Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Prove this equality $$\sqrt[3]{2+11i}+\sqrt[3]{2-11i}=4$$
This is how I did it:
$$\sqrt[3]{2+11i}+\sqrt[3]{2-11i}=4 \;\;\; |^{3}$$
$$2+11i+3\sqrt[3]{(2+11i)^2(2-11i)}+3\sqrt[3]{(2+11i)(2-11i)^2}+2-11i=64$$
$$4+3\sqrt[3]{(2+11i)(2-11i)} \cdot (\sqrt[3]{2+11i}+\sqrt[3]{2-11i})=64$$
$$3\sqrt[3]{4+121} \cdot (\sqrt[3]{2+11... | Close. As lemon pointed out, you started by assuming what was needed to be proven. This is easily fixed, however.
$$\sqrt[3]{2+11i}+\sqrt[3]{2-11i}=x \;\;\; |^{3}$$
$$2+11i+3\sqrt[3]{(2+11i)^2(2-11i)}+3\sqrt[3]{(2+11i)(2-11i)^2}+2-11i=x^3$$
$$4+3\sqrt[3]{(2+11i)(2-11i)} \cdot (\sqrt[3]{2+11i}+\sqrt[3]{2-11i})=x^3$$
$$3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/991516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Primality Criterion for Specific Class of Proth Numbers Is this proof acceptable ?
Theorem :
Let $N = k\cdot 2^n+1$ with $n>1$ , $k<2^n$ , $3 \mid k $ , and
$\begin{cases}
k \equiv 3 \pmod {30} , & \text{with }n \equiv 1,2 \pmod 4 \\
k \equiv 9 \pmod {30} , & \text{with }n \equiv 2,3 \pmod 4 \\
k \equiv 21 \pmo... | You just took two (true) theorems and emerged them into one, and based your proof, just by assuming both theorems are true.
For me it is unacceptable.
Proving first part is just too easy, but it is always nice to also show why you used or took as granted the given formulas. Not many people heard about Legendre symbol.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/993281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Polynomial Factoring over a finite field Ok, so I'm trying to figure out how to factor polynomials over a finite field.
My polynomial is x^5 + x^2 + x + 1 and I have to factor over GF(2)
I know the answer is (x+1)^2 * (x^3 + x + 1), because both (x+1)^2 and (x^3 + x + 1) are irreducible. (x+1)^2 is an irreducible polyn... | If you multiply $(x+1)^2$ and $x^3+x+1$ you get $x^5+2 x^3+x^2+x+1$, and if you multiply $x+1$ and $x^4+2x^3+x^2+1$ you get $x^5+3x^4+3x^3+x^2+x+1$. You don't get $x^5+x^2+x+1$. So what's happening here?
The Galois field $GF(p)$ for a prime $p$ is the integers modulo $p$. When you factor a polynomial over $GF(p)$, you ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/997422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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tough factorisation problem How would you factorise this equation given that $x=7$ is a root of this equation
$$x^3 - 67x + 126 = 0.$$
Any help would be thoroughly appreciated.
| $x^3-67x+126$ and the root is $x=7$. One writes $x=7$ as $x-7=0$ and convert the above equation in quadratic, by dividing $x^3-67x+126$ by $x-7$; so the quotient is $x^2+7x-18$. Now one factorises this $x^2+7x-18=x^2+9x-2x-18=x(x+9)-2(x+9)=(x-2)(x+9)$ then $\displaystyle(x-2)(x+9)=\frac{x^3-67x+126}{x-7}$, so $(x-2)(x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/998585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Tricky Decomposition Decompose the following polynomial:
$A=x^4(y^2+z^2) + y^4(z^2+x^2) + z^4(x^2+y^2) +2x^2 y^2 z^2 $
(by the way I'm not quite sure about the tags and the title feel free to edit them if you wish)
| Define the polynomial:
$$
f(a,b,c) = a^2(b + c) + b^2(c + a) + c^2(a + b) + 2abc
$$
Notice that $f$ is cyclic:
$$
f(a,b,c) = f(b,c,a) = f(c,a,b)
$$
Observe that since:
\begin{align*}
f(a,-a,c)
&= a^2(-a + c) + a^2(c + a) + 0 - 2a^2c \\
&= a^2(-a + c + c + a - 2c) \\
&= 0
\end{align*}
it follows by the Factor Theorem th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/998667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Evaluate this infinite sum $\displaystyle \sum_{n=1}^{\infty} \left(x^{3n-2} - x^{3n+1}\right)$
We can't simplify anything any more. Except.
$\displaystyle \sum_{n=1}^{\infty} \left( \frac{x^{3n}}{x^2} - \frac{x^{3n}x^3}{x^2} \right)$
$\displaystyle \sum_{n=1}^{\infty} \frac{x^{3n}(1 - x^3)}{x^2}$
| $\displaystyle S=\sum_{n=1}^{\infty} x^{3n-2} - x^{3n+1}=\dfrac{1-x^3}{x^2}\sum_{n=1}^\infty(x^3)^n$
If $|x^3|\le1,$ using Infinite Geometric Series Formula Derivation,
$\displaystyle S=\frac{1-x^3}{x^2}\cdot\dfrac{x^3}{1-x^3}=x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1002860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Zero as an Eigenvalue of stress tensor Given, $\sigma=\left[ \begin{array}{ccc}
0 & 0 & \alpha\beta^2 \\
0 & 0 & -\alpha\beta \\
\alpha\beta^2 & -\alpha\beta & 0 \end{array} \right]$, I am interested in the eigenvalues as well as eigenvectors of the stress tensor. $\lambda_1=0, \lambda_2=\alpha\beta\sqrt{1+\beta^2}, \l... | Assuming $\alpha\beta\ne 0$ we have:
$$\left( \begin{array}{ccc}
0 & 0 & \alpha\beta^2 \\
0 & 0 & -\alpha\beta \\
\alpha\beta^2 & -\alpha\beta & 0 \end{array} \right)\left( \begin{array}{c}
\frac{1}{\beta} \\
1 \\
0 \end{array} \right) =0 \left( \begin{array}{c}
\frac{1}{\beta} \\
1 \\
0 \end{array} \right)$$
$$\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1002950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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What are the integer solutions of the system $a^2+b^2=c^2$, $a^3+b^3+c^3=d^3$? How to solve these equations to find the integer numbers (a, b, c, and d)?
$$a^2+b^2=c^2\tag{1}$$
$$a^3+b^3+c^3=d^3\tag{2}$$
I know one of solutions which is $a=3, b=4, c=5, d=6$, but I don't know if there are others... | $(3,4,5,6)$ is the only non-trivial solution involving strictly positive integers (by non-trivial I mean where all common factors are removed).
Proof:
$(a, b, c)$ is a Pythagorean triple (EDIT: and WLOG is a primitive triple since $k|a,b,c\implies k|d$), so it can be parametrised as $a = x^2 - y^2$, $b = 2xy$, $c = x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1003333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Evaluating $\int_0^2(\tan^{-1}(\pi x)-\tan^{-1} x)\,\mathrm{d}x$ Hint given: Write the integrand as an integral.
I'm supposed to do this as double integration.
My attempt:
$$\int_0^2 [\tan^{-1}y]^{\pi x}_{x}$$
$$= \int_0^2 \int_x^{\pi x} \frac { \mathrm{d}y \mathrm{d}x} {y^2+1}$$
$$= \int_2^{2\pi} \int_{\frac{y}{\pi}... | A more straight forward approach uses integration by parts.
Define:
\begin{align}
& I(c)=\int_{a}^{b}dx(1 \times \arctan{c x})=\int_{ac}^{bc}\frac{dy}{c} (1 \times\arctan{ y})=\\&\frac{1}{c}y \arctan(y)|_{ac}^{bc}-\frac{1}{2c}\int_{ac}^{bc}\frac{y}{1+y^2}
\end{align}
using partial fraction this reads:
\begin{align}
I(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1005422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 1
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Why is $2x-1=7$ not $x=-4 \text{ or } x=4$ How would you explain why $3(2x-1)^2=147$, is $2x-1=7 \text{ or } 2x-1=-7$. But not $2x=8 \text{ so } x=4 \text{ or } x=-4$?
| If $3(2x-1)^2=147=3\times7^2$, then you have an equation of the form $A^2=B^2$ with $A=2x-1$ and $B=7$, hence $A=\pm B$:
You can write that $A^2=B^2$ is equivalent to $A^2-B^2=0$, but $A^2-B^2=(A-B)\cdot(A+B)$, hence it's also equivalent to $(A-B)\cdot(A+B)=0$. But a product is zero if and only if at least one of the f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1005733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Show sequance is monotonic Let $x>0$ (fixed) and $n$ be natural. Show that $$\displaystyle (x^n+x^{n-1}+...+1)^{\frac{1}{n}}$$ is monotonic.
I tried by induction but didn't work but intuition tells me it's decreasing.
| For $x=1$, $\sqrt[n]{n}$ is incsreasing since $\frac{d}{{dy}}\left( {\sqrt[y]{y}} \right) = \sqrt[y]{y}\frac{{1 - \ln y}}{{{y^2}}} \leqslant 0$ for $e \geqslant y \geqslant 1$ and is decreasing for $y > e$. Indeed, the first four members of the sequence are 1, 1.41421, 1.44225 and 1.41421, so the sequence is not monoto... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1012071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Decide whether the $\sqrt{\sqrt{5}+3}+\sqrt{\sqrt{5}-2}$ is rational Working needs to be shown
$\sqrt{\sqrt{5}+3}+\sqrt{\sqrt{5}-2}$
My guess is to multiply by $\sqrt{\sqrt{5}+3}-\sqrt{\sqrt{5}-2}$ then we have a rational number but is it enough to prove the rationality of a number?
| If $\sqrt{\sqrt{5}+3}+\sqrt{\sqrt{5}-2}$ is rational, then so is $\dfrac{5}{\sqrt{\sqrt{5}+3}-\sqrt{\sqrt{5}-2}}$ and consequently $\sqrt{\sqrt{5}+3}-\sqrt{\sqrt{5}-2}\in \mathbb Q$.
The sum of two rational numbers is a rational number, thus $\left(\sqrt{\sqrt{5}+3}+\sqrt{\sqrt{5}-2}\right)+\left(\sqrt{\sqrt{5}+3}-\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1012376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I find the determinant of a 4x4 matrix when the diagonal is made up of variables? Evaluate: $\det(A)$, where $A=
\begin{bmatrix}
a & 1 & 1 & 1 \\
1 & a & 1 & 1 \\
1 & 1 & a & 1 \\
1 & 1 & 1 & a\end{bmatrix}$
| By adding appropriate multiples of row 4 to the other 3 rows, you can see that the determinant of your matrix is
$$
\left|
\begin{array}{cccc}
0 &1-a &1-a &1-a^2\\
0 & a-1 & 0 & 1-a\\
0 & 0 & a-1 & 1-a\\
1& 1 & 1 & a
\end{array}
\right|
$$
Then, using cofactor expansion along the first column, the determinant is
$$
-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1016943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Does this system has a solution? Solve the system of equations
$$\begin{cases}2x_1+x_3+5x_4=0 \\x_1+2x_2-x_3=0\\x_1+x_2+2x_4=0\end{cases}$$
I created the matrix for this system and I found that it has many solution but I'm not sure if that true of not.
| As you said, it has many solutions.
Working with the matrix we get
$ \left( \begin{array}{cccc}
2 & 0 & 1 & 5\\
1 & 2 & -1 & 0 \\
1 & 2 & 0 &2 \end{array} \right) \to$ $ \left( \begin{array}{cccc}
1 & 0 & 1/2 & 5/2\\
0 & 2 & -3/2 & -5/2 \\
0 & 2 & -1/2 & -1/2 \end{array} \right) \to$ $ \left( \begin{array}{cccc}
1 & 0 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1017600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $\dfrac{x^2-6x+9}{\sqrt{x^2-6x+13}-2}=\sqrt{(x-3)^2+4}+2$ and hence find the limit as x tends to 3. Show that $\dfrac{x^2-6x+9}{\sqrt{x^2-6x+13}-2}=\sqrt{(x-3)^2+4}+2$ and hence find the limit as x tends to 3.
How do I start with this question? Rationalising the denominator? But how would I expand it?
| Yes, you are right. You want to use the identity $$a^2-b^2=(a+b)(a-b)$$ In your case $a:=\sqrt{x^2-6x+13}$ and $b=2$. The expansion is $$\left(\sqrt{x^2-6x+13}-2\right)\cdot\left(\sqrt{x^2-6x+13}+2\right)=x^2-6x+13-4=x^2-6x+9$$ Now, this simplifies significantly with the term in the numerator (but do not forget to mult... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1019072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate the integral $\int_{0}^{4}\sqrt{9t^2+t^4}dt$
Evaluate the integral $\displaystyle\int_{0}^{4}\sqrt{9t^2+t^4}dt$.
I know I have to substitute $u=9t^2+t^4$ and $\displaystyle\frac{du}{dt}=18t+4t^3$
But what should I do hereafter?
| $\int{\sqrt{9t^2+t^4}}dt = \int{t\sqrt{9+t^2}}dt$
$u = 9+t^2 \implies du = 2t \cdot dt \implies t \cdot dt = \frac{1}{2}du$
$\frac{1}{2}\int{\sqrt{u}}du = \frac{1}{3}u^{\frac{3}{2}}+c = \frac{1}{3}(9+t^2)^{\frac{3}{2}}+c$
$\frac{1}{3}((9+16)^{\frac{3}{2}}-9^{\frac{3}{2}})=\frac{98}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1019911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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summation of $\sum^n_{k=0} (n-k)^2$ I'm trying to find the recurrence of
$$ T(n) = T (n-1) + n^2$$
After following the steps,
$$T (n) = T (n-1) + n^2 = T (n-2) + (n-1)^2 + n^2 $$
$$T (n) = T (n-2) + (n-1)^2 + n^2 = T(n-3) + (n-3)^2 + (n-1)^2 + n^2 $$
$$T (n) = T (n-3) + (n-3)^2 + (n-1)^2 + n^2 =T (n-4) + (n-4)^2 +... | Hint: It is the same as $\sum_{k=0}^n k^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1020993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Chain rule with triple composition We are supposed to apply the chain rule on the following function $f$:
$$
f(x) = \sqrt{x+\sqrt{2x+\sqrt{3x}}}
$$
I assumed we could rewrite this as $$ f(x) = g(h(j(x))) $$
However, I was not sure how to define the functions $$g(x), h(x), j(x) $$
Any help would be appreciated.
| 1) $j(x) = 2x + \sqrt{3x}$
$j'(x) = 2 + \frac{\sqrt{3}}{2\sqrt{x}}$
2) $h(x) = x + \sqrt{2x + \sqrt{3x}} = x + \sqrt{j(x)}$
$h'(x) = 1 + \frac{1}{2*\sqrt{j(x)}} * j'(x)$ $=1 + \frac{1}{2*\sqrt{2x + \sqrt{3x}}} * (2 + \frac{\sqrt{3}}{2\sqrt{x}})$
3) $g(x) = \sqrt{x + \sqrt{2x + \sqrt{3x}}} = \sqrt{h(x)}$
$g'(x) = \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1021755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
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Find bases for eigenspaces of A $$A = \begin{pmatrix} 6 & 4 \\ -3 & -1\end{pmatrix}$$
Find the bases for eigenspaces $E_{\lambda_1}$ and $E_{\lambda_2}$ of $A$.
I don't really know where to start on this problem.
| Consider the matrix $$\lambda I - A = \begin{pmatrix}\lambda-6 & -4 \\ 3 & \lambda +1\end{pmatrix}$$
Now, our eiegnavlues of $A$, are the solution to the equation $\det{(\lambda I - A)}=0$
\begin{align}\det{(\lambda I - A)}&=0 \\ \implies(\lambda -6)(\lambda+1)-(3)(-4) &=0 \\ \implies \lambda^2-5\lambda +6 &= 0 \\\impl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1025200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Does $a\ln(x^2 +y^ 2 )+b$ satisfy Laplace’s equation? I can't verify that $F(x,y) = a\ln(x^2 +y^ 2 )+b$ satisfies Laplace’s equation ($F_{xx}+F_{yy}=0$). Here is what I did:
\begin{align*}
F_x &= \frac{2ax}{x^2 + y^2}
&F_y &= \frac{2ay}{x^2+y^2}
\\
F_{xx} &= \frac{2a}{x^2+y^2} - \frac{4ax^2}{(x^2+y^2)^2}
&
F_{yy} &= \f... | Let $F(x,y) = a \ln(x^2 + y^2) + b $. Then, Using quotient rule:
$$ F_x = \frac{ 2ax}{x^2 + y^2} \iff F_{xx} = \frac{2a(x^2+y^2)-2x(2ax)}{(x^2+y^2)^2} = \frac{2ay^2 - 2ax^2}{(x^2+y^2)^2}$$
Similarly,
$$ F_y = \frac{ 2ay}{x^2+y^2} \implies F_{yy} = \frac{2a(x^2+y^2) -2ay(2y)}{(x^2+y^2)^2} = \frac{2ax^2 - 2ay^2}{(x^2+y^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1025542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Modular arithmetic and using in well-ordering principle I need to prove the following, but I do not know how to go about it.
If $$ (*)\:\:\: x^{3} - y^{3}= 3^{n} $$
Then $$ x \equiv 0 (mod 3) \:\: and \:\:\: y \equiv 0 (mod 3)$$
In addition, to prove that the equation above (*), there is no solution in $ N^{+} $ (in... | Let $x$ and $y$ be positive integers such that $x^3-y^3=3^n$, where $n\ge 3$. We first show that $3$ divides $x$ and $3$ divides $y$. Then later we use this to show that in fact there are no positive solutions.
In the OP, it is observed that $x\equiv y\pmod{3}$, so $x-y$ is divisible by $3$. Now we use the fact that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1026035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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How to integrate $\int_0^1 \sqrt{-x^6+x^4-x^2+1}\:dx$ How do I integrate
\begin{equation}
\int_0^1 \sqrt{-x^6+x^4-x^2+1}\:dx,
\end{equation}
which has arisen from a problem I'm working on? I've noticed I can do the following:
\begin{align}
\int_0^1\sqrt{-x^6+x^4-x^2+1}\:dx & =\int_0^1\sqrt{\left(-x^4\right)\left(x^2-1\... | I think I can simplify it some, but I can't run it to ground.
Let $\theta = \cos^{-1} x$, so $x = \cos \theta$ and $dx = -\sin \theta \:d\theta$. Also $1 - x^2 = 1 - \cos^2 \theta = \sin^2 \theta$.
$$\begin{align}
\int_0^1 \sqrt{-x^6+x^4-x^2+1}\:dx & =
\int_{\frac{\pi}2}^0 -\sin \theta\sqrt{-\cos^6 \theta + \cos^4 \th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1026654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Prove that $\sec^2(x) - \tan^2(x) = 1$ given that $\sin^2(x) + \cos^2(x) = 1$ I need to prove from $\sin^2(x) +\cos^2(x) = 1$ that $\sec^2(x) - \tan^2(x) = 1$.
| $$ \sec^2(x)-\tan^2(x)=1 $$
$$ \frac{1}{\cos^2(x)}-\frac{\sin^2(x)}{\cos^2(x)}=1 $$
$$ \frac{1-\sin^2(x)}{\cos^2(x)}=1 $$
$$ 1-\sin^2(x)= \cos^2(x) $$
$$ 1= \sin^2(x)+\cos^2(x) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1026894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Find the possible values of |A + B + C | $ |A |= |B | = |C | = 1 $ ,where A B and C are complex nos
$$ \frac{A^2}{BC}+ \ \frac{B^2}{ \ {CA}} \ +\ \frac{C^2}{ \ {AB}} + 1 = 0$$
Find the possible values of $ |A + B + C |$
Tried substituting cos(theta) + i sin(theta)
| $$A=\cos a+i\sin a, B=\cos b+i\sin b, C=\cos c+i\sin c, a, b, c \in [0,2\pi).$$
From the given condition results:
$$\cos (2a-b-c)+\cos (2b-c-a)+\cos (2c-a-b)+1=0 $$
$$\sin (2a-b-c)+\sin (2b-c-a)+\sin (2c-a-b)=0 $$
or $$\cos x+\cos y+\cos z+1=0 $$
$$\sin x+\sin y+\sin z=0 $$ where
$$2a-b-c=x, 2b-c-... | {
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"question_score": "4",
"answer_count": 1,
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Proving the inequality $a^4+b^4+c^4+2abc(a+b+c)\ge \frac{(a+b+c)^4}9$ If $a,b,c$ are non-negative real numbers prove the following inequality $a^4+b^4+c^4+2abc(a+b+c)\ge \frac{(a+b+c)^4}9$.
| Expanding, this is equivalent to (with $\sum$ denoting cyclic sums):
$$4\sum a^4 +3 abc\sum a \ge 2\sum ab(a^2+b^2)+3\sum a^2b^2 $$
which follows from combining Schur's inequality of fourth degree
$$3\sum a^4+3abc\sum a \ge 3\sum ab(a^2+b^2)$$
with the AM GMs:
$$\sum a^4 \ge \sum a^2b^2, \quad \sum ab(a^2+b^2) \ge 2\s... | {
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"source": "stackexchange",
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Exponential Function of Quaternion - Derivation The equation for the exponential function of a quaternion $q = a + b i + c j + dk$ is supposed to be $$e^{q} = e^a \left(\cos(\sqrt{b^2+c^2+d^2})+\frac{(b i + c j + dk)}{\sqrt{b^2+c^2+d^2}} \sin(\sqrt{b^2+c^2+d^2})\right)$$
I'm having a difficult time finding a derivation... | The definition of quaternionic exponential is given by the absolutely convergent series
$$
e^z=\sum_{k=0}^\infty\dfrac{z^k}{k!}
$$
It is well known that, from this definition, if $x, y$ commute we have $e^xe^y=e^ye^x=e^{x+y}$.
Since real quaternions commute with all other quaternions, for $a \in \mathbb{R}$ we have $e... | {
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"source": "stackexchange",
"question_score": "34",
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Simplify - Help I am supposed to simplify this:
$$(x^2-1)^2 (x^3+1) (3x^2) + (x^3+1)^2 (x^2-1) (2x)$$
The answer is supposed to be this, but I can not seem to get to it:
$$x(x^2-1)(x^3+1)(5x^3-3x+2)$$
Thanks
| $$(x^2-1)^2 (x^3+1) (3x^2) + (x^3+1)^2 (x^2-1) (2x)$$
Factor out the $(x^2-1)$ term to get
$$(x^2-1)\left((x^2-1) (x^3+1) (3x^2) + (x^3+1)^2(2x)\right)$$
then factor out the $(x^3+1)$ term to get
$$(x^2-1)(x^3+1)\left((x^2-1) (3x^2) + (x^3+1)(2x)\right)$$
then factor out the $x$ term to get
$$x(x^2-1)(x^3+1)\left((x^2-... | {
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"source": "stackexchange",
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Where did I go wrong in this integration $\int\frac{\ln(1-e^x)}{e^{2x}}\,dx$ Here is the closest I've come to the answer
Link to Wolfram equality not giving true as output
NB! I marked where I'm unsure in $\color{red}{red}$ color. And please don't get startled because of the wall of equation, it's simply a more tedious... | $$I=\int\frac{\ln(1-u)}{u^3}\,du$$
$$I=\frac{\ln(1-u)}{-2u^2}-\int\frac{1}{2u^2(1-u)}$$
Now, $$\frac{1}{u^2(1-u)}=\frac{1}{u^2}+\frac1{u}+\frac1{(1-u)}$$
So:
$$I=\frac{1}{2}\left(-\frac{\ln(1-u)}{u^2}-\left(-\frac1u+\ln u-\ln|1-u|\right)\right)+c\\
I=\frac{1}{2e^{2x}}\left(-\ln(1-e^x)+e^{x}-e^{2x}x+e^{2x}\ln|1-e^x|\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1032789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding Fourier cosine series of sine function I am trying to find Fourier cosine series of following function, but think that I am messing up somewhere.
$$
f(x)=\sin \bigg ( \frac{\pi x}{l} \bigg )
$$
Fourier cosine series can be written as
$$
f(x)=\frac{a_0}{2} + \sum\limits_{n=1}^\infty a_n \cos \bigg ( \frac{n \pi ... | With
$$f(x)=\sin\left(\dfrac{\pi x}{l}\right),$$
you want to find $a_n$. The formula says:
$$a_n=\dfrac{2}{l}\int_0^l \sin\left(\dfrac{\pi x}{l}\right) \cos\left(\dfrac{n\pi x}{l}\right)dx.$$
With $u=\pi x/l$, you get:
$$a_n=\dfrac{2}{\pi}\int_0^\pi \sin(u) \cos(nu)\, du.$$
One of these identities can help you:
$$\int\... | {
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Find the area of a triangle if its two sides measure $6 in.$ and $9 in.$, and the bisector of the angle between the sides is $4\sqrt{3}$ in. Find the area of a triangle if its two sides measure $6$ in. and $9$ in., and the bisector of the angle between the sides is $4\sqrt{3}$ in. I'm thinking of using the formula $A$=... | Let $|BC|=a$, $|AC|=b$, $|AB|=c$,
angle bisector $|AD|=\beta_a$ in $\triangle ABC$.
Given $b=6,\ c=9,\ \beta_a=4\sqrt3$, find the area $S_{ABC}$.
Using known expression for the length of the angle bisector,
\begin{align}
\beta_a&=bc\left(
1-\frac{a^2}{(b+c)^2}
\right)
\tag{1}\label{1}
,
\end{align}
the missing side le... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Get the numbers from (0-30) by using the number $2$ four times How can I get the numbers from (0-30) by using the number $2$ four times.Use any common mathematical function and the (+,-,*,/,^)
I tried to solve this puzzle, but I couldn't solve it completely. Some of my results were:
$$2/2-2/2=0$$ $$(2*2)/(2*2)=1$$ $$2/... | We don't seem to have 15 yet:
${2\cdot 2 + 2 \choose 2} = 15$
If we're going to use $\Gamma$ and the like, we can get $30$ too:
$2 \cdot {\Gamma(2+2) \choose 2} = 30$
No $19$ or $27$ or $29$ yet either (though this keeps getting more absurd):
$\lfloor{2+\sqrt{2}}\rfloor^{\lfloor{2+\sqrt{2}}\rfloor} = 27$
$\lfloor \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1034122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 11,
"answer_id": 9
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Indefinite Integral of Reciprocal of Trigonometric Functions How to evaluate following integral
$$\int \frac{\mathrm dx}{\sin^4x+\cos^4x\:+\sin^2(x) \cos^2(x)}$$
Can you please also give me the steps of solving it?
| $$\int \frac{1}{\sin^4x+\cos^4x+\sin^2\left(x\right)\cos^2\left(x\right)}dx$$
$$=\int \frac{1}{\sin^4x+\cos^4x\:+2\sin^2\left(x\right)\cos^2\left(x\right)-\sin^2\left(x\right)\cos^2\left(x\right)}dx$$
$$=\int \frac{1}{(\sin^2x+\cos^2x)^2-\sin^2\left(x\right)\cos^2\left(x\right)}dx$$
$$=\int \frac{1}{1-\frac{1}{4}\sin^2... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Why $(x-5)^2-4$ can be factorised as $(x-5-2)(x-5+2)$ I would like to understand why $(x-5)^2-4$ can be factorised as $(x-5-2)(x-5+2)$
I am particularly concerned with the term, $-4$.
| Here is another way to look at it: Use the FOIL method on the right hand side.
$(x-5-2)(x-5+2)$ becomes $x^2-5x+2x-5x+25-10-2x+10-4$
Combine and cancel your like terms and you're left with $x^2-10x+25-4$, which of course can be rewritten as $(x-5)^2-4$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving $\displaystyle\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \csc A + \cot A$ I got this question from a paper but can't solve it and the question paper has no solutions section.How do you prove this?
$$\displaystyle\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \csc A + \cot A$$
Thanks in advance.
| $$\mathrm{cosec} A+\mathrm{cotan} A=\frac{1}{\sin A}+\frac{\cos A}{\sin A}=\frac{1+\cos A}{\sin A}\\=\frac{1+\cos A}{\sin A}\frac{\cos A+\sin A-1}{\cos A+\sin A-1}\\=\frac{\cos A+\sin A-1+\cos^2 A+\sin A\cos A-\cos A}{\sin A\cos A+\sin^2 A-\sin A}\\\underbrace{=}_{(1)}\frac{\sin A-1+\cos^2 A+\sin A\cos A}{\sin A\cos A+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1039642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Using mathematical induction to prove $\frac{1}1+\frac{1}4+\frac{1}9+\cdots+\frac{1}{n^2}<\frac{4n}{2n+1}$ This induction problem is giving me a pretty hard time:
$$\frac{1}1+\frac{1}4+\frac{1}9+\cdots+\frac{1}{n^2}<\frac{4n}{2n+1}$$
I am struggling because my math teacher explained us that in this case ($n^2$) when... | $$\frac{1}1+\frac{1}4+\frac{1}9+\ldots+\frac{1}{n^2}<\frac{4n}{2n+1}$$
fist proof for one: $1< \frac{4}{3}$,
for k-> $$\frac{1}1+\frac{1}4+\frac{1}9+\ldots+\frac{1}{k^2}<\frac{4k}{2k+1}$$
for k+1-> $$\frac{1}1+\frac{1}4+\frac{1}9+\ldots+\frac{1}{n^2}+\frac{1}{(k+1)^2}<\frac{4k+4}{2k+3}$$
Now,
$$\frac{4k}{2k+1}+\frac{... | {
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"url": "https://math.stackexchange.com/questions/1040312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving $x^2 - y^2 + z^2 \gt (x - y + z)^2$ Prove that
$$x^2 - y^2 + z^2 > (x - y + z)^2$$
where: $x < y <z$ for all natural numbers.
Thank for help.
| Hints:
$(x-y+z)^2-z^2=(x-y)(x-y+2z) > (x-y)(x-y+2y)=(x-y)(x+y)\\=x^2-y^2$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Other challenging logarithmic integral $\int_0^1 \frac{\log^2(x)\log(1-x)\log(1+x)}{x}dx$ How can we prove that:
$$\int_0^1\frac{\log^2(x)\log(1-x)\log(1+x)}{x}dx=\frac{\pi^2}{8}\zeta(3)-\frac{27}{16}\zeta(5) $$
| \begin{align}J&=\int_0^1\frac{\log^2(x)\log(1-x)\log(1+x)}{x}dx\\
U&=\int_0^1\frac{\log(1+x)\log^3 x}{1-x}dx,V=\int_0^1\frac{\log(1-x)\log^3 x}{1+x}dx\\
J&\overset{\text{IBP}}=\frac{1}{3}\Big[\ln^3 x\ln(1-x)\ln(1+x)\Big]_0^1-\frac{1}{3}\int_0^1 \left(\frac{1}{1+x}-\frac{1}{1-x}\right)\ln^3 x\,dx\\
&=\frac{1}{3}\Big(U-V... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1043771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
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"answer_id": 3
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Integrate the function Find the following integral,
$$I_n=\displaystyle\int (\arctan \theta)^n\ d\theta$$
Thinking that maybe there is some reduction formula, I tried using integration by parts but unable to derive it. Any help will be appreciated.
Actually the problem that I have been struggling over for quite some t... | The basic method for obtaining $\mathcal{I}_{1}$ is integration by parts:
$$\begin{align}
\mathcal{I}_{1}
&=\int\arctan{(x)}\,\mathrm{d}x\\
&=x\arctan{(x)}-\int\frac{x}{x^2+1}\,\mathrm{d}x\\
&=x\arctan{(x)}-\frac12\ln{\left(x^2+1\right)}+\color{grey}{constant}.\\
\end{align}$$
The strategy for obtaining $\mathcal{I}_{... | {
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Show $ 5 + \frac{a}{b} + \frac{b}{c} + \frac{c}{a} ≥ (1+a)(1 +b)(1+c)$ when $abc =1$ Just an inequality that I couldn't solve:
For $a,b,c$ positive reals such that $abc = 1$, show
$$5 + \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \ge (1 + a)(1 + b)(1 + c).$$
WhatI have done so far: I have shown this is equivalent, after ... | Substitute $a = u/v, b = v/w, c = w/ u$ to get rid of the constraint $abc=1$, and you have the equivalent inequality
$$\sum_{cyc} u^3v^3 + 3u^2v^2w^2 \ge uvw \sum_{cyc} uv(u+v)$$
Now let $x = uv, y = vw, z = wu$ and we get in these new variables the inequality:
$$\sum_{cyc} x^3 + 3xyz \ge \sum_{cyc} xy(x+y)$$
which is... | {
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Give $x,y,z: \begin{cases}x^2+y^2+z^2=5\\x-y+z=3\end{cases}$. Find Min, Max of $P=\frac{x+y-2}{z+2}$ Give $x,y,z: \begin{cases}x^2+y^2+z^2=5\\x-y+z=3\end{cases}$
Find Min, Max of $P=\frac{x+y-2}{z+2}$
Please help me ?
| edit 1:
It is easy to go directly:
$x^2+(x+z)^2+z^2=5 \implies x= \pm \dfrac{\sqrt{1+6z-3z^2}-z+3}{2},y=\pm \dfrac{\sqrt{1+6x-3x^2}+z-3}{2},1-\dfrac{2}{\sqrt{3}}\le z\le 1+\dfrac{2}{\sqrt{3}}$
$P= \dfrac{\pm2\sqrt{1+6z-3z^2}-2}{z+2}$
edit2:
$P=-\dfrac{2(1+\sqrt{1+6z-3z^2})}{z+2},P'=-\dfrac{2(\sqrt{-3z^2+6z+1}+9z-5) }{\... | {
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"url": "https://math.stackexchange.com/questions/1046054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do I prove this seemingly simple trigonometric identity $$a = \sin\theta+\sin\phi\\b=\tan\theta+\tan\phi\\c=\sec\theta+\sec\phi$$
Show that, $8bc=a[4b^2 + (b^2-c^2)^2]$
I tried to solve this for hours and have gotten no-where. Here's what I've got so far :
$$
\\a= 2\sin(\frac{\theta+\phi}{2})\cos(\frac{\theta-\ph... | $\color{red}{c^2=\sec^2\theta+\sec^2\phi+2\sec\theta\sec\phi\tag{1}}$
$\color{blue}{b^2=\tan^2\theta+\tan^2\phi+2\tan\theta\tan\phi\tag{2}}$
$(1)-(2)$ gives,
$\begin{align}\left(c^2-b^2\right) & =\sec^2\theta+\sec^2\phi+2\sec\theta\sec\phi-\tan^2\theta-\tan^2\phi-2\tan\theta\tan\phi\\ &=2(1+\sec\theta\sec\phi-\tan\thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1048023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 1
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Find the sum to n terms of the series $\frac{1}{1.2.3}+\frac{3}{2.3.4}+\frac{5}{3.4.5}+\frac{7}{4.5.6}+\cdots$.. Question :
Find the sum to n terms of the series $\frac{1}{1.2.3}+\frac{3}{2.3.4}+\frac{5}{3.4.5}+\frac{7}{4.5.6}+\cdots$
What I have done :
nth term of numerator and denominator is $2r-1$ and $r(r+1)(r+2... | You can get the result to pop out if you play around with the indices of the infinite series and expand a couple terms. $$\sum_{r=1}^\infty\frac{2r-1}{r(r+1)(r+2)} = \sum_{r=1}^\infty\frac{1}{2r}-\frac{1}{r+1}+\frac{1}{2(r+2)} \\ = \frac{1}{2}\sum_{r=1}^\infty \frac{1}{r}-\sum_{r=1}^\infty\frac{1}{r+1}+\frac{1}{2}\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1048308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
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How do I prove that $\lim _{n\to \infty} (\sqrt{n^2 + 1} - n) = 0$? As you can see from the title, I need help proving that $\lim _{n\to \infty} (\sqrt{n^2 + 1} - n) = 0$. I first looked for $N$ by using $|\sqrt{n^2 + 1} - n - 0| = \sqrt{n^2 + 1} - n < \varepsilon$ and solving for $n$. I got $n > \frac{1 - \varepsilon ... | For any positive $c$,
$(n+\frac{c}{2n})^2
=n^2+c+\frac{c^2}{4n^2}
> n^2+c
$,
so
$\sqrt{n^2+c}
<n+\frac{c}{2n}
$.
Therefore,
if $c > 0$,
$n
<\sqrt{n^2+c}
<n+\frac{c}{2n}
$
so
$\lim_{n \to \infty} \sqrt{n^2+c}-n
= 0
$.
You can prove this when $c < 0$
similarly.
| {
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"url": "https://math.stackexchange.com/questions/1050741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Prove that $1 + 4 + 7 + · · · + 3n − 2 = \frac {n(3n − 1)}{2}$ Prove that
$$1 + 4 + 7 + · · · + 3n − 2 =
\frac{n(3n − 1)}
2$$
for all positive integers $n$.
Proof: $$1+4+7+\ldots +3(k+1)-2= \frac{(k + 1)[3(k+1)+1]}2$$
$$\frac{(k + 1)[3(k+1)+1]}2 + 3(k+1)-2$$
Along my proof I am stuck at the above section where it w... | The solution must be a quadratic polynomial (because its first order difference is a linear polynomial) so it is obtained with the Lagrangian formula on three points. Taking $(0,0),(1,1),(2,5)$,
$$0\frac{(n-1)(n-2)}{(0-1)(0-2)}+1\frac{(n-0)(n-2)}{(1-0)(1-2)}+5\frac{(n-0)(n-1)}{(2-0)(2-1)}=\frac{n(3n-1)}2.$$
| {
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"url": "https://math.stackexchange.com/questions/1050814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 11,
"answer_id": 8
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How to evaluate the sum Please help me to evaluate the following
$$\sum\limits_{n=1}^\infty \frac{\left(\frac{3-\sqrt{5}}{2}\right)^n}{n^3}$$
Have no idea how to evaluate. Any trick please?
Thanks
| Differentiating: $\displaystyle f(z) = \operatorname{Li}_{3}(z) + \operatorname{Li}_{3}(1-z)+ \operatorname{Li}_{3}\left(1 - \frac{1}{z}\right)$
We have, $\displaystyle f'(z) = \frac{\operatorname{Li}_2(z)}{z} - \frac{\operatorname{Li}_2(1-z)}{1-z} +\frac{\operatorname{Li}_2\left(1 - \frac{1}{z}\right)}{1 - \frac{1}{z}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1052776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Simplifying $\frac{x^6-1}{x-1}$ I have this:
$$\frac{x^6-1}{x-1}$$
I know it can be simplified to $1 + x + x^2 + x^3 + x^4 + x^5$
Edit : I was wondering how to do this if I didn't know that it was the same as that.
| Just apply the following reduction formula inductively.
$$\frac{x^n-1}{x-1} = x^{n-1}+\frac{x^{n-1}-1}{x-1}$$
Derivation
$$\begin{array}{lll}
\frac{x^n-1}{x-1}&=&\frac{x^n - x^{n-1}+x^{n-1}-1}{x-1}\\
&=&\frac{x^n - x^{n-1}+x^{n-1}-1}{x-1}\\
&=&\frac{x^{n-1}(x-1)+x^{n-1}-1}{x-1}\\
&=&x^{n-1}+\frac{x^{n-1}-1}{x-1}\\
\e... | {
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"url": "https://math.stackexchange.com/questions/1055671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 10
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Prove using induction $2^{3n}-1$ is divisble by $7$ for all $n$ $\in \mathbb N$
Show that
$2^{3n}-1$ is divisble by $7$ for all $n$ $\in \mathbb N$
I'm not really sure how to get started on this problem, but here is what I have done so far:
Base case $n(1)$:
$\frac{2^{3(1)}-1}{7} = \frac{8-1}{7} = \frac{7}{7}$
But n... | For simplicity, please note that $2^{3n}=8^n$.
First, show that this is true for $n=1$:
*
*$\frac{8^1-1}{7}=1\in\mathbb{N}$
Second, assume that this is true for $n$:
*
*$\frac{8^n-1}{7}=k\in\mathbb{N}$
Third, prove that this is true for $n+1$:
*
*$\frac{8^{n+1}-1}{7}=\frac{8\cdot8^n-1}{7}$
*$\frac{8\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1056045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Why is $\sum \frac{\sqrt{n^3+n}-\sqrt{n^3}}{\sqrt{n(n+1)}-1}$ convergent? We have the following series:
$$\sum_{n=1}^{\infty} \frac{\sqrt{n^3+n}-\sqrt{n^3}}{\sqrt{n(n+1)}-1}$$
According to WolframAlpha it is convergent but I don't see why. Intuitively, the expression under the sum must be assymptotically similair to so... | You may write, as $n$ is great:
$$
\begin{align}
\frac{\sqrt{n^3+n}-\sqrt{n^3}}{\sqrt{n(n+1)}-1}&=\frac{(\sqrt{n^3+n}-\sqrt{n^3})(\sqrt{n(n+1)}+1)}{n(n+1)-1}\\\\
&=\frac{n^{5/2} }{n^2}\frac{\left(\sqrt{1+\frac{1}{n^2}}-1\right)\left(\sqrt{1+\frac{1}{n^2}}+\frac1n\right)}{1+\frac{1}{n}-\frac{1}{n^2}}\\\\
&=\frac{n^{5/2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1056131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Sum of squares and $5\cdot2^n$ Does anyone know of a proof of the result that $5\cdot2^n$ where $n$ is a nonnegative integer is always the sum of two squares?
That is, nonzero integers $x,y$ must always exist where:
$x^2+y^2=5\cdot2^n$
when $n$ is $0$ or a positive integer?
For example:
$$1^2+2^2=5\cdot 2^0$$
$$1^2+3... | Consider the following two cases:
$$n=2k\implies 5\cdot 2^{2k}=(1+4)\cdot 2^{2k}=2^{2k}+2^{2k+2}=\left(2^{k}\right)^2+\left(2^{k+1}\right)^2$$
$$n=2k+1\implies 5\cdot 2^{2k+1}=10\cdot 2^{2k}=(1+9)\cdot 2^{2k}=2^{2k}+9\cdot 2^{2k}=\left(2^{k}\right)^2+\left(3\cdot 2^{k}\right)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1058568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Help with $\int \frac{1}{(\sin x + \cos x)}$ Kindly solve this question
$$\int \frac{1}{(\sin x + \cos x)} dx$$
I reached up to
$$\frac{(1+\tan^2x)}{1-\tan^2x + 2\tan x}$$
| Hint :
\begin{align}
\int \frac{\mathrm dx}{\sin x + \cos x} &=\int \frac{\mathrm dx}{\sin x + \cos x}\cdot \frac{\sin x - \cos x}{\sin x - \cos x}\\
&=\int \frac{\sin x - \cos x}{\sin^2 x - \cos^2 x}\mathrm dx\\
&=\int \frac{\sin x}{1 - 2\cos^2 x}\mathrm dx-\int \frac{\cos x}{2\sin^2 x -1}\mathrm dx\\
&=\int \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Evaluating $ \int \frac{1}{\sin x} dx $ Verify the identity
$$\sin x = \frac {2 \tan\frac{x}{2}}{1 + \tan^2\frac{x}{2}}$$
Use this identity and the substitution $t = \tan\frac{x}{2}$ to evaluate the integral of $$ \int \frac{1}{\sin x} dx $$
My answer:
$$ A = \frac{x}{2} \Rightarrow \sin 2A= \frac{2\tan A}{1+\tan^2 A} ... | $$I=\int\frac{dx}{\sin x}=\int\frac{1+\cos x}{\sin x}\frac{dx}{1+\cos x}=\int\frac{1+\cos x}{\sin x}d\left(\frac{\sin x}{1+\cos x}\right)$$
$$=\ln\left|\frac{\sin x}{1+\cos x}\right|+C=\ln\left|\tan {\frac{x}{2}}\right|+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Just a proof of algebra If $a+b+c=0$,
Show that
$\left[\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}\right]\left[\dfrac{b-c}{a}+\dfrac{c-a}{b}+\dfrac{a-b}{c}\right]=9.$
I am struck with this problem but can't find a solution. Please help me.
| Assuming $a,b,c$ are non-zero & distinct
$$F=\dfrac{c}{a-b}\left[\dfrac{b-c}{a}+\dfrac{c-a}{b}+\dfrac{a-b}{c}\right]$$
$$=1+\frac c{a-b}\cdot\frac{b^2-bc+ca-a^2}{ab}$$
$$=1+\frac c{a-b}\cdot\frac{(b-a)(b+a)-c(b-a)}{ab}$$
$$=1+\frac c{a-b}\cdot\frac{(b-a)(b+a-c)}{ab}$$
$$=1-\frac{c(b+a-c)}{ab}$$
As $b+a=-c,$
$$F=1-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1060560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Prove by induction that an expression is divisible by 11
Prove, by induction that $2^{3n-1}+5\cdot3^n$ is divisible by $11$ for any even number $n\in\Bbb N$.
I am rather confused by this question. This is my attempt so far:
For $n = 2$
$2^5 + 5\cdot 9 = 77$
$77/11 = 7$
We assume that there is a value $n = k$ such th... | $\begin{align}2^{3}\equiv -3\pmod {11} & \implies 3^n\equiv 2^{3n}\pmod {11} \ \color{blue}{(\text{since}\ n\ \text{is even})}\\& \implies 2^{3n-1}+5\cdot 3^n\equiv 2^{3n-1}+5\cdot 2^{3n}\pmod {11}\\&\implies2^{3n-1}+5\cdot 3^n\equiv 2^{3n-1}(1+5\cdot 2)\pmod {11}\\&\implies 2^{3n-1}+5\cdot 3^n\equiv 0\pmod {11}\end{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1061477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 9,
"answer_id": 6
} |
If $\left(1+\sin \phi\right)\cdot \left(1+\cos \phi\right) = \frac{5}{4}\;,$ Then $\left(1-\sin \phi\right)\cdot \left(1-\cos \phi\right)$ If $\displaystyle \left(1+\sin \phi\right)\cdot \left(1+\cos \phi\right) = \frac{5}{4}\;,$ Then $\left(1-\sin \phi\right)\cdot \left(1-\cos \phi\right) = $
$\bf{My\; Try::}$ Given $... | Let $$A = (1+\sin(\theta))(1+\cos(\theta))$$ and $$B = (1-\sin(\theta))(1-\cos(\theta))$$ Observe that $$A+B = 2+2\sin(\theta)\cos(\theta)$$ $$AB = (\sin(\theta)\cos(\theta))^2$$
Substituting $A = 5/4$, we now have a quadratic in $B$ that is easily solved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1061928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
If both $a,b>0$, then $a^ab^b \ge a^bb^a$ Prove that $a^a \ b^b \ge a^b \ b^a$, if both $a$ and $b$ are positive.
| $$a^a \ b^b \;?\; a^b \ b^a \\
\frac{a^a}{b^a} \;?\; \frac{a^b}{b^b} \\
\left(\frac{a}{b}\right)^a \;?\; \left(\frac{a}{b}\right)^b \\
\left(\frac{a}{b}\right)^{a-b} \;?\; 1 $$
if $a \ge b$, then $c = \frac{a}{b} \ge 1$, and $d = a-b \ge 0$. Thus $c^d \ge 1$, so $?$ is $\ge$.
if $a < b$, then:
$$\left(\frac{a}{b}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1063223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 10,
"answer_id": 1
} |
Evaluate this Trigonometric Expression: $\sqrt[3]{\cos \frac{2\pi}{7}} + \sqrt[3]{\cos \frac{4\pi}{7}} + \sqrt[3]{\cos \frac{6\pi}{7}}$
Evaluate
$$ \sqrt[3]{\cos \frac{2\pi}{7}} + \sqrt[3]{\cos \frac{4\pi}{7}} + \sqrt[3]{\cos \frac{6\pi}{7}}$$
I found the following
*
*$\large{\cos \frac{2\pi}{7}+\cos \frac... |
let $$x=\sqrt[3]{\cos{\dfrac{2\pi}{7}}},y=\sqrt[3]{\cos{\dfrac{4\pi}{7}}},z=\sqrt[3]{\cos{\dfrac{6\pi}{7}}},$$
then we have
$$\begin{cases}
x^3+y^3+z^3=-\dfrac{1}{2}\\
(xy)^3+(yz)^3+(xz)^3=-\dfrac{1}{2}\\
(xyz)^3=\dfrac{1}{8}
\end{cases}$$
use this identity
$$a^3+b^3+c^3=(a+b+c)^3-3(a+b+c)(ab+bc+ac)+3abc$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1065431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
A reduction formula for $\int_0^1 x^n/\sqrt{9 - x^2}\,\mathrm dx$
Let $$I_n = \int_0^1 \frac{x^n}{\sqrt{9 - x^2}}\,\mathrm dx$$
Using integration, show that $$nI_n = 9(n - 1)nI_{n - 2} - 2\sqrt2$$
I've found that $\displaystyle I_0 = \sin^{-1}\left(\frac{1}{3}\right)$, but that's it.
I'm struggling to go any further.... | $$I_n = \int_0^1 \frac{x^n}{\sqrt{9 - x^2}}\,\mathrm dx= \int_0^1 x^{n-1}\cdot\frac{x}{\sqrt{9 - x^2}}\,\mathrm dx$$
Using Integration By Parts
$$\frac{x}{\sqrt{9 - x^2}}\,\mathrm dx=\,\mathrm dv\iff-\sqrt{9-x^2}=v$$
$$x^{n-1}=u\iff (n-1)x^{n-2}\,\mathrm dx=\,\mathrm du$$
$$\begin{align}
I_n
&= -x^{n-1}\cdot\sqrt{9-x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1067022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to evaluate $\int_{0}^{\infty}\frac{(x^2-1)\ln{x}}{1+x^4}dx$? How to evaluate the following integral
$$I=\int_{0}^{\infty}\dfrac{(x^2-1)\ln{x}}{1+x^4}dx=\dfrac{\pi^2}{4\sqrt{2}}$$
without using residue or complex analysis methods?
| \begin{align}
&\int_{0}^{\infty}\frac{(x^2-1)\ln{x}}{1+x^4}dx
=2\int_{0}^{1}\frac{(x^2-1)\ln{x}}{1+x^4}dx \\
=&\ \sqrt2 \int_{0}^{1}\ln x \ d\left( -\tanh^{-1}\frac{\sqrt2x}{1+x^2}\right)
\overset{ibp}={\sqrt2} \int_0^1 \frac1x \tanh^{-1}\frac{\sqrt2x}{1+x^2}dx\\
=& \ {\sqrt2} \int_0^1 \int_0^{\pi/4}
\frac{2(x^2+1)\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1067499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Find $x$ such that $x \equiv7\pmod {37}$ and $x^2 \equiv 12\pmod {37^2}$ Find $x$ such that $x \equiv7 \pmod {37}$ and $x^2 \equiv 12\pmod {37^2})$
My attempt: Given $x \equiv7\pmod {37}$
so $37|(x-7)$ so $37^2|(x-7)^2$
so $x^2-14x+49 \equiv 0\pmod {37^2}$
as $12-14x+49 \equiv 0\pmod{37^2}$ [as $x^2 \equiv 12\pmod {37... | I would do $$(37n+7)^2=12\pmod{37^2}\\14n*37+49=12\pmod{37^2}\\14n+1=0\pmod{37}$$
which has a smaller modulo
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1071441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What's $\int \frac{1}{\sqrt{25-x^2}}$ What is $$\int \frac{1}{\sqrt{25-x^2}}$$ WolframAlpha says $\sin^{-1}(\frac{x}{5})$ while I got $\frac{1}{5}\sin^{-1}(\frac{x}{5})$. What is correct?
Thanks in advance.
| Let $x = 5u \to dx = 5du \to \displaystyle \int \dfrac{dx}{\sqrt{25-x^2}} = \displaystyle \int \dfrac{5du}{\sqrt{25-25u^2}} = \displaystyle \int \dfrac{du}{\sqrt{1-u^2}} =\arcsin u + C = \arcsin \left(\frac{x}{5}\right) + C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1072925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Inequality $\frac{x^3+y^3}{x-y}>4$ Let $x>y>0$ and $xy\geq 1$. Prove that $$\frac{x^3+y^3}{x-y}>4.$$
Of course we can factor $(x^3+y^3)=(x+y)(x^2-xy+y^2)$, but it is not very useful. For fixed $x-y$, we can try to find the minimum value of $x,y$ such that $xy\geq 1$, and show that $\frac{x^3+y^3}{x-y}>4$ for those valu... | $$x^3+y^3 - 4(x-y) \ge 2(x^2 - xy +y^2) - 4(x-y) $$
$$= 2(x^2 - 2xy +y^2 - 2(x-y) + xy) \ge 2(x-y-1)^2$$
Where, we used $x+y \ge 2\sqrt{xy} \ge 2$ and $xy \ge 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1074023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Trigonometric equation, find $\sin \theta $ Find $\sin \theta $ if $a$ and $c$ are constants
$$ 1-\left(c-a\tan\theta\right)^2=\frac{\sin^2\theta\cos^4\theta }{a^2-\cos^4\theta } $$
| after simpßlification we find this here
$$2\,\sin \left( \theta \right) \left( \cos \left( \theta \right)
\right) ^{5}ac- \left( \cos \left( \theta \right) \right) ^{8}+{a}^{
2} \left( \cos \left( \theta \right) \right) ^{6}- \left( \cos
\left( \theta \right) \right) ^{6}{c}^{2}+2\, \left( \cos \left(
\theta \r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1074396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
How can I write this power series as a power series representation? How can I write this power series ($1+x+2x^2+2x^3+3x^4+3x^5+4x^6+4x^7+5x^8....$) as a power series representation (like $\dfrac{1}{1-x}$ or something neat like that)?
| Hint: using $y=x^2$ and derivative in $y$: $$(1+x)(1+2x^2+3x^4+\ldots) $$
$$ =(1+x)(1+2y+3y^2+4y^3 +\ldots)$$
$$= (1+x)(y+y^2+y^3+y^4+\ldots)'$$
$$ = (1+x)\left( \frac{y}{1-y}\right)'$$
Edit:
$$ = (1+x) \frac{1}{(1-y)^2} $$
$$ = \frac{1+x}{(1-x^2)^2} $$
$$ = \frac{1}{(1-x)(1+x^2)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1075054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Diophantine equation : two products of linear factors differ by a constant Recently, I was asked the following question by a friend : find
all $a,b,c,a',b',c',k \in {\mathbb Z}$ with $k\neq 0$ such that the identity
$$
(X-a)(X-b)(X-c)+k=(X-a')(X-b')(X-c')
$$
holds in ${\mathbb Z}[X]$. Does anyone have an idea ?
Note ... | These are not all solutions, but one infinite family.
For example, for any integers $x,y,z$, take $a = xyz$, $b = -x(x+y)z$, $c = -y(x+y)z$, $k = x^2 y^2 (x+y)^2 z^3$. Note that $ab + ac + bc = 0$. If $p(X) = (X-a)(X-b)(X-c)$, we have
$p(0) = -x^2 y^2 (x+y)^2 z^3$ and $p'(0) = 0$, and then
$$p(X) + k = X^2 (X + (x^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1076358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Explanation solution partial-fraction of $\frac{x^2 + 2}{x^2 - 1}$ The partial fraction of $\dfrac{x^2+2}{x^2-1}$ is $1 + \dfrac{3}{2}\cdot(\dfrac{1}{x-1}-\dfrac{1}{x+1})$.
I understand how you get $\dfrac{3}{2}\cdot(\dfrac{1}{x-1}-\dfrac{1}{x+1})$ but from where does the $1 +$ come?
| 1 comes from long division of $x^2+2$ by $x^2-1.$ when you divide $x^2 + 2$ by $x^2 - 1$ you get the quotient $1$ and the remainder $3.$ that is $$x^2 + 2 = 1(x^2-1)+3 \mbox{ or }{x^2 + 2 \over x^2 -1} = 1 + {3 \over x^2 - 1}$$
this is very similar to how you do division in integers. for example when you divide $123$ b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1076437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove $4^k - 1$ is divisible by $3$ for $k = 1, 2, 3, \dots$ For example:
$$\begin{align}
4^{1} - 1 \mod 3 &=
\\
4 -1 \mod 3 &=
\\
3 \mod 3 &=
\\3*1 \mod 3 &=0
\\
\\
4^{2} - 1 \mod 3 &=
\\
16 -1 \mod 3 &=
\\
15 \mod 3 &=
\\3*5 \mod 3 &= 0
\\
\\
4^{3} - 1 \mod 3 &=
\\
64 -1 \mod 3 &=
\\
21 \mod 3 &=
\\3*7 \mod 3 &=
... | Note that $x^k - y^k$ is always divisible by $x-y$. (In fact, $x^k - y^k = (x - y)(x^{k-1} + x^{k-2}y + x^{k-3}y^2 + \cdots + y^{k-1})$ Specialize for $x=4, y=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1076946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 7
} |
Generalisation of $(n+3)^2-(n+2)^2-(n+1)^2+n^2=4$ After seeing the neat little identity $(n+3)^2-(n+2)^2-(n+1)^2+n^2=4$ somewhere on MSE, I tried generalising this to higher consecutive powers in the form
$\sum_{k=0}^a\epsilon_k(n+k)^p=C$, where $C$ is a constant and $\epsilon_k=\pm1$. I discovered a relatively simple ... | Note that since $R$ and $1$ commute,
$$
R^{2^k}-1=\left[\sum\limits_{j=0}^{2^k-1}R^j\right](R-1)\tag{1}
$$
Therefore,
$$
\begin{align}
\prod_{k=0}^{n-1}\left(R^{2^k}-1\right)x^n
&=\left[\prod_{k=0}^{n-1}\sum_{j=0}^{2^k-1}R^j\right](R-1)^nx^n\tag{2a}\\
&=\left[\prod_{k=0}^{n-1}\sum_{j=0}^{2^k-1}R^j\right]n!\tag{2b}\\
&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1079575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 2,
"answer_id": 0
} |
How to integrate $f(x) = \frac{1}{a + b \cos x + c \sin x }$ over $x \in (0,\pi/2)$
Conjecture 1
$$ \begin{align*} I_{T}=\int_0^{2\pi} \frac{\mathrm{d}x}{a + b \cos x + c
\sin x} & = \frac{2\pi}{\rho} \tag{1} \\ I_{T/4} = \int_0^{\pi/2} \frac{\mathrm{d}x}{a +
b \cos x + c \sin x} & = \frac{2}{\rho} \arctan \lef... | Assuming $b^2+c^2<a^2$ and expressing $\sin(x)$ and $\cos(x)$ in terms of $\tan(x/2)$ we have:
$$I_{T/4}=2\int_{0}^{1}\frac{dt}{a(1+t^2)+b(1-t^2)+c(2t)}=2\int_{0}^{1}\frac{dt}{(a-b)t^2+2ct+(a+b)}.$$
Since the discriminant of $p(t)=(a-b)t^2+2ct+(a+b)$ is negative by the initial assumptions, $p(t)$ does not vanish on $[0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Find the value of : $\lim_{ x \to \infty} \left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)$ I need to calculate the limit of the function below:
$$\lim_{ x \to \infty} \left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)$$
I tried multiplying by the conjugate, substituting $x=\frac{1}{t^4}$, and both led to nothing.
| Set $\dfrac1x=h^2$ where $h>0$ to get
$$\lim_{ x \to \infty} \sqrt{x+\sqrt{x+\sqrt{x}}}$$
$$=\sqrt{\frac1{h^2}+\sqrt{\frac1{h^2}+\sqrt{\frac1{h^2}}}}=\sqrt{\frac1{h^2}+\sqrt{\frac1{h^2}+\frac1h}}=\sqrt{\frac1{h^2}+\sqrt{\frac{h+1}{h^2}}}$$
$$=\sqrt{\frac1{h^2}+\frac{\sqrt{h+1}}h}=\sqrt{\frac{1+h\sqrt{h+1}}{h^2}}=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1082649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 5
} |
Why is $\cos(x/2)+2\sin(x/2)=\sqrt5 \sin(x/2+\tan^{-1}(1/2))$ true? According to Wolfram Alpha the following equality holds:
$$\cos\left(\frac{x}{2}\right)+2\sin\left(\frac{x}{2}\right)=\sqrt5 \sin\left(\frac{x}{2}+\tan^{-1}\left(\frac{1}{2}\right)\right)$$
I also checked it numerically. Why is it true?
| Using $\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$,
\begin{align}
\sin\left(x + \tan^{-1} \left(\frac 1 2\right) \right)
&= \sin x \cos \left(\tan^{-1}\left(\frac 1 2\right)\right) +
\cos x \sin \left(\tan^{-1}\left(\frac 1 2\right)\right) \\
&= \sin x \frac 2 {\sqrt 5} + \cos x \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1082979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Value of $\lim\limits_{z \to 0}\bigl(\frac{\sin z}{z}\bigr)^{1/z^2}$ Find the value of $$\lim\limits_{z \to 0}\left(\dfrac{\sin z}{z}\right)^{1/z^2}$$
So I took a log: $$\frac{1}{z^2}\log\left(\frac{\sin z}z\right)$$ If I could expand it something like $\log(1+x)$ .. any hints ?
| $\sin{z}=z-\frac{z^3}{3!}+\frac{z^5}{5!}-......$
Therefore, we have
$\lim_{z \to 0}({\frac{\sin{z}}{z}})^{1/z^2}=\lim_{z \to 0}(\frac{z-\frac{z^3}{3!}+\frac{z^5}{5!}-......}{Z})^{1/z^2}$
=$\lim_{z \to 0}(1-\frac{z^2}{3!}+\frac{z^4}{5!}-......)^{1/z^2}$
=$(\lim_{z \to 0}(1-\frac{z^2}{3!}+\frac{z^4}{5!}-......))^{1/z^2}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1085225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
General Term of a Sequence What would be the best way in finding a general term $a_n, n>3$ for the recursive sequence?
$$a_n = \dfrac{6a_{n-1}^2a_{n-3} -8 a_{n-1}a_{n-2}^2}{a_{n-2}a_{n-3}}$$
where $a_1 = 1 ; a_2 = 2 ; a_3 = 24$ ;
| Given the recursive sequence:
$$a_n = \dfrac{6a_{n-1}^2a_{n-3} -8 a_{n-1}a_{n-2}^2}{a_{n-2}a_{n-3}}\tag{E}$$
$(\bf E)$ can be rearranged to:
$$\frac{a_n}{a_{n-1}}=6\frac{a_{n-1}}{a_{n-2}}-8\frac{a_{n-2}}{a_{n-3}}\tag{1}$$
Seeing a pattern in $(1)$ let a new term $t_n$ be: $$t_n=\frac{a_n}{a_{n-1}}\tag{2}$$
Now $t_n$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1085493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Arithmetic Error in Calculation of the Limit of a Given Function I consider a function $f(x)$ which is equal to $\dfrac{\frac{3}{x}-\frac{1}{2}}{x-6}$
While trying to evaluate the $\lim_{x \to 6} f(x)$
It is true that $\dfrac{\frac{3}{x}-\frac{1}{2}}{x-6} = \dfrac{\frac{3}{x}}{x-6}-\dfrac{\frac{1}{2}}{x-6}$
It is also ... | Note: $$\frac a{\frac bc} = \frac{ac}b$$
You arrive at only $ac$.
Also, $$\dfrac{\frac ab}{c} = \frac{a}{bc}$$
So $$ \dfrac{\frac{3}{x}-\frac{1}{2}}{x-6} = \dfrac{\frac{3}{x}}{x-6}-\dfrac{\frac{1}{2}}{x-6} = \frac{3}{x(x-6)} - \frac 1{2(x-6)} = \frac{2\cdot 3}{2x(x-6)}-\frac{x}{2x(x-6)} = \dfrac{6-x}{2x(x-6)}$$
$$= \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1086930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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A hint to help me integrate $ \int {x^5+1\over x^4+x^3+x^2} \, dx $ Numerator is of a higher degree than the denominator, so after division I get the following:
$$ \int {x^5+1\over x^4+x^3+x^2} \, dx = \frac 12x^2-x + \int {1\over x^2+x+1} \, dx $$
| \begin{align}
& \underbrace{x^2+x+1=\left(x^2+x+\frac 1 4 \right)+\frac 3 4}_{\text{completing the square}} = \left( x+\frac 1 2 \right)^2 + \frac 3 4 \\[10pt]
= {} & \frac 3 4 \left( \frac 4 3 \left( x+\frac 1 2 \right)^2 + 1 \right) = \frac 3 4 \left( \left( \frac{2x+1}{\sqrt{3}} \right)^2 +1 \right) = \frac 3 4 (u^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1088027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Pythagorean type diophantine equation.
How to find all solutions to
$$ a^2+b^2+c^2+d^2=e^2+2$$
where all variables $a$ to $e$ are positive integers and $e^2 \equiv 1 \mod 8$
I tried using parameterization similar to pythagoras equation, but no success so far. Any help will be appreciated.
Thanks!
| For such equations, you can use the standard approach.
One approach is to use equations Pell. For the beginning will talk about a more simple way.
I. Eq.1
$$a^2+b^2+c^2+d^2=(2q+1)^2+2$$
Solutions have the form:
$$a=x + k$$
$$b=x + p$$
$$c=x + s$$
$$d=x - z+1$$
$$q=x$$
where,
$$x = -1 -z+ (z^2 - k p - k s - p s),\qua... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1088695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Area of circle and parabola I shall determine the surface area of the intersection of a circle with Radius $R= 1 $ and the set of points above the parabola $f(x) = 2x^2$. Only the positive $x,y$-plane is of interest.
My approach is to parameterize it with polar coordinates as $x=R\cos(\phi)$ and $y=R\sin(\phi)$, thus ... | I will assume the circle is centered at the origin since you did not state.
We will have the equation of the circle: $x^2+y^2=1$
For the parabola, we have $y=2x^2$
To get the points of intersection, we just have to set y=$2x^2$ in the equation of the parabola.
$x^2+4x^4=1$
Solving the equation, we get $x=0.68,-0.68$ as... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1090958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How to compute $\int_{0}^{+\infty} \frac{dx}{e^{x+1} + e^{3-x}}$? How to compute $\int_{0}^{+\infty} \frac{dx}{e^{x+1} + e^{3-x}}$?
My partial solution:
$$
\int_{0}^{+\infty} \frac{dx}{e^{x+1} + e^{3-x}} = \int_{0}^{+\infty} \frac{dx}{e^{3-x}(1 + e^{2x-2})} \\
= \int_{0}^{+\infty} \frac{e^{x-3}dx}{1 + e^{2x-2}}.
$... | It is natural to let $y=e^x$. Then $dy=e^x\,dx$, so $dx=\frac{1}{y}\,dy$. Thus we want
$$\int_1^\infty \frac{1}{y}\cdot \frac{1}{e\cdot y+e^3 \cdot \frac{1}{y}}\,dy,$$
which simplifies to
$$\int_1^\infty \frac{1}{e}\cdot\frac{1}{y^2+e^2}\,dy.$$
Let $y=et$, and we are nearly finished.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1091549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How does polynomial long division work?
I get normal long division but this doesn't make sense. How can doing it by only dividing by the leading term work? The problem is $$\dfrac{3x^3 - 2x^2 + 4x - 3 } {x^2 + 3x + 3},$$ not $$\dfrac{3x^3 - 2x^2 + 4x - 3} {x^2}.$$
| Some trick on how to do this quickly: $3x^3 - 2x^2 + 4x - 3 = 3x(x^2+3x+3) - 11x^2 - 5x - 3 = 3x(x^2+3x+3) - 11(x^2+3x+3) + 28x + 30 = (3x-11)(x^2+3x+3) + 28x+30$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1092503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 0
} |
Why does the limit behavior of this function take over at 35? I've been working with this function on an semi-related question:
$$f(N)=\left\lfloor \frac{10N}{\lceil \frac{3}{4} N \rceil} \right\rfloor$$
It's clear that $10\leq f(N) \leq 13$, and that $\displaystyle\lim_{n\rightarrow \infty} f(N)=13$.
It's also true th... | Let
$N = 4n+k$
where $0 \le k \le 3$.
I will show that
$f(N)
= 13+\left\lfloor \frac{1-\frac{3k}{n}}{3+\frac{k}{n}} \right\rfloor
$.
Therefore,
if $n \ge 9$,
$f(N) = 13$.
Since
$\lceil \frac{a}{b} \rceil
= \lfloor \frac{a+b-1}{b} \rfloor
$
if $a$ and $b$
are integers with
$a \ge 0$
and $b \ge 1$,
$\begin{array}\\
f(N)
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1093809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Calculate limit on series with nested sum I want to calculate the limit of following series:
$$\sum_{n=0}^{\infty} \sum_{k=0}^{n} \frac{1}{3^k} \cdot \frac{1}{2^{n-k}}$$
As far I could simply the series to:
$$\sum_{n=0}^{\infty} (\sum_{k=0}^{n} (\frac{1}{3})^k) \cdot (\sum_{k=0}^{n} 2^{n-k})$$
which would then allow me... | $$\sum^{\infty}_{n=0}\sum^{n}_{k=0}\frac{1}{3^k}\cdot\frac{1}{2^{n-k}}=\sum^{\infty}_{n=0}\sum^{n}_{k=0}\frac{1}{3^k}\cdot\frac{2^k}{2^{n}}=\sum^{\infty}_{n=0}\frac{1}{2^{n}}\sum^{n}_{k=0}\Big(\frac{2}{3}\Big)^k$$
Using the geometric series result for a convergent geometric series yields
$$\sum^{\infty}_{n=0}\frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1096394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
How to show $I_p(a,b) = \sum_{j=a}^{a+b-1}{a+b-1 \choose j} p^j(1-p)^{a+b-1-j}$
Show that $$I_p(a,b) = \frac{1}{B(a,b)}\int_0^p u^{a-1}(1-u)^{b-1}~du\\= \sum_{j=a}^{a+b-1}{a+b-1 \choose j} p^j(1-p)^{a+b-1-j}$$ when $a,b$ are positive integers.
I have no idea how to proceed. Please help.
| Hint: Use $\frac{1}{B(a,b)}=\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}=\frac{(a+b-1)!}{(a-1)!(b-1)!}$ and integration by parts to evaluate the integral:
$$I=\int_{0}^{1}u^{a-1}(1-u)^{b-1} = \left.\frac{1}{a}u^a(1-u)^{b-1}\right|_{0}^{p}+\frac{b-1}{a}\int_{0}^{p}u^{a}(1-u)^{b-2}\,du, $$
$$ I= \frac{1}{a}p^a(1-p)^{b-1}+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1096636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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How to solve $\int_0^\pi \frac{x\sin x}{1+ \sin^2 x}dx$? I have solved
$$\int_0^\pi \frac{x\sin x}{1+ \cos^2 x}dx=\pi^2/4$$. But I cannot solve it with the same way that I used for the upper problem. Can you help me to solve this problem?
$$\int_0^\pi \frac{x\sin x}{1+ \sin^2 x}dx$$
| We are going to evaluate the integral
$\displaystyle \int_{0}^{\pi} \frac{x \sin x}{1+\sin ^{2} x} d x \tag*{} $
by substitution, odd and even functions.
Let $\displaystyle y=x-\frac{\pi}{2},$ then
$\displaystyle \quad \int_{0}^{\pi} \frac{x \sin x}{1+\sin ^{2} x} d x \\$
$\displaystyle=\int_{-\frac{\pi}{2}}^{\frac{\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1100566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Limit $\lim \limits_{x \to 2}{\frac{\sqrt{x^3 - 3x^2 + 4}-x +2}{x^2 - 4}}$ I try to calculate $\lim \limits_{x \to 2}{\frac{\sqrt{x^3 - 3x^2 + 4}-x +2}{x^2 - 4}}$. So, $\frac{\sqrt{x^3 - 3x^2 + 4}-x +2}{x^2 - 4} = \frac{(x-2)x}{(x+2)(x+\sqrt{(x-2)^2(x+1)}-2)}$ but I don't know what to do next.
| No L'Hospital, too messy. You are almost there, rewrite the expression as
$$
\frac{(x-2)x}{(x+2)(x-2)(1 +\sqrt{\frac{(x-2)^2(x+1)}{(x-2)^2}}}
$$
what do you get?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1102589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
$(3/4) \cdot (8/9) \cdot (15/16) \cdot \ldots$ How to prove that limit is $1/2$ How can one prove $$\frac34 \cdot \frac89 \cdot \cdots \cdot \frac{n^2-1}{n^2}$$ has a limit $1/2$?
I tried some manipulations with the terms with no success and have just no idea,
Thanks
| Rewrite each $1-\frac{1}{k^2}$ as $\frac{(k-1)(k+1)}{k^2}$ and observe the mass cancellations.
$$\left(\frac{3}{2}\frac{4}{3}\frac{5}{4}\cdots\frac{n}{n-1}\frac{n+1}{n}\right)\cdot\left(\frac{1}{2}\frac{2}{3}\frac{3}{4}\cdots\frac{n-2}{n-1}\frac{n-1}{n}\right)$$
After mass cancellations, pull the $$\frac{n+1}{2}\text{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1105294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How can we say two algebraic expressions are "equal" if one is undefined at certain points and the other isn't? I'm trying to understand why it is that we can say $\frac{x^2-1}{x-1} = \frac{(x-1)(x+1)}{(x-1)} = x+1$ but then have it also be the case that the two functions $f(x) = \frac{x^2-1}{x-1}$ and $g(x)=x+1$ are n... | Let $f(x) = \frac{x^2-1}{x-1}$. The domain of $f(x)$ is $x \in \mathbb{R}-\{1\}$.
Actually, $$\frac{x^2 -1}{x-1} = x+1,$$ for all $x \neq 1$.
Hence, $f(x) \neq x+1$, because $1$ is in the domain of $x+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1108186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Invert a $2\times 2$ Matrix containing trig functions Invert the $2\times 2$ matrix:
\begin{bmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix}
My thought was to append the $2\times 2$ identity matrix to the right of the trig matrix and use row operations to get the answer.
I have to show... | Several people have posted ways to do this, but none did it the way the original posted proposed, i.e. row operations. Nor any with trigonometric identities. So here are those two methods:
\begin{align}
& \begin{bmatrix} \cos\theta & -\sin\theta & 1 & 0 \\ \sin\theta & \cos\theta & 0 & 1
\end{bmatrix} \\
\text{Now pu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1111399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Rewrite a circle's equation to easily see centre and radius $$x^{2}+y^{2}-5x-15y+30=0$$
I'm supposed to rewrite this equation so that you can easily see the centre and radius of the circle. I don't even know where to start. According to Mathematica the centre is $(5/2, 15/2)$ and the radius is $\sqrt{65/2}$.
| Answer:
$$x^2 +y^2 -5x-15y+30 = 0$$
$$x^2 -5x +(\frac{5}{2})^2 +y^2 -15x +(\frac{15}{2})^2 -(\frac{5}{2})^2-(\frac{15}{2})^2+30 = 0$$
$$(x-\frac{5}{2})^2+(y-\frac{15}{2})^2 = \frac{225+25}{4}-30$$
$$(x-\frac{5}{2})^2+(y-\frac{15}{2})^2 =(\sqrt{\frac{65}{2}})^2$$
Center$$ (\frac{5}{2},\frac{15}{2})$$and the radius $$= \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1111856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
convergence of $\sum_{n=1}^\infty \int_0^{\frac {1}{n}} \dfrac {\sqrt x}{1+x^2} dx$ Test convergence of $\sum_{n=1}^\infty \int_0^{\frac {1}{n}} \dfrac {\sqrt x}{1+x^2} dx$
Attempt: Since, $\lim_{n \rightarrow \infty} \dfrac {1}{n} \in (0,1) \implies \sum_{n=1}^\infty \int_0^{\frac {1}{n}} \dfrac {\sqrt x}{1+x^2} dx \g... | First note that the denominator is there just to confuse you. For sufficiently small $x$,
$$1\ge\frac1{1+x^2}\ge\frac12,$$
so we can disregard that altogether. Now
$$\sum_{n=1}^\infty\int_0^{1/n}\sqrt x\,dx=\frac23\sum_{n=1}^\infty \frac 1{n^{3/2}},$$
which converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1116307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Why it's true? $\arcsin(x) +\arccos(x) = \frac{\pi}{2}$ The following identity is true for any given $x \in [-1,1]$:
$$\arcsin(x) + \arccos(x) = \frac{\pi}{2}$$
But I don't know how to explain it.
I understand that the derivative of the equation is a truth clause, but why would the following be true, intuitively?
$$\in... | More simple....
From $\cos \alpha=\sin \left(\dfrac{\pi}{2}-\alpha\right)$ we have:
$$
\cos y=x \Rightarrow \sin\left(\dfrac{\pi}{2}-y\right)=x \Rightarrow
$$
$$
\Rightarrow
\begin{cases}
\arccos x=y \\
\arcsin x= \dfrac{\pi}{2}-y
\end{cases}
\Rightarrow
$$
$$
\Rightarrow
\arccos x+\arcsin x=\dfrac{\pi}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1116974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 0
} |
Convergence of $ \sum_{n=1} ^\infty \frac {1}{n(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n} )}$ Convergence of $$ \sum_{n=1} ^\infty \dfrac {1}{n(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n} )}$$
Attempt: I believe not a nice attempt: $ n(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n} ) \leq n( 1+1+... | The other way to look at it is, denoting $\displaystyle H_n = \sum\limits_{k=1}^{n} \frac{1}{k}$
The tail of the series: $$\displaystyle \frac{\frac{1}{n+1}}{H_{n+1}}+\frac{\frac{1}{n+2}}{H_{n+2}}+\cdots +\frac{\frac{1}{n+r}}{H_{n+r}} \ge \frac{\sum\limits_{k=1}^{r}\frac{1}{n+k}}{H_{n+r}} = \frac{H_{n+r}-H_{n}}{H_{n+r}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1119733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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If $a^2+b^2+c^2=1$ then prove the following. If $a^2+b^2+c^2=1$, prove that $\frac{-1}{2}\le\ ab+bc+ca\le 1$.
I was able to prove that $ ab+bc+ca\le 1$. But I am unable to gain an equation to prove that
$ \frac{-1}{2}\le\ ab+bc+ca$ .
Thanks in advance !
| If $ab+bc+ca < \dfrac{-1}{2} \Rightarrow 2(ab+bc+ca) + 1 < 0 \Rightarrow 2(ab+bc+ca) + a^2+b^2+c^2 < 0 \Rightarrow (a+b+c)^2 < 0$. Contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1119879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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Find the value of $ ( ab + bc + ca )^2 $ If $a,b,c$ are real numbers which satisfy
$a^2+b^2+ab = 9$
$b^2+c^2+bc = 16$
$c^2+a^2+ca = 25$
find the value of $ ( ab + bc + ca )^2 $
| let
since
$$\begin{cases}a^2-2a\cdot b\cos{\dfrac{2\pi}{3}}+b^2=9\\
b^2-2bc\cos{\dfrac{2\pi}{3}}+c^2=16\\
c^2-2ca\cos{\dfrac{2\pi}{3}}+a^2=25
\end{cases}
$$
then we let
$$PA=a,PB=b,PC=c,\angle APB=\angle BPC=\angle APC=\dfrac{2\pi}{3}$$
By cosine
$$|AB|^2=25,|BC|^2=16,|AC|^2=9$$
so $\angle C=\dfrac{\pi}{2}$,
then we ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 0
} |
Antiderivative of $\frac{\sqrt{4-x}}{x\sqrt{x}}$ I need help to find the antiderivative of the function $\displaystyle x \, \mapsto \, \frac{\sqrt{4-x}}{x\sqrt{x}}$ on $]0,4[$. I have tried the change of variables $u = \sqrt{4-x}$ but it didn't help. Which change of variables (or trick) am I not thinking of ?
| Let us try this:
$$\frac{4-x}x=u^2\implies x=\frac4{u^2+1}\implies dx=-\frac{8u}{(u^2+1)^2}\,du$$
and then
$$\int\frac1x\sqrt\frac{4-x}x\;dx=\int\frac{u^2+1}4\cdot u\cdot \left(-\frac{8u}{(u^2+1)^2}du\right)=-2\int\frac{u^2}{(u^2+1)^2}du=$$
$$=-\arctan u+\frac u{u^2+1}+C\longrightarrow-\arctan\sqrt\frac{4-x}x+\frac{\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1121559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
If $a^2+ab+b^2=c^2+cd+d^2$ then $a+b+c+d$ is not prime. Let $a,b,c,d$ be positive integers such that $a^2+ab+b^2=c^2+cd+d^2$. Show that $a+b+c+d$ is not prime.
My proof looks like this:
$(a+b)^2 - ab=(c+d)^2-cd$
$(a+b)^2 - (c+d)^2=ab-cd$
$(a+b+c+d)(a+b-c-d)=ab-cd$
$a+b+c+d=\frac{ab-cd}{a+b-c-d}$
I'd like to have $a+b+c... | It is necessary to use the formula. Diophantine equation $a^2+b^2=c^2+d^2$
For example this.
You can write a similar equation and solutions: $$a^2+ac+c^2=x^2+xy+y^2$$
Solutions have the form: $$a=q^2+k^2-p^2+kq$$
$$c=q^2+k^2+2p^2+kq-3pk-3pq$$ $$x=q^2-2k^2-p^2+3pk-2qk$$ $$y=k^2-2q^2-p^2+3pq-2qk$$
This means... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1122457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Solve the PDE by the method of characteristics. I am trying to figure out where my solution went wrong. I am off by a factor of two.
$$ u_x + u_y + u = e^{x+2y}$$
I first found that the characteristic curves are determined by $$\frac{dy}{dx} = 1 \implies y-x = C.$$
I then solved the ODE $$\frac{du}{dx} + u = e^{x+2y}$$... | $$\begin{align}
u_x + u_y + u &= e^{x + 2y} \\
\implies u_x + u_y &= e^{x + 2y} - u \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (*)\\
\end{align} $$
Setting $u = u(x(s),y(s))$ we find
$$\begin{align}
\frac{d}{ds} u &= \frac{\partial u}{\partial x} \cdot \frac{dx}{ds} + \frac{\partial u}{\partial y} \cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1124412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $a^2 + b^2 + c^2 \geq ab + bc + ca$ for all positive integers $a$, $b$, and $c$ Show that $a^2 + b^2 + c^2 \geq ab + bc + ca$ for all positive integers $a$, $b$, and $c$.
I am not sure how to approach this problem. Should I divide this problem into multiple cases based on whether $a$,$b$,$c$ and odd/even or i... | $0\le\dfrac12\left((a-b)^2+(b-c)^2+(c-a)^2\right)=\dfrac12\left(a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2\right)=a^2+b^2+c^2-ab-bc-ca\implies a^2+b^2+c^2\ge ab+bc+ca$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1125709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solving integral $\int \arcsin x \cos x dx$ Can anyone give me a hint how to solve $$\int \arcsin(x)\cos(x)dx ~?$$
| $\int\sin^{-1}x\cos x~dx$
$=\int\sin^{-1}x~d(\sin x)$
$=\sin^{-1}x\sin x-\int\sin x~d(\sin^{-1}x)$
$=\sin^{-1}x\sin x-\int\dfrac{\sin x}{\sqrt{1-x^2}}dx$
Let $x=\sin u$ ,
Then $dx=\cos u~du$
$\therefore\sin^{-1}x\sin x-\int\dfrac{\sin x}{\sqrt{1-x^2}}dx$
$=\sin^{-1}x\sin x-\int\sin\sin u~du$
$=\sin^{-1}x\sin x-\int\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1128772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find $\int_0^1 \frac{dx}{(1+x^n)^2\sqrt[n]{1+x^n}}$ Find $$\int_0^1 \frac{dx}{(1+x^n)^2\sqrt[n]{1+x^n}}$$
with $n \in \mathbb{N}$.
My tried:
I think that, I need to find the value of
$$I_1=\int_{0}^1 \frac{dx}{(1+x^n)\sqrt[n]{1+x^n}}$$
because:
$$\begin{align} I_1=\int_{0}^1 \frac{dx}{(1+x^n)\sqrt[n]{1+x^n}} &= \frac{... | Observe that
$$
\left(\frac{1}{(1+x^n)^{1/n}}\right)'=-\frac{x^{n-1}}{(1+x^n)^{1+1/n}} \tag1
$$
giving
$$
\begin{align}
\left(\frac{x}{(1+x^n)^{1/n}}\right)'=\left(x \times\frac{1}{(1+x^n)^{1/n}}\right)'=\frac{1}{(1+x^n)^{1/n}}-\frac{x \times x^{n-1}}{(1+x^n)^{1+1/n}}=\frac{1}{(1+x^n)^{1+1/n}}
\end{align}
$$
then
$$
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1128875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
no. of real roots of the equation $ 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7} = 0$
The no. of real roots of the equation $\displaystyle 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7} = 0 $
$\bf{My\; Try::}$ First we will find nature of graph of function $\displaystyle ... | Let
$$f(x)=1+\frac x1+\cdots+\frac{x^7}{7}$$
Assume that there are two different roots $a_1$ and $a_2, (a_1<a_2)$ of the equation $f(x)=0$ so by the mean value theorem we get for $a_3\in(a_1,a_2)$
$$0=f(a_1)-f(a_2)=(a_1-a_2)f'(a_3)\implies f'(a_3)=1+a_3+a_3^2+\cdots+a_3^6=0$$
and using the geometric sum
$$1+x+\cdots+x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1128932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Finding the argument $\theta$ of a complex number I want to find the Argument of $z = -\sqrt{2 - \sqrt{3}} + i\sqrt{2 + \sqrt{3}}$ where $z$ is a complex number of the form $z = a + bi$.
I find that the modulus is $2$, but am having trouble simplifying $\theta = \arctan\left(\frac{\sqrt{2 + \sqrt{3}}}{-\sqrt{2 - \sqrt{... | Note
$$\frac{1}{2}\sqrt{2 - \sqrt{3}} = \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{1 - \cos \frac{\pi}{6}}{2}} = \sin \frac{\pi}{12} $$
and similarly $$\frac{1}{2}\sqrt{2 + \sqrt{3}} = \cos \frac{\pi}{12}.$$ Therefore $$z = 2\left(-\sin \frac{\pi}{12} + i\cos \frac{\pi}{12}\right) = 2e^{i(\frac{\pi}{2} + \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1130834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
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.