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How to change the value of $a$ in a Taylor series yet still get the correct approximation? Set up
For example, if $f(x) = x^3 + 4x + 6$, then the second degree Taylor polynomial
is
$$
P_2(a) = (a^3 + 4a + 6) + (3a + 4)( x - a) + \left(\frac{6a}{2!}(x - a)^2\right)
$$
If I want to approximate $f(1.2)$ then I can use $P_2 (1.2)$.
You can see that this is close by graphing the functions, and noting that about
$f(1.2)$ they're very close.
Question
From other questions I get the idea that I'm able to change the value of $a$ ,
by setting it to zero or whatever, and still approximate the function as needed.
But looking at this example, I don't know how I could change the value of $a$ ,
and still approximate the original function at $x = 1.2$.
Sometimes people say that you're able to change the value which the function is being evaluated about, but I don't understand how this is so.
definitions
Taylors formula is $f(x) = P_n(x) + R_n(x)$ where $P_n(x)$ is
\begin{equation}
\begin{aligned}
P_n(x) = f(a)
+ \frac{f'(a)}{1!}(x - a)
+ \frac{f''(a)}{2!}(x - a)^2
& + \ldots
+ \frac{f^{(n)}(a)}{n!}(x - a)^n \\
\end{aligned}
\end{equation}
And $R_n (x) $ is (where $\xi$ is between $a$ and $x$)
\begin{equation}
\begin{aligned}
R_n(x) = \frac{f^{(n + 1)}(\xi)}{(n + 1)!}(x - a)^{(n + 1)}
\end{aligned}
\end{equation}
| I think you've gotten a little confused between your $x$ and your $a$. The Taylor series is a function of $x$, parameterised by $a$. So, for example, the 2nd-order Taylor polynomial of your cubic is:
$$T_{2;a}(x) = (a^3 + 4a + 6) + (x - a)(3a^2 + 4) + 6a\frac{(x - a)^2}{2}$$
If we evaluate this series at $a = 0$, we get the (linear) polynomial:
$$\begin{eqnarray}T_{2;0}(x) & = & (0^3 + 4\cdot0 + 6) + (x - 0)(3\cdot0^2 + 4) + 6\cdot 0 \frac{(x - a)^2}{2} \\
& = & 6 + 4x\end{eqnarray}$$
Using this function, the approximation for $f(1.2)$ is $T_{2;0}(1.2) = 6 + 4\cdot 1.2 = 10.8$.
If we evaluate the series at $a = 1$, we get the polynomial:
$$\begin{eqnarray}T_{2;1}(x) & = & (1^3 + 4 \cdot 1 + 6) + (x - 1)(3 \cdot 1^2 + 4) + 6 \cdot 1 \frac{(x - 1)^2}{2} \\
& = & 11 + 7(x - 1) + 3(x - 1)^2 \\
& = & 7 + x + 3x^2\end{eqnarray}$$
Using this function, the approximation for $f(1.2)$ is $T_{2;1}(1.2) = 11 + 7(0.2) + 3(0.2^2) = 12.52$. Note that this is a lot closer to the true value of $f(1.2) = 12.528$.
If we take $a = 1.2$, then we'll get a Taylor polynomial that is exactly right for $f(1.2)$, but whose approximation gets worse further away from it.
| {
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"url": "https://math.stackexchange.com/questions/2369458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
if $x+y+z=1$ then find the maximum of the form
Let $x>0$, $y>0$ and $z>0$ such that $x+y+z=1$. Find the maximal value of
$$x\sqrt{y}+y\sqrt{z}$$
I think $x=y=\dfrac{4}{9},z=\dfrac{1}{9}$ then the maximum $\dfrac{4}{9}$,but how to use AM-GM prove it?
| Without AM-GM, the Lagrange multiplier method will do:
$$L=x\sqrt{y}+y\sqrt{z}+t(1-x-y-z).$$
$$\begin{cases} L_x=\sqrt{y}-t=0 \\ L_y=\frac{x}{2\sqrt{y}}+\sqrt{z}-t=0 \\ L_z=\frac{y}{2\sqrt{z}}-t=0 \\ L_t=1-(x+y+z)=0 \end{cases} \Rightarrow x=y=\frac{4}{9}, z=\frac{1}{9}.$$
Note: To solve the system of equations, square $(1)$ divide it by $(3)$ to find $t=2\sqrt{z}$. Then substitute this into $(1)$ to find $y=4z$. Then substitute these into $(2)$ to find $x=4z$. And finally substitute all into $(4)$ to find $z=\frac{1}{9}$, hence the rest.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2374389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
$x^2 + y^2 = 10 , \,\, \frac{1}{x} + \frac{1}{y} = \frac{4}{3}$ I couldn't solve the above equation. Below, I describe my attempt at solving it.
$$x^2 + y^2 = 10 \tag{1}$$
$$\frac{1}{x} + \frac{1}{y} = \frac{4}{3} \tag{2}$$
Make the denominator common in the RHS of $(2)$.
$$\frac{y + x}{xy} = \frac{4}{3} \tag{2.1}$$
Multiply $(2.1)$ through by $3$:
$$\frac{3(y + x)}{xy} = 4 \tag{2.2}$$
Let $y = mx$
Substitute $y = mx$ into $(1)$ and $(2.2)$:
$$x^2 + m^2x^2 = 10 \tag{1.1}$$
$$\frac{3(mx + x)}{mx^2} = 4$$
Factorise:
$$\frac{3x(m + 1)}{x(mx)} = 4$$
$$\frac{3(m + 1)}{(mx)} = 4$$
Cross multiply:
$$3(m+1) = 4mx$$
Collect like terms:
$$4mx - 3m = 3$$
Factorise:
$$m(4x - 3) = 3$$
Divide through by $(4x - 3)$:
$$m = \frac{3}{4x - 3} \tag{3}$$
Substitute $(3)$ into $(1.1)$
$$x^2 + \left(\frac{3}{4x - 3}\right)^2x^2 = 10$$
$$x^2 + \frac{9x^2}{16x^2 - 24x + 9} = 10$$
Multiply through by ${16x^2 - 24x + 9}$:
$$16x^2 - 24x^3 + 9x^2 + 9x^2 = 10(16x^2 - 24x + 9)$$
Divide through by $2$:
$$8x^4 - 12x^3 + 9x^2 = 80x^2 - 120x + 45$$
$$8x^4 - 12x^3 - 71x^2 +120x - 45 = 0$$.
$x = 0$ is not a solution of the above equation, so divide through by $x^2$:
$$8x^2 - 12x - 71 + \frac{120}{x} - \frac{45}{x^2} = 0 \tag{1.2}$$
Let $v = x - \frac{k}{x}$.
$$v^2 = x^2 - 2k + \frac{k^2}{x^2}$$
Rewriting $(1.2)$:
$$\left(8x^2 - 71 - \frac{45}{x^2}\right) + \left(-12x + \frac{120}{x}\right)$$
$$\left(8x^2 - 71 - \frac{45}{x^2}\right) + \left(-12(x - \frac{10}{x}\right) \tag{1.3}$$
Putting $k = 10$, works for $v$, but not for $v^2$.
I don't know any other approach to solving the equation from $(1.3)$.
| Let $s=x+y$ and let $p=xy$. Then you know that $10=x^2+y^2=(x+y)^2-2xy=s^2-2p$ and that $\frac43=\frac1x+\frac1y=\frac sp$. So let us solve the system$$\left\{\begin{array}{l}s^2-2p=10\\p=\frac34s.\end{array}\right.$$Replacing $p$ by $\frac34s$ in the first equation gives $s^2-\frac32s=10$. This equation has two solutions: $s=4$ (for which $p=3$) and $s=-\frac52$ (for which $p=-\frac{15}{8}$). Now, all that's left is to solve the systems$$\left\{\begin{array}{l}x+y=4\\xy=3\end{array}\right.\text{ and }\left\{\begin{array}{l}x+y=-\frac52\\xy=-\frac{15}8.\end{array}\right.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2374762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Calculate root interval of $x^5+x^4+x^3+x^2+1$ So I have to find an interval (in the real numbers) such that it contains all roots of the following function:
$$f(x)=x^5+x^4+x^3+x^2+1$$
I've tried to work with the derivatives of the function but it doesn't give any information about the interval, only how many possible roots the function might have.
| Let $f(x)=x^5+x^4+x^3+x^2+1$.
Hence, $f'(x)=x(5x^3+4x^2+3x+2)$ and since $(5x^3+4x^2+3x+2)'=15x^2+8x+3>0$,
we see that the polynomial $5x^3+4x^2+3x+2$ has one real root $x_1$ and this root is negative.
Thus, $x_{min}=0$ and $x_{max}=x_1$.
But $f(0)>0$, which say that $f$ has an unique real root and since
$$f(-1.25)f(-1.24)<0,$$
we get that this root placed in $(-1.25,-1.24)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2375023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 2
} |
Show that the following matrix has $4$ as an eigen value of multiplicity $3$ Show that the following matrix has $4$ as an eigen value of multiplicity $3$
$\begin{bmatrix} 5&1&1&1&1&1\\1&5&1&1&1&1\\1&1&5&1&1&1\\1&1&1&4&0&1\\1&1&1&0&3&0\\1&1&1&1&0&4\end{bmatrix}$.
How should I try this?Isn't this equivalent to finding the entire spectrum of the above matrix?
Or there are some tricks involved here?Please help.
| Hint:$A=\begin{bmatrix}
4 & 0 & 0 &0&0&0\\
0&4&0&0&0&0\\
0&0&4&0&0&0\\
0&0&0&3&0&0\\
0&0&0&0&2&0\\
0&0&0&0&0&3
\end{bmatrix}+\begin{bmatrix}
1 & 1 & 1 &1&1&1\\
1 & 1 & 1 &1&1&1\\
1 & 1 & 1 &1&1&1\\
1 & 1 & 1 &1&1&1\\
1 & 1 & 1 &1&1&1\\
1 & 1 & 1 &1&1&1
\end{bmatrix}$
$$\text{Both matrices commute, eigenvalues for the first are} ~~4,4,4,3,2,3$$
$$\text{eigenvalues for thesecond matrix are}~~0,0,0,0,0,6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2375649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Convert a equation with fractions into whole numbers So I have this equation:
$$\frac{2}{3}a^2-\frac{4}{9}a^2 = 8a$$
So this is a really easy problem, I could just multiply
$$\frac{2}{3}*\frac{3}{3} = \frac{6}{9}$$
Then subtract
$$\frac{6}{9}a^2 - \frac{4}{9}a^2 = \frac{2}{9}a^2=8a$$
$$36a=a^2$$
$$36=a$$
However, I want to solve the equation by getting rid of the fractions right at the beginning:
$$\frac{2}{3}a^2-\frac{4}{9}a^2 = 8a$$
So I thought it'd be much simpler if I could get rid of these fractions by multiplying everything by a single value. Therefore, I thought what value can I multiply 2 and 4 so it gives me a divisible value by 3 and 9? It took me some time but I came up with 9
$$\frac{2}{3}a^2-\frac{4}{9}a^2 = 8a$$
$$9(\frac{2}{3}a^2-\frac{4}{9}a^2) = 9(8a)$$
$$6a^2-9a^2 = 8a$$
My question is if there is an easier way to find this value that when multiplied it eliminates the fractions? It took a few valuable seconds to figure out it was 9, so I was wondering if this process has a name or any rules to find it quicker?
| You're simply finding a common denominator to work with. If you have an expression $\frac{n_1}{d_1}+\frac{n_2}{d_2}+...+\frac{n_k}{d_k}$, then you want the denominator to be $\text{lcm}(d_1,d_2,...,d_k)$.
This now gives for your expression
$$\frac{\frac{n_1}{d_1}\text{lcm}(d_1,d_2,...,d_k)+\frac{n_2}{d_2}\text{lcm}(d_1,d_2,...,d_k)+...+\frac{n_k}{d_k}\text{lcm}(d_1,d_2,...,d_k)}{\text{lcm}(d_1,d_2,...,d_k)}$$
The terms in the numerator are all integers which is what you're after.
In your case, the lcm is $9$, so you have
$$\frac{2}{3}a^2-\frac{4}{9}a^2 = 8a$$
$\implies$
$$\frac{1}{9}\left(9\times\frac{2}{3}a^2-9\times\frac{4}{9}a^2\right) = 8a$$
$\implies$
$$\left(6a^2-4a^2\right) = 9\times8a$$
(Also, as pointed out, you have a typo on the bottom line)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2376720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
$\int_{0}^{\frac{\pi}{2}}\frac{d\theta}{a + \sin^2(\theta)} = \frac{\pi}{2\sqrt{a(a+1)}}$, for $a > 0$ I want to show that
$$\int_{0}^{\frac{\pi}{2}}\frac{d\theta}{a + \sin^2(\theta)} = \frac{\pi}{2\sqrt{a(a+1)}}$$
for $a > 0$.
I try several methods
*
*Substitutions to rationalize
*The famous $U = \tan(\frac{\theta}{2})$
*Multiplying by conjugates
and other calculus techniques and still, I can not prove the equality.
Anyone can give me a hint of how should I begin to work this integral?
I'm very ashamed that I can not solve this problem.
|
\begin{align} \int_{0}^{\frac{\pi}{2}}\frac{d\theta}{a +
> \sin^2(\theta)} &= \frac{\pi}{2[a(a+1)]^\frac{1}{2}} \end{align}
\begin{align}
I=
\int_{0}^{\frac{\pi}{2}}\frac{d\theta}{a + \sin^2(\theta)}
&=
\int_{0}^{\frac{\pi}{2}}\frac{d\theta}{(\sqrt{a+1} + \cos\theta)
(\sqrt{a+1} - \cos\theta}
\\
&=\frac1{2\sqrt{a+1}}
\left(
\int_0^{\tfrac\pi2}
\frac{d\theta}{\sqrt{a+1} + \cos\theta}
+
\int_0^{\tfrac\pi2}
\frac{d\theta}{\sqrt{a+1} - \cos\theta}
\right)
.
\end{align}
Now we can use a table integral
\begin{align}
\int \frac{dx}{u+v\cos x}
&=
\frac2{\sqrt{u^2-v^2}}
\arctan\left( \frac{(u-v)\tan\tfrac{x}2}{\sqrt{u^2-v^2}} \right)
,\quad u^2>v^2
\end{align}
for $u=\sqrt{a+1}$, $v=\pm 1$.
\begin{align}
I&=
\frac1{2\,u}
\frac2{\sqrt{u^2-1}}
\left(
\arctan\left(
\sqrt{\frac{u-1}{u+1}}\tan\tfrac\pi4
\right)
+
\arctan\left(
\sqrt{\frac{u+1}{u-1}}\tan\tfrac\pi4
\right)
\right)
,\\
&=
\frac1{u\,\sqrt{u^2-1}}
\left(
\arctan\left(
\sqrt{\frac{u-1}{u+1}}
\right)
+
\arctan\left(
\sqrt{\frac{u+1}{u-1}}
\right)
\right)
.
\end{align}
And since $\arctan x+\arctan \tfrac1x=\tfrac\pi2$ for $x>0$, we have
\begin{align}
I&=
\frac{\pi}{2\,\sqrt{a+1}\,\sqrt{a}}
.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
} |
Find the maximum positive integer that divides $n^7+n^6-n^5-n^4$
Find the maximum positive integer that divides all the numbers of the form $$n^7+n^6-n^5-n^4 \ \ \ \mbox{with} \ n\in\mathbb{N}-\left\{0\right\}.$$
My attempt
I can factor the polynomial
$n^7+n^6-n^5-n^4=n^4(n-1)(n+1)^2\ \ \ \forall n\in\mathbb{N}.$
If $n$ is even then exists $\,k\in\mathbb{N}-\left\{0\right\}\,$ such that $n=2k.\;$ so:
$n^4(n-1)(n+1)^2=2^4 k^4(2k-1)(1+2k)^2$
If $n$ is odd then exists $k\in\mathbb{N}$ such that $n=2k+1$ so
$$n^4(n-1)(n+1)^3=2^3(2k+1)^4(k+1)^2$$
Can I conclude that the maximum positive integer that divides all these numbers is $N=2^3?$ (Please, help me to improve my english too, thanks!)
Note: I correct my "solution" after a correction... I made a mistake :\
| we can use some common sense.
If $n$ is even $16|n^4$. But if $n$ is odd then $2\not \mid n^4$. However $n+1$ and $n-1$ are both even so $2^3|(n-1)(n+1)^2$. But if $n$ is odd then either $n + 1$ or $n -1$ is divisible by $4$ so $2^4|(n-1)(n+1)^2$. So $2^4$ will divide all such numbers.
Can any higher powers of $2$ divide it? Why should they? Example: if $n=2$ then then number is $16*1*3^2$ which is not divisible by any higher power.
But are there any other factors that must divide it? Either $n$ or $n-1$ or $n+1$ is divisible by $3$ so $3|n^4(n-1)(n+1)^2$. But if $3|n-1$ then $1$ is the only guaranteed power of $3$ that divides. (if $n-1 = 3$ then $9 \not \mid n^4(n-1)(n+1)^2$.
So $3*2^4$ will always be a factor.
Are there any other prime factors that must occur? Well, if $p$ is a prime $p > 3$ and if we let $n-1 = p+1$ we have $n^4(n-1)(n+1)^2 = (p+2)^2(p+1)*(p+3)^2$ and $p$ is not a factor of that.
So $48 = 3*2^4$ the largest integer that must divide all $n^4(n-1)(n+1)^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2377112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
Maximum value of $a^3 + b^3 + c^3$ If $a+b+c=5$ and $ \frac{1}{a} +\frac{1}{b} +\frac{1}{c} =\frac{1}{5}$ then find the maximum value of $a^3 + b^3 +c^3$ where $a,b,c $ is real numbers.
My Attempt
writing $a+b=5-c$ and $ \frac{1}{a} +\frac{1}{b} =\frac{1}{5} -\frac{1}{c}$ and after algebraic manipulation I obtain $(a+b)(\frac{1}{ab} + \frac{1}{5c})=0$. If $(a+b)=0$ we get $c=5$ and $a^3 + b^3 +c^3 =125$. But I can't analyze the case $(\frac{1}{ab} + \frac{1}{5c})=0$. Can anyone help me from this point. Thanks in advance.
| Let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative, and $abc=w^3$.
Hence, our conditions give $u=\frac{5}{3}$ and $w^3=15v^2$ and since
$$0\leq(a-b)^2(a-c)^2(b-c)^2=27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6),$$
we obtain:
$$v^2(3v^2+25)^2\leq0$$ or $v^2\leq0$.
Thus, $$a^3+b^3+c^3=27u^3-27uv^2+3w^3=125+30v^2\leq125.$$
The equality does not occur because for this we need $abc=0$, which is impossible,
but $125$ is a supremum, of course.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2377466",
"timestamp": "2023-03-29T00:00:00",
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How can one prove $-\arctan\left(\frac{x}{y}\right)+\pi H(y)=2\arctan\left(\frac{y}{\sqrt{x^2+y^2}+x}\right)+\frac{\pi}{2}$? How can one prove the following?
$$
-\arctan\left(\frac{x}{y}\right)+\pi H(y)=2\arctan\left(\frac{y}{\sqrt{x^2+y^2}+x}\right)+\frac{\pi}{2}$$
with $H(x)$ the Heaviside function.
I have the impression that this might follow from the following
$$H(y)=H(-x)+2H(x)H(y)-H(xy)$$
and
$$\arctan\left(\frac{x}{y}\right)+\arctan\left(\frac{y}{x}\right)=\pi H(xy)-\frac{\pi}{2}$$
but I can't quite get it right...
| For $y>0$ you'd like to prove that
$$
\frac{\pi}{2}-\arctan\frac{x}{y}=2\arctan\frac{y}{\sqrt{x^2+y^2}+x}
$$
Note that the left-hand side is a number in the interval $[0,\pi]$ and the same for the right-hand side. The identity is true for $x=0$.
Suppose $x>0$, so both sides are in $(0,\pi/2)$. Then take the tangent of both sides; if $z=y/(\sqrt{x^2+y^2}+x)$, the identity to prove is
$$
\frac{y}{x}=\frac{2z}{1-z^2}
$$
For simplicity, set $D=\sqrt{x^2+y^2}$; then
$$
1-z^2=1-\frac{y^2}{2x^2+y^2+2xD}=\frac{2x^2+2xD}{2x^2+y^2+2xD}=\frac{2x(x+D)}{(D+x)^2}=
\frac{2x}{D+x}
$$
Hence
$$
\frac{2z}{1-z^2}=\frac{2y}{D+x}\frac{D+x}{2x}=\frac{y}{x}
$$
Suppose now $x<0$ and set $x=-t$; then the identity to prove is
$$
\frac{\pi}{2}+\arctan\frac{t}{y}=2\arctan\frac{y}{\sqrt{t^2+y^2}-t}=
2\arctan\frac{\sqrt{t^2+y^2}+t}{y}
$$
Recall that $\arctan\frac{1}{u}=\pi/2-\arctan u$ (for $u>0$), so the identity to prove is
$$
\frac{\pi}{2}+\arctan\frac{t}{y}=\pi-\arctan\frac{y}{\sqrt{t^2+y^2}+t}
$$
that's already been proved.
For $y<0$ you want to prove that
$$
-\frac{\pi}{2}-\arctan\frac{x}{y}=2\arctan\frac{y}{\sqrt{x^2+y^2}+x}
$$
Changing $y=-z$, the identity becomes
$$
-\frac{\pi}{2}+\arctan\frac{x}{z}=-2\arctan\frac{z}{\sqrt{x^2+z^2}+x}
$$
that has already been proved.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Why does rotation preserve angles and distance in the Euclidean plane? An angle in radians is a distance moved around a unit circle. A rotation of a point around a center of rotation, moves the point a distance around a circle around the center that goes through the point. The distance is given by the angle of the rotation multiplied by the distance from center.
So why do rotations preserve distance and angles when the distance measure is Pythagorean.
| A linear algebra approach:
Any linear transformation of $\mathbb{R}^n$, such as a rotation about the origin, can be represented with an $n \times n$ matrix $A$, where $A\mathbf{v}$ is the output when the transformation acts on a vector $\mathbf{v}$. Any counterclockwise rotation of $\mathbb{R}^2$ about the origin looks like $\left[ \begin{array}{ccc}
\cos(\theta) & -\sin(\theta) \\
\sin(\theta) & \cos(\theta) \end{array} \right]$ where $\theta$ is the angle of rotation. See my post here for a justification of this fact.
First, let us investigate $\|\mathbf{v}\|$ and $\|A \mathbf{v}\|$. If $\mathbf{v} = \left[ \begin{array}{ccc}
a \\
b \end{array} \right]$, then $\|\mathbf{v}\| = \sqrt{a^2 + b^2}$. You can check that $A \mathbf{v} = \left[ \begin{array}{ccc}
\cos(\theta)a - \sin(\theta)b \\
\sin(\theta)a + \cos(\theta)b \end{array} \right]$. We have:
$$\|A \mathbf{v}\| = \sqrt{\Big( \cos(\theta)a - \sin(\theta)b \Big)^2 + \Big( \sin(\theta)a + \cos(\theta)b \Big)^2}$$
$$ \quad \quad \quad \ = \sqrt{\Big( \sin^2(\theta) + \cos^2(\theta) \Big)a^2 + \Big(\sin^2(\theta) + \cos^2(\theta) \Big)b^2}$$
And this reduces to $\sqrt{a^2 + b^2} = \|\mathbf{v}\|$ due to the Pythagorean identity: $\sin^2(\theta) + \cos^2(\theta) = 1$. Thus, the length of any vector is preserved under any rotation.
To see why rotations preserve angles, use the fact that the cosine of the angle between unit vectors $\mathbf{v}$ and $\mathbf{w}$ is equal to their dot product $\mathbf{v} \cdot \mathbf{w}$. So let $\mathbf{v} = \left[ \begin{array}{ccc}
a \\
b \end{array} \right]$ and $\mathbf{w} = \left[ \begin{array}{ccc}
c \\
d \end{array} \right]$ be generic unit vectors; you will find that $\mathbf{v} \cdot \mathbf{w}$ and $A\mathbf{v} \cdot A \mathbf{w}$ are equal (following reduction of the latter using trig identities).
| {
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"source": "stackexchange",
"question_score": "4",
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Examining the convergence with parameter $a$ For $a \in R$, let $x_1=a$ and $x_{n+1}=\frac{1}{4}(x_{n}^2+3)$ for all $n≥2$. Examine the convergence of the sequence ${x_n}$ for different values of $a$. Also, find $\lim x_n$, whenever it exists.
I am having problems on how to take $a$ as a parameter. I am unable to think which all values of $a$ would give different results.
Hints are appreciated. Thanks.
| *
*Let $a>3$.
Hence, $$x_{n+1}-3=\frac{(x_n-3)(x_n+3)}{4}>0$$ and by induction we obtain here $x_n>3$ for all $n\in\mathbb N$.
In another hand, $$x_{n+1}-3=\frac{(x_{n-1}-3)(x_{n-1}+3)}{4}>$$
$$>\frac{3}{2}(x_{n-1}-3)>\left(\frac{3}{2}\right)^2(x_{n-2}-3)>...>\left(\frac{3}{2}\right)^{n-1}(x_1-3),$$
which says that for $a>3$ our sequence does not converge.
*$a=3$.
In this case $x_n=3$ and our sequence converges.
*$1<a<3$.
$$x_{n+1}-1=\frac{(x_{n}-1)(x_{n}+1)}{4},$$ which says that $1<x_n<3$ for all $n$ by induction and
$$x_{n+1}-x_n=\frac{(x_n-3)(x_n-1)}{4}<0,$$ which says that $\{x_n\}$ decreasing.
Thus, there is $\lim\limits_{n\rightarrow+\infty}x_n$ and let this limit equal to $x$.
Thus, $1\leq x<3$ and $x=\frac{1}{4}(x^2+3),$ which gives $x=1$.
*$a=1$.
In this case $x_n=1$ and our sequence converges.
*$-3<a<1$.
Here $|x_n+1|\leq|a-1|<4$ for all $n$ and $$|x_{n}-1|=|\frac{(x_{n-1}-1)(x_{n-1}+1)}{4}|\leq|x_{n-1}-1|\cdot|\frac{a-1}{4}|\leq$$
$$\leq|x_{n-2}-1|\cdot|\frac{a-1}{4}|^2\leq|a-1|\cdot|\frac{a-1}{4}|^{n-1},$$
which says that $\lim\limits_{n\rightarrow+\infty}x_n=1$ by the limit definition.
*$a=-3$.
In this case $x_2=3$ and $x_n=3$ for all $n\geq2$, which says that $x$ converges.
*$a<-3$.
Since $x_{n+1}-3=\frac{(x_n-3)(x_n+3)}{4}>0$ by induction and $x_{n+1}-x_n=\frac{(x_n-3)(x_n-1)}{4}>0$,
we see that our sequence has no a limit.
Done!
| {
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Two positive integers $a$ and $b$ are such that $a+b=\frac{a}{b} +\frac{b}{a}$ . What is the value of $a^2 + b^2$? Two positive integers $a$ and $b$ are such that $a+b=\frac{a}{b} +\frac{b}{a}$ . What is the value of $a^2 + b^2$?
Got this in a sample question paper for a contest but i have no idea how to solve it. I tried to apply inequalities but it did it not work. Please help
| Interesting problem and elegant solution of @quasi.
Alternatively: Without loss of generality, assume $1\le a\le b$. Then:
$$a+b=\frac{a}{b}+\frac{b}{a}\le \frac{b}{b}+\frac{b}{1}=1+b\Rightarrow a\le 1 \Rightarrow a=1.$$
| {
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"source": "stackexchange",
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Matrix to power 30 using eigen values issue
$$ \text{If matrix } A = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \text{ then find } A^{30}.$$
I tried to approach through diagonalization using eigen values method.
I got eigen values as $-1, 1, 1$
As per diagonalization $ A = P*D*P^{-1}.$ So $ A^{30} = P*D^{30}*P^{-1}. $
But $ D^30 = I.$
So, $ A^{30} = P*I*P^{-1} = I $
But $ A^{30} $ is not equal to I. If we do general multiplication without all these.
Where is the mistake in my approach?
| You have computed the eigenvalues of $A$ to be $\{-1, 1, 1\}$. The repeated eigenvalue may be an obstacle to diagonalization. In this case, the geometric and algebraic multiplicities of the eigenvalue $1$ are different, so A is not diagonalizable. You continued as if $A$ were diagonalizable, so this is the mistake in your approach.
You have added a follow-on question in comments (instead of to your question, so you should not be surprised if other answers do not address it). I see that others have demonstrated Jordan normal form and induction. Another method is binary decomposition of the exponent and repeated squaring to get power-of-$2$ powers of $A$: \begin{align*}
A^1 &= \begin{pmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \text{,} \\
A^2 &= A^1 \cdot A^1 = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix} \text{,} \\
A^4 &= A^2 \cdot A^2 = \begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 2 & 0 & 1 \end{pmatrix} \text{,} \\
A^8 &= A^4 \cdot A^4 = \begin{pmatrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 4 & 0 & 1 \end{pmatrix} \text{,} \\
A^{16} &= A^8 \cdot A^8 = \begin{pmatrix} 1 & 0 & 0 \\ 8 & 1 & 0 \\ 8 & 0 & 1 \end{pmatrix} \text{,} \\
A^{30} = A^{11110_{\,2}} &= A^{16}\cdot A^8 \cdot A^4 \cdot A^2 = \begin{pmatrix} 1 & 0 & 0 \\ 15 & 1 & 0 \\ 15 & 0 & 1 \end{pmatrix} \text{.}
\end{align*}
Of course, after computing $A^4$ or $A^8$, perhaps one would notice the pattern...
There is a slightly faster way to get there using the above method. $A^{-1}$ is easy enough to compute (using the minors method, for example). $$
A^{-1} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ -1 & 1 & 0 \end{pmatrix}
$$ Then, $A^{30} = (A^{15})^2 = (A^{16} \cdot A^{-1})^2$. This replaces three multiplies with a multiply and an inverse. Continuing to think about this leads to the computationally intractable problem of optimal addition chains.
| {
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"url": "https://math.stackexchange.com/questions/2384523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Please prove $\frac{1 + \sin\theta - \cos\theta}{1 + \sin\theta + \cos\theta} = \tan \left(\frac \theta 2\right)$
Prove that $\dfrac{1 + \sin\theta - \cos\theta}{1 + \sin\theta + \cos\theta} = \tan\left(\dfrac{\theta}{2}\right)$
Also it is a question of S.L. Loney's Plane Trignonometry
What I've tried by now:
\begin{align}
& =\frac{1+\sin\theta-\sin(90-\theta)}{1+\cos(90-\theta)+\cos\theta} \\[10pt]
& =\frac{1+2\cos45^\circ \sin(\theta-45^\circ)}{1+2\cos45^\circ \cos(45-\theta)} \end{align}
Cause I do know
\begin{align} & \sin c + \sin d = 2\sin\left(\frac{c+d}{2}\right)\cos\left(\frac{c-d}{2}\right) \\[10pt] \text{and } & \cos c + \cos d = 2\cos\left(\frac{c+d}{2}\right)\sin\left(\frac{c-d}{2}\right)
\end{align}
I can't think of what to do next..
| HINT: use the tan-half angle substution
$$\sin(x)=2\,{\frac {\tan \left( x/2 \right) }{1+ \left( \tan \left( x/2
\right) \right) ^{2}}}$$
$$\cos(x)={\frac {1- \left( \tan \left( x/2 \right) \right) ^{2}}{1+ \left(
\tan \left( x/2 \right) \right) ^{2}}}
$$
with $t=\tan(x/2)$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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An inequality with $\cos$ and triangle sides Here is the problem:
Let $ABC$ be a triangle with sides $a, b, c$. Show that $\dfrac{\cos A}{a^3}+\dfrac{\cos B}{b^3}+\dfrac{\cos C}{c^3}\geq\dfrac{3}{2abc}.$
Here's my attempt:
By the cosine formula, we have $\cos A = \dfrac{b^2+c^2-a^2}{2bc}$ etc, which the left hand side can be transformed into:
\begin{equation*}
\dfrac{a^2+b^2-c^2}{2abc^3}+\dfrac{a^2+c^2-b^2}{2ab^3c}+\dfrac{b^2+c^2-a^2}{2a^3bc}
\end{equation*}
And then I'm stuck. Can someone help me? Thanks.
| You're almost done. If you factor out $\frac{1}{2abc}$ from the expression you obtained, you get
$$\frac{1}{2abc}\left(\frac{a^2+b^2-c^2}{c^2}+\frac{a^2+c^2-b^2}{b^2} + \frac{b^2+c^2-a^2}{a^2}\right).$$
So you just need to prove $\frac{a^2+b^2-c^2}{c^2}+\frac{a^2+c^2-b^2}{b^2} + \frac{b^2+c^2-a^2}{a^2}\ge 3$. Writing
\begin{align}&\frac{a^2+b^2-c^2}{c^2}+\frac{a^2+c^2-b^2}{b^2} + \frac{b^2+c^2-a^2}{a^2} \\
&= \left(\frac{a^2}{b^2}+\frac{b^2}{a^2}\right) + \left(\frac{a^2}{c^2}+\frac{c^2}{a^2}\right) + \left(\frac{b^2}{c^2}+\frac{c^2}{b^2}\right) - 3\end{align}
it is enough to show that each of the terms in the parentheses is at least $2$. Why is that true?
| {
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Inequality - Maximize k such that I'm not sure how to solve this problem I came across online. Here it is:
Let $a, b, c$ be nonnegative reals. Maximize $k$ such that
$$(a+b+c)\left(\dfrac 1a + \dfrac 1b + \dfrac 1c\right) + \dfrac{k(ab + bc+ca)}{a^2+b^2+c^2} \geq 9 + k.$$
What I know is this:
$a, b, c$ must be positive, because otherwise we'd have division by zero. I also know that $(a+b+c)\left(\dfrac 1a + \dfrac 1b + \dfrac 1c\right) \geq 9$ for positive $a,b,c$.
| Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, our inequality it's $f(w^3)\geq0$, where $f$ is a decreasing function.
Thus, it remains to prove our inequality for a maximal value of $w^3$.
We know that $a$, $b$ and $c$ are positive roots of the following equation
$$(x-a)(x-b)(x-c)=0$$ or
$$x^3-3ux^2+3v^2x-w^3=0$$ or
$$x^3-3ux^2+3v^2x=w^3,$$
which says that a graph of $f(x)=x^3-3ux^2+3v^2x$ and the line $y=w^3$
have three common points: $(a,f(a))$, $(b,f(b))$ and $(c,f(c))$.
Now, let $a\leq b\leq c$ and since $u^2\geq v^2$, we see that $$f'(x)=3x^2-6xu+3v^2=3(x^2-2ux+v^2)=3(x-x_1)(x-x_2).$$
Let $x_1\leq x_2$.
We see that $x_1>0$ and $\left(x_1,f(x_1)\right)$ is a maximum point and $\left(x_2,f(x_2)\right)$ is a minimum point.
Now, let $w^3$ increases and $u$ and $v^2$ are constants.
Thus, $a$, $b$ and $c$ are change and $w^3$ will get a maximal value,
when the line $y=w^3$ will touch to the graph of $f$,
which happens for equality case of two variables (for $a=b$ in our case).
Since our inequality is homogeneous, we can assume $b=c=1$, which gives
$$(a+2)\left(\frac{1}{a}+2\right)-9\geq k\left(1-\frac{2a+1}{a^2+2}\right)$$ or
$$\frac{2(a^2+2)}{a}\geq k.$$
Now, by AM-GM
$$\frac{2(a^2+2)}{a}=2\left(a+\frac{2}{a}\right)\geq2\cdot2\sqrt2=4\sqrt2,$$
which gives the answer $4\sqrt2$.
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Given $m$, find three numbers $a,b,c$ such that $c-b=m(b-a)$ and $a+b,a+c,b+c$ are all squares I've worked with the next problem
*
*Given an integer $m$, find three integers $a,b,c$, with $b\neq a$, such that $c-b=m(b-a)$ and
$$a+b=x^2, \ \ a+c=y^2, \ \ b+c=z^2,$$
where $x,y, z \in \mathbb{Z}$.
I've not been able to give a result that guarantees the existence of solutions of this problem for some $m$ values.
Can you help me, please?
| The condition requires $$y^2-x^2=m(z^2-y^2)$$ or $$x^2-(m+1)y^2=-mz^2.$$
You also require $x+y+z$ even, and then you get $$a=\frac{x^2+y^2-z^2}{2},b=\frac{x^2+z^2-y^2}{2},c=\frac{y^2+z^2-x^2}{2}.$$
But there is a general approach to finding all rational solutions to:
$$aX^2+bY^2=c$$
if you have one rational solution.
Here, we have $(a,b,c)=(1,-(m+1),-m)$. and base solution $(X_0,Y_0)=(1,1)$.
Then if $p,q$ are integers, not both zero, then there is another rational solution $(X,Y)=(X_0+pt,Y_0+qt)$ with $t\neq 0$.
Solving, we get $$(1+pt)^2-(m+1)(1+qt)^2=-m\\
2pt + p^2t^2 -2(m+1)qt-(m+1)q^2t^2=0\\
2t\left[p-q(m+1)\right] + t^2\left[p^2-(m+1)q^2\right]=0$$
Assuming $t\neq 0$, you have:
$$t=\frac{2q(m+1)-2p}{p^2-(m+1)q^2}$$
This obviously has problems if $p^2-(m+1)q^2=0$, which can happen when $m+1$ is a perfect square.
Also, if $q=1,p=m+1$ then the numerator is zero, so doesn't give a new solution, but that's okay.
Now $$1+pt = \frac{(m+1)q(2p-q)-p^2}{p^2-(m+1)q^2}\\
1+qt=\frac{(m+1)q^2+p(p-2q)}{p^2-(m+1)q^2}
$$
Then any time $(m+1)q+p$ is even, and $p^2-(m+1)q^2\neq 0$ you get the solution:
$$(x,y,z)=\left((m+1)q(2p-q)-p^2,(m+1)q^2+p(p-2q),p^2-(m+1)q^2\right)$$
Trying $(p,q)=(4,2)$ you get:
$$(x,y,z)=(6m+2,2m+2,6-2m)\\(a,b,c)=(18m^2+4m-14,18m^2-28m+18,-14m^2+4m+18)$$
The general formula gives (I think, still needs checking):
$$a=2\left[(q-2p)^2q^2(m+1)^2-(4q^2-6qp+4p^2)pq(m+1)+p^2(p-2q)^2\right]\\
b=2\left[(q-2p)^2q^2(m+1)^2-(4p^2-2pq+4q^2)pq(m+1)+(p^2+4pq-4q^2)p^2\right]\\
c=2\left[(q^2+4pq-4p^2)q^2(m+1)^2+(4p^2+6pq+4q^2)pq(m+1)+p^2(p+2q)^2\right]
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Identify if some special name is given to polynomials Is there some special name given to the two polynomials below:
$${n \choose 1}x - {n \choose 3}x^3 + \ldots \label{} \tag{1}$$ and $$1 - {n \choose 2}x^2 + {n \choose 4}x^4 - \ldots \label{} \tag{2}$$.
| HInt:try $$(1+ix)^n$$ wher $i^2=-1$
$$(1+ix)^n=\\\left(\begin{array}{c}n\\ 0\end{array}\right)1^{n}(ix)^{0}+\left(\begin{array}{c}n\\ 1\end{array}\right)1^{n-1}(ix)^{1}+
\left(\begin{array}{c}n\\ 2\end{array}\right)1^{n-2}(ix)^{2}+
\left(\begin{array}{c}n\\ 3\end{array}\right)1^{n-3}(ix)^{3}+
\left(\begin{array}{c}n\\ 4\end{array}\right)1^{n-4}(ix)^{4}+...=\\\left(\begin{array}{c}n\\ 0\end{array}\right)+\left(\begin{array}{c}n\\ 1\end{array}\right)(ix)+\left(\begin{array}{c}n\\ 20\end{array}\right)(ix)^{2}+\left(\begin{array}{c}n\\ 3\end{array}\right)(ix)^{3}+\left(\begin{array}{c}n\\ 4\end{array}\right)(ix)^{4}+...=\\(
\left(\begin{array}{c}n\\ 0\end{array}\right)-\left(\begin{array}{c}n\\ 2\end{array}\right)x^2+\left(\begin{array}{c}n\\ 4\end{array}\right)x^4-...)+\\i(
\left(\begin{array}{c}n\\ 1\end{array}\right)x-\left(\begin{array}{c}n\\ 3\end{array}\right)x^3+...)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2386396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Induction on Fermat Numbers: $F_n = \prod_{j=0}^{n-1}F_j+2$ Is the Following Proof Correct?
Theorem. Given that $\forall n\in\mathbf{N}(F_n = 2^{2^n}+1)$ show that the following is true
$$\forall n\in{1,2,3...}\left(F_n = \prod_{j=0}^{n-1}F_j+2\right)$$
Proof. We construct the proof by recourse to Mathematical-Induction.
Basis-Step: For $n=1$ we have $F_1=F_0+2=(2^{2^0}+1)+2 = (2+1)+2=5$.
Inductive-Step: Assume now for an arbitrary $k\in\{1,2,3,...\}$ the following
$$F_k = \left(\prod_{j=0}^{k-1}F_j+2\right)\tag{1}$$
consequently using $(1)$ we have
$$\left(\prod_{j=0}^{k}F_j+2\right) = \left(\prod_{j=0}^{k-1}F_j\cdot F_k+2\right)=(F_k-2)\cdot F_k=(F_k)^{2}-2F_k+2\tag{2}$$
thus $(F_k)^{2}-2F_k+2 = (2^{2^k}+1)^2-2(2^k+1)+2$ expanding all terms we have $(2^{2^k}+1)^2-2(2^k+1)+2 = 2^{2^{k+1}}+2\cdot 2^{2^k}+1-2\cdot 2^{2^k}-2+2 = 2^{2^{k+1}}+1 = F_{k+1}$
thus
$$F_{k+1}=\left(\prod_{j=0}^{k}F_j+2\right)\tag{3}$$
$\blacksquare$
| Looks good to me :)
NB - those are Fermat numbers, not Fermat primes.
| {
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Inequality : $\sqrt[3]{\frac{x^3+y^3+z^3}{xyz}} + \sqrt{\frac{xy+yz+zx}{x^2+y^2+z^2}} \geq 1+\sqrt[3]{3}$
Let $x>0$, $y>0$ and $z > 0$. Prove that $$\sqrt[3]{\frac{x^3+y^3+z^3}{xyz}} + \sqrt{\frac{xy+yz+zx}{x^2+y^2+z^2}} \geq
1+\sqrt[3]{3}.$$
From Micheal Rozenberg's answer :
$(x+2)\sqrt{x^2+2}\left(\sqrt{x^2+2}+\sqrt{2x+1}\right)\geq\sqrt[3]x\left(\sqrt[3]{(x^3+2)^2}+\sqrt[3]{3x(x^3+2)}+\sqrt[3]{9x^2}\right)$,
Prove that $(x+2)^2(x+2\sqrt{x}+3)\geq9\sqrt[3]{x(x^3+2)^2}$,
LHS :
$(x+2)\sqrt{x^2+2}\left(\sqrt{x^2+2}+\sqrt{2x+1}\right)\geq (x+2) \frac{x+2}
{\sqrt{3}}\left(\frac{x+2}{\sqrt{3}}+\frac{2\sqrt{x}+1}{\sqrt{3}}\right)= \frac{(x+2)^2}{3}(x+2\sqrt{x}+3)$
RHS :
$\sqrt[3]{3x(x^3+2)}\leq \sqrt[3]{(x^3+2)^2}$
$\sqrt[3]{9x^2}\leq \sqrt[3]{(x^3+2)^2}$
so $\sqrt[3]{(x^3+2)^2}+\sqrt[3]{3x(x^3+2)}+\sqrt[3]{9x^2}\leq 3\sqrt[3]{(x^3+2)^2}$
$\sqrt[3]{x}(\sqrt[3]{(x^3+2)^2}+\sqrt[3]{3x(x^3+2)}+\sqrt[3]{9x^2})\leq 3\sqrt[3]{x(x^3+2)^2}$
Thus,
$\frac{(x+2)^2}{3}(x+2\sqrt{x}+3) \geq 3\sqrt[3]{x(x^3+2)^2}$
$(x+2)^2(x+2\sqrt{x}+3) \geq 9\sqrt[3]{x(x^3+2)^2}$
| Here is my, as Michael Rozenberg rightly dubs it, ugly proof of the last step in Sergic Primazon's proof that is $g(t)$ decreases for $t\in(0,1]$.
$$g'(t) = -3^{-\frac23}\Big(\frac1{t^2}+\frac1t-1\Big)^{-\frac23}\Big(\frac2{t^3}+\frac1{t^2}\Big)+\frac12\Big(\frac1t\Big)^{\frac12}\ .$$
We want to show $g'(t)<0$ or
$$3^{\frac23}\Big(\frac1t\Big)^{\frac12}\Big(\frac1{t^2}+\frac1t-1\Big)^{\frac23} < 2\Big(\frac2{t^3}+\frac1{t^2}\Big). $$
Let $s:=\frac1t-1\in[0,\infty)$. The above inequality is equivalent to it being raised to the power of $6$, because it is easy to see its both sides are positive after making the substitution of $t$ by $s$. The transformed inequality is
$$2^6(s+1)^9(2s+3)^6-3^4((1+s)^2+s)^4>0 \tag1$$
The left hand side of the above inequality is
$$4096 s^{15} + 73728 s^{14} + 617472 s^{13} + 3191808 s^{12} + 11388672 s^{11} + 29712384 s^{10} + 58555200 s^9 + 88764399 s^8 + 104358708 s^7 + 95156902 s^6 + 66743280 s^5 + 35363277 s^4 + 13701744 s^3 + 3665574 s^2 + 605556 s + 46575 $$
All the coefficients of the polynomial are positive, so Inequality (1) holds for $s\ge 0$ implying $g'(t)<0$ for $t\in(0,1]$.
| {
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Let $a_{1}>0,a_{2}>0$ and $a_{n}=\frac{2a_{n-1}a_{n-2}}{a_{n-1}+a_{n-2}}, n>2$, then $\{ a_{n}\}$ converges to $\frac{3a_{1}a_{2}}{a_{1}+a_{2}}$. Let $a_{1}>0,a_{2}>0$ and $a_{n}=\dfrac{2a_{n-1}a_{n-2}}{a_{n-1}+a_{n-2}}, n>2$, then $\{ a_{n}\}$ converges to $\dfrac{3a_{1}a_{2}}{a_{1}+a_{2}}$.
My attempt:
\begin{align}
a_{n} &= \frac{2a_{n-1}a_{n-2}}{a_{n-1}+a_{n-2}} \\
&= \frac{2}{\dfrac{1}{a_{n-1}}+\dfrac{1}{a_{n-2}}} \\
& \le \frac{1}{\sqrt{a_{n-1}a_{n-2}}}
\end{align}
I used AM- GM inequality here. I am not able to proceed further. How to solve the question? Please help me.
| Following in the footsteps of @martycohen elsewhere on this page, a generalized version of the sequence in the OP can be constructed. Thus, consider
$$g_n=\frac{ag_{n-1}g_{n-2}}{bg_{n-1}+cg_{n-2}}=\frac{a}{\frac{b}{g_{n-2}}+\frac{c}{g_{n-1}}}$$
The let $f_n=1/g_n$ to find that
$$f_n=\left(\frac{c}{a} \right)f_{n-1}+\left(\frac{b}{a} \right)f_{n-2}$$
which I call a generalized Fibonacci sequence and have demonstrated solutions for arbitrary constants and initial conditions in various pages in MSE. See here, for example.
| {
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Is there a geometric method to show $\sin x \sim x- \frac{x^3}{6}$ I found a geometric method to show $$\text{when}\; x\to 0 \space , \space \cos x\sim 1-\frac{x^2}{2}$$ like below :
Suppose $R_{circle}=1 \to \overline{AB} =2$ in $\Delta AMB$ we have $$\overline{AM}^2=\overline{AB}\cdot\overline{AH} \tag{*}$$and
$$\overline{AH}=\overline{OA}-\overline{OH}=1-\cos x$$ when $x$ is very small
w.r.t. * we can conclude $$x^2 \approx 2(1-\cos x) \\\to \cos x \sim 1- \frac{x^2}{2}$$
Now I have two question :
$\bf{1}:$ Is there other idea(s) to prove (except Taylor series) $x\to 0 , \space \cos x\sim 1-\frac{x^2}{2}\\$
$\bf{2}:$ How can show $\sin x \sim x- \frac{x^3}{6}$ with a geometric concept ?
Thanks in advance.
| Let assume that for small $x$, $$\sin x = x + a x^2 + bx^3 + O(x^4)\tag{1}$$
In this trigonometric circle we can compute $BC$ in two ways:
First, considering the $ABC$ triangle:
$$BC= \sqrt{AB^2+AC^2}=\sqrt{\sin^2 x+\left(1-\sqrt{1-\sin^2 x}\right)^2}=\\= x+ax^2+\left(b+\frac18\right)x^3+O(x^4) \tag{2}$$
[*]
Second, bisecting the angle $x$:
$$BC = 2 \sin \frac{x}{2}=x+\frac{a}{2} x^2 + \frac{b}{4}x^3 + O(x^4)\tag{3}$$
Equating (2) and (3) we get $a=0$ and $b=-\frac{1}{6}$, that is: $\sin x = x - \frac16 x^3 + O(x^4)$
I'm not sure if this a geometric proof, though...
[*] To prove (2) we only need to know the expansion, for $t\to 0$: $$\sqrt{1+t}=1+\frac{t}{2}+\cdots \tag{4}$$
(here and in the following expansions the dots means a remainder $O(t^{k+1})$ where $k$ is the highest exponent in the preceeding terms).
Then
$$\sqrt{1-t^2}=1-\frac{t^2}{2}+\cdots \\
(1-\sqrt{1-t^2})^2=\frac{t^4}{4}+\cdots\\
t^2+(1-\sqrt{1-t^2})^2=t^2(1+\frac{t^2}{4}+\dots)\\
\sqrt{t^2+(1-\sqrt{1-t^2})^2}=t (1+\frac{t^2}{8}+\cdots)
=t +\frac{t^3}{8}+\cdots
$$
Letting $t=x+ax^2+bx^3+O(x^4)$ that gives
$$ x+ax^2+bx^3+\frac{x^3}{8}+O(x^4)$$
With some more work (I've verified it with Maxima), doing higher order expansions we can get more terms ($\sin x = x -\frac16 x^3 + \frac{1}{120}x^5 +\cdots$).
| {
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Solve Diophantine equation: $xy+5y=2x+(y+2)^2$
Solve Diophantine equation: $xy+5y=2x+(y+2)^2$ for positive integers $x,y$.
Rearranging the equation gives: $y^2=xy-2x+y-4$.So the RHS must be a perfect square,how can it be more restricted?
| It is equivalent to
$$x(y-2)=y^2-y+4\iff x=\frac{y^2-y+4}{y-2}=y+1+\frac6{y-2}$$
for $y\neq2$.
EDIT
As @lab bhattacharjee suggested, we may also arrange the equation with $y$ as the subject.
$$y^2-(1+x)y+(2x+4)=0\tag1$$
For integer $y$, we must have some integer $z$ such that
$$\Delta=z^2=(1+x)^2-4(2x+4)=x^2-6x-15=(x-3)^2-24$$
$$\iff(x-3-z)(x-3+z)=24$$
So there exists some integer $a=x-3-z$ and $\frac{24}{a}=x-3+z$.
Thus we have $$a+\frac{24}a=2x-6\iff x=\frac a2+\frac{12}a+3$$
So $a=2,4,6,12\iff x=8,10$. Substitute both values of $x$ into $(1)$ and we get
$$(x,y)=(8,4),(8,5),(10,3),(10,8)$$
| {
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Maclaurin series with zero denominator when evaluating derivative I have following function:
$\displaystyle f(x) = \frac{\ln(1+x^2) - x^2}{\sqrt{1+x^4} - 1}$
As you can see, when doing the quotient rule for the denominator in your head, the derivative of this function results to $0$ in the denominator when evaluating for $x = 0$. My question is, how is it possible that there exists a Maclaurin series according to Wolfram Alpha?
Do I overlook something?
| Technically speaking, even the function itself is undefined at zero, as the denominator of $f(x)$ turns into zero when we try to plug in $x=0$. But this is a removable discontinuity, as the graph of $f(x)$ (available on the page that you linked to) shows. And moreover, with this discontinuity "removed" (or patched, so to speak), the function becomes smooth and analytic.
To understand better what's going on here, let's set up the Maclaurin expansions of the numerator and denominator separately:
$$\ln(1+x^2)-x^2=\left(x^2-\frac{x^4}{2}+\frac{x^6}{3}-\frac{x^8}{4}+\cdots\right)-x^2=-\frac{x^4}{2}+\frac{x^6}{3}-\frac{x^8}{4}+\cdots;\\
\sqrt{1+x^4}-1=\left(1+\frac{x^4}{2}-\frac{x^8}{8}+\cdots\right)-1=\frac{x^4}{2}-\frac{x^8}{8}+\cdots.$$
You can see that both expansions start with $x^4$, so when we put them into a fraction, this $x^4$ can be factored out in both of them and canceled out:
$$\frac{\ln(1+x^2)-x^2}{\sqrt{1+x^4}-1}=\frac{-\frac{x^4}{2}+\frac{x^6}{3}-\frac{x^8}{4}+\cdots}{\frac{x^4}{2}-\frac{x^8}{8}+\cdots}=\frac{x^4\left(-\frac{1}{2}+\frac{x^2}{3}-\frac{x^4}{4}+\cdots\right)}{x^4\left(\frac{1}{2}-\frac{x^4}{8}+\cdots\right)}=\frac{-\frac{1}{2}+\frac{x^2}{3}-\frac{x^4}{4}+\cdots}{\frac{1}{2}-\frac{x^4}{8}+\cdots}=-1+\cdots,$$
see WolframAlpha for the rest. :-)
| {
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First four nonzero terms of the McLaurin expansion of $\frac{xe^x}{\sin x}$ at $x_0=0$
Define if necessary the given function so as to be continuous at $x_0=0$ and find the first four nonzero terms of its MacLaurin series.
$$ \frac{xe^x}{\sin x}$$
Given $f(x)= \frac{h(x)}{g(x)}$ where $h(x)= xe^x$, and $g(x)=\sin x$
We have the following MacLaurin series
$$f(x) = \sum_{n=0}^{\infty} a_n x^n = a_0 + a_1 x +a_2 x^2 + a_3 x^3 +a_4 x^4 + ...$$
$$g(x) = \sum_{n=0}^{\infty} b_n x^n = 0 + x+ 0 -\frac{1}{6} x^3 + 0 + \frac{1}{120} x^5 +...$$
$$h(x) = xe^x = \sum_{n=0}^{\infty} c_n x^n = \sum_{n=0}^{\infty} \frac{n}{(n)!}x^n = 0 + x + x^2 + \frac{1}{2} x^3 + \frac{1}{6} x^4 + \frac{1}{24} x^5$$
.
The Maclaurin coefficients of $f(x)g(x)$
$$ c_0 = a_0b_0 = 0$$
$$ c_1 = a_0b_1 + a_1b_0 = a_0$$
$$ c_2 = a_0b_2 + a_1b_1 +a_2b_0 = a_1 $$
$$ c_3 = a_0b_3 +\cdots a_3b_0 = - \frac{1}{6} a_0 + a_2$$
$$ c_4 = a_0b_4 +\cdots + a_4b_0 = - \frac{1}{6} a_1 + a_3 $$
$$ \implies f(x)g(x) = a_0 x^1 + a_1 x^2 + \left( - \frac{1}{6} a_0 + a_2\right)x^3 + \left(- \frac{1}{6} a_1 + a_3\right) x^4 + \cdots$$
.
Equating coefficients of $f(x)g(x)$ with coefficients of $h(x)$
$$ c_0 = 0$$
$$ c_1 = a_0 = 1 $$
$$ c_2 = a_1 = 1 $$
$$ c_3 = - \frac{1}{6} a_0 + a_2 = 1/2 \implies a_2 = 2/3 $$
$$ c_4 = - \frac{1}{6} a_1 + a_3 = 1/6 \implies a_3 = 1/3 $$
.
it follows that the first 4 nonzero terms are:
$$f(x) = \frac{x e^x}{\sin(x)} = 1 + x + \frac{2}{3} x^2 + \frac{1}{3} x^3$$
.
I would like to know if I am going in the right direction?
My main question regards the original question "Define if necessary the given function so as to be continuous at $x_0=0$" . I am not sure how to handle this question, if a definition is needed how? and why? if a definition is not needed how? and why?
Thx for your input/help.
| I didn't check the computations but, yes, you are doing it right.
As for the other question, note that the expression $\frac{xe^x}{\sin x}$ is undefined if $x=0$. In order to make $f$ continuous at $0$; you must define $f(0)$ as $\lim_{x\to0}\frac{xe^x}{\sin x}$, which is equal to $1$.
| {
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Solving the equation $2x^4-3x^2-9=0$ How would I go about solving $2x^4-3x^2-9=0$?
I started by taking $x^2$ out, thus getting $x^2(2x^2-3) = 9$ but I don't know if that did anything useful, usually if this equation is equal to $0$ it's easy but this time it's equal to $9$ so I'm lost.
| We can add $3x^2-3x^2$ to the left-hand side and factor by grouping.$$\begin{align*}2x^4-6x^2+3x^2-9=0 & \iff 2x^2(x^2-3)+3(x^2-3)=0\\ & \iff (x^2-3)(2x^3+3)=0\end{align*}$$However, if you did not see that sneaky trick, we can always make the substitution $X=x^2$ to get your polynomial into a quadratic.$$2X^2-3X-9=0\iff (X-3)(2X+3)=0\iff (x^2-3)(2x^3+3)=0$$In fact, fourth degree polynomials of the form $ax^4+bx^2+c$ are called biquadratics.
| {
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Prove $\sum_{n=0}^\infty\frac{2^n}{2^{2^n}}\geq\frac{1}{4(\ln 2)^2}$
Prove that:
$$\sum_{n=0}^\infty\frac{2^n}{2^{2^n}}\geq\frac{1}{4(\ln 2)^2}$$
I computed the indefinite integral:
$$\int\frac{2^x}{2^{2^x}}dx=-\frac{1}{2^{2^x}(\ln 2)^2}+C$$
How can I continue from here?
| You are on the right track. Note that $2^x-x$ is increasing for $x\geq 1$. Therefore $f(x)=\frac{1}{2^{2^x-x}}$ is decreasing and
$$\sum_{n=1}^\infty\frac{2^n}{2^{2^n}}=\sum_{n=1}^\infty f(n)\geq \sum_{n=1}^\infty \int_{n}^{n+1}f(x) dx=\int_1^{\infty}f(x)dx=\left[-\frac{1}{2^{2^x}(\ln 2)^2}\right]_1^{\infty}=\frac{1}{4(\ln 2)^2}.$$
This is a stronger inequality because the sum starts form the index $1$ and $f(0)=1/2>0$.
| {
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How to evaluate $\binom{2}{2}\binom{10}{3} + \binom{3}{2}\binom{9}{3} + \binom{4}{2}\binom{8}{3} + \ldots + \binom{9}{2}\binom{3}{3}$ I really can't understand how to approach exercise 41.
I tried by comparing the 5th term of binomial expansion and something like that, but getting 12C5, I think the answer should be 13C6. How to approach?
| Numerical method.
The sum is equal to $1716$.
Now we want to find $x,y$ in:
$$\frac{x!}{y!(x-y)!}=1716=2^2\cdot 3\cdot 11\cdot 13=\frac{2^3\cdot 3\cdot 11\cdot 13}{2!}=\frac{2^3\cdot 9\cdot 11\cdot 13}{3!}=\frac{2^5\cdot 9\cdot 11\cdot 13}{4!}=$$
$$\frac{2^4\cdot 9\cdot 10\cdot 11\cdot 13}{5!}=\frac{8\cdot 9\cdot 10\cdot 11\cdot 12\cdot 13}{6!}=\frac{13!}{6!\cdot 7!}=\frac{13!}{6!\cdot (13-6)!}.$$
Hence: $x+y=13+6=19$.
| {
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Show that: $\sin ^2 (a+b)+\sin ^2 (a-b)=4(1-\cos ^2 a \cos ^2 b)$ Show that: $$\sin ^2 (a+b)+\sin ^2 (a-b)=4(1-\cos ^2 a \cos ^2 b)$$
I tried two approaches.
Approach 1: $$\sin ^2 (a \pm b)=\sin a\cos b \pm \sin b\cos a$$
This reduced to: $$2(\cos ^2 a +\cos ^2 b -2\cos ^2 a\cos ^2b)$$
I can't see where to go from here.
Approach 2: $$\sin ^2 (a \pm b)=1-\cos ^2 (a \pm b)$$
This reduced to: $$2(\cos ^2 a +\cos ^2 b -2\cos ^2 a\cos ^2b)$$
Both approaches reduced to the same expression which is a good sign, but I still can't figure out how to make the final manipulation.
| $$S=\sin ^2 (a+b)+\sin ^2 (a-b)=[\sin (a+b)+\sin (a-b)]^2-2\sin(a+b)\cdot\sin(a-b)$$
Now use that sum-product relations:
$$\sin (a+b)+\sin (a-b)=2\sin (a)\cos (b)$$
$$2\sin(a+b)\cdot\sin(a-b)=\cos(2b)-\cos(2a)=2(\cos^2(b)-\cos^2(a))$$
So,
$$S=4(1-\cos^2(a))\cos^2(b)+2\cos^2(a)-2\cos^2(b)$$
$$S=4(1-\cos^2(a))\cos^2(b)+2\cos^2(a)-2\cos^2(b)$$
$$S=2[\cos^2(a)+\cos^2(b)]-4\cos^2(a)\cos^2(b)$$
and then the statement is wrong.
| {
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Inequality : $ (a_1a_2+a_2a_3+\ldots+a_na_1)\left(\frac{a_1}{a^2_2+a_2}+\frac{a_2}{a^2_3+a_3}+ \ldots+\frac{a_n}{a^2_1+a_1}\right)\geq \frac{n}{n+1} $ Let $a_1, a_2, \ldots, a_n $ be positive real numbers such that $\displaystyle\sum^n_{i=1} a_i=1$. Prove that
$$ (a_1a_2+a_2a_3+\ldots+a_na_1)\left(\frac{a_1}{a^2_2+a_2}+\frac{a_2}{a^2_3+a_3}+ \ldots+\frac{a_n}{a^2_1+a_1}\right) \geq \frac{n}{n+1}$$
My attempt :
By Holder inequality,
$\left(\displaystyle\sum_{cyc} a_1a_2\right)\left(\displaystyle\sum_{cyc} \frac{a_1}{a^2_2+a_2}\right)\left(\displaystyle\sum_{cyc} a_1(a_2+1)\right)\geq \left(\displaystyle\sum_{cyc} a_1\right)^3$
As $\displaystyle\sum^n_{i=1} a_i=1$,
$\Leftrightarrow \left(\displaystyle\sum_{cyc}a_1a_2\right)\left(\displaystyle\sum_{cyc} \frac{a_1}{a^2_2+a_2}\right)\left(\displaystyle\sum_{cyc} a_1a_2+1\right)\geq 1$
Please suggest on how to proceed.
| We consider two cases:
1) Let's first consider the case when $a_1,...,a_n$ fulfill $\sum a_ia_{i+1}\leq \frac{1}{n}.$
By CS we have
\begin{equation}
\left(\sum a_ia_{i+1} \right)\left(\sum\frac{a_i}{a_{i+1}^2+a_{i+1}} \right)\geq \left(\sum a_i \sqrt{\frac{a_{i+1}}{a_{i+1}^2+a_{i+1}}} \right)^2=\left(\sum a_i \frac{1}{\sqrt{a_{i+1}+1}} \right)^2.
\end{equation}
Now observe that the function $f(x)=\frac{1}{\sqrt{x+1}}$ is clearly convex. So by Jensen
\begin{equation}\sum_{i=1}^m\lambda_i f(x_i)\geq f(\sum_{i=1}^m \lambda_i x_i)
\end{equation}
holds for all $x_1,...,x_m>-1$ and all positive $\lambda_1,...,\lambda_m$ with $\sum_{i=1}^m\lambda_i=1.$ Applying this to $\lambda_i=a_i$ and $x_i=a_{i+1}$ we get
\begin{equation}
\left(\sum a_i \frac{1}{\sqrt{a_{i+1}+1}}\right)^2\geq \left(\frac{1}{\sqrt{1+\sum a_ia_{i+1}}}\right)^2=\frac{1}{1+\sum a_ia_{i+1}}.
\end{equation}
So to finish the proof of the inequality all we have to show is
\begin{equation}
n+1\geq n\left(1+\sum a_ia_{i+1} \right)=n+n\sum a_ia_{i+1},
\end{equation}which is true because $\sum a_ia_{i+1}\leq \frac{1}{n}.$
2) Now consider the case where $\sum a_i a_{i+1}\geq \frac{1}{n}.$ Because $a_1,...,a_n$ and $\frac{1}{a_1^2+a_1},...,\frac{1}{a_n^2+a_n}$ are reversly ordered, we obtain by the main theorem on ordered sequences (I hope it's called like that in english) and AM-HM
\begin{equation}
\sum \frac{a_i}{a_{i+1}^2+a_{i+1}}\geq \sum \frac{a_i}{a_i^2+a_i}=\sum \frac{1}{a_i+1}\geq \frac{n^2}{\sum (a_i +1)}=\frac{n^2}{n+1}.
\end{equation}
So in this second case
\begin{equation}
\left( \sum a_ia_{i+1}\right)\left( \sum \frac{a_i}{a_{i+1}^2+a_{i+1}}\right)\geq \frac{1}{n}\frac{n^2}{n+1}=\frac{n}{n+1}.
\end{equation}
Observe that we don't have to prove $\sum a_ia_{i+1}\geq \frac{1}{n}$ or $\leq \frac{1}{n}$ for any such $n$ numbers $a_1,...,a_n$ with $\sum a_i=1$, but that depending on how large $\sum a_ia_{i+1}$ is, we find a proof for the inequality.
| {
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Simplify $\sqrt{6-\sqrt{20}}$ My first try was to set the whole expression equal to $a$ and square both sides. $$\sqrt{6-\sqrt{20}}=a \Longleftrightarrow a^2=6-\sqrt{20}=6-\sqrt{4\cdot5}=6-2\sqrt{5}.$$
Multiplying by conjugate I get $$a^2=\frac{(6-2\sqrt{5})(6+2\sqrt{5})}{6+2\sqrt{5}}=\frac{16}{2+\sqrt{5}}.$$
But I still end up with an ugly radical expression.
| $(\sqrt{5}-\sqrt{1})^2= 6-\sqrt{20}$
| {
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Show that for any natural number n between $n^2$ and$(n+1)^2$ there exist 3 distinct natural numbers a, b, c, so that $a^2+b^2$ is divisible by c Show that for any natural number n ,one can find 3 distinct natural numbers a, b, c, between $n^2$ and$(n+1)^2$, so that $a^2+b^2$ is divisible by c.
It's easy to prove that such three distinct numbers exist, by supposing the contrary and coming to contradiction(i.e."suppose $(n+1)^2-n^2=0$ -->$n=-1$, $-1$ is not a natural number, and so on.."), but how to show divisibility?
(The task is from 1998 St. Petersburg City Mathematical Olympiad)
| Since $n^2+2n<(n+1)^2$ and since $n^2$ is included, one obvious answer is
$$\frac{\left(n^2+n \right)^2 +\left(n^2+2n \right)^2 }{n^2}$$
Extended exposition
Since $n^2+2n<(n+1)^2=n^2+2n+1$, let $a=(n^2+n)$, $b=(n^2+2n)$ and $c=n^2$ then
$$\frac{a^2+b^2}{c}=\frac{\left(n^2+n \right)^2 +\left(n^2+2n \right)^2 }{n^2}$$
$n^2$ factors out of the numerator and we have
$$\frac{a^2+b^2}{c}=\frac{n^2\left(\left(n+1 \right)^2 +\left(n+2 \right)^2\right) }{n^2}=\left(n+1 \right)^2 +\left(n+2 \right)^2$$
Since we now think $n^2$ is not included this simple answer is precluded and left for reference only.
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Show that if $a+b+c=0$, $2(a^4 + b^4+ c^4)$ is a perfect square
Show that for $\{a,b,c\}\subset\Bbb Z$ if $a+b+c=0$ then $2(a^4 + b^4+ c^4)$ is a perfect square.
This question is from a math olympiad contest.
I started developing the expression $(a^2+b^2+c^2)^2=a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2$ but was not able to find any useful direction after that.
Note: After getting 6 answers here, another user pointed out other question in the site with similar but not identical content (see above), but the 7 answers presented include more comprehensive approaches to similar problems (e.g. newton identities and other methods) that I found more useful, as compared with the 3 answers provided to the other question.
| We can simplify the algebra by noticing that the expressions are homogenous:
$
2(a^4+b^4+c^4)
\\\quad=2(a^4+b^4+(a+b)^4)
\\\quad=2b^4(x^4+1+(x+1)^4),
\quad x=a/b
$
$
2(x^4+1+(x+1)^4)
\\\quad=4 x^4 + 8 x^3 + 12 x^2 + 8 x + 4
\\\quad=4 (x^2 + x)^2 + 8 (x^2 + x) + 4
\\\quad=4(y^2+2y+1),
\quad y=x^2+x
\\\quad=4(y+1)^2
\\\quad=(2(y+1))^2
\\\quad=(2(x^2+x+1))^2
$
$
2(a^4+b^4+c^4)
\\\quad=b^4(2(x^2+x+1))^2
\\\quad=(2b^2(x^2+x+1))^2
\\\quad=(2(a^2+ab+b^2)^2
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2401281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 6
} |
Reverse Logarithmic Inequality Given
$$\frac{1}{\log_{4}\left(\frac{x+1}{x+2}\right)}<\frac{1}{\log_{4}(x+3)}.$$
Then what is the range of values of $x$ for which this inequality is satisfied.
My Try On simplification, I got $x$ from $(-\infty,-2)$. That however is not the correct answer. Can someone tell me each and every step and every individual domains and inequalities I need to consider in this?
| The arguments of the logarithms should be positive. Hence $x+3>0$ and $\frac{x+1}{x+2}>0$, that is $x\in (-3,-2)\cup (-1,+\infty).$
If $x\in (-3,-2)$ then $0<x+3<1$
and $\frac{x+1}{x+2}>1$. Therefore
$$\log_4(x+3)<0\quad\mbox{and}\quad\log_{4}\left(\frac{x+1}{x+2}\right)>0$$
and the inequality does not hold.
If $x\in (-1,+\infty)$ then then $x+3>1$
and $0<\frac{x+1}{x+2}<1$. Therefore
$$\log_4(x+3)>0\quad\mbox{and}\quad\log_{4}\left(\frac{x+1}{x+2}\right)<0$$
and the inequality holds.
So the inequality holds iff $x>-1$. Note that if we replace the base $4$ with another number greater than $1$ then the result is the same.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2401413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Compute $\int_0^{\pi /2}\frac x {\tan x} \, dx$ using contour integration How can I calculate the integral $$\int_0^{\pi /2}\frac x {\tan x} \, dx$$ with complex integration? (Contours, residue theorem, etc.) I was thinking on using the fact that $\displaystyle \tan x=\frac{e^{ix}-e^{-ix}}{i(e^{ix}+e^{-ix})}$, which implies $e^{ix}=z$. I still have not been succesful.
| A complex-analytic solution. Here is a combination of calculus and a basic complex analysis: Perform integration by parts to sanitize the integrand:
$$ \int_{0}^{\frac{\pi}{2}} \frac{x}{\tan x} \, dx
= \underbrace{ \left[ \vphantom{\int} x \log\sin x \right]_{0}^{\frac{\pi}{2}} }_{=0}
- \int_{0}^{\frac{\pi}{2}} \log \sin x \, dx. $$
Also notice that, if $x \in (0,\frac{\pi}{2}]$ then $\log|\sin x| = -\log 2 + \operatorname{Re}\log(1 - e^{2ix})$. So
\begin{align*}
\int_{0}^{\frac{\pi}{2}} \frac{x}{\tan x} \, dx
&= - \frac{1}{2} \int_{0}^{\pi} \log |\sin x| \, dx \\
&= \frac{\pi}{2}\log 2 - \frac{1}{2} \operatorname{Re} \left( \int_{0}^{\pi} \log(1 - e^{2ix}) \, dx \right) \\
(z = e^{2ix}) \quad&= \frac{\pi}{2}\log 2 - \frac{1}{4} \operatorname{Re} \left( \oint_{|z|=1} \frac{\log(1 - z)}{iz} \, dz \right),
\end{align*}
Since $z \mapsto \frac{\log(1 - z)}{iz}$ is holomorphic in the unit disc $\mathbb{D}$ and has only logarithmic singularity on the boundary $\partial \mathbb{D}$, we can still apply the Cauchy integral theorem to conclude that the integral vanishes. Therefore we have
$$ \int_{0}^{\frac{\pi}{2}} \frac{x}{\tan x} \, dx = \frac{\pi}{2}\log 2. $$
Alternative complex-analytic solution. Use the substitution $x = \arctan u$ to write
$$ \int_{0}^{\frac{\pi}{2}} \frac{x}{\tan x} \, dx = \frac{1}{2}\int_{-\infty}^{\infty} \frac{\arctan u}{u(u^2+1)} \, du. $$
This hints us that we may use some contour integration technique, but we need to resolve the branch cut of $\arctan$. Here we give one such trick. Notice that
$$ \frac{\arctan u}{u} = \int_{0}^{1} \frac{dt}{1+u^2 t^2}. $$
Plugging this back and interchanging the order of integration, we get
$$ \int_{0}^{\frac{\pi}{2}} \frac{x}{\tan x} \, dx = \frac{1}{2}\int_{0}^{1} \left( \int_{-\infty}^{\infty} \frac{du}{(u^2 t^2+1)(u^2+1)} \right) \, dt. $$
Now we can perform contour integration to obtain that
\begin{align*}
\int_{-\infty}^{\infty} \frac{du}{(u^2 t^2+1)(u^2+1)}
&= 2\pi i \left( \underset{u = i}{\mathrm{Res}} \, \frac{1}{(u^2 t^2+1)(u^2+1)} + \underset{u = i/t}{\mathrm{Res}} \, \frac{1}{(u^2 t^2+1)(u^2+1)} \right) \\
&= \frac{\pi}{t+1}.
\end{align*}
So we have
$$ \int_{0}^{\frac{\pi}{2}} \frac{x}{\tan x} \, dx = \frac{1}{2}\int_{0}^{1} \frac{\pi}{t+1} \, dt = \frac{\pi}{2} \log 2. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2401761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
find a and b such that the limit will exist and find the limit I am given this question $$\lim_{x\rightarrow 0} \frac{e^{\sqrt{1+x^2}}-a-bx^2}{x^4}$$
And asked to find the $a$ and $b$ such that the limit exists and also to compute the limit
My solution:
Let $t$=$\sqrt{1+x}$.
Then the Maclaurin polynomial is : $$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+\mathcal{O}(x^3)$$
Now plugging in $x^2$ for x we get $$\sqrt{1+x^2}=1+\frac{x^2}{2}-\frac{x^4}{8}+\mathcal{O}(x^6)$$
and from the common Maclaurin polynomial we have that $e^t=1+t+\frac{t^2}{2}+ \mathcal{O}(t^3)$. Plugging in $\sqrt{x^2+1}$ for $t$ we get:
$$e^{1+\frac{x^2}{2}-\frac{x^4}{8}+\mathcal{O}(x^6)
}$$
which in turn is:
$$e^{1+\frac{x^2}{2}-\frac{x^4}{8}+\mathcal{O}(x^6)
}=1+\left(1+\frac{x^2}{2}-\frac{x^4}{8}+\mathcal{O}(x^6)\right)+\mathcal{O}(x^6)$$
hence we have:
$$\lim_{x\rightarrow 0} \frac{1+\left(1+\frac{x^2}{2}-\frac{x^4}{8}+\mathcal{O}(x^6)\right)+\mathcal{}O(x^6)-a-bx^2}{x^4}$$
And i argue that $a=2, b=1/2$ and the $\lim=-1/8 $
However the book disagrees with me and argues that the they should be $a=e, b=e/2$ and limit $=0$
i mean i can see how theyd done it, by not expanding $e^t$ but by only expanding $\sqrt{x^2+1}$ however what i dont understand how come we didnt get the same values or at least the same value for the limit?
| First rewrite the equation replacing $x$ by $X$: Thus we require $L=\lim_{X\rightarrow 0}\dfrac{e^{\sqrt{1+X^2}}-a-bX^2}{X^4}$
Now let $x:=\sqrt{1+X^2}$, then $X^2=x^2-1$ with $X^4=(x^2-1)^2$ and $x\rightarrow 1$ as $X\rightarrow 0$.
Now, we have $L=\lim_{x\rightarrow 1}\dfrac{e^x-a-b(x^2-1)}{(x^2-1)^2}=\dfrac{\lim_{x\rightarrow 1}e^x-a-b(x^2-1)}{\lim_{x\rightarrow 1}(x^2-1)^2}=\dfrac{e-a}{0}$
Thus $L$ is not defined (finite) unless $e-a=0$. Now if $a=e$ we have $L=\frac{0}{0}$, so we may use L'Hopital.
Consider then $L=\dfrac{\lim_{x\rightarrow 1}e^x-2bx}{\lim_{x\rightarrow 1}4(x^2-1)x}=\dfrac{e-2b}{0}$
which is not defined unless $e-2b=0$.
Thus for a finite limit it is necessary that $b=\frac{e}{2}$ and $a=e$.
Again use L'Hopital:
$L=\dfrac{\lim_{x\rightarrow 1}e^x-e}{\lim_{x\rightarrow 1}12x^2-4}=\dfrac{e-e}{8}=0$ as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2402304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Wrong result for a $1^\infty$ limit I wanna know the $$\lim_{x\rightarrow\infty}(\sqrt{x^2+2x+4}-x)^x .$$
I applied $1^\infty$ algorithm and I have in last sentence $$\lim_{x\rightarrow\infty} \left(\frac{2x+4-\sqrt{x^2+2x+4}}{\sqrt{x^2+2x+4}+x}\right)^x.$$ After that I used $x=e^{\ln x}$ and got this: $$\lim_{x\rightarrow\infty} \left(\frac{x^2+4x-x\sqrt{x^2+2x+4}}{\sqrt{x^2+2x+4}+x}\right).$$ The result for this, I think is $2$, but I'm not really sure of this, Wolfram-Alpha says that it's $\frac{3}{2}$. Can explain somebody why, but without L'Hospital?
Thks a lot!
| Want
$\lim_{x\rightarrow\infty}(\sqrt{x^2+2x+4}-x)^x
$.
Since,
for small $z$,
$\sqrt{1+z}
=1+\frac12 z + O(z^2)
$
and
$\ln(1+z)
=z+O(z^2)
$,
$\begin{array}\\
\sqrt{x^2+2x+4}
&=\sqrt{(x+1)^2+3}\\
&=(x+1)\sqrt{1+\frac{3}{(x+1)^2}}\\
&=(x+1)(1+\frac{3}{2(x+1)^2}+O(x^{-4}))\\
&=x+1+\frac{3}{2(x+1)}+O(x^{-3})\\
\text{so}\\
\sqrt{x^2+2x+4}-x
&=1+\frac{3}{2(x+1)}+O(x^{-3})\\
\end{array}
$
Therefore
$\ln(\sqrt{x^2+2x+4}-x)
=\ln(1+\frac{3}{2(x+1)}+O(x^{-3}))
=\frac{3}{2(x+1)}+O(x^{-2})
$
so
$\begin{array}\\
x\ln(\sqrt{x^2+2x+4}-x)
&=\frac{3x}{2(x+1)}+O(x^{-1})\\
&=\frac{3(x+1)-3}{2(x+1)}+O(x^{-1})\\
&=\frac32-\frac{3}{2(x+1)}+O(x^{-1})\\
&=\frac32+O(x^{-1})\\
\end{array}
$
so that
$\lim_{x\rightarrow\infty}(\sqrt{x^2+2x+4}-x)^x
=e^{3/2}
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2403985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Volume made be a Revolving Pentagon I was looking to the volume created if I revolve the area of the pentagon having vertices
(1,0), (2,2), (0,4), (-2, 2), (-1,0) about the x-axis.
My work:
I used the Second theorem of Pappus, which states that the volume created is equal to the product
of circumference of the circle described by the centroid and the area. In short: $V = (2\pi \bar y)( A)$
I need to get the centroid of the pentagon.
$$\bar x = \frac{1+2+0+(-2)+(-1)}{5} = 0$$
$$\bar y = \frac{0+2+4+2+0}{5} = \frac{8}{5}$$
The area of the polygon laid out in Cartesian Coordinates is:
$$A = \frac{1}{2} \left[ \begin{matrix}
x_1 & x_2 & x_3 & ... & x_n & x_1\\
y_1 & y_2 & y_3 & ... & y_n & y_1\\
\end{matrix} \right] $$
Turns out, the area described by the pentagon above is 10 square units.
We can now get the volume of solid of revolution of pentagon:
$$V = 2 \pi \bar y A$$
$$V = 2 \pi \left(\frac{8}{5}\right) (10)$$
We now get $32 \pi$ cubic units.
But my book said the volume is $\frac{104}{3} \pi$
Where did I go wrong?
| It's
$$2\left(\frac{4\cdot\pi\cdot4^2}{3}-\frac{1\cdot\pi\cdot2^2}{3}-\frac{2\cdot\pi\cdot2^2}{3}\right)=\frac{104\pi}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2404832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Solve $\lfloor \frac{2x-1}{3} \rfloor + \lfloor \frac{4x+1}{6} \rfloor=5x-4$ I came across this problem and it doesn't seem so tricky, but I didn't do it.
Solve the equation
$$\lfloor {\frac{2x-1}{3}}\rfloor + \lfloor {\frac{4x+1}{6}}\rfloor=5x-4.$$
My thoughts so far:
Trying to use the inequality $k-1 < \lfloor k\rfloor \leq k$
$5x-4$ is integer, so $x=\frac{a}{5},$ where $a$ is from $\mathbb Z,$ then replace in the original equation, to get
$$\lfloor \frac{4a-10}{30}\rfloor + \lfloor \frac{4a+5}{30}\rfloor =a-4$$
Let $k= \frac{4a-10}{30}$
so $\lfloor k\rfloor +\lfloor k+\frac{1}{2}\rfloor =a-4$
but $\lfloor k\rfloor =\lfloor k+\frac{1}{2}\rfloor $ or $\lfloor k\rfloor +1=\lfloor k+\frac{1}{2}\rfloor $, so we have 2 cases
I strongly think that a shorter solution exists, I would appreciate a hint!
|
The graph of the function $g(x) = \lfloor\frac{2x-1}{3}\rfloor + \lfloor\frac{4x+1}{6}\rfloor - (5x-4)$
Once we know $0 < x < 1$, we have $\lfloor\frac{2x-1}{3} \rfloor = 0$ and $\lfloor \frac{4x+1}{6} \rfloor = 0$. Thus $x = \frac{4}{5}$ is the only solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2405495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Review: Prove by induction Nicomachus's theorem Please help me out reviewing the way I wrote this proof:
Prove by induction: $1^3+2^3+3^3+...+n^3=\left(\frac{n(n+1)}{2}\right)^2$ with $n\geqslant1$
Proof:
Lets define the set, $S=\left \{n\in N:n\geqslant1, 1^3+2^3+3^3+...+n^3=\left(\frac{n(n+1)}{2}\right)^2 \right \}$
If $n=1$ then, $1\in S$
Lets asume that $k\in S$ with $k\geqslant1$, then
$\begin{gather*} 1^3+2^3+3^3+...+k^3=\left(\frac{k(k+1)}{2}\right)^2\\
\end{gather*}$
Now lets proof that $k+1\in S$,
$\begin{align*}1^3+2^3+3^3+...+k^3+(k+1)^3=&\left(\frac{k(k+1)}{2}\right)^2+(k+1)^3\\ =&(k+1)^2\left (\frac{k^2}{4}+(k+1)\right)\\=&(k+1)^2 \left(\frac{(k^2+4k+4)}{4} \right)\\ =&(k+1)^2 \left(\frac{(k+2)^2}{4}\right)\\ =&\left(\frac{(k+1)(k+2)}{2}\right)^2 \end{align*}$
Which is true.
| Yes, the proof seems okay to me.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2407633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Expectation of minimum of two random numbers Let's assume we picked chose two numbers in the range $[1,n]$ where $n\in \mathbb N$ independently, such that each one of them is distributed randomly.
How can I find the expectation of the minimum of these two numbers?
I know that we have $2n-1$ possibilities of getting $1$ as minimum, $2(n-1)-1$ possibilities of getting $2$ as minimum, and so on!
But how do I formalize this and calculate the expectation?
| For $1 \le k \le n$, let $p_k$ be the probability that the minimum of the two numbers is $k$, and let $e$ be the expected minimum value.
\begin{align*}
\text{Then}\;\;p_k &=
2
\left(
{\small{\frac{1}{n}}}
\right)
\left(
{\small{\frac{n-k+1}{n}}}
\right)
-
{\small{\frac{1}{n^2}}}
\\[6pt]
&=\frac{2n-2k+1}{n^2}\\[8pt]
\text{Hence}\;\;e &=\sum_{k=1}^n kp_k\\[4pt]
&=\sum_{k=1}^n k\left(\frac{2n-2k+1}{n^2}\right)\\[4pt]
&=\frac{1}{n^2}\left(\sum_{k=1}^n k(2n-2k+1)\right)\\[4pt]
&=\frac{1}{n^2}\left(\sum_{k=1}^n (2n+1)k-2k^2)\right)\\[4pt]
&=\frac{1}{n^2}\left(\left(\sum_{k=1}^n (2n+1)k\right)-\left(\sum_{k=1}^n 2k^2)\right)\right)\\[4pt]
&=\frac{1}{n^2}\left(
\left((2n+1)\sum_{k=1}^n k\right)
-
\left(2\sum_{k=1}^n k^2)\right)
\right)\\[4pt]
&=\frac{1}{n^2}
\left(
(2n+1)
\left({\small{\frac{n(n+1)}{2}}}\right)
-2
\left({\small{\frac{n(n+1)(2n+1)}{6}}}\right)
\right)
\\[4pt]
&=\frac{(n+1)(2n+1)}{6n}\\[4pt]
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2407805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Conversion to base $9$ Where am I going wrong in converting $397$ into a number with base $9$?
$397$ with base $10$ $$= 3 \cdot 10^2 + 9 \cdot 10^1 + 7 \cdot 10^0$$ To convert it into a number with base $9$
$$3 \cdot 9^2 + 9 \cdot 9^1 + 7 \cdot 9^0 = 243 + 81 + 7 = 331$$
But answer is $481$?????
| You cannot simply keep the same coefficients but change the $10$'s to $9$'s. What you must do is something like the following:
$$\begin{align}397&=3\times \color{red}{10}^2+9\times\color{red} {10}^1+7\times \color{red}{10}^0\\&=3\times(9+1)^2+9\times(9+1)+7\times 1\\&=3\times\color{blue}9^2+(3\times18+3)+\color{blue}9^2+\color{blue}9^1+(7)\\&=4\times \color{blue}9^2+(6\times9)+(10)+\color{blue}9^1\\&=4\times \color{blue}9^2+8\times \color{blue}9^1+1\times \color{blue}9^0\end{align}$$
What you did was simply change the tens in red to nines without doing any intermediate steps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2408902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Quadratic Diophantine Equation$(x^2+x)(y^2-1)=240$ For whole numbers $x$ and $y$, $$x,y | (x^2+x)(y^2-1)= 240$$
Find the biggest and smallest value for $x-y$.
How do you proceed with such a question?
Are their formulas or something for that type of equation?
I'd appreciate any help.
| Here is a table of whole numbers $x$ so that $x^2+x\mid240$:
$$
\begin{array}{c|c}
x&1&2&3&4&5&15\\\hline
x^2+x&\color{#090}{2}&6&12&20&\color{#090}{30}&240
\end{array}
$$
For $x$ to work, we need $\frac{240}{x^2+x}+1=y^2$:
$\frac{240}{2}+1=11^2$
$\frac{240}{6}+1=41$
$\frac{240}{12}+1=21$
$\frac{240}{20}+1=13$
$\frac{240}{30}+1=3^2$
$\frac{240}{240}+1=2$
The choices we have are $\{x=1,y=11\}$ and $\{x=5,y=3\}$. Thus, the smallest is $x-y=-10$ and the largest is $x-y=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2411174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Calculate the rest of the division of $(4^{103} + 2(5^{104}))^{102}$ by $13$ I already know some techniques to solve big exponents such as, Euler/Fermat Theorem,Euler/Carmichael,successive squaring.
But these problems seem to be more dificult as they involve operations insted of a single number.
How could I solve them ?
*
*$(4^{103} + 2(5^{104}))^{102}$ by $13$
*$53^{103}+103^{53}$ by $39$
| $$
(4^{103}+2 \cdot 5^{104})^{102}\overset{13}{\equiv}
(4^{3 \cdot 34+1}+2 \cdot 5^{2 \cdot 52})^{102}\overset{13}{\equiv}
(4^{3 \cdot 34}4^1+2 \cdot 5^{2 \cdot 52})^{102}\overset{13}{\equiv}
\\
\left((4^{3})^{34} \cdot 4^1+2 \cdot (5^{2})^{52}\right)^{102}\overset{13}{\equiv}
\left(64^{34} \cdot 4^1+2 \cdot 25^{52}\right)^{102}\overset{13}{\equiv}
\\
\left((-1)^{34}4^1+2(-1)^{52}\right)^{102}\overset{13}{\equiv}
6^{102}\overset{13}{\equiv}
6^{6}\overset{13}{\equiv}
36^{3}\overset{13}{\equiv}
10^{3}\overset{13}{\equiv}
100.10\overset{13}{\equiv}
9.10\overset{13}{\equiv}
-1
$$
$$
53^{103}+103^{53}\overset{13}{\equiv}
1^{103}+(-1)^{53}\overset{13}{\equiv}
1-1\overset{13}{\equiv} 0;
$$
$$
53^{103}+103^{53}\overset{3}{\equiv}
(-1)^{103}+1^{53}\overset{3}{\equiv}
-1+1\overset{3}{\equiv} 0;
$$
which implies that $53^{103}+103^{53}\overset{39}{\equiv} 0$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2411403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that inequality $\sum _{cyc}\frac{a^2+b^2}{a+b}\le \frac{3\left(a^2+b^2+c^2\right)}{a+b+c}$
Let $a>0$,$b>0$ and $c>0$. Prove that:
$$\dfrac{a^2+b^2}{a+b}+\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}\le \dfrac{3\left(a^2+b^2+c^2\right)}{a+b+c}.$$
$$\Leftrightarrow \frac{(a^2+b^2)(a+b+c)}{a+b}+\frac{(b^2+c^2)(a+b+c)}{b+c}+\frac{(c^2+a^2)(a+b+c)}{c+a}\leq 3(a^2+b^2+c^2)$$
$\Leftrightarrow 2(a^2+b^2+c^2)+\frac{c(a^2+b^2)}{a+b}+\frac{a(b^2+c^2)}{b+c}+\frac{b(a^2+c^2)}{a+c}\leq 3(a^2+b^2+c^2)$
$\Leftrightarrow \frac{c(a^2+b^2)}{a+b}+\frac{a(b^2+c^2)}{b+c}+\frac{b(a^2+c^2)}{a+c}\leq a^2+b^2+c^2$
$\Leftrightarrow \frac{c(a+b)^2-2abc}{a+b}+\frac{a(b+c)^2-2abc}{b+c}+\frac{b(a+c)^2-2abc}{a+c}\leq a^2+b^2+c^2$
$\Leftrightarrow 2(ab+bc+ac)\leq a^2+b^2+c^2+2abc\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)$
I can't continue. Help me
| your inequality is equivalent to
$$ab(a-b)(a^2-b^2)+ac(a-c)(a^2-c^2)+bc(b-c)(b^2-c^2)\geq 0$$ which is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2413136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
For what value of $k$ is one root of the equation
For which value of $k$ is one root of the equation $x^2+3x-6=k(x-1)^2$ double the other?
My Attempt:
$$x^2+3x-6=k(x-1)^2$$
$$x^2+3x-6=k(x^2-2x+1)$$
$$x^2+3x-6=kx^2-2kx+k$$
$$(1-k)x^2+(3+2k)x-(6+k)=0$$
| If $ax^2+bx+c$ has two roots, then the sum of the roots is equal to $-b/a$ and the product of the roots is equal to $c/a$. In your case, if one root is $p$, then the other one would be $2p$; applying the formulas to:
$$(1-k)x^2+(3+2k)x-(6+k)=0$$
$$\frac{3+2k}{k-1}=3p \quad\mbox{ and }\quad \frac{6+k}{k-1}=2p^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2413338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
asymptotic behavior of $\displaystyle{\sum_{r = 0}^k} { 2k - r \choose k} x^r $ for large $k$ Problem
This problem comes from the partition function $Z( n, r )$ in this question. I want to analyze the large $k$ behavior of
\begin{equation}
f(k) = \sum_{r = 0}^k { 2k - r \choose k} x^r
\end{equation}
For example,
\begin{equation}
\sum_{r =0}^k { 2k - r \choose k} 2^r = 4^k
\end{equation}
A more specific question would be the limit as a function of $x$
\begin{equation}
\lim_{k\rightarrow \infty} \frac{\ln f(k)}{k}
\end{equation}
One can assume $x > 0$ or even $x > 1$. I'm interested at the point $x = 3$.
A preliminary analysis shows that
\begin{equation}
\sum_{r=0}^{\infty} { 2k -r \choose k } x^r = { 2k \choose k} {}_2F_1( 1, -2k;-k; x )
\end{equation}
then the problem reduces to the analysis of the Gauss hypergeometric function.
Update 1: numerical data
I calculate the results for $x = 0.1$ to $10$ and $k$ up to $25$
$\ln f( x, k )$ vs $k$ is pretty much on the straight line even for $k$ of order $10$.
Then I fit the curve (in range $k \in [15, 25]$) and compare the slope with the result from @ThomasAndrews
Based on this finite size numerical calculation, we can see that for $x> 4$, $2\ln x - \ln( x- 1)$ is a very accurate estimation. It is actually the exact answer( see the error estimation by @ThomasAndrews ).
However, the error term can not be improved, because the data in range $[2,4]$ is slightly above $2\ln x - \ln( x- 1)$. The estimation doesn't work for $k < 2$.
[2017-09-07]Update
Following @ThomasAndrews, we should analyze the scaling behavior of
\begin{equation}
f( k, x ) = x^k \sum_{r=0}^{k} { k + r \choose r } x^{-r} \quad x \ge 2
\end{equation}
I here present a slightly different break up.
Using Pascal's rule for binomial coefficients
\begin{equation}
{ k + r \choose r} = { k - 1 \choose 0} + { k \choose 1 } + \cdots + { k+r - 1 \choose r }
\end{equation}
we have
\begin{equation}
\begin{aligned}
f(k,x) = x^k\Bigg\{ &\left[ { k - 1 \choose 0} + { k \choose 1 } + \cdots + { 2k - 1 \choose k } \right] \frac{1}{x^k} \\
+&\left[ { k - 1 \choose 0} + { k \choose 1 } + \cdots + { 2k - 2 \choose k-1 } \right] \frac{1}{x^{k-1}}\\
+\vdots &\\
+&\left[ { k - 1 \choose 0}\right] \frac{1}{x^0} \Bigg\} \\
\end{aligned}
\end{equation}
Summing along the vertical lines to get a recursive relation
\begin{equation}
\begin{aligned}
\frac{f(k,x)}{x^k} &= \frac{x}{x - 1}\left[ (1 - x^{-k-1}){ k - 1 \choose 0} + x^{-1} ( 1 - x^{-k}){ k \choose 1 } + \cdots + x^{-k}( 1 -x^{-1} ){ 2k - 1 \choose k } \right] \\
&= \frac{x}{x - 1} \left[ \sum_{r =0 }^{k-1} { k - 1 + r \choose r } x^{-r} + x^{-k}{2k - 1 \choose k } - x^{-k-1} { 2k \choose k } \right]\\
&= \frac{x}{x-1} \frac{f( k-1, x ) }{x^{k-1}} + \frac{x - 2}{x-1} x^{-k} { 2k - 1 \choose k } \quad k \ge 1\\
&= \left( \frac{x}{x - 1}\right)^k \left[ f(0,x) + \frac{x - 2}{x-1} \sum_{r=1}^{k} \left( \frac{x^2}{x - 1}\right)^{-r} { 2r - 1 \choose r } \right]
\end{aligned}
\end{equation}
Therefore
\begin{equation}
f(k,x) = \left( \frac{x^2}{x - 1}\right)^k + \frac{x - 2}{x-1} \left( \frac{x^2}{x - 1}\right)^k \sum_{r=1}^{k} \left( \frac{x^2}{x - 1}\right)^{-r} { 2r - 1 \choose r }
\end{equation}
A consistency check is
\begin{equation}
f( k, 2 ) = 4^k \rightarrow \text{correct}
\end{equation}
For general $x > 1, x \ne 2$, we have $\frac{x^2}{x - 1} > 4$, which is in the convergence range of the series
\begin{equation}
\sum_{r=0}^{\infty} { 2r \choose r } x^r = \frac{1}{\sqrt{ 1- 4x} } \quad |x| < \frac{1}{4}
\end{equation}
Consequently, we can use this generating function to approximate the second term of $f$ (the error is $o(1)$ because of the convergence of the series),
\begin{equation}
\begin{aligned}
f( k ,x ) &=\left( \frac{x^2}{x - 1}\right)^k + \frac{1}{2} \frac{x - 2}{x-1} \left( \frac{x^2}{x - 1}\right)^k \sum_{r=1}^{k} \left( \frac{x^2}{x - 1}\right)^{-r} { 2r \choose r } \\
&\sim \left( \frac{x^2}{x - 1}\right)^k + \frac{1}{2} \frac{x - 2}{x-1} \left( \frac{x^2}{x - 1}\right)^k \left( \frac{x}{|x-2|} - 1 \right) \\
&= \left( \frac{x^2}{x - 1}\right)^k + \frac{1}{x-1} \left( \frac{x^2}{x - 1}\right)^k \quad x > 2 \\
&= \frac{1}{x} \left( \frac{x^2}{x - 1}\right)^{k+1}
\end{aligned}
\end{equation}
We conclude that
\begin{equation}
\label{eq:Z_N-asymp}
\lim_{k \rightarrow \infty} \frac{\ln f(k,x)}{k} = 2\ln x - \ln ( x+ 1 ) \quad x \ge 2
\end{equation}
I'm not sure why the fitting result is a bit off the theoretical line in $[2,4]$.
Thanks for your attention.
| An upper bound, and a lower bound when $x>4$.
Assuming you mean $x^r$ rather than $x^{k}$, then you get:
$$\sum_{r=0}^{k}\binom{2k-r}{k}x^r = \sum_{s=0}^{k}\binom{k+s}{k}x^{k-s}=x^k\sum_{s=0}^{k}\binom{k+s}{k}x^{-s}$$
Now, $$\sum_{s=0}^{\infty}\binom{k+s}{k}y^s =\frac{1}{(1-y)^{k+1}}$$
for $|y|<1$.
So, when $x>1$, letting $y=\frac{1}{x}$, you get $f(k)\leq x^k\frac{1}{\left(1-\frac{1}{x}\right)^{k+1}}= \frac{x^{2k+1}}{(x-1)^{k+1}}$, or:
$\log f(k)\leq (2k+1)\log x - (k+1)\log(x-1)$ and thus:
$$\limsup_{k\to\infty}\frac{\log f(k)}{k}\leq 2\log(x)-\log(x-1)$$
The error function:
$$E_k=\frac{x^{2k+1}}{(x-1)^{k+1}}-f(k)=\sum_{s=1}^{\infty}\binom{2k+s}{k}x^{-s}$$
We can note that $\binom{2k+s}{k}<\binom{2(k+s)}{k+s}$ and get:
$$E_k<x^{k}\sum_{t=k+1}^{\infty} \binom{2t}{t}x^{-t}<x^{k}\left(\frac{1}{\sqrt{1-4/x}}-1\right)=\frac{4x^{k}}{\sqrt{x^2-4x}+x-4}=O(x^{k-1})$$
when $x>4$.
I think that means, when $x>4$, that $\lim_{k\to\infty}\frac{\log f(k)}{k}=2\log(x)-\log(x-1)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
How many positive integer solutions are there to the equation $^2 + 2^2 = 4^2$?
How many positive integer solutions are there to the equation $^2 + 2^2 = 4^2$?
I realised that this looks a lot like the Pythagorean theorem -- it could be written as $^2 + (\sqrt{2}y)^2 = (2z)^2$ as well. Then wouldn't there be an infinite number of solutions that are positive integers? For some reason they don't have the answer to this one, so I just wanted to check that I got this right.
| Observe $x^2$ divisible by $2$, thus $x$ is divisible by $2$. Write
$$ \begin{align*} x = 2n &\implies 4n^2 + 2y^2 = 4z^2 \\
&\implies y^2+2n^2=2z^2 \\
&\implies y = 2m \\
&\implies 4m^2+2n^2=2z^2 \\
&\implies n^2+2m^2=z^2 \\
&\implies n = k(a^2-2b^2), \end{align*}$$
where $m = 2kab, z = k(a^2+2b^2), a,b,k \in \mathbb{Z}^{+}$ (see a solution of this equation by member Ivan Loh also at MSE in 2013 post). So $x = 2n = 2k(a^2-2b^2), y = 2m = 4kab, z = k(a^2+2b^2)$. I hope this helps.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
} |
Calculate the Limit of $f_n:= (\frac{2}{3}+\frac{i}{2})^n+\sqrt[n]{2n}-\frac{n^3}{2^n}$ $\lim_{n \rightarrow \infty}(\frac{2}{3}+\frac{i}{2})^n+\sqrt[n]{2n}-\frac{n^3}{2^n} = \lim_{n \rightarrow \infty}(\frac{2}{3}+\frac{i}{2})^n+1-0$
Now, how to evaluate $(\frac{2}{3}+\frac{i}{2})^n$?
| With $||.||_2$ we will denote the complex absolute value.
We see that: $$||\frac{2}{3}+\frac{i}{2}||_2=\sqrt{\frac{4}{9}+\frac{1}{4}}=\sqrt{\frac{25}{36}}=\frac{5}{6} <1$$
Thus $||\frac{2}{3}+\frac{i}{2}||_2^n=(\frac{5}{6})^n \to 0$
Now we have that: $$||f_n-1||_2=||(\frac{2}{3}+\frac{i}{2})^n+(\sqrt[n]{2n}-1)+\frac{n^3}{2^n}||_2 \leq ||\frac{2}{3}+\frac{i}{2}||_2^n+|(\sqrt[n]{2n}-1)|+\frac{n^3}{2^n} \to 0$$
Thus $f_n \to 1$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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} |
Show that $x^6 - x^4 +2 >0$ for all $x$
Show that $x^6 - x^4 +2 >0$ for all $x$.
I can prove this using calculus and I can prove it by factorising the expression.
Is there any way to prove it another way, perhaps by considering cases of $x$? I need a simpler way to explain to a student that cannot factorise degree $6$ polynomials and use calculus at this stage. They can factorise degree $2$ though, if that helps!
Thanks.
| Use sum of squares! $x^6>x^4$ for $|x|>1$
Otherwise $|x|<1$ and we have $x^6-x^4+2=(x^3-\frac{1}{2}x)^2-\frac{1}{4}x^2+2>0$ since $2>\frac{1}{4}x^2$
Or, magically, we have
$x^6-x^4+2=(x^3-x)^2+(x^2-1)^2+x^2+1>0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2418707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
Find $\iiint (x^2+y^2+z^2)~dV$, above the cone $z=\sqrt{3(x^2+y^2)}$, inside the sphere $x^2+y^2+z^2\leq a^2$.
Find
$$\iiint (x^2+y^2+z^2)~dV$$
above the cone $z=\sqrt{3(x^2+y^2)}$, and inside the sphere $x^2+y^2+z^2\leq a^2$.
I think that the solution for this problem is following:
Use cylindrical coordinates:
$x = r\cdot \cos(\theta)$
$y = r\cdot \sin(\theta)$
$z = z$
and we will get:
$$\int_0^a \int_0^{\pi} \int_0^{a/2} (r^2+z^2)r dr d\theta dz$$
Is this approach correct?
| Hint: Alternative approach is using the spherical coordinates for this problem:
$$\int_0^a \int_0^{2\pi} \int_{0}^{\frac{\pi}{3}} r^4\sin\varphi d\varphi d\theta dr=\color{blue}{\dfrac{\pi}{5}a^5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2418966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $ \lim_{x\to \pi/2} \frac{\sqrt{1+\cos(2x)}}{\sqrt{\pi}-\sqrt{2x}}$
Evaluate
$$ \lim_{x\to \pi/2} \frac{\sqrt{1+\cos(2x)}}{\sqrt{\pi}-\sqrt{2x}}$$
I tried to solve this by L'Hospital's rule..but that doesn't give a solution..appreciate if you can give a clue.
| Alternatively:
$$\lim_{x\to \pi/2} \frac{\sqrt{1+\cos(2x)}}{\sqrt{\pi}-\sqrt{2x}}=\lim_{x\to \pi/2} \frac{\sqrt{1+\cos(2x)}}{\sqrt{\pi}-\sqrt{2x}}\cdot \frac{\sqrt{1-\cos(2x)}}{\sqrt{1-\cos(2x)}}\cdot \frac{\sqrt{\pi}+\sqrt{2x}}{\sqrt{\pi}+\sqrt{2x}}=$$
$$\lim_{x\to \pi/2} \frac{|\sin{(2x)|}\cdot(\sqrt{\pi}+\sqrt{2x})}{(\pi-2x)\cdot\sqrt{1-\cos{(2x)}}}=\lim_{x\to \pi/2} \frac{|\sin{(\pi-2x)|}\cdot\sqrt{2\pi}}{(\pi-2x)}$$
Note:
$$\lim_{x\to \pi/2^+} \frac{|\sin{(\pi-2x)|}\cdot\sqrt{2\pi}}{(\pi-2x)}=\lim_{x\to \pi/2^+} \frac{-\sin{(\pi-2x)}\cdot\sqrt{2\pi}}{(\pi-2x)}=-\sqrt{2\pi}.$$
$$\lim_{x\to \pi/2^-} \frac{|\sin{(\pi-2x)|}\cdot\sqrt{2\pi}}{(\pi-2x)}=\lim_{x\to \pi/2^-} \frac{\sin{(\pi-2x)}\cdot\sqrt{2\pi}}{(\pi-2x)}=\sqrt{2\pi}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that for $n\to \infty, u_n - \sqrt{n} - \frac{1}{2} \sim \frac{1}{8\sqrt{n}}$ $\forall n\in\mathbb{N}^*, u_n = \sqrt{n+u_{n-1}}$ with $u_1 = 1$.
I've already shown that $ u_n \sim \sqrt{n}$ and that $\underset{n\rightarrow \infty}{\lim} u_n - \sqrt{n} = \dfrac{1}{2}$
How to show that $n\rightarrow \infty, u_n - \sqrt{n} - \dfrac{1}{2} \sim \dfrac{1}{8\sqrt{n}}$ ? (Or it might not be this, maybe something else ? I dont really know... I intuited this result with the asymptotic exampsion of $\sqrt{n}$.
I know that I have to use the asymptotic expansion, but I don't know how to use it correctly (I've some difficulties to keep the $o(1)$).
What I've done :
$u_n = \sqrt{n+u_{n-1}}= \sqrt{n}(1+\dfrac{1}{2\sqrt{n}} + o(1)) = \sqrt{n} + \dfrac{1}{2} + o(1)$
| Write $y_n = u_n - \sqrt{n} - \frac{1}{2}$. Then the recurrence becomes
$$y_n = \sqrt{n + \sqrt{n} + \frac{1}{2} + y_{n-1}} - \sqrt{n + \sqrt{n} + \frac{1}{4}} = \frac{\frac{1}{4} + y_{n-1}}{\sqrt{n + \sqrt{n} + \frac{1}{2} + y_{n-1}} + \sqrt{n} + \frac{1}{2}}.$$
Now look at $8\sqrt{n}\cdot y_n$ and use that you already know $y_n \to 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integrating $\int x+1 \, dx$ using $u$-substition gives wrong answer. $\int (x+1) dx$
Substituting:
$u = x + 1$ and $du = dx$
$$\int u \,du = \frac{u^2}{2}$$
Which gives:
$$\int (x+1) \, dx = \frac{(x+1)^2}{2}$$
This is completely wrong as the correct answer is $\frac{x^2}{2}+x$
Why did this happen?
| You forgot the constant C. It is very important:
$$
\int u du = \frac{u^2}{2} + C = \frac{(x+1)^2}{2} + C = \frac{x^2}{2} + x + \frac{1}{2}+ C = \frac{x^2}{2} + x + C'
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2421968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
for $abc = 1$ and $a \le b \le c$ prove that $(a+1)(c+1)>3$ This inequality has been given to me by my teacher to keep me occupied and after hours of fumbling around with it, and later trying to google it. I found nothing at all.
For any three positive real numbers $a$, $b$ and $c$, where $abc = 1$ and $a\le b \le c$, prove that:
$$(a+1)(c+1)>3.$$
| Our conditions give $c\geq1$ and $c\geq b$.
Thus, by AM-GM we obtain:
$$(a+1)(c+1)=ac+a+3\cdot\frac{c}{3}+1\geq5\sqrt[5]{ac\cdot a\cdot\left(\frac{c}{3}\right)^3}+1=$$
$$=\frac{5}{\sqrt[5]{27}}\sqrt[5]{a^2c^4}+1\geq\frac{5}{\sqrt[5]{27}}\sqrt[5]{a^2b^2c^2}+1=\frac{5}{\sqrt[5]{27}}+1>3.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\lim_{x\to 1} \frac{x+2}{x^2+1}=\frac{3}{2}$ The title is quite clear. I am required to prove $\lim_{x\to 1} \frac{x+2}{x^2+1}=\frac{3}{2}$ using the epsilon-delta definition of the limit.
Given any $\varepsilon \gt 0$, there exists a $\delta =$
Such that $0 \lt \lvert x-1 \rvert \lt \delta \Rightarrow \lvert\frac{x+2}{x^2+1} -\frac{3}{2} \rvert \lt\varepsilon$
$\frac{x+2}{x^2+1} -\frac{3}{2}= \frac{-3x^2+2x+1}{2(x^2+1)}$
I have managed to express both the numerator and the denominator as such
$-3x^2+2x+1 = -(x-1)(3x+1)\\
2(x^2+1) = 2x^2+2= 2(x-1)^2 + 4(x-1) +4$
Returning back to the fraction $\frac{-(x-1)(3x+1)}{2(x-1)^2 + 4(x-1) +4}$
I am unsure on how to continue and would really appreciate some guidance.
| Your inequality can be written
$$
\left|\frac{(x-1)(3x+1)}{x^2+1}\right|<2\varepsilon\tag{*}
$$
If $x>0$, then $x^2+1>1$, so $\frac{1}{x^2+1}<1$. The condition that $x>0$ is satisfied as soon as $\delta<1$, when you take $0<|x-1|<\delta$.
In this case, we have
$$
\left|\frac{(x-1)(3x+1)}{x^2+1}\right|<
\lvert(x-1)(3x+1)\rvert
$$
Since also $3x+1<7$, because $x<2$, we have
$$
\lvert(x-1)(3x+1)\rvert<7\lvert x-1\rvert<7\delta
$$
If we take $\delta=\min\{2\varepsilon/7,1\}$, we achieve what we want: for $0<|x-1|<\delta$, we have
$$
\left|\frac{(x-1)(3x+1)}{x^2+1}\right|<
\lvert(x-1)(3x+1)\rvert
<7\lvert x-1\rvert<7\delta\le7\frac{2\varepsilon}{7}=2\varepsilon
$$
The solution set of (*) may well be larger than the set described by
$$
0<\lvert x-1\rvert<\min\!\left\{\frac{2\varepsilon}{7},1\right\}
$$
but this is unimportant: what's needed for checking the correctness of a limit is that to every $\varepsilon>0$ there corresponds a $\delta>0$ et cetera: once you find a value for $\delta$ all numbers $\delta'$ with $0<\delta'<\delta$ would be good as well.
In some applications related to approximations, getting “the largest $\delta$” can be important, but not for the proof of correctness.
| {
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If $a,b$ are positive integers and $x^2+y^2\leq 1$ then find the maximum of $ax+by$ without differentiation. If $x^2+y^2\leq 1$ then maximum of $ax+by$
Here what I have done so far.
Let $ax+by=k$ . Thus $by=k-ax$.
So we can have that $$b^2x^2+(k-ax)^2 \leq b^2$$
$$b^2x^2+k^2-2akx +a^2x^2-b^2\leq 0 $$
By re-writing as a quadratic of $x$ ,
$$(a^2+b^2)x^2-2akx +k^2-b^2\leq 0 $$
Since $a^2+b^2$ is positive , above quadratic has a minimum. Thus it to be negeative it must have roots.
So $$(-2ak)^2-4(a^2+b^2)(k^2-b^2) \geq 0$$
$$a^2k^2-(a^2+b^2)k^2+(a^2+b^2)b^2 \geq 0$$
$$(a^2+b^2) \geq k^2$$
So maximum of $k$ is $\sqrt{a^2+b^2}$
Is this correct ?
If it is correct any shorter method ? Thanks in advance.
| You can consider $$(ax+by)^2+(bx-ay)^2=(a^2+b^2)(x^2+y^2)\le a^2+b^2$$ so that $$(ax+by)^2\le a^2+b^2-(bx-ay)^2$$
Equality is attained when $x^2+y^2=1$ and $bx=ay$
How did I get that second square? Well I wanted to work with the square of the target expression and to get a term in $x^2+y^2$, so I needed to swap $a$ and $b$ around so I had equivalent terms in $x$ and $y$. Then I wanted to eliminate the cross term, so I put the minus sign in to achieve that.
It is sometimes useful to pick out the cross term and eliminate the squares by doing something like $$(ax+by)^2-(ax-by)^2=4abxy$$ - for example in working with an area represented by the cross term.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Show $k\mid 12$ with $2^k=1\bmod 13$ Let $k$ be the smallest positive integer such that $2^k=1\bmod 13$. Show that $k\mid 12$.
I'm not very good at proofs and I'm confused as how to prove this. I started by saying $2^k-1=13n$. But I don't know where to go from there.
| It seems quicker just to show directly that $k=12$.
Powers of $2 \bmod 13$:
$\begin{array}{c|rl|l}
k &2^k &\bmod 13 & \text{using -ve}\\ \hline
1 && 2 \\ 2 && 4 \\ 3 && 8 \\
4 & 16\equiv\! & 3 \\
5&& 6 \\ 6&& 12 &\equiv -1\\
7 & 24\equiv\! & 11 &\equiv -2 \leftarrow 2^6\cdot 2^1\\
8 & 22\equiv\! &9 &\equiv -4 \leftarrow 2^6\cdot 2^2 \\
9 & 18\equiv\! &5 &\equiv -8 \leftarrow 2^6\cdot 2^3 \\
10 && 10 &\equiv -3 \leftarrow 2^6\cdot 2^4 \\
11 & 20\equiv\! & 7 &\equiv -6\\
12 & 14\equiv\! &1 &\equiv -(-1)
\end{array}$
The shortcut for negative equivalence values in the second half of the cycle also shown. This demonstrates that finding $p{-}1$ must indicate the halfway point of the cycle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2427362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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For $x: (1 \to b)$ where $a$ & $b$ are co prime, why is $ax \bmod b$ distinct So, out of curiosity, I was wondering why $x (1 \to b)$ where $a$ & $b$ are co prime, why is $ax \bmod b$ distinct.
For example, let $a = 5$, and $b= 8$:
$\begin{array}{c|c|c}
x & 5x & 5x\bmod 8 \\ \hline
1 & 5 & 5 \\
2 & 10 & 2 \\
3 & 15 & 7 \\
4 & 20 & 4 \\
5 & 25 & 1 \\
6 & 30 & 6 \\
7 & 35 & 3 \\
8 & 40 & 0 \\
\end{array}$
As you can see $5x \bmod 8$ is always distinct. Why?
| Suppose $ax\equiv ay\pmod{b}$, with $1\le x\le b$ and $1\le y\le b$. Without loss of generality, we can assume $x\ge y$. Then
$$
a(x-y)\equiv0\pmod{b}
$$
that is, $b\mid a(x-y)$; since $a$ and $b$ are coprime, we conclude that
$$
b\mid (x-y)
$$
but $0\le x-y<b$, so we only have one possibility, namely $x-y=0$.
In a slightly different way, we can use Bézout’s identity: find $u$ and $v$ such that $ua+vb=1$. Then
$$
uax=(1-vb)x=x-vbx\equiv x\pmod{b}
$$
Since from $ax\equiv ay\pmod{b}$ we get $uax\equiv uay\pmod{b}$, we conclude $x\equiv y\pmod{b}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2429728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the remainder when $p!$ is divided by $p+1$?
Let $p$ be a prime larger than $7$. What is the remainder when $p!$ is divided by $p+1$?
I tried plugging in the next prime (11), which doesn't help with such big numbers. Then I tried writing $\frac{p!}{p+1} = p(p-1)(p-2).../(p+1)$. Dividing each of $p$, $(p-1)$ etc individually by $(p+1)$, I always get $(p+1)$ as a remainder, and multiplying all those remainders together and dividing again by $p+1$ will give $0$, which I'm not sure is right. How can I solve this?
| Remember that if $p > 7$, then $p! = 1 \times 2 \times 3 \times \ldots \times (p - 1) \times p$.
The smallest prime factor of $p + 1$ is less than $\sqrt{p + 1}$ and clearly $\lceil \sqrt{p + 1} \rceil < p$. The largest prime factor of $p + 1$ is at most $\frac{p + 1}{2}$, which is also less than $p$.
This means that all the prime factors of $p + 1$ are included among the prime factors of $p!$, and in greater numbers. Therefore $(p + 1) \mid p!$.
For example, given $p = 11$, we have $p! = 39916800 = 2^8 \times 3^4 \times 5^2 \times 7 \times 11$. By contrast, $p + 1 = 12 = 2^2 \times 3^{(1)}$, so $$\frac{39916800}{12} = 3326400 = 2^6 \times 3^3 \times 5^2 \times 7 \times 11.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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Compute: $\arctan{\frac{1}{7}}+\arctan{\frac{3}{4}}.$
Compute: $\arctan{\frac{1}{7}}+\arctan{\frac{3}{4}}.$
I want to compute this sum by computing one term at a time. It's clear that
$$\arctan{\frac{1}{7}}=A \Longleftrightarrow\tan{A}=\frac{1}{7}\Longrightarrow A\in\left(0,\frac{\pi}{2}\right).$$
Drawing a right triangle with sides $1$, $7$ and $5\sqrt{2},$ I get that $$\begin{array}{lcl}
\sin{A} & = & \frac{1}{5\sqrt{2}}\Longleftrightarrow A= \arcsin{\frac{1}{5\sqrt{2}}} \\
\cos{A} & = & \frac{7}{5\sqrt{2}}\Longleftrightarrow A= \arccos{\frac{7}{5\sqrt{2}}} \\
\end{array}$$
But this will not get me standard angles for $A.$
| $$\tan\left(\arctan\frac{1}{7}+\arctan\frac{3}{4}\right)=\frac{\frac{1}{7}+\frac{3}{4}}{1-\frac{1}{7}\cdot\frac{3}{4}}=\frac{4+21}{28-3}=1$$
and since $0^{\circ}<\arctan\frac{1}{7}+\arctan\frac{3}{4}<90^{\circ}$, we get the answer:
$$\arctan\frac{1}{7}+\arctan\frac{3}{4}=45^{\circ}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving an equation about exponential of matrices
Let $A, B$ be $n\times n$ matrices over $\mathbb{C}$. Assume that
$$[A,[A,B]]=[B,[A,B]] = 0$$
Show
$$e^{\epsilon A}Be^{-\epsilon A} = B + \epsilon[A,B]$$
Then, I have to show the equation below. Here $[A, B]$ is equal to $AB-BA$ and $\epsilon$ is a real parameter. I can't find a way to show the equation below... Could anyone explain to me in detail?
| $\textbf{Proof}$. Since
$$
e^{\varepsilon A} = \sum_{k=0}^{\infty} \dfrac{(\varepsilon A)^k}{k!}
= 1 + \varepsilon A + \frac{\varepsilon^2 A^2}{2!} +
\frac{\varepsilon^3 A^3}{3!} + \ldots,
$$
we have
$$
e^{-\varepsilon A} = \sum_{k=0}^{\infty} \dfrac{(-\varepsilon A)^k}{k!}
= 1 - \varepsilon A + \frac{\varepsilon^2 A^2}{2!} -
\frac{\varepsilon^3 A^3}{3!} + \ldots.
$$
So
\begin{align*}
e^{\varepsilon A} B e^{-\varepsilon A}
&= \left(1 + \varepsilon A + \frac{\varepsilon^2 A^2}{2!} +
\frac{\varepsilon^3 A^3}{3!} + \ldots\right)B \left( 1 - \varepsilon A + \frac{\varepsilon^2 A^2}{2!} -
\frac{\varepsilon^3 A^3}{3!} + \ldots\right) \\
&= \left(1 + \varepsilon A + \frac{\varepsilon^2 A^2}{2!} +
\frac{\varepsilon^3 A^3}{3!} + \ldots\right) \left( B - \varepsilon B A + \frac{\varepsilon^2 B A^2}{2!} -
\frac{\varepsilon^3 B A^3}{3!} + \ldots \right) \\
&= B + \varepsilon (AB-BA) + 2! \varepsilon^2 (A^2B-2ABA + BA^2)
+3! \varepsilon^3[A, [A,[A,B]] ]
+ \ldots \\
&= B + \varepsilon [A,B] + 2! \varepsilon^2 [A,[A,B]] +3! \varepsilon^3[A, [A,[A,B]] ]
+ \ldots \\
&= B + \varepsilon [A,B]
\end{align*}
since $[A,[A,B]]= A(AB-BA)-(AB-BA)A=A^2B-2ABA + BA^2=0$. $\hspace{2cm}\square$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Characteristic $2$ ring $A$ with $x^{2^k +1}=x$ for some $k$ for every $x \in A$. Show that $x^2 = x$.
Let $A$ be a ring of characteristic $2$ with $x^{2^k +1}=x$ for some $k$ for every $x \in A$ (i.e. $k$ is specific to $x$). Show that $x^2 = x$ for every $x \in A$.
I have been fiddling about for a while now but have not had much luck. I have to tried to use the fact $A$ is characteristic 2 e.g. by looking at different bracket expansions, however since we don't know any properties about $A$ (e.g. domain?) I can't seem to equate these expressions to anything well known or nice. Perhaps I am missing an obvious short rearrangement trick.
Any hints/ideas would be appreciated.
| We start with a lemma: By Theorem 2 of Binomial coefficients modulo a prime (Fine, 1947), we have that if the binary expansion of the positive integer $N$ has $m$ ones on it, then the number of odd entries of the $N$-th row of Pascal's Triangle is $2^m$. In particular, if $N=2^k$ then $m=1$ and $2^m=2$, so the only odd entries are the $1$s from the extremes. Similarly, if $N=2^k+1$ (which is odd) then $m=2$ and $2^m=4$, so the only odd entries are $1, N, N, 1$ at the extremes.
Now let $k$ and $n$ be such that $x^{2^k+1}=x$ and $(x+1)^{2^n+1}=x+1$.
By the lemma we have $$x+1=(x+1)^{2^n+1} = x^{2^n+1}+x^{2^n}+x+1,$$ which implies $x^{2^n+1}+x^{2^n}=0$, or $$x^{2^n+1}=x^{2^n}.$$
By the lemma we also have $$(x+1)^{2^k+1} = x^{2^k+1}+x^{2^k}+x+1 = x+x^{2^k}+x+1 = x^{2^k}+1 = (x+1)^{2^k},$$ the last step again by our lemma.
We have to put these results together. We will use that, for a big enough power of $x$, the chain of powers of $x$ stabilizes, while for a bigger power $M$ of $x$, the "cycle is closed" and we get $x^M=x$.
Suppose $n<k$, so that $2^n<2^k+1$. Since $x^{2^n}=x^{2^n+1}$, we infer $x^{2^n}=x^s$ for every $s>2^n$, hence
$$x=x^{2^k+1}=x^{2^n}=x^{2\cdot 2^n}=(x^{2^n})^2=x^2.$$
If on the contrary $n\geq k$, then as $(x+1)^{2^k}=(x+1)^{2^k+1}$, we infer $(x+1)^{2^k}=(x+1)^s$ for every $s>2^k$, so that
$$x+1=(x+1)^{2^n+1}=(x+1)^{2^k}=(x+1)^{2\cdot 2^k} = ((x+1)^{2^k})^2 = (x+1)^2 = x^2+1,$$
which clearly implies $x=x^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $m = \sqrt {n +\sqrt {n +\sqrt {n + \cdots}}}$ is an integer if and only if $n$ is equal to twice a triangular number. $m = \sqrt {n +\sqrt {n +\sqrt {n + \cdots}}}$
Prove that $m$ is an integer if and only if $n$ is equal to twice a triangular number.
Supposing that $n$ is twice a triangular number. Then I have that $m = \sqrt {2t +m}$ for some traingular number $t$ thus $m^2-m-2t=0$ so, by the quadratic formula $m=\dfrac{1+\sqrt{8t+1}}{2}$
So I need to show that $8t +1$ is a square for any $t$. If so than it's square root is clearly an odd number so $1+\sqrt{8t+1}$ is clearly even and therefore divisible by $2$. So $m$ is an integer.
But I'm unsure how to show this.
| We have
\begin{eqnarray*}
m= \sqrt{ n+ \sqrt{n+ \sqrt{n+ \cdots}}} \\
\end{eqnarray*}
So $m= \sqrt{n+m}$. Square this and solve the quadratic
\begin{eqnarray*}
m= \frac{-1 + \sqrt{1+4n}}{2}
\end{eqnarray*}
So $1+4n$ will need to be a perfect square to deal with the square root; indeed a square of an odd value to deal with the division by $2$. So $1+4n =(2k+1)^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find limits of sequences Prove that
$$a) \lim\limits_{n \to \infty} (\frac{1^p+2^p+...+n^p}{n^p} - \frac{n}{p+1})=\frac{1}{2},$$
$$b) \lim\limits_{n \to \infty} \frac{1^p+3^p+...+(2n-1)^p}{n^{p+1}}=\frac{2^p}{p+1},$$
where is $p \in \Bbb N $.
Thanks to Stolz–Cesàro theorem in $a)$ I went to $$\lim\limits_{n \to \infty} \frac{(n+1)^p}{(n+1)^p-n^p}$$ which after dividing by $n^p$ goes to
$$\lim\limits_{n \to \infty} \frac{1+\frac{p-1}{n}+\frac{p(p-1)}{2n^2}+...+\frac{1}{n^p}}{\frac{p-1}{n}+\frac{p(p-1)}{2n^2}+...+\frac{1}{n^p}} = \frac{1}{\infty}$$
At the same time $$\lim\limits_{n \to \infty} \frac{n}{p+1}=\infty,$$
so initial limit should be $\frac{1}{\infty}-\infty = ?$
I can't figure out what I'm missing ( $b)$ seems to complicated to me so I didn't even try it).
| Solution for (b):
$$\lim_n\frac{1^p+3^p+\cdots +(2n-1)^p}{n^{p+1}}=\lim_n \frac{1^p+2^p+\cdots +(2n)^p}{n^{p+1}}-\lim_n\frac{2^p+4^p+\cdots+(2n)^p}{n^{p+1}}$$
$$=2^{p+1}\lim_n\frac{1^p+2^p+\cdots +n^p}{n^{p+1}}-2^p\lim_n\frac{1^p+2^p+\cdots+n^p}{n^{p+1}}$$
$$=(2^{p+1}-2^p).\lim_n\frac{1^p+2^p+\cdots+n^p}{n^{p+1}}=2^p.\lim_n\frac{1}{n}\sum_{r=1}^n \left(\frac{r}{n}\right)^p$$
$$=2^p.\int_0^1x^p\,dx=\frac{2^p}{p+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2434998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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The asymptotic expansion of $x^x$ at $0$. If we put $x^x$ in Wlofram Alpha, we get the following:
$$
x^x = 1 + x \log(x) + \frac12 x^2 \log^2(x) + \frac16 x^3 \log^3(x) + \frac1{24} x^4 \log^4(x) + \frac1{120} x^5 \log^5(x) + O(x^6)
$$
Is there a name of this form of expansion and is there a systematic way to find such expansions?
| We know that
$$x^x=e^{x\log x}$$
Also by Taylor's Series,
$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots$$
So
$$\begin{align}
x^x&=e^{x\log x}\\
&=1+x\log x+\frac{(x\log x)^2}{2!}+\frac{(x\log x)^3}{3!}+\frac{(x\log x)^4}{4!}+\frac{(x\log x)^5}{5!}+\cdots\\
&=1+x\log x+\frac{1}{2}x^2\log^2 x+\frac{1}{6}x^3\log^3x+\frac{1}{24}x^4\log^4x+\frac{1}{120}x^5\log^5x+O(x^6)
\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Inverting a $2 \times 2$ matrix $\mod 26$ I am trying to invert the matrix (in mod 26)
\begin{bmatrix}
19 & 7\\
19 & 0 \\
\end{bmatrix}
I compute the determinant
((19*0)-(19*7)=23 (Reducing in mod 26)
\begin{bmatrix}
0 & -7\\
-19 & 19 \\
\end{bmatrix}
Switch a & d and negate b & c
Multiply by 23^-1 = 17
To get
\begin{bmatrix}
0 & 11\\
15 & 25 \\
\end{bmatrix}
But this can't be right
since wolfram gives me
\begin{bmatrix}
0 & 15\\
11 & 11\\
\end{bmatrix}
What could I be doing wrong?
| Let $A$ be the given matrix. Then $AA^{-1}=I_2$. Hence, we must row reduce the augmented system:
$$\begin{bmatrix}19&7&|&1&0\\19&0&|&0&1\end{bmatrix}\to\begin{bmatrix}19&7&|&1&0\\0&-7&|&-1&1\end{bmatrix}\to \begin{bmatrix}19&0&|&0&1\\0&-7&|&-1&1\end{bmatrix}\to\begin{bmatrix}1&0&|&0&11\\0&1&|&-11&11\end{bmatrix}$$
using $19^{-1}=11$ and $-7=19$ in $\Bbb Z_{26}$.
Hence, $$A^{-1}=\begin{bmatrix}0&11\\-11&11\end{bmatrix}=\begin{bmatrix}0&11\\15&11\end{bmatrix}\qquad\text{in $\Bbb Z_{26}$ since $-11=15$}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Draw the function $y=\arctan{\frac{e^x-1}{\sqrt{3}}}-\arctan{\frac{e^x-4}{e^x\sqrt{3}}}$ I'm not allowed to use differentiation for this. I know the domain and range of the arctan function. The function $\arctan:\mathbb{R}\rightarrow \left(-\frac{\pi}{2},\frac{\pi}{2}\right).$ Is my job to see what happens to the arguments $\frac{e^x-1}{\sqrt{3}}$ and $\frac{e^x-4}{e^x\sqrt{3}}$ in this range? I'm sorry for not having a better question but I'm stuck.
| Note:If you find $f'$ when $f(x)=y=\arctan{\frac{e^x-1}{\sqrt{3}}}-\arctan{\frac{e^x-4}{e^x\sqrt{3}}}$ , you will see
$f'(x)=0$ so $f(x)$ is a constant function
$$f(x)=\arctan{\frac{e^x-1}{\sqrt{3}}}-\arctan{\frac{e^x-4}{e^x\sqrt{3}}}=\\
\arctan{\frac{e^x-1}{\sqrt{3}}}-\arctan{\frac{e^{-x}(e^x-4)}{\sqrt{3}}}=\\
\arctan{\frac{e^x-1}{\sqrt{3}}}-\arctan{\frac{1-4e^{-x}}{\sqrt{3}}}\\$$I tis easier to work .
$$f'(x)=\frac{\bigg(\dfrac{e^x-1}{\sqrt{3}}\bigg)'}{1+\bigg(\dfrac{e^x-1}{\sqrt{3}}\bigg)^2}-\frac{\bigg(\dfrac{1-4e^{-x}}{\sqrt{3}}\bigg)'}{1+\bigg(\dfrac{1-4e^{-x}}{\sqrt{3}}\bigg)^2}\\=
\frac{\frac{e^x}{\sqrt{3}}}{1+\frac{(e^x-1)^2}{3}}-\frac{\frac{4e^{-x}}{\sqrt{3}}}{1+\frac{(1-4e^{-x})^2}{3}}
$$ to simplify $\frac{\frac{4e^{-x}}{\sqrt{3}}}{1+\frac{(1-4e^{-x})^2}{3}} $ multiply by $\frac{e^{2x}}{e^{2x}}$ hence:
$$f'=\frac{\sqrt3 e^x}{4+e^{2x}-2e^x}-\frac{4\sqrt3 e^{-x}}{4+16e^{-2x}-8e^{-x}}\\=
\frac{\sqrt3 e^x}{4+e^{2x}-2e^x}-\frac{4\sqrt3 e^{-x}}{4(1+4e^{-2x}-2e^{-x}}.\frac{e^{2x}}{e^{2x}}\\=
\frac{\sqrt3 e^x}{4+e^{2x}-2e^x}-\frac{\not4\sqrt3 e^{+x}}{\not4(1.e^{2x}+4e^{-2x+2x}-2e^{-x+2x}}.\frac{e^{2x}}{e^{2x}}\\=
\dfrac{\sqrt3 e^x}{4+e^{2x}-2e^x}-\frac{\sqrt3 e^x}{4+e^{2x}-2e^x} \\=0 \checkmark$$ so $\forall x \in \mathbb{R} :f'=0\implies f(x)=const$ now check a number in $f(x)$
forexample $x=0$
$$f(0)=\arctan{\frac{1-1}{\sqrt{3}}}-\arctan{\frac{1-4}{1.\sqrt{3}}}=0-(-\frac{\pi}{3})=\frac{\pi}{3}$$
A trig way $$\arctan{\frac{e^x-1}{\sqrt{3}}}-\arctan{\frac{1-4e^{-x}}{\sqrt{3}}}=a-b \\ \tan(a-b)=\frac{\tan a -\tan b}{1+\tan a . \tan b}=\\\frac{\frac{e^x-1}{\sqrt{3}} -\frac{1-4e^{-x}}{\sqrt{3}}}{1+\frac{e^x-1}{\sqrt{3}} . \frac{1-4e^{-x}}{\sqrt{3}}}=\sqrt{3}\underbrace{\frac{e^x-1-1+4e^{-x}}{3+(e^x-1).(1-4e^{-x})}}_{1}=\\\sqrt 3 \implies \tan(a-b)=\arctan \sqrt3=\frac{\pi}{3}|$$
| {
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In how many ways can $5$ students be chosen such that at least one student is chosen from each state? There are $3$ states and $3$ students representing each state. In how many ways can $5$ students be chosen such that at least one student is chosen from each state?
According to me, the answer should be $3C1 \cdot 3C1 \cdot 3C1 \cdot 6C2$, but this answer is wrong.
Can someone please help me?
| Your answer over counts.
Suppose the representatives of state $A$ are $a, b, c$, the representatives of state $B$ are $d, e, f$, and the representatives of state $C$ are $g, h, i$. You count the selection $\{a, b, c, d, g\}$ three times, once for each of the ways we could choose a representative of state $A$.
$a, d, g, \qquad b, c$
$b, d, g, \qquad a, c$
$c, d, g, \qquad a, b$
and the selection $\{a, b, d, e, g\}$ four times, once for each of the ways we could pick a representative of state $A$ and once for each of the ways we could pick a representative of state $B$.
$a, d, g, \qquad b, e$
$a, e, g, \qquad b, d$
$b, d, g, \qquad a, e$
$b, e, g, \qquad a, d$
If we do a direct count, the number of permissible selections is
$$\binom{3}{3}\binom{3}{1}\binom{3}{1} + \binom{3}{1}\binom{3}{3}\binom{3}{1} + \binom{3}{1}\binom{3}{1}\binom{3}{3} + \binom{3}{2}\binom{3}{2}\binom{3}{1} + \binom{3}{2}\binom{3}{1}\binom{3}{2} + \binom{3}{1}\binom{3}{2}\binom{3}{2} = 3\binom{3}{3}\binom{3}{1}^2 + 3\binom{3}{2}^2\binom{3}{1}$$
Your answer $\binom{3}{1}\binom{3}{1}\binom{3}{1}\binom{6}{2}$ counts each of the first three terms three times and each of the last three terms four times. Notice that
$$3 \cdot 3\binom{3}{3}\binom{3}{1}^2 + 4 \cdot 3 \binom{3}{2}^2\binom{3}{1} = \binom{3}{1}^3\binom{6}{2}$$
As RobertZ and Muralidharan indicated, it is easier to subtract the number of cases in which exactly two states are represented from the total number of ways of selecting five representatives from the nine available people.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2438640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Compute $\lim_{x\rightarrow -\infty}\left(\sqrt{x^2-6x+9}+x-1\right)$ I note that $\sqrt{x^2-6x+9}=|x-3|$. Splitting upp the limit into cases gives
*
*$x\geq 3:$
$$\lim_{x\rightarrow -\infty}\left(\sqrt{x^2-6x+9}+x-1\right)=\lim_{x\rightarrow -\infty}(|x-3|+x-1)=2\lim_{x\rightarrow -\infty}(x-2)=-\infty.$$
*
*$x< 3:$
$$\lim_{x\rightarrow -\infty}\left(\sqrt{x^2-6x+9}+x-1\right)=\lim_{x\rightarrow -\infty}(-x+3+x-1)=\lim_{x\rightarrow -\infty}2=2.$$
I get two different values of the limit, but plotting the function clearly shows that the answer should be $2$.
| Since $\lim_{x \rightarrow - \infty}$ it suffices to consider 'small' negative $x$.
$|x-3| = -x + 3$.
Example: $x= -7$:
$|x-3| = |-7-3| = -x +3.$
Hence:
$\lim_{x \rightarrow -\infty} (|x-3| + x-1) = $
$\lim_{x \rightarrow -\infty}( -x +3 +x -1 )= 2 $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2440429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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In how many ways can a person score a total of $130$ marks in an exam consisting of $3$ sections of $50$ marks each? In how many ways can a person score a total of $130$ marks in an exam consisting of $3$ sections of $50$ marks each?
What I did:-
$$a+b+c=130$$
So applying partition rule
$n=130$ and $130$ marks has to be divided into three groups so $r=$
$132C2$
Now we will subtract the case when $a$, $b$ or $c$ is $51$, i.e. more than $50$. Suppose $a$ is $51$.
\begin{align*}
51+b+c & = 130\\
b+c & = 79
\end{align*}
Now $n=79$ and $r=2$.
$80C1$
Similarly, for $b$ and $c$ also $= 80C1$
Final answer $= 132C2-3 \cdot (80C1)$
But actual answer is $132C2-3 \cdot (81C2)+3 \cdot (30C2)$
Please someone explain me this. What is wrong with my answer? Why are we adding?
| We wish to solve the equation
$$a + b + c = 130 \tag{1}$$
in the nonnegative integers subject to the restrictions that $a, b, c \leq 50$.
As you observed, equation 1 has
$$\binom{130 + 2}{2} = \binom{132}{2}$$
solutions in the nonnegative integers.
From these, we must exclude those in which or more of the variables exceeds $50$. Notice that at most two variables can exceed $50$ since $3 \cdot 51 = 153 > 130$.
Suppose $a > 50$. Then $a' = a - 51$ is a nonnegative integer. Substituting $a' + 51$ for $a$ in equation 1 yields
\begin{align*}
a' + 51 + b + c & = 130\\
a' + b + c & = 79 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers with
$$\binom{79 + 2}{2} = \binom{81}{2}$$
solutions. By symmetry, there are the same number of solutions in which $b > 50$ or $c > 50$. Hence, there are
$$\binom{3}{1}\binom{81}{2}$$
solutions in which a variable exceeds $50$.
However, if we subtract $\binom{3}{1}\binom{81}{2}$ from the total, we will have subtracted too much since we have counted those cases in which two variables exceed $50$ twice, once for each way we could designate one of the variables as the one that exceeds $50$. We only want to subtract those cases once, so we must add them back.
Suppose $a > 50$ and $b > 50$. Let $a' = a - 51$ and let $b' = b - 51$. Then $a'$ and $b'$ are nonnegative integers. Substituting $a' + 51$ for $a$ and $b' + 51$ for $b$ in equation 1 yields
\begin{align*}
a' + 51 + b' + 51 + c & = 130\\
a' + b' + c & = 28 \tag{3}
\end{align*}
Equation 3 is an equation in the nonnegative integers with
$$\binom{28 + 2}{2} = \binom{30}{2}$$
solutions. By symmetry, there are an equal number of solutions in which $a$ and $c$ both exceed $50$ and in which $b$ and $c$ both exceed $50$. Thus, there are
$$\binom{3}{2}\binom{30}{2}$$
solutions in which two of the variables exceed $50$.
By the Inclusion-Exclusion Principle, the number of ways a person could obtain a score of $130$ marks on an examination consisting of three parts worth $50$ marks each is
$$\binom{132}{2} - \binom{3}{1}\binom{81}{2} + \binom{3}{2}\binom{30}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2441521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A median of a triangle is the geometric mean of the adjacent sides; find the cosine of one angle in terms of the others
$AD$ is a median of $\triangle ABC$. $|AD|$ is the geometric mean of $|AB|$ and $|AC|$.
Find $1+\cos A$ in terms of $\cos B$ and $\cos C$.
Edit
This is the second part of the question
Also prove that 1+cosA=√2|cosB-cosC|.
| $$\frac{1}{2}\sqrt{2b^2+2c^2-a^2}=\sqrt{bc},$$ which gives
$$a^2=2b^2+2c^2-4bc.$$
Thus, $b\neq c$ and
$$\cos\beta\cos\gamma=\frac{a^2+c^2-b^2}{2ac}\cdot\frac{a^2+b^2-c^2}{2ab}=$$
$$=\frac{(2b^2+2c^2-4bc+c^2-b^2)(2b^2+2c^2-4bc+b^2-c^2)}{4(2b^2+2c^2-4bc)bc}=$$
$$=\frac{(3c^2-4bc+b^2)(3b^2-4bc+c^2)}{8(b-c)^2bc}=\frac{(b-c)^2(3b-c)(b-3c)}{8bc(b-c)^2}=$$
$$=\frac{3(b^2+c^2)-10bc}{8bc}=\frac{3}{8}\left(\frac{b}{c}+\frac{c}{b}\right)-\frac{5}{4}.$$
Thus, $$\frac{b}{c}+\frac{c}{b}=\frac{10+8\cos\beta\cos\gamma}{3}.$$
Id est, $$1+\cos\alpha=1+\frac{b^2+c^2-a^2}{2bc}=1+\frac{b^2+c^2-2b^2-2c^2+4bc}{2bc}=$$
$$=3-\frac{1}{2}\left(\frac{b}{c}+\frac{c}{b}\right)=3-\frac{5+4\cos\beta\cos\gamma}{3}=\frac{4}{3}(1-\cos\beta\cos\gamma).$$
About your second problem.
$$1+\cos\alpha=1+\frac{b^2+c^2-a^2}{2bc}=$$
$$=1+\frac{b^2+c^2-2b^2-2c^2+4bc}{2bc}=3-\frac{1}{2}\left(\frac{b}{c}+\frac{c}{a}\right).$$
Also, we see that $3-\frac{1}{2}\left(\frac{b}{c}+\frac{c}{a}\right)\geq0$.
In another hand,
$$\sqrt2|\cos\beta-\cos\gamma|=\sqrt2\left|\frac{a^2+c^2-b^2}{2ac}-\frac{a^2+b^2-c^2}{2ab}\right|=$$
$$=\sqrt2\left|\frac{2b^2+2c^2-4bc+c^2-b^2}{2ac}-\frac{2b^2+2c^2-4bc+b^2-c^2}{2ab}\right|=$$
$$=\sqrt2\left|\frac{b^2-4bc+3c^2}{2ac}-\frac{3b^2-4bc+c^2}{2ab}\right|=$$
$$=\sqrt2|b-c|\cdot\left|\frac{b-3c}{2ac}-\frac{3b-c}{2ab}\right|=$$
$$=\frac{\sqrt2|b-c|}{2abc}\cdot\left|(b-3c)b-(3b-c)c\right|=$$
$$=\frac{\sqrt2|b-c|}{2\sqrt2|b-c|bc}\cdot\left|b^2+c^2-6bc\right|=$$
$$=3-\frac{1}{2}\left(\frac{b}{c}+\frac{c}{a}\right)$$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2442259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finding equation of circle with the given restrictions
Tangents are drawn to circle $x^2+y^2-6x-4y-11=0$ from point $P(1,8)$ touching circle at $A$ and $B$. Let there be a circle whose radius passes through point of intersection of circles $x^2+y^2-2x-6y+6=0$ and $x^2+y^2+2x-6y+6=0$ and intersect the circumcircle of $PAB$ orthogonally. Find minimum radius of such a circle.
My attempt
Circumcircle would be $(x-1)(x-3)+(y-2)(y-8)=0$ and centre would lie on radical axis of the given circles ie $x=0$ .
therefore equation of circle will be $x^2+y^2+2fx+c=0$. Applying orthogonality condition , $c=-19-10f$ .
I am unable to find minimum radius .
Please suggest any ways to minimise the radius or any better solution if possible .
| So here's how I tried to solve it :
Since the circle is passing through the point of intersection of the two circles.We can write the family of circles passing through them by S1 + $\omega $ S2, where $\omega $ is a parameter which produces the infinite circle passing through the point of intersection of the two circles.
$\ x^2+y^2-2x+6y+6 $ + $\omega $( x^2+y^2+2x-6y+6 ) = 0
Simplifying we get :
$\ x^2(1+\omega) $+ $y^2$(1+$\omega $) +x(2$\omega $ - 2) +y(-6$\omega $ - 6) + 6 + 6$\omega = 0 $
Since this is a circle it has its centre at $\ (1-\omega),(3\omega + 3) $
Now it is given in the question that a circle from this family is orthogonal to the given circle:
Note: Here I mention first circle as the one you wrote in the diameter form.
$\ r_1^2 + r_2^2 = d^2 $ where $\ r_1 $ is the radius of the first circle and $\ r_2 $ is the radius of the second circle, d is the distance between the two centres of the circles.
First circle has centre (2,5) and radius $\sqrt 10 $
Putting in the values in the formula we get :
$\ 10 $ + (1 - $\omega $)^2 + (3 $\omega $ + 3 - 5)^2 - 6 - 6$\omega $ = (1- $\omega $ - 2)^2 + (3 $\omega $ + 3 - 5) ^2:
Simplifying it all we get $\omega $ = -1/4.
Which gives us the circle when we put the value of $\omega $ in the family of circle to get the required circle.
We get the radius (root of 77 )/4
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find a quadratic equation with integral coefficients whose roots are $\frac{α}{β}$ and $\frac{β}{α}$ The roots of the equation $2x^2-3x+6=0$ are α and β. Find a quadratic equation with integral coefficients whose roots are $\frac{α}{β}$ and $\frac{β}{α}$.
The answer is $4x^2+5x+4=0$
I don't know how to get to the answer. Could someone explain the steps?
| Rename $a= \alpha $ and $b=\beta$
So this is $(x-{a\over b})(x-{b\over a})=0$ thus
$$x^2-{a^2+b^2\over ab}x+1=0$$
Since $ab = {6\over 2}=3$ and $a^2+b^2 = (a+b)^2-2ab = ({3\over 2})^2-6 = {-15\over 4} $ we get:
$$x^2+{5\over 4}x+1=0$$
and thus the conclusion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2446243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\frac{a^3}{b^2} + \frac{b^3}{c^2} + \frac{c^3}{a^2} \geq 3 \, \frac{a^2 + b^2 + c^2}{a + b + c}$ Proposition
For any positive numbers $a$, $b$, and $c$,
\begin{equation*}
\frac{a^3}{b^2} + \frac{b^3}{c^2} + \frac{c^3}{a^2}
\geq 3 \, \frac{a^2 + b^2 + c^2}{a + b + c} .
\end{equation*}
I am requesting an elementary, algebraic explanation to this inequality. (I suppose the condition for equality is that $a = b = c$.) I am not familiar with symmetric inequalities in three variables. I would appreciate any references.
| I found an AM-GM proof$:$
$$\text{LHS}-\text{RHS}=\frac{1}{a+b+c} \sum \Big({\dfrac {{a}^{3}}{b}}+{\dfrac {{a}^{4}}{14{b}^{2}}}+{
\dfrac {{a}^{3}c}{{b}^{2}}}+{\dfrac {11{b}^{4}}{14{c}^{2}}
}+{\dfrac {{c}^{4}}{7{a}^{2}}}-3{a}^{2}\Big)\geqslant0,$$
which is obvious by AM-GM.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2446312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
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The area of curve $y^3=x$ bounded by lines. Find the area of region on $xy$ plane shaded by curve $y^3=x$ and lines $y=1$ and $x=8$
My solution: The line $x=8$ intersect curve at point $y=2$ and line $y=1$ intersect at point x=1. So the intended area is $$=\left(1-\int \limits_{0}^{1}y^3dy\right)+7=8-\frac{1}{4}=7,75$$
However, the answer in the book is $4,25$.
Can anyone point out where is my mistake with graphs? I have rechecked many times and did not detected it.
| Referring to the graph:
1-method:
$$S=\int_{x=1}^{x=8} x^{1/3}dx-7\cdot 1=\left(\frac{3}{4} x^{4/3}\right) \bigg|_1^8-7=\left(12-\frac34\right)-7=4\frac14.$$
2-method:
$$S=8\cdot 1-\int_{y=1}^{y=2} y^3dy=8-\left(\frac{1}{4} y^4\right) \bigg|_1^2=8-\left(4-\frac14\right)=4\frac14.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $\sum_{k=1}^\infty \frac{2^k a^{2^k}}{1+a^{2^k}}=\frac{a}{1-a}$ for $-1How to prove $$\sum_{k=1}^\infty \frac{2^k a^{2^k}}{1+a^{2^k}}=\frac{a}{1-a}$$ for $-1<a<1$?
It gives a hint: $2^{k+1}-1=1+2+\cdots+2^k$. How can we use this hint to cancel some terms?
| Here is a brutal-force computation: Expand everything and group according to the exponent of $a$.
$$
\sum_{k=0}^{\infty} \frac{2^k a^{2^k}}{1+a^{2^k}}
= \sum_{k=0}^{\infty} 2^k \sum_{j=1}^{\infty} (-1)^{j-1} a^{j \cdot 2^k}
= \sum_{n=1}^{\infty} \left( \sum_{(j,k) \ : \ j \cdot 2^k = n} (-1)^{j-1} 2^k \right) a^{n}.
$$
Now write $n = m \cdot 2^r$, where $m$ is odd and $r \geq 0$. Then
*
*If $0 \leq k < r$, then $j = m \cdot 2^{r-k}$ is even and hence $(-1)^{j-1} = -1$.
*If $k = r$, then $j = m$ is odd and hence $(-1)^{j-1} = 1$.
So we have
$$ \sum_{(j,k) \ : \ j \cdot 2^k = n} (-1)^{j-1} 2^k
= 2^r - \sum_{j=0}^{r-1} 2^j
= 1 $$
and hence
$$
\sum_{k=0}^{\infty} \frac{2^k a^{2^k}}{1+a^{2^k}}
= \sum_{n=1}^{\infty} a^n
= \frac{a}{1-a}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Number of Trailing Zeros of Binomial Coefficient If $x+2=18181818...$ $n$ $digits$, find the number zeros at the end of ${x \choose x/2}$. I have tried using Legendre's formula for factorials, but I have got nowhere because of the strange value of $x$.
| We want to find how many Trailing zeros for $\binom{2n}{n}$.
First Write the binomial as factorials $\frac{(2n)!}{(n!)^2}$
And apply Legendre formula for every term separately.
For $(2n)!$ there are $\sum \limits_{k=1}^{\infty} \lfloor \frac{2n}{2^k} \rfloor $ $2$'s and $\sum \limits_{k=1}^{\infty} \lfloor \frac{2n}{5^k} \rfloor $ $5$'s
Now for $(n!)^2$ it have the same powers of primes as $n!$ but they are multiplied by $2$ (simple powers rule), so there are $\sum \limits_{k=1}^{\infty} 2\lfloor \frac{n}{2^k} \rfloor $ $2$'s and $\sum \limits_{k=1}^{\infty} 2\lfloor \frac{n}{5^k} \rfloor $ $5$'s
So for the expression $\frac{(2n)!}{(n!)^2}$ there are $\sum \limits_{k=1}^{\infty} \lfloor \frac{2n}{2^k} \rfloor-2 \lfloor \frac{n}{2^k} \rfloor $ $2$'s and $\sum \limits_{k=1}^{\infty} \lfloor \frac{2n}{5^k} \rfloor -2\lfloor \frac{n}{5^k} \rfloor$ $5$'s (powers in denominator are subtracted ,simple powers rule).
So $\frac{(2n)!}{(n!)^2} $ there are $Min(\sum \limits_{k=1}^{\infty} \lfloor \frac{2n}{2^k} \rfloor-2 \lfloor \frac{n}{2^k} \rfloor ,\sum \limits_{k=1}^{\infty} \lfloor \frac{2n}{5^k} \rfloor-2 \lfloor \frac{n}{5^k} \rfloor )$ Trailing Zeros,
Checked for first $1000$ numbers and its correct.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the inequality for $x$: $\log_4 (x^2 − 2x + 1) < \log_2 3$ Solve the inequality for $x$: $$\log_4 (x^2 − 2x + 1) < \log_2 3$$
I got two answers and I'm not sure if I did it correctly.
1st ans: $(-2,1)\cup (1,4)$
2nd ans: $x \neq -2,4$
| \begin{array}{c}
\log_4 (x^2 − 2x + 1) < \log_2 3 \\
x^2 − 2x + 1 < 4^{\log_2 3} \\
0<(x-1)^2 < 2^{2\log_2 3} \\
0<(x-1)^2 < 2^{\log_2 3^2} \\
0<(x-1)^2 < 3^2 \\
\text{$-3 < x-1 < 3$ and $x \ne 1$} \\
\text{$-2 < x < 4$ and $x \ne 1$} \\
x \in (-2,1) \cup (1,4)
\end{array}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2451064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the matrix $X$ in $AX$
Given that $A$ is a $3$ x $3$ matrix such that $$ A \begin{pmatrix}
1 \\
0 \\
0 \\
\end{pmatrix} = \begin{pmatrix}
1 \\
1 \\
1 \\
\end{pmatrix} $$ and $$ A \begin{pmatrix}
0 \\
0 \\
1 \\
\end{pmatrix} = \begin{pmatrix}
0 \\
0 \\
1 \\
\end{pmatrix} $$
Find a matrix $X$ such that
$$A X = \begin{pmatrix}
1 & 0 & 4 \\
1 & 0 & 4 \\
1 & 0 & 7 \\
\end{pmatrix}$$
How do I approach this question? From the two matrix equations given above, I can find the first and the third columns, but I don't think I have the second column to find matrix $A$. Thanks.
| You can take$$X=\begin{pmatrix}1&0&4\\0&0&0\\0&0&3\end{pmatrix}.$$
And how do I know that? Note that$$X.\begin{pmatrix}1\\0\\0\end{pmatrix}=\begin{pmatrix}1\\0\\0\end{pmatrix}\implies AX\begin{pmatrix}1\\0\\0\end{pmatrix}=\begin{pmatrix}1\\1\\1\end{pmatrix},$$that$$X.\begin{pmatrix}0\\1\\0\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}\implies AX.\begin{pmatrix}0\\1\\0\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix},$$and that$$X.\begin{pmatrix}0\\0\\1\end{pmatrix}=4\begin{pmatrix}1\\0\\0\end{pmatrix}+3\begin{pmatrix}0\\0\\1\end{pmatrix}\implies AX.\begin{pmatrix}0\\0\\1\end{pmatrix}=4\begin{pmatrix}1\\1\\1\end{pmatrix}+3\begin{pmatrix}0\\0\\1\end{pmatrix}=\begin{pmatrix}4\\4\\7\end{pmatrix}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $\sin x - \cos x = \frac{1}{2}$ then determine: $\sin^4 x + \cos^4 x$ If $$\sin x - \cos x = \frac{1}{2}$$ then determine: $$\sin^4 x + \cos^4 x$$
I tried making it $(\sin^2 x)^2+(\cos^2 x)^2$ but then I get nothing that can help. What is the trick to this?
| I'm using the identity $2(u^2+v^2)=(u+v)^2+(u-v)^2$ repeatedly:
$$(\cos x+\sin x)^2=2(\cos^2 x+\sin^2 x)-(\cos x-\sin x)^2=2-\frac14=\frac74,$$ so
\begin{align}2(\cos^4 x+\sin^4 x)&=(\cos^2 x+\sin^2 x)^2+(\cos^2 x-\sin^2 x)^2\\&=(\cos^2 x+\sin^2 x)^2+(\cos x+\sin x)^2\,(\cos x-\sin x)^2\\&=1+\frac74\cdot\frac14=\frac{23}{16}\end{align} We get $$\cos^4 x+\sin^4 x=\frac{23}{32}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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showing $f$ to be a bijection but stuck trying to show that $f$ is surjective Recall that the closed interval $[a,b]$={$x \in \mathbb{R} \vert a\le x \le b$}
prove that $\vert [2,5] \vert = \vert [-2,3] \vert$
(cardinality of [a,b] not absolute value)
proof: define $f:[2,5] \rightarrow [-2,3]$ by $f(x)=c+ \frac{d-c}{b-a} (x-a)=-2+ \frac{5}{3}(x-2)$
show that $f$ is injective:
let $a,b \in [2,5]$ such that $f(a)=f(b)$
then
$-2+\frac{5}{3}(a-2)=-2+\frac{5}{3}(b-2)$
$\frac{5a-10}{3}=\frac{5b-10}{3}$
$\frac{5a}{3}=\frac{5b}{3}$ $\rightarrow$ $a=b$
so f is injective.
now when i try to show that f is surjective, i let $y\in [-2,3]$
then if $x\in [2,5]$ such that $x=\frac{3(y+2)+10)}{15}$
we see that $f(x)=-2+\frac{5}{3}((\frac{3(y+2)+10}{15})-2)$
basically trying to show that this simplifies to $f(x)=y$ and hence is surjective, was gettng way too tedious and wasnt looking very promising. Can someone please show me where I went wrong here?
| You have
$$f(x) = \frac{5}{3}(x - 2) - 2$$
Let $y = f(x)$, then solve for $x$.
\begin{align*}
y & = \frac{5}{3}(x - 2) - 2\\
y + 2 & = \frac{5}{3}(x - 2)\\
\frac{3}{5}(y + 2) & = x - 2\\
\frac{3}{5}(y + 2) + 2 & = x
\end{align*}
Substituting this value for $x$ in the function $f$ yields
\begin{align*}
f\left(\frac{3}{5}(y + 2) + 2\right) & = \frac{5}{3}\left[\frac{3}{5}(y + 2) + 2 - 2\right] - 2\\
& = \frac{5}{3}\left[\frac{3}{5}(y + 2)\right] - 2\\
& = y + 2 - 2\\
& = y
\end{align*}
so $f$ is surjective.
Since you did not show your work, I am not sure how you obtained a $15$ in your denominator. However, that was your error.
Note: Another approach to showing that a bijection exists would be to solve for
$$g(x) = \frac{3}{5}(x + 2) + 2$$
then showing that $g = f^{-1}$ by demonstrating that
\begin{align*}
(g \circ f)(x) & = x && \text{for each $x \in \text{Dom}_f = [2, 5]$}\\
(f \circ g)(x) & = x && \text{for each $x \in \text{Dom}_g = [-2, 3]$}
\end{align*}
| {
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"url": "https://math.stackexchange.com/questions/2452226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving the equation $c=\dfrac{x^2+y^2-1}{x^2+(y+1)^2}$ Solving the equation $c=\dfrac{x^2+y^2-1}{x^2+(y+1)^2}$
$$c=\dfrac{x^2+y^2-1}{x^2+(y+1)^2}$$
$$c{x^2+c(y+1)^2}={x^2+y^2-1}$$
$$c{x^2+cy^2+2cy+c}={x^2+y^2-1}\text{ [expanded]}$$
$$1+c=x^2-cx^2+y^2-cy^2-2cy\text{ [moved to other side]}$$
$$1+c=(1-c)x^2+\color{red}{(1-c)y^2-2cy}\text{ [factor, will complete square of red]}$$
$$1+c=(1-c)x^2+\color{red}{(1-c)(y^2-\dfrac{2cy}{c-1}+(\dfrac{2c}{c-1})^2)-(\dfrac{2c}{c-1})^2}\text{ [completed square]}$$
$$1+c+\color{red}{(\dfrac{2c}{c-1})^2}=(1-c)x^2+\color{red}{(1-c)(y-\dfrac{2c}{c-1})^2}\text{ [done]}$$
So we see that this is an ellipse, stretched in the $y$ direction by a factor of $1-c$, and translated both in $x$ and $y$.
$c\neq 1$.
Consider $c=2$
$19=-x^2-(y+16)^2$
$-19=x^2+(y+16)^2$
But this not a circle nor an ellipse. It's supposed to be a circle? What is the issue here?
| $$
-(1-c)y^2-2cy = -(1-c)\left[y^2+\frac{2c}{1-c}y\right]=-(1-c)\left[\left(y+\frac{c}{1-c}\right)^2-\left(\frac{c}{1-c}\right)^2\right]
$$
Not what you have above.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Solve: $4 \log_5 x- \log_5y =1 \quad \&\quad 5\log_5 x+ 3\log_5 y =14$ Find $x,y$ given:
$$4 \log_5 x- \log_5y =1 \quad \&\quad
5\log_5 x+ 3\log_5 y =14$$
How do I help my son this assignment? Not so good in maths myself
| $$
\begin{align}
4 \log_5 x- \log_5y &=1\tag 1 \\
5\log_5 x+ 3\log_5 y &=14\tag 2
\end{align}
$$
Multiplying $(1)$ by $3$ and adding $(1)+(2)$ we have
$$17\log_5 x=17\quad\Longrightarrow\quad \log_5 x=1\quad\Longrightarrow\quad x=5$$
Multiplying $(1)$ by $-5$ and $(2)$ by $4$ and adding $(1)+(2)$ we have
$$17\log_5 y=51\quad\Longrightarrow\quad \log_5 y=51/17=3\quad\Longrightarrow\quad y=5^3=125$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2453373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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what is the image of the set $ \ A=\{(x,y)| x^2+y^2 \leq 1 \} \ $ under the linear transformation what is the image of the set $ \ A=\{(x,y)| x^2+y^2 \leq 1 \} \ $ under the linear transformation $ \ T=\begin{pmatrix}1 & -1 \\ 1 & 1 \end{pmatrix} \ $
Answer:
From the given matrix , we can write as
$ T(x,y)=(x-y,x+y) \ $
But what would be the image region or set $ \ T(A) \ $ ?
| \begin{align}T(x,y) &= (x-y, x+y)\\
&=\left(\begin{bmatrix} 1 & -1 \\ 1 & 1\end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix} \right)^T \\
&=\sqrt{2} \left(\begin{bmatrix} \cos \left( \frac{\pi}{4}\right) & -\sin \left( \frac{\pi}{4}\right) \\ \sin \left( \frac{\pi}{4}\right) & \cos \left( \frac{\pi}{4}\right)\end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix} \right)^T\end{align}
Hence, the image will be $ T(A)=\{(x,y)| x^2+y^2 \leq 2 \} \ $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2454179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How did x become -sqrt(x^2)?
In computing the limit as $x \to -\infty$, we must remember that for $x<0$, we have $\sqrt{x^2} = |x| = -x$. So when we divide the numerator by $x<0$, we get
$$
\frac{\sqrt{2x^2+1}}{x}
= \frac{\sqrt{2x^2+1}}{-\sqrt{x^2}}
= - \sqrt{\frac{2x^2+1}{x^2}}
= - \sqrt{2 + \frac{1}{x^2}}
$$
How did the denominator change from $x$ to $-\sqrt{x^2}$ ? I can't seem to understand.
I'm mainly confused by where the negative came from.
| Because we are looking at the limit $x\rightarrow-\infty,$ our target is negative x from the limit's perspective. That means since the value $\sqrt{x^2}=|x|$ (you can confirm this by doing casework on positive and negative values - note that square roots are always positive!), $\sqrt{x^2}=|x|=-x$ since $|x|=-x$ for negative x.
(since $-x=\sqrt{x^2}$ we can negate both sides to get $x=-\sqrt{x^2}$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2454812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integrate $\frac{\arctan x}{x^2}\,\mathrm dx$ $$\int \frac{\arctan
(x)}{x^2}\,\mathrm dx$$
(I have a constant in front of the integral, but I figure it just distributes to each of the terms in the end answer, and simplified it for now).
I figured it would be a chance for using integration by parts. I let $u=\arctan x, du=\frac{1}{x^2+1}, dv=\frac{1}{x^2}, v=\frac{-1}{x}$.
I get then $$\frac{-\arctan (x)}{x} -\int\left( -\frac{1}{1+x^2}\right)\,\mathrm dx$$ I cancel out the negatives in front of the integral and what's in front of the 1.
So I do partial fractions, and find that $\frac{1}{x(x^2+1)}=\frac{A}{x}+\frac{B}{(x^2+1)}$. I get that A=1 and B=-1 by letting x=0 and x=1, respectively.
So I have the $\arctan (x)$, and $\int\left( -\frac{1}{1+x^2} ,dx\right)$, simple enough, I thought, which generates $\ln |x|$ and $-\arctan (x)$. Somewhere, this must be wrong, but I can't see where. Please help!
| $$\int { \frac { \arctan { x } }{ { x }^{ 2 } } dx } =-\int { \arctan { x } d\left( \frac { 1 }{ x } \right) } =-\frac { \arctan { x } }{ x } +\int { \frac { dx }{ x\left( { x }^{ 2 }+1 \right) } } =\\ =-\frac { \arctan { x } }{ x } +\int { \left[ \frac { 1 }{ x } -\frac { x }{ { x }^{ 2 }+1 } \right] dx } =-\frac { \arctan { x } }{ x } +\int { \frac { dx }{ x } - } \frac { 1 }{ 2 } \int { \frac { d\left( { x }^{ 2 }+1 \right) }{ { x }^{ 2 }+1 } } =\\ =-\frac { \arctan { x } }{ x } +\ln { \left| x \right| - } \frac { 1 }{ 2 } \ln { \left( { x }^{ 2 }+1 \right) +C } $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $\lim_{(x,y) \to (0,0)} \frac{x^3+y^3}{x^2+y^2}$ assuming it exists.
Find $\lim_{(x,y) \to (0,0)} \frac{x^3+y^3}{x^2+y^2}$ assuming it
exists.
Since limit exists, we can approach from any curve to get the limit...
if we approach (0,0) from y=x
$\lim_{(x,y) \to (0,0)} \frac{x^3+y^3}{x^2+y^2} \Rightarrow \lim_{x \to 0} \frac{x^3+x^3}{x^2+x^2} = x = 0$
is this method correct?
| A better way to solve this problem is to use polar coordinates
$$x=r\cos \phi$$
$$y=r\sin \phi$$
$$\lim_{(x,y) \to (0,0)} \frac{x^3+y^3}{x^2+y^2}=\lim_{r\to 0}\frac{r^3(\cos^3 \phi+\sin^3 \phi)}{r^2}=\lim_{r\to 0}\left[r(\cos^3 \phi+\sin^3 \phi)\right].$$
Because $|\cos^3 \phi + \sin^3\phi|\leq 1$ the limit exists (by the sandwich theorem) and is zero.
Remark: If you assume that the limit exists then you could also choose $y=0$ and let $x \to 0$ (because it does not matter how you approach $(0,0)$).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Need hint for solving the following problem
If $f(x^{500}-1)=5x^{1015}+3x^{244}+7x+10$. Find the sum of coefficients of $f(x^5+1)$.
Let $x^{500}-1=y$, then $f(y)=5(y+1)^\frac{1015}{500}+3(y+1)^\frac{244}{500}+7(y+1)^\frac{1}{500}+10$.
But, I don't think that it is a right approach as each term of $f(y)$ gets transformed into an infinite series via binomial expansion, in that case it is not possible to sum the coefficients.
Another approach(This may be wrong!!,Please check)
Let,$x^{500}-1=x^5+1\implies x^{500}-x^5-1=1$.So,$f(x^5+1)=5(x^{500}-x^5-1)x^{1015}+3(x^{500}-x^5-1)x^{244}+7(x^{500}-x^5-1)x+10(x^{500}-x^5-1)$
So,the sum of coeffecients$=5-5-5+3-3-3+7-7-7+10-10-10=-25$
Please give some hints.
| The sum of the coefficients of $x$ in $g(x)=f\!\left(x^5+1\right)$ is $g(1)=f(2)$. However, given the formula for $f\!\left(x^{500}-1\right)$, we plug in $x=3^{1/500}$ to get
$$
f(2)=45\cdot3^{3/100}+3\cdot3^{61/125}+7\cdot3^{1/500}+10=68.651334833\dots
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Finding minimal polynomial of $x=a+b\sqrt[3]{2}+c\sqrt[3]{4}$ I want to find the minimal polynomial of $x=a+b\sqrt[3]{2}+c\sqrt[3]{4}$
For simple case like $x=a+b\sqrt[3]{2}$, I can find
\begin{align}
(x-a)^3 = 2b^3 \qquad \Rightarrow \qquad (x-a)^3-2b^3 =0
\end{align}
I tried to do similar tings such as
\begin{align}
(x-a)^3 = (\sqrt[3]{2}(b+c\sqrt[3]{2}))^3 = 2 (b+c\sqrt[3]{2})^3
\end{align}
but this does not seems good.
From the argument of
\begin{align}
x^3-2 = (x-\sqrt[3]{2})(x-\sqrt[3]{2}w)(x-\sqrt[3]{2}w^2)
\end{align}
where $w^2+w+1=0$,
I came up with some idea, and do the computation via mathematica. And i figure out the minimal polynomial for that is
\begin{align}
&(x-(a+b\sqrt[3]{2}+c\sqrt[3]{4}))(x-(a+bw\sqrt[3]{2}+cw^2\sqrt[3]{4}))(x-(a+bw^2\sqrt[3]{2}+cw\sqrt[3]{4})) \\
&=x^3 +3 a^2 x-3 a x^2-6 b c x-2 b^3-4 c^3 +6 a b c-a^3
\end{align}
The question is how to obtain this polynomial from the starting point.
| Since $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-yz)$, we obtain:
$$a-x+b\sqrt[3]{2}+c\sqrt[3]{4}=0$$ or
$$(a-x)^3+2b^3+4c^3-6(a-x)bc=0,$$
which is third degree.
I hope now it's clear.
By the way, we got your polynomial exactly.
| {
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"timestamp": "2023-03-29T00:00:00",
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Third order convergence of an iteration scheme Consider the iteration scheme
$x_{n+1}=\alpha x_n(3-\frac{x_n^2}{a})+\beta x_n(1+\frac{a}{x_n^2})$
For third order convergence to $\sqrt 2$, the values of $\alpha$ and $\beta$ are ......
I tried it by plugging $x_{n+1}=x_n=\sqrt a$ as $n\rightarrow\infty$ and got $\alpha +\beta =1$.
After this I tried to put it in the form
$Lim_{n\rightarrow\infty}$$\frac{|x_{n+1}-\sqrt a|}{|x_n-\sqrt a|^3}=M$ where $M\ne 0$
but couldn't succeed. Any suggestion and hints please!
The given answer is $\alpha=1/8$,$\beta=3/8$.
| Let $x_n = \sqrt{2}+\epsilon$ (and of course $a=2$). Then
$$
x_{n+1} = \alpha (\sqrt{2}+\epsilon)(3-\frac12(\sqrt{2}+\epsilon)^2
+\beta (\sqrt{2}+\epsilon) \left( 1+\frac{2}{(\sqrt{2}+\epsilon)^2}\right)
$$
Expanding in $\epsilon$, to one order more than we need in the first line,
$$
x_{n+1} = (\sqrt{2}+\epsilon) \left( \alpha (2-\sqrt{2}\epsilon-\epsilon^2/2)+\beta\left(1+1-\sqrt{2}\epsilon +\frac32 \epsilon^2 -\sqrt{2} \epsilon^3 + O(\epsilon^4)\right) \right)
\\
x_{n+1} = (2\sqrt{2}\alpha + 2\sqrt{2}\beta) + \left( -2\alpha +2\alpha -2\beta + 2\beta\right) \epsilon \\+ \left(-\frac12\sqrt{2}\alpha-\sqrt{2}\alpha +\frac32\sqrt{2}\beta-\sqrt{2}\beta\right)\epsilon^2+O(\epsilon^3)
$$
and since this needs to equal $\sqrt{2} + O(\epsilon^3)$ we have
$$
\alpha+\beta = \frac12\\ \frac12 \beta - \frac32 \alpha = 0
$$
the solution to which is $\alpha = \frac18, \beta = \frac38$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to prove that $21\mid a^2+b^2\Rightarrow 441 \mid a^2 + b^2$ How to prove that $441 \mid a^2 + b^2$ if it is known that $21 \mid a^2 + b^2$.
I've tried to present $441$ as $21 \cdot 21$, but it is not sufficient.
| If $3\mid(a^2+b^2)$ then $3$ divides both $a$ and $b$, since $-1$ is not a quadratic residue $\!\!\pmod{3}$.
The same applies $\!\!\pmod{7}$. If $21\mid(a^2+b^2)$, from the CRT we get that $3$ and $7$ divide both $a$ and $b$, hence $3^2$ and $7^2$ divide both $a^2$ and $b^2$ and $3^2\cdot 7^2\mid (a^2+b^2)$ as wanted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2467327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that if $r\overrightarrow {X}=\overrightarrow {0}$, then either $r=0$ or $\overrightarrow {X}=\overrightarrow{0}$
Suppose $\overrightarrow{X}\in{\bf R}^2$ and $r\in{\bf R}$. Show that if $r\overrightarrow {X}=\overrightarrow{0}$, then either $r=0$ or $\overrightarrow {X}=\overrightarrow{0}$.
[Attempt:]
Let $\overrightarrow {X}=\left( \begin{matrix} a\\ b\end{matrix} \right)$. Assume $r\neq 0$. Then,
$r\overrightarrow {X}=\overrightarrow {0}$,
$\overrightarrow {X}=\dfrac {1} {r }\overrightarrow {0}$,
$\left( \begin{matrix} a\\ b\end{matrix} \right)=\dfrac {1} {r }\left( \begin{matrix} 0\\ 0\end{matrix} \right)$,
$\left( \begin{matrix} a\\ b\end{matrix} \right)=\left( \begin{matrix} \dfrac {1} {r }0\\ \dfrac {1} {r }0\end{matrix} \right)$,
$\left( \begin{matrix} a\\ b\end{matrix} \right)=\left( \begin{matrix} 0\\ 0\end{matrix} \right)$.
Thus, we have $a=0$ and $b=0$, that is, $\overrightarrow {X}=\overrightarrow {0}$.
Now, assume $\overrightarrow {X}\neq\overrightarrow {0}$. We will show that $r=0$. Then,
$r\overrightarrow {X}=\overrightarrow {0}$,
$\left( \begin{matrix} r a\\ r b\end{matrix} \right)=\left( \begin{matrix} 0\\ 0\end{matrix} \right)$. Since $a\neq 0$ and $b\neq 0$, we obtain $r=0$.
So, we are done.
Can you check my proof?
$$
\newcommand{\Vec}[1]{\overrightarrow{#1}}
$$
| Your proof is correct. As long as you have shown that $\vec{X}=\vec{0}$ assuming $r\neq 0$, all the rest of your work are redundant.
| {
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Proving that: $ (a^3+b^3)^2\le (a^2+b^2)(a^4+b^4)$ This problem is from Challenge and Thrill of Pre-College Mathematics:
Prove that $$ (a^3+b^3)^2\le (a^2+b^2)(a^4+b^4)$$
It would be really great if somebody could come up with a solution to this problem.
| $$\begin{array}{rrcl}
& (a^3+b^3)^2 &\le& (a^2+b^2)(a^4+b^4) \\
\iff& a^6 + 2a^3b^3 + b^6 &\le& a^6+a^2b^4+b^2a^4+b^6 \\
\iff& 2a^3b^3 &\le& a^2b^4+b^2a^4 \\
\iff& 2ab &\le& b^2+a^2 \\
\iff& 0 &\le& b^2-2ab+a^2 \\
\iff& 0 &\le& (b-a)^2 \\
\end{array}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solving for $k$: $\sqrt{k-\sqrt{k+x}}-x = 0$
$$\sqrt{k-\sqrt{k+x}}-x = 0$$
Solve for $k$ in terms of $x$
I got all the way to
$$x^{4}-2kx^{2}-x+k^{2}-x^{2}$$
but could not factor afterwards. My teacher mentioned that there was grouping involved
Thanks Guys!
Edit 1 : The exact problem was solve for $x$ given that $$\sqrt{4-\sqrt{4+x}}-x = 0$$ with a hint of substitute 4 with k
| In this kind of problem, you have to be very careful about the domain of definition. Squaring the equation an find an equivalent polynomial equations is not enough, you need to verify if the solutions found are effective solutions of the original equation.
First two remarks :
*
*$x=\sqrt{\cdots}\quad$ thus $x\ge 0$
*If $k<0$ then $k-\sqrt{k+x}<0$ and we cannot take the square root of this, so $k\ge 0$
In particular: with $x\ge 0$ and $k\ge 0$
*
*$\sqrt{x+k}$ is also well defined
*$\sqrt{(x+a)^2}=x+a$ for any $a\ge 0$, we will use that property later.
The equation squared twice becomes $(x^2-k)^2=x+k$
$\iff k^2-k(2x^2+1)+(x^4-x)=0$
with $\Delta=(2x^2+1)^2-4(x^4-x)=(2x+1)^2$
So $k=\frac 12(2x^2+1\pm(2x+1))=x(x-1)$ or $(x^2+x+1)$
Let's no substitute back in the original problem to eliminate superfluous solutions.
*
*$k=x(x-1)$
$\sqrt{k-\sqrt{k+x}}=\sqrt{x^2-x-\sqrt{x^2}}=\sqrt{x^2-x-x}=\sqrt{x^2-2x}$
This can be equal to $x$ if and only if $x=0\qquad$ [$x^2-2x=x^2\iff x=0$]
*
*$k=x^2+x+1$
$\sqrt{k-\sqrt{k+x}}=\sqrt{x^2+x+1-\sqrt{(x+1)^2}}=\sqrt{x^2+x+1-(x+1)}=\sqrt{x^2}=x$
So the equation is always verified
Finally the solutions are $(0,0)$ and $(x\ge 0,k=x^2+x+1)$
| {
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Number of ways for a 6 digit license plate such that each digit is unique except for one The problem asks how many different six digit license plates can be made if there are three digits that occur once and one digit that occurs thrice?
My idea is that there can be 10 numbers for the reoccurring digits, and $6 \choose 3$ ways to lay them out. Then there are $9\cdot8\cdot7$ ways to lay out the remaining unique digits. This gives $10\cdot{6\choose3}\cdot9\cdot8\cdot7$ possible plates.
Is this thought process correct? I'm having some trouble on these problems.
Thanks
|
Then there are $9\cdot8\cdot7$ ways to lay out the remaining unique digits.
This is not about the number of ways of 'laying them out', but of picking digits for the three remaining digits.
So yes, you have indeed $6 \choose 3$ ways to lay out the recurring digit, and you have $10$ choices for what that digit is.
But then for the other three digits, you can pick $9$ choices for the 'first' (i.e. the first one you encounter from left to right in the digit string) digit, $8$ for the next, and $7$ for the last, giving you a total of:
$$10 \cdot {6 \choose 3} \cdot 9 \cdot 8 \cdot 7$$
Alternatively:
There are $9 \choose 3$ ways to pick the $3$ remaining digits, and $3!$ ways to lay them out:
$$10 \cdot {6 \choose 3} \cdot {9 \choose 3} \cdot 3!$$
which works out to be the same (of course!):
$$10 \cdot {6 \choose 3} \cdot {9 \choose 3} \cdot 3! =$$
$$ 10 \cdot {6 \choose 3} \cdot \frac{9!}{3!\cdot 6!} \cdot 3! =$$
$$10 \cdot {6 \choose 3} \cdot \frac{9!}{6!} =$$
$$10 \cdot {6 \choose 3} \cdot 9 \cdot 8 \cdot 7$$
| {
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} |
$|\int_{0}^{a+bi}\cos(z^2)\,dz|\leq(a^2+b^2)^{1/2}\sinh(2ab)/(2ab)$ Assuming that $a>0$ and $b>0$, derive the estimate
$$\left|\int_{0}^{a+bi}\cos(z^2)\,dz\right|\leq \frac{(a^2+b^2)^{1/2}\sinh(2ab)}{2ab}$$
I know $|\int_{\gamma}f(z)\,dz|\leq \int_{\gamma}|f(z)|\,dz$, but I do not know how to apply it here, I have tried the following:
$$\left|\int_{0}^{a+bi}\cos\left(z^2\right)\,dz\right|\leq \int_{0}^{a+bi}\left|\frac{e^{iz^2}+e^{-iz^2}}{2}\right|\,dz,$$
but I do not know what else to do. Could anyone help me with this please? Thank you very much.
| First note that \begin{align}
\int_0^{a+bi} \cos (z^2) dz&=\int_0^1\cos((a+bi)^2t^2)\cdot (a+bi)dt\quad (z=(a+bi)t,\, 0\le t\le 1)\\
&=(a+bi)\int_0^1\cos((a^2-b^2)t^2+2abt^2i)dt.
\end{align}
Using the inequality \begin{align}
|\cos (\alpha +\beta i)|&=\left|\frac{\exp(i(\alpha +\beta i))+\exp(-i(\alpha +\beta i))}{2}\right|\le \frac{|\exp (i\alpha -\beta )|+|\exp(-i\alpha +\beta )|}{2}\\
&=\frac{\exp(-\beta )+\exp(\beta )}{2},
\end{align}
we have \begin{align}
\left|\int_0^{a+bi} \cos (z^2) dz\right|&\le \sqrt{a^2+b^2}\int_0^1\left|\cos((a^2-b^2)t^2+2abt^2i)\right|dt\\
&\le \frac{\sqrt{a^2+b^2}}{2}\int_0^1 \left(\exp(2abt^2)+\exp(-2abt^2)\right)dt.
\end{align}
Now we use the inequality $$\exp(2abt^2)+\exp(-2abt^2)\le \exp(2abt)+\exp(-2abt)\quad (0\le t\le 1).$$
This follows the fact that $f(x)=e^x+e^{-x}$ is increasing for $x>0$.
Thus we have \begin{align}
\left|\int_0^{a+bi} \cos (z^2) dz\right|&\le \frac{\sqrt{a^2+b^2}}{2}\int_0^1 \left(\exp(2abt)+\exp(-2abt)\right)dt\\
&=\frac{\sqrt{a^2+b^2}}{2}\cdot \frac{\exp(2ab)-\exp(-2ab)}{2ab}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove $\lim_{x\rightarrow 0}\frac{\sqrt{9-x}-3}{x}=-\frac{1}{6}$ with epsilon-delta I am looking to show that $$f:(0,1)\rightarrow \mathbb{R} \lim_{x\rightarrow 0}\frac{\sqrt{9-x}-3}{x}=-\frac{1}{6}$$ Normally for these types of problems I have been doing epsilon-delta proofs but I cannot figure out how to define my $\delta$.
Essentially I need with finding an appropriate $\delta$. I am not looking for an application of L'Hopital's Rule.
So far with determining a $\delta$ from $\varepsilon$ I have $$\left| \frac{\sqrt{9-x}-3}{x}+\frac{1}{6} \right|<\varepsilon $$
$$\left| \frac{-1}{\sqrt{9-x}-3}+\frac{1}{6} \right|<\varepsilon $$
$$\left| \frac{-1}{\sqrt{9-x}-3}\right|+\frac{1}{6} \leq\varepsilon $$
by the triangle inequality
$$\left| \frac{-1}{\sqrt{9-x}-3}\right| \leq\varepsilon -\frac{1}{6}$$
$$ \frac{1}{\left|\sqrt{9-x}-3\right|} \leq\varepsilon -\frac{1}{6}$$
$$ \frac{1}{\varepsilon -\frac{1}{6}} \leq\left|\sqrt{9-x}-3\right|$$
$$ \frac{1}{\varepsilon -\frac{1}{6}} \leq\left|\sqrt{9-x}\right|+3$$
$$ \left(\frac{1}{\varepsilon -\frac{1}{6}}\right)-3 \leq\left|\sqrt{9-x}\right|$$ because the square root is always positive
$$ \left(\frac{1}{\varepsilon -\frac{1}{6}}\right)-3 \leq\sqrt{9-x}$$
$$ \left(\left(\frac{1}{\varepsilon -\frac{1}{6}}\right)-3\right)^2 \leq9-x$$
$$ 9-\left(\left(\frac{1}{\varepsilon -\frac{1}{6}}\right)-3\right)^2 \geq x$$
| Consider that $$\frac{\sqrt{9-x}-3}{x}+\frac{1}{6} = \frac{x+6\sqrt{9-x}-18}{6x} = -\frac{\left(\sqrt{9-x}-3\right)^2}{6x}$$ If $\lim_{x\to 0^+} \frac{f(x)}{x} = c$, then for $\epsilon > 0$, there is a $\delta > 0$ such that $$\left\lvert \frac{f(x)}{x}-c\right\rvert = \left\lvert \frac{f(x)-cx}{x}\right\rvert < \epsilon \Leftrightarrow \lvert f(x)-cx\rvert < \epsilon x$$ for $0 < x < \delta$. Therefore, for $0 < x < \delta$, \begin{align*} \left\lvert \frac{f(x)^2}{x}\right\rvert &= \left\lvert \frac{((f(x)-cx)+cx)^2}{x}\right\rvert \\ &= \left\lvert \frac{(f(x)-cx)^2+2(f(x)-cx)cx+c^2x^2}{x}\right\rvert \\ &\leq \frac{\lvert f(x)-cx\rvert^2}{\lvert x\rvert}+\frac{2c\lvert f(x)-cx\rvert x}{x}+c^2x \\ &< (c+\epsilon)^2x \end{align*} so $\lim_{x\to 0^+} \frac{f(x)^2}{x} = 0$. We get the same result for $\lim_{x\to 0^-} \frac{f(x)}{x} = d$, so if the directional limits of $\frac{f(x)}{x}$ exist as $x\to 0$, then $\lim_{x\to 0} \frac{f(x)^2}{x} = 0$. Now, if you can reason that the directional limits of $\frac{\sqrt{9-x}-3}{x}$ exist as $x\to 0$,$^1$ then you will have $$\lim_{x\to 0} \frac{\sqrt{9-x}-3}{x} = -\frac{1}{6}$$
$^1$This follows from the facts that $f(x) = \frac{\sqrt{9-x}-3}{x}$ is monotonic on $(-\infty, 0)$ and $(0, 9)$, as $f'(x) = -\frac{\left(\sqrt{9-x}-3\right)^2}{2x^2\sqrt{9-x}} < 0$ for $x\neq 0, 9$, and that $f(x)$ is bounded, i.e. $-\frac{1}{3}\leq f(x) < 0$. The directional limits of monotonic, bounded functions always exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2477969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Inequality question: If $a + b + c =1$, what is the minimum value of $\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}$. If $a + b + c =1$, what is the minimum value of $\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}$. I've tried AM-HM but it gave $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geq 9$ which gives $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} + 2 (\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} )\geq 81$
| By Holder
$$\sum_{cyc}\frac{1}{ab}=\sum_{cyc}\frac{1}{ab}\sum_{cyc}a\sum_{cyc}b\geq\left(\sum_{cyc}\sqrt[3]{\frac{1}{ab}\cdot a\cdot b}\right)^3=\left(\sum_{cyc}1\right)^3=27.$$
The equality occurs for $a=b=c=\frac{1}{3}$, which says that $27$ is a minimal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2478189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
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