Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Prove that an expression is divisible by a polynomial Question: Show that the polynomial $f(x)=(x+1)^{2n} +(x+2)^n - 1$
is divisible by $g(x) = x^2+3x+2$, where $n$ is an integer.
I have tried to use mathematical induction. The basis case wasn't that difficult, but when it comes to the inductive step itself, I got a ... | Putting the zeros of $x^2+3x+2=0$ i.e., $-1,-2$ one by one, in $f(x)=(x+1)^{2n}+(x+2)^n-1$
we get, $f(-1)=(-1+1)^{2n}+(-1+2)^n-1=0$
and $f(-2)=(-2+1)^{2n}+(-2+2)^n-1=0$
(i)So, using Remainder Theorem,
$(x+1)\mid f(x)$ and $(x+2)\mid f(x)\implies lcm(x+1,x+2)\mid f(x)$
(ii) Alternatively,
$\frac{(x+1)^{2n}+(x+2)^n-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/228332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Induction: $\sum_{k=0}^n \binom nk k^2 = n(1+n)2^{n-2}$ I found crazy (for me at least) induction example, in fact it just would be nice to prove. (Even have problems with starting) Any hints are highly valued: $$0^2\binom{n}{0}+1^2\binom{n}{1}+2^2\binom{n}{2}+\cdots+n^2\binom{n}{n}=n(1+n)2^{n-2} $$
| Although a year late, here is a proof using the properties
$$
k\binom{n}{k}=n\binom{n-1}{k-1}\tag{1}
$$
and
$$
\sum_{k=0}^n\binom{n}{k}=2^n\tag{2}
$$
Thus,
$$
\begin{align}
\sum_{k=0}^nk^2\binom{n}{k}
&=\sum_{k=0}^nnk\binom{n-1}{k-1}\tag{3}\\
&=\sum_{k=0}^nn\binom{n-1}{k-1}+\sum_{k=0}^nn(k-1)\binom{n-1}{k-1}\tag{4}\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/231596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 7,
"answer_id": 0
} |
Methods to solve a system of many Ax=B equations using least-squares I am working with a force measurement instrument which needs calibration via a calibration matrix. For each of a set of controlled measurements I have a vector $k$ of three known, independent values, and a corresponding vector $m$ containing three mea... | Let's denote $C$ as
\[ C = \begin{pmatrix} c_{11} & c_{12} & c_{13}\\ c_{21} & c_{22} & c_{23} \\ c_{31} & c_{32} & c_{33} \end{pmatrix} \]
We can write the equation $Cm_i = k_i$ as
\begin{align*}
m_{i1} c_{11} + m_{i2}c_{12} + m_{i3} c_{13} &= k_{i1}\\
m_{i1} c_{21} + m_{i2}c_{22} + m_{i3} c_{23} &= k_{i2}\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/232118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
absolute value integrated For example I have the function $f(x) = |x^2-1| = \sqrt{(x^2-1)^2}$
$\int \sqrt{(x^2-1)^2}dx = \frac{x(x^2-3)\sqrt{(x^2-1)^2)}}{3 (x^2-1)}+constant$
But plotted it looks like this: Plot 1. There are values < 0.
I have to take the absolute value again to get the correct function.
Plot 2 Why do ... | $x^2-1$ is positive outside of the interval $[-1,1]$.
Therefore,
$$|x^2-1| = \left\{ \begin{array}{rl} -x^2+1, & -1 \le x \le 1, \\ x^2-1, & \textrm{ otherwise}.\end{array}\right.$$
Or more completely
$$|x^2-1| = \left\{ \begin{array}{rl} x^2-1, & \textrm{ if } x < -1, \\
-x^2+1, & \textrm{ if } -1 \le x \le 1, \\ x^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/232451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Decompose a Matrix into a Sum of Tensor Products Given the matrix
$$\phantom{-}
\pmatrix{
0 & a+b& -b+a&0\\
\phantom{-}a+c& a+d&\phantom{-} b+c&b+d\\
-c+a&c+b&-d+a&d+b\\
0&c+d&\phantom{-}d-c&0\\
}.
$$
How could one decompose it into the (smallest) sum of tensor product, like
$M\otimes E_{k}$ and $E_{n}\otimes M$, whe... | You cannot decompose $$A=\phantom{-}
\pmatrix{
0 & a+b& -b+a&0\\
\phantom{-}a+c& a+d&\phantom{-} b+c&b+d\\
-c+a&c+b&-d+a&d+b\\
0&c+d&\phantom{-}d-c&0\\
}
$$
into $X\otimes M+M\otimes Y$. The system $A=X\otimes M+M\otimes Y$ for $X=\pmatrix{x_1&x_2\\x_3&x_4}, \ Y=\pmatrix{y_1&y_2\\y_3&y_4}$ does not have a solution in $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/233792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
A couple of formulas for $\pi$ While I was studying sums of polygonal numbers I discovered a couple of formulas for $\pi$. Most of the formulas I found were already known, but I can't seem to find any references to the following four:
$$\pi=\sum_{n=1}^\infty \frac{3}{n(2n-1)(4n-3)},$$
$$\pi=\sum_{n=1}^\infty \frac{3\sq... | The first formula may be rewritten by changing the lower limit of the summation:
$$\begin{align}
\frac{\pi}{3}
&=\sum_{n=1}^\infty \frac{1}{n(2n-1)(4n-3)} \\
&=\sum_{n=0}^\infty \frac{1}{(n+1)(2(n+1)-1)(4(n+1)-3)} \\
&=\sum_{n=0}^\infty \frac{1}{(n+1)(2n+1)(4n+1)} \\
\end{align}$$
This last equation is given by Lehmer ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/235657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
} |
Chebyshev polynomials of first kind I know the chebyshev polynomials of the first kind can be approximated using the cosine function, where $T_n(\cos \theta)=\cos(n \theta)$ and I know that chebyshev polynomials are a family of orthogonal polynomials. How would I prove that $$\int_{-1}^1 \frac{T_h(x)T_i(x)}{\sqrt{1-x^2... | Using Tschebyscheff differential equation in it's autoadjoint form we have, for $T_n$ and $T_m$
\begin{align}
\frac{d}{dx}\left\{\sqrt{1-x^2} T_n'\right\} + \frac{n^2}{\sqrt{1-x^2}} T_n &= 0 \\
\frac{d}{dx}\left\{\sqrt{1-x^2} T_m'\right\} + \frac{m^2}{\sqrt{1-x^2}} T_m &= 0
\end{align}
Multiplying the first equation by... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/238358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
If $(a, b, c)$ is a Pythagorean triple, then $(a+1, b+1, c+1)$ is not a Pythagorean triple. Pythagoras stated that there exist positive natural numbers, $a$, $b$ and $c$ such that $a^2+b^2=c^2$. These three numbers, $a$, $b$ and $c$ are collectively known as a
Pythagorean triple. For example, $(8, 15, 17)$ is one of th... | Suppose $a^2+b^2=c^2$ and $(a+1)^2+(b+1)^2=(c+1)^2$ both hold.
Simplifying the second equation and subtracting the first gives $$2a + 2b = 2c - 1$$ but this is impossible sso they can't both hold!
hint: think about even and odd numbers to complete the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/239312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Solve the recurrence $T(n) = 2T(n-1) + n$ Solve the recurrence $T(n) = 2T(n-1) + n$
where $T(1) = 1$ and $n\ge 2$.
The final answer is $2^{n+1}-n-2$
Can anyone arrive at the solution?
| By subtraction
$$
\begin{array}{rrcl}
& T(n)-2T(n-1) &=& n \\
& T(n-1) - 2T(n-2) &=& n-1 \\
\hline
& T(n)-3T(n-1)+2T(n-2) &=& 1
\end{array}$$
Again, by subtraction
$$
\begin{array}{rrcl}
& T(n)-3T(n-1)+2T(n-2) &=& 1 \\
& T(n-1)-3T(n-2)+ 2T(n-3) &=& 1 \\
\hline
& T(n)-4T(n-1)+5T(n-2)-2T(n-3) &=& 0
\end{array}$$
Characte... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/239974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 9,
"answer_id": 6
} |
Proving an equation with invertible matrices I need to prove that if $A$, $B$ and $(A + B)$ are invertible then $(A^{-1} + B^{-1})^{-1} = A(A + B)^{-1}B$
I'm a bit lost with this one,
I can't find a way to make any assumptions about $(A^{-1} + B^{-1})$,
Neither by using $A^{-1}$, $B^{-1}$ and $(A + B)^{-1}$.
If someone... | How about a purposedly complicated proof, but without matrix inversion?
Let $X = A(A + B)^{-1}B$. By writing the rightmost $B$ as $(A+B)-B$, we get
$$X = A(A + B)^{-1}[(A+B)-A] = \underbrace{A - A(A + B)^{-1}A}_{X_1}.$$
Similarly, if we write the leftmost $A$ as $(A+B)-B$ instead, we get
$$X = [(A+B)-B](A + B)^{-1}B = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/241287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Solve $x^2 + 10 = 15$ How do I solve the following equation?
$$x^2 + 10 = 15$$
Here's how I think this should be solved.
\begin{align*}
x^2 + 10 - 10 & = 15 - 10 \\
x^2 & = 15 - 10 \\
x^2 & = 5 \\
x & = \sqrt{5}
\end{align*}
I was thinking that the square root of 5 is iregular repeating 2.23606797749979 number. 2.236 m... | The idea is there, however be aware of the fact that $x^2 = a$ is equivalent to $x = \sqrt{a}$ OR $x = -\sqrt{a}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/246071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How do you prove a Quadratic Residue? Help Someone please.
Prove or disprove: 28 is a quadratic residue (mod 65).
| Recall that if $a$ and $m$ are relatively prime, then $a$ is a QR of $m$ if the congruence
$$x^2\equiv a \pmod{m}$$
has a solution.
If the congruence $x^2\equiv 28\pmod{65}$ has a solution, then so do the congruences $x^2\equiv 28\pmod{5}$ and $x^2\equiv 28\pmod{13}$. Indeed the converse also holds, by the Chinese Re... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/246317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What am I doing wrong in calculating this determinant? I have matrix:
$$
A = \begin{bmatrix}
1 & 2 & 3 & 4 \\
2 & 3 & 3 & 3 \\
0 & 1 & 2 & 3 \\
0 & 0 & 1 & 2
\end{bmatrix}
$$
And I want to calculate $\det{A}$, so I have written:
$$
\begin{array}{|cccc|ccc}
1 & 2 & 3 & 4 & 1 & 2 & 3 \\
2 & 3 & 3 & 3 & 2 & 3 & 3 \\
0 & ... | Sarrus's rule works only for $3\times 3$-determinants. So you have to find another way to compute $\det A$, for example you can apply elementary transformations not changing the determinant, that is e. g. adding the multiple of one row to another:
\begin{align*}
\det \begin{bmatrix}
1 & 2 & 3 & 4 \\
2 & 3 & 3 & 3 \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/246606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
p(n) is count of all n-digit numbers... Let $n$ be an arbitrary positive integer, and $p(n)$ be the number of $n$-digit numbers which consist only of the digits $1,2,3,4,5$, and in which each two neighbour digits differ by $2$ or more.
My task is to prove that this inequality is true for any positive integer $n$:
$$\ 5... | Just an idea (not a complete answer). Let $p_k(n)$ be the number of $n$-digit numbers of the requested form that end in the digit $k$. Then $p_k(1) = 1$ for all $k \in \{1, \dotsc, 5\}$ and $p_k(n+1)$ can be computed recursively by
$$
\begin{pmatrix}
p_1(n+1) \\
p_2(n+1) \\
p_3(n+1) \\
p_4(n+1) \\
p_5(n+1)
\end{pmatr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/249342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Why does $\frac{a}{\frac{b}{x}} = x \times \frac{a}{b}$? As much as it embarasses me to say it, but I always had a hard time understanding the following equality:
$$
\frac{a}{\frac{b}{x}} = x \times \frac{a}{b}
$$
I always thought that the left-hand side of the above equation was equivalent to
$$
\frac{a}{\frac{b}{x}} ... | $\frac{a}{\frac{b}{x}}$ means a divided by $\frac{b}{x}$.
Note that
$$\frac{b}{x}\frac{x}{b} =1$$
This means that
$$a \frac{b}{x}\frac{x}{b} = a$$
Now divide both sides by $\frac{b}{x}$ and you get
$$a\frac{x}{b} = \frac{a}{ \frac{b}{x}}$$
The mistake you make is confusing $\frac{a}{\frac{b}{x}}$ with $\frac{\frac{a}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/251317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 3
} |
Show $2(x+y+z)-xyz\leq 10$ if $x^2+y^2+z^2=9$ If $x,y,z$ are real and $x^2+y^2+z^2=9$, how can we prove that $2(x+y+z)-xyz\leq 10$?
Please provide a solution without the use of calculus. I know the solution in that way.
| without loss of generality we can assume that $|x|\le |y|\le |z|$,and Note that $3z^2\ge x^2+y^2+z^2$, which implies that $z^2\ge 3$
we have
\begin{align}
[2(x+y+z)-xyz]^2&=[2(x+y)+z(2-xy)]^2\\
&\le [(x+y)^2+z^2][4+(2-xy)^2]\\
&=(9+2xy)(8-4xy+(xy)^2))\\
&=2(xy)^3+(xy)^2-20(xy)+72\\
&=(xy+2)^2(2xy-7)+100
\end{align}
fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/252178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Property of Fejer kernel Let
$$
F_n(x) = \frac{1}{n} \left( \frac{ \sin(\frac{1}{2} n x ) } { \sin(\frac{1}{2} x ) } \right)^2
$$
be the n-th Fejer-Kernel. Then
$$
\forall \epsilon > 0, r < \pi : \exists N \in \mathbb{N} : \forall n \ge N : \int_{[-\pi,\pi] \setminus [-r, r]} F_n(t) \, \mathrm{d} t < \epsilon
$$
The... | Note that you're integrating over the set $r < |x| \le \pi$.
Let $f(x) = \sin^2(\frac{1}{2}x)$. Differentiating, we get that $f'(x) = \frac{\sin(x)}{2}$. Now if $x \in (0,\pi]$, then $\sin(x) > 0$, and so $f$ is increasing on $(0,\pi]$. Thus $\sin^2(\frac{1}{2} r) < \sin^2(\frac{1}{2}x)$ for $0 < r < x \le \pi$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/255002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Can't argue with success? Looking for "bad math" that "gets away with it" I'm looking for cases of invalid math operations producing (in spite of it all) correct results (aka "every math teacher's nightmare").
One example would be "cancelling" the 6's in
$$\frac{64}{16}.$$
Another one would be something like
$$\frac{9}... | The problem : solve the equation $$e^{(x-2)}+e^{(x+8)}=e^{(4-x)}+e^{(3x+2)}$$
Incorrect method
As everybody knows $$e^a+e^b=e^{ab}$$
So let's apply it to solve the equation :
$\begin{array}{lll}
e^{(x-2)}+e^{(x+8)}=e^{(4-x)}+e^{(3x+2)} & \iff & e^{(x-2)(x+8)}=e^{(4-x)(3x+2)} \\
& \iff &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/260656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "289",
"answer_count": 42,
"answer_id": 26
} |
What kind of software can be used to solves systems of equations? For example, I have to solve the following equations:
$$\left\{\begin{align*}
&x^2 + y^2 + z^2 = 1\\
&Ax + By + Cz = 0
\end{align*}\right.$$
for $y$ and $z$, where $x$, $A$, $B$ and $C$ are known values. I need this for production. I guess I could solve ... | Hot off the Mathematica presses:
> Solve[{x^2 + y^2 + z^2 - 1 == 0, A x + B y + C z == 0}, {y, z}] // Simplify
$$\begin{align}
\Bigg\{\Bigg\{y & \to -\frac{A B x+\sqrt{-C^2 \left(A^2 x^2+B^2 \left(-1+x^2\right)+C^2 \left(-1+x^2\right)\right)}}{B^2+C^2}, \\
z & \to \frac{-A C^2 x+B \sqrt{-C^2 \left(A^2 x^2+B^2 \left(-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/261299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Fourier transform of $f(x)=\frac{1}{x^2+6x+13}$ How to find the Fourier transform of the following function:
$$f(x)=\frac{1}{x^2+6x+13}$$
| Breaking down to partial fractions yields
$$
\begin{align}
\frac1{x^2+6x+13}
&=\frac1{(x+3)^2+4}\\
&=\frac1{(x+3-2i)(x+3+2i)}\\
&=\frac1{4i}\left(\frac1{x+3-2i}-\frac1{x+3+2i}\right)
\end{align}
$$
Therefore, for $\xi\ge0$,
$$
\begin{align}
\int_{-\infty}^\infty\frac1{x^2+6x+13}e^{-i2\pi x\xi}\,\mathrm{d}x
&=\frac1{4i}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/262321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Prove every odd integer is the difference of two squares I know that I should use the definition of an odd integer ($2k+1$), but that's about it.
Thanks in advance!
| Proof for odd numbers: Divide any odd number, $x$, in half, then let $C$ be the half rounded up and $B$ be the half rounded down. Then, $C^2 - B^2 = x$. Let's examine this algebraically:
$$C = \frac{x+1}{2}$$
$$B = \frac{x-1}{2}$$
$$C^2 - B^2 = \biggl(\frac{x+1}{2}\biggr)^2 - \biggl(\frac{x-1}{2}\biggr)^2 = \frac{x^2+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/263101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "64",
"answer_count": 9,
"answer_id": 8
} |
What is the proper convergence proof for this sequence? Limit of $n^{1/2}(n^{1/n}-1)=0$, as $n$ approaches infinity.
I need a strict math proof that this sequence converges to zero, but without using L'Hôpital's rule, because I am not allowed to use it yet. Thanks in advance.
| Denote $a_n=n^\tfrac{1}{n}-1, \;\; b_n={\sqrt{n}\left(n^\tfrac{1}{n}-1\right)}.$
Obviously, $a_n \geqslant {0}.$ Rewrite $n^{\tfrac{1}{n}}=1+a_n $ and apply the binomial formula
\begin{gather}n=(1+a_n)^n= \\
=1+na_n+\dfrac{n(n-1)}{2!}a_n^2+\dfrac{n(n-1)(n-2)}{3!}a_n^3+\ldots+a_n^n \geqslant \\
\geqslant \dfrac{n(n-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/264743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\cot^2{(\pi/7)} + \cot^2{(2\pi/7)} + \cot^2{(3\pi/7)} = 5$ Prove that $\cot^2{(\pi/7)} + \cot^2{(2\pi/7)} + \cot^2{(3\pi/7)} = 5$ .
I am sure this is derived from using roots of unity and Euler's complex number function, but I am very uncomfortable in these areas so some help would be great. It is evident t... | The following approach is intimately related to Ofir's; you can think of this as the linear algebraic variation of his route.
Consider the perturbed Toeplitz tridiagonal matrix
$$\begin{pmatrix}2&-1&&&\\-1&2&-1&&\\&-1&\ddots&\ddots&\\&&\ddots&2&-1\\&&&-1&1\end{pmatrix}$$
which has the characteristic polynomial
$$(-1)^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/265229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
where is wrong in the sum of series $\frac{1}{3}+\frac{1}{4}\cdot\frac{1}{2!}+\frac{1}{5}\cdot\frac{1}{3!}+\cdots$ I came through two types of solutions of the series $\frac{1}{3}+\frac{1}{4}\cdot\frac{1}{2!}+\frac{1}{5}\cdot\frac{1}{3!}+\cdots$
$$\begin{align*}
\frac{1}{3}+\frac{1}{4}\cdot\frac{1}{2!}+\frac{1}{5}\cdot... | You have $$\int^{1}_{0}x(e^{x}-1)dx=[e^{x}(x-1)-\frac{x^{2}}{2}]|^{1}_{0}=\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/266562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to compute the determinant of a tridiagonal matrix with constant diagonals? How to show that the determinant of the following $(n\times n)$ matrix
$$\begin{pmatrix}
5 & 2 & 0 & 0 & 0 & \cdots & 0 \\
2 & 5 & 2 & 0 & 0 & \cdots & 0 \\
0 & 2 & 5 & 2 & 0 & \cdots & 0 \\
\vdots & \vdots& \vdots& \vdots & \vdots & \vdots... | We will generalize Calvin Lin's answer a bit. Let
$$A_n = \begin{bmatrix} a & b & 0 & 0 & \cdots & 0\\ c & a & b & 0 & \cdots & 0\\ 0 & c & a & b & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \cdots & 0\\ \vdots & \vdots & \vdots & \vdots & a & b\\ 0 & 0 & 0 & \cdots & c & a \end{bmatrix}.$$
We then have, by using... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/266998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 5,
"answer_id": 2
} |
$\frac{n}{2} < 1+\frac{1}{2} + \frac{1}{3}+ \ldots +\frac{1}{2^n-1}How to show that for $n\geqslant 2$
$$\frac{n}{2} < 1+\frac{1}{2} + \frac{1}{3}+ \ldots +\frac{1}{2^n-1}<n$$
| HINT: $$\frac12<2^n\left(\frac1{2^{n+1}-1}\right)<\sum_{k=2^n}^{2^{n+1}-1}\frac1k<2^n\left(\frac1{2^n}\right)=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/267191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Smart demonstration to the formula $ \sum_{n=1} ^{N} \frac{n}{2^n} = \frac{-N + 2^{N+1}-2}{2^n}$ Someone could give me a smart and simple solution to show the folowing identity?
$$ \sum_{n=1} ^{N} \frac{n}{2^n} = \frac{-N + 2^{N+1}-2}{2^n}$$
| Proceed by induction.
Base case is true with $N=1$.
Inductive (I replaced the actual sum with $\sum$):
$$ \frac{N+1}{2^{N+1}} + \sum^N = \sum^{N+1} $$
$$ \frac{N+1}{2^{N+1}} + \frac{-N+2^{N+1}-2}{2^N} = \frac{-(N+1) + 2^{N+2}-2}{2^{N+1}} $$
$$ \frac{N + 1 - 2N + 2^{N+2} - 4}{2^{N+1}} = \frac{-N-3+2^{N+2}}{2^{N+1}} $$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/268434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Alternate proof that for every natural number $n,\ \left\lfloor\left(\frac{7+\sqrt{37}}{2}\right)^n\right\rfloor$ is divisible by $3$ Original Problem:
Prove that for every natural number $n$,$$\left\lfloor\left(\frac{7+\sqrt{37}}{2}\right)^n\right\rfloor$$ is divisible by $3$.
I found the problem in the book Winning... | Considering the number $\frac{7+\sqrt{37}}{2}$ and its quadratic conjugate $\frac{7-\sqrt{37}}{2}$:
$$
\left(\frac{7+\sqrt{37}}{2}\right)\left(\frac{7-\sqrt{37}}{2}\right)=\frac{49-37}{4}=3\tag{1}
$$
$$
\left(\frac{7+\sqrt{37}}{2}\right)+\left(\frac{7-\sqrt{37}}{2}\right)=7\tag{2}
$$
Thus, we get that both of these con... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/270157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 1
} |
Evaluating $\int^{\infty}_{0}{\frac{\ln x}{(1+x^2)^2}dx}$ $$\int^{\infty}_{0}{\frac{\ln x}{(1+x^2)^2}dx}$$
I've done a similar one: $\int^{\infty}_{1}{\frac{\ln^n x}{x^2}dx}$ using IBP, but in this case I've tried IBP multiple times, differentiating all of the possible choices, and it's only getting more convoluted. Al... | Let $x=1/y$. We then get
$$I = \int_0^{\infty} \dfrac{\ln(x)}{(1+x^2)^2} dx = \int_0^{1} \dfrac{\ln(x)}{(1+x^2)^2} dx + \int_1^{\infty} \dfrac{\ln(x)}{(1+x^2)^2} dx\\
= \int_0^{1} \dfrac{\ln(x)}{(1+x^2)^2} dx - \underbrace{\int_0^{1} \dfrac{x^2\ln(x)}{(1+x^2)^2} dx}_{x \to 1/x}\\
= \int_0^{1} (1-x^2)\left(\sum_{k=0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/271097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 0
} |
Algebra question involving fractions How would I perform the indicated operation.
$$\frac{t+2}{t^2+5t+6}+\frac{t-1}{t^2+7t+12}-\frac{2}{t+4}.$$
I simplified it to
$$
\frac{t+2}{(t+3)(t+2)} + \frac{t-1}{(t+4)(t+3)}-\frac{2}{t+4}.
$$
Then I did the lowest common denominator but I still have problems.
According to my boo... | Note that $\dfrac{t+2}{t^2+5t+6} = \dfrac{t+2}{(t+2)(t+3)} = \dfrac1{t+3}$ for $t \neq -2$.
Hence, $\dfrac{t+2}{t^2+5t+6} - \dfrac2{t+4} = \dfrac1{t+3} - \dfrac2{t+4} = \dfrac{t+4-2t-6}{(t+3)(t+4)} = - \dfrac{t+2}{(t+3)(t+4)}$.
Hence, $$\dfrac{t+2}{t^2+5t+6} - \dfrac2{t+4} + \dfrac{t-1}{t^2+7t+12} = \dfrac{t-1}{t^2+7t+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/274912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Understanding the solution of $\int\left(1-x^{p}\right)^{\frac{n-1}{p}}\log\left(1-x^{p}\right)dx$ I would like to solve the integral
$$\int\left(1-x^{p}\right)^{\frac{n-1}{p}}\log\left(1-x^{p}\right)dx.$$
My problem is that I arrived at a solution via wolfram alpha, but I would like to understand how one would arrive... | You made a substitution $u=1-x^p$, transforming the integral as
$$
\int \left(1-x^p\right)^{(n-1)/p} \log\left(1-x^p\right) \mathrm{d} x = - \frac{1}{p} \int \left(1-u\right)^a u^b \log\left(u\right) \mathrm{d} u
$$
where $a = \frac{1}{p}-1$ and $b=\frac{n-1}{p}$. In order to integrate this we shall take advantage... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/275158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
a question on $O$- notation I'm trying to understand $O$-notationbetter. I've found a really helpful answer : Big O notation, $1/(1-x)$ series
But I have trouble with quotient. Let me discuss about this example function:
$$f(x)=a+bx+\frac{c+dx}{e+fx^2}e^{ax}$$
If I calculate for the zero order error $O(x^0)$, I'll get
... | Taylor series get you a long way in practice. Using the Taylor series for the exponential function and $\frac{1}{1 + x}$ you can get as many order terms exactly as you need
$$\begin{align} f(x) &=a+bx+\frac{c+dx}{e+fx^2}e^{ax} \\
&= a + bx + \frac{1}{e} \cdot \frac{c+dx}{1 + \frac{f}{e}x^2} \cdot \left(1 + ax + \frac{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/277264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
What will be the one's digit of the remainder in: $\left|5555^{2222} + 2222^{5555}\right|\div 7=?$ What will be the ones digit of the remainder in: $$\frac{\left|5555^{2222} + 2222^{5555}\right|} {7}$$
| Hint: $ 5555 \equiv 4 \pmod{7}$, $2222 \equiv 3 \pmod{7}$.
Edit: Calculate that $4^3 \equiv 1, 3^6 \equiv 1 \pmod{7}$. Hence, this implies that $4^{3k} \equiv (4^3)^k \equiv 1^k \equiv 1 \pmod{7}, 3^{6j} \equiv (3^6)^j \equiv 1^j \equiv \pmod{7}$.
Now, $2222 \equiv 2 \pmod{3}$, and $5555 \equiv 5 \pmod{6}$. Hence,
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/279333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
find the sum of the alternating series How to find the sum of the infinite series
$$\frac{1}{12}-\frac{1\cdot 4}{12 \cdot 18 } + \frac{1\cdot 4\cdot 7}{12\cdot 18\cdot 24} - \frac{1 \cdot 4 \cdot 7\cdot 10}{12 \cdot 18 \cdot 24 \cdot 30}+...$$
I understood the answer posted in Yahoo Answer till the last but one step:
T... | We can rewrite the series as
$$
\begin{align}
-3\sum_{n=2}^\infty\binom{2/3}{n}(1/2)^n
&=-3\left((1+1/2)^{2/3}-1-1/3\right)\\
&=4-3\ (3/2)^{2/3}
\end{align}
$$
A Bit of Explanation
By the Generalized Binomial Theorem, we have
$$
\sum_{n=0}^\infty\binom{2/3}{n}(1/2)^n=(1+1/2)^{2/3}
$$
The first two terms are $\binom{2/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/280849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Green's function for $\frac{d^2}{dx^2} + \frac{1}{4}$ I am trying to solve the following differential equation $$y''+\frac{1}{4}y=\sin(2x)~~~y(0)=y(\pi)=0,$$ using Green's function. I have found the Green's function for the operator $y''+\frac{1}{4}$ to be $$G(x,\xi)= \sin\left(\frac{1}{2}(x-\xi)\right).$$ However, whe... | Let's start from the beginning. You want to express the problem as solving for a function $G(x,y)$ that satisfies, for $x \in [0,\pi]$
$$\frac{d}{dx^2} G(x,x') + \frac{1}{4} G(x,x') = \delta(x-x') $$
where $\delta(x-x') = 0 \, \forall \, x \ne x'$ and $\int_0^{\pi} dx' \: \delta(x-x') = 1 $. The solution you seek, $y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/284870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
limit of the sum $\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n} $ Prove that : $\displaystyle \lim_{n\to \infty} \frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n}=\ln 2$
the only thing I could think of is that it can be written like this :
$$ \lim_{n\to \infty} \sum_{k=1}^n \frac{1}{k+n} =\lim_{n\to \inf... | Set $a_n
= 1/(n+1) + 1/(n+2)+···+ 1/(2n)
=\sum_{k=1}^n \dfrac1{n+k}
$.
$\begin{array}\\
a_{n+1}-a_n
&=\sum_{k=1}^{n+1} \dfrac1{n+1+k}-\sum_{k=1}^n \dfrac1{n+k}\\
&=\sum_{k=2}^{n+2} \dfrac1{n+k}-\sum_{k=1}^n \dfrac1{n+k}\\
&=\dfrac1{n+n+2}+\dfrac1{n+n+1}-\dfrac1{n+1}\\
&=\dfrac1{2n+2}+\dfrac1{2n+1}-\dfrac1{n+1}\\
&=-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/285308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 7,
"answer_id": 6
} |
How to transform $2^{n-2}\frac{(2n-5)(2n-7)...(3)(1)}{(n-1)(n-2)...(3)(2)(1)}$ into $\frac{1}{n-1}\binom{2n-4}{n-2}$? Just an algebraic step within the well known solution for the number of triangulations of a convex polygon!
| First, rewrite the denominator as $(n-1)\cdot(n-2)!$, then multiply numerator and denominator by $(n-2)!$ :
$$2^{n-2}\frac{(2n-5)(2n-7)\ldots(3)(1)(n-2)(n-3)\ldots(2)(1)}{(n-1)(n-2)!(n-2)!}$$
Next, distribute the $n-2$ factors of $2$ over the $n-2$ 'elements' of $(n-2)!$ in the numerator:
$$\frac{(2n-5)(2n-7)\ldots(3)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/286884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Integration by parts $ \int \sqrt{x-x^2} dx $ I have to integrate $$ \int \sqrt{x-x^2} dx $$
The answer on my textbook is $ \frac 14 \left( \arcsin(\sqrt x)-(1-2x)\sqrt{x-x^2} \right) $ but I want to solve this by myself so can you please give me a clue ? :) Thank you.
| for calculation of Integral $\displaystyle \int \sqrt{a^2-x^2}dx$
We will Use Integration by parts method
Let $\displaystyle \mathbb{I} = \int \sqrt{a^2-x^2}.xdx = \sqrt{a^2-x^2}.x-\int\frac{1}{2\sqrt{a^2-x^2}}.-2x.xdx$
$\displaystyle \mathbb{I}= x.\sqrt{a^2-x^2}-\int\frac{(a^2-x^2)-a^2}{\sqrt{a^2-x^2}}dx$
$\displaysty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/287238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Calculating the limit $\lim \limits_{n \to \infty} \sqrt[n]{4n + \sin \sqrt{n} + \cos (\frac{1}{n^2}) + 17}$ I am trying to calculate this limit:
$$
\lim \limits_{n \to \infty} \sqrt[n]{4n + \sin \sqrt{n} + \cos (\tfrac{1}{n^2}) + 17}$$
I understand I should use squeeze theorem but I am having some trouble applying it ... | $\lim \limits_{n \to +\infty} \sqrt[n]{4n + \sin \sqrt{n} + \cos (\tfrac{1}{n^2}) + 17}$
$4n+15 \le 4n + \sin \sqrt{n} + \cos (\tfrac{1}{n^2}) + 17 \le 4n+19$
$\sqrt[n]{4n+15} \le \sqrt[n]{4n + \sin \sqrt{n} + \cos (\tfrac{1}{n^2}) + 17} \le \sqrt[n]{4n+19}$
$\lim \limits_{n \to +\infty} \sqrt[n]{4n+15} \le \lim \limit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/287358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Is the sum always bigger than $n^2$? Let $s(n)$ an arithmetical function defined as
$$s(n)=(p_1+1)^{e_1} (p_2+1)^{e_2} \cdots (p_m+1)^{e_m}$$
where prime factorization of $n$ is $n=p_1^ {e_1} p_2 ^{e_2} \cdots p_m^{e_m}$.
(For example, $s(49)=s(7^2)=8^2=64$, $s(60)=s(2^2 \times 3^1 \times 5^1)=3^2 \times 4^1 \times 6^1... | This is true. If $\nu_p(n)$ is maximal such that $p^{\nu_p(n)}\mid n$, then
$$
s(n)\ge (\frac32)^{\nu_2(n)} (\frac43)^{\nu_3(n)}n.
$$
Therefore, for all $i=1$, $\dots$, $2^5 3^3$,
$$
s(2^5 3^3n+i)\ge (\frac32)^{\min(\nu_2(i),5)} (\frac43)^{\min(\nu_3(i),3)} (2^5 3^3 n+i).
$$
Summing over $i=1$, $\dots$, $2^5 3^3$ give... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/287862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Bell Numbers: How to put EGF $e^{e^x-1}$ into a series? I'm working on exponential generating functions, especially on the EGF for the Bell numbers $B_n$.
I found on the internet the EGF $f(x)=e^{e^x-1}$ for Bell numbers. Now I tried to use this EGF to compute $B_3$ (should be 15). I know that I have to put the EGF int... | You probably don't want to use the series expansion for the exponential as you've done, because you will receive contributions to each coefficient from all orders of this expansion, so you'll still have infinite sums left to evaluate. Instead, you should use the fact that term $n$ in a sequence is given by $F^{(n)}(0)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/289097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
primes of the form $a^2+b^2=x^2-xy+y^2$? Let $a,b,x,y$ be strict positive integers.
Im intrested in primes $p$ such that $p=a^2+b^2=x^2-xy+y^2$.
What is the analogue PNT for these type of primes ? I think these primes are all the primes $p \equiv 1 \pmod{12}$.
| By Fermat's theorem on sums of two squares, $p$ can be written as $a^2+b^2$ iff $p \equiv 1 \pmod 4$ or $p = 2$. By this question, $p$ can be written as $x^2-xy+y^2$ iff $p \equiv 1 \pmod 3$ or $p = 3$. A prime $p$ satisfies both if and only if $p \equiv 1 \pmod {12}$, by the Chinese Remainder Theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/291696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proving $ (1)(2) + (2)(3) + (3)(4) + \cdots+ (n) (n+1) = \frac{(n)(n+1)(n+2)}{3} $ by induction I have the Following Proof By Induction Question:
$$ (1)(2) + (2)(3) + (3)(4) + \cdots+ (n) (n+1) = \frac{(n)(n+1)(n+2)}{3} $$
Can Anybody Tell Me What I'm Missing.
This is where I've Gone So Far.
Show Truth for N = 1
LHS =... | Your proof is fine, but you should show clearly how you got to the last expression.
$\dfrac{k(k+1)(k+2)}{3}+(k+1)(k+2)$
$=\dfrac{k}{3}(k+1)(k+2)+(k+1)(k+2)$
$=(\dfrac{k}{3}+1)(k+1)(k+2)$
$=\dfrac{k+3}{3}(k+1)(k+2)$
$=\dfrac{(k+1)(k+2)(k+3)}{3}$.
You should also word your proof clearly. For example, you can say "Let $P(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/292775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find the spanning set of the range of the linear transformation $T(x)=Ax$. Let
$$
A=
\begin{bmatrix}
-4 & -4 & 12 & 0 \\
-4 & -4 & 12 & 0 \\
4 & -2 & 0 &-6 \\
1 &-4 &7 &-5 \\
\end{bmatrix}
$$
Find the spanning set of the range of the linear transformation $T(x)=A... | It $T(v)=Av,~~~v\in\mathbb K^4$, then $$
T(v)=
\begin{pmatrix}
1 & 0 & -1 & -1 \\
0 & 1 & -2 & 1 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{pmatrix}\begin{pmatrix}
x \\
y \\
z \\
t \\
\end{pmatrix}=\begin{pmatrix}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/294344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find the maximum and minimum of $\cos x\sin y\cos z$.
Given $x\geq y\geq z\geq\pi/12$, $x+y+z=\pi/2$, find the maximum and minimum of $\cos x\sin y\cos z$.
I tried using turn $\sin y$ to $\cos(x+z)$, and Jensen Inequality, but filed. Please help. Thank you.
*p.s. I'm seeking for a solution without calculus.
| Let $$P=\cos x\sin y\cos z=\frac{\cos z}{2}\bigg[2\cos x \sin y\bigg] = \frac{\cos z}{2}\bigg[\sin(x+y)-\sin(x-y)\bigg]\leq \frac{\cos z}{2}\cdot \sin (x+y)$$
So $$P\leq \frac{\cos z \cdot \cos z}{2}=\frac{1}{4}(1+\cos 2z)\leq \frac{1}{4}\bigg[1+\cos 2\cdot \frac{\pi}{12}\bigg] = \frac{2+\sqrt{3}}{8}$$
Above equality h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/294402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Deriving $e^{ix} = 1 + ix + \frac{(ix)^2}{2!}+...$ I know how to prove equation:
$$e = \lim \limits_{x \rightarrow \infty} \left( 1 + \frac{1}{x} \right)^x$$
How can I now derive the series:
$$e^{ix} = 1 + ix + \frac{(ix)^2}{2!}+\frac{(ix)^3}{3!}+...$$
Those two seem very similar to me...
| Use the binomial theorem on the following limit:
$$e^z = \lim_{x \rightarrow \infty} \left( 1 + \frac{z}{x} \right)^x$$
i.e.,
$$\begin{align} \left( 1 + \frac{z}{x} \right)^x &= 1 + \frac{x}{1!} \frac{z}{x} + \frac{x (x-1)}{2!} \left (\frac{z}{x} \right )^2 + \frac{x (x-1) (x-2)}{3!} \left (\frac{z}{x} \right )^3 + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/294855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Show $\mathbb{Q}[\sqrt[3]{2}]$ is a field by rationalizing I need to rationalize $\displaystyle\frac{1}{a+b\sqrt[3]2 + c(\sqrt[3]2)^2}$
I'm given what I need to rationalize it, namely $\displaystyle\frac{(a^2-2bc)+(-ab+2c^2)\sqrt[3]2+(b^2-ac)\sqrt[3]2^2}{(a^2-2bc)+(-ab+2c^2)\sqrt[3]2+(b^2-ac)\sqrt[3]2^2}$
But I fail to... | You can multiply it by $1+d\sqrt[3]2+e\sqrt[3]4$ and insist the product not have any factors of $\sqrt[3]2$ or $\sqrt[3]4$. So
$$(a+b\sqrt[3]2+c\sqrt[3]4)(1+d\sqrt[3]2+e\sqrt[3]4)=\\a+ebe+2cd+\sqrt[3]2(ad+b+2ce)+\sqrt[3]4(ae+bd+c)$$
So we need $ad+b+2ce=0, ae+bd+c=0$, giving $$e=-\frac {ad+b}{2c}\\-a^2d-ab+2cbd+2c^2=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/294993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How do I integrate $\int \frac{x+1}{2\sqrt{x+1}} \mathrm dx$? I know how to do the bottom part, but I can't figure it out how to get $x+1$ on top
$$\int \frac{x+1}{2\sqrt{x+1}} \mathrm dx$$
| $$\int \frac {x+1}{2(x+1)^{1/2}} = \int\frac{x+1}{2\sqrt{x+1}}\,dx.$$ Let $u=(x+1)^{1/2}$, so $du=\dfrac{1}{2(x+1)^{1/2}}\,dx.\;$ Substitution gives us $$\int u^2\,du\;\;=\;\;\frac{u^3}{3}+C.$$
Back substitute, and our answer is $$\frac {((x+1)^{1/2})^3}{3} \;=\;\frac{(x+1)^{3/2}}{3}+C.$$
This can be approached in a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/295815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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probability distribution $X, Y$ and $X+Y$ A box contains $5$ ticket, $\{ 0 , 0 , 0 , 4 , 4\}$.
Drawing two tickets at random w/o replacement.
$X$ be the sum of the first two draws and Y be the outcome of the first draw.
Question: Find distribution of $X + Y$?
*
*What I did --
I found the distribution of $X$. The... | For the distribution of $X$, I do not think there is a substantially better way than yours.
For the distribution of $X+Y$, we do something similar. If it helps to keep track of things, draw a tree diagram. Let $W=X+Y$.
Case $1$: Maybe the first pick was $0$. This has probability $\frac{3}{5}$. Given that this happene... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplifying $\frac{1}{x} + \frac{5+x}{x+1} - \frac{7x^2 + 3}{(x+2)^2}$ I'm having trouble simplifying this expression:
$$\frac{1}{x} + \frac{5+x}{(x+1)} - \frac{7x^2 + 3}{(x+2)^2}$$
Would you first do the addition or subtraction?
What's the steps to solve this?
The final answer is
$$\frac{-6x^4 + 3x^3 + 26x^2 + 25x + 4... | HINT
The guiding idea is the same as when you're evaluating $\frac{1}{3} + \frac{3}{4}$, which is to say that you find a common denominator. In my example, it would be $3\cdot 4$. In yours, it would be...
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many solutions to prime = $a^3+b^3+c^3 - 3abc$ Let $a,b,c$ be integers.
Let $p$ be a given prime.
How to find the number of solutions to $p = a^3+b^3+c^3 - 3abc$ ?
Another question is ; let $w$ be a positive integer. Let $f(w)$ be the number of primes of type prime = $a^3+b^3+c^3 - 3abc$ below $w$. How does the fu... | We may also prove that there are infinitely many primes when $a+b+c=1$. This answer along with Andre Nicolas' answer and Will Jagy's answer allows us to calculate the exact number of representations $a^3+b^3+c^3-abc=p$ for each prime $p$.
We begin with the factorization $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Fractions in binary? How would you write a fraction in binary numbers; for example, $1/4$, which is $.25$? I know how to write binary of whole numbers such as 6, being $110$, but how would one write fractions?
| Not sure how helpful this is but whenever I work with binary I always use a table.
So if you want 0.25 or 15.75 (base 10) in binary:
\begin{array}{r|rrrr|rrrr}
& 8 & 4 & 2 & 1 & \frac{1}{2} & \frac{1}{4} & \frac{1}{8} & \frac{1}{16} \\
& 8 & 4 & 2 & 1 & 0.5 & 0.25 & 0.125 & 0.0625 \\
\hline
0.25 & 0 & 0 & 0 & 0 & 0 & 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/301435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 8,
"answer_id": 6
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Question about Geometry involving angles and lines
The answer is C however if angle ACD is 110 degrees and angle AB is 110 degrees how does it equal 180?
| Since it's an isosceles triangle, angles $B, C$ (the angles opposite segments $AC, AB$ respectively, which are equal in length) are equal in measure:
$$ \angle B = \angle C$$
We also know the the sum of the measures of the angles of a triangle sum to $180^\circ$.
$$\angle A + \angle B + \angle C = \angle A + 2 \angle... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Non-induction proof of $2\sqrt{n+1}-2<\sum_{k=1}^{n}{\frac{1}{\sqrt{k}}}<2\sqrt{n}-1$
Prove that $$2\sqrt{n+1}-2<\sum_{k=1}^{n}{\frac{1}{\sqrt{k}}}<2\sqrt{n}-1.$$
After playing around with the sum, I couldn't get anywhere so I proved inequalities by induction. I'm however interested in solutions that don't use induct... | Mean Value Theorem can also be used,
Let $\displaystyle f(x)=\sqrt{x}$
$\displaystyle f'(x)=\frac{1}{2}\frac{1}{\sqrt{x}}$
Using mean value theorem we have:
$\displaystyle \frac{f(n+1)-f(n)}{(n+1)-n}=f'(c)$ for some $c\in(n,n+1)$
$\displaystyle \Rightarrow \frac{\sqrt{n+1}-\sqrt{n}}{1}=\frac{1}{2}\frac{1}{\sqrt{c}}$..... | {
"language": "en",
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If $a,b \in \mathbb R$ satisfy $a^2+2ab+2b^2=7,$ then find the largest possible value of $|a-b|$ I came across the following problem that says:
If $a,b \in \mathbb R$ satisfy $a^2+2ab+2b^2=7,$ then the largest possible value of $|a-b|$ is which of the following?
$(1)\sqrt 7$, $(2)\sqrt{7/2}$ , $(3)\sqrt {35}$ $(... | As the distance between points is invariant 1,2 under Rotation,
using Rotation of axes by putting $a=x\cos\theta-y\sin\theta$ and $b=x\sin\theta+y\cos\theta$ in $a^2+2ab+2b^2=7$
$(x\cos\theta-y\sin\theta)^2+2(x\cos\theta-y\sin\theta)(x\sin\theta+y\cos\theta)+2(x\sin\theta+y\cos\theta)^2=7$
or, $x^2(\cos^2\theta+2\cos\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/310348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Problem with graphing linear equations Well, I can understand how to graph basic liner equations, for example:
$$y=2x-4$$
The y-intercept would be -4 and the slope would be 2. The coordinates could then be (0,-4)(1, -2)
However, how would I solve a linear equation like this: $$y = \frac{2x}{4}$$
What are the steps to ... | Strong Hint:
The equation $y = \dfrac{2x}{4}$ is read as the linear relationship described by multiplying the $x$-value by $\dfrac{2}{4}$.
Since the numerator $2$ and the denominator $4$ are both even, we can simplify this fraction, meaning to reduce the fraction to lowest terms. To do that, we divide the numerator an... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\sum_{j = 0}^{n} (-\frac{1}{2})^j = \frac{2^{n+1} + (-1)^n}{3 \times 2^n}$ whenever $n$ is a nonnegative integer. I'm having a really hard time with the algebra in this proof. I'm supposed to use mathematical induction (which is simple enough), but I just don't see how to make the algebra work.
$\sum_{j = 0... | Try writing your sum
$$\sum_{j = 0}^{k} \left(-\frac{1}{2}\right)^k + \left(-\frac{1}{2}\right)^{k + 1} $$
$$ =\frac{2^{k+1} + (-1)^k}{3 \times 2^k} + \frac{(-1)^{k+1}}{2^{k+1}} $$
$$ = \frac{2^{k+2} + 2(-1)^k}{3 \times 2^{k+1}} + \frac{(-1)^{k+1}}{2^{k+1}} \;$$
$$ = \frac{2^{k+2} + 2(-1)^k}{3 \times 2^{k+1}} + \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/311709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find Maclaurin series of $(\sin(x^3))^{1/3}$ How do I find Maclaurin series for the function:
$$\sqrt[3]{\sin(x^3)}$$
The answer should be:
$$ x - \frac {x^7}{18} - \frac {{x}^{13}}{3240} + o(x^{13})$$
I tried:
$$\sin x = x - \frac {x^3}{3!} + \frac {x^5}{5!} - \frac {x^7}{7!} + ...$$
So, I changed $x$ to $x^3$ to get... | If you factor out $x^3$ from cubic root you'll get
$$
x\sqrt[3]{1-\left (\frac{x^6}{3!}-\frac{x^{12}}{5!} +o(x^{12})\right )}
$$
Now, use power series expansion for cubic root
$$
\sqrt[3]{1-x} = 1-\frac x3-\frac {x^2}9+o(x^2) \\
\sqrt[3]{1-\left (\frac{x^6}{3!}-\frac{x^{12}}{5!} +o(x^{12})\right )} = 1-\frac 13\left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/311836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Integrating :$\int\sqrt{\sin x} \cos^{\frac{3}{2}}x dx$ How to integrate :
$$\int\sqrt{\sin x} \cos^{\frac{3}{2}}x dx$$
| Let's make a change of variables $u = \sin^2(x)$. Formally, $\sqrt{\sin(x)} = u^{1/4}$, $\cos^{3/2}(x) = (1-u)^{3/4}$, and $\mathrm{d}x = \frac{\mathrm{d}u}{2 \sqrt{u} \sqrt{1-u}}$.
Thus:
$$
\int \sqrt{\sin( x)} \cos^{3/2}( x) \, \mathrm{d}x = \frac{1}{2}\int u^{-1/4} (1-u)^{1/4} \mathrm{d} u
$$
In another answer o... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Need to prove $\frac{3}{5}(2^{\frac{1}{3}}-1)\le\int_0^1\frac{x^4}{(1+x^6)^{\frac{2}{3}}}dx\le1$ I need to show that
$$\frac{3}{5}(2^{\frac{1}{3}}-1)\le\int_0^1\frac{x^4}{(1+x^6)^{\frac{2}{3}}}dx\le1$$
I just know that if in $[a,b]$, $f(x)\le g(x)\le h(x)$, then
$$\int_a^bf(x)dx\le\int_a^bg(x)dx\le\int_a^bh(x)dx$$
How... | Consider left part first. You need to get integral there, something like $f(1)-f(0)$
$$
\frac 35 \left ( 2^{\frac 13}-1\right ) = \frac 35 \left ( (1+\fbox 1)^{\frac 13}-(1+\fbox 0)^{\frac 13}\right ) =\frac 35 \left . \left ( 1+x^n\right )^{\frac 13} \right |_0^1
$$
Now, you can imagine that values inside of the boxes... | {
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"timestamp": "2023-03-29T00:00:00",
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How do I completely solve the equation $z^4 - 2z^3 + 9z^2 - 14z + 14 = 0$ where there is a root with the real part of $1$. I would please like some help with solving the following equation:
$$z^4 - 2z^3 + 9z^2 - 14z + 14 = 0$$
All I know about the equation is that there is a root with the real part of $1$.
My approach ... | As the coefficients of the different powers of $z$ are real,
if one of the four roots is $1+yi,$ the other must be its conjugate $1-yi$ .
If the other two roots are $x,w,$
then using Vieta's Formulas for the coefficient of $z^3$, $1+yi+1-yi+x+w=2\implies w=-x$
So, $$(z-x)(z+x)\{z-(1+yi)\}\{z-(1-yi)\}=0$$
$$\implies ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/312791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Rewrite so that the denominator does not have any root expressions: $\frac{\sqrt[3]{49} +\sqrt[3]{7x} + \sqrt[3]{x^2}}{\sqrt[3]{x} -\sqrt[3]{7}}$ I am struggling with rewriting the following so that the denominator does not have any root expressions:
$$\frac{\sqrt[3]{49} +\sqrt[3]{7x} + \sqrt[3]{x^2}}{\sqrt[3]{x} -\sq... | Hint:$a^3-b^3=(a-b)(a^2+ab+b^2)$
Let $a=x^{1/3},b=7^{1/3}$
$x-7=a^3-b^3=(a-b)(a^2+ab+b^2)=(x^{1/3}-7^{1/3})(7^{2/3}+(7x)^{1/3}+x^{2/3})$
$\displaystyle \frac{(7^{2/3}+(7x)^{1/3}+x^{2/3})}{(x^{1/3}-7^{1/3})}=\frac{(7^{2/3}+(7x)^{1/3}+x^{2/3})}{(x^{1/3}-7^{1/3})}\frac{(7^{2/3}+(7x)^{1/3}+x^{2/3})}{(7^{2/3}+(7x)^{1/3}+x^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/312981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Factorization of cyclic polynomial
Factorize $$a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)$$
Since this is a cyclic polynomial, factors are also cyclic
$$f(a) = a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)$$
$$f(b) = b(b^2-c^2)+b(c^2-b^2)+c(b^2-b^2) = 0 \Rightarrow a-b$$
is a factor of the given expression. Therefore, other factors are $... | \begin{align}
a (b^2-c^2)+b (c^2-a^2)+c (a^2-b^2)
&=ab^2-ac^2+bc^2-ba^2+ca^2-cb^2\\
&=a^2c-a^2b+ab^2-ac^2+bc^2-cb^2\\
&=a^2 (c-b)-a (c^2-b^2)+bc (c-b)\\
&=(c-b)(a^2-ac-ab+bc)\\
&=(c-b)(bc-ab-ac-a^2)\\
&=(c-b)(b (c-a)-a (c-a))\\
&=(c-b)(b-a)(c-a)\\
&=(a-b)(b-c)(c-a)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/314105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Obtaining the pdf of the sum of two iid random variables with pdf $f_{X}(x)= \sqrt{\frac{1}{2\pi x}e^{-x/2}}$ X1, and X2 are independent with the pdf: $f_{X}(x)= \sqrt{\frac{1}{2\pi x}e^{-x/2}}$ defined for x>0
Y=X1+X2
What is the pdf of Y?
This is what I did so far:
$$f_{Y}(y) = \int_{0}^{\infty } f_{X_{1}}(y-x_{2})f_... | Effectively $x_2$ only goes to $y$, not to $\infty$. (The density is $0$ at negatives.)
For the work that remains, complete the square. After making the right substitution, you should end up needing to find $\int\frac{du}{\sqrt{1-u^2}}$, which is $\arcsin u +C$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/314727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find of $\int e^x \cos(2x) dx$ I did the following.
Using the LIATE rule:
$$\begin{align*}
u &=& \cos(2x)\\
u\prime &=& -2 \sin(2x)\\
v &=& e^x\\
v\prime&=&e^x
\end{align*}$$
We get:
$$\int e^x \cos(2x)dx = e^x \cos(2x) +2 \int e^x \sin(2x)dx $$
Now we do the second part.
$$\begin{align*}
u &=& \sin(2x)\\
u\prime &=& ... | We know that $\cos(2x)$ is the real part of $e^{2ix}$, so we'll calculate $\int e^xe^{2ix}dx $, then we take its real part. We have
$$\int e^xe^{2ix}dx=\frac{1}{2i+1}e^{(2i+1)x}+C=\frac{e^x}{5}(1-2i)(\cos(2x)+i\sin(2x))+C\\=\frac{1}{5}(e^x\cos2x+2e^x\sin2x)+C+i(\text{imaginary part}).$$
Now, you can conclude.
| {
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"url": "https://math.stackexchange.com/questions/316639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Please, help me to find where is a mistake in the solutions of the equation. I have this equation and I will be very thankful to anyone who can provide me any help with the one discrepancy in my solution and the solution from the self-learning website:
$$
\frac{1+\tan(x) + \tan^2(x) + ... + \tan^n(x) + ...}{1-\tan(x) +... | If $|\tan x|\ge 1$ both numerator and denominator of the left-hand side are not finite, because the geometric series
$$N=1+\tan(x) + \tan^2(x) + ... + \tan^n(x) + ...$$
and
$$D=1-\tan(x) + \tan^2(x) - ... +(-1)^n\tan^n(x)+...$$
don't converge. If $|\tan x|< 1$, we have
$$N=\lim_{n\rightarrow \infty }\frac{\tan ^{n}x-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/317149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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equation $\displaystyle a_{n+1}.x^2-2x\sqrt{a^2_{1}+a^2_{2}+.......+a^2_{n}+a^2_{n+1}}+\left(a_{1}+a_{2}+.......+a_{n}\right) = 0$ has real roots The Largest positive integer $n$ such that for all real no. $a_{1},a_{2},.......,a_{n},a_{n+1}.$ The equation
$\displaystyle a_{n+1}.x^2-2x\sqrt{a^2_{1}+a^2_{2}+.......+a^2_... | Let $x=a_{n+1}$ then the last inequality is $x^2-xT+S \ge 0$ where $T=a_1+a_2+\cdots+a_n$ and $S=a_1^2+a_2^2+\cdots+a_n^2$. This requires its own discriminant $\Delta=T^2-4s\le 0$. As you try increasing the number of $a_i$'s you see at 4 elements the inequality can be made to work but not after that. Here is how it tu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/318374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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values of basic trig function, given values in terms of trig functions Get the function value if $\displaystyle \sin x = \frac{2}{\sqrt{13}}$ and $\displaystyle \cos x = -\frac{3}{\sqrt{13}}$ and the function value is going to be calculated from $\displaystyle \sin\left(2x-\frac{7\pi}{6}\right)$.
| $$\sin(2x-\frac{7\pi}{6})=\sin2x\cos(7\pi/6)-\cos2x\sin(7\pi/6)=$$
$$=2\sin x\cos x\cos(7\pi/6)-(\cos^2x-\sin^2x)\sin(7\pi/6)=$$
$$=2(2/\sqrt 13)(-3/\sqrt13)\cos(7\pi/6)-(4/13-9/13)\sin(7\pi/6)=$$
$$=-12/13\cos(7\pi/6)+5/13\sin(7\pi/6)=$$
$$=-12/13(-\sqrt 3/2)+5/13(-1/2)=$$
$$=12\sqrt 3/26-5/26=(12\sqrt 3-5)/26$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/319617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Partial Fraction Decomposition $\frac{2x^2+3x+3}{(x+1)(x^2+1)}$ How would one go about decomposing this fraction?
$$
2x^2+3x+3\over
(x+1)(x^2+1)
$$
Here is what I have so far:
$$
{2x^2+3x+3\over
(x+1)(x^2+1)
} =
{ A\over x+1 } + { Bx+C\over x^2+1} $$
$$
{2x^2+3x+3\over
(x+1)(x^2+1)
} =
{ A(x^2+1) } + (Bx+C)(x+1) $$
... | $$
\begin{align}
2x^2 + 3x +3 &= A(x^2 + 1) + (Bx+C)(x+1)\\
&= x^2(A+B) + x(B+C) + A+C
\end{align}
$$
By comparing the corresponding co-efficients :
$$
A+B = 2;\\
B+C = 3;\\
C+A = 3;
$$
By solving, $A = 1$, $B = 1$, $C = 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/320260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find $E(f(f(f(\ldots f(x)))$ I have a random function $f(x)$ which returns one of the integers in the range $[0, x-1]$ with equal probability and $f(0) = 0$.
What is the expected value $E(f(f(f(\ldots f(x)))$ ($n$-times $f(x)$)? The answer should be a function of $x$ and $n$.
| I fail to find a general pattern, but here are some results.
Write $E_n(x) = E(f^n(x))$.
It is easily seen that $E_n(x) = 0$ if $x \leq n$ and $E_n(n+1) = \frac{1}{(n+1)!}$.
Furthermore, if $x \geq 1$, we have
$$
E_1(x) = \frac{x-1}{x} + \dots + \frac{1}{x} = \frac{x-1}{2}
$$
$$
E_2(x) = \frac{1}{x}\sum_{y=1}^{x-1}\fra... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
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Does there exists a polynomial $Q(x,y)$ such that $x-1=Q(x^2-1,x^3-1)$. I have a question: Does there exists a polynomial $Q(x,y)$ such that $x-1=Q(x^2-1,x^3-1)$.
I did as following: Let $Q(x,y)=\sum\limits_{k=0}^n\sum\limits_{i+j=k}a_{ij}x^iy^j$. Then I could find some coefficients: the coefficient of $(x^2-1)$ is $0$... | The polynomials $x^2-1,x^3-1,(x^2-1)^2,(x^2-1)(x^3-1),(x^2-1)^3,(x^3-1)^2,(x^2-1)^2(x^3-1),(x^2-1)^4,(x^2-1)(x^3-1)^2$ are $9$ in number and all of degree at most $8$. I suspect they span the $8$-dimensional space of polynomials of degree at most $8$ vanishing at $x=1$. If that's right, then $x-1$ will be a linear comb... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Integers with 15 divisors (from brilliant.org) How many integers from 1 to 19999 have exactly 15 divisors?
Note: This is a past question from brilliant.org.
| If $n = p_1^{a_1}p_2^{a_2} \ldots p_k^{a_k}$, then the number of divisors is given by $d(n) = (1+a_1)(1+a_2)\cdots(1+a_k)$. We are given that $d(n) = 15$. From this, show that there cannot be $3$ distinct primes diving $n$. Hence, $n=p_1^{a_1}$ or $n=p_1^{a_1} p_2^{a_2}$.
$1$. If $n=p_1^{a_1}$, then $(1+a_1) = 15 \impl... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Showing $(x,y)$ pairs exist for $\sqrt{\quad\mathstrut}$ If we were to show that there exists infinitely many $(x,y)$ pairs in $\mathbb{Q}^2$ for which both $\sqrt{x^2+y^4}$ and $\sqrt{x^4+y^2}$ are rational. If the power root for $x$ and $y$ vary but never the same, how can we prove that we can always represent the ou... | Let $x=y=t^2-1$. Then $x^2+y^3=x^3+y^2=t^2(t^2-1)^2$ so both squareroots are $t(t^2-1)$, which is rational provided $t$ is rational. This is infinitely many pairs $(x,y).$
With a little work and Pell's equation, one can find pairs $(x,y)$ of form $(2t,t)$ which work, if it's a concern that $x=y$ in the above example.
E... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Please help me to find the value of $ABCDE$ (step by step) $ABCD\times E = DCBA$ with $A,B,C,D$, and $E$ distinct decimal digits (and $ABCD$ representing the concatenation of those digits). How can I find the value each of them?
| 2178*4 = 8712
Consider $E$ cannot be 1. otherwise $D=A$.If $E\ge 5$, then $A=1$, $D=E$, contradiction!
So $E\ge 2$, thus we have
*
*$E = 2$, then $A$ can only be $1,2,3,4$, since the product should be
4-digit number.
$DE$ has end digit as $A$, then $A$ is even, $A$ can be 2,4, since $A\neq E$, therefore $A=4$. The... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/324018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to integrate $\int \frac 1 {\begin{pmatrix}\frac1 3 t - 8\end{pmatrix}^5} \,dt$ by substitution? I am required to solve this question via integration by an appropriate substitution. This is the first time I am doing it with no help on the substitution part. Hence, could somebody help me check this working?
$$
\begi... | Thank you to all who have helped, my proposed solution has been verified and is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/324694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Definite Integral $\int_0^{\pi/2} \frac{\log(\cos x)}{x^2+\log^2(\cos x)}dx = \frac{\pi}{2}\left(1-\frac{1}{\log 2}\right)$ I want to prove that
$$\int_0^{\pi/2} \frac{\log(\cos x)}{x^2+\log^2(\cos x)}dx = \frac{\pi}{2}\left(1-\frac{1}{\log 2}\right)$$
| Consider $$f(z) = \frac{1}{\log(1-iz)} \frac{1}{1+z^{2}}$$ where the branch cut for $\log (1-iz)$ runs down the imaginary axis from $z=-i$.
Then integrate around a contour that consists of the line segment $[-R,R]$ and the upper half of the circle $|z|=R$, and is indented around the simple pole at the origin.
Since $ z... | {
"language": "en",
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"source": "stackexchange",
"question_score": "22",
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Prove that: $(1+a_1)(1+a_2)...(1+a_n) \le 1 + S_n + (S_n)^2/2! + ... + (S_n)^n/n!$ I am currently working on this problem from Hardy's Course of Pure Mathematics and have gotten stuck near the end. I was wondering if someone could help me determine what to go next.
Question
If $a_1, a_2, ...,a_n$ are all positive and $... | Continue the proof using MI. Assume P(n) is true.
Let $y=a_{n+1}, S=S_n$
Let $f(y)=1+(S+y)+\frac{(S+y)^2}{2!}+...+\frac{(S+y)^{(n+1)}}{(n+1)!}-(1+a_1)(1+a_2)...(1+a_n)(1+y) \quad y>0$
$f'(y)=1+(S+y)+\frac{(S+y)^2}{2!}+...+\frac{(S+y)^{n}}{n!}-(1+a_1)(1+a_2)...(1+a_n)$
$ > 1+S+\frac{S^2}{2!}+...+\frac{S^{n}}{n!}-(1+a_1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/326721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 2
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Range of $\frac{1}{2\cos x-1}$ How can we find the range of $$f(x) =\frac{1}{2\cos x-1}$$
Since range of $\cos x$ can be given as : $-1 \leq \cos x \leq 1$
therefore we can proceed as :$$\begin{array}{rcl}
-2 \leq & 2\cos x & \leq 2 \\
-2-1 \leq & 2\cos x -1 & \leq 2-1\\
-3 \... | No, you are fine down to $-3 \le 2\cos x -1 \le 1$, but you need to change the sense of the inequality when you invert: Think of a positive $x \le 1$, then $\frac 1x \ge 1$. The idea is that if the denominator is small (and you have trapped it near $0$) the inverse is large.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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How to evaluate $\int_0^{\pi/2}x^2\ln(\sin x)\ln(\cos x)\ \mathrm dx$ Find the value of
$$I=\displaystyle\int_0^{\pi/2}x^2\ln(\sin x)\ln(\cos x)\ \mathrm dx$$
We have the information that
$$J=\displaystyle\int_0^{\pi/2}x\ln(\sin x)\ln(\cos x)\ \mathrm dx=\dfrac{\pi^2}{8}\ln^2(2)-\dfrac{\pi^4}{192}$$
| Presented below is a complete solution that evaluates the integral to the following close-form
\begin{align}
&\int_0^{\pi/2}x^2\ln(\sin x)\ln(\cos x)\ dx\\
=& -\frac\pi2 \text{Li}_4(\frac12)-\frac{3\pi}8\ln2\ \zeta(3)
+\frac{\pi^5}{320}+\frac{\pi^3}{16}\ln^22-\frac\pi{48}\ln^42\tag1
\end{align}
To derive (1), evaluate ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "60",
"answer_count": 6,
"answer_id": 5
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Computing $\int \frac{6x}{\sqrt{x^2+4x+8}} dx$ I am trying to compute an indefinite integral. Thanks!
$$\int \frac{6x}{\sqrt{x^2+4x+8}} dx.$$
| $$\begin{align}\int \dfrac{6x}{\sqrt{x^2 + 4x + 8}}\, dx\;
& = \;\int \dfrac{6x+12 -12}{\sqrt{x^2 + 4x + 8}} \,dx \\ \\ \\ \\
& = \int \dfrac {6(x + 2) - 12}{\sqrt {\color{blue}{\bf (x^2+ 4x + 4) + 4}}}\,dx
\end{align}$$
split the integral into the difference of two integrals:
$$ = \;\int \dfrac {6(x + 2)}{\sqrt {(x^2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$ \sqrt{2-\sqrt{2}} $ simplified So I have a (nested?) square root $ \sqrt{2-\sqrt{2}} $. I know that $ \sqrt{2-\sqrt{3}} = \frac{\sqrt{6}-\sqrt{2}}{2} $. I know how to turn the simplified version into the complex one, but not vice versa. What is $ \sqrt{2-\sqrt{2}} $ simplified in this fashion and what are the steps?
| this is a general way to do it:
$\sqrt{2-\sqrt{2}}=\sqrt x+\sqrt y $
then $2-\sqrt 2=2 \sqrt{xy}+x+y$
and then make the rational parts equal to the rational parts as such:
$2=x+y$
now we know $y=2-x$. substitute this into the irrational part:
$-\sqrt 2 =2\sqrt{(x)}\sqrt{(2-x)}$ square both sides to get
$2=4(x)(2-x) \ri... | {
"language": "en",
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"source": "stackexchange",
"question_score": "13",
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Solving recursion with generating function I am trying to solve a recursion with generating function, but somehow I ended up with mess.....
$$y_n=y_{n-1}-2y_{n-2}+4^{n-2}, y_0=2,y_1=1 $$
\begin{eqnarray*}
g(x)&=&y_0+y_1x+\sum_2^{\infty}(y_{n-1}-2y_{n-2}+4^{n-2})x^n\\
&=&2+x+\sum^{\infty}_2y_{n-1}x^n-2\sum_2^{\infty}y_{... | Use the technique in Wilf's "generatingfunctionology" Define the generating function $A(z) = \sum_{n \ge 0} y_n z^n$, and write:
$$
y_{n + 2} = y_{n + 1} - 2 y_n + 4^n \quad y_0 = 2, y_1 = 1
$$
Using properties of generating functions:
$$
\frac{A(z) - y_0 - y_1 z}{z^2}
= \frac{A(z) - y_0}{z} - 2 A(z) + \frac{1}{1 - 4... | {
"language": "en",
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"source": "stackexchange",
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Prove $\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6$
If $a,b,c$ are non-negative numbers and $a+b+c=3$, prove that:
$$\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6.$$
Here's what I've tried:
Using Cauchy-Schawrz I proved that:
$$(3a + b^3)(3 + 1) \ge (3\sqrt{a} + \sqrt{b^3})^2$$
$$\sqrt{(3a ... | Proof by KaiRain.
By squaring and collecting in terms, the original inequality is equivalent with:
\begin{align*} \sum_{cyc} \sqrt{ \left(3a+b^3 \right) \left(3b+c^3 \right)} \ge \frac{3 \left(a+b \right) \left(b+c \right) \left(c+a \right)}{2} \end{align*}
We will prove that:
\begin{align} \sqrt{ \left(3a+b^3 \right) ... | {
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"source": "stackexchange",
"question_score": "17",
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Solve the equation by putting new variable I want to solve the equation
$$x^2+(2 x+3) \sqrt{3 x^2+6 x+2}=6 x+5.$$
I tried. Put $t = \sqrt{3 x^2+6 x+2}.$ We have
$$x^2+(2 x+3)t - 6x - 5 = 0.$$
Discriminant of this equation (with unknown $x$) is $56-36t+4t^2$. It is not a square number.
| Just put it like this:
$(2 x + 3) \sqrt{(3 x^2 + 6 x + 2)} = (6 x + 5 - x^2)$
then square both sides:
$(2 x + 3)^2 (3 x^2 + 6 x + 2) = (6 x + 5 - x^2)^2$
and collect the terms:
$(-1 + x (5 + x)) (7 + x (17 + 11 x))=0$
The answer is straightforward. Will this help?:)
P.S.: the solutions are
$\left\{x\to \frac{1}{22} \l... | {
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"source": "stackexchange",
"question_score": "2",
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Why $\sum_{k=1}^{\infty} \frac{k}{2^k} = 2$? Can you please explain why
$$
\sum_{k=1}^{\infty} \dfrac{k}{2^k} =
\dfrac{1}{2} +\dfrac{ 2}{4} + \dfrac{3}{8}+ \dfrac{4}{16} +\dfrac{5}{32} + \dots =
2
$$
I know $1 + 2 + 3 + ... + n = \dfrac{n(n+1)}{2}$
| Start with the geometric series
$$s(x) = \sum_{k=0}^{\infty} x^k = \frac{1}{1-x}$$
Then
$$x \frac{d}{dx} s(x) = \sum_{k=1}^{\infty} k x^k = \frac{x}{(1-x)^2}$$
Your case has $x=1/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/337937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 12,
"answer_id": 3
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Evaluate $\lim\limits_{n\rightarrow \infty}\left\{\frac{1^3}{n^4}+\frac{2^3}{n^4}+\frac{3^3}{n^4}+\dots +\frac{n^3}{n^4}\right\}$
*
*Evaluate $\displaystyle\lim_{n\rightarrow \infty}\left\{\frac{1^3}{n^4}+\frac{2^3}{n^4}+\frac{3^3}{n^4}+\dots +\frac{n^3}{n^4}\right\}$.
*Examine whether $x^{1/x}$ possesses a maximum ... | It is happens that there is an attractive closed form for $1^3+2^3+\cdots+n^3$, which can readily be proved by induction:
$$1^3+2^3+3^3+\cdots +n^3=\left(\frac{n(n+1)}{2}\right)^2.$$
Divide by $n^4$. Fairly quickly we find that the desired limit is $\dfrac{1}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/338121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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generating functions / combinatorics Calculate number of solutions of the following equations:
$$ x_1 + x_2 + x_3 + x_4 = 15 $$
where $ 0 \le x_i < i + 4 $
I try to solve it using generating functions/enumerators :
$$ (1+x+x^2+x^3+x^4)(1+x+x^2+x^3+x^4+x^5)(1+x+x^2+x^3+x^4+x^5+x^6)(1+x+x^2+x^3+x^4+x^5+x^6+x^7)$$
and ta... | To take a slightly different route from i707107's solution, once you obtain the product of rational functions, you can use the identity:
and
$$
\frac{1}{(1-x)^n} = 1 + \binom{1 + n -1}{1}x + \binom{2 + n -1}{2}x^2 + \dots + \binom{r + n -1}{r}x^r + \dots.
$$
to expand the last term of the simplified product (2nd li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/339230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Given the graphic sequence of a simple graph, how to construct the adjacency matrix? Suppose I have the graphic sequence (7,6,6,6,5,4,4,2), how do I get the adjacency matrix out of it?
| The graph has $8$ vertices, and $\tfrac{1}{2}(7+6+6+6+5+4+4+2)=20$ edges by the Handshaking Lemma. We can exhaustively generate the non-isomorphic $8$-vertex $20$-edge graphs with vertex degrees between $2$ and $7$ using geng which comes with nauty using:
geng 8 20:20 -d2 -D7
I wrote a script that filters out the one... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/339381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to find the value of $\sin{\dfrac{\pi}{14}}+6\sin^2{\dfrac{\pi}{14}}-8\sin^4{\dfrac{\pi}{14}}$
Determine
$$
\sin\left(\pi \over 14\right) + 6\sin^{2}\left(\pi \over 14\right)
-8\sin^{4}\left(\pi \over 14\right)
$$
My idea: Let $\displaystyle{\sin\left(\pi \over 14\right)} = x$.
| A non-trivial result is that $\sin \dfrac{\pi}{14}$ is a root to the equation $^{\color{blue}{(**)}}$
$$1 - 4x - 4x^2 + 8x^3 = 0. \tag{1}$$
If you believe this, then you can simply apply polynomial division to obtain
$$x + 6x^2 - 8x^4 = -\left(x + \frac{1}{2}\right)(8x^3 - 4x^2 - 4x + 1) + \frac{1}{2}.$$
So then, fil... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/346664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 2,
"answer_id": 0
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Sum of $\sum_{n=0}^\infty \frac{(x+2)^{n+2}}{3^n} $ Calculate the sum of the next series and for which values of $x$ it converges:
$$\sum_{n=0}^\infty \frac{(x+2)^{n+2}}{3^n}$$
I used D'Alembert and found that the limit is less than 1, so: $-5 < x < 1$ (because the fraction must be less than 1).
and then I assigned the... | Your sum is equal to
$$
\sum_{n=0}^\infty\frac{(x+2)^{n+2}}{3^n}=(x+2)^2\sum_{n=0}^\infty\left(\frac{x+2}{3}\right)^n,
$$
which you recognize as a geometric series $\sum_n q^n$ with sum $1/(1-q)$. Thus,
$$
\sum_{n=0}^\infty\frac{(x+2)^{n+2}}{3^n}=(x+2)^2\frac{1}{1-\frac{x+2}{3}} =\frac{3(x+2)^2}{1-x}
$$
as long as $|(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/348079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Find the value of $\alpha $ given $2\sin\theta -\sqrt 5 \cos \theta \equiv - 3\cos (\theta + \alpha )$ Given:
$$2\sin\theta -\sqrt 5 \cos \theta \equiv - 3\cos (\theta + \alpha ),$$
where $$0 <\alpha < 90^\circ, $$
find $α.$
The issue I have with this question is the $-3$ on the right hand side, it really complicat... | $$\Rightarrow {\rm{2sin}}\theta {\rm{ - }}\sqrt 5 \cos \theta = - 3\cos (\theta + \alpha )$$
$$\Rightarrow -\frac{2}{3}\sin \theta+\frac{\sqrt 5 }{3}\cos \theta= \cos (\theta + \alpha ) $$
$$\Rightarrow -\frac{2}{3}\sin \theta+\frac{\sqrt 5 }{3}\cos \theta= \cos \theta \cos \alpha - \sin \theta \sin \alpha$$
$$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/349185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $x \neq 0,y \neq 0,$ then $x^2+xy+y^2$ is ..... I came across the following problem that says:
If $x \neq 0,y \neq 0,$ then $x^2+xy+y^2$ is
1.Always positive
2.Always negative
3.zero
4.Sometimes positive and sometimes negative.
I have to determine which of the aforementioned options is right.
Now ... | $x^2 + xy + y^2$ is a quadratic form and can be written
$$x^2 + xy + y^2 = \begin{bmatrix} x & y\end{bmatrix}\begin{bmatrix} 1 & \frac{1}{2} \\ \frac{1}{2} & 1 \end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix}$$
A matrix $\mathbf{A}$ is positive-definite if $\mathbf{z'}\mathbf{A}\mathbf{z}>0 $ for all $\mathbf{z}\neq 0$... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 4
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Generating function for the solutions of equation $ 2x_1 + 4x_2 = n$ Let's denote $h_n$ as the number of soulutions of the following equation:
$$ 2x_1 + 4x_2 = n$$
where $x_i \in \mathbb N$.
Find generating function of the sequence $h_n$ and calculate $h_{2000}$.
I've found the generating function:
$$\frac{1}{1-x^2}\cd... | As you observed, if the generating function for $h_n$ is $$f(x) = h_0 + h_1x + h_2x^2 + h_3x^3 + \dots + h_nx^n + \dots,$$
then $$f(x) = (1 + x^2 + x^4 + x^6 + \dots)(1 + x^4 + x^8 + x^{12} + \dots) = \frac{1}{1-x^2}\frac{1}{1-x^4}$$
To actually get the coefficients of $x^n$ in the above, we resort to partial fractions... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
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Help please on complex polynomials I wanted to know if there's any good approaches to these questions
a)By considering $z^9-1$ as a difference of two cubes, write $1+z+z^2+z^3+z^4+z^5+z^6+z^7+z^8$
as a product of two real factors one of which is a quadratic.
b) Solve $z^9-1=0$ and hence write down the 6 solutions of $z... | (a)
\begin{align*}
(z - 1)(z^8 + z^7 + \ldots + 1) = z^9 - 1 & = (z^3 - 1)(z^6 + z^3 + 1) \\
& = (z - 1)(z^2 + z + 1)(z^6 + z^3 + 1) \\
\therefore \qquad z^8 + z^7 + \ldots + 1 & = (z^2 + z + 1)(z^6 + z^3 + 1).
\end{align*}
(b)
The roots of $z^9 - 1$ are $r_n = e^{\frac{2\pi}9ni}$, for $n = 0, \ldots, 8$. Since $r_0 = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/350879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Simplify $\sum_{i=0}^n (i+1)\binom ni$ Simplifying this expression$$1\cdot\binom{n}{0}+ 2\cdot\binom{n}{1}+3\cdot\binom{n}{2}+ \cdots+(n+1)\cdot\binom{n}{n}= ?$$
$$\text{Hint: } \binom{n}{k}= \frac{n}{k}\cdot\binom{n-1}{k-1} $$
| We have
$$\sum_{k=0}^n\dbinom{n}k x^k = (1+x)^n$$
Hence, we get that
$$\sum_{k=0}^n\dbinom{n}k x^{k+1} = x(1+x)^n$$
Differentiating the above, we get
$$\sum_{k=0}^n(k+1) \dbinom{n}k x^{k} = \dfrac{d(x(1+x)^n)}{dx}$$
Now set $x=1$ to get the answer.
$$\sum_{k=0}^n(k+1) \dbinom{n}k x^{k} = \dfrac{d(x(1+x)^n)}{dx} = (1+x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/351289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
Really Confused on a surface area integral can't seem to finish the integral off. Basically the question asks to compute $\int \int_{S} ( x^{2}+y^{2}) dA$ where S is the portion of the sphere $x^{2} + y^{2}+ z^{2}= 4$ and $z \in [1,2]$ we start with a chnage of variables
$x=x $
$y=y$
$ z= 2 \cdot(4-(x^{2} + y^{2}... | Sometimes cylindrical coordinates is at least as easy when there is axial symmetry. The integral to be evaluated becomes
$$2 \pi \int_1^2 dz \, r(z)^3 \sqrt{1+\left ( \frac{d r}{dz} \right)^2} $$
where $r(z) = \sqrt{4 - z^2}$ and $r$ is the distance from the axis to the sphere. This reduces to, upon evaluation of the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/356385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Constructing a degree 4 rational polynomial satisfying $f(\sqrt{2}+\sqrt{3}) = 0$ Goal: Find $f \in \mathbb{Q}[x]$ such that $f(\sqrt{2}+\sqrt{3}) = 0$.
A direct approach is to look at the following
$$
\begin{align}
(\sqrt{2}+\sqrt{3})^2 &= 5+2\sqrt{6} \\
(\sqrt{2}+\sqrt{3})^4 &= (5+2\sqrt{6})^2 = 49+20\sqrt{6} \\
\end... | A 'mechanical' approach follows. Let $x=\sqrt{2}+\sqrt{3}$. Then $x^2=5+2\sqrt{6}$ which means $x^2-5=2\sqrt{6}$. Now $$(x^2-5)^2=24\Longrightarrow(x^2-5)^2-24=0.$$ By construction, one of the roots of $f(x)=(x^2-5)^2-24$ is $\sqrt{2}+\sqrt{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/359054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 3
} |
How do I evaluate $\lim_{n \rightarrow \infty} \frac{1\cdot2+2\cdot3 +3\cdot4 +4\cdot5+\ldots}{n^3}$ How to find this limit : $$\lim_{n \rightarrow \infty} \frac{1\cdot2+2\cdot3 +3\cdot4 +4\cdot5+\ldots+n(n+1)}{n^3}$$ As, if we look this limit problem viz. $\lim_{x \rightarrow \infty} \frac{1+2+3+\ldots+n}{n^2}$ then w... | Or we can apply Stolz–Cesàro theorem to obtain $$\lim_{n \rightarrow \infty} \frac{1\cdot2+2\cdot3 +3\cdot4 +4\cdot5+\ldots+n(n+1)}{n^3}=\lim_{n \to \infty}\frac{(n+2)(n+1)}{(n+1)^3-n^3}=\lim_{n\to \infty}\frac{(n+2)(n+1)}{3n^2+3n+1}=\frac{1}{3} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/364284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Solving a 2nd order differential equation by the Frobenius method Can you, please, help me to solve this equation:
$$(x+1)^2y''+(x+1)y'-y=0$$
Here, for me the problem is, I am finding relationship among 3 members: $a_n, a_{n+1}, a_{n+2}$, not between 2 members: $a_n$ and $a_{n+1}$
I would like to solve it using the Fro... | Let $y=\sum\limits_{n=0}^\infty a_n(x+1)^{n+r}$ ,
Then $y'=\sum\limits_{n=0}^\infty(n+r)a_n(x+1)^{n+r-1}$
$y''=\sum\limits_{n=0}^\infty(n+r)(n+r-1)a_n(x+1)^{n+r-2}$
$\therefore(x+1)^2\sum\limits_{n=0}^\infty(n+r)(n+r-1)a_n(x+1)^{n+r-2}+(x+1)\sum\limits_{n=0}^\infty(n+r)a_n(x+1)^{n+r-1}-\sum\limits_{n=0}^\infty a_n(x+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/366233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find $(a, b)$ such that $\lim_{x \to 0} \frac{ax -1 + e^{bx}}{x^2} = 1$ Find $(a, b)$ such that $\lim_{x \to 0} \frac{ax -1 + e^{bx}}{x^2} = 1$
I am able to find $b = \pm \sqrt{2}$ using L'Hopitals Rule, but unable to do anything for $a$.
| As the limit in question is of the indeterminate form "$\frac{0}{0}$", we use L'Hôpital's rule and consider the limit $$\lim_{x \to 0} \frac{a +be^{bx}}{2x}.$$ Note that $\lim_{x\to 0}a+be^{bx} = a+be^0 = a+b$ and $\lim_{x\to 0} 2x = 0$. The only way the overall limit can be $1$ is if we have a limit in the indetermina... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/366384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"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.