 \cdot 9^0 = 243 + 81 + 7 = 331$$
But answer is $481$?????
| $481_{[9]} = 4\,\cdot\,81 +8\,\cdot9+1\,\cdot\,1 = 324 + 72 + 1 = 397_{[10]}$
The conversion you presented is incorrect, you converted $397_{[9]}$ into $331_{[10]}$, which is not what you want.
To convert it properly:
$397/9 = 44$ (remainder = $1$)
$44/9 = 4$ (remainder = $8$)
$4/9 = 0$ (remainder = $4$)
The remainders... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2408902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Interesting way to evaluate $ \int \cos^3 x\ dx$ I have read these days a nice way for integrating $\cos^3x$:
First differentiate:
$$f=\cos^3(x)$$$$f'=-3\cos^2(x)\sin(x)$$$$f''=6\cos(x)\sin^2(x)-3\cos^3(x)$$$$f''=6\cos(x)(1-\cos^2(x))-3f$$$$f''=6\cos(x)-6\cos^3(x)-3f$$$$f''=6\cos(x)-9f$$
Then integrate:
$$f'= 6\sin(x) ... | \begin{eqnarray*}
\int \cos^3(x) dx &=& \int (1-\sin^2(x)) \cos(x) dx = \int \cos(x) dx - \int \sin^2(x) \cos(x) dx \\
\int \cos(x) dx &=& \sin(x) + C_1
\end{eqnarray*}
Now, we need to integrate $\sin^2(x) \cos(x)$. We do this with the substitution $u = \sin(x)$.
\begin{eqnarray*}
\int \sin^2(x) \cos(x) dx &=& \int u^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2410578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 2
} |
Quadratic Diophantine Equation$(x^2+x)(y^2-1)=240$ For whole numbers $x$ and $y$, $$x,y | (x^2+x)(y^2-1)= 240$$
Find the biggest and smallest value for $x-y$.
How do you proceed with such a question?
Are their formulas or something for that type of equation?
I'd appreciate any help.
| Notice that $y^2-1$ is adivisor of $240$;
also we know that $-1 \leq y^2-1$,
by checking throuh all divisors of $240$,
we get the folowing possibilities for $y^2-1$:
*
*$y^2-1=-1$ but there is no pair in this case.
*$y^2-1=0$ but there is no pair in this case.
*$y^2-1=3$ and $x^2+x=80$;
but notice that the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2411174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove that inequality $\sum _{cyc}\frac{a^2+b^2}{a+b}\le \frac{3\left(a^2+b^2+c^2\right)}{a+b+c}$
Let $a>0$,$b>0$ and $c>0$. Prove that:
$$\dfrac{a^2+b^2}{a+b}+\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}\le \dfrac{3\left(a^2+b^2+c^2\right)}{a+b+c}.$$
$$\Leftrightarrow \frac{(a^2+b^2)(a+b+c)}{a+b}+\frac{(b^2+c^2)(a+... | The following quantity is clearly positive
\begin{eqnarray*}
\sum_{perms} a^2 b (a-b)^2 \geq 0
\end{eqnarray*}
This can be rearranged to
\begin{eqnarray*}
3(a^2+b^2+c^2)(a+b)(b+c)(c+a) \geq (a+b+c) \left( \sum_{cyc} (a^2 +b^2)(c+a)(c+b) \right)
\end{eqnarray*}
Now divide this by $(a+b+c)(a+b)(b+c)(c+a)$ and we have
\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2413136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
For what value of $k$ is one root of the equation
For which value of $k$ is one root of the equation $x^2+3x-6=k(x-1)^2$ double the other?
My Attempt:
$$x^2+3x-6=k(x-1)^2$$
$$x^2+3x-6=k(x^2-2x+1)$$
$$x^2+3x-6=kx^2-2kx+k$$
$$(1-k)x^2+(3+2k)x-(6+k)=0$$
| Hint: Let $a$ and $2a$ be the two roots.
Then
$$
-\dfrac{3+2k}{1-k}=3a,
\qquad
-\dfrac{6+k}{1-k}=2a^2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2413338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Max and min of $f(x)=\sin^2{x}+\cos{x}+2$. My first try was to look for when $\cos{x}$ and $\sin{x}$ attain their min and max respectively, which is easy using the unit circle. So the minimum of $\sin{x}+\cos{x}$ has maximum at $\sqrt{2}$ and minimum at $-\sqrt{2}.$ But if the sine-term is squared, how do I find the mi... | Credit to @iamwhoiam for suggesting this solution.
There's nothing wrong with your method - you just missed a root. When you find the roots of the first derivative, either $2 \cos x - 1$ or $\sin x$ should equal $0$, so we have:
First case:
$2 \cos x - 1 = 0$, $\cos x = \frac{1}{2}$.
Since $\cos(x)$ is an even functio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Find $\lim\limits_{n \to \infty} \frac{\sqrt 1 + \sqrt 2 + \dots + \sqrt{n}}{n\sqrt{n}}$.
$$\lim_{n \to \infty} \dfrac{\sqrt 1 + \sqrt 2 + \dots + \sqrt{n}}{n\sqrt{n}}$$
$$\lim_{n \to \infty} \dfrac{\sqrt 1 + \sqrt 2 + \dots + \sqrt{n}}{n\sqrt{n}} =\lim_{n\to \infty} \dfrac1{n}\sum^{n}_{k = 1} \sqrt{\dfrac k n} $$
Wh... | picture......................................
$$ \frac{2}{3} n \sqrt n < \mbox{SUM} < \frac{2}{3} \left( \; (n+1) \sqrt {n+1} \; - \; 1 \right) \; < \; \frac{2}{3} \left( \; (n+1) ( 1 +\sqrt n) \; - \; 1 \right) = \frac{2}{3} \left( \; n \sqrt n + n + \sqrt n \right) $$
$$ \frac{2}{3} n \sqrt n < \mbox{SUM} < ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2415683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 1
} |
A.P.: Sum of numbers in a particular group Question: A series of odd positive integers are divided into the following groups $(1), (3,5), (7,9,11),\dotsm$ . Prove that the sum of numbers in the $n^{\text{th}}$ group is $n^3$.
My attempt:
It is observed that the number of terms in the $n^{\text{th}}$ group is equal to $... | There are $\color{red}n$ terms in the $n$-th group of numbers.
The "middle" term (for groups with even terms, this refers to the average of the two centre-most terms) of each group is $\color{red}{n^2}$*, i.e.
$$\overset{\overset{\;1^2}{\downarrow}}{\underbrace{(1)}_{1\text{ term}}}, \overset{\overset{\;2^2}{\downarr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2415775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
complex number relations I can structure a solution using geometry but I am struggling to find on using algebra.
In an Argand diagram, the complex numbers $z, w$ and $z+w$ are represented by the points $P, Q$ and $S$ respectively.
Show that:
$$| z + w | ≥ | z | − | w |$$
Thanks for any help
| Let $z=x+yi$ and $w=a+bi$, where $\{x,y,a,b\}\subset\mathbb R$.
Thus, we need to prove that
$$\sqrt{(x+a)^2+(y+b)^2}+\sqrt{a^2+b^2}\geq\sqrt{x^2+y^2}$$ or
$$\sqrt{(a^2+b^2)((x+a)^2+(y+b)^2)}\geq-a^2-b^2-ax-by.$$
If $-a^2-b^2-ax-by<0$ then the inequality is obviously true, but for
$-a^2-b^2-ax-by\geq0$ it's enough to pr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2416587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find $\int \frac{\sqrt{2-3x}\,dx}{x^2+1}$ Find $$\int \frac{\sqrt{2-3x}\,dx}{x^2+1}$$
I used substitution $x=\frac{2}{3} \sin^2 y$ So we get $dx=\frac{4}{3} \sin y \cos y dy$
hence
$$I= \frac{36\sqrt{2}}{3} \int \frac{\sin y \cos^2 y\, dy}{4 \sin^4 y+9}$$
Now if again i use substitution $\cos y=t$ we get
$$I=-12\sqrt{2... | hint: let $u = \sqrt{2-3x}\implies 2-3x=u^2\implies x = \dfrac{2-u^2}{3}, dx = -\dfrac{2udu}{3}$. Then do a fraction decomposition. It can be done but its not short !
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2417128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Finding all $a,b,c$ such that $(ax)^2+(b+cy)^2 \leq 1$ whenever $x^2+y^2 \leq 1$. Suppose $x,y,a,b,c \in \mathbb R$ and suppose that $$(ax)^2+(b+cy)^2 \leq 1$$ for all $x,y$ such that $x^2+y^2 \leq 1$. What are the allowed values of $a,b,c$?
Plugging in $x=1$, $y=0$ we get $a^2+b^2 \leq 1$. Similarly for $x=0,y=1$ we ... | I played around for a bit
and came to a point
where I didn't want
to go further.
I'll show what I did
in the hope that
someone else
might find this useful.
Assume that
$x$ and $y$
are on the boundary,
so
$x^2+y^2 = 1
$.
This becomes
$\begin{array}\\
1
&\ge a^2x^2+b^2+2bcy+c^2y^2\\
&=a^2(1-y^2)+b^2+2bcy+c^2y^2\\
&=a^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2418043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find maximum of $f(x)=3\cos{2x}+4\sin{2x}.$ Find maximum of $f(x)=3\cos{2x}+4\sin{2x}.$ The shortest way to do this is to only consider $2x\in[0,2\pi)$, set $t=2x$ and note that
*
*Max$(\cos{t})=1$ for $t=0.$
*Max$(\sin{t})=1$ for $t=\frac{\pi}{2}.$
Max of these two functions added is when $t$ equals the angle ex... | With $\tan t=\dfrac43$ then $\cos t=\dfrac{1}{\sqrt{1+(\dfrac43)^2}}=\dfrac35$ so
\begin{align}
3\cos{2x}+4\sin{2x}
&=3(\cos{2x}+\tan t\sin{2x})\\
&=\dfrac{3}{\cos t}(\cos t\cos{2x}+\sin t\sin{2x})\\
&=\dfrac{3}{\cos t}\cos(t-2x)\\
&=5\cos(t-2x)\\
&\leqslant5
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2420674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Finding $\int\frac{x}{\sqrt{3-2x-x^2}} dx$. I was looking for the integral of
$$\int\frac{x}{\sqrt{3-2x-x^2}} dx$$
My work:
$$\int \frac{x}{\sqrt{3-2x-x^2}} dx = \int \frac{x}{\sqrt{(3)+(-2x-x^2)}} dx $$
$$ = \int \frac{x}{\sqrt{(3)-(2x+x^2)}} dx $$
$$ = \int \frac{x}{\sqrt{(3)-(1+2x+x^2) +1}} dx $$
$$\int \frac{x}{... | Integration strategy:
By adding/subtracting a constant to the numerator, you can let the derivative of the polynomial under the radical appear and you can integrate the ratio.
Then remains the inverse of the square root of this quadratic polynomial. By completing the square, you see that by translating by $1$ ($\to4-u^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Prove that for any integer $n, n^2+4$ is not divisible by $7$. The question tells you to use the Division Theorem, here is my attempt:
Every integer can be expressed in the form $7q+r$ where $r$ is one of $0,1,2,3,4,5$ or $6$ and $q$ is an integer $\geq0$.
$n=7q+r$
$n^2=(7q+r)^2=49q^2+14rq+r^2$
$n^2=7(7q^2+2rq)+r^2$
$n... | Your proof is correct (fleablood's comment). Let me rephrase a bit:
$A(n):= n^2 +4$; $B(r,q) = 7 q^2 + 2rq;$
$A(n) = 7B(r,q) + (r^2 +4)$.
Fairly simple to show that:
$A(n)$ is divisible by $7 \iff $
$(r^2 +4)$ is divisible $ n.$
By inspection
$(r^2 +4) , r = 0,1,2,3,4,5,6,$
is not divisible by $7$.
$\Rightarrow$: $A(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 1
} |
How to split $\frac{x^3}{(x^2+5)^2}$ into partial fractions?
Split $\frac{x^3}{(x^2+5)^2}$ into partial fractions.
I have tried letting $$\frac{x^3}{(x^2+5)^2}=\frac{ax+b}{x^2+5}+\frac{cx^3+dx^2+ex+f}{(x^2+5)^2},$$ but this yields a system of only four simultaneous equations.
I have also tried looking at $$\frac{x^3... | If we format it into the partial fractions 'template':
$$\frac{x^3}{(x^2+5)^2}=\frac{Ax+B}{(x^2+5)}+\frac{Cx+D}{(x^2+5)^2}$$
Multiplying both sides by the denominator,
$$\frac{x^3\left(x^2+5\right)^2}{\left(x^2+5\right)^2}=\frac{\left(B+Ax\right)\left(x^2+5\right)^2}{x^2+5}+\frac{\left(D+Cx\right)\left(x^2+5\right)^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to do this Orthogonality Integral? I am solving a question, where
$$ p_1(x) = x , Q_0(x) = \frac {\ln \frac {1+x} {1-x}}{2}$$
are solutions of legedre's differential equation corresponding to different eigen values.
I have to evaluate their orthogonality integral
$$ \int \limits_{-1}^{1} x \frac {\ln \frac {1+x} {1... | Ignoring limits,
$$
\int x\ln\left(\frac{1+x}{1-x}\right) dx \\
= \frac{1}{2}\int \frac{d}{dx}(x^2-1)\ln\left(\frac{1+x}{1-x}\right)dx \\
= \frac{1}{2}(x^2-1)\ln\left(\frac{1+x}{1-x}\right)-\frac{1}{2}\int(x^2-1)\frac{d}{dx}\ln\left(\frac{1+x}{1-x}\right)dx \\
= \frac{1}{2}(x^2-1)\ln\left(\frac{1+x}{1-x}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $\lim_{x\to 1} \frac{x+2}{x^2+1}=\frac{3}{2}$ The title is quite clear. I am required to prove $\lim_{x\to 1} \frac{x+2}{x^2+1}=\frac{3}{2}$ using the epsilon-delta definition of the limit.
Given any $\varepsilon \gt 0$, there exists a $\delta =$
Such that $0 \lt \lvert x-1 \rvert \lt \delta \Rightarrow \lvert\f... | Your fraction, in absolute value, equals
$$\left|\frac{(x-1)(3x+1)}{2(x^2+1)} \right|<\left|\frac{(x-1)(3x+1)}{x^2} \right|.$$
Now, if $|x-1|<\delta$, then $|x|<\delta+1$. So if $\delta\leq1$, then $|x|<2$ and $|x^2|<4$. So if we choose $\delta=\min\{1,\frac{7\epsilon}{4}\}$, then the above inequality is less than $\ep... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2424030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Prove the sequence $\frac{n^2}{n^2-n-5}$ converges I know the basic procedure for proving convergence of sequences, but I'm having difficulty reducing this sequence. I believe the sequence converges to 1, so I've set up the following two approaches:
1:
$\left|\frac{n^2}{n^2 - n - 5} - 1\right| = \left|\frac{n^2}{n^2 - ... | Your first approach should ring a bell since the nominator is first degree while the denominator is second degree.
You can do the formals by applying an estimate for the nominator and denominator. For $n>5$ we have that $0 < n+5 < 2n$ and $n^2 - n - 5 > n^2 - n^2/5 - n^2/5 = 3n^2/5$. This means that for $n>5$ we have
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2425132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Is it possible to have a 3rd number system based on division by zero? Is it possible in mathematics to use a third number line based on division by zero; in addition to the real and imaginary number lines?
This is because some solutions blow up when there is a division by zero. Would it be possible to solve them with t... | One way to approach this problem could be to imagine zero as a very small number.
So to start with, let us define,
$$small \to 0$$
$$\therefore$$
instead of $$ (2 \cdot 0) \cdot p = 1 $$
as above @Jim H. Use:
$$2 \cdot small \cdot \frac{1}{small} = 2$$
As
$$(2 \cdot small) \cdot \frac{1}{small} \neq small \cdot \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2425242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
If $a,b$ are positive integers and $x^2+y^2\leq 1$ then find the maximum of $ax+by$ without differentiation. If $x^2+y^2\leq 1$ then maximum of $ax+by$
Here what I have done so far.
Let $ax+by=k$ . Thus $by=k-ax$.
So we can have that $$b^2x^2+(k-ax)^2 \leq b^2$$
$$b^2x^2+k^2-2akx +a^2x^2-b^2\leq 0 $$
By re-writing as a... | We can solve this by geometry,
$$x^2+y^2\le 1$$ is area bounded by circle with centre at origin and radius 1
&
$$ax+by=c$$
Is a line
We want to find max value of c,
Note: a and b are positive so slope is negative.
We want to find (x,y) such that it lies in circle and on line but also that line gives us maximum ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2426188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Eavaluating the roots of quadratic equation If $b>a , c>0$
Determine the intervals that the roots of the equation
$(x-a)(x-b) -c =0$ belong to
My work is to get the values of the roots in terms of a , b and c using the general form but i couldn't determine those intervals
| The given equation is $x^2-(a+b)x+ab-c=0$. But $$\Delta=(a+b)^2-4(ab-c)=(a-b)^2+4c>0.$$ So the equation has two distinct real roots. The two roots are $\frac{(a+b)+\sqrt{(a-b)^2+4c}}{2}$ and $\frac{(a+b)-\sqrt{(a-b)^2+4c}}{2}$. Hence the roots are in the interval $[\frac{(a+b)-\sqrt{(a-b)^2+4c}}{2},\frac{(a+b)+\sqrt{(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2428040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integration changing limits - does the question have an error? I am asking about changing the limits of integration.
I have the following integral to evaluate -
$$\int_2^{3}\frac{1}{(x^2-1)^{\frac{3}{2}}}dx$$ using the substitution $x = sec \theta$.
The problem states
Use the substitution to change the limits into t... | For simplicity, write $a=\arccos\frac{1}{2}$ and $b=\arccos\frac{1}{3}$. One can take care of them at the end.
It is true that $a=\pi/3$, but it's definitely wrong that $b=\pi/3$. Actually, $b\approx1.230959$, but you won't need that.
The integral becomes
$$
\int_a^b-\frac{1}{(\sec^2\theta-1)^{3/2}}\sec^2\theta\sin\the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Probability, without replacement An urn contains 11 balls, of which 4 are white. Three players $A$, $B$, and $C$, successively draw from the urn. $A$ first, then $B$, then $C$, then $A$ again, and so on. The winner is the first one to draw a white ball. Find the probability of winning for each player, assuming the ball... | Assume $4$ white and $7$ red balls in the box. Then:
$$P(A)=P(W_A)+P(R_AR_BR_CW_A)+P(R_AR_BR_CR_AR_BR_CW_A)=\frac{4}{11}+\frac{7\cdot 6\cdot 5\cdot 4}{11\cdot 10\cdot 9\cdot 8}+\frac{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 4}{11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5}\approx 0.4818.$$
Similarly for $P(B), P(C)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove $ \forall n \ge 4$, $n^{3} + n < 3^{n}$
Prove that $\forall n \ge 4$, $n^{3} + n < 3^{n}$
My attempt: Base case is trivial.
Suppose $ \ n \ge 4$, $n^{3} + n < 3^{n}$. Then,
$$ (n+1)^{3} + (n+1) = n^3 + n + 3n^2 + 3n + 1 +1 < 3^{n} + 3n^2 + 3n + 2 \\< 3\cdot 3^{n} + 3n^2 + 3n +2.$$
Not sure how to get rid of $3... | By the binomial theorem, $3^n = (1+2)^n \ge \binom{n}{1}+ \binom{n}{4}2^4$ for $n \ge 4$.
So, it's enough to prove that $\binom{n}{1}+ \binom{n}{4}2^4 > n^3+n$ (*).
This simplifies to $2 n^3 - 15 n^2 + 22 n - 12 > 0$.
The largest critical point of $2 x^3 - 15 x^2 + 22 x - 12$ is at $x \approx 4.1$. Also (*) is true for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
$1/\cos x$ integration Question: The aim of this question is to integrate $\frac{1}{\cos x}$
a. Write an integral as $\int \frac{\cos x}{\cos^2(x)}dx$ [DONE]
b. Use $u= \sin x$ [DONE]
c. Use partial fraction to integrate [DONE]
d. Multiply a fraction top and bottom by $1+\sin x$ to simplify it.
My answer to part c is $... | From your answer in part (c)
$ A =\frac{1}{2} \ln(1+\sin x ) - \frac{1}{2}\ln (1-\sin x) $
$ = \ln (\sqrt{1+\sin x}) - \ln (\sqrt{1-\sin x}) $
$ = \ln\left(\frac{\sqrt{1+\sin x }}{\sqrt{1-\sin x}}\right) $
$= \ln\left(\sqrt{\frac{1+\sin x }{1-\sin x} \cdot \frac{1+\sin x}{1+\sin x}}\right) $
$= \ln\left(\frac{1+\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Cantor's "original" proof of $\mathbb{R}$ uncountable Found this problem in a set of notes I'm self studying from. I believe there might be an error.
"Consider $\{x_n\}_{n∈N}$ a sequence in $[0, 1]$. Construct new sequences $a_n$ and $b_n$ as follows: $a_0$ and $b_0$ are the first two elements in the sequence ${x_n}$ ... | Nope, it is correct as stated. Your intersection $\bigcap_{n \geq 3} \left( \frac{1}{2} - \frac{1}{n}, \frac{1}{2} + \frac{1}{n} \right)$ is actually not empty, since it contains $\frac{1}{2}$. Indeed, for any $n\geq 3$, $\frac{1}{2}-\frac{1}{n}<\frac{1}{2}<\frac{1}{2}+\frac{1}{n}$.
(That said, the proof would work j... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2436779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that for $n \ge 2$ the follow inequality holds $\frac {4^n}{n+1} \lt \frac{(2n)!}{(n!)^2}$. As already shown above I need to prove that $$\frac {4^n}{n+1} \lt \frac{(2n)!}{(n!)^2}\qquad \forall n \ge 2.$$
What I've come up with is the following:
$\underline{n=2:}\qquad$ $$\frac {16}{3} \lt 6$$
$\underline{n=k:}\... | As your first steps of mathematical induction are right, I'll just perform a proof of inductive thesis for $n=k+1$.
First notice, that for $k>1$
$$\frac{4(k+1)}{(k+2)} < \frac{(2k+2)(2k+1)}{(k+1)^2}$$
proof:
$$(k-1)k(k+1)^2(k+2)>0\\
\frac{k(k-1)}{(k+1)^2(k+2)}>0\\
\frac{4k^2-4k}{(k+1)^2(k+2)}>0\\
\frac{-4k^2+4k}{(k+1)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2437448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find all the points that lie $ 3$ units from each of the points $ (2,0,0), (0,2,0), \text{ and }(0,0,2)$
I calculated the result $$ \{(x,y,z) \in\mathbb{R^3}: x=y=z\}.$$
I'm wondering whether I did this problem correctly and if I did how to draw the set of solutions.
I used the euclidean distance formula with the squ... | To avoid forgetting solutions it is best to keep the whole system together.
$\begin{cases}
(x-2)^2+y^2+z^2=9\\
x^2+(y-2)^2+z^2=9\\
x^2+y^2+(z-2)^2=9\\
\end{cases}\iff\begin{cases}
(-4x+4)+x^2+y^2+z^2=9\\
(-4y+4)+x^2+y^2+z^2=9\\
(-4z+4)+x^2+y^2+z^2=9\\
\end{cases}$
Now replacing row $(2)$ by $(2)-(1)$ you get $4x=4y$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2439824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove the limit $\lim_{x\to 1+}\frac{1}{\sqrt{x}}=1$, using epsilon-delta definition. $$\lim_{x\to 1+}\frac{1}{\sqrt{x}}=1$$
The proof that I have:
Let $\varepsilon > 0$, we must show that
$$\exists \delta >0: 0<x-1<\delta \Rightarrow \left | \frac{1}{\sqrt{x}}-1\right|<\epsilon$$
So usually, when doing these $\varep... | Coming from the right, you just have to show that $1-\frac{1}{\sqrt x} \longrightarrow$ as $x \longrightarrow 1$. To show the limit, you could define $y=x-1$.
Then $$1-\frac{1}{\sqrt x} = \frac{\sqrt{1+y} - 1}{\sqrt{1+y}}$$.
Note that when $y < \delta$
$$\frac{\sqrt{1+\delta} - 1}{\sqrt{1+\delta}} \leq \sqrt{1+\delt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2441520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Probability of a target being hit I don't have the answer of the following question so I wanted to cross check my solution.
$A$ can hit a target 3 times in 5 shots, $B$ 2 times in 5 shots and $C$ 3 times in 4 shots. Find the probability of the target being hit at all when all of them try.
my method
P(target being hit... | There is also a way we can use we take all the probable situations and subtract the ones that we don't succeed hitting;
For $A$ it is $\dfrac{2}{5}$
For $B$ it is $\dfrac{3}{5}$
And for $C$ it is $\dfrac{1}{4}$
$$1-\frac{2}{5}\frac{3}{5}\frac{1}{4}=\frac{47}{50}$$
I accidentally wrote this answer without seeing the fi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2442777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$\frac{a^3}{b^2} + \frac{b^3}{c^2} + \frac{c^3}{a^2} \geq 3 \, \frac{a^2 + b^2 + c^2}{a + b + c}$ Proposition
For any positive numbers $a$, $b$, and $c$,
\begin{equation*}
\frac{a^3}{b^2} + \frac{b^3}{c^2} + \frac{c^3}{a^2}
\geq 3 \, \frac{a^2 + b^2 + c^2}{a + b + c} .
\end{equation*}
I am requesting an elementary, alg... | Actually this is not a symmetric inequality, since the LHS is cyclic but not symmetric.
Anyway, here's the proof I've come up with. From Cauchy-Schwarz on $\displaystyle \left(\frac{a^2}{b\sqrt{a}}, \: \frac{b^2}{c\sqrt{b}}, \: \frac{c^2}{a\sqrt{c}}\right)$ and $\displaystyle \left(b\sqrt{a}, \: c\sqrt{b}, \: a\sqrt{c}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2446312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Find all $4 \times 4$ real matrices such that $A^3=I$ Find all $4 \times 4$ real matrices such that $A^3=I$.
The minimal polynomial must divide $x^3-1$. Since the matrix is real, the minimal polynomial must be either $x-1$ or $x^3-1$ (i.e., if it contains one of the complex roots of unity, it must contain the conjugat... | Since $p(A) = 0$ where $p(x) = x^3 - 1 = (x-1) (x^2 + x + 1)$, and $x-1$ and $x^2 + x + 1$ are irreducible over $\mathbb{R}[x]$, then $\mathbb{R}^4$ can be decomposed as a direct sum of cyclic subspaces for $A$ corresponding to the polynomials $x-1$ and $x^2 + x + 1$. Now, we know that a cyclic subspace with annihilat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2446574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{ab}{a+b} + \frac{cd}{c+d} \leq \frac{(a+c)(b+d)}{a+b+c+d}$ $$\frac{ab}{a+b} + \frac{cd}{c+d} \leq \frac{(a+c)(b+d)}{a+b+c+d}$$
I tried applying a.m. g.m inequality to l.h.s and tried to find upper bound for l.h.s and lower bound for r.h.s but i am not getting answer .
| I suppose that $a,b,c,d$ are $>0$. An idea is to put
$$F(x)=\frac{(x+c)(b+d)}{x+b+c+d}-\frac{xb}{x+b}-\frac{cd}{c+d}$$
and to compute the derivative:
$$F^{\prime}(x)=\frac{(b+d)^2}{(x+b+c+d)^2}-\frac{b^2}{(x+b)^2}$$
This show that the minimum of $F$ on $]0,+\infty[$ is obtained for $x=bc/d$, and to finish you have to c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2448601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
$a^2b^2 + b^2c^2 + c^2d^2 + d^2a^2$ can be expressed as the sum of two squares in at least two different ways The question: Every expression of the form $a^2b^2 + b^2c^2 + c^2d^2 + d^2a^2$ can be expressed as the sum of two squares in at least two different ways. Find any one of the three possible ordered pairs of
posi... | $$\text{First way} $$
$$a^2b^2 + b^2c^2 + c^2d^2 + d^2a^2=\color{red}{a^2b^2 + c^2d^2 + b^2c^2+ d^2a^2+2(ab)(cd) - 2(bc)(da)}=\color{blue}{(ab+cd)^2+(bc-da)^2} $$
$$\text{Second way} $$
$$a^2b^2 + b^2c^2 + c^2d^2 + d^2a^2= \color{red}{a^2b^2+c^2d^2 -2(ab)(cd)+b^2c^2+ d^2a^2+2(bc)(da)}=\color{blue}{(ab-cd)^2+(bc+da)^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2449930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do you solve quadratic inequalities/functions?
Find the least value of $n$ such that $(n-2)x^2+8x+n+4 > 0$, where $x \in \mathbb R$ and $n \in \mathbb N$.
What I did -
As $x$ is real, $D>0$
$64-4(n+4)(n-2)>0$
$(n-4)(n-6)<0$
$n \epsilon (-6,4)$
But as $n \epsilon N$
The least value is 3
According to the book's... | the case $n=0$ gives $$x^2-4x-2<0$$ which is not fulfilled for all real $x$
$$n=1$$ gives $$x^2-8x-5<0$$
$$n=2$$ gives $$8x+6>0$$
for $$n>2$$ we get
$$x^2+2\left(\frac{4}{n-2}\right)x+\left(\frac{4}{n-2}\right)^2+\frac{n+4}{n-2}-\left(\frac{4}{n-2}\right)^2>0$$ and we get
$$\left(x+\frac{4}{n-2}\right)^2+\frac{n^2+2n-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2451981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $\sin x - \cos x = \frac{1}{2}$ then determine: $\sin^4 x + \cos^4 x$ If $$\sin x - \cos x = \frac{1}{2}$$ then determine: $$\sin^4 x + \cos^4 x$$
I tried making it $(\sin^2 x)^2+(\cos^2 x)^2$ but then I get nothing that can help. What is the trick to this?
| You have
$$
\frac{1}{4}=(\sin x-\cos x)^2=\sin^2x-2\sin x\cos x+\cos^2x
$$
so
$$
\sin x\cos x=\frac{1}{2}\left(1-\frac{1}{4}\right)=\frac{3}{8}
$$
Then
$$
\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-\dotsb
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2452142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\int \frac{dx}{x\sqrt{4x^2+1}}$ I was evaluating $\int \frac{dx}{x\sqrt{4x^2+1}}$ using Table of Integrals.
My work
I was evaluating $\int \frac{dx}{x\sqrt{4x^2+1}}$ using Table of Integrals. I found in the Table of Integrals an integral that is akin to
$\int \frac{dx}{x\sqrt{4x^2+1}}$. The integral I was ta... | After "Now getting the integral", when you use the formula, you introduce a $2$ that should not be there. You would get
$$
-\ln\left(\frac{1+\sqrt{4x^2+1}}{x}\right)
=\ln\left(\frac{x}{1+\sqrt{4x^2+1}}\right).
$$
You also need to review the rules for logarithms. It is not true that $\ln 1/x=1/\ln x$ (you used it twice,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2453001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Solving the equation $c=\dfrac{x^2+y^2-1}{x^2+(y+1)^2}$ Solving the equation $c=\dfrac{x^2+y^2-1}{x^2+(y+1)^2}$
$$c=\dfrac{x^2+y^2-1}{x^2+(y+1)^2}$$
$$c{x^2+c(y+1)^2}={x^2+y^2-1}$$
$$c{x^2+cy^2+2cy+c}={x^2+y^2-1}\text{ [expanded]}$$
$$1+c=x^2-cx^2+y^2-cy^2-2cy\text{ [moved to other side]}$$
$$1+c=(1-c)x^2+\color{red}{(... | You do not need all those calculations!
Starting from $c=\dfrac{x^2+y^2-1}{x^2+(y+1)^2}$
you get
$\left(x^2+y^2-1\right)-c \left(x^2+(y+1)^2\right)=0$
expand
$-c x^2-c y^2-2 c y-c+x^2+y^2-1=0$
collect
$(1-c) x^2+(1-c) y^2-2 c y-c-1=0$
and divide each term by $c-1$ provided that $c\ne 1$
$x^2+y^2-\dfrac{2 c }{1-c}\,y-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2453143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
The intersection of $5$ planes If the intersection of planes $a^{k}x+2^{k}y+3^{k}z+d^{k}=0(0\leq k\leq 4)$ is a line, then $a=?$ and $d=?$.
| The matrix of coefficients is
$$A=\left(
\begin{array}{cccc}
1 & 1 & 1 & 1 \\
a & 2 & 3 & d \\
a^2 & 4 & 9 & d^2 \\
a^3 & 8 & 27 & d^3 \\
a^4 & 16 & 81 & d^4 \\
\end{array}
\right)$$
The system has One and only one solution, that is one point, if $\text{rank }A=4$
We know that the planes intersect along a line, a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2454173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find numbers $a\leq b$ that maximize the value of the integral $\int_a^ba-x-x^2dx$ I'm trying to find numbers $a\leq b$ that maximize the value of the integral $\int_a^ba-x-x^2dx$. I tried computing the integral and i got the function in two variables
$$f(a,b)=\int_a^ba-x-x^2dx=ab-\frac{b^2}2-\frac{b^3}3-\frac{a^2}2+\f... | We have that $$ab - \frac{b^2}{2} - \frac{b^3}{3} - \frac{a^2}{2} + \frac{a^3}{3} \leq 0 \iff ab + \frac{a^3}{3} \leq \frac{a^2}{2} + \frac{b^2}{2} + \frac{b^3}{3} \ \ \ \ \ \ \ (I)$$
To prove this inequality holds, note that
$$0 \leq \left( \frac{a}{\sqrt{2}} - \frac{b}{\sqrt{2}} \right)^2 = \frac{a^2}{2} - ab + \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2458197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding minimal polynomial of $x=a+b\sqrt[3]{2}+c\sqrt[3]{4}$ I want to find the minimal polynomial of $x=a+b\sqrt[3]{2}+c\sqrt[3]{4}$
For simple case like $x=a+b\sqrt[3]{2}$, I can find
\begin{align}
(x-a)^3 = 2b^3 \qquad \Rightarrow \qquad (x-a)^3-2b^3 =0
\end{align}
I tried to do similar tings such as
\begin{alig... | An alternative approach.
Unless $b=c=0$, $\alpha=a+b\sqrt[3]2+c\sqrt[3]4$ has degree $3$ and so
a cubic minimum polynomial. Let's assume this. Observe that
$$\pmatrix{a&b&c\\2c&a&b\\2b&2c&a}\pmatrix{1\\\sqrt[3]2\\\sqrt[3]4}=\alpha\pmatrix{1\\\sqrt[3]2\\\sqrt[3]4}.$$
Therefore $\alpha$ is an eigenvalue of
$\pmatrix{a&b&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2461150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Derivation of fourth-order accurate formula for the second derivative I am trying to derive / prove the fourth order accurate formula for the second derivative:
$f''(x) = \frac{-f(x + 2h) + 16f(x + h) - 30f(x) + 16f(x - h) - f(x -2h)}{12h^2}$.
I know that in order to do this I need to take some linear combination for t... | $$
f(x+h) =
f(x) + h f'(x) + \frac{h^2}{2} f''(x)
+ \frac{h^3}{6} f'''(x)
+ O(h^4)
$$
$$
f(x-h) =
f(x) - h f'(x) + \frac{h^2}{2} f''(x)
- \frac{h^3}{6} f'''(x)
+ O(h^4)
$$
$$
f(x+2h) =
f(x) + 2h f'(x) + 2 h^2 f''(x)
+ \frac{4 h^3}{3} f'''(x) + O(h^4)
$$
$$
f(x-2h) =
f(x) - 2h f'(x) + 2 h^2 f''(x)
- \frac{4 h^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2461506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\int_{0}^{\infty} \frac{\ln x}{x^2+6x+10}dx$ Evaluate $$\int_{0}^{\infty} \frac{\ln x}{x^2+6x+10}dx$$
The given answer is $0.370429$. Is there any method to solve this? Thanks in advance.
| Note
$$ \int_0^\infty\frac{x^a}{x+1}dx=-\frac{\pi}{\sin(a\pi)}$$
and hence
$$ \int_0^\infty\frac{x^a}{x+b}dx=-\frac{b^a\pi}{\sin(a\pi)}. $$
So
\begin{eqnarray}
&&\int_{0}^{\infty} \frac{\ln x}{x^2+6x+10}dx\\
&=&\lim_{a\to0}\frac{d}{da}\int_{0}^{\infty} \frac{x^a}{x^2+6x+10}dx\\
&=&\lim_{a\to0}\frac{d}{da}\int_{0}^{\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2463122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
How to prove that $21\mid a^2+b^2\Rightarrow 441 \mid a^2 + b^2$ How to prove that $441 \mid a^2 + b^2$ if it is known that $21 \mid a^2 + b^2$.
I've tried to present $441$ as $21 \cdot 21$, but it is not sufficient.
|
Lemma. If a prime $p \equiv 3\pmod{4}$ and $p$ divides $a^2+b^2$ then $p$ divides $a,b$.
Proof. Assume the contrary that $p \nmid a,p \nmid b$.
By Fermat's little theorem, we have $a^{p-1} \equiv 1 \pmod{p}, b^{p-1} \equiv 1 \pmod{p}$ so $a^{p-1}+b^{p-1} \equiv 2 \pmod{p}$.
On the other hand, note that $x+y$ divides ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2467327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find the $n^{th}$ partial sum of a telescoping series $$
a_k = \frac{6^k}{(3^{k+1}+2^{k+1})(3^k + 2^k)}
$$
Been stuck on this for a while, I started by looking at values of $a_k$ for varying values of $k$ and looking for a pattern, but haven't come to anything useful yet. The next step, I think, is to try and find a pa... | This is just a supplement on how to obtain hint given by Qing Zhang.
Consider the following
$$\frac{6^k}{(3^{k+1}+2^{k+1})(3^k+2^k)} = \frac{A \cdot 3^k + B \cdot 2^k}{3^{k+1} +2^{k+1}} + \frac{C\cdot 3^k + D\cdot 2^k}{3^k+2^k}$$
where $A,B,C,D$ are constants to be found.
By making the same denominator on right side... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2469969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solving for $k$: $\sqrt{k-\sqrt{k+x}}-x = 0$
$$\sqrt{k-\sqrt{k+x}}-x = 0$$
Solve for $k$ in terms of $x$
I got all the way to
$$x^{4}-2kx^{2}-x+k^{2}-x^{2}$$
but could not factor afterwards. My teacher mentioned that there was grouping involved
Thanks Guys!
Edit 1 : The exact problem was solve for $x$ given that $$\s... | As discovered $\sqrt{k - \sqrt{k+x}} = x$ leads to the equation $k^2 - (2 \, x^2 + 1) \, k + (x^4 - x) = 0$. Now,
\begin{align}
k &= \frac{1}{2} \, \left[(2 \, x^2 + 1) \pm \sqrt{ (2 \, x^2 + 1)^2 - 4 \, (x^4 - x) } \right] \\
&= \frac{1}{2} \, [(2 \, x^2 + 1) \pm (2 x + 1)] \\
&= \begin{cases}{ x^2 + x + 1 = \frac{1 -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2470062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
If $z = \tan(x/2)$, what is $\sin(x)$ and $\cos(x)$? While reading mathematical gazette, I noticed an interesting "theorem". If $z = \tan(x/2)$, then $\sin(x) = \frac{2z}{1+z^2}$ and $\cos(x) = \frac{1-z^2}{1+z^2}$.
How can I derive these so I don't have to remember them?
| \begin{align}
z & = \tan \frac x 2 & & \frac z 1 = \tan = \frac{\text{opposite}}{\text{adjacent}} \\[10pt]
x & = 2\arctan z \\[10pt]
\sin x & = \sin(2\arctan z) \\
& = \sin(2\theta) = 2\sin\theta\cos\theta \\[10pt]
& = 2\sin(\arctan z)\cos(\arctan z) \\[10pt]
& = 2 \frac{\text{opposite}}{\text{hypotenuse}} \cdot \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
In how many ways can $m$ employees be assigned to $n$ projects if every project is assigned to at least one employee? In how many ways can we assign $m$ employees to $n$ projects so that every employee is assigned to exactly one project and every project is assigned to at least one employee? Also, $m>n$.
My answer:
Let... | For $m=5$, $n=4$ . . .
Exactly one project gets two people, the others get one.
*
Choose the project with $2$ people: $\binom{4}{1}=4$ choices.
Choose the $2$ people for that project: $\binom{5}{2}=10$ choices.
Assign the remaining $3$ people to the remaining $3$ projects: $3!=6$ choices.
Hence the number of vali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2477051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove $\lim_{x\rightarrow 0}\frac{\sqrt{9-x}-3}{x}=-\frac{1}{6}$ with epsilon-delta I am looking to show that $$f:(0,1)\rightarrow \mathbb{R} \lim_{x\rightarrow 0}\frac{\sqrt{9-x}-3}{x}=-\frac{1}{6}$$ Normally for these types of problems I have been doing epsilon-delta proofs but I cannot figure out how to define my $\... | With
$$\frac{\sqrt{9-x}-3}{x}=\frac{-1}{\sqrt{9-x}+3}$$
then
\begin{align}
\left|\frac{\sqrt{9-x}-3}{x}+\frac{1}{6}\right|
&= \left|\frac{-1}{\sqrt{9-x}+3}+\frac{1}{6}\right|\\
&= \left|\frac{\sqrt{9-x}-3}{6(\sqrt{9-x}+3)}\right|\\
&= \left|\frac{-x}{6(\sqrt{9-x}+3)^2}\right|\\
&\leq\dfrac{1}{54}|x|
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2477969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Inequality question: If $a + b + c =1$, what is the minimum value of $\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}$. If $a + b + c =1$, what is the minimum value of $\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}$. I've tried AM-HM but it gave $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geq 9$ which gives $\frac{1}{a^2} + \f... | Assuming $a,b,c$ are positive reals,
$$\sum\limits_{\textrm{cyc}}\frac 1{ab}=\frac 1{abc}\sum\limits_{\textrm{cyc}}a=\frac 1{abc}\geq \frac 1{\left(\frac{a+b+c}3\right)^3}=27$$
with equality when $a=b=c=1/3$ giving $3\cdot\dfrac 1{(1/3)^3}=27$
where the penultimate step follows from the AM-GM inequality
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2478189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Sum of trigonometric secants I don't know what formula to use to solve the next sum
$\textrm{sec}\left ( \frac{2\pi }{7} \right )+\textrm{sec}\left ( \frac{4\pi }{7} \right )+\textrm{sec}\left ( \frac{6\pi }{7} \right )$
Please, give me some advice. Thanks for your help.
| $$\textrm{sec}\left ( \frac{2\pi }{7} \right )+\textrm{sec}\left ( \frac{4\pi }{7} \right )+\textrm{sec}\left ( \frac{6\pi }{7} \right )=\frac{\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}+\cos\frac{2\pi}{7}\cos\frac{6\pi}{7}+\cos\frac{4\pi}{7}\cos\frac{6\pi}{7}}{\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{6\pi}{7}}=$$
$$=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2478907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Use induction to prove that $3|(4^n − 1) $ for any integer $n \geq 0$ Use induction to prove that :
$ 3 |(4^n − 1)$ for any integer $n ≥ 0$
Hint: If $k \geq 0$ is an integer then
$4^{(k+1)} = 4\cdot4^k = 3 \cdot4^k + 4^k$.
Honestly have no idea how to even start this one
| The case $k = 0$ is self-evident: $3 \mid 0$.
Consider that
$3 \mid 4 - 1, \tag 1$
which is the case $k = 1$;. then since we have
$4(4^k - 1) + 3 = 4^{k + 1} - 4 + 3 = 4^{k + 1} - 1, \tag 2$
if we assume
$3 \mid 4^k - 1, \tag 3$
and recall as in (1) that
$3 \mid 3, \tag 4$
we see that
$3 \mid 4(4^k - 1) + 3 \tag 5$
d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2479652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Showing that $\frac{\sin(a_{n-1}) + 1}{2}$ is a Cauchy sequence In my homework set, I have the following question:
Show that $$ a_n = \frac{\sin(a_{n-1}) + 1}{2}, \quad a_1=0 $$
satisfies the definition of Cauchy sequence.
As we went over the concept of Cauchy sequences a bit too quickly in class, I'm puzzled abou... | I think it is not necessary to define such a function defined by Robert Z.
$\displaystyle |a_{n+1}-a_n|\le \frac{1}{2}|a_n-a_{n-1}|$ for all $n=1,2,\cdots$
So , $\displaystyle |a_{n+1}-a_n|\le \frac{1}{2}.|a_n-a_{n-1}|\le \frac{1}{2^2}.|a_{n-1}-a_{n-2}|\le \cdots \le \frac{1}{2^{n-1}}.|a_2-a_1| \text{ for all } n=1,2,\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2481450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
The maximum possible value of $x^2+y^2-4x-6y$ subject to the condition $|x+y|+|x-y|$ The maximum possible value of
$x^2+y^2-4x-6y$
subject to the condition $|x+y|+|x-y|$=4
My workout...
Now if we add 13 to the equation we get
$x^2+y^2-4x-6y+13-13$
or,$x^2+y^2-4x-6y+4+9-13$
or,$(x-2)^2+(y-3)^2-13$
Are there any methods... |
Solution is x=-2, y=-2.
Let us call $S(x,y)=x^2+y^2-4x-6y$.
Let us analyze our condition $|x+y|+|x-y|=4$. If we now rise this to quadrat :
$$
(|x+y|+|x-y|)^2=4^2
$$
$$
x^2+y^2+2xy+x^2+y^2-2xy+2x^2-2y^2=4x^2=4^2
$$
$$
x=+2,-2
$$
2 Solutions, or
$$
x^2+y^2+2xy+x^2+y^2-2xy-2x^2+2y^2=4y^2=4^2
$$
$$
y=+2,-2
$$
For x=2 or x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2484326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Parametric equation of the intersection between $x^2+y^2+z^2=6$ and $x+y+z=0$ I'm trying to find the parametric equation for the curve of intersection between $x^2+y^2+z^2=6$ and $x+y+z=0$. By substitution of $z=-x-y$, I see that $x^2+y^2+z^2=6$ becomes $\frac{(x+y)^2}{3}=1$, but where should I go from here?
| When you substitute $z=-(x+y)$ into $x^2+y^2+z^2=6$ we get
\begin{eqnarray*}
x^2+y^2+(x+y)^2=6 \\
x^2+xy+y^2 =3.
\end{eqnarray*}
Now multiply this by $4$ & complete the square
\begin{eqnarray*}
(2x+y)^2+3y^2=12.
\end{eqnarray*}
This is easily parameterised by $2x+y = \sqrt{12} \cos ( \theta)$ and $y = 2 \sin ( \theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2484914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Addition of two piecewise function Suppose I have function
$$y(x)=\begin{cases}
x+1\qquad & 0\leq x\leq1 \\
2-x\qquad & 1<x \leq2 \\
0\qquad & \mathrm{elsewhere}
\end{cases}$$
I need to find the function $g(x)=y(x+2)+2y(x+1)$
$$y(x+2)=\begin{cases}
x+3\qquad &-2\leq x\leq-1 \\
-x\qquad & -1<x \leq0 \\
0\qqu... | In your calculation of $g(x)$, the first line, $x+3$, was obtained by adding $y(x+2)=x+3$ and $2y(x+1)=0$. The first of these two equations, $y(x+2)=x+3$, is valid for $x$ in the range $-2\leq x\leq-1$. The second, $2y(x+1)=0$, is valid when $x<-1$ and also when $x>1$. In particular, it is not valid when $x=-1$. So t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2485305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Simplest way to get the lower bound $\pi > 3.14$ Inspired from this answer and my comment to it, I seek alternative ways to establish $\pi>3.14$. The goal is to achieve simpler/easy to understand approaches as well as to minimize the calculations involved. The method in my comment is based on Ramanujan's series $$\frac... | If we consider the Beuker-like integral
$$ 0<\int_{0}^{1}\frac{x^8(1-x)^8}{1+x^2}\,dx = 4\pi-\frac{188684}{15015} $$
we get, through partial fraction decomposition and few operations in $\mathbb{Q}$,
$$ \pi > \frac{47171}{15015} > 3.14159.$$
Inspired by Professor Vector's brilliant approach, I am adding a further appr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2485558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "55",
"answer_count": 11,
"answer_id": 4
} |
Find all posible $x$ given the congruence $(x+y)^p-y^p \equiv 1 \pmod{q}$ for both $pA few examples.
$$\begin{array}{c|c|l}
\text{prime } p & \text{prime } q & \text{possible values } x \\ \hline
3 & 5 & [1, 2, 4] \\
5 & 13 & [1, 3, 5, 8, 9, 10] \\
13 & 17 & [1, 4, 7, 10, 14, 15, 16] \\
23 & 37 & [1, 2, 3, 6, 11, 12,... | If $z = x+y$, you are solving $z^p \equiv 1 + y^p \mod q$.
If $\gcd(p, q-1) = 1$, the map $t \mapsto t^p$ is one-to-one on $\mathbb Z/q \mathbb Z$, so there is always one $z$. If $\gcd(p, q-1) = g > 1$, the map is neither one-to-one nor onto, so for some $y$ there is no solution and for others there are several. T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2486569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Prove the following inequality by induction for all $n\in \Bbb N$ I know how to prove the base case for this. But how would I continue from there?
| Set $$S_n =\sum_{i=1}^n \frac{1}{\sqrt{i}} \geq \sqrt{n}$$
then $$S_{n+1} =\sum_{i=1}^{n+1} \frac{1}{\sqrt{i}} \geq \sqrt{n} =S_n +\frac{1}{\sqrt{n+1}}$$
Obviously $S_1\ge 1 =\sqrt1.$
If we assume that $$S_n\ge \sqrt{n}$$
then
we have that
$$S_{n+1} =S_n +\frac{1}{\sqrt{n+1}} \ge \sqrt{n}+\frac{1}{\sqrt{n+1}} =\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2487195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to show that $\int_{0}^{\infty}e^{-x}(x^2+3x+3)\cos(x){\mathrm dx\over (1+x)^3}=1?$ How to show that $(1)$ is
$$\int_{0}^{\infty}e^{-x}(x^2+3x+3)\cos(x){\mathrm dx\over (1+x)^3}=1?\tag1$$
$$x^2+3x+3=\left(x+{3\over 2}\right)^2+{3\over 4}$$
$$\int_{0}^{\infty}e^{-x}\cos(x){\mathrm dx\over (1+x)^3}+\int_{0}^{\infty}... | By partial integration, we have
\begin{align}
&\int_0^{+\infty}\frac{e^{-x}\cos x}{1+x}dx
=\int_0^{+\infty}\frac{e^{-x}}{1+x}d\sin x\\
&=\left.\frac{e^{-x}\sin x}{1+x}\right|_0^{+\infty}
+\int_0^{+\infty}e^{-x}\sin x\left[\frac{1}{1+x}+\frac1{(1+x)^2}\right]dx\\
&=0-\int_0^{+\infty}e^{-x}\left[\frac{1}{1+x}+\frac1{(1+x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2488730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
For how many positive integers $n$ is $4^n − 1$ a prime number? This was the initial question and one of the methods to work it out include this:
The second method uses the fact when $n$ is a positive integer $x − 1$ is a factor of $x^n − 1$,
since, for $n ≥ 2$,
$$x^n − 1 = (x − 1)(x^{n−1} + x^{n−2} + \ldots + x + 1).$... | For the question at hand, if $P(x)$ is a polynomial and $P(r)=0$, then $P(x)=(x-r)Q(x)$ for some polynomial $Q$. If the coefficients of $P$ and $r$ are all integers, then the coefficients of $Q$ are also all integers.
The specific case of $x^n-1$ might be best understood by example:
$$\begin{align}
(x-1)(x^3+x^2+x+1)
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2490762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Conditional Joint Distribution Concern Random variables $X$ and $Y$ are jointly distributed with density
$$f(x,y)=
\begin{cases}
3x & 0 <y<2x<2 \\
0 & \text{otherwise}
\end{cases}
$$
(i) Find $P\left(X<\frac{1}{2} \Big| Y < \frac{1}{2}\right)$
$$
P\left(X<\frac{1}{2} \Big| Y < \frac{1}{2}\right) = \frac{P\left(X < \f... | In the denominator you should have
$\int_{0}^{\frac{1}{2}} \int_{\frac{y}{2}}^{1} 3x dx dy$
and in the numerator you should have
$\int_{0}^{\frac{1}{2}} \int_{\frac{y}{2}}^{\frac{1}{2}} 3x dy dx$
In other words, you should have $$\dfrac{\int_{0}^{\frac{1}{2}} \int_{\frac{y}{2}}^{\frac{1}{2}}3x dx dy}{\int_{0}^{\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2493844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$x+y+z=3$, prove the inequality For $x,y,z>0$ and $x+y+z=3$ prove that $\frac{x^3}{(y+2z)^2}+\frac{y^3}{(z+2x)^2}+\frac{z^3}{(x+2y)^2}\ge \frac{1}{3}$.
QM, AM, GM, HM suggested ;)
| $$\sum_{\text{cyc}}(y+2z)=3(x+y+z)=9$$
By Holder inequality:
$$\sum_{\text{cyc}}\frac{x^3}{(y+2z)^2}=$$
$$=\frac{\sum_{\text{cyc}}(y+2z)\sum_{\text{cyc}}(y+2z)\sum_{\text{cyc}}\frac{x^3}{(y+2z)^2}}{81}\ge$$
$$\ge \frac{(x+y+z)^3}{81}=\frac{1}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2497192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Binomial summation problem
Show that
$$ \frac{\binom{n}{0}}{1} - \frac{\binom{n}{1}}{4} +\dots + (-1)^n \frac{\binom{n}{n}}{3n+1} = \frac{3^n \cdot n!}{ 1\cdot 4\cdot 7\cdots(3n+1)}.$$
I don't know how to proceed in such type of problems. Any help or hint will be much appreciated.
| Note $$(1-x^{3})^{n} = \binom{n}{0} x^{0} - \binom{n}{1}x^{3} + \binom{n}{2}x^{6} + \cdots +(-1)^{n}\binom{n}{n}x^{3n}$$ Then what you need is $$\int_{0}^{1}(1-x^{3})^{n} \ dx = \frac{\Gamma\frac{4}{3} \cdot n!}{\Gamma\left(n+\frac{4}{3}\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2498077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the number of terms of a product to make it 1/2 How to find $k$ such that:
$$\prod_{j=1}^{k-1} \Big(1 - \frac jn \Big) = \frac 12$$
?
According to a book I'm reading, this should hold approximatively when $k \approx \sqrt{n}$ (no proof is given).
Indeed, it seems to work. The equation is equivalent to
$$\begin{ali... | Starting with
$\sum_{j=1}^{k-1} -\log\Big(1 - \frac jn \Big)
= \log(2)
$
we can use,
for $0< x < 1$,
$-\log(1-x)
=\sum_{m=1}^{\infty} \dfrac{x^m}{m}
$.
Then,
using the first $h$ terms,
$-\log(1-x)
\gt\sum_{m=1}^{h} \dfrac{x^m}{m}
$.
Looking at the remaining terms,
$\sum_{m=h+1}^{\infty} \dfrac{x^m}{m}
\lt \sum_{m=h+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2500187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solving $a \sin\theta + b \cos\theta = c$ Could someone help me with the steps for solving the below equation $$a \sin\theta + b \cos\theta = c$$
I know that the solution is $$\theta = \tan^{-1} \frac{c}{^+_-\sqrt{a^2 + b^2 - c^2}} - \tan^{-1} \frac{a}{b} $$
I just can't figure out the right steps to arrive at this sol... | here is a trick:
write
$$\frac{a}{\sqrt{a^2+b^2}}\sin(\theta)+\frac{b}{\sqrt{a^2+b^2}}\cos(\theta)=\frac{c}{\sqrt{a^2+b^2}}$$
Setting $$\cos(\phi)=\frac{a}{\sqrt{a^2+b^2}}$$ and $$\sin(\phi)=\frac{b}{\sqrt{a^2+b^2}}$$ then you will get
$$\sin(\phi+\theta)=\frac{c}{\sqrt{a^2+b^2}}$$
so $$\theta=\arcsin\left(\frac{c}{\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2502976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Star operator in matrices and new math to find inverse? Example Calculation
I realized I could make some progress on an open problem if I introduced a new operation: $f(*) x = f(x)$, where $f$ is a function and $x$ is a variable, then I could make progress. It seems initially just like some new notation but its not as ... | Lets take a 'general case' of an mxn matrix 'multiplying' a n element column vector, that is
$$
\left(
\begin{array}{cccc}
f_{11}(\star) & f_{12}(\star) & \ldots & f_{1n}(\star) \\
\vdots \\
f_{m1}(\star) & f_{m2}(\star) & \ldots & f_{mn}(\star)
\end{array}
\right)
\left(
\begin{array}{c}
x_1\\
\vdots \\
x_n
\end{ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2504894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$a_1 = a_2 = 1$ and $a_n = \frac{1}{2} \cdot (a_{n-1} + \frac{2}{a_{n-2}})$. Prove that $1 \le a_n \le 2: \forall n \in \mathbb{N} $
Let $a_n$ be a sequence satisfying $a_1 = a_2 = 1$ and $a_n = \frac{1}{2} \cdot (a_{n-1} + \frac{2}{a_{n-2}})$. Prove that $1 \le a_n \le 2: \forall n \in \mathbb{N} $
Attempt at soluti... | $$a_{n+1} = \frac{1}{2}a_{n} + \frac1{a_{n-1}}$$
Assume that for all $k \leq n$ we have $1 \leq a_k \leq 2$. Under those constraints it's easy to see that the above formula for $a_{n+1}$ takes on maximal value of $\frac{1}{2}\cdot 2 + \frac{1}{1} = 2$ and minimal value of $\frac{1}{2} \cdot 1 + \frac{1}{2} = 1$. But bo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2506658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
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.