Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Calculating $ \int \frac{1}{x} \sqrt{\frac{x+a}{x-a}}\mathrm dx $ I want to evaluate $ \int \frac{1}{x} \sqrt{\frac{x+a}{x-a}}\mathrm dx $.
$ x=a\cosh(2t), \int \frac{1}{x} \sqrt{\frac{x+a}{x-a}}dx= \int \frac{2\tanh(2t)}{\tanh(t)}dt= \int \frac{4}{1+\tanh^2(t)}\mathrm dt $
$ u=\tanh(t), \int \frac{4}{1+\tanh^2(t)}dt=2 \int (\frac{1}{1+u^2}+\frac{1}{1-u^2})\mathrm du=2 \mathrm{artanh}(u)+2\arctan(u)+C= $
$ \mathrm{arcosh}(x/a)+2\arctan(\sqrt{\frac{x^2-a^2}{x^2+a^2}})+C $
However, I could not manage to show that the derivative of this function is $ \frac{1}{x} \sqrt{\frac{x+a}{x-a}} $.
| After substitutions back into the correct anti-derivative in terms of $u$, I get different result:
$$
\cosh^{-1}\left(\frac{x}{a}\right) + 2 \arctan\left( \sqrt{ \frac{x-a}{x+a} }\right) + C
$$
Differentiating this, I get $\dfrac{1}{x} \left( \dfrac{x-a}{x+a} \right)^{-\frac12}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/60164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
finding the coefficient of $x^{14}$ in the expression: $\frac{5x^2-x^4}{(1-x)^3}$ I have a homework question which requires me to find the coefficient of $x^{14}$ in the expression: $\dfrac{5x^2-x^4}{(1-x)^3}$
I have not figured out a way to do this (I believe this is because my algebra is weak).
This is a question in a combinatorics class and has to do with Generating Functions if that is any help.
Help is greatly appreciated. Thanks, Jason
| To complement @Bill's answer, I will explain one way to quickly obtain a partial fraction expansion in this example.
Use substitution. Set $x = 1 -y$, so that the given expression becomes:
$$
\frac{5(1-y)^2 - (1-y)^4}{y^3} = \frac{4 - 6y - y^2 + 4y^3 - y^4}{y^3} =
\frac{4}{y^3} - \frac{6}{y^2} - \frac{1}{y} + 4 - y.
$$
Substituting back $1-x$ for $y$, we get:
$$
\frac{4}{(1-x)^3} - \frac{6}{(1-x)^2} - \frac{1}{(1-x)} + 3 + x.
$$
Using the binomial series1, we can calculate the coefficient of $x^{14}$ to be:
$$
4 \binom{14+3-1}{14} - 6 \binom{14+2-1}{14} - \binom{14+1-1}{14} + 0 + 0 = 4 \binom{16}{2} - 6 \times 15 - 1 = 389.
$$
1As I remarked in a comment above, the binomial series for $(1-x)^{\beta+1}$ has a particularly simple form.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/61879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Prove using mathematical induction: $1 \cdot 2 \cdot 3 + \ldots + n \cdot (n+1) \cdot (n+2) = \frac{n(n+1)(n+2)(n+3)}{4}$ So I have the easy stuff done. However, I'm not sure how to go about doing the inductive step with such an awkward proof. :/
Statement:
$$ 1 \cdot 2 \cdot 3 + … + n \cdot (n+1) \cdot (n+ 2) = \frac{n \cdot (n + 1) \cdot (n + 2) \cdot (n +3)}{4} .$$
Base step: $n = 1$:
$$ 0 \cdot (0 + 1) \cdot (0 + 2) + 1 \cdot (1 + 1) \cdot (1 +2) = \frac{ 1 \cdot (1+ 1) \cdot (1 + 2) \cdot (1 + 3) }{4} .$$
$$ 0 \cdot (0 + 1) \cdot (0 + 2) +1 \cdot (1 + 1) \cdot (1 +2) = 0 + 6 = 6$$
$$ \frac{ 1 \cdot (1+ 1) \cdot (1 + 2) \cdot (1 + 3) }{4} = \frac{24}{ 4} = 6 .$$
$$6 = 6.$$
Inductive step:
$n = n + 1$
| Recommendation 1. Never, ever, write "$n=n+1$". It's just wrong.
Recommendation 2. It's easier if you state the inductive step in two parts: state the induction hypothesis explicitly, then state what you need to prove.
(Recommendation 2 is my strongest recommendation, especially if you are beginning induction and are having some difficulty; it helps organize your thoughts, it helps you keep straight what you are assuming vs. what you want to prove, it helps prevent confusion between the inductive step and the theorem as a whole. It's a chore, perhaps, but I really do strongly recommend doing it).
Recommendation 3. It is often useful to prevent confusion by stating the inductive step using a letter different from what you are trying to prove, so that your "induction hypothesis" doesn't get confused with the theorem to be proven; using a letter that is not being used for anything else can help. (It also prevents writing things that bring such horror to a trained mathematician like "$n=n+1$"...) Here, the statement we are trying to prove uses $n$ for the variable and nothing else. To avoid getting confused, use $k$ for the induction hypothesis and $k+1$ for what you need to prove in the induction hypothesis.
This is purely aesthetical; it plays no mathematical role. It's just a way to prevent possible confusion, and nothing more.
Here, try this:
Inductive step.
Induction hypothesis: The result holds for $k$. That is, we assume that:
$$(1)(2)(3) + (2)(3)(4) + \cdots + (k)(k+1)(k+2) = \frac{k(k+1)(k+2)(k+3)}{4}$$
is true.
To be proven: The result holds for $k+1$; that is, we need to prove that:
$$\small (1)(2)(3) + (2)(3)(4) + \cdots + (k+1)\bigl((k+1)+1\bigr)\bigl((k+1)+2\bigr) = \frac{(k+1)(k+2)(k+3)(k+4)}{4}.$$
is true.
(The above is what I mean by "state the induction hypothesis explicitly, then state what you need to prove)
Okay let's try to prove that the second highlighted statement is true. With induction, we want to bring the Induction Hypothesis into play somehow; usually, by trying to think of the $k+1$ case as being "do the $k$-step and then do something else"; then we can use the induction hypothesis to simplify the "$k$-step". This works very well here.
We have:
$$(1)(2)(3) + (2)(3)(4) + \cdots + (k+1)(k+2)(k+3).$$
In order to think about it as "the $k$-step and then a bit more", let's write the the $k$th summand as well as the $k+1$st:
$$(1)(2)(3) + (2)(3)(4) + \cdots + k(k+1)(k+2)+ (k+1)(k+2)(k+3).$$
Now, we can do the sum by first adding the first $k$ summands, and then adding $(k+1)(k+2)(k+3)$ to the result of that; so the entire sum we have is equal to:
$$
\Biggl( (1)(2)(3)+(2)(3)(4)+\cdots+k(k+1)(k+2)\Biggr) + (k+1)(k+2)(k+3).$$
Now, we may realize that the stuff inside the big parenthesis is precisely
the sum about which the Induction Hypothesis tells us something. The Induction Hypothesis says that:
$$(1)(2)(3) + (2)(3)(4) + \cdots + (k)(k+1)(k+2) = \frac{k(k+1)(k+2)(k+3)}{4},$$
so we can replace that first parenthetical expression with $\frac{k(k+1)(k+2)(k+3)}{4}$. So, in summary, we have:
$$\begin{align*}
(1)(2)(3) &+ (2)(3)(4) + \cdots + k(k+1)(k+2)+ (k+1)(k+2)(k+3)\\
&=\Biggl( (1)(2)(3)+(2)(3)(4)+\cdots+k(k+1)(k+2)\Biggr) + (k+1)(k+2)(k+3)\\
&= \Biggl(\frac{k(k+1)(k+2)(k+3)}{4}\Biggr) + (k+1)(k+2)(k+3)
\end{align*}$$
with the first equality being just by associativity of the sum; and the second equality justified because we have applied the induction hypothesis.
That is, we have simplified the sum of the first $k$ terms through the use of the induction hypothesis.
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/64987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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} |
An integral about Bessel function Is there somebody who knows the solution for the integral
$$\int_0^\infty\frac{J^3_1(ax)J_0(bx)}{x^2} dx$$
where $a>0,b>0$ and $J(\cdot)$ the bessel function of the first kind with integer order?
Reference, or solution from computer programs all are welcome. Thanks!
| The integral can be approached by using Mellin convolution technique. Let's first start with a simple case of $a=b$. In this case we compute Mellin transform of $J_1(x)^2$ and $J_0(x) J_1(x)$:
$$
\mathcal{M}_s( J_1(x)^2 ) = \int_0^\infty x^{s} J_1(x)^2 \frac{\mathrm{d} x}{x} = \frac{\Gamma \left(\frac{1}{2}-\frac{s}{2}\right) \Gamma
\left(\frac{s}{2}+1\right)}{2 \sqrt{\pi } \Gamma \left(1-\frac{s}{2}\right) \Gamma
\left(2-\frac{s}{2}\right)} \qquad \text{for} \qquad -2 < \mathrm{Re}(s) < 1
$$
and
$$
\mathcal{M}_s( J_1(x) J_0(x) ) = \frac{\Gamma \left(1-\frac{s}{2}\right) \Gamma \left(\frac{s+1}{2}\right)}{2 \sqrt{\pi } \Gamma \left(\frac{3}{2}-\frac{s}{2}\right)^2} \qquad \text{for} \qquad -1 < \mathrm{Re}(s) < 2
$$
Now
$$
\begin{eqnarray}
\int_0^\infty x^{-2} J_1(a x)^3 J_0(a x) \mathrm{d} x &=& \frac{a}{2} \frac{1}{2 \pi i} \int_{-1 - i \infty}^{-1 + i \infty} \frac{\Gamma \left(\frac{1}{2}-\frac{s}{2}\right) \Gamma \left(\frac{s}{2}+1\right)
\Gamma \left(\frac{s}{2}+\frac{3}{2}\right) \Gamma \left(-\frac{s}{2}\right)}{2
\pi \Gamma \left(1-\frac{s}{2}\right) \Gamma \left(2-\frac{s}{2}\right) \Gamma
\left(\frac{s}{2}+2\right)^2} \mathrm{d} s \\
&=& \frac{a}{2 \pi i} \int_{-\frac{1}{2} - i \infty}^{-\frac{1}{2} + i \infty} \frac{\Gamma \left(\frac{1}{2}-s\right) \Gamma \left(s+1\right)
\Gamma \left(s+\frac{3}{2}\right) \Gamma \left(-s\right)}{2
\pi \Gamma \left(1-s\right) \Gamma \left(2-s\right) \Gamma
\left(s+2\right)^2} \mathrm{d} s \\
&=& \frac{a}{2 \pi} G_{4,4}^{2,2}\left(1\left|
\begin{array}{c}
\frac{1}{2},1,2,2 \\
1,\frac{3}{2},0,-1 \\
\end{array}
\right.\right) = a \left( \frac{3 }{16} - \frac{1}{\pi^2} \right)
\end{eqnarray}
$$
Here $G_{4,4}^{2,2}(1)$ denotes Meijer's G-function.
Now when $a \not= b$, Mellin transform of $J_1(a x) J_0(b x)$ is no longer a ratio of $\Gamma$-functions:
$$
\left. \mathcal{M}_s( J_1( a x) J_0(b x)) \right\vert_{-1 < \mathrm{Re}(s) <2} = \left\{
\begin{array}{cc}
\frac{a 2^{s-1} b^{-s-1} \Gamma \left(\frac{s+1}{2}\right) \,
_2F_1\left(\frac{s+1}{2},\frac{s+1}{2};2;\frac{a^2}{b^2}\right)}{\Gamma
\left(\frac{1}{2}-\frac{s}{2}\right)} & a < b \\
\frac{2^{s-1} a^{-s} \Gamma \left(\frac{s+1}{2}\right) \,
_2F_1\left(\frac{s-1}{2},\frac{s+1}{2};1;\frac{b^2}{a^2}\right)}{\Gamma
\left(\frac{3}{2}-\frac{s}{2}\right)} & a > b
\end{array}
\right.
$$
Notice that Gauss hypergeometric function $ {}_2F_1$ can be represented by its defining sum, whose terms are ratios of Gamma functions themselves. Continuing this way will produce sum representation of the integral.
Integrals of this form has been discussed by W.N. Bailey in 1936, see link.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/66661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 0
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What is the probability that $XIf two dice are rolled repeatedly, and $X$ is the number of tosses until $3$ is rolled, and $Y$ is the number of tosses until a $5$ is rolled, what is the probability$(X < Y)$? Also, if $Z$ is the number of rolls until a $10$ is rolled, what the probability$(X < Y < Z)$?
Is $P(X < Y) = \sum_{n=2}^{\infty}\sum_{k=1}^{\infty}((32/36)^{(n-1)}(4/36)-(34/36)^{(k-1)}(2/36)]$?
| (a) $P(X<Y)$
Suppose $q = P(3 \text{ before } 5)$. When will we see a $3$ before we see a $5$? That is if we first see a $3$ on the first roll (we are done), or if we don't see a $3$ or a $5$ on the first roll and then see a $3$ before a $5$ after that. In other words:
$$q = P(3) \cdot 1 + P(5) \cdot 0 + P(\text{neither } 3 \text{ nor } 5) \cdot q$$
Since we know that $P(3) = \frac{2}{36}$, $P(5) = \frac{4}{36}$ we get $P(\text{neither } 3 \text{ nor } 5) = 1 - \frac{2}{36} - \frac{4}{36} = \frac{30}{36}$, so
$$q = \frac{2}{36} + \frac{30}{36} q \quad \Longleftrightarrow \quad q = \frac{1}{3}$$
Note that we can also rewrite the first equation as
$$q = P(3) \cdot 1 + P(5) \cdot 0 + P(\text{neither } 3 \text{ nor } 5) \cdot q \quad \Longleftrightarrow \quad q = \frac{P(3)}{P(3) + P(5)} = \frac{P(3)}{P(3 \text{ or } 5)} = P(3\ |\ 3 \text{ or } 5)$$
This is in accordance with a comment made by Dilip: We can simply ignore all throws which are neither a $3$ or a $5$, as they are irrelevant. Then the probability of seeing a $3$ first is simply the conditional probability of throwing a $3$, given that it's either a $3$ or a $5$.
(b) $P(X<Y<Z)$
We can use the same approach as above. First, let us ignore all throws which are not $3$, $5$ or $10$. We now need that the first event is a $3$, which happens with probability
$$P(X<Y,X<Z) = \frac{P(X)}{P(X \text{ or } Y \text{ or } Z)} = \frac{P(X)}{P(X) + P(Y) + P(Z)} = \frac{2}{2+4+3} = \frac{2}{9}$$
Now if this happens, we only need that in all successive events of $3,5,10$ we first see some number of $3$s, and then see a $5$ (and not a $10$ yet). But that means that after this first event, we can again ignore all events $X$, and look at the first throw resulting in either a $5$ or a $10$. Then
$$P(Y<Z | X<Y,X<Z) = \frac{P(Y)}{P(Y \text{ or } Z)} = \frac{P(Y)}{P(Y)+P(Z)} = \frac{4}{4+3} = \frac{4}{7}$$
So the combined probability is then
$$P(X<Y<Z) = P(X<Y,X<Z) \cdot P(Y<Z | X<Y, X<Z) = \frac{2}{9} \cdot \frac{4}{7} = \frac{8}{63}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/66996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Evaluating $ \int {\frac{1}{2+\sqrt{1-x}+\sqrt{1+x}}}\mathrm dx $ I would like to evaluate: $$ \int {\frac{1}{2+\sqrt{1-x}+\sqrt{1+x}}}\mathrm dx $$
$$ \frac{1}{2+\sqrt{1-x}+\sqrt{1+x}}=\frac{\sqrt{1-x}+\sqrt{1+x}-2}{2(\sqrt{1-x^2}-1)} $$
The substitution $ x \rightarrow \sin(x) $ or $ \cos(x) $ can only simplify the denominator, and $ x \rightarrow \sqrt{1+x}$ or $ \sqrt{1-x} $ is also useless...
Can you help me find a useful substitution?
$$ x=\cos(2t) $$
$$ \int {\frac{1}{2+\sqrt{1-x}+\sqrt{1+x}}}\mathrm dx=-\int {\frac{\sqrt{2}\sin(t)\cos(t)}{\sqrt{2}+\sin(t)+\cos(t)}}\mathrm dt $$
$$ u=\tan(t/2) $$
$$ -4\sqrt{2}\int \frac{u(1-u^2)}{(1+u^2)^2((\sqrt{2}-1)u^2+2u+1+\sqrt{2})}\mathrm du $$
But now it looks even more complicated... ?
| Integral=
$$\int\frac{dx}{2+\sqrt{1-x}+\sqrt{1+x}}$$
$$=\int\frac{\sqrt{1-x}-\sqrt{1+x}}{(2+\sqrt{1-x}+\sqrt{1+x})(\sqrt{1-x}-\sqrt{1+x})}dx$$
Substitution:
$z=2+\sqrt{1-x}+\sqrt{1+x} $
$$dz=\frac{\sqrt{1-x}-\sqrt{1+x}}{2\sqrt{1-x^2}}dx $$
$\sqrt{1-x^2}=(1/2)(z^2-4z+2)$
$\sqrt{1-x}-\sqrt{1+x}=\sqrt{4z-z^2}$
Integral=
$$=\int\frac{z^2-4z+2}{z\sqrt{4z-z^2}}dz$$
$$=\int\frac{z-4}{\sqrt{4z-z^2}}dz+\int\frac{2dz}{z\sqrt{4z-z^2}}$$
$$=\int\frac{-(1/2)(-2z+4)}{\sqrt{4z-z^2}}dz-\int\frac{2dz}{\sqrt{4z-z^2}}+\int\frac{2dz}{z\sqrt{4z-z^2}}$$
For the third integral you may use the substitution $z=1/t$
We have,Integral
$$=-\sqrt{4z-z^2}-4\sin^{-1}\frac{\sqrt{z}}{2}-\sqrt{\frac{4-z}{z}}+C$$
Where $z=2+\sqrt{1-x}+\sqrt{1+x}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/67172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 2
} |
Find the limit as $x$ tends to $-\infty$ Find the limit as $x$ tends to $-\infty$ of
$$ f(x)=\frac{\sqrt{x^2+1}}{x+1} $$
I did
$$ f(x) = \frac{\sqrt{1+1/x^2}}{1+1/x}\to \frac{\sqrt{1+0}}{1-0} =1 $$
(as $x$ tends to -infinity, $1/x^2$ tends to $0$ and $1/x$ tends to $0, 0$ gets negative sign fro negative infinity)
But the correct answer is $-1$, where did I go wrong?
| Here's how I show the numerator manipulations to students:
$$\sqrt{x^{2} + 1} \;\;=\;\; \sqrt{x^{2}\cdot \left(1 + \frac{1}{x^{2}}\right)} \;\;=\;\; \sqrt{x^2} \cdot \sqrt{1 + \frac{1}{x^{2}}\;}$$
$$= \;\;|x| \cdot \sqrt{1 + \frac{1}{x^{2}}\;}\;\; =\;\; (-x) \cdot \sqrt{1 + \frac{1}{x^{2}}\;}$$
In the above, I've used two "precalculus facts". One is the fact that $\sqrt{x^2} = |x|.$ The other is from the definition of absolute value: If $x < 0,$ then $|x| = -x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/71742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Integral solutions of $x^2+y^2+1=z^2$ I am interested in integral solutions of $$x^2+y^2+1=z^2.$$ Is there a complete theory comparable to the one for $x^2+y^2=z^2?$
| the equation: $X^2+Y^2=Z^2-1$
Solutions can be written using the solutions of Pell's equation: $p^2-2k(k-1)s^2=1$
$k$ - given by us.
Solutions have form:
$X=2kps-2(k-1)s^2$
$Y=2(k-1)ps+2ks^2$
$Z=p^2+2(k^2-k+1)s^2$
And more:
$X=2p^2-2(3k-2)ps+2(2k-1)(k-1)s^2$
$Y=2p^2-2(3k-1)ps+2k(2k-1)s^2$
$Z=3p^2-4(2k-1)ps+2(3k^2-3k+1)s^2$
Although it should be a more general solution to record.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/74931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
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"answer_id": 2
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Where is the mistake in this logarithmic equation? The problem is:
Given this equation find the possible values of $x$.
$$\log_{2}\left ( \frac{x+2}{x-1} \right )+\log_{2}\left [ (x+2)(x-1) \right ]=2$$
First I defined the domain of the function.
*
*The denominator $x-1$ must not be zero: $x-1\neq0$.
*The fraction $\frac{x+2}{x-1}$ must be bigger than zero: $\frac{x+2}{x-1}>0$.
*The product $(x+2)(x-1)$ must be bigger than zero: $(x+2)(x-1)>0$.
In the end I concluded that the domain should be $\left ]-\infty,-2\right [\cup \left ]1,+\infty\right[$.
Then I sarted to solve:
$$\begin{align*}
\log_{2}\left ( \frac{x+2}{x-1} \right )&+\log_{2}\left [ (x+2)(x-1) \right ]=2
\\
&\iff
\log_{2}(x+2)-\log_{2}(x-1)+\log_{2}(x+2)+\log_{2}(x-1)=2\\
&\iff
2 \cdot \log_{2}(x+2)=2\\
&\iff \log_{2}(x+2)=1 \\
&\iff x+2=2^1\\
&\iff x=2-2\\
&\iff x=0.
\end{align*}$$
This result is not in the domain!
Where I miss?
| One problem is that you can have both the product and the fraction be valid arguments for $\log_2$, but the individual factors/denominator/numerator not be valid arguments. Specifically, if $x\lt -2$, then neither $x-1$ nor $x+2$ are valid arguments, so you cannot take the first step in general.
Instead, why not work those identities in the other direction?
$$\begin{align*}
\log_2\left(\frac{x+2}{x-1}\right) + \log_2\Bigl( (x+2)(x-1)\Bigr)
&= \log_2\left(\left( \frac{x+2}{x-1}\right)(x+2)(x-1)\right)\\
&= \log_2\left(\bigl(x+2\bigr)^2\right).
\end{align*}$$
Now, if $x+2\gt0$ (and hence, if $x\gt 1$, then this is equal to $2\log_2(x+2)$, from which you conclude that $\log_2(x+2) = 1$, hence $x+2=2$, hence $x=0$, but that is not in the domain.
What does that mean? It means that you cannot find a solution with $x+2\gt 0$.
What if $x+2\lt 0$, though? Then simply go from $\log_2((x+2)^2) = 2$ to $(x+2)^2 = 2^2 = 4$, and hence $-(x+2) = 2$, or $x+2=-2$, hence $x=-4$. This is in the domain, and satisfies $x+2\lt 0$.
Indeed, note that if $x=-4$, then $\frac{x+2}{x-1} = \frac{-2}{-5} = \frac{2}{5}$, and $(x+2)(x-1) = (-2)(-5) = 10$, so you have that the left hand side is
$$\log_2(2/5) + \log_2(10) = \log_2(20/5) = \log_2(4) = 2,$$
so $x=-4$ is a solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/77040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Angle addition formula for cotangent when three angles are being added $$\cot(x+y+z) = \frac{\cot (x) \cot (y) \cot (z) - (\cot (x)+\cot (y)+\cot (z))}{\cot (x)\cot(y)+\cot(y)\cot(z) + \cot(x)\cot(z) - 1 }$$
I know the addition theorem for two variables, but what to do when you have 3 variables in the argument.
| Set $a=x+y$ and $b=z$ and apply$^1$
$$
\cot (a+b)=\frac{\cot a\cot b-1}{\cot b+\cot a}
.$$
--
$^1$From
$$
\cos (a+b)=\cos a\cos b-\sin a\sin b
$$
and
$$
\sin (a+b)=\sin a\cos b+\cos a\sin b
$$
we get
$$
\begin{eqnarray*}
\cot (a+b) &=&\frac{\cos (a+b)}{\sin (a+b)}=\frac{\cos a\cos b-\sin a\sin b}{
\sin a\cos b+\cos a\sin b} \\
&=&\dfrac{\dfrac{\cos a\cos b-\sin a\sin b}{\sin a\sin b}}{\dfrac{\sin a\cos
b+\cos a\sin b}{\sin a\sin b}} \\
&=&\frac{\cot a\cot b-1}{\cot b+\cot a}.
\end{eqnarray*}
$$
Set $a=x+y$ and $b=z$. Then
$$
\begin{eqnarray*}
\cot \left( x+y+z\right) &=&\cot (a+b)=\dfrac{\cot a\cot b-1}{\cot b+\cot a}
\\
&=&\dfrac{\cot \left( x+y\right) \cot z-1}{\cot z+\cot \left( x+y\right) }=
\dfrac{\dfrac{\cot x\cot y-1}{\cot x+\cot y}\cot z-1}{\cot z+\dfrac{\cot x\cot
y-1}{\cot x+\cot y}} \\
&=&\dfrac{\dfrac{\cot x\cot y\cot z-\cot z-\cot x-\cot y}{\cot x+\cot y}
}{\dfrac{\cot x\cot z+\cot y\cot z+\cot x\cot y-1}{\cot x+\cot y}} \\
&=&\dfrac{\cot x\cot y\cot z-\cot x-\cot y-\cot z}{\cot x\cot y+\cot x\cot
z+\cot y\cot z-1}.
\end{eqnarray*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/77586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Sum of a rearranged alternating harmonic series, with three positive terms followed by one negative term The series is this:
$$ 1 + 1/3 + 1/5 - 1/2 + 1/7 + 1/9 + 1/11 - 1/4 + 1/13 + 1/15 + 1/17 - 1/6 ...$$
The hint is to consider partial sums to $4n$ terms. I did that, and got the summation
$$ \sum 1/(6x-5) + 1/(6x-3) + 1/(6x-1) - 1/(2x) $$
But then I got stuck. The terms don't seem to simplify to anything... Wolfram alpha gives the result of this summation as $\log(12)/2$, but I have no idea how to get there.
Any help greatly appreciated.
| Hint:
Try writing the sum as
$$
\sum_{k=0}^\infty \frac{1}{6k+1}+\frac{1}{6k+3}+\frac{1}{6k+5}-\frac{3}{6k+6}\tag{1}
$$
Then use the fact that
$$
\begin{align}
\sum_{k=0}^n\frac{1}{2k+1}
&=\log(2n+1)+\gamma-\tfrac{1}{2}(\log(n)+\gamma)+O\left(\tfrac{1}{n}\right)\\
&=\log(2)+\tfrac{1}{2}(\log(n)+\gamma)+O\left(\tfrac{1}{n}\right)\tag{2}
\end{align}
$$
and that
$$
\begin{align}
\sum_{k=0}^n\frac{1}{2k+2}
&=\tfrac{1}{2}(\log(n+1)+\gamma)+O\left(\tfrac{1}{n}\right)\\
&=\tfrac{1}{2}(\log(n)+\gamma)+O\left(\tfrac{1}{n}\right)\tag{3}
\end{align}
$$
Addendum:
Both formulas $(2)$ and $(3)$ follow from the asymptotic expansion for the partial harmonic series:
$$
\sum_{k=1}^n\frac{1}{k}=\log(n)+\gamma+O\left(\tfrac{1}{n}\right)
$$
where $\gamma$ is the Euler-Mascheroni Constant. Thus,
$$
\begin{align}
\sum_{k=0}^n\frac{1}{2k+1}
&=\sum_{k=1}^{2n+1}\frac{1}{k}-\sum_{k=1}^{n}\frac{1}{2k}\\
&=\log(2n+1)+\gamma-\tfrac{1}{2}(\log(n)+\gamma)+O\left(\tfrac{1}{n}\right)\\
&=\log(2)+\tfrac{1}{2}(\log(n)+\gamma)+O\left(\tfrac{1}{n}\right)
\end{align}
$$
and
$$
\begin{align}
\sum_{k=0}^n\frac{1}{2k+2}
&=\sum_{k=1}^{n+1}\frac{1}{2k}\\
&=\tfrac{1}{2}(\log(n+1)+\gamma)+O\left(\tfrac{1}{n}\right)\\
&=\tfrac{1}{2}(\log(n)+\gamma)+O\left(\tfrac{1}{n}\right)
\end{align}
$$
Using $(2)$, we get
$$
\begin{align}
\sum_{k=0}^n \frac{1}{6k+1}+\frac{1}{6k+3}+\frac{1}{6k+5}
&=\log(2)+\tfrac{1}{2}(\log(3n+2)+\gamma)+O\left(\tfrac{1}{n}\right)\\
&=\tfrac{1}{2}(\log(n)+\log(12)+\gamma)+O\left(\tfrac{1}{n}\right)\tag{4}
\end{align}
$$
and using $(3)$, we get
$$
\begin{align}
\sum_{k=0}^n \frac{3}{6k+6}
&=\tfrac{1}{2}(\log(n)+\gamma)+O\left(\tfrac{1}{n}\right)\tag{5}
\end{align}
$$
Combining $(4)$ and $(5)$, we get
$$
\sum_{k=0}^n \frac{1}{6k+1}+\frac{1}{6k+3}+\frac{1}{6k+5}-\frac{3}{6k+6}=\tfrac{1}{2}\log(12)+O\left(\tfrac{1}{n}\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/78002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
Prove that if $a$ and $b$ are odd, coprime numbers, then $\gcd(2^a +1, 2^b +1) = 3$
Prove that if $a$ and $b$ are odd, coprime numbers, then $\gcd(2^a +1, 2^b +1) = 3$.
I was thinking among the lines of:
Since $a$ and $b$ are coprime numbers, $\gcd(a,b)=1$. Then there exist integers $x$ and $y$ such that, $ax+by=1$.
Then, $a=(1-by)/x$, $b=(1-ax)/y$
So if I write $2^a+1$ as:
$2^{(1-by)/x}+1$
Then can I say that the above expression is equivalent to 0 (mod 3)?
| One of the possible form of odd coprime numbers is:
$a=2m-n$ and $b=m$ , (see this page)
So we may write following equalities:
$2^{a}+1=2^{2m-n}-2+3=2(2^{2m-n-1}-1)+3$
$2^{b}+1=2^{m}-2+3=2(2^{m-1}-1)+3$
So in order to prove that $\gcd(2^{a}+1,2^{b}+1)=3$ we need to prove that:
$\gcd(2^{2m-n-1}-1,2^{m-1}-1)=2^{\gcd(2m-n-1,m-1)}-1=3\cdot p $ for some natural number $p$
The last equality is true only if: $\gcd(2m-n-1,m-1)=2\cdot k$ (even number) for all $m,n$.
Since $m$ and $n$ are odd numbers you can easily show that this is always true.
Similarly you can show for other two forms of $a$ and $b$ that $\gcd(2^{a}+1,2^{b}+1)=3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/78239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Solving simultaneous congruences Trying to figure out how to solve linear congruence by following through the sample solution to the following problem:
$x \equiv 3$ (mod $7$) $x \equiv 2$ (mod $5$) $x \equiv 1$
(mod $3$)
Let:
$n_1$ = 7
$n_2 = 5$
$n_3 = 3$
$N = n_1 \cdot n_2 \cdot n_3 = 105$
$m_1 = \frac{N}{n_1} = 15$
$m_2 = \frac{N}{n_2} = 21$
$m_3 = \frac{N}{n_3} = 35$
$gcd(m_1,n_1)$ = $gcd(15,7) = 1 = 15 \times 1 - 7 \times 2$ so $y_1 = 1$ and $x_1 = 15$
$gcd(m_2,n_2)$ = $gcd(21,5) = 1 = 21 \times 1 - 5 \times 4$ so $y_2 = 1$ and $x_2 = 21$
$gcd(m_3,n_3)$ = $gcd(35,3) = 1 = -35 \times 1 + 3 \times 12$ so $y_3 = -1$ and $x_3 = -35$
I understand up to this point, but the next line I don't get:
So $x = 15 \times 3 + 21 \times 2 - 35 \times 1 \equiv 52$ (mod $105$)
Where is the $\times 3$, $\times 2$, $\times 1$ from? Is it just because there are 3 terms, so it starts from 3 then 2 then 1? And where is the 52 coming from?
| HINT $\rm\quad y\: :=\: x-2 \equiv 0\pmod 5\ $ so $\rm\: y = 5\:j\:.\ $ Therefore
$\rm\ mod\ 3\!:\ {-}1\equiv y = 5\:j \equiv -j\ \Rightarrow\ j\equiv 1\ \Rightarrow\ y = 5\:(1+3\:k) = 5 + 15\:k$
$\rm\ mod\ 7\!:\ \ \ \ 1 \equiv y = 5 + 15\:k\equiv -2 + k\ \Rightarrow\ k\equiv 3\ \Rightarrow\ y = 5+15\:(3+7\:n) = 50 + 105\:n$
NOTE $\ $ Such exploitation of the "law of small numbers" often yields quicker soltuions than rote application of algorithms (Chinese Remainder Theorem, extended Euclidean algorithm, Hermite-Smith normal forms, etc).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/79282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Geometric series? Common ratio of $\sum_{n=0}^\infty 2^{-n}z^{n^2}$ I am trying to find the common ratio of $\sum_{n=0}^\infty 2^{-n}z^{n^2}$. Writing it out in full I got: $\frac{1}{2^1}+(z^1)^1+\frac{1}{2^2}+(z^2)^2+\frac{1}{2^3}+(z^3)^3+...$
So the common ratio is $\frac{1}{2}z^{something?}$
Thanks.
| Let us be systematic about how we find the ratio of consecutive terms. The terms are given by $ a_n = \frac{z^{n^2} }{2^n} $ so the ratio is $$ \frac{a_{n+1} }{a_n } = \frac{ z^{(n+1)^2} }{2^{n+1} } \cdot \frac{2^n}{z^{n^2} }= \frac{ z^{(n+1)^2-n^2} }{2} = \frac{z^{2n+1 }}{2}. $$
Note that this value changes with each $n$ so it is not "common", and this series is not a geometric series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/81025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Ax=b problem solving Can you please tell how to solve this:
Ax=b
$$A=\begin{pmatrix}
2 & 4 & -8 & 4 & 1\\
4 & 10 &-16 & 8 &-8\\
-4 & -12 & 17 & -8 & 20\\
-2 & -10 & 10 & -3 & 34 \end{pmatrix}$$
$$b=\begin{pmatrix}-33\\-36\\4\\-62\end{pmatrix}$$
I have to give a fundamental and
particular solutions.
I have tried so:
But I do not know whether it is right and what to do next ...
Please, help me! Thanks.
| $$\left(\begin{array}{ccccc|c}2&4&-8&4&1&-33\\ 4&10&-16&8&-8&-36\\ -4&-12&17&-8&20&4\\-2&-10&10&-3&34&-62\end{array}\right)\to\left(\begin{array}{ccccc|c}2&4&-8&4&1&-33\\0&2&0&0&-10&30\\ 0&-2&1&0&12&-32\\0&-6&2&1&35&-95\end{array}\right)\to$$
$$\left(\begin{array}{ccccc|c}2&4&-8&4&1&-33\\0&1&0&0&-5&15\\ 0&0&1&0&2&-2\\0&0&2&1&5&-5\end{array}\right)\to\left(\begin{array}{ccccc|c}2&4&-8&4&1&-33\\0&1&0&0&-5&15\\ 0&0&1&0&2&-2\\0&0&0&1&1&-9\end{array}\right)\to$$
$$\left(\begin{array}{ccccc|c}2&0&0&0&33&-73\\0&1&0&0&-5&15\\ 0&0&1&0&2&-2\\0&0&0&1&1&-9\end{array}\right)$$ So your general solution is $(-\frac{73}{2},15,-2,-9,0)+x(-\frac{33}{2},5,-2,-1,1)$. Choosing a particular value of $x$ (e.g., $x=0$) you will get a particular solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/82763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Square Roots of Complex Number $3-4i$
What I did
$z^2=3-4i$
$(a+bi)^2 = 3-4i$
$a^2-b^2+2abi = 3-4i$
Then got 2 simultaneous equations
$a^2-b^2=3$ and $2ab=-4$
Solve for $a^2$ in 1st equation: $a^2=3+b^2$
Subbed into 2nd equation to power of 2
$(3+b^2)b^2=4$
$b^4+3b^2-4=0$
Is there a better way than below? Solving power 4 equation then cubic?
Then solved solved power 4 equation ...
$(b-1)(Ab^2+Bb^2+Cb+D)=b^4+3b^2-4$
$(b-1)(b^3+b^2+4b+4)=b^4+3b^2-4$
then solved for cubic equation, getting ...
$(b-1)(b+1)(b^2+4)=0$
So
$b=\pm 1 \text{ or } 2i$
What did the book do now?
I did: (subbing into $a^2=3+b^2$)
When $b=1, a^2=3+1^2 \Longrightarrow a = 2$
When $b=-1, a^2=3+(-1)^2 \Longrightarrow a=2$
When $b=2i, a^2=3+(2i)^2=3+4(-1)=-1, a=i$
So I will have 3 equations
*
*$2+i$ // why is this not in the book?
*$2-i$
*$i-2$
UPDATE (In response to @David Mitra)
Ok. I let $x=a^2$ and $b=b^2$ giving the quadratic equation: $x=3+y$ and $xy=4$. Then (after subbing) $(y-1)(y+4)=0$
Then when $y=1, x=4$. $y=-4, x=-1$
Then $b=\pm 1 or \pm 2i$ and $a=\pm 2i or \pm i$
Finally testing equations:
$(2+i)^2 = 3+4i$ (rej)
$(2-i)^2=5$ what to do?
$(i-2)^2=3$
$(-2-i)^2=3$
... 8 equations (but 4 unique)
| The value of $a$ is calculated incorrectly when $b=1$. From $2ab=-4$ there is a unique value of $a$ for every value of $b$. When $b=1$, $a$ can only have the value $-2$ (not $+2$ as stated in the question), and then $z = i - 2$.
The quartic equation for $b$, although derived by squaring, is correct and is not responsible for introducing any extra solutions. It could have been derived without squaring, using $ab=−2$. The error is in the step "subbing into $a^2=3+b^2$ ". That quadratic equation derives two values of $a$ from each value of $b$, when we know from $2ab=−4$ that there is only one value of $a$ for each value of $b$. If $2ab=-4$ had been used for the determination of $a$ from $b$ no extra solutions would appear, and every value of $b$ would determine a correct value of $a$.
What did the book do now?
The book does not say, but I guess it meant: eliminate one of the variables using $xy = -2$, obtaining a quartic (degree 4) equation in the other variable that is a quadratic equation in $x^2$ (or $y^2$, whichever is the surviving variable). The quartic has two real and two imaginary roots. Where the book says there are two solutions, that means solutions with real values of $x$ and $y$. These can be found by taking the two real solutions of the quartic and solving for the other variable through $xy=-2$; if one variable is real, the value found for the other variable will also be real.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/84161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Find x in $4^{\sin^2x}+4^{\cos^2x}=8$ $$4^{\sin^2x}+4^{\cos^2x}=8$$
I solved like this:
\begin{align*}4^{\sin^2x}+4^{\cos^2x}=8&\Rightarrow4^{\sin^2x}+4^{1-\sin^2x}=8\\
&\Rightarrow4^{\sin^2x}+\frac{4}{4^{\sin^2x}}=8 |\cdot4^{\sin^2x}\\
&\Rightarrow4^{2\sin^2x}-8\cdot4^{\sin^2x}+4=0\\
y=4^{\sin^2x}&\Rightarrow y^2-8y+4=0\\
&\Rightarrow\Delta=64-16=48\\
&\Rightarrow y_{1,2}=\frac{8\pm 4\sqrt{3}}{2}\\
&\Rightarrow y_{1,2}=4\pm 2\sqrt{3}\\
&\Rightarrow 4^{\sin^2x}=4 \pm 2\sqrt{3}
\end{align*}
But now I'm stuck.
| If you do want complex roots, one of the possible values of $x$ is $\arcsin(\sqrt{\log_4(4+2\sqrt{3})}) = \pi/2 + i t$
where
$$ t = \frac{1}{2} \ \ln \left( {\frac {\ln \left( 2 \right) }{\ln \left( 2+\sqrt {3} \right) +\sqrt {\ln \left( 4+2\,\sqrt {3} \right) \ln \left( 1+1/
2\,\sqrt {3} \right) }}} \right) \approx -.6285882035
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/84641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to get from $2^{99} \cdot (\cos{(99\times \frac{5\pi}{6}) + i\cdot \sin{(99\times \frac{5\pi}{6})}})$ to $0+2^{99}i$?
How do I get from 2nd last to last step? How did they simplify cos & sin $99\times \frac{5\pi}{6}$?
| $99\cdot \frac {5 \pi}{6}=82.5\pi $ and you can ignore the multiples of $2 \pi$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/84785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Show $\frac{d}{dx} \tan^3{x}-3 \tan{x}+3x = 3 \tan^4{x}$ How do I go about doing this?
Show $\frac{d}{dx} \tan^3{x}-3 \tan{x}+3x = 3 \tan^4{x}$.
My work.
$$
\begin{align*}
\frac{d}{dx} \tan^3 x -3 \tan x+3x
&= 3 \tan^2 x \sec^2 x - 3 \sec^2 x + 3
\\ &= 3 \frac{\sin^2 x}{\cos^2 x} \frac{1}{\cos^2 x} - 3 \frac{1}{\cos^2 x} + 3
\\ &= 3 \frac{\sin^2 x}{\cos^4 x} - 3 \frac{1}{\cos^2 x} + 3
\\ &= \frac{3\sin^2 x - 3\cos^2 x + 3\cos^4 x}{\cos^4 x}
\\ &= \frac{3(\sin^2 x - \cos^2 x + \cos^4 x)}{\cos^4 x}.
\end{align*}
$$
I got here so far, did I make a mistake or something?
| One of the outcomes of the Mean Value Theorem is that if two functions have the same derivative, then the differ by a constant. Or, in mathy speak, if $f'(x)=g'(x)$ then there is a number $C$ so that $f(x)=g(x)+C$. If you take a derivative of $f(x)=3\tan^2(x)\sec^2(x)-3\sec^2(x)+3$, you will get
$$
f'(x)=6(\tan(x)\sec^4(x)+\tan^3(x)\sec^2(x)-\sec^2(x)\tan(x).
$$
Getting common denominators and using $\cos^2(x)+\sin^2(x)=1$ this becomes
$$
\frac{12\sin^3(x)}{\cos^5(x)}=12\tan^3(x)\sec^2(x).
$$
But, the right hand side is the derivative of $3\tan^4(x)$. So, there is some $C$ that makes $f(x)=3\tan^4(x)+C$. Just plug in $x=0$ to see that $C=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/85253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
} |
The sum of powers of $2$ between $2^0$ and $2^n$ Lately, I was wondering if there exists a closed expression for $2^0+2^1+\cdots+2^n$ for any $n$?
| As the sum of a geometric sequence, $1+2^1+2^2+\ldots+2^n = \frac{2^{n+1}-1}{2-1}=2^{n+1}-1$.
You can see it by computing $(1+2^1+2^2+\ldots+2^n)(2-1)$ and distribute. All the terms except for $2 \cdot 2^n - 1$ will cancel each other out.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/86838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is there a less ad hoc way to find the degree of this extension? I want to find the degree of $\mathbb{Q}(\sqrt{3+2\sqrt{2}})$ over $\mathbb{Q}$. I observe that
$3+2\sqrt{2}=2+2\sqrt{2}+1=(\sqrt{2}+1)^2$ so
$$
\mathbb{Q}(\sqrt{3+2\sqrt{2}})=\mathbb{Q}(\sqrt{2}+1)=\mathbb{Q}(\sqrt{2})
$$
so the degree is 2.
Is there a more mechanical way to show this without noticing the factorization?
| We can immediately see that $\sqrt{3+2\sqrt{2}}$ is a root of
$$(X^2 -3)^2 -8 = X^4 -6X^2 + 1$$
So we can ask, whether this polynomial is irreducible or not.
The polynomial has not roots in $\mathbb Q$, since for a root $\frac rs \in \mathbb Q$ with $(r,s) = 1$, we would need to have $r|1$, $s|1$, but $\pm1$ is not a root.
So let's try to find factors of degree 2:
$$
\begin{align}
X^4 - 6X^2 + 1 &= (X^2 + aX \pm 1)(X^2 + cX \pm 1) \\
&= X^4 + (a+c)X^3 + (ac\pm 2)X^2 \pm (a+c)X + 1
\end{align}
$$
so $a = -c$ and $-6 = ac \pm 2 = -a^2 \pm 2$. This implies that we must choose $a = 2$, $c=-2$ and the minus-sign for $\pm 1$. We have thus arrived at the factorization
$$X^4 - 6X^2 + 1 = (X^2 + 2X - 1)(X^2 - 2X - 1)$$
Both the factors on the right must be irreducible over $\mathbb Q$ (because they can't have a root in $\mathbb Q$) and $\sqrt{3+2\sqrt{2}}$ must be a root of one of them (we need not even figure out which one).
From this it follows that $[\mathbb Q(\sqrt{3+2\sqrt{2}}):\mathbb Q] = 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/90459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
$\sum \limits_{n=1}^{\infty}n(\frac{2}{3})^n$ Evalute Sum
Possible Duplicate:
How can I evaluate $\sum_{n=1}^\infty \frac{2n}{3^{n+1}}$
How can you compute the limit of
$\sum \limits_{n=1}^{\infty} n(2/3)^n$
Evidently it is equal to 6 by wolfram alpha but how could you compute such a sum analytically?
| Toss a coin that has probability $1/3$ of landing "heads" until we get a head. Let $X$ be the number of tosses required. We find the mean of $X$ in two different ways. Let our sum be $S$.
Note that $P(X=1)=1/3$, $P(X=2)=(2/3)(1/3)$, $P(X=3)=(2/3)^2(1/3)$, and so on. It follows that
$$E(X)=1\cdot\left(\frac{1}{3}\right)+ 2\cdot\left(\frac{2}{3}\right)\left(\frac{1}{3}\right)+ 3\cdot\left(\frac{2}{3}\right)^2\left(\frac{1}{3}\right)+ 4\cdot\left(\frac{2}{3}\right)^3\left(\frac{1}{3}\right)+\cdots.$$
Thus
$$E(X)=\frac{1}{2}\left[1\cdot\left(\frac{2}{3}\right)+ 2\cdot\left(\frac{2}{3}\right)^2+ 3\cdot\left(\frac{2}{3}\right)^3+ 4\cdot\left(\frac{2}{3}\right)^4+\cdots\right]=\frac{S}{2}.$$
If the first toss is a head, then $X=1$. If that the first toss is a tail, we have used up $1$ toss, and the game begins again. By the Law of Total Expectation,
$$E(X)=1\cdot\frac{1}{3} +(1+E(X))\cdot\frac{2}{3}.$$
Solve for $E(X)$. We get $E(X)=3$, and therefore $S=6$.
Comment: There is a very nice book on bijective arguments called Proofs that Really Count. Maybe one should start collecting Mean Proofs.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "9",
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Solve recurrence equations-homework extras Extras from my homework. The first one should be easier, but still hard enough.
1) $a_{n+3}-(3/2)a_{n+2}-a_{n+1}-(1/4)a_n=0$
2) $a_{n+3}-3a_{n+2}-3a_{n+1}+a_n=n^2+2^n$
| First problem
Hint:
For the first one, you can use the same trick of the fibonacci numbers.
The characteristic polynomial for this question is:
$$q^3-\frac{3q^2}{2}-q-\frac{1}{4}=0\;.$$
The second Mathematica says:
Solution:
$$\begin{align*}
a(n)&= c_1 \left(2-\sqrt{3}\right)^n+c_2
\left(2+\sqrt{3}\right)^n+c_3 (-1)^n\\\\
&+\frac{-24 \sqrt{3} \zeta
\left(-2,\frac{n}{2}\right)+48 \zeta \left(-2,\frac{n}{2}\right)+24
\sqrt{3} \zeta \left(-2,\frac{n+1}{2}\right)-48 \zeta
\left(-2,\frac{n+1}{2}\right)}{9 \left(\sqrt{3}-1\right)^3
\left(1+\sqrt{3}\right)}\\\\
&+\frac{12 \sqrt{3} n^2-24 n^2-30 \sqrt{3} n+60
n-2^{n+3}+\sqrt{3} 2^{n+2}+27 \sqrt{3}-54}{9 \left(\sqrt{3}-1\right)^3
\left(1+\sqrt{3}\right)}
\end{align*}$$
| {
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Squeeze an integral Would you have any idea about this problem ?
Prove that for all nonnegative integers $n$, the following inequalities hold:
$$\frac{e^2}{n+3}\leq \int_{1}^{e} x (\ln x)^n \,dx \leq \frac{e^2}{n+2} ,$$
where $\ln$ is the natural logarithm.
Thank you.
| As bleh notes already, if $I_n = \int_1^e x \cdot (\ln x)^n \, dx$, then integration by parts gives
$$
I_n = \frac{e^2}{2} - \frac{n}{2} I_{n-1}.
$$
Now, multiplying the above equation by $\frac{(-2)^n}{n!}$, we get
$$
\frac{(-2)^n I_n}{n!} = \frac{e^2}{2} \cdot \frac{(-2)^n}{n!} + \frac{(-2)^{n-1}}{(n-1)!} I_{n-1}.
$$
Plugging in $n = 1, 2, \ldots, N$, and adding the $N$ equations, we get
$$
\frac{(-2)^N I_N}{N!} = \frac{e^2}{2} \cdot \sum_{n=1}^N \frac{(-2)^n}{n!} + \frac{(-2)^{0}}{0!} I_{0}, \tag{$\ast$}
$$
after a telescopic cancellation. Now $I_0 = \int_1^e x \, dx = \frac{e^2}{2} - \frac{1}{2} = \frac{e^2}{2} (1 - e^{-2})$. Plugging this in $(\ast)$, we get
$$
\frac{(-2)^N I_N}{N!} = \frac{e^2}{2} \cdot \left( \sum_{n=0}^N \frac{(-2)^n}{n!} - e^{-2} \right). \tag{$\ast\ast$}
$$
Using the series expansion $e^{-2} = \sum \limits_{n=0}^{\infty} \frac{(-2)^n}{n!}$, we can rewrite $(\ast\ast)$ as
$$
\frac{(-2)^N I_N}{N!} = - \frac{e^2}{2} \cdot \sum_{n=N+1}^{\infty} \frac{(-2)^n}{n!} ,
$$
or
$$
I_N = e^2 \left( \frac{1}{N+1} - \frac{2}{(N+1)(N+2)} + \frac{4}{(N+1)(N+2)(N+3)} - \cdots \right),
$$
which is an alternating series whose terms are monotone decreasing (in magnitude). Therefore, the odd and even partial sums provide upper and lower bounds on $I_N$.
| {
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Prove $2^{1/3}$ is irrational. Please correct any mistakes in this proof and, if you're feeling inclined, please provide a better one where "better" is defined by whatever criteria you prefer.
*
*Assume $2^{1/2}$ is irrational.
*$2^{1/3} * 2^{x} = 2^{1/2} \Rightarrow x = 1/6$.
*$2^{1/3} * {2^{1/2}}^{1/3} = 2^{1/2}$.
*if $2^{1/2}$ is irrational, then ${2^{1/2}}^{1/3}$ is irrational.
*$2^{1/3} = 2^{1/2} / {2^{1/2}}^{1/3}$.
*$2^{1/3}$ equals an irrational number divided by an irrational number.
*$2^{1/3}$ is an irrational number.
| Let $x=2^{1/3}$ be a rational $\frac{p}{q}$ where $p$ and $q$ are natural numbers having no common factors.
Then $x^3 = 2$, and $x = \frac{x^3}{x^2} = \frac{2}{(2^{1/3})^2} = \frac{2}{\big(\frac{p}{q}\big)^2}$
Hence $x = \frac{p}{q} = \frac{2q^2}{p^2}$.
Since $\frac{p}{q}$ is in its lowest terms, then the second denominator $p^2$ is a multiple of $q$, which is a contradiction unless $q=1$.
But if $q=1$, then $x=2^{1/3}=p$, a natural number, so $x$ is a natural number as well. But $x^3=2$, and no natural number is equal to 2 when cubed. Hence, we have a contradiction, and so $x$ must be irrational, as required.
| {
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} |
are all $n!$ ($n>3$) the difference of two squares? For the small values of n I have been able to check, it seems that for $n>3$, there exist whole numbers $x,y$ s.t. $n! = x^2 - y^2$. For example ..
$4! = 5^2 - 1^2$
$5! = 11^2 - 1^2$
$6! = 27^2 - 3^2$
$7! = 71^2 - 1^2$
$8! = 201^2 - 9^2$
$9! = 603^2 - 27^2$
$10! = 1905^2 - 15^2$
$11! = 6318^2 - 18^2$
$12! = 21888^2 - 288^2$
In most of the cases above, the $x$ value is just the next integer larger than $\sqrt{n!}$, though at $n=12$ and $n=17$ it's the one following that. With the tools at hand I've only been able to check this as far as $n=17$.
I expect there's probably already a name for this, but not knowing that name, googling was coming up dry.
| If $n >3$, then $n!$ is divisible by $4$.
So $n!=4k=(2)(2k)$ for some integer $k$.
Note now that
$$4k=(k+1)^2-(k-1)^2.$$
If $n$ is large, there are many representations of $n!$ as a difference of two squares. For let $2a$ and $2b$ be any two even numbers whose product is $n!$. Then
$$n!=4ab=(a+b)^2-(a-b)^2.$$
Comment: Let $a$ be an odd integer. Then $a+1$ and $a-1$ are even, and therefore $(a+1)/2$ and $(a-1)/2$ are integers. We have
$$a=\left(\frac{a+1}{2}\right)^2-\left(\frac{a-1}{2}\right)^2,$$
so $a$ is a difference of two squares.
If $a$ is divisible by $4$, the argument we gave above shows that $a$ is a difference of two squares.
If $a$ is even but not divisible by $4$, then $a$ is not a difference of two squares. For a difference of two even squares is divisible by $4$, and a difference of two odd squares is divisible by $8$.
| {
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} |
limit of a sequence and principle of induction I try to evaluate this limit:
$$\lim_{n\to+\infty} \frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}$$
I considered this inequality
$$\frac{1}{4n}\le\left [ \frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)} \right]^2\le \frac{1}{2n+1}$$
and so
$$\lim_{n\to+\infty} \frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}=0$$
my questions are:
1)- how do I prove the inequality with the principle of induction?
2)- there is another way to solve this limit?
| I will only show the upper inequality as @JavaMan's comment shows. For $n=1$, it's only $\frac 14\leq\frac 13$, and if it's true for $n$ then
\begin{align*}\left(\prod_{j=1}^{n+1}\frac{2j-1}{2j}\right)^2&=\left(\frac{2n+1}{2(n+1)}\right)^2\left(\prod_{j=1}^n\frac{2j-1}{2j}\right)^2\\
&\leq \left(\frac{2n+1}{2(n+1)}\right)^2\frac 1{2n+1}\\
&=\frac{2n+1}{4(n+1)^2},
\end{align*}
which is $\leq \frac 1{2n+3}$ since $\frac{(2n+1)(2n+3)}{4(n+1)^2}=\frac{4n^2+5n+3}{4n^2+8n+4}\leq 1$.
An other way: put $a_n:=\prod_{j=1}^n\frac{2j-1}{2j}$. Then by $\ln(1+x)\leq x$ for $x\geq -1$ we have
$$\ln a_n =\sum_{j=1}^n\ln\left(1-\frac 1{2j}\right)\leq -\frac 12\sum_{j=1}^n\frac 1j,$$
so $a_n=\exp\left(-\frac 12\sum_{j=1}^n\frac 1j\right)$ and we can conclude since the harmonic series diverges.
| {
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About one generating function Initially, I have the following problem: find $$\sum_{k=0}^{n+1}(−1)^{n−k}4^k{n+k+1 \choose 2k}.$$ I thought, if I found the function $g_n(x) = \sum_{k=0}^{n}{n+k \choose 2k}x^k$, the answer would be $(-1)^ng_{n+1}(-4)$. I tried to factorize ${n + k \choose 2k}$ to ${n+k \choose k}{n \choose k}/{2k \choose k}$. It is known that $$(1-x)^{-n-1} = \sum_{k=0}^{\infty}{n + k \choose k}x^k,$$ $$(1-4x)^{-1/2} = \sum_{k=0}^{\infty}{2k \choose k}x^k,$$ $$(1+x)^n = \sum_{k=0}^n{n \choose k}x^k.$$ I have tried to combine these formulae, but nothing interesting has been found yet.
| Your sum can be rewritten like so:
$$\begin{align}
\sum_{k=0}^{n+1} (-1)^{n-k} 4^k \binom{n+k+1}{2k} &= \sum_{k=0}^{n+1} (-1)^{n-k} 4^k \binom{n+1+k}{n+1-k} \\
&= \sum_{j \geq 0} (-1)^{j-1} 4^{n+1-j} \binom{2(n+1)-j}{j} \\
&= \sum_{j \geq 0} (-1)^{j-1} 4^{m/2-j} \binom{m-j}{j} \\
&= - \frac{1}{2^m}\sum_{j \geq 0} (-1)^j 4^{m-j} \binom{m-j}{j}, \\
\end{align}$$
where we switched indices twice via $k = n+1-j$ and $m = 2n+2$. So what we need is the bivariate generating function of the sum
$$G_m(x,y) = \sum_{j \geq 0} \binom{m-j}{j} x^j y^{m-j}.$$
(Note that when $x = y= 1$ this gives $F_{m+1}$, the $m+1$ Fibonacci number.)
I considered this generating function in this answer. In particular, $G(x,y)$ satisfies the recurrence $$G_m(x,y) = y G_{m-1}(x,y) + xy G_{m-2}(x,y)$$ with initial conditions $G_0(x,y) = 1$, $G_1(x,y) = y$. With $x = -1, y = 4$, we need to solve $$G_m = 4 G_{m-1} -4 G_{m-2},$$
which has auxiliary equation $z^2 - 4z + 4 = 0$. There is a double root at $z = 2$, and so the general solution is $G_m = A2^m + Bm 2^m$. Subbing in the initial conditions, we have $$G_m = 2^m + m 2^m.$$
Therefore,
$$\frac{-1}{2^m} \sum_{j \geq 0} (-1)^j 4^{m-j} \binom{m-j}{j} = \frac{-1}{2^m} (2^m + m 2^m) = -1 -m.$$
Since $m = 2n+2$, subbing back in terms of $n$ gives us the answer you're after:
$$\sum_{k=0}^{n+1} (-1)^{n-k} 4^k \binom{n+k+1}{2k} = -1 -(2n+2) = -2n-3.$$
More generally, the bivariate generating function is $$G_m(x,y) = \frac{\left(y + \sqrt{y^2+4xy}\right)^{m+1} - \left(y - \sqrt{y^2+4xy}\right)^{m+1}}{2^{m+1}\sqrt{y^2+4xy}},$$
provided, of course, that $y^2 + 4xy \neq 0$. However, this is precisely the case that the OP's question falls in, and so we have to go to the recurrence relation to find the solution. (Binet's formula for the Fibonacci numbers is obtained by setting $x = y = 1$, so this is a generalization of that formula.)
| {
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The length of curves. I can't see what I'm doing wrong here, it's very simple.
The length of the curve $x^2$ between $0$ and $4$ is $\int_0^4 \sqrt{(1+4x^2)}dx$ isn't it?
I don't know how to calculate this but wolframalpha tells me that it's about $16.8$ units. That can't be right: maybe it's an optical illusion but I have it drawn up on my board and the curve doesn't look any longer than 8 units.
What am I doing wrong?
EDIT: You know what? I realised my mistake: my y-axis is scaled. Doh!
Feel free to delete this question :)
| Integrating by parts:
$$
\int \sqrt{1+4x^2} \mathrm{d x} = x \sqrt{1+4x^2} - \int \frac{4 x^2}{\sqrt{1+4x^2}} \mathrm{d} x = x \sqrt{1+4x^2} - \int \sqrt{1+4x^2} \mathrm{d} x + \int \frac{\mathrm{d}x }{\sqrt{1+4x^2}}
$$
Solving for $\int \sqrt{1+4x^2} \mathrm{d} x$:
$$
\int \sqrt{1+4x^2} \mathrm{d} x = \frac{x}{2} \sqrt{1+4x^2} + \frac{1}{2} \int \frac{\mathrm{d}x }{\sqrt{1+4x^2}} = \frac{x}{2} \sqrt{1+4x^2} + \frac{1}{4} \operatorname{arcsinh}(2x)
$$
This gives
$$
\int_0^4 \sqrt{1+4x^2} \mathrm{d} x = 2 \sqrt{65} + \frac{1}{4} \operatorname{arcsinh}(8)
\approx 16.8186
$$
| {
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incremental computation of standard deviation How can I compute the standard deviation in an incremental way (using the new value and the last computed mean and/or std deviation) ?
for the non incremental way, I just do something like:
$$S_N=\sqrt{\frac1N\sum_{i=1}^N(x_i-\overline{x})^2}.$$
mean = Mean(list)
for i = 0 to list.size
stdev = stdev + (list[i] - mean)^2
stdev = sqrRoot( stdev / list.size )
| Forgive my poor math background, what I need is detail!
I added my progress here for someone like me.
$$
s_n^2=\frac {\sum_{i=1}^{n}(x_i-\bar{x}_n)^2}{n-1} \\
= \frac {\sum_{i=1}^n(x_i - \bar{x}_{n-1} + \bar{x}_{n-1} - \bar{x}_n)^2}{n-1} \\
= \frac {\sum_{i=1}^{n}(x_i - \bar{x}_{n-1})^2 + 2\sum_{i=1}^n(x_i - \bar{x}_{n-1})(\bar{x}_{n-1} - \bar{x}_n) + \sum_{i=1}^n(\bar{x}_{n-1} - \bar{x}_n)^2} {n-1} \\
= \frac {(\sum_{i=1}^{n-1}(x_i - \bar{x}_{n-1})^2 + (x_n - \bar{x}_{n-1})^2) + (2\sum_{i=1}^{n-1}(x_i - \bar{x}_{n-1})(\bar{x}_{n-1} - \bar{x}_n) + 2(x_n - \bar{x}_{n-1})(\bar{x}_{n-1} - \bar{x}_n)) + \sum_{i=1}^n(\bar{x}_{n-1} - \bar{x}_n)^2} {n-1} \\
= \frac {(n-2)s_{n-1}^2 + (x_n - \bar{x}_{n-1})^2 + 0 + 2(x_n - \bar{x}_{n-1})(\bar{x}_{n-1} - \bar{x}_n)) + n(\bar{x}_{n-1} - \bar{x}_n)^2} {n-1} \\
= \frac {(n-2)s_{n-1}^2 + x_n^2 - 2x_n\bar{x}_{n-1} + \bar{x}_{n-1}^2 + 2x_n \bar{x}_{n-1} - 2x_n \bar{x}_n - 2\bar{x}_{n-1}^2 + 2\bar{x}_{n-1}\bar{x}_n + n\bar{x}_{n-1}^2 - 2n\bar{x}_{n-1}\bar{x}_n + n\bar{x}_n^2} {n-1} \\
= \frac {(n-2)s_{n-1}^2 + x_n^2 + \bar{x}_{n-1}^2 - 2x_n \bar{x}_n - 2\bar{x}_{n-1}^2 + 2\bar{x}_{n-1}\bar{x}_n + n\bar{x}_{n-1}^2 - 2n\bar{x}_{n-1}\bar{x}_n + n\bar{x}_n^2} {n-1} \\
= \frac {(n-2)s_{n-1}^2 + x_n^2 - 2x_n\bar{x}_{n-1} + \bar{x}_{n-1}^2 + 2x_n \bar{x}_{n-1} - 2x_n \bar{x}_n - 2\bar{x}_{n-1}^2 + 2\bar{x}_{n-1}\bar{x}_n + n\bar{x}_{n-1}^2 - 2n\bar{x}_{n-1}\bar{x}_n + n\bar{x}_n^2} {n-1} \\
= \frac {(n-2)s_{n-1}^2 + (x_n^2 - 2x_n\bar{x}_n + \bar{x}_n^2) + (n-1)(\bar{x}_{n-1}^2 - 2\bar{x}_{n-1}\bar{x}_n + \bar{x}_n^2)} {n-1} \\
= \frac {(n-2)s_{n-1}^2 + (n-1)(\bar{x}_{n-1} - \bar{x}_n)^2 + (x_n - \bar{x}_n)^2} {n-1} \\
= \frac {n-2}{n-1}s_{n-1}^2 + (\bar{x}_{n-1} - \bar{x}_n)^2 + \frac {(x_n - \bar{x}_n)^2}{n-1}
$$
and
$$
(\bar{x}_{n-1} - \bar{x}_n)^2 + \frac {(x_n - \bar{x}_n)^2}{n-1} \\
= (\bar{x}_{n-1} - \frac {x_n + (n-1)\bar{x}_{n-1}}{n})^2 + \frac {(x_n - \frac {x_n + (n-1)\bar{x}_{n-1}}{n})^2}{n-1} \\
= \frac {1}{n} (x_n - \bar{x}_{n-1})^2
$$
so
$$
s_n^2 = \frac{n-2}{n-1}s_{n-1}^2 + \frac{1}{n}(x_n - \bar{x}_{n-1})^2
$$
| {
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Sum of n squares $(x_1^2+x_2^2 + \dots + x_n^2)^2 (y_1^2+y_2^2 + \dots + y_n^2) = z_1^2+z_2^2 + \dots + z_n^2$ Consider this 5-Square Identity,
$(x_1^2+x_2^2+x_3^2+x_4^2+x_5^2)^2 (y_1^2+y_2^2+y_3^2+y_4^2+y_5^2) = z_1^2+z_2^2+z_3^2+z_4^2+z_5^2$
where,
$\begin{align}
z_1 &= (-x_1^2+x_2^2+x_3^2+x_4^2+x_5^2)y_1 - 2x_1(0x_1 y_1+x_2 y_2+x_3 y_3+x_4 y_4 + x_5 y_5)\\
z_2 &= (x_1^2-x_2^2+x_3^2+x_4^2+x_5^2)y_2 - 2x_2(x_1 y_1+0x_2 y_2+x_3 y_3+x_4 y_4 + x_5 y_5)\\
z_3 &= (x_1^2+x_2^2-x_3^2+x_4^2+x_5^2)y_3 - 2x_3(x_1 y_1+x_2 y_2+0x_3 y_3+x_4 y_4 + x_5 y_5)\\
z_4 &= (x_1^2+x_2^2+x_3^2-x_4^2+x_5^2)y_4 - 2x_4(x_1 y_1+x_2 y_2+x_3 y_3+0x_4 y_4 + x_5 y_5)\\
z_5 &= (x_1^2+x_2^2+x_3^2+x_4^2-x_5^2)y_5 - 2x_5(x_1 y_1+x_2 y_2+x_3 y_3+x_4 y_4 + 0x_5 y_5)
\end{align}$
The pattern is easily seen for,
$(x_1^2+x_2^2 + \dots + x_n^2)^2 (y_1^2+y_2^2 + \dots + y_n^2) = z_1^2+z_2^2 + \dots + z_n^2$
The case n = 4 is used in Pfister’s 8-square Identity. How to prove the pattern indeed holds true for ALL positive integer n?
| We can write
\begin{align*}
z_k&=y_k\left(\sum_ix_i^2-2x_k^2\right)-2x_k\sum_{i\neq k}x_iy_i\\
&=y_k\sum_ix_i^2-2x_k\left(\sum_{i\neq k}x_iy_i+x_ky_k\right)\\
&=y_k\sum_ix_i^2-2x_k\sum_ix_iy_i
\end{align*}
so hopping we are working in a commutative ring
$$z_k^2=y_k^2\left(\sum_ix_i^2\right)^2-4x_ky_k\left(\sum_ix_i^2\right)\left(\sum_ix_iy_i\right)+4x_k^2\left(\sum_ix_iy_i\right)^2$$
and finally
\begin{align*}
\sum_{k=1}^nz_k^2&=\sum_{k=1}^ny_k^2\left(\sum_ix_i^2\right)^2-4x_ky_k\left(\sum_ix_i^2\right)\left(\sum_ix_iy_i\right)+4x_k^2\left(\sum_ix_iy_i\right)^2\\
&=\left(\sum_ix_i^2\right)^2\sum_{k=1}^ny_k^2-4\left(\sum_kx_ky_k\right)\left(\sum_ix_i^2\right)\left(\sum_ix_iy_i\right)\\
&+4\left(\sum_kx_k^2\right)\left(\sum_ix_iy_i\right)^2\\
&=\left(\sum_ix_i^2\right)^2\sum_{k=1}^ny_k^2.
\end{align*}
| {
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Another symmetric inequality How would one show that for positive $a,b,c,d$ and $a+b+c+d = 4$ that
$$ \frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} \leq \frac{4}{abcd} $$
| Consider
$$\left(\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a}\right)abcd = a^2cd + b^2ad + c^2ab + d^2bc = ac(ad + bc) + bd(ab + cd)$$
Since there is cyclic symmetry, we can assume that $ad + bc \le ab + cd$.
So
$$ac(ad + bc) + bd(ab + cd) \le (ac + bd)(ab + cd)$$
Now $xy \le \left(\frac{x+y}{2}\right)^2$
and so
$$(ac + bd)(ab + cd) \le \left(\frac{ac + bd + ab + cd}{2}\right)^2 = \left(\frac{(a+d)(b+c)}{2}\right)^2$$
Applying $xy \le \left(\frac{x+y}{2}\right)^2$ again we get
$$\left(\frac{(a+d)(b+c)}{2}\right)^2 \le \left(\frac{\left(\frac{a+b+c+d}{2}\right)^2}{2}\right)^2 = 4$$
Thus
$$\left(\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a}\right)abcd \le 4$$
and so
$$\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} \le \frac{4}{abcd}$$
What we have shown is that, for four positive numbers,
$$ \left(\frac{a+b+c+d}{4}\right)^4 \ge abcd\frac{\left(\frac{a}{b} + \frac{b}{c} + \frac{c}{d}+ \frac{d}{a}\right)}{4}$$
and since $\frac{a}{b} + \frac{b}{c} + \frac{c}{d}+ \frac{d}{a} \ge 4$, this inequality is stronger than $\text{AM} \ge \text{GM}$ for $4$ numbers.
Somewhat surprisingly, we only used $\text{AM} \ge \text{GM}$ (twice) to prove it! And for two numbers, a similar inequality is actually false!
| {
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"url": "https://math.stackexchange.com/questions/105065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Prove: $\int_0^\infty \sin (x^2) \, dx$ converges. $\sin x^2$ does not converge as $x \to \infty$, yet its integral from $0$ to $\infty$ does.
I'm trying to understand why and would like some help in working towards a formal proof.
| Just with elementary tools: see Here,How to prove only by Transformation that: $ \int_0^\infty \cos(x^2) dx = \int_0^\infty \sin(x^2) dx $
In fact we have: \begin{split} \int_0^\infty \sin(x^2) dx=\frac{1}{2\sqrt{2}} \int^\infty_0\frac{\sin^2 x}{x^{3/2}}\,dx<\infty\end{split}
See below
Employing the change of variables $2u =x^2$ after integration by parts we get \begin{split} \int_0^\infty \sin(x^2) dx&=&\frac{1}{\sqrt{2}}\int^\infty_0\frac{\sin(2x)}{\sqrt{x}}\,dx\\& =&\frac{1}{\sqrt{2}} \int^\infty_0\frac{\sin(2x)}{x^{1/2}}\,dx\\&=&\frac{1}{\sqrt{2}}\underbrace{\left[\frac{\sin^2 x}{x^{1/2}}\right]_0^\infty}_{=0} +\frac{1}{2\sqrt{2}} \int^\infty_0\frac{\sin^2 x}{x^{3/2}}\,dx \\&= &\frac{1}{2\sqrt{2}} \int^\infty_0\frac{\sin^2 x}{x^{3/2}}\,dx\end{split}
Given that $ \sin 2x =(\sin^2x)'$ and $$\lim_{x\to 0}\frac{\sin x}{x}=1$$
However, $$ \int^\infty_1\frac{\sin^2 x}{x^{3/2}}\,dx\le \int^\infty_1\frac{1}{x^{3/2}}\,dx<\infty$$
since $|\sin x|\le |x|$ we have,
$$\int^1_0\frac{\sin^2 x}{x^{3/2}}\,dx \le \int^1_0\frac{\ x^2}{x^{3/2}}\,dx = \int^1_0\sqrt x\,dx<\infty$$
| {
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"source": "stackexchange",
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Show that $\lim_{n\rightarrow \infty}n \big[1-\frac{(n+1)^{n}}{en^{n}}\big]=\frac{1}{2}$ This is exercise from Guidorizzi's book Cálculo (In Portuguese).
Show that $\lim_{n\rightarrow \infty}n \big[1-\frac{(n+1)^{n}}{en^{n}}\big]=\frac{1}{2}.$
All I managed to do is rewrite the equation as $n \big[1-\frac{(1+\frac{1}{n})^{n}}{e}\big]$.
What to do from now on?
| $$\log\left(1 + \frac{1}{n}\right)^n = n \log \left(1+\frac{1}{n}\right) = n\left(\frac{1}{n}-\frac{1}{2n^2} + O\left(\frac{1}{n^3}\right)\right)$$
Thus
$$\left(1 + \frac{1}{n}\right)^n =e^{1 - \frac{1}{2n} + O\left(\frac{1}{n^2}\right)} $$
Substituting, we need to find
$$\lim_{n \to \infty} n\left(1 - \frac{e^{1 - \frac{1}{2n} + O\left(\frac{1}{n^2}\right)}}{e}\right) = n\left(1-e^{\frac{-1}{2n} + O\left(\frac{1}{n^2}\right)}\right)$$
Now using the expansion of $e^x$,
you get
$$n\left(1-e^{\frac{-1}{2n} + O\left(\frac{1}{n^2}\right)}\right) = n\left(\frac{1}{2n} + O\left(\frac{1}{n^2}\right)\right)$$
which tends to $\frac{1}{2}$ and $n \to \infty$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove by induction (formula for $\sum^n(3i-1)^2$) Anyone knows how to do this? The answer I'm getting is not correct.
Prove by induction that, for all integers $n\ge1$, $$\sum_{i=1}^n (3i-1)^2 = \frac12 n(6n^2 + 3n - 1). $$
Thanks
This Is what I have managed to get. After this I think I'm doing something wrong:
$$ \frac12 n(6n^2 + 3n - 1 + 6n + 4) (3n +2). $$
| You prove it the usual way:
Base. Establish the result for $n=1$.
If $n=1$, then the left hand side is
$$\sum_{i=1}^1(3i-1)^2 = (3(1)-1)^2 = 4,$$
and the right hand side is
$$\frac{1}{2}(1)\left(6(1)^2 + 3(1) - 1\right) = \frac{1}{2}(8) = 4.$$
So we have equality.
Inductive Step.
Induction Hypothesis. The result holds for $k$; that is,
$$\sum_{i=1}^k(3i-1)^2 = \frac{1}{2}k(6k^2 + 3k - 1).$$
To prove. The result holds for $k+1$.
$$\begin{align*}
\sum_{i=1}^{k+1}(3i-1)^2 &= \left(\sum_{i=1}^k(3i-1)^2\right) + (3(k+1)-1)^2\\
&= \frac{1}{2}k(6k^2+3k-1) + (3(k+1)-1)^2 &\text{(by the Induction Hypothesis)}\\
&=\frac{1}{2}k(6k^2+3k-1) + (3k+2)^2\\
&=\frac{1}{2}k(6k^2+3k-1) + (9k^2 + 12k+4)\\
&= \frac{1}{2}\left(6k^3 + 3k^2 - k + 18k^2 + 24k + 8\right)\\
&= \frac{1}{2}\left(6k^3 + 21k^2 + 23k + 8\right)\\
&=\frac{1}{2}\left(6(k^3 + 3k^2 + 3k + 1) + 3k^2 + 5k + 2\right)\\
&=\frac{1}{2}\left(6(k+1)^3 + 3(k^2 + 2k + 1) -k - 1\right)\\
&=\frac{1}{2}\left(6(k+1)^3 + 3(k+1)^2 - (k+1)\right)\\
&=\frac{1}{2}(k+1)\left(6(k+1)^2 + 3(k+1) - 1\right).
\end{align*}$$
As this is exactly what we needed to show for $k+1$, this proves the inductive step.
Conclusion. The given formula holds for $n=1$; and if it holds for $k$, then it also holds for $k+1$. By mathematical induction, the given formula holds for all positive integers $n$. $\Box$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving the inequality $\frac{x}{\sqrt{x+12}} - \frac{x-2}{\sqrt{x}} > 0$ I'm having troubles to solve the following inequality..
$$\frac{x}{\sqrt{x+12}} - \frac{x-2}{\sqrt{x}} > 0$$
I know that the result is $x>0$ and $x<4$ but I cannot find a way to the result..
Thanks in advance!
| First write the inequality as
$$
{x\sqrt x-(x-2)\sqrt{x+12}\over \sqrt x\sqrt{x+12} }>0
$$
This holds if and only if $x>0$ and $x\sqrt x-(x-2)\sqrt{x+12}>0$.
We have
$$\eqalign{
&x\sqrt x-(x-2)\sqrt{x+12}>0\cr
\iff&x\sqrt x>(x-2)\sqrt{x+12}\cr
}
$$
For $x>2$, squaring both sides of the above gives
$$x^3>(x-2)^2 (x+12)$$
or
$$
0>4(2x^2-11x+12)
$$
or
$$
0>(2x-3)(x-4).
$$
The solution to the above is $3/2<x<4$.
Thus for $x>2$, the only solutions to the original inequality are $2<x<4$.
For $0<x\le2$, we obviously have a solution to the original inequality (since both terms will be positive for $0<x<2$, and for $x=2$, we have the inequality ${2\over\sqrt{14}}>0$).
So the solution set is $(0,4)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find surface area of $x=\sqrt{a^2-y^2}$ I still hard time to find surface area of function...
I have
The given curve is rotated about the $y$-axis. Find the area of the resulting surface. $$x= \sqrt{a^2-y^2},\quad y\in\left[0,\frac{a}{2}\right]$$
If you have any idea, would you post your idea ? Thank you :)
| We have formula
$$\text{Surface area } S = \int 2\pi x \sqrt{1+\left(\frac{dx}{dy}\right)^2} dy$$
we have $x= \sqrt{a^2-y^2}$ and $\frac{dx}{dy}= \frac{-y}{\sqrt{a^2-y^2}}$. Hence we have,
$$S=\int_0^{\frac{a}{2}}2\pi \sqrt{a^2-y^2}\sqrt{1+ \frac{y^2}{{a^2-y^2}}} dy$$
$$=\int_0^{\frac{a}{2}}2\pi \sqrt{a^2-y^2} \frac{a}{\sqrt{a^2-y^2}}dy $$
$$=\int_0^{\frac{a}{2}}2\pi .a dy$$
$$=\pi .a^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/113858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $a^2 + b^2 + c^2 $ is not a prime number I am having difficulty solving this problem:
Let $a, b, c \in\mathbb{Z}$, $abc \neq 0$ and $a\neq c$ be such that
$$\frac{a}{c} = \frac{a^2+b^2}{c^2+b^2}.$$
Prove that $a^2 + b^2 + c^2$ is not a prime number.
Thanks in advance!
| $\rm\ a^2+b^2+c^2 = (a+c)^2-b^2 + 2\: (b^2-ac)\ $ so $\rm\:b^2\! = ac\:\Rightarrow\:$ it factors (difference of squares)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Proving that $\binom{n}{k}\binom{\smash{k}}{m}\binom{m}{r} = \binom{n}{r}\binom{n-r}{n-m}\binom{n-m}{n-k}$ How would you show that
$$\binom{n}{k}\binom{k}{m}\binom{m}{r} = \binom{n}{r}\binom{n-r}{n-m}\binom{n-m}{k-m}$$
for $n\geq k\geq m\geq r$ ?
| $$
\begin{align}
& \binom{n}{k}\binom{k}{m}\binom{m}{r} \\
= & \frac{n!}{(n-k)! \cdot k!} \cdot \frac{k!}{(k-m)! \cdot m!} \cdot \frac{m!}{(m-r)! \cdot r!} \\
= & \frac{n! \cdot k! \cdot m!}{(n-k)! \cdot k! \cdot (k-m)! \cdot m!\cdot (m-r)! \cdot r!} \\
= & \frac{n!}{(n-k)! \cdot (k-m)! \cdot (m-r)! \cdot r!} \cdot \frac{(n-r)!}{(n-r)!} \\
= & \frac{n!}{r! \cdot (n-r)!} \cdot \frac{(n-r)!}{(n-k)! \cdot (k-m)! \cdot (m-r)!} \\
= & \frac{n!}{r! \cdot (n-r)!} \cdot \frac{(n-r)!}{(n-k)! \cdot (k-m)! \cdot (m-r)!} \cdot \frac{(n-m)!}{(n-m)!} \\
= & \frac{n!}{r! \cdot (n-r)!} \cdot \frac{(n-r)!}{(n-m)! \cdot (m-r)!} \cdot \frac{(n-m)!}{(k-m)! \cdot (n-k)!} \\
= & \frac{n!}{r! \cdot (n-r)!} \cdot \frac{(n-r)!}{(n-m)! \cdot ((n-r)-(n-m))!} \cdot \frac{(n-m)!}{(k-m)! \cdot ((n-m)-(k-m))!} \\
= & \binom{n}{r}\binom{n-r}{n-m}\binom{n-m}{k-m}
\end{align}
$$
Q.E.D.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/114311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Elementary Geometry The side of the square measures $1\ \mathrm{cm}$ , and $AC = 1\ \mathrm{cm}$, find the value of $AB$
| First try:
There are 3 rectangular triangles here, where we get 3 Pythagoras from:
$$
\begin{eqnarray*}
1^2+(1-d)^2&=&c^2\\
a^2+d^2&=&1^2\\
(1+a)^2+1^2&=&(1+c)^2,
\end{eqnarray*}
$$
with $a=\overline{AB}$,$d=\overline{BC}$ and $c=\overline{0C}$, where $0$ is the upper left point. With this, we ask Wolfram to get
$$
d=\overline{AB}\approx 0.883204 = \frac{1}{2}\left( \sqrt{2}-1+\sqrt{2\sqrt{2}-1} \right)
$$
Second try:
We have $\sin \alpha= \frac{a}{1}= \frac{1+a}{1+c}$ and $1+(1-d)^2=c^2$.
Further, since $d=\cos( \arcsin (a) )=\sqrt{1-a^2}$, we get
$$
a=\frac{1+a}{1+\sqrt{1+(1-d)^2}}=\frac{1+a}{1+\sqrt{1+(1-\sqrt{1-a^2})^2}}, \tag{*}
$$
with the same solution as above.
$(*)$ can be reformulated to
$$
a^8-2a^6+7a^4-6a^2+1=A^4-2A^3+7A^2-6A+1=0,
$$
where the roots of polynomial of $4$th order can be found here. But because it works out nice, let's go one step further and substitute $a=u+1/2$, so we get:
$$
u^4+\frac{11}{2}u^2-\frac{7}{16}=0.
$$
From here on, the rest is left to you. Nice Question!
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove: $\frac{n^5}5 + \frac{n^4}2 + \frac{n^3}3 - \frac n {30}$ is an integer for $n \ge 0$ I am attempting to prove the following problem:
Prove that $\frac{n^5}5 + \frac{n^4}2 + \frac{n^3}3 - \frac n {30}$ is an integer for all integers $n = 0,1,2,...$
I attempted to solve it by induction, but when proving for $n= x+1$ the algebra gets very messy very fast. I was wondering if this is the only way or if there is a quicker way to prove this. I guess I am a little unsure as to how to prove something is an integer.
I also noticed that letting $f(x) = \frac{x^5}5 + \frac{x^4}2 + \frac{x^3}3 - \frac x{30}$ and deriving $f(x)$ yields a fairly clean result, but I don't know if this helps me at all. Any help would be great.
| First write over a common denominator:
$$\frac{n^5}5 + \frac{n^4}2 + \frac{n^3}3 - \frac n{30} = \frac{1}{30}(6n^5+15n^4+10n^3-n)$$
Thus we want to show that 30 always divides expression in parenthesis. Since $\gcd(5,6)=1$, it suffices to prove divisibility by 5 and 6 individually. My first attempt was to factor this polynomial (using a computer, as I'm lazy):
$$6n^5+15n^4+10n^3-n=n(n+1)(2n+1)(3n(n+1)-1).$$
Observe that $6\sum_{k=1}^n k^2=n(n+1)(2n+1)$. Hence the expression is divisible by 6. You can check that 5 does not divide the remaining factor $3n(n+1)-1$, so we need a different approach.
Instead, we reduce the original expression mod 5, that is,
$$6n^5+15n^4+10n^3-n \equiv n^5-n \;\text{mod}\;5.$$
The computer tells me that the expression on the right appears to be divisible by 5. Can you prove this by induction (or even easier Fermat's little theorem)?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Can $f(x)=\sin(x) + \cos^2(x)$ take the value $\sqrt{2}$? Can $f(x)$ take the value of $\sqrt{2}$, where
$$f(x) = \sin(x) + (\cos(x))^2\quad ?$$
When equating the value of $f(x)$ to $\sqrt{2}$, it gives imaginary values of $\sin(x)$.
Thanks in advance.
| Since $\cos^2x=1-\sin^2 x$, our equation becomes $\sin^2 x-\sin x+\sqrt{2}-1=0$.
Solve the equation $w^2-w+\sqrt{2}-1=0$. The roots are
$$w=\frac{1\pm \sqrt{1-4(\sqrt{2}-1})}{2}.$$
The discriminant $1-4(\sqrt{2}-1)$ is negative, so our quadratic in $w$ has no real roots. It follows that there cannot be a real number $x$ such that $\sin x+\cos^2 x=\sqrt{2}$.
This does not fully answer the question. We can ask whether there are complex non-real solutions. For that, look at the two non-real values of $w$ obtained above. Call them $w_1$ and $w_2$. Recall that if $z$ is a complex number, then
$$\sin z=\frac{e^{iz}-e^{-iz}}{2i}.$$
So we want to solve the equations
$$e^{iz}-e^{-iz}=2iw_j \quad (j=1,2).$$
Multiply both sides by $e^{iz}$, and simplify. We get
$$e^{2iz}-2iw_je^{iz}-1=0.$$
Make the substitution $u=e^{iz}$. We arrive at the equations
$$u^2-2iw_j u-1=0 \quad (j=1,2).$$
These are a quadratic equation with complex coefficients. One can write down the solutions in more or less the usual way, and from them obtain the values of $z$ that work. There are infinitely many of them, just like there are, for example, infinitely many $x$ such that $\sin x=1/2$.
Remark To show that there are no real solutions, one does not need to recall the Quadratic Formula. We are interested in the equation $\sin^2 x-\sin x-1+\sqrt{2}=0$.
If we complete the square, we get
$$\sin^2 x-\sin x-1=\left(\sin x-\frac{1}{2}\right)^2-\frac{5}{4}.$$
Since $\left(\sin x-\frac{1}{2}\right)^2\ge 0$, it follows that
$\sin^2 x-\sin x-1$ must always be $-\frac{5}{4}$. Since $\sqrt{2}-\frac{5}{4}$ is positive, we conclude that we cannot have $\sin^2 x-\sin x-1+\sqrt{2}=0$.
.
| {
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Proving $x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$ In Spivak's Calculus 3rd Edition, there is an exercise to prove the following:
$$x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$$
I can't seem to get the answer. Either I've gone wrong somewhere, I'm overlooking something, or both. Here's my (non) proof:
$$\begin{align*}
x^n - y^n &= (x - y)(x^{n-1} + x^{n-2}y +\cdots+ xy^{n-2} + y^{n-1}) \\
&= x \cdot x^{n-1} + x \cdot x^{n-2} \cdot y + \cdots + x \cdot x \cdot y^{n-2} + x \cdot y^{n-1}\\
&\qquad + (-y) \cdot x^{n-1} + (-y) \cdot x^{n-2} \cdot y + \cdots + (-y) \cdot x \cdot y^{n-2} + (-y) \cdot y^{n-1}\\
&= x^n + x^{n-1} y + \cdots + x^2 y^{n-2} + x y^{n-1} - x^{n-1}y - y^2 x^{n-2} - \cdots- x y^{n-1} - y^n \\
&= x^n + x^2 y^{n-2} - x^{n-2} y^2 - y^n \\
&\neq x^n - y^n
\end{align*}$$
Is there something I can do with $x^n + x^2 y^{n-2} - x^{n-2} y^2 - y^n$ that I'm not seeing, or did I make a mistake early on?
EDIT:
I should have pointed out that this exercise is meant to be done using nine of the twelve basic properties of numbers that Spivak outlines in his book:
*
*Associate law for addition
*Existence of an additive identity
*Existence of additive inverses
*Commutative law for additions
*Associative law for multiplication
*Existence of a multiplicative identity
*Existence of multiplicative inverses
*Commutative law for multiplication
*Distibutive law
| Your method is sound, you just made a sort of arithmetic mistake. When cancelling or otherwise combining two sequences, try explicitly lining things up to make sure you do it right:
$$ \begin{align} x^n &+& x^{n-1} y &+& x^{n-2} y^2 &+& \cdots + x y^{n-1} &
\\ &-& x^{n-1} y &-& x^{n-2} y^2 &+& \cdots - x y^{n-1} &+& y^n
\end{align}
$$
I've found that, when shorthand starts becoming awkward and/or error prone, that it really is helpful to switch to summation notation. So, you are trying to prove
$$ x^n - y^n = (x-y) \sum_{k=0}^{n-1} x^k y^{n-1-k} $$
and the first stem of your work would be
$$ \cdots = \left( \sum_{k=0}^{n-1} x^{k+1} y^{n-1-k} \right)
- \left( \sum_{k=0}^{n-1} x^{k} y^{n-k} \right)$$
and now, we can change the index to line things up: I'm substituting k = j-1:
$$ \cdots = \left( \sum_{(j-1)=0}^{n-1} x^{(j-1)+1} y^{n-1-(j-1)} \right)
- \left( \sum_{k=0}^{n-1} x^{k} y^{n-k} \right)$$
and simplifying
$$ \cdots = \left( \sum_{j=1}^{n} x^{j} y^{n-j} \right)
- \left( \sum_{k=0}^{n-1} x^{k} y^{n-k} \right)$$
and now replacing $j$ with $k$.
$$ \cdots = \left( \sum_{k=1}^{n} x^{k} y^{n-k} \right)
- \left( \sum_{k=0}^{n-1} x^{k} y^{n-k} \right)$$
(can you take it from here?)
| {
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convergence of $\sum \limits_{n=1}^{\infty }\bigl\{ \frac {1\cdot3 \cdots (2n-1)} {2\cdot 4\cdots (2n)}\cdot \frac {4n+3} {2n+2}\bigr\} ^{2}$ I am investigating the convergence of
$$\begin{split}\sum _{n=1}^{\infty }\left\{ \dfrac {1\cdot 3\cdots (2n-1)} {2\cdot 4\cdots (2n)}\cdot \dfrac {4n+3} {2n+2}\right\} ^{2} &= \sum _{n=1}^{\infty }\left\{ \dfrac {\prod _{t=1}^n (2t-1)} {\prod _{t=1}^n (2t)}\cdot \dfrac {4n+3} {2n+2}\right\} ^{2} \\
&=\sum _{n=1}^{\infty }\left\{ \prod _{t=1}^n\left( 1-\dfrac {1} {2t}\right) \dfrac {4n+3} {2n+2}\right\} ^{2}
\end{split}$$
which after some manipulations I have reduced to
$$\sum _{n=1}^{\infty }e^ \left\{ 2\ln \left(2 -\dfrac {1} {2n+2}\right) +2\cdot \sum _{t=1}^{n}\ln \left( 1-\dfrac {1}{2t}\right) \right\} $$
and from an alternative approach I was able to reduce it to
$$\sum _{n=1}^{\infty } \dfrac{\left( 4n+3\right) ^{2}}{4\left(n+1\right)^{2}} \prod _{t=1}^n\left( 2+\dfrac{1}{2t^{2}}-\dfrac{2}{t}\right)$$
I am unsure how to proceed from here in either of the two cases. Any help would be much appreciated.
| Denote by $a_n$ the general term, which is positive. We can rewrite it as $\left(\frac{(2n)!}{4^nn!n!}\right)^2\left(\frac{4n+3}{2n+2}\right)^2$, which is equivalent to $b_n:=4\left(\frac{(2n)!}{4^nn!n!}\right)^2$. Now we use Stirling's formula, which states that $n!\overset{+\infty}{\sim}\left(\frac ne\right)^n\sqrt{2\pi n}$. We get
\begin{align*}
b_n&\overset{+\infty}{\sim} 4\left(\frac{\left(\frac{2n}e\right)^{2n}\sqrt{4n\pi}}{4^n\left(\frac ne\right)^{2n}2\pi n}\right)^2\\
&=\frac 4{n\pi},
\end{align*}
and using the fact that the harmonic series diverges, we get that the series $\sum_n a_n$ is divergent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/118383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
Given that $a+b\sqrt[3]{2} +c\sqrt[3]{4} =0$, where $a,b,c$ are integers. Show $a=b=c=0$ Given that $\displaystyle{a+b\sqrt[3]{2} +c\sqrt[3]{4} =0}$, where $a,b,c$ are integers. Show $a=b=c=0$
Do I use modular arithmetic?
| Start with $b \sqrt[3]{2} + c \sqrt[3]{4} = -a$. Cube this relation to find another
equation of the form $B \sqrt[3]{2} + C \sqrt[3]{4} = -A$ for rationals A, B, C. Eliminating
the cube root of 4 from these equations will tell you that the cube root of 2 is rational.
This contradiction shows that $a = b = c = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/120489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 0
} |
N Choose K Confusion $$
\binom{kn}{2} = k\binom{n}{2} + \binom{k}{2} n^2,n,k ≥ 2
$$
Im not sure how to prove algebraically this. Can someone help me?
| Since $\binom{m}{2} = \frac{m(m-1)}{2}$, you are seeking to prove:
$$
\frac{k n (k n -1)}{2} = k \frac{n(n-1)}{2} + n^2 \frac{k(k-1)}{2} \stackrel{\text{expand}}{=} {\color\red{ k \frac{n^2}{2}}} - k \frac{n}{2} + n^2 \frac{k^2}{2} {\color\red{- n^2 \frac{k}{2}}} \stackrel{\text{factor}}{=} k n \frac{k n -1}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/120795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $\lim_{n\rightarrow \infty} {a^n \over n!}=0$
Possible Duplicate:
How to prove that $\lim\limits_{n \to \infty} \frac{k^n}{n!} = 0$
As the topics, how to prove $\lim\limits_{n\rightarrow \infty} \dfrac{a^n}{n!}=0$
$\forall a \in \mathbb{R^+}$
| Let's suppose $a$ is... I don't know, $5$ for a moment. And let's look at the sequence.
$\dfrac{5}{1}, \dfrac{5^2}{2\cdot 1}, \dfrac{5^3}{3\cdot 2}, \dfrac{5^4}{4 \cdot 3 \cdot 2}, \dfrac{(5) \cdot 5^4}{(5) \cdot 4 \cdot 3 \cdot 2}, \dfrac{(5^2) \cdot 5^4}{(6 \cdot 5) \cdot 4!}, \dfrac{(5^3) \cdot 5^4}{(7 \cdot 6 \cdot 5) 4!}, \ldots$
So in particular, after $n = 5$, we have a constant multiplied by something bounded by $\left(\frac{5}{6}\right)^n$, and thus it goes to 0.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/121386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
} |
Convergence of $\sum_{n=1}^{\infty} \frac{n+5}{\sqrt[3]{n^7+n^2-n}}$ Convergence of
$$\sum_{n=1}^{\infty} \frac{n+5}{\sqrt[3]{n^7+n^2-n}}$$
I think ratio test will be very tedious, root test too since its not a exponential equation. So I tried limit comparison test, noting that
$$\frac{n+5}{\sqrt[3]{n^7+n^2-n}} \approx \frac{n}{n^{7/3}} \approx \frac{1}{n^{4/3}} \approx \frac{1}{n^2}, \qquad \frac{1}{n^2} \text{ diverges}$$
$$\lim_{n\to\infty} \frac{n+5}{\sqrt[3]{n^7 + n^2 - n}} \cdot n^2 = \lim_{n\to\infty} \frac{n^3+5n^2}{\sqrt[3]{n^7 + n^2 - n}}$$
Now, should I multiply top & bottom by $\frac{1}{n^3}$? But will the bottom become
$$= \lim_{n\to\infty} \frac{1+\frac{5}{n}}{\sqrt[3]{\frac{n^7 + n^2 - n}{n^{27}}}} = \frac{1}{0}$$
Which seems wrong?
| Seems a straightforward comparison test will also work.
$$
0\le \sum_1^\infty \frac{n+5}{\sqrt[3]{n^7 +n^2 - n}}
\le \sum_1^\infty \frac{n+5n}{\sqrt[3]{n^7}} = 6\sum_1^\infty \frac{1}{n^\frac43}.
$$
On the right we have a convergent p-series. The second inequality is valid because $5\le 5n$ and $n^2-n\ge 0$ for $n\ge 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/121700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Induction Proof: Proof of Strict Inequality involving Exponents of 3 in the Denominator. I have been trying to do this problem by using induction but I became stuck halfway through:
Use induction to show that
$$2\left(1+ \frac{1}{8} + \frac{1}{27} + \cdots + \frac{1}{n^3}\right) \lt 3 - \frac{1}{n^2}$$
for $n\geq 2$.
Does the series $$\sum_{n=1}^{\infty}\frac{1}{n^3}$$ converge? Justify your conclusions.
So far I have this:
Base Case of Induction, $n=2$
$$\begin{align*}
\frac{1}{n^3} &\lt 3- \frac{1}{n^2}\\
\frac{1}{8} &\lt 3 - \frac{1}{4}\\
\frac{1}{8} &\lt \frac{11}{4}
\end{align*}$$
Induction Step: Assume true for some $k \geq 2 :$
$$
2\left(1+ \frac{1}{8} + \frac{1}{27} + \cdots + \frac{1}{k^3}\right) \lt 3 - \frac{1}{k^2}$$
Show true with $n= k+1$
$$ 2 + 4 + \frac{2}{27} + \cdots + \frac{2}{k^3} + \frac{2}{(k+1)^3} \lt 3-\frac{1}{(k+1)^2}.$$
From there, I have no idea what to do.
I was planning to have
$$3 - \frac{1}{k^2} + \frac{2}{(k+1)^3} \lt 3 - \frac{1}{(k+1)^2}$$
but I feel like it won't work since
$$2\left(1 + \frac{1}{8} + \frac{1}{27}+\cdots+\frac{1}{k^3}\right)$$
does not equal
$$3 - \frac{1}{k^2}.$$
| When you go from
$$2\left(1+ \frac{1}{8} + \frac{1}{27} + \cdots + \frac{1}{k^3}\right) \lt 3 - \frac{1}{k^2}\tag{1}$$
to
$$2\left(1+ \frac{1}{8} + \frac{1}{27} + \cdots + \frac{1}{k^3}+\frac1{(k+1)^3}\right) \lt 3 - \frac{1}{(k+1)^2}\;,\tag{2}$$
the righthand side increases by
$$\left(3-\frac1{(k+1)^2}\right)-\left(3-\frac1{k^2}\right)=\frac1{k^2}-\frac1{(k+1)^2}\;,$$
while the lefthand side increases by $\dfrac2{(k+1)^3}$. If you can show that $$\frac2{(k+1)^3}\le\frac1{k^2}-\frac1{(k+1)^2}\;,\tag{3}$$ you’ve shown that $(1)$ implies $(2)$, which is your induction step; do you see why?
Now $(3)$ can by multiplied through by $k^2(k+1)^3$ to yield the equivalent inequality $$2k^2\le (k+1)^3-k^2(k+1)\;,$$ which in turn simplifies to $0\le 3k+1$. This is certainly true for $k\ge 2$, and it’s equivalent to $(3)$, so $(3)$ is true for $k\ge 2$, and your induction step is done.
For the second part of the question, note that the $n$-th partial sum of the infinite series is $$\sum_{k=1}^n\frac1{k^3}\;,$$ and from the first part of the question you know that $$2\sum_{k=1}^n\frac1{k^3}<3-\frac1{n^2}\;,$$ so $$\sum_{k=1}^n\frac1{k^3}<\frac32-\frac1{2n^2}\;.$$ Thus, the partial sums form an increasing sequence bounded above by ... what? And what can you conclude from that about the convergence of the infinite series?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Cauchy by Definition I'm a bit confused as to how to show a sequence is Cauchy by definition. For example let $a_n = 1 - \frac{2}{\sqrt{n}}$.
$$\left|\left( 1 - \frac{2}{\sqrt{n}}\right) - \left( 1 - \frac{2}{\sqrt{m}}\right)\right| = \left|\frac{2}{\sqrt{m}} - \frac{2}{\sqrt{n}}\right|$$
And let $m \geq n$. Then this is where I got lost in class: $|\frac{2}{\sqrt{m}} - \frac{2}{\sqrt{n}}| \leq |\frac{2}{\sqrt{n}} + \frac{2}{\sqrt{n}}|$.
Can someone explain why the last step is justified? Is it because $m \geq n \implies \frac{2}{\sqrt{m}} \leq \frac{2}{\sqrt{n}}$ and $\frac{2}{\sqrt{m}} - \frac{2}{\sqrt{n}} \leq 0 \leq \frac{2}{\sqrt{n}} + \frac{2}{\sqrt{n}}$? But the inequality may not hold when taking absolute values of both sides...
| In the situation you have, with $m\geq n$, we have
$$\left|\frac{2}{\sqrt{m}} - \frac{2}{\sqrt{n}}\right| = \frac{2}{\sqrt{n}} - \frac{2}{\sqrt{m}} \leq \frac{2}{\sqrt{n}} + \frac{2}{\sqrt{n}} = \left|\frac{2}{\sqrt{n}} +\frac{2}{\sqrt{n}}\right|$$
because
$$-\frac{2}{\sqrt{m}} \leq 0 \leq \frac{2}{\sqrt{n}},$$
so adding $\frac{2}{\sqrt{n}}$ to both sides leads to the last inequality.
Alternatively,
$$\begin{align*}
\left|\frac{2}{\sqrt{m}} - \frac{2}{\sqrt{n}}\right| &\leq \left|\frac{2}{\sqrt{m}} \right| + \left|\frac{2}{\sqrt{n}}\right| &\text{(triangle inequality)}\\
&=\frac{2}{\sqrt{m}} + \frac{2}{\sqrt{n}}\\
&\leq \frac{2}{\sqrt{n}} + \frac{2}{\sqrt{n}}\\
&= \left|\frac{2}{\sqrt{n}} + \frac{2}{\sqrt{n}}\right|.
\end{align*}$$
You prove a sequence $\{a_n\}$ is Cauchy "by definition" by showing that for every $\epsilon\gt 0$ there exists an $N$ (which will depend on $\epsilon$) such that for all $n,m\geq N$ we have $|a_n-a_m|\lt \epsilon$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
computing a trignometric limit I am trying to show that $$\lim _{n\rightarrow \infty }\dfrac {1+\cos \dfrac {x} {n}+\cos \dfrac {2x} {n}+\ldots +\cos\dfrac {\left( n-1\right) x} {n}} {n } = \dfrac{\sin x}{x}$$
I have attempted a number of approaches such as trigonometric tricks and the most recent with substitution of $\cos \dfrac {x} {n}=\dfrac {e^{i\frac {x} {n}}+e^{-\frac {ix} {n}}} {2}$
Starting from the left hand side i ended up with an expression such as $$\lim _{n\rightarrow \infty }\frac {\frac {1-e^{ix}} {1-e^{\frac {ix} {n}}}+\frac {1-e^{-ix}} {1-e^{\frac {-ix} {n}}}} {2n}$$ I am unsure how i could possibly rearrange this so upon taking the limit i can convert to RHS.
Any help with this method or even an alternative proof strategy would be much appreciated.
| Here is slightly long winded solution using trigonometric identities only.
$$\lim _{n\rightarrow \infty }\dfrac {1+\cos \dfrac {x} {n}+\cos \dfrac {2x} {n}+\ldots +\cos\dfrac {\left( n-1\right) x} {n}} {n } $$
using the trignometric identity $$\sum _{n=1}^{N}\cos n\theta =\dfrac {-1} {2}+\dfrac {\sin \left( N+\dfrac {1} {2}\right) \theta } {2\sin \dfrac {\theta } {2}}$$
our equation can be rewritten as
$$\lim _{n\rightarrow \infty }\dfrac {1-\dfrac {1} {2}+\dfrac {\sin \left( n-1+\dfrac {1} {2}\right) \dfrac {x} {n}} {2\sin x / 2n}} {n}$$
$$\lim _{n\rightarrow \infty }\dfrac {1} {2n}\left( \dfrac {\sin\dfrac {x} {2n}+\sin \left( x-\dfrac {x} {2n}\right) } {\sin \dfrac {x} {2n}}\right) $$
$$\lim _{n\rightarrow \infty }\dfrac {1} {2n}\left( \dfrac {2\sin\dfrac {x} {2}\cos \left(\dfrac {\dfrac {x} {n}-x} {2}\right)} {\sin \dfrac {x} {2n}}\right) $$
The following restructure of the denominator was pointed out to me by Aryabhata. :-)
$$\lim _{n\rightarrow \infty }\dfrac {\sin \dfrac {x} {2}\cos \dfrac {x} {2}\left( \dfrac {1} {n}-1\right) } {\dfrac {x\sin \dfrac {x} {2n}} {2\dfrac {x} {2n}}}$$
Taking the limit we get
$$\dfrac {\sin \dfrac {x} {2}\cos -\dfrac {x} {2}} {\dfrac {x} {2}}$$
which is the same as
$$\dfrac {\sin \dfrac {x} {2}\cos \dfrac {x} {2}} {\dfrac {x} {2}}$$
which gives RHS using the trigonometric identity once again.
$$\dfrac{\sin x}{x}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Showing that $\lceil (\sqrt{3} + 1)^{2n} \rceil$ is divisible by $2^{n+1}$. I have a question which has fluxommed me and my pals for the past few days. Any help or solution is welcome
Show using Binomial theorem that the integer just after $(3^{1/2} + 1)^{2n}$ is divisble by $2^{n+1}$.
here n belongs to natural numbers (positive integers).
| This is a classic, the standard proof is to consider
$$ (\sqrt{3} + 1)^{2n} = x_n + y_n \sqrt{3}$$
for integer $x_n$, $y_n$, (which you can see using Binomial theorem) and show that
$$ (\sqrt{3} - 1)^{2n} = x_n - y_n \sqrt{3}$$
again using Binomial theorem.
Now use the fact that $\sqrt{3} - 1 \lt 1$ and that
$$(\sqrt{3} + 1)^{2n} + (\sqrt{3} - 1)^{2n} = 2x_n$$
Thus the integer you are looking for is $2x_n$.
Get a recurrence for $x_n$ and $y_n$ and use induction.
To get the recurrence:
we have that $(\sqrt{3} + 1)^2 = 4 + 2\sqrt{3}$ and so
$$ x_{n+1} + y_{n+1} = (x_n + y_n \sqrt{3})(4 + 2 \sqrt{3}) = (4x_n + 6y_n) + (2x_n + 4y_n) \sqrt{3}$$
Thus
$$ x_{n+1} = 4x_n + 6y_n$$
$$ y_{n+1} = 2x_n + 4y_n$$
Write $$x_{n+2} = 4x_{n+1} + 6y_{n+1}$$ and eliminate $y_n$ and $y_{n+1}$
(I believe it comes to $x_{n+2} = 8x_{n+1} - 4 x_n$)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Basketball Team Combinatorial Question: A basketball team has 5 players, 3 in forward position (which includes a center) and 2 in guard position. In how many ways can we make a team if there are 6 forwards, 4 guards and 2 people who can play forward or guard?
The way I am approaching this question is as follows:
We need to pick 3 people out of 6 for the forward position: $\binom{6}{3}$.
We also need to pick 2 guards out of possible 4: $\binom{4}{2}$.
So far, the answer is: $\binom{6}{3} \times \binom{4}{2}$. (assuming that the 2 people who can play forward or guard statement is disregarded).
If two people can play forward or guard, I am thinking of assuming both play forward, add that to assuming one is forward one is guard, plus assuming two are guards. So:
$\binom{8}{3}\binom{4}{2} + \binom{7}{3}\binom{5}{2} + \binom{6}{3}\binom{6}{2}$
Would this yield the right answer? If not, why?
| I would do this carefully, the long way, and then see whether I have missed any shortcuts. A procedure that does not involve crossing the fingers has high priority!
You counted correctly the number $\binom{6}{3}\binom{4}{2}$ of teams made up of specialists. So there are $120$ such teams. Now we count separately the teams that contain $1$ versatile player, and $2$ versatile players.
If we are going to have $1$ versatile player, she can be chosen in $\binom{2}{1}$ ways. She can replace a forward, leaving $\binom{6}{2}\binom{4}{2}$ choices for the rest of the team, or she can replace a guard, leaving $\binom{6}{3}\binom{4}{1}$ choices for the rest of the team. Thus there are
$$\binom{2}{1}\left(\binom{6}{2}\binom{4}{2}+\binom{6}{3}\binom{4}{1}\right)$$
teams with exactly $1$ versatile player. So there are $340$ such teams.
If we are going to use $2$ versatile players, they can be both replace forwards, leaving $\binom{6}{1}\binom{4}{2}$ choices, or both replace guards, leaving $\binom{6}{3}$ choices, or do one of each, leaving $\binom{6}{2}\binom{4}{1}$ choices, for a total of
$$\binom{6}{1}\binom{4}{2}+\binom{6}{3}+\binom{6}{2}\binom{4}{1}$$
teams with exactly $2$ versatile players. So there are $96$ such teams.
Finally, add up.
Remark: I interpreted team to mean a set of $5$ people. If by team we mean a set of $5$, together with a specification of what positions (forward or guard) they are playing, the answer would be different, since one versatile player in each position would have to be counted twice, once for X playing forward and Y playing guard, and once for the reverse. And one can complicate things further.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove $n^5 - n$ is divisible by $30$ without reduction How can I prove that prove $n^5 - n$ is divisible by $30$?
I took $n^5 - n$ and got $n(n-1)(n+1)(n^2+1)$
Now, $n(n-1)(n+1)$ is divisible by $6$.
Next I need to show that $n(n-1)(n+1)(n^2+1)$ is divisible by $5$.
My guess is using Fermat's little theorem but I don't know how.
| By the Chinese remainder theorem, you just have to prove that $n^5\equiv n\mod 5$, $\bmod 3$ and $\bmod2$.
The case $n=5$ is exactly the statement of Little Fermat's theorem.
Also, the same theorem says that $n^3\equiv n\mod 3$, so $\;n^5=n^3\cdot n^2\equiv n\cdot n^2=n^3\equiv n$.
Similarly, modulo $2$, $\;n^5=\bigl(n^2\bigr)^2n\equiv n^2n\equiv n\cdot n=n^2\equiv n.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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Prove $\sin{2\theta} = \frac{2 \tan\theta}{1+\tan^{2}\theta}$ How to prove: $$\sin{2\theta} = \frac{2 \tan\theta}{1+\tan^{2}\theta}$$ Help please. Don't know where to start.
| Set $a=b=\theta $ in the identity
$$\begin{equation*}
\sin (a+b)=\sin a\cdot \cos b+\cos a\cdot \sin b
\end{equation*}$$
to get this one
$$\begin{equation*}
\sin 2\theta =2\sin \theta \cdot \cos \theta .
\end{equation*}$$
Then divide the RHS by $\sin ^{2}\theta +\cos ^{2}\theta =1$ and afterwards both
numerator and denominator by $\cos ^{2}\theta \neq 0$
$$\begin{equation*}
\sin 2\theta =\dfrac{2\sin \theta \cdot \cos \theta }{\sin ^{2}\theta +\cos
^{2}\theta }=\dfrac{2\dfrac{\sin \theta \cdot \cos \theta }{\cos ^{2}\theta }}{
\dfrac{\sin ^{2}\theta +\cos ^{2}\theta }{\cos ^{2}\theta }},
\end{equation*}$$
and simplify.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Calculating a real integral using complex integration $$\int^\infty_0 \frac{dx}{x^6 + 1}$$
Does someone know how to calculate this integral using complex integrals? I don't know how to deal with the $x^6$ in the denominator.
| Thankfully the integrand is even, so we have
$$
\int^\infty_0 \frac{dx}{x^6 + 1} = \frac{1}{2}\int^\infty_{-\infty} \frac{dx}{x^6 + 1}.
\tag{1}
$$
To find this, we will calculate the integral
$$
\int_{\Gamma_R} \frac{dz}{z^6+1},
$$
where $\Gamma_R$ is the semicircle of radius $R$ in the upper half-plane, $C_R$, together with the line segment between $z=-R$ and $z=R$ on the real axis.
(Image courtesy of Paul Scott.)
Then
$$
\int_{\Gamma_R} \frac{dz}{z^6+1} = \int_{-R}^{R} \frac{dx}{x^6+1} + \int_{C_R} \frac{dz}{z^6+1}.
$$
We need to show that the integral over $C_R$ vanishes as $R \to \infty$. Indeed, the triangle inequality gives
$$\begin{align}
\left| \int_{C_R} \frac{dz}{z^6+1} \right| &\leq L(C_R) \cdot \max_{C_R} \left| \frac{1}{z^6+1} \right| \\
&\leq \frac{\pi R}{R^6 - 1},
\end{align}$$
where $L(C_R)$ is the length of $C_R$. From this we may conclude that
$$
\lim_{R \to \infty} \int_{\Gamma_R} \frac{dz}{z^6+1} = \int_{-\infty}^{\infty} \frac{dx}{x^6+1}.
\tag{2}
$$
The integral on the left is evaluated by the residue theorem. For $R > 1$ we have
$$
\int_{\Gamma_R} \frac{dz}{z^6+1} = 2\pi i \sum_{k=0}^{2} \operatorname{Res}\left(\frac{1}{z^6+1},\zeta^k \omega\right),
$$
where $\zeta$ is the primitive sixth root of unity and $\omega = e^{i\pi/6}$. Note that this is because $\omega$, $\zeta\omega$, and $\zeta^2 \omega$ are the only poles of the integrand inside $\Gamma_R$. The sum of the residues can be calculated directly, and we find that
$$
\int_{\Gamma_R} \frac{dz}{z^6+1} = 2\pi i \sum_{k=0}^{2} \operatorname{Res}\left(\frac{1}{z^6+1},\zeta^k \omega\right) = \frac{\pi}{3 \sin(\pi/6)} = \frac{2\pi}{3}.
$$
Thus, from $(1)$ and $(2)$ we conclude that
$$
\int_{0}^{\infty} \frac{dx}{x^6+1} = \frac{\pi}{3}.
$$
In general,
$$
\int_{0}^{\infty} \frac{dx}{x^{2n}+1} = \frac{\pi}{2 n \sin\left(\frac{\pi}{2n}\right)}
$$
for $n \geq 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to $\int \sqrt{10x-x^2}dx$ Theres a hint to use $x=5+5\sin{t}$. Ok, but how do I know what substitution to use if a hint wasn't given? Is it "trivial" or perhaps, its very unlikely that that will appear?
Anyways, I did:
$\int \sqrt{10(5+5\sin{t}) - (5+2\sin{t})^2} dx \\
= \int \sqrt{50+50\sin{t} - (25+50\sin{t} + 25\sin^2{t})} dx\\
= \int \sqrt{ 25-25\sin^2{t} } dx \\
= 5 \int \sqrt{1-\sin^2{t}} dx \\
= 5\sin^{-1}{\sin{t}} \\
= 5t \\
= 5 \sin^{-1}{\frac{x-5}{5}}$
But the answer was:
$$\frac{25}{2}\sin^{-1}{\frac{x-5}{5}}+\frac{x-5}{2}\sqrt{10x-x^2}+c$$
What did I do wrong? Or is the answer wrong perhaps?
| What you "did wrong" in making the subsitution was forget to take into account the
$$\frac{dx}{dt} = 5\cos t.$$
Continuing,
$$\begin{align}
\int \sqrt{10x-x^2}dx & = \int \sqrt{10(5 + 5\sin{t}) + (5+5\sin{t})^2}5\cos{t}dt\\
& = 5\int \sqrt{25 - 25\sin^2t}\cos{t}dt\\
& = 25\int \cos^2tdt\\
& = \frac{25}{2}\sin{t}\cos{t} + \frac{25}{2}t\\
& = \frac{25}{2}\frac{(x-5)}{5}\sqrt{1-(\frac{x-5}{5})^2} + \frac{25}{2}\sin^{-1}(\frac{x-5}{5})\\
& = \frac{x-5}{2}\sqrt{25 - (x-5)^2} + \frac{25}{2}\sin^{-1}(\frac{x-5}{5})\\
& = \frac{x-5}{2}\sqrt{10x-x^2} + \frac{25}{2}\sin^{-1}(\frac{x-5}{5}).
\end{align}$$
Not the easiest method, but it does work.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/136175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Calculus optimisation This meant to be a relatively easy problem but I cannot get my head around it. It is from Burkill's "First course in Analysis", book $4$(f), $10$.
An open bowl is in the form of a segment of a sphere of metal of negligible thickness. Find the shape of the bowl if its volume is the greatest for a given area of metal. (Solution: Hemisphere)
Could anyone help me with the solution of the problem?
Here is one of my attempts. I assumed that the problem is circumferentially symmetric so I considered the planar problem instead. I took the area of the segment of a disk with radius R, central angle $\theta$, and area A which I calculated as follows:
$$A = \text{sector area} - \text{area of triangle} = \frac{R^2\pi}{2\pi} \theta - 2 \frac{1}{2} R \cos\left(\frac{\theta}{2}\right) R\sin\left(\frac{\theta}{2}\right) = \frac{R^2\theta}{2}-\frac{R^2}{2}\sin\theta.$$
This is constrained by the area that is the length of material we have say $L$:
$$L = R\theta.$$
Substituting in for $R$:
$$A = \frac{L^2}{2\theta} - \frac{L^2}{2\theta^2} \sin\theta.$$
Differentiate to find turning value:
$$\frac{dA}{d\theta} = \frac{L^2}{2}\left(-\frac{1}{\theta^2} + \frac{2}{\theta^3} \sin\theta + \frac{1}{\theta^2} \cos\theta\right) = 0.$$
I am bit stuck now how to get $\theta$ out of this and I am questioning whether my method is really correct. Could anyone help me out? Thank you!
| I think the problem implies that the bowl is obtained by cutting a sphere with a plane, so that the bowl is indeed rotationally symmetric. Let $2 \theta$ be the aperture of the bowl, with $0 < \theta < \pi$. Then the area of the metal is $A(R,\theta) = 4 \pi R^2 \sin^2\left(\frac{\theta}{2}\right) $.
The volume of liquid such a bowl could hold is given by an integral, obtained by shell method:
$$ \begin{eqnarray}
V_\theta &=& \int_{R \cos \theta}^R A(\rho, \arccos\left( \frac{R \cos\theta}{\rho} \right)) \mathrm{d} \rho = \int_{R \cos \theta}^R 4 \pi \rho^2 \sin^2\left(\frac{1}{2} \arccos\left( \frac{R \cos\theta}{\rho} \right)\right) \rho \mathrm{d} \rho \\
&=& \int_{R \cos\theta}^R 2 \pi \rho \left( \rho - R \cos(\theta) \right) \mathrm{d} \rho = \frac{4}{3} \pi R^3 \left( 3 - 2 \sin^2\left(\frac{\theta}{2}\right) \right) \sin^4\left(\frac{\theta}{2}\right) = \frac{4}{3} \pi R^3 \left( 3 - 2 \frac{A}{4 \pi R^2} \right) \frac{A^2}{16 \pi^2 R^4}
\end{eqnarray}
$$
The above expression is maximal for $A = 4 \pi R^2$, meaning $\theta = \pi$, i.e. exactly the hemisphere.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/139594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find $b_3$ in a $3\times 1$ range vector For the matrix
$$A=\left(\begin{array}{rrrr}
2 & 1 & 5 & 0\\
1&2&4& -3\\
1 & 1 & 3 & -1
\end{array}\right)$$
I needed to find a range vector $\mathbf{b}$ such that $b_1=5$, $b_2=2$.
I'm not very clear on what algebraic characteristics this vector should have, so as to allow $A\mathbf{x}=\mathbf{b}$ to have solutions. I understand geometrically that all the equations should provide at least non-contradictory inputs.
I used a few steps of elimination by pivoting on $[A|\mathbf{b}]$:
$$\begin{align*}
R_1&:=R_1/2;\\
R_2&:=-R_1;\\
R_3&:=-R_1;
\end{align*}$$
At this point I noticed that $R_2$ was just 3 times $R_3$ (if only the matrix $A$, not vector b is considered), so I proceeded to solve
$-.5=3(b_3-2.5) \implies b_3=2\frac{1}{3}$, which extended this relationship to the rows of vector $\mathbf{b}$. Even though this would mean the third equation was denoting a parallel space, it just seemed easy to solve the equality. When testing with a calculator 2 1/3 for b3 does work and several random values didn't. However, most importantly, I'm not clear about the linear algebra behind this. Please explain this simply and concisely.
Because I was unsure if that's the right way to solve this, I also proceeded with elimination to reach the form:
$$\left(\begin{array}{rrrr|r}
1&0&2&1&2 \frac{2}{3}\\
0&1&1&-2 & -\frac{1}{3}
\end{array}\right)$$
That didn't seem helpful because the only variable I was concerned with, b3, disappeared as the third row became 0.
I apologize for the lengthy post because the question is rather simple, but I was trying to be detailed to help the reader understand what I'm talking about.
| You start with the Augmented Matrix
$$A=\left(\begin{array}{rrrr|r}
2 & 1 & 5 & 0&5\\
1&2&4& -3&2\\
1 & 1 & 3 & -1&b_3
\end{array}\right)$$
and you want to find a value of $b_3$ for which the matrix corresponds to a system with a solution; that is, you want to make sure that after you row reduce, you don't have a line which has all $0$s before the vertical bar equal to something with a nonzero thing after it (that would correspond to an equation that reads "$0=\text{something nonzero}$", which is bad).
If we divide the first row by $2$, we get (avoid mixed fractions! They lead to errors)
$$\left(\begin{array}{rrrr|r}
1 & \frac{1}{2} & \frac{5}2 & 0&\frac{5}{2}\\
1&2&4& -3&2\\
1 & 1 & 3 & -1&b_3
\end{array}\right)$$
Then you seem to say that you subtracted Row 1 from both Row 2 and Row 3; if you do that, we get:
$$\left(\begin{array}{rrrr|r}
1 & \frac{1}{2} & \frac{5}{2} & 0 & \frac{5}{2}\\
0 & \frac{3}{2} & \frac{3}{2} & -3 & -\frac{1}{2}\\
0 & \frac{1}{2} & \frac{1}{2} & -1 & b_3-\frac{5}{2}
\end{array}\right).$$
(Aside: it would have been easier, and avoided fractions, to first exchange rows 1 and 2).
Your next observation was that if you divide the second row by $3$, you get something whose first four entries equal the third row:
$$\left(\begin{array}{rrrr|r}
1 & \frac{1}{2} & \frac{5}{2} & 0 & \frac{5}{2}\\
0 & \frac{1}{2} & \frac{1}{2} & -1 & -\frac{1}{6}\\
0 & \frac{1}{2} & \frac{1}{2} & -1 & b_3-\frac{5}{2}
\end{array}\right).$$
Now, if we subract the second row from the third, we would get:
$$\left(\begin{array}{rrrr|r}
1 & \frac{1}{2} & \frac{5}{2} & 0 & \frac{5}{2}\\
0 & \frac{1}{2} & \frac{1}{2} & -1 & -\frac{1}{6}\\
0 & 0 & 0 & 0 & b_3-\frac{5}{2}+\frac{1}{6}
\end{array}\right).$$
Now remember: for the system to have a solution, we cannot have a row that has $0\ 0\ 0\ 0|x$ with $x\neq 0$. So that final row tells us that we must have:
$$b_3 - \frac{5}{2}+\frac{1}{6} = 0.$$
This tells us what $b_3$ needs to be:
$$b_3 = \frac{5}{2}-\frac{1}{6} = \frac{14}{6}=\frac{7}{3}.$$
This is the same answer you got. And indeed, what we did above shows that $b_3=\frac{7}{3}$ is the only value that can work: any other value for $b_3$ would lead to a system that is inconsistent, and so has no solutions.
Explanation. From the perspective of systems of linear equations, all we are doing is transforming the system into simpler and simpler systems, just like we usually do with row reduction. The key is that the third equation is a linear combination of the first two: the first equation reads
$$2x+y+5z + 0w = b_1$$
and the second reads
$$x+2y+4z-3w = b_2;$$
if you add them, we get
$$3x + 3y + 9z - 3w = (b_1+b_2),$$
and you can note that the third equation is
$$x+y+3z-w = b_3.$$
If the system is consistent, then we must have
$$\begin{align*}
\frac{1}{3}(b_1+b_2) &= \frac{1}{3}(3x+3y+9z-3w)\\
&= x+y+3z-w\\
&= b_3,
\end{align*}$$
which is why we have $b_3 = \frac{7}{3} = \frac{1}{3}(5+2) = \frac{1}{3}(b_1+b_2)$. The linear dependency between the first two rows of $A$ and the third forces the same linear dependency between the first two entries of $\mathbf{b}$ and the third. The row reduction just takes the guessing and luck out of figuring out the linear dependencies between the rows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/139784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Calculating stationary distribution of markov chain I am asked to compute the stationary distribution of the markov chain with state space $E=\{0\dots,n\}$ and transition matrix below:
\begin{bmatrix}
0 & 1 \\
\frac{1}{n} & 0 & \frac{n-1}{n} \\
& \frac{2}{n} & 0 & \frac{n-2}{n} & \\
& & \ddots & \ddots & \ddots \\
& & & \frac{n-1}{n} & 0 & \frac{1}{n} \\
& & & & 1 & 0 & \\
\end{bmatrix}
What I did was use $\pi P = \pi$. And I got:
$\pi_0=\frac{1}{n}\pi_1 \Rightarrow \pi_1 =n\pi_0 \\
\pi_1 =\pi_0 +\frac{2}{n}\pi_2 \Rightarrow \pi_2 =\frac{n(n-1)}{2}\pi_0 \Rightarrow \pi_2=\frac{n-1}{2}\pi_1 \\$
I tried fiddling with it here and there but I cant seem to get anywhere to finish this problem. i.e. I can't seem to find $\pi_k$ for all $k \in E=\{0,\dots,n\}$. How would I finish this problem?
| The $\pi = P \pi$ equation, component-wise, reads:
$$ \begin{eqnarray}
\sum_{m=0}^n \pi_m \left(\frac{m}{n} \delta_{m,k+1} + \frac{n-m}{n} \delta_{m,k-1}\right) &=& \pi_k \\
(n-k+1) \pi_{k-1} + \pi_{k+1} (k+1) &=& n \pi_k
\end{eqnarray}
$$
It is easiest to solve this using probability generating function $g(x) = \sum_{k=0}^n x^k \pi_k$. Multiplying the equation with $x^k$
$$
(n-k+1) x^{k} \pi_{k-1} + x^{k} \pi_{k+1} (k+1) = n x^k \pi_k
$$
and summing over $k$:
$$
\begin{eqnarray}
\sum_{k=1}^n (n-k+1) x^{k} \pi_{k-1} + \sum_{k=0}^{n-1} \pi_{k+1} (k+1) x^{k} &=& n \sum_{k=0}^n x^k \pi_k \\
\sum_{k=1}^{n+1} (n-k+1) x^{k} \pi_{k-1} + \sum_{k=-1}^{n-1} \pi_{k+1} (k+1) x^{k} &=& n g(x) \\
\sum_{k=0}^{n} (n-k) x^{k+1} \pi_{k} + \sum_{k=0}^{n} k x^{k-1} \pi_{k} &=& n g(x) \\
n x g(x) - x^2 g^\prime(x) + g^\prime(x) &=& n g(x) \\
(1-x^2) g^\prime(x) &=& n (1-x) g(x) \\
(1+x) g^\prime(x) &=& n g(x) \\
g(x) &=& C (1+x)^n
\end{eqnarray}
$$
Normalization is determined from $g(1) = 1$ requirement, since $g(x)$ is the probability generating function. Hence
$$
\pi_k = \frac{1}{2^n} \binom{n}{k}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/140436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Understanding congruence and using the notation correctly I am working on the following problem presented in a text:
Find the remainder when $3^{1000}$ is divided by 13. I have just been introduced to the notation for saying that $a\equiv b \pmod{n}$. From what I understand of this it is stating that $a$ is congruent to $b$ when divided by $n$, that is $a$ will have the same remainder as $b$ when $b$ is divided by $n$ and vice versa.
From this I am assuming that $a \equiv a \pmod{n}$, where $a$ and $n$ are any integers?
The following property has also been introduced:
If $a \equiv b \pmod{n}$, then $a^k \equiv b^k \pmod{n}$ (*).
We have
$3^1 \equiv 3 \pmod{13}$, $3^2 \equiv 9 \pmod{13}$, and $3^3\equiv 27 \equiv 1 \pmod{13}$.
The above line bothers me because essentially it looks like we are just writing
$3 \equiv 3 \pmod{13}$, $9 \equiv 9 \pmod{13}$
however the first part I don't get is $3^3\equiv 27 \equiv 1 \pmod{13}$ is the extra 1 present here. Assuming my conclusion above is true then I would not have a problem with writing
$3^1 \equiv 3 \pmod{13}$, $3^2 \equiv 9 \pmod{13}$, and $3^3\equiv 27 \pmod{13}$,
but the extra 1 included is really bothering me as I do not understand its significance.
Secondly the next line in the solution states that:
"We repeatedly multiply by 3 and use property (*):
$3^4 \equiv 3 \pmod{13}$, $3^5 \equiv 9 \pmod{13}$, $3^6 \equiv 1 \pmod{13}$.
Thus the remainders form the repeating pattern 3,9,1,3,9,1,... .
I am not following why we have stopped writing that $3^n$ is congruent to the result of $3^n$. So where before we had $3^2 \equiv 9 \pmod{13}$, why do we now have $3^5 \equiv 9 \pmod{13}$ instead of $3^5 \equiv 243 \pmod {13}$?
| Fermat's Theorem tells us that $\,\,3^{13}\equiv 3\pmod{13}\Longrightarrow 3^{12}\equiv 1\pmod{13}\,$ , and since $\,1,000=12\cdot 83+4\,$ , we get $$3^{1,000}=\left(3^{12}\right)^{83}3^4\equiv81\equiv 3\pmod{13}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/145009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculate sums of inverses of binomial coefficients How to calculate the sum of sequence $$\frac{1}{\binom{n}{1}}+\frac{1}{\binom{n}{2}}+\frac{1}{\binom{n}{3}}+\cdots+\frac{1}{\binom{n}{n}}=？$$ How about its limit?
| Assuming $C_n^k$ stands for $\binom{n}{k} = \frac{n!}{(n-k)! \cdot k!}$, and using, for $k>0$:
$$
\frac{1}{\binom{n}{k}} = k \operatorname{Beta}(k,n-k+1) = k \int_0^1 (1-x)^{k-1} x^{n-k} \mathrm{d} x
$$
Hence
$$
S = \sum_{k=1}^n \frac{1}{\binom{n}{k}} = \int_0^1 \sum_{k=0}^n k (1-x)^{k-1} x^{n-k} \mathrm{d} x = \int_0^1 \frac{x^{n+1} -(1-x)^n ((2n+1)x-n)}{(1-2x)^2} \mathrm{d} x
$$
Now changing integration variable $x = \frac{1}{2} + u$:
$$\begin{eqnarray}
S &=& \int_{-1/2}^{1/2} \frac{\mathrm{d} u}{4 u^2} \left( \left(\frac{1}{2}+u\right)^{n+1} - \frac{1}{2} \left(\frac{1}{2}-u\right)^n \left( 1 + 2 (2n+1) u\right)\right) \\
&=& \int_{-1/2}^{1/2} \frac{\mathrm{d} u}{4 u^2} \sum_{k=2}^{n+1} 2^{k-n-1} \left((-1)^k \left((2 n+1) \binom{n}{k-1}-\binom{n}{k}\right)+\binom{n+1}{k}\right) u^{k} \\
&=& \sum_{k=1}^{\lfloor\frac{n+1}{2}\rfloor} 2^{2k-n-1} \left(\left((2 n+1) \binom{n}{2k-1}-\binom{n}{2k}\right)+\binom{n+1}{2k}\right) \underbrace{\int_{-1/2}^{1/2} \frac{\mathrm{d} u}{4 u^2} u^{2 k}}_{\frac{1}{4^k} \frac{1}{2k-1}} \\
&=& \sum_{k=1}^{\lfloor\frac{n+1}{2}\rfloor} \frac{1}{2^{n}} \frac{1}{2k-1}\frac{(n+1)!}{ (2k-1)! (n-2k+1)!} \stackrel{\ast}{=} \sum_{k=1}^n \frac{n+1}{n+1-k} \frac{1}{2^k}
\end{eqnarray}
$$
Thus
$$
\sum_{k=0}^n \frac{1}{\binom{n}{k}} = \sum_{k=0}^n \frac{n+1}{n+1-k} \frac{1}{2^k}
$$
For large $n$ the sum approaches the value of $2$ from above:
I am hoping this sum has a nice probabilistic underpinnings to it.
Added: Derivation of equality $\stackrel{\ast}{=}$ above
$$ \begin{eqnarray}
S &=& \sum_{k=1}^{\lfloor\frac{n+1}{2}\rfloor} \frac{1}{2^{n}} \frac{1}{2k-1}\frac{(n+1)!}{ (2k-1)! (n-2k+1)!} = \frac{n+1}{2^n} \underbrace{\sum_{k=0}^{\lfloor\frac{n-1}{2}\rfloor} \frac{1}{2k+1} \binom{n}{2k+1}}_{A_n}
\end{eqnarray}
$$
We will now establish that $A_{n+1} - A_{n} = \frac{2^{n}}{n+1}$, which would imply $$A_n = \sum_{k=1}^n \frac{2^{k-1}}{k} \stackrel{k \to n+1-k}{=} \sum_{k=1}^n \frac{2^{n-k}}{n+1-k}$$
We thus proceed, firstly for even, $n=2m$:
$$ \begin{eqnarray}
A_{2m+1} -A_{2m} &=& \sum_{k=0}^{m} \frac{1}{2k+1} \left( \underbrace{\binom{2m+1}{2k+1} - \binom{2m}{2k+1}}_{\binom{2m}{2k}}\right) \\
&=& \sum_{k=0}^{m} \frac{1}{2k+1} \binom{2m}{2k} = \frac{1}{2}\int_0^1 \left( (1+x)^{2m} + (1-x)^{2m} \right) \mathrm{d} x = \frac{2^{n}}{n+1}
\end{eqnarray}
$$
and then, similarly, for odd, $n=2m+1$:
$$\begin{eqnarray}
A_{2m+2}-A_{2m+1} &=& \sum_{k=0}^m \frac{1}{2k+1} \binom{2m+1}{2k} \\
&=& \frac{1}{2}\int_0^1 \left( (1+x)^{2m+1} + (1-x)^{2m+1} \right) \mathrm{d} x = \frac{2^n}{n+1}
\end{eqnarray}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/151441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "42",
"answer_count": 5,
"answer_id": 2
} |
How to find the antiderivative of $\sqrt{25-x^2}$? How to find the antiderivative of $\sqrt{25-x^2}$?
How would I do it? Integration by substitution doesn't seem to work in this case.
| A trigonometric substitution can do the trick. Note that $1-\sin^2\theta = \cos^2\theta$.
Let $x = 5\sin\theta$, with $-\frac{\pi}{2}\leq \theta\leq\frac{\pi}{2}$. Then
$$\sqrt{25-x^2} = \sqrt{25-25\sin^2\theta} = 5\sqrt{\cos^2\theta} = 5|\cos\theta| = 5\cos\theta$$
(with the last equality because with $-\frac{\pi}{2}\leq \theta\leq\frac{\pi}{2}$, we have $\cos\theta\geq 0$.
We also have $dx = 5\cos\theta d\theta$. So
$$\begin{align*}
\int\sqrt{25-x^2}\,dx &= \int5\cos\theta 5\cos\theta\,d\theta\\
&= 25\int\cos^2\theta\,d\theta.
\end{align*}$$
The integral of $\cos^2\theta$ can be done by using Integration by Parts or a reduction formula. Then convert back to $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/151747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to find the error in a proof? (that $1=0$) So I devised this proof that $1=0$. Of course it is false, but I don't know why. Why?
$$\begin{align*}
x+1&=y\\
\frac{x+1}{y}&=1\\
\frac{x+1}{y}-1&=0\\
\frac{x+1}{y}-\frac{y}{y}&=0\\
\frac{x-y+1}{y}&=0\\
x-y+1&=0\\
x-y+1&=\frac{x-y+1}{y}\\
y(x-y+1)&=x-y+1\\
y&=1\\
x+1&=1\\
x&=0\qquad * * * *\\
y-1&=x\\
\frac{y-1}{x}&=1\\
\frac{y-1}{x}-1&=0\\
\frac{y-1}{x}-\frac{x}{x}&=0\\
\frac{y-x-1}{x}&=0\\
y-x-1&=0\\
y-x-1&=\frac{y-x-1}{x}\\
x(y-x-1)&=y-x-1\\
x&=1\qquad * * * *\\
1&=0\\
\end{align*}$$
| You have a problem already by going from
$$x-y+1=0= y(x-y+1)$$
to
$$y=1$$
because you are dividing by zero. The fact that you are deducing $y=1$ from a starting place, namely $$x+1=y$$
that does not assume anything about $x$ or $y$, should have been a signal that something was wrong.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/152742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
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how to find this limit $\lim_{x\rightarrow 0} \frac{\sin x^2}{ \ln ( \cos x^2 \cos x + \sin x^2 \sin x)} = -2$ without using L'Hôpital's rule I am looking for simple trigonometric or algebraic manipulation so that this limit can be solved without using L'Hôpital's rule
$$ \lim_{x\rightarrow 0} \frac{\sin x^2}{ \ln ( \cos x^2 \cos x + \sin x^2 \sin x)} = -2$$
link on wolframalpha. Thank you for help!!
| $$\lim_{x\rightarrow 0} \frac{\sin x^2}{ \ln ( \cos x^2 \cos x + \sin x^2 \sin x)} = ?$$
Hint :
$$\cos (x-y)=\cos(x)\cos(y)+\sin(x)\sin(y)$$
So:
$$(\cos x^2 \cos x + \sin x^2 \sin x)=\cos(x^2-x)$$
$$\lim_{x\rightarrow 0} \frac{\sin x^2}{ \ln (\cos(x^2-x))} = \lim_{x\rightarrow 0} \frac{\dfrac{\sin x^2}{\cos(x^2-x)-1}}{\dfrac{ \ln (1+(\cos(x^2-x)-1))}{\cos(x^2-x)-1}} = ? $$
$$
\lim_{x\rightarrow 0} \dfrac{\sin x^2}{\cos(x^2-x)-1}=\dfrac{\sin x^2}{x^2}\times \dfrac{x^2}{\cos(x^2-x)-1}\times \dfrac{\cos(x^2-x)+1}{\cos(x^2-x)+1} =-2$$
thus :
$$\lim_{x\rightarrow 0} \frac{\sin x^2}{ \ln ( \cos x^2 \cos x + \sin x^2 \sin x)} = -2$$
note that :
$$\lim_{x \to 0} \dfrac{\sin x}{x}=1=\lim_{x \to 0}\dfrac{\ln(1+x)}{x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/153918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
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} |
Solve an absolute value equation simultaneously My question is :
Solve simultaneously
$$\left\{\begin{align*}&|x-1|-|y-2|=1\\&y = 3-|x-1|\end{align*}\right.$$
What I did :
$y=3 - |x-1|$ is given.
Thus $y = 3-(x-1)$ or $y = 3-\left(-(x-1)\right),$ and so $$y = 4-x\qquad\mbox{ or } \qquad y = 2+x.$$
If $y = 2+x$, then $$x - 1 = 4-y \quad(1)\qquad\mbox{ or } \qquad y - 2 = x \quad(2).$$
Substituting 1), we get $$|4-y| - |y-2| = 1 \qquad\mbox{ or } \qquad |y-2-1|-|y-2| = 1.$$
I got here but I am not getting how to get the final solution. Any help would be greatly appreciated.
| It’s entirely possible to solve this system by an exhaustive analysis of cases along the lines on which you’ve begun, but there’s an easier way. Notice that both equations have a term $|x-1|$. Let’s let $u=|x-1|$ and see whether the resulting system is a little easier to solve. After the substitution the original system
$$\left\{\begin{align*}&|x-1|-|y-2|=1\\&y = 3-|x-1|\end{align*}\right.\tag{1}$$ becomes
$$\left\{\begin{align*}&u-|y-2|=1\\&y=3-u\end{align*}\right.\;.\tag{2}$$
The second equation of $(2)$ implies that $y-2=1-u$; substituting this into the first equation yields $$u-|1-u|=1\;.\tag{3}$$ If $u<1$, then $1-u>0$, and therefore $|1-u|=1-u$, and $(3)$ reduces to $u-(1-u)=1$, which is easily solved to get $u=1$ and hence $y=3-u=2$. Recall that $u=|x-1|$; thus, $|x-1|=1$, $x-1=\pm 1$, and $x=0$ or $x=2$. You can check that if $y=2$ and $x=0$ or $x=2$, then $(1)$ is satisfied.
If $u\ge1$, then $|1-u|=u-1$, and $(3)$ reduces to $u-(u-1)=1$, or $1=1$. This is always true, so every value of $u>1$ leads to a solution. Let’s check that. Let $u$ be any number greater than $1$, and let $y=3-u$. Then $y-2=1-u<0$, so $|y-2|=2-y=u-1$, and $$u-|y-2|=u-(2-y)=u-(u-1)=1\;,$$ so $(2)$ is indeed satisfied. All we need to do now is solve for $x$ in this case. We have $|x-1|=u$, so $x-1=\pm u$, and $x=1\pm u$. In other words, for each $u\ge 1$ there are two solutions,
$$\left\{\begin{align*}&x=1+u\\&y=3-u=3-(x-1)=4-x\end{align*}\right.\tag{4}$$ and
$$\left\{\begin{align*}&x=1-u\\&y=3-u=3-(1-x)=2+x\;.\end{align*}\right.\tag{5}$$
Finally, $(4)$ and $(5)$ can be simplified by getting rid of $u$. Since $u\ge 1$, $(4)$ gives us every value of $x\ge 2$, and $(5)$ gives us every value of $x\le 0$. Thus, $(4)$ and $(5)$ reduce to saying that the solutions to $(1)$ are
$$\begin{align*}
&y=4-x\text{ for any }x\ge 2\\
&y=2+x\text{ for any }x\le 0\;,
\end{align*}\tag{6}$$
with no solutions having $0<x<2$. You can improve a little on $(6)$ by recasting it in terms of $y$: note that both branches of the solution give you all values of $y\le 2$, so you can rephrase $(6)$ by saying that the solutions are all pairs such that
$$y\le 2,\text{ and }x=4-y\text{ or }x=y-2\;.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Primes of the form $p=a^2-2b^2$. I've stumbled upon this and I was wondering if anyone here could come up with a simple proof:
Let $p$ be a prime such that $p\equiv 1 \bmod 8$, and let $a,b\geq 1$ such that
$$p=a^2-2b^2.$$
Question: Is $b$ necessarily a square modulo $p$?
I have plenty of numerical data to support an affirmative answer, but the proof eludes me so far. For instance:
\begin{align*}
17 & = 5^2 - 2\cdot 2^2\\
&= 7^2 - 2\cdot 4^2\\
& = 23^2 - 2\cdot 16^2\\
& = 37^2 - 2\cdot 26^2\\
& = 133^2 - 2\cdot 94^2\\
\end{align*}
and $2\equiv 36$, $4$, $16$, $26\equiv 9$, $94\equiv 9 \bmod 17$ are squares.
Thanks!
| Since $p\equiv 1\pmod 8$, $2$ is a square modulo $p$. It will therefore be enough to show that any odd prime divisor of $b$ is also a square modulo $p$. Then any prime divisor of $b$ will be a square modulo $p$, therefore $b$ itself will be.
Let $q$ be an odd prime divisor of $b$, and consider your equation modulo $q$. You find that $p\equiv a^2 \pmod{q}$, so that $p$ is a square modulo $q$. By quadratic reciprocity (using that $p\equiv 1\pmod 4$), $q$ is a square modulo $p$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
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Evaluating $\int \sqrt{5 + 4x - x^2}dx$ $$\int \sqrt{5 + 4x - x^2}dx$$
I am pretty certain what I need to do to this problem is complete the square and turn it into a trig subsitution but I have no idea how to complete the square with a $-x^2$ or really with this problem at all, I just can't make it work.
I tried to see if I could make the problem be the same in any way by just pulling out a negative but that didn't seem to work.
I got the problem up to
$$\int \sqrt{ -1(x-2)^2 - 1}dx$$
But I do not think that does me any good. What I think I need to do is have a difference of squares with a square in it or something, I just have to get rid of the 4x term somehow.
| Our integral can be written as,
\begin{equation}
\begin{split}
\int\sqrt{5+4x-x^{2}}dx&=\int\sqrt{9-4+4x-x^{2}}dx\\
&=\int\sqrt{3^{2}-(x-2)^{2}}dx.\
\end{split}
\end{equation}
Now by trigonometric substitution,
take $$x-2=3\sin \theta. (\because \text{If we have} \sqrt{a^{2}-x^{2}} \text{ then we have to substitute} x=a\sin\theta.)$$
Thus,
\begin{equation}
\begin{split}
\int\sqrt{3^{2}-(x-2)^{2}}dx&=\int\sqrt{3^{2}-3^{2}\sin^{2}\theta} 3\cos \theta d\theta\\
&=\int 3\sqrt{1-\sin^{2}\theta} 3\cos \theta d\theta\\
&=9\int\cos ^{2}\theta d\theta\\
&=9\int\frac{1+\cos 2\theta}{2}d\theta\\
&=\frac{9}{2}\int 1 d\theta+\frac{9}{2}\int \cos 2\theta d\theta\\
&=\frac{9}{2}\left[\theta+\frac{\sin 2\theta}{2}\right]+C=\frac{9}{2}\left[\theta +\frac{2\sin \theta\cos \theta}{2}\right]+C\\
&=\frac{9}{2}\left[\sin^{-1}\left(\frac{x-2}{3}\right)+\frac{1}{9}(x-2) \sqrt{9-(x-2)^{2}}\right]+C
\end{split}
\end{equation}
Thus,
$$\int \sqrt{5+4x-x^{2}} dx=\frac{9}{2}\sin^{-1}\left(\frac{x-2}{3}\right)+\frac{(x-2) \sqrt{5+4x-x^{2}}}{2}+C$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Question Regarding Cardano's Formula In Cardano's derivation of a root of the cubic polynomial $f(X)=X^3+bX+c$ he splits the variable $X$ into two variables $u$ and $v$ together with the relationship that $u+v=X$. From this he finds that $x=u+v$, where
$$u=\sqrt[3]{\frac{-c}{2}+\sqrt{\frac{c^2}{4}+\frac{b^3}{27}}}$$ and $$v=\sqrt[3]{\frac{-c}{2}-\sqrt{\frac{c^2}{4}+\frac{b^3}{27}}}$$
is a root of $f(X)$.
Is there an intuitive explanation for why Cardano splits the variable $X$ into two parts, $u$ and $v$?
| Cardano knew that any quadratic equation of the form $$x^2+bx+c=0\tag{1}$$ can be written as $$x^2-(u+v)x+uv=0,\tag{2}$$ where $u$ and $v$ are the roots of the equation. Since by setting $t=u+v$ in the reduced cubic $$t^3+pt+q=0\tag{3}$$ we get $$(u^3+v^3+q)+(3uv+p)(u+v)=0,\tag{4}$$ then every root of the system $$u^3+v^3+q=0\tag{5a}$$ $$3uv+p=0\tag{5b}$$ is a root of $(4)$ as well, and based on the property of the quadratic equation indicated in $(2)$ it's now easy to find a formula for $t$ satisfying equation $(3)$.
Added. We just need to find two numbers $u^3$ and $v^3$ such that their sum is $-q$ and their product is $-p^3/27$, which we know from $(1)-(2)$ are the roots of the quadratic equation $$Y^2+qY-\frac{p^3}{27}=0.\tag{6}$$
Consequently,
$t=u+v=\sqrt[3]{u^3}+\sqrt[3]{v^3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/157746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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How to graph this? I have a non-right triangle. I will call
the bottom or base edge $b$,
the top left edge $a$,
the top right edge $c$.
Let $ a=c+2 $ and $b=10$. How do I graph a curve where the graph $ x $ and $ y$ coordinates represent the vertex where edges $ a $ and $c$ meet?
| Taking the origin as the intersection of $a$ and $b$, with the equation in terms of the angle $\theta$ at that vertex:
Through the cosine law we have:
$$(a-2)^2=a^2+100-20a\cos\theta$$
$$-4a+4=100-20a\cos\theta$$
$$a=\frac {96}{20\cos\theta-4}$$
With our particular setup, $a$ is the radius. So this is a polar equation, where
$$r(\theta)=\frac{96}{20\cos\theta-4}=\frac{24}{5\cos\theta-1}$$
we can convert to cartesian coordinates if you want:
$$5r\cos\theta-r=24$$
$$(5r\cos\theta-24)^2=r^2$$
$$(5x-24)^2=x^2+y^2$$
$$y^2=24x^2-240x+576$$
$$y^2=24(x-5)^2-24$$
$$(x-5)^2-\frac{y^2}{24}=1$$
which is a hyperbola. You want the rightmost branch (so $a>0$), which is valid when $\cos\theta<\frac 1 5$.
EDIT: picture time!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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the solution set of $\left | \frac{2x - 3}{2x + 3} \right |< 1$ what is the solution set of $\left | \frac{2x - 3}{2x + 3} \right |< 1$ ?
I solved it by first assuming: $-1 < \frac{2x - 3}{2x + 3 } < 1$
ended with: $x > 0 > -3/2$
Is that a correct approach?
And how to derive the solution set from the last inequality?
Is it $(-\frac{3}{2}, \infty)$ or $(0, \infty )$ ?
Thanks in advance.
| Equivalently, you are indeed trying to solve the
inequalities
$$-1\lt \frac{2x-3}{2x+3}\lt 1,$$
so it is reasonable to start by saying so. Then there are two cases to consider, $2x+3>0$ and $2x+3\lt 0$. In the first case we obtain the equivalent inequalities
$$-(2x+3) \lt 2x-3\lt 2x+3.$$
The inequality on the right always holds. The inequality on the left holds iff $-2x-3\lt 2x-3$, that is, iff $x\gt 0$.
In the case $2x+3 \lt 0$, multiplying through by $2x+3$ switches the direction of the inequalities, so we want
$$-(2x+3) \gt 2x-3 \gt 2x+3.$$
But the inequality $2x-3\gt 2x+3$ can never hold. We conclude that our inequality holds precisely if $x$ is in the interval $(0,\infty)$.
I would recommend that for more complicated problems you consider the following sort of approach. The expression $\frac{2x-3}{2x+3}$ changes sign "at" $x=-3/2$ and at $x=3/2$. Consider the three cases (i) $x\lt -3/2$, (ii) $-3/2\lt x\lt 3/2$, and (iii) $x \gt 3/2$ separately.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/158906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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How to solve this recurrence $T(n) = 2T(n/2) + n\log n$ How can I solve the recurrence relation $T(n) = 2T(n/2) + n\log n$? It almost matches the Master Theorem except for the $n\log n$ part.
| Let us take $n = 2^m$. Then we have the recurrence $$T(2^m) = 2T(2^{m-1}) + 2^m \log_2(2^m) = 2T(2^{m-1}) + m 2^m$$
Calling $T(2^m)$ as $f(m)$, we get that
\begin{align}
f(m) & = 2 f(m-1) + m 2^m\\
& = 2(2f(m-2) + (m-1)2^{m-1}) + m2^m\\
& = 4f(m-2) + (m-1)2^m + m2^m\\
& = 4(2f(m-3) +(m-2)2^{m-2}) + (m-1)2^m + m2^m\\
& = 8f(m-3) +(m-2)2^m + (m-1)2^m + m2^m\\
\end{align}
Proceeding on these lines, we get that
\begin{align}
f(m) &= 2^m f(0) + 2^m (1+2+3+\cdots+m) = 2^m f(0) + \dfrac{m(m+1)}{2}2^m\\
& = 2^m f(0) + m(m+1)2^{m-1}
\end{align}
Hence, $T(n) = n T(1) + n \left(\dfrac{\log_2(n) (1+\log_2(n))}{2} \right) = \mathcal{\Theta}(n \log^2 n)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/159720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 2,
"answer_id": 1
} |
Evaluate the series: $ \sum_{k=1}^{\infty}\frac{1}{k(k+1)^2k!}$ Evaluate the series:
$$ \sum_{k=1}^{\infty}\frac{1}{k(k+1)^2k!}$$
| Partial fraction decomposition gives
$$\frac{1}{k(k+1)^2}=\left(\frac{1}{k}-\frac{1}{k+1}\right)\frac{1}{k+1}=\frac{1}{k}-\frac{1}{k+1}-\frac{1}{(k+1)^2}$$
Hence this series is
$$\sum_{k=1}^\infty\left(\frac{1}{k}-\frac{1}{k+1}-\frac{1}{(k+1)^2}\right)\frac{1}{k!}$$
$$=\left(\sum_{k=1}^\infty\frac{1}{k \cdot k!}\right)-\left(\sum_{k=1}^\infty\frac{1}{(k+1)!}\right)-\left(\sum_{r=2}^\infty\frac{1}{r\cdot r!}\right).$$
Notice how in the third sum we set $r=k+1$ so $(k+1)^2k!=(k+1)\cdot(k+1)!=r\cdot r!$. The middle term is clearly $e-2$, and the difference between the outside series is $\frac{1}{1\cdot 1!}$, hence we obtain $3-e$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/160219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
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How can I calculate $\lim_{x \to 0} \log(\cos(x))/\log(\cos(3x))$ without l'Hopital? How can I calculate the following limit without using, as Wolfram Alpha does, without using l'Hôpital?
$$
\lim_{x\to 0}\frac{\log\cos x}{\log\cos 3x}
$$
| Note that since $x \to 0$ both $\cos x$ and $\cos 3x$ tend to $1$ and then using $\lim\limits_{t \to 1}\dfrac{\log t}{t - 1} = 1$ we get $$\lim_{x \to 0}\frac{\log\cos x}{\cos x - 1} = 1 = \lim_{x \to 0}\frac{\log \cos 3x}{\cos 3x - 1}$$ We can now easily see that $$\begin{aligned}L &= \lim_{x \to 0}\frac{\log \cos x}{\log \cos 3x}\\
&= \lim_{x \to 0}\frac{\log \cos x}{\cos x - 1}\cdot\frac{\cos x - 1}{\cos 3x - 1}\cdot\frac{\cos 3x - 1}{\log \cos 3x}\\
&= \lim_{x \to 0}\frac{\cos x - 1}{\cos 3x - 1}\\
&= \lim_{x \to 0}\frac{(\cos x - 1)(\cos x + 1)}{\cos x + 1}\cdot\frac{\cos 3x + 1}{(\cos 3x - 1)(\cos 3x + 1)}\\
&= \lim_{x \to 0}\frac{1 + \cos 3x}{1 + \cos x}\cdot\frac{\sin^{2}x}{\sin^{2}3x}\\
&= \lim_{x \to 0}\frac{\sin^{2}x}{x^{2}}\cdot \frac{x^{2}}{(3x)^{2}}\cdot\frac{(3x)^{2}}{\sin^{2}3x}\\
&= \frac{1}{9}\end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/163568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Which of these statements about $Q=\frac{1}{100}+\frac{1}{101}+\cdots+\frac{1}{1000}$ is correct? Pick the correct option regarding $Q$.
$$Q=\frac{1}{100}+\frac{1}{101}+\cdots+\frac{1}{1000}$$
Pick one option:
*
*$Q>1\qquad$ 2. $Q\leq \frac{1}{3}\qquad$ 3. $\frac{1}{3}<Q\leq \frac{2}{3}\qquad$ 4. $\frac{2}{3}<Q\leq 1$
My approach :
$$Q = \frac{1}{100}+\frac{1}{101}+\cdots+\frac{1}{1000} > \frac{1}{1000}+\frac{1}{1000}+\cdots+\frac{1}{1000} = \frac{901}{1000}$$
$$\implies \frac{1}{100}+\frac{1}{101}+\cdots+\frac{1}{1000} > \frac{9}{10}$$
So, $Q > 1$. Option 1 is correct.
Now, my question is: can this be proved with some other approach?
| Divide the sum into two parts:
$$\frac1{100}+\ldots+\frac{1}{500}+\frac1{501}+\ldots+\frac1{1000}$$
The first part, $100$ to $500$ is such that:
$$\frac1{100}+\ldots+\frac{1}{500}>\frac1{500}+\ldots+\frac{1}{500}=\frac{401}{500}$$
The second part, $501$ to $1000$ is such that:
$$\frac1{501}+\ldots+\frac{1}{1000}>\frac1{1000}+\ldots+\frac{1}{1000}=\frac{500}{1000}$$
Therefore the sum of $100$ to $1000$ is larger than $\frac{401}{500}+\frac{500}{1000}$, which itself is larger than $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/163927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
How to evalutate this exponential integral Is there an easy way to compute $$\int_{-\infty}^\infty\exp(-x^2+2x)\mathrm{d}x$$
without using a computer package?
| In general
$$
\begin{align}
\int_{x=0}^\infty e^{-(ax^2+bx)}\,dx&=\int_{x=0}^\infty \exp\left(-a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}\right)\right)\,dx\\
&=\exp\left(\frac{b^2}{4a}\right)\int_{x=0}^\infty \exp\left(-a\left(x+\frac{b}{2a}\right)^2\right)\,dx\\
\end{align}
$$
Let $u=x+\frac{b}{2a}\;\rightarrow\;du=dx$, then
$$
\begin{align}
\int_{x=0}^\infty e^{-(ax^2+bx)}\,dx&=\exp\left(\frac{b^2}{4a}\right)\int_{x=0}^\infty \exp\left(-a\left(x+\frac{b}{2a}\right)^2\right)\,dx\\
&=\exp\left(\frac{b^2}{4a}\right)\int_{u=0}^\infty e^{-au^2}\,du.\\
\end{align}
$$
The last form integral is Gaussian integral that equals to $\frac{1}{2}\sqrt{\frac{\pi}{a}}$. Hence
$$
\int_{x=0}^\infty e^{-(ax^2+bx)}\,dx=\frac{1}{2}\sqrt{\frac{\pi}{a}}\exp\left(\frac{b^2}{4a}\right).
$$
In your case,
$$
\begin{align}
\int_{-\infty}^\infty e^{-x^2+2x}\,dx&=2\int_0^\infty e^{-(x^2-2x)}\,dx\\
&=2\cdot\frac{1}{2}\sqrt{\frac{\pi}{1}}\exp\left(\frac{(-2)^2}{4\cdot1}\right)\\
&=e\sqrt{\pi}.
\end{align}
$$
$$\text{# }\mathbb{Q.E.D.}\text{ #}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "9",
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For prime $p>2: 1^23^25^2\cdot\cdot\cdot(p-2)^2 \equiv (-1)^{\frac{p+1}{2}} \pmod p$
Possible Duplicate:
Why is the square of all odds less than an odd prime $p$ congruent to $(-1)^{(p+1)/(2)}\pmod p$?
If p is an odd prime, prove that $1^2 \times 3^2 \times 5^2 \cdots \times (p-2)^2 \equiv (-1)^{(p+1)/2}\pmod{p}$
I'd love your help with proving the following claim:
For prime $p>2$:
$$1^23^25^2\cdot\cdot\cdot(p-2)^2 \equiv (-1)^{\frac{p+1}{2}} \pmod p.$$
I instantly thought of Wilson Theorem which says that $1\cdot2\cdot3\cdot\cdot\cdot\cdot(p-1) \equiv (-1) \pmod p$, but I can't see how to use it.
I also tried to divide it to two cases, for $p \equiv 1 \pmod4$, and $p \equiv 3 \pmod4$, but again I didn't reach the conclusion.
Thanks a lot!
| Since $p$ is odd, the negatives of the odd residues listed in your product are the even residues: $-1=p-1\pmod{p}$, $-3=p-3\pmod{p}$, $-5=p-5\pmod{p}$, etc.
By Wilson's Theorem, the product of all of these non-zero residues is $(p-1)!=-1\pmod{p}$.
Your product can be changed to $(p-1)!$ by
$$
\begin{align}
\color{#C00000}{1^2}\cdot\color{#00A000}{3^2}\cdot\color{#0000FF}{5^2}\cdots(p-2)^2
&=(-1)^{\frac{p-1}{2}}\color{#C00000}{1\cdot(p-1)}\cdot\color{#00A000}{3\cdot(p-3)}\cdot\color{#0000FF}{5\cdot(p-5)}\cdots(p-2)\cdot2\\
&=(-1)^{\frac{p-1}{2}}(-1)\\
&=(-1)^{\frac{p+1}{2}}\pmod{p}
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Are there rngs whose rngs of matrices are commutative? If $R$ is a unital ring and $M_{2\times 2}(R)$ is a commutative ring, then $R$ is a trivial ring because if $$\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}=\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}\begin{pmatrix}0 & 1 \\ 0 & 1 \end{pmatrix}=\begin{pmatrix}0 & 1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}=\begin{pmatrix}0 & 0\\0 & 0\end{pmatrix}, $$ then $0=1.$
However, $M_{2\times 2}(R)$ can be commutative for possibly non-unital rings of any size. That is, for any cardinal number $\kappa$ (finite or not), there exists an abelian group of order $\kappa.$ Equipping this group with the zero multiplication gives a rng $R$ such that $M_{2\times 2}(R)$ has zero multiplication and so is commutative. The trivial ring is also in this class.
Are there any examples of rngs $R$ whose multiplication is non-zero and such that $M_{2\times 2}(R)$ is commutative?
| No. Expanding the first cell of the product of arbitrary matrices, we would have to have $aa'+bc'=a'a+b'c$ for any $a,b,c,a',b',c'$, so it suffices to look at the particular case $a=a'=b'=c=0$:
$$\begin{pmatrix}0&x\\0&0\end{pmatrix}\begin{pmatrix}0&0\\y&0\end{pmatrix}=\begin{pmatrix}xy&0\\0&0\end{pmatrix}\\
\begin{pmatrix}0&0\\y&0\end{pmatrix}\begin{pmatrix}0&x\\0&0\end{pmatrix}=\begin{pmatrix}0&0\\0&yx\end{pmatrix}$$
Therefore $xy=0$ for any $x,y\in R$, and the family of such rngs is exactly the abelian additive groups equipped with zero multiplication.
| {
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"url": "https://math.stackexchange.com/questions/164905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Largest number that divides $n^2(n^2 - 1)(n^2 - n - 2)$ for all $n$ Obtain the greatest natural that divides $n^2(n^2 - 1)(n^2 - n - 2)$ for all natural numbers $n$.
What should be the approach in these type of questions?
Should I equate with prime factorization $2^a 3^b 5^c \cdots$ etc?
| \begin{align}
f(n) & = n^2(n^2-1)(n^2-n-2)\\
& = n^2(n+1)(n-1)(n+1)(n-2)\\
& = n^2(n+1)^2(n-1)(n-2)
\end{align}
$n=3$ gives us $9 \times 16 \times 2 = 288$. So the natural number dividing $n^2(n+1)^2(n-1)(n-2)$ for all $n$ should be a divisor of $288$.
Also, note that $24$ has to divide $n^2(n+1)^2(n-1)(n-2)$ since $24$ divides $(n-2)(n-1)n(n+1)$ since it is a product of $4$ consecutive integers.
Further, $2$ divides $n(n+1)$. Hence, we know that at-least $48$ divides $f(n)$. Hence, the desired number should be a multiple of $48$ and must divide $288$.
Can you finish it from here by looking at $f(n)$ for couple of other values?
Move the mouse over the gray area for the answer.
Since the largest number must be a multiple of $48$ and must divide $288$. The possible options are $$\{48, 96, 144, 288\}$$
$$\text{$f(4) = 2400$. And $144,288$ doesn't divide $2400$. Hence, the largest number cannot be $144$ or $288$.}$$
$$\text{$f(5) = 10,800$. And $96$ doesn't divide $10,800$. Hence, the largest number cannot be $96$. Hence, the largest number is $48$.}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Convergence of the sequence $\sqrt{1+2\sqrt{1}},\sqrt{1+2\sqrt{1+3\sqrt{1}}},\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1}}}},\cdots$ I recently came across this problem
Q1 Show that $\lim\limits_{n \rightarrow \infty} \underbrace{{\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+n\sqrt{1}}}}}}}_{n \textrm{ times }} = 3$
After trying it I looked at the solution from that book which was very ingenious but it was incomplete because it assumed that the limit already exists.
So my question is
Q2 Prove that the sequence$$\sqrt{1+2\sqrt{1}},\sqrt{1+2\sqrt{1+3\sqrt{1}}},\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1}}}},\cdots,\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+n\sqrt{1}}}}}$$ converges.
Though I only need solution for Q2, if you happen to know any complete solution for Q1 it would be a great help .
If the solution from that book is required I can post it but it is not complete as I mentioned.
Edit: I see that a similar question was asked before on this site but it was not proved that limit should exist.
| Let $f_n(0)=\sqrt{1+n}$ and $f_n(k)=\sqrt{1+(n-k)f_n(k-1)}$. Then $0<f_n(0)<n+1$ when $n>0$. Assume that $f_n(k)<n+1-k$ and we can show by induction that
$$
f_n(k+1) < \sqrt{1+(n-k-1)(n-k+1)} = \sqrt{1+(n-k)^2-1} = n+1-(k+1)
$$
for all k. Your expression is $f_n(n-2)$ which is increasing in $n$ and bounded above by $3$, so converges.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Number of solutions, $a+b+c=n$, $a\gt b\gt c\ge0$ Number of non negative integral solutions for $a + b + c = n$
Where $n$ is a positive integer are
$$\binom{n + 3 - 1}{3 - 1}$$
But if a condition is there $a > b > c$
Is there any direct method by which we can find out required number of solutions.
I believe that we should multiply original number of solutions by $1 / 4$ as there are four following cases possible
$a = b > c$
$a = b = c$
$a > b = c$
$a > b > c$
| When $c=0$, there are $[(n-1)/2]$ solutions, since $b$ can be any positive integer not exceeding $(n-1)/2$.
When $c\ne0$, let $d=a-c$, $e=b-c$, then $d\gt e\gt0$ and $d+e=n-3c$, so the number of solutions (by the previous argument) is $[(n-3c-1)/2]$.
$c$ can be anything from 0 to $[n/3]-1$, so the answer is $$f(n)=\sum_{c=0}^m\left[{n-3c-1\over2}\right]$$ where $m=[n/3]-1$. Now we need a closed form for $f(n)$. Here's one way to go. Note $$f(n+6)=\sum_{c=-2}^{m}\left[{n-3c-1\over2}\right],\qquad f(n+12)=\sum_{c=-4}^{m}\left[{n-3c-1\over2}\right]$$ so $$f(n+12)-2f(n+6)+f(n)=\left[{n+11\over2}\right]+\left[{n+8\over2}\right]-\left[{n+5\over2}\right]-\left[{n+2\over2}\right]=6$$ Thus on any arithmetic progression with common difference 6, $f(n) $ is a quadratic polynomial. So we just have to find the 6 quadratics.
We calculate $f(3)=1$, $f(9)=7$, $f(15)=19$, so (by standard techniques) $f(6n+3)=3n^2+3n+1$; the whole deal looks like this: $$\eqalign{f(6n)&=3n^2\cr f(6n+1)&=3n^2+n\cr f(6n+2)&=3n^2+2n\cr f(6n+3)&=3n^2+3n+1\cr f(6n+4)&=3n^2+4n+1\cr f(6n+5)&=3n^2+5n+2\cr}$$ A simpler formula is $$f(n)=\left[{n^2+6\over12}\right]$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding Asymptotes of Hyperbolas To find a asymptote its either b2/a2 or a2/b2 depending on the way the equation is written.
With the problem
$$\frac{(x+1)^2}{16} - \frac{(y-2)^2}{9} = 1$$
The solutions the sheet I have is giving me is $3/4x - 3/4$ and $3/4 x + 5/4$
I thought it was just supposed to be $\pm 3/4x$.
| Hint: Your hyperbola is the standard hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$, moved one unit to the left and $2$ units up. So write down the asymptotes of $\frac{x^2}{16}-\frac{y^2}{9}=1$, and move them in the same way.
Added: The asymptotes of $\frac{x^2}{16}-\frac{y^2}{9}=1$ are, as you know, $\frac{y}{3}=\pm \frac{x}{4}$. So the asymptotes in your case are
$$\frac{y-2}{3}=\pm\frac{x+1}{4}.$$
The "plus" case simplifies to $y-2=\frac{3}{4}x+\frac{3}{4}$, then to $y=\frac{3}{4}x+\frac{11}{4}$.
The "minus" case simplifies to $y=-\frac{3}{4}x+\frac{5}{4}$.
| {
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Solve: $|-(x + 1)^2+1|\geq 1$ I tried to figure out this question. Can any one help? I know the answer but I keep getting the wrong one. Thanks in advance.
Find all values of x that satisfy $|-(x + 1)^2+1|\geq 1$
| $$|-(x + 1)^2+1|\geq 1 >0$$
Since both sides of the inequlity are positive squaring the inequality we get,
$$((x + 1)^2-1)^2 \geq 1 $$
$$\Leftrightarrow ((x + 1)^2-1)^2-1^2\geq 0 \tag {difference of squares} $$
$$\Leftrightarrow ((x + 1)^2-1-1)((x + 1)^2-1+1)\geq 0 $$
$$\Leftrightarrow ((x + 1)^2-2)(x+1)^2 \geq 0 \tag {with $(x+1)^2>0$ iff $x \neq -1$} $$
Hence,
Case 1 $x = -1$
Then, $((x + 1)^2-2)(x+1)^2 =0 \geq 0$ as required
Case 2 $x \neq -1$
Then $(x + 1)^2-2\geq 0$ (dividing the inequality by $(x + 1)^2>0$)
$$\Leftrightarrow(x + 1- \sqrt{2})(x + 1+ \sqrt{2})\geq 0$$
$$ \Leftrightarrow x \leq -1-\sqrt{2} \ or \ x \geq -1+\sqrt{2}$$
Therefore the final solution is $$x \in (- \infty,-1-\sqrt{2} ] \cup \{-1 \} \cup [-1+\sqrt{2}, +\infty)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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A misbehaved Power Series (contradiction with ln(2))
Possible Duplicate:
Explain why calculating this series could cause paradox?
Using the power series expansion $\ln(x+1)=\sum _{k=1}^\infty \frac{(-1)^{k+1}}{k}x^k$ we have $\ln(2)=(1-\frac{1}{2})+(\frac{1}{3}-\frac{1}{4})+\cdots+(\frac{1}{2k-1}-\frac{1}{2k})+\cdots=\sum _{k=1}^\infty (\frac{1}{2k-1}-\frac{1}{2k})$. Let's rearrange the terms of this sequence so that two negative terms follow each positive term:
$\ln(2)=(1-\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{6}-\frac{1}{8})+\cdots+(\frac{1}{2k-1}-\frac{1}{4k-2}-\frac{1}{4k})+\cdots=\sum _{k=1}^\infty (\frac{1}{2k-1}-\frac{1}{4k-2}-\frac{1}{4k})$. However, $\frac{1}{2k-1}-\frac{1}{4k-2}-\frac{1}{4k}=\frac{1}{2}(\frac{1}{2k-1}-\frac{1}{2k})$; therefore, $\sum _{k=1}^\infty (\frac{1}{2k-1}-\frac{1}{4k-2}-\frac{1}{4k})=\frac{1}{2}\sum _{k=1}^\infty (\frac{1}{2k-1}-\frac{1}{2k})$, but how can the latter inequality hold if $\sum _{k=1}^\infty (\frac{1}{2k-1}-\frac{1}{2k})$ and $\sum _{k=1}^\infty (\frac{1}{2k-1}-\frac{1}{4k-2}-\frac{1}{4k})$ are equal to $\ln(2)$?
Feel free to come up with a better title...
| You have rediscovered the difference between absolute convergence and conditional convergence, and the fact that rearrangement of conditionally convergent series can alter the sum.
If the sum of the positive terms and the sum of the negative terms are both finite, then the series converges absolutely and rearrangement won't change the sum.
| {
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"timestamp": "2023-03-29T00:00:00",
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proving :$\frac{(ab+b)(2b+1)}{(ab+a)(5b+1)}+\frac{(bc+c)(2c+1)}{(bc+b)(5c+1)}+\frac{(ca+a)(2a+1)}{(ca+c)(5a+1)}\ge\frac{3}{2}$ Let $a,b,c, >0$ be real numbers such that $$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge3$$
How to prove that :
$$\frac{(ab+b)(2b+1)}{(ab+a)(5b+1)}+\frac{(bc+c)(2c+1)}{(bc+b)(5c+1)}+\frac{(ca+a)(2a+1)}{(ca+c)(5a+1)}\ge\frac{3}{2}$$
| Let $a=\frac{1}{x} ,b=\frac{1}{y} ,c=\frac{1}{z} $
$ \frac{(x+1)(y+2)}{(y+1)(y+5)} + \frac{(y+1)(z+2)}{(z+1)(z+5)}+\frac{(z+1)(x+2)}{(x+1)(x+5)} \geq \frac{3}{2} $
note that $\frac{y+2}{(y+1)(y+5)}\ge \frac{3}{4(y+2)} $
then is enough to prove that $\frac{x+1}{y+2} + \frac{y+1}{z+2} + \frac{z+1}{x+2} \geq 2$
by Cauchy-Schwarz
$\frac{x+1}{y+2} + \frac{y+1}{z+2} + \frac{z+1}{x+2} =$
$\frac{(x+1)^2}{ (x+1)(y+2) }+\frac{(y+1)^2}{(y+1)(z+2)}+\frac{(z+1)^2}{(z+1)(x+2)}\geq \frac{((x+1)+(y+1)+(z+1))^2}{(x+1)(y+2) +(y+1)(z+2) +(z+1)(x+2) } \geq 2$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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question related to radical sign My question is-
Let $p(x)= \sqrt{x + 2 + 3\sqrt{2x-5}} - \sqrt{x - 2 + \sqrt{2x-5}}$. Then $p(2012)^6$ equals?
Any solution for this question would be greatly appreciated.
Thank you,
| This problem can be solved using the method that is used to denest radicals of the form $\sqrt{a+c\sqrt{b}}$ so that $\sqrt{a+c\sqrt{b}}=\sqrt{d}+\sqrt{e}$. See Stack Exchange question number 220818 for an example of this. This leads to setting each of the nested radicals above to equal the sum of two square roots so that $$\sqrt{x + 2 + 3\sqrt{2x-5}} =(\sqrt{a}+\sqrt{b})$$ and $$\sqrt{x - 2 + \sqrt{2x-5}}=(\sqrt{e}+\sqrt{f})$$ Let's denest the first nested radical you can do the second. I will give the result only for the second, which is below. Set $$a+b = (x+2)$$ and $$2\sqrt{a}\sqrt{b} = 3\sqrt{2x-5}$$ Square both sides of the second equation to get $4ab=18x-45$ and solve for $b$ to get $b=\frac{18x-45}{4a}$then substitute $b$ into the first equation to get $$a+\frac{18x-45}{4a}=(x+2)$$ then multiply both sides by $a$ to get the quadratic equation $$a^2-(x+2)a+\frac{18x-45}{4}=0$$ We can now solve this using the quadratic formula to get $$\frac{x+2+\sqrt{(x+2)^2-\frac{4(18x-45)}{4}}}{2}$$ and $$\frac{x+2-\sqrt{(x+2)^2-\frac{4(18x-45)}{4}}}{2}$$ or $$\frac{x+2+\sqrt{(x-7)(x-7)}}{2}$$
and $$\frac{x+2-\sqrt{(x-7)(x-7)}}{2}$$ The two roots therefore turn out to be $\frac{2x-5}{2}$ and $\frac{9}{2}$. Because we are looking for the sum of two square roots $\sqrt{a}$ and $\sqrt{b}$ we have to take the square roots of each of the roots of the quadratic and this yields the sum $$\frac{\sqrt{4x-10}}{2}+\frac{3\sqrt{2}}{2}$$ for the first nested radical in your question.
Now use the identical procedure to denest the second nested radical $$\sqrt{x - 2 + \sqrt{2x-5}}$$ Going through the above denesting procedure on this yields $$\frac{\sqrt{4x-10}}{2}+\frac{\sqrt{2}}{2}$$ Combining the two results we get $$\left(\frac{\sqrt{4x-10}}{2}+\frac{3\sqrt{2}}{2}\right)
- \left(\frac{\sqrt{4x-10}}{2}+\frac{\sqrt{2}}{2}\right) = \sqrt{2}$$ So you can see that the result is $p(x) = \sqrt{2}$ for all $x$.
| {
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Evaluate $\int^1_0 \frac{dx}{x^2+x+1}$ Here is an integral which I don't know the answer:
$$\int^1_0 \frac{dx}{x^2+x+1}$$
I tried to use complex number to solve i.e. the root of $x^2+x+1$ is $(-1/2 + \sqrt {3}i/2)$.
Let w=$(-1/2 + \sqrt {3}i/2)$ , then it becomes $ \int_{0}^{1} 1/(x-w)^2\,\mathrm{d} (x-w)$ , the answer is in terms of complex number. I wanna ask whether my method is correct and seek for another method. Thank you.
| $$I := \int^1_0 \frac{dx}{x^2+x+1}$$
This looks like a job for . . . $\arctan$!
Recall that
$$\int\frac{du}{u^2+a^2} = \frac { \arctan \left( \frac {u}{a} \right)} {a} + C$$
Complete the square in the bottom to get
$$I = \int^1_0 \frac{dx} {{\left(x+\frac{1}{2} \right)}^2 + \frac{3}{4}}$$
Let $u = x + \frac{1}{2}, du = dx$.
$$ \frac { 2\arctan \left( \frac{2u}{\sqrt{3}} \right) } {\sqrt{3}}$$
Back-substitute $u=x+\frac{1}{2}$ to get:
$$ \frac { 2\arctan \left( \frac{2x+1}{\sqrt{3}}\right) } {\sqrt{3}}$$
Now just evaluate at your endpoints to get
$$I = \frac{\pi}{3\sqrt{3}} \approx .605$$
| {
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A hyperbola passing through integer lattice points Prove that for any $n\geq 0$, there is a hyperbola that passes through exactly $n$ lattice points (= points with integer coordinates) and find an example.
For example it is easy to see that the hyperbola $xy=1$ passes throught exactly two lattice points. It is also easy to see that the hyperbola
$$
x^2-2y^2=1
$$
passes through infinitely many lattice points, because this is a Pell equation and its integer solutions are related to the units of the ring of algebraic integers. The intermediate case of a prescribed number of lattice points requires another idea.
| Note that the hyperbola $2x^2-2y^2=1$ has no points on the integer lattice.
We next deal with the case where $n \gt 0$ is even, even though the proof we give later for $n$ odd actually works for all non-zero $n$.
Look at the hyperbola with equation $x^2-y^2=3^k$, or equivalently $(x-y)(x+y)=3^k$. We obtain all the solutions by putting $x-y=u$, $x+y=v$, where $u$ and $v$ are integers and $uv=3^{k}$. The number of possibilities where $u$ and $v$ are positive is $k+1$, since $u$ can take on all values from $3^0$ to $3^k$. Double this to include the negative possibilities. So the number of lattice points is $2(k+1)$. Now let $k=\frac{n}{2}-1$.
Finally we deal with the case where $n$ is odd. Actually, the oddness of $n$ will be irrelevant.
Consider the hyperbola $(4x+1)^2-y^2=5^k$, or equivalently $(4x+1 -y)(4x+1+y)=5^{k}$. The total number of ordered pairs $(u,v)$ of integers (possibly both negative) such that $uv=5^k$ is $2(k+1)$.
So exactly as earlier, the equation $z^2-y^2=5^k$ has precisely $2(k+1)$ integer solutions. In all of these solutions, $z$ is odd.
Note that exactly one of the odd integers $z$ and $-z$ is congruent to $1$ modulo $4$. So the number of solutions of $(4x+1)^2-y^2=5^k$ is $\frac{2(k+1)}{2}$, that is, $k+1$. Now let $k=n-1$.
| {
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Compute the trigonometric integrals How would you compute the following integrals?
$$ I_{n} = \int_0^\pi \frac{1-\cos nx}{1- \cos x} dx $$
$$ J_{n,m} = \int_0^\pi \frac{x^m(1-\cos nx)}{1- \cos x} dx $$
For instance, i noticed that the first integral is convergent for any value of $n$ since $\lim_{x\to0} \frac{1- \cos nx}{1 - \cos x}= n^2$. This fact may also be extended to the second integral, as well.
| Here's a way to find $I_n$ with the residue theorem: first, by symmetry, $I_n = \frac{1}{2} \int_0^{2\pi} \frac{1 - \cos(nx)}{1- \cos(x)}dx$. Consider $z = e^{ix}$ such that the integral with now be on the edge of the unit circle $C$, and use the fact that $$\cos(nx) = \frac{e^{inx} + e^{-inx}}{2}.$$ as well as $\frac{dz}{dx} = i e^{ix}$. So $dx = \frac{1}{iz} dz$ and
$$\int_0^{2\pi} \frac{1 - \cos(nx)}{1 - \cos(x)} dx = \int_C \frac{1}{iz} \frac{1 - \frac{1}{2}(z^{n} + z^{-n})}{1 - \frac{1}{2}(z + z^{-1})}dz$$ $$= \frac{1}{i}\int_C \frac{1}{z^{n}} \left(\frac{z^n - 1}{z - 1}\right)^2 dz$$ $$= \frac{2\pi i}{i} \mathrm{Res}_{z = 0} \frac{1}{z^{n}} (1 + ... + z^{n-1})^2 = 2\pi n$$
and so $I_n = \pi n$.
| {
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A binomial identity I was wandering if someone knows an elementary proof of the following identity:
$$
\frac{(a)_n (b)_n}{(n!)^2} = \sum_{k=0}^n (-1)^k {1-a-b \choose k}
\frac{(1-a)_{n-k}(1-b)_{n-k}}{((n-k)!)^2}\ ,
$$
where $a,b$ are arbitrary real numbers,
$$(x)_0=1,\quad
(x)_n:=x(x+1)\cdots(x+n-1) \quad
\mbox{for $n\geq 1$}
$$
is the Pochhammer's symbol, and
$$
{x\choose 0}=1,\quad
{x\choose k} = \frac{x(x-1)\cdots (x-k+1)}{k!} \quad
\mbox{for $k\geq 1$}$$
is a binomial coefficient.
The proof that I know uses the Hypergeometric differential equation. One has to continue analytically the solutions along a path connecting two singular points. This could be done by some well known integral representation of the Hypergeometric function.
I think that there should be a combinatorial proof. Since this is an identity between polynomials in $a$ and $b$, it is enough to prove it for $a$ and $b$ negative integers, i.e., we may assume that $a=-p$ and $b=-q$ where $p,q\in \mathbb{Z}_{\geq 0}$. The identity then turns into
$$
{p\choose n}{q\choose n} =
\sum_{k=0}^n (-1)^k {1+p+q\choose k}{p+n-k\choose n-k}{q+n-k\choose n-k}.
$$
I did not try very hard to proof the above identity and I did not search the literature that much, but since it comes from an interesting subject I think that it is worth finding an alternative proof.
| Suppose we seek to evaluate
$$\sum_{k=0}^n (-1)^k
{p+q+1\choose k} {p+n-k\choose n-k} {q+n-k\choose n-k}$$ which is
claimed to be $${p\choose n}{q\choose n}.$$
Introduce $${p+n-k\choose n-k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{p+n-k}}{z^{n-k+1}}
\; dz$$
and $${q+n-k\choose n-k}
= \frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{(1+w)^{q+n-k}}{w^{n-k+1}}
\; dw.$$
Observe that these integrals vanish when $k\gt n$ and we may extend
$k$ to infinity.
We thus obtain for the sum $$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{p+n}}{z^{n+1}}
\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{(1+w)^{q+n}}{w^{n+1}}
\\ \times \sum_{k\ge 0} {p+q+1\choose k} (-1)^k
\frac{z^k w^k}{(1+z)^k (1+w)^k}
\; dw\; dz.$$
Note that while there is no restriction on $k$ the sum only contains a
finite number of terms. Continuing,
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{p+n}}{z^{n+1}}
\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{(1+w)^{q+n}}{w^{n+1}}
\\ \times
\left(1-\frac{z w}{(1+z)(1+w)}\right)^{p+q+1}
\; dw\; dz$$
or $$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n-q-1}}{z^{n+1}}
\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{(1+w)^{n-p-1}}{w^{n+1}}
(1+ z + w)^{p+q+1}
\; dw\; dz$$
Supposing that $p\ge n$ and $q\ge n$
and $\epsilon \ll 1$ and $\gamma \ll 1$ this may be re-written as
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1} (1+z)^{q+1-n}}
\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{n+1} (1+w)^{p+1-n}}
\\ \times (1+ z + w)^{p+q+1}
\; dw\; dz$$
Put $w = (1+z) u$ so that $dw = (1+z) \; du$ to get
with $\delta \lt \gamma/(1+\epsilon)$
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1} (1+z)^{q+1-n}}
\frac{1}{2\pi i}
\int_{|u|=\delta}
\frac{1}{(1+z)^{n+1} u^{n+1} (1+(1+z)u)^{p+1-n}}
\\ \times (1+ z)^{p+q+1} (1+u)^{p+q+1}
\; (1+z) \; du\; dz$$
Note that the pole at $u=-1/(1+z)$ has norm $\delta/\gamma \gt \delta$
so it is not inside the contour in $u$. This yields
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^p}{z^{n+1}}
\frac{1}{2\pi i}
\int_{|u|=\delta}
\frac{1}{u^{n+1} (1+ u + z u)^{p+1-n}}
\\ \times (1+u)^{p+q+1}
\; du\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^p}{z^{n+1}}
\frac{1}{2\pi i}
\int_{|u|=\delta}
\frac{(1+u)^{n+q}}{u^{n+1} (1 + z u /(1+u))^{p+1-n}}
\; du\; dz$$
Extracting the residue for $z$ first we obtain
$$\sum_{k=0}^n {p\choose n-k}
\frac{(1+u)^{n+q}}{u^{n+1}}
{k+p-n\choose k}
(-1)^k \frac{u^k}{(1+u)^k}.$$
The residue for $u$ then yields $$\sum_{k=0}^n (-1)^k {p\choose n-k}
{k+p-n\choose k} {n-k+q\choose n-k}.$$
The sum term here is $$\frac{p!\times (p+k-n)!\times (q+n-k)!}
{(n-k)! (p+k-n)! \times k! (p-n)! \times (n-k)! q!}$$ which simplifies
to $$\frac{p!\times n! \times (q+n-k)!}
{(n-k)! \times n!\times k! (p-n)! \times (n-k)! q!}$$ which is
$${n\choose k} {p\choose n}{q+n-k\choose q}$$ so we have for the sum
$${p\choose n}
\sum_{k=0}^n {n\choose k} (-1)^k {q+n-k\choose q}.$$
To evaluae the remaining sum we introduce $${q+n-k\choose q}
= \frac{1}{2\pi i}
\int_{|v|=\epsilon} \frac{(1+v)^{q+n-k}}{v^{q+1}} \; dv$$
getting for the sum $${p\choose n}
\frac{1}{2\pi i}
\int_{|v|=\epsilon}
\frac{(1+v)^{q+n}}{v^{q+1}}
\sum_{k=0}^n {n\choose k} (-1)^k \frac{1}{(1+v)^k} \; dv
\\ = {p\choose n}
\frac{1}{2\pi i}
\int_{|v|=\epsilon}
\frac{(1+v)^{q+n}}{v^{q+1}}
\left(1-\frac{1}{1+v}\right)^n \; dv
\\ = {p\choose n}
\frac{1}{2\pi i}
\int_{|v|=\epsilon}
\frac{(1+v)^{q}}{v^{q-n+1}} \; dv
= {p\choose n} {q\choose q-n}$$
which is $${p\choose n} {q\choose n}.$$
This concludes the argument.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/174054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 1
} |
Maximum Profit at which combination A dealer deals only in colour TVs and VCRs . He wants to spend up to Rs.12 lakhs to buy 100 pieces. He can purchase colour TV at Rs. 10,000 and a VCR at Rs. 15,000. He can sell a colour TV at Rs. 12,000 and a VCR at Rs. 17,500. His objective is to maximiz profits
Fpr the maximum profit number of colour TVs and VCRs that he should stock is
I have found answer using hit and trial
But is there a mathmatical way to guarantee that maximum profit happens for 60 colour TV and 40 VCR
| assuming total number of pieces of CTVs and VCRs together is $c$ which is a constant once the purchase is made.
let, $x$ is the number of CTVs sold(purchased), and $(c-x)$ is the number of VCRs sold(purchased)
profit function, $f(x)=(x)(2)(10^3)+(c-x)(\frac{5}{2})(10^3)$ -------(1)
cost price, $g(x)=(x)(10)(10^3) + (c-x)(15)(10^3)\le(12)(10^5)$ -------(2)
integer, $c = 100$ -------(3)
maximum number of CTVs which can be purchased, $x = \frac{(12)(10^5)}{(10)(10^3)}=120$
maximum number of VCRs which can be purchased, $(c-x) = \frac{(12)(10^5)}{(15)(10^3)}$
i.e. $0 \le x\le 100$ -------(4)
we have, $0 \le (c-x)\le 80$ -------(5)
we always need to check the boundary conditions which means consider the following equation
$g(x_b)=(x_b)(10)(10^3) + (c-x_b)(15)(10^3)=(12)(10^5)$ -------(6)
also, $f(x_b)=(x_b)(2)(10^3)+(c-x_b)(\frac{5}{2})(10^3)$ -------(7)
from equation (6),
$x_b = (3c-240)=60$
$f(x)=(2x+\frac{5}{2}(c-x))(10^3)$
$f(x)=(\frac{5c}{2}-\frac{x}{2})(10^3)$ -------(8)
if $x$ is the number of CTVs sold, then on solving, (1) and (2) we have
$f(x)\le(c+120)(10^3)$ -------(9)
i.e. profit $f(x)\le(220)(10^3)$, where $x$ is the number of CTVs sold(purchased)
also, from (8) and (9) we have $x_m$ the maximum number of CTVs sold(purchased) from the following equation
$(\frac{5c}{2}-\frac{x_m}{2})(10^3)=(c+120)(10^3)$
i.e. $x_m=60$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/178689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Expressing $\sin^4x-\sin^6x$ in another way I am slightly confused on how one would subtract $\sin^4x-\sin^6x$.
I know that $\sin^2x=(1/2)(1-\cos2x)$,
so $\sin^4x$ would logically be $[(1/2)(1-\cos2x)]^2=(1/4)(1-2\cos(2x)+\cos^2(2x)$
However the value of $\sin^6x$ eludes me. Would it be $(1-\cos2x)^3$? I did that and got $1-\cos2x-2\cos2x-2\cos^2(2x)+\cos^2(2x)-\cos^3(2x)$
I am not sure if that is correct.
| I have noticed the other answers either don't answer your question (and get many upvotes none-the-less) or are much more complicated than necessary (even more complicated than you understand based on the comment by Raymond). So, here is the answer to your actual question.
You dropped the 1/2.
$$\sin^6 x = (\sin^2 x)^3 = \left(\frac{1 - \cos{2x}}{2}\right)^3 = \frac{1}{8} (1 - \cos{2x})^3$$
so you end up missing $1/8$ in the end. Also, you messed up the sign on one term. In general
$$(A + B)^3 = A^3 + 3A^2B + 3AB^2 + B^3$$
so here we have
\begin{align*}
(1 - \cos(2x))^3 &= \bigg(1 + (-\cos(2x))\bigg)^3 \\
&= 1 + 3(-\cos(2x)) + 3(-\cos(2x))^2 + (-\cos(2x))^3 \\
&= 1 - 3\cos(2x) + 3\cos^2(2x) - \cos^3(2x).
\end{align*}
Therefore, your final answer is
\begin{align*}
\sin^4x - \sin^6x &= \frac{1}{4}\bigg(1 - 2\cos(2x) + \cos^2(2x)\bigg) \\
&- \frac{1}{8}\bigg(1 - 3\cos(2x) + 3\cos^2(2x) - \cos^3(2x)\bigg) \\
&= \frac{1}{8}\bigg(1 - \cos(2x) - \cos^2(2x) + \cos^3(2x)\bigg)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/180348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
A problem on indefinite integral: $\int(\cos x)^m\sin(nx) \mathrm dx$
If
$$I(m,n)=\int(\cos x)^m\sin(nx) ~\mathrm dx,$$
How do I get $7I(4,3)-4I(3,2)$?
| A general solution
$$\begin{align}
I(m,n)
&=\int(\cos x)^m \sin(nx) ~\mathrm{d}x \\
&=\int(\cos x)^m\sin((n-1)x+x) ~\mathrm{d}x \\
&=\int(\cos x)^{m+1}\sin((n-1)x) ~\mathrm{d}x+\int(\cos x)^m\cos((n-1)x)(\sin x)~\mathrm{d}x \\
&\stackrel{\color{red}{\text{IBP}}}{=}-\dfrac{\cos((n-1)x)}{n-1}(\cos x)^{m+1}-\left(\dfrac{m+1}{n-1}-1\right)\int(\cos x)^m\cos((n-1)x)(\sin x)~\mathrm{d}x \\
&=-\dfrac{\cos((n-1)x)}{n-1}(\cos x)^{m+1}-\left(\dfrac{m-n+2}{n-1}\right)\int(\cos x)^m\cos((n-1)x)(\sin x)~\mathrm{d}x
\end{align}$$
Now
$$\begin{align}
&\int(\cos x)^m\cos((n-1)x)(\sin x)~\mathrm{d}x \\
&\stackrel{\color{red}{\text{IBP}}}{=} -\dfrac{(\cos x)^{m+1}\cos((n-1)x)}{m+1}+\left(\dfrac{n-1}{m+1}\right)\int (\cos x)^{m+1}\sin((n-1)x) ~\mathrm{d}x \\
&=-\dfrac{(\cos x)^{m+1}\cos((n-1)x)}{m+1}+\left(\dfrac{n-1}{m+1}\right) I(m+1,n-1)
\end{align}$$
This is tedious but now everything can be evaluated by successively putting values into it. It would be great if someone shows a nice way of evaluating the given value by some manipulation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/180662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Inequality. $a^2+b^2+c^2 \geq a+b+c$ Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that
$a^2+b^2+c^2 \geq a+b+c$.
Thanks
| We will use the following form of Cauchy-Schwarz inequality:
From Cauchy-Schwarz inequality applied on the vectors $\displaystyle{ \left( \frac{x_1}{\sqrt{a_1}} , \frac{x_2}{\sqrt{a_2}} , \cdots , \frac{x_n}{\sqrt{a_n}} \right)}$ and $ \displaystyle{ \left( \sqrt{a_1} ,\sqrt{a_2} , \cdots , \sqrt{a_n} \right) }$ where $ x_1 ,x_2 \cdots ,x_n \in \mathbb R $ and $ a_1, a_2, \cdots ,a_n >0 $ we get that
$ \displaystyle{ \frac{x_1^{2}}{a_1} +\frac{x_2^{2}}{a_2} + \cdots + \frac{x_n^{2}}{a_n} \geq \frac{\left(x_1 + x_2 + \cdots + x_n \right)^{2}}{a_1 +a_2 + \cdots + a_n} }$
Back to our problem now the given inequality can be written equivelant in the form
$ \displaystyle{ \frac{a^2}{abc} + \frac{b^2}{abc} +\frac{c^2}{abc} \geq a+ b+c \quad (\star)}$
Using now the inequality we stated in the begging we get that the left-hand-side of $(\star)$ is :
$ \displaystyle{ \frac{a^2}{abc} + \frac{b^2}{abc} +\frac{c^2}{abc} \geq \frac{ \left(a+b+c \right)^2}{3abc} = \frac{ \left(a+b+c \right)^2}{3} \geq \left(a+b+c \right) \cdot \frac{ 3 \sqrt[3]{abc}}{3} = a+b+c }$
which is what we need to prove. Q.E.D
P.S. I the last step we use the AM-GM inequality: $ \displaystyle{ a+b+c \geq 3
\sqrt[3]{abc} }$.
P.S The above form of Cauchy-Schwarz inequality is called Cauchy-Schwarz in Engel form.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/181626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 8,
"answer_id": 1
} |
Solve the Diophantine equation $ 3x^2 - 2y^2 =1 $ Solve $$ 3x^2 - 2y^2 =1 $$ in $ \mathbb{Z}$.
How can we do it?
( All of answers gave me a great help. Thanks a lot kind stackexchangers.)
| This is pretty much the same information as Mark Bennet. Not the entire earthly existence of Mark Bennet, just his answer. We begin with an automorph of your indefinite quadratic form. I like to call the matrix that goes on the right the automorph, so for me it is
$$ A \; = \;
\left( \begin{array}{rr}
5 & 4 \\
6 & 5
\end{array}
\right)
$$
which solves
$$
\left( \begin{array}{rr}
5 & 6 \\
4 & 5
\end{array}
\right) \cdot
\left( \begin{array}{rr}
3 & 0 \\
0 & -2
\end{array}
\right) \cdot
\left( \begin{array}{rr}
5 & 4 \\
6 & 5
\end{array}
\right)
=
\left( \begin{array}{rr}
3 & 0 \\
0 & -2
\end{array}
\right)
$$
Note that we have
$$ A^{-1} \; = \;
\left( \begin{array}{rr}
5 & -4 \\
-6 & 5
\end{array}
\right)
$$
We take the convention
$$ A^{0} \; = \; I = \;
\left( \begin{array}{rr}
1 & 0 \\
0 & 1
\end{array}
\right)
$$
The key thing is that any power of $A,$ positive or negative or $0,$ is also an automorph, and takes a solution of your equation as a column vector to another solution. In this case, all solutions can be generated in this manner as
$$
A^n \cdot
\left( \begin{array}{r}
1 \\
1
\end{array}
\right).
$$
For example,
$$
\left( \begin{array}{rr}
5 & 4 \\
6 & 5
\end{array}
\right) \cdot
\left( \begin{array}{r}
1 \\
1
\end{array}
\right)
=
\left( \begin{array}{r}
9 \\
11
\end{array}
\right),
$$
$$
\left( \begin{array}{rr}
5 & 4 \\
6 & 5
\end{array}
\right) \cdot
\left( \begin{array}{r}
9 \\
11
\end{array}
\right)
=
\left( \begin{array}{r}
89 \\
109
\end{array}
\right),
$$
meanwhile
$$
\left( \begin{array}{rr}
5 & -4 \\
-6 & 5
\end{array}
\right) \cdot
\left( \begin{array}{r}
1 \\
1
\end{array}
\right)
=
\left( \begin{array}{r}
1 \\
-1
\end{array}
\right),
$$
$$
\left( \begin{array}{rr}
5 & -4 \\
-6 & 5
\end{array}
\right) \cdot
\left( \begin{array}{r}
1 \\
-1
\end{array}
\right)
=
\left( \begin{array}{r}
9 \\
-11
\end{array}
\right).
$$
Furthermore, you can still negate both entries. So all solutions are
$$ \pm A^n \cdot
\left( \begin{array}{r}
1 \\
1
\end{array}
\right) $$
with $n \in \mathbb Z.$
This is from pages 21-34 of Duncan A. Buell, Binary Quadratic Forms: Classical Theory and Modern Computations. If you had an oversized target or other difficulty, as in $3 x^2 - 2 y^2 = 25,$ the automorph idea would still work, but more "seed" vectors would be required. This is discussed in John H. Conway, The Sensual Quadratic Form.
I give an example of Conway's "river" method, including the necessary diagram, at Another quadratic Diophantine equation: How do I proceed?
I recently purchased a home scanner, it does just one page at a time but fairly well. Here, with any luck, is Conway's "River" diagram for this problem:
Good, it is there and clear. Note how the ways to express $1$ stay close to the river, and there is just one way per cycle. I have written in the $x,y$ values for each $3x^2 - 2 y^2 = n$ as a column vector in that position. In comparison, if I wanted $3x^2 - 2y^2 = 25,$ i would have several solutions per cycle, for two reasons: for any solution of $3x^2 - 2y^2 = 1,$ I could just multiply through by $5,$ giving one imprimitive solution per cycle. There are also two primitive solutions per cycle, beginning with the two $3 \cdot 3^2 - 2 \cdot 1^2 = 3 \cdot 11^2 - 2 \cdot 13^2 = 25.$
In case it was not clear, the business with automorphs works along with the river construction, all that is really needed is correct identification of the solutions in a single cycle. So, all solutions to $3x^2 - 2 y^2 = 25$ are
$$ \pm A^n \cdot
\left( \begin{array}{r}
5 \\
5
\end{array}
\right), \; \; \;
\pm A^n \cdot
\left( \begin{array}{r}
3 \\
1
\end{array}
\right), \; \; \;
\pm A^n \cdot
\left( \begin{array}{r}
11 \\
13
\end{array}
\right),
$$
with $n \in \mathbb Z.$
In case this makes any sense, this diagram is a small section of a parametrization of $PSL_2 \mathbb Z.$ All the solutions we give have $x \geq 0.$ You get the other solutions by negating both $x,y.$ Well, pages 1-33 in Conway's book.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/181758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Inequality. $\frac{1}{16}(a+b+c+d)^3 \geq abc+bcd+cda+dab$ I want to prove the following inequality :
$$\frac{1}{16}(a+b+c+d)^3 \geq abc+bcd+cda+dab, $$
$a,b,c,d \in \mathbb{R}_{+} .$
In my book, at the answers chapter the author uses AM $\geq$ GM, but I haven't any idea how I can use that.
Thanks :)
| Also notice that:
$$(a+b+c+d)^3 - 16(abc+abd+acd+bcd) = (a+b+c+d)(a+b-c-d)^2 + 4(c-d)^2(a+b) + 4(a-b)^2(c+d) \ge 0$$
Or
$$(a+b)[(a+b-c-d)^2 + 4(c-d)^2] + (c+d)[(a+b-c-d)^2 + 4(a-b)^2]\ge0$$
This way is suggested by a friend of mine.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/184029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 5,
"answer_id": 0
} |
A divisibility question involving Fermat's theorem use How I am supposed to solve questions like
if $\gcd (a,42) =1$ and $168= 3\times 7\times8$ then show that $168$ divides $a^6-1$.
| From $(a,42) =1$ and using Fermat's little theorem :
*
*$(a,7) =1$ so that $\ a^{7-1}\equiv 1\pmod{7}\quad$ ($7$ divides $a^6-1$)
*$(a,3) =1$ so that $\ a^{3-1}\equiv 1\pmod{3}\quad$ ($3|(a^2-1)$ and $(a^2-1)|(a^6-1)$)
*$(a,2) =1$ so that $a$ is odd but for odd $a$ we get $\ (\pm 1)^2\equiv (\pm 3)^2\equiv 1\pmod{8}$ and $a^6\equiv 1\pmod{8}$ (i.e. 8 divides $a^6-1$).
We conclude that $8\cdot 3\cdot 7$ divides $a^6-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/185759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
The functional equation $f(x)+f(y)+f(z)+f(x+y+z)=f(x+y)+f(y+z)+f(z+x)$ How to find the all continuous functions $f\colon \mathbb{R}\to\mathbb{R}$ such that $f(x)+f(y)+f(z)+f(x+y+z)=f(x+y)+f(y+z)+f(z+x)$
| All continuous solutions are of the form $f(x) = a x + b x^2$.
For given $z$, let $g(x) = f(x+z) - f(x) - f(z)$. The functional equation becomes
$g(x+y) = g(x) + g(y)$. This is the Cauchy functional equation, and it is known that its
only continuous solutions are $g(x) = c x$. Of course $c$ can depend on $z$. Now we need to solve
$f(x+z) - f(x) - f(z) = c(z) x$. Since the left side is symmetric in $x$ and $z$, $c(z) x = c(x) z$, so $c(z) = k z$ for some constant $k$.
Taking $z=-x$, and using $f(0)=0$, we get $- f(x) - f(-x) = - k x^2$, or $f(x) + f(-x) = k x^2$.
Note that if $f(x)$ is a solution of our equation, so is $f(-x)$, and by linearity so are the even and odd parts $ (f(x) + f(-x))/2$ and $(f(x) - f(-x))/2$. Thus it suffices to consider the two cases $f$ even and $f$ odd.
If $f$ is even,
$f(x) + f(-x) = 2 f(x) = k x^2$, so $f(x) = (k/2) x^2$.
If $f$ is odd, $f(x) + f(-x) = 0$ so $k=0$. now we have $f(x+z) - f(x) - f(z) = 0$, which is again Cauchy's functional equation, and so $f(x) = a x$ for some constant $a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/186011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
} |
Inequality. $\frac{ab+c}{a+b}+\frac{ac+b}{a+c}+\frac{bc+a}{b+c} \geq 2.$ Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that (using rearrangements inequalities, you can also view this exercise here, exercise number 3.1.8 )
$$\frac{ab+c}{a+b}+\frac{ac+b}{a+c}+\frac{bc+a}{b+c} \geq 2.$$
thanks.
| The inequality follows $$\frac{(b+c)(c+a)}{a+b}+\frac{(a+c)(a+b)}{b+c}+\frac{(a+b)(b+c)}{a+c}\geq 2$$
Let $a+b=z,b+c=x,c+a=y$ then $x+y+z=2$. Therefore $$x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}\geq xyz(x+y+z)$$
This is true by AM-GM.
P/s: Sorry for my bad English.
| {
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"url": "https://math.stackexchange.com/questions/191605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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When $2^n + m$ is a perfect square This problem showed up when I was doing some recreative math, but the problem was interesting itself and I found very difficult to solve.
Given $n\in\mathbb{N}$ and $m\in\mathbb{Z}$, the question is: There is a method to know if $2^n + m$ is a perfect square?
Thank you very much.
| Assume $2^n+m=a^2$ with $a,m,n\in\mathbb N$, $0<m\le\sqrt{2^n}$.
If $n=2k$ is even, then $(2^k)^2=2^n<a$ and $(2^k+1)^2=2^n+2\cdot 2^k+1>2^n+m$, contradiction.
If $n=2k+1$ is odd, then $\left(\frac a{2^k}\right)^2=2+\frac m{2^{2k}}$, i.e. $\frac a{2^k}$ is a relatively good approximation for $\sqrt 2$.
In fact with $\frac a{2^k}>\sqrt 2$ we find that
$$\frac m{2^{2k}}=\left(\frac a{2^k}\right)^2-2=\left(\frac a{2^k}+\sqrt2\right)\left(\frac a{2^k}-\sqrt2\right)>2\sqrt 2 \cdot\left(\frac a{2^k}-\sqrt2\right),$$
hence
$$0<\frac a{2^k}-\sqrt2<\frac{m}{2^n\sqrt 2}\le \frac{\sqrt{2^n}}{2^n\sqrt 2}=\frac1{2^{k+1}}.$$
Letting $a=\lceil2^k\sqrt 2\rceil$ will therefore lead to an instance of $2^n+m=a^2$ about 50% of the time.
If we relax the restriction on $m$ to $m\le 2\cdot\sqrt{ 2^n}$, this rate will go up to 100%.
For example $2^{59}+9092137 = 759250125^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/194745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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