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Non-Linear System. Find the conditional expectation. I've had my test for this course and I think I failed it again. The hardest part for me is findig the correct distributions. This is a test exercise I couldn't figure out or at least, I probably failed the question. I hope I can get your help, so next time I won't fail my test.
Given:
$$
\begin{align}
X_{k+1} &= X_k\\
Y_k &= X_k^2 - V_k^2
\end{align}
$$
Also $X_0$ and $V_k$ are $\sim U(0,1)$. With $V_k$ white noise, in this case means always independent of $X_k$. I did the test and I came up with several approaches, but the reason I think they failed is since the outcome of the different ways were not the same.
First: Determine $E[X_0|Y_0]$, for simplicity $X_0 = X$ and $Y_0 = Y$.
The approach I did that I thought was most promising:
$$
\begin{align}
E[X|Y] &= \int x f_{X|Y} dx = \int x \left(\frac{d}{dx}P(X\leq x|Y =y)\right)dx\\
&= \int x \left(\frac{d}{dx}P(V_0^2\leq x^2 - y|Y =y)\right)dx\\
&= \int x \left(\frac{x}{\sqrt{x^2 - y}} \mathbb{1}_{x^2-1<y<x^2}\mathbb{1}_{0<x<1} \right)dx\\
&= \left(\mathbb{1}_{-1<y<0}\int_{0}^{\sqrt{y+1}} + \mathbb{1}_{0<y<1}\int_{\sqrt{y}}^{1}\right)\frac{x^2}{\sqrt{x^2 - y}}dx
\end{align}
$$
I couldn't get any further than this. What should I have done? Are those steps done right? I also thought of computing ${f_{X,Y}(x,y)}{f_Y(y)}$ and using a parametrization $X_0 = r \cosh(\theta)$ and $V_0 = r \sinh(\theta)$. But how I should do that in this case is beyond me at the moment--although I guess this can't be done either since $\cosh(\theta) \geq 1$.
Second: coming up...
|
There are two ways you could tackle the problem:
(a) Use "change of variables" from $X$ and $V$ to $X$ and $Y$ and then find the required conditional distribution and expectation.
(b) Use the "direct" method, which is what you've done.
For this particular problem, I think (a) is a slightly simpler but I will use (b) here since it is in line with your approach.
I'll assume for $E(X\mid Y=y)$ that $y\geq 0$. The working is similar in the $y\lt 0$ case.
\begin{eqnarray*}
E(X\mid Y=y) &=& \int_x xP(X=x\mid X^2-V^2 = y)\;dx \\
&=& \int_x xP(X=x\cap V^2 = x^2-y)\;dx \;/\; P(X^2-V^2 = y) \\
&=& \int_x xP(X=x)P(V^2 = x^2-y)\;dx \;/\; P(X^2-V^2 = y) \\
&=& \int_{x=\sqrt{y}}^{1} x\cdot 1\cdot \dfrac{1}{2\sqrt{x^2-y}}\;dx \;/\; P(X^2-V^2 = y) \\
&& \qquad\qquad\qquad\qquad\text{(if $U=V^2,\;$ then $\frac{dv}{du}=\frac{1}{2\sqrt{u}}\;$ so $f_U(u)=\frac{1}{2\sqrt{u}}$)} \\
&=& \left[\frac{1}{2}\sqrt{x^2-y}\right]_{x=\sqrt{y}}^{1} \bigg/ P(X^2-V^2 = y) \\
&=& \frac{1}{2}\sqrt{1-y} \bigg/ P(X^2-V^2 = y). \qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{(1)}
\end{eqnarray*}
We now find $P(X^2-V^2 = y)$ by way of $P(X^2-V^2 \leq y)$.
\begin{eqnarray*}
P(X^2-V^2 \leq y) &=& 1 - \int_{x=\sqrt{y}}^{1} \sqrt{x^2-y}\;dx \\
&=& 1 - \frac{1}{2} \left[ x\sqrt{x^2-y} - y\ln \left(\sqrt{x^2-y} + x\right) \right]_{x=\sqrt{y}}^{1} \\
&=& 1 - \frac{1}{2} \left[ \sqrt{1-y} - y\ln \left(\sqrt{1-y} + 1\right) +y\ln\left(\sqrt{y}\right) \right]. \\
\end{eqnarray*}
Differenting wrt $y$, we have
\begin{eqnarray*}
P(X^2-V^2 = y) &=& - \frac{1}{2} \left[ \frac{-1}{2\sqrt{1-y}} - \ln \left(\sqrt{1-y} + 1\right) +\frac{y}{\sqrt{1-y}+1}\cdot \frac{1}{2\sqrt{1-y}} + \ln\left(\sqrt{y}\right) + \frac{1}{2} \right] \\
&=& \frac{1}{2}\ln \left(\sqrt{1-y} + 1\right) - \frac{1}{2}\ln\left(\sqrt{y}\right) \qquad\qquad\qquad\qquad\qquad\text{after simplification.}
\end{eqnarray*}
Substituting into $(1)$ we have, for $y\geq 0$
\begin{eqnarray*}
E(X\mid Y=y) &=& \frac{\sqrt{1-y}}{\ln \left(\sqrt{1-y} + 1\right) - \ln\left(\sqrt{y}\right)}.
\end{eqnarray*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
$(x+y+z)^3-(y+z-x)^3-(z+x-y)^3-(x+y-z)^3=24xyz$? The question given is
Show that $(x+y+z)^3-(y+z-x)^3-(z+x-y)^3-(x+y-z)^3=24xyz$.
What I tried is suppose $a=(y+z-x),\ b=(z+x-y)$ and $c=(x+y-z)$ and then noted that $a+b+c=x+y+z$. So the question statement reduced to $(a+b+c)^3-(a^3+b^3+c^3)$. Then I tried to invoke the identity $(a^3+b^3+c^3-3abc)=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$ by adding and subtracting $3abc$ in the question statement. After doing all this when I substituted back the values of $a,b$ and $c$, I ended up with the initial question statement.
Any hints will be appreciated.
|
Pascal's (or Tartaglia's) Tetrahedron: the left outline is a binomial expansion of $(x+y)^3$, while the right outline is a binomial expansion of $(x+z)^3$ and the bottom outline is a binomial expansion of $(y+z)^3$. Each of them is highlighted in yellow for identification purposes. Furthermore, you may notice that the terms with the highest powers are situated at the vertices, because the coefficients of those terms are “1”. Imagine to rewrite this equation in "graphically", treating expressions in triangles as the components of a matrix (which add-term contracts):
Hence the claim.
|
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|
Proving inequalities using Calculus In general how do you prove inequalities using calculus, I believe it is using maxima or minima right? For example
$$a^2b+b^2c+c^2a \le 3, \qquad a,b,c \ge 0,\quad a+b+c=3.$$
How would you use calculus, just a sketch, I am interested in the method itself.
|
$f(a,b) = a^2b + b^2(3-a-b) + (3-a-b)^2a = a^2b + 3b^2-ab^2-b^3+9a+a^3+ab^2-6a^2-6ab+2a^2b=3a^2b+3b^2-b^3+9a+a^3-6a^2-6ab$. Thus: $f_a = 6ab+9+3a^2-12a-6b = 0 = f_b = 3a^2+6b-3b^2-6a\Rightarrow 3(a^2-b^2) = 6(a-b)\Rightarrow (a-b)(a+b-2)=0$. There are $2$ cases:
Case $1$: $a = b \Rightarrow 6a^2+9+3a^2-12a-6a = 0\Rightarrow 9a^2-18a+9 = 0\rightarrow 9(a-1)^2 = 0\to a = 1=b, c = 1$, since $a+b+c=3$.
Case $2$: $a+b = 2 \Rightarrow 6a(2-a)+9+3a^2-12a-6(2-a) = 0\Rightarrow 12a-6a^2+9+3a^2 - 12a-12+6a=0\to -3a^2-3+6a=0\to -3(a-1)^2=0 \to a = 1, b = 1, c = 1$.
Either case gives $f_{max} = f(1,1,1) = 3\to a^2b+b^2c+c^2a \leq 3$
|
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|
How can I calculate this limit? $\lim _{ m\rightarrow \infty }{ \left( \lim _{ n\rightarrow \infty }{ \cos ^{ 2n }{ \left( \pi m!x \right) } } \right) } $
Attempt : since $\cos ^{ 2 }{ x=\frac { 1+cos2x }{ 2 } } $ so we can write $$\lim _{ m\rightarrow \infty }{ \left( \lim _{ n\rightarrow \infty }{ \cos ^{ 2n }{ \left( \pi m!x \right) } } \right) } =\lim _{ m\rightarrow \infty }{ \left( \lim _{ n\rightarrow \infty }{ \frac { \left( 1+cos\left( 2\pi m!x \right) \right) ^{ n } }{ 2^{ n } } } \right) } $$
then i applied Teylor expansion for $cos\left( 2\pi m!x \right)$
$$cox\left( 2\pi m!x \right) =1-\frac { \left( 2\pi m!x \right) ^{ 2 } }{ 2! } +o\left( \frac { \left( 2\pi m!x \right) ^{ 4 } }{ 4! } \right) $$
but this way seems me very long,how can i proceed?
Thanks in advance!
|
Let
\begin{equation*}
f_{m}(x)=\lim_{n\rightarrow \infty }(\cos m!\pi x)^{2n}.
\end{equation*}
Note that for $m$ fixed, $f_{m}(x)$ is the limit of geometric sequence, $%
r^{n}$ with
\begin{equation*}
r=(\cos m!\pi x)^{2}\in \lbrack 0,1].
\end{equation*}
If $r=1,$ the limit of the geometric sequence $(r^{n})$ is $1,$ otherwise,
the limit is $0.$ So when $m!x$ is an integer $f_{m}(x)=1,$ and otherwise
(when $m!x$ is not integer) $f_{m}(x)=0.$ Now let
\begin{equation*}
f(x)=\lim_{m\rightarrow \infty }f_{m}(x).
\end{equation*}
Let us discuss when $x$ is rational or is irrational. If $x$ is irrational,
then
\begin{equation*}
(\cos m!\pi x)^{2}\neq 1
\end{equation*}
and then $f(x)=0.$ If $x$ is rational, say $x=\frac{a}{b}$ with $a$ and $b$
integers. We have
\begin{equation*}
m!x=\frac{m!a}{b}=\frac{1\times 2\times \cdots \times m\times a}{b}.
\end{equation*}
If $m\geq b$ then $f(x)=1.$ It follows that
\begin{equation*}
\lim_{m\rightarrow \infty }\lim_{n\rightarrow \infty }(\cos m!\pi x)^{2n}=0
\end{equation*}
if $x$ is irrational
\begin{equation*}
\lim_{m\rightarrow \infty }\lim_{n\rightarrow \infty }(\cos m!\pi x)^{2n}=1
\end{equation*}
if $x$ is instead rational.
|
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|
Solving the differential equation $x^2y''+xy'-y=x^2$
$$x^2y''+xy'-y=x^2$$
My attempt:
Divided by $x^2$:
$$y''+\frac{y'}{x}-\frac{y}{x^2}=1$$
Now to solve the homogenous equation using Euler's method
$$y''+\frac{y'}{x}-\frac{y}{x^2}=0$$
To look for solution $y=x^r$
so $y'=rx^{r-1}$
$y''=r(r-1)x^{r-2}$
So:
$$r(r-1)x^{r-2}+\frac{rx^{r-1}}{x}-\frac{x^r}{x^2}=0$$
Divided by $x^r$:
$$r(r-1)x^{-2}+\frac{rx^{-1}}{x}-\frac{1}{x^2}=0$$
Is it correct so far?
My problem:
I don't know how to find $r_1,r_2$
|
As per ChickenP's answer,
$y_c = Ax + \dfrac{B}{x}$
Now since the D.E. is linear, also account for a particular solution, i.e.
$y_p = s \cdot x^2$ so $y'' + \dfrac{y'}{x} - \dfrac{y}{x^2} = 2s + 2s -s = 3s = 1$ giving $s = \frac{1}{3}$, so $y_p = \frac{1}{3}x^2$
Then the full solution is
$y = Ax + \dfrac{B}{x} + \frac{1}{3}x^2$
|
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|
Do more generalizations of Schur's inequality exist?
I meet this following problem
If $$n\ge 3,\sum_{i=1}^{n}\left(\prod_{j\neq i}(a_{i}-a_{j})\right)\ge 0$$
where $a_{i}$ are real numbers.
when $n=3$, it is Schur's inequality
so which $n$ such this inequality?
but for more generalization form of Schur'.s Inequality exists?
|
Now I have solve it, this inequality it is only true for $n=3,n=5$ holds.
For $n=3$ then inequality can write as
$$(a_{1}-a_{2})(a_{1}-a_{3})+(a_{2}-a_{1})(a_{2}-a_{3})+(a_{3}-a_{1})(a_{3}-a_{2})=a^2_{1}+a^2_{2}+a^2_{3}-a_{1}a_{2}-a_{1}a_{3}-a_{2}a_{3}\ge 0$$
it is clear true
for $n=5$,WLOG, we assume that
$$a_{1}\ge a_{2}\ge a_{3}\ge a_{4}\ge a_{5}$$
Lemma 1:
$$I=(a_{1}-a_{2})(a_{1}-a_{3})(a_{1}-a_{4})(a_{1}-a_{5})+(a_{2}-a_{1})(a_{2}-a_{3})(a_{2}-a_{4})(a_{2}-a_{5})\ge 0$$
Proof: because
$$I=(a_{1}-a_{2})[(a_{1}-a_{3})(a_{1}-a_{4})(a_{1}-a_{5})-(a_{2}-a_{3})(a_{2}-a_{4})(a_{2}-a_{5})]$$
since
$$a_{1}-a_{3}\ge a_{2}-a_{3}\ge 0$$
$$a_{1}-a_{4}\ge a_{2}-a_{4}\ge 0$$
$$a_{1}-a_{5}\ge a_{2}-a_{5}\ge 0$$
$$a_{1}-a_{2}\ge 0$$
so
$$I\ge 0$$
Use same methods we have
$$(a_{5}-a_{1})(a_{5}-a_{2})(a_{5}-a_{3})(a_{5}-a_{4})+(a_{4}-a_{1})(a_{4}-a_{2})(a_{4}-a_{3})(a_{4}-a_{5})\ge 0$$
and
$$(a_{3}-a_{1})(a_{3}-a_{2})(a_{3}-a_{4})(a_{3}-a_{5})\ge 0$$
so for $n=5$ is hold
for $n=4$, you can take $a_{1}=0,a_{2}=a_{3}=a_{4}=1$
for $n\ge 6$,you can take
$$a_{1}=0,a_{2}=a_{3}=a_{4}=1,a_{5}=a_{6}=\cdots=a_{n}=-1$$
then $$\sum_{i=1}^{n}\left(\prod_{j\neq i}(a_{i}-a_{j})\right)=(-1)^3<0$$
|
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|
Solving a system of non-linear equations Let
$$(\star)\begin{cases}
\begin{vmatrix}
x&y\\
z&x\\
\end{vmatrix}=1, \\
\begin{vmatrix}
y&z\\
x&y\\
\end{vmatrix}=2, \\
\begin{vmatrix}
z&x\\
y&z\\
\end{vmatrix}=3.
\end{cases}$$
Solving the above system of three non-linear equations with three unknowns.
I have a try.
Let$$A=\begin{bmatrix}
1& 1/2& -1/2\\
1/2& 1& -1/2\\
-1/2& -1/2& -1
\end{bmatrix}$$
We have $$(x,y,z)A\begin{pmatrix}
x\\
y\\
z
\end{pmatrix}=0.$$
There must be a orthogonal matrix $T$,such that $T^{-1}A T=diag \begin{Bmatrix}
\frac{1}{2},\frac{\sqrt{33}+1}{4},-\frac{\sqrt{33}-1}{4}
\end{Bmatrix}.$
$$\begin{pmatrix}
x\\
y\\
z
\end{pmatrix}=T\begin{pmatrix}
x^{'}\\
y^{'}\\
z^{'}
\end{pmatrix}\Longrightarrow\frac{1}{2} {x'}^{2}+\frac{\sqrt{33}+1}{4} {y'}^{2}-\frac{\sqrt{33}-1}{4}{z'}^{2}=0.$$
But even if we find a $\begin{pmatrix}
x_0^{'}\\
y_0^{'}\\
z_0^{'}
\end{pmatrix} $ satisfying $\frac{1}{2} {x_0'}^{2}+\frac{\sqrt{33}+1}{4} {y_0'}^{2}-\frac{\sqrt{33}-1}{4}{z_0'}^{2}=0,\begin{pmatrix}
x_0\\
y_0\\
z_0
\end{pmatrix}=T\begin{pmatrix}
x_0^{'}\\
y_0^{'}\\
z_0^{'}
\end{pmatrix}$ may not be the solution of $(\star)$
If you have some good ideas,please give me some hints. Any help would be appreciated!
|
Given $x^2-yz = 1, \quad y^2-xz = 2, \quad z^2-xy = 3$, we can sum all of these to get
$$(x-y)^2+(y-z)^2+(z-x)^2 = 12 \tag{1}$$
OTOH, subtracting gives $(y^2-x^2)+z(y-x)=1 \implies (x+y+z)(y-x) = 1$
and similarly $(x+y+z)(z-y) = 1$, so we must have $y-x = z - y = a$, say. Using this in $(1)$,
$$a^2+a^2+4a^2=12 \implies a = \pm \sqrt2$$
So we have $y = x \pm \sqrt2, \quad z = x \pm 2\sqrt2$. Using these in say the first equation, you should be able to solve for $x$ and then $y, z$.
|
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|
Find the area of triangle, given an angle and the length of the segments cut by the projection of the incenter on the opposite side.
In a triangle $ABC$, one of the angles (say $\widehat{C}$) equals $60^\circ$.
Given that the incircle touches the opposite side ($AB$) in a point that splits it in two segments having length $a$ and $b$, what is the area of $ABC$?
|
An alternative approach: let we just find the inradius $r$. Assuming that $\widehat{C}=60^\circ$, $I_C$ is the projection of the incenter on the $AB$-side, $AI_C=a,BI_C=b$, we have:
$$\widehat{A}=2\arctan\frac{r}{a},\qquad \widehat{B}=2\arctan\frac{r}{b}\tag{1}$$
hence:
$$ \frac{\pi}{3} = \arctan\frac{r}{a}+\arctan\frac{r}{b}\tag{2} $$
must hold, from which:
$$ \sqrt{3} = \frac{\frac{r}{a}+\frac{r}{b}}{1-\frac{r^2}{ab}}\tag{3}$$
follows. Since:
$$ \Delta = \frac{1}{2}(a+\sqrt{3}\,r)(b+\sqrt{3}\,r)\sin 60^\circ =\frac{\sqrt{3}}{4}ab\left(1+\sqrt{3}\left(\frac{r}{a}+\frac{r}{b}\right)+3\frac{r^2}{ab}\right)\tag{4}$$
by exploiting $(3)$ we have:
$$ \Delta = \color{red}{\sqrt{3}\,ab},\tag{5}$$
pretty nice, don't you think?
|
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|
Help with Definite integral question Anyone please help with this question:
(a) Show that:
\begin{align}
\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx
\end{align}
(b) Hence show that:
\begin{align}
\int_{0}^{\frac{\pi}{4}} \frac{1-\sin(2x)}{1+\sin(2x)} dx = \int_{0}^{\frac{\pi}{4}} \tan^2{x} dx
\end{align}
And evaluate the integral.
I'd done part (a) and for part (b) I tried:
\begin{align}
\int_{0}^{\frac{\pi}{4}} \frac{1-\sin(2(\frac{\pi}{4} - x))}{1+\sin(2(\frac{\pi}{4} - x))}
\end{align}
\begin{align}
\int_{0}^{\frac{\pi}{4}} \frac{1-\sin\left(\frac{\pi}{2} - 2x\right)}{1+\sin \left(\frac{\pi}{2} - 2x\right)}
\end{align}
Which give:
\begin{align}
\int_{0}^{\frac{\pi}{4}} \frac{1-\cos(2x)}{1+\cos(2x)}
\end{align}
But I was lost after that. I'm not sure if I'm on the right track...
Thanks for your time.
This is a H/W question from P. 173 of Arnold and Arnold. 4 Unit Mathematics. Melbourne 1993.
|
Using the double angle identities to express $\cos 2x$ in terms of $\cos x$ and $\sin x$, we get:
$$\frac{1-\cos 2x}{1+\cos 2x} = \frac{1- (1 - 2 \sin^2 x)}{1 + (2 \cos^2 x - 1)} = \frac{\sin^2 x}{\cos^2 x} = \tan^2 x$$
|
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|
Evaluating a function at a point where $x =$ matrix. Given $A=\left(
\begin{array} {lcr}
1 & -1\\
2 & 3
\end{array}
\right)$
and $f(x)=x^2-3x+3$ calculate $f(A)$.
I tried to consider the constant $3$ as $3$ times the identity matrix ($3I$) but the answer is wrong. Appreciate any ideas.
|
$f(A) = \begin{pmatrix}
1 & -1 \\ 2 & 3
\end{pmatrix}^2 - 3\begin{pmatrix} 1 & -1 \\ 2 & 3 \end{pmatrix} + 3\begin{pmatrix}
1 & 0 \\ 0 & 1
\end{pmatrix} $
$f(A) = \begin{pmatrix}
-1 & -4 \\ 8 & 7
\end{pmatrix} + \begin{pmatrix}
-3 & 3 \\ -6 & -9
\end{pmatrix} + \begin{pmatrix}
3 & 0 \\ 0 & 3
\end{pmatrix}$
$f(A) = \begin{pmatrix}
-1 & -1 \\ 2 & 1
\end{pmatrix}$
|
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|
According to Stewart Calculus Early Transcendentals 5th Edition on page 140, in example 5, how does he simplify this problem? In Stewart's Calculus: Early Transcendentals 5th Edition on page 140, in example 5, how does
$$\lim\limits_{x \to \infty} \frac{\dfrac{1}{x}}{\dfrac{\sqrt{x^2 + 1} + x}{x}}$$
simplify to
$$\lim\limits_{x \to \infty} \frac{\dfrac{1}{x}}{\sqrt{1 + \dfrac{1}{x^2}} + 1}$$
I understand how he simplifies $\frac{x}{x}$ to 1 but how does he simplify $\frac{\sqrt{x^2 + 1}}{x}$?
|
Since you are calculating the limit at $x\to\infty$ you may assume that $x>0$. Then
$$\frac{\sqrt{x^2+1}+x}x=\frac{\sqrt{x^2+1}}{\sqrt{x^2}}+\frac xx=\sqrt{\frac{x^2+1}{x^2}}+1=\sqrt{1+\frac{1}{x^2}}+1$$
|
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|
particular solution of a difference equation I am unable to find a particular solution of the following difference equation
$$
y[k-1]-5y[k]+6y[k+1]=-u[k-1]+4u[k]
$$
with $u[k]=\big(\frac{1}{2}\big)^k$.
This is what I tried so far. Because $u[k]=\big(\frac{1}{2}\big)^k$ we try $y_p[k]=\alpha \big(\frac{1}{2}\big)^k$. Substituting this is in the difference equation gives us
$$
\begin{align}
\alpha \bigg(\frac{1}{2}\bigg)^{k-1}-5\alpha \bigg(\frac{1}{2}\bigg)^k + 6\alpha \bigg(\frac{1}{2}\bigg)^{k+1} &= -\bigg(\frac{1}{2}\bigg)^{k-1} + 4 \bigg(\frac{1}{2}\bigg)^k\\
\bigg(\frac{1}{2}\bigg)^{k-1}\bigg(\alpha -\frac{5\alpha}{2}+\frac{6\alpha}{4}\bigg)&= \bigg(\frac{1}{2}\bigg)^{k-1}
\end{align}
$$
Thus $\Big(\alpha -\frac{5\alpha}{2}+\frac{6\alpha}{4}\Big)=1$ which results in $0\alpha=2$ which has no possible solution.
The problem is that I have no idea which particular solution I could try that solves this problem, namely the fact that $1-5/2+6/4=0$.
Any help would be greatly appreciated.
|
We change notation slightly. Note that $-u_{k-1}+4u_k=\frac{4}{2^k}-\frac{2}{2^k}=\frac{2}{2^k}$. We look for a solution of the shape $y_i=\alpha\frac{i}{2^i}$.
Substitute in $6y_{k+1}-5y_k+y_{k-1}$, forgetting about the $\alpha$ for a while. We get
$$\frac{6(k+1)}{2^{k+1}}-\frac{5k}{2^k}+\frac{k-1}{2^{k-1}}.$$
This simplifies to
$$\frac{1}{2^{k+1}}\left(6k+6 -10k+4k+4k-4\right).$$
I am sure you can finish from here.
|
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|
Calculus - Solving limits with square roots I am having trouble understanding how to solve this limit by rationalizing. I have the problem correct (I used Wolfram Alpha of course), but I still don't understand how it is completed. I was trying to solve this by multiplying both the numerator and the denominator by $\sqrt{x^2+11}+6$, but I am stuck. Is that the way to solve this? If so, could you walk me through it so I can solve others?
Problem:
|
A possible step-by-step solution: write $x=y+5$ (so that you are looking for a limit as $y\to 0$), and the denominator is $x-5=y$
$$\begin{align}
\sqrt{x^2+11} &= \sqrt{(y+5)^2+11} = \sqrt{y^2+10y + 36} = \sqrt{36}\sqrt{1+\frac{10}{36}y+\frac{y^2}{36}} \\
&= 6 \sqrt{1+\frac{5}{18}y+\frac{y^2}{36}}
\end{align}
$$
From there,
$$\begin{align}
\sqrt{x^2+11} - 6 &= 6 \left( \sqrt{1+\frac{5}{18}y+\frac{y^2}{36}} - 1\right)
\end{align}
$$
Now, if you know Taylor series, you can immediately apply the fact that $\sqrt{1+u} = 1+\frac{u}{2} + o(u)$ when $u\to 0$ to get that
$$\begin{align}
\sqrt{x^2+11} - 6 &= 6 \left( 1+\frac{5}{36}y+ o(y) - 1\right)
= 6 \left( \frac{5}{36}y+ o(y)\right)
= \frac{5}{6}y+ o(y)
\end{align}
$$
and now
$$\frac{\sqrt{x^2+11} - 6}{y} = \frac{\frac{5}{6}y+ o(y)}{y} = \frac{5}{6} + o(1) \xrightarrow[y\to 0]{} \frac{5}{6}.$$
|
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|
Finding bases such that the matrix representation is a block matrix where one submatrix is the identity matrix Problem: Let $L: \mathbb{R}^2 \rightarrow \mathbb{R}^3$ be a linear map with \begin{align*} [L]_{\alpha}^{\beta} = \begin{pmatrix} 2 & 3 \\ 4 & 6 \\ 6 & 9 \end{pmatrix} \end{align*} as the matrix representation with respect to the standard bases $\alpha$ for $\mathbb{R}^2$ and $\beta$ for $\mathbb{R}^3$. Now find a basis $\mathcal{V}$ for $\mathbb{R}^2$ and a basis $\mathcal{W}$ for $\mathbb{R}^3$ such that \begin{align*} [L]_{\mathcal{V}}^{\mathcal{W}} = \begin{pmatrix} \mathbb{I}_r & O \\ O & O \end{pmatrix} \end{align*} where each $O$ represents a block matrix with all zeroes and/or does not appear.
Attempt at solution: I'm not sure if I understand what's being asked here. Since the matrix $[L]_{\alpha}^{\beta}$ is with respect to the standard bases, we have $L(1,0) = (2,4,6)$ and $L(0,1) = (3,6,9)$. From this I determined the general formula of $L$ as $L(x,y) = (2x + 3y, 4x + 6y, 6x + 9y)$.
Now, I assume the condition $\begin{pmatrix} \mathbb{I}_r & O \\ O & O \end{pmatrix}$ means we want a matrix representation of the form $\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{pmatrix}$. Let $\mathcal{V} = \left\{v_1, v_2\right\}$ and $\mathcal{W} = \left\{w_1, w_2, w_3\right\}$ be the other two bases we seek. So we want \begin{align*} L(v_1) = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = 1 w_1 + 0 w_2 + 0 w_3 \qquad \text{and} \qquad L(v_2) = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = 0 w_1 + 1 w_2 + 0 w_3 \end{align*} Now I'm stuck, and I don't know how to find a concrete example of $\mathcal{V}$ and $\mathcal{W}$ that would fit with the explicit formula for $L$ I found earlier.
|
so as I mentioned in the comment, we first choose a proper basis $B$ in $\mathbb{R}^2$, we take
$$
B=\left\{\begin{pmatrix} 1 \\ 0 \end{pmatrix},\begin{pmatrix} -1.5 \\ 1 \end{pmatrix}\right\}
$$
and first keep the standard basis $A$ in $\mathbb{R}^3$, therefore our matrix looks like
$$
L^{B}_A=\begin{pmatrix} 2 & 0 \\ 4 & 0 \\ 6 & 0 \end{pmatrix}
$$
Next thing we do is to find the proper basis $C$ in $\mathbb{R}^3$, this is rather obvious and we take
$$
C=\left\{\begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix},\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \right\}
$$
and therefore our linear maps L looks with respect to the basis $B$ and $C$
$$
L^B_C=\begin{pmatrix} 1 & 0 \\ 0 & 0 \\ 0 & 0 \end{pmatrix}
$$
and that's it. By the way, the choice of
$ \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$ is arbitrary, you just need to fill up $\left\{\begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix}\right\}$ to a basis of $\mathbb{R}^3$.
The complete line of computation looks like
$$
\begin{pmatrix} 1 & 0 \\ 0 & 0 \\ 0 & 0 \end{pmatrix}=
\begin{pmatrix} 2 & 0 & 0\\ 4 & 1 & 0\\ 6 & 0 & 1 \end{pmatrix}^{-1}
\begin{pmatrix} 2 & 3 \\ 4 & 6 \\ 6 & 9 \end{pmatrix}
\begin{pmatrix} 1 & -1.5 \\ 0 & 1 \end{pmatrix}
$$
bests
|
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|
Question on proving quadratic inequality
Let $ax^2+bx+c$ = 0 be a quadratic equation and $\alpha$,$\beta$ are real roots. Condition for $\alpha < -1$ and $\beta > 1$. Show that $1 +\frac{c}{a}$ + $\left|\frac{b}{a}\right| < 0$.
I have tried but could not prove this inequality. I want to know how to solve this. Show that I can get an idea and solve further questions myself.
|
Without loss of generality, let $a \gt 0$.
Now firstly, Since $\alpha$ is a root of $f(x)$, hence $a\alpha^2+b\alpha+c=0$.
$\Rightarrow$ $\alpha = \frac {-b - \sqrt{b^2-4ac}}{2a} \lt -1$ $\Rightarrow$ $-b - \sqrt{b^2-4ac} \lt -2a$ $\Rightarrow$ $\sqrt{b^2-4ac} \gt 2a-b$ $\Rightarrow$ $b^2-4ac \gt 4a^2-4ab+b^2$ $\Rightarrow$ $a^2+ab-ac \lt 0$ $\Rightarrow$ $1-\frac ba+\frac ca \lt 0$.
Similarly, $\beta=\frac {-b+\sqrt {b^2-4ac}}{2a} \gt 1$ $\Rightarrow$ $\sqrt {b^2-4ac} \gt b+2a$ $\Rightarrow$ $b^2-4ac \gt b^2+4ab+4a^2$ $\Rightarrow$ $a^2+ab+ac \lt 0$ $\Rightarrow$ $ 1+\frac ba+\frac ca \lt 0$.
Hence, $1+\frac ca+|\frac ba| \lt 0$.
|
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|
Arrangements of Chairs in a Circle
Ten chairs are arranged in a circle. Find the number of subsets of this set of chairs that contain at least three adjacent chairs.
Hints only please!
This is a confusing worded-problem.
We could break it into, $7$ cases but that would take very long?
Case 1: 3 chairs adjacent.
Ways to do this: $$\binom{10}{3} \cdot \binom{7}{7} = 120$$
But I see that, for the next, $\binom{10}{4}$.
So it will be:
$$1 + \binom{10}{3} + \binom{10}{4} + \binom{10}{5} + \binom{10}{6} + \cdots + \binom{10}{9}$$
$$= 1 + 120 + 210 + 252+\cdots$$
But this isnt a legit method it looks like.
I am not sure how to use PIE/anything else here?
|
The number of sets of exactly three adjacent chairs is $10$. Either they're chairs $1$, $2$, and $3$, or $2,3,4$, or $3,4,5$, etc. After $8,9,10$ comes $9,10,1$ and then $10,1,2$.
The number of subsets of the remaining set of seven chairs is $2^7=128$.
However, if one just picks $10\times128=1280$, then one is counting some combinations more than once. For example, if one has $1,2,3,4,6,7,8$, then one counts each of these:
\begin{align}
& \underbrace{1,2,3,}\ 4, 6, 7, 8 \\[10pt]
& 1,\ \underbrace{2,3,4,}\ 6,7,8 \\[10pt]
& 1,2,3,4,\ \underbrace{6,7,8}
\end{align}
But they're really all three the same combination and should be counted only once. One should therefore expect a smaller number than $1280$.
So I would search for combinations in which no chairs are adjacent or in which exactly two chairs are adjacent, and subtract that from the total number of subsets.
|
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|
How prove $\lim x_n = \sqrt{\frac{a}{b}}$ if $b \neq 0$ for $x_1=c$ and $x_{n+1}=\frac{x_n^2 + \frac{a}{b}}{2x_n}$? Let $x_1=c$ and $x_{n+1}=\frac{x_n^2 + \frac{a}{b}}{2x_n}$. How prove $\lim x_n = \sqrt{\frac{a}{b}}$ if $b \neq 0$ ?
|
Let $r=\frac{a}{b}$ and assume that $r>0$ and $c>0$. To prove existence of the limit, rewrite the equation as follows:
$$
2x_{n+1}=x_n + \frac{r}{x_n}\quad (\star)
$$
Since $x_1=c>0$ and $r>0$, it follows by induction that all $x_n$ are positive. Applying the AM-GM Inequality, we see that $x_{n+1}\geq \sqrt{r}$ (for $n\geq 1$). Squaring both sides of $(\star)$ yields
$$
\begin{align*}
4x_{n+1}^2&=x_n^2 + 2r + \frac{r^2}{x_n^2}\\
\implies 4x_{n+1}^2-4r&=\left(x_n-\frac{r}{x_n}\right)^2\\
\implies \left|\frac{x_{n+1}^2-r}{x_n^2-r}\right|&=\frac{\left|1-\frac{r}{x_n^2}\right|}{4}.
\end{align*}
$$
Since $x_n\geq \sqrt{r}$ for $n\geq 2$, it follows that $\left|1-\frac{r}{x_n^2}\right|\leq 1$ for $n\geq 2$. Consequently
$$
\left|x_{n+1}^2-r\right|\leq \left(\frac{1}{4}\right)\cdots \left(\frac{1}{4}\right)\left(\frac{\left|1-\frac{r}{c^2}\right|}{4}\right)\left|c^2-r\right|=\frac{C}{4^n},
$$
for some positive constant $C$.
Taking the limit as $n\to\infty$, it follows that $\lim_{n\to\infty} x_{n+1}^2=r$, and therefore $\lim_{n\to\infty}x_n=\sqrt{r}$ as desired.
|
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|
Evaluating a Summation with a binomial Problem:
Evaluate for $n=11$$$\begin{align} \sin^{4n}\left(\frac{\pi}{4n}\right) + \cos^{4n}\left( \frac{\pi}{4n}\right) = \frac{1}{4^{2n-1}} \left[ \sum_{r=0}^{n-1} \binom{4n}{2r} \cos\left(1 - \frac{r}{n} \right) \pi \, + \frac{1}{2} \binom{4n}{2n} \right]. \end{align} $$
Sorry for this odd question. I saw this formula here on MSE which is quite helpful for a question I need to solve ($\sin^{4n}\frac{\pi}{4n} + \cos^{4n} \frac{\pi}{4n})$ for given values of $n$. Unfortunately I don't know how to evaluate this formula manually. I was thus hoping to use Wolfram Alpha to evaluate this for different values of $n$ but was unable to enter it correctly. I would be truly grateful if somebody would kindly show me how to input this formula into Wolfram Alpha or solve it for $n=11$. Many thanks in advance.
|
With some trivial manipulation it straighforward to check that we just have to compute:
$$ \sum_{r=0}^{2n}\binom{4n}{2r}\cos\frac{4\pi r}{n}=\text{Re}\sum_{r=0}^{2n}\binom{4n}{2r}\exp\left(8r\cdot\frac{2\pi i}{4n}\right).\tag{1} $$
Since:
$$ \sum_{r=0}^{2n}\binom{4n}{2r} z^{2r} = \frac{1}{2}\sum_{k=0}^{4n}\binom{4n}{k}\left(z^k+(-z)^k\right)=\frac{(1+z)^{4n}+(1-z)^{4n}}{2}\tag{2}$$
by taking $\omega=\exp\frac{2\pi i}{n}$ we have:
$$ \sum_{r=0}^{2n}\binom{4n}{2r}\cos\frac{4\pi r}{n}=\text{Re}\left(\frac{(1+\omega)^{4n}+(1-\omega)^{4n}}{2}\right)\tag{3}$$
then, since $1\pm \omega = 2\cos\left(\frac{\pi}{n}\right)\exp\left(\pm\frac{\pi i}{n}\right)$,
$$ \sum_{r=0}^{2n}\binom{4n}{2r}\cos\frac{4\pi r}{n} = \frac{1}{2}\left(2\cos\frac{\pi}{n}\right)^{4n}\left(\cos(4\pi)+\cos(-4\pi)\right)=\color{red}{\left(2\cos\frac{\pi}{n}\right)^{4n}}.\tag{4}$$
|
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|
is there a general formula for cases like $\sqrt{2}$ = $\frac {2}{\sqrt{2}}$? I just noticed that $\sqrt{2}$ is equal to $\frac {2}{\sqrt{2}}$:
$\sqrt{2} = 1.414213562$
$\frac {2}{\sqrt{2}} = 1.414213562$
It is confirmed by a hand-calculator.
I tried to proof this as follows:
$\sqrt{2}$ = $\frac {2}{\sqrt{2}}$
$2^{\frac 12} = \frac {2}{2^{\frac 12}}$
$2 = 2^{\frac 12} \cdot 2^{\frac 12}$
$2 = 2^{\frac 12 + \frac 12} $
$ 2 = 2^1 $
$ 2 = 2$
It is also true for:
$-\frac {2}{\sqrt5} = \frac {-2}{5}\sqrt 5$
I didn't know this relationship beforehand and it was new for me, my question is: Is there a general rule for this?
|
Yes. $$\frac{x^a}{x^b} = x^{a-b}$$ In this case we have $$\frac{2^1}{\sqrt{2}} = \frac{2^1}{2^{1/2}} = 2^{1- 1/2} = 2^{1/2} = \sqrt{2}$$
|
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|
Generalize multiples of $999...9$ using digits $(0,1,2)$ The smallest $n$ such that $9n$ uses only the three digits $(0,1,2)$ is $1358$, giving a product $12222$. For $99n$ this is $11335578$, giving $1122222222$. Similarly,
$999(111333555778)=111222222222222,$
$9999(1111333355557778) = 11112222222222222222, ...$
The product seems to be $k$ 1's followed $4k$ 2's for $k$ digit $999...9$. For $n$, this seems to be $1..3...5...7...8$, with $1, 3, 5$ $k$ times and $7$ $k-1$ or $0$ times. Can this pattern be proven and extended for any $k$?
|
It is readily verified that
$$\tag1(10^k-1)\cdot\frac{10^{4k}+2\cdot 10^{3k}+2\cdot 10^{2k}+2\cdot 10^k+2}{9} =\frac{10^{5k}-1}9+\frac{10^{4k}-1}{9}$$
and that the two fractions are in fact integers and the number on the right, being the sum of two repunits, is written with $1$ and $2$ only.
Remains to show that the numbers in $(1)$ are minimal.
Note that $(10^k-1)n=10^kn-n$ is the difference of two numbers of different length, so that $10^kn$ (as well as $n$) must start with at least $k-1$ digits $\in\{0,1,2\}$ and the $k$th digit $\in\{0,1,2,3\}$; actually, a $3$ as $k$th digit is only possible if it is followed by a digit $\le 2$ as otherwise no carray (or rather: borrow) orccurs. Also, $(10^k-1)n\equiv -n\pmod{10^k}$ so that the last $k$ digits of $n-1$ must be $\in\{7,8,9\}$. Note that
$$\left\lfloor\frac{(10^k-1)n}{10^k}\right\rfloor=n-\left\lceil\frac{n}{10^k} \right\rceil=(n-1)-\left\lfloor\frac{n}{10^k} \right\rfloor$$
so that we conclude about the next $k$ digits: We subtract someting from digits $7,8,9$ and obtain digits $0,1,2$; this is only possible if no borrow occurs, hence the next $k$ digits of $n-1$ are $\in\{5,6,7,8,9\}$. By repeating the argument, the next $k$ digits of $n-1$ are in $\{3,4,5,6,7,8,9\}$. Therefore, $n$ either has more than $4k$ digits, or $n-1$ is described by the regular expression [12][012]{k-1}[3-9]{k}[5-9]{k}[7-9]{k}. We can exclude the case of more than $4k$ digits as $(1)$ already beats that.
Thus we are left with the second case that can also be formulated as:
$$ n=\alpha\cdot 4^{3k}+\beta\cdot10^{2k}+\gamma\cdot 10^k+\delta+1$$
where $\alpha,\beta,\gamma,\delta$ are $k$-digit number with digits $\in\{0,1,2\}$, $\in\{3,\ldots,9\}$, $\in\{5,\ldots,9\}$, $\in\{7,\ldots,9\}$, respectively. Then
$$(10^k-1)n=\alpha\cdot 10^{4k}+(\beta-\alpha)\cdot 10^{3k}+(\gamma-\beta)\cdot 10^{2k} +(\delta-\gamma)\cdot10^k+(10^k-1-\delta)$$
A stated above, the subtractions $\delta-\gamma,\gamma-\beta,\beta-\alpha$ involve no borrows. Therefore any zero occuring in $\alpha$ would produce a digit $\ge 3$ in $\beta-\alpha$ and hence also in $(10^k-1)n$. We conclude that $\alpha$ contains no zeroes. But then $\alpha \ge 11\ldots 1$, $\beta\ge 33\ldots 3$, $\gamma\ge 55\ldots 5$, $\delta\ge 77\ldots 7$ and ultimately the solution $(1)$ is inded the minimal solution.
|
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|
Euler's proof of sum of natural numbers I've to recheck Euler's proof of the sum of the natural numbers, but I don't know exactly what it is? It has something to do with the $\zeta(s)$?
|
Euler was assuming the following things:
$$0+1+x+x^2+x^3+x^4+...=\sum_{k=0}^{\infty} x^k=\frac{1}{1-x},|x|<1$$
If we take the $\frac{d}{dx}$ (derivative) of that sum we get:
$$0+1+2x+3x^2+4x^3+5x^4+...=\sum_{k=0}^{\infty} (k+1)x^k=\frac{1}{(1-x)^2},|x|<1$$
Second thing he assumed was:
$$\sum_{k=0}^{\infty} (k+1)x^k=1-2+3-4+5-6+...=\frac{1}{4}$$
Eulers proof of the sum of all natural numbers:
$$\zeta(s)=\sum_{n=1}^{\infty} \frac{1}{n^s}=\frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+...\Longleftrightarrow$$
$$2^{-s}\cdot \zeta(s)=2^{-s}\cdot \left(\frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+...\right)\Longleftrightarrow$$
$$2^{-s}\cdot \zeta(s)=2^{-s}+4^{-s}+6^{-s}+8^{-s}+...\Longleftrightarrow$$
$$(1-2\cdot 2^{-s})\zeta(s)=(1-2\cdot 2^{-s})\left(\frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+...\right)\Longleftrightarrow$$
$$(1-2\cdot 2^{-s})\zeta(s)=1+2^{-s}+3^{-s}+...-2(2^{-s}+4^{-s}+6^{-s})\Longleftrightarrow$$
$$(1-2\cdot 2^{-s})\zeta(s)=1-2^{-s}+3^{-s}-4^{-s}+...$$
Now we choose $s=-1$:
$$(1-2\cdot 2^{--1})\zeta(-1)=1-2^{--1}+3^{--1}-4^{--1}+...\Longleftrightarrow$$
$$(1-2\cdot 2^1)\zeta(-1)=1-2^1+3^1-4^1+...\Longleftrightarrow$$
$$(1-2\cdot 2)\zeta(-1)=1-2+3-4+...\Longleftrightarrow$$
$$-3\cdot\left(\sum_{n=1}^{\infty} \frac{1}{n^{-1}}\right)=1-2+3-4+...\Longleftrightarrow$$
$$-3\cdot\left(\sum_{n=1}^{\infty} n\right)=1-2+3-4+...\Longleftrightarrow$$
$$-3\cdot\left(1+2+3+4+5+...\right)=1-2+3-4+...\Longleftrightarrow$$
$$-3\cdot\left(1+2+3+4+5+...\right)=\frac{1}{4}\Longleftrightarrow$$
$$1+2+3+4+5+...=\frac{\frac{1}{4}}{-3}\Longleftrightarrow$$
$$1+2+3+4+5+...=-\frac{1}{12}\Longleftrightarrow$$
$$\sum_{n=1}^{\infty} n=-\frac{1}{12}$$
|
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|
Find lim$_{x \to 0}\left(\frac{1}{x} - \frac{\cos x}{\sin x}\right).$ $$\lim_{x \to 0} \left(\frac{1}{x} - \frac{\cos x}{\sin x}\right) = \frac{\sin x - x\cos x}{x\sin x}= \frac{0}{0}.$$
L'Hopital's: $$\lim_{x \to 0} \frac{f'(x)}{g'(x)} = -\frac{1}{x^2} + \frac{1}{\sin ^2x} = \frac{0}{0}.$$
Once again, using L'Hopital's: $$\lim_{x \to 0} \frac{f''(x)}{g''(x)} = \frac{2}{x^3}- \frac{2\cos x}{\sin ^3x} = \frac{0}{0}\,\ldots$$ The terms are getting endless here. Any help? Thanks.
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$$\begin{align}\lim_{x\to 0}\frac{\sin x-x\cos x}{x\sin x}&=\lim_{x\to 0}\frac{\cos x-(\cos x-x\sin x)}{\sin x+x\cos x}\\&=\lim_{x\to 0}\frac{x\sin x}{\sin x+x\cos x}\\&=\lim_{x\to 0}\frac{x}{1+\frac{x}{\sin x}\cdot \cos x}\\&=\frac{0}{1+1\cdot 1}\end{align}$$
|
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Solution verification for $y'$ when $y=\sin^{-1}(\frac{2x}{1+x^2})$ I was required to find $y'$ when $y=\sin^{-1}(\frac{2x}{1+x^2})$
This is my solution.
Above when I put $\sqrt{x^4-2x+1}=\sqrt{(1-x^2)^2}$ then I get the correct answer but when I put $\sqrt{x^4-2x+1}=\sqrt{(x^2-1)^2}$ then I get something else.
Now my question is that which one is correct and why? How you will come to know that we should put $\sqrt{x^4-2x+1}=\sqrt{(1-x^2)^2}$ and not $\sqrt{x^4-2x+1}=\sqrt{(x^2-1)^2}$
Kindly help me.
|
HINT:
For real $a,$ $$\sqrt{a^2}=|a|$$
$=+a$ if $a\ge0$ and $=-a$ if $a<0$
Things will be clearer with the following :
Using my answer here: showing $\arctan(\frac{2}{3}) = \frac{1}{2} \arctan(\frac{12}{5})$ and $\arctan y=\arcsin\dfrac y{\sqrt{y^2+1}}$
$$2\arctan x=\begin{cases} \arcsin\frac{2x}{1+x^2} &\mbox{if } x^2<1\\ \pi+\arcsin\frac{2x}{1+x^2} & \mbox{if } x^2>1\end{cases} $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Definite Integration with Trigonometric Substitution I'm working on a question that involves using trigonometric substitution on a definite integral that will later use u substitution but I am not sure how to go ahead with this.
$$\int_1^2\frac1{x^2\sqrt{4x^2+9}}dx$$
My first step was to use $\sqrt{a^2+x^2}$ as $x=a\tan\theta$ to get...
$$2x=3\tan\theta :x=\frac32\tan\theta$$
$$dx=\frac32\sec^2\theta$$
Substituting:
$$\int\frac{\frac32\sec^2\theta}{\frac94\tan^2\theta\sqrt{9\tan^2\theta+9}}$$
The problem here is how do I change the limit it goes to?
$$\frac43=\tan\theta$$
and
$$\frac23=\tan\theta$$
Following DR.MV's answer so far..
$$\frac29\int_{\tan^-1(\frac23)}^{\tan^-1(\frac43)}\frac{\sec^2\theta}{\tan^2\theta\sqrt{9\sec^2\theta}}d\theta$$
$$=\frac29\int_{\tan^-1(\frac23)}^{\tan^-1(\frac43)}\frac{\sec\theta}{\tan^2\theta}d\theta$$
$$=\frac29\int_{\tan^-1(\frac23)}^{\tan^-1(\frac43)}\frac{\cos\theta}{\sin^2\theta}d\theta$$
Now $u=\sin\theta$ so $du=\cos\theta d\theta$
$$=\frac29\int_{?}^{?}\frac{1}{u^2}du$$
This is where I am stuck now...
|
$$\int_1^2\frac1{x^2\sqrt{4x^2+9}}dx = \int_a^b\frac{\frac{2}{3}\sec^2\theta}{\frac{9}{4}\tan^2\theta\sqrt{9\tan^2\theta+9}}d\theta$$
where $a = \tan^{-1}\frac{2}{3}$ and $b = \tan^{-1}\frac{4}{3}$
$$ \int_a^b\frac{2\sec\theta}{9\tan^2\theta}d\theta = \left(-\frac{2\text{cosec}\theta}{9} \right)_a^b$$
Hint: $\text{cosec}\left(\tan^{-1}\frac{4}{3}\right) = \frac{5}{4}$, $\text{cosec}\left(\tan^{-1}\frac{2}{3}\right) = \frac{\sqrt{13}}{2}$
EDIT: let $\tan^{-1}\frac{4}{3}=d $
$$ \frac{4}{3} = \tan d$$
$$\frac{4}{3} = \frac{\sin d}{\cos d} $$
$$ \frac{1}{\sin d} = \text{cosec}\ d = \frac{3}{4}\sec d$$
$$ \sec^2d = \tan^2 d + 1 = 1+ \left(\frac{4}{3}\right)^2 = \frac{25}{9}$$
$$\sec d =\frac{5}{3}$$
$$ \text{cosec}\ d = \frac{3}{4}\times \frac{5}{3} =\frac{5}{4}$$
You can do the same for the second one.
|
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|
Prove the inequality $\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$ with the constraint $abc=1$ If $a,b,c$ are positive reals such that $abc=1$, then prove that $$\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$$ I tried substituting $x/y,y/z,z/x$, but it didn't help(I got the reverse inequality). Need some stronger inequality. Thanks.
|
Let $a=\frac{x^2}{yz}$ and $b=\frac{y^2}{xz}$, where $x$, $y$ and $z$ are positives.
Hence, $c=\frac{z^2}{xy}$ and we need to prove that:
$$\sum_{cyc}\frac{x}{\sqrt{x^2+8yz}}\geq1.$$
Now, by Holder
$$\left(\sum_{cyc}\frac{x}{\sqrt{x^2+8yz}}\right)^2\sum_{cyc}x(x^2+8yz)\geq(x+y+z)^3.$$
Thus, it remains to prove that
$$(x+y+z)^3\geq\sum_{cyc}x(x^2+8yz)$$ or
$$\sum_{cyc}z(x-y)^2\geq0.$$
Done!
Another way:
$$\sum_{cyc}\sqrt{\frac{a}{a+8}}=\sum_{cyc}\frac{x}{\sqrt{x^2+8xy}}\geq\sum_{cyc}\frac{x^{\frac{4}{3}}}{x^{\frac{4}{3}}+y^{\frac{4}{3}}+z^{\frac{4}{3}}}=1$$
Also we can use the Contradiction method.
Let $\frac{a}{a+8}=\frac{p^2}{9}$, $\frac{b}{b+8}=\frac{q^2}{9}$ and $\frac{c}{c+8}=\frac{q^2}{9}$, where $p$, $q$ and $r$ are positives.
Hence, we need to prove that $p+q+r\geq3.$
But the condition $abc=1$, gives
$$8^3=\prod_{cyc}\left(\frac{9}{p^2}-1\right)$$ or
$$81=57p^2q^2r^2-p^2q^2-p^2r^2-q^2r^2+9(p^2+q^2+r^2).$$
Now, let $p+q+r<3$, $p=kx$, $q=y$ and $r=z$, where $k>0$ and $x+y+z=3$.
Hence, $kx+y+z<3=x+y+z$, which gives $0<k<1$.
Thus, since $9-p^2-q^2>(p+q)^2-p^2-q^2>0$, we obtain:
$$81=57p^2q^2r^2-p^2q^2-p^2r^2-q^2r^2+9(p^2+q^2+r^2)=$$
$$=k^2x^2(57y^2z^2-y^2-z^2+9)-y^2z^2+9(y^2+z^2)<$$
$$<x^2(57y^2z^2-y^2-z^2+9)-y^2z^2+9(y^2+z^2),$$
which is contradiction because we'll prove now that
$$57x^2y^2z^2-x^2y^2-x^2z^2-y^2z^2+9(x^2+y^2+z^2)\leq81.$$
Indeed, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, it's obvious that our inequality is equivalent to $f(w^3)\leq0$, where $f$ is a convex function.
Id est, it's enough to prove the last inequality for an extremal value of $w^3$,
which happens in the following cases.
*
*$y=x$, $z=3-2x$, where $0<x<1.5$, which gives
$$x(x-1)^2(9+15x+19x^2-19x^3)\geq0,$$
which is true for $0<x<\frac{3}{2}$;
*$w^3=0$.
Let $z=0$ and $y=3-x$, where $0<x<3$.
We obtain, $x(4x^3-12x^2-9x+54)\geq0$, which is obvious.
Done again!
|
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|
Number theory with binary quadratic 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.
Given $$ \frac {x^2-y^2+2y-1}{y^2-x^2+2x-1} = 2$$ find $x-y$
I'm not sure if given choices is right... (A)2 (B)3 (C)4 (D)5 (E)6
I've tried to move them $$x^2-y^2+2y-1 = 2y^2-2x^2+4x-2$$
$$x^2-y^2+2y-1 - 2y^2+2x^2-4x+2 = 0$$
$$3x^2-3y^2+2y-4x+1=0$$
$$(3x-1)(x-1)-(3y-1)(y+1)+1=0$$
I've stuck in here, not sure if I've found x and y, or not...
EDIT: I've move other questions to other posts, thanks for helping me identifying the questions category.
|
\begin{align} 2&= \frac {x^2-y^2+2y-1}{y^2-x^2+2x-1} \\
&=\frac {x^2-(y^2-2y+1)}{y^2-(x^2-2x+1)} \\
&= \frac {x^2-(y-1)^2}{y^2-(x-1)^2} \\
&= \frac {(x-y+1)(x+y-1)}{(y+x-1)(y-x+1)}\end{align}
$$\implies {x-y+1}=2y-2x+2\\
3(x-y)=1 $$
Thus, $x-y=\dfrac13$
|
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|
Polynomial of $11^{th}$ degree Let $f(x)$ be a polynomial of degree $11$ such that $f(x)=\frac{1}{x+1}$,for $x=0,1,2,3.......,11$.Then what is the value of $f(12)?$
My attempt at this is:
Let $f(x)=a_0+a_1x+a_2x^2+a_3x^3+......+a_{11}x^{11}$
$f(0)=\frac{1}{0+1}=1=a_0$
$f(1)=\frac{1}{1+1}=\frac{1}{2}=a_0+a_1+a_2+a_3+......+a_{11} $
$f(2)=\frac{1}{2+1}=\frac{1}{3}=a_0+2a_1+4a_2+8a_3+......+2^{11}a_{11} $
.
.
.
$f(11)=\frac{1}{11+1}=\frac{1}{12}=a_0+11a_1+11^2a_2+11^3a_3+......+11^{11}a_{11} $
for calculating $f(12)$, I need to calculate $a_0,a_1,a_2,....,a_11$ but I could solve further.Is my approach right,how can I solve further or there is another right way to solve it.
$(A)\frac{1}{13}$
$(B)\frac{1}{12}$
$(C)0 $
$(D)\frac{1}{7}$
which one is correct answer?
|
Somewhat less vague...
$(x+1)f(x)-1$ is a polynomial of degree 12 with roots at every integer in $[0,11]$, so could be $$(x+1)f(x)-1 = A \prod_{c=0}^{11} x-c$$
for some/any (nonzero) constant $A$.
When $x=-1$, we have $(0)f(-1)-1 = A (-1)^{12} 12!$, or $-1 = A \, 12!$ and discover only $A = \frac{-1}{12!}$ is consistent with the givens. (Why $-1$? Because it is the only choice we haven't already used (we have used the integers 0, ... 11) that makes some expression containing $x$s zero.)
Hence, $13 f(12) - 1 = \frac{-1}{12!} 12!$ and $f(12) = \frac{-1+1}{13} = 0$.
|
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|
Solve $10x^4-7x^2(x^2+x+1)+(x^2+x+1)^2=0$ How to solve the following equation?
$$10x^4-7x^2(x^2+x+1)+(x^2+x+1)^2=0$$
My attempt:
$$ 10x^4 - (7x^2+1)(x^2+x+1)=0$$
Thats all i can
Update
Tried to open brakets and simplify:
$$(7x^2+1)(x^2+x+1) = 7x^4+7x^3+7x^2+x^2+x+1=7x^4+7x^3+8x^2+1 $$
$$10x^4 - (7x^2+1)(x^2+x+1)= 3x^4-7x^3-8x^2-1=0 $$
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$10x^4-7x^2(x^2+x+1)+(x^2+x+1)^2=0$
Divide by $x^4$ on both sides,
$10-\frac{7x^2(x^2+x+1)}{x^4}+\frac{(x^2+x+1)^2}{x^4}=0$
$10-7(1+\frac{1}{x}+\frac{1}{x^2})+(1+\frac{1}{x}+\frac{1}{x^2})^2=0$
Put $(1+\frac{1}{x}+\frac{1}{x^2})=t$
$10-7t+t^2=0$,solving we get $t=2,5$
when $t=2$
$1+\frac{1}{x}+\frac{1}{x^2}=2$
simplify we get,$x^2-x-1=0$.............(1)
when $t=5$
$1+\frac{1}{x}+\frac{1}{x^2}=5$
simplify we get,$4x^2-x-1=0...........(2)$
Solve (1) and (2) and get the answer.
|
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|
Am I getting the right answer for the integral $I_n= \int_0^1 \frac{x^n}{\sqrt {x^3+1}}\, dx$?
Let $I_n= \int_0^1 \dfrac{x^n}{\sqrt {x^3+1}}\, dx$. Show that $(2n-1)I_n+2(n-2)I_{n-3}=2 \sqrt 2$ for all $n \ge 3$. Then compute $I_8$.
I get an answer for $I_8={{2 \sqrt 2} \over 135}(25-16 \sqrt 2)$, could somebody please check against my answer and see if I made a mistake.
|
If we set
$$
I_n=\int_0^1\frac{x^n}{\sqrt{x^3+1}}\,\mathrm{d}x\tag{1}
$$
Then, integration by parts gives
$$
\begin{align}
I_{n+3}+I_n
&=\int_0^1\frac{x^n(x^3+1)}{\sqrt{x^3+1}}\,\mathrm{d}x\\
&=\int_0^1x^n\sqrt{x^3+1}\ \mathrm{d}x\\
&=\frac1{n+1}\int_0^1\sqrt{x^3+1}\ \mathrm{d}x^{n+1}\\
&=\frac1{n+1}\left[\sqrt2-\int_0^1\frac32\frac{x^{n+3}}{\sqrt{x^3+1}}\,\mathrm{d}x\right]\tag{2}
\end{align}
$$
Applying a bit of algebra to $(2)$ yields
$$
\left(2n+5\right)I_{n+3}+(2n+2)I_n=2\sqrt2\tag{3}
$$
which, after substituting $n\mapsto n-3$, gives
$$
\bbox[5px,border:2px solid #C0A000]{(2n-1)I_n+(2n-4)I_{n-3}=2\sqrt2}\tag{4}
$$
Next, substitute $x\mapsto(x-1)^{1/3}$:
$$
\begin{align}
I_8
&=\int_0^1\frac{x^8}{\sqrt{x^3+1}}\,\mathrm{d}x\\
&=\frac13\int_1^2\frac{(x-1)^2}{\sqrt{x}}\,\mathrm{d}x\\
&=\frac13\left[\frac25x^{5/2}-\frac43x^{3/2}+2x^{1/2}\right]_1^2\\
&=\bbox[5px,border:2px solid #C0A000]{\frac{14\sqrt2-16}{45}}\tag{5}
\end{align}
$$
|
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|
Common solutions of two inequations Find the real values of $a$ for which the inequations $x^2-4x-6a\leq 0$ and $x^2+2x+a\leq0$ have only one real solution common.
My attempt:
Let $\alpha$ be one real common root of two inequations.
then $ \alpha^2-4\alpha-6a\leq 0$.......(1)
$\alpha^2+2\alpha+a\leq 0$.......(2)
Subtract the two inequations,we get
$-6\alpha-7a\leq0$ but could not solve further for values of $a$.
$\alpha^2-4\alpha-6a\leq 0$ can be written as $(\alpha-2)^2-6a-4\leq0$
$\alpha^2+2\alpha+a\leq 0$ can be written as $(\alpha+1)^2+a-1\leq0$
Any hint will be helpful for me.
|
Hint:
$$x^2-4x-6a \le 0 \iff 2-\sqrt{4+6a} \le x \le 2+\sqrt{4+6a} \tag{A}$$
$$x^2+2x+a \le 0 \iff -1-\sqrt{1-a} \le x \le -1+\sqrt{1-a} \tag{B}$$
Now find the situations when $A \cap B$ is exactly one point, to get $a \in \{0, 1\}$.
|
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|
Derivation of Gradshteyn and Ryzhik integral 3.876.1 (in question) In the Gradshteyn and Ryzhik Table of Integrals, the following integral appears (3.876.1, page 486 in the 8th edition):
\begin{equation}
\int_0^{\infty} \frac{\sin (p \sqrt{x^2 + a^2})}{\sqrt{x^2 + a^2}} \cos (bx) dx =
\begin{cases}
\frac{\pi}{2} J_0 \left( a \sqrt{p^2 - b^2} \right) & 0 < b < p \\
0 & b > p > 0
\end{cases}
\end{equation}
for $a > 0$, where $J_0$ is the Bessel function.
I am interested how this result could be derived (not necessarily rigorously proved), especially the range of $p$ such that the value is $0$.
Thank you.
|
$$I(a,b,p)\equiv\int_0^{\infty}\frac{\sin\left(p\sqrt{x^2+a^2}\right)}{\sqrt{x^2+a^2}}\cos(bx)\,dx\\\\
$$
Enforcing the substitution $x\to a\sinh x$ and assuming that $a>0$ yields
$$\begin{align}
I(a,b,p)&=\int_0^{\infty}\sin\left(pa\cosh x\right)\cos(ba\sinh x)\,dx\\\\
&=\frac12\int_0^{\infty}\left(\sin\left(pa\cosh x+ba\sinh x\right)+\sin\left(pa\cosh x-ba\sinh x\right)\right)\,dx\\\\
&=\frac12\int_{-\infty}^{\infty}\sin\left(pa\cosh x+ba\sinh x\right)\,dx
\end{align}$$
Recalling that
$$A\cosh x+B\sinh x=
\begin{cases}
\sqrt{A^2-B^2}\cosh(x-\text{artanh}(B/A))&,0<B<A\\\\
\sqrt{B^2-A^2}\sinh(x-\text{artanh}(A/B))&,B>A>0
\end{cases}
$$
we have
$$I(a,b,p)=
\begin{cases}
\frac12\int_{-\infty}^{\infty}\sin\left(a\sqrt{p^2-b^2}\cosh x\right)\,dx&,0<b<p\\\\
\frac12\int_{-\infty}^{\infty}\sin\left(a\sqrt{b^2-p^2}\sinh x\right)\,dx&,b>p>0
\end{cases}
$$
Noting that the integrand of the first integral is an even function of $x$, while the integrand of the second integral is an odd function of $x$ reveals
$$I(a,b,p)=
\begin{cases}
\int_{0}^{\infty}\sin\left(a\sqrt{p^2-b^2}\cosh x\right)\,dx&,0<b<p\\\\
0&,b>p>0
\end{cases}
$$
From Equation $(10.9.9)$ HERE, we see that
$$\int_{0}^{\infty}\sin\left(a\sqrt{p^2-b^2}\cosh x\right)\,dx=\frac{\pi}{2}J_0\left(a\sqrt{p^2-b^2}\right)$$
for $a\sqrt{p^2-b^2}>0$ and $p>b>0$
whereupon we have the final result
$$I(a,b,p)=
\begin{cases}
\frac{\pi}{2}J_0\left(a\sqrt{p^2-b^2}\right)\,dx&,0<b<p\\\\
0&,b>p>0
\end{cases}
$$
for $a>0$.
|
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|
Calculate the limit $\lim_{x\to 0} \left(\frac 1{x^2}-\cot^2x\right)$ The answer of the given limit is $2/3$, but I cannot reach it. I have tried to use the L'Hospital rule, but I couldn't drive it to the end. Please give a detailed solution!
$$\lim_{x\to 0} \left(\dfrac 1{x^2}-\cot^2x\right)$$
|
Here's a non-series solution.
$$\lim_{x \rightarrow 0} \left(\frac{1}{x^2} - \cot^2(x)\right) = \lim_{x \rightarrow 0} \frac{\sin^2(x) - x^2\cos^2(x)}{x^2\sin^2(x)}$$
Instead of directly doing l'Hospital, we can make our life easier using $\lim_{x \rightarrow 0}\frac{\sin(x)}{x} = 1$ Or rather, $\lim_{x \rightarrow 0}\frac{x}{\sin(x)} = 1$:
$$\lim_{x \rightarrow 0} \frac{\sin^2(x) - x^2\cos^2(x)}{x^2\sin^2(x)} = \lim_{x \rightarrow 0}\left(\frac{x}{\sin(x)}\right)^2\left(\frac{\sin^2(x) - x^2\cos^2(x)}{x^4}\right) = \lim_{x \rightarrow 0} \frac{\sin^2(x) - x^2\cos^2(x)}{x^4}$$
Now l'Hospital. Note that we can freely factor out constants, and also $\cos(x)$ (as it approaches 1 as $x \rightarrow 0$):
\begin{align*} \lim_{x \rightarrow 0} \frac{\sin^2(x) - x^2\cos^2(x)}{x^4} &= \lim_{x \rightarrow 0} \frac{2\sin(x)\cos(x) - 2x\cos^2(x) + 2x^2\sin(x)\cos(x)}{4x^3} \\
&= \frac{1}{2}\lim_{x \rightarrow 0}\frac{\sin(x) - x\cos(x) + x^2\sin(x)}{x^3} \\
&= \frac{1}{2}\lim_{x \rightarrow 0}\frac{\cos(x) - \cos(x) + x\sin(x) + 2x\sin(x) + x^2\cos(x)}{3x^2} \\
&= \frac{1}{6}\lim_{x \rightarrow 0}\frac{3x\sin(x) + x^2\cos(x)}{x^2} \\
&= \frac{1}{6}\left(\lim_{x \rightarrow 0}3\frac{\sin(x)}{x} + \lim_{x \rightarrow 0} \cos(x)\right) \\
&= \frac{1}{6}(3 + 1) = \frac{2}{3}\end{align*}
|
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|
how can i prove this trigonometry equation I need help on proving the following:
$$\frac{\cos {7x} - \cos {x} + \sin {3x}}{ \sin {7x} + \sin {x} - \cos {3x} }= -\tan {3x}$$
So far I've only gotten to this step:
$$\frac{-2 \sin {4x} \sin {3x} + \sin {x}}{ 2 \sin {4x} \cos {3x} - \cos {x}}$$
Any help would be appreciated as trigonometry is not my forte.
|
You are almost there (I think you have a typo in the expression you got) :
Using
$$\cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}$$
$$\sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$$
we have
$$\begin{align}\frac{\cos(7x)-\cos x+\sin(3x)}{\sin(7x)+\sin x-\cos(3x)}&=\frac{-2\sin(4x)\sin(3x)+\sin(3x)}{2\sin(4x)\cos(3x)-\cos(3x)}\\&=\frac{\sin(3x)(-2\sin(4x)+1)}{\cos(3x)(2\sin(4x)-1)}\\&=\frac{\sin(3x)}{\cos(3x)}\cdot \frac{-2\sin(4x)+1}{-(-2\sin(4x)+1)}\end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove using mathematical induction that $x^{2n} - y^{2n}$ is divisible by $x+y$ Prove using mathematical induction that
$(x^{2n} - y^{2n})$ is divisible by $(x+y)$.
Step 1: Proving that the equation is true for $n=1 $
$(x^{2\cdot 1} - y^{2\cdot 1})$ is divisible by $(x+y)$
Step 2: Taking $n=k$
$(x^{2k} - y^{2k})$ is divisible by $(x+y)$
Step 3: proving that the above equation is also true for $(k+1)$
$(x^{2k+2} - y^{2k+2})$ is divisible by $(x+y)$.
Can anyone assist me what would be the next step? Thank You in advance!
|
Hint: Rewrite:
$$x^{2k+2}-y^{2k+2}=(x^{2k+2}-y^{2k}x^2)+(y^{2k}x^2-y^{2k+2})=x^2(x^{2k}-y^{2k})+y^{2k}(x^2-y^2).$$
Added: Note it can be proved without induction:
$$x^{2n}-y^{2n}=(x^2)^n-(y^2)^n=(x^2-y^2)(x^{2(n-1)}+x^{2(n-2)}y^2+\dots+x^2y^{2(n-2)}+y^{2(n-1)}),$$
and the first factor is divisible by $x+y$.
|
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|
How to integrate ${x^3}/(x^2+1)^{3/2}$? How to integrate $$\frac{x^3}{(x^2+1)^{3/2}}\ \text{?}$$
I tried substituting $x^2+1$ as t, but it's not working
|
$x^2+1=t^2\Rightarrow x\,dx=t\,dt\;,\;x^2=t^2-1$
$$\int\frac{x^3}{(x^2+1)^{\frac{3}{2}}}dx=\int\frac{t^2-1}{t^2}dt=\int(1-t^{-2})dt=t+\frac{1}{t}+C$$
$t=\sqrt{1+x^2}\Rightarrow$ answer$=\sqrt{1+x^2}+\frac{1}{\sqrt{1+x^2}}+C=\frac{x^2+2}{\sqrt{x^2+1}}+C$
|
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|
Is $\left(45+29\sqrt{2}\right)^{1/3} + \left(45-29\sqrt{2}\right)^{1/3}$ an integer? The problem is the following:
Prove that this number
$$x = \left(45+29\sqrt{2}\right)^{1/3} + \left(45-29\sqrt{2}\right)^{1/3}$$
is an integer. Show which integer it is.
I thought that it has some relations with something like complex numbers such as the set $N(\sqrt{2})$ or the some kind of integer polynomial which has root $x$ and therefore show that it has integer solutions.
|
Given the hint that $(45+29\sqrt{2})^{1/3} + (45-29\sqrt{2})^{1/3}$ is an integer.
One should first investigate whether
$$(45 + 29\sqrt{2})^{1/3} = a + b\sqrt{2}$$
for some integers $a, b$.
Notice $$(a+b\sqrt{2})^3 = a ( a^2 + 6 b^2 ) + (3a^2 + 2b^2)b\sqrt{2}$$
This suggest us to look for integer solutions for following equation:
$$\begin{cases}
a( a^2 + 6b^2) &= 45\\
b(3a^2 + 2b^2) &= 29
\end{cases}
$$
Since $b | 29$ and $29$ is a prime, the most natural guess is $b = 1$. Substitute $b = 1$ in the second equation, we find
$a = \sqrt{\frac{29 - 2}{3}} = 3$ which also solves the first equation. This leads to
$$(45+29\sqrt{2})^{1/3} = 3 + \sqrt{2} \implies
(45-29\sqrt{2})^{1/3} = 3 - \sqrt{2}\\ \implies
(45+29\sqrt{2})^{1/3} + (45-29\sqrt{2})^{1/3} = 6$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Find all values that solve the equation For which values a, the equation
$$ a\sin{x}+(a+1)\sin^2{\frac{x}{2}} + (a-1)\cos^2{\frac{x}{2}} =1 $$
has a solution?
My idea: I think it's possible to factorize equation or reduce equation to the form like: $a(\sin^2{\frac{x}{2}} + \cos^2{\frac{x}{2}}) =1 $
Let's go:
$$ 2a\sin{\frac{x}{2}}\cos{\frac{x}{2}} + asin^2{\frac{x}{2}} + sin^2{\frac{x}{2}} + a\cos^2{\frac{x}{2}} - \cos^2{\frac{x}{2}} = 1$$
$$ a\left(sin{\frac{x}{2}}+\cos{\frac{x}{2}}\right)^2 = 1 - \sin^2{\frac{x}{2}} + \cos^2{\frac{x}{2}}$$
$$ a\left(\sin{\frac{x}{2}}+\cos{\frac{x}{2}}\right)^2 =2\cos^2{\frac{x}{2}}$$
I can't finish...
|
Note that your equation reduces to
\begin{align}
1
&= a\sin{x}+(a+1)\sin^2{\frac{x}{2}} + (a-1)\cos^2{\frac{x}{2}} \\
&= a \sin x +a \left(\sin^2{\frac{x}{2}} + \cos^2{\frac{x}{2}}\right) +
\sin^2{\frac{x}{2}}- \cos^2{\frac{x}{2}} \\
&= a \sin x +a + \sin^2{\frac{x}{2}}- \cos^2{\frac{x}{2}} \\
&= a \sin x +a - \cos(x) \\
\end{align}
And this could be rewritten as
$$
\frac{1 + \cos x}{1 + \sin x} = a
$$
for $\sin x \ne -1$ or
$$
\cos x = -1
$$
for $\sin x = -1$ which can not happen.
So we assume $\sin x \ne -1$.
|
{
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|
Find the equation of the circle. Find the equation of the circle whose radius is $5$ which touches the circle $x^2 + y^2 - 2x -4y - 20 = 0$ externally at the point $(5,5)$
|
Express that the circle is through the point $(5,5)$
$$(5-x_c)^2+(5-y_c)^2=5^2,$$
and that the gradients are collinear a this point
$$(x_c-5)(2\cdot5-4)-(y_c-5)(2\cdot 5-2)=0.$$
From the second equation,
$$y_c=\frac{3x_c+5}4,$$
and plugging in the first,
$$x_c^2-10x_c+9=0.$$
The solutions are $(x_c,y_c)=(1,2)$ and $(9,8)$ and we choose the second.
$$\color{green}{(x-9)^2+(y-8)^2=5^5}.$$
|
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|
The value of the definite integral The value of the definite integral $\displaystyle\int\limits_0^\infty \frac{\ln x}{x^2+4} \, dx$ is
(A) $\dfrac{\pi \ln3}{2}$ (B) $\dfrac{\pi \ln2}{3}$ (C) $\dfrac{\pi \ln2}{4}$ (D) $\dfrac{\pi \ln4}{3}$
I tried using integration by parts,
\begin{align}
& \int_0^\infty \frac{\ln x}{x^2+4}dx = \ln x\int_0^\infty \frac{1}{x^2+4}dx-\int_0^\infty \left(\frac{d}{dx}\ln x\int_0^\infty \frac{1}{x^2+4}\right) \, dx \\[10pt]
= {} & \left[\ln x \frac{1}{2}\tan^{-1}\frac{x}{2}\right]-\int_0^\infty \frac{1}{x}\frac{1}{2}\tan^{-1}\frac{x}{2} \, dx
\end{align}
I could not move ahead. Can someone help me to get final answer?
|
Let $$\displaystyle I = \int_{0}^{\infty}\frac{\ln x}{x^2+4}dx = \int_{0}^{2}\frac{\ln x}{x^2+4}dx+\underbrace{\int_{2}^{\infty}\frac{\ln x}{x^2+4}dx}_{J}$$
Now for Calculation of $$\displaystyle J = \int_{2}^{\infty}\frac{\ln x}{x^2+4}dx\;,$$ Now put $\displaystyle x=\frac{4}{y}\;,$ Then $\displaystyle dx = -\frac{4}{y^2}dy$
So we get $$\displaystyle J = -\int_{2}^{0}\left[\frac{\ln 4-\ln y}{y^2+4}\right]dy = \int_{0}^{2}\frac{\ln 4}{x^2+4}dy-\int_{0}^{\infty}\frac{\ln y}{x^2+4}dx$$
Above we have used $$\displaystyle \bullet\; \int_{a}^{b}f(y)dy = \int_{a}^{b}f(x)dx$$
And we also used $$\displaystyle \bullet\; \int_{a}^{b}f(x)dx = -\int_{b}^{a}f(x)dx$$
Now Put value of $\displaystyle J$ in $\displaystyle I\;,$ So we get
$$\displaystyle I = \int_{0}^{2}\frac{\ln x}{x^2+4}dx+\int_{0}^{2}\frac{\ln 4}{x^2+4}dx-\int_{0}^{2}\frac{\ln x}{x^2+4}dx$$
So we get $$\displaystyle I = \ln 4\cdot \left[\tan^{-1}\left(\frac{x}{2}\right)\right]_{0}^{2} = \ln (4)\cdot \frac{1}{2}\cdot \frac{\pi}{4}= \frac{\ln (2)\cdot \pi}{4}$$
|
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|
How to factor intricate polynomial $ ab^3 - a^3b + a^3c -ac^3 +bc^3 - b^3c $ I would like to know how to factor the following polynomial.
$$ ab^3 - a^3b + a^3c -ac^3 +bc^3 - b^3c $$
What is the method I should use to factor it? If anyone could help.. Thanks in advance.
|
$ab^3 - a^3b + a^3c -ac^3 +bc^3 - b^3c =$
$ab^3 - a^3b + a^3c -b^3c -ac^3 + bc^3=$
$ab(b^2-a^2)+c(a^3-b^3)-c^3(a-b)=$
$ab(a+b)(b-a)+c(a-b)(a^2+ab+b^2)-c^3(a-b)=$
$(a-b)[-ab(a+b)+c(a^2+ab+b^2)-c^3]=$
$(a-b)(-a^2b-ab^2+ca^2+cab+cb^2-c^3)=$
$(a-b)(ca^2-a^2b+cab-ab^2+cb^2-c^3)=$
$(a-b)[a^2(c-b)+ab(c-b)+c(b^2-c^2)]= $
$(a-b)(c-b)[a^2+ab-c(b+c)]=$
$(a-b)(c-b)(a^2+ab-bc-c^2)=$
$(a-b)(c-b)(a^2-c^2+ab-bc)=$
$(a-b)(c-b)[(a-c)(a+c)+b(a-c)]=$
$(a-b)(c-b)(a-c)(a+b+c)$
The suggestion of replacing $b$ with $a$ was the following: if in such an expression when you replace $b$ with $a$ you get a $0$ then one of the factors may be $(a-b)$.
In the future if you want an idea of the factors that appear you could write on wolframalpha factor$(ab^3 - a^3b + a^3c -ac^3 +bc^3 - b^3c)$
|
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|
Show by induction that $F_n \geq 2^{0.5 \cdot n}$, for $n \geq 6$ I have the following problem:
Show by induction that $F_n \geq 2^{0.5 \cdot n}$, for $n \geq 6$
Where $F_n$ is the $nth$ Fibonacci number.
Proof
Basis
$n = 6$.
$F_6 = 8 \geq 2^{0.5 \cdot 6} = 2^{\frac{6}{2}} = 2^3 = 8$
Induction hypothesis
Assume $F_n \geq 2^{\frac{n}{2}}$, for some $n \geq 6$.
Inductive step
Lets shows that $F_{n+1} \geq 2^{\frac{n + 1}{2}}$.
We know that
*
*$F_{n + 1} = F_n + F_{n - 1}$
*$2^{\frac{n + 1}{2}} > 2^{\frac{n}{2}}$
*$2^{\frac{n + 1}{2}} = 2^{\frac{n}{2} + \frac{1}{2}} = 2^{\frac{n}{2}} \cdot 2^{\frac{1}{2}} \leq 2^{\frac{n}{2}} \cdot 2 = 2^{\frac{n}{2}} + 2^{\frac{n}{2}}$
Since we have assumed that $F_n \geq 2^{\frac{n}{2}}$, then $$F_n + F_{n - 1} = F_{n + 1} \geq 2^{\frac{n}{2}} + F_{n - 1} \geq 2^{\frac{n}{2}} + F_{n - 1} + 2^{\frac{n}{2}} - F_n = 2^{\frac{n}{2}} + 2^{\frac{n}{2}} + F_{n - 1} - F_n$$
The last inequality is true because $2^{\frac{n}{2}} - F_n$ is negative or $0$, since $F_n \geq 2^{\frac{n}{2}}$.
I have tried a lot of things, but I cannot figure out how to proceed and how to conclude that indeed $F_{n + 1} \geq 2^{\frac{n}{2}} \cdot 2^{\frac{1}{2}}$. I feel really stupid after trying for a long time to do this alone and not managing to do it.
|
Suppose we knew for 2 values of n $F_n \ge 2^{\frac n2}$ i.e for n = 6 and n = 7.
We know this holds for n=6 and n=7.
We also know that $F_{n+1} = F_{n} + F_{n-1}$
So we assume for some k and k-1 (7 and 6) $F_{k-1} \ge 2^{\frac {k-1}{2}}$ and $F_{k+1} \ge 2^{\frac {k+1}{2}}$
We know $F_{k} \ge F_{n-1}$
so $F_{k+1} \ge 2 \cdot F_{n-1}$
Using the assumption $F_{k+1} \ge 2 \cdot 2^{\frac {k-1}{2}} = 2^{\frac{k+1}{2}}$ as required.
EDIT: If you want a phrasing in the language of induction (propositional)
$$P(k) = F_{n} \ge 2^{\frac{n}{2}} \mbox{ & } F_{n-1} \ge 2^{\frac{n-1}{2}}\\$$
We then prove:
$$
P(k+1) = F_{n+1} \ge 2^{\frac{n+1}{2}} \mbox{ & } F_{n} \ge 2^{\frac{n}{2}}$$
Above I proved the second from the first.
|
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|
Showing that $\frac{1}{2^n +1} + \frac{1}{2^n +2} + \cdots + \frac{1}{2^{n+1}}\geq \frac{1}{2}$ for all $n\geq 1$
Show that $$\frac{1}{2^n +1} + \frac{1}{2^n +2} + \cdots + \frac{1}{2^{n+1}}\geq \frac{1}{2}$$ for all $n\geq 1$
I need this in order to complete my proof that $1 + \frac{n}{2} \leq H_{2^n}$, but I don't have any ideas.
I know it is true at least for the first $20$ cases, but I can't prove it. Any suggestions?
|
Hint for every $1\leq k\leq 2^n$
$$\frac{1}{2^n+k}\geq \frac{1}{2^{n+1}} $$
do every term in your sum is greater then $\frac{1}{2^{n+1}} $ and in the sum there is $2^n$ terms so your sum is greater then $2^n$ times $\frac{1}{2^{n+1}} $ $\cdots$
|
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|
How many divisors of the combination of numbers?
Find the number of positive integers that are divisors of at least one of $A=10^{10}, B=15^7, C=18^{11}$
Instead of the PIE formula, I would like to use intuition.
$10^{10}$ has $121$ divisors, and $15^7$ has $64$ divisors, and $18^{11}$ has $276$ divisors.
Number of divisors total with no restriction is: $461$. $A,B \to $ there are $5^{0} \to 5^{7} = 8$ divisors. $B, C \to$ there are: $3^{0} \to 3^{7} = 8$ divisors. $A, C \to$ there are $2^{0} \to 2^{10} = 11$ divisors.
So far: $461 - 27 = 434$. I took out: $15$ divisors from $B$, $19$ from $A$, and $19$ from $C$. So in total:
$$434 + 53 = 487$$
But this isnt right.
|
Each divisor is of the form $2^i3^j5^k$, where the exponents are nonnegative.
If $i=0$ and $j=0$, then $0\le k \le 10$, which accounts for $11$ possibilities.
If $i=0$, $j>0$, and $k=0$, then $1\le j \le 22$, which accounts for $22$ possibilities.
If $i=0$, $j>0$, and $k>0$, then $1\le j,k \le 7$, which accounts for $49$ possibilities.
If $i>0$, $j=0$, and $k=0$, then $1\le i\le11$, for $11$ more.
If $i>0$, $j=0$, and $k>0$, then $1\le i\le10$ and $1\le k \le10$, for $100$ more.
If $i>0$ and $j>0$, then $k=0$, $1\le i\le11$, and $1\le j \le22$, for $242$ more.
This analysis by cases is exhaustive, and there are $11+22+49+11+100+242=435$ factors.
|
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|
How to solve $\lfloor \sqrt{(k + 1)\cdot2009} \rfloor = \lfloor \sqrt{k\cdot2009} \rfloor$ Is there any way to solve this equation (or to tell how many solutions are there), other than checking all 2009 possibilities? $\lfloor \sqrt{(k + 1)\cdot2009} \rfloor = \lfloor \sqrt{k\cdot2009} \rfloor, 0 \le k \le 2009, k \in \Bbb Z$
|
As shown in N.S.'s answer, there are no solutions for $k \le 501$ since $\sqrt{2009(k+1)} > \sqrt{2009k}+1$ for all $k \le 501$. Also, for all $502 \le k \le 2009$, we have $\sqrt{2009(k+1)} < \sqrt{2009k}+1$.
Thus, for each $502 \le k \le 2009$, we have either $\lfloor\sqrt{2009(k+1)}\rfloor - \lfloor\sqrt{2009k}\rfloor = 0$ or $\lfloor\sqrt{2009(k+1)}\rfloor - \lfloor\sqrt{2009k}\rfloor = 1$.
Since $\displaystyle\sum_{k = 502}^{2009}\left(\lfloor\sqrt{2009(k+1)}\rfloor - \lfloor\sqrt{2009k}\rfloor\right)$ $= \lfloor\sqrt{2009 \cdot 2010}\rfloor - \lfloor\sqrt{2009 \cdot 502}\rfloor$ $= 2009 - 1004 = 1005$, we have that $\lfloor\sqrt{2009(k+1)}\rfloor - \lfloor\sqrt{2009k}\rfloor = 1$ for exactly $1005$ values of $k$ in the range $502 \le k \le 2009$.
Since there are $2009-502+1 = 1508$ values of $k$ in the range $502 \le k \le 2009$, we have that $\lfloor\sqrt{2009(k+1)}\rfloor - \lfloor\sqrt{2009k}\rfloor = 0$ for exactly $1508 - 1005 = 503$ values of $k$.
|
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|
$p,q,r$ primes, $\sqrt{p}+\sqrt{q}+\sqrt{r}$ is irrational. I want to prove that for $p,q,r$ different primes, $\sqrt{p}+\sqrt{q}+\sqrt{r}$ is irrational.
Is the following proof correct?
If $\sqrt{p}+\sqrt{q}+\sqrt{r}$ is rational, then $(\sqrt{p}+\sqrt{q}+\sqrt{r})^2$ is rational, thus $p+q+r+2\sqrt{pq}+2\sqrt{pr}+2\sqrt{qr}$ is rational, therefore $\sqrt{pq}+\sqrt{pr}+\sqrt{qr}$ is rational.
If $\sqrt{pq}+\sqrt{pr}+\sqrt{qr}$ is rational, then $(\sqrt{pq}+\sqrt{pr}+\sqrt{qr})^2$ is rational, therefore $pq+qr+pr+\sqrt{p^2qr}+\sqrt{pq^2r}+\sqrt{pqr^2}$ is rational, therefore $p\sqrt{qr}+q\sqrt{pr}+r\sqrt{pq}$ is rational.
Now suppose $p<q<r$. If $p\sqrt{qr}+q\sqrt{pr}+r\sqrt{pq}$ and $\sqrt{pq}+\sqrt{pr}+\sqrt{qr}$ are rational, then $$p\sqrt{qr}+q\sqrt{pr}+r\sqrt{pq}-p(\sqrt{pq}+\sqrt{pr}+\sqrt{qr})$$
is rational, therefore $(q-p)\sqrt{pr}+(r-p)\sqrt{pq}$ is rational.
If $(q-p)\sqrt{pr}+(r-p)\sqrt{pq}$ is rational, then $((q-p)\sqrt{pr}+(r-p)\sqrt{pq})^2$ is rational, thus $(q-p)^2pr+2(q-p)(r-p)\sqrt{p^2qr}+(r-p)^2pq$ is rational, thus $\sqrt{qr}$ is rational.
But $q,r$ are distinct primes, thus $qr$ can't be a square. Thus $\sqrt{qr}$ is irrational. Contradiction.
Also, is there an easier proof?
|
Suppose $\sqrt{p}+\sqrt{q}+\sqrt{r} = a \to (a-\sqrt{p})^2 = (\sqrt{q}+\sqrt{r})^2\to a^2-2a\sqrt{p} +p =q+r+2\sqrt{qr}\to a^2-q+p-r=2a\sqrt{p}+2\sqrt{qr}\to (a^2-q+p-r)^2=4a^2p+4qr+8a\sqrt{pqr}\to \sqrt{pqr} $ is a rational number, and this is not possible.
|
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|
Prove the series has positive integer coefficients How can I show that the Maclaurin series for
$$
\mu(x) = (x^4+12x^3+14x^2-12x+1)^{-1/4}
\\
= 1+3\,x+19\,{x}^{2}+147\,{x}^{3}+1251\,{x}^{4}+11193\,{x}^{5}+103279\,
{x}^{6}+973167\,{x}^{7}+9311071\,{x}^{8}+\cdots
$$
has positive integer coefficients? (I have others to do, too, but this one will be a start.)
possibilities
(a) The coefficients $Q(n)$ satisfy the recurrence
$$
(n+1)Q(n)+(12n+21)Q(n+1)+(14n+35)Q(n+2)+(-12n-39)Q(n+3)+(n+4)Q(n+4) = 0
$$
(b) $\mu(x)$ satisfies the differential equation
$$
(x^3+9x^2+7x-3)\mu(x)+(x^4+12x^3+14x^2-12x+1)\mu'(x)=0
$$
(c) factorization of $x^4+12x^3+14x^2-12x+1$ is
$$
\left( x-\sqrt {5}+3+\sqrt {15-6\,\sqrt {5}} \right) \left( x-\sqrt
{5}+3-\sqrt {15-6\,\sqrt {5}} \right) \left( x+\sqrt {5}+3-\sqrt {15+
6\,\sqrt {5}} \right) \left( x+\sqrt {5}+3+\sqrt {15+6\,\sqrt {5}}
\right)
$$
(d) $(1-8X)^{-1/4}$ has positive integer coefficients, But if $X$ is defined by $1-8X=x^4+12x^3+14x^2-12x+1$, then $X$ does not have integer coefficients.
(e) Can we compute the series for $\log\mu(x)$
$$3\,x+{\frac {29}{2}}{x}^{2}+99\,{x}^{3}+{\frac {3121}{4}}{x}^{4}+{
\frac {32943}{5}}{x}^{5}+{\frac {348029}{6}}{x}^{6}+\dots
$$
and then recognize that its exponential has integer coefficients?
|
I can prove the integer property, but not (yet) positivity.
Lemma: The coefficients of Taylor series of $\mu(x)$ are integers.
Proof. We can write
$$x^4+12x^3+14x^2-12x+1=\left(1-x\right)^4\Bigl(1-8\nu(x)\Bigr),$$
where $\nu(x)$ is given by
$$\nu(x)=\frac{2}{1-x}-\frac{7}{\left(1-x\right)^2}+\frac{7}{\left(1-x\right)^3}-\frac{2}{\left(1-x\right)^4}.$$
The coefficients of Taylor series of $\nu(x)$ are clearly integer and we have $\nu(0)=0$. Since the coefficients of $(1-x)^{-1}$ and $(1-8x)^{-\frac14}$ are integers, the statement follows. $\qquad\square$
Another option is to rewrite $\mu(x)$ in terms of $y=\frac{x}{1-x}$:
$$\mu(x)\mapsto \frac{1+y}{\left[1-8y\left(1-y^2\right)\left(1+2y\right)\right]^{\frac14}}.$$
It is interesting to note that even the coefficients of different powers of $y$ seem to be positive.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Find the value below
Let $a,b,c$ be the roots of the equation $$8x^{3}-4x^{2}-4x+1=0$$
Find $$\frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}$$
It's just for sharing a new ideas, thanks:)
|
A systematic way to compute sums of the form $a^{-k} + b^{-k} + c^{-k}$ is
using Newton's identities.
Notice
$$\begin{align}
a, b, c \text{ roots of } p(x) &= 8x^3 - 4x^2 - 4x + 1\\
\implies\quad
\frac{1}{a}, \frac{1}{b}, \frac{1}{c}
\text{ roots of } q(x) &= x^3 p(\frac{1}{x}) = x^3 - \color{red}{4} x^2 - \color{green}{4}x + \color{blue}{8}
\end{align}
$$
Let $p_k = a^{-k} + b^{-k} + c^{-k}$ for $k = 1,2,3$. Newton's identities tell us:
$$\begin{cases}
p_1 - \color{red}{4} \times 1 &= 0\\
p_2 - \color{red}{4} p_1 - \color{green}{4}\times 2 &= 0\\
p_3 - \color{red}{4} p_2 - \color{green}{4} p_1 + \color{blue}{8}\times 3 &= 0
\end{cases}
\quad\implies\quad
\begin{cases}
p_1 &= 4\\
&\;\Downarrow\\
p_2 &= 4\times \underbrace{4}_{p_1} + 4\times 2 = 24\\
&\;\Downarrow\\
p_3 &= 4\times \underbrace{24}_{p_2} + 4\times \underbrace{4}_{p_1} - 8\times 3 = 88
\end{cases}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1387198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Proving that $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}$ I have written the left side of the equation as $$\left(1+\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2n-1}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n}\right).$$ I don't know how to find the sums of these sequences. I know the sums for odd and even integers, but I can't figure this out.
|
Try using $\Sigma$-notation to make your problem more manageable in terms of its algebraic expressions and the like. To this end, note that
$$
1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}\tag{1}
$$
becomes
$$
\sum_{i=1}^n\frac{1}{2i-1}-\sum_{i=1}^n\frac{1}{2i}=\sum_{i=1}^n\frac{1}{i+n}.\tag{2}
$$
Before moving on to the induction proof, you should observe that
$$
\sum_{i=1}^{k+1}\frac{1}{i+k+1}=\sum_{i=2}^{k+2}\frac{1}{i+k},\tag{3}
$$
a simple $\Sigma$-manipulation that will be of use to us in a moment. Now let's prove $(2)$ (and hence $(1)$) by induction.
Claim: For $n\geq1$, let $S(n)$ denote the statement
$$
S(n) : \sum_{i=1}^n\frac{1}{2i-1}-\sum_{i=1}^n\frac{1}{2i}=\sum_{i=1}^n\frac{1}{i+n}.
$$
Base step ($n=1$): $S(1)$ says that $\sum_{i=1}^1\frac{1}{2i-1}-\sum_{i=1}^1\frac{1}{2i}=\sum_{i=1}^1\frac{1}{i+1}$ and this is true because $1-\frac{1}{2}=\frac{1}{2}$.
Inductive step $S(k)\to S(k+1)$: Fix some $k\geq 1$ and assume that
$$
S(k) : \sum_{i=1}^k\frac{1}{2i-1}-\sum_{i=1}^k\frac{1}{2i}=\sum_{i=1}^k\frac{1}{i+k}
$$
holds. To be proved is that
$$
S(k+1) : \sum_{i=1}^{k+1}\frac{1}{2i-1}-\sum_{i=1}^{k+1}\frac{1}{2i}=\sum_{i=1}^{k+1}\frac{1}{i+k+1}
$$
follows. Beginning with the left-hand side of $S(k+1)$,
\begin{align}
\sum_{i=1}^{k+1}\frac{1}{2i-1}-\sum_{i=1}^{k+1}\frac{1}{2i} &= \left(\sum_{i=1}^k\frac{1}{2i-1}-\sum_{i=1}^k\frac{1}{2i}\right)+\frac{1}{2k+1}-\frac{1}{2k+2}\tag{by defn.}\\[1em]
&= \sum_{i=1}^k\frac{1}{i+k}+\frac{1}{2k+1}-\frac{1}{2k+2}\tag{by $S(k)$}\\[1em]
&= \sum_{i=1}^{k+1}\frac{1}{i+k}-\frac{1}{2k+2}\tag{by defn.}\\[1em]
&= \sum_{i=2}^{k+2}\frac{1}{i+k}-\frac{1}{2k+2}-\frac{1}{2k+2}+\frac{1}{k+1}\tag{by defn.}\\[1em]
&= \sum_{i=2}^{k+2}\frac{1}{i+k}-\frac{1}{k+1}+\frac{1}{k+1}\tag{like terms}\\[1em]
&= \sum_{i=1}^{k+1}\frac{1}{i+k+1},\tag{by $(3)$}
\end{align}
one arrives at the right-hand side of $S(k+1)$, thereby showing $S(k+1)$ is also true, completing the inductive step.
Thus, by mathematical induction, the claim $S(n)$ is true for all $n\geq 1$. $\blacksquare$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How do I find the remainder of $4^0+4^1+4^2+4^3+ \cdots + 4^{40}$ divided by 17? Recently I came across a question,
Find the remainder of $4^0+4^1+4^2+4^3+ \cdots + 4^{40}$ divided by 17?
At first I applied sum of G.P. formula but ended up with the expression $1\cdot \dfrac{4^{41}-1}{4-1}$. I couldn't figure out how to proceed further. Secondly I thought of using the fact $(a+b+\cdots) \pmod {17} = (r_a+r_b\dots) \pmod {17}$ but it is getting more messier.
Please explain in detail. And also mention the formula being used.
|
HINT:
Observe that for any non-negative integer $a,$ $$1+4+4^2+4^3=(1+4)(1+4^2)\equiv0\pmod{17}$$
$$\implies\sum_{a=m}^n4^{4a}(1+4+4^2+4^3)\equiv0\pmod{17}$$
Here $m=0,n=9$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
Show that $\sqrt{4+2\sqrt{3}}-\sqrt{3}$ is rational using the rational zeros theorem What I've done so far:
Let $$r = \sqrt{4+2\sqrt{3}}-\sqrt{3}.$$
Thus, $$r^2 = 2\sqrt{3}-2\sqrt{3}\sqrt{4+2\sqrt{3}}+7$$
and $$r^4=52\sqrt{3}-28\sqrt{3}\sqrt{4+2\sqrt{3}}-24\sqrt{4+2\sqrt{3}}+109.$$
I did this because in a similar example in class, we related $r^2$ and $r^4$ to find a polynomial such that $mr^4+nr^2 = 0$ for some integers $m,n$. However, I cannot find such relation here. Am I doing this right or is there another approach to these type of problems.
|
In terms of your method, I think the first thing I'd do is rewrite it as,
$$r + \sqrt{3} = \sqrt{4 + 2\sqrt{3}}.$$
If you square both sides,
$$r^2 + 3 + 2\sqrt{3}r = 4 + 2\sqrt{3}.$$
Isolating $\sqrt{3}$,
$$2\sqrt{3}(r - 1) = 1 - r^2.$$
Squaring again,
$$12(r-1)^2 = r^4 - 2r^2 + 1.$$
Expanding and rearranging,
$$(r-1)^2\left(r^2+2r-11\right)=r^4 - 14r^2 + 24r - 11 = 0.$$
Therefore, if $r$ is rational, then $r = \pm 1, \pm 11$, which narrows things down. The only rational root of this polynomial is $1$, which narrows it down again. So, this method tells us that we only need to verify that $r$ is or is not $1$ (it is $1$), then we'll definitely know whether or not it is irrational.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Find the sum of binomial coefficients
Calculate the value of the sum
$$
\sum_{i = 1}^{100} i\binom{100}{i} = 1\binom{100}{1} +
2\binom{100}{2} +
3\binom{100}{3} +
\dotsb +
100\binom{100}{100}
$$
What I have tried:
$$\begin{align}
S &= 0\binom{100}{0}+1\binom{100}{1}+ \dotsb +99\binom{100}{99}+100\binom{100}{100} \\ \\
&=100\binom{100}{100}+99\binom{100}{99}+ \dotsb +1\binom{100}{1}+0\binom{100}{0}
\end{align}$$
and I'm stuck here, I don't know if it's true or not, any help will be appreciated.
|
$$\begin{align}
\sum_{i=1}^ni\binom ni
&=\sum_{i=0}^{n-1}(i+1)\binom n{i+1}
\color{lightgray}{=\sum_{i=0}^{n-1}(i+1)\frac {n(n-1)^{\underline{i}}}{(i+1)i!}=\sum_{i=0}^{n-1}n\frac {(n-1)^{\underline{i}}}{i!}}\\
&=\sum_{i=0}^{n-1}n\binom {n-1}i\\
&=n\sum_{i=0}^{n-1}\binom {n-1}i\\
&=n(1+1)^{n-1}\\
&=n\cdot 2^{n-1}
\end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Card Draw question What are the ways to draw 13 cards from a pack of 52 cards such that
(a) the hand is void in at least one suit, (b) the hand is not void in any suit.(“void in a suit” means having no cards of that suit)
Current approach
I am thinking on following lines:
Case-1: All cards from exactly 1 suit $= {4 \choose 1} * {13 \choose 13} = 4$ ways
Case-2: Cards from exactly 2 suits = ${4 \choose 2} * [{26 \choose 13} - {2 \choose 1} * {13 \choose 13}] = 62,403,588$ ways
Case-3: Cards from exactly 3 suits = ${4 \choose 3} * [{39 \choose 13} - {3 \choose 2} * [{26 \choose 13} - {2 \choose 1} * {13 \choose 13}] - {3 \choose 1} * {13 \choose 13}] = 32,364,894,588$ ways
Case-4: Cards from exactly 4 suits = ${4 \choose 4} * [{52 \choose 13} - {4 \choose 3} * [{39 \choose 13} - {3 \choose 2} * [{26 \choose 13} - {2 \choose 1} * {13 \choose 13}]] - {4 \choose 2} * [{26 \choose 13} - {2 \choose 1} * {13 \choose 13}] - {4 \choose 1} * {13 \choose 13}] = 602,586,261,420$ ways
I have no ways to verify answer for Case-4. However as expected, answer for (a) matches sum case-1 + case-2 + case-3. I am sure there must be a simpler and more direct way to arrive at these numbers. Any insights. Many Thanks
|
For (a), select the 3 suits represented out of 4, and then select 13 cards out of the remaining $52 - 13 = 39$, for a total of $\binom{4}{3} \cdot \binom{39}{13} = 32489701776$. For (b), it is all posibilities (taking 13 card out of 52) less the ones missing a suit. Part (a) uses at most 3 suits, by a similar reasoning:
*
*At most one suit: $\binom{4}{1} \binom{13}{13}$
*At most two suits: $\binom{4}{2} \binom{2 \cdot 13}{13}$
*At most three suits: $\binom{4}{3} \binom{3 \cdot 13}{13}$
Then exactly 2 suits is (at most 3) - (at most 2), i.e.,
$$
\binom{4}{1} \binom{13}{13} - \binom{4}{2} \binom{2 \cdot 13}{13} = 32427298176
$$
This is essentially inclusion-exclusion.
Thanks to JMoravitz for catching a dumb mistake. I plead not enough coffee.
|
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|
$\int\limits_{1/2}^{2}\frac{1}{x}\sin (x-\frac{1}{x})dx$ $\int\limits_{1/2}^{2}\frac{1}{x}\sin (x-\frac{1}{x})dx$ has value equal to
$(A)0\hspace{1cm}(B)\frac{3}{4}\hspace{1cm}(C)\frac{3}{4}\hspace{1cm}(D)2 $
I tried to solve this question by putting $x-\frac{1}{x}=t$ and limits have changed to $\frac{-3}{2}$ to $\frac{3}{2}$ but what to do with remaining $\frac{1}{x}$ and $dx$ in the integration.Please help....
|
Let $$I(a) = \int^{a}_{\frac{1}{a}}\frac{1}{x}\sin \left(x-\frac{1}{x}\right)dx$$
Where $a>0$
Now $$I'(a) = \frac{1}{a}\sin \left(a-\frac{1}{a}\right)+\sin \left(\frac{1}{a}-a\right)\cdot a \cdot \frac{1}{a^2} = 0$$
So $$I'(a) = 0\Rightarrow I(a) = \mathcal{C}$$
Now put $a=1\;,$ in first equation, We get $I(1) = 0$
So put $a = 1$ in $I(a)= \mathcal{C}\;,$ We get $I(1) = 0 = \mathcal{C}$
So we get $\displaystyle I(a) = 0\;,$ Put $a=2\;,$ We get $$I(2) = \int^{2}_{\frac{1}{2}}\frac{1}{x}\sin \left(x-\frac{1}{x}\right)dx = 0$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Coefficient of $x^9$ in $(1+x)\cdot (1+x^2)\cdot (1+x^3)\cdots(1+x^9)$
The Coefficient of $x^9$ in $(1+x)\cdot (1+x^2)\cdot (1+x^3)\cdots(1+x^9)$ is
My Try We Know that Coefficient of $x^9$ occur (which is $ = 1$) when we
multiply $x^0\cdot x^9$ and $x^1\cdot x^8$ and $x^2\cdot x^7$ and $x^3\cdot x^6$ and $x^4\cdot x^5$
Now I did not understand How can I calculate it bcz answer given as $ = 8$
Help me, Thanks
|
This is the same as the number of ways to express $9$ as the sum of distinct positive integers $\le 9$.
$9 = 9 = 8+1 = 7+2 = 6+3 = 6+2+1 = 5+4 = 5+3+1 = 4+3+2$
So there are eight ways.
|
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|
Finding $\lim_{(x,y)\rightarrow (0,0)}\frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4}$ I want to find $\lim_{(x,y)\rightarrow 0}\frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4}$.
What I tried:
Denote $x=r\cos(\theta)$ and $y=r\sin(\theta)$. So the limit is:
$$\lim_{r\rightarrow 0}\frac{e^{-\frac{1}{r^2(\cos^2(\theta)+\sin^2(\theta))}}}{r^4(\cos^4(\theta)+\sin^4(\theta))} = \lim_{r\rightarrow 0}\frac{e^{-\frac{1}{r^2}}}{r^4(\cos^4(\theta)+\sin^4(\theta))}$$
Am I on the right track? If so, how do I continue?
|
Let
$1/r^2
= z
$.
Then
$U(r, \theta)
=\frac{e^{-\frac{1}{r^2}}}{r^4(\cos^4(\theta)+\sin^4(\theta))}
=\frac{z^2e^{-z}}{\cos^4(\theta)+\sin^4(\theta)}
$.
Also,
$\begin{array}\\
\cos^4(\theta)+\sin^4(\theta)
&=\cos^4(\theta)+\sin^4(\theta)+2\cos^22(\theta)\sin^2(\theta)-2\cos^2(\theta)\sin^2(\theta)\\
&=(\cos^2(\theta)+\sin^2(\theta))^2-2\cos^2(\theta)\sin^2(\theta)\\
&=1-\frac12(2\cos(\theta)\sin(\theta))^2\\
&=1-\frac12(\sin(2\theta))^2\\
&\ge 1-\frac12\\
&=\frac12
\end{array}
$
so
$|U(r, \theta)|
\le\frac12 z^2e^{-z}
$.
Since
$\lim_{z \to \infty} z^2e^{-z}
= 0
$,
as has been shown here many times,
$\lim_{r \to 0}|U(r, \theta)|
= 0
$.
(Added later)
To show
$\lim_{z \to \infty} z^2e^{-z}
= 0
$,
from the power series
for $e^z$,
$e^z \ge \frac{z^3}{6}
$,
so
$\frac{z^2}{e^z}
\le \frac{6}{z}
\to 0
$.
Note that this easily generalizes:
To show
$\lim_{z \to \infty} z^ne^{-z}
= 0
$,
from the power series
for $e^z$,
$e^z \ge \frac{z^{n+1}}{(n+1)!}
$,
so
$\frac{z^n}{e^z}
\le \frac{(n+1)!}{z}
\to 0
$.
|
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|
How to factor $4x^2 + 2x + 1$? I want to know how to factor $4x^2 + 2x + 1$? I found the roots using quadratic equation and got $-1 + \sqrt{-3}$ and $-1 - \sqrt{-3}$, so I thought the factors would be $(x - (-1 + \sqrt{-3}))$ and $(x - (-1 - \sqrt{-3}))$
However, according to MIT's course notes, the factors are $(1 - (-1 + \sqrt{-3})x)$ and $(1 - (-1 - \sqrt{-3})x)$
Course Notes (pg. 30)
Are these two expressions equivalent, or am I simply not factoring correctly? Thanks.
|
The solution is $\frac{-1 \pm \sqrt {-3}}{4}$ (suppose we know what we mean here by $\sqrt{-3}$).
Then the expansion would be
$\begin{align*}
&4x^2+2x+1=4\left(x-\frac{-1+ \sqrt {-3}}{4}\right)\left(x-\frac{-1- \sqrt {-3}}{4}\right)\\
& = 4\left(x-\frac{1}{-1- \sqrt {-3}}\right)\left(x-\frac{1}{-1+ \sqrt {-3}}\right)\\
&=[-1+(-1- \sqrt {-3})x][-1+(-1+ \sqrt {-3})x] \\
&=[1-(-1- \sqrt {-3})x][1-(-1+ \sqrt {-3})x]
\end{align*}$
as written in the lecture note.
Note that here we've used the fact that $(-1+ \sqrt {-3})(-1- \sqrt {-3})=4$.
(Thus $\frac{-1+ \sqrt {-3}}{4}=\frac{1}{-1- \sqrt {-3}}$
And
$\frac{-1- \sqrt {-3}}{4}=\frac{1}{-1+\sqrt {-3}}$)
|
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"timestamp": "2023-03-29T00:00:00",
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|
Counting integral solutions Suppose $a + b + c = 15$
Using stars and bars method, number of non-negative integral solutions for the above equation can be found out as $15+3-1\choose15$ $ =$ $17\choose15$
How to extend this principle for finding number of positive integral solutions of
$a + b + 3c = 15$?
I tried to do it by substituting $3c$ with another variable $d$. But could not succeed.
|
The generating function for the number of non-negative integral solutions to $a+b+3c=n$ is
$$
f(x)=\frac1{1-x}\frac1{1-x}\frac1{1-x^3}
$$
Since
$$
\begin{align}
(1-x)^{-2}
&=\sum_{k=0}^\infty\binom{-2}{k}(-x)^k\\
&=\sum_{k=0}^\infty\binom{k+1}{k}x^k\\
&=\sum_{k=0}^\infty(k+1)x^k
\end{align}
$$
and
$$
\begin{align}
\left(1-x^3\right)^{-1}
=\sum_{k=0}^\infty x^{3k}
\end{align}
$$
the coefficient of $x^n$ in $f(x)$ is
$$
\begin{align}
\sum_{k=0}^{\lfloor n/3\rfloor}(n-3k+1)
&=(n+1)\left(\lfloor n/3\rfloor+1\right)-\frac32\left(\lfloor n/3\rfloor+1\right)\lfloor n/3\rfloor\\
&=\left(n+1-\frac32\lfloor n/3\rfloor\right)\left(\lfloor n/3\rfloor+1\right)
\end{align}
$$
For $n=15$, we get $51$ ways.
|
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|
Find $\int \frac{1}{\sin\frac{x}{2}\sqrt{\cos^3\frac{x}{2}}}dx$ $\int \frac{1}{\sin\frac{x}{2}\sqrt{\cos^3\frac{x}{2}}}dx$
I tried to solve it.I put $\frac{x}{2}=t$ then $\int \frac{1}{\sin\frac{x}{2}\sqrt{\cos^3\frac{x}{2}}}dx=$
$\int \frac{2 dt}{\sin t\sqrt{\cos^3 t} }=\int \frac{2 dt}{\sin t \cos t\sqrt{\cos t} }$ and then i could not solve.Please help.
|
$$\int\frac{dx}{\sin\frac{x}{2}\sqrt{\cos^3\frac{x}{2}}}$$
$$=\int\frac{\sin\frac{x}{2}dx}{\sin^2\frac{x}{2}\sqrt{\cos^3\frac{x}{2}}}$$
$$=\int\frac{\sin\frac{x}{2}dx}{\left(1-\cos^2\frac{x}{2}\right)\sqrt{\cos^3\frac{x}{2}}}$$
Let $\cos\frac{x}{2}=t\implies -\frac{1}{2}\sin\frac{x}{2}=dt$
$$=-\frac{1}{2}\int\frac{dt}{\left(1-t^2\right)t^{3/2}}$$
Can you solve further?
|
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|
Number of non-decreasing functions?
Let $A = \{1,2,3,\dots,10\}$ and $B = \{1,2,3,\dots,20\}$.
Find the number of non-decreasing functions from $A$ to $B$.
What I tried:
Number of non-decreasing functions = (Total functions) - (Number of decreasing functions)
Total functions are $20^{10}$. And I think there are ${20 \choose 10}$ decreasing functions. Since you choose any $10$ codomain numbers and there's just one way for them to be arranged so that the resultant is a decreasing function.
However my answer doesn't match. Where am I going wrong?
How can I directly compute the non-decreasing functions like without subtracting from total?
|
Means Here function satisfy the following Condition.
$f(1),f(2),f(3),.......f(10)\in \left\{1,2,3,4,5,...,20\right\}$
$\; \bullet f(1)<f(2)<..........<f(10)\;,$ Total no. of possibilities $ = \displaystyle \binom{20}{10}$
$\;\bullet f(1)\leq f(2)<f(3)<.......<f(10)\;,$ Total no. of possibilities $\displaystyle \binom{20}{9}$
There are $9$ Such cases. So Total Possibilities for $\bf{2^{nd}}$ cases $ \displaystyle 9 \times \binom{20}{9}$
$\;\bullet\; f(1)\leq f(2)\leq f(3)<......,f(10)\;,$ Total cases $\displaystyle \binom{20}{8}$
These are $8$ such cases . So Total $\displaystyle 8\times \binom{20}{8}$
Similarly Calculate in that manner.
may be it helps to you.
Now We can add all possibilities , we get
$\displaystyle = \binom{20}{10}+9\times \binom{20}{9}+8\times \binom{20}{8}+........+1\times \binom{20}{1}$
$\displaystyle = \underbrace{\binom{20}{10}+\binom{20}{9}}+8\left[\underbrace{\binom{20}{9}+\binom{20}{8}}\right]+..........+\binom{20}{1}$
$\displaystyle = \binom{21}{10}+8\times \binom{21}{9}+.....+\binom{20}{1}$
Above we have uese the formula $\displaystyle \binom{n}{r}+\binom{n}{r-1} = \binom{n+1}{r}$
Now Adding in similar manner, We get final answer $\displaystyle = \binom{29}{10}$
|
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|
Prove that the equation $\ 5x^4 + x − 3 = 0\ $ has no rational solutions. I'm locked at $\ x\left(5x^3 + 1\right) = 3$. Not too sure where to go from there but I'm getting the feeling it's really really obvious..
|
Let us suppose $x = a / b$ is a rational solution, where a, b are integers, mutually prime. Then:
$$
5(a/b)^4 + a/b -3 = 0
$$
and after multiplying by $b^4$:
$$
5a^4 + ab^3 -3b^4=0
$$
or equivalent
$$
a(5a^3 + b^3) = 3b^4 (*)
$$
The last equality implies that $a$ divides $3b^4$. Because $a$ and $b$ are coprime, then $a$ must divide $3$, so $a \in \{-1, 1, -3, 3\}$.
Case 1
If $a = 1$ then the last equality becomes:
$$
5 + b^3 = 3b^4
$$
or equivalent
$$
5 = b^3(3b - 1)
$$
Therefore $b^3$ divides $5$, but the only cube numbers to divide $5$ are $-1$ and $1$ so $b \in \{-1, 1\}$. Both values for $b$ don't satisfy the last equality, so this case doesn't lead to a solution.
Case 2
If $a=3$ then $(*)$ becomes:
$$
3(135 + b^3)=3b^4
$$
or equivalent
$$
135 = b^3(b-1)
$$
The last equality is not possible because $b^3(b-1)=b^2b(b-1)$ and $b(b-1)$ is always an even number, therefore $b^3(b-1)$ is even, while $135$ is odd number.
The other cases ($a=-1$ and $a=-3$) are pretty similar to the above ones.
|
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|
Evaluate $\int e^x \sin^2 x \mathrm{d}x$ Is the following evaluation of correct?
\begin{align*} \int e^x \sin^2 x \mathrm{d}x &= e^x \sin^2 x -2\int e^x \sin x \cos x \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int e^x (\cos^2 x - \sin^2x) \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int e^x (1 - 2\sin^2x) \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int -2 e^x \sin^2x \mathrm{d}x + 2 e^x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 e^x -4 \int e^x \sin^2x \mathrm{d}x \end{align*}
First two steps use integration by parts. In the first step we differentiate $\sin^2 x$. In the second step we differentiate $\sin x \cos x$.
Using this, we reach $$5\int e^x \sin^2 x \mathrm{d}x = e^x \sin^2 x -2e^x \sin x \cos x + 2 e^x$$
$$\int e^x \sin^2 x \mathrm{d}x = \frac{e^x \sin^2 x -2e^x \sin x \cos x + 2 e^x}{5}+C$$
I can't reach the form that most integral calculators give, which has terms $\cos(2x)$ and $\sin(2x)$ by just using trig identities, so I wonder whether the result is correct. I would also be interested in a method that immediately gives the form $$-\frac{e^x[2 \sin(2x)+ \cos(2x)-5]}{10}+C$$
|
Notice
$$\int e^x\sin^2x\mathrm{d}x=$$ $$=\int e^x\left(\frac{1-\cos 2x}{2}\right)\mathrm{d}x$$
$$=\frac{1}{2}\int e^xdx-\frac{1}{2}\int e^x \cos 2x \mathrm{d}x$$
Using $\displaystyle \int e^{ax}\cos (bx) \mathrm{d}x=\frac{e^{ax}}{a^2+b^2}(a\cos bx+b\sin bx)$, we get
$$=\frac{1}{2}e^x-\frac{1}{2}\frac{e^x}{1^2+2^2}(\cos 2x+2\sin 2x)+C$$
$$=\frac{1}{2}e^x-\frac{1}{10}e^x(\cos 2x+2\sin 2x)+C$$
$$=-\frac{e^x(2\sin 2x+\cos 2x-5)}{10}+C$$
|
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|
How to get to this equality $\prod_{m=1}^{\infty} \frac{m+1}{m}\times\frac{m+x}{m+x+1}=x+1$?
How to get to this equality $$\prod_{m=1}^{\infty} \frac{m+1}{m}\times\frac{m+x}{m+x+1}=x+1?$$
I was studying the Euler Gamma function as it gave at the beginning of its history, and need to solve the following product operator, tried a few things, but I could not ...
$$\prod_{m=1}^{\infty} \frac{m+1}{m}\times\frac{m+x}{m+x+1}=\prod_{m=1}^{\infty} \frac{m^2+mx+m+x}{m^2+mx+m} =\prod_{m=1}^{\infty} \frac{m(m+x+1)+x}{m(m+x+1)}$$
WolframAlpha checked in and the result really is correct.
http://www.wolframalpha.com/input/?i=simplify+prod+m%3D1+to+inf+%5B%28m%2B1%29%2Fm+*+%28m%2Bx%29%2F%28m%2B1%2Bx%29%5D
|
$$\begin{align}
\prod_{n=1}^N\frac{m+1}{m}\frac{m+x}{m+x+1}&=\left(\prod_{n=1}^N\frac{m+1}{m}\right)\left(\prod_{n=1}^N\frac{m+x}{m+x+1}\right)\\\\
&=\left(\frac{2}{1}\frac{3}{2}\frac{4}{3}\cdots \frac{N}{N-1}\frac{N+1}{N}\right)\left(\frac{1+x}{2+x}\frac{2+x}{3+x}\frac{3+x}{4+x}\cdots \frac{N-1+X}{N+x}\frac{N+x}{N+1+X}\right)\\\\
&=\frac{N+1}{1}\frac{1+x}{N+1+x}\\\\
&\to 1+x \,\,\text{as}\,\,N\to \infty
\end{align}$$
|
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|
Find $\int_0^\frac{\pi}{2} \frac {\theta \cos \theta } { \sin \theta + \sin ^ 3 \theta }\:d\theta$ My Calc 2 teacher wasn't able to solve this:
$$\int_0^\frac{\pi}{2} \frac {\theta \cos \theta } { \sin \theta + \sin ^ 3 \theta }\:d\theta$$
Can someone help me solve this?
|
You can transform the integral by parts to get
$$\int_0^{\pi/2} d\theta \frac{\theta \cos{\theta}}{\sin{\theta} (1+\sin^2{\theta})} = \int_0^{\pi/2} d(\sin{\theta}) \left (\frac1{\sin{\theta}}-\frac{\sin{\theta}}{1+\sin^2{\theta}} \right )\theta \\ = -\frac{\pi}{4} \log{2} - \int_0^{\pi/2} d\theta \left [\log{(\sin{\theta})} - \frac12 \log{(1+\sin^2{\theta})} \right ]$$
Now,
$$\int_0^{\pi/2} d\theta \log{(\sin{\theta})} = - \frac{\pi}{2} \log{2}$$
$$\int_0^{\pi/2} d\theta \,\log{(1+\sin^2{\theta})} = \frac{\pi}{2} \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k 2^{2 k}} \binom{2 k}{k} $$
Defining
$$f(x) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k 2^{2 k}} \binom{2 k}{k} x^k$$
we find that
$$f'(x) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{2^{2 k}} \binom{2 k}{k} x^{k-1} = \frac1{x} \left (\frac1{\sqrt{1+x}}-1 \right )$$
so that
$$f(x) = \log{\left [\frac{\sqrt{1+x}-1}{ x \left (\sqrt{1+x}+1\right )} \right ]}+C$$
$$\lim_{x \to 0} f(x) = 0 \implies C=\log{4} $$
Thus, putting this all together, we get
$$\int_0^{\pi/2} d\theta \frac{\cos{\theta}}{\sin{\theta} (1+\sin^2{\theta})} = \frac{\pi}{2} \log{(1+\sqrt{2})} - \frac{\pi}{4} \log{2} = \frac{\pi}{4} \log{\left (\frac{3}{2} + \sqrt{2} \right )}$$
|
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|
Find Solution to an infinite Nested Radicals How do I find the solution to the following:
$$ \sqrt{ 7 - \sqrt{\frac{7}{2} + \sqrt{\frac{7}{4} - \sqrt{\frac{7}{16} + \sqrt{\frac{7}{256} - \ldots}}}}}$$
I first tried looking for a pattern for the denominators, but the $16$ seems to be throwing me off. Can we use calculus in this maybe, to find limits or something?
|
Hint 1:
$$X = \sqrt{7 - \dfrac{A}{\sqrt{2}}}$$
where $A = \displaystyle\sqrt{7 + \sqrt{7 - \sqrt{7 + \sqrt{7 - ....}}}}$
Hint 2:
$$(A^{2} - 7)^{2} = 7 - A$$
|
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|
LU-factorization: why can't I get a unit lower triangular matrix? I want to find an $LU$-factorization of the following matrix: \begin{align*} A = \begin{pmatrix} 3 & -6 & 9 \\ -2 & 7 & -2 \\ 0 & 1 & 5 \end{pmatrix} \end{align*} This matrix is invertible (the determinant is $33$), so I should be getting a $LU$ decomposition where $L$ is a lower triangular matrix with only $1s$ on the diagonal. But that's not what I'm getting. Here is what I did: \begin{align*} \begin{pmatrix} 3 & -6 & 9 \\ -2 & 7 & -2 \\ 0 & 1 & 5 \end{pmatrix} & \begin{matrix} \xrightarrow{R_1 \rightarrow 1/3 R_1} \end{matrix} \begin{pmatrix} 1 & -2 & 3 \\ -2 & 7 & -2 \\ 0 & 1 & 5 \end{pmatrix} \begin{matrix} \xrightarrow{R_2 \rightarrow R_2 + 2R_1} \end{matrix} \begin{pmatrix} 1 & -2 & 3 \\ 0 & 3 & 4 \\ 0 & 1 & 5 \end{pmatrix} \\ & \begin{matrix} \xrightarrow{R_2 \rightarrow 1/3 R_2} \end{matrix} \begin{pmatrix} 1 & -2 & 3 \\ 0 & 1 & \frac{4}{3} \\ 0 & 1 & 5 \end{pmatrix} \begin{matrix} \xrightarrow{R_3 \rightarrow R_3 - R_2} \end{matrix} \begin{pmatrix} 1 & -2 & 3 \\ 0 & 1 & \frac{4}{3} \\ 0 & 0 & \frac{11}{3} \end{pmatrix} = U. \end{align*} This matrix is now in echelon form. The elementary matrices corresponding to the operations are: \begin{align*} E_1 & = \begin{pmatrix} \frac{1}{3} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \qquad (R_1 \rightarrow 1/3 R_1) \quad E_2 = \begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \qquad (R_2 \rightarrow R_2 + 2R_1) \\ E_3 &= \begin{pmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & 1 \end{pmatrix} \qquad (R_2 \rightarrow 1/3 R_2) \quad E_4 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -1 & 1 \end{pmatrix} \qquad (R_3 \rightarrow R_3 - R_2) \end{align*} Hence we have $E_4 E_3 E_2 E_1 A = U$. Now \begin{align*} E_4 E_3 E_2 E_1 = M = \begin{pmatrix} \frac{1}{3} & 0 & 0 \\ \frac{2}{9} & \frac{1}{3} & 0 \\ - \frac{2}{9} & - \frac{1}{3} & 1 \end{pmatrix} \end{align*} is a lower triangular matrix. Now we can write \begin{align*} A = (E_4 E_3 E_2 E_1)^{-1} U = M^{-1} U = LU, \end{align*} with \begin{align*} M^{-1} = L = \begin{pmatrix} 3 & 0 & 0 \\ -2 & 3 & 0 \\ 0 & 1 & 1 \end{pmatrix} \end{align*} But why does my matrix $L$ not have $1s$ on the diagonal? I thought an $LU$-factorization of an invertible matrix is unique, and that in that case $L$ is an unit lower triangular matrix? Did I overlook something or made a mistake? I haven't interchanged any rows. Please help!
|
In the framework of the Gaussian elimination procedure, you get $L$ with all $1$s on the diagonal by never truly rescaling rows. You rescale them in the intermediate step before you add them to another row, but then you "scale them back" so that you're only changing the row that you actually added to. You can always perform Gaussian elimination in this way. (By contrast, you may not always be able to perform Gaussian elimination without exchanging rows to obtain your next pivot, so an $LU$ decomposition may not always exist.)
|
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|
Solve for $v$ - simplify as much as possible Solve for $v$. Simplify the answer.
$$-3 = -\frac{8}{v-1}$$
Here is what I tried:
$$-3 = \frac{-8}{v-1} $$
$$(-8) \cdot (-3) = \frac{-8}{v-1} \cdot (-8) $$
$$24 = v-1$$
$$25 = v$$
|
It may be instructive to, as an exercise, try a few problems similar to this one without skipping any steps (as follows).
$$\begin{array}{lll}
-3&=&-\displaystyle\frac{8}{v-1}\\
-3&=&(-1)\cdot\displaystyle\frac{8}{v-1}\\
-3&=&\displaystyle\frac{-1}{1}\cdot\displaystyle\frac{8}{v-1}\\
-3&=&\displaystyle\frac{(-1)\cdot8}{1\cdot(v-1)}\\
-3&=&\displaystyle\frac{-8}{v-1}\\
\displaystyle\frac{-3}{1}&=&\displaystyle\frac{-8}{v-1}\\
\displaystyle\frac{-3}{1}\cdot\frac{v-1}{1}&=&\displaystyle\frac{-8}{v-1}\cdot\frac{v-1}{1}\\
\displaystyle\frac{-3(v-1)}{1\cdot 1}&=&\displaystyle\frac{-8(v-1)}{(v-1)\cdot 1}\\
\displaystyle\frac{-3(v-1)}{1}&=&\displaystyle\frac{-8(v-1)}{1\cdot(v-1)}\\
-3(v-1)&=&\displaystyle\frac{-8}{1}\cdot\frac{v-1}{v-1}\\
-3(v-1)&=&\displaystyle\frac{-8}{1}\cdot 1\\
-3(v-1)&=&\displaystyle\frac{-8}{1}\\
-3(v-1)&=&-8\\
\displaystyle\frac{-3(v-1)}{-3}&=&\displaystyle\frac{-8}{-3}\\
\displaystyle\frac{-3(v-1)}{-3\cdot 1}&=&\displaystyle\frac{(-1)\cdot8}{(-1)\cdot 3}\\
\displaystyle\frac{-3}{-3}\cdot \frac{v-1}{1}&=&\displaystyle\frac{(-1)}{(-1)}\cdot\frac{8}{3}\\
\displaystyle 1\cdot \frac{v-1}{1}&=&\displaystyle 1\cdot\frac{8}{3}\\
\displaystyle \frac{v-1}{1}&=&\displaystyle \frac{8}{3}\\
\displaystyle v-1&=&\displaystyle \frac{8}{3}\\
\displaystyle (v-1)+1&=&\displaystyle \frac{8}{3}+1\\
\displaystyle (v-1)+1&=&\displaystyle \frac{8}{3}+\frac{3}{3}\\
\displaystyle (v+(-1))+1&=&\displaystyle \frac{8+3}{3}\\
\displaystyle v+((-1)+1)&=&\displaystyle \frac{11}{3}\\
\displaystyle v+0&=&\displaystyle \frac{11}{3}\\
\displaystyle v&=&\displaystyle \frac{11}{3}\\
\end{array}$$
|
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|
Find $\int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx$ $$\int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx=\frac{u}{v}$$ where $u$ and $v$ are in their lowest form. Find the value of $\dfrac{1000u}{v}$
$$\int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx=\int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^2(x^2-1)+1}}dx$$ I put $x^2-1=t$ but no benefit. Please guide me.
|
Let $$\displaystyle I = \int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx = \int \frac{\left(1-x^{-2}\right)\cdot x^{-3}}{\sqrt{2-2x^{-2}+x^{-2}}}dx\;,$$
Now Let $x^{-2} = u\;,$ Then $-2x^{-3}dx = du$ and Changing Limit, We get
$$\displaystyle I = -\frac{1}{2}\int_{1}^{\frac{1}{4}}\frac{1-u}{\sqrt{u^2-2u+2}}du = -\frac{1}{2}\int_{1}^{\frac{1}{4}}\frac{(1-u)}{\sqrt{(1-u)^2+1}}du$$
Now Let $(1-u) = t\;,$ Then $du = -dt$ and Changing Limit, We get
$$\displaystyle I = \frac{1}{2}\int_{0}^{\frac{3}{4}}\frac{t}{\sqrt{1+t^2}}dt=\frac{1}{2}\left[\sqrt{1+t^2}\right]_{0}^{\frac{3}{4}} =\frac{1}{8}$$
So we get $\displaystyle I = \frac{1}{8} = \frac{u}{v},$ So $u=1,v=8$
So we get $\displaystyle 1000 \times \frac{u}{v} = 1000 \times \frac{1}{8} = 125$
|
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|
Triangle area inequalities; semiperimeter I've got stuck on this problem :
Proof that for every triangle of
sides $a$, $b$ and $c$ and area
$S$, the following
inequalities are true :
$4S \le a^2 + b^2$
$4S \le b^2 + c^2$
$4S \le a^2 + c^2$
$6S \le a^2 + b^2 + c^2$
The first thing that came to my mind was the inequality $S \le \frac 12 ab$. That is derived from the fact that $S = \frac 12ab \cdot \sin(\angle ACB)$ and $0 < \sin(\angle ACB) \le 1$.
Anyway, this wasn't enough to solve the problem. Some help would be well received.
Thanks!
|
Those inequalities are very weak. It is trivial that
$$ 2S\leq ab\leq \left(\frac{a+b}{2}\right)^2\leq \frac{a^2+b^2}{2}$$
by the GM-AM-QM inequality, hence $S\leq\frac{a^2+b^2+c^2}{6}$ by adding the first three inequalities.
However, they cannot all hold as equalities, since a (Euclidean) triangle cannot have three right angles, so it is reasonable to expect that $S\leq K(a^2+b^2+c^2)$ always holds with a constant $K<\frac{1}{6}$. By Heron's formula:
$$ 4S=\sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)} $$
but:
$$ a^4+b^4+c^4 = 3\cdot QM(a^2,b^2,c^2)^2 \geq 3\cdot AM(a^2,b^2,c^2)^2 = \frac{1}{3}(a^2+b^2+c^2)^2 $$
hence the optimal constant (achieved by the equilateral triangle) is $K=\color{red}{\frac{1}{4\sqrt{3}}}.$
|
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|
Prove the trigonometric identity $\cos(x) + \sin(x)\tan(\frac{x}{2}) = 1$ While solving an equation i came up with the identity $\cos(x) + \sin(x)\tan(\frac{x}{2}) = 1$.
Prove whether this is really true or not. I can add that $$\tan\left(\frac{x}{2}\right) = \sqrt{\frac{1-\cos x}{1+\cos x}}$$
|
Yup! We have $$\cos x + \sin x \tan \frac{x}{2} = (\cos ^2 \frac{x}{2} - \sin ^2 \frac{x}{2}) + (2 \sin \frac{x}{2} \cos \frac{x}{2}) \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} = \cos ^2 \frac{x}{2} + \sin ^2 \frac{x}{2} = 1$$
|
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|
If $ a + b + c \mid a^2 + b^2 + c^2$ then $ a + b + c \mid a^n + b^n + c^n$ for infinitely many $n$
Let $ a,b,c$ positive integer such that $ a + b + c \mid a^2 + b^2 + c^2$.
Show that $ a + b + c \mid a^n + b^n + c^n$ for infinitely many positive integer $ n$.
(problem composed by Laurentiu Panaitopol)
So far no idea.
|
Here's a more intuitive way to get the idea of considering powers of $2$.
Added (below): in the same way we can prove that any $n=6k\pm1$ works.
Note that $a+b+c\mid(a+b+c)^2-(a^2+b^2+c^2)=2(ab+bc+ca)$. By The Fundamental Theorem of Symmetric Polynomials (FTSP), $a^n+b^n+c^n$ is an integer polynomial in $a+b+c$, $ab+bc+ca$ and $abc$. If $3\nmid n$, no term has degree divisible by $3$ so each term has at least one factor $a+b+c$ or $ab+bc+ca$. If we can find infinitely many $n$ such that the terms without a factor $a+b+c$ have a coefficient that is divisible by $2$, then we are done because $a+b+c\mid2(ab+bc+ca)$. This suggests taking a look to the polynomial $a^n+b^n+c^n$ over $\Bbb F_2$. Note that over $\Bbb F_2$, $a^{2^n}+b^{2^n}+c^{2^n}=(a+b)^{2^n}+c^{2^n}=(a+b+c)^{2^n}$ is divisible by $(a+b+c)$. Because the polynomial given by FTSP over $\Bbb F_2$ is the reduction modulo $2$ of that polynomial over $\mathbb Z$ (this is a consequence of the uniqueness given by the FTSP), this shows that the coefficients of those terms that have no factor $a+b+c$ is divisible by $2$, and we are done because $3\nmid2^n$. (In fact all coefficients except that of $(a+b+c)^{2^n}$ are divisible by $2$.)
Added later
Ievgen's answer inspired me to generalise the above approach to $n=6k\pm1$. Consider again $a^n+b^n+c^n$ as an integer polynomial in $abc,ab+bc+ca,a+b+c$ (which we can do by FTSP). Because $3\nmid n$, no term has the form $(abc)^k$. It remains to handle the terms of the form $m\cdot(ab+bc+ca)^k(abc)^l$. If $a+b+c$ is odd, then $a+b+c\mid ab+bc+ca$ and we're done. If $a+b+c$ is even, at least one of $a,b,c$ is even so $2\mid abc$, and hence $a+b+c\mid m\cdot(ab+bc+ca)^k(abc)^l$. (Note that $l>0$ because $n$ is odd.)
|
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|
Find the least positive residue of $10^{515}\pmod 7$. I tried it, but being a big number unable to calculate it.
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$10^{6} \equiv 1 $ mod $7$ By Fermat's little theorem. Notice $515=6 \times 85+5$ so $10^{515} \equiv 10^{5}$ mod $7$. Notice $10 \equiv 3$ mod 7, so $10^{2} \equiv 3^{2} \equiv 2$ and $10^{3} \equiv 20 \equiv -1\equiv 6$ mod $7$, $10^{4} \equiv 60 \equiv 4$ mod $7$, so $10^{5} \equiv 40 \equiv 5$ mod $7$. So $10^{515}\equiv 5$ mod $7$
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"timestamp": "2023-03-29T00:00:00",
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|
Find min of $M=\frac{1}{2+\cos2A}+\frac{1}{2+\cos2B}+\frac{1}{2-\cos2C}$
Find min of $$M=\frac{1}{2+\cos2A}+\frac{1}{2+\cos2B}+\frac{1}{2-\cos2C}$$, where $A, B, C$ are three angle of triangle $ABC$
Using Cauchy-Schwarz, we obtain: \begin{align*}
M &= \frac{1}{2+\cos2A}+\frac{1}{2+\cos2B}+\frac{1}{2-\cos2C}\\
&\ge \frac{9}{6+2\cos(A+B)\cos(A-B)+1-2\cos^2C}\\
& \ge \frac{9}{7-2\cos C-2\cos^2C}\ge \frac{9}{\frac{15}{2}}=\frac{6}{5}
\end{align*}
The equality holds if and only if $\left\{\begin{matrix}A=B\\\cos C=-\frac{1}{2} \end{matrix}\right.\iff \left\{\begin{matrix}A=B=30^{\circ}\\ C=120^{\circ}\end{matrix}\right.$
But the change $\cos(A+B)\cos(A-B)\le-\cos C $ may be not correct. :(
|
After using C-S we need to prove that \begin{align*}
&\displaystyle\frac{9}{6-2\cos\gamma\cos\left(\alpha-\beta\right)+1-2\cos^2\gamma}\geq\frac{6}{5} \\
\iff&\displaystyle4\cos^2\gamma+4\cos\gamma\cos\left(\alpha-\beta\right)+1\geq0 \\
\iff&\displaystyle\left(2\cos\gamma+\cos\left(\alpha-\beta\right)\right)^2+\sin^2\left(\alpha-\beta\right)\geq0
\end{align*}Done!
|
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|
Vector Functions of One Variable Question
A particle moves along the curve of the intersection of the cylinders $y=-x^2$ and $z=x^2$ in the direction in which $x$ increases. (All distances are in cm.) At the instant when the particle is at the point $(1,\,-1,\,1)$ its speed is $9$ cm/s, and the speed is increasing at a rate of $3$ cm/s$^2$. Find the velocity and acceleration of particle at that instant.
Solution (Well at least my attempt at one)
$\mathbf{r}(t) = x\mathbf{i} - x^2\mathbf{j} + x^2\mathbf{k}$
$\mathbf{v}(t) = \left(\mathbf{i} - 2x\mathbf{j} + 2x\mathbf{k}\right)\dfrac{dx}{dt}$
\begin{align*}
\implies ||\mathbf{v}|| &=\left(\sqrt{1^2 +4x^2+4x^2}\right) \left|\dfrac{d x}{dt}\right|\\
&=(\sqrt{1+8x^2})\dfrac{dx}{dt}
\end{align*}
$|\frac{dx}{dt}|=\frac{dx}{dt}$ since the particle is always increasing
$\implies$ at the point $(1,\,-1,\,1),\,\, ||\mathbf{v}|| = 9$, so $dx/dt = 3$ and the velocity at that point is
$\mathbf{v}(t) = 3\mathbf{i} - 6\mathbf{j} + 6\mathbf{k}$
Now acceleration is where I have a bit of trouble
$\mathbf{a}(t) =\dfrac{d^2x}{dt^2}(\mathbf{i} - 2x\mathbf{j} + 2x\mathbf{k}) + (\dfrac{dx}{dt})^2(- 2\mathbf{j} + 2\mathbf{k})$
But I don't exactly know what to do from here any help would be appreciated
|
As I mentioned in my comment above, you can figure out $\dfrac{d^2x}{dt^2}$ from your equation in $|\mathbf{v}|$, given that you know that the change of speed is 3cm/s/s.
\begin{eqnarray}
|\mathbf{v}| &=& (\sqrt{1+8x^2})\dfrac{dx}{dt}\\
\dfrac{d|\mathbf{v}|}{dt} &=& (\sqrt{1+8^2})\dfrac{d^2x}{dt^2} + \frac{8x}{\sqrt{1+8x^2}}\left(\dfrac{dx}{dt}\right)^2 = 3 \\
\dfrac{d^2x}{dt^2} &=& -7
\end{eqnarray}
Plugging this into your equation for $\mathbf{a}$ gives
\begin{equation}\mathbf{a} = -7\mathbf{i} - 4\mathbf{j} + 4\mathbf{k}\end{equation}
and it can easily be checked that $|\mathbf{a}|=3$
|
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|
Does the limit $\lim\limits_{x\to0}\left(\frac{1}{x\tan^{-1}x}-\frac{1}{x^2}\right)$ exist? Does the limit: $$\lim\limits_{x\to0}\frac{1}{x\tan^{-1}x}-\frac{1}{x^2}$$ exist?
|
Using Why is $\arctan(x)=x-x^3/3+x^5/5-x^7/7+\dots$? on,
$$\lim_{x\to0}\dfrac{x-\tan^{-1}x}{x^2\tan^{-1}x}=\lim_{x\to0}\dfrac{\dfrac {x^3}3+O(x^5)}{x^2\tan^{-1}x} =\dfrac13\lim_{x\to0}\dfrac{x+O(x^3)}{\tan^{-1}x}$$ as $x\ne0$ as $x\to0$
Now set $\tan^{-1}x=y\implies x=\tan y,x\to0\implies y\to0$
$$\lim_{x\to0}\dfrac{x+O(x^3)}{\tan^{-1}x}=\lim_{y\to0}\dfrac{\tan y+O(\tan^3y)}{y}=1$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Simulataneous equations Suppose you have the following system of linear congruence
$2x+5y$ is congruent to 1 (mod6)
$x+y$ is congruent to 5 (mod6)
where $x,y \in \mathbb{Z}$
How would you obtain a general solution for this system. Also is there a way to determine whether the system is solvable or not?
|
The system
\begin{align*}
x+y & \equiv 5 \pmod{6}\\
2x+5y & \equiv 1 \pmod{6}
\end{align*}
can be rewritten as
$$
\begin{pmatrix}
1 & 1\\
2 & 5
\end{pmatrix}
\begin{pmatrix}
x\\y
\end{pmatrix}
\equiv \begin{pmatrix}
5\\1
\end{pmatrix} \pmod{6}
$$
Now consider using row-operations (modulo $6$ of course) to get
$$
\left[
\begin{array}{cc|c}
1 & 1 & 5\\
2 & 5 & 1\\
\end{array}
\right]
\longrightarrow \left[
\begin{array}{cc|c}
1 & 1 & 5\\
0 & 3 & 3\\
\end{array}
\right]
$$
The second congruence is $3y \equiv 3 \pmod{6}$. But $\gcd(3,6) \neq 1$ so you cannot simply cancel out $3$. However this gives us $y \equiv 1,3,5 \pmod{6}$. Now the first congruence gives
$$x \equiv 5-y \equiv 4, 2, 0 \pmod{6}.$$
Thus
$$(x,y) \in \{(4,1), (2,3),(0,5)\} \subset \mathbb{Z}_6 \times \mathbb{Z}_6.$$
Remark: the solvability of such system of congruences $Ax \equiv b \pmod{n}$ is linked to the $\gcd(D, n)$, where $D=\text{det}A$. In particluar, if $\gcd=1$, then the system has a unique solution.
|
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|
For $n\times n$ matrices, is it true that $AB=CD\implies AEB=CED$?
If $A,B,C,D,E$ are $n\times n$ matrices, does $AB=CD$ imply $AEB=CED$?
I only know that $AB=CD \implies ABE=CDE$, but I don't see how you can sandwhich $E$ within it.
Also, if $AB=CD=0$, does $\det(AB)=\det(CD)=0$?
I think this should be true because $AB$ and $CD$ are the same matrices and $\det(0)=0$
|
I don't think this works by sandwiching.....
$$A=\begin{pmatrix}
1 & 3 \\
0 & 1 \\
\end{pmatrix},
B=\begin{pmatrix}
1 & 2 \\
0 & 1 \\
\end{pmatrix},
C=\begin{pmatrix}
1 & 1 \\
0 & 1 \\
\end{pmatrix},
D=\begin{pmatrix}
1 & 4 \\
0 & 1 \\
\end{pmatrix},$$
Then we have that $AB=CD$ by matrix multiplication, however if
$$E=\begin{pmatrix}
1 & 1 \\
1 & 1 \\
\end{pmatrix}$$
Then $AEB=\begin{pmatrix} 4 & 12\\ 1 & 3 \\ \end{pmatrix}\neq \begin{pmatrix} 2 & 10\\ 1 & 5 \\ \end{pmatrix}=CED$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving that $\frac{1}{2}<\frac{2}{3}<\frac{3}{4}<$...$<\frac{n-1}{n}$ In an attempt to find a pattern, I did this:
Let a,b,c,d be non-zero consecutive numbers. Then we have:
$a=a$
$b=a+1$
$c=a+2$
$d=a+3$
This implies:
$\frac{a}{b}=\frac{a}{a+1}$
$\frac{b}{c}=\frac{a+1}{a+2}$
$\frac{c}{d}=\frac{a+2}{a+3}$
I don't know how that helps. I'm greatly seeking your help. Thank you very much.
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Now that I understand better, another simple way to prove the inequality is:
$$\begin{align}& n^{2}-1<n^{2}\\
& \implies\dfrac{n^{2}-1}{n(n+1)}<\dfrac{n^2}{n(n+1)}\\
&\implies \dfrac{(n-1)(n+1)}{n(n+1)}<\dfrac{n^{2}}{n(n+1)}\\
&\implies \dfrac{n-1}{n}<\dfrac{n}{n+1}\\
\end {align}$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Closed form for $\sum_{k=1}^\infty(\zeta(4k+1)-1)$ Wikipedia gives
$$\sum_{k=2}^\infty(\zeta(k)-1)=1,\quad\sum_{k=1}^\infty(\zeta(2k)-1)=\frac34,\quad\sum_{k=1}^\infty(\zeta(4k)-1)=\frac78-\frac\pi4\left(\frac{e^{2\pi}+1}{e^{2\pi}-1}\right)$$
from which we can easily find $\sum_{k=1}^\infty(\zeta(2k+1)-1)$ and $\sum_{k=1}^\infty(\zeta(4k+2)-1)$. From here it's natural to ask the following
Question: Is there a known closed form for $\sum_{k=1}^\infty(\zeta(4k+1)-1)$?
Related: Closed form for $\sum\limits_{k=1}^{\infty}\zeta(4k-2)-\zeta(4k)$ and Closed form for $\sum_{k=1}^{\infty} \zeta(2k)-\zeta(2k+1)$.
Progress
Note that we are done once we have a closed form for $\sum_{k=1}^\infty(\zeta(4k+1)-\zeta(4k+3))$. So I tried the same approach as in one of the questions above and this is what I got:
We have $$\zeta(4k+1)-\zeta(4k+3)=\sum_{n\geq2}^\infty\left(1-\frac1{n^2}\right)\frac1{n^{4k+1}}$$
Hence
$$\sum_{k\geq1}\zeta(4k+1)-\zeta(4k+3)=\sum_{n\geq2}\left(1-\frac1{n^2}\right)\sum_{k\geq1}\frac1{n^{4k+1}}=\sum_{n\geq2}\left(1-\frac1{n^2}\right)\frac1{n^5}\frac1{1-\frac1{n^4}}\\=\sum_{n\geq2}\frac1{n^3+n^5}.$$
In the same way I get $$\sum_{k\geq1}\zeta(4k+1)-\zeta(4k)=-\sum_{n\geq2}\frac1{n+n^2+n^3+n^4}$$ and $$\sum_{k\geq1}\zeta(4k+1)-\zeta(4k+2)=\sum_{n\geq2}\frac1{n^2+n^3+n^4+n^5}.$$
It suffices (in fact it is equally hard) to evaluate any of these series.
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By expressing the infinite sum as a double sum and then switching the order of summation, we get
$$\begin{align} \sum_{n=1}^{\infty} \left[\zeta(4n+1) - 1 \right] &= \sum_{n=1}^{\infty} \sum_{m=2}^{\infty} \frac{1}{m^{4n+1}} = \sum_{m=2}^{\infty} \frac{1}{m}\sum_{n=1}^{\infty} \left( \frac{1}{m^{4}} \right)^{n} \\ &= \sum_{m=2}^{\infty} \frac{1}{m} \frac{\frac{1}{m^{4}}}{1-\frac{1}{m^{4}}} = \sum_{m=2}^{\infty} \frac{1}{m} \frac{1}{m^{4}-1}. \end{align}$$
One can use the series representation $$\psi(2+z) = - \gamma +1 + \sum_{n=2}^{\infty} \frac{z}{n(n+z)}, \tag{1}$$ where $\psi(x)$ is the digamma function, to show
$$\psi(2+z) + \psi(2-z) + \psi(2+iz) + \psi(2-iz) = - 4 \gamma +4 -4z^{4} \sum_{n=2}^{\infty} \frac{1}{n(n^{4}-z^{4})}.$$
Letting $z=1$, we find
$$ \begin{align} \sum_{m=2}^{\infty} \frac{1}{m(m^{4}-1)} &= 1-\gamma - \frac{1}{4} \Big(\psi(3) + \psi(1) + \psi(2+i) + \psi(2-i) \Big) \\ &= 1-\gamma - \frac{1}{4} \left(\frac{3}{2} - \gamma - \gamma + \psi(2+i)+\psi(2-i) \right) \\ &= \frac{5}{8}-\frac{\gamma}{2} - \frac{1}{4} \Big(\psi(2-i)+\psi(2+i) \Big) \\ &\approx 0.0390670072. \end{align}$$
EDIT:
$(1)$ I find it easier to prove that series representation for $\psi(2+z)$ using the difference equation $$\psi(N+1+z) - \psi(z) = \sum_{n=0}^{N} \frac{1}{n+z} $$ as opposed to manipulating the series representation for $\psi(1+z)$.
$$ \begin{align} \sum_{n=2}^{N} \frac{z}{n(n+z)} &= \sum_{n=2}^{N} \frac{1}{n} - \sum_{n=2}^{N} \frac{1}{n+z} \\ &= \Big( H_{N}-1 \Big) - \Big(\psi(N+1+z) - \psi(z) - \frac{1}{z} - \frac{1}{1+z} \Big)\\ &= \Big( \psi(N+1) + \gamma -1 \Big) - \Big(\psi(N+1+z) - \psi(2+z) \Big) \end{align}$$
Letting $N \to \infty$ leads to the result since $\psi(x+a) = \ln(x) + \mathcal{O} \left(\frac{1}{x} \right)$ as $x \to \infty$.
|
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|
How to prove $\sum _{k=1}^{\infty} \frac{k-1}{2 k (1+k) (1+2 k)}=\log_e 8-2$? How to prove:
$$\sum _{k=1}^{\infty} \frac{k-1}{2 k (1+k) (1+2 k)}=\log_e 8-2$$
Is it possible to convert it into a finite integral?
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Let me try.
Note that the $$RHS = 3\ln 2 + 2 = 2 + 3 \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} = 2 + 3\sum_{k=1}^\infty \frac{1}{2k-1}-\frac{1}{2k}$$
We have $$\begin{eqnarray} \sum_{k=1}^\infty \frac{k-1}{2k(k+1)(2k+1)} &=& \sum_{k=1}^\infty \frac{2k - (k+1)}{2k(k+1)(2k+1)} \\ &=&\sum_{k=1}^\infty \left(\frac{1}{(k+1)(2k+1)} - \frac{1}{2k(2k+1)} \right)\\
&=& \sum_{k=1}^\infty \left(\frac{2}{2k+1} - \frac{1}{k+1} - \frac{1}{2k} + \frac{1}{2k+1}\right) \\ &=& \sum_{k=1}^\infty \left(\frac{3}{2k+1} - \frac{1}{k+1} - \frac{1}{2k}\right)\\
&=& 2 + 3\sum_{k=1}^\infty \frac{1}{2k-1}-\frac{1}{2k} \\&=& RHS.\end{eqnarray}$$
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|
How to calculate the limit that seems very complex.. Someone gives me a limit about trigonometric function and combinatorial numbers.
$I=\displaystyle \lim_{n\to\infty}\left(\frac{\sin\frac{1}{n^2}+\binom{n}{1}\sin\frac{2}{n^2}+\binom{n}{2}\sin\frac{3}{n^2}\cdots\binom{n}{n}\sin\frac{n+1}{n^2}}{\cos\frac{1}{n^2}+\binom{n}{1}\cos\frac{2}{n^2}+\binom{n}{2}\cos\frac{3}{n^2}\cdots\binom{n}{n}\cos\frac{n+1}{n^2}}+1\right)^n$
when $n$ is big enough, $\displaystyle 0\leqslant\frac{n+1}{n^2}\leqslant \frac\pi 2$ I tried $\displaystyle \frac{x}{1+x}\leqslant\sin x\leqslant x$
Use $\sin x\leqslant x$, I got $I\leqslant e$, But use another inequality I got nothing.
I don't know the answer is $e$ or not. Who can help me. Thanks.
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Consider $$\lim_{n\to\infty} n\frac{\displaystyle\sum\limits_{k=0}^n {n \choose k}\sin\frac{k+1}{n^2}}{\displaystyle\sum\limits_{k=0}^n {n \choose k}\cos\frac{k+1}{n^2}}.$$ If this limit equals $\alpha$, your limit equals $e^\alpha$. In order to simplify a bit, thanks to the Taylor expansions of $\sin$ and $\cos$ near $0$, we may multiply top and bottom by $n^2$: $$\begin{align} \lim_{n\to\infty} n\frac{\displaystyle\sum\limits_{k=0}^n {n \choose k}n^2\sin\frac{k+1}{n^2}}{\displaystyle\sum\limits_{k=0}^n {n \choose k}n^2\cos\frac{k+1}{n^2}}&=\lim_{n\to\infty} {n}\frac{\displaystyle\sum\limits_{k=0}^n {n \choose k}(k+1)}{\displaystyle\sum\limits_{k=0}^n {n \choose k}\left(n^2-\frac{1}{2}\frac{(k+1)^2}{n^2}\right)} \\ &= \lim_{n\to\infty} {n}\frac{\displaystyle\sum\limits_{k=0}^n k{n \choose k} + \displaystyle\sum\limits_{k=0}^n {n \choose k}}{n^2\displaystyle\sum\limits_{k=0}^n {n \choose k} - \frac{1}{2n^2}\displaystyle\sum\limits_{k=0}^n (k+1)^2{n \choose k}} \\ &= \lim_{n\to\infty} \frac{2^{n-1}n(n+2)}{n^2 2^n-2^{n-3}(1+5/n+4/n^2)}=\frac{1}{2}.\end{align}$$ Thus your expression approaches $\sqrt{e}$.
|
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|
Find the value of $\sum_{n=1}^\infty \frac{1}{n(2n-1)}$ I will show two solutions of this problem.
First solution :
$$\sum_{n=1}^\infty \frac{1}{n(2n-1)}=\sum_{n=1}^\infty \left(\frac{2}{2n-1}-\frac{1}{n}\right)$$
$$=\sum_{n=1}^\infty \left(\int_0^1 (2x^{2n-2}-x^{n-1})dx\right)$$
$$=\int_0^1 \left( \sum_{n=1}^\infty(2x^{2n-2}-x^{n-1})\right)dx$$
$$=\int_0^1 \left( \frac{2}{1-x^2}-\frac{1}{1-x} \right) dx $$
$$=\int_0^1 \frac{1}{1+x}dx $$
$$=\ln 2$$
Second solution :
$$\sum_{n=1}^\infty \frac{1}{n(2n-1)}=\sum_{n=1}^\infty \left(\frac{2}{2n-1}-\frac{1}{n}\right)$$
$$=\sum_{n=1}^\infty \left(\int_0^1 (2x^{2n-2}-2x^{2n-1})dx\right)$$
$$=\int_0^1 \left( \sum_{n=1}^\infty(2x^{2n-2}-2x^{2n-1})\right)dx$$
$$=\int_0^1 \left( \frac{2}{1-x^2}-\frac{2x}{1-x^2} \right) dx $$
$$=\int_0^1 \frac{2}{1+x}dx $$
$$=2\ln 2$$
In fact, $\displaystyle\sum_{n=1}^\infty \frac{1}{n(2n-1)}=2\ln 2$.
Why the first solution is false but the second is true ?
I have known that uniformly convergent series can be integrated term by term.
Is this mean that $\sum_{n=1}^\infty(2x^{2n-2}-2x^{2n-1})$ ; second solution, is uniformly convergent but $\sum_{n=1}^\infty(2x^{2n-2}-x^{n-1})$ ; first solution, is not ?
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Here is an even more baffling calculation:
\begin{align*}
0 &= \sum_{n=1}^{\infty} \left( \frac{2}{2n} - \frac{1}{n} \right) \\
&= \sum_{n=1}^{\infty} \int_{0}^{1} (2 x^{2n-1} - x^{n-1}) \, dx \\
&= \int_{0}^{1} \sum_{n=1}^{\infty} (2 x^{2n-1} - x^{n-1}) \, dx \\
&= \int_{0}^{1} \left( \frac{2x}{1-x^2} - \frac{1}{1-x} \right) \, dx \\
&= - \int_{0}^{1} \frac{dx}{x+1} \\
&= -\log 2.
\end{align*}
So where did we mess up? It is the third step, where we interchanged the order of integration and summation. In this step, we are actually dealing with $\infty - \infty$ type indeterminate, which are cleverly hidden, and we failed to manage it properly. This is more evident if we plot the graph of the partial sum
$$y = \sum_{k=1}^{n} (2x^{2k-1} - x^{k-1})$$
for $n = 1, \cdots, 20$ as follows. (Color changes from red to green to blue as $n$ increases.)
The mass of $\log 2$ is concentrated around the high spike at $x = 1$, which vanishes as $n \to \infty$. This is why in both calculation we end up losing $\log 2$.
Speaking differently, the common mistake we have done is essentially the same as in the following bogus argument:
\begin{align*}
0
&= \left( 1 + \frac{1}{2} + \frac{1}{3} + \cdots \right) - \left( 1 + \frac{1}{2} + \frac{1}{3} + \cdots \right) \\
&= \left( \frac{2}{2} + \frac{2}{4} + \frac{2}{6} + \cdots \right) - \left( 1 + \frac{1}{2} + \frac{1}{3} + \cdots \right) \\
&= -1 + \left(\frac{2}{2} - \frac{1}{2}\right) - \frac{1}{3} + \left( \frac{2}{4} - \frac{1}{4}\right) - \cdots \\
&= -1 + \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \cdots \\
&= -\log 2.
\end{align*}
In our calculation, shifting the order of the summand is achieved by doubling the exponent; that is, by changing $\int_{0}^{1} x^{n-1} \,dx$ by $\int_{0}^{1} 2x^{2n-1} \,dx$.
|
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|
Solve $\vert x-2\vert+2\vert x-4\vert\leq \vert x+1\vert$ I was helping someone with abolute values and inequalities and found this question.
What is the easiest way to solve this?
The only thing I thought of is to add the L.H.S and graph it with the R.H.S to answer the questoin is there simpler way to deal with this?
Thank you
|
You have:
$$
x-2\ge 0 \iff x\ge 2
$$
$$
x-4\ge 0 \iff x\ge 4
$$
$$
x+1\ge 0 \iff x\ge -1
$$
so we can split the inequality $|x-2|+2|x-4|-|x+1|\le 0$ in four systems:
$$
\begin{cases}
x<-1\\
2-x+2(4-x)+x+1\le 0
\end{cases}
$$
$$
\begin{cases}
-1\le x<2\\
2-x+2(4-x)-x-1\le 0
\end{cases}
$$
$$
\begin{cases}
2\le x<4\\
x-2+2(4-x)-x-1\le 0
\end{cases}
$$
$$
\begin{cases}
4\le x\\
x-2+2(x-4)-x-1\le 0
\end{cases}
$$
and the solution of the inequality is the union of the solutions of these systems.
With a bit of algebra you can see that the first two systems have no solutions, and the solutions of the other two are: $ \dfrac{5}{2}\le x<4$ and $4\le x \le \dfrac{11}{2}$, so the final solution is the union: $\dfrac{5}{2}\le x\le\dfrac{11}{2}$.
|
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|
Calculating the $n^\text{th}$ derivative How do we calculate the $n^{\text{th}}$ derivative for
$$
\frac{x^3}{(x-a)(x-b)(x-c)}?
$$
How can I obtain the partial fraction for the given term?
|
Using Partial fraction
Above Given fraction is a Improper fraction then by actual division, Since same degree in above and below.
I.e $\displaystyle \frac{x^3}{x\cdot x \cdot x}$
Then $$\displaystyle \frac{x^3}{(x-a)(x-b)(x-c)}= 1+\frac{f(x)}{(x-a)(x-b)(x-c)}$$
Where $f(x)$ is a degree of polynomial $2$
So $$\displaystyle \frac{x^3}{(x-a)(x-b)(x-c)}= 1+\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}....................(1)$$
So $$x^3 = (x-a)(x-b)(x-c)+A(x-b)(x-c)+B(x-c)(x-a)+C(x-a)(x-b)$$
Now Put $x=a\;,$ We get $\displaystyle a^3 = A(a-b)(a-c)\Rightarrow A = \frac{a^3}{(a-b)(a-c)}$
Similarly Put $x=b\;,$ We get $\displaystyle B = \frac{b^3}{(b-c)(b-a)}$
Similarly Put $x=c\;,$ We get $\displaystyle C = \frac{c^3}{(c-a)(c-b)}$
Now Let $$\displaystyle f(x) = \frac{x^3}{(x-a)(x-b)(x-c)} = 1+\frac{A}{(x-a)}+\frac{B}{(x-b)}+\frac{C}{(x-c)}$$
So $$\displaystyle f'(x) = -\frac{A}{(x-a)^2}-\frac{B}{(x-b)^2}-\frac{C}{(x-c)^2}$$
So $$\displaystyle f''(x) = \frac{2A}{(x-a)^3}+\frac{2B}{(x-b)^3}+\frac{2C}{(x-c)^3}$$
In a Similar fashion, We get $$\displaystyle f^{n}(x) = (-1)^n\frac{n!\cdot A}{(x-a)^{n+1}}+(-1)^n\frac{n!\cdot B}{(x-b)^{n+1}}+(-1)^n\frac{n!\cdot C}{(x-c)^{n+1}}$$
and put values of $A,B,C$ from above.
|
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|
Solve complex equation $z^4=a^{16}$ Let $a$ be some complex number. I have to solve equation
$$z^4=a^{16}$$
One is tempted to "simplify" it to $z=a^4$, so it is the solutions. But somebody told me, there are more solutions than that. Is it true and why? It seems counterintuitive.
|
Note that every $a \in \mathbb{C}$ can be written as $a = re^{i\varphi}$, hence your equation delivers
$$z^4 = r^{16} e^{16i\varphi}$$
The complex roots are given by
$$z^n = re^{i\varphi} \implies z_k = r^{\frac{1}{n}}e^{i\left( \frac{\varphi}{n} + \frac{2\pi k}{n}\right)} \qquad k=0,\dots,n-1$$
So you have $n$ solutions for the $n^{\text{th}}$ complex root. In your case
\begin{align*}
z_0 &= r^4 e^{i\left( \frac{16\varphi}{4} + 0 \right)} = \left( r, \varphi \right) = \left( r^4, 4\varphi \right)\\
z_1 &= r^4 e^{i\left( \frac{16\varphi}{4} + \frac{2\pi}{4} \right)} = \left( r, \varphi \right) = \left( r^4, 4\varphi + \frac{\pi}{2} \right)\\
z_2 &= r^4 e^{i\left( \frac{16\varphi}{4} + \frac{4\pi}{4} \right)} = \left( r, \varphi \right) = \left( r^4, 4\varphi + \pi \right)\\
z_3 &= r^4 e^{i\left( \frac{16\varphi}{4} + \frac{6\pi}{4} \right)} = \left( r, \varphi \right) = \left( r^4, 4\varphi + \frac{3\pi}{2} \right)\\
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
All real solution of the equation $2^x+3^x+6^x = x^2$
Find all real solution of the equation $2^x+3^x+6^x = x^2$
$\bf{My\; Try::}$ Let $$f(x) = 2^x+3^x+6^x-x^2\;,$$ Now Using first Derivative
$$f'(x) = 2^x\cdot \ln 2+3^x\cdot \ln 3+6^x\cdot \ln 3-2x$$
Now for $x<0\;,$ We get $f'(x)>0,$ So function $f(x)$ is strictly Increasing function.
so it will cut $\bf{X-}$ axis at exactly one point.
So $f(x)=0$ has exactly one root for $x=-1$ for $x<0$
Now How can I calculate no. of real roots for $x\geq 0$
Thanks
|
The derivative $f'(x)$ is certainly greater than $h(x)=6^x \ln 6 - 2x.$ [I assume you had a typo when you wrote the term $6^x \ln 3,$ as the derivative of $6^x$ is $6^x \ln 6.$]
Now assume $x \ge 0$ and note $h(0)=\ln 6 >0.$ Also we have
$$h'(x)=6^x (\ln 6)^2 -2 > (3.2)\cdot 6^x -2\ge 1.2 >0,$$
using the underestimate $3.2$ for $(\ln 6)^2$ and the fact that for nonnegative $x$ one has $6^x \ge 1.$
Thus we have both $h(0)>0$ and $h$ increasing on $[0,\infty),$ and can conclude as desired that $h(x) \ge 0$ for $x \ge 0.$ Combining this with your result for negative $x$ one gets $f$ increasing on $\mathbb{R}$ so the only zero of $f$ is at $x=-1.$
|
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|
Evaluating $\lim _{x\to 1}\left(\frac{\left(2x^2-1\right)^{\frac{1}{3}}-x^{\frac{1}{2}}}{x-1}\right)$ Without L'Hopital or Calculus? What is: $\lim _{x\to 1}\left(\frac{\left(2x^2-1\right)^{\frac{1}{3}}-x^{\frac{1}{2}}}{x-1}\right)$?
Thanks in advance
Much appreciated!
|
It mostly depends on what you're allowed to use. In order to do it without recurring to derivatives, I'd split it into two parts:
\begin{align}
\lim _{x\to 1}\biggl(\frac{(2x^2-1)^{1/3}-x^{1/2}}{x-1}\biggr)
&=\lim _{x\to 1}\biggl(\frac{(2x^2-1)^{1/3}-1+1-x^{1/2}}{x-1}\biggr)
\\[6px]
&=\lim_{x\to 1}\left(\frac{(2x^2-1)^{1/3}-1}{x-1}-
\frac{x^{1/2}-1}{x-1}\right) \\[6px]
&=\lim_{x\to 1}\frac{(2x^2-1)^{1/3}-1}{x-1}-
\lim_{x\to 1}\frac{x^{1/2}-1}{x-1}
\end{align}
provided both limits in the last expression exist (they do). The second one is easy: multiply by $x^{1/2}+1$ to get
$$
\lim_{x\to 1}\frac{x^{1/2}-1}{x-1}=
\lim_{x\to 1}\frac{x-1}{(x-1)(x^{1/2}+1)}=
\lim_{x\to 1}\frac{1}{x^{1/2}+1}=\frac{1}{2}
$$
The first one can be treated in a similar way recalling that
$$
a^3-b^3=(a-b)(a^2+ab+b^2)
$$
where $a=(2x^2-1)^{1/3}$ and $b=1$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to find the equations whose roots are equal to the following numbers? How to find the equations whose roots are equal to the following numbers ?
(a) $\sin^2\frac{\pi}{2n+1}$,$\sin^2\frac{2\pi}{2n+1}$,$\sin^2\frac{3\pi}{2n+1}$,...,$\sin^2\frac{n\pi}{2n+1}$
(b)$\cot^2\frac{\pi}{2n+1}$,$\cot^2\frac{2\pi}{2n+1}$,$\cot^2\frac{3\pi}{2n+1}$,...,$\cot^2\frac{n\pi}{2n+1}$
I got stuck while solving this problem.This was from a complex number chapter practice problem exercise. I tried to fit in Demoivre's theorem considering roots of unity.But no idea how to bring about the squared sine terms. Any suggestions?
|
For $(b),$ see Trig sum: $\tan ^21^\circ+\tan ^22^\circ+...+\tan^2 89^\circ = ?$
For $(a),$
Method $\#1:$ $$\cot^2y=\dfrac{1-\sin^2y}{\sin^2y}$$
Method $\#2:$
Using De Moivre's formula,
$$\cos(2m+1)x+i\sin(2m+1)x=(\cos y+i\sin y)^{2m+1}=\cdots$$
Equating the imaginary parts (See this),
$$\sin(2m+1)x=(2n+1)\sin x-\cdots+(-1)^{m-1}2^{2m-2}(2m+1)\sin^{2m-1}x+(-1)^m2^{2m}\sin^{2m+1}x$$
Now if $\sin(2m+1)x=0,(2m+1)x=r\pi$ where $r$ is any integer
$x=\dfrac{r\pi}{2m+1}$ where $r\equiv0,\pm1,\pm2,\cdots,\pm m\pmod{2m+1}$
Can you take it from here?
|
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|
Why the maximum value of $|AB|+\sqrt{2}|CD|$ is $10$?
Assume a Convex quadrilateral $ABCD$ a,such $|AC|=1,|BC|=3\sqrt{2}$,and $|AD|=|BD|=\frac{\sqrt{2}}{2}|AB|$, find the maximum of the value $|AB|+\sqrt{2}|CD|$
The Book only have answer: $10$.
I wanted to solve this by using law of cosines in triangles $ABC$and $BDC$ to determine $AB$ and $CD$,
Assume that $\angle ACB=x$,then $AB=\sqrt{1+18-6\sqrt{2}\cos{x}}=\sqrt{19-6\sqrt{2}\cos{x}}$，But I can't determine $CD$ .
|
Let $\angle ADC=x$, then $\angle BDC=90^\circ-x$. Apply the law of cosine to $\triangle ADC$ we have
$$2AD\cdot CD\cos x=AD^2+CD^2-1\tag{1}$$
and to $\triangle BCD$ we have
$$2BD\cdot CD \cos (90^\circ-x) = BD^2+CD^2-18\tag{2}.$$
Square both (1) and (2) and add them up we get
$$4AD^2\cdot CD^2 = (AD^2+CD^2-1)^2+(AD^2+CD^2-18)^2.$$
so
$$0=2AD^4+2CD^4-26(AD^2+CD^2)+145\ge (AD^2+CD^2)^2-38(AD^2+CD^2)+(1+18^2).$$
It follows that
$$AD^2+CD^2\le 19+\sqrt{19^2-1-18^2}=19+6=25.$$
Thus
$$(AD+CD)^2\le 2(AD^2+CD^2)=50$$
or equivalently
$$\frac{AB+\sqrt{2}CD}{\sqrt2}\le \sqrt{2(AD^2+CD^2)}\le \sqrt{50}.$$
So
$$AB+\sqrt{2}CD\le 10.$$
The equality happens when $AD=CD=BD$, equivalently $\angle ACB=135^\circ$.
|
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|
What is row reduced echelon form? How to row reduce this matrix? I'm not being able to grasp the concept of row reduced echelon form. Please, explain how to row reduce one of the the following matrices.
$A =
\begin{bmatrix}
1&3&4&5\\3&9&12&9\\1&3&4&1
\end{bmatrix}$
$B=
\begin{bmatrix}
1&2&1&2\\0&1&0&1\\-1&2&0&3
\end{bmatrix}$
|
You begin to put the matrix in echelon form, with the pivots equal to $1$, going downwards row after row. When that is done, all coefficients under a pivot (in the same column) are equal to $0$.Then you make the coefficients above the pivots equal to $0$, starting fom the last pivot and going upwards:
\begin{align*}
&\begin{bmatrix}
1&3&4&5\\3&9&12&9\\1&3&4&1
\end{bmatrix}
\rightsquigarrow
\begin{bmatrix}
1&3&4&5\\0&0&0&-6\\0&0&0&-4\\
\end{bmatrix}
\rightsquigarrow
\begin{bmatrix}
1&3&4&5\\0&0&0&1\\0&0&0&1
\end{bmatrix}\\
\rightsquigarrow
&\begin{bmatrix}
1&3&4&5\\0&0&0&1\\0&0&0&0
\end{bmatrix}\rightsquigarrow
\begin{bmatrix}
1&3&4&0\\0&0&0&1\\0&0&0&0
\end{bmatrix}
\end{align*}
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Maclaurin expansion of $y=\frac{1+x+x^2}{1-x+x^2}$ to $x^4$
Maclaurin expansion of $$\displaystyle y=\frac{1+x+x^2}{1-x+x^2}\,\,\text{to } x^4$$
I have tried by using Maclaurin expansion of $\frac1{1-x}=1+x+x^2+\cdots +x^n+o(x^n)$, but it seems not lead me to anything.
|
The sortest way is division of numerator by the denominator along the increasing powers of $x$:
This division tell us that
\begin{align*}1+x+x^2&=(1-x+x^2)(1+2x+2x^2-2x^4)+x^5(-2+2x)\\
\text{so that}\quad\frac{1+x+x^2}{1-x+x^2}&=1+2x+2x^2-2x^4+\frac{x^5(-2+2x)}{1-x+x^2}\\&=1+2x+2x^2-2x^4+O(x^5).\end{align*}
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
How is it possible that $\int\frac{dy}{(1+y^2)(2+y)}$ = $\frac{1}{5}\int\frac{dy}{y+2}-\int\frac{ydy}{1+y^2}+\int\frac{2dy}{1+y^2} $? Suppose we have a fraction
$$I=\int\frac{dy}{(1+y^2)(2+y)}$$
How is it possible that
$$5I = \int\frac{dy}{y+2}-\int\frac{ydy}{1+y^2}+\int\frac{2dy}{1+y^2} ?$$
How are they using partial fractions to do this?
|
$$\begin{align*}
\frac1{y+2}-\frac{y}{1+y^2}+\frac2{1+y^2}&=\frac1{y+2}+\frac{2-y}{1+y^2}\\
&=\frac{(1+y^2)+(4-y^2)}{(y+2)(1+y^2)}\\
&=\frac5{(y+2)(1+y^2)}
\end{align*}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Complex numbers - locus of a point Question:
If $z \neq 1$ and ${z^2} \over {z-1}$ is real, then find the locus of the point represented by the complex number $z$.
I'm not sure how to approach this question. I attempted to substitute $z = x + iy$, however, that didn't solve the problem. It's quite clear, by observation, that any point on the real axis would satisfy the equation. However, there also have to be some other points. How would I approach this question?
|
Let $\displaystyle z=x+iy$ in $\displaystyle \frac{z^2}{z-1} = \frac{(x+iy)^2}{x+iy-1} = \frac{x^2-y^2+2ixy}{(x-1)+iy}\times \frac{(x-1)-iy}{(x-1)-iy}$
So we get $\displaystyle \frac{(x^2-y^2)(x-1)+2xy^2+i\left[-y(x^2-y^2)++2xy(x-1)\right]}{(x-1)^2+y^2}$
$\displaystyle = \frac{(x^2-y^2)(x-1)+2xy^2}{(x-1)^2+y^2}+i\frac{2xy(x-1)-y(x^2-y^2)}{(x-1)^2+y^2}$
so put $\displaystyle \frac{2xy(x-1)-y(x^2-y^2)}{(x-1)^2+y^2} = 0\Rightarrow 2xy(x-1)-y(x^2-y^2) =0$
So we get $\displaystyle y\left[2x^2-2x-x^2+y^2\right] = 0\Rightarrow y[x^2+y^2-2x] =0$
So $y=0$ or $x^2+y^2-2x=0$
|
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|
Solving $\int_0^{\frac{1}{2}} x^{\frac{3}{2}} (1-2x)^{\frac{3}{2}}dx$? How to solve the definite integral $\int_0^{\frac{1}{2}} x^{\frac{3}{2}} (1-2x)^{\frac{3}{2}}~dx$?
Funnily enough, this is actually a problem that showed up during my real analysis course. At first glance, the problem seemed solvable by using knowledge from basic elementary calculus, but it turned out not to be so simple. I initially tried the substitution $u=\sqrt{x}$, but that lead me nowhere...
Let $u=x^{\frac{1}{2}}$, $du=\frac{1}{2}x^{-\frac{1}{2}}$.
$\int_0^{\frac{1}{2}} x^{\frac{3}{2}} (1-2x)^{\frac{3}{2}}~dx$ =
$2\int_0^{\frac{1}{\sqrt{2}}} u^4 (1-2u^2)^{\frac{3}{2}}~du$.
EDIT: I have tried an additional substitution to the above, namely $u= \frac{\sin t}{\sqrt{2}}$.
$$
\begin{align*}
2\int_0^{\frac{1}{\sqrt{2}}} u^4 (1-2u^2)^{\frac{3}{2}}~du&=
2\int_0^{\frac{\pi}{2}} \frac{\sin^4 t}{4} (1-\sin^2 t)^{\frac{3}{2}}\cdot \frac{\cos t}{\sqrt{2}}~dt\\
&=\frac{1}{2\sqrt{2}}\int_0^{\frac{\pi}{2}}\sin^4t\cos^4t~dt.
\end{align*}
$$
Hopefully my calculations aren't wrong...
|
$$\int_0^{\frac{1}{2}} x^{\frac{3}{2}} (1-2x)^{\frac{3}{2}}dx$$
Substitute $u = \sqrt{x}$
$$ = 2\int_0^{\frac{1}{\sqrt{2}}} u^4 (1-2u^2)^{\frac{3}{2}}dx$$
Substitute u = $\frac{\sin(s)}{\sqrt{2}}$, $du = \frac{\cos(s)}{\sqrt{2}}$ (this will be valid for $0 < s < \frac{\pi}{2}$)
$$ = \sqrt{2}\int_0^{\frac{\pi}{2}} \frac{1}{4}\sin^4(s)\cos(s)cos^2(s)^{3/2}ds$$
$$ = \sqrt{2}\int_0^{\frac{\pi}{2}} \frac{1}{4}\sin^4(s)\cos^4(s)ds$$
$$ = \frac{1}{2\sqrt{2}}\int_0^{\frac{\pi}{2}} \sin^4(s)\cos^4(s)ds$$
We then apply the following reduction formula with $m=4$ and $n=4$ (which is really messy... I'll leave out a lot of intermediate stuff):
$$\int \cos^m(s)\sin^m(s)ds = -\frac{\cos^{m+1}(s)\sin^{m+1}(s)}{m+n} + \frac{n-1}{m+n}\int \cos^m(s) \sin^{n-2}(s)ds$$
$$\Rightarrow \frac{3}{16\sqrt{2}}\int_0^{\frac{\pi}{2}} \sin^2(s)\cos^4(s)ds$$
$$=\frac{3}{16\sqrt{2}}\int_0^{\frac{\pi}{2}} (1-\cos^2(s))\cos^4(s)ds$$
$$=\frac{3}{16\sqrt{2}}\int_0^{\frac{\pi}{2}} \big(\cos^4(s) - \cos^6(s)\big)ds$$
$$=\frac{3}{16\sqrt{2}}\int_0^{\frac{\pi}{2}}\cos^4(s)ds - \frac{3}{16\sqrt{2}}\int_0^{\frac{\pi}{2}}\cos^6(s)ds$$
We then apply the following reduction formula for $m=6$:
$$\int \cos^m(s) = \frac{\sin(s)\cos^{m-1}(s)}{m} + \frac{m-1}{m}\int \cos^{m-2}(s)ds$$
$$\Rightarrow \frac{1}{32\sqrt{2}}\int_0^{\frac{\pi}{2}}\cos^4(s)ds$$
Apply the cosine reduction formula again for $m=4$:
$$\Rightarrow \frac{3}{128\sqrt{2}}\int_0^{\frac{\pi}{2}}\cos^2(s)ds$$
We then make the substitution $cos^2(s) = \frac{1}{2}\cos(2s)+\frac{1}{2}$, and after a few very basic calculus and algebra steps arrive at the solution.
$$= \frac{3\pi}{512\sqrt{2}}$$
|
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|
If $\frac{x^2+y^2}{x+y}=4$, then what are the possible values of $x-y$?
If $\frac{x^2+y^2}{x+y}=4$,then all possible values of $(x-y)$ are given by
$(A)\left[-2\sqrt2,2\sqrt2\right]\hspace{1cm}(B)\left\{-4,4\right\}\hspace{1cm}(C)\left[-4,4\right]\hspace{1cm}(D)\left[-2,2\right]$
I tried this question.
$\frac{x^2+y^2}{x+y}=4\Rightarrow x+y-\frac{2xy}{x+y}=4\Rightarrow x+y=\frac{2xy}{x+y}+4$
$x-y=\sqrt{(\frac{2xy}{x+y}+4)^2-4xy}$, but I am not able to proceed. I am stuck here. Is my method wrong?
|
Given $$\displaystyle \frac{x^2+y^2}{x+y} = 4\Rightarrow x^2+y^2 = 4x+4y$$
So we get $$x^2-4x+4+y^2-4y+4 = 8\Rightarrow (x-2)^2+(y-2)^2 = (2\sqrt{2})^2$$
Now Put $$x-2 = 2\sqrt{2}\cos \phi\Rightarrow x = 2+2\sqrt{2}\cos \phi$$
and $$y-2 = 2\sqrt{2}\sin \phi\Rightarrow y = 2+2\sqrt{2}\sin \phi$$
So $$\displaystyle x-y = 2\sqrt{2}\left(\cos\phi-\sin \phi\right) = 4\cdot \left[\cos \phi \cdot \frac{1}{\sqrt{2}}-\sin \phi\cdot \frac{1}{\sqrt{2}}\right] = 4\cos\left(\phi+\frac{\pi}{4}\right)$$
So we know that $$\displaystyle -4 \leq 4\cos\left(\phi+\frac{\pi}{4}\right)\leq 4$$
So we get $$-4 \leq x-y\leq 4\Rightarrow x-y\in \left[-4,4\right]$$
|
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|
Can't understand this pseudo-inverse relation. In the Answer to a different Question, a curious matrix relation came up:
M is symmetric and non-singular, G is non-symmetric and singular.
Theorem:
When $M$ is positive/negative definite, or more generally, if $G$ and $(G^\dagger G) M (G^\dagger G)$ have the same ranks then
$$G^{T}\left(GMG^{T}\right)^{\dagger}G=\left((G^\dagger G)M(G^\dagger G)\right)^\dagger$$
I have no idea how to prove this or why the rank relationships asserted would be true.
An example:
$M=\left(
\begin{array}{ccc}
1. & 2. & 3. \\
2. & 4. & 5. \\
3. & 5. & 6. \\
\end{array}
\right)$ $G=\left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 0 \\
\end{array}
\right)$ $GMG^T=\left(
\begin{array}{ccc}
1. & 4. & 0. \\
4. & 16. & 0. \\
0. & 0. & 0. \\
\end{array}
\right)$
$GMG^T$ has a rank of 1 and the relation does not hold. But if we add a small number times the Identity Matrix to M, $GMG^T$ will have rank 2 and the relation will hold.
What is going on there? Can anyone help with an explanation or proof? And under exactly what circumstances is this true or not true?
|
Primary matrices
The matrix
$$
\mathbf{M} =
\left[
\begin{array}{ccc}
1 & 2 & 3 \\
2 & 4 & 5 \\
3 & 5 & 6 \\
\end{array}
\right]
$$
has rank $\rho_{M} = 3$. The matrix
$$
\mathbf{G} =
\left[
\begin{array}{ccc}
1 & 0 & 0 \\
2 & 0 & 0 \\
0 & 0 & 0 \\
\end{array}
\right]
$$
has rank $\rho_{G} = 2$.
To make manipulation easier, introduce the submatrices
$$
m =
\left[
\begin{array}{ccc}
1 & 2 \\
2 & 4
\end{array}
\right], \quad
g =
\left[
\begin{array}{ccc}
1 & 0 \\
2 & 0
\end{array}
\right].
$$
We see a stencil matrix
$$
%
\mathbf{G} =
%
\left[
\begin{array}{ccc}
g & \mathbf{0} \\
\mathbf{0} & 0
\end{array}
\right], \quad
%
\mathbf{G}^{\dagger} =
%
\left[
\begin{array}{ccc}
g^{-1} & \mathbf{0} \\
\mathbf{0} & 0
\end{array}
\right], \quad
%
%
$$
Product matrices
The product matrix
$$
\mathbf{G} \mathbf{M} \mathbf{G}^{\mathrm{T}} =
%
\left[
\begin{array}{cc}
gmg^{\mathrm{T}} & \mathbf{0} \\
\mathbf{0} & 0 \\
\end{array}
\right]
%
=
%
\left[
\begin{array}{rrr}
1 & 2 & 0 \\
2 & 4 & 0 \\
0 & 0 & 0 \\
\end{array}
\right]
$$
has rank $\rho_{GMG^{T}} = 1$.
The pseudoinverse is
$$
\left( \mathbf{G} \mathbf{M} \mathbf{G}^{\mathrm{T}} \right)^{\dagger} =
%
\frac{1}{25}
\left[
\begin{array}{rrr}
1 & 2 & 0 \\
2 & 4 & 0 \\
0 & 0 & 0 \\
\end{array}
\right]
%
=
%
\frac{1}{25}
\mathbf{G} \mathbf{M} \mathbf{G}^{\mathrm{T}}.
$$
The final product matrix is
$$
\mathbf{G}^{\dagger}\mathbf{G} \mathbf{M} \mathbf{G}^{\dagger}\mathbf{G} =
\left[
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{array}
\right].
$$
LHS
The left hand side of the identity is
$$
\mathbf{G}^{\mathrm{T}} \left( \mathbf{G} \mathbf{M} \mathbf{G}^{\mathrm{T}} \right)^{\dagger} \mathbf{G} =
%
\left[
\begin{array}{rrr}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{array}
\right]
$$
RHS
The right hand side of the identity is
$$
\left( \mathbf{G}^{\dagger}\mathbf{G} \mathbf{M} \mathbf{G}^{\dagger}\mathbf{G} \right)^{\dagger} =
\left[
\begin{array}{rrr}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{array}
\right]
$$
The equality holds in this case:
$$
\mathbf{G}^{\mathrm{T}} \left( \mathbf{G} \mathbf{M} \mathbf{G}^{\mathrm{T}} \right)^{\dagger} \mathbf{G}
=
\left( \mathbf{G}^{\dagger}\mathbf{G} \mathbf{M} \mathbf{G}^{\dagger}\mathbf{G} \right)^{\dagger}
$$
Postscript Asymmetric $\mathbf{G}$ used after comments by @Dohleman.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1444777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
$\int\frac{x dx}{\sqrt{x^4+4x^3-6x^2+4x+1}}$ $\int\frac{x dx}{\sqrt{x^4+4x^3-6x^2+4x+1}}$
I was given this question by my senior.I tried to solve it but could not reach the answer.
Let $I= \int\frac{x dx}{\sqrt{x^4+4x^3-6x^2+4x+1}} $
$I=\int\frac{dx}{\sqrt{x^2+4x-6+\frac{4}{x}+\frac{1}{x^2}}}$
Then after repeated attempts, i could not solve further.
I think this function is not integrable.Am i correct?If not,how should i move ahead.Please help.
|
Note that:
$$(x+1)^4=x^4+4x^3+6x^2+4x+1$$
Then
$$(x+1)^4-12x^2=x^4+4x^3-6x^2+4x+1$$
$$(x+1)^4-12x^2=12x^2((\frac{(x+1)^2}{\sqrt{12}x})^2-1)$$
Let us define
$$f(x)=\frac{(x+1)^2}{\sqrt{12}x}$$
And derive it:
$$f'(x)=\frac{2(x+1)x-(x+1)^2}{12x^2}=\frac{x^2-1}{12x^2}$$
Now lets have a look at
$$[arcsin(f(x))]'=\frac{f'(x)}{\sqrt{f(^2x)-1}}=\frac{x^2-1}{12x^2\sqrt{(\frac{(x+1)^2}{\sqrt{12}x})^2-1}}=\frac{x^2-1}{12x\sqrt{x^4+4x^3-6x^2+4x+1}}$$
Simplifying a bit:
$$[arcsin(f(x))]'=\frac{x}{12\sqrt{x^4+4x^3-6x^2+4x+1}}-\frac{1}{12x\sqrt{x^4+4x^3-6x^2+4x+1}}$$
Thus:
$$\int{\frac{xdx}{\sqrt{x^4+4x^3-6x^2+4x+1}}}=12\arcsin(\frac{(x+1)^2}{\sqrt{12}x})+\int{\frac{dx}{x\sqrt{x^4+4x^3-6x^2+4x+1}}}$$
Now, that's not what you were looking for, but that's my best attempt
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1445369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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|
Show that $5\cos^2x - 2\sqrt{3}\sin x\cos x + 3\sin^2x = \cos 2x - \sqrt{3}\sin2x + 4$ Show that $5\cos^2x - 2\sqrt{3}\sin x\cos x + 3\sin^2x = \cos 2x - \sqrt{3}\sin2x + 4$
Attempt.
I'm preparing for a trigonometry examination tomorrow. So I'm solving as many questions as possible. This is confusing. I've tried solving it but its not proofing. I've tried replacing with trigonometric identities and simplifying. Please someone should help even if it is how to start?
|
Notice, the following formula
$$\cos^2A+\sin^2 A=1$$
$$2\sin A\cos A=\sin 2A$$
$$\cos^2 A-\sin^2 A=\cos 2A$$
Now, we have
$$LHS=5\cos^2 x-2\sqrt 3\sin x\cdot \cos x+3\sin^2x$$
$$=4\cos^2 x+\cos^2 x-\sqrt 3(2\sin x\cdot \cos x)+4\sin^2x-\sin^2 x$$
$$=4(\cos^2 x+\sin^2x)-\sqrt 3\sin 2x+(\cos^2 x-\sin^2 x)$$
$$=4(1)-\sqrt 3\sin 2x+(\cos 2x)$$
$$=\cos 2x-\sqrt 3\sin 2x+4=RHS$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1456576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Calculating $\lim_{x\to 0}\frac{x^2+x\cdot \sin x}{-1+\cos x}$ Without L'Hopital,
$$\lim_{x\to 0}\frac{x^2+x\cdot \sin x}{-1+\cos x}$$
That's
$$\frac{x^2+x\cdot \sin x}{-1+\left(1-2\sin^2\frac{x}{2}\right)} = \frac{x^2+x\cdot \sin x}{-2\sin^2\frac{x}{2}}$$
Let's split this
$$\frac{x\cdot x}{-2\cdot\sin\frac{x}{2}\cdot \sin\frac{x}{2}} + \frac{x\cdot \sin x}{-2\cdot\sin\frac{x}{2}\cdot \sin\frac{x}{2}}$$
On the left one, it would be great to have $\frac{x}{2} \cdot \frac{x}{2}$. To do so, I will multiply and divide by $\frac{1}{4}$:
$$\frac{\frac{x}{2}\cdot \frac{x}{2}}{\frac{1}{4}\cdot-2\cdot\sin\frac{x}{2}\cdot \sin\frac{x}{2}} + \frac{x\cdot \sin x}{-2\cdot\sin\frac{x}{2}\cdot \sin\frac{x}{2}}$$
Then we use the identity $\frac{x}{\sin x} = 1$:
$$-2 + \frac{x\cdot \sin x}{-2\cdot\sin\frac{x}{2}\cdot \sin\frac{x}{2}}$$
On the right side, I'd like to have $\frac{x}{2}$ there, so I will multiply and divide by $\frac{1}{2}$:
$$-2 + \frac{\frac{x}{2}\cdot \sin x}{-2\cdot\sin\frac{x}{2}\cdot \sin\frac{x}{2}\cdot\frac{1}{2}}$$
We use the identity $\frac{x}{\sin x} = 1$:
$$-2 + \frac{\sin x}{-2\cdot\sin\frac{x}{2}\cdot\frac{1}{2}}$$
Hmm... we could perform the addition I suppose:
$$\frac{\sin x - 2(-2\cdot\sin\frac{x}{2}\cdot\frac{1}{2})}{-2\cdot\sin\frac{x}{2}\cdot\frac{1}{2}}$$
Simplify:
$$\frac{\sin x + (4\cdot-2\sin\frac{x}{2}\cdot-2\frac{1}{2})}{-2\cdot\sin\frac{x}{2}\cdot\frac{1}{2}} = \frac{\sin x +8\sin\frac{x}{2}}{\sin\frac{x}{2}}$$
Then
$$\frac{\sin x}{\sin\frac{x}{2}} + \frac{8\sin\frac{x}{2}}{\sin\frac{x}{2}} = \frac{\sin x}{\sin\frac{x}{2}} + 8$$
I'm close! The answer should be $-4$, but I don't know what to do with
$$\frac{\sin x}{\sin\frac{x}{2}} + 8$$
So basically I need to know what to do with $\frac{\sin x}{\sin\frac{x}{2}}$, but I also included my whole procedure just in case I've been doing it wrong all along.
|
You did this in a needlessly elaborate way.
$$\frac{x^{2}+x\sin x}{-1+\cos x}\approx\frac{x^{2}+x\left(x-x^{3}/6\right)}{-1+\left(1-x^{2}/2\right)}=\frac{2x^{2}-x^{4}/6}{-x^{2}/2}=-4$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1457217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Hyperbolic Trig Inequality The following hyperbolic trig inequality came up.
$$0 \leq y \leq x \leq 2 \implies \sinh(x)-\sinh(y) \leq \sinh(x-y)\cdot e^{xy/2}.$$
I spent many hours trying to prove it. The first few terms of the Taylor series work out, but I couldn't get a general proof. I passes the usual sanity checks ($y=0$, $y=x$, $x=2$) and I also verified it numerically by plotting and checking about $10^9$ random $x$ and $y$ values.
Any ideas?
|
Completing Brevan Ellefsen's solution.
First, since
$\sinh x - \sinh y = 2 \sinh \frac{x-y}{2} \cosh \frac{x+y}{2}$,
and
$\sinh (x-y) = 2 \sinh \frac{x-y}{2} \cosh \frac{x-y}{2}$,
we have
$$
r\equiv\frac{\sinh x - \sinh y}{\sinh (x-y)}
= \frac{\cosh \frac{x+y}{2} } {\cosh \frac{x-y}{2}}
= \frac{1 + \tanh\frac x 2 \tanh \frac y 2 } { 1 - \tanh\frac x 2 \tanh\frac y 2}.
$$
For $0 \le \dfrac{x}{2}, \dfrac{y}{2} \le 1$, we have
$$
0
\le
\tanh\frac{x}{2} \cdot \tanh\frac{y}{2}
\le
\frac{x}{2} \cdot \tanh\frac{y}{2}
\le
\tanh\left(\frac{x}{2} \cdot \frac{y}{2}\right) \le 1.
$$
Since $\dfrac{1+t}{1-t}$ is an increasing function of $t$ for $t \in (0, 1)$, we get
$$
r \le \frac{1 + \tanh\frac{xy}{4}}{1-\tanh\frac{xy}{4}}
=\exp\frac{xy}{2}.
$$
This completes the proof.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1460482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
guess the color of the next card, what is the payoff of this game? There are 26 red cards and 26 black cards on the table which are randomly shuffled and are facing down onto the table. The host turns up the cards one at a time. You can stop the game any time (even at the beginning of the game). Once you stops the game, the next card is turned up: if it is red, you get $1; otherwise you pay the host one dollar. What is the payoff of this game?
|
[Thanks to Brian Scott for pointing out my original misinterpretation of this problem.]
The value of the game is zero.
Let $f(r,b)$ denote the value of the game if there are $r$ red cards and $b$ black cards remaining. If we stop at that point, the expected value is $(r-b)/(r+b)$; but as we also have the option of continuing, we have the recurrence
$$f(r,b)=\max\left((r-b)/(r+b),\frac{r}{r+b}f(r-1,b)+\frac{b}{r+b}f(r,b-1)\right).$$
The boundary conditions are $f(0,0)=0$, $f(1,0)=1$, $f(0,1)=-1$, and $f(r,b)=0$ if either $r$ or $b$ is negative. Calculating individual values of $f(r,b)$ is easy using dynamic programming. Here are values of $f(r,b)$ for small $r$ and $b$:
$$
\begin{array}{c|cccccc}
r\,\backslash b & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline
0 & 0 & -1 & -1 & -1 & -1 & -1 \\
1 & 1 & 0 & -\frac{1}{3} & -\frac{1}{2} & -\frac{3}{5} & -\frac{2}{3} \\
2 & 1 & \frac{1}{3} & 0 & -\frac{1}{5} & -\frac{1}{3} & -\frac{3}{7} \\
3 & 1 & \frac{1}{2} & \frac{1}{5} & 0 & -\frac{1}{7} & -\frac{1}{4} \\
4 & 1 & \frac{3}{5} & \frac{1}{3} & \frac{1}{7} & 0 & -\frac{1}{9} \\
5 & 1 & \frac{2}{3} & \frac{3}{7} & \frac{1}{4} & \frac{1}{9} & 0 \\
\end{array}
$$
Apparently $f(r,b)=-f(b,r)$; this is easy to prove from the recurrence. Accordingly $f(26,26)=0$.
[Actually it follows from the recurrence (and is evident from the table] that $f(r,b)=(r-b)/(r+b)$ -- in other words, it is never optimal to continue playing.]
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1461915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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.