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Determinants using elementary row operations Let matrix $A$ be defined as \begin{pmatrix}
a_{11} & a_{12} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots & \vdots & \vdots & \vdots \\
a_{n1} & a_{n2} & \cdots & a_{nn}\\
\end{pmatrix}
And $det(A)=d$
Let $B=$
\begin{pmatrix}
2a_{11} & 2a_{12} & \cdots & 2a_{1n} \\
2a_{21} & 2a_{22} & \cdots & 2a_{2n} \\
\vdots & \vdots & \vdots & \vdots \\
2a_{n1} & 2a_{n2} & \cdots & 2a_{nn}\\
\end{pmatrix} $$+$$
\begin{pmatrix}
3a_{11} & 3a_{12} & \cdots & 3a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots & \vdots & \vdots & \vdots \\
a_{n1} & a_{n2} & \cdots & a_{nn}\\
\end{pmatrix}
Find $det(B)$ using elementary row operations. I think the answer is $3^{(n-1)}*5*d$, but I'm not sure.
|
$$\begin{pmatrix}
2a_{11} & 2a_{12} & \cdots & 2a_{1n} \\
2a_{21} & 2a_{22} & \cdots & 2a_{2n} \\
\vdots & \vdots & \vdots & \vdots \\
2a_{n1} & 2a_{n2} & \cdots & 2a_{nn}\\
\end{pmatrix} +
\begin{pmatrix}
3a_{11} & 3a_{12} & \cdots & 3a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots & \vdots & \vdots & \vdots \\
a_{n1} & a_{n2} & \cdots & a_{nn}\\
\end{pmatrix}=\begin{pmatrix}
5a_{11} & 5a_{12} & \cdots & 5a_{1n} \\
3a_{21} & 3a_{22} & \cdots & 3a_{2n} \\
\vdots & \vdots & \vdots & \vdots \\
3a_{n1} & 3a_{n2} & \cdots & 3a_{nn}\\
\end{pmatrix}$$
Since the determinant is a $n$-linear function (considering that each line is an element of $\mathbb R^n$), it follows that, since the first line of $B$ is the first line of $A$ multiplied by $5$ and since all the other $n-1$ lines of $B$ are the same $n-1$ lines of $A$ multiplied by $3$, it follows that
$\det(B)=5.3^{n-1}.det(A)=5.3^{n-1}.d$
Your answer is correct.
|
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"url": "https://math.stackexchange.com/questions/1314087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Expansion of Generating Functions If you roll $10$ dice, how many ways can you get a total sum of top faces of $25$?
I understand how to write the generating function of $(x+x^2+ \dots +x^6)^{10}$ and the fact that you need to find the coefficient of $x^{25}$, but how do you do this? The conventional binomial theorem only works for binomial expansion....
|
I should mention, this answer assumes the dice are distinct. (e.g. you have a red die, a blue die, a green die, etc... or a "first die" a "second die" a "third die", etc...)
We know that each result on the die is minimum 1, so let us ask a similar question instead to simplify the arithmetic involved: What is the coefficient of $x^{15}$ in the expansion of $(1+x+x^2+\dots+x^5)^{10}$.
Find how many ways you can add to 15 using 1's, 2's, 3's, etc in decreasing order using at most 10 positive integers less than or equal to 5.
$$\begin{array}{l}5+5+5\\
5+5+4+1\\
5+5+3+2\\
5+5+3+1+1\\
5+5+2+2+1\\
5+5+2+1+1+1\\
5+5+1+1+1+1+1\\
5+4+4+2\\
\color{red}{5+4+4+1+1}\\
5+4+3+3\\
5+4+3+2+1\\
5+4+3+1+1+1\\
5+4+2+1+1+1+1\\
5+4+1+1+1+1+1+1\\
\vdots\\
2+2+2+2+2+1+1+1+1+1\end{array}$$
For each of these, you can calculate their specific contribution. The line colored red for example occurs when you use one occurrence of $x^5$, two occurrences of $x^4$, two occurrences of $x^1$, and the remaining five occurrences as $x^0$. If you were to reimagine this as finding the coefficient of $A^1B^2C^2D^5$ in the expansion of $(A+B+C+D+E+F)^{10}$, it will be $\binom{10}{1,2,2,5}=\frac{10!}{1!2!2!5!}$.
I.e., the red line corresponds to rolling one 6, two 5s, two 2s, and five 1s.
Find the contribution of each case separately and add them together.
Of course, this is a very tedious process to do by hand, and so it is much easier to simply ask a calculator to expand $(1+x+x^2+x^3+x^4+x^5)^{10}$ and find the coefficient that way.
Wolfram shows that it is $831204$
If you are curious in the number of ways you can do so where the dice are indistinct, just count how many ways are in the list above.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1315009",
"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\lim\limits_{x\to\ 0}(\frac{1}{\sin(x)\arctan(x)}-\frac{1}{\tan(x)\arcsin(x)})$
Evaluate $$\lim\limits_{x\to\ 0}\left(\frac{1}{\sin(x)\arctan(x)}-\frac{1}{\tan(x)\arcsin(x)}\right)$$
It is easy with L'Hospital's rule, but takes too much time to calculate derivatives. What are some shorter methods?
|
Here is a solution to the EDITED problem asked by OP. I will make use of
some standard limits only, without l'Hospital rule nor Taylor series.
First, some simple transformations are required:
\begin{eqnarray*}
\frac{1}{\sin x\arctan x}-\frac{1}{\tan x\arcsin x} &=&\frac{1}{\sin
x\arctan x}-\frac{\cos x}{\sin x\arcsin x} \\
&=&\frac{1}{\sin x}\left( \frac{1}{\arctan x}-\frac{\cos x}{\arcsin x}%
\right) \\
&=&\frac{1}{\sin x}\left( \frac{\arcsin x-\cos x\arctan x}{\arctan x\arcsin x%
}\right) .
\end{eqnarray*}
Now, we re-write this expression using the expressions involved in standard
limits as follows:
\begin{eqnarray*}
&=&\frac{1}{\sin x}\left( \frac{\arcsin x-x+x-\arctan x+\arctan x-\cos
x\arctan x}{\arctan x\arcsin x}\right) \\
&=&\frac{1}{\sin x}\left( \frac{\left( \arcsin x-x\right) +\left( x-\arctan
x\right) +\arctan x\left( 1-\cos x\right) }{\arctan x\arcsin x}\right) \\
&=&\frac{x}{\sin x}\frac{x}{\arctan x}\frac{x}{\arcsin x}\left( \frac{\left(
\arcsin x-x\right) +\left( x-\arctan x\right) +\arctan x\left( 1-\cos
x\right) }{x^{3}}\right) \\
&=&\frac{x}{\sin x}\frac{x}{\arctan x}\frac{x}{\arcsin x}\left( \frac{%
\arcsin x-x}{x^{3}}+\frac{x-\arctan x}{x^{3}}+\frac{\arctan x}{x}\left(
\frac{1-\cos x}{x^{2}}\right) \right)
\end{eqnarray*}
Standard limits used are
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{\sin x}{x} &=&1,\ \ \ \ \lim_{x\rightarrow 0}%
\frac{\arcsin x}{x}=1,\ \ \ \ \lim_{x\rightarrow 0}\frac{\arctan x}{x}=1 \\
\lim_{x\rightarrow 0}\frac{1-\cos x}{x^{2}} &=&\frac{1}{2},\ \ \ \
\lim_{x\rightarrow 0}\frac{\arcsin x-x}{x^{3}}=\frac{1}{6},\ \ \ \ \
\lim_{x\rightarrow 0}\frac{x-\arctan x}{x^{3}}=\frac{1}{3}
\end{eqnarray*}
Therefore,
\begin{equation*}
\lim_{x\rightarrow 0}\left( \frac{1}{\sin x\arctan x}-\frac{1}{\tan x\arcsin
x}\right) =1\cdot 1\cdot 1\left( \frac{1}{6}+\frac{1}{3}+1\cdot \left( \frac{%
1}{2}\right) \right) =1.
\end{equation*}
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
with this inequality $\ln{x}\ln{(1-x)}<\sqrt{x(1-x)}$
If $0<x<1$, show that
$$\ln{x}\ln{(1-x)}<\sqrt{x(1-x)}$$
use derivative it's not easy, such
$$
f(x)=(\ln{x}\ln{(1-x)})^2-x(1-x),
$$
$$
f'(x)=2x-1+\dfrac{2\ln{x}\ln^2{(1-x)}}{x}+\dfrac{2\ln^2{x}\ln{(1-x)}}{x-1}
$$
and we $f(x)=f(1-x)$,then we prove inequality hold in $x\in(0,1/2]$.
can you someone have brief solution?
|
Let
\begin{equation*}
f(x)=\ln x\ln (1-x)-\sqrt{x(1-x)},\ for\ 0<x\leq \frac{1}{2}.
\end{equation*}
\begin{eqnarray*}
f^{\prime }(x) &=&\frac{\ln (1-x)}{x}-\frac{\ln x}{1-x}+\frac{x-\frac{1}{2}}{%
\sqrt{x(1-x)}} \\
&=&\frac{(1-x)\ln (1-x)}{x(1-x)}-\frac{x\ln x}{x(1-x)}+\frac{(x-\frac{1}{2})%
\sqrt{x(1-x)}}{x(1-x)} \\
f^{\prime }(x) &=&\frac{g(x)}{x(1-x)},\ where
\end{eqnarray*}
\begin{eqnarray*}
g(x) &=&(1-x)\ln (1-x)-x\ln (x)+(x-\frac{1}{2})\sqrt{x(1-x)} \\
g^{\prime }(x) &=&\sqrt{x\left( 1-x\right) }-\ln \left( 1-x\right) -\ln x+%
\frac{\left( -\frac{1}{4}\right) \left( 2x-1\right) ^{2}}{\sqrt{x\left(
1-x\right) }}-2 \\
g^{\prime \prime }(x) &=&\frac{1}{1-x}-\frac{1}{x}+3\frac{\left( \frac{1}{2}%
-x\right) }{\sqrt{x\left( 1-x\right) }}+\frac{\left( -\frac{1}{8}\right)
\left( 2x-1\right) ^{3}}{x\left( 1-x\right) \sqrt{x(1-x)}} \\
&=&h(x)\frac{\left( 2x-1\right) }{x\left( x-1\right) \sqrt{x(1-x)}},\ where\
\end{eqnarray*}
\begin{eqnarray*}
h(x) &=&x-x^{2}-\sqrt{x-x^{2}}+\frac{1}{8} \\
h^{\prime }(x) &=&k(x)\frac{\left( \frac{1}{2}-x\right) }{\sqrt{x(1-x)}},\ \
where
\end{eqnarray*}
\begin{eqnarray*}
k(x) &=&\left( 2\sqrt{x-x^{2}}-1\right) \\
k^{\prime }(x) &=&\frac{2(\frac{1}{2}-x)}{\sqrt{x\left( 1-x\right) }}>0,\
for\ 0<x<\frac{1}{2}.
\end{eqnarray*}
Can you take it from here?
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $f(x) = \frac{\sin^{-1} x}{\sqrt{1- x ^2}}$, then evaluate $(1-x^2)f''(x) - xf(x)$
$f(x) = \dfrac{\sin^{-1} x}{\sqrt{1- x ^2}}$
Differentiating the given function, we get
$f'(x) = \dfrac{1 + \dfrac{x\sin^{-1}x}{\sqrt{1-x^2}}}{1-x^2}$
which can also be written as
$f'(x) = \dfrac{1 + xf(x)}{1-x^2}$
Differentiating this one more time
$f''(x) = \dfrac{(1-x^2)[xf'(x) + f(x)] - [1 + xf(x)](-2x)}{(1-x^2)^2}$
Now the next obvious step is to substitute for $f'(x)$ since we have no $f'(x)$ term in the question.
Which further simplify the equation
$f''(x) = \dfrac{3x[1 + xf(x)] + (1-x^2)f(x)}{(1-x^2)^2}$
Head scratch, I have no idea how to proceed. Possibilities are I have made a mistake above, if so please point out and help me take this ahead.
|
we will differentiate using the implicit definition of $y.$ we have $$y = \frac{\sin^{-1} x}{\sqrt {1- x^2}} \to y^2(1-x^2) = (\sin^{-1} x)^2 \tag 1$$ differentiating once, we have $$ -2xy^2 +2yy'(1-x^2) =\frac 2{\sqrt{1-x^2}}\sin^{-1} x = 2y$$ diving out by $y, $ you get $$1+xy =y'(1-x^2) \tag 2 $$ differentiating $(2),$ we have $$xy' + y = -2xy' +y''(1-x^2) \to (1-x^2)y''-x'y =y+2xy'=y+\frac{2x(1+xy)}{1-x^2}\\=\frac{y(1+x^2)+2x}{1-x^2} $$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1317246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Find this closed form $\sum_{k=1}^{n}\left(\lfloor a_{k}\rfloor +\lfloor a_{k}+\frac{1}{2}\rfloor \right)$ Let $\dfrac{1}{a_{k}}=\dfrac{1}{k^2}+\dfrac{1}{k^2+1}+\cdots+\dfrac{1}{(k+1)^2-1}$
I need some ideas to exploit for finding the closed form of
$$\sum_{k=1}^{n}\left(\lfloor a_{k}\rfloor +\lfloor a_{k}+\dfrac{1}{2}\rfloor \right)$$
we get
$$\dfrac{2k+1}{k^2+2k}<\dfrac{1}{a_{k}}<\dfrac{2k+1}{k^2}$$
Therefore
$$\dfrac{k^2}{2k+1}<a_{k}<\dfrac{k^2+2k}{2k+1}$$
|
Let
$$b_{n}=\dfrac{1}{n^2}+\dfrac{1}{n^2+1}+\cdots+\dfrac{1}{n^2+n-1}+\dfrac{1}{n^2+n}+\dfrac{1}{n^2+n+1}+\cdots+\dfrac{1}{n^2+2n}$$
so
$$\dfrac{2}{n+1}<\dfrac{1}{n}+\dfrac{1}{n+2}=\dfrac{n+1}{n^2+n}+\dfrac{n}{n^2+2n}<b_{n}<\dfrac{n}{n^2}+\dfrac{n+1}{n^2+n}=\dfrac{2}{n}$$
so we have
$$\dfrac{2}{k+1}<\dfrac{1}{a_{k}}<\dfrac{2}{k}$$
so
$$k<2a_{k}<k+1$$
so
$$\lfloor 2a_{k}\rfloor =k$$
so use indenity
$$\lfloor a_{k}\rfloor +\lfloor a_{k}+\dfrac{1}{2}\rfloor=\lfloor 2a_{k}\rfloor=k$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1318072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
How do you solve the summation of $2-4+8-16+32- \dots 2^{48}$? This is a summation problem but I can't seem to figure out how to solve this with the mix of subtraction and addition.
|
$$S=2-4+8-16+32- \dots -2^{48} = (-1)(-2+4-8+16-32+ \dots +2^{48}) $$
$$ = (-1)\left((-2)^1+(-2)^2+(-2)^3+ \dots +(-2)^{48}\right)$$
which is a geometric series with first term $a=-2$ and common ratio $r=-2$ and $n=48$ terms. $S$ then becomes
$$S = (-1)\times \frac{a(1-r^{n+1})}{1-r} = -1\times \frac{-2(1-(-2)^{49})}{1-(-2) }= \frac{2(2^{49}+1)}{3}$$
Another way to look at this is to split the overall sum up into a positive part and a negative part:
$$S=2-4+8-16+...-2^{48} = (2+8+32+\dots +2^{47})-(4+16+64+\dots + 2^{48})$$ which are both geometric series with all positive terms
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1319964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
What is wrong with this integration of $ \int_0^{2\pi}\sin x /(1 + A \sin x)$ A worked solution of the integral has been provided as an answer to a previous question.
But I am still unclear why I get the wrong answer from the following method which uses a formula for the definite integral provided by Wolfram Alpha.
Wolfram Alpha indicates the following solution formula:-
$$
\int_0^{2\pi} \frac{\sin x}{1 + A \sin x} dx
=
(1/A)\left( x - \frac{2 \tan^{-1} \left( \frac{A + \tan{(x/2)}}{\sqrt{(1-A^2)}}\right)}{\sqrt{(1-A^2)}} + constant\right)^{2\pi}_0
$$
My thinking is that substituting for $x$ by $2 \pi$ and by $0$ leads to the corresponding values of $\tan(x/2)$ being the same, namely $\tan(\pi) = \tan(0) = 0$. It follows therefore that the fractional term:-
$$
\frac{2 \tan^{-1} \left( \frac{A + \tan{(x/2)}}{\sqrt{(1-A^2)}}\right)}{\sqrt{(1-A^2)}}
$$
has the same value ( let us call it $Q$ ) for $x=2\pi$ and $x=0.$ Therefore the definite integral becomes
$$
\int_0^{2\pi} \frac{\sin x}{1 + A \sin x} dx
=
(1/A)\left( (2\pi - Q + constant) - (0 - Q + constant) \right)
= 2\pi/A.
$$
For $A = 0.2$ we derive the result
$$
\int_0^{2\pi} \frac{\sin x}{1 + 0.2 \sin x} dx
= 10 \pi.
$$
However Wolfram Alpha gives the answer as $-0.64782$ (approx.) which is reasonable looking at the graph and agrees with the formula provided in @abel's answer to my previous question.
So what is wrong with my reasoning? I realize that the $\tan$ function is not continuous over the range $0,2\pi$ but I don't know what that implies for the analysis of the definite integral. What rules or tricks should I apply in order to use the Wolfram Alpha formula? Must I always beware the appearance of $\tan()$ functions in integrand formulae?
|
Using the guidance provided by Chappers and Hans I have come up with this approach.
The Wolfram Alpha solution is:-
$$
\int_0^{2\pi} \frac{\sin x}{1 + A \sin x} dx
=
(1/A)\left( x - \frac{2 \tan^{-1} \left( \frac{A + \tan{(x/2)}}{\sqrt{(1-A^2)}}\right)}{\sqrt{(1-A^2)}} + C\right)^{2\pi}_0
$$
The antiderivative (on the RHS of the equation above) cannot be evaluated by plugging in values for the limiting values of $x=0$ and $x=2\pi$ because the $\tan^{-1}$ term in the anti-derivative function is not continuous over the range $0..2\pi$
See this plot.
However the $\tan^{-1}$ term is cyclic with a period of $\pi$. Therefore the definite integral over $0..2\pi$ is the same as the definite integral over any other range of length $2\pi$ or two times the definite integral over any range of length $\pi$. We proceed by selecting one particular range: $-\pi/2..\pi/2$ over which the $\tan^{-1}$ term is continuous. We will need to multiply the expression by two to compensate for the difference in ranges.
Proceeding thus...
$$
\int_0^{2\pi} \frac{\sin x}{1 + A \sin x} dx
=
Q =
(2*1/A)\left( x - \frac{2 \tan^{-1} \left( \frac{A + \tan{(x/2)}}{\sqrt{(1-A^2)}}\right)}{\sqrt{(1-A^2)}} + C\right)_{-\pi/2}^{+\pi/2}
$$
$$Q =
(2/A) \left[
\left( \pi/2- \frac{2 \tan^{-1} \left( \frac{A + \tan{(+\pi/4)}}{\sqrt{(1-A^2)}}\right)}{\sqrt{(1-A^2)}} + C\right)
-
\left( (-\pi/2) - \frac{2 \tan^{-1} \left( \frac{A + \tan{(-\pi/4)}}{\sqrt{(1-A^2)}}\right)}{\sqrt{(1-A^2)}} + C\right)
\right]
$$
$$Q =
(2/A) \left[
\pi - \frac{ 2\tan^{-1} \left( \frac{A + 1}{\sqrt{(1-A^2)}}\right)}{\sqrt{(1-A^2)}}
+
\frac{ 2\tan^{-1} \left( \frac{A -1}{\sqrt{(1-A^2)}}\right)}{\sqrt{(1-A^2)}}
\right]
$$
$$Q =
\frac{2}{A} \left[
\pi
+
\frac{ 2\tan^{-1} \left( \frac{A - 1}{\sqrt{(1-A^2)}}\right)
-2\tan^{-1} \left( \frac{A + 1}{\sqrt{(1-A^2)}}\right)
}
{\sqrt{(1-A^2)}}
\right]
$$
It happens that (for $0<A<1$) (tested in Excel, see Note at bottom of answer)
$$
\tan^{-1} \left( \frac{A - 1}{\sqrt{(1-A^2)}}\right)
-\tan^{-1} \left( \frac{A + 1}{\sqrt{(1-A^2)}}\right)
=\tan^{-1} (-\infty)
= - \pi/2
$$
and so we get
$$
\int_0^{2\pi} \frac{\sin x}{1 + A \sin x} dx
= Q
=
\frac{2\pi}{A} \left[
1- \frac{1}{\sqrt{(1-A^2)}}
\right]
$$
which is the same as the answer derived by @abel in answer to the previous question and for which, with $A=0.2$ we get the result $Q=-0.64782$ (approx.).
============================================================
Note
No proof is provided here for the assertion that ( for $0<A<1$ )
$$
\tan^{-1} \left( \frac{A - 1}{\sqrt{(1-A^2)}}\right)
-\tan^{-1} \left( \frac{A + 1}{\sqrt{(1-A^2)}}\right)
=\tan^{-1} (-\infty)
= - \pi/2
$$
However it can be checked (a) by testing in a spreadsheet such as Excel or (b) using the inverse-tangent addition theorem:-
$$
\tan^{-1} x \pm \tan^{-1} y = \tan^{-1} \left( \frac{x \pm y}{1 \mp xy} \right).
$$
|
{
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"url": "https://math.stackexchange.com/questions/1321158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Prove (or provide a counterexample): no pair of primitive Pythagorean triples (a,b,c) and (2a,k,c) exists. A primitive Pythagorean triple is an ordered set of coprime integers (a,b,c) such that $a^2+b^2=c^2$. Show that the system of Diophantine equations
$$a^2+b^2=c^2$$
$$4a^2+k^2=c^2$$
have no solutions.
|
$$c^2=a^2+b^2=4a^2+k^2$$
$$k^2+3a^2=b^2$$
All solutions of this Diophantine equation are defined parameterization.
$$a=2ps$$
$$k=p^2-3s^2$$
$$b=p^2+3s^2$$
Then the amount of;
$$a^2+b^2=4p^2s^2+p^4+6p^2s^2+9s^4=p^4+10p^2s^2+9s^4$$
A square can not be.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1321478",
"timestamp": "2023-03-29T00:00:00",
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|
Integral using contour integration Here is the integral I want to evaluate:
$$\int_{0}^{2\pi} \frac{dx}{a+b \cos x }, \quad a>b >0$$
Apparently there are limitations as to what values $a, b$ are supposed to take but let us not concern about this. Since using the sub $u =\tan \frac{x}{2}$ (the Weiersstrass sub) results that the integral is $0$ (as it should, since it is not $1-1$ function in this interval) I got down down the way of contour integration. Hence:
$$\begin{aligned}
\int_{0}^{2\pi}\frac{dx}{a+ b\cos x} &\overset{x=i \ln u}{=\! =\! =\!} \oint \limits_{|z|=1} \frac{dz}{iz \left [ a + \frac{b}{2}\left ( z+z^{-1} \right ) \right ]} \\
&= \frac{1}{i} \oint \limits_{|z|=1} \frac{dz}{za + \frac{bz^2}{2}+\frac{b}{2}}\\
&=\frac{2}{i} \oint \limits_{|z|=1} \frac{dz}{bz^2 +b +2za} \\
&= \frac{2}{i} 2\pi i \sum_\text{residues} f(z)\\
&= \frac{4\pi}{1+ \sqrt{1-8ab}+2a}
\end{aligned}$$
However, judging by intuition this must not be the result. Because this one restricts the integral too much. What i mean is, that for $a=6, b=3$ we have that:
$$\int_0^{2\pi} \frac{dx}{6+3\cos x}= \frac{2\pi}{3\sqrt{3}}$$
My formula cannot derive the result because then radical would be negative. What am I doing wrong here?
|
$$\frac 1i \frac{2}{2 a z+b z^2+b}$$
This has a residue at
$$\frac{\sqrt{a^2-b^2}-a}{b}$$
Which is evaluated as
$$\frac{1}{i\sqrt{a^2-b^2}}$$
And the final answer is
$$\frac{2 \pi}{\sqrt{a^2-b^2}}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
inequalities with fraction problem x $\frac{1}{x} > \frac{2x} {x^2 +2}$ solving this inequalities:
My long solution (wrong) :
multiplying $(x^2 + 2)^2 (x)^2 \dots$ (multiplying square of each denominator, getting rid of the > or < 0)
$x (x^2 +2)^2 > 2x(x^2) ( x^2 +2)$
$x(x^4+4x^2 +4)>2x^2(x^2+2)$
$x^5+4x^3+4x>2x^4+4x^2$
$x^5+4x^3+4x-2x^4-4x^2>0$
$x ( x^4 + 4x^2 + 4 +2x^3 -4x) >0 $
$x>0$ or ^ ...
This solving method doesn't look right
|
As both sides of the inequation have the same sign, it is equivalent to:
$$\frac{x^2+2}{x}=x + \frac2x> 2x\iff x<\frac2 x\iff\begin{cases}x^2<2&\text{if}\enspace x>0,\\x^2>2&\text{if}\enspace x<0\end{cases}$$
Thus the solutions are $\,(0< x<\sqrt 2)\,$ or $\,(x<-\sqrt 2)$.
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Asymptotic expansion computation I found a paper in which the following expression
$$\log\left(1 - \frac{1}{x}\right) + \frac{1}{x} + \frac{A}{x^2}+\epsilon x = 0$$
Where $\epsilon$ is a constant of order $10^{-2}$ to $10^{-5}$, $A$ is a constant of order unity, is approximated for $\frac{1}{x}<<1$ as
$$\epsilon x^4 + \left(A-\frac{1}{2}\right) x - \frac{1}{3} + O\left(\frac{1}{x}\right) = 0$$
I am unable to derive this result.
Going back to the first equation, I denote $\frac{1}{x} = y$.
Then by using the Taylor expansion
$$\log(1-y) = y -\frac{1}{2}y^2 +\frac{2}{3}y^3 +...$$
And substitute I obtain
$$2y +\left(a-\frac{1}{2}\right)y^2 + \frac{2}{3}Y^3 + \frac{\epsilon}{y} +O(x^4)= 0$$
And it is not even close. Any hint on my error would be very appreciated, thanks
|
$$\begin{align}0 &=\ln\left(1 - \frac{1}{x}\right) + \frac{1}{x} + \frac{A}{x^2}+\epsilon x \\
&= -\frac1x - \frac1{2x^2}-\frac1{3x^3} +\cdots+\frac1x+\frac a{x^2} + \epsilon x \\
&=\epsilon x^4+(a-1/2)x -1/3+\cdots \end{align} $$
|
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|
Evaluate $\lim_{x\to 0}\frac{\sqrt[m]{\cos\alpha x}-\sqrt[m]{\cos\beta x}}{\sin^2x},m\in \mathbb{N}$
Evaluate $$\lim_{x\to 0}\frac{\sqrt[m]{\cos\alpha x}-\sqrt[m]{\cos\beta x}}{\sin^2x},m\in \mathbb{N}$$
I used L'Hospital's rule, but that didn't work. Could Taylor series be used? I don't know how to use it with irrational functions.
|
This can be done via the use of following standard limits $$\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1},\,\lim_{x \to 0}\frac{1 - \cos x}{x^{2}} = \frac{1}{2},\,\lim_{x \to 0}\frac{\sin x}{x} = 1$$ Let us proceed as follows
\begin{align}
L &= \lim_{x \to 0}\frac{\sqrt[m]{\cos \alpha x} - \sqrt[m]{\cos \beta x}}{\sin^{2}x}\notag\\
&= \lim_{x \to 0}\frac{\sqrt[m]{\cos \alpha x} - \sqrt[m]{\cos \beta x}}{x^{2}}\cdot\frac{x^{2}}{\sin^{2}x}\notag\\
&= \lim_{x \to 0}\frac{\sqrt[m]{\cos \alpha x} - \sqrt[m]{\cos \beta x}}{x^{2}}\cdot 1\notag\\
&= \lim_{x \to 0}\frac{(1 - \sqrt[m]{\cos \beta x}) - (1 - \sqrt[m]{\cos \alpha x})}{x^{2}}\notag\\
&= \lim_{x \to 0}\frac{(1 - \sqrt[m]{\cos \beta x})}{x^{2}} - \frac{(1 - \sqrt[m]{\cos \alpha x})}{x^{2}}\notag\\
&= \lim_{x \to 0}\frac{1 - \sqrt[m]{\cos \beta x}}{1 - \cos \beta x}\cdot\frac{1 - \cos \beta x}{x^{2}} - \frac{1 - \sqrt[m]{\cos \alpha x}}{1 - \cos \alpha x}\cdot\frac{1 - \cos \alpha x}{x^{2}}\notag\\
&= \lim_{t \to 1}\frac{1 - t^{1/m}}{1 - t}\cdot\lim_{x \to 0}\frac{1 - \cos \beta x}{x^{2}} - \lim_{t \to 1}\frac{1 - t^{1/m}}{1 - t}\cdot\lim_{x \to 0}\frac{1 - \cos \alpha x}{x^{2}}\notag\\
&= \frac{1}{m}\left(\lim_{x \to 0}\frac{1 - \cos \beta x}{x^{2}} - \lim_{x \to 0}\frac{1 - \cos \alpha x}{x^{2}}\right)\notag\\
&= \frac{1}{m}\left(\lim_{x \to 0}\beta^{2}\cdot\frac{1 - \cos (\beta x)}{(\beta x)^{2}} - \lim_{x \to 0}\alpha^{2}\cdot\frac{1 - \cos (\alpha x)}{(\alpha x)^{2}}\right)\notag\\
&= \frac{1}{m}\left(\frac{\beta^{2}}{2} - \frac{\alpha^{2}}{2}\right)\notag\\
&= \frac{\beta^{2} - \alpha^{2}}{2m}\notag
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solution of a quartic equation. Suppose that the equation $x^4-2x^3+4x^2+6x-21=0$ is known to have two roots that are equal in magnitude but opposite in sign. Solve the equation.
This is what I have been thinking. Suppose $\zeta_1$ $\zeta_2$ are roots. Such that $|\zeta_1|=|\zeta_2|$. Then $(x-\zeta_1)(x-\zeta_2)$ divides the polynomial. I don't know where to go from here. Any advice?
|
Let $P(x)= x^4−2x^3+4x^2+6x−21=0$ Let $a,-a $ are the two roots given. So, $P(a)=P(-a)=0$
\begin{align}
a^4−2a^3+4a^2+6a−21 &= &a^4+2a^3+4a^2-6a−21\\
\implies 4a^3-12a =0\\
\implies a(a^2-3)=0\\
a=0,a=\pm\sqrt3
\end{align}
But $a=0$ is not a root of the polynomial. Hence $\pm \sqrt3$ are the two roots of $P(x)$ because at two points the functional values are zero and the functional values are same only at three points. Now by the long division or Synthetic Division we can factor $P(x)$. $P(x)=(x^2-3)(x^2-2x+7)$. Now we have to find the roots of the quadratic equation $(x^2-2x+7).$ The roots of $x^2-2x+7$ is $1\pm \sqrt6 i.$ So, the roots of the polynomial $P(x)$ are $\pm \sqrt3 ,1\pm i \sqrt6.$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Solve $x^8 \equiv 3 \pmod {13}$ I need to find all solutions to $x^8 \equiv 3 \pmod {13}$.
What I've tried: I know $2$ is a primitive root modulo $13$.
So it is equivalent to solve
$2^{8t} \equiv 2^4 \pmod {13}$
Then I get $t = 2 + 3k$.
I think I'm wrong... and if not, what are the final solutions??
|
In $\mathbb{F}_{13}$,
$$ x^8-3 = x^8-16 = (x^4-4)(x^4+4)=(x^2-2)(x^2+2)(x^2+2x+2)(x^2-2x+2) $$
hence we just have to check which of the numbers $-2,2,-1$ are quadratic residues. Since $13$ is a prime of the form $8k+5$, the only quadratic residue among the previous ones is $-1$, so $x^8-3$ splits as:
$$ x^8-3 = (x-4)(x+4)(x-6)(x+6)(x^2-2)(x^2+2) $$
and the solutions of $x^8\equiv 3\pmod{13}$ are given by $x\in\{\pm 4,\pm 6\}\pmod{13}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1332767",
"timestamp": "2023-03-29T00:00:00",
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|
Could translate/explain this for me? I have this problem:
$$
10x^2 - 7x - 12 = 0
$$
And apparently the method to factoring it is to find two numbers whose product is the same as the product of the coefficient of $x^2$ and the constant term, and whose sum is the same as the coefficient of $x$.
However, I can barely understand what that even means, and as far as I'm concerned it seems impossible with the expression given.
Could someone possibly just walk me through this method?
|
to find two numbers whose product is the same as the product of the coefficient of $x^2$ and the constant term, and whose sum is the same as the coefficient of $x$.
Take the smallest, innermost noun phrases first:
to find two numbers whose product is the same as the product of the coefficient of $x^2$ and the constant term, and whose sum is the same as the coefficient of $x$.
Identify those pieces in the given polynomial:
$$ \underbrace{10}_{\text{coefficient of $x^2$}}x^2 + \overbrace{(-7)}^{\text{coefficient of $x$}}x + \underbrace{(-12)}_{\text{constant term}} $$
Plug those in to the question, to simplify it:
to find two numbers whose product is the same as the product of $10$ and $-12$, and whose sum is the same as $-7$.
"The product of $10$ and $-12$" is $-120$. Let's plug that in.
to find two numbers whose product is the same as $-120$, and whose sum is the same as $-7$.
Now the words "the same as" seem pointless, so let's get rid of them:
to find two numbers whose product is $-120$, and whose sum is $-7$.
An obvious way to proceed here is just to list all the pairs of numbers whose product is $-120$, and to see which of them have sum $-7$. (We can list them all the ones where both numbers are integers, at least... hopefully that'll be enough.) Now, $-120$ has lots of factors, so there's lots of possibilities to consider. Any way of splitting up the factors on the RHS of
$$ -120 = -1\cdot 2\cdot 2 \cdot 2 \cdot 3\cdot 5 $$
into two groups will make a factorization of $-120$ which might work. For example, we might try
$$ -1\cdot 2\cdot 3 \qquad\text{and}\qquad 2\cdot 2\cdot 5 $$
Those are $-6$ and $20$; their product is $-120$, but their sum is $14$, not $-7$ as we wished.
By playing around for a while, or by systematically checking every possibility, you'll eventually find that
$$ 2\cdot 2\cdot 2 \qquad\text{and}\qquad -1\cdot 3\cdot 5 $$
that is, $8$ and $-15$, have product $-120$ and sum $-7$.
The rest of the method, which you didn't mention, is to break up the $x$ term using these two numbers,
\begin{align*}
10x^2 - 7x - 12
&= 10x^2 - 15x + 8x - 12
\end{align*}
then to take out common factors from the left two terms and from the right two terms,
\begin{align*}
10x^2 - 7x - 12
&= 10x^2 - 15x + 8x - 12 \\
&= 5x(2x - 3) + 4(2x - 3)
\end{align*}
and then, if we did everything right, we'll find, as we did here, that there's now a common factor between the left and the right, which we take out:
\begin{align*}
10x^2 - 7x - 12
&= 10x^2 - 15x + 8x - 12 \\
&= 5x(2x - 3) + 4(2x - 3) \\
&= (5x+4)(2x-3)
\end{align*}
Factorization complete.
|
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|
How to derive FWHM of sinc function So this is probably a simple question, but I am unable to get my head around it. If we have $\operatorname{sinc}(2 \pi v L)$, what is the width of that $\operatorname{sinc}$ in terms of $v$ at half the maximum point.
Normally I'd just say that the maximum is $1$, thus half point is.... $1/2$
$$
\operatorname{sinc}(2\pi v L) = \frac{1}{2}
\\
\frac{ \sin(2 \pi vL) } { 2 \pi v L} = \frac{1}{2}
\\
\sin(2\pi vL) = \pi v L
$$
EDIT: How can you analytically find this?
Thanks.
|
I cannot resist the pleasure of providing an approximate solution for $Lv \leq 1$. This approximation was made by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician. $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ So $$\frac{\sin(x)}x\simeq \frac{16 (\pi -x) }{5 \pi ^2-4 (\pi -x) x}$$ and for a value equal to $\frac 12$, the solution is given by $$x=\frac{\pi }{2}+\sqrt{16-(\pi -4) \pi }-4\approx 1.89477$$ while the exact solution would be $\approx 1.89549426$.
Equations which mix polynomial and trigonometric functions do not show explicit solutions and numerical methods (or sophisticated approximations) should be used.
The solution being close to $\frac{3\pi}5$, we can build the simplest Pade approximant of the left hand side and using exact values for the trigonometric function of the angle get $$x=\frac{250 \left(5+\sqrt{5}\right)+3 \pi \left(100 \sqrt{10-2 \sqrt{5}}-3 \pi
\left(-55+5 \sqrt{5}+3 \sqrt{2 \left(5+\sqrt{5}\right)} \pi \right)\right)}{500
\sqrt{5+2 \sqrt{5}}+75 \left(9+\sqrt{5}\right) \pi -45 \sqrt{2
\left(5+\sqrt{5}\right)} \pi ^2}$$ which is $\approx 1.89549416$
Edit
In a more general manner, if we expand $\frac{\sin(x)}x$ as its simplest approximant at $x=\theta$, the solution of $\frac{\sin(x)}x=\frac 12$ is given by $$x=\frac{-3 \theta ^2-\left(\theta ^2-2\right) \cos (2 \theta )+\theta \sin (\theta )
\left(\theta ^2+8 \cos (\theta )\right)-2}{\left(\theta ^2-2\right) \sin (\theta
)-3 \theta +2 \sin (2 \theta )+2 \theta \cos (\theta )-\theta \cos (2 \theta )}$$ $\theta=\frac{29 \pi}{48}$ would have been a very good choice since the values of its trigonometric functions are known (with many radicals).
Using it, the result would have been $\approx 1.89549427$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Simplify a Combinatorial Sum $\sum_{k=0}^\infty {a\choose k}{b\choose c-k}{d-k\choose e}$ Is there a way to simplify
$$\sum_{k=0}^\infty {a\choose k}{b\choose c-k}{d-k\choose e}$$
where $a,b,c,d,e$ are natural numbers?
In particular, I would like to see the case for $a=45, b=3,c=4,d=48,e=4$.
|
The series can be cast into hypergeometric form and is given by
\begin{align}
\sum_{k=0}^\infty \binom{a}{k} \binom{b}{c-k} \binom{d-k}{e} = \binom{b}{c} \binom{d}{e} \, {}_{3}F_{2}(-a, -c, e-d; -d, b-c+1; 1)
\end{align}
Further reduction may be possible depending on the values of $\{a,b,c,d,e\}$.
Proof of result
\begin{align}
S &= \sum_{k=0}^\infty \binom{a}{k} \binom{b}{c-k} \binom{d-k}{e} \\
&= \sum_{k=0}^{\infty} \frac{b! \, \Gamma(a+1) \Gamma(d+1-k)}{k! \, \Gamma(a+1-k) \Gamma(c+1-k) \Gamma(b-c+1+k) \, e! \, \Gamma(d-e+1-k)} \\
&= \frac{b! \, d!}{e! \, c! \, (b-c)!} \, \sum_{k=0}^{\infty} \frac{(-a)_{k} (-c)_{k} (e-d)_{k}}{k! \, (-d)_{k} (b-c+1)_{k}} \\
&= \binom{b}{c} \binom{d}{e} \, {}_{3}F_{2}(-a, -c, e-d; -d, b-c+1; 1).
\end{align}
It is to be noted that $b-c+1 \neq 0,-1,-2, \cdots$ and $d \leq 0$.
As given by the example values $a=45, b=3,c=4,d=48,e=4$ the series is best handled in a straight forward manor.
\begin{align}
\sum_{k=0}^\infty \binom{45}{k} \binom{3}{4-k} \binom{48-k}{4} &= \sum_{k=1}^{4} \binom{45}{k} \binom{3}{4-k} \binom{48-k}{4} \\
&= 45 \, \binom{47}{4} + 3 \cdot 45 \cdot 22 \, \binom{46}{4} + 3 \, \binom{45}{3} \binom{45}{4} + \binom{44}{4} \binom{45}{4} \\
&= 27,061, 623, 270.
\end{align}
|
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|
Find the volume formed by rotating the region bounded by $y = e^{-x} \sin x$, $x\ge 0$ about $y =0$.
Find the volume formed by rotating the region bounded by $y = e^{-x} \sin x$, $x\ge 0$ about $y =0$.
I tried to graph this using Wolfram Alpha, but it didn't help. I don't know how to start or graph this.
|
$$f(x) = e^{-x} \sin x$$
Your volume can be found using cylindrical coordinates :
\begin{align}
\mathcal{V}\equiv\int\limits_0^\infty \int\limits_0^{2\pi}\int\limits_0^{f(x)} r dr d\theta dx &= 2\pi \int\limits_0^\infty\int\limits_0^{f(x)} r dr dx\\
&=\pi \int\limits_0^\infty f^2(x) dx =\pi \int\limits_0^\infty e^{-2x}\sin^2x dx = \pi \int\limits_0^\infty e^{-2x}\dfrac{1-\cos 2x}{2}dx\\
&= \dfrac{\pi}{2} \left\lbrace \int\limits_0^\infty e^{-2x}dx -\dfrac{1}{2} \int\limits_0^\infty e^{-2x}\left[e^{2ix} + e^{-2ix}\right]dx\right\rbrace\\
&= \dfrac{\pi}{2} \left\lbrace \dfrac{1}{2} -\dfrac{1}{2} \left[\dfrac{e^{-2(1-i)x}}{-2(1-i)} + \dfrac{e^{-2(1+i)x}}{-2(1+i)}\right]_0^\infty\right\rbrace\\
&= \dfrac{\pi}{2} \left\lbrace \dfrac{1}{2} +\left[\dfrac{e^{-2x}}{4} \left(\dfrac{e^{2ix}}{1-i} + \dfrac{e^{-2ix}}{1+i}\right)\right]_0^\infty\right\rbrace\\
&= \dfrac{\pi}{2} \left\lbrace \dfrac{1}{2} +\left[\dfrac{e^{-2x}}{4} \left(\dfrac{e^{2ix}}{1-i} + \dfrac{e^{-2ix}}{1+i}\right)\right]_0^\infty\right\rbrace\\
&=\dfrac{\pi}{2} \left\lbrace \dfrac{1}{2} +\left[\dfrac{e^{-2x}}{8} \left((1+i)e^{2ix} + (1-i)e^{-2ix}\right)\right]_0^\infty\right\rbrace\\
&=\dfrac{\pi}{2} \left\lbrace \dfrac{1}{2} +\left[\dfrac{e^{-2x}}{4} \left(\cos 2x - \sin 2x\right)\right]_0^\infty\right\rbrace\\
&= \dfrac{\pi}{2} \left\lbrace \dfrac{1}{2} - \dfrac{1}{4}\right\rbrace\\
&= \dfrac{\pi}{8}
\end{align}
|
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|
Checking if a matrix is in the span of other matrices Problem: Expand the following set matrices \begin{align*} \left\{ \begin{pmatrix} 1 & -1 \\ 2 & 3 \end{pmatrix}, \begin{pmatrix} 2 & 1 \\ -1 & 4 \end{pmatrix}, \begin{pmatrix} 0 & -3 \\ 5 & 4 \end{pmatrix}\right\} \end{align*} to a basis of $\mathbb{R}^{2 \times 2}$.
What is the best way to do this? Should I just pick an arbitrary matrix and check for linear dependence?
I picked the matrix $\begin{pmatrix} 3 & 1 \\ 2 & 1 \end{pmatrix}$. So now we need to know if there are scalars $a,b$ and $c$ such that \begin{align*} \begin{pmatrix} 3 & 1 \\ 2 & 1 \end{pmatrix} = a \begin{pmatrix} 1 & -1 \\ 2 & 3 \end{pmatrix} + b \begin{pmatrix} 2 & 1 \\ -1 & 4 \end{pmatrix} + c \begin{pmatrix} 0 & -3 \\ 5 & 4 \end{pmatrix}. \end{align*} This gives rise to the following system of equations: \begin{align*} \begin{cases} a + 2b &= 3 \\ -a + b -c &= 1 \\ 2a - b + 5c &= 2 \\ 3a + 4b + 4c &= 1 \end{cases} \end{align*} I put this in augmented matrix-form as follows: \begin{align*}\left(\begin{array}{ccc|c} 1 & 2 & 0 & 3 \\ -1 & 1 & -3 & 1 \\ 2 & -1 & 5 & 2 \\ 3 & 4 & 4 & 1 \end{array}\right) \end{align*} I eventually row-reduced it to \begin{align*}\left(\begin{array}{ccc|c} 1 & 2 & 0 & 3 \\ 0 & 1 & -1 & 4/3 \\ 0 & 0 & 0 & 8/3 \\ 0 & 0 & 2 & -16/3 \end{array}\right) \end{align*} Since the third row implies a contradiction, can I conclude on the basis of this that my four matrices are linearly independent?
|
Instead of a blind guess you could check which standard basis element is not in the span of your set. (You should also check that your set is linearly independent if it is not already assumed so).
i.e. check that these systems have solutions
\begin{align*} \begin{cases} a + 2b &= 1 \\ -a + b -c &= 0 \\ 2a - b + 5c &= 0 \\ 3a + 4b + 4c &= 0 \end{cases} \end{align*}
\begin{align*} \begin{cases} a + 2b &= 0 \\ -a + b -c &= 1 \\ 2a - b + 5c &= 0 \\ 3a + 4b + 4c &= 0 \end{cases} \end{align*}
\begin{align*} \begin{cases} a + 2b &= 0 \\ -a + b -c &= 0 \\ 2a - b + 5c &= 1 \\ 3a + 4b + 4c &= 0 \end{cases} \end{align*}
\begin{align*} \begin{cases} a + 2b &= 0 \\ -a + b -c &= 0 \\ 2a - b + 5c &= 0 \\ 3a + 4b + 4c &= 1 \end{cases} \end{align*}
I think only the last one doesn't work. (You should check) So just add in the matrix
\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}
|
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|
Derivative of the function $y = 2^{\sqrt{\tan x}}$ How to find derivative of the following function: $y = 2^{\sqrt{\tan x}}$ , $y' = ?$ I did the following $$\frac{d}{dx}2^{\sqrt{\tan x}} = 2^{\sqrt{\tan x}}\ln{2}(\sqrt{\tan}x)'$$ and stopped here. Can you guide me?
The solution is as follows in the book:
$$2^{\sqrt{\tan x}}\frac{\ln{2}}{2\cos^{2}{x}\sqrt{\tan{x}}}$$
|
If $y = 2^{\sqrt{\tan x}}$ then $\displaystyle y = e^{\sqrt{\tan x} \ln 2}$. So, using the chain rule yields $$y' = \frac{\mathrm{d}}{\mathrm{d}x} \left( \sqrt{\tan x} \cdot \ln 2\right) \cdot e^{\sqrt{\tan x} \ln 2}$$
But, using the chain rule again yields $$\frac{\mathrm{d}}{\mathrm{d}x} \left( \sqrt{\tan x} \cdot \ln 2\right) = \frac{\sec^2 x \ln 2}{2\sqrt{\tan x}}$$
So $$y' = e^{\sqrt{\tan \left(x\right)}\cdot \ln 2}\cdot \frac{\sec ^2\left(x\right)\ln \left(2\right)}{2\sqrt{\tan x}}$$
Changing back to base 2 and simplifying gives $$y'=2^{\sqrt{\tan x}}\frac{\ln{2}}{2\sqrt{\tan{x}}} \cdot \frac{1}{\cos^2 x} \\
=2^{\sqrt{\tan x}}\dfrac{\ln 2}{2\cos^2 x\sqrt{\tan x}}$$
Edit: For added clarification, let's look at $\dfrac{\mathrm{d}}{\mathrm{d}x} \sqrt{\tan x}$. Let $g(x) = \tan x$ and $f(x) = \sqrt{x}$. So $\sqrt{\tan x}$ is the composite function $f(g(x))$. It's derivative is hence $$f'(g(x)) \cdot g'(x) = \frac{1}{2\sqrt{\tan x}} \cdot \sec^2 x$$
|
{
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"url": "https://math.stackexchange.com/questions/1343813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How do i find $\tan(\theta)$ such that : $\frac{16}{\sin^6(\theta)} + \frac{81}{\cos^6(\theta)}=625$?? How do i find $\tan(\theta)$ such that :$$\frac{16}{\sin^6(\theta)} + \frac{81}{\cos^6(\theta)}=625$$?
Note : i used some trigono-form but sorry i didn't succed .
Thank you for any help.
|
I cannot say that this is an elegant solution but oh well,
i will use $s$ for $sin$, $c$ for $cos$ and $t$ for tan
we know that $\frac{s}{c} = t$
so $\frac{16}{c^{6}*t^{6}} + \frac{81}{c^{6}} = 625$
$16 + 81t^{6} = 625c^{6}*t^{6}$
we know that $ 1 + t^{2} = \frac{1}{c^{2}}$
or $16 + 81t^{6} = \frac{625*t^{6}}{(t^{2} +1)^{3}}$
simplifying gives
$81t^{12} + 243t^{10} + 243t^{8} + - 528t^{6} + 48t^{4} + 48t^{2} +16 = 0$
let $y = t^{2}$
$81y^{6} + 243y^{5} + 243y^{4} + - 528y^{3} + 48y^{2} + 48y^{1} +16 = 0$
using the rational root theorem, we find that $\frac{2}{3}$ works
so $\sqrt{\frac{2}{3}} = \boxed{\tan(x) = \frac{\sqrt{6}}{3}}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How do I solve this trigonometric equation?
Solve the equation $$\sin\left(\frac{3\pi}{10}-\frac{x}{2}\right)=\frac{1}{2}\sin\left(\frac{\pi}{10}+\frac{3x}{2}\right).$$
I tried applying some Ratio properties, but they just made the equation nasty. Some hints please. Thanks.
|
HINT:
Method $\#1:$
Let $\dfrac{3\pi}{10}-\dfrac x2=y\iff\dfrac x2=\dfrac{3\pi}{10}-y$
$\implies\dfrac\pi{10}+\dfrac{3x}2=\dfrac\pi{10}+3\left(\dfrac{3\pi}{10}-y\right)=\pi-3y$
So, we have $\sin y=\dfrac12\sin(\pi-3y)\iff2\sin y=\sin3y$
Now use $\sin3y=3\sin y-4\sin^3y$ and $\cos2B=1-2\sin^2B$
Method $\#2:$
As $\sin(180^\circ-B)=\sin B,$
$$\sin(18^\circ+3x/2)=\sin\left[180^\circ-(18^\circ+3x/2)\right]=\sin3(54^\circ-x/2)$$
Write $\sin(54^\circ-x/2)=u$ and use $\sin3y=3\sin y-4\sin^3y$
Method $\#3:$
$$\sin(54^\circ-x/2)=\sin[90^\circ-(36^\circ+x/2)]=\cos(36^\circ+x/2)$$
As $\cos(90^\circ+A)=-\sin A,$
$$\sin(18^\circ+3x/2)=-\cos(90^\circ+18^\circ+3x/2)=-\cos3(36^\circ+x/2)$$
Write $\cos(36^\circ+x/2)=u$ and use $\cos3y=4\cos^3y-3\cos y$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Calculate Infinite Limit I'm trying to calculate the limit and when I get to the last step I plug in infinity for $\frac 8x$ and that divided by -4 I get - infinity for my answer but the book says 0. Where did I go wrong?
$$
\frac {8x^3-x^2}{7+11x-4x^4}
$$
Divide everything by $x^4$
$$
\frac {\frac{8x^3}{x^4}-\frac{x^2}{x^4}}{\frac{7}{x^4}+\frac{11x}{x^4}-\frac{4x^4}{x^4}}
$$
Results
$$
\frac {\frac{8}{x}-\frac{1}{x^2}}{\frac{7}{x^4}+\frac{11x}{x^4}-4} = \infty
$$
|
Hint
When you have expressions which are ratios of polynomials and that $x\to \infty$, the best is to factor the highest powers in numerator and denominator. So, in your case $$A=\frac {8x^3-x^2}{7+11x-4x^4}=\frac{x^3\big(8-\frac 1x\big)}{x^4\big(\frac 7{x^4}+\frac {11}{x^3}-4\big)}$$ So, since $x$ is large, we can "ignore" $\frac 1x$ (very small if compared to $8$), as well as $\frac 7{x^4}+\frac {11}{x^3}$ (very small if compared to $-4$). All of that arrives to $$A\approx \frac {8x^3}{-4x^4}=-\frac 2x$$ from which you can conclude.
If you apply to $$B=\frac{a_0+a_1x+a_2x^2+\cdots+a_nx^n}{b_0+b_1x+b_2x^2+\cdots+b_mx^m}$$ you could easily show that, for an infinite value of $x$, $$B\approx \frac{a_nx^n}{b_mx^m}=\frac{a_n}{b_m}x^{n-m}$$ Now, look what happens if $n>m$, $n=m$, $n<m$.
|
{
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"url": "https://math.stackexchange.com/questions/1352293",
"timestamp": "2023-03-29T00:00:00",
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|
How to prove that $7^{31} > 8^{29}$ How can I prove that $7^{31}$ is bigger than $8^{29}$?
I tried to write exponents as multiplication, $2\cdot 15 + 1$, and $2\cdot 14+1$, then to write this inequality as $7^{2\cdot 15}\cdot 7 > 8^{2\cdot 14}\cdot 8$. I also tried to write the right hand side as $\frac{8^{31}}{8^2}$.
|
First, $7^5=16807$ and $2^{14}=16384$. And we have $\dfrac{7^{31}}{8^{29}}=\left(\dfrac{7^5}{2^{14}}\right)^6\times\dfrac78$.
Now, for $x\ge0$ and integer $n\ge0$, we have $(1+x)^n\ge1+nx$, hence
$$\left(\frac{7^5}{2^{14}}\right)^6\ge1+6\left(\frac{7^5}{2^{14}}-1\right)=1+\frac{6\times423}{2^{14}}=1+\frac{3\times423}{2^{13}}=\frac{9461}{8192}$$
Finally,
$$\frac{9461}{8192}\times\frac78=\frac{66227}{65536}>1$$
Therefore,
$$\frac{7^{31}}{8^{29}}>1$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Solving Riccati equation $v'=av^2+bv+c$ I am trying to solve Riccati equation
$$
v(t)'=av(t)^2+bv(t)+c
$$
where $a$,$b$,$c$ are real-valued constants. Wolfram|Alpha gives the solution (here), but I am not able to reproduce the result.
Thank you in advance.
|
$$v(t)'=av(t)^2+bv(t)+c\Longleftrightarrow$$
$$\frac{dv(t)}{dt}=av(t)^2+bv(t)+c\Longleftrightarrow$$
$$\frac{\frac{dv(t)}{dt}}{av(t)^2+bv(t)+c}=1\Longleftrightarrow$$
$$\int\left(\frac{\frac{dv(t)}{dt}}{av(t)^2+bv(t)+c}\right)dt=\int 1dt\Longleftrightarrow$$
$$\frac{2\tan^{-1}\left(\frac{b+2av(t)}{\sqrt{-b^2+4ac}}\right)}{\sqrt{-b^2+4ac}}=t+k_1\Longleftrightarrow$$
$$\left(t+k_1\right)\left(\sqrt{-b^2+4ac}\right)=2\tan^{-1}\left(\frac{b+2av(t)}{\sqrt{-b^2+4ac}}\right)\Longleftrightarrow$$
$$\frac{\left(t+k_1\right)\left(\sqrt{-b^2+4ac}\right)}{2}=\tan^{-1}\left(\frac{b+2av(t)}{\sqrt{-b^2+4ac}}\right)\Longleftrightarrow$$
$$\tan\left(\frac{\left(t+k_1\right)\left(\sqrt{-b^2+4ac}\right)}{2}\right)=\tan\left(\tan^{-1}\left(\frac{b+2av(t)}{\sqrt{-b^2+4ac}}\right)\right)\Longleftrightarrow$$
$$\tan\left(\frac{\left(t+k_1\right)\left(\sqrt{-b^2+4ac}\right)}{2}\right)=\frac{b+2av(t)}{\sqrt{-b^2+4ac}}\Longleftrightarrow$$
$$\tan\left(\frac{\left(t+k_1\right)\left(\sqrt{-b^2+4ac}\right)}{2}\right)\left(\sqrt{-b^2+4ac}\right)=b+2av(t)\Longleftrightarrow$$
$$\tan\left(\frac{\left(t+k_1\right)\left(\sqrt{-b^2+4ac}\right)}{2}\right)\left(\sqrt{-b^2+4ac}\right)-b=2av(t)\Longleftrightarrow$$
$$\frac{\tan\left(\frac{\left(t+k_1\right)\left(\sqrt{-b^2+4ac}\right)}{2}\right)\left(\sqrt{-b^2+4ac}\right)-b}{2a}=v(t)\Longleftrightarrow$$
$$v(t)=\frac{1}{2a}\left(\sqrt{4ac-b^2}\tan\left(\frac{1}{2}\left(k_1\sqrt{4ac-b^2}+t\sqrt{4ac-b^2}\right)\right)-b\right)$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
A golden trigonometric diophantine equation After answering this question I reflected on the identity $$\cos\frac{\pi}{5}=\phi\cos\frac{\pi}{3}$$ and thought of looking for all the quadruplets of positive integers $(a,b,c,d)$ satisfying $$\cos \frac{a}{b}\pi=\phi\cos\frac{c}{d}\pi,\tag{1}$$ where both fractions are in lowest terms and smaller than $2$. Of course $a/b=c/d=1/2,3/2$, $a/b=3c/d=3/2$ and $c/d=3a/b=3/2$ work, but let us focus on non-trivial solutions, and assume none of the cosines is $0$. At first I tried using properties of $\cos$ and $\phi$, but eventually I turned to infnite products. However, it didn't work either, but I'll go through my reasoning. Hopefully someone will be able to go further, if not come up with something else; and perhaps simpler.
Using the infinite product formula for $\cos x$ we can rewrite $(1)$ as $$\require\cancel \prod_{n=1}^\infty\left(1-\frac{a^2\cancel{\pi^2}}{b^2\cancel{\pi^2}(n-1/2)^2}\right)=\phi\prod_{n=1}^\infty\left(1-\frac{c^2\cancel{\pi^2}}{d^2\cancel{\pi^2}(n-1/2)^2}\right) \\ \prod_{n=1}^\infty\frac{b^2(n-1/2)^2-a^2}{b^2(n-1/2)^2}=\phi\prod_{n=1}^\infty\frac{d^2(n-1/2)^2-c^2}{d^2(n-1/2)^2},$$ and isolating $\phi$, $$\prod_{n=1}^\infty\frac{d^2}{b^2}\frac{b^2(n-1/2)^2-a^2}{d^2(n-1/2)^2-c^2}=\phi $$ $$\prod_{n=1}^\infty\frac{\left(n-\displaystyle\frac{1}{2}\right)^2-\displaystyle\frac{a^2}{b^2}}{\left(n-\displaystyle\frac{1}{2}\right)^2-\displaystyle\frac{c^2}{d^2}}=\phi \\ \frac{\left(\displaystyle\frac{1}{2}-\displaystyle\frac{a}{b}\right)_\infty \left(\displaystyle\frac{1}{2}+\displaystyle\frac{a}{b}\right)_\infty}{\left(\displaystyle\frac{1}{2}-\displaystyle\frac{c}{d}\right)_\infty \left(\displaystyle\frac{1}{2}+\displaystyle\frac{c}{d}\right)_\infty}=\phi \tag{2} $$ $$ \lim_{n\to\infty}\frac{\Gamma\left(n+\displaystyle\frac{1}{2}-\displaystyle\frac{a}{b}\right)\Gamma\left(n+\displaystyle\frac{1}{2}-\displaystyle\frac{a}{b}\right)}{\Gamma\left(n+\displaystyle\frac{1}{2}-\displaystyle\frac{c}{d}\right)\Gamma\left(n+\displaystyle\frac{1}{2}+\displaystyle\frac{c}{d}\right)}\frac{\Gamma\left(\displaystyle\frac{1}{2}-\displaystyle\frac{c}{d}\right)\Gamma\left(\displaystyle\frac{1}{2}+\displaystyle\frac{c}{d}\right)}{\Gamma\left(\displaystyle\frac{1}{2}-\displaystyle\frac{a}{b}\right)\Gamma\left(\displaystyle\frac{1}{2}+\displaystyle\frac{a}{b}\right)}=\phi \\ \frac{\Gamma\left(\displaystyle\frac{1}{2}-\displaystyle\frac{c}{d}\right)\Gamma\left(\displaystyle\frac{1}{2}+\displaystyle\frac{c}{d}\right)}{\Gamma\left(\displaystyle\frac{1}{2}-\displaystyle\frac{a}{b}\right)\Gamma\left(\displaystyle\frac{1}{2}+\displaystyle\frac{a}{b}\right)}=\phi \tag{3}.$$ Note that $(x)_n$ is the Pochhammer symbol.
Now, unfortunately I doubt $(3)$ is more tractable than $(1)$, therefore I looked up an infinite product to substitute for $\phi$ in $(2)$, and found it here: $(2)$ is equivalent to $$\prod_{n=1}^\infty\frac{\left(n-\displaystyle\frac{1}{2}\right)^2-\displaystyle\frac{a^2}{b^2}}{\left(n-\displaystyle\frac{1}{2}\right)^2-\displaystyle\frac{c^2}{d^2}}=\frac{1}{2}\prod_{n=0}^\infty\frac{100n(n+1)+25}{100n(n+1)+9} \\ \prod_{n=1}^\infty\frac{\left(n-\displaystyle\frac{1}{2}\right)^2-\displaystyle\frac{a^2}{b^2}}{\left(n-\displaystyle\frac{1}{2}\right)^2-\displaystyle\frac{c^2}{d^2}}\frac{100n(n+1)+9}{100n(n+1)+25}= \frac{25}{18},\tag{4}$$
which at least contains no irrational numbers.
|
Suppose you put $a/b$ and $c/d$ in terms of a common denominator as $m/n$ and $k/n$, and let $\omega = \exp(\pi i/n)$, so
$$\eqalign{\cos(a\pi/b) &= \cos(m \pi/n) = (\omega^m + \omega^{-m})/2\cr
\cos(c\pi/d) &= \cos(k \pi/n) = (\omega^k + \omega^{-k})/2\cr}$$
The equation $\phi^2 = \phi + 1$ then becomes $$
\left(\dfrac{\omega^m + \omega^{-m}}{\omega^k + \omega^{-k}}\right)^2 = \dfrac{\omega^m + \omega^{-m}}{\omega^k + \omega^{-k}} + 1$$
or $$(\omega^{2m} + 1)^2 \omega^{2k}- (\omega^{2m} + 1)(\omega^{2k} + 1)\omega^{m+k} - (\omega^{2k} + 1)^2 \omega^{2m} = 0$$
$$ {\omega}^{2\,k+4\,m}+{\omega}^{2\,k}-{\omega}^{3\,k+3\,m}-{\omega}^{k+
3\,m}-{\omega}^{3\,k+m}-{\omega}^{k+m}-{\omega}^{2\,m+4\,k}-{\omega}^{
2\,m} = 0$$
Now the minimal polynomial of $\omega$ is the cyclotomic polynomial
$C_n(X)$, so this will work only if $C_n(X)$ divides
$$ {X}^{2\,k+4\,m}+{X}^{2\,k}-{X}^{3\,k+3\,m}-{X}^{k+3\,m}-{X}^{3\,k+m}-{
X}^{k+m}-{X}^{2\,m+4\,k}-{X}^{2\,m}
$$
That's not necessarily going to mean $\cos(m\pi/n)/\cos(k\pi/n) = \phi$: the left side could actually be $\cos(jm\pi/n)/\cos(jk\pi/n)$ for some $j$ coprime to $n$, and the right side could be either $\phi$ or $-1/\phi$.
But it's a necessary condition, and we can filter out the ones we want.
No surprises: searching through $1 \le m,k \le 100$ I get the following results:
$$ \phi = \dfrac{\cos(\pi/3)}{\cos(2\pi/5)} =\dfrac{\cos(2\pi/3)}{\cos(3\pi/5)}$$
(together with others related by the symmetries of $\cos$)
|
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|
If $n = 4k + 1$, does $4$ divide $n^2 -1$? How would I show that $4$ divides $n^2 -1$ if $n = 4k+1$?
Is there more than one way to solve this?
|
For $k = 0$, $n = 1$, $n^2-1 = 0 = 4\cdot 0$.
Assume statement is true for some $k = h$. Let $(4h+1)^2-1 = 4a$.
For $k = h+1$,
$$\begin{align*}
(4h+5)^2 - 1 &= (4 + 4h+1)^2 - 1\\
&= 4^2 + 2\cdot4(4h+1) +(4h+1)^2-1\\
&= 4^2 + 2\cdot4(4h+1) + 4a
\end{align*}$$
which is divisible by $4$.
Similarly for $k=h-1$.
|
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|
weighing avg by columns vs rows outputs different results First of all, I just know the basics of math, so please be patient.
I have an overall score for a company with the product1 and product2. However when I do the overall for each criteria like A, B, C, D, E, the sum of these criteria is not equal to the overall score.
Example A
Data
$\begin{array}{c|ccccc}
\;&A&B&C&D&E\\
\hline
\text{product1}&1&2&3&4&5\\
\text{product2}&2&2&3&3&3
\end{array}$
Weighing
$\begin{array}{c|ccccc}
\;&AA&BB&CC&DD&EE\\
\hline
\text{product1}&2&3&2&3&5\\
\text{product2}&2&4&3&3&4
\end{array}$
Formula for row $i$: $(A_i \times AA_i + B_i \times BB_i + \dots ) /sum(AA_i:EE_i)$
product1: $\dfrac{2+6+6+12+25}{15}=3.4$
product2: $\dfrac{4+8+9+9+12}{16}=2.625$
Total AVG: $\dfrac{3.4+2.625}{2}=3.0125$
If I try to find what is the weighing avg for each column, the total is different. Both examples should output the same weighing avg? If not, why? What is the most adequate way to represent the avg for this type of need? the first example or the second?
Example B
Formula for column $J$: $(J_1 \times JJ_1 + J_2 \times JJ_2)/sum(JJ_1:JJ_2)$
$\begin{array}{c|ccccc}
\;&A&B&C&D&E\\
\hline
\text{average}&\dfrac{2+4}{6}=1.5
&\dfrac{6+8}{7}=2
&\dfrac{6+9}{5}=3
&\dfrac{12+9}{6}=3.5
&\dfrac{25+12}{9}=4.111
\end{array}$
Total AVG: $\dfrac{1.5+2+3+3.5+4.111}{5}=2.822$
|
Your $2.8$ is not correct. If you do the weighted average of $1,5\ \ 2 \ \ 3 \ \ 3,5\ \ 4$ with the weights you are using you get $3.0333333$
|
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|
Proof of $\sum^{2N}_{n=1} \frac{(-1)^{n-1}}{n} = \sum^{N}_{n=1} \frac{1}{N+n}$ The title pretty much summarizes my question. I am trying to prove the following:
$$\displaystyle \forall N \in \mathbb{N}: \sum^{2N}_{n=1} \frac{(-1)^{n-1}}{n} = \sum^{N}_{n=1} \frac{1}{N+n}.$$
I tried proving this using induction. Starting with the base case $N = 1$:
$$\displaystyle \sum^{2}_{n=1} \frac{(-1)^{n-1}}{n} = \frac{1}{1} + \frac{-1}{2} = \frac{1}{2} = \frac{1}{1+1} = \sum^{1}_{n=1} \frac{1}{N+n}.$$
My problem is the inductive step for $N+1$, starting with
$$\displaystyle \sum^{2(N+1)}_{n=1} \frac{(-1)^{n-1}}{n} = \sum^{N+1}_{n=1} \frac{1}{(N+1)+n}.$$
And now my problem:
\begin{align}
\sum^{2(N+1)}_{n=1} \frac{(-1)^{n-1}}{n} &\Leftrightarrow \sum^{2N}_{n=1} \frac{(-1)^{n-1}}{n} + \sum^{2}_{n=1} \frac{(-1)^{n-1}}{n} \\[1em]
&\underset{\Leftrightarrow}{\text{ind. hyp.}} \sum^{N}_{n=1} \frac{1}{N + n} + \sum^{2}_{n=1} \frac{(-1)^{n-1}}{n}
\end{align}
Is this the correct start, and if so, how do I continue?
|
you may find it easier to note that
$$
S=\sum_{n=1}^{2N} (-1)^{n-1}\frac1{n} = \sum_{n=1}^{N} \left(\frac1{2n-1} - \frac1{2n} \right)
$$
for then:
$$
S+\sum_{n=1}^{N} \frac1{n} = \sum_{n=1}^{2N} \frac1{n} = \sum_{n=1}^{N} \left( \frac1{n}+\frac1{N+n} \right)
$$
|
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|
Is this matrix diagonalizable over $\mathbb{R}$ or $\mathbb{C}$? Problem: Let $A = \begin{pmatrix} 6 & 0 \\ -2 & 2 \end{pmatrix}$. Is this matrix diagonalizable over $\mathbb{R}$? If not, is it diagonalizable over $\mathbb{C}$? Compute the eigenvalues $\lambda$ and the corresponding eigenspaces $E_{\lambda}$.
Attempt at solution: The characteristic polynomial of this matrix is $\det(A - x \mathbb{I}_2)$: \begin{align*} \det \begin{pmatrix} 6 - x & 0 \\ -2 & 2-x \end{pmatrix} = x^2 - 6x +10 \end{align*} Since this polynomial has no roots over $\mathbb{R}$, it is not diagonalizable over $\mathbb{R}$. The complex roots of this polynomial are $\lambda_1 = 3+i$ and $\lambda_2 = 3-i$. This are the eigenvalues of $A$. The eigenspace corresponding to $\lambda_1$ is the nullspace of the matrix $(A - \lambda_1 \mathbb{I}_2)$: \begin{align*} E_{\lambda_1} = N(A - \lambda_1 \mathbb{I}_2) = \left\{ \begin{pmatrix} x \\ y \end{pmatrix} \in \mathbb{C}^2 : \begin{pmatrix} -i +3 & 0 \\ -2 & -i -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\right\} \end{align*}
Now I was wondering what the geometric multiplicity is of $\lambda_1$, i.e. the dimension of the corresponding eigenspace. If this is not equal to the algebraic multiplicity (which is $1$), then the matrix is not diagonalizable over $\mathbb{C}$. How many free variables are there in the matrix \begin{align*} \begin{pmatrix} -i +3 & 0 \\ -2 & -i -1 \end{pmatrix} ? \end{align*} I thought I could row reduce this to $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$. But then I have $\dim(E_{\lambda_1}) = 0$, which is impossible since it must always be greater than or equal to $1$? Some help would be appreciated.
|
You miscalculated the determinant. You should have ended up with the characteristic polynomial
$$
p(x) = (x-6)(x-2)
$$
check your work, note that $0 \times (-2) = 0$.
|
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|
3 Questions on number theory. We have to everything using number congruences, and I am just a beginner, I know a few theorems, and we have to solve these using basics.
1) If $n=a^4$ where $a \in \mathbb Z$ then prove that $n \equiv 0,1,5$ or $6(mod \ 10)$.
My work: $a^4 \equiv x(mod \ 10)$
We have to find $x$.
But how to simplify the above congruence ?
2) Prove that $4n^2+4 \equiv 0(\mod \ 19)$ for any $n$.
But I think this question is wrong because it doesn't work for $1$ or $2$. I think it may be a misprint. It may be $\not\equiv$ in place of $\equiv$
3) Solve for $n$, $5n \equiv 3(mod \ 8)$.
Sol: $n \equiv -1(mod \ 8)$
So $n=8k-1$
Solution set becomes $\{...,-17,-9,-1,7,15,...\}$
|
Question 1: Because we can write any number $a \in \mathbb{Z}$ as $a=10\cdot b + c$ with $b,c \in \mathbb{N}$ we see that $a^4 = (10 \cdot b + c)^4 \mod 10$. It is easy to see that this forms a polynomial with all terms except for the $c^4$ part is a multiple of $10$. Since we are working in $\mod 10$ we see that $a^4=(10 \cdot b + c)^4 = c^4 \mod 10$. This means we only have to look at the last digit of any number in $\mathbb{Z}$ and we are left with 10 cases:
*
*Numbers ending on $0: 0^4=0$ so last digit is $0$.
*Numbers ending on $1: 1^4=1$ so last digit is $1$.
*Numbers ending on $2: 2^4=16$ so last digit is $6$.
*Numbers ending on $3: 3^4 =81$ so last digit is $1$.
*Numbers ending on $4: 4^4 =256$ so last digit is $6$.
*Numbers ending on $5: 5^4 =625$ so last digit is $5$.
*Numbers ending on $6: 6^4 =1296$ so last digit is $6$.
*Numbers ending on $7: 7^4 =2401$ so last digit is $1$.
*Numbers ending on $8: 8^4 =4096$ so last digit is $6$.
*Numbers ending on $9: 9^4 =6561$ so last digit is $1$.
Thus we have the numbers $0,1,5,6$.
Question 2:
$4n^2+4 = 0 \mod 19$ is not true, take $n=1$ and the result is trivial.
If we instead look at $4n^2+4 \neq \mod 19$ we see that we get a repeating pattern every 19 numbers. This can be seen by looking at \begin{align*}
4(n+19)^2+4 \mod 19 &= 4(n^2 + 38n + 19^2) +4 \mod 19 \\
&= 4 n^2 +4 \mod 19 + 38n \mod 19 + 19^2 \mod 19\end{align*} where the last 2 terms are obviously $0 \mod 19$.
You now have to compute the first 19 cases by hand (or excel/program it like I did), which shows indeed that $4n^2+4 \neq \mod 19$
Question 3: You have done this correctly.
|
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|
Find numbers $\overline{abcd}$ so that $\overline{abcd}+\overline{bcd}+\overline{cd}+d+1=\overline{dcba}$ Find the numbers $\overline{abcd}$, with digits not null that satisfy the equality
\begin{equation}\overline{abcd}+\overline{bcd}+\overline{cd}+d+1=\overline{dcba}\end{equation}
where
\begin{equation}\overline{abcd}=1000a+100b+10c+d\end{equation}
I see that $4d+1=\overline{..a},\ d\neq0,\ a\neq 0,\ a<d$ but that gives too many pairs to start with $(d,a)\in\{(5,1),(6,5),(8,3),(9,7)\}$
|
As you observed, $$(d,a) \in \{(5,1), (8,3), (9,7), (6, 5)\}$$ but notice also that $d = a+k_3$, where $k_3$ is the carry resulting from $bcd+cd+d+1$. This quantity is at most $999+99+9+1 = 1108$ so the carry is either $0$ or $1$ and thus $d = a$ or $d=a+1$. Combined with the foregoing, this means that $(d,a) = (6, 5)$.
Now we have $$\overline{5bc6} + \overline{bc6} + \overline{c6} + 6 + 1 = \overline {6cb5}$$ and at this point anything we do will solve the problem as there are only 9 possibilities to check. Remembering we need $10\le 2b+k_2 \le 19$ where $k_2$ is the carry resulting from $\overline{c6} + 7$, so $b\ge 5$, we do trial and error on $b\in \{5,6,7,8,9\}$ and immediately find $b=5, c=7$.
Computer search confirms the uniqueness of the answer.
|
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|
Using equation to find value of $1/x - 1/y$ $$\left(\frac{48}{10}\right)^x=\left(\frac{8}{10}\right)^y=1000$$
What is the value of $\frac{1}{x}-\frac{1}{y}$?
I have already used that when $48$ divided by $10$ then it becomes $4.8$ and when $8$ divided by $10$ then it becomes $0.8$ by getting $10^x$ and $10^y$ to make it simpler but I cannot proceed further.
|
$$\left(\frac{48}{10}\right)^x=\left(\frac{8}{10}\right)^y=1000$$
$$48^x = 10^{x+3} \; \; \; \& \; \; \; 8^y = 10^{y+3}$$
$$x+3 = x \log {48} \Rightarrow \; \; \; 1 + \frac{3}{x} = \log {48}$$
$$y+3 = y \log {8} \Rightarrow \; \; \; 1 + \frac{3}{y} = \log {8}$$
$$3 \times (\frac{1}{x} - \frac{1}{y}) = \log {48} - \log {8} = \frac{\log_2 {48}}{3}$$
$$\frac{1}{x} - \frac{1}{y} = \frac{\log_2 {48}}{9} \approx 0.62055$$
|
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|
Limits. Finding positive value Find a real number k such that the limit $$\lim_{n\to\infty}\ \left(\frac{1^4 + 2^4 + 3^4 +....+ n^4}{n^k}\right)$$ has as positive value.
If I am not mistaken every even $k$ can be the answer. But the answer is 5.
|
Do you know Riemann Sum?
$\begin{eqnarray}
&& \lim_{n\to\infty} \left( \frac{1^4+2^4+3^4+\ldots+n^4 }{n^k } \right) \\
&=& \lim_{n\to\infty}n^{4-k} \left( \left(\frac1n\right)^4 +\left(\frac2n\right)^4 + \left(\frac3n\right)^4 + \ldots + \left(\frac nn\right)^4 \right) \\
\end{eqnarray} $
Should it converge to a finite non-zero value, it should be in the form of $\displaystyle \lim_{n\to\infty} \frac1n \displaystyle\sum_{k=1}^n f\left( \frac kn \right) $
So $4-k = -1 \Rightarrow k = 5 $.
If you want to evaluate the limit: it becomes $ \displaystyle \int_0^1 x^4 \, dx = \frac15 $.
Alternatively, you can apply Stolz Cesaro Theorem: $a_n= 1^4 + 2^4 + \ldots+n^4 $, $ b_n = n^k $.
Show that $ \displaystyle \lim_{n\to\infty} = \frac{a_{n+1} - a_n}{b_{n+1} - b_n } = \lim_{n\to\infty} \frac{(n+1)^4}{(k+1)n^{k-1}} $, then we have $k-1 =4 $ again.
|
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|
Find the matrix $\mathbf{A}$ if $A\binom{7}{-1} = \binom{6}{2}.$ Find the $2\times2$ matrix $A$ where $A^2=A$ and
$$A\begin{pmatrix} 7 \\ -1 \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \end{pmatrix}.$$
I tried plugging in: $A= \begin{pmatrix}a&b\\c&d\end{pmatrix}$ but that became messy very quickly. I got the equations:
$7a-b = 6$
$7c-d = 2$
$a^2+bc = a$
$ab+bd = b$
$ac + cd = c$
$bc + d^2 = d$ from trying that method. What should I do?
|
As $A^2=A$, you also know that $A\begin{pmatrix}6\\2\end{pmatrix}=\begin{pmatrix}6\\2\end{pmatrix}$. Since $\begin{pmatrix}7\\-1\end{pmatrix},\begin{pmatrix}6\\2\end{pmatrix}$ are linearly independant, this determines $A$ completely. We conclude that both columns of $A$ are multiples of $\begin{pmatrix}6\\2\end{pmatrix}$, so
$$ A=\begin{pmatrix}6x&6y\\2x&2y\end{pmatrix}$$
for certain $x,y$. Now $A\begin{pmatrix}6\\2\end{pmatrix}=\begin{pmatrix}6\\2\end{pmatrix}$ just means that $6x+2y=1$, and $A\begin{pmatrix}7\\-1\end{pmatrix}=\begin{pmatrix}6\\2\end{pmatrix}$ means that $7x-y=1$. Can you find $x,y$ now?
|
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|
Prove by induction that $\frac{n^3}{3}+\frac{2n}{3}$ is an integer. The question that I am working on is:
Prove that $\dfrac{n^3}{3}+\dfrac{2n}{3} \in \mathbb Z \ \forall \ n \in \mathbb N$
The method that I think would be will work for this question is that I say that $3|(n^3+2n)$ and prove that.Would this be a good way to do this question?
So far I have done the following:
1) Base Case
n = 1
$3|n^3+2n = 3|3 \checkmark$
2) Assume, $n^3+2n$ is divisible by $3$ for $n =k, k \in \mathbb N$
$k^3+2k = 3m , m \in \mathbb N$
Let $n = k + 1$ ; Then:
$(k+1)^3 + 2(k+1)$
= $k^3+2k+3k^2+3k+3$
Since $k^3+2k = 3m$
$3m + 3k^2+3k+3$
Now, I'm stuck I don't know what else to do further.
|
Look at this:
assuming
$\dfrac{k^3 + 2k}{3} \in \Bbb Z, \tag{1}$
we have
$\dfrac{(k + 1)^3 + 2(k + 1)}{3} = \dfrac{k^3 + 3k^2 + 3k + 1 + 2k + 2}{3} = \dfrac{(k^3 + 2k) + 3(k^2 + k +1)}{3}$
$= \dfrac{k^3 + 2k}{3} + k^2 + k + 1 \in \Bbb Z, \tag{2}$
by (1).
QED!!!
|
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|
Solving $6 \cos x - 5 \sin x = 8$ My attempt:
Using the formula for linear combinations of sine and cosine:
$$A \cos x+B \sin x=C \sin (x+\phi)$$
$$
\sqrt{51} \left(\frac{6}{\sqrt{51}} \cos x - \frac{5}{\sqrt{51}}\sin x\right) = 8
$$
$$
\frac{6}{\sqrt{51}} \cos x - \frac{5}{\sqrt{51}}\sin x = \frac{8}{\sqrt{51}}
$$
And then assume:
$$
\frac{6}{\sqrt{51}}= \cos \psi ; \frac{5}{\sqrt{51}}= \sin\psi ;
$$
$$
\cos \psi \cos x - \sin \psi \sin x = \cos (x+ \psi) = \cos(x + \arccos ( \frac{6}{\sqrt{51}}))
$$
$$
x + \arccos\left(\frac{6}{\sqrt{51}}\right) = \arcsin\left( \frac{8}{\sqrt{51}}\right)
$$
$$
x \approx 12^\circ
$$
But answer is:
$$
-\frac{\pi}{4} + (-1)^n \frac{\pi}{4} + \pi n , n\in\Bbb Z
$$
|
Edit: In general, the maximum value of $a\cos A+b\sin A$ is $\sqrt{a^2+b^2}$
Hence the maximum value of $6\cos x-5\sin x$ is $\sqrt{6^2+(-5)^2}=\sqrt{61}$
But $RHS=8$ $\implies \sqrt{61}<8$ i.e. equality does not hold true for any value of $x$
Hence there is no solution of the given equation: $6\cos x-5\sin x=8$
|
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|
Baby Rudin claim: $1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}...$ converges This sequence is a rearrangement of the series $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}...$. Note that at this point in the text we do not have any theorem about the convergence of rearrangements.
Let $\{s_n\}$ be the sequence of partials sums of the series then for $n \ge 0$
$s_{3(n+1)} = \sum ^n _ {k=0} \frac{1}{4k+1} + \frac{1}{4k+3} - \frac{2}{4k+4}$
We can view it as the sequence(on $n$) of partials sums of
$\sum_0 a_n = \sum_0 \frac{1}{4n+1} + \frac{1}{4n+3} - \frac{2}{4n+4}$
Where $|a_n| = a_n = \frac{1}{4n+4}\{\frac{3}{4n+1}+\frac{1}{4n+3}\} \le \frac{1}{4n^2}$.
By the comparison test $s_{3(n+1)}$ converges to some real $\alpha$.
But $s_{3(n+1)+1} = s_{3(n+1)}+ \frac{1}{4n+5}$ and $s_{3(n+1)+2} = s_{3(n+1)}+ \frac{1}{4n+5}+\frac{1}{4n+7} $hence we have a partition of $\{s_n\}$ into subsequences which tend to $\alpha$ and this implies $s_n \rightarrow \alpha$.
Is my proof correct? Any alternative solutions are appreciated.
|
Showing Convergence
Breaking the series into chunks of $3$ terms, which is okay since the terms tend to $0$, we get
$$
\begin{align}
\sum_{k=0}^\infty\left[\frac1{4k+1}+\frac1{4k+3}-\frac2{4k+4}\right]
&=\sum_{k=0}^\infty\left[\left(\frac1{4k+1}-\frac1{4k+4}\right)+\left(\frac1{4k+3}-\frac1{4k+4}\right)\right]\\
&=\sum_{k=0}^\infty\left[\frac3{(4k+1)(4k+4)}+\frac1{(4k+3)(4k+4)}\right]
\end{align}
$$
Which can be compared to
$$
\left[\frac34+\frac3{16}\sum_{k=1}^\infty\frac1{k^2}\right]
+\left[\frac1{12}+\frac1{16}\sum_{k=1}^\infty\frac1{k^2}\right]
=\frac56+\frac14\sum_{k=1}^\infty\frac1{k^2}
$$
which converges by the $p$-test.
One Approach to Evaluation
If curious about the actual sum, it can be computed, using $(11)$ from this answer, as
$$
\begin{align}
&\frac14\sum_{k=1}^\infty\left[-\left(\frac1k-\frac1{k-\frac34}\right)-\left(\frac1k-\frac1{k-\frac14}\right)\right]\\
&=\frac14\left[-H_{-3/4}-H_{-1/4}\right]\\
&=\frac14\left[-(-\pi/2-3\log(2))-(\pi/2-3\log(2))\right]\\
&=\frac32\log(2)
\end{align}
$$
Another Approach to Evaluation
Using the fact that the alternating Harmonic Series converges to $\log(2)$, we get
$$
\begin{align}
\sum_{k=0}^\infty\left[\frac1{4k+1}+\frac1{4k+3}-\frac2{4k+4}\right]
&=\sum_{k=0}^\infty\left[\frac1{4k+1}-\frac1{4k+2}+\frac1{4k+3}-\frac1{4k+4}\right]\\
&+\sum_{k=0}^\infty\left[\hphantom{\frac1{4k+1}}+\frac1{4k+2}\hphantom{\ \!+\frac1{4k+3}}-\frac1{4k+4}\right]\\
&=\log(2)+\frac12\log(2)\\[6pt]
&=\frac32\log(2)
\end{align}
$$
|
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|
Differentiate the Function: $y=2x \log_{10}\sqrt{x}$ $y=2x\log_{10}\sqrt{x}$
Solve using: Product Rule $\left(f(x)\cdot g(x)\right)'= f(x)\cdot\frac{d}{dx}g(x)+g(x)\cdot \frac{d}{dx}f(x)$
and $\frac{d}{dx}(\log_ax)= \frac{1}{x\ \ln\ a}$
$(2x)\cdot [\log_{10}\sqrt{x}]'+(\log_{10}\sqrt{x})\cdot [2x]'$
$y'=2x\frac{1}{\sqrt{x}\ln 10}+\log_{10}\sqrt{x}\cdot 2$
Answer in book is $y'= \frac{1}{\ln10}+\log_{10}x$
|
$$\log\sqrt{x} = \frac{1}{2}\log_{10} x= \frac{\ln x}{2\ln 10}$$
$$\frac{d}{dx}\ln x= \frac{1}{x}$$
using chain rule,
$$\frac{d}{dx}\left(2x\log\sqrt{x}\right)= 2x \frac{dx}{dx}\frac{\ln x}{2\ln 10} + \frac{\ln x}{2\ln 10}\frac{d}{dx} 2x $$
$$=\frac{2x}{2x\ln 10} + \frac{2\ln x}{2\ln 10} $$
$$ =\frac{1}{\ln 10}+ \log_{10} x$$
Note the change of base: $$\log_{10}x = \frac{\ln x}{\ln 10} $$
|
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|
Wanted : for more formulas to find the area of a triangle? I know some formulas to find a triangle's area, like the ones below.
*
*Is there any reference containing most triangle area formulas?
*If you know more, please add them as an answer
$$s=\sqrt{p(p-a)(p-b)(p-c)} ,p=\frac{a+b+c}{2}\\s=\frac{h_a*a}{2}\\s=\frac{1}{2}bc\sin(A)\\s=2R^2\sin A \sin B \sin C$$
Another symmetrical form is given by :$$(4s)^2=\begin{bmatrix}
a^2 & b^2 & c^2
\end{bmatrix}\begin{bmatrix}
-1 & 1 & 1\\
1 & -1 & 1\\
1 & 1 & -1
\end{bmatrix} \begin{bmatrix}
a^2\\
b^2\\
c^2
\end{bmatrix}$$
Expressing the side lengths $a$, $b$ & $c$ in terms of the radii $a'$, $b'$ & $c'$ of the mutually tangent circles centered on the triangle's vertices (which define the Soddy circles)
$$a=b'+c'\\b=a'+c'\\c=a'+b'$$gives the paticularly pretty form $$s=\sqrt{a'b'c'(a'+b'+c')}$$
If the triangle is embedded in three dimensional space with the coordinates of the vertices given by $(x_i,y_i,z_i)$ then $$s=\frac{1}{2}\sqrt{\begin{vmatrix}
y_1 &z_1 &1 \\
y_2&z_2 &1 \\
y_3 &z_3 &1
\end{vmatrix}^2+\begin{vmatrix}
z_1 &x_1 &1 \\
z_2&x_2 &1 \\
z_3 &x_3 &1
\end{vmatrix}^2+\begin{vmatrix}
x_1 &y_1 &1 \\
x_2&y_2 &1 \\
x_3 &y_3 &1
\end{vmatrix}^2}$$
When we have 2-d coordinate $$ s=\frac{1}{2}\begin{vmatrix}
x_a &y_a &1 \\
x_b &y_b &1 \\
x_c &y_c & 1
\end{vmatrix}$$
In the above figure, let the circumcircle passing through a triangle's vertices have radius $R$, and denote the central angles from the first point to the second $q$, and to the third point by $p$ then the area of the triangle is given by:
$$ s=2R^2|\sin(\frac{p}{2})\sin(\frac{q}{2})\sin(\frac{p-q}{2})|$$
|
$\require{cancel}$
Some errors
spotted in
Baker's papers, referenced in
the accepted answer.
Expression N20
is definitely wrong (even the dimension):
\begin{align}
20.\ &
\cancel{\color{blue}{\frac{R\,r}{\beta_a\beta_b\beta_c}\,
\left(\frac1a+\frac1b\right)
\left(\frac1b+\frac1c\right)
\left(\frac1c+\frac1a\right)}}
,
\end{align}
and most likely,
the term $\beta_a\beta_b\beta_c$
must be in the numerator
and a constant $\tfrac12$ is missing,
so the correct form
should be
\begin{align}
20.\ &
\tfrac12\,{R\,r}{\beta_a\beta_b\beta_c}\,
\left(\frac1a+\frac1b\right)
\left(\frac1b+\frac1c\right)
\left(\frac1c+\frac1a\right)
.
\end{align}
In the second
Baker's paper,
expressions N63, N80
are wrong :
\begin{align}
63.\ &
\cancel{\color{blue}{R^2\sin 2A (1+\cos C)}}
\end{align}
\begin{align}
80.\ &
\cancel{\color{blue}{a\,(-\beta_a\,\sin \tfrac12\,A+\beta_b\,\sin \tfrac12\,B+\beta_c\,\sin \tfrac12\,C)}}
\\
\quad&=
\cancel{\color{blue}{2\,s\,(\beta_a\,\sin \tfrac12\,A+\beta_b\,\sin \tfrac12\,B+\beta_c\,\sin \tfrac12\,C)}}
.
\end{align}
Also, in N94 the two first expressions are correct:
\begin{align}
94.\ &
r^2\cot\tfrac12\,A+2\,R\,r\,\sin A
\\
&=
r_a^2\cot\tfrac12\,A-2\,R\,r_a\,\sin A
,
\end{align}
but there are obvious typos in the third and fourth.
They should be
\begin{align}
&=r_b^2\cot\tfrac12\,B-2\,R\,r_b\,\sin B
\\
&=r_c^2\cot\tfrac12\,C-2\,R\,r_c\,\sin C
.
\end{align}
|
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|
Show that $ \tan (A + \theta) $ can be simplified to $- \cot \theta$ as A tends to $\frac{\pi}{2}$ So far I have used the identity,
$$\tan\left(\frac{\pi}{2} + \theta\right) = \frac{\tan A + \tan \theta} {1 - \tan A \tan \theta}$$
As $A \to \frac{\pi}{2}$, $\tan A \to \infty$, so my reasoning is, $\infty + \tan \theta = \infty$, which gives:
$$\frac{\infty} {- \infty \tan \theta}$$
$\infty$ cancels to $-1$ leaving $\frac{-1}{\tan \theta}$ as the answer, or $-\cot \theta$
My approach seems a bit iffy, and I was hoping someone could confirm if this is right or not.
Thanks
|
Using:
$$\lim_{A \to \pi/2} \frac{- \tan A \tan \theta}{1 - \tan A \tan \theta} = 1:$$
$$\lim_{A \to \pi /2} \frac{\tan A + \tan \theta}{1 - \tan A \tan \theta} = \lim_{A \to \pi/2} \frac{\tan A}{1 - \tan A \tan \theta} + \frac{\tan \theta}{1 - \tan A \tan \theta} = \lim_{A \to \pi /2} \frac{\tan A}{-\tan A \tan \theta} + \frac{\tan \theta}{-\tan A \tan \theta} = \lim_{A \to \pi/2} \frac{-1}{\tan \theta} + \frac{-1}{\tan A} = -\frac{1}{\tan \theta} = -\cot \theta$$
|
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|
Limit of the sequence $\sin \left( {2\pi \sqrt {{n^2} + n} } \right)$ I would like to calculate the following limit: ${\lim _{n \to \infty }}\sin \left( {2\pi \sqrt {{n^2} + n} } \right)$
I am not sure if this limit exists...
|
Hint: Note that
$$\sqrt{n^2+n}-\left(n+\frac{1}{2}\right)=\frac{\left(\sqrt{n^2+n}-\left(n+\frac{1}{2}\right)\right)\left(\sqrt{n^2+n}+\left(n+\frac{1}{2}\right)\right)}{\sqrt{n^2+n}+\left(n+\frac{1}{2}\right)}=\frac{-1/4}{\sqrt{n^2+n}+\left(n+\frac{1}{2}\right)}.$$
Thus
$$\sin\left(2\pi\sqrt{n^2+n}\right)=\sin\left(2\pi n+\pi - \frac{\pi/2}{\sqrt{n^2+n}+\left(n+\frac{1}{2} \right)}\right).$$
|
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|
Confusion about the integral $\int dT/(1-T^2)$ From some reference on the internet we have the following real valued function and its derivative:
$$
M(T) = \frac{\sqrt{1-T^2}}{1+T}
\quad \Longrightarrow \quad \frac{dM}{dT} = - M/(1-T^2)
$$
The reverse of differentiation is integration:
$$
\frac{dM}{dT} = - M/(1-T^2) \quad \Longrightarrow \quad \int\frac{dM}{M} = - \int \frac{dT}{1-T^2}
= \int \frac{dT}{T^2-1}\quad \Longrightarrow \\
\int\frac{dM}{M} = \frac{1}{2} \left[\, \int \frac{dT}{T-1} -\int \frac{dT}{T+1}\,\right]
\quad \Longrightarrow \\ \ln|M| = \frac{1}{2}\left[\,\ln|T-1|-\ln|T+1|\,\right] =
\ln\left(\sqrt{\left|\frac{T-1}{T+1}\right|}\right) \quad \Longrightarrow \\
M = \sqrt{\frac{1-T}{1+T}} = \frac{\sqrt{1-T^2}}{1+T}
$$
Here is where the confusion starts, because we also could have proceeded in the following way:
$$
\frac{dM}{dT} = - M/(1-T^2) \quad \Longrightarrow \quad \int\frac{dM}{M} = - \int \frac{dT}{1-T^2}
\quad \Longrightarrow \\
\int\frac{dM}{M} = - \frac{1}{2} \left[\, \int \frac{dT}{1-T} + \int \frac{dT}{1+T}\,\right]
\quad \Longrightarrow \\ \ln|M| = - \frac{1}{2}\left[\,\ln|1-T|+\ln|1+T|\,\right] =
\ln\left(\frac{1}{\sqrt{\left|1-T^2\right|}}\right) \quad \Longrightarrow \\
M = \frac{1}{\sqrt{1-T^2}}
$$
Which is clearly wrong. But .. where is the error?
|
$$
\int \frac{dT}{1-T} = -\ln(1-T)+\text{constant}.
$$
The initial minus sign comes from the chain rule. You omitted it.
(No absolute value sign is needed since $\sqrt{1-T^2}$ exists only when $1-T$ and $1+T$ are both $\ge0$.)
|
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|
New Idea to prove $1+2x+3x^2+\cdots=(1-x)^{-2}$
Given $|x|<1 $ prove that $\\1+2x+3x^2+4x^3+5x^4+...=\frac{1}{(1-x)^2}$.
1st Proof: Let $s$ be defined as
$$
s=1+2x+3x^2+4x^3+5x^4+\cdots
$$
Then we have
$$
\begin{align}
xs&=x+2x^2+3x^3+4x^4+5x^5+\cdots\\
s-xs&=1+(2x-x)+(3x^2-2x^2)+\cdots\\
s-xs&=1+x+x^2+x^3+\cdots\\
s-xs&=\frac{1}{1-x}\\
s(1-x)&=\frac{1}{1-x}\\
s&= \frac{1}{(1-x)^2}
\end{align}
$$
2nd proof:
$$
\begin{align}
s&=1+2x+3x^2+4x^3+5x^4+\cdots\\
&=\left(1+x+x^2+x^3+\cdots\right)'\\
&=\left(\frac{1}{1-x}\right)'\\
&=\frac{0-(-1)}{(1-x)^2}\\
&=\frac{1}{(1-x)^2}
\end{align}
$$
3rd Proof:
$$
\begin{align}
s=&1+2x+3x^2+4x^3+5x^4+\cdots\\
=&1+x+x^2+x^3+x^4+x^5+\cdots\\
&+0+x+x^2+x^3+x^4+x^5+\cdots\\
&+0+0+x^2+x^3+x^4+x^5+\cdots\\
&+0+0+0+x^3+x^4+x^5+\cdots\\
&+\cdots
\end{align}
$$
$$
\begin{align}
s&=\frac{1}{1-x}+\frac{x}{1-x}+\frac{x^2}{1-x}+\frac{x^3}{1-x}+\cdots\\
&=\frac{1+x+x^2+x^3+x^4+x^5+...}{1-x}\\
&=\frac{\frac{1}{1-x}}{1-x}\\
&=\frac{1}{(1-x)^2}
\end{align}
$$
These are my three proofs to date. I'm looking for more ways to prove the statement.
|
Problem: For a given integer $N$, how many integers $n_1$ and $n_2$ larger than or equal to zero are there that satisfy the equation:
$$n_1 + n_2 = N$$
We note that it is the coefficient of $x^N$ in the expansion of
$$\left(\sum_{k=0}^{\infty}x^k\right)^2 = \frac{1}{(1-x)^2}$$
We can also solve the problem by noting that it is the number of ways you can color N objects with 2 colors. The count the number of solutions, we note that there is aone to one correspondence between colorings and a string consisting of N 0's and one 1. The 0's to the left of the 1 represent to objects with color 1 the 0's to the right represent the objects with color 2. The total number of such strings is equal to $\binom{N+1}{1} = N+1$. This means that the coefficient of $x^N$ in $\frac{1}{(1-x)^2}$ is equal to $N+1$.
|
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|
partial fraction derivative question So I have this partial fraction derivative question. I know how to solve it, but for some reason I keep swapping two numbers. Here is the problem:
$$\int\frac{3-4x}{x^2+x}= \frac{A}{x}+\frac{B}{x+1}$$
$$(3-4x)=A(x)+B(x+1)$$
Let $x=-1$
$$(3-4(-1))=A(-1)+B(0)$$
$$-7=A$$
Let $x=0$
$$(3-4(0))=A(0)+B(1)$$
$$3=B$$
Therefore,
$$\int\frac{-7}{x}+\int\frac{3}{x+1}$$
Answer is: $$-7ln|x|+3|x+1|+C$$
shouldn't the answer actually be $$3ln|x|-7|x+1|+C$$
Where am I going wrong?
|
$$\frac{3-4x}{x^2+x}= \frac{A}{x}+\frac{B}{x+1}$$
$$\frac{A}{x}+\frac{B}{x+1}=\frac {A(x+1)}{x(x+1)}+\frac{Bx}{x(x+1)}$$
then we have to set:$$3-4x=(A+B)x+A$$
Then $A=3$ and $B=-7$ so our integral will be:
$$\int\frac{3-4x}{x^2+x}dx=\int\frac{3}{x}dx+\int \frac{-7}{x+1}dx $$
$=3\ln\mid x\mid-7\ln\mid(x+1)\mid+c$
|
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|
Is this problem wrongly built? Or is there a solution which I don't know how to arrive at? I was solving a Cauchy-Schwarz's inequality based problem. Given that $x^2+y^2+z^2=1$ I am supposed to show that $x+y+z \le 6$. After struggling for a while I realised that I could solve this inequality had the condition been $x^2+y^2+z^2=12$. This was my solution:
$(x.1+y.1+z.1)^2 \le (x^2+y^2+z^2)(1^2+1^2+1^2)$
$\implies (x+y+z)^2 \le 12*3$
$\implies (x+y+z)^2 \le 12*3 \le 36$
Taking square root on both sides, I got
$(x+y+z) \le 6$
Now I am wondering whether I detected a typo in the problem ($x^2+y^2+z^2=12$, not $x^2+y^2+z^2=1$) or is there a real way to reach $(x+y+z) \le 6$ starting from $x^2+y^2+z^2=1$ based on Cauchy-Schwarz principles!
The main reason for this doubt is I am stuck on an extension of this problem in which I need to prove $x^3+y^3+z^3 \ge 24$
Any help is appreciated.
|
In the first part of your problem, assuming $x^2+y^2+z^2=1$ is actually no problem, because $x^2+y^2+z^2=1$ implies $x,y,z\le 1$, so $x+y+z\le 3<6$.
Also, using Cauchy-Schwarz you can slightly improve this bound to $x+y+z\le \sqrt{3}$.
Looking however also at your next problem, showing that $x^3+y^3+z^3\ge 24$, it seems that you are right. The assumption $x^2+y^2+z^2=1$ does not imply this, because clearly $1^2+0^2+0^2=1$, but $1^3+0^3+0^3=1<24$.
Taking however, your modified assumption $x^2+y^2+z^2=12$ we obtain by Cauchy-Schwarz:
$$144=(x^\frac32 x^\frac12+y^\frac32 y^\frac12+z^\frac32 z^\frac12)^2 \le (x^3+y^3+z^3)(x+y+z)\le 6(x^3+y^3+z^3),$$
as required. Therefore, you are right in saying, that this is most likely a typo in the problem statement.
|
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|
Using Lagrange's Method in Finding Extreme Values of $x^2 + y^2 + z^2$ for $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$ (New to This Method) Did I do this hw question correctly (at least in theory, I do not expect anyone to check my algebra work)? In particular, did I solve for lambda and plug lambda back into my equations for x,y, and z correctly, or is there a better way? Something does not feel right about it. Thanks!
Apply Lagrange's method in finding the extreme values $x^2 + y^2 + z^2$ subject to the constraint $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$, where $a > b > c > 0$.
Here is my work thus far:
Let $f(x,y,z) = x^2 + y^2 + z^2$, and let $g(x,y,z) - k = \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} - 1$.
Then let $u=f(x,y,z) + \lambda(g(x,y,z) - k)$. This gives us: $$u=(x^2 + y^2 + z^2) + \lambda\frac{x^2}{a^2} + \lambda\frac{y^2}{b^2} + \lambda\frac{z^2}{c^2} - \lambda$$
Taking the partial of u with respect to each variable (including lambda) and setting it equal to zero, we get:
$u_x = 2x + \frac{2\lambda x}{a^2} = 0$
$u_y = 2y + \frac{2\lambda y}{b^2} = 0$
$u_z = 2z + \frac{2\lambda z}{c^2} = 0$
$u_\lambda = \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$
Solving for x, y, and z, we get $x = -\frac{\lambda}{a^2},\ y = -\frac{\lambda}{b^2},\ z = -\frac{\lambda}{c^2}$, which we can plug into our partial, $u_\lambda$, for x, y, and z. In our partial, we get $$\frac{\lambda^2}{a^6} + \frac{\lambda^2}{b^6} + \frac{\lambda^2}{c^6} = 1$$
Solving for $\lambda$ gives us the difficult solution of $$\lambda = \pm \frac{a^3b^3c^3}{\sqrt{a^6b^6 + a^6c^6 + b^6c^6}}$$
Since $a,\ b,\ c>0$, the plugging in the positive solution for $\lambda$ will give us our max and vice versa for the min (if the function were odd, see below). Plugging $\lambda$ into our x, y, and z equations yield:
$x = \pm \frac{ab^3c^3}{\sqrt{a^6b^6 + a^6c^6 + b^6c^6}}$
$y = \pm \frac{a^3bc^3}{\sqrt{a^6b^6 + a^6c^6 + b^6c^6}}$
$z = \pm \frac{a^3b^3c}{\sqrt{a^6b^6 + a^6c^6 + b^6c^6}}$
Since f is even (each variable is raised to the second power), then either the positive or negative solution will give a positive result.
|
Setting $f_x=\lambda g_x, \;f_y=\lambda g_y, \;f_z=\lambda g_z$ gives
$\;\;\displaystyle2x=\lambda\cdot\frac{2x}{a^2}, \;\;2y=\lambda\cdot\frac{2}{b^2},\;\;2 z=\lambda\cdot\frac{2z}{c^2}$.
Therefore $\textbf{1)}$ $x=0$ or $\lambda=a^2\;\;\;$$\textbf{2)}$ $y=0$ or $\lambda=b^2\;\;\;$ $\textbf{3)}$ $z=0$ or $\lambda=c^2$
Since $\displaystyle\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$ and $a>b>c$, this implies that either
$\textbf{A)}$ $x=0, y=0, z^2=c^2\;\;$ or$\;\;$ $\textbf{B)}$ $x=0, z=0, y^2=b^2\;\;$ or$\;\;$ $\textbf{C)}$ $y=0, z=0, x^2=a^2$.
Therefore $x^2+y^2+z^2$ has largest value $a^2$ and smallest value $c^2$.
|
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|
Prove that if $x \neq 0$, then if $ y = \frac{3x^2+2y}{x^2+2}$ then $y=3$ The problem is the following (Velleman's exercise 3.2.10):
Suppose that $x$ and $y$ are real numbers. Prove that if $x \neq 0$, then if $ y = \frac{3x^2+2y}{x^2+2}$ then $y=3$.
My approach so far: Suppose $x \neq 0$. Suppose $ y = \frac{3x^2+2y}{x^2+2}$. Suppose $y \neq 3$. Suppose $y=4$. Then $4x^2=3x^2 \iff x=0$. Contradiction to $x \neq 0$.
Is it correct? I am a noob so feel free to bash it as good as you can. Thanks in advance.
|
$y = \frac{3x^2+2y}{x^2+2}$, multiplying both sides of the equation by $x^2+2$ results in an equivalent equation because that term is never $0$ (in the reals at least).
You end up with $yx^2 + 2y = 3x^2+2y$
subtract $2y$ from both sides (always legitimate).
$yx^2=3x^2$
Since $x\neq 0$ we can divide both sides by $x^2$ and get $y=3$
|
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|
Solve the equation $(m^2-m-2)x=m^2+4m+3$ Here's how I solve it
I think that m is the variable (am I right?).
Then
$$m^2x-mx-2x-m^2-4m-3=0$$
$$m^2(x-1)-m(x+4)-(2x+3)=0$$
$$D=x^2+8x+16+4(x-1)(2x+3)$$
$$=x^2+8x+16+4(2x^2-2x+3x-3)$$
$$=9x^2+12x+4$$
$$=(3x+2)^2$$
$$m=\frac{x+4\pm (3x-2))}{2(x-1)}$$
$$m_1=\frac{4x+2}{2x-2}$$
$$m_2=\frac{-2x+6}{2x-2}$$
Is this right?
I don't know if the tag is right, so please don't be based on it.
|
$m^2-m-2=(m-2)(m+1)$ and $m^2+4m+3=(m+1)(m+3)$
Clearly $m+1=0$ is a solution
Else $(m-2)x=m+3\implies x=\dfrac{m+3}{m-2}=1+\dfrac5{m-2}$
If $x$ is an integer, $(m-2)$ must divide $5$
|
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|
coefficient of $x^{17}$ in the expansion of $(1+x^5+x^7)^{20}$ I found this questions from past year maths competition in my country, I've tried any possible way to find it, but it is just way too hard.
find the coefficient of $x^{17}$ in the expansion of $(1+x^5+x^7)^{20}$
(A)3400 (B)3410 (C)3420 (D)3430 (E)3440
so it would be $$x^{140} + ...... + 1$$
This requires binomial theorem and Multinomial theorem, but I'm not sure how to calculate it. Any tips or formula would be appreciate.
|
So if you think about
$$ (1 + x^5 + x^7)^{20} $$
That intuitively is just
$$ ((1 + x^5) + x^7) \times ((1 + x^5) + x^7) \times ((1 + x^5) + x^7) ... $$
Which can be expanded out term by term. By the Binomial Theorem as
$$ (1 + x^5)^{20} (x^7)^0 + \begin{pmatrix} 20 \\ 1\end{pmatrix}(1 + x^5)^{19}x^7 + \begin{pmatrix} 20 \\ 2\end{pmatrix}(1 + x^5)^{18}x^{14} ...$$
Now we can take each of the $$(1 + x^5)$$ terms and expand them as well.
Note that the number 17 can be expressed as sum in terms of multiples of 5 and 7 as
$$ 10 + 7$$
If we consider say $15$ in the sum it's too big, and same for $14$ likewise arguments can be made to show that $5$ by itself won't add to a positive multiple of 7 to make 17.
So that means every power of 17 is contained in the term
$$\begin{pmatrix} 20 \\ 1\end{pmatrix}(1 + x^5)^{19}x^7 $$
Of our binomial expression. We need to find the coefficient of $x^{10}$ in
$$ (1 + x^5)^{19} $$
Call that C. Then
$$ C \begin{pmatrix} 20 \\ 1\end{pmatrix}$$
Is the answer.
By Binomial Theorem:
$$ (1 + x^5)^{19} = 1 + \begin{pmatrix} 19 \\ 1\end{pmatrix}x^5 + \begin{pmatrix} 19 \\ 2\end{pmatrix}x^{10} ... $$
So then $C = \begin{pmatrix} 19 \\ 2\end{pmatrix} $
So the answer then is
$$\begin{pmatrix} 19 \\ 2\end{pmatrix} \begin{pmatrix} 20 \\ 1\end{pmatrix} $$
To get more advanced techniques (it is worthwhile to take a look at the multinomial theorems). They generalize these ideas for arbitrarily large sums raised to integer powers like the binomial theorem for 2 elements raised to a power.
Note that
$$\begin{pmatrix} 19 \\ 2\end{pmatrix} \begin{pmatrix} 20 \\ 1\end{pmatrix} = \frac{19!}{2!17!} 20 = 19 \times 9 \times 20 = 3420$$
|
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|
Digital Roots of Square Numbers Can anyone offer a proof of the following:
The digital root of a square number is always $1$, $4$, $7$ or $9$. (It is never $2$, $3$, $5$, $6$ or $8$.)
Digital root : Add the digits of a number until you get a single digit.
examples: The digital root of $144$ is $1+4+4 = 9$.
The digital root of $14289$ is $1+4+2+8+9 = 24 2+4 = 6$.
The digital root of $1428842$ is $1+4+2+8+8+4+2 = 29 2+9 = 11 1+1 = 2$.
Square number: A number whose square root is an integer.
Examples: $25$ $36$ $144$ $400$
$116$ is not a square number.
|
Mod $9$: $0^2 \equiv 3^2 \equiv 6^2 \equiv 0$, $1^2 \equiv 8^2 \equiv 1$, $2^2 \equiv 7^2 \equiv 4$, $4^2 \equiv 5^2 \equiv 7$. That's all!
|
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Complex integration by Cauchy's residue theorem
Evaluate the following integral by Cauchy's Residue Theorem
$$\int_C\frac{2z^2-z+1}{(2z-1)(z+1)^2}\,dz$$where , $C:r=2\cos \theta$ , $0\le \theta \le \pi.$
I have problem about the contour $C$.
Here, $r^2=4\cos^2 \theta=\frac{4x^2}{x^2+y^2}$ , as $\tan \theta =y/x$.
Then , $x^2+y^2=\frac{4x^2}{x^2+y^2}\implies x^2+y^2=\pm 2x\implies (x\pm 1)^2+y^2=1$. Thus we get two semicircles, which we take for the integration and why?
|
$$\int_{C}\frac{2z^2-z+1}{(2z-1)(z+1)^2}dz$$$$=\frac{1}{2}\int_{C}\frac{2z^2-z+1}{\left(z-\frac{1}{2}\right)(z+1)^2}dz$$
We see that the singularities $z=\frac{1}{2}$ & $z=-1$ are on the real axis inside the curve, $C: r=2\cos \theta$ hence there are the poles of first & second order respectively. Let's find out the residues at these poles as follows
$\color{red}{\text{Residue at}\ z=\frac{1}{2}}$
$$Res\left(f(1/2)\right)=\lim_{z\to \frac{1}{2}} \frac{2z^2-z+1}{(z+1)^2}$$ $$=\frac{2\left(\frac{1}{2}\right)^2-\frac{1}{2}+1}{\left(\frac{1}{2}+1\right)^2}=\frac{4}{9}$$ $\color{red}{\text{Residue at}\ z=-1}$
$$Res\left(f(-1)\right)=\lim_{z\to {-1}}\frac{d}{dz}\left(\frac{2z^2-z+1}{z-\frac{1}{2}}\right)$$ $$=\lim_{z\to {-1}}\left(\frac{\left(z-\frac{1}{2}\right)(4z-1)-(2z^2-z+1)(1)}{\left(z-\frac{1}{2}\right)^2}\right)$$ $$=\frac{\left(-1-\frac{1}{2}\right)(4(-1)-1)-(2(-1)^2-(-1)+1)}{\left(-1-\frac{1}{2}\right)^2}=\frac{2}{3}$$
Hence, using Cauchy Residue theorem, we get $$\frac{1}{2}\int_{C}\frac{2z^2-z+1}{\left(z-\frac{1}{2}\right)(z+1)^2}dz=\frac{1}{2}(2\pi i)\left(Res(f(1/2))+Res(f(-1))\right)$$ $$=\pi i\left(\frac{4}{9}+\frac{2}{3}\right)=\color{}{\frac{10\pi i}{9}}$$ $$\bbox[5px, border:2px solid#C0A000]{\int_{C}\frac{2z^2-z+1}{\left(2z-1\right)(z+1)^2}dz=\color{blue}{\frac{10\pi i}{9}}}$$
|
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|
Solve $x,y\in \mathbb{Z}$ Solve for $x,y\in \mathbb{Z}$
$$x^{6}=y^{2}+53$$
I tried but I couldn't complete
|
$$x^6 = y^2 + 53$$
$$x^6-y^2 = 53$$
From $x^6-y^2 = (x^3+y)(x^3-y)$
$$(x^3+y)(x^3-y) = 53$$
The only factors of $53$ are $1$ and $53$ so let: $$x^3-y = 1$$$$x^3+y = 53$$
Adding the $2$ equations we get: $$2x^3 = 54$$ $$x = 3$$$$y = 3^3-1 = 26$$
Our solutions are $x = 3$, $y = 26$.
But notice that in the original equation we are raising $x$ to the $6^\text{th}$ power and $y$ to the $2^\text{nd}$ power so the signs on $x$ and $y$ can be changed so we get $$\boxed{x = \pm3, y = \pm26}$$
|
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|
Find the least $N$ so there is no square
Find the least positive integer $N$ such that the set of $1000$ consecutive integers beginning with $1000 \cdot N$ contains no square of an integer.
Let $x^2$ appear before $1000N$ so:
$(x+1)^2 - x^2 > 1000 \implies x \ge 500$
Let $A = [1000N, 1000(N+1)]$
So I let $x=500$ then, $x^2 = 250000$, obviously this is impossible since the set has a square already.
So now I need to set some number, $k^2 \ge 501^2 \equiv 0 \pmod{1000}$
So I need to solve the quadratic residue:
$k^2 \equiv 0 \pmod{1000}$ with constraints for $k$.
Lets see: $500^2 = 250000$ and $501^2 = 251001$ and $502^2 = 252004$
$501^2 = 500^2 + 1000 + \sum_{k=1}^{1} 2k - 1$
$502^2 = 500^2 + 2(1000) + \sum_{k=1}^{2} 2k - 1$
The next $n$th square after $500$ is generalized:
$p = 500^2 + n(1000) + \sum_{k=1}^{n} 2k - 1$
Take $p \pmod{1000}$
$p \equiv n + \sum_{k=1}^{n} 2k - 1 \pmod{1000}$
$p \equiv n + n^2 \pmod{1000}$
$n(n+1) \equiv 0 \pmod{1000}$
$n \equiv 0$ and $n \equiv -1 \pmod{1000}$
But this method doesnt work properly. The final answer is: $N = 282$.
Please some hints only.
|
HINT :
As you wrote, we have $x\ge 500$.
Observe that
$$501^2=251001,\quad 502^2=252004,\quad 503^2=253009,\cdots$$
$$251\times 1000=251000,\quad 252\times 1000=252000,\quad 253\times 1000=253000,\cdots$$
Now we need to have
$$(500+a)^2-(250+a)\times 1000\gt 1000\Rightarrow a\gt 31$$
where $N=250+a$.
|
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Proving the inequality $\frac{a+b}{2} - \sqrt{ab} \geq \sqrt{\frac{a^2+b^2}{2}} - \frac{a+b}{2}$
Show that for any two positive real numbers $a$ and $b$, $\frac{a+b}{2} - \sqrt{ab} \geq \sqrt{\frac{a^2+b^2}{2}} - \frac{a+b}{2}$
My attempt: $(\sqrt a-\sqrt b)^2\geq0\\\frac{a+b}{2}\geq \sqrt{ab}$
$(a-b)^2\geq0\\a^2+b^2\geq2ab \\\sqrt{\frac{a^2+b^2}{2}}\geq\sqrt{ab}$
I do not know how to link them together. Appreciate any tips.
|
Run this backwards:
$$\frac{a+b}2-\sqrt{ab}\geq\sqrt{\frac{a^2+b^2}2}-\frac{a+b}2\\
a+b\geq\sqrt{ab}+\sqrt{\frac{a^2+b^2}2}\\
a^2+2ab+b^2\geq ab+\frac{a^2+b^2}2+2\sqrt{ab\frac{a^2+b^2}2}\\
a^2+2ab+b^2\geq4\sqrt{ab\frac{a^2+b^2}2}\\
(a+b)^4\geq 8(a^3b+ab^3)\\
(a-b)^4\geq0$$
|
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|
Find min & max of $f(x,y) = x + y + x^2 + y^2$ when $x^2 + y^2 = 1$ Problem: Find the maximum and minimal value of $f(x,y) = x + y + x^2 + y^2$ when $x^2 + y^2 = 1$.
Since $x^2 > x$ (edit $x^2 \geq x$) for all $x \in \mathbb{R}$, $f$ is bowl-ish with a minimal value in the bottom.
This is a critical point which means that we can set partial derivatives of $f$ equal to $0$ and try to solve for $x$ and $y$
$\nabla f = (1+2x, 1+2y) = (0,0) \implies (x,y) = (\frac{-1}{2}, \frac{-1}{2})$
So we get the minimal value $f(\frac{-1}{2}, \frac{-1}{2}) = \frac{-1}{2} + \frac{-1}{2} + (\frac{-1}{2})^2 \frac{-1}{2})^2 = -\frac{1}{2}$
But how about the maximal value? How does $x^2 + y^2 = 1$ restrict $f$?
|
By $MA\geq MG$ we have $2xy\leq x^2+y^2=1$. Note that $(x+y)^2-(x^2+y^2)=2xy$ and this implies that $(x+y)^2\leq 2$ . So, $-\sqrt{2}\leq x+y\leq \sqrt{2}$. Then $f(x,y)\leq 1+\sqrt{2}$ with equality for $x=y=\frac{\sqrt{2}}{2}$ and $f(x,y)\geq 1-\sqrt{2}$ with equality for $x=y=-\frac{\sqrt{2}}{2}$.
|
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|
Prove $ \ \frac{a}{x^3 + 2x^2 - 1} + \frac{b}{x^3 + x - 2} \ = \ 0 \ $ has a solution in $ \ (-1,1) $ If $a$ and $b$ are positive numbers, prove that the equation
$$\frac{a}{x^3 + 2x^2 - 1} + \frac{b}{x^3 + x - 2} = 0$$
has at least one solution in the interval $ \ (-1,1) \ $ .
The question is from the exercises section of a textbook chapter on limits/continuity.
I've been stumped on this one for a couple of days. I've been trying to calculate $\lim _{x \to -1}$ and $\lim _{x \to 1}$ and then show the function is continuous to show a root must lie in the interval. Factorising the denominators gives...
$$\frac{a}{(x+1)(x^2+x-1)} + \frac{b}{(x-1)(x^2+x+2)} = 0$$
So of course $x = 1$ and $x = -1$ are undefined and so the limits will be one-sided. Playing around with equation I haven't been able to find an equivalent function across $x \neq -1, x \neq 1$.
The only thing I have been able to show is
$$\frac{a}{b} = - \frac{(x+1)(x^2+x-1)}{(x-1)(x^2+x+2)}$$
and so
$$\lim _{x \to -1} \frac{a}{b} = 0, \lim _{x \to 1} \frac{b}{a} = 0$$
but I'm not sure if this is significant or I'm overthinking things. Could anyone point me in the right direction?
|
In the interval $(-1,1)$ your equation is equivalent to
$f(x):= a(x-1)(x^2+x+2)+b(x+1)(x^2+x-1)=0$
Now you can see that $f(x)$ is continuous and
$\lim\limits_{x\to -1}f(x)<0, \quad
\lim\limits_{x\to 1}f(x)>0$
Thus $f(x)=0$ has at least one solution (in the interval $(-1,1)$).
|
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|
Compare $A=\frac{1.0\,000\,004}{(1.0\,000\,006)^2}$ and $ B=\frac{(0.9\,999\,995)^2}{0.9\,999\,998}$ My work:
*
*$1.0\,000\,004 = 1+\frac{4}{10^7}=1+\frac{1}{125\cdot 10^6}$
*$ (1.0\,000\,006)^2=(1+\frac{6}{10^7})^2=(1+\frac{3}{5\cdot 10^6})^2$
*$ (0.9\,999\,995)^2=(\frac{9\,999\,995}{10^7})^2=(\frac{1\,999\,999}{2\cdot10^6})^2$
*$ 0.9\,999\,998=\frac{9.999\,998}{10^7}=\frac{4\,999\,999}{5\cdot10^6}$
The two fractions become:
$A=\dfrac{1+\frac{1}{125\cdot 10^6}}{(1+\frac{3}{5\cdot 10^6})^2}$ and $B=\dfrac{(\frac{1\,999\,999}{2\cdot10^6})^2}{\frac{4\,999\,999}{5\cdot10^6}}$
At this stage, I don't know how to continue.
Any help is appreciated. Thank you.
|
Let $a=\frac{1}{10^7}$.
Then, $$A=\frac{1+4a}{(1+6a)^2},\quad B=\frac{(1-5a)^2}{1-2a}$$
Now,
$$\begin{align}A-B&=\frac{1+4a}{(1+6a)^2}-\frac{(1-5a)^2}{1-2a}\\&=\frac{(1+4a)(1-2a)-(1-5a)^2(1+6a)^2}{(1+6a)^2(1-2a)}\\&=\frac{3 a^2 (20a (1-15 a)+17)}{(1+6a)^2(1-2a)}\end{align}$$
This is positive, so $A\gt B$.
|
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|
Finding $\frac{ \mathrm{d}y}{ \mathrm{d}x}$ when given that $\sqrt{1-x^2} + \sqrt{1-y^2} ={a}(x-y)$ Finding $\frac{ \mathrm{d}y}{ \mathrm{d}x}$ when given that $\sqrt{1-x^2} + \sqrt{1-y^2} = a(x-y)$. $a$ is a constant.
I have the final answer, which is $$\frac{ \mathrm{d}y}{ \mathrm{d}x} = \sqrt{\frac{1-y^2}{1-x^2}}.$$
But I've only been able to get till $$\frac{ \mathrm{d}y}{ \mathrm{d}x} = \frac{a\sqrt{1-x^2}+x}{a\sqrt{1-y^2}-y}.\sqrt{\frac{1-y^2}{1-x^2}}.$$
How can the two terms be cancelled off? Thanks.
|
start with the defining equation and use difference of squares ...
$$ a(x-y) = \sqrt{1-y^2}+\sqrt{1-x^2} $$
$$ a \frac{x^2-y^2}{x+y} = \frac{x^2-y^2}{\sqrt{1-y^2}-\sqrt{1-x^2}} $$
$$ a (\sqrt{1-y^2}-\sqrt{1-x^2}) = x+y $$
$$ a \sqrt{1-y^2}-y =a\sqrt{1-x^2}+x $$
so
$$ \frac{a\sqrt{1-x^2}+x}{ a \sqrt{1-y^2}-y }=1 $$
|
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|
Evaluate $\int \frac{a}{x(x^2-a^2)^{1/2}} dx$ I was trying to integrate $\int \frac{a}{x(x^2-a^2)^{1/2}} dx$ and by applying on what i saw on the formula of inverse trigonometric functions, there is formula like $\frac{1}{a}Arcsec\frac{u}{a}$ = $\int \frac{du}{u\sqrt(u^2-a^) }$ so my answer is $\frac{1}{a}Arcsec\frac{x}{a}$ but my friend said it should be something like $arcsin?$
|
To obtain the solution that the OP quotes
$$
\int \frac{a}{x\sqrt{x^2-a^2}}dx
$$
we then have
$$
\int \frac{a}{x^2\sqrt{1-\left(\frac{a}{x}\right)^2}}dx
$$
Here I used the fact
$$
\sqrt{x^2-a^2} = \sqrt{x^2\left(1-\frac{a^2}{x^2}\right)} = \sqrt{x^2}\sqrt{\left(1-\frac{a^2}{x^2}\right)} = x\sqrt{1-\left(\frac{a}{x}\right)^2}
$$
let $u = \frac{a}{x}$
we have
$$
-\int \frac{du}{\sqrt{1-u^2}} = -\arcsin(u) + C
$$
|
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|
Quadratic formula does not work If I put the equation: $5x^2-x-4 =0$ in the quadratic formula, than I get
$x = 1$ or $x = \frac{-4}{5}$
but the real zeros are: $x = -1$ or $x = \frac{4}{5}$
Can somebody explain me if the quadratic formula fails or me?
|
$5x^2-x-4$ does indeed have roots $x = 1, -\frac{4}{5}$, so you did it correctly.
To illustrate:
$$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
$$=\frac{1 \pm \sqrt{1^2-4*5*(-4)}}{2(5)}$$
$$=\frac{1 \pm \sqrt{81}}{10}$$
$$= \frac{1\pm 9}{10}$$
$$x = 1, -\frac{4}{5}$$
As Peter said if you had meant $5x^2+x-4$ then your roots would have been $x = -1, +\frac{4}{5}$.
|
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|
$\int\limits_{-1/2}^{1/2}(\sin^{-1}(3x-4x^3)-\cos^{-1}(4x^3-3x))dx=$ $\int\limits_{-1/2}^{1/2}(\sin^{-1}(3x-4x^3)-\cos^{-1}(4x^3-3x))dx=$
$(A)0\hspace{1cm}(B)\frac{-\pi}{2}\hspace{1cm}(C)\frac{\pi}{2}\hspace{1cm}(D)\frac{7\pi}{2}$
I tried and got the answer but my answer is not matching the options given.Is my method not correct?
$\int\limits_{-1/2}^{1/2}(\sin^{-1}(3x-4x^3)-\cos^{-1}(4x^3-3x))dx=\int\limits_{-1/2}^{1/2}(3\sin^{-1}(x)-3\cos^{-1}(x))dx$$=3\int\limits_{-1/2}^{1/2}(\sin^{-1}(x)-\cos^{-1}(x))dx=3\int\limits_{-1/2}^{1/2}(\frac{\pi}{2}-2\cos^{-1}(x))dx=\frac{3\pi}{2}-6\int\limits_{-1/2}^{1/2}\cos^{-1}(x)dx$
$=\frac{3\pi}{2}+6\int\limits_{2\pi/3}^{\pi/3}t \sin t dt=\frac{-3\pi}{2}$
|
We use the facts that
$$\arccos(-x)=\pi-\arccos(x)$$
and
$$\arccos(x)+\arcsin(x)=\pi/2.$$
We have
\begin{align}
& \int_{-1/2}^{1/2}\left(\arcsin\left(3x-4x^3\right)-\arccos\left(4x^3-3x\right)\right)\mathrm{d}x \\
& \quad = \int_{-1/2}^{1/2}\left(\arcsin\left(3x-4x^3\right)-\pi+\arccos\left(3x-4x^3\right)\right)\mathrm{d}x \\
& \quad = \int_{-1/2}^{1/2}\left(-\pi+\pi/2\right)\mathrm{d}x=-\pi/2.
\end{align}
|
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|
Proving $\frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \cdots + \frac{1}{n\cdot(n+2)} = \frac{3}{4} - \frac{(2n+3)}{2(n+1)(n+2)}$ by induction for $n\geq 1$ I'm having an issue solving this problem using induction. If possible, could someone add in a very brief explanation of how they did it so it's easier for me to understand?
$$\frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \cdots + \frac{1}{n\cdot(n+2)} = \frac{3}{4} - \frac{(2n+3)}{2(n+1)(n+2)}$$
How do I prove the above equation for all integers where $n\geq1$?
|
When dealing with a sum like you are here (summing $n$ terms for some general expression), I would almost always recommend that you use $\Sigma$-notation, for it tidies up a lot of the algebraic mess you have to deal with in your induction proof. With that in mind, you may write your claim as follows.
Claim: For any $n\geq 1$, the statement
$$
S(n) : \sum_{i=1}^n\frac{1}{i(i+2)}=\frac{3}{4}-\frac{2n+3}{2(n+1)(n+2)}
$$
is true.
Base step ($n=1$): $S(1)$ says that $\sum_{i=1}^1\frac{1}{i(i+2)}=\frac{1}{3}=\frac{3}{4}-\frac{5}{12}$, and this is true.
Induction step ($S(k)\to S(k+1)$): Fix some $k\geq 1$, and assume that
$$
S(k) : \sum_{i=1}^k\frac{1}{i(i+2)}=\frac{3}{4}-\frac{2k+3}{2(k+1)(k+2)}
$$
is true. To be proved is that
$$
S(k+1) : \sum_{i=1}^{k+1}\frac{1}{i(i+2)}=\frac{3}{4}-\frac{2k+5}{2(k+2)(k+3)}
$$
follows. Beginning with the left side of $S(k+1)$,
\begin{align}
\sum_{i=1}^{k+1}\frac{1}{i(i+2)}&= \sum_{i=1}^k\frac{1}{i(i+2)}+\frac{1}{(k+1)(k+3)}\tag{by defn.}\\[1em]
&= \left(\frac{3}{4}-\frac{2k+3}{2(k+1)(k+2)}\right)+\frac{1}{(k+1)(k+3)}\tag{by $S(k)$}\\[1em]
&= \frac{3}{4}-\frac{2k^2+7k+5}{2(k+1)(k+2)(k+3)}\tag{common denom.}\\[1em]
&= \frac{3}{4}-\frac{(2k+5)(k+1)}{2(k+1)(k+2)(k+3)}\tag{factor}\\[1em]
&= \frac{3}{4}-\frac{2k+5}{2(k+2)(k+3)},\tag{simplify}
\end{align}
one arrives at the right side of $S(k+1)$, thereby showing $S(k+1)$ is also true, completing the inductive step.
By mathematical induction, the claim $S(n)$ is true for all $n\geq 1$. $\blacksquare$
|
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|
Computing the limit of a series involving trigonometric identities Let $x_n\in(n\pi,(n+1)\pi)$ with $\tan(x_n)=x_n$. Can
$$s:=\sum_{n=1}^\infty \frac{1}{x_n^2}$$
be determined explicitly?
My ideas so far: First, I wrote $x_n$ as
$$x_n= \pi (n+z_n)$$
with $z_n\in(0,\frac{1}{2})$.
Using $\zeta(2)=\frac{\pi^2}{6}$ gives with replacing $z_n=\frac{1}{2}$ and $z_n=0$ lower and upper bound on $s$:
$$\frac{1}{2}-\frac{4}{\pi^2} <s< \frac{1}{6}.$$
Maybe it's useful to use the (not absolutely converging) identity
$$\pi\cot(\pi z) = \sum_{k\in\mathbb Z} \frac{1}{z+k}$$
since we have
$$s = \sum_{n=1}^\infty \cot(\pi z_n)^2.$$
|
At each $x_n$, the function $f(z) = \tan z - z$ has a simple zero. Therefore
$$g(z) = \frac{f'(z)}{z^2\cdot f(z)}$$
has a simple pole with residue $\frac{1}{x_n^2}$ there. Also, the negative solutions to $\tan z = z$ are the points $-x_n, \, n \in \mathbb{N}\setminus \{0\}$. The equation $\tan z = z$ has no non-real solutions. To see that, use the addition theorems of $\sin$ and $\cos$ to obtain
$$\tan (x+iy) = \frac{\sin x \cos x + i \sinh y\cosh y}{\cos^2 x + \sinh^2 y}.$$
For $x = 0$, we have $\tan (iy) = i\tanh y$, so $\tan (iy) = iy \iff y = 0$ since $\lvert\tanh y\rvert < \lvert y\rvert$ for $y\in \mathbb{R}\setminus \{0\}$, and for $x,y \in \mathbb{R}\setminus \{0\}$ we have
$$\biggl\lvert \frac{\sin x\cos x}{x}\biggr\rvert < 1 < \frac{\sinh y\cosh y}{y},$$
but $\tan z = z$ would imply
$$\frac{\sin x\cos x}{x} = \cos^2 x + \sinh^2 y = \frac{\sinh y\cosh y}{y},$$
which is incompatible with the above chain of inequalities. At the points $\bigl(k+\frac{1}{2}\bigr)\pi$ for $k\in \mathbb{Z}$, $f$ has a simple pole, hence $g(z)$ has a simple pole with residue $-\frac{1}{\bigl(k+\frac{1}{2}\bigr)^2\pi^2}$ there.
For $N \in \mathbb{N}$, let $C_N$ be the rectangular path with vertices $(\pm 1\pm i)(N+1)\pi$ in the plane, traversed in positive orientation. Then we have
$$\frac{1}{2\pi i} \int_{C_N} g(z)\,dz = \operatorname{Res}\bigl(g(z);0\bigr) + 2\sum_{n = 1}^N \frac{1}{x_n^2} - \frac{8}{\pi^2}\sum_{n = 0}^N \frac{1}{(2n+1)^2}.$$
By the standard estimate, we have
$$\lim_{N\to \infty} \int_{C_N} g(z)\,dz = 0$$
since $\frac{f'(z)}{f(z)} = \frac{\tan^2 z}{\tan z - z}$ is bounded on $C_N$ independently of $N$: $\lvert\tan z\rvert \to 1$ uniformly in $\operatorname{Re} z$ for $\lvert \operatorname{Im} z\rvert \to +\infty$, and for $\operatorname{Re} z = k\pi$, we have $\tan z = \tanh (\operatorname{Im} z)$ hence $\lvert \tan z\rvert < 1$. Thus we find
$$\sum_{n = 1}^\infty \frac{1}{x_n^2} = \frac{4}{\pi^2}\sum_{n = 0}^\infty \frac{1}{(2n+1)^2} -\frac{1}{2} \operatorname{Res}\bigl(g(z);0\bigr) = \frac{1}{2} -\frac{1}{2} \operatorname{Res}\bigl(g(z);0\bigr).$$
The MacLaurin expansion of $\tan$ yields
\begin{align}
\frac{\tan^2 z}{z^2(\tan z - z)} &= \frac{\bigl(1 + \frac{z^2}{3} + O(z^4)\bigr)^2}{\frac{z^3}{3}\bigl(1 + \frac{2z^2}{5} + O(z^4)\bigr)}\\
&= \frac{3}{z^3}\biggl( 1 + \frac{2z^2}{3} + O(z^4)\biggr)\biggl(1 - \frac{2z^2}{5} + O(z^4)\biggr)\\
&= \frac{3}{z^3}\biggl(1 +\frac{4z^2}{15} + O(z^4)\biggr),
\end{align}
so we have
$$\operatorname{Res}\bigl(g(z);0\bigr) = \frac{4}{5}$$
and therefore
$$\sum_{n = 1}^\infty \frac{1}{x_n^2} = \frac{1}{10}.$$
|
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|
Solve trigonometric inequality $ \sin x \sin 2x - \cos x \cos 2x > \sin 6x $ Solve this trigonometric inequality:
$$ \sin x \sin 2x - \cos x \cos 2x > \sin 6x $$
My steps:
$$ \cos x \cos 2x - \sin x \sin 2x < - \sin 6x $$
$$ \cos 3x < \sin (-6x)$$
$$ \cos 3x < \cos (\frac{\pi}{2}+6x) $$
From this we get:
$$ 3x > \dfrac{\pi}{2}+6x+2k\pi$$
$$ -3x > \dfrac{\pi}{2}+2k\pi$$
$$ x < -\dfrac{\pi}{6}+ \dfrac{2k\pi}{3}$$
and
$$ 3x < -\dfrac{\pi}{2} - 6x + 2k\pi$$
$$ 9x < -\dfrac{\pi}{2} + 2k\pi$$
$$ x < - \dfrac{\pi}{18} + \dfrac{2k\pi}{9} $$
as you can see the solution is not correct
|
Your error is that in general $\cos A < \cos B$ does not imply that $A > B$. (If you don't see this right away, try $A = \pi$ and $B = \dfrac{3\pi}{2}$).
Instead, try the following approach:
$ \cos x \cos 2x - \sin x \sin 2x < -\sin 6x $
$\cos 3x < -\sin 6x$
$\cos 3x + \sin 6x < 0$
$\cos 3x + 2\sin 3x\cos 3x < 0$
$\cos 3x(1+2\sin 3x) < 0$
So we need exactly one of $\cos 3x$ and $1+2\sin 3x$ to be negative.
We know that $\cos 3x < 0$ when $\dfrac{\pi}{2}+2\pi k < 3x < \dfrac{3\pi}{2}+2\pi k$ for some integer $k$. Also, $1+2\sin 3x < 0$ when $\sin 3x < -\dfrac{1}{2}$, i.e. $\dfrac{7\pi}{6}+2\pi k < 3x < \dfrac{11\pi}{6}+2\pi k$ for some integer $k$.
Can you finish the problem from here?
|
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|
Limit of $\frac{\sin^2(x^p)}{x^q+x^r}$ as $x \to 0^+$
Suppose $p,q,r \in \mathbb{R}$ with $p > 0$. Write (without proof) the limit of the following expression: $$\lim_{x \to 0^+} \frac{\sin^2(x^p)}{x^q+x^r}$$ In particular, find and prove the limit of the expression if $p=3$, $q=6$, $r=7$.
I got (through countless graphing on Desmos):
$$\lim_{x \to 0^+} \frac{\sin^2(x^p)}{x^q+x^r}=\begin{cases} 0 & \text{if either $q < 2p$ or $r < 2p$} \\ \frac 12 & \text{if $q=r=2p$} \\ 1 & \text{if either $q=2p$, $r > 2p$ or $q > 2p$, $r=2p$} \\ \infty & \text{if $q > 2p$, $r > 2p$} \end{cases}$$
Also, in the case of $p=3$, $q=6$, $r=7$, the expression becomes $$\lim_{x \to 0^+} \frac{\sin^2(x^3)}{x^6+x^7}.$$
I was able to prove that $$\lim_{x \to 0^+} \frac{\sin^2(x^3)}{x^6+x^7} = \lim_{x \to 0^+} \frac{(\sin(x^3))^2}{x^6(1+x)} \le \lim_{x \to 0^+} \frac{(x^3)^2}{x^6(1+x)}=\lim_{x \to 0^+}\frac 1{1+x} = 1.$$ Now, the graph of this function on Desmos tells me that, in fact, $$\lim_{x \to 0^+} \frac{\sin^2(x^3)}{x^6+x^7} = 1.$$ How can I prove the other inequality, to achieve this result? Namely, I have trouble with proving $$\lim_{x \to 0^+} \frac{\sin^2(x^3)}{x^6+x^7} \ge 1.$$
|
It is well known that $$\frac{\sin\left(x\right)}{x}\longrightarrow1
$$ at $x\rightarrow0
$, then $$\lim_{x\rightarrow0^{+}}\frac{\sin^{2}\left(x^{3}\right)}{x^{6}+x^{7}}=\lim_{x\rightarrow0^{+}}\frac{x^{6}}{x^{6}+x^{7}}=1
$$ no need to split in two inequalities.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Completing the square of $(x+a)(x+b)$ The problem is simple, to complete the square of $(x+a)(x+b)$. My calculations yield
$$\left(x+\frac{a+b}{2}\right)^2-\frac{(a+b)^2}{4}+ab,$$
But the textbook's answer is different ("problem 361", at the bottom of the page):
$$\left(x+\frac{a+b}{2}\right)^2-\frac{(a-b)^2}{4}$$
Did I do anything the wrong way?
$$(x+a)(x+b)=x^2+xb+ax+ab=x^2+x(a+b)+ab=$$
$$=\left(x^2+2*\frac{a+b}{2}*x+\left(\frac{a+b}{2}\right)^2\right)-\left(\frac{a+b}{2}\right)^2+ab=$$
$$=\left(x+\frac{a+b}{2}\right)^2-\frac{(a+b)^2}{4}+ab$$
|
$\displaystyle \bf{(x+a)\cdot (x+b)} = x^2+(a+b)x+ab$
$\displaystyle = \underbrace{x^2+(a+b)x+\left(\frac{a+b}{2}\right)^2}+\underbrace{ab-\left(\frac{a+b}{2}\right)^2} = \left[x+\frac{a+b}{2}\right]^2-\left(\frac{a-b}{2}\right)^2$
So your answer is Right.
|
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|
Being careful with terms of infinite sums $ cos(x): = \sum_{k=0}^\infty \frac{(-1)^nx^{2n}}{2n!}$
$=1- \frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}...$
I would like to show that for $x \in [0,2]$
$cos(x) \leq 1- \frac{x^2}{2!}+\frac{x^4}{4!}$
This means the stuff from $-\frac{x^6}{6!}$ onwards must sum to something negative.
So I reckon we can group that stuff into pairs : $(-\frac{x^6}{6!}+\frac{x^8}{8!})+ (-\frac{x^{10}}{10!}+\frac{x^{12}}{12!})...$ and show each bracket is negative.
Indeed : $(-\frac{x^{2n}}{(2n)!}+\frac{x^{2n+2}}{(2n+2)!})= \frac{x^{2n}}{(2n)!}[-1+\frac{x^{2}}{(2n+1)(2n+2)}]$ and the stuff in the square bracket is certainly negative for $x \in [0,2] $ .
But is this legal? Can I pair terms like this? Why? I am sure I have seen a case where grouping terms has given different answers to the evaluation of an infinite sum. If you know of a case I am talking about could you please give it as an example. It would help me understand better if I saw when it was not allowed and when it was.
|
The thing you are talking about is called rearrangement of terms of a series. For example
$$
1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5} + \dots = \sum_{k=1}^\infty (-1)^{k+1} \frac{1}{k} = \log2 \text{.}
$$
On the other hand
$$
1-\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{6}-\frac{1}{8}+\frac{1}{5}-\frac{1}{10}-\frac{1}{12}+\dots=\\
(1-\frac{1}{2})-\frac{1}{4}+(\frac{1}{3}-\frac{1}{6})-\frac{1}{8}+(\frac{1}{5}-\frac{1}{10})-\frac{1}{12}+\dots=\\
\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{10}-\frac{1}{12} + \dots = \frac{1}{2} \log 2.$$
Now the operation you have made with your series does not change its arrangement, so it does not change its sum. Hope that helps.
|
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|
Find $\int \frac{5x^4+4x^5}{(x^5+x+1)^2}$ $\displaystyle \int \frac{5x^4+4x^5}{(x^5+x+1)^2}$
Since in the denominator of the integrand,$(x^5+x+1)^2$ is there.So the answer must be in the form $\displaystyle \frac{f(x)}{(x^5+x+1)}$, but i could not figure out what $f(x)$ i should suppose.I could not imagine any other way out.Please help...
|
Let $\displaystyle \frac{5x^4+4x^5}{(x^5+x+1)^2}dx = \frac{d}{dx}\left[\frac{ax+b}{(x^5+x+1)}\right] = \frac{(x^5+x+1)\cdot a-(ax+b)\cdot (x^5+x+1)}{(x^5+x+1)^2}$
Now Equate above equation, and get the value of $a$ and $b$
|
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|
Calculate $\lim_{(x,y,z)\to(0,0,0)}\frac{xy+yz+zx}{x^2+y^2+z^4}$ $$\lim_{(x,y,z)\to(0,0,0)}\frac{xy+yz+zx}{x^2+y^2+z^4}$$
In order to calculate this limit, I did:
$$\lim_{(x,y,z)\to(0,0,0)}\left(y\frac{x}{x^2+y^2+z^4}+z\frac{y}{x^2+y^2+z^4}+x\frac{z}{x^2+y^2+z^4}\right)$$
then, in each of these terms, one part is limited by $1$, and the other goes to $0$, then the entire thing goes to $0$
Am I rigth?
|
Let $y=nx$, $z=mx$. So,
$$
\frac{xy+yz+xz}{x^2 + y^2 + z^4} = \frac{n+nm+m}{1 + n^2 + m^4x^2}.
$$
Limit depends on $n$ and $m$; so, it doesn't exists.
|
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|
To compute improper integral $\int_3^{5}\frac{x^{2}\, dx}{\sqrt{x-3}{\sqrt{5-x}}}$ I am given improper integral as
$$\int_3^{5}\frac{x^{2}}{\sqrt{x-3}{\sqrt{5-x}}}dx$$
DOUBT
I see that problem is at both the end points, so i need to split up the integral. But problem seems to me that when i split up integral one of terms in denominator which is ${5-x}$ becomes negative. So how do i split up the integral ? Kindly help
Thanks
|
Notice, the following
$$x^2=A(x-3)(5-x)+B(8-2x)+C$$
On solving we get
$$x^2=-(x-3)(5-x)-4(8-2x)+17$$
Now, we have
$$\int_{3}^{5}\frac{x^2dx}{\sqrt{(x-3)}\sqrt{(5-x)}}$$
$$=\int_{3}^{5}\frac{(-(x-3)(5-x)-4(8-2x)+17)dx}{\sqrt{(x-3)}\sqrt{(5-x)}}$$
$$=-\int_{3}^{5}\frac{(x-3)(5-x)dx}{\sqrt{(x-3)}\sqrt{(5-x)}}-4\int_{3}^{5}\frac{(8-2x)dx}{\sqrt{(x-3)}\sqrt{(5-x)}}+17\int_{3}^{5}\frac{dx}{\sqrt{(x-3)}\sqrt{(5-x)}}$$
$$=-\int_{3}^{5}\sqrt{(x-3)}\sqrt{(5-x)}-4\int_{3}^{5}\frac{(8-2x)dx}{\sqrt{8x-x^2-15}}+17\int_{3}^{5}\frac{dx}{\sqrt{1-(x-4)^2}}$$
$$=-\int_{3}^{5}\sqrt{1-(x-4)^2}-4\int_{3}^{5}\frac{(8-2x)dx}{\sqrt{8x-x^2-15}}+17\int_{3}^{5}\frac{dx}{\sqrt{1-(x-4)^2}}$$
$$=-\frac{1}{2}\left[(x-4)\sqrt{1-(x-4)^2}+\sin^{-1}\left(x-4\right)\right]_{3}^{5}-4\left[2\sqrt{1-(x-4)^2}\right]_{3}^{5}+17\left[\sin^{-1}(x-4)\right]_{3}^{5}$$
$$=-\frac{1}{2}(\pi)-8(0)+17(\pi)=\frac{33\pi}{2}$$
|
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|
Modular maths: How do I find the remainder? How do I find the remainder of $5^{22} \pmod{25}$?
And also how do I find the remainder of $3^{16} + 7 \pmod{5}$?
|
Well $5^3 = 125 \equiv -1 \pmod {7}$ because $7 \mid 125 - (-1)$
And since ${5}^{3} \equiv -1 \pmod {7}$ then ${(5^3)}^{7} \equiv (-1)^7 \pmod{7}$ and so $5^{21} \equiv -1 \pmod{7}$ and so $5^{22} \equiv -5 \pmod 7$. But since $-5 \equiv 2 \pmod 7$ then $5^{22} \equiv 2 \pmod 7$ as well
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
How is $\left|\frac{xy}{\sqrt{x^2+y^2}}\right| \leq \frac{\sqrt{|xy|}}{\sqrt{2}}$ $$\left|\frac{xy}{\sqrt{x^2+y^2}}\right| \leq \frac{\sqrt{|xy|}}{\sqrt{2}}$$
Does this apply in general, because here in this example i have $x \to 0, y \to 0$. Some inequality is used I believe to prove this one, but I do not see which.
|
$$\begin{align*}
(|x|-|y|)^2 &\ge 0\\
x^2 - 2|xy| + y^2 &\ge 0\\
x^2 + y^2 &\ge 2|xy|\\
\sqrt{x^2+y^2} &\ge \sqrt{2|xy|}\\
\frac{1}{\sqrt2} &\ge \frac{\sqrt{|xy|}}{\sqrt{x^2+y^2}}\\
\frac{\sqrt{|xy|}}{\sqrt2} &\ge \frac{|xy|}{\sqrt{x^2+y^2}}\\
\end{align*}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Factoring Quadratic equation I am trying to factor $9x^2-6x+1$ after finding the roots, I am using the following formula $a(x-x_1)(x-x_2)$ in this case there is just one root ($\frac{1}{3}$)
How do I know that the answer is $(3x-1)^2$ and not just $(3x-1)$?
|
Using the quadratic formula for $ax^2+bx+c=0$ we get: $$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
In your case we have $x_1=\frac{1}{3}$ and $x_2=\frac{1}{3}$, hence we can write $$9x^2-6x+1=9(x-x_1)(x-x_2)=9\left(x-\frac{1}{3}\right)\cdot\left(x-\frac{1}{3}\right)=9\left(x-\frac{1}{3}\right)^2.$$
|
{
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|
Finding Volume of Rugby Ball I am asked to find the volume of rugby ball whose surface is given by the ellipsoid:
$$\frac{x^2}{4} + \frac{y^2}{4} + \frac{z^2}{9} = 1$$
I am having trouble figuring out which coordinate system I should use. Is it possible to solve the triple integral of the volume by just using cartesian co-ordinates, without making conversions to the spherical or cylindrical coordinate system?
|
Notice,
In general, the volume of ellipsoid $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$ is $$=\frac{4\pi}{3}(abc)$$
Hence, the volume of ellipsoid $\frac{x^2}{4}+\frac{y^2}{4}+\frac{z^2}{9}=1\iff \frac{x^2}{2^2}+\frac{y^2}{2^2}+\frac{1^2}{3^2}=1$ is $$=\frac{4\pi}{3}(2\cdot2\cdot 3)$$ $$=\frac{48\pi}{3}$$ $$=\color{blue}{16\pi}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluation of $\int\frac{x^7+2}{\left(x^2+x+1\right)^2}dx$
Evaluation of $$\int\frac{x^7+2}{\left(x^2+x+1\right)^2}dx$$
$\bf{My\; Try::}$ Let $$\displaystyle \mathop{I = \int\frac{x^7+2}{(x^2+x+1)^2}}dx = \int\frac{(x^7-1)+3}{(x^2+x+1)^2}dx$$
$$\mathop{\displaystyle = \int\frac{x^7-1}{(x^2+x+1)^2}}+\displaystyle \int\frac{3}{(x^2+x+1)^2}dx$$
Now Using the formula $$\bullet x^7-1 = (x-1)\cdot \left[x^6+x^5+x^4+x^3+x^2+x+1\right]$$
So $$\bullet \; (x^7-1) = (x-1)\left[(x^4+x)\cdot (x^2+x+1)+1\right]$$
So we get $$\displaystyle I = \int\frac{(x-1)\cdot (x^4+x)\cdot (x^2+x+1)}{(x^2+x+1)^2}dx+\frac{1}{2}\int\frac{2x-2}{(x^2+x+1)^2}+3\int\frac{1}{(x^2+x+1)^2}dx$$
$$\displaystyle I = \underbrace{\int\frac{(x-1)(x^4+x)}{x^2+x+1}dx}_{J}+\underbrace{\frac{1}{2}\int\frac{(2x+1)}{(x^2+x+1)^2}dx}_{K}+\underbrace{\frac{3}{2}\int\frac{1}{(x^2+x+1)^2}dx}_{L}$$
Now $$\displaystyle J = \int\frac{(x-1)(x^4-x)+2x(x-1)}{(x^2+x+1)}dx = \int(x^3-2x^2+x)dx+\int\frac{2x^2-2x}{x^2+x+1}dx$$
Now we can solve the integral Using the formulae.
My question is can we solve it any other way, If yes then plz explain here
Thanks
|
HINT:
Notice, we have $$\int\frac{x^7+2}{(x^2+x+1)^2}dx$$
$$=\int\frac{x^7+2}{\left(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right)^2}dx$$
Let, $$x+\frac{1}{2}=t\implies dx=dt$$
$$=\int\frac{\left(t-\frac{1}{2}\right)^7+2}{\left(t^2+\frac{3}{4}\right)^2}dt$$
Use reduction to solve further
|
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|
$\arctan (x) + \arctan(1/x) = \frac{\pi}{2}$
How can I show that $\arctan (x) + \arctan(1/x) =\frac{\pi}{2}$?
I tried to let $x = \tan(u)$. Then
$$ \arctan(\tan(u)) + \arctan(\tan(\frac{\pi}{2} - x)) = \frac{\pi}{2}$$
but it does not seem useful.
I'd appreciate most a proof that gives intuition and / or uses geometric
insight.
|
Let $f(x)=\arctan(x)+\arctan(\frac1x)$. Then $f'(x)=\frac{1}{1+x^2}+\frac{1}{1+1/x^2}\frac{-1}{x^2}=\frac{1}{1+x^2}-\frac{1}{1+x^2}=0$ so $f(x)$ is constant for $x>0$. Then note that $f(1)=\frac{\pi}{2}$ and thus $f(x)=\frac{\pi}{2}$.
There is a discontinuity at $x=0$, so the derivative only makes sense for $x\neq 0$. When $x<0$, you can check that $f(-1)=-\frac{\pi}{2}$. $f(x)$ has the same derivative so for $x<0$, $f(x) -\frac{\pi}{2}$
|
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|
How to find $\#\{1\le x\le 5^k:5^k|(x^4-1)\}$? Find $\#\{1\le x\le 5^k:5^k|(x^4-1)\}$. I am not so sure how it is done, nor am I completely sure if it is about any specific $k$ or all of them together. What I did arrive at, not being really sure, is:
If $k=0$, the answer is $1$, I guess, so suppose $k>1$.
I need to have $x^4\equiv 1 \pmod{5^k}$ satisfied. I know it is for $x\equiv 1 \pmod{5^k}$. Since $5^k=5^{k-1}\cdot 4+5^{k-1}=4\cdot5^{k-1}+4\cdot5^{k-2}+5^{k-3}=4(5^{k-1}+...+5^2+5)+5\equiv 1 \pmod4$, then by a theorem taught in class, $x\equiv -1$ is a primitive root modulo $5^k$. That is, $x=5^k-1\in \{1\le x\le 5^k:5^k|(x^4-1)\}$. So $\#\{1\le x\le 5^k:5^k|(x^4-1)\}\ge 2$. I am stuck at the moment. I don't know how to find more $x$ solutions such that $x^4\equiv 1 \pmod{5^k}$. How have my argues been so far? I could really use your guiding.
|
First note that if $x$ works for $k+1$ then $x$ (reduced modulo $5^k$) also works for $k$, so every solution for $k+1$ comes from a solution for $k$.
Now suppose we have a solution for $k$, that is, $a^4-1\equiv0\bmod{5^k}$. Let $b=a+5^kt$; then there is exactly one value of $t$ in $\{\,0,1,2,3,4\,\}$ such that $b$ is a solution for $k+1$. We see this as follows.
$b^4-1=a^4+4a^35^kt+5^{2k}c-1$ for some $c$, so dividing $b^4-1\equiv0\bmod{5^{k+1}}$ through by $5^k$ we get $${a^4-1\over5^k}+4a^3t\equiv0\bmod5$$ Since $a\not\equiv0\bmod5$, this equation has a unique solution $t$, and we are done.
All I have done here is I have gone through the standard proof of Hensel's Lemma for this particular congruence.
|
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|
Is $\sum_{n=1}^\infty \frac{m}{(n+m)^2}$ bounded for all $m\in\mathbb{N}$? I'm trying to figure out if there is a finite constant $C$ such that $\sum_{n=1}^\infty \frac{m}{(n+m)^2}\leq C$ for all $m\in\mathbb{N}$.
I can see that $\sum_{n=1}^\infty \frac{m}{(n+m)^2}\leq\sum_{n=1}^\infty \frac{m}{n^2}=mc$ for some finite constant $c$, but this is a weaker statement.
If I naively replace integers by reals and sums by integrals then I feel like the original series should be bounded.
|
By the integral comparison test,
$\sum_{n=1}^\infty \frac{1}{(n+m)^2}
=\sum_{n=m+1}^\infty \frac{1}{n^2}
\approx \int_m^{\infty} \frac{dx}{x^2}
=\frac{-1}{x}\big|_m^{\infty}
=\frac{1}{m}
$,
so
$\sum_{n=1}^\infty \frac{m}{(n+m)^2}
\approx 1
$.
To be more precise,
$\int_m^{\infty} \frac{dx}{x^2}
> \sum_{n=m+1}^\infty \frac{1}{n^2}
>\int_{m+1}^{\infty} \frac{dx}{x^2}
$,
so
$\frac{1}{m}
> \sum_{n=m+1}^\infty \frac{1}{n^2}
>\frac{1}{m+1}
$
or
$1
>\sum_{n=m+1}^\infty \frac{m}{n^2}
=\sum_{n=1}^\infty \frac{m}{(n+m)^2}
>(1-\frac1{m+1})
$.
|
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|
Solve in $\mathbb{Q}$ the equation $x^2-(\sqrt{2}+1)x+\sqrt{2}=0$. Solve in $\mathbb{Q}$ the equation $x^2-(\sqrt{2}+1)x+\sqrt{2}=0$
Somebody can help me? I dont remember how to do.
|
Using Quadratic Formula
If $ax^2+bx+c=0\;,$ Then $\displaystyle x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$
Here $a = 1$ and $b=-\left(\sqrt{2}+1\right)$ and $c=\sqrt{2}$
So $$\displaystyle x = \frac{(\sqrt{2}+1)\pm \sqrt{2+1+2\sqrt{2}-4\sqrt{2}}}{2}$$
So $$\displaystyle x = \frac{(\sqrt{2}+1)\pm (\sqrt{2}-1)}{2} = \sqrt{2}\;\;,1$$
|
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|
Solving a rational equation with multiple and nested fractions
This is the equation to solve: $\dfrac{\dfrac{x+\dfrac{1}{2}} {\dfrac{1}{2}+\dfrac{x}{3}}}{\dfrac{1}{4}+\dfrac{x}{5}}=3$
What I did:
$x+\dfrac{1}{2}=\dfrac{2x+1}{2}$
$\dfrac{x}{3}+\dfrac{1}{2}=\dfrac{2x+3}{6}$
$\dfrac{x}{5}+\dfrac{1}{4}=\dfrac{4x+5}{20}$
$\dfrac{2x+1}{2}\div \dfrac{2x+3}{6} = \dfrac{2x+1}{2}\times \dfrac{6}{2x+3}= \dfrac{6x+3}{2x+3}$
$\dfrac{6x+3}{2x+3}\div\dfrac{4x+5}{20}=\dfrac{6x+3}{2x+3}\times\dfrac{20}{4x+5}=3$
$\implies\dfrac{120x+60}{(2x+3)(4x+5)}=3$
I know how to solve the equation. But right now I tried several times and I got wrong answers.
So I appreciate your help. Thank you.
|
$$\frac { \frac { 2x+1 }{ \frac { 2 }{ \frac { 3+2x }{ 6 } } } }{ \frac { 5+4x }{ 20 } } =3\quad \Rightarrow \frac { \frac { 2x+1 }{ 2 } \cdot \frac { 6 }{ 2x+3 } }{ \frac { 5+4x }{ 20 } } =3\quad \Rightarrow \frac { \frac { 2x+1 }{ 2x+3 } }{ \frac { 5+4x }{ 20 } } =1\Rightarrow \frac { 2x+1 }{ 2x+3 } =\frac { 5+4x }{ 20 } \Rightarrow \\ \Rightarrow 8{ x }^{ 2 }-18x-5=0\Rightarrow x=\frac { 9\pm 11 }{ 8 } $$
|
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|
Prove that $(1)\frac{1}{2}\leq\int_{0}^{2}\frac{dx}{2+x^2}\leq\frac{5}{6}$ $(2)2e^{-1/4}<\int_{0}^{2}e^{x^2-x}dx<2e^2$ Prove that $(1)\frac{1}{2}\leq\int_{0}^{2}\frac{dx}{2+x^2}\leq\frac{5}{6}$
$(2)2e^{-1/4}<\int_{0}^{2}e^{x^2-x}dx<2e^2$
I tried to prove it but my answer is not correct.
For first part,As $0\leq x\leq2\Rightarrow 2\leq x^2+2\leq6\Rightarrow\frac{1}{6}\leq\frac{1}{x^2+2}\leq\frac{1}{2}\Rightarrow\frac{1}{6}\int_{0}^{2}1dx\leq\int_{0}^{2}\frac{dx}{2+x^2}\leq\frac{1}{2}\int_{0}^{2}1dx$$\Rightarrow\frac{1}{3}\leq\int_{0}^{2}\frac{dx}{2+x^2}\leq1$
In the second part,$0\leq x\leq2\Rightarrow -2<x^2-x<4\Rightarrow e^{-2}<e^{x^2-x}<e^4\Rightarrow2e^{-2}<\int_{0}^{2}e^{x^2-x}dx<2e^4$
Where have i gone wrong?What is the correct method to solve it.
|
$(i)$ Let $$\displaystyle I = \int_{0}^{2}\frac{1}{2+x^2}dx = \int_{0}^{1}\frac{1}{2+x^2}dx+\int_{1}^{2}\frac{1}{2+x^2}dx\geq \int_{1}^{2}\frac{1}{3}dx+\int_{0}^{1}\frac{1}{6}dx \geq \frac{1}{2}$$
And $$\displaystyle I = \int_{0}^{2}\frac{1}{2+x^2}dx = \int_{0}^{1}\frac{1}{2+x^2}dx+\int_{1}^{2}\frac{1}{2+x^2}dx\leq \int_{0}^{1}\frac{1}{2}dx+\int_{1}^{2}\frac{1}{3}dx \leq\frac{5}{6}$$
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"timestamp": "2023-03-29T00:00:00",
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|
If $x^4+7x^2y^2+9y^4=24xy^3$,show that $\frac{dy}{dx}=\frac{y}{x}$ If $x^4+7x^2y^2+9y^4=24xy^3$,show that $\frac{dy}{dx}=\frac{y}{x}$
I tried to solve it.But i got stuck after some steps.
$x^4+7x^2y^2+9y^4=24xy^3$
$4x^3+7x^2.2y\frac{dy}{dx}+7y^2.2x+36y^3.\frac{dy}{dx}=24x.3y^2\frac{dy}{dx}+24y^3$
$\dfrac{dy}{dx}=\dfrac{24y^3-4x^3-14y^2x}{14x^2y+36y^3-72xy^2}$
How to move ahead?Or there is some other elegant way to solve it.
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Another way to view this is that the differential equation $ \ \frac{dy}{dx} \ = \ \frac{y}{x} \ $ is separable and has the general solution given by
$$ \ \int \ \frac{dy}{y} \ = \ \int \ \frac{dx}{x} \ \ \Rightarrow \ \ \ln |y| \ = \ \ln |x| \ + \ C \ \ \Rightarrow \ \ y \ = \ A \ x \ \ , $$
with $ \ A \ $ being any real number. This is reasonable, since the differential equation can be interpreted as saying that the slope of the solution curve(s) at any point is $ \ \frac{y}{x} \ $ . The origin $ \ ( 0 , \ 0 ) \ $ is a point on the curve of the given equation, so the slope of the curve everywhere must be constant.
We can solve for the constant(s) by inserting our general solution into the curve equation to obtain
$$x^4 \ + \ 7x^2y^2 \ + \ 9y^4 \ - \ 24xy^3 \ = \ 0 \ \ \rightarrow \ \ x^4 \ + \ 7x^2 \ (Ax)^2 \ + \ 9 \ (Ax)^4 \ - \ 24x \ (Ax)^3 \ = \ 0 $$
$$ \Rightarrow \ \ x^4 \ ( \ 9A^4 \ - \ 24A^3 \ + \ 7 A^2 \ + 1 \ ) \ = \ 0 \ \ . $$
Other than at the origin ( $ \ x = 0 \ $ ) , the curve has two real slopes, $ \ A \ \approx \ 2.3230 \ $ and $ \ A \ \approx \ 0.54349 \ $ (the other two solutions for $ \ A \ $ are a complex conjugate pair). If we plot the equation $ \ x^4 \ + \ 7x^2y^2 \ + \ 9y^4 \ - \ 24xy^3 \ = \ 0 \ $ , we find that we in fact have a degenerate quartic curve, appearing as a pair of straight lines (the "wiggles" near the origin are a graphing artifact, as they are present at all plotting scales in the same way).
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|
Find the value of function at $\,x=5$ If $\,f(x)\,$ is a non-constant polynomial of $\,x\,$ such that $\,f\left(x^3\right)-f\left(x^3-2\right)=f^2\left(x\right)+12\,$ is true for all $\,x\,$ then find the value of $\,f\left(5\right).\,$
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Given
$$
f(x^3) - f(x^3-2) = f(x)^2 + 12. \tag 1
$$
Let $f(x) = P_n(x)$, where $P_n(x)$ is a polynomial of degree $n$.
LHS is of degree $3(n-1)$ and RHS is of degree $2n$.
Whence
$$
\bbox[16px,border:2px solid #800000] { f(x) = P_3(x) }. \tag 2
$$
We can differentiate $(1)$ and we obtain
$$
\begin{eqnarray}
3 x^2 \Big[ f'(x^3) - f'(x^3-2) \Big] &=& 2 f(x) f'(x)\\\\
3 \Big[ 2 x + 3 x^4 \Big] \Big[ f'(x^3) - f'(x^3-2) \Big]
&=& 2 f(x) f''(x) + 2f'(x)^2\\\\
3 \Big[ 2 + 18 x^3 + 9 x^6 \Big] \Big[ f'(x^3) - f'(x^3-2) \Big]
&=& 2 f(x) f'''(x) + 6 f'(x) f''(x)
\end{eqnarray}
$$
Put in $x=0$ so we get
$$
\begin{eqnarray}
f(0) f'(0) &=& 0\\\\
f(0) f''(0) + f'(0)^2 &=& 0\\\\
f(0) f'''(0) + 3 f'(0) f''(0) &=& 3 \Big[ f'(0) - f'(-2) \Big]
\end{eqnarray}
$$
However, $f'(0) \ne 0$ gives a contradiction,
as $f(0)=0$ and then $f(0) f''(0) + f'(0)^2=0$.
So
$$
\bbox[16px,border:2px solid #800000] { f'(0) = 0 } \tag 3
$$
If $f(0)=0$ we get $f'(0) = f'(-2) = 0$, which implies that $f(x) = c$ - which is to be excluded.
So
$$
\bbox[16px,border:2px solid #800000] { f(0) \ne 0 } \tag 4
$$
As $f(0) f''(0) + f'(0)^2 = 0$.
So
$$
\bbox[16px,border:2px solid #800000] { f''(0) = 0 } \tag 5
$$
And we also get
$$
f'''(0) = - 3\frac{ f'(-2) }{ f(0) } \tag 6
$$
From $(2)$ $(3)$ $(4)$ and $(5)$ follows that
$$
f(x) = p x^3 + q
\tag 7
$$
From $(7)$ follows
$$
6 p = - \frac{36p}{q}, \quad \Rightarrow \quad q = -6. \tag 8
$$
Put in $x=0$ in $(1)$ and we get
$$
f(0) - f(-2) = f(0)^2 + 12.
$$
Whence
$$
- 6 + 8 p + 6 = 6^2 + 12, \quad \Rightarrow \quad p = 6. \tag 9
$$
So we obtain
$$
\bbox[16px,border:2px solid #800000] { f(x) = 6 x^3 - 6 }
\quad \Rightarrow \quad
\bbox[16px,border:2px solid #800000] {f(5) = 6 \times 5^3 - 6 = 744 }
$$
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|
Proving uniqueness of solutions to $\sin^2A + \sin^2B = \sin (A+B)$ without using multivariable calculus In the course of solving a trigonometric problem (see $a^2+b^2=2Rc$,where $R$ is the circumradius of the triangle.Then prove that $ABC$ is a right triangle), in one approach the following equation needed to be solved:
$$\boxed{\sin^2A + \sin^2B = \sin (A+B)}$$
subject to $A\in(0,\frac{\pi}{2}),B\in(0,\frac{\pi}{2}),\pi-(A+B) > \max(A,B)$, i.e. $A$ and $B$ are the two angles of a triangle not opposite the longest side.
Clearly, any $A,B\:|\:A+B=\frac{\pi}{2}$ is a family of solutions. Since multivariable calculus is presumably beyond the level of the original problem:
How to prove that there are no other solutions $\underline{\text{without}}$ using multivariable calculus?
[I don't think a trig identity will suffice as there are other solutions if the restrictions on $A,B$ are relaxed.]
(For completeness - using multivariable calculus)
Part 1
Proof that $\sin^2A + \sin^2B < \sin (A+B)$ over region $R_1=\{0<A,B\land A+B<\frac{\pi}{2}\}$.
Consider $$f(x,y)=\sin^2x + \sin^2y - \sin (x+y)$$
Then
$$\begin{align}
f_x &= \sin 2x - \cos(x+y) \\
f_y &= \sin 2y - \cos(x+y) \\
f_{xx} &= 2\cos 2x + \sin(x+y) \\
f_{yy} &= 2\cos 2y + \sin(x+y) \\
f_{xy} &= \sin(x+y) &= f_{yx} \\
\end{align}$$
For local extrema we require $f_x=f_y=0$. But
$$f_x=f_y=0 \implies \sin2x=\sin2y \implies y=x \lor y=\frac{\pi}{2}-x$$
Exclude $y=\frac{\pi}{2}-x$ as it violates the restriction that $x+y<\frac{\pi}{2}$.
If $y=x$, $f_x=0 \implies \sin2x=\sin2y \implies x=y=\frac{\pi}{8}$.
The determinant of the Hessian is
$$D(x,y) = \begin{vmatrix} f_{xx} & f_{xy}\\ f_{yx} & f_{yy} \end{vmatrix} = f_{xx}f_{yy} - f_{xy}f_{yx} = f_{xx}f_{yy} - (f_{xy})^2$$
At $P(\frac{\pi}{8},\frac{\pi}{8})$, we have
$$\begin{align}
f_{xx}&=2\cos\frac{\pi}{4}+\sin\frac{\pi}{4}=\frac{3}{\sqrt2} \\
f_{yy}&=2\cos\frac{\pi}{4}+\sin\frac{\pi}{4}=\frac{3}{\sqrt2} \\
f_{xy}&=\sin\frac{\pi}{4}=\frac{1}{\sqrt2} \\
D(x,y)&=\frac{9}{2}-\frac{1}{2}=4
\end{align}$$
Since both $f_{xx}$ and $D$ are positive at $P$, this is a local minimum (with $f(x,y)|_P=1-\sqrt2$).
On the boundaries:
*
*$f(0,y)=\sin^2y-\sin y<0$ on $x=0,y\in(0,\frac{\pi}{2})$
*$f(x,0)=\sin^2x-\sin x<0$ on $x\in(0,\frac{\pi}{2}),y=0$
*$f(0,0)=0$
*$f(x,y)=0$ for $x,y\geq0,x+y=\frac{\pi}{2}$
Since $(0,0)$ is not part of the domain, and there are no other local extrema, we have $f(x,y)<0$ over $R_1$.
Part 2
It will be sufficient to prove that $\sin^2A + \sin^2B > \sin (A+B)$ over region $R_2=\{0<A,B<\frac{\pi}{2} \land A+B>\frac{\pi}{2} \}$.
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Suppose $a,b\in(0,\frac\pi2)$ and
\begin{align*}
\sin^2 a + \sin^2 b &= \sin(a+b) \\
&= \sin a\cos b + \cos a \sin b \tag{1}
\end{align*}
Squaring both sides, we get
\begin{align*}
\sin^4 a + 2\sin^2 a\sin^2 b + \sin^4 b
&= \sin^2 a \cos^2 b + 2\sin a\sin b\cos a\cos b + \cos^2 a\sin^2 b \\
&= \sin^2 a (1-\sin^2 b) + 2\sin a\sin b\cos a\cos b + (1-\sin^2 a)\sin^2 b \\
&= \sin^2 a - 2\sin^2 a\sin^2 b + \sin^2 b + 2\sin a\sin b\cos a\cos b
\end{align*}
Rearranging,
\begin{align*}
4\sin^2 a\sin^2 b
&= \sin^2 a - \sin^4 a + \sin^2 b - \sin^4 b + 2\sin a\sin b\cos a\cos b \\
&= \sin^2 a(1-\sin^2 a) + \sin^2 b(1-\sin^2 b) + 2\sin a\sin b\cos a\cos b \\
&= \sin^2 a\cos^2 a + \sin^2 b\cos^2 b + 2\sin a\sin b\cos a\cos b \\
&= (\sin a\cos a + \sin b\cos b)^2
\end{align*}
By our assumptions, everything is positive, so we can take square roots, getting
$$ 2\sin a\sin b = \sin a\cos a + \sin b\cos b $$
Adding this to (1) yields
$$ \sin^2 a + 2\sin a\sin b + \sin^2 b
= \sin a\cos a + \sin a\cos b + \sin b\cos b + \sin b\cos a $$
that is,
$$ (\sin a + \sin b)^2 = (\sin a + \sin b)(\cos a + \cos b) $$
Therefore (again, everything being positive)
$$ \sin a + \sin b = \cos a + \cos b $$
Rearranging,
$$ \sin a - \cos a = \cos b - \sin b $$
and squaring,
$$ \sin^2 a - 2\sin a\cos a + \cos^2 a = \sin^2 b - 2\sin b\cos b + \cos^2 b $$
Simplifying,
$$ \sin 2a = \sin 2b $$
Since $a,b\in(0,\frac\pi2)$, we conclude that either $a=b$ or $a+b=\frac\pi2$. In the case $a=b$, (1) reduces to $\sin 2a = 2\sin^2 a = 1 - \cos 2a$, which is equivalent to $\sin 4a = 0$, giving (in our interval) only the solution $a=\frac\pi4$, which is also part of the second case. So $a+b=\frac\pi2$.
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|
How Euler get $\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$. I saw on wikipedia that he consider $$\frac{\sin(x)}{x}=1-\frac{x^2}{3!}+\frac{x^4}{5!}+...$$
The roots are given by $x=\pm n\pi$ and thus (to me)
$$\frac{\sin x}{x}=(x-\pi)(x+\pi)(x-2\pi)(x+2\pi)(x-3\pi)(x+3\pi)...$$
How can he get
$$\frac{\sin x}{x}=\left(1-\frac{x}{\pi}\right)\left(1+\frac{x}{\pi}\right)\left(1-\frac{x}{2\pi}\right)\left(1+\frac{x}{2\pi}\right)\left(1-\frac{x}{3\pi}\right)\left(1+\frac{x}{3\pi}\right)...\ \ \ ??$$
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Euler's proof is a bit adhoc. He uses the fact that for finite polynomials if:
$$(x-r_1)\dots(x-r_n)=x^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0$$
then $$\dfrac{1}{r_1}+\dots+\dfrac{1}{r_n}=\dfrac{-a_1}{a_0}$$
Then he assumes the result is true for power series, and uses:
$$\dfrac{\sin(\sqrt x)}{\sqrt x}=1-\dfrac{x}{3!}+\dfrac{x^2}{5!}+\dots$$
so $r_k=(k\pi)^2$ and $a_0=1$ and $a_1=\dfrac{-1}{6}$.
Reference
|
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|
Trying to solve the trig equation $\sqrt{3+4\cos^2(x)}=\frac{\sin(x)}{\sqrt 3}+3\cos(x)$ The equation is
$$\sqrt{3+4\cos^2(x)}=\frac{\sin(x)}{\sqrt 3}+3\cos(x)$$
My solution goes like this
$$
\begin{cases}
3+4\cos^2(x)=\frac{\sin^2(x)}{3}+\frac{6}{\sqrt 3}\sin(x)\cos(x)+9\cos^2(x) \\
\frac{\sin(x)}{\sqrt 3}+3\cos(x) \ge 0
\end{cases}
$$
$$3(\sin^2(x)+\cos^2(x))+4\cos^2(x)=\frac{\sin^2(x)}{3}+\frac{6}{\sqrt 3}\sin(x)\cos(x)+9\cos^2(x)$$
$$2\cos^2(x)+\frac{\sin^2(x)}{3}-3\sin^2(x)+\frac{6}{\sqrt 3}\sin(x)\cos(x)=0$$
I multiply by 3 and divide by $\cos^2(x)$:
$$8\tan^2(x)-6\sqrt{3}\tan(x)-6=0$$
Let $t=\tan(x)$, then
$$4t^2-3\sqrt{3}t-3=0$$
$$t_1=\frac{7\sqrt 3}{2}$$
$$t_2=\frac{11\sqrt 3}{4}$$
The solutions for $x$ would be arc-tangents of these values.
But the textbook's answer is
$$\color{green}{x_1=\frac{\pi}{3}+2\pi n; x_2=-\arctan\left(\frac{\sqrt 3}{4}\right)+2\pi n}$$
Where did I make a mistake?
P.S. From the textbook
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The roots of $4 t^2-3 \sqrt{3} t-3$ are $-\frac{\sqrt{3}}{4}$ and $\sqrt{3}$.
A slightly different approach to this problem could be the following:
\begin{align}
3+4\cos^2(x)-\frac{\sin^2(x)}{3}-\frac{6}{\sqrt 3}\sin(x)\cos(x)-9\cos^2(x)=\frac{1}{3} \left(-3 \sqrt{3} \sin (2 x)-7 \cos (2 x)+1\right)
\end{align}
With this we have $$3 \sqrt{3} \sin (2 x)+7 \cos (2 x)=2\sqrt{19}\sin(2x+y)$$ where $y=\arcsin\frac{7}{2\sqrt{19}}$. Hence the roots can be found as \begin{align}
x&=\frac12\Big(\arcsin\frac{1}{2\sqrt{19}}-\arcsin\frac{7}{2\sqrt{19}}\Big)+2\pi n\\
x&=\arcsin\frac{1}{2\sqrt{19}}+\arcsin\frac{7}{2\sqrt{19}}+2\pi n\\
\end{align}
now note that \begin{align}
\arcsin\frac{1}{2\sqrt{19}}+\arcsin\frac{7}{2\sqrt{19}}&=\arcsin( \frac{1}{2\sqrt{19}}\sqrt{1-\frac{7}{2\sqrt{19}}^2} + \frac{7}{2\sqrt{19}}\sqrt{1-\frac{1}{2\sqrt{19}}^2})\\
&=\arcsin\frac{\sqrt3}{2}\\
&=\frac{\pi}{3}
\end{align}
Further using $\tan(\alpha+\beta)$ and $\tan(\arcsin x)=\frac{x}{\sqrt{1-x^2}}$ we obtain $$\arcsin\frac{1}{2\sqrt{19}}-\arcsin\frac{7}{2\sqrt{19}}=-2\arctan \frac{\sqrt3}{4}.$$
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|
Integration of the square root of a quadratic I am in the tricky situation of trying to integrate the following.
$$\sqrt{4 a^2 (y-b)^2+c^4}$$
$a, b$ and $c$ are all known constants.
Can anybody provide insight as to how to do this?
I have tried to rearrange to fit the form:
$$\int (ax+b)^{\alpha}dx = \dfrac1a \cdot \dfrac{(ax+b)^{\alpha+1}}{\alpha+1} + \text{ constant}$$
But do not seem able to do so. Maybe excessive toiling has hidden an obvious answer from my eyes.
Thanks for the help.
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Let $$\displaystyle I = \int \sqrt{4a^2(y-b)^2+c^4}dy\;,$$ Let $(y-b) = t\;,$ Then $dy = dt$
So Integral $$\displaystyle I = \int \sqrt{4a^2t^2+c^4}dt = 2a\underbrace{\int \sqrt{t^2+k^2}dt}_{J}\;,$$ Where $\displaystyle k= \frac{c^2}{2a}$
For calculation of Integral $J\;,$ We Use Integration by parts.
Now Let $$\displaystyle J = \int \sqrt{t^2+k^2}\cdot tdt = \sqrt{t^2+k^2}\cdot t-\int\frac{t^2}{\sqrt{t^2+k^2}}dt$$
So $$\displaystyle J= \sqrt{t^2+k^2}\cdot t-\int\frac{(t^2+k^2)-k^2}{\sqrt{t^2+k^2}}dt = \sqrt{t^2+k^2}\cdot t-J+k^2\ln |t+\sqrt{t^2+k^2}|$$
So we Get $$\displaystyle J = \frac{1}{2}\sqrt{t^2+k^2}\cdot t+\frac{k^2}{2}\ln|t+\sqrt{t^2+k^2}|$$
So we get $$\displaystyle I = 2a\cdot J = a\cdot \sqrt{t^2+k^2}\cdot t+ak^2\cdot \ln|t+\sqrt{t^2+k^2}|$$
Where $\displaystyle k = \frac{c^2}{2a}$
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|
Finding the maximum of a function on $ \Bbb{S}^{7} $. I'm trying to find the maximum of the function
$$2 a^2 h+\sqrt{3} a d f+\sqrt{3} a e g+2 b^2 h-\sqrt{3} b d g+\sqrt{3} b e f\\+2 c^2
h+\sqrt{3} c d^2+\sqrt{3} c e^2-\sqrt{3} c f^2-\sqrt{3} c g^2\\-d^2 h-e^2 h-f^2
h-g^2 h-2 h^3$$
on the sphere $$a^2+b^2+c^2+d^2+e^2+f^2+g^2+h^2=1 \text{.}$$
I suspect it is $2$, but I can't come up with a proof. Lagrange multipliers don't work, since the resulting equations don't have an analytic solution.
The value $2$ is attained for $h=-1$.
My current idea is that terms of the form $x^2y$ can't be bigger than $\frac{2}{3\sqrt{3}}$ and terms of the form $xyz$ can't be bigger than $\frac{1}{3\sqrt{3}}$, but I'm not sure this helps.
I would also be satisfied with a slightly higher bound than $2$.
Additional information:
The function is proportional to $\sum_{ijk=1}^8 d_{ijk}p^i p^j p^k$ (the constraint is just $\sum_i (p^i)^2 = 1$), where $ \tau_i\tau_j = -\frac{2}{3}\delta_{ij} +(f_{ijk}-i\, d_{ijk})\tau_k$ holds for the generators $\tau_i$ of $SU(3)$ with $[\tau_i,\tau_j]=f_{ijk} \tau_k$ and the normalization condition $\mathrm{Tr}(\tau_i\tau_j) = -2\delta_{ij}$ is imposed.
|
Let $$a=p\cos\chi,\, b=p\sin\chi\\
d=m\cos\phi,\,e=m\sin\phi\\
f=k\cos\theta,\,g=k\sin\theta\\
q^2=c^2+q^2$$
Then replace $h^3$ by $h(1-p^2-c^2-m^2-k^2)$
$$\text{Maximise }h\left[4p^2+4c^2+m^2+k^2-2\right]+\sqrt{3}pmk\cos(\chi-\phi+\theta)+\sqrt{3}c(m^2-k^2)$$
Let the angle be zero, $m=n\cos\alpha,k=n\sin\alpha$
$$\text{Maximise } h[4p^2+4c^2+n^2-2]+\sqrt{\frac34}\left[pn^2\sin2\alpha+2cn^2\cos2\alpha\right]$$
Let $p=q\cos\beta,c=q\sin\beta$,
$$\text{Maximise }h[4q^2+n^2-2]+\sqrt{\frac34}qn^2[\sin2\alpha\cos\beta+2\cos2\alpha\sin\beta]$$
Let $\alpha=0,\beta=\pi/2$
$$\text{Maximize }h[4q^2+n^2-2]+\sqrt{3}qn^2\\
q^2+n^2+h^2=1$$
Use Lagrange Multipliers, I found that either $n=0$, in which case we maximize $h(2-4h^2)$, which is maximized at $h=-1$; or else $h=0$, in which case we maximize $\sqrt{3}(1-q^2)q$, whose maximum is $2/3$.
|
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|
Intermediate digits of 34! Problem: Given that $34!=295232799cd96041408476186096435ab000000$. Find $a, b, c, d$.
$a, b, c, d$ are single digits.
I am able to find $a$ and $b$ but cant find $c, d$.
I did the prime factorisation of $34!$ using De-Polignac's formula.
I got $\frac{34!}{10^7}=2^{25}\times3^{15}\times7^4\times11^3\times13^2\times17^2\times19\times23\times29\times31$
So, i got $b=0$ and $a=2$ since $2^{25}\times3^{15}\times7^4\times11^3\times13^2\times17^2\times19\times23\times29\times31 \equiv 2 \mod 10$.
So, please help me find $c$ and $d$
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Here $34!$ contain $2^{32}$ and $5^7$. So $2^{25}$ remaining and $2^{7}\cdot 5^7 = (10)^7$ form $7$ zeros at the end.
So $34! = 295232799cd9604140809643ab \times 10^7$ now after deleting $7$ zero,s ,
So we get $34! = 295232799cd9604140809643ab$
Now we use divisibility test for last $7$ digits using $2^7$.
Here $(a,b)\in \{0,1,2,3,4,5,6,7,8,9\}$
So Divisiblilty by $2^2$, we get
$00,04,08,12,16,20,24,28,32,36,40,44,48,52,56,60,64,68,72,76,80,84,88,92,96$ ect
So Divisibility by $2^3$
$304,312,320,328,336,344,352,368,376,384$
In a similar manner
at last when divisible by $2^{7},$ we get $ab = 52$
Now, since these are digits, $0 \leq c, d \leq 9$, so $-9 \leq c-d \leq 9$ and $0 \leq c+d \leq 18$.
Use the fact that $34!$ is a multiple of $9,$ to tell you the value of $c+d$. We get that $c+d = 3$ or $12$.
Use the fact that $34!$ is a multiply of $11,$ to tell you the value of $c-d$.
We get that $c-d = -3$ or $8$. Since $2c$ is an even number from $0$ to $18,$
we conclude that $c=0, d=3$.
So $(a,b,c,d) = (5,2,0,3)$
|
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|
How I can evaluate $\lim_{(x,y) \rightarrow (0,0)} xy(\frac{1+xy}{x^3+y^3})^{1/3}$ I don't have idea how I can evaluate this double limit
$$\lim_{(x,y) \rightarrow (0,0)} xy \left(\frac{1+xy}{x^3+y^3} \right) ^{1/3}$$ could you help me please!
I try prove that $f$ is continuous: $f(x,y)=xy \left(\frac{1+xy}{x^3+y^3} \right) ^{1/3}$ if $x\not=-y$ and $f(x,y)=0$ if $x=-y$
|
This is prossible?
$ 0\leq|xy\left(\frac{1+xy}{x^3+y^3}\right)^{1/3}|=|\left(\frac{x^3y^3+x^4y^4}{x^3+y^3}\right)^{1/3}|=|\left(\frac{x^3y^3}{x^3+y^3}+\frac{x^3y^3(xy)}{x^3+y^3}\right)^{1/3}|=|\left(\frac{x^3}{x^3+y^3}y^3+\frac{x^3}{x^3+y^3}y^3(xy)\right)^{1/3}|\leq|(y^3+y^3(xy))^{1/3}|\leq (y^2|y|+y^4|x|)^{1/3}\leq(|y|+|x|)^{1/3}$
so like the function is bounded by two function whose limits tends to $0$ then $f(x,y)$ tends to $0$¿?
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
How do you factor $3x^{3/2} -9x^{1/2}+6x^{-1/2}$? How do you factor $3x^{3/2} -9x^{1/2}+6x^{-1/2}$ ?
I factored out a 3 to get:
$3(x^{3/2} -3x^{1/2}+2x^{-1/2})$, but it seems this can be factored further.
|
$$ 3 x^{\frac{3}{2}} -9 x^{\frac{1}{2}} + 6x^{-\frac{1}{2}}$$
Factor out $\sqrt{x}$ to get
$$\sqrt{x} (3 x - 9 + 6x^{-1})$$
Furthermore you get
$$\sqrt{x} 3(x - 3 + 2x^{-1})$$
and then
$$\sqrt{x} 3 \frac{(x-2)(x-1)}{x}$$
Simplify:
$$3 \frac{(x-2)(x-1)}{\sqrt{x}}$$
|
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|
Minimum value of $\cos x+\cos y+\cos(x-y)$ What is the minimum value of $$ \cos x+\cos y+\cos(x-y). $$ Here $x,y$ are arbitrary real numbers. Mathematica gives (with NMinimize) $-3/2$. But I don't know if this is correct and if so, how to prove it.
|
Let $$\displaystyle z=\cos x+\cos y+\cos(x-y) = 2\cos\left(\frac{x+y}{2}\right)\cdot \cos\left(\frac{x-y}{2}\right)+2\cos^2 \left(\frac{x-y}{2}\right)-1$$
so we get $$\displaystyle 2\cos^2 \left(\frac{x-y}{2}\right)-2\cos\left(\frac{x+y}{2}\right)\cdot \cos \left(\frac{x-y}{2}\right)-(1+z) =0$$
Now Let $$\displaystyle \cos\left(\frac{x-y}{2}\right) = t\;,$$ and $$\displaystyle \bullet \; -1\le \cos \left(\frac{x\pm y}{2}\right)\le 1$$
So we get $$\displaystyle 2t^2-2\cos \left(\frac{x+y}{2}\right)-(1+z) =0\;,$$ Now for real roots, its $\bf{Discriminant\geq 0}$
So we get $$\displaystyle 4\cos \left(\frac{x+y}{2}\right)+8(1+z)\geq 0$$
So we get $$\displaystyle \cos \left(\frac{x+y}{2}\right)+2(1+z)\geq 0\Rightarrow 2(1+z)\geq \cos \left(\frac{x+y}{2}\right)\geq -1$$
So we get $$\displaystyle (1+z)\geq -\frac{1}{2}\Rightarrow z\geq -\frac{3}{2}$$
and equality hold when $\displaystyle \cos \left(\frac{x+y}{2}\right) = -1=\cos \pi\Rightarrow x+y=2\pi$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1419019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Is the matrix $A$ positive (negative) (semi-) definite? Given, $$A = \begin{bmatrix}
2 &-1 & -1\\
-1&2 & -1\\
-1& -1& 2
\end{bmatrix}.$$
I want to see if the matrix $A$ positive (negative) (semi-) definite.
Define the quadratic form as $Q(x)=x'Ax$.
Let $x \in \mathbb{R}^{3}$, with $x \neq 0$.
So, $Q(x)=x'Ax = \begin{bmatrix}
x_{1} &x_{2} &x_{3}
\end{bmatrix} \begin{bmatrix}
2 &-1 & -1\\
-1&2 & -1\\
-1& -1& 2
\end{bmatrix} \begin{bmatrix}
x_{1}\\x_{2}
\\x_{3}
\end{bmatrix}$.
After multiplying out the matrices I am left with $$Q(x) = 2(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-x_{1}x_{2} - x_{1}x_{3}-x_{2}x_{3}).$$
Not sure what I can do with this result. Any suggestions on how to proceed would be appreciated.
|
If you want to proceed with this solution, you should complete the square. It is important that you "complete one variable completely every time". We write
$$\begin{aligned}
x_1^2+x_2^2+x_3^2-x_1x_2-x_1x_3-x_2x_3&=\Bigl(x_1-\frac{1}{2}x_2-\frac{1}{2}x_3\Bigr)^2+\frac{3}{4}x_2^2+\frac{3}{4}x_3^2-\frac{3}{2}x_2x_3\\
&=\Bigl(x_1-\frac{1}{2}x_2-\frac{1}{2}x_3\Bigr)^2+\frac{3}{4}\bigl(x_2-x_3\bigr)^2.
\end{aligned}
$$
Can you conclude from here?
|
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|
How can this expression be simplified? How do I factorize $$a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)?$$
I've tried it in different ways but failed. Wish some one could help solving it out.
|
Another proof. By Laplace expansion (along the last row):
$$ -\sum_{cyc}a^2(b-c) = \det\begin{pmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{pmatrix},\tag{1}$$
But the RHS is a Vandermonde matrix, whose determinant is well-known.
Gaussian elimination gives:
$$ D=\det\begin{pmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{pmatrix}=\det\begin{pmatrix} 1 & 0 & 0 \\ a & b-a & c-a \\ a^2 & b^2-a^2 & c^2-a^2 \end{pmatrix} \tag{2}$$
and by expanding along the first row and factoring out $(b-a)$ and $(c-a)$:
$$ D = (b-a)(c-a)\cdot \det\begin{pmatrix} 1 & 1 \\ b+a & c+a\end{pmatrix}=(b-a)(c-a)(c-b).\tag{3}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
A trick for calculating $n^6$ that I don't understand I was doing a math exercise and it asked to find what are the possible units digits of $n^6$ knowing that $n\in\mathbb Z$. The solution said that because we are concerned only with finding what the units digits of $n^6$ could be, it is sufficient to take the sixth power of numbers $0,1,2,\dots,9$ and the results would be our answers (which are $0,1,4,5,6,9$). How come this is true? Thanks for your help!
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What is the units digit of $340274513\times 384759374\,{}$?
$$
\begin{array}{ccccccccccccc}
& & & & 3 & 4 & 0 & 2 & 7 & 4 & 5 & 1 & 3 \\
& & & \times & 3 & 8 & 4 & 7 & 5 & 9 & 3 & 7 & 4 \\
\hline
& & & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & 2 \\
& & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\
& \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\
& & & \vdots \\
& & & \vdots \\
\hline
\cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & 2
\end{array}
$$
The method of multiplying that you learned as a child shows you that the units digit in the product is determined by the units digits, and not any of the other digits, in the numbers that you're multiplying.
Moreover you can write the number $340274513$ as $\Big(34027451\times10\Big)+3$ and $384759374$ as $\Big(38475937\times 10\Big)+4$, and then you have
\begin{align}
& 340274513\times384759374 \\[10pt]
= {} & \Big(\big(34027451\times10\big)+3\Big)\times\Big(\big(384759374\times10\big)+4\Big) \\[10pt]
= {} & \Big(\underbrace{\text{something}\times 10}_{\begin{smallmatrix} \text{This contributes} \\ \text{nothing to the} \\ \text{units digit.} \end{smallmatrix}}\Big) + (3\times 4)
\end{align}
|
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|
if $\sin(x)=\frac{1}{3}$ and $\sec(y)=\frac{5}{4} $, what is $\sin(x+y)?$ if $\sin(x)=\frac{1}{3}$ and $\sec(y)=\frac{5}{4} $, what is $\sin(x+y)?$
Here is my thought process:
the identity says:
$\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)$
We know $\sin(x)=1/3$ and $\sec(y)=\frac{1}{\cos(y)} = \frac{4}{5}$, so
$\sin(x+y)=\frac{1}{3}\times\frac{4}{5}+\cos(x)\sin(y)$
But I'm stuck here because I don't know how to find $\cos(x)$ or $\sin(y)$
|
You have started off well. To get the values of $\cos x$ and $\sin y$, you should use the identity
$$\sin^2\theta + \cos^2\theta \equiv 1$$
I will show you how to find $\cos x$.
You can use the same method to find $\sin y$ yourself.
For example, if $\sin x = \frac{1}{3}$ then we have
\begin{eqnarray*}
\sin^2 x + \cos^2x &\equiv& 1 \\ \\
\left(\frac{1}{3}\right)^{\! 2} + \cos^2x &=& 1 \\ \\
\frac{1}{9} + \cos^2x &=& 1 \\ \\
\cos^2x &=& \frac{8}{9} \\ \\
\cos x&=& \pm\frac{2}{3}\sqrt{2}
\end{eqnarray*}
Next you need to decide on the $\pm$. Looking at the graph of $y=\cos x$ we see that $\cos x > 0$ if $0^{\circ} \le x < 90^{\circ}$ or $270^{\circ} < x \le 360^{\circ}$. So, for example: if we know that $x$ is acute then
$$\cos x = +\frac{2}{3}\sqrt{2}$$
If we know that $x$ is obtuse then
$$\cos x = -\frac{2}{3}\sqrt{2}$$
If $x$ is reflex and less that $270^{\circ}$ then
$$\cos x = +\frac{2}{3}\sqrt{2}$$
If $x$ is reflex and more than $270^{\circ}$ then
$$\cos x = -\frac{2}{3}\sqrt{2}$$
The same is true modulo $360^{\circ}$.
|
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"url": "https://math.stackexchange.com/questions/1422492",
"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\int \frac{\sin ^3(\theta/2)}{\cos(\theta/2)\sqrt{\cos^3\theta+\cos^2\theta+\cos \theta}}d\theta$ I came about the following practice problem in a book of integration:-
$Q.$ Evaluate $$I=\int \frac{\sin ^3(\theta/2)}{\cos(\theta/2)\sqrt{\cos^3\theta+\cos^2\theta+\cos \theta}}d\theta$$ To do this, first I substituted $\cos(\theta/2)=u \implies \frac {-1}{2}\sin (\frac{\theta}2)\ d\theta=du \implies \sin^3 (\frac{\theta}2)d\theta=-2(1-u^2)\ du$. This gives $$\begin{align}I&=\int \frac{2(u^2-1)\ du}{u\sqrt{(2u^2-1)^3+(2u^2-1)^2+(2u^2-1)}}\\&=\frac 12\int \frac {(u^2-1)(4u\ du)}{u^2\sqrt{(2u^2-1)^3+(2u^2-1)^2+(2u^2-1)}}\end{align}$$ Now substitute $z=2u^2-1 \implies dz=4u\ du$. We have $u^2=\frac {z+1}2 \implies u^2-1=\frac {z-1}2$. Hence $$\begin{align}I&=\int \frac {{{z-1}\over2}\ dz}{(\frac {z+1}2)\sqrt{z^3+z^2+z}}\\&=\int \frac{(z-1)\ dz}{(z+1)\sqrt{z^3+z^2+z}}\\&=\int \left[\frac{1}{\sqrt{z^3+z^2+z}}-\frac2{(z+1)\sqrt{z^3+z^2+z}}\right]\ dz\end{align}$$ I don't know how to proceed further. Some hints would be appreciated. Is there any easier way to integrate the given expression? I also tried substituting $\tan (\theta/2)=u$, but it becomes even more messier than what I have shown here.
|
Let $$\displaystyle I = \int \frac{\sin ^3(\theta/2)}{\cos(\theta/2)\sqrt{\cos^3\theta+\cos^2\theta+\cos \theta}}d\theta = \frac{1}{2}\int\frac{2\sin^2 \frac{\theta}{2}\cdot 2\sin \frac{\theta}{2}\cdot \cos \frac{\theta}{2}}{2\cos^2 \frac{\theta}{2}\sqrt{\cos^3 \theta+\cos^2 \theta+\cos \theta}}d\theta$$
So we get $$\displaystyle I = \frac{1}{2}\int\frac{(1-\cos \theta)\cdot \sin \theta}{(1+\cos \theta)\sqrt{\cos^3 \theta+\cos^2 \theta+\cos \theta}}d\theta$$
Now Put $\cos \theta = t\;,$ Then $\sin \theta d\theta = -dt$
So Integral $$\displaystyle I = -\frac{1}{2}\int\frac{(1-t)}{(1+t)\sqrt{t^3+t^2+t}}dt = -\frac{1}{2}\int\frac{(1-t^2)}{(1+t)^2\sqrt{t^3+t^2+t}}dt$$
So we get $$\displaystyle I = \frac{1}{2}\int\frac{\left(1-\frac{1}{t^2}\right)}{\left(t+\frac{1}{t}+2\right)\sqrt{t+\frac{1}{t}+1}}dt$$
Now Let $\displaystyle \left(t+\frac{1}{t}+1\right) = u^2\;,$ Then $\left(1-\frac{1}{t^2}\right)dt = 2udu$
So Integral $$\displaystyle I = \frac{1}{2}\int\frac{2u}{u^2+1}\cdot \frac{1}{u}du = \tan^{-1}(u)+\mathcal{C}$$
So we get $$\displaystyle I = \tan^{-1}\sqrt{\left(t+\frac{1}{t}+1\right)}+\mathcal{C}$$
So we get $$\displaystyle \displaystyle I = \tan^{-1}\sqrt{\left(\cos \theta+\sec \theta+1\right)}+\mathcal{C}$$
|
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|
There are 30 tokens numbers from 0 to 30. Find the number of ways of choosing 3 tickets such that the sum of the numbers on the tokens is 30 Then the number of solutions is divisible by
(A) 2 (B) 3 (C) 5 (D) 7
One way to solve its by finding the number of unequal integral solutions of $x+y+z=30$.
The possible cases are $x<y<z, y<x<z,..$ (3! Ways)
If $x<y<z$, let $x=a, y=a+b$ and $z=a+b+c$
Then $3a+2b+c=30$ where $0\le a,b,c\le 30$
The number of solutions is the coefficient of $x^{30}$ in $(1-x^3)^{-1}(1-x^2)^{-1}(1-x)^{-1}$ which is as difficult as counting the numbers manually.
Total number of ways would be $3!$ times the number obtained from above. Hence divisible by 2 and 3. Is this correct?
|
start|range|combos of 2 to get 30 (Gauss method)
0 .....1-29 ...$\lfloor\frac{29}{2}\rfloor = 14$
1 .....2-27 ...$\lfloor\frac{26}{2}\rfloor = 13$
2 .....3-25 ..$\lfloor\frac{23}{2}\rfloor = 11$
3 .....4-23 ...$\lfloor\frac{20}{2}\rfloor = 10$
4 .....5-21 ...$\lfloor\frac{17}{2}\rfloor = 8$
5 .....6-19 ...$\lfloor\frac{14}{2}\rfloor = 7$
6 .....7-17 ...$\lfloor\frac{11}{2}\rfloor = 5$
7 .....8-15 ...$\lfloor\frac{8}{2}\rfloor = 4$
8 .....9-13 ...$\lfloor\frac{5}{2}\rfloor = 2$
9 .....10-11 ...$\lfloor\frac{2}{2}\rfloor = 1$
Total 75
Ans:(B) and (C)
|
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.