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Trace constraints and rank-one positive semi-definite matrices. Let $C_1, C_2, \dots, C_N$ be $M \times M$ Hermitian matrices and $c > 0$ be a given constant. Let $W$ be a positive (possibly semi) definite matrix such that
\begin{align}
\text{trace}\{WC_1\} & \geq c \\
\text{trace}\{WC_2\} & \geq c \\
\vdots & \\
\text{trace}\{WC_N\} & \geq c \\
\end{align}
**Does this guarantee that there exists a vector $w$ such that
\begin{align}
w^{H}C_1w &\geq c \\
w^{H}C_2w &\geq c \\
\vdots & \\
w^{H}C_Nw &\geq c \\
\end{align}
NOTE: $w^{H}C_iw=\operatorname{trace}\{ww^HC_i\}$. Thus, if the given $W$ is a rank-one matrix, the solution is straightforward.
| Here is a counterexample: $$C_1=\begin{pmatrix}2&3\\3&-1\end{pmatrix} \quad C_2=\begin{pmatrix}2&-3\\-3&-1\end{pmatrix} \quad C_3=\begin{pmatrix}2&-3i\\3i&-1\end{pmatrix}\quad C_4=\begin{pmatrix}2&3i\\-3i&-1\end{pmatrix} \\
C_5=\begin{pmatrix}-1&3\\3&2\end{pmatrix} \quad C_6=\begin{pmatrix}-1&-3\\-3&2\end{pmatrix} \quad C_7=\begin{pmatrix}-1&-3i\\3i&2\end{pmatrix}\quad C_8=\begin{pmatrix}-1&3i\\-3i&2\end{pmatrix}
\\
\\ W=\begin{pmatrix}1&0\\0&1\end{pmatrix} $$
The assumptions hold with $c=1$. Suppose that $w=\begin{pmatrix}a \\ b \end{pmatrix}\in\mathbb C^2$ is a vector such that $$w^HC_kw>0\ \text{ for all }\ k=1,\dots,8\tag1$$ This will lead to a contradiction.
Using (1) with $k=1,\dots,4$ we find that
$$2|a|^2-|b|^2 > 6\max(|\operatorname{Re}(a\bar b)|, |\operatorname{Im}(a\bar b)|)\tag2$$
while using (1) with $k=5,\dots,8$ yields
$$-|a|^2+2|b|^2 > 6\max(|\operatorname{Re}(a\bar b)|, |\operatorname{Im}(a\bar b)|)\tag3$$
Since $a$ and $b$ are interchangeable in (2)-(3), we assume $|a|\le |b|$ from now on.
For any complex number $z$ we have $\max (|\operatorname{Re}z|, |\operatorname{Im} z|)\ge |z|/\sqrt{2}$. Therefore, (2) implies
$$2|a|^2-|b|^2 > \frac{6}{\sqrt{2}}|ab| \tag4$$
Recalling that $|a|\le |b|$ and using (4), we arrive at
$$|a|^2\ge 2|a|^2-|b|^2 > \frac{6}{\sqrt{2}}|ab| \ge \frac{6}{\sqrt{2}}|a|^2$$ which is a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/361433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Coefficient of Taylor Series of $\sqrt{1+x}$ The coefficient of $x^3$ in the Taylor series of the function $f(x) = \sqrt{1+x}$ about the point $a = 0$ is $$1\over 16$$
Can someone show me how to get this value?
| Another way to see this is to expand in Binomial series:
$$
(1+x)^{\frac{1}{2}} = \sum_{k=0}^{\infty}\binom{\frac{1}{2}}{k}x^k
$$
Although $\binom{\frac{1}{2}}{k}$ looks intimidating, it's nothing when you replace $\frac{1}{2} = \alpha \Rightarrow \binom{\alpha}{k} = \alpha\cdot (\alpha-1)\cdots(\alpha-k+1) \cdot \frac{1}{k!} = \frac{1}{2} \cdots (-\frac{1}{2}) \cdots \frac{1}{2}-(k-1)\cdot \frac{1}{k!} = \frac{\sqrt{\pi}}{k! 2 \Gamma(\frac{3}{2}-k)}$
This is easy since you only need to plug in $k=3$ and $\Gamma(-\frac{3}{2})=\frac{4 \sqrt{\pi}}{3}$ and therefore $\binom{\frac{1}{2}}{3} = \frac{1}{16}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does $y(y+1) \leq (x+1)^2$ imply $y(y-1) \leq x^2$? Can anyone see how to prove the following?
If $x$ and $y$ are real numbers with $y\geq 0$ and $y(y+1) \leq (x+1)^2$ then $y(y-1) \leq x^2$.
It seems it is true at least according to Mathematica.
| Given $y^2 + y \le x^2 + 2x + 1$, if possible, let $x^2 < y(y-1)$. Clearly $y > 1$.
Then $x^2 + (2x + 1) < y^2 - y + (2x + 1)$
So $y^2 + y < y^2 - y + 2x + 1$, which resolves to $y < x + \frac{1}{2}$.
Hence we also have $y - 1 < x - \frac{1}{2}$.
As $y > 1$, the LHS is positive, and we can multiply the last two to get
$y(y-1) < x^2 - \frac{1}{4} \implies y(y-1) < x^2$, a contradiction.
| {
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"timestamp": "2023-03-29T00:00:00",
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In how many ways i can write 12? In how many ways i can write 12 as an ordered sum of integers where
the smallest of that integers is 2? for example 2+10 ; 10+2 ; 2+5+2+3 ; 5+2+2+3;
2+2+2+2+2+2;2+4+6; and many more
| Let's split the question into a few pieces.
First part: how many ways to sum if all numbers are even? In this case, it's a simple partition problem, and the answer is equally simple. If you imagine the digits like this:
..|..|..|..|..|.. = 2+2+2+2+2+2
Then each of the five dividers can either be there or not, giving exactly $2^5=32$ ways of summing when they're all even.
Second part: Having the odd part summing to six, how many ways? In this case, there's just 3+3, and then you place three 2s and two 3s, which is equivalent to choosing two places in a sequence of five to place the 3s. But you can also place a 2 and a 4, a 4 and a 2, or a 6. So that's $\binom{5}{2}+2\binom{4}{2}+\binom{3}{2}=10+2\cdot6+3=25$. So far, we have 57.
Third part: Having the odd part summing to eight, how many ways? In this case, there's 3+5 and 5+3, and either two 2s or a 4. So that's $2\binom{4}{2}+2\binom{3}{2}=2\cdot 6+2\cdot 3 = 18$. We're up to 75.
Fourth part: Having the odd part summing to ten, how many ways? In this case, there's 3+7, 5+5, and 7+3, and then a 2 in each case. So there's three ways of placing the 2, and then three ways of choosing the remaining number pair, for a total of 9. That brings us to 84.
Final part: Having the odd part summing to twelve, how many ways? In this case, it's easiest to count - there's 3+3+3+3, 3+9, 5+7, 7+5, and 9+3, for a total of 5. So our answer is 89.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving a set of 3 Nonlinear Equations In the following 3 equations:
$$
k_1\cos^2(\theta)+k_2\sin^2(\theta) = c_1
$$
$$
2(k_2-k_1)\cos(\theta)\sin(\theta)=c_2
$$
$$
k_1\sin^2(\theta)+k_2\cos^2(\theta) = c_3
$$
$c_1$, $c_2$ and $c_3$ are given, and $k_1$, $k_2$ and $\theta$ are the unknowns. What is the best way to solve for the unknowns? Specifically, I need to solve many independent instances of this system in an algorithm. Therefore, ideally the solution method should be fast.
| Re-write in terms of double-angle expressions, and define values $A$, $B$, $C$:
$$\begin{align}
k_1 \cos^2\theta + k_2 \sin^2\theta = c_1 &\quad\implies\quad \cos 2\theta = \frac{2 c_1-k_1-k_2}{k_1-k_2} & =: A \\[5pt]
2 \left(k_1-k_2\right) \sin\theta \cos\theta = c_2 &\quad\implies\quad \sin 2\theta = \frac{c_2}{k_1-k_2} &=: B \\[5pt]
k_1 \sin^2\theta + k_2 \cos^2\theta = c_3 &\quad\implies\quad \cos 2\theta = \frac{k_1+k_2-2c_3}{k_1-k_2} &=: C
\end{align}$$
Now, $A=C$ implies $k_1 + k_2 = c_1 + c_3$ (but we'd know that simply by adding your first and third equations together), so that $k_2 = c_1+c_3-k_1$ and
$$
A = \frac{c_1-c_3}{2k_1-c_1-c_3} \qquad\qquad
B = \frac{c_2}{2k_1-c_1-c_3}
$$
Observe that you can already write
$$\tan 2\theta = \frac{c_2}{c_1-c_3}$$
which gives you $\theta$. For $k_1$ and $k_2$, note that $A^2+B^2=1$ implies
$$(c_1-c_3)^2 + c_2^2 = \left(2k_1-c_1-c_3\right)^2$$
so that
$$\begin{align}
k_1 &= \frac{1}{2}\left( c_1+c_3 \pm \sqrt{(c_1-c_3)^2+c_2^2} \right) \\
k_2 &= \frac{1}{2}\left( c_1+c_3 \mp \sqrt{(c_1-c_3)^2+c_2^2} \right)
\end{align}$$
(I would guess that the $k_1$ and $k_2$ are simply the two roots of the quadratic ---so that, say, $k_1$'s "$\pm$" is "$+$", while $k_2$'s "$\mp$" is "$-$"--- but this ambiguity is really for you to resolve based on your circumstances.)
| {
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"timestamp": "2023-03-29T00:00:00",
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Summing $ \sum _{k=1}^{n} k\cos(k\theta) $ and $ \sum _{k=1}^{n} k\sin(k\theta) $ I'm trying to find
$$\sum _{k=1}^{n} k\cos(k\theta)\qquad\text{and}\qquad\sum _{k=1}^{n} k\sin(k\theta)$$
I tried working with complex numbers, defining $z=\cos(\theta)+ i \sin(\theta)$ and using De Movire's, but so far nothing has come up.
| HINT:
$$\sum_{1\le k\le n}\left(\cos k\theta+i\sin k\theta\right)=\sum_{1\le k\le n}e^{i k\theta}=e^{i\theta}\cdot\frac{e^{in\theta}-1}
{e^{i\theta}-1}$$
Differentiate wrt $\theta$ and equate the real & the imaginary parts
$$\text{On differentiation, the left hand becomes }\sum_{1\le k\le n}k\left(-\sin k\theta+i\cos k\theta\right)$$
Alternatively, if $S=\sum_{1\le k\le n}k\cdot a^k=a+2\cdot a^2+\cdots+(n-1)\cdot a^{n-1}+n\cdot a^n$
$a\cdot S=a^2+2\cdot a^3+\cdots+(n-1)\cdot a^n+n\cdot a^{n+1}$
$$\text{On subtraction,}(a-1)S=n\cdot a^{n+1}-(a+a^2+\cdots+a^{n-1}+a^n)=n\cdot a^{n+1}-\frac{a(a^n-1)}{a-1}$$
$$\implies S=\sum_{1\le k\le n}k\cdot a^k=\frac{n\cdot a^{n+1}}{a-1}-\frac{a(a^n-1)}{(a-1)^2}$$
Putting $a=e^{i\theta},$
$$S=\sum_{1\le k\le n}k\cdot e^{ik\theta}=\frac{n\cdot e^{i(n+1)\theta}}{e^{i\theta}-1}-\frac{e^{i\theta}(e^{in\theta}-1)}{(e^{i\theta}-1)^2}$$ (The derivative method should bring us here,too)
$$S=\frac{ne^{i(n+1)\theta}}{e^{i\frac\theta2}(e^{i\frac\theta2}-e^{-i\frac\theta2})}-\frac{e^{i\theta}(e^{in\theta}-1)}{\{e^{i\frac\theta2}(e^{i\frac\theta2}-e^{-i\frac\theta2})\}^2}$$
As $e^{ix}-e^{-ix}=2i\sin x,$ $$S=\frac{ne^{i\frac{(2n+1)\theta}2}}{(e^{i\frac\theta2}-e^{-i\frac\theta2})}-\frac{(e^{in\theta}-1)}{(e^{i\frac\theta2}-e^{-i\frac\theta2})^2}$$
$$S=\frac{n\left(\cos\frac{(2n+1)\theta}2+i\sin\frac{(2n+1)\theta}2\right)}{2i\sin\frac\theta2}-\frac{(\cos n\theta+i\sin n\theta-1)}{(2i\sin\frac\theta2)^2}$$
$$=\frac{\cos n\theta-1}{4\sin^2\frac\theta2}+\frac{n\sin \frac{(2n+1)\theta}2}{2\sin\frac\theta2}+i\left(\frac{\sin n\theta}{4\sin^2\frac\theta2}-\frac{n(\cos\frac{(2n+1)\theta}2}{2\sin\frac\theta2}\right)$$
Again, $$S=\sum_{1\le k\le n}k\cdot e^{ik\theta}=\sum_{1\le k\le n}k(\cos k\theta+i\sin k\theta)=\sum_{1\le k\le n}k\cdot\cos k\theta+i\sum_{1\le k\le n}k\cdot\sin k\theta$$
Now, equate the real & the imaginary parts
| {
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"timestamp": "2023-03-29T00:00:00",
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Vectors and subspace For which real numbers $w, x, y, z$ do the vectors: $(w, 1, 1, 1)$, $(1, x, 1, 1)$, $(1, 1, y, 1)$, $(1, 1, 1, z)$ NOT form a basis of $\mathbb{R}^4$? For each of the values that you found, what is the dimension of the subspace of $\mathbb{R}^4$ that they span?
| The fact that the vectors does not form a basis of $\mathbb{R}^4$ means that they are not all linearly independent, which in turn is equivalent to:
$$det(A) = \begin{vmatrix} w & 1 & 1 & 1 \\ 1 & x & 1 & 1 \\ 1 & 1 & y & 1 \\ 1 & 1 & 1 & z \end{vmatrix} = 0$$
The determinant can be caluclated by expanding it using cofactors:
$$det(A) = w\begin{vmatrix} x & 1 & 1 \\ 1 & y & 1 \\ 1 & 1 & z \end{vmatrix} - \begin{vmatrix} 1 & 1 & 1 \\ 1 & y & 1 \\ 1 & 1 & z \end{vmatrix} + \begin{vmatrix} 1 & x & 1 \\ 1 & 1 & 1 \\ 1 & 1 & z \end{vmatrix} - \begin{vmatrix} 1 & x & 1 \\ 1 & 1 & y \\ 1 & 1 & 1 \end{vmatrix}$$
You can easily compute these determinants and see that:
$$det(A)=w(xyz+2-x-y-z)-(yz+2-y-z-1)+(x+z+1-xz-2)-(xy+2-x-y-1)=wxyz-wx-wy-wz-xy-xz-yz+2(w+x+y+z)-3$$
So all numbers $(w,x,y,z)$ satisfying the condition are solution to the problem:
$$wxyz-wx-wy-wz-xy-xz-yz+2(w+x+y+z)-3 = 0$$
The dimensionality of the subspace spanned by this vector is equal to the number of linearly indepndant vectors of the 4, or in other words the rank of the matrix $A$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Maximal Smallest Number Ran into a interesting problem which I have no idea how to solve but have the desire to.
Let a and b be two positive real numbers and let $m$(a,b) be the smallest of the three numbers $a,$ $1/b$ and $1/a + b.$ For which pairs (a,b) is $m$(a,b) maximal? Prove!
| It is not too hard to see that the maximum can only occur for $a=\frac{1}{b}=\frac{1}{a}+b$.
Suppose for instance that $a<\min(\frac{1}{b},b+\frac{1}{a})$, then we can increase $a$ a little and improve our $m(a,b)$.
For $\frac{1}{b}$ and $b+\frac{1}{a}$ similar things happen.
Thus $a=\frac{1}{b} = b+\frac{1}{a}$ implies $a = \sqrt{2}$ and $b = \frac{1}{2}\sqrt{2}$.
Checking the answer is quite easy:
Suppose that $m(a,b)>\sqrt{2}$, then $a>\sqrt{2}$ and $\frac{1}{b}>\sqrt{2}$. But then $\frac{1}{a}<\frac{1}{2}\sqrt{2}$ and $b<\frac{1}{2}\sqrt{2}$, so $b+\frac{1}{a}<\sqrt{2}$. A contradiction.
Thus $m(a,b)\leq\sqrt{2}$ for all $a$ and $b$. We already know that $m(\sqrt{2},\frac{1}{2}\sqrt{2})=\sqrt{2}$, thus this is optimal.
| {
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"timestamp": "2023-03-29T00:00:00",
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Symmetry properties of $\sin$ and $\cos$. Why does $\cos\left(\frac{3\pi}{2} - x\right) = \cos\left(-\frac{\pi}{2} - x\right)$? For a question such as:
If $\sin(x) = 0.34$, find the value of $\cos\left(\frac{3\pi}{2} - x\right)$.
The solution says that:
\begin{align*}
\cos\left(\frac{3\pi}{2} - x\right) &= \cos\left(-\frac{\pi}{2} - x\right)\\
&= \cos\left(\frac{\pi}{2} + x\right)\\
&= -\sin(x)
\end{align*}
Why does $\cos\left(\frac{3\pi}{2} - x\right) = \cos\left(-\frac{\pi}{2} - x\right)$?
Similarly, if $\cos(x) = 0.6$, find $\sin\left(\frac{3\pi}{2} + x\right)$, the solution says $\sin\left(\frac{3\pi}{2} + x\right)$ is equal to $\cos(\pi+x)$. How did they get $\cos(\pi+x)$?
| Recall your elementary trigonometric identity $\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$
So we may begin by expanding $\cos\left(\frac\pi2-x\right)$.$$\begin{align*}\cos\left(\frac\pi2-x\right)&=\cos\frac\pi2\cos(-x)-\sin\frac\pi2\sin(-x)\\&=0(\cos x)-1(-\sin x)\text{ knowing that }\cos x,\sin x\text{ are even and odd, respectively}\\&=\sin x\end{align*}$$
This can be seen graphically as the two sinusoidal waves are identical apart from a phase shift of $\frac\pi2$ rad:
Additionally, our functions are also periodic with period $2\pi$, i.e. for any integer $k$ we find that adding said multiple of $2\pi$ does not affect our result: $$\cos(x+2\pi k)=\cos x\\\sin(x+2\pi k)=\sin x$$
With this information, we may easily finish off our solution; recognize that by subtracting $2\pi$ from $\frac{3\pi}2-x$ yields our familiar $-\frac\pi2-x$. Recalling evenness of $\cos x$, i.e. $\cos(-x)=\cos x$, we rewrite $\cos\left(-\frac\pi2-x\right)=\cos\left(\frac\pi2+x\right)$. All we have left to do is recognize our phase shift.
| {
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A question on a proof of $y(y+1) \le (x+1)^2 \implies y(y-1) \le x^2$ I was reading the question posted here:Does $y(y+1) \leq (x+1)^2$ imply $y(y-1) \leq x^2$?. The solution posted was:
"Given $y^2+y≤x^2+2x+1$, if possible, let $x2<y(y−1)$. Clearly $y>1.$
Then $x^2+(2x+1)<y^2−y+(2x+1)$
So $y^2+y<y^2−y+2x+1$, which resolves to $y<x+1/2.$
Hence we also have $y−1<x−1/2.$
As $y>1$, the LHS is positive, and we can multiply the last two to get
$y(y−1)<x^2−1/4⟹y(y−1)<x^2$, a contradiction."
However, I don't think this is right because he had $y<x+1/2$, and in the last step he multiplied $y−1$ by $y$ and $x−1/2 by x+1/2$. However, this inequality is not in the same proportion anymore, because y does not equal x+1/2. It seems like he increased the value of the right side compared to the left, and then concluding that the right side is bigger/ It's like if I try to prove that $a \lt b$, and to solve this I say $a \lt b+5$. While this is true, this changes the inequality.
| To reword the proof given there:
We start with real numbers $x,y$ such that
$$\tag1 y\ge0 $$
$$\tag2 y^2+y\le x^2+2x+1$$
$$\tag3 x^2<y(y-1)$$
If we had $y\le1$, then the right hand side in $(3)$ would be the product of a nonnegative (according to $(1)$) and a nonpositve (by assumtion) factor, hence nonnegative, contradicting $x^2\ge0$. Therefore
$$\tag4 y>1.$$
Adding $2x+1$ to $(3)$ and combining with $(2)$ we find
$y^2+y\le x^2+2x+1<y^2-y+2x+1$, i.e.
$$\tag5 y\le x+\frac12.$$
By subtracting $1$, this becomes
$$\tag6 y-1\le x-\frac12.$$
Because of $(4)$ and $(6)$, the number $x-\frac12$ is positive, hence we are allowed to multiply $(5)$ with $x-\frac12$. Also, we can multiply $(6)$ with the (again by $(4)$) positive number $y$ and combine this to find
$$ x^2\stackrel{(3)}<y(y-1)\stackrel{y\cdot(6)}\le y\left(x-\frac12\right)\stackrel{(x-\frac12)\cdot(5)}\le \left(x+\frac12\right)\left(x-\frac12\right)=x^2-\frac14<x^2,$$
contradiction.
| {
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Find the values of $x,y$ and $z$ in these equations I'm stuck with these equations. Can somebody help me with solving it?
If $$(a+b) = \frac{x(y+z)}{x+y+z},\quad (b+c) = \frac{y(x+z)}{x+y+z} ,\quad (a+c) = \frac{z(x+y)}{x+y+z}$$ then find $x,y,z$.
| On adding all the 3 we get,
$$a+b+c=\frac{xy+yz+zx}{x+y+z}$$
Subtracting from the given equations we get $$c=\frac{yz}{x+y+z}$$ $$b=\frac{xz}{x+y+z}$$ $$a=\frac{xy}{x+y+z}$$
$$\implies \frac{y}{x}=\frac{c}{b}$$
$$\implies \frac{z}{x}=\frac{c}{a}$$
$$\implies \frac{z}{y}=\frac{b}{a}$$
$$a=\frac{y}{1+\frac{y}{x}+\frac{z}{x}}$$
$$\implies y = a(1+\frac{y}{x}+\frac{z}{x})=a(1+\frac{c}{b}+\frac{c}{a})=\frac{ab+bc+ca}{b}$$
I guess you can do the same for $x$ and $z$
| {
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Sums that are pythagorean and normal I noticed that
$3^2+4^2+15^2=9^2+13^2$
and also
$3+4+15=9+13$
Is there an easy way to find all pairs of disjoint sets of positive integers whose sum are the same and whose sum of squares are the same? How common are they?
| set of equations:
$\left\{\begin{aligned}&x+y+z=a+b\\&x^2+y^2+z^2=a^2+b^2\end{aligned}\right.$
Solutions and can be written as follows.
$x=t+qk+qt+k$
$y=t$
$z=qk$
$a=t+qk+k$
$b=t+qk+qt$
And more:
$z=kq$
$y=(k+q+t)^2$
$x=t^2+kq+tk+qt$
$a=k^2+2qk+t^2+qt+2kt$
$b=q^2+2qk+t^2+2qt+kt$
$k,t,q$ - are integers and can be any mark.
| {
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How to estimate the following integral: $\int_0^1 \frac{1-\cos x}{x}\,dx$
How to estimate the following integral?
$$
\int_0^1 \frac{1-\cos x}{x}\,dx
$$
| First note that the integral exists since $$0 \leq \dfrac{1-\cos(x)}x = \dfrac{2 \sin^2(x/2)}x \leq \dfrac{x}2$$ Hence, the integral is between $0$ and $1/4$. To compute the integral, proceed as follows. We have
$$\cos(x) = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} \mp = \sum_{k=0}^{\infty}(-1)^k \dfrac{x^{2k}}{(2k)!}$$
Hence,
$$1-\cos(x) = \dfrac{x^2}{2!} - \dfrac{x^4}{4!} + \dfrac{x^6}{6!} \pm = \sum_{k=1}^{\infty} (-1)^{k-1} \dfrac{x^{2k}}{(2k)!} $$
This gives us
$$\dfrac{1-\cos(x)}x = \sum_{k=1}^{\infty} (-1)^{k-1} \dfrac{x^{2k-1}}{(2k)!} $$
Now lets get back to the integral.
\begin{align}
\int_0^1 \dfrac{1-\cos(x)}x dx & = \int_0^1 \sum_{k=1}^{\infty} (-1)^{k-1}\dfrac{x^{2k-1}}{(2k)!} dx = \sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{(2k)!}\int_0^1 x^{2k-1} dx\\
& = \sum_{k=1}^{\infty}\dfrac{(-1)^{k-1}}{(2k)!} \cdot \dfrac1{2k}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/368988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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} |
Find $E[X^2 + Y^2]$ for the given joint density function. Let X and Y have joint probability density function
$f(x,y) = \frac{3}{2}(x^2 +y^2)$, with $0<x<1, 0<y<1$.
Find $E[X^2+Y^2]$.
I came up with $\frac{14}{15}$ by integrating $\int_0^1 \int_0^1 (x^2 +y^2)(x^2+y^2)$, but I've
never had to do anything but E[X].
| $\mathbf{E}[\varphi(X,Y)] = \int \int_{D} \varphi(x,y) f(x,y) dx dy = \frac{3}{2} \int _{0}^{1} \int _{0}^{1}(x^4 + 2 x^2 y^2 + y^4)dx dy = \frac{3}{2}\int_{0}^{1}[\int_{0}^{1}(x^4 + 2 x^2 y^2 + y^4)dy]dx$
Are you familiar with double integration? Can you handle from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/369540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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An application of contraction mapping theorem Suppose $\mathbb{R}$ is equipped with the standard metric and I want to show that the equation
$$
3x^3-3x^2+x-2=0
$$
has a unique root $c\geq1$. The equation is equivalent to
$$
f(x)=x,
$$
where $f(x)=1+\frac{1}{1+3x^2}$, so by showing that $f:[1,+\infty)\to \mathbb{R}$ is a contraction mapping I can prove the uniqueness of the root $c$, which will be the fixed point of $f$. Let $x,y\geq 1$, then
$$
|f(x)-f(y)|=\left|\frac{1}{1+3x^2}-\frac{1}{1+3y^2} \right|=3|x-y|\left| \frac{x+y}{(1+3x^2)(1+3y^2)}\right|
$$
So I guess I need to find a bound on $\left| \frac{x+y}{(1+3x^2)(1+3y^2)}\right|$, this is where I am stuck.
| Note that $x^2-2x+1=(x-1)^2\ge0$, hence $x^2+1\ge2x$ and similarly $y^2+1\ge2y$, hence indeed (very wastefully)
$$\begin{align}(1+3x^2)(1+3y^2)&=9x^2y^2+3x^2+3y^2+1\\&\ge 9+3x^2+3y^2+1\\&=3(x^2+1)+3(y^2+1)+3\\&\ge 6x+6y+3\\&>6( x+y) \end{align}$$
and you get a contraction factor of $\frac12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/370712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
What is the polar form of $ z = 1- \sin (\alpha) + i \cos (\alpha) $? How do I change $ z = 1- \sin (\alpha) + i \cos (\alpha) $ to polar? I got $r = (2(1-\sin(\alpha))^{\frac{1}{2}} $. I have problems with the exponential part. What should I do now?
| We have
\begin{eqnarray}
z&=&1-\sin\alpha+i\cos\alpha=1+i(\cos\alpha+i\sin\alpha)=1+e^{i\pi/2}e^{i\alpha}=1+e^{i\frac{\pi+2\alpha}{2}}\\
&=&\left(e^{-i\frac{\pi+2\alpha}{4}}+e^{i\frac{\pi+2\alpha}{4}}\right)e^{i\frac{\pi+2\alpha}{4}}=2\cos\left(\frac{2\alpha+\pi}{4}\right)e^{i\frac{\pi+2\alpha}{4}}.
\end{eqnarray}
For $\alpha \in \frac{\pi}{2}+2\pi\mathbb{Z}$ we have $z=0$.
For $\alpha \in (-\frac{3\pi}{2},\frac{\pi}{2})+4\pi\mathbb{Z}$ we have $|z|=2\cos\left(\frac{2\alpha+\pi}{4}\right),\ \arg z=\frac{\pi+2\alpha}{4}$.
For $\alpha \in (\frac{\pi}{2},\frac{5\pi}{2})+4\pi\mathbb{Z}$ we have $|z|=-2\cos\left(\frac{2\alpha+\pi}{4}\right),\ \arg z=\frac{5\pi+2\alpha}{4}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Prove That $x=y=z$ If $x, y,z \in \mathbb{R}$,
and if
$$ \left ( \frac{x}{y} \right )^2+\left ( \frac{y}{z} \right )^2+\left ( \frac{z}{x} \right )^2=\left ( \frac{x}{y} \right )+\left ( \frac{y}{z} \right )+\left ( \frac{z}{x} \right ) $$
Prove that $$x=y=z$$
| Let $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$, and let
$$
G=\sqrt[3]{abc}, A = \frac{a+b+c}{3}, Q=\sqrt{\frac{a^2+b^2+c^2}{3}}
$$
Then the given condition is $Q^2=A$, but by the power mean inequality
$$
Q^2\ge Q \ge A \ge G = 1
$$
with equality in each case only if $a=b=c=1$, i.e. if $x=y=z$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/372143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Prove that for any $n \in \mathbb{N}, 2^{n+2} 3^{n}+5n-4$ is divisible by $25$? I have question
Q
Prove that for any $n \in \mathbb{N}, 2^{n+2} 3^{n}+5n-4$ is divisible by $25$?
by using induction
Thanks
| Let $f(n)=2^{n+2}3^n+5n-4=4\cdot6^n+5n-4$
$f(m+1)-f(m)=4\cdot6^{m+1}+5(m+1)-4-(4\cdot6^m+5m-4)$
$=5\{5+4(6^m-1)\}$
Now, we know $(6^m-1)$ is divisible by $6-1=5$
$\implies f(m+1)\equiv f(m)\pmod {25}$
Now, $f(1)=2^3\cdot3+5\cdot1-4=25\implies f(1)$ divisible by $25$
Alternatively,
$2^{n+2}3^n+5n-4=4\cdot6^n+5n-4$
$=4(1+5)^n-5n-4$
$=4\left(1+\binom n1 5+\binom n2 5^2+\cdots +\binom n{n-1}5^{n-1}+5^n\right)+5n-4$ (Using Binomial Expansion)
$= 4\{1+5n+5^2\left(\binom n2 +\binom n35 +\binom n45^2+\cdots+\binom n{n-1}5^{n-3}+5^{n-2}\right)\}+5n-4$
As the Binomial coefficients are all integers, all the terms after $\binom n1 5$ is divisible by $5^2=25,$
$5^2\left(\binom n25^2 +\binom n35 +\binom n45^2+\cdots+\binom n{n-1}5^{n-3}+5^{n-2}\right)$ can bewritten as $25k$ where $k$ is an integer
So, $2^{n+2}3^n+5n-4=4\{1+5n+25k\}+5n-4=25(n+4k)$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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sum of squares of the roots of equation The equation is $$x^2-7[x]+5=0.$$
Here $[x]$ the greatest integer less than or equal to $x$. Some other method other than brute forcing. I tried a method of putting $[x]=q$ and $x=q+r$ which gives an equation: $$(q+r)^2-7q+5=0.$$
| HINT:
$0=x^2-7[x]+5\ge x^2-7x+5$
So, $x^2-7x+5\le0$
If $a,b(>a)$ are the roots of $x^2-7x+5=0, a\le x\le b$
Now, $a,b$ are $\approx .8,6.19$
Now, testing for the cases
As $x^2=7[x]-5$
If $0<x<1,[x]=0\implies x^2+5=0$ which does not have any real solution
If $1<x<2,[x]=1\implies x^2=7-5=2$
If $2<x<3,[x]=2\implies x^2=14-5=9$ but $x$ can not be integer greater than $[x]$
If $3<x<4,[x]=3\implies x^2=7\cdot3-5=16$ but $x$ can not be integer greater than $[x]$
If $4<x<5,[x]=4\implies x^2=7\cdot4-5=23$
If $5<x<6,[x]=5\implies x^2=7\cdot5-5=30$
If $6<x<7,[x]=6\implies x^2=7\cdot6-5=37$
If $7<x<8,[x]=7\implies x^2=7\cdot7-5=44<7^2$
So, the values of $x^2$ are $2,23,30,37$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Differentiating $\;y = x a^x$ My attempt:
$$\eqalign{
y &= x{a^x} \cr
\ln y &= \ln x + \ln {a^x} \cr
\ln y &= \ln x + x\ln a \cr
{1 \over y}{{dy} \over {dx}} &= {1 \over x} + \left(x \times {1 \over a} + \ln a \times 1\right) \cr
{1 \over y}{{dy} \over {dx}} &= {1 \over x} + \left({x \over a} + \ln a\right) \cr
{1 \over y}{{dy} \over {dx}} &= {1 \over x} + \left({{x + a\ln a} \over a}\right) \cr
{1 \over y}{{dy} \over {dx}} &= {{a + {x^2} + ax\ln a} \over {ax}} \cr
{{dy} \over {dx}} &= {{a + {x^2} + ax\ln a} \over {ax}} \times x{a^x} \cr
{{dy} \over {dx}} &= {{x{a^{x + 1}} + {x^3}{a^x} + {a^{x + 1}}{x^2}\ln a} \over {ax}} \cr
{{dy} \over {dx}} &= {a^x} + {x^2}{a^{x - 1}} + {a^x}x\ln a \cr
{{dy} \over {dx}} &= {a^x}\left(1 + {x^2}{a^{ - 1}} + x\ln a\right) \cr
{{dy} \over {dx}} &= {a^x}\left({{a + {x^2} + x\ln a} \over a}\right) \cr} $$
However the answer in the back of the book is:
$${dy \over dx} = a^x (1 + x\ln a)$$
What have I done wrong?
| $$
\frac{d}{dx} \ln a = \frac 1 a \frac{da}{dx}.
$$
If $\dfrac{da}{dx} = 0$, then you need to use that fact.
But here's a quicker way: Since $a$ is constant, $\ln a$ is constant, so its derivative is $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/376642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Number of circular placements of $n$ identical letters such that no two letters are adjacent. Suppose I have to place $3$ identical letters on a circular table which has $7$ slots in such a way that no two letters are in consecutive slots. In how many ways can I do this?
Can this be generalized into $n$ identical letters on a circular table with m slots?
| Whenever you have a circular table, it's easiest to fix the position of one thing that you're placing, then place the rest relative to that position.
So, let's say your letters are $\color{red}{\operatorname{A}}$, $\color{blue}{\operatorname{B}}$, and $\color{green}{\operatorname{C}}$. Let's fix $\color{red}{\operatorname{A}}$ in position $1$. There are $6$ remaining positions.
$$\begin{array}{|ccccccc|}
1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline
\color{red}{\operatorname{A}} & \times & \circ & \circ & \circ & \circ & \times \\
\end{array}$$
Note that $1$ is adjacent to $2$ and $7$. So, neither $\color{blue}{\operatorname{B}}$ nor $\color{green}{\operatorname{C}}$ can go there. If we place $\color{blue}{\operatorname{B}}$ in position $3$ or position $6$, we see that there are two places available for $\color{green}{\operatorname{C}}$.
$$\begin{array}{|ccccccc|}
1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline
\color{red}{\operatorname{A}} & \times & \color{blue}{\operatorname{B}} & \times & \circ & \circ & \times \\ \hline \color{red}{\operatorname{A}} & \times & \color{blue}{\operatorname{B}} & \times & \color{green}{\operatorname{C}} & \times & \times \\\color{red}{\operatorname{A}} & \times & \color{blue}{\operatorname{B}} & \times & \times & \color{green}{\operatorname{C}} & \times \\
\end{array}\hspace{30pt}\begin{array}{|ccccccc|}
1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline
\color{red}{\operatorname{A}} & \times & \circ & \circ & \times & \color{blue}{\operatorname{B}} & \times \\ \hline
\color{red}{\operatorname{A}} & \times & \color{green}{\operatorname{C}} & \times & \times & \color{blue}{\operatorname{B}} & \times \\ \color{red}{\operatorname{A}} & \times & \times & \color{green}{\operatorname{C}}& \times & \color{blue}{\operatorname{B}} & \times \\
\end{array}$$
On the other hand, if $\color{blue}{\operatorname{B}}$ goes into positions $4$ or $5$, then there is only $1$ position left for $\color{green}{\operatorname{C}}$.
$$\begin{array}{|ccccccc|}
1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline
\color{red}{\operatorname{A}} & \times & \times & \color{blue}{\operatorname{B}} & \times & \circ & \times \\ \hline
\color{red}{\operatorname{A}} & \times & \times & \color{blue}{\operatorname{B}} & \times & \color{green}{\operatorname{C}} & \times \\
\end{array}\hspace{30pt}\begin{array}{|ccccccc|}
1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline
\color{red}{\operatorname{A}} & \times & \circ & \times & \color{blue}{\operatorname{B}} & \times & \times \\ \hline
\color{red}{\operatorname{A}} & \times & \color{green}{\operatorname{C}}& \times & \color{blue}{\operatorname{B}} & \times & \times \\
\end{array}$$
In each of these final positions, $\color{blue}{\operatorname{B}}$ and $\color{green}{\operatorname{C}}$ can be permuted (i.e. $\color{green}{\operatorname{C}}$ placed before $\color{blue}{\operatorname{B}}$), but doing so only moves us to a solution we already obtained. (This also holds for the general case: the order of placement does not matter. Why?) Thus we are left with $$2\times 2 + 2\times 1=6\text{ ways.}$$
Can you extend this counting technique to higher numbers of slots?
| {
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"timestamp": "2023-03-29T00:00:00",
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A Permutation problem with sum restrictions In how many permutations of digits 1, 2, 3,...,9 are the following two conditions satisfied:
*
*Sum of digits between 1 and 2 (including both) is 12.
*Sum of digits between 2 and 3 (including both) is 23.
My Attempt
Clearly 3 cannot lie between 1 and 2. So the sequence will be 1..2..3 or 3..2..1
or 2..1..3 or 3..1..2.
For case 1, I started counting the possible sub cases.
Similarly, I might count all cases, but I am sure there is a more elegant way.
Just a hint would suffice.
| Properly between $1$ and $2$ we need a total of $9$.
As we cannot use $1,2,3$ to produce this $9$, we find thet the $9$ can only be the digit $9$ itself or $4+5$.
If $1$ is between $2$ and $3$, we need a total of $23-3-12=8$ between $2$ and $3$. This can only be $8$ itself (as $1,2,3$ are already used). This gives us $(2,9,1,8,3)$ or reversed as possibilities; or we have $(2,4,5,1,8,3)$ or the latter with $4,5$ swapped or either reversed. In the first cases, this block and the remaining four digits can occur in $5!$ orders, in the second case, the block and remaining three digits can occur in $4!$ orders. This gives us $2\cdot 5!+4\cdot 4!=336$ possibilities.
If $1$ is not between $2$ and $3$, we need a total of $18$ between $2$ and $3$, where we must avoid $1,2,3$. Note that
$$18=9+5+4=8+6+4=7+6+5$$
are the only possibilities we have. However, we have used either $9$ or $4$ and $5$ between $1$ and $2$. This strikes out the first sum and allows only $9$ between $1$ and $2$ in the other two cases.
The three summands between $2$ and $3$ can occur in $3!$ orders. The whole block $(1,9,2,?,?,?,3)$ can be flipped and the block and the remaining two digits can occur in $3!$ orders. This gives us $2\cdot3!\cdot 3!=72$ possibilities.
In total we count $336+72=408$ possibilities.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that for any odd $n$ it follows that $n^2 \equiv 1 \mod 8$ and for uneven primes $p\neq 3$ we have $p^2 \equiv 1\mod 24$. Show that for any odd $n$ it follows that $n^2 \equiv 1 \mod 8$ and for uneven primes $p\neq 3$ we have $p^2 \equiv 1\mod 24$.
My workings so far: I proceeded by induction. Obviously $1^2 \equiv 1 \mod 8$. Then assume that for a certain uneven number $k$ we have $k^2 \equiv 1 \mod 8$. Then the the next uneven number is $k+2$ and $(k+2)^2 = k^2 + 4k + 4 \equiv 4k + 5 \mod 8$. by the induction hypothesis. Now since $k$ is uneven we can write it as $k=2j+1$ and thus $4k+5 =8j+9 \equiv 1 \mod 8$ and we have shown what was asked for by induction.
However, in the case of a prime number $n$ I am not so certain how to proceed because I can't use induction. It comes down to proving that $24|(p^2-1)$. This is certainly the case for the first uneven prime $p \neq 3$, namely $p=5$ such that $p^2-1 =24$. How would I proceed from here, or how should I approach the problem differently? Many thanks in advance!
| If $n$ is odd, $n = 2 k + 1$ and $n^2 = 4 k^2 + 4 k + 1 = 4 k (k + 1) + 1$, and $4 k (k + 1)$ is divisible by 8. Thus $n^2 \equiv 1 \pmod{8}$.
If $p > 3$ is prime, it is odd, and so by the above $p^2 \equiv 1 \pmod{8}$. As it is prime, $p \equiv 1 \pmod{3}$ or $p \equiv 2 \pmod{3}$, so $p^2 \equiv 1 \pmod{3}$. Now, as $\gcd(3, 8) = 1$, combining the above congruences gives $p^2 \equiv 1 \pmod{24}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/380696",
"timestamp": "2023-03-29T00:00:00",
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Variance for the number of trials before success in an urn problem without replacement This question is asked as an extension of:
Expectation of number of trials before success in an urn problem without replacement
(Note: I am not the author of the original question.)
We have $b$ blue balls and $r$ red balls in an urn. Sampling the urn sequentially and without replacement, we remove red balls until we select a blue ball. In the previous question, Byron Schmuland (and others) calculated that the expected number of balls drawn before drawing a blue ball should be:
$\mathbb{E}(\mbox{number of balls drawn})=1+\sum_{i=1}^r \mathbb{E}(Z_i)=1+r\left({1\over b+1}\right)$
And therefore, we have that:
$\mathbb{E}(\mbox{number of red balls drawn})=1+\sum_{i=1}^r \mathbb{E}(Z_i)-1=1+r\left({1\over b+1}\right)-1$
My question is:
What is the variance for the number of red balls drawn, i.e. $Var[\mbox{number of red balls drawn}]$?
| The probability of drawing $n$ red balls then a blue ball is
$$
\frac{r}{b+r}\frac{r-1}{b+r-1}\frac{r-2}{b+r-2}\cdots\frac{r-n+1}{b+r-n+1}\frac{b}{b+r-n}
=\frac{\binom{\vphantom{b+}r}{n}}{\binom{b+r-1}{n}}\frac{b}{b+r}\qquad\tag{1}
$$
Since a blue ball must be drawn eventually, $(1)$ indicates that
$$
\sum_{n=0}^r\frac{\binom{\vphantom{b+}r}{n}}{\binom{b+r-1}{n}}=\frac{b+r}{b}\tag{2}
$$
which can be proven by induction on $r$.
$(2)$ is true for $r=0$. Assume $(2)$ is true for $r-1$, then
$$
\begin{align}
\sum_{n=0}^r\frac{\binom{\vphantom{b+}r}{n}}{\binom{b+r-1}{n}}
&=1+\sum_{n=1}^r\frac{\frac{\vphantom{b+}r}{n}\binom{r-1}{n-1}}{\frac{b+r-1}{n}\binom{b+r-2}{n-1}}\\
&=1+\frac{r}{b+r-1}\sum_{n=0}^{r-1}\frac{\binom{r-1}{n}}{\binom{b+r-2}{n}}\\
&=1+\frac{r}{b+r-1}\frac{b+r-1}{b}\\
&=\frac{b+r}{b}
\end{align}
$$
There is a more direct proof of this identity at the end of this answer.
Using $(2)$, the expected number of red balls drawn is
$$
\begin{align}
\sum_{n=0}^r\frac{\binom{\vphantom{b+}r}{n}}{\binom{b+r-1}{n}}\frac{b}{b+r}n
&=\frac{br}{b+r}\sum_{n=0}^r\frac{\binom{r-1}{n-1}}{\binom{b+r-1}{n}}\\
&=\frac{br}{b+r}\sum_{n=0}^r\frac{\binom{r\vphantom{+1}}{n}-\binom{r-1}{n}}{\binom{b+r-1}{n}}\\
&=\frac{br}{b+r}\left(\frac{b+r}{b}-\frac{b+r}{b+1}\right)\\
&=\frac{r}{b+1}\tag{3}
\end{align}
$$
which agrees with the result in the question.
Using $(2)$ the expected value of $n(n-1)$ is
$$
\begin{align}
\sum_{n=0}^r\frac{\binom{\vphantom{b+}r}{n}}{\binom{b+r-1}{n}}\frac{b}{b+r}n(n-1)
&=\frac{br(r-1)}{b+r}\sum_{n=0}^r\frac{\binom{r-2}{n-2}}{\binom{b+r-1}{n}}\\
&=\frac{br(r-1)}{b+r}\sum_{n=0}^r\frac{\binom{r\vphantom{-1}}{n}-2\binom{r-1}{n}+\binom{r-2}{n}}{\binom{b+r-1}{n}}\\
&=\frac{br(r-1)}{b+r}\left(\frac{b+r}{b}-2\frac{b+r}{b+1}+\frac{b+r}{b+2}\right)\\
&=\frac{2r(r-1)}{(b+1)(b+2)}\tag{4}
\end{align}
$$
Thus, the variance, which is the mean of the squares minus the square of the mean is
$$
\frac{2r(r-1)}{(b+1)(b+2)}+\frac{r}{b+1}-\left(\frac{r}{b+1}\right)^2
=\frac{br(b+r+1)}{(b+1)^2(b+2)}\tag{5}
$$
| {
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} |
Binomial theorem for prime exponent Could you explain to me why for prime $p$ we have the following?
$$(x+y)^p - (x^p + y^p)= x^p + \binom{p}{1}x^{p-1}y + \binom{p}{2}x^{p-2}y^2 + \binom{p}{p-1}xy^{p-1} + y^p.$$
I found it here: Abstract Algebra - Solutions to Homework (pdf),
solution 5d, page 3.
I would really appreciate your help.
Thank you.
| As written, what's in the notes is incorrect. We should have
\begin{align*}
(x + y)^p - (x^p + y^p) &= x^p + \begin{pmatrix} p\\ 1\end{pmatrix}x^{p - 1}y + \begin{pmatrix} p\\ 2\end{pmatrix} x^{p - 2}y^2 + \ldots + \begin{pmatrix} p\\ p - 1\end{pmatrix}x y^{p - 1} + y^p - (x^p + y^p)\\
&= \begin{pmatrix} p\\ 1\end{pmatrix}x^{p - 1}y + \begin{pmatrix} p\\ 2\end{pmatrix} x^{p - 2}y^2 + \ldots + \begin{pmatrix} p\\ p - 1\end{pmatrix}x y^{p - 1}\\
&= \sum_{k = 1}^{p-1}\begin{pmatrix}p\\k\end{pmatrix}x^{p - k} y^k.
\end{align*}
I can only assume the writer of the notes got hungry and left in the middle of writing the solutions, and forgot to write a few terms ($\ldots$) upon returning.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/385884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find the integral of $\int {{2 \over {\sqrt x (x - 4)}}dx} $ given the substitution: $u = \sqrt x $ My stab at it:
$\eqalign{
& \int {{2 \over {\sqrt x (x - 4)}}dx} \cr
& u = \sqrt x \cr
& {{du} \over {dx}} = {1 \over 2}{x^{ - {1 \over 2}}} = {1 \over {2\sqrt x }} \cr
& {{dx} \over {du}} = 2\sqrt x \cr
& \int {f(x)dx} = \int {{2 \over {\sqrt x (x - 4)}} \times 2\sqrt x {\rm{ }}du} \cr
& = \int {{{4\sqrt x } \over {\sqrt x (x - 4)}}du} \cr
& = \int {{{4u} \over {u({u^2} - 4)}}du} \cr
& = \int {{4 \over {({u^2} - 4)}}du} \cr
& = \int {{4 \over {({u^2} - 4)}}du} \cr
& = 4 \times {1 \over {2u}} \times \ln |({u^2} - 4)| + C \cr
& = {{2\ln |(x - 4)|} \over {\sqrt x }} + C \cr} $
My problem is the answer is:
$\ln |{{\sqrt x - 2} \over {\sqrt x + 2}}| + C$
I dont know how to simplify to reach this, this is most likely because I've probably made a mistake integrating. I am aware that you can reproduce the above answer using partial fractions to simplify the expression that is in terms of U and then integrate, but i was wondering if it was possible to duplicate the answer with my way of doing things..
Thank you for your help!
| You did just fine up to $\;\int \frac{4}{u^2-4}\,du\;\;$
Note that $$\int \frac{4}{u^2-4}du = \int \frac{4}{(u - 2)(u+2)}\,du = \int \left(\frac {A}{u - 2} + \frac{B}{u + 2} \right)\,du$$
That is, we can use partial fractions to break our integral into the sum of two nicely integrable functions by simply solving for the numerators $A, B$.
In this case, $A$ and $B$ turn out to be "nice": To solve for $A, B$, note that $A(u + 2) + B(u - 2) = 4$.
Letting $u = 2 \implies 4A = 4 \implies A = 1$
Letting $u = -2 \implies -4B = 4 \implies B = -1$.
So our integral $$\int \left(\frac {A}{u - 2} + \frac{B}{u + 2} \right)\,du = \int \left(\frac 1{u-1} + \frac {-1}{u + 2} \right) \,du$$
Also note that for the solution you are given:
$$\ln \left|{{\sqrt x - 2} \over {\sqrt x + 2}}\right| + C = \ln\left|\sqrt{x} - 2\right| - \ln\left| \sqrt x + 2\right| + C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/386691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Are all limits solvable without L'Hôpital Rule or Series Expansion Is it always possible to find the limit of a function without using L'Hôpital Rule or Series Expansion?
For example,
$$\lim_{x\to0}\frac{\tan x-x}{x^3}$$
$$\lim_{x\to0}\frac{\sin x-x}{x^3}$$
$$\lim_{x\to0}\frac{\ln(1+x)-x}{x^2}$$
$$\lim_{x\to0}\frac{e^x-x-1}{x^2}$$
$$\lim_{x\to0}\frac{\sin^{-1}x-x}{x^3}$$
$$\lim_{x\to0}\frac{\tan^{-1}x-x}{x^3}$$
| Using only trigonometric identities, in this answer, it is shown that
$$
\lim_{x\to0}\frac{x-\sin(x)}{x-\tan(x)}=-\frac12\tag{1}
$$
Therefore, if we subtract from $1$, we get
$$
\lim_{x\to0}\frac{\tan(x)-\sin(x)}{\tan(x)-x}=\frac32\tag{2}
$$
Using the limits proven geometrically in this answer, we can derive
$$
\begin{align}
\lim_{x\to0}\frac{\tan(x)-\sin(x)}{x^3}
&=\lim_{x\to0}\frac{\tan(x)(1-\cos(x))}{x^3}\\
&=\lim_{x\to0}\frac{\tan(x)}x\frac{\sin^2(x)}{x^2}\frac1{1+\cos(x)}\\
&=\frac12\tag{3}
\end{align}
$$
we can divide $(3)$ by $(2)$ to get
$$
\bbox[5px,border:2px solid #C0A000]{\lim_{x\to0}\frac{\tan(x)-x}{x^3}=\frac13}\tag{4}
$$
and we can multiply $(1)$ by $(4)$ to get
$$
\bbox[5px,border:2px solid #C0A000]{\lim_{x\to0}\frac{\sin(x)-x}{x^3}=-\frac16}\tag{5}
$$
Note that $(4)$ implies
$$
\begin{align}
\lim_{x\to0}\frac{\tan(x)-x}{\tan^3(x)}
&=\lim_{x\to0}\frac{\tan(x)-x}{x^3}\lim_{x\to0}\frac{x^3}{\tan^3(x)}\\
&=\frac13\cdot1\tag{6}
\end{align}
$$
Therefore, substituting $x\mapsto\tan^{-1}(x)$,
$$
\bbox[5px,border:2px solid #C0A000]{\lim_{x\to0}\frac{\tan^{-1}(x)-x}{x^3}=-\frac13}\tag{7}
$$
Similarly, $(5)$ implies
$$
\begin{align}
\lim_{x\to0}\frac{\sin(x)-x}{\sin^3(x)}
&=\lim_{x\to0}\frac{\sin(x)-x}{x^3}\lim_{x\to0}\frac{x^3}{\sin^3(x)}\\
&=-\frac16\cdot1\tag{8}
\end{align}
$$
Therefore, substituting $x\mapsto\sin^{-1}(x)$,
$$
\bbox[5px,border:2px solid #C0A000]{\lim_{x\to0}\frac{\sin^{-1}(x)-x}{x^3}=\frac16}\tag{9}
$$
Using the Binomial Theorem, we have
$$
\left(1+\frac xn\right)^n-1-x
=\frac{n-1}{2n}x^2+\sum_{k=3}^n\binom{n}{k}\frac{x^k}{n^k}\tag{10}
$$
and for $|x|\le1$,
$$
\begin{align}
\left|\sum_{k=3}^n\binom{n}{k}\frac{x^k}{n^k}\right|
&=|x|^3\left|\sum_{k=3}^n\binom{n}{k}\frac{x^{k-3}}{n^k}\right|\\
&\le |x|^3\sum_{k=3}^\infty\frac1{k!}\\[6pt]
&=|x|^3\left(e-\tfrac52\right)\tag{11}
\end{align}
$$
Combining $(10)$ and $(11)$ and taking the limit as $n\to\infty$ yields
$$
\frac{e^x-1-x}{x^2}=\frac12+O(|x|)\tag{12}
$$
and therefore,
$$
\bbox[5px,border:2px solid #C0A000]{\lim_{x\to0}\frac{e^x-1-x}{x^2}=\frac12}\tag{13}
$$
A simple corollary of $(13)$ is
$$
\lim_{x\to0}\frac{e^x-1}x=1\tag{14}
$$
Therefore, it follows that
$$
\begin{align}
\lim_{x\to0}\frac{e^x-1-x}{(e^x-1)^2}
&=\lim_{x\to0}\frac{e^x-1-x}{x^2}\lim_{x\to0}\frac{x^2}{(e^x-1)^2}\\
&=\frac12\tag{15}
\end{align}
$$
If we substitute $x\mapsto\log(1+x)$ in $(15)$, we get
$$
\lim_{x\to0}\frac{x-\log(1+x)}{x^2}=\frac12\tag{16}
$$
Therefore,
$$
\bbox[5px,border:2px solid #C0A000]{\lim_{x\to0}\frac{\log(1+x)-x}{x^2}=-\frac12}\tag{17}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/387333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "121",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\int {\sin 2\theta \over 1 + \cos \theta} \, d\theta $, using the substitution $u = 1 + \cos \theta $
Evaluate $$\int_0^{\pi \over 2} {{\sin 2\theta } \over {1 + \cos \theta }} \, d\theta$$ using the substitution $u = 1 + \cos \theta $
Using
$$\begin{align}
u &= 1 + \cos \theta \\
\frac{du}{d\theta} &= -\sin\theta \\
d\theta &= \frac{du}{-\sin\theta}\\
\end{align}$$
$$\begin{align}
\int_0^{\pi \over 2} {{\sin 2\theta } \over {1 + \cos \theta }} &= \int_0^{\pi \over 2} {{\sin 2\theta } \over {1 + \cos \theta }} {{du} \over { - \sin \theta }} \\
& = \int_0^{\pi \over 2} {{2\sin \theta \cos \theta } \over {1 + \cos \theta }} {{du} \over { - \sin \theta }} \\
& = \int_1^2 {{2(u - 1)} \over { - u}}\, du \\
& = \int_1^2 {{2 - 2u} \over u} \,du \\
&= \int_1^2\left( {2 \over u} - 2\right) \,du \\
& = \left[ 2\ln |u| - 2u \right]_1^2 \\
& = (2\ln 2 - 2(2)) - (2\ln1 - 2(1)) \\
& = (2\ln 2 - 4) - (2\ln 1 - 2) \\
& = 2\ln 2 - 2 \\
\end{align} $$
This answer is incorrect, the answer in the book is
$$2 - 2\ln 2$$
Could someone tell me where and how I went wrong? I'm not sure, but I think it may have been when I distributed the minus sign from the denominator to the numerator.
| HINT:
$$\frac{\sin2\theta}{1+\cos\theta}=\frac{2\sin\theta\cos\theta}{1+\cos\theta}=\frac{2\sin\theta(1+\cos\theta-1)}{1+\cos\theta}=2\sin\theta -\frac{2\sin\theta}{1+\cos\theta}$$
Now put $1+\cos\theta=u$ in the second integral.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
What have I done wrong in solving this problem with indices rules? The question asks to simplify:
$$\left(\dfrac{25x^4}{4}\right)^{-\frac{1}{2}}.$$
So I used $(a^m)^n=a^{mn}$ to get
$$\dfrac{25}{4}x^{-2} = \dfrac{25}{4} \times \dfrac{1}{x^2} = \frac{25}{4x^2} = \frac{25}{4}x^{-2}$$
However, this isn't the answer, and I can't see what I've done wrong.
This is what the mark scheme says:
$$\left(\dfrac{25x^4}{4}\right)^{-\frac{1}{2}} = \left[\left(\frac{4}{25x^4}\right)^{\frac{1}{2}} \text{ or } \left(\frac{5x^2}{2}\right)^{-1} \text{ or } \frac{1}{\left(\dfrac{25x^4}{4}\right)^{\frac{1}{2}}}\right] = \frac{2}{5}x^{-2}$$
To me, my answer looks more simple than theirs, and I can't see what I've done wrong.
| $$(\dfrac{25x^4}{4})^{-\dfrac{1}{2}}$$
$$(\dfrac{25}{4})^{-\dfrac{1}{2}}(x^{-2})$$
$$(\dfrac{4}{25})^{\dfrac{1}{2}}(x^{-2})$$
$$(\dfrac{2}{5})(x^{-2})$$
$$\dfrac {2}{5x^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/388748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
If $f(x)=\sqrt{1-\sqrt{1-x^2}}$, then prove that $f(x)$ is continuous on $[-1,1]$ and differentiable on $(-1,0) \cup (0,1)$.
If $f(x)=\sqrt{1-\sqrt{1-x^2}}$, then prove that $f(x)$ is continuous on $[-1,1]$ and differentiable on $(-1,0) \cup (0,1)$.
Please prove using $$\lim_{x\to c^+}f(x)=\lim_{x\to c^-}f(x)=f(c)$$
and
$$\lim_{x\to c^+}\frac{f(x)-f(c)}{x-c}=\lim_{x\to c^-}\frac{f(x)-f(c)}{x-c}$$
Thankyou
| Using $\lim_{x\to c}f(x)=f(c)$ to check continuity, in this case, is just the same as checking that you have a composition of continuous functions, which is continuous wherever it is defined.
Since $1-x^2\ge0$ means $|x|<1$ and implies $\sqrt{1-x^2}\le1$, your function is defined (and continuous) in $[-1,1]$.
The derivative can be computed with the chain rule, keeping in mind that the derivative of $x\mapsto\sqrt{x}$ exists only for $x>0$:
$$
f'(x)=\frac{1}{2\sqrt{1-\sqrt{1-x^2}}}\frac{x}{\sqrt{1-x^2}}
$$
You have to be careful where $1-\sqrt{1-x^2}=0$ and $\sqrt{1-x^2}=0$, that is at $x=0$, $x=1$ and $x=-1$, where the formula doesn't apply.
However, since the function is continuous, you can compute the limit of the derivative; it's easy to show that
\begin{align}
\lim_{x\to-1^+}f'(x)&=-\infty\\
\lim_{x\to 1^-}f'(x)&=\infty
\end{align}
so your function is not differentiable in $-1$ and $1$. For the limit at $0$ you can write the derivative as
\begin{align}
f'(x)&=
\frac{1}{2\sqrt{1-x^2}}\frac{x}{\sqrt{1-\sqrt{1-x^2}}}\\
&=
\frac{1}{2\sqrt{1-x^2}}\frac{x\sqrt{1+\sqrt{1-x^2}}}{\sqrt{1-(1-x^2)}}\\
&=
\frac{\sqrt{1+\sqrt{1-x^2}}}{2\sqrt{1-x^2}}\frac{x}{\sqrt{x^2}}\\
&=
\frac{\sqrt{1+\sqrt{1-x^2}}}{2\sqrt{1-x^2}}\frac{x}{\lvert x\rvert}\\
\end{align}
so that
$$
\lim_{x\to0^-}f'(x)=-\frac{\sqrt{2}}{2},\qquad
\lim_{x\to0^+}f'(x)=\frac{\sqrt{2}}{2}
$$
which proves that $f$ is not differentiable at $0$.
Here's the graph
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
I need help to evaluate this integral with a constant $c$ Find the value of the constant $c$ for which the integral converges, and evaluate the integral:
$$\int_0^\infty \left(\frac{9x}{x^2+1}-\frac{9c}{2x+1}\right)dx$$
| We have
\begin{align}
\int_0^a \left(\dfrac{9x}{x^2+1} - \dfrac{9c}{2x+1}\right) dx & = \left. \left(\dfrac92 \log\left(x^2+1 \right) - \dfrac{9c}2 \log(2x+1) \right) \right \vert_0^a\\
& = \dfrac{9}2 \log \left(\dfrac{a^2+1}{(2a+1)^c}\right)
\end{align}
Given this, as $a \to \infty$, what happens if $c<2$ or $c>2$? Once you get the $c$ from this, use this $c$ and let $a \to \infty$, to get the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/389778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Reducibility of $x^{2n} + x^{2n-2} + \cdots + x^{2} + 1$ Just for fun I am experimenting with irreducibility of certain polynomials over the integers. Since $x^4+x^2+1=(x^2-x+1)(x^2+x+1)$, I thought perhaps $x^6+x^4+x^2+1$ is also reducible. Indeed:
$$x^6+x^4+x^2+1=(x^2+1)(x^4+1)$$
Let $f_n(x)=x^{2n}+x^{2n-2}+\cdots + x^2+1$. Using Macaulay2 (powerful software package) I checked that:
$$f_4(x) = (x^4-x^3+x^2-x+1)(x^4+x^3+x^2+x+1)$$
We get that
$$ f_5(x)=(x^2+1)(x^2-x+1)(x^2+x+1)(x^4-x^2+1)$$
The polynomial is reducible for $n=6, 7, 8, 9$ as far as I checked. I suspect that $f_n(x)$ is reducible over integers for all $n\ge 2$. Is this true?
Thanks!
| There is a closed quotient form for the polynomial using the geometric series formula:
$$\sum_{k=0}^n x^{2k}=\frac{x^{2n+2}-1}{x^2-1}.$$
The theory of cyclotomic polynomials tells us how to reduce polynomials of the form $x^m-1$. In particular, the irreducible factors of $x^m-1$ are $\Phi_d(x)$ for each divisor $d\mid m$ (no repeated factors are present in this factorization). See references for $\Phi_n(x)$ for further details.
Thus,
$$\frac{x^{2n+2}-1}{x^2-1}=\prod_{\substack{d\mid (n+1) \\ d\ne1}}\Phi_d(x^2).$$
If $n+1$ is composite then the above factorization involves more than one cyclotomic polynomial hence $f_n(x)$ is reducible. On the other hand, if $n>1$ is odd we have
$$\frac{x^{2(n+1)}-1}{x^2-1}=\frac{x^n+1}{x+1}\frac{x^n-1}{x-1}.$$
Both quotients are in fact integer-coefficient polynomials:
$$\frac{x^n-1}{x-1}=x^{n-1}+\cdots+x+1, \quad \frac{x^n+1}{x+1}=\frac{(-x)^n-1}{(-x)-1}=(-x)^{n-1}+\cdots+(-x)+1.$$
Note the latter depended on $n$ being odd so that $(-x)^n=-x^n$.
The only exception is $n=1$, and in that particular case we know that $x^2+1$ is irreducible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/391086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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"answer_id": 1
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Complex integral help involving $\sin^{2n}(x)$ Show that $$\int_0^\pi \sin^{2n} \theta d\theta=\dfrac{\pi(2n)!}{(2^n n!)^2} $$
So far I have came up with:
$$\sin^{2n} \theta = \left(\dfrac {z-z^{-1}}{2i} \right)^{2n}$$ and I know I should be using:
$$(a+b)^n=\sum_{k=0}^n \dfrac{n!}{(n-k)!k!}a^kb^{n-k}$$
but I'm not sure how to get the conclusion. Any help will be greatly appreciated.
| This is my favorite example of integration by parts.
Let $I_n = \displaystyle \int_{0}^{\frac{\pi}{2}} \sin^n(x) dx$.
$I_n = \displaystyle \int_{0}^{\frac{\pi}{2}} \sin^{n-1}(x) d(-\cos(x)) = -\sin^{n-1}(x) \cos(x) |_{0}^{\frac{\pi}{2}} + \int_{0}^{\frac{\pi}{2}} (n-1) \sin^{n-2}(x) \cos^2(x) dx$
The first expression on the right hand side is zero since $\sin(0) = 0$ and $\cos(\frac{\pi}{2}) = 0$.
Now rewrite $\cos^2(x) = 1 - \sin^2(x)$ to get
$$I_n = (n-1) \left(\displaystyle \int_{0}^{\frac{\pi}{2}} \sin^{n-2}(x) dx - \int_{0}^{\frac{\pi}{2}} \sin^{n}(x) dx \right) = (n-1) I_{n-2} - (n-1) I_n$$
Rearranging we get $n I_n = (n-1) I_{n-2}$, $I_n = \frac{n-1}{n}I_{n-2}$.
Using this recurrence we get
$$I_{2k+1} = \frac{2k}{2k+1}\frac{2k-2}{2k-1} \cdots \frac{2}{3} I_1$$
$$I_{2k} = \frac{2k-1}{2k}\frac{2k-3}{2k-2} \cdots \frac{1}{2} I_0$$
$I_1$ and $I_0$ can be directly evaluated to be $1$ and $\frac{\pi}{2}$ respectively and hence,
$$I_{2k+1} = \frac{2k}{2k+1}\frac{2k-2}{2k-1} \cdots \frac{2}{3} = \dfrac{4^k (k!)^2}{(2k+1)!}$$
$$I_{2k} = \frac{2k-1}{2k}\frac{2k-3}{2k-2} \cdots \frac{1}{2} \frac{\pi}{2} = \dfrac{(2k)!}{(k!)^2}\dfrac{\pi}{2^{2k+1}}$$
Your integral is twice the above integral, since the sine function is symmetric about $x=\pi/2$. Hence, the value of your integral is
$$J_{2k+1} = \dfrac{2^{2k+1} (k!)^2}{(2k+1)!}$$
$$J_{2k} = \frac{2k-1}{2k}\frac{2k-3}{2k-2} \cdots \frac{1}{2} \frac{\pi}{2} = \dfrac{(2k)!}{(k!)^2}\dfrac{\pi}{2^{2k}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How prove this $\left(\sqrt{a^2+b^4}-a\right)\left(\sqrt{b^2+a^4}-b\right)\le a^2b^2$ let $a,b\in R$,and such that
$$\left(\sqrt{a^2+b^4}-a\right)\left(\sqrt{b^2+a^4}-b\right)\le a^2b^2$$
prove that $$a+b\ge 0$$
I think this is very beatifull problem, have you nice methods? Thank you,
I have see this problem
$$\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1$$
then we have $$x+y=0$$
This problem have some nice methods,
let $f(x)=\ln\left({x+\sqrt{x^2+1}}\right)$, then $f(x)$ is increasing and is odd funciotion, and
$$\ln\left({y+\sqrt{y^2+1}}\right)=-\ln{-y+\sqrt{y^2+1}}=-f(-y)$$
so
$$f(x)+f(-y)=0\Longrightarrow f(x)=f(-y)$$
then $x+y=0$
and I have see this problem
$$\left(x+\sqrt{y^2+1}\right)\left(y+\sqrt{x^2+1}\right)=1$$
then we have $$x+y=0$$
solution: let $$x=\dfrac{1}{2}\left(u-\dfrac{1}{u}\right),y=\dfrac{1}{2}\left(v-\dfrac{1}{v}\right)$$
then we have
$$\dfrac{(uv-1)((u+v)^2uv+(u-v)^2)}{u^2v^2}=0$$
since $$(u+v)^2uv+(u-v)^2\ge 0$$
so $$uv=1$$
then $$x+y=\dfrac{1}{2}\left(u+v-\dfrac{1}{u}-\dfrac{1}{v}\right)=0$$
| Step 1: Multiply by the conjugate, we get that
$$ b^4 a^4 \leq a^2 b^2 ( \sqrt{a^2 + b^4} +a ) ( \sqrt{ b^2 + a^4} + b), $$
$$ a^2 b^2 \leq ( \sqrt{a^2 + b^4} +a ) ( \sqrt{ b^2 + a^4} + b)$$.
This gives us the chain of inequalities
$$ ( \sqrt{a^2 + b^4} -a ) ( \sqrt{ b^2 + a^4} - b) \leq a^2 b^2 \leq ( \sqrt{a^2 + b^4} +a ) ( \sqrt{ b^2 + a^4} + b)$$
Take the extreme ends, we get $0 \leq a \sqrt{b^2 + a^4} + b \sqrt{a^2 + b^4} $.
Step 2: Consider cases
If $a, b <0$, then the terms in the LHS are clearly greater than $b^2 \times a^2$, hence the inequality is not true.
If $a,b \geq 0$, then clearly $a+b \geq 0.$
Hence, we may assume that $ a \leq 0 \leq b$, and we want to show that $ -a \leq b$. Now, because we are used to dealing with positive (non-negative reals), let me replace $a$ with $-a$ (not necessary, but simplifies considerations later)
Step 3: With this substitution, the inequality in step 1 gives us
$$ a\sqrt{b^2 + a^4} \leq b \sqrt{a^2 + b^4}$$
Since the LHS is non-negative, we may square it to obtain
$$a^2 (b^2 + a^4) \leq b^2 (a^2 + b^4 \Rightarrow a^6 \leq b^6 \Rightarrow a \leq b.$$
But this is what we want to show in Step 2, hence we are done. (remember we substituted $a$ for $-a$.)
Step 1 gives you another way to prove your equality case. Namely, we get that
$$ 0 = y \sqrt{x^2+1} + x \sqrt{y^2+1} $$
Hence, we have $ - y \sqrt{x^2 + 1} = x \sqrt{y^2+1} \Rightarrow y^2(x^2+1)=x^2(y^2+1) \Rightarrow y^2=x^2$. Then check that $y=x$ is not a valid solution (unless $y=x=0$), hence we must have $y=-x$.
This seems much more direct than your approach, and is motivated by considering conjugates.
The inequality in step 1 can also be obtained directly, by expanding and showing that
$$ 0 \leq \sqrt{a^2+b^4} \sqrt{b^2+a^4} + ab - a^2b^2 \leq a \sqrt{b^2 + a^4} + b \sqrt{a^2 + b^4}$$
However, this is not immediately obvious from the question. It's more of 'on hindsight'.
I'd be interested in seeing how the equality case can be obtained through direct expansion. (I don't see how to do this as yet.)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Solve $\sqrt{2x-5} - \sqrt{x-1} = 1$ Although this is a simple question I for the life of me can not figure out why one would get a 2 in front of the second square root when expanding. Can someone please explain that to me?
Example: solve $\sqrt{(2x-5)} - \sqrt{(x-1)} = 1$
Isolate one of the square roots: $\sqrt{(2x-5)} = 1 + \sqrt{(x-1)}$
Square both sides: $2x-5 = (1 + \sqrt{(x-1)})^{2}$
We have removed one square root.
Expand right hand side: $2x-5 = 1 + 2\sqrt{(x-1)} + (x-1)$-- I don't understand?
Simplify: $2x-5 = 2\sqrt{(x-1)} + x$
Simplify more: $x-5 = 2\sqrt{(x-1)}$
Now do the "square root" thing again:
Isolate the square root: $\sqrt{(x-1)} = \frac{(x-5)}{2}$
Square both sides: $x-1 = (\frac{(x-5)}{2})^{2}$
Square root removed
Thank you in advance for your help
| $$2x-5 = (1 + \sqrt{x-1})^2$$
to expand RHS use this formula or simple mulipty it with itself(to do square).
formula is:
$(a+b)^2=a^2+b^2+2\times a\times b$
so your expansion will be
$$2x-5 = (1^2 + (\sqrt{x-1})^2+2\times1\times \sqrt{x-1})$$
$$2x-5 = (1 + {x-1}+2\times \sqrt{x-1})$$
$$2x-5 = x+2\sqrt{x-1}$$
$$x-5 = 2\sqrt{x-1}$$
now you have your way.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Knowing that $n= 3598057$ is a product of two different prime numbers and that 20779 a square root of $1$ mod $n$, find prime factorization of $n$.
Knowing that $n= 3598057$ is a product of two different prime numbers and that 20779 is a square root of $1$ mod $n$, find prime factorization of $n$.
What I have done so far:
$n = p \cdot q$
$x^2 \equiv 1\pmod{n}$
$x^2 -1 \equiv 0\pmod{n}$
$(x-1)(x+1) \equiv 0\pmod{n}$
$x-1 = 20779 \lor x + 1=20779$
I have also noticed that:
$(x-1)(x+1) \equiv 0\pmod{p \cdot q}$
$(x-1)(x+1) \equiv 0\pmod{p} \land (x-1)(x+1) \equiv 0\pmod{q}$
But I have no idea what to do next. Any hints?
| From $(x-1)(x+1)\equiv 0 \pmod n$ you cannot conclude that $x-1 = 20779 \lor x + 1=20779$ when $n$ is not prime. You can conclude that $x-1 \equiv 0 \pmod p$ and $x+1 \equiv 0 \pmod q$ (or the other way-we could swap $p,q$) when you know that $n$ has only two prime factors unless $pq$ divides one of $x+1, x-1$. So factor $20778$ and $20780$ looking for factors that will multiply to make $3598057$. The other factors are small enough to find by hand.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Comparing $\sqrt{1001}+\sqrt{999}\ , \ 2\sqrt{1000}$ Without the use of a calculator, how can we tell which of these are larger (higher in numerical value)?
$$\sqrt{1001}+\sqrt{999}\ , \ 2\sqrt{1000}$$
Using the calculator I can see that the first one is 63.2455453 and the second one is 63.2455532, but can we tell without touching our calculators?
| And one more... I'm reflecting at the concavity and monotonicty of the curve of the sqrt-function and conclude, that whatever greater or smaller the relation is, it should be the same if I decrease the arguments down to zero. Since we do not yet know whether the relation is $\gt$ or $\lt$ I introduce the "indeterminate comparision symbol" $\mathcal C$ which can assume the greater or smaller-relation and we rewrite the original question
$$ \sqrt{1001} + \sqrt{ 999} \qquad \mathcal C \qquad 2\sqrt{1000} \tag 1 $$
as
$$ \sqrt{1001} - \sqrt{1000} \qquad \mathcal C \qquad \sqrt{1000} - \sqrt{999} \tag 2$$
Then we use the monotonicity and concavity of the sqrt-function (from x=1001 down to x=0) and write the consecutive differences
$ \displaystyle \qquad \begin{array} {lll}
&\sqrt{1001} & - & \sqrt{1000} & \mathcal C & \sqrt{1000} &-& \sqrt{999} \\
&\sqrt{1000} & - & \sqrt{ 999} & \mathcal C& \sqrt{ 999} &-& \sqrt{998} \\
&\sqrt{ 999} & - & \sqrt{ 998} & \mathcal C& \sqrt{ 998} &-& \sqrt{997} \\
&\cdots \\
&\sqrt{ 2} & - & \sqrt{ 1} & \mathcal C & \sqrt{ 1} &-& \sqrt{ 0} \\
\end{array} $
and compare the whole sums which are nicely telescoping
$ \displaystyle \qquad \begin{array} {lll}
\sum =&\sqrt{1001} &-&1 & \mathcal C & \sqrt{1000} &&& \end{array} $ $ \tag 3$
which can then be rewritten by
$ \displaystyle \qquad \begin{array} {rrrrrrr}
&\sqrt{1001} &-&\sqrt{1000} & \mathcal C & 1 && & // * (\sqrt{1001} + \sqrt{1000})\\
& 1001 & - & 1000 & \mathcal C & \sqrt{1001} &+& \sqrt{1000} \\
& & & 1 & \mathcal C & \sqrt{1001} &+& \sqrt{1000} &\sim 2\sqrt{1000} \\
\end{array} $
Here, in the last comparision, the geater/smaller-relation is obvious and thus our operator $ \mathcal C = "\lt" $
and we have the result
$ \displaystyle \qquad \begin{array} {lll}
&\sqrt{1001} & + & \sqrt{ 999} & \lt & 2 \cdot \sqrt{1000}
\end{array} $ $ \tag 4$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
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linear Transformation of polynomial with degrees less than or equal to 2 I would like to determine if the following map $T$ is a linear transformation:
\begin{align*}
T: P_{2} &\to P_{2}\\
A_{0} + A_{1}x + A_{2}x^{2} &\mapsto A_{0} + A_{1}(x+1) + A_{2}(x+1)^{2}
\end{align*}
My attempt at solving:
\begin{align}
T(p + q) &= p(x+1) + q(x+1)\\
&= \left[A_{0} + A_{1}(x+1) + A_{2}(x+1)^2\right] + \left[b_{0} + b_{1}(x+1) + b_{2}(x+1)^2\right]\\
&= \left(A_{0} + b_{0}\right) + \left(A_{1} + b_{1}\right)(x+1) + \left(A_{2} + b_{2}\right)(x+1)^2\\
&= T(p) + T(q)
\end{align}
Is this right so far? If not, what am I doing wrong?
| Why is $T(p+q)=p(x+1)+q(x+1)$? That's basically what you're trying to show.
You could try to just expand $T(p)$ and see that $T$ can be easily expressed as a matrix in base $1,x,x^2$.
| {
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Finding invariant factors of finitely generated Abelian group There is this question that I wasn't sure how to do but somehow got the answers partially correct (maybe).
Suppose that the abelian group $M$ is generated by three elements $x,y,z$ subject to the relations $4x+y+2z=0, 5x+2y+z=0,6y-6z=0$. Determine the invariant factors of $M$.
My attempt:
I wasn't sure what theorem or reasoning to use, but I have a feeling to write the equations in matrix form and find the equivalent matrix. My question is normally if we write in matrix form, we should be priorly given a map or homomorphism, but this question did not, or whether somehow we can interpret the equations as map? if so, what is the map?
Let me try to make one. Let $M$ be the $\mathbb{Z}$-module $F/N$ where $F=\mathbb{Z}^3$ and $N=\langle(4,5,0),(1,2,6),(2,1,-6)\rangle\le\mathbb{Z}^3$. Consider the homomorphism $\varphi:\mathbb{Z}^3\to\mathbb{Z}^3$ given by $(x,y,z)\to x(4,5,0)+y(1,2,6)+z(2,1,-6)$. Am I correct? Please correct me if I am wrong.
Then I find the equivalent matrix:
$\begin{bmatrix}4 & 1 & 2\\5 & 2 & 1\\0 & 6 & -6\end{bmatrix}$ $4R_2-5R_1$ $\begin{bmatrix}4 & 1 & 2\\0 & 3 & -6\\0 & 6 & -6\end{bmatrix}$ $R_3-2R_2$ $\begin{bmatrix}4 & 1 & 2\\0 & 3 & -6\\0 & 0 & 6\end{bmatrix}$ $3R_1+R_2$ $\begin{bmatrix}12 & 6 & 0\\0 & 3 & -6\\0 & 0 & 6\end{bmatrix}$ $R_2+R_3$ $\begin{bmatrix}12 & 6 & 0\\0 & 3 & 0\\0 & 0 & 6\end{bmatrix}$ $R_1-2R_2$ $\begin{bmatrix}12 & 0 & 0\\0 & 3 & 0\\0 & 0 & 6\end{bmatrix}$. Then by performing column and row swaps, the matrix becomes $\begin{bmatrix}3 & 0 & 0\\0 & 6 & 0\\0 & 0 & 12\end{bmatrix}$. Since 3,6,12 are non-units in $\mathbb{Z}$, then the invariant factors of M are 3, 6, 12.
However, the solution says that the invariant factors are 1,3,6.
Was my method correct? Where did I make the mistake?
Thanks!
| When you do the operation $4 R_2 - 5 R_1$, you are altering the determinant by a factor of $4$, since you are multiplying your matrix on the left by
$$
\begin{bmatrix}1 & 0 & 0\\-5&4&0\\0&0&1\end{bmatrix}.
$$
Ditto for $3 R_1 + R_2$, which alters the determinant by a factor of $3$. These two operations have introduced the spurious factor $12$.
I don't have the time to post the correct calculations (which are a lengthier alternative to the use of the determinant in this case), perhaps later.
So here's how you do it:
$$\begin{align}
&\begin{bmatrix}4 & 1 & 2\\5 & 2 & 1\\0 & 6 & -6\end{bmatrix} C_1 \leftrightarrow C_2 \begin{bmatrix}1 & 4 & 2\\2 & 5 & 1\\6 & 0 & -6\end{bmatrix} R_2 - 2 R_1 \begin{bmatrix}1 & 4 & 2\\0 & -3 & -3\\6 & 0 & -6\end{bmatrix} R_3 - 6 R_1\\& \begin{bmatrix}1 & 4 & 2\\0 & -3 & -3\\0 & -24 & -18\end{bmatrix} C_2 - 4 C_1, C_3 - 2 C_1 \begin{bmatrix}1 & 0 & 0\\0 & -3 & -3\\0 & -24 & -18\end{bmatrix} R_3 - 8 R_2\\& \begin{bmatrix}1 & 0 & 0\\0 & -3 & -3\\0 & 0 & 6\end{bmatrix} C_3 - C_2 \begin{bmatrix}1 & 0 & 0\\0 & -3 & 0\\0 & 0 & 6\end{bmatrix} {-R_2} \begin{bmatrix}1 & 0 & 0\\0 & 3 & 0\\0 & 0 & 6\end{bmatrix}
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve equations $\sqrt{t +9} - \sqrt{t} = 1$ Solve equation: $\sqrt{t +9} - \sqrt{t} = 1$
I moved - √t to the left side of the equation $\sqrt{t +9} = 1 -\sqrt{t}$
I squared both sides $(\sqrt{t+9})^2 = (1)^2 (\sqrt{t})^2$
Then I got $t + 9 = 1+ t$
Can't figure it out after that point.
The answer is $16$
| From $\sqrt{t +9} - \sqrt{t} = 1$ you don't get $\sqrt{t +9} = 1 -\sqrt{t}$ but $$\sqrt{t +9} = 1 +\sqrt{t}.$$
You can solve as follows, using the algebraic identity $(a+b)^2=a^2+2ab+b^2$:
$$
\begin{eqnarray*}
\sqrt{t+9}-\sqrt{t} &=&1\Leftrightarrow \sqrt{t+9}=1+\sqrt{t}\tag{1} \\
\text{Square both sides of $(1)$} &\Rightarrow &\left( \sqrt{t+9}\right) ^{2}=\left(
1+\sqrt{t}\right) ^{2} \\
\text{Compute and simplify} &\Leftrightarrow &t+9=1+2\sqrt{t}+t\Leftrightarrow 9=1+2\sqrt{t} \\
\text{Simplify} &\Leftrightarrow &9-1=2\sqrt{t}\Leftrightarrow 8=2\sqrt{t} \\
\text{Simplify} &\Leftrightarrow &\frac{8}{2}=\sqrt{t}\Leftrightarrow 4=
\sqrt{t}\tag{2} \\
\text{Square both sides of $(2)$} &\Rightarrow &4^{2}=\left( \sqrt{t}\right) ^{2} \\
&\Leftrightarrow &16=t.\tag{3}
\end{eqnarray*}
$$
Final comment. When we square both sides of an equation we get a new equation which has the same solutions of the original equation, but can have additional solutions. However in this case we got only the solution $t=16$, which is a solution of $(1)$ too.
ADDED. In your recent question solve the equation $\sqrt{3x-2}+2-x=0$, we get two solutions after squaring
$$
\begin{equation*}
\sqrt{3x-2}+2-x=0\Rightarrow 3x-2=x^{2}-4x+4\Leftrightarrow x\in \{1,6\}
\end{equation*}
$$
but only $x=6$ is a solution of $\sqrt{3x-2}+2-x=0$, as explained in
this comment by Glen O.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the sum of a Taylor expansion I want to find the following sum:
$$
\sum\limits_{k=0}^\infty (-1)^k \frac{(\ln{4})^k}{k!}
$$
I decided to substitute $x = \ln{4}$:
$$
\sum\limits_{k=0}^\infty (-1)^k \frac{x^k}{k!}
$$
The first thing I noticed is that this looks an awful lot like the series expansion of $e^x$:
$$
e^x = \sum\limits_{k=0}^\infty \frac{x^k}{k!}
$$
The only obstacle is the $(-1)^k$ term. I tried getting rid of it by rewriting:
\begin{align*}
\sum\limits_{k=0}^\infty (-1)^k \frac{x^k}{k!} &= 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} - \frac{x^5}{5!} + \dots\\
&= (1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \dots) - (x + \frac{x^3}{3!} + \frac{x^5}{5!} + \dots)\\
&= \sum\limits_{k=0}^\infty \frac{x^{2k}}{(2k)!} - \sum\limits_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}
\end{align*}
These sums look a lot like the series expansions of $\sin(x)$ and $\cos(x)$:
\begin{align*}
\sin(x) &= \sum\limits_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!}\\
\cos(x) &= \sum\limits_{k=0}^\infty (-1)^k \frac{x^{2k}}{(2k)!}
\end{align*}
However, these sums do have a $(-1)^k$ term, just when I got rid of it! So now I'm stuck. Can someone help me in the right direction?
| Note that $$e^{-x}=\sum_{k=0}^\infty \frac{(-x)^k}{k!}=\sum_{k=0}^\infty (-1)^k\frac{x^k}{k!}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Trigonometric function integration: $\int_0^{\pi/2}\frac{dx}{(a^2\cos^2x+b^2 \sin^2x)^2}$ How to integrate $$\int_0^{\pi/2}\dfrac{dx}{(a^2\cos^2x+b^2 \sin^2x)^2}$$
What's the approach to it?
Being a high school student , I don't know things like counter integration.(Atleast not taught in India in high school education ).I just know simple elementary results of definite and indefinite integration. Substitutions and all those works good. :)
| Put $t=\tan x\implies dt=\sec^2x.dx$
$$
\begin{align}
&\int_0^{\pi/2}\frac{dx}{(a^2\cos^2x+b^2\sin^2x)^2}=\int_0^{\pi/2}\frac{\sec^4x.dx}{(a^2+b^2\tan^2x)^2}\\
&=\int_0^\infty\frac{1+t^2}{(a^2+b^2t^2)^2}dt=\frac{1}{b^2}\int_0^\infty\frac{dt}{a^2+b^2t^2}+\frac{b^2-a^2}{b^2}\int^\infty_0\frac{dt}{(a^2+b^2t^2)^2}=I_1+I_2
\end{align}
$$
$$
I_1=\frac{1}{b^2}\int_0^\infty\frac{dt}{a^2+b^2t^2}=\bigg[\frac{1}{b^2}.\frac{1}{ab}.\tan^{-1}\frac{bt}{a}\bigg]_0^\infty=\frac{\pi}{2ab^3}
$$
Put $t=\frac{a}{b}\tan\theta\implies dt=\frac{a}{b}\sec^2\theta.d\theta$
$$
\begin{align}
I_2&=\frac{b^2-a^2}{b^2}\int^\infty_0\frac{dt}{(a^2+b^2t^2)^2}=\frac{b^2-a^2}{b^2}\int_0^{\pi/2}\frac{\frac{a}{b}\sec^2\theta.d\theta}{(a^2+a^2\tan^2\theta)^2}\\
&=\frac{b^2-a^2}{b^2}\frac{a}{b.a^4}\int_0^{\pi/2}\frac{\sec^2\theta}{\sec^4\theta}d\theta=\frac{b^2-a^2}{a^3b^3}\int_0^{\pi/2}\cos^2\theta.d\theta\\
&=\frac{b^2-a^2}{a^3b^3}\int_0^{\pi/2}\bigg(\frac{1+\cos2\theta}{2}\bigg)d\theta=\frac{b^2-a^2}{2a^3b^3}\bigg[\theta+\frac{\sin2\theta}{2}\bigg]_0^{\pi/2}=\frac{\pi}{4}\frac{b^2-a^2}{a^3b^3}
\end{align}
$$
$$
\begin{align}
\int_0^{\pi/2}\frac{dx}{(a^2\cos^2x+b^2\sin^2x)^2}&=I_1+I_2=\frac{\pi}{2ab^3}+\frac{\pi}{4}\frac{(b^2-a^2)}{a^3b^3}\\
&=\frac{2\pi.a^2+\pi.b^2-\pi.a^2}{4a^3b^3}=\frac{\pi}{4}.\frac{a^2+b^2}{a^3b^3}
\end{align}
$$
| {
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"url": "https://math.stackexchange.com/questions/402223",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "7",
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Simplified form of $\left(6-\frac{2}{x}\right)\div\left(9-\frac{1}{x^2}\right)$. Tried this one a couple of times but can't seem to figure it out.
I am trying to simplify the expression:
$$\left(6-\frac{2}{x}\right)\div\left(9-\frac{1}{x^2}\right)$$
So my attempt at this is:
$$=\bigg(\dfrac{6x}{x}-\dfrac{2}{x}\bigg)\div\bigg(\dfrac{9x^2}{x^2}-\dfrac{1}{x^2}\bigg)$$
$$=\bigg(\dfrac{6x-2}{x}\bigg)\div\bigg(\dfrac{9x^2-1}{x^2}\bigg)$$
$$=\dfrac{6x-2}{x}\cdot\dfrac{x^2}{9x^2-1}$$
$$=\dfrac{(6x-2)(x^2)}{(x)(9x^2-1)}$$
$$=\dfrac{6x^3-2x^2}{9x^3-x}$$
This is the part that I get stuck at. I can't decide what to factor out:
$$=\dfrac{x(6x^3-2x^2)}{x(9x^3-x)}$$
$$=\dfrac{(6x^2-2x)}{(9x^2-1)}$$
Edit, missed a difference of squares:
$$=\dfrac{2x^2(6x^3-2x^2)}{x(9x^3-x)}$$
$$=\dfrac{2x^2(3x-1)}{x(3x-1)(3x+1)}$$
Giving a final answer of:
$$=\boxed{\dfrac{2x}{3x+1}}$$
| You can factor a $2x^2$ out of the top to get $6x^3-2x^2=2 x^2 (3x-1 )$. After this, you can factor an $x$ out of the bottom to get $(9x^2-1)$, which then factors into $(3x-1) (3x+1)$ as it is a difference of squares. So, putting that all together,
$$\frac{\left(6x^3-2x^2\right)}{\left(9x^3-x\right)}=\frac{2 x^2 (3x-1 )}{x (3x-1) (3x+1)}=\hspace{2pt}\ldots\hspace{2pt}\Large{?}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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$R_n = 3(2^n)-4(5^n)$, $n \geq0$, prove $R_n$ satisfies $R_n = 7R_{n-1}-10R_{n-2}$ So the question is:
$R_n=3(2^n)-4(5^n)$ for $n\ge 0$; prove that $R_n$ satisfies $R_n=7R_{n-1}-10R_{n-2}$.
I don't really know what to do from here. If I substitute
$$R_n = 3(2^n)-4(5^n)$$
into
$$Rn = 7R_{n-1}-10R_{n-2}$$
I end up getting
$$R_n = 7\Big(3(2^{n-1})-4(5^{n-1})\Big)-10\Big(3(2^{n-2})-4(5^{n-2})\Big)$$
Dont know what to do...
EDIT: Thanks to Zev, what I did was:
$$3(2^n)-4(5^n)=7\bigg[3(2^{n-1})-4(5^{n-1})\bigg]-10\bigg[3(2^{n-2})-4(5^{n-2})\bigg].$$
$$\begin{align*}
3(2^n)-4(5^n)&=21(2^{n-1})-28(5^{n-1})-30(2^{n-2})+40(5^{n-2})\\\\
3(2^n)-4(5^n)&=21(2^{n})(2^{-1})-28(5^{n})(5^{-1})-30(2^{n})(2^{-2})+40(5^{n})(5^{-2})\\\\
3(2^n)-4(5^n)&=21/2(2^{n})-28/5(5^{n})-30/2(2^{n})+40/5(5^{n})\\\\
3(2^n)-4(5^n)&=(2^{n})[21/2-30/4]+(5^{n})[40/25-28/25]\\\\
3(2^n)-4(5^n)&=(2^{n})[3]+(5^{n})[-4]\\\\
3(2^n)-4(5^n)&=3(2^{n})-4(5^{n})
\end{align*}$$
| We need to eliminate $2^n,3^n$
$$R_n=3\cdot 2^n-4\cdot 5^n \ \ \ \ (1)$$
$$(1)\implies 3\cdot2^n-5\cdot 5^n-R_n=0\ \ \ \ (2)$$
$$(1)\implies R_{n+1}=3\cdot 2^{n+1}-4\cdot 5^{n+1}=6\cdot2^n-20\cdot 5^n$$
$$\implies 6\cdot 2^n-20\cdot 5^n-R_{n+1}=0 \ \ \ \ (3)$$
Solving $(2),(3)$ for $2^n,3^n$ we get $$2^n=\frac{5R_n-R_{n+1}}9\text{ and } 3^n=\frac{2R_n-R_{n+1}}{12}$$
$$(1)\implies R_{n+2}=3\cdot 2^{n+2}-4\cdot 5^{n+2}=12\cdot2^n-100\cdot 5^n\ \ \ \ (4)$$
Put the values of $2^n,3^n$ in $(4)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/405384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
$1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1 = 12$ or $1$? Does the expression $1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1$ equal $1$ or $12$ ?
I thought it would be 12 this as per pemdas rule:
$$(1+1+1+1+1+1+1+1+1+1+1+1)\cdot (0+1) = 12 \cdot 1 = 12$$
Wanted to confirm the right answer from you guys. Thanks for your help.
| Your answer of $12$ is numerically correct, but you’ve applied the precedence rules incorrectly. In the absence of parentheses multiplication is done before addition, so
$$1+1+1+1+1+1+1+1+1+1+1+1\cdot 0+1$$
is to be evaluated as
$$1+1+1+1+1+1+1+1+1+1+1+(1\cdot 0)+1\;,$$
not as
$$(1+1+1+1+1+1+1+1+1+1+1+1)\cdot(0+1)\;.$$
Since $1\cdot0=0$, this simplifies to
$$1+1+1+1+1+1+1+1+1+1+1+0+1=12\;.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/405543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Trigonometric identities using $\sin x$ and $\cos x$ definition as infinite series Can someone show the way to proof that $$\cos(x+y) = \cos x\cdot\cos y - \sin x\cdot\sin y$$ and $$\cos^2x+\sin^2 x = 1$$ using the definition of $\sin x$ and $\cos x$ with infinite series.
thanks...
| Let me do a different one. Begin with
$$
\sin x = \sum_{k=0}^\infty \frac{(-1)^k x^{2k+1}}{(2k+1)!},
\qquad
\cos x = \sum_{k=0}^\infty \frac{(-1)^k x^{2k}}{(2k)!}
$$
So compute
$$\begin{align}
\sin(x+y) &= \sum_{k=0}^\infty \frac{(-1)^k(x+y)^{2k+1}}{(2k+1)!}
\\ &=
\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}\sum_{j=0}^{2k+1} \binom{2k+1}{j} x^j y^{2k+1-j}
\\ &=
\sum_{k=0}^\infty (-1)^k\sum_{j=0}^{2k+1} \frac{1}{j!(2k+1-j)!} x^j y^{2k+1-j}
\\ &=
\sum_{j=0}^\infty \frac{x^j}{j!}\sum_k \frac{(-1)^k}{(2k+1-j)!} y^{2k+1-j}
\end{align}$$
where the inner sum is over all $k$ such that $2k+1 \ge j$. Consider two cases for the inner sum, $j$ even and $j$ odd.
If $j=2n$, then $2k+1 \ge j$ iff $2k+1 \ge 2n$ iff $k \ge n$. So the $k$-sum is:
$$
\sum_{k=n}^\infty \frac{(-1)^k}{(2k+1-2n)!} y^{2k+1-2n}
$$
Use change of variables $i=k-n$ to get
$$
\sum_{i=0}^\infty \frac{(-1)^{i+n}}{(2i+1)!} y^{2i+1} = (-1)^n \sin y .
$$
If $j=2n+1$, then $2k+1 \ge j$ iff $2k+1 \ge 2n+1$ iff $k \ge n$. So the $k$-sum is
$$
\sum_{k=n}^\infty \frac{(-1)^k}{(2k+1-2n-1)!}y^{2k+1-2n-1}
$$
Again use change of variables $i=k-n$ to get
$$
\sum_{i=0}^\infty \frac{(-1)^{i+n}}{(2i)!} y^{2i} = (-1)^n \cos y.
$$
So finally we have
$$
\sin(x+y) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!} \sin y
+\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}\cos y
= \cos x \sin y + \sin x \cos y.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/405628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Deriving Reduction formula - Indefinite integration using integration by parts I was working on finding the reduction formula for :
$\int \frac{dx}{(x^2+a^2)^n}$
By using integration by parts formula ( $\int f(x) g(x) dx = f(x) \int g(x)dx -\int (f'(x)\int g(x))$ considering 1.dx as second function :
Let $I_n$ = $\int \frac{dx}{(x^2+a^2)^n} = \frac{dx}{(x^2+a^2)^n}.\int dx -\int \{\frac{dx}{(x^2+a^2)^n}.\int \}dx$
=$\frac{dx}{(x^2+a^2)^n}-\int (x^2+a^2)^{n-1}.2x^2dx$
=$\frac{dx}{(x^2+a^2)^n} +2n\int \frac{x^2}{(x^2+a^2)^{n+1}}dx$
=$\frac{dx}{(x^2+a^2)^n} + \frac{2n}{(x^2+a^2)}\int \frac{dx}{(x^2+a^2)^{n-1}}-\frac{dx}{(x^2+a^2)^n}$
Therefore , $I = \frac{x}{(x^2+a^2)^n}+2nI_n-2n^2I_{n+1}$
Please guide further how to proceed and conclude..Thanks
| If $$I_n=\int \frac{dx}{(x^2+a^2)^n}$$ then setting $$u=\frac{1}{(x^2+a^2)^n},~~dv=dx$$ leads us to have $$du=\frac{-2nxdx}{(x^2+a^2)^{n+1}},~~v=x$$ and the method of integrating by parts gives us $$I_n=\frac{x}{(x^2+a^2)^n}+2n\int\frac{x^2dx}{(x^2+a^2)^{n+1}}=...=\frac{x}{(x^2+a^2)^n}+2nI_n-2na^2I_{n+1}$$ Now we see the integrating of $I_{n+1}$ will be dependent to integrating of $I_n$ and....
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/405886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
What does $\lim\limits_{x\to\pi/6}\frac{1-\sqrt{3}\tan x}{\pi-6x}$ evaluate to? What does $$\lim_{x\to\pi/6}\frac{1-\sqrt{3}\tan x}{\pi-6x}$$ evaluate to? This very likely needs substitution.
| $$\frac{1-\sqrt{3}\tan x}{\pi-6x}=\frac{\sqrt3}6\cdot\frac{\tan \frac\pi6-\tan x }{\frac\pi6-x}\left(\text{ as }\tan\frac\pi6=\frac1{\sqrt3}\right)$$
$$=\frac{\sqrt3}6\cdot\frac{\sin\left(\frac\pi6-x\right)}{\left(\frac\pi6-x\right)\cos x\cos \frac\pi6}$$
$$\lim_{x\to\frac\pi6}\frac{1-\sqrt{3}\tan x}{\pi-6x}=\frac{\sqrt3}{6\cos \frac\pi6}\cdot\lim_{x\to\frac\pi6}\frac{\sin\left(\frac\pi6-x\right)}{\left(\frac\pi6-x\right)}\frac1{\lim_{x\to\frac\pi6}\cos x}$$
$$=\frac{\sqrt3}{6\cos\frac\pi6}\cdot\lim_{y\to0}\frac{\sin y}y\cdot\frac1{\cos\frac\pi6}$$ (Putting $\frac\pi6-x=y$ in the first limit)
$$\lim_{x\to\frac\pi6}\frac{1-\sqrt{3}\tan x}{\pi-6x}= \frac{\sqrt3}{6\cos^2\frac\pi6}=\frac2{3\sqrt3}$$
$$\text{In fact, }\lim_{x\to a}\frac {f(x)-f(a)}{x-a}=f'(a)$$
$$\implies \lim_{x\to a}\left(\frac {\tan x-\tan a}{x-a}\right)=\left(\frac {d\tan x}{dx}\right)_{x=a}=\sec^2a$$
$$\implies \lim_{x\to \frac\pi6}\left(\frac {\tan x-\frac1{\sqrt3}}{x-\frac\pi6}\right)=\sec^2\frac\pi6=\frac43$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/408315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Why dividing by zero still works Today, I was at a class. There was a question:
If $x = 2 +i$, find the value of $x^3 - 3x^2 + 2x - 1$.
What my teacher did was this:
$x = 2 + i \;\Rightarrow \; x - 2 = i \; \Rightarrow \; (x - 2)^2 = -1 \; \Rightarrow \; x^2 - 4x + 4 = -1 \; \Rightarrow \; x^2 - 4x + 5 = 0 $. Now he divided $x^3 - 3x^2 + 2x - 1$ by $x^2 - 4x + 5$.
The quotient was $x + 1$ and the remainder $x - 6$. Now since $\rm dividend = quotient\cdot divisor + remainder$, he concluded that $x^3 - 3x^2 + 2x - 1 = x-6$ since the divisor is $0$.
Plugging $2 + i$ into $x - 6$, we get $-4 + i$.
But my question is, how was he able to divide by zero in the first place? Why did dividing by zero work in this case?
| because he's only dividing by zero when $x$ takes that value (or its conjugate) otherwise he's really just figuring out what the next thing he'll need to multiply is. Yes, it may look a little dodgy, think of this as a sort of back-of-the-envelope calculation. It's only a means to find out the next step and since in the final expression we no longer have this factor in the denominator, everything works out fine.
Think for example of Factorizing $x^2+7x+12$. We can similarly divide by $x+4$ and get a result of $x+3$ to conclude that $x^2+7x+12=(x+4)(x+3)$. While we can only technically divide by $x+4$ if $x\not=-4$, the process still works out fine.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/408527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 4,
"answer_id": 3
} |
If $\frac{\cos^4\theta}{\cos^2\phi}+\frac{\sin^4\theta}{\sin^2\phi}=1$, show $\frac{\cos^4\phi}{\cos^2\theta} +\frac{\sin^4\phi}{\sin^2\theta}=1$
If $\dfrac{\cos^4\theta}{\cos^2\phi}+\dfrac{\sin^4\theta}{\sin^2\phi}=1$, prove that $\dfrac{\cos^4\phi}{\cos^2\theta} +\dfrac{\sin^4\phi}{\sin^2\theta}=1$.
Unable to move further ...request you to please suggest how to proceed ..Thanks..
| We can Write
$$\frac{Cos^4\theta}{Cos^2\phi}+\frac{Sin^4\theta}{Sin^2\phi}=Cos^2\phi+Sin^2\phi$$ $\implies$
$$ \left(Cos^2\phi-\frac{Cos^4\theta}{Cos^2\phi}\right)+\left(Sin^2\phi-\frac{Sin^4\theta}{Sin^2\phi}\right)=0$$ $\implies$
$$ Cos^4\phi+Sin^4\phi=Cos^4\theta+Sin^4\theta$$ $\implies$
$$ Sin\phi Cos\phi=Sin\theta Cos\theta \implies Sin(2\phi)=Sin(2\theta)$$ So
$$\phi=k\pi+\theta ,\: k \in \mathbb{Z}$$ So
$$ Cos\phi=(-1)^k Cos\theta, Sin\phi=(-1)^k Sin\theta$$ Finally the Result follows..
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/410304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Determine the base of $\ker\phi$ and fibre of phi of a polynom function Good evening,
given is a base $B$ defined as
$$B:=(x\mapsto1,x\mapsto x,x\mapsto x^2,x\mapsto x^3 ,x\mapsto x^4)$$ a set $V$ defined as $$V:= \{ f: \mathbb{R} \mapsto \mathbb{R}\ |\ \exists\ {a_0},...{a_4} \in \mathbb{R}\ : f(x)=\sum_{i=0}^{4}{a_ix^i} \ \forall \ x \in \mathbb{R}\}$$ a function $\phi$ defined as $$\phi(f)(x)=f''(x)+x \cdot f'(x) - f(x+1)$$ (where $f'(x)$ is the first derivative and $f''(x)$ the second derivative)
and an element $g, g \in V$ $$g(x)=3x^4+2x^3-x+1$$
The task is to find a base of the kernel of phi and to determine $\phi^{-1}(\{g\})$.
To determine a base of the $\ker\phi$ I need to find all functions that map to $0$. But before that I calculated the mapping rule (? not sure on the english word).
$$\phi(\begin{pmatrix}a\\b\\c\\d\\e\end{pmatrix}) = \phi(\lambda_1\cdot\begin{pmatrix}1\\0\\0\\0\\0\end{pmatrix}+\lambda_2\cdot\begin{pmatrix}0\\1\\0\\0\\0\end{pmatrix}+\lambda_3\cdot\begin{pmatrix}0\\0\\1\\0\\0\end{pmatrix}+\lambda_4\cdot\begin{pmatrix}0\\0\\0\\1\\0\end{pmatrix}+\lambda_5\cdot\begin{pmatrix}0\\0\\0\\0\\1\end{pmatrix})$$
The next step was to determine the coefficients $\lambda_1, ... ,\lambda_5$ which directly result of this matrix: $$\begin{pmatrix} 1 & 0 & 0 & 0 & 0&| & a \\ 0 & 1 & 0 & 0 & 0&| & b\\ 0 & 0 & 1 & 0 & 0&| & c \\ 0 & 0 & 0 & 1 & 0&|& d \\ 0 & 0 & 0 & 0 & 1&|& e \end{pmatrix}$$
$$\Rightarrow \lambda_1 = a, \lambda_2 = b, \lambda_3 = c, \lambda_4 = d, \lambda_5 = e$$
Knowing this we can continue finding the mapping rule:
$$\overset{\phi\ linear}= a\cdot\phi(\begin{pmatrix}1\\0\\0\\0\\0\end{pmatrix})+b\cdot\phi(\begin{pmatrix}0\\1\\0\\0\\0\end{pmatrix})+c\cdot\phi(\begin{pmatrix}0\\0\\1\\0\\0\end{pmatrix})+d\cdot\phi(\begin{pmatrix}0\\0\\0\\1\\0\end{pmatrix})+e\cdot\phi(\begin{pmatrix}0\\0\\0\\0\\1\end{pmatrix})$$
$$=a\cdot\begin{pmatrix}-1\\0\\0\\0\\0\end{pmatrix}+b\cdot\begin{pmatrix}-1\\0\\0\\0\\0\end{pmatrix}+c\cdot\begin{pmatrix}1\\-2\\1\\0\\0\end{pmatrix}+d\cdot\begin{pmatrix}-1\\3\\-3\\2\\0\end{pmatrix}+e\cdot\begin{pmatrix}-1\\-4\\6\\-4\\3\end{pmatrix}$$
$$=\begin{pmatrix}-a-b+c-d-e\\-2c+3d-4e\\c-3d+6e\\2d-4e\\3e\end{pmatrix}$$
So far so good, now using the definition: $\ker\phi:=\{v \in V:\phi(v)=0\}.$
$$\begin{pmatrix}-a-b+c-d-e\\-2c+3d-4e\\c-3d+6e\\2d-4e\\3e\end{pmatrix}=\begin{pmatrix}0\\0\\0\\0\\0\end{pmatrix}$$
$$\Rightarrow e=0,d=0,c=0,-a=b$$
Now I've got a problem choosing a base due to $-a=b$. As far as I can tell a base of $\ker\phi$ would have 2 elements in this case which aren't linear independent. I thought of: $$\{a\cdot\begin{pmatrix}-1\\0\\0\\0\\0\end{pmatrix}+b\cdot\begin{pmatrix}-1\\0\\0\\0\\0\end{pmatrix} | a,b \in \mathbb{R}\}$$
For the other question I need to know the $\ker\phi$ since for $\forall v \in V, g\in \phi(g): $$\phi^{-1}({g})= v + \ker\phi$$
Anyway I calculated $v$ by solving this equation:
$$\begin{pmatrix}-a-b+c-d-e\\-2c+3d-4e\\c-3d+6e\\2d-4e\\3e\end{pmatrix}=\begin{pmatrix}1\\-1\\0\\2\\3\end{pmatrix}$$
$$\Rightarrow v =x^4+3x^3+3x^2-2$$
If you need further information simply ask in this thread. I hope I described the problem detailed enough.
| Imo it'd be easier (or at least slightly less messier) as follows:
$$f(x)=\sum_{k=0}^4a_kx^k\implies $$
$$f(x)\in\ker\phi\iff f''(x)+xf'(x)=f(x+1)\iff$$
$$2a_2+(6a_3+a_1)x+(12a_4+2a_2)x^2+3a_3x^3+4a_4x^4=$$
$$=\sum_{k=0}^4a_k+\left(\sum_{k=1}^4ka_k\right)x+\left(a_2+3a_3+6a_4\right)x^2+\left(a_3+4a_4\right)x^3+a_4x^4$$
Comparing corresponding powers of $\;x\;$ we get:
$$4a_4=a_4\iff a_4=0\\3a_3=a_3\iff a_3=0\\2a_2=a_2\iff a_2=0\\a_1=a_1\\a_0=-a_1$$
We thus get that
$$f(x)\in\ker\phi\iff f(x)=a-ax\;,\;\;a\in\Bbb R\implies \dim\ker\phi=1\;,\;\;\ker\phi=\text{Span}\{1-x\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/411269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Is there a contradiction is this exercise? The following exercise was a resolution to this problem
Let $\displaystyle\frac{2x+5}{(x-3)(x-7)}=\frac{A}{(x-7)}+\frac{B}{(x-3)}\space \forall \space x \in \mathbb{R}$. Find the values for $A$ and $B$
The propose resolution was:
In order to isolate $A$ on the right side, multiply all the equation by $x-7$
$\displaystyle\frac{(2x+5)(x-7)}{(x-3)(x-7)}=\frac{A(x-7)}{(x-7)}+\frac{B(x-7)}{(x-3)}$
Now is my doubt. The resolution suggests that $x-7$ cancel out.
$\displaystyle\frac{(2x+5)}{(x-3)}=A+\frac{B(x-7)}{(x-3)}$
But, $x-7$ can be equal zero. In this situation, is allowed to perform this operation? One can say "for every $x\neq7$", but on the next step the resolution says "for $x=7$ we have".
$\displaystyle\frac{(14+5)}{(7-3)}=A+\frac{B(0)}{(7-3)} \Leftrightarrow A=\frac{19}{4}$
I think there is a contradiction is this resolution.
| If $\frac{2x+5}{x-3}=A+B\frac{x-7}{x-3}$ for all $x\ne 7$ (or $3$), then we can take the limit as $x\to 7$ on both sides to get the given result, and since both sides are continuous, it is equivalent to just plugging in $x=7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/411658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
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Interestingly restricted compositions of $n$ Let $n$ be a non-negative integer. How many compositions of $n$ are there where the $i$-th part has the same parity as $i$?
The main problem I'm having with this problem is that I can't really formulate a single generating function for the set of all tuples of natural numbers where the $i$-th element has the same parity as $i$. How should I proceed?
| Let $a_n$ be the number of compositions of $n$ whose $i$-the part has the same parity as $i$, and let $b_n$ be the number of compositions of $n$ whose $i$-the part has the opposite parity to $i$. Considering the possible values of the first part of a composition of $n$, we see that
$$a_{2n+1}=\sum_{k=0}^nb_{2k}\quad\text{ and }\quad a_{2n}=\sum_{k=0}^{n-1}b_{2k+1}$$
and
$$b_{2n+1}=\sum_{k=0}^{n-1}a_{2k+1}\quad\text{ and }\quad b_{2n}=\sum_{k=0}^{n-1}a_{2k}\;,$$
and hence that
$$\left\{\begin{align*}a_n&=a_{n-2}+b_{n-1}\\
b_n&=a_{n-2}+b_{n-2}\;.
\end{align*}\right.\tag{1}$$
If we make the blanket assumption that $a_n=b_n=0$ for all $n<0$ and set $a_0=0$ and $b_0=1$, $(1)$ is valid for all $n\in\Bbb Z$ except in the case of $b_0$, adding an Iverson bracket term to get
$$\left\{\begin{align*}a_n&=a_{n-2}+b_{n-1}\\
b_n&=a_{n-2}+b_{n-2}+[n=0]\;.
\end{align*}\right.\tag{2}$$
fixes this.
Let
$$A(x)=\sum_{n\ge 0}a_nx^n\quad\text{and}\quad B(x)=\sum_{n\ge 0}b_nx^n$$
be the generating functions for the two sequences. Then multiplying the recurrences $(2)$ by $x^n$ and summing over $n\ge 0$ yields
$$\left\{\begin{align*}
A(x)&=x^2A(x)+xB(x)\\
B(x)&=x^2A(x)+x^2B(x)+1\;.
\end{align*}\right.$$
Then
$$B(x)=\frac{x^2}{1-x^2}A(x)+\frac1{1-x^2}$$
and
$$A(x)=\frac{x}{1-x^2}B(x)=\frac{x^3}{(1-x^2)^2}A(x)+\frac{x}{(1-x^2)^2}\;,$$
so
$$\begin{align*}
A(x)&=\frac{x}{1-x^2}\left(1-\frac{x^3}{(1-x^2)^2}\right)^{-1}\\\\
&=\frac{x}{1-x^2}\cdot\frac{(1-x^2)^2}{1-2x^2-x^3+x^4}\\\\
&=\frac{x(1-x)}{1-2x^2-x^3+x^4}
\end{align*}$$
and
$$\begin{align*}
B(x)&=\frac{1-x^2}xA(x)\\\\
&=\frac{(1-x)^2(1+x)}{1-2x^2-x^3+x^4}\;.
\end{align*}$$
Some actual numbers:
$$\begin{array}{rccc}
n:&0&1&2&3&4&5&6&7&8&9&10&11&12&13\\
a_n:&0&1&0&2&1&3&4&5&10&11&21&27&43&64\\
b_n:&1&0&1&1&1&3&2&6&6&11&16&22&37&43
\end{array}$$
OEIS has $\langle a_n+b_n:n\ge 0\rangle=\langle b_{n+2}:n\ge 0\rangle$ as OEIS A062200 and $\langle a_n:n\ge 0\rangle$ as OEIS A122514. It has little more information beyond the recurrence
$$b_n=2b_{n-2}+b_{n-3}-b_{n-4}\;,$$
which is easily derived from $(1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/412222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that for every $n\in \mathbb{N}^{+}$, there exist a unique $x_{n}\in[\frac{2}{3},1]$ such that $f_{n}(x_{n})=0$ Let $f_{n}(x)=-1+x+\dfrac{x^2}{2^2}+\dfrac{x^3}{3^2}+\cdots+\dfrac{x^n}{n^2}$,
(1) Prove that for every $n\in \mathbb{N}^{+}$, then there exist unique $x_{n}\in[\frac{2}{3},1]$ such that
$f_{n}(x_{n})=0$
(2) Show that the sequence $(x_{n})$ of (1) is such that
$$0<x_{n}-x_{n+p}<\dfrac{1}{n}$$
for all $p\in \mathbb{N}$.
(3):$x_{n}=A+\dfrac{B}{n}+\dfrac{C}{n^2}+O(\dfrac{1}{n^2})$, find $A,B,C$?
My attempts :
For $(1)$, I have prove it :
$$f_{n}(1)=-1+1+\dfrac{1}{2^2}+\cdots+\dfrac{1}{n^2}>0,$$
\begin{align}
f_{n}(\frac{2}{3})&=-1+\dfrac{2}{3}+\cdots+\dfrac{(\dfrac{2}{3})^n}{n^2}\\
&\le-\dfrac{1}{3}+\dfrac{1}{4}\sum_{k=2}^{n}\left(\dfrac{2}{3}\right)^k\\
&=-\dfrac{1}{3}\cdot\left(\dfrac{2}{3}\right)^{n-1}<0
\end{align}
and
$$f'_{n}(x)=1+\dfrac{x}{2}+\cdots+\dfrac{x^{n-1}}{n}>0$$
But for $(2)$, I have methods:
since
$$x>0, f_{n+1}(x)=f_{n}(x)+\dfrac{x^{n+1}}{(n+1)^2}>f_{n}(x)$$
so $$f_{n+1}(x_{n})>f_{n}(x_{n})=f_{n+1}(x_{n+1})=0$$
since $f_{n+1}(x)$ is increasing, so $x_{n+1}<x_{n}$
since
$$f_{n}(x_{n})=-1+x_{n}+\dfrac{x^2_{n}}{2^2}+\cdots+\dfrac{x^n_{n}}{n^2}=0$$
$$f_{n+p}(x_{n})=-1+x_{n+p}+\dfrac{x^2_{n+p}}{2^2}+\cdots+\dfrac{x^{n+p}_{n+p}}{(n+p)^2}=0$$
and use $0<x_{n+p}<x_{n}\le 1$ we have
\begin{align}
x_{n}-x_{n+p}&=\sum_{k=2}^{n}\dfrac{x^k_{n+p}-x^k_{n}}{k^2}+\sum_{k=n+1}^{n+p}\dfrac{x^k_{n+p}}{k^2}\le\sum_{k=n+1}^{n+p}\dfrac{x^k_{n+p}}{k^2}\\
&\le\sum_{k=n+1}^{n+p}\dfrac{1}{k^2}<\sum_{k=n+1}^{n+p}\dfrac{1}{k(k-1)}=\dfrac{1}{n}-\dfrac{1}{n+p}<\dfrac{1}{n}
\end{align}
But for part $(3)$ I can't prove it,Thank you everyone can help,and for part $(2)$ have other nice methods?
| $x_{n}-x_{n+p}>0$ trivial, since $f_{n+p}(x_{n})>f_{n}(x_{n})=0$ and these $f$ always increase.
For the other inequality, I couldn't find a neat way to do it, so pardon some of the messy calculation. The idea is to use mean value theorem:
If $n\leq 3$ then trivial. So assume $n\geq 4$:
Now a bit of number crunching go in. You could use a Ferrari's quartic formula, or you could show that $f_{4}(\frac{4}{5})>0$. Either way you get that $x_{4}<\frac{4}{5}$ and thus $x_{n}<\frac{4}{5}$ for all $n\geq 4$.
Now we have $n(\frac{(x_{n})^{n+1}}{(n+1)^{2}}+\frac{(x_{n})^{n+2}}{(n+2)^{2}}+...+\frac{(x_{n})^{n+p}}{(n+p)^{2}})<\frac{(x_{n})^{n+1}}{n+1}+\frac{(x_{n})^{n+2}}{n+2}+...+\frac{(x_{n})^{n+p}}{n+p}<\frac{(x_{n})^{n+1}}{n+1}+\frac{(x_{n})^{n+2}}{n+1}+...+\frac{(x_{n})^{n+p}}{n+1}<\frac{(x_{n})^{n+1}}{n+1}(1+x_{n}+(x_{n})^{2}+...)=\frac{(x_{n})^{n+1}}{n+1}\frac{1}{1-x_{n}}$.
Use the earlier bound $n\geq 4,x_{n}<\frac{4}{5}$ give $\frac{(x_{n})^{n+1}}{n+1}\frac{1}{1-x_{n}}<\frac{(\frac{4}{5})^{5}}{5}\frac{1}{1-\frac{4}{5}}<1$
Notice that $f'_{n+p}(x)=1+\frac{x}{2}+...+\frac{x^{n+p-1}}{n+p}>1$ so $f'_{n+p}(x)>n(\frac{(x_{n})^{n+1}}{(n+1)^{2}}+\frac{(x_{n})^{n+2}}{(n+2)^{2}}+...+\frac{(x_{n})^{n+p}}{(n+p)^{2}})=nf_{n}(x_{n})+n(\frac{(x_{n})^{n+1}}{(n+1)^{2}}+\frac{(x_{n})^{n+2}}{(n+2)^{2}}+...+\frac{(x_{n})^{n+p}}{(n+p)^{2}})=n(-1+x_{n}+\frac{(x_{n})^{2}}{2^{2}}+...+\frac{(x_{n})^{n+p}}{(n+p)^{2}})=nf_{n+p}(x_{n})$
Now if $x_{n}-x_{n+p}\geq\frac{1}{n}$ then by applying mean value theorem there exist some $x\in[x_{n+p},x_{n}]$ such that $f'_{n+p}(x)=\frac{f_{n+p}(x_{n})-f_{n+p}(x_{n+p})}{x_{n}-x_{n+p}}=\frac{f_{n+p}(x_{n})}{x_{n}-x_{n+p}}\leq\frac{f_{n+p}(x_{n})}{(\frac{1}{n})}=nf_{n+p}(x_{n})$ which contradict the inequality just derived above.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is $S=\sum_{r=1}^\infty \tan^{-1}\frac{2r}{2+r^2+r^4}$ finite? Problem:
If $$S=\sum_{r=1}^\infty \tan^{-1}\left(\frac{2r}{2+r^2+r^4}\right)$$ Then find S ??
Solution:
I know that $\tan^{-1} x + \tan^{-1} y= \tan^{-1} \frac {x +y} {1-xy} $
But I have no idea how to such complicated question with it.
| HINT:
As $1+a^2+a^4=(1+a^2)^2-a^2=(1+a^2-a)(1+a^2+a),$
$$\frac{2\cdot a}{2+a^2+a^4}=\frac{(1+a^2+a)-(1+a^2-a)}{1+(1+a^2-a)(1+a^2+a)}$$
$$\implies \arctan \left(\frac{2\cdot a}{2+a^2+a^4}\right)=\arctan(1+a^2+a)-\arctan(1+a^2-a)$$
Can you recognize the Telescoping series?
So, $$\sum_{1\le r\le n}\arctan \left(\frac{2\cdot r}{2+r^2+r^4}\right)=\arctan (1+n^2+n)-\arctan 1$$
$$=\arctan\left(\frac{n^2+n}{n^2+n+2}\right)=\arctan\left(\frac{1+\frac1n}{1+\frac1n+\frac2{n^2}}\right)$$
$$\implies \lim_{n\to\infty}\arctan\left(\frac{1+\frac1n}{1+\frac1n+\frac2{n^2}}\right)=\arctan1=\frac\pi4$$
| {
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The limit of $\lim\limits_{x \to \infty}\sqrt{x^2+3x-4}-x$ I tried all I know and I always get to $\infty$, Wolfram Alpha says $\frac{3}{2}$. How should I simplify it?
$$\lim\limits_{x \to \infty}\sqrt{(x^2+3x+4)}-x$$
I tried multiplying by its conjugate, taking the squared root out of the limit, dividing everything by $\sqrt{x^2}$, etc.
Obs.: Without using l'Hôpital's.
| $$ \lim_{x \rightarrow \infty} \left(x^2 + 3x + 4\right)^{ \frac{1}{2}} - x $$
$$ = \lim_{x \rightarrow \infty} \left(x^2\left(1+\frac{3}{x}+\frac{4}{x^2}\right)\right)^{ \frac{1}{2}} - x $$
$$ = \lim_{x \rightarrow \infty} x\left(1+\frac{3}{x}+\frac{4}{x^2}\right)^{ \frac{1}{2}} - x $$
Then by Taylor expansion, we get that
$$ = \lim_{x \rightarrow \infty} x\left(1+\frac{1}{2}\left(\frac{3}{x}+\frac{4}{x^2}\right)+\operatorname{o}\left(\frac{1}{x^2}\right)\right) - x $$
$$ = \lim_{x \rightarrow \infty} \frac{3}{2} + \operatorname{o}\left(\frac{1}{x}\right) = \frac{3}{2} $$
as required.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find $x^{2000}+x^{-2000}$ when $x + x^{-1} = \frac{1}{2}(1 + \sqrt{5})$
Let $x+x^{-1}=\dfrac{1+\sqrt{5}}{2}$. Find $x^{2000}+x^{-2000}$.
How many nice methods do you know for solving this problem? Thank you everyone.
My method: because $x+\dfrac{1}{x}=2\cos{\dfrac{2\pi}{5}}$, so $$x^{2000}+\dfrac{1}{x^{2000}}=2\cos{\dfrac{2000\pi}{5}}=2.$$
Can you think of other nice methods? Or this problem has not used Euler's theorem: $(\cos{x}+i\sin{x})^n=\cos{nx}+i\sin{nx}$
| Here is an algebraic way avoiding trig functions: note that your number $x$ satisfies
$$x^2-(\frac{1+\sqrt{5}}{2})x+1=0 \quad \implies \quad \text{by multiplying by the conjugate} \quad x^4-x^3+x^2-x+1=0$$ and then use the factorization
$$x^{10}-1=(x^6+x^5-x-1)(x^4-x^3+x^2-x+1)$$ to see that $x^{10}=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find maximum value of $f(x)=2\cos 2x + 4 \sin x$ where $0 < x <\pi$ Find the maximum value of $f(x)$ where
\begin{equation}
f(x)=2\cos 2x + 4 \sin x \ \
\text{for} \ \ 0<x<\pi
\end{equation}
| We can utilize algebra to avoid calculus as follows
$$f(x)=2\cos2x+4\sin x$$
$$=2(1-2\sin^2x)+4\sin x$$
$$=2-(4\sin^2x-4\sin x)$$
$$=2-(2\sin x-1)^2+1\le 3$$ as for real $x,(2\sin x-1)^2\ge 0$
the equality occurs if $2\sin x-1=0\iff \sin x=\frac12\implies x=n\pi+(-1)^n\frac\pi6$
For $0<x<\pi, x=\frac\pi6$ or $\pi-\frac\pi6$
| {
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Does $x^2>x^3+1 \implies x < -{1\over P}$? How could one prove that:
$$x^2>x^3+1 \implies x < -{1\over P}$$
where $P$ denotes the plastic constant, the unique real root of $x^3-x-1=0$?
| The function $x^2 - x^3 - 1$ is monotone decreasing on $\{x < 0\}$ so we need to look at the (unique) $x$ where $$x^2 = x^3 + 1.$$ But for this $x$, dividing by $x^3$, $$-\frac{1}{x} = -1 - \frac{1}{x^3},$$ so $-\frac{1}{x}$ solves the defining equation for the plastic constant; so $-\frac{1}{x} = P$ and $x = \frac{-1}{P}$.
| {
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Simplify a expression with nested radical signs Simplify :
$\sqrt{10+6 \sqrt{2}+5 \sqrt{3}+4 \sqrt{6}}$
I have tried completing square but failed, Can anyone help me please? Thanks.
| $$10+6 \sqrt{2}+5 \sqrt{3}+4 \sqrt{6}=5(2+\sqrt3)+2\sqrt6(2+\sqrt3)=(2+\sqrt3)(5+2\sqrt6)$$
Now, $$5+2\sqrt6=3+2+2\cdot\sqrt2\cdot\sqrt3=(\sqrt3+\sqrt2)^2$$
and $$2+\sqrt3=\frac{4+2\sqrt3}2=\frac{3+1+2\cdot\sqrt3\cdot1}2=\frac{(\sqrt3+1)^2}2$$
Can you take it from here?
| {
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What does an expression $[x^n](1-x)^{-1}(1-x^2)^{-1}(1-x^3)^{-1}(1-x^4)^{-1}...$ mean? I came across the function that describes number of partitions of $n$ (I mean partitions like $5=4+1=3+2=3+1+1$ and so on.
There was defined a Cartesian product: $$\{0,1,1+1,1+1+1,...\}\times\{0,2,2+2,2+2+2,...\}\times\{0,3,3+3,...\}\times...$$ so that each partition is an element of finite sum (sum over components is finite) of this product. For example $1+2+2=(1,2+2,0,0,...)$.
And then followed something I didn't understand: author introduced polynomials in the following way:
$$\begin{align}1:1+x+x^2+x^3+...&=(1-x)^{-1}\\2:1+x^2+x^4+x^6+...&=(1-x^2)^{-1}\\3:1+x^3+x^6+x^9+...&=(1-x^3)^{-1}\end{align}\\...$$
(powers are corresponding elements of product), then multiplied all the sums and stated, that number of partitions of $n$ is:
$$[x^n](1-x)^{-1}(1-x^2)^{-1}(1-x^3)^{-1}...$$
Later he showed an example, but it was not numeric – $x$ was left in the expression.
So how do I understand this?
Thanks in advance!
| Often the notation
$$
[x^n](a_0+a_1x+a_2x^2+a_3x^3+\dots+a_nx^n+\dots)
$$
is used to mean the coefficient of $x^n$ in the given polynomial or power series ($a_n$ in the series I've given).
The formulas presented by the author are not polynomials, but power series. Polynomials are finite versions of power series. That is,
$$
\frac1{1-x}=1+x+x^2+x^3+x^4+\dots
$$
never terminates; there is no highest power of $x$.
For example,
$$
\left[x^6\right]\frac1{1-x}\frac1{1-x^2}\frac1{1-x^3}\frac1{1-x^4}\cdots=11
$$
| {
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Let $p_n$ be the sequence defined by $p_n=\sum_{k=1}^n\frac{1}{k}$. Show that $p_n$ diverges even though $\lim_{n\to\infty}(p_n-p_{n-1})=0$ I have tried this as :
$$p_n=\sum_{k=1}^n\frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n-1}+\frac{1}{n}$$
$$p_{n-1}=\sum_{k=1}^{n-1}\frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n-1}$$
$$(p_n-p_{n-1})=\frac{1}{n}$$
$$\lim_{n\to\infty}(p_n-p_{n-1})=\lim_{n\to\infty}\frac{1}{n}=0$$
But dont know how to show $p_n$ diverges?
| Recall the inequality $e^x>1+x \forall x>0$
Now Consider $$e^{p_n}=e^{\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)}$$ $$=e\cdot e^{\frac{1}{2}}\cdot e^{\frac{1}{3}}\cdots e^{\frac{1}{n}}$$ $$>\left(1+1\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\cdots\left(1+\frac{1}{n}\right)$$ $$=(2)\left(\frac{3}{2}\right)\left(\frac{4}{3}\right)\cdots\left(\frac{n+1}{n}\right)$$ $$=n+1$$
Thus, taking $\ln$ on both sides (inequality maintains as $\ln$ is an increasing function),
$$p_n>\ln(n+1)$$ and $\ln(n+1)$ is unbounded as $n\uparrow$ and hence $p_n$ diverges.
| {
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$a+b+c =0$; $a^2+b^2+c^2=1$. Prove that: $a^2 b^2 c^2 \le \frac{1}{54}$ If a,b,c are real numbers satisfying $a+b+c =0; a^2+b^2+c^2=1$.
Prove that $a^2 b^2 c^2 \le \frac{1}{54}$.
| So, $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$
$\implies ab+bc+ca=-\frac12.$
So $a,b,c$ are the roots of the equation $t^3-\frac12t-u=0$ where $u=abc.$
Using discriminant of the cubic equation fro real roots, $$\triangle= -4\cdot1\cdot\left(-\frac12\right)^3-27(-u)^2=\frac12-27u^2\ge0\iff u^2\le \frac1{54}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $10^{20} +20^{10}$ is divided by 4 then what would be its remainder? If $$10^{20} +20^{10}$$ is divided with 4 then what would be its remainder?
| Since
$$
\begin{eqnarray*}
10^{20}+20^{10} &=&\left( 10^{10}\right) ^{2}+2^{10}10^{10} \\
&=&10^{10}\left( 10^{10}+2^{10}\right) \\
&=&2^{10}5^{10}\left( 2^{10}5^{10}+2^{10}\right) \\
&=&2^{10}2^{10}5^{10}\left( 5^{10}+1\right) \\
&=&4^{10}5^{10}\left( 5^{10}+1\right) \\
&=&4\left( 4^{9}5^{10}\left( 5^{10}+1\right) \right) ,
\end{eqnarray*}
$$
the remainder would be is $0$, because
$$\frac{10^{20}+20^{10}}{4}=4^{9}5^{10}\left( 5^{10}+1\right).$$
| {
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Using partial fractions to find explicit formulae for coefficients? The set of binary string whose integer representations are multiples of 3 have the generating function
$$\Phi_S(x)={1-x-x^2 \over 1-x-2x^2}$$
Let $a_n=[x^n]\Phi_s(x)$ represent the number of strings in $S$ with length $n$. Using partial fraction expansion to determine an explicit formula for $a_n$ for all integers $n\geq 0$. You may use the initial conditions $a_0=1$, $a_1=0$.
Hmm. What are the initial conditions for? How are partial fractions and generating functions related to recurrences? The terms do seem to follow $a_n=a_{n-1}+2a_{n-2}$ but how to prove it?
| Note that we get very lucky, since $1-x-2x^2=(1-2x)(1+x)$.
The first step, since the numerator has degree $\ge $ the degree of the denominator, is to divide. We get
$$\frac{1-x-x^2}{1-x-2x^2}=\frac{1}{2}+\frac{1}{2}\frac{1-x}{1-x-2x^2}.$$
We now work for a while with the simpler $\frac{1-x}{1-x-2x^2}$, We try to find $A$ and $B$ such that
$$\frac{1-x}{1-x-2x^2}=\frac{A}{1+x}+\frac{B}{1-2x}=\frac{A(1+x)+B(1-2x)}{1-x-2x^2}.$$
So we want to have
$$1-x \quad\text{identically equal to}\quad A(1+x)+B(1-2x).$$
Put $x=\frac{1}{2}$. We get $A=\frac{1}{3}$. Putting $x=-1$, or otherwise, we find that $B=\frac{2}{3}$.
Now use the fact that
$$\frac{1}{1-t}=1+t+t^2+t^3+\cdots,$$
putting in turn $t=-x$ and $t=2x$, to find the power series expansions of $\frac{1}{1+x}$ and $\frac{1}{1-2x}$.
Finally, use our previous calculations to find the power series expansion of our original function. From this we can read off the coefficient of $x^k$ for any $k$. Note that our original function is equal to
$$\frac{1}{2}+\frac{1}{6}\cdot\frac{1}{1+x}+\frac{1}{3}\cdot\frac{1}{1-2x}.$$
| {
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Prove $\frac{\cos^2 A}{1 - \sin A} = 1 + \sin A$ by the Pythagorean theorem. How do I use the Pythagorean Theorem to prove that $$\frac{\cos^2 A}{1 - \sin A} = 1 + \sin A?$$
| Let's simplify matters:
In a triangle with hypotenuse equal to $1$ (think of the unit circle, an angle $A$ between the $x$ axis and the hypotenuse, we know that $$\sin A=\frac{\text{opposite}}{\text{hypotenuse}}\quad \text{and}\quad \cos A=\frac{\text{adjacent}}{\text{hypotenuse}}$$ then, since hypotenuse $=1$, we have the leg opposite the angle $A$ given by $\sin A=\text{opposite}/1$ and the leg along the x-axis of length $\cos A=\text{adjacent}/1$. Now, by the Pythagorean Theorem, and substitution, we have that
$$\begin{align}\text{opposite}^2 + \text{adjacent}^2 &= \text{hypotenuse}^2 = 1^2 \\ \sin^2A +\cos^2 A & = 1\end{align}$$
This gives us the well-known identity: $$\sin^2A + \cos^2 A = 1\tag{1}$$
We can express this identity in terms of $\cos^2 A$ by subtracting $\sin^2 A$ from both sides of the identity to get $$\begin{align}\cos^2 A & = 1 - \sin^2 A \\ &= (1)^2 - (\sin A)^2\tag{2}\end{align}$$
Now, we know that for any difference of squares, we can factor as follows: $$(x^2 - y^2) = (x +y)(x - y)\tag{3}$$
Since equation $(2)$ is a difference of squares, we have that $$\begin{align}\cos^2 A &= 1 - \sin^2 A \\ & = (1)^2 - (\sin A)^2 \\ &= (1 +\sin A)(1 - \sin A)\tag{4}\end{align}$$
Substituting gives us:
$$\begin{align}\frac{\cos^2 A}{1 -\sin A} & = \frac{1 - \sin^2 A}{1 -\sin A} \\ \\ &= \frac{(1 + \sin A)(\color{blue}{\bf 1 - \sin A})}{\color{blue}{\bf 1 -\sin A}}\\ \\ & = 1 + \sin A\end{align}$$
| {
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Show that $7 \mid( 1^{47} +2^{47}+3^{47}+4^{47}+5^{47}+6^{47})$ I am solving this one using the fermat's little theorem but I got stuck up with some manipulations and there is no way I could tell that the residue of the sum of each term is still divisible by $7$. what could be a better approach or am I on the right track? Thanks
| One more solution.
By Fermat's Little Theorem,
$a^{p - 1} \equiv 1 \pmod{p}$ for $a \not\equiv 0$. Thus,
$$
a^{48} = \left(a^6\right)^8 \equiv 1^8 = 1 \pmod{7}
$$
for each $a \in \{1, 2, \ldots, 6\}$.
As a consequence, $a^{47} \equiv a^{-1}$, so
$$
\begin{align}
1^{47} + 2^{47} + 3^{47} + 4^{47} + 5^{47} + 6^{47} &\equiv 1^{-1} + 2^{-1} + 3^{-1} + 4^{-1} + 5^{-1} + 6^{-1} \\
&\equiv 1 + 4 + 5 + 2 + 3 + 6 \\
&\equiv 0 \pmod{7}.
\end{align}
$$
| {
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If $a,b$ are roots for $x^2+3x+1=0$.Calculating $(\frac{a}{b+1})^2 +(\frac{b}{a+1})^2$ If $a,b$ are roots for the equation $x^2+3x+1=0$.How to calculate $$\left(\frac{a}{b+1}\right)^2 +\left(\frac{b}{a+1}\right)^2$$
| HINT:
As $ab=1, \frac a{b+1}=\frac{a}{\frac1a+1}=\frac{a^2}{a+1}=y$(say)
So, $a^2=y(a+1)$
and again $a^2+3a+1=0\implies a^2=-3a-1$
So, $ay+y=-3a-1\implies a=-\frac{y+1}{y+3} $
As $a$ is root of the given eqaution $$\left(-\frac{y+1}{y+3} \right)^2+3\left(-\frac{y+1}{y+3} \right)+1=0$$
Simply to get $y^2+4y-1=0$
Using Vieta's Formula, the required sum will be $(-4)^2-2(-1)=18$
| {
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Find $S$ where $S=\sqrt[3] {5+2 \sqrt {13}}+\sqrt[3]{5-2 \sqrt {13}}$, why am I getting an imaginary number? $\large S=\sqrt[3] {5+2 \sqrt {13}}+\sqrt[3]{5-2 \sqrt {13}}$
Multiplying by conjugate:
$\large S=\dfrac {-3}{\sqrt[3] {5+2 \sqrt {13}}-\sqrt[3]{5-2 \sqrt {13}}}$
From the original:
$\large S-2\sqrt[3]{5-2 \sqrt {13}}
=\sqrt[3] {5+2 \sqrt {13}}-\sqrt[3]{5-2 \sqrt {13}}$
Substituting:
$\large S=\dfrac{-3}{S-2\sqrt[3]{5-2 \sqrt {13}}}$
This leads to a quadratic equation in $\large S$ which I checked in wolframalpha and I got imaginary solutions. Why does this happen? I am not looking for an answer telling me how to solve this problem, I just want to know why this is wrong. Thanks.
| For simplicity, we'll let $S = a+b$ where $a=\sqrt[3]{5+2\sqrt{13}}$ and $b=\sqrt[3]{5-2\sqrt{13}}$.
Note that $S^3=a^3+3a^2b+3ab^2+b^3=5+2\sqrt{13}+5-2\sqrt{13}+3a^2b+3ab^2$.
$S^3+9S-10=3a^2b+3ab^2+9a+9b=3a(ab+3)+3b(ab+3)=3(a+b)(ab+3)=3S(ab+3)$.
Now, $ab=\sqrt[3]{(5+2\sqrt{13})(5-2\sqrt{13})}=\sqrt[3]{25-52}=\sqrt[3]{-27}=-3$. So $ab+3=0$. Thus, we see that $S$ is a root of the cubic equation $x^3+9x-10$. It is clear that $1$ is a root so the cubic is equal to $(x-1)(x^2+x+10)$. By the Quadratic Formula, the other two roots are imaginary. Thus $S=1$.
| {
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"source": "stackexchange",
"question_score": "3",
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Approximation of DE It depends on my previous question. Closed form solution of DE
I don't want to deal with Airy functions. How can I approximate this DE in continous domain $[0,1]$?
$$y''(x)+(x+1)y(x)=0\quad\text{ with the initial conditions}\quad y(0)=0\quad y'(0)=1$$
What if the conditions change to
$$y''(x)+(x+1)y(x)=0\quad\text{ with the initial conditions}\quad y(0)=0\quad y'(1)=1$$
May I use the same methods?
| The simplest "brute-force" approach would be to attempt a power series solution. The basic idea is to guess that the solution is of the form $y(x) = \sum_{n=0}^{\infty}a_{n}x^n$ and our task is to find the $a_{n}$. In principle this can be used to solve the DE to arbitrary precision by simply computing more and more terms of the series. We have:
$$
y(x) = \sum_{n=0}^{\infty}a_{n}x^n\implies y'(x) = \sum_{n=0}^{\infty}na_{n}x^{n-1}\implies y''(x) = \sum_{n=0}^{\infty}n(n-1)a_{n}x^{n-2}
$$
Note that the first two terms in the series of $y''(x)$ are zero, so we can rewrite the series as
$$
y''(x) = \sum_{n=2}^{\infty}n(n-1)a_{n}x^{n-2}= \sum_{n=0}^{\infty}(n+1)(n+2)a_{n+2}x^{n}
$$
Sub these into the DE:
$$
\sum_{n=0}^{\infty}(n+1)(n+2)a_{n+2}x^{n} + (x+1)\sum_{n=0}^{\infty}a_{n}x^n=0 \\
\sum_{n=0}^{\infty}(n+1)(n+2)a_{n+2}x^{n} + \sum_{n=0}^{\infty}a_{n}x^{n+1}+\sum_{n=0}^{\infty}a_{n}x^{n}=0
$$
We need to get all the powers of $x$ to be the same to match up the coefficients. By pulling the first term of the the first and third sum, we have:
$$
2a_{2}+\sum_{n=1}^{\infty}(n+1)(n+2)a_{n+2}x^{n} + \sum_{n=0}^{\infty}a_{n}x^{n+1}+a_{0}+\sum_{n=1}^{\infty}a_{n}x^{n}=0 \\
a_{0}+2a_{2}+\sum_{n=0}^{\infty}(n+2)(n+3)a_{n+3}x^{n+1} + \sum_{n=0}^{\infty}a_{n}x^{n+1}+a_{0}+\sum_{n=0}^{\infty}a_{n+1}x^{n+1}=0 \\
a_{0}+2a_{2}+\sum_{n=0}^{\infty}\big((n+2)(n+3)a_{n+3} + a_{n}+a_{n+1}\big)x^{n+1}=0
$$
Now, in order for ths to be zero, we need $a_{0}+2a_{2}=0$ and $(n+2)(n+3)a_{n+3} + a_{n}+a_{n+1} = 0$ for all $n$. Your initial conditions immediately imply that $a_{0}=0$ and $a_{1}=1$, so we must have $a_{2}=0$. Now, consider the recurrance relation with $n=0$, this gives:
$$
6a_{3} + a_{0}+a_{1} = 0\implies a_{3}=-1/6
$$
For $n=1$: $12a_{4} + a_{1}+a_{2} = 0\implies a_{4}=-1/12$. This procedure can be carried out indefinately. The fourth-order approximation (at zero) is thus $y(x)\approx x -\frac{1}{6}x^3 - \frac{1}{12}x^4$ (I would probably take a few more terms to make sure the approximation is still decent at $x=1$.)
In response to your edit:
$y(0) = 0\implies a_{0}=0\implies a_{2}=0$ (as above) and $y'(1) = 1 \implies \sum_{n=1}^{\infty}na_{n} = 1.$ Say we want a fourth-order approximation (as above.) Then we have $a_{1} + 2a_{2} + 3a_{3} + 4a_{4}=1\implies a_{1}+3a_{3}+4a_{4}=1$. Taking $n=0$ and $n=1$ gives the following set of equations:
$$
a_{1} + 3a_{3} + 4a_{4}=1 \\
a_{1} + 6a_{3} = 0 \\
a_{1} + 12a_{4} = 0
$$
Solving gives $a_{1} =6, a_{3} = -1, a_{4} = -1/2$. The new approximation is thus $y(x) \approx 6x - x^{3} - \frac{1}{2}x^4$.
Disclaimer: This approximation is bad. Like, really bad. You'd need to take many more terms to get accuracy comparable to the first situation.
| {
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} |
Integral $\int_0^\frac{\pi}{2} \sin^7x \cos^5x\, dx $ im asked to find the limited integral here but unfortunately im floundering can someone please point me in the right direction?
$$\int_0^\frac{\pi}{2} \sin^7x \cos^5x\, dx $$
step 1 brake up sin and cos so that i can use substitution
$$\int_0^\frac{\pi}{2} \sin^7(x) \cos^4(x) \cos(x) \, dx$$
step 2 apply trig identity
$$\int_0^\frac{\pi}{2} \sin^7x\ (1-\sin^2 x)^2 \, dx$$
step 3 use $u$-substitution
$$ \text{let}\,\,\, u= \sin(x)\ du=\cos(x) $$
step 4 apply use substitution
$$\int_0^\frac{\pi}{2} u^7 (1-u^2)^2 du $$
step 5 expand and distribute and change limits of integration
$$\int_0^1 u^7-2u^9+u^{11}\ du $$
step 6 integrate
$$(1^7-2(1)^9+1^{11})-0$$
i would just end up with $1$ however the book answer is $$\frac {1}{120}$$
how can i be so far off?
| $$\int_0^\frac\pi2\sin^7x\cos^5xdx=\int_0^\frac\pi2\sin^7x\cos^4x\cos xdx=\int_0^\frac\pi2\sin^7x(1-\sin^2x)^2\cos xdx$$
$$=\int_0^1 u^7(1-u^2)^2 du (\text{ Putting }\sin x=u)$$
$$=\int_0^1 (u^7-2u^9+u^{11}) du$$
$$=\left(\frac{u^8}8-2\frac{u^{10}}{10}+\frac{u^{12}}{12}\right)_0^1$$
$$=\frac18-\frac15+\frac1{12}$$
$$=\frac{15-24+10}{120}=\frac1{120} $$
| {
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How to integrate these integrals $$\int^{\frac {\pi}2}_0 \frac {dx}{1+ \cos x}$$
$$\int^{\frac {\pi}2}_0 \frac {dx}{1+ \sin x}$$
It seems that substitutions make things worse:
$$\int \frac {dx}{1+ \cos x} ; t = 1 + \cos x; dt = -\sin x dx ; \sin x = \sqrt{1 - \cos^2 x} = \sqrt{1 - (t-1)^2} $$
$$ \Rightarrow
\int \frac {-\sqrt{1 - (t-1)^2}}{t} = \int \frac {-\sqrt{t^2 + 2t}}{t} = \int \frac {-\sqrt t \cdot \sqrt t \cdot \sqrt{t +
2}}{\sqrt t \cdot t} $$
$$= \int \frac {- \sqrt{t + 2}}{\sqrt t } = \int - \sqrt{1 + \frac 2t} = ? $$
What next? I don’t know. Also, I’ve tried another “substitution”, namely $1 + \cos x = 2 \cos^2 \frac x2) $
$$ \int \frac {dx}{1+ \cos x} = \int \frac {dx}{2 \cos^2 \frac x2} = \int \frac 12 \cdot \sec^2 \frac x2 = ? $$
And failed again. Help me, please.
| You were on the right track in the last line
$$ \frac{1}{2} \int \sec^2 \frac{x}{2} dx = \tan \frac{x}{2} + C $$
That's it. Put the limits in and you're done.
For the second one, substitute $t = \frac{\pi}{2} - x $ and you're back to the first integral
| {
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How find this value $\frac{a^2+b^2-c^2}{2ab}+\frac{a^2+c^2-b^2}{2ac}+\frac{b^2+c^2-a^2}{2bc}$ let $a,b,c$ such that
$$\left(\dfrac{a^2+b^2-c^2}{2ab}\right)^2+\left(\dfrac{b^2+c^2-a^2}{2bc}\right)^2+\left(\dfrac{a^2+c^2-b^2}{2ac}\right)^2=3,$$
find the value
$$\dfrac{a^2+b^2-c^2}{2ab}+\dfrac{a^2+c^2-b^2}{2ac}+\dfrac{b^2+c^2-a^2}{2bc}$$
is true?
Yes, I tink this problem can prove
$$(a+b+c)(a+b-c)(a+c-b)(b+c-a)=0$$
so $$\dfrac{a^2+b^2-c^2}{2ab}+\dfrac{a^2+c^2-b^2}{2ac}+\dfrac{b^2+c^2-a^2}{2bc}=1or -3$$
How many nice methods prove $$(a+b+c)(a+b-c)(a+c-b)(b+c-a)=0$$ ?
and I see this easy problem
http://zhidao.baidu.com/question/260913315.html
| First, we simplify the initial equation by multiplying out by the denominator. Let
$$f(a,b,c) = c^2(a^2+b^2-c^2)^2 + a^2(b^2+c^2-a^2)^2 + b^2(c^2+a^2-b^2)^2 - 12a^2b^2c^2$$
In Ron Gordon's deleted post, he realized that $f(a,b,c) = 0 $ if $a+b=c$, $b+c=a$, $c+a=b$. This strongly suggests that $a+b-c, b+c-a, c+a-b$ are factors (sort of Remainder Factor Theorem). And indeed they are. We have
$$f(a,b,c) = -(a+b-c)(a-b+c)(-a+b+c)(a+b+c)(a^2+b^2+c^2)$$
which you can check in Wolfram.
Since the denominators are non-zero, we have $(a^2+b^2+c^2)>0$. Thus $$f(a,b,c) = 0 \Leftrightarrow (a+b-c)(a-b+c)(-a+b+c)(a+b+c) = 0 $$
We now split into cases.
Case 1. $(a+b-c)(a-b+c)(-a+b+c) = 0$
(Once again, multiply by denominators, and using Ron's observation.) Defining
$$g(a,b,c) = c(a^2+b^2-c^2) + a(b^2+c^2-a^2) + b(c^2+a^2-b^2) - 2abc $$
gives us
$$g(a,b,c) = - (a+b-c)(a-b+c)(-a+b+c) = 0 $$
Hence, the answer is 1.
Case 2. $a+b+c = 0$.
Then each term can be simplified in the form $\frac{ (a+b)^2 - c^2 - 2ab}{2ab} = -1$, hence the answer is -3.
| {
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How to solve the given initial-value problem? Solve the given problem which the input function $g(x)$ is discontinuous?
$y''+4y = g(x)$, $y(0) = 1$, $y'(0) = 2$, where
$$g(x) = \begin{cases} \sin x, & 0\leq x\leq\frac{\pi}{2}\\
0,& x>\frac{\pi}{2} \end{cases}$$
And the given answer is,
$$y = \begin{cases} \cos 2x+\frac56\sin2x+\frac13\sin x, & 0\leq x\leq\frac{\pi}{2}\\
\frac23\cos 2x+\frac56\sin2x,& x>\frac{\pi}{2} \end{cases}$$
| First let $g(x)=\sin(x)$.
$$
\begin{align*}
y'' + 4y &= \sin(x)
\end{align*}
$$
Applying the Laplace transform to both sides gives
\begin{align*}
s^2Y(s)-s-2+4Y(s) &= \dfrac{1}{s^2+1}\\
Y(s)(s^2+4) &= \dfrac{1}{s^2+1}+s+2\\
Y(s) &= \dfrac{1}{(s^2+1)(s^2+4)}+\dfrac{s}{s^2+4}+\dfrac{2}{s^2+4}\\
\end{align*}
Via partial fraction decomposition we get
\begin{align*}
Y(s) &= \dfrac{1}{3(s^2+1)}-\dfrac{1}{3(s^2+4)}+\dfrac{s}{s^2+4}+\dfrac{2}{s^2+4}\\
\end{align*}
Thus the inverse Laplace transform gives the desired solution:
$$y(x) = \frac{1}{3}\sin(x) - \frac{5}{6}\sin(2x)+\cos(2x)$$
For $g(x)=0$, we have
$$\begin{align*}y'' + 4y &= 0\\
\end{align*}$$
Let us make the ansatz that $y(x)=e^{\lambda x}$. Thus,
$$\begin{align*}
\lambda^2e^{\lambda x} + 4e^{\lambda x} &= 0\\
\lambda^2 + 4 &= 0\\
\end{align*}$$
Hence $\lambda = \pm 2i$, and the general solution is:
$$\begin{align*}
y(x) &= c_1 e^{2ix} + c_2e^{-2ix}\\
&=c_1(\cos(2x)+i\sin(x))+c_2(\cos(2x)-i\sin(2x))\\
&=(c_1+c_2)\cos(2x)+i(c_1-c_2)\sin(2x)\\
&=k_1\cos(2x)+k_2\sin(2x)
\end{align*}$$
where the last line simply redefines our constants, since they are already arbitrary. To satisfy initial conditions, $k_1=k_2=1$, which gives the specific solution
$$ y(x) = \cos(2x)+\sin(2x) $$
In conclusion, the piecewise function is
$$y(x) =
\begin{cases}
\frac{1}{3}\sin(x) - \frac{5}{6}\sin(2x)+\cos(2x) & \text{if } 0 \leq x \leq \frac{\pi}{2} \\
\cos(2x)+\sin(2x) & \text{if } x > \frac{\pi}{2}
\end{cases}$$
| {
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Prove that $\frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}+\frac{1}{(1+d)^2}\geq 1$ Let $abcd=1$ and $a,b,c$ and $d$ are all positive.
Prove that
$\dfrac{1}{(1+a)^2}+\dfrac{1}{(1+b)^2}+\dfrac{1}{(1+c)^2}+\dfrac{1}{(1+d)^2}\geq 1$
I am probably able to do this by assuming $a\geq b\geq c\geq d$
and by using derivative show that
$\dfrac{1}{(1+a)^2}+\dfrac{1}{(1+b)^2}\geq \dfrac{2}{(1+\sqrt{ab})^2}$
In this way I reduce the number of variables.
However this approach seems to be tedious.
Is there any other methods of proving this inequality?
I hope to learn the proof of the inequality in the title.
| This inequality I have nice methods:use
$$\dfrac{1}{(1+a)^2}+\dfrac{1}{(1+b)^2}\ge\dfrac{1}{1+ab}$$
pf:use $Cauchy-Schwarz$ inequality
we have
$$(a+1)^2\le\left(\dfrac{a}{b}+1\right)(ab+1)$$
$$(b+1)^2\le\left(\dfrac{b}{a}+1\right)(ab+1)$$
so
$$\dfrac{1}{(1+a)^2}+\dfrac{1}{(1+b)^2}\ge\dfrac{1}{\left(\dfrac{a}{b}+1\right)(ab+1)}+\dfrac{1}{\left(\dfrac{b}{a}+1\right)(ab+1)}=\dfrac{1}{ab+1}$$
$$\dfrac{1}{(1+c)^2}+\dfrac{1}{(1+d)^2}\ge\dfrac{1}{1+cd}$$
since $abcd=1$ then we have
$$\dfrac{1}{1+ab}+\dfrac{1}{1+cd}=\dfrac{1}{1+\dfrac{1}{cd}}+\dfrac{1}{1+cd}=\dfrac{cd}{1+cd}+\dfrac{1}{1+cd}=1$$
| {
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Substitution for $\int \frac {dx} {ax^2 + bx + c}$ I'm looking for the substitution that makes easier to solve integral containing quadratic polynomial in denominator (!) when such polynomial cannot be broken into parts (if it can, then it's possible to use partial fraction decomposition). Example:
$$\int \frac {dx} {5x^2 + x - 2}$$
Formulas suggested by wikipedia are hard to remember. So I hope there is some kind of substitution.
link: http://en.wikipedia.org/wiki/List_of_integrals_of_rational_functions#Integrands_of_the_form_xm_.2F_.28a_x2_.2B_b_x_.2B_c.29n
| HINT:
$$\frac1{ax^2+bx+c}=\frac{4a}{(2ax+b)^2+4ca-b^2}$$
Case $1:$ If $4ca-b^2=0,$ $$\int \frac{dx}{ax^2+bx+c}=4a\int\frac{dx}{(2ax+b)^2} $$ Put $y=2ax+b$
Case $2:$ If $4ca-b^2>0, 4ca-b^2=d^2$(say)
$$\int \frac{dx}{ax^2+bx+c}=4a\int\frac{dx}{(2ax+b)^2+d^2} $$
Put $2ax+b=d\tan\theta$ as $\tan^2\theta+1=\sec^2\theta$
Case $3:$ If $4ca-b^2<0, 4ca-b^2=-d^2$(say)
$$\int \frac{dx}{ax^2+bx+c}=4a\int\frac{dx}{(2ax+b)^2-d^2} $$
Put $2ax+b=d\sec\theta$ as $\sec^2\theta-1=\tan^2\theta$
| {
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Factor Equations Please check my answer in factoring this equations:
Question 1. Factor $(x+1)^4+(x+3)^4-272$.
Solution: $$\begin{eqnarray}&=&(x+1)^4+(x+3)^4-272\\&=&(x+1)^4+(x+3)^4-272+16-16\\
&=&(x+1)^4+(x+3)^4-256-16\\
&=&\left[(x+1)^4-16\right]+\left[(x+3)^4-256\right]\\
&=&\left[(x+1)^2+4\right]\left[(x+1)^2-4\right]+\left[(x+3)^2+16\right]\left[(x+3)^2-16\right]\\
&=&\left[(x+1)^2+4\right]\left[(x+1)^2-4\right]+\left[(x+3)^2+16\right]\left[(x+3)-4\right]\left[(x+3)+4\right]\end{eqnarray}.$$
Question 2. Factor $x^4+(x+y)^4+y^4$
Solution: $$\begin{eqnarray}&=&(x^4+y^4)+(x+y)^4\\
&=&(x^4+y^4)+(x+y)^4+2x^2y^2-2x^2y^2\\
&=&(x^4+2x^2y^2+y^4)+(x+y)^4-2x^2y^2\\
&=&(x^2+y^2)^2+(x+y)^4-2x^2y^2
\end{eqnarray}$$
I am stuck in question number 2, I dont know what is next after that line.
| For the first, I will put $y=\frac{x+1+x+3}2=x+2$
so that $x +1=y-1, x+3=y+1$ and the odd powers of $y$ vanish in $(y-1)^4+(y+1)^4$
$$\implies (x+1)^4+(x+3)^4-272=(y-1)^4+(y+1)^4-272$$
$$=2\{y^4+6y^2+1\}-272=2(y^4+6y^2-135)$$
$$=2\{y^4+(15-9)y^2-135\}=2(y^2+15)(y^2-9) =2(y^2-15)(y+3)(y-3)$$
$$=2\{(x+2)^2-15\}(x+5)(x-1)$$
$$\text{Now, }(x+2)^2-15=x^2+4x+4-15=x^2+4x-11\text{ which is not reducible}$$
| {
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How prove that $xyz+\sqrt{x^2y^2+y^2z^2+x^2z^2}\ge \frac{4}{3}\sqrt{xyz(x+y+z)}$ let $x,y,z>0$,and such that
$x^2+y^2+z^2=1$,prove that
$$xyz+\sqrt{x^2y^2+y^2z^2+x^2z^2}\ge \dfrac{4}{3}\sqrt{xyz(x+y+z)}$$
Does this have a nice solution? Thank you everyone.
| We need to prove that
$$xyz+\sqrt{(x^2+y^2+z^2)(x^2y^2+x^2z^2+y^2z^2)}\geq\frac{4}{3}\sqrt{xyz(x+y+z)(x^2+y^2+z^2)}.$$
Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, we need to prove that $f(w^3)\geq0,$ where
$$f(w^3)=w^3+3\sqrt{(3u^2-2v^2)(3v^4-2uw^3)}-4\sqrt{u(3u^2-2v^2)w^3}.$$
But by AM-GM
$$f'(w^3)=1-\frac{3u\sqrt{3u^2-2v^2}}{\sqrt{3v^4-2uw^3}}-2\sqrt{\frac{u(3u^2-2v^2)}{w^3}}\leq1-\sqrt{\frac{u(3u^2-2v^2)}{w^3}}\leq0,$$
which says that $f$ is a decreasing function and it's enough to prove our inequality
for a maximal value of $w^3$, which happens for equality case of two variables.
Since our inequality is homogeneous, it's enough to do it for $y=z=1$, which gives
$$x+\sqrt{(x^2+2)(2x^2+1)}\geq\frac{4}{3}\sqrt{x(x+2)(x^2+2)}$$ or after squaring of the both sides
$$x^4-16x^3+11x^2-32x+9+9x\sqrt{(x^2+2)(2x^2+1)}\geq0.$$
Now, by C-S $$\sqrt{(x^2+2)(2x^2+1)}=\sqrt{(x^2+1+1)(x^2+x^2+1)}\geq x^2+x+1.$$
Id est, it remains to prove that
$$x^4-16x^3+11x^2-32x+9+9x(x^2+x+1)\geq0$$ or
$$(x-1)^2(x^2-5x+9)\geq0.$$
Done!
| {
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Induction of inequality involving AP Prove by induction that
$$(a_{1}+a_{2}+\cdots+a_{n})\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}\right)\geq n^{2}$$
where $n$ is a positive integer and $a_1, a_2,\dots, a_n$ are real positive numbers
Hence, show that
$$\csc^{2}\theta +\sec^{2}\theta +\cot^{2}\theta \geq 9\cos^{2}\theta$$
Please help me.
Thank you!
| For the first part, we need a base case and an inductive step.
Base Case: suppose we have only one number, $a_1$. Then $$(a_1)\left(\frac1{a_1}\right)=1\leq 1$$
Inductive Step: By Isomorphism's work, this amounts to showing that
$$
a_n\left(\frac{1}{a_1} + \frac{1}{a_2} \cdots + \frac{1}{a_{n-1}}\right) + \frac1{a_n}(a_1 + a_2 \cdots + a_{n-1})\geq 2n-2
$$
Start by multiplying through to get
$$
\begin{align}
a_n\left(\frac{1}{a_1} + \frac{1}{a_2} \cdots + \frac{1}{a_{n-1}}\right) + \frac1{a_n}(a_1 + a_2 \cdots + a_{n-1}) & = \\
\left(\frac{a_n}{a_1} + \frac{a_n}{a_2} \cdots + \frac{a_n}{a_{n-1}}\right) + \left(\frac{a_1}{a_n} + \frac{a_2}{a_n} \cdots + \frac{a_{n-1}}{a_n}\right) &= \\
\left(\frac{a_n}{a_1} + \frac{a_1}{a_n}\right) +
\left(\frac{a_n}{a_2} + \frac{a_2}{a_n}\right) + \dots +
\left(\frac{a_n}{a_{n-1}} + \frac{a_{n-1}}{a_n}\right) &\geq \\
2+2+\dots+2
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/440761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Demonstrate this triangle! Help Give a triangle ABC with
$$\sin{\left(\frac{3A}{2}\right)}+\sin{\left(\frac{3B}{2}\right)}=2\cos{\left(\frac{(A-B)}{2}\right)}$$
Demonstrate that triangle ABC is equilateral triangle!!
Thank all!
P/S: I'm sorry. Because I speak English not good!
| $\sin{\left(\dfrac{3A}{2}\right)}+\sin{\left(\dfrac{3B}{2}\right)}=2\sin{\left(\dfrac{3A+3B}{4}\right)}\cos{\left(\dfrac{3A-3B}{4}\right)}$
let $x=\cos{\dfrac{A-B}{4}},p=\sin{\left(\dfrac{3A+3B}{4}\right)} \implies 1\ge x > \dfrac{\sqrt{2}}{2}, 1 \ge p >0 $,then we have:
$p(4x^3-3x)=2x^2-1$, it is trivial LHS< RHS when $\dfrac{\sqrt{3}}{2} >x > \dfrac{\sqrt{2}}{2}$, now we prove then $1 \ge x \ge \dfrac{\sqrt{3}}{2}$, LHS $\le$ RHS
LHS $\le 4x^3-3x \implies 4x^3-3x \le 2x^2-1 \iff 4x^3-3x - 2x^2+1\le 0 \iff (x-1)(4x^2+2x-1) \le 0$
it is true. so the "=" will hold when $p=1$ and $x=1$
so we have $\dfrac{3A+3B}{4}=\dfrac{\pi}{2},A=B \implies A=B=C=\dfrac{\pi}{3}$
QED
Edit: this can be more simple way:
WOLG, let $A \ge B $,
$x=\dfrac{A-B}{2} \ge 0 ,y=\dfrac{3(A+B)}{4} < \dfrac{3\pi}{4}, x<\dfrac{3\pi}{4} $, now it becomes:
$\sin{y}\cos{\dfrac{3x}{2}} = \cos{x}$
$\cos{x}$ is mono decreasing function on [$0,\pi)$, if $x \ge 0, \dfrac{3x}{2} \ge x, \cos{\dfrac{3x}{2}} \le \cos{x}, 1 \ge \sin{y} >0, \implies \sin{y} \cos{\dfrac{3x}{2}} \le \cos{x} \implies \sin{y}=1,\cos{\dfrac{3x}{2}}=\cos{x} \implies y=\dfrac{\pi}{2},x=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/441937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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} |
Find the point where $3x-4y+25=0$ is a tangent to the circle $x^2+y^2=25$ Find the coordinates of the point where the line $3x-4y+25=0$ is a tangent to the circle $x^2+y^2=25$.
Can someone please show me the first few steps of solving the simultaneous equation?
| $$
3x - 4y + 25 = 0\\
4y = 3x + 25\\
m = \frac34
$$
Differentiate circle equation $ x^2 + y^2 =25 $
$ \dfrac{dy}{dx}= -\dfrac xy = m = \dfrac 34 $
Solve for x and y to get, $ x = \mp 3, y =\pm 4 $
Only the first set $ (-3,4)$ satisfies the given equation of straight line tangent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/443136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Simple limit problem: $\lim_{x\to2}(\frac{1}{x-2}-\frac{4}{x^2-4})$ While trying to help my sister with her homework she gave me the next limit: $$\lim_{x\to2}(\frac{1}{x-2}-\frac{4}{x^2-4})$$
I know the conventional way of solving it would be (That's what i showed her):
$$\lim_{x\to2}(\frac{1}{x-2}-\frac{4}{x^2-4})=\lim_{x\to2}\left(\frac{x+2-4}{x^2-4}\right)=\lim_{x\to2}\left(\frac{x-2}{(x+2)(x-2)}\right)=\lim_{x\to2}\left(\frac{1}{x+2}\right)=\frac14$$
But she gave me the next answer:
$$\begin{align}
\lim_{x\to2}(\frac{1}{x-2}-\frac{4}{x^2-4})&=\lim_{x\to2}\frac{1}{x-2}-4\lim\frac{1}{x^2-4}\\
&=\lim_{x\to2}\frac{1}{x-2}-4\lim_{x\to2}\frac{1}{x+2}\lim_{x\to2}\frac{1}{x-2}\\
&=\lim_{x\to2}\frac{1}{x-2}-4\frac14\lim_{x\to2}\frac{1}{x-2}\\
&=\lim_{x\to2}\frac{1}{x-2}-\lim_{x\to2}\frac{1}{x-2}\\
&=0
\end{align}$$
I actually couldn't explain her why is she wrong. Cause technically it looks fine. What am i missing?
| If $\lim_{x \to 2} \frac{1}{x-2}-\frac 4{x^2-4}=0$, that means $$\lim_{x \to 2} \frac{1}{x-2}- \lim_{x \to 2} \frac 4{x^2-4}=0 \implies \lim_{x \to 2} \frac{1}{x-2}=\lim_{x \to 2} \frac 4{x^2-4}$$.
But $\frac1{x-2} \neq \frac 4{x^2-4}$ or $\frac 1{x-2} \neq \frac1{x-2} \frac4{x+2}$
Thus $\lim_{x \to 2} \frac{1}{x-2}-\frac 4{x^2-4}=0$ do not true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/443556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
How to prove two trigonometric identities I want to show that
$${\sin}^2 \alpha + 4{\sin}^4\frac{\alpha}{2} = 4{\sin}^2 \frac{\alpha}{2}$$
and
$${\sin}^2 \alpha + 4{\cos}^4\frac{\alpha}{2} = 4{\cos}^2 \frac{\alpha}{2}$$
They should be true, as Wolfram Alpha says so. However, I want to prove them, and I have no idea how to proceed here. Any ideas?
| $${\sin}^2 \alpha + 4{\sin}^4\frac{\alpha}{2}$$
$$\left(2\sin\frac {\alpha} {2}\cos \frac{\alpha}{2}\right)^2 + 4{\sin}^4\frac{\alpha}{2}$$
$$4\sin^2\frac {\alpha} {2}\cos^2 \frac{\alpha}{2} + 4{\sin}^4\frac{\alpha}{2}$$
$$4\sin^2\frac {\alpha} {2}(\cos^2 \frac{\alpha}{2} + {\sin}^2\frac{\alpha}{2})$$
$$4\sin^2\frac {\alpha} {2}\cdot 1$$
$$4\sin^2\frac {\alpha} {2}$$
same way second part can be prove
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/450962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
If $\int f(x) \sin{x} \cos{x}\,\mathrm dx = \frac {1}{2(b^2 - a^2)} \log f(x) +c $. Find $f(x)$
Problem: If $\int f(x) \sin{x} \cos{x}\,\mathrm dx = \frac {1}{2(b^2 - a^2)} \log f(x) +c $. Find $f(x)$
Solution: $\int f(x) \sin{x} \cos{x}\,\mathrm dx = \frac {1}{2(b^2 - a^2)} \log f(x) +c $
Differenting both sides,we get
$ f(x) \sin{x} \cos{x} = \frac {f'(x)}{2(b^2 - a^2)f(x)} $
Am I doing right ?
| It is a good start. For simplicity write $y$ for $f(x)$. We can rewrite the result you got as
$$\frac{y'}{y^2}=2(b^2-a^2)\sin x\cos x.$$
Integrate both sides. It may be handy to note that $2\sin x\cos x=\sin(2x)$. Or not, since it is clear that $2\sin x\cos x$ is the derivative of $\sin^2 x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/451131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Need help calculating this determinant using induction This is the determinant of a matrix of ($n \times n$) that needs to be calculated:
\begin{pmatrix}
3 &2 &0 &0 &\cdots &0 &0 &0 &0\\
1 &3 &2 &0 &\cdots &0 &0 &0 &0\\
0 &1 &3 &2 &\cdots &0 &0 &0 &0\\
0 &0 &1 &3 &\cdots &0 &0 &0 &0\\
\vdots &\vdots &\vdots&\ddots &\ddots &\ddots&\vdots &\vdots&\vdots\\
0 &0 &0 &0 &\cdots &3 &2 &0 &0\\
0 &0 &0 &0 &\cdots &1 &3 &2 &0\\
0 &0 &0 &0 &\cdots &0 &1 &3 &2\\
0 &0 &0 &0 &\cdots &0 &0 &1 &3\\
\end{pmatrix}
The matrix follows the pattern as showed.
I have to calculate it using induction (we haven't learnt recursion so far).
Thanks
| Let $D_n$ be the determinant of our matrix of size $n$. We can calculate $D_n$ by expansion of the first column: $D_n = 3 D_{n - 1} - 1 \cdot 2 \cdot D_{n - 2}$. For the second term we expanded again by the first row. We can see: $D_1 = 3$, $D_2 = 7$. By our recurence, we can count more terms: 3, 7, 15, 31, 63, …. Now we can guess that $D_n = 2^{n + 1} - 1$. The formula works for first terms. We will prove the formula by induction. Induction step is: if it works for all $k ≤ n - 1$, then it works for n. We have $D_n = 3 D_{n - 1} - 2 D_{n - 2} = 3 (2^n - 1) - 2 (2^{n - 1} - 1) = 2^{n + 1} - 1$. And we're done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/451941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
n is+ve integer, how many solutions $(x,y)$ exist for $\frac{1}{x} + \frac{1}{y} = \frac{1}{n}$ with $x$, $y$ being positive integers and $(x \neq y)$ I wanted to know, how can i solve this.
For a given positive integer n, how many solutions $(x,y)$ exist for $\frac{1}{x} + \frac{1}{y} = \frac{1}{n}$ with $x$ and $y$ being positive integers and $(x \neq y)$.
| It is clear that the two positive fractions on the left have to be less than $\frac 1n$, so we have $x,y \gt n$.
Let $x=n+a, y=n+b$ and clear fractions then $$n(n+b)+n(n+a)=(n+a)(n+b)$$ which simplifies to $$ab=n^2$$
Any such $a,b$ give a pair of fractions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/452137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Differentiability of function involving absolute values I need to check the differentiability at $x=-10$ of the following function.
$f(x)=\cos|x-5|+\sin|x-3|+|x+10|^3-(|x|+4)^2$
Now,
\begin{align*}
\text{LHD} &= \lim_{x\to -10^-}\frac{f(x)-f(-10)}{x+10}\\
&= \lim_{x\to -10^-}\frac{\cos(x-5)-\sin(x-3)-(x+10)^3-(4-x)^2-(\cos15-\sin13-14^2)}{x+10}\\
&= \lim_{x\to -10^-}\frac{2\sin(\frac{x+10}{2})\sin(\frac{20-x}{2})+2\cos(\frac{x+10}{2})\sin(\frac{16-x}{2})-(x+10)^3+(10+x)(18-x)}{x+10}\\ \\
&=-\infty
\end{align*}
Similarly,
RHD = $+\infty$. So, not differentiable. But the answer says that $f(x)$ is differentiable at $-10$.
Can anyone please help?
| f(x)=cos(x-5)- sin (x-3)-(x-3)^3-(-x+4)^2 when xis less than -10 or = -10
=cos(x-5)-sin(x-3)+(x-3)^3-(-x+4)^2 whenis between-10 and 0
LHD at x=-10
=sin 15- c0s 13 -196
RHD AT x=-10
=sin15-cos 13 -196
icomputed the lrft hand derivative and right hand derivative using the definition and l`hopital rule
so the derivative at x=-10 exists and the function f is differentiable at x=-10
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/452699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to factor $8xy^3+8x^2-8x^3y-8y^2$ How can I factor $8xy^3+8x^2-8x^3y-8y^2$ or the different form $2x(4y^3+4x)-2y(4x^3+4y)$
Is there any general methods that work?
A possible solution should be $8(x^2-y^2)(1-xy)$ But please do not start from here as in the general case I will not know the answer...
Thanks!
Alexander
| $$ 8xy^3+8x^2-8x^3y-8y^2= 8xy^3-8x^3y+8x^2-8y^2$$
$$= 8xy(y^2-x^2)-8(y^2-x^2)= (8xy-8)(y^2-x^2) $$
$$ = 8(1-xy)(x^2-y^2). $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/452978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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show that $\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$ show that
$$\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$$
using different ways
thanks for all
| $$
\begin{aligned}
\int_{0}^{\infty} \frac{\sin ^{3} x}{x^{3}} d x
=& \frac{1}{4} \int_{0}^{\infty} \frac{3 \sin x-\sin 3 x}{x^{3}} d x \\
\stackrel{IBP }{=} & \frac{1}{8} \int_{0}^{\infty} \frac{(3 \sin x-\sin 3 x)^{(2)}}{x} d x\\
=&\frac{1}{8} \int_{0}^{\infty} \frac{-3 \sin x+9 \sin 3 x}{x} d x \\
=&\frac{1}{8}\left(-\frac{3 \pi}{2}+\frac{9 \pi}{2}\right) \\
=&\frac{3 \pi}{8}
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/453198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 8,
"answer_id": 3
} |
For what $n$ is $x^2 + x+ 1\mid x^{2n} + x^n + 1$? For what $n \in \mathbb{N}$ is $x^2 + x+ 1\mid x^{2n} + x^n + 1$?
The only obvious thing that I could see was noticing that $(x^3 - 1) = (x-1)(x^2 + x+1)$. So, if $x^3 - 1\mid x^{2n} + x^n + 1$. I don't even know if this will help.
| HINT:
If $\omega$ is a root of $x^2+x+1=0, \omega^2+\omega+1=0\implies \omega^3=1$
So if $3|n, n=3m$(say),
$$\omega^n=\omega^{3m}=(\omega^3)^m=1 \text{ and } \omega^{2n}=\omega^{6m}=(\omega^3)^{2m}=1$$
$$\implies \omega^{2n}+\omega^n+1=3\ne0$$
$\implies \omega$ is not a root of $x^{2n}+x^n+1=0$
If $n=3m+1,\omega^{2n}+\omega^n+1=\omega^{2(3m+1)}+\omega^{3m+1}+1=\omega^2+\omega+1=0$
$\implies \omega$ is a root of $x^{2n}+x^n+1=0$
$\implies (x-\omega)|(x^{2n}+x^n+1)$
Similarly, $(x-\omega^2)|(x^{2n}+x^n+1)$
and as $(x-\omega)(x-\omega^2)=x^2+x+1\implies (x^2+x+1)|(x^{2n}+x^n+1)$
Similarly, for $n=3m+2$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Differentiation, an issue with an exercise I'm currently working on an exercise that involves quite a few fractional exponents. This is it: $$y = \frac {(x^4 + a)^\frac {1}{3}} {(x^3 + a)^ \frac {1}{2}} $$
I take the multiplication route by doing:
$$(x^4 + a)^ \frac{1}{3}(x^3 + a) ^ \frac {-1}{2} $$
I eventually get $$(x^4 + a)^\frac {1}{3} \frac{-3}{2}x^2(x^3 + a)^ \frac{-3}{2} + (x^3 + a)^ \frac{-1}{2}\frac{4}{3}x^3(x^4 + a)^\frac{-2}{3}$$
This is what I can't get past at the moment. I'm not sure of the next step I should take to ensure I get the correct answer. Any help on this is much appreciated as always! There may be some formatting issues with the negative fractional exponents as they do look a little odd.
| Logarithmic differentiation is useful here.
$$\log{y} = \frac13 \log{(x^4+a)} - \frac12 \log{(x^3+a)}$$
Then
$$\frac{y'}{y} = \frac13 \frac{4 x^3}{x^4+a} - \frac12 \frac{3 x^2}{x^3+a}$$
Multiply by the original $y$ and you are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/454902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
If a,b,c are sides of a triangle, prove: $ \sqrt{a+b-c} + \sqrt{b+c-a} + \sqrt{c+a-b} \le \sqrt{a} + \sqrt{b} + \sqrt{c} $ I did substitute $a=x+y, b=x+z, c=y+z$ and I arrived at $\sqrt{2x} + \sqrt{2y} + \sqrt{2z} \le \sqrt{x+y} + \sqrt{x+z} + \sqrt{y+z}$.
However, after this, I tried various methods like AM-GM and Cauchy-Schwarz inequality for hours and I still can't prove it. Can someone help please? Thanks.
| Since $\sqrt{x}$ is concave down, Jensen's inequality tells us that
$ \dfrac 12 ( \sqrt{2x} + \sqrt{2y}) \leq \sqrt{ \dfrac{ 2x + 2y } 2 } = \sqrt{x+y}$.
Summing cyclically gives the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/456541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Prove: $\frac{1}{1^2} +\frac{1}{2^2} + \cdots + \frac{1}{n^2} + \cdots = \sum_{n=1}^\infty \frac{1}{n^2} < 2$ While I don't doubt that this question is covered somewhere else I can't seem to find it, or anything close enough to which I can springboard. I however am trying to prove $$\frac{1}{1^2} +\frac{1}{2^2} + \cdots + \frac{1}{n^2} + \cdots = \sum_{n=1}^\infty \frac{1}{n^2} < 2$$ by induction.
I have seen it many times and proved it before but can't remember what it was I did. I see that for the first two terms $n = 1, n=2$ I get:
for $n = 1$, $\frac{1}{1^2} = 1 < 2$
for $n = 2$, $\frac{1}{1^2} + \frac{1}{2^2} = \frac{5}{4} < 2$
Now I am stumped, I know I want to show this works for the $n+1$ term and am thinking, let the series $\sum_{n=1}^\infty \frac{1}{n^2} = A(n)$ Then look to show the series holds for $A(n+1)$ But $A(n+1) = A(n) + \frac{1}{(n+1)^2}$ But now what? If I tried $A(n+1) - A(n) = \frac{1}{(n+1)^2}$ , but would have to show that this is less than $2 - A(n)$. I am stuck.
Thanks for your thoughts,
Brian
| Hint Prove instead that
$$\frac{1}{1^2} +\frac{1}{2^2} + \cdots + \frac{1}{n^2} < 2-\frac{1}{n}$$
Interesting, this stronger exercise is an easy induction problem, while the weaker result you mentioned cannot be proven directly by induction for obvious reasons (LHS increases, RHS is constant < --- in the stronger version both sides are increasing, this issue is fixed).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/456595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
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Find $a,b\in\mathbb{Z}^{+}$ such that $\large (\sqrt[3]{a}+\sqrt[3]{b}-1)^2=49+20\sqrt[3]{6}$ find positive intergers $a,b$ such that
$\large (\sqrt[3]{a}+\sqrt[3]{b}-1)^2=49+20\sqrt[3]{6}$
Here i tried plugging
$x^3=a,y^3=b$
$(x+y-1)^2=x^2+y^2+1+2(xy-x-y)=49+20\sqrt[3]{6} $
the right hand part is a square hence can be written as $(p+q)^2$
| I did an Excel search and found $(288,48)=(6^2*2^3,6*2^3)$ and the reverse as solutions. This is confirmed by Alpha
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/456681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Convergence of the Jacobi iteration method I think I am not quite understanding the Jacobi Method or some related concept for indirectly solving linear systems of equations of the form $Ax=b$. We need the norm
$||I-Q^{-1}A||_\infty < 1$ and the Jacobi Method states to let $Q$ be the matrix with the same diagonal entries as $A$. Then we have:
\begin{equation}
||I-Q^{-1}A||_\infty = \max_{1\leq i\leq n}\sum\limits_{j=1,j\neq i}^n \left|\frac{a_{ij}}{a_{ii}}\right|
\end{equation}
Let $A$ be a diagonally dominant matrix. We see that $Q^{-1}A$ has $1$'s on the diagonal. the matrix $I-Q^{-1}A$ has $0$s on the diagonal and the sum of each row is less
than 1. My book then concludes that $||I-Q^{-1}A||_\infty < 1$. It is not obvious to me from the above equations that this is true. I think I constructed a
counter-example:
\begin{align}
A=&\begin{pmatrix}\frac{1}{2}&\frac{1}{4}&\frac{1}{8}\\
0&\frac{2}{3}&\frac{1}{2}\\
3&\frac{1}{2}&4\end{pmatrix}\\
Q^{-1}A =& \begin{pmatrix}1&\frac{1}{2}&\frac{1}{4}\\
0&1&\frac{3}{4}\\
\frac{3}{4}&\frac{1}{8}&1\end{pmatrix}\\
I-Q^{-1}A =& \begin{pmatrix}0&\frac{-1}{2}&\frac{-1}{4}\\
0&0&\frac{-3}{4}\\
\frac{-3}{4}&\frac{-1}{8}&0\end{pmatrix}
\end{align}
$A$ is diagonally dominant but $||I-Q^{-1}A||_\infty$ which is the sum of the absolute value of the maximum entry in each column $=\frac{3}{4}+\frac{1}{2}+\frac{3}{4}>1$
Is this correct? If not, what did I do wrong?
| Your understanding of the infinity norm for matrices is wrong:
$$ \Vert I - Q^{-1}A \Vert_\infty = \max \bigg \{0+\bigg|-\frac 12\bigg| + \bigg|-\frac14\bigg|, 0 +0+\bigg|-\frac34\bigg|, \bigg|-\frac34\bigg| + \bigg|-\frac18\bigg| + 0 \bigg\} = \frac 7 8 < 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/457914",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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"answer_id": 0
} |
Prove this result related to progressions I'm stuck with the following problem from Higher Algebra by Hall and Knight:
If $r<1$ and positive, and $m$ is a positive integer, show that $(2m+1)r^m(1-r) < 1-r^{2m+1}$. Hence show that $nr^n$ is indefinitely small when $n$ is indefinitely great.
I have no clue about the second part of the problem, but for now I've spent the whole afternoon trying to work out the first part -- no success.
Here are my two attempts:
1) Consider the sum of the sequence
$S = 1 + r + r^2 + ... + r^{2m}$
$\Rightarrow S = \dfrac{1-r^{2m+1}}{1-r}$
Now I can claim that the sum of these terms is greater than any of them ($m$ and $r$ being positive), so $S > r^m$. This leaves me with:
$\dfrac{1-r^{2m+1}}{1-r} > r^m$
Which is incomplete.
2) This time we consider the sequence:
$S = 1 + 3r + 5r^2 + ... + (2m-1)r^{m-1} + (2m+1)r^m$
$\Rightarrow rS = r + 3r^2 + ... + (2m-3)r^{m-1} + (2m-1)r^{m} + (2m+1)r^{m+1}$
$\Rightarrow (1-r)S = 1 + (2r + 2r^2 + ... + 2r^m) - r(2m+1)r^m$
(after some simplification)
$\Rightarrow (1-r)^2S = 1 + r -3r^{m+1} -2mr^{m+1}$
Now, I can claim that the LHS is greater than zero, which gives:
$2mr^{m+1} < 1 + r - 3r^{m+1}$
$\Rightarrow (2m+1)r^{m+1} < 1 + r - 2r^{m+1}$
etc. However, this too is of different form.
Please provide some help!
| You have $0 < r < 1$, so you can divide by $r^m(1-r) > 0$ and get the equivalent inequality
$$\begin{align}2m+1 &< \frac{1}{r^m} \frac{1-r^{2m+1}}{1-r} = \frac{1}{r^m}\left(1 + r + \dotsb + r^{2m-1} + r^{2m}\right)\\
&= \left(\frac{1}{r^m} + \frac{1}{r^{m-1}} + \dotsb + \frac1r + 1 + r + \dotsb + r^{m-1} + r^m \right)
\end{align}$$
to prove.
In that form, we can easily deduce it from
$$\left(x + \frac1x \geqslant 2\right) \land \left(x+\frac1x = 2 \iff x = 1\right),$$
since we can write
$$\frac{1}{r^m} + \dotsb \frac1r + 1 + r + \dotsb r^m = 1 + \sum_{k=1}^m \left(r^k + \frac{1}{r^k}\right) > 1 + m\cdot 2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/457976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Closed form for this continued fraction Is there a closed form for this continued fraction?
$$x+\frac{1}{x+\frac{1}{x+\frac{1}{...}}}$$
| Here's a cute, handwavy way to do it:
$$ f(x) = x + \cfrac{1}{x+\cfrac{1}{x+\cfrac{1}{x+\dots}}}$$
Notice then that
$$ \frac{1}{f(x)} = \cfrac{1}{x+\cfrac{1}{x+\cfrac{1}{x+\dots}}} $$
Thus
\begin{align}
x+\frac{1}{f(x)} &= x+\cfrac{1}{x+\cfrac{1}{x+\cfrac{1}{x+\dots}}} \\
&=f(x)
\end{align}
So, we have a functional relationship
$$ x + \frac{1}{f(x)} = f(x) $$
or equivalently
$$ xf(x) + 1 = f(x)^{2} $$
This quadratic equation may be solved quite simply, we have candidate solutions:
$$ f_{\pm}(x) = \frac{x\pm\sqrt{x^{2}+4}}{2}$$
We should note that $f(1)$ as a continued fraction is precisely the golden ratio. Hence we deduce
$$ f(x) = \frac{x+\sqrt{x^{2}+4}}{2}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/459938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Compare the sum of the squares of the median of a triangle to the sum of the squares of sides You have to compare the sum of the squares of the median of a triangle to the sum of the squares of sides?
| Let $ABC$ be a triangle and let $I,J,K$ be the midpoints of $[BC],[AC],[AB]$ respectively.
$\overrightarrow{AI}^2=\frac{1}{4}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)^2=\frac{1}{4}\left(AB^2+AC^2+2\overrightarrow{AB}\cdot\overrightarrow{AC}\right)$
The same for $BJ^2$ and $CK^2$. Adding the three equations, we obtain
$$AI^2+BJ^2+CK^2=\frac{1}{4}\left(2AB^2+2AC^2+2BC^2+(\overrightarrow{AB}\cdot\overrightarrow{AC}+\overrightarrow{BA}\cdot\overrightarrow{BC})+(\overrightarrow{AB}\cdot\overrightarrow{AC}+\overrightarrow{CA}\cdot\overrightarrow{CB})+(\overrightarrow{BA}\cdot\overrightarrow{BC}+\overrightarrow{CA}\cdot\overrightarrow{CB})\right)=\frac{3}{4}\left(AB^2+AC^2+BC^2\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/460588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Prove the inequality $\,\frac{1}{\sqrt{1}+ \sqrt{3}} +\frac{1}{\sqrt{5}+ \sqrt{7} }+\ldots+\frac{1}{\sqrt{9997}+\sqrt{9999}}\gt 24$
Prove the inequality $$\frac{1}{\sqrt{1}+ \sqrt{3}}
+\frac{1}{\sqrt{5}+ \sqrt{7} }+......... +\frac{1}{\sqrt{9997}+\sqrt{9999}} > 24$$
My work:
Rationalizing the denominator gives
$$\frac{\sqrt{3}-1}{2} +\frac{\sqrt{7}-\sqrt{5}}{2}+......+\frac{\sqrt{9999}-\sqrt{9997}}{2} .$$
Now by taking two as common and separating the positive and negative terms gives
$$\frac{1}{2} [ \{\sqrt{3} +\sqrt{7}+\dots +\sqrt{9999}\} - \{1+\sqrt{5} +\dots+\sqrt{9997}\}].$$
Can we do like this please suggest. Thanks.
| Another idea: by considering concavity and a left endpoint approximation, the desired sum is an overestimate of the following integral:
$$\frac{1}{2}\int_0^{2500}\sqrt{4x+3}-\sqrt{4x+1}\approx 24.6528$$
More explicitly, notice that your sum is:
$$\frac{1}{2}\sum_{n=0}^{2499}\sqrt{4n+3}-\sqrt{4n+1}$$
We can think of this as one half of the sum of the areas of $2500$ rectangles of width $1$ and height $\sqrt{4n+3}-\sqrt{4n+1}$. These rectangles can be visualized in the plane as follows: consider the two curves $f(x)=\sqrt{4x+3}$ and $g(x)=\sqrt{4x+1}$. The $n$th rectangle (starting the count from $0$) is then formed by the $4$ points:
$$(n,f(n)),(n,g(n)),(n+1,f(n)),(n+1,g(n))$$
Notice that the base has length $1$, and the height is exactly $\sqrt{4n+3}-\sqrt{4n+1}$. Also, notice that the area of the rectangle is well-approximated by the area between the two curves, and in fact is an overestimate if you consider the fact that the upper curve always has a shallower slope. This means that the area between these two curves from $x=0$ to $x=2500$ is an underestimate of your desired sum. This is what the integral above calculates, the area between the two curves.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/462118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Factorize $(x+1)(x+2)(x+3)(x+6)- 3x^2$ I'm preparing for an exam and was solving a few sample questions when I got this question -
Factorize : $$(x+1)(x+2)(x+3)(x+6)- 3x^2$$
I don't really know where to start, but I expanded everything to get :
$$x^4 + 12x^3 + 44x^2 + 72x + 36$$
I used rational roots test and Descartes rule of signs to get guesses for the roots. I tried them all and it appears that this polynomial has no rational roots.So, what should I do to factorize this polynomial ?
(I used wolfram alpha and got the factorization : $(x^2 + 4x + 6) (x^2 + 8x + 6)$ But can someone explain how to get there ?)
| $$(x+1)(x+2)(x+3)(x+6)-3x²$$
$$(x+1)(x+6)(x+3)(x+2)-3x²$$
$$(x²+6x+1x+6)(x²+3x+2x+6)-3x²$$
$$(x²+6x+6+1x)(x²+5x+6+x-x)-3x²$$
$$(x²+6x+6+1x)(x²+5x+x+6-x)-3x²$$
$$(x²+6x+6+1x)(x²+6x+6-x)-3x²$$
$$(x²+6x+6)²(+1x)(-x)-3x²$$
$$(x²+6x+6)²-x²-3x²$$
$$(x²+6x+6)²-4x²$$
$$(x²+6x+6)²-(2x)²$$
Apply $a²-b²=(a+b)(a-b)$
$$(x²+6x+6+2x)(x²+6x+6-2x)$$
$$(x²+8x+6)(x²+6x-2x+6)$$
$$(x²+8x+6)(x²+4x+6)$$
$$x(x+8+6/x)x(x+4+6/x)$$
Answer: $$x²(x+8+6/x)(x+4+6/x)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/463506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 4,
"answer_id": 2
} |
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