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Derivative of $\sec^{-1}x$ and integral of $\frac{1}{x\sqrt{x^2-1}}$ My attempt is as follows:-
*
*Derivative of $\sec^{-1}x$
Let $\theta=\sec^{-1}x,$ where $\theta\in [0,\pi]-{\dfrac{\pi}{2}}.$
$$\sec\theta=x.$$
Differentiating both sides with respect to $x:$
$$\sec\theta\cdot\tan\theta\cdot\dfrac{\mathrm d\theta}{\mathrm dx}=1\\
\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{1}{x\sqrt{x^2-1}}.$$
As $\sec^{-1}x$ is a strictly increasing function, its derivative should be positive, hence we write $x$ as $|x|$ to ensure that $\dfrac{\mathrm d\theta}{\mathrm dx}$ will not be negative if $x$ is negative. But I wonder why I didn't get $\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{1}{|x|\sqrt{x^2-1}}$ in the above calculation?
*Integral of $\dfrac{1}{x\sqrt{x^2-1}}$
Case $1:x>0$
Then the integral is definitely $\sec^{-1}x.$
Case $2: x<0$
Then the integral is $-\sec^{-1}x.$
But many textbooks write that $\displaystyle\int\frac{1}{x\sqrt{x^2-1}}\,\mathrm dx=\sec^{-1}x+C.$
Shouldn't $\displaystyle\int\frac{1}{|x|\sqrt{x^2-1}}\,\mathrm dx=\sec^{-1}x+C\:?$ What am I missing here?
|
To avoid $\pm$ signs, proceed as follows:
*
*Derivative of $\sec^{-1}x$. Let $\theta=\sec^{-1}x,$ with $\theta\in [0,\pi]-{\dfrac{\pi}{2}}$. Then, $\sin \theta >0$
$$\cos\theta=\frac1{\sec\theta}=\frac1x\implies \sin\theta\cdot\dfrac{\mathrm d\theta}{\mathrm dx}=\frac1{x^2}\\
\dfrac{\mathrm d\theta}{\mathrm dx}=\frac{1}{x^2\sin\theta} =\frac{1}{x^2\sqrt{1-\frac1{x^2}}}= \frac{1}{\sqrt{x^2(x^2-1)}} $$
*Integral of $\dfrac{1}{x\sqrt{x^2-1}}$. For integration valid for the whole domain $|x|>1$
$$\int \frac{1}{x\sqrt{x^2-1}}dx
= \int \frac{d(\sqrt{x^2-1})}{(x^2-1)+1}=\tan^{-1}\sqrt{x^2-1}+C
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How many five-digit numbers with distinct digits can be formed from the digits $0, 1, 2, 3, 4, 5, 6, 7$ under the given conditions?
You can use the digits $0,1,2,3,4,5,6,7$.
You have to make a number of $5$ digits, above $9999$ (no $0$ in the start).
All digits should be different from each other, the number must have $3$ EVEN digits and $2$ ODD digits ($0$ is even), and the number should be divisible by $25$.
I need help in understanding the principle of one of the solving techniques which is:
letting everything roll as if $0$ can be in the start (let's call this I) - that time when $0$ is in the start (let's call this II):
What I don't understand is when do you use or don't use the other conditions ($3$ even and $2$ odd, divisible by $25$, ...). Is it in II and I or just one of them?
If someone can tell me what's wrong with this, I'll be glad:
Divide to $3$ cases: $25$, $50$, $75$.
first one is:
_ _ _ 2 5, (I) >>>(pick 1 odd out of 3)*(pick 2 even out of 3)*3!
minus
0 _ _ 2 5, (II) >>>(pick 1 odd out of 3)*(pick 1 even out of 3)*2!
the second one is 50
_ _ _ 5 0 which is special >>> (pick 2 even out of 3)*(pick 1 odd out of 3)*3!
and the third one is
_ _ _ 7 5 >>>(pick 2 even out of 3)*3!
minus
0 _ _ 7 5, (II) >>>(pick 2 even out of 3)*2!
The answer in the book is $138$, and this doesn't add up to it. Thanks for your answers.
|
You have to take the conditions into account both when you count the five-digit strings and when you count the five-digit strings that begin with $0$.
The answer in the text is incorrect.
Method 1: We use your method.
Five-digit numbers of the form _ _ _ $25$ that use three even and two odd digits, all of which are distinct: Once we have used $2$ and $5$, we have three even and three odd digits remaining. To form a five-digit string with three even digits and two odd digits, we must choose two of the three remaining even digits and one of the three remaining odd digits, then arrange the three selected numbers in the indicated positions, which can be done in
$$\binom{3}{2}\binom{3}{1}3! = 54$$
ways.
From these, we must subtract those five-digit strings of the form $0$ _ _ $25$ that use three even and two odd digits, all of which are distinct. Once we have used $0$, $2$, and $5$, we have two even and three odd digits remaining. We must choose one of the two remaining even digits and one of the three remaining odd digits, then arrange them in the indicated positions, which can be done in
$$\binom{2}{1}\binom{3}{1}2! = 12$$
ways.
Hence, we have
$$\binom{3}{2}\binom{3}{1}3! - \binom{2}{1}\binom{3}{1}2! = 42$$
admissible numbers of the form _ _ _$25$.
Five-digit numbers of the form _ _ _$50$ that use three even and two odd digits, all of which are distinct: Once we have used $5$ and $0$, we have three even and three odd digits remaining. As shown above, there are
$$\binom{3}{2}\binom{3}{1}3! = 54$$
strings of this form, all of which are admissible.
Five-digit numbers of the form _ _ _$75$ that use three even and two odd digits, all of which are distinct: Once we have used $5$ and $7$, we have four even and two odd digits remaining. We must use three of the four even digits in the remaining slots. We must choose three of the four even digits, then arrange them in those slots, which can be done in
$$\binom{4}{3}3! = 24$$
ways.
From these, we must subtract those strings of the form $0$_ _$75$ that use three even and two odd digits, all of which are distinct. Once we have used $0$, $5$, and $7$, we are left with three even and two odd digits. We must select two of the three even numbers and arrange them in the indicated positions, which can be done in
$$\binom{3}{2}2! = 6$$
ways.
Hence, there are
$$\binom{4}{3}3! - \binom{3}{2}2! = 18$$
admissible numbers of the form _ _ _$75$.
Total: Since the three cases are mutually exclusive and exhaustive, the number of five-digit numbers with distinct digits which are divisible by $25$ composed from the set $\{0, 1, 2, 3, 4, 5, 6, 7\}$ is
$$42 + 54 + 18 = 114$$
Method 2: We do a direct count.
Five-digit numbers of the form _ _ _ $25$ that use three even and two odd digits, all of which are distinct: We consider two cases, depending on whether the leading digit is even or odd.
Leading digit is even: Since we cannot use $0$ or $2$ for the leading digit, the leading digit can be chosen in two ways from the two remaining even numbers. We must use one of the two remaining even numbers (which include $0$ and the other unused even digit) and one of the remaining three odd numbers, then arrange them in the thousands and hundreds places, which can be done in
$$\binom{2}{1}\binom{2}{1}\binom{3}{1}2! = 24$$
ways.
Leading digit is odd: We must place one of the three remaining odd numbers in the first position. We must choose two of the remaining even numbers for the remaining two slots, then arrange them in those slots, which can be done in
$$\binom{3}{1}\binom{3}{2}2! = 18$$
Hence, there are
$$\binom{2}{1}\binom{2}{1}\binom{3}{1}2! + \binom{2}{1}\binom{2}{1}\binom{3}{1}2! = 42$$
admissible numbers of this form.
Five-digit numbers of the form _ _ _$50$ that use three even and two odd digits, all of which are distinct: We consider two cases, depending on whether the leading digit is even or odd.
Leading digit is even: Since we cannot use $0$ for the leading digit, we can choose the leading digit in three ways from the remaining even digits. We must choose one even number from the remaining two even numbers and one odd number from the remaining three odd numbers, then arrange them in the remaining two positions. There are
$$\binom{3}{1}\binom{2}{1}\binom{3}{1}2! = 36$$
such arrangements.
Leading digit is odd: We can choose the leading digit in three ways from the remaining odd digits. We must choose two of the remaining three even numbers then arrange them in the thousands and hundreds places. There are
$$\binom{3}{1}\binom{3}{2}2! = 18$$
such arrangements.
Hence, there are
$$\binom{3}{1}\binom{2}{1}\binom{3}{1}2! + \binom{3}{1}\binom{3}{2}2!$$
admissible numbers of this form.
Five-digit numbers of the form _ _ _$75$ that use three even and two odd digits, all of which are distinct: The three remaining digits must be even. Since the leading digit cannot be $0$, it can be chosen in three ways. To fill the remaining two slots, we must choose two of the remaining three even numbers (which include $0$ and the other two unused even numbers), then arrange them in those slots, which can be done in
$$\binom{3}{1}\binom{3}{2}2! = 18$$
ways.
Total: Since the three cases are mutually exclusive and exhaustive, the number of five-digit numbers with distinct digits which are divisible by $25$ composed from the set $\{0, 1, 2, 3, 4, 5, 6, 7\}$ is
$$42 + 54 + 18 = 114$$
|
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|
Tricky complex contour integration Show that $$\int_{0}^{2\pi} \frac{\cos^2{(\theta)}}{13-5\cos{(2\theta)}} \; d\theta = \frac{\pi}{10}$$ using complex contour integration.
I let $z = e^{i\theta}$ which means $dz = ie^{i\theta} \; d\theta = iz \; d\theta$ (meaning we know integrate over the unit circle $C$). Hence,
$$\int_{0}^{2\pi} \frac{\cos^2{(\theta)}}{13-5\cos{(2\theta)}} \; d\theta = \oint_{C} \frac{1/4\left(z+1/z\right)^2}{13-5/2\left(z^2+1/z^2\right)} \; \frac{dz}{iz} = -\frac{i}{2}\oint_{C}\frac{z(z^2+1)^2}{(z^3-5)(5z^3-1)} \; dz.$$
Not sure what to do next. Is it to find the poles that are in the unit circle? Thanks!!!
|
On the last step, you should have
$$I=\frac i2\oint_C\frac{(z^2+1)^2}{\color{red}z(z^{\color{red}2}-5)(5z^{\color{red}2}-1)}~\mathrm dz$$
Now use the two roots in the unit circle from $5z^2-1=0\Rightarrow z=\pm\sqrt5/5$ and $z=0$, which are first order poles, to get
$$I=\frac\pi{10}$$
|
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|
Evaluate $\int\frac{dx}{(1+\sqrt{x})(x-x^2)}$
$$
\int\frac{dx}{(1+\sqrt{x})(x-x^2)}
$$
Set $\sqrt{x}=\cos2a\implies\dfrac{dx}{2\sqrt{x}}=-2\sin2a.da$
$$
\int\frac{dx}{(1+\sqrt{x})(x-x^2)}=\int\frac{dx}{(1+\sqrt{x})x(1-x)}=\int\frac{-4\sin2a\cos2a.da}{2\cos^2a.\cos^22a.\sin^22a}\\
=\int\frac{-2.\sec^2a.da}{\sin2a\cos2a}
$$
I think I am getting stuck here, is there a better substitution that I can chose so that the integral becomes more simple to evaluate ?
Solution as per my reference: $\dfrac{2(\sqrt{x}-1)}{\sqrt{1-x}}$
Note: I'd prefer to choose a substitution which does not make use of partial fractions, as there seems to be 4 terms for the substitution $\sqrt{x}=y\implies \frac{dx}{2\sqrt{x}}=dy$.
$$
I=\int\frac{dx}{(1+\sqrt{x})(x-x^2)}=\int\frac{dx}{(1+\sqrt{x})x(1-x)}\\
=\int\frac{2dy}{y(1+y)(1-y^2)}=\int\frac{2dy}{y(1+y)^2(1-y)}\\
\frac{2}{y(1+y)^2(1-y)}=\frac{A}{y}+\frac{B}{1-y}+\frac{C}{1+y}+\frac{D}{(1+y)^2}
$$
|
Note that there is an inconsistency. The solution you listed is not for the integral posted. The integral instead should be
$$
I=\int\frac{dx}{(1+\sqrt{x})\sqrt{x-x^2}}
$$
If so, first use the substitution $t=\sqrt x$ to rewrite it as
$$I=\int\frac{2dx}{(1+t)\sqrt{1-t^2}}$$
Then, let $u=\frac{1-t}{1+t}$ and the integral simplifies to
$$I = - \int \frac{du}{\sqrt u} = -2\sqrt u +C $$
|
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|
Evaluate $\int \frac{\cos^2x}{1+\tan x}dx$ Evaluate $\int \dfrac{\cos^2x}{1+\tan x}dx$
Here are my various unsuccessful attempts:-
Attempt $1$:
$$\tan x=t$$
$$\sec^2 x=\dfrac{dt}{dx}$$
$$\int \dfrac{dt}{(1+t^2)^2(1+t)}$$
$$\ln(1+t)=y$$
$$\dfrac{1}{1+t}=\dfrac{dy}{dt}$$
$$\int \dfrac{dy}{\left(1+(e^y-1)^2\right)^2}$$
$$\int \dfrac{dy}{(1+e^{2y}+1-2e^y)^2}$$
$$\int \dfrac{dy}{(e^{2y}+2-2e^y)^2}$$
$$\int \dfrac{dy}{e^4y+4+4e^{2y}+4e^{2y}-4e^y(e^{2y}+2)}$$
$$\int \dfrac{dy}{e^4y-4e^{3y}+8e^{2y}-8e^y+4}$$
From here I gave up on this method.
Attempt $2$:
$$\int \dfrac{\cos^3x}{\cos x+\sin x}$$
$$\int\dfrac{3\cos x+\cos 3x}{4(\cos x+\sin x)}$$
$$\dfrac{1}{4}\left(3\cdot\int\dfrac{\cos x}{\cos x+\sin x}+\int \dfrac{\cos 3x}{\cos x+\sin x}\right)$$
Solving the first integral
$$\cos x=A(\cos x+\sin x)+B(-\sin x+\cos x)+C$$
$$A+B=1$$
$$A-B=0$$
$$A=\dfrac{1}{2},B=\dfrac{1}{2}$$
$$\dfrac{1}{4}\left(\dfrac{3}{2}\left(x+\ln|\sin x+\cos x|\right)+\int \dfrac{\cos 3x}{\cos x+\sin x}\right)$$
I was not understanding how to proceed for second integration.
Attempt $3$:
For @mvpq
$$\tan x=t$$
$$\sec^2 x=\dfrac{dt}{dx}$$
$$\int \dfrac{dt}{(1+t^2)^2(1+t)}$$
Trying to write expression as
$$\dfrac{1}{(1+t^2)^2(1+t)}=\dfrac{A}{1+t}+\dfrac{B}{1+t^2}+\dfrac{C}{(1+t^2)^2}$$
$$\dfrac{1}{(1+t^2)^2(1+t)}=\dfrac{A(1+t^2)^2+B(1+t^2)+C(1+t)}{(1+t)(1+t^2)^2}$$
$$\dfrac{1}{(1+t^2)^2(1+t)}=(A+B+C)+Ct+(B+2A)t^2+At^4$$
$$A+B+C=1\tag{1}$$
$$A=0$$
$$A+2B=0$$
$$B=0$$
$$C=0$$
But $A+B+C=1$, hence no solution, so cannot be solved by partial fractions.
Attempt $4$:
For @heropup
$$\tan x=t$$
$$\sec^2 x=\dfrac{dt}{dx}$$
$$\int \dfrac{dt}{(1+t^2)^2(1+t)}$$
Trying to write expression as
$$\dfrac{1}{(1+t^2)^2(1+t)}=\dfrac{A}{1+t}+\dfrac{Bt+C}{1+t^2}+\dfrac{Dt+E}{(1+t^2)^2}$$
$$\dfrac{1}{(1+t^2)^2(1+t)}=\dfrac{A(1+t^2)^2+(Bt+C)(1+t^2)(1+t)+(Dt+E)(1+t)}{(1+t)(1+t^2)^2}$$
$$1=A(1+t^2)^2+(Bt+C)(1+t^2)(1+t)+(Dt+E)(1+t)$$
Placing $t=-1$ in the equation
$$4A=1, A=\dfrac{1}{4}$$
$$1=(Bt+C)(1+t+t^2+t^3)+(Dt+Dt^2+E+Et)$$
$$1=Bt+Bt^2+Bt^3+Bt^4+C+Ct+Ct^2+Ct^3+Dt+Dt^2+E+Et$$
$$1=Bt^4+(B+C)t^3+(B+C+D)t^2+(B+C+D+E)t+(C+E)$$
$$B=0$$
$$B+C=0$$
$$C=0$$
$$B+C+D=0$$
$$D=0$$
$$B+C+D+E=0$$
$$E=0$$
$$C+E=1$$
$$E=1$$
This is contradiction right?
Any hints?
|
You may continue and finish the second approach as follows,
$$I=\dfrac{1}{4}\int\dfrac{3\cos x+\cos3x}{\cos x+\sin x}dx$$
Rewrite the term $\cos3x $ in the integrand as
$\cos3x = (\cos2x-\sin2x)(\sin x+ \cos x)+\sin x$ and the integral becomes
$$I=\dfrac{1}{4}\int\left(2+\cos2x-\sin2x+\dfrac{\cos x-\sin x}{\cos x+\sin x}\right)dx$$
$$=\frac x2+\frac18(\sin2x+\cos2x)+\frac14\ln|\sin x+\cos x|+C$$
|
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|
How many ways can we get a number by addition if each part of the addition has to be smaller or equal to a set value? For example, if we need to get 5 with the largest number we can use being 3, we can use:
*
*3 + 2
*3 + 1 + 1
*2 + 2 + 1
*2 + 1 + 1 + 1
*1 + 1 + 1 + 1 + 1
Is there any way to find out the solution with any two numbers without calculating every one of them?
|
If you wish to find the number of ways to express a positive integer $n$ as the sum of $m$ positive integers ($m\le n$) and each of the positive integers must be less than or equal to a certain value $\alpha$ then the number of such integers will be the number of positive integral solutions to these equations:
$x_1+x_2=n$
$x_1+x_2+x_3=n$
$x_1+x_2+x_3+\cdots+x_m=n$
such that $1\le x_i \le \alpha$
for the equation $x_1+x_2=n$
the number of positive integral solutions is the same as the coefficient of $x^n$ in the expansion $(x+x^2+x^3+\cdots+x^\alpha)^2$
for the equation $x_1+x_2+x_3=n$
the number of positive integral solutions is the same as the coefficient of $x^n$ in the expansion $(x+x^2+x^3+\cdots+x^\alpha)^3$
for the equation $x_1+x_2+x_3+\cdots+x_m=n$
the number of positive integral solutions is the same as the coefficient of $x^n$ in the expansion $(x+x^2+x^3+\cdots+x^\alpha)^m$
So the total number of the solutions of all of these equations is the desired number.
This method of finding the coefficient in the expansion works because the term $x^n$ is formed only if in the multiplication, the powers of some terms get added to give the specified $n$.
For example in your question, $5$ has to be expressed as the sum of positive integers not greater than $3$
$x_1+x_2=5$
Coefficient of $x^5$ in the expansion of $(x+x^2+x^3)^2$
$=(x+x^2+x^3)(x+x^2+x^3)$
$=x^2+2x^3+3x^4+2x^5+x^6$
Thus,$[x^5]=2$
Hence, 5 can be expressed in 2 ways as the sum of 2 positive integers lesser than or equal 3, which are
$(2,3)$ and $(3,2)$ ( yes, this method gives us the ordered pairs)
You can do the same for rest of the equations.
In general as well, if you allow integers from a certain interval (not only positive) say, $ \alpha \le x_i \le \beta$ then the number of ways to express $n$ as sum of integers (order is considered) is
($[x^n]$ is the coefficient of $x^n$)
$$\sum_{m=2}^n [x^n] (x^ \alpha+x^{\alpha+1}+x^{\alpha+2}+\cdots+x^ \beta)^m$$
|
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|
How to solve this quadratic equation (with $x$ represented by a fraction containing a square root)? If $x-\frac{4}{5} = \pm \frac{\sqrt{31}}{5}$, how can I find the values of a and b in the following equation?
$$5x^2+ax+b=0?$$
I've tried substituting $(\frac{4}{5} \pm \frac{\sqrt{31}}{5})$ for $x$ and got totally confused there.
|
Since the roots are $4/5\pm\sqrt{31}/5$, we have that $x^2+(a/5)x+b/5=(x-(4/5+\sqrt{31}/5))(x-(4/5-\sqrt{31}/5))$.
So, $x^2+(a/5)x+b/5=x^2-(8/5)x+(16/25-31/25)=x^2-(8/5)x-3/5$.
Hence $a=-8, b=-3$.
|
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|
Calculate $2^{5104} \bmod 10$ using mental arithmetic I am practising interview for Jane Street's trader internship and I found the following question.
Question: Calculate $2^{5104} \bmod 10$ using mental arithmetic.
I know that $2^5 \bmod 10 \equiv 2 \bmod 10.$
So,
\begin{align*}
2^{5104} & = (2^5)^{1020} 2^4 \\
& \equiv 2^{1020}2^4 \\
& = (2^5)^{204}2^4 \\
& \equiv(2^5)^{40}2^8 \\
& \equiv (2^5)^8 2^8 \\
& \equiv (2^5)^3 2 \\
& \equiv 6 \bmod 10.
\end{align*}
However, I find the calculations above very taxing if I use mental arithmetic. I believe there should be a faster way to answer the question but I am not able to find one.
|
The Chinese Remainder Theorem says:
$$2^{5104}\equiv x \pmod{2\cdot 5} \iff \begin{cases}2^{5104}\equiv x\pmod 2 \\ 2^{5104}\equiv x\pmod 5\end{cases}$$
It follows that $x\equiv 0\pmod 2$ and $x\equiv 1\pmod 5$.
Consequently $x\equiv 6\pmod{10}$.
|
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|
Factoring $1 - y-x^2-y^2-yx^2+y^3$ I was struggling to factorize this expression, I need it cleaned to bracket form but I was struggling to understand the process to get the answer? Is there a special formula or technique for factorizing these sorts of expressions?
$$1 - y-x^2-y^2-yx^2+y^3$$
Answer: $$ \ ( 1+y)[(1-y)^2-x^2] $$
|
Here is how I factored it:
$$1-y-x^2-y^2-yx^2+y^3=$$
$$(1-y^2)-(y-y^3)-(x^2+yx^2)=$$
$$(1-y^2)-y(1-y^2)-x^2(1+y)=$$
$$(1-y^2)(1-y)-x^2(1+y)=$$
$$(1-y)(1+y)(1-y)-x^2(1+y)=$$
$$(1+y)(1-y)^2-(1+y)x^2=$$
$$\boxed{(1+y)[(1-y)^2-x^2]}$$
You can factorize it further:
$$(1+y)[(1-y)^2-x^2]=$$
$$(1+y)[(1-y-x)(1-y+x)]=$$
$$\boxed{(1+y)(1-y-x)(1-y+x)}$$
We can expand $(1+y)(1-y-x)(1-y+x)$ to make sure we got it:
$$\require{cancel} (1+y)(1-y-x)(1-y+x)=$$
$$(1\cancel{-y}-x\cancel{+y}-y^2-xy)(1-y+x)=$$
$$(1-x-y^2-xy)(1-y+x)=$$
$$(1-x-y^2-xy)-(y-xy-y^3-xy^2)+(x-x^2-xy^2-x^2y)=$$
$$1\cancel{-x}-y^2\cancel{-xy}-y\cancel{+xy}+y^3\cancel{+xy^2}\cancel{+x}-x^2-\cancel{xy^2}-x^2y=$$
$$1-y^2-y+y^3-x^2-x^2y=$$
$$\boxed{1-y-x^2-y^2-yx^2+y^3}$$
|
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|
Solve the irrational radical equation Solve equation:$$\frac{x + \sqrt{3}}{\sqrt{x} + \sqrt{x + \sqrt{3}}} + \frac{x - \sqrt{3}}{\sqrt{x} - \sqrt{x - \sqrt{3}}} = \sqrt{x}$$
answer: $S=\{2\}$
Attemp:
$$\frac{x+\sqrt{3}}{\sqrt{x}+\sqrt{x+\sqrt{3}}}+\frac{x-\sqrt{3}}{\sqrt{x}-\sqrt{x-\sqrt{3}}}=\sqrt{x}\Leftrightarrow \frac{\left( \sqrt{x}-\sqrt{x+\sqrt{3}} \right)\left( x+\sqrt{3} \right)}{-\sqrt{3}}+\frac{\left( \sqrt{x}+\sqrt{x+\sqrt{3}} \right)\left( x-\sqrt{3} \right)}{\sqrt{3}}=\sqrt{x}$$
$$\frac{2\sqrt{x}\left( \sqrt{x}\sqrt{x+\sqrt{3}}-\sqrt{3} \right)}{\sqrt{3}}=\sqrt{x}\Leftrightarrow 2\left( \sqrt{x}\sqrt{x+\sqrt{3}}-\sqrt{3} \right)=\sqrt{3}\Leftrightarrow \sqrt{x}\sqrt{x+\sqrt{3}}=3\sqrt{3}$$
|
When you inserted factors to obtain a common denominator, you made the error
$$\left( \sqrt{x}+\sqrt{x+\sqrt{3}} \right) \left( \sqrt{x}-\sqrt{x-\sqrt{3}} \right) \neq -\sqrt{3} \text{.} $$
That you did not get the same denominator in both fractions should have highlighted a problem.
Since we require $x - \sqrt{3} \geq 0$ for the original expression to be defined, we may assume $x \geq \sqrt{3}$, which fact we will use in combining radicals.
$$\left( \sqrt{x}+\sqrt{x+\sqrt{3}} \right) \left( \sqrt{x}-\sqrt{x-\sqrt{3}} \right) \\
= x + \sqrt{x^2+x\sqrt{3}} - \sqrt{x^2-x\sqrt{3}} - \sqrt{x^2 - 3} \text{.} $$
(This is not an instance of the difference of two squares factorization because $\sqrt{x + \sqrt{3}} \neq \sqrt{x - \sqrt{3}}$.)
|
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"timestamp": "2023-03-29T00:00:00",
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|
Vector, Matrix multiplication notation Given a vector $v=\begin{bmatrix} 1 \\ 2 \\ 3\end{bmatrix}$ and matrix $M=\begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6 \\ 7 & 8 & 9\end{bmatrix}$ is there a standard notation for scaling each row of $M$ by the corresponding element of $v$. e.g.
$$\begin{bmatrix} 1 \\ 2 \\ 3\end{bmatrix} \circ_{Row} \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6 \\ 7 & 8 & 9\end{bmatrix} = \begin{bmatrix} 1 & 2 & 3\\ 8 & 10 & 12 \\ 21 & 24 & 27\end{bmatrix} = A$$
I could define it using $A_{ij} = v_i \cdot M_{ij}$ but would rather use an operator if there is one.
|
You could do it by
$$\begin{bmatrix} 1&0&0 \\0& 2&0 \\0&0& 3\end{bmatrix}
\begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6 \\ 7 & 8 & 9\end{bmatrix} = \begin{bmatrix} 1 & 2 & 3\\ 8 & 10 & 12 \\ 21 & 24 & 27\end{bmatrix}$$
easily.
|
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|
Find real $a$, $b$, $c$, $d$ satisfying $(1-a)^2+(a-b)^2+(b-c)^2+c^2=\frac{1}{4}$
If real numbers $a$, $b$, $c$, $d$ satisfy
$$(1-a)^2+(a-b)^2+(b-c)^2+c^2=\frac{1}{4}$$
then find $(a,b,c,d)$.
What I try:
$$1+2a^2+2b^2+2c^2-2a-2ab-2bc=\frac{1}{4}$$
$$8a^2+8b^2+8c^2-8a-8ab-8bc+7=0$$
How do I solve it? Help me, please.
|
You can observe this as quadratic equation on $a$ :
$$8a^2-8a(1+b) +(8b^2+8c^2 -8bc+7)=0$$ which has to have a solution so the discriminant must be nonnegative:
$$64(1+b)^2-32(8b^2+8c^2 -8bc+7)\geq 0$$
so $$6b^2-4b(1+2c)+8c^2+5\leq 0$$
Now again, the discriminat must $\geq 0$ since that inequality has a solution...
|
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|
evaluate the integral $\int x^3(5x^2+3)^5 dx$ How to evaluate for the integral:
$$\int x^3(5x^2+3)^5 dx$$
What I've tried:
For example,
(1) $u = x^2$ whence $du = 2x$ $dx$
(2)$u = ax^2+b$ whence $du = 2ax$ $dx$
Therefore,
(3) $u = 5x^2+3$ whence $du = 10x$ $dx$
$$I = \int \frac{du}{10}u^5 = \frac{1}{10} \int \frac{u^6}{6} du= \frac{1}{10}\frac{1}{6}u^6$$
However, this does not lead me to the answer:
$$\frac{(10x^2-1)(5x^2+3)^6}{700}$$
Where did I go wrong and how do I correct this?
|
$$\int x^3(5x^2+3)^5 dx=\frac{1}{5}\int(5x^3+3x-3x)(5x^2+3)^5 dx=$$
$$=\frac{(5x^2+3)^7}{5\cdot7\cdot10}-\frac{3(5x^2+3)^6}{5\cdot6\cdot10}+C.$$
|
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|
Lagrange multipliers... Three variables. Let $f:\mathbb{R}^3\mapsto \mathbb{R}$ be defined by
\begin{equation*}
f(x,y,z) := x+3y+5z.
\end{equation*}
Compute the minimum of $f$ subject to the constraint that
\begin{equation*}
g(x,y,z) = x^2+y^2+z^2-1 = 0.
\end{equation*}
The method of Lagrange multipliers is as follows: We first solve $\nabla f(x,y,z) = \lambda \nabla g(x,y,z)$ and $g(x,y,z) = k$ then plug in all solutions, $(x,y,z)$, into $f(x,y,z)$ and identify the minimum and maximum values, provided they exist and $\nabla g\neq 0$ at the point. Let's apply this to our problem.
We have $1 = 2\lambda x$, $3 = 2\lambda y$ and $5 = 2\lambda z$ along with $x^2+y^2+z^2 = 1$. Solving for $x$, $y$ and $z$ we get
\begin{equation*}
x = \frac{1}{2\lambda}, \quad y = \frac{3}{2\lambda}, \quad z = \frac{5}{2\lambda}
\end{equation*}
and plugging into our constraint gives
\begin{equation*}
x^2+y^2+z^2 = \frac{1}{4\lambda^2}+\frac{9}{4\lambda^2}+\frac{25}{4\lambda^2} = 1 \Longleftrightarrow \lambda^2 = \frac{35}{4},
\end{equation*}
or $\lambda = \pm\frac{\sqrt{35}}{2}$. For $\lambda = \frac{\sqrt{35}}{2}$, $(x,y,z) = \left(\frac{1}{\sqrt{35}},\frac{3}{\sqrt{35}},\frac{5}{\sqrt{35}}\right)$ and for $\lambda = -\frac{\sqrt{35}}{2}$, $(x,y,z) = \left(-\frac{1}{\sqrt{35}},-\frac{3}{\sqrt{35}},-\frac{5}{\sqrt{35}}\right)$. Substituting these into $f(x,y,z)$ gives
\begin{equation*}
f\left(\frac{1}{\sqrt{35}},\frac{3}{\sqrt{35}},\frac{5}{\sqrt{35}}\right) = \frac{1}{\sqrt{35}}+\frac{9}{\sqrt{35}}+\frac{25}{\sqrt{35}} = \frac{35}{\sqrt{35}} = \sqrt{35}
\end{equation*}
which is the maximum value of $f$, and
\begin{equation*}
f\left(-\frac{1}{\sqrt{35}},-\frac{3}{\sqrt{35}},-\frac{5}{\sqrt{35}}\right) = -\frac{1}{\sqrt{35}}-\frac{9}{\sqrt{35}}-\frac{25}{\sqrt{35}} = -\frac{35}{\sqrt{35}} = -\sqrt{35}
\end{equation*}
which is the minimum value of $f$ as required.
Is this all good? Thanks!
|
Yes, your computations and your result are correct. Using the Cauchy-Schwarz inequality, we can also obtain the same result very quickly:
Note that (where $\langle\cdot,\cdot\rangle$ is the euclidean inner product and $\|\cdot\|$ is the euclidean norm) $$f(x,y,z)=\left\langle(x,y,z),(1,3,5)\right\rangle \text{ and that } g(x,y,z)=0 \text{ means that }\|(x,y,z)\|=1.$$
Hence, by Cauchy-Schwarz, $$\lvert f(x,y,z)\rvert\le\|(x,y,z)\|\cdot\|(1,3,5)\|=1\cdot\|(1,3,5)\|=\sqrt{35}.$$
Equality is achieved if and only if $(x,y,z)$ is a scalar multiple of $(1,3,5)$, which, together with $\|(x,y,z)\|=1$, implies that we have the two global minimizers/maximizers $$(x,y,z)=\pm\frac1{\sqrt{35}}(1,3,5)$$
|
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|
Nested radical of Ramanujan and cyclic inequality Inspired by a nested radical of Ramanujan I propose this :
Let $a,b,c\in (2\sqrt[4\,]{5},6-2\sqrt[4\,]{5})$ such that $a+b+c=9$ then we have :
$$\sqrt[4\,]{\frac{a+2\sqrt[4\,]{5}}{a-2\sqrt[4\,]{5}}}+\sqrt[4\,]{\frac{b+2\sqrt[4\,]{5}}{b-2\sqrt[4\,]{5}}}+\sqrt[4\,]{\frac{c+2\sqrt[4\,]{5}}{c-2\sqrt[4\,]{5}}}\leq\sqrt[4\,]{\frac{a+2\sqrt[4\,]{5}}{b-2\sqrt[4\,]{5}}}+\sqrt[4\,]{\frac{b+2\sqrt[4\,]{5}}{c-2\sqrt[4\,]{5}}}+\sqrt[4\,]{\frac{c+2\sqrt[4\,]{5}}{a-2\sqrt[4\,]{5}}}$$
I try to prove it term by term like this :
$$\sqrt[4\,]{\frac{a+2\sqrt[4\,]{5}}{a-2\sqrt[4\,]{5}}}\leq \sqrt[4\,]{\frac{a+2\sqrt[4\,]{5}}{b-2\sqrt[4\,]{5}}}$$
But it's cyclic so we are in the wrong way.
On the other hand I can prove that :
$$3\sqrt[4\,]{\frac{3+2\sqrt[4\,]{5}}{3-2\sqrt[4\,]{5}}}\leq\sqrt[4\,]{\frac{a+2\sqrt[4\,]{5}}{a-2\sqrt[4\,]{5}}}+\sqrt[4\,]{\frac{b+2\sqrt[4\,]{5}}{b-2\sqrt[4\,]{5}}}+\sqrt[4\,]{\frac{c+2\sqrt[4\,]{5}}{c-2\sqrt[4\,]{5}}}$$
Because it's not hard to show that the function $f(x)=\sqrt[4\,]{\frac{x+2\sqrt[4\,]{5}}{x-2\sqrt[4\,]{5}}}$ is convex on the interval $(2\sqrt[4\,]{5},6-2\sqrt[4\,]{5})$ and after apply Jensen's inequality.
I tried also majorization ,C-S,AM-GM and more but without success.
If you have nice idea I let you play with this .
Thanks a lot for sharing your time and knowledge .
|
Remark: For applications of the Symmetric Function Theorem, also see
Prove that $\sum\limits_{cyc}\frac{1}{\sqrt{a^2+ab+b^2}}\geq\frac{2}{\sqrt{ab+ac+bc}}+\sqrt{\frac{a+b+c}{3(a^3+b^3+c^3)}}$
How prove this inequality $\sum\limits_{cyc}\frac{x+y}{\sqrt{x^2+xy+y^2+yz}}\ge 2+\sqrt{\frac{xy+yz+xz}{x^2+y^2+z^2}}$
Proof: We will apply Ji Chen's Symmetric Function Theorem for $n=3$
(see https://artofproblemsolving.com/community/c6h194103p1065812):
Symmetric Function Theorem: Let $d\in (0,1)$. Let $x, y, z, u, v, w$ be non-negative real numbers satisfying
$$x+y+z \ge u+v+w, \quad
xy+yz+zx \ge uv+vw+wu, \quad
xyz \ge uvw.$$
Then $x^d + y^d+z^d \ge u^d + v^d+w^d$.
Now let us prove the inequality. We need to prove that
$$\sqrt[4]{X} + \sqrt[4]{Y} + \sqrt[4]{Z} \le \sqrt[4]{U} + \sqrt[4]{V} + \sqrt[4]{W}.$$
Clearly, $XYZ = UVW$. According to the Symmetric Function Theorem, it suffices to prove that
$$X + Y + Z \le U + V + W$$
and
$$XY + YZ + ZX \le UV + VW + WU.$$
After some manipulations, it suffices to prove that
$$2(-a^2+ab+ac-b^2+bc-c^2)\sqrt[4]{5} + a^2c+ab^2-3abc+bc^2\ge 0$$
and
$$2(a^2-ab-ac+b^2-bc+c^2)\sqrt[4]{5} +a^2b-3abc+ac^2+b^2c \ge 0.$$
The latter is obvious.
I used Mathematica Resolve to verify the former. I believe that the proof of the former is not very hard (omitted here).
We are done.
|
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|
Is this solution correct for equation $3x^2-4y^2=13$? Prove equation $3x^2-4y^2=13$ has no integer solution.
Solution: Suppose (y, 3)=1, We have:
$y≡( 1, 2) \mod (3)$
⇒ $4y^2≡( 1, 2) \ mod(3)$
$3x^2≡0 \mod (3)$
The common remainder between $3x^2$ and $4y^2$ is 0 , so we may write:
⇒ $3x^2-4y^2≡ 0 \ mod(3)$
But, $13≡1 \mod (3)$
Hence this equation has no integer solution. Similar result comes out when we consider remainder of nomials on 4.
Is this solution correct?
|
Here is another method:
$3x^2-4y^2=13$
Clearly x must be odd; let $x=2k+1$, then:
$3x^2-4y^2=12k^2+12k+3-4y^2=4(3k^2+3k-y^2)=10$
⇒$2(3k^2+3k-y^2)=5$
LHS is even but RHS is odd, so the equation can not have integer solution.
|
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|
$\begin{cases} x^3-y^3=19(x-y) \\ x^3+y^3=7(x+y) \end{cases}$
I should solve the following system: $$\begin{cases} x^3-y^3=19(x-y) \\
x^3+y^3=7(x+y) \end{cases}$$
by reducing the system to a system of second degree.
We can factor:
$$\begin{cases} (x-y)(x^2+xy+y^2)=19(x-y) \\
(x+y)(x^2-xy+y^2)=7(x+y) \end{cases}$$
I really don't want to divide the equations by $x-y$ and $x+y$, respectively. I am taught to divide by expressions containing variables only in special cases. Is there any other way here?
|
You can divide by variables if you ensure they are not zero. Here, you can consider the cases $x=y$ and $x=-y$ first. If $x=y$ the first equation is trivial and the second becomes $2x^3=14x$ or $x=0,\pm \sqrt 7$. You can do the same for $x=-y$ and find a pair of solutions.
Then decree that $x+y \neq 0, x-y \neq 0$ and divide away. Once you do that, you can subtract the two equations to get $2xy=12$ and use that to get expressions for $(x+y)^2, (x-y)^2$
|
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|
number of real solution in exp equation
Find number of real values of $x$ in
$$3^x (3^x-1)=|3^x-1|+|3^x-2|$$
What i try
Put $\displaystyle 3^{x}-\frac{1}{2}=t.$ Then
$$\bigg(t-\frac{1}{2}\bigg)\cdot \bigg(t+\frac{1}{2}\bigg)= \bigg|t-\frac{1}{2}\bigg|+\bigg|t-\frac{3}{2}\bigg|$$
So $$t^2-\frac{1}{4}=|t-0.5|+|t-1.5|$$
How do i solve it Help me please
|
If $3^x > 0$ and $|A| + |B| \ge 0$ so $3^x(3^x-1) \ge 0$ so $3^x-1 \ge 0$. So $x \ge 0$.
If $x=0$ then LHS $= 0$ and RHS $=0 + 1= 1$. So $x = 0$ is not a solution.
Case 1: If $0 < x < \log_3 2$ then $3^x -1 > 0$ but $3^x - 2 < 0$ so
$3^x(3^x-1) = (3^x -1) + (2 - 3^x)$. Let $a = 3^x$ and solve for $a(a-1)=(a-1)+(2-a)$ and when you know what $a$ is (watch out! $a \ge 1$ so don't include any negative solutions or any solutions less than $1$) then solve for $3^x =a$.
Case 2: If $x \ge \log_3 2$ then $3^x -1 > 0$ and $3^x -2 \ge 0$ so
$3^x(3^-1) = (3^x-1) + (3^x -2)$. Let $b = 3^x$ and solve for $b(b-1)=(b-1)+(b-2)$ ane when you know what $b$ is (watch out! $b \ge 2$ so don't include any solutions less than $2$) then solve for $3^x = a$.
|
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|
Show $\frac{\sqrt{\sqrt[4]8-\sqrt{\sqrt2+1}\;}}{\sqrt{\sqrt[4]8+\sqrt{\sqrt2-1}\;} -\sqrt{\sqrt[4]8-\sqrt{\sqrt2-1}\;}}=\frac1{\sqrt2}$ Days ago, I tried to demonstrate this equality, reducing radicals, multiplying by the conjugate of the denominator, etc. But, I did not reach anything similar to the right side.
$$
\frac{\sqrt{\sqrt[4]{8}-\sqrt{\sqrt{2}+1}\;}}{\sqrt{\sqrt[4]{8}+\sqrt{\sqrt{2}-1}\;} -\sqrt{\sqrt[4]{8}-\sqrt{\sqrt{2}-1}\;}}=\frac{1}{\sqrt{2}}
$$
Can you help? I lack vision for this problem. You may have fantastic tricks that I have not heard about.
Good 2020!
|
Apply the denest formula $\sqrt{a-\sqrt c}=\sqrt{\frac{a+\sqrt{a^2-c}}2}-\sqrt{\frac{a-\sqrt{a^2-c}}2} $ to the numerator
$$N=\sqrt{\sqrt[4]8-\sqrt{\sqrt2+1}}=\sqrt{\frac{\sqrt[4]8+\sqrt{\sqrt2-1}}2}-\sqrt{\frac{\sqrt[4]8-\sqrt{\sqrt2-1}}2}=\frac1{\sqrt2}D$$
where $D$ is the denominator.
Alternatively, evaluate
$$D^2 = \left(\sqrt{\sqrt[4]8+\sqrt{\sqrt2-1}}-\sqrt{\sqrt[4]8-\sqrt{\sqrt2-1}}\right)^2$$
$$=\sqrt[4]8+\sqrt{\sqrt2-1}+\sqrt[4]8-\sqrt{\sqrt2-1}-2\sqrt{\sqrt8-(\sqrt2-1)}$$
$$=2\sqrt[4]8-2\sqrt{\sqrt2+1}=2N^2$$
Thus, $\frac ND = \frac1{\sqrt2}$.
|
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|
show that the following recursive series is bounded and monotonic I need to show that the following recursive series $$a_1=1, \quad a_{n+1}=\frac{a_n+3}{5}$$ is bounded and monotonic.
At first, I want to show that the series is bounded from below with $\frac{3}{4}$. By induction:
For $n=1$ we have $a_1=1>\frac{3}{4}$. We assume it is true for $n$, i.e. $a_n>\frac{3}{4}$ and we check for $a_{n+1}$:
$$a_{n+1}=\frac{a_n+3}{5}>\frac{\frac{3}{4}+3}{5}=15>\frac{3}{4}.$$
We conclude that $a_n>\frac{3}{4}$ for all n.
The series is increasing because
$a_{n+1}-a_{n}=\frac{-4a_n+3}{5}=\frac{-4(a_n-\frac{3}{4})}{5}<0$$
so we have also $a_n \leq a_1=1$.
Is my solution ok? Thank you for your help.
|
$$
\begin{align}
a_{n+1}-\frac34
&=\frac{a_n+3}5-\frac34\\
&=\frac{a_n-\frac34}5
\end{align}
$$
Thus, if $a_n\ge\frac34$, then $a_{n+1}\ge\frac34$. Therefore, since $a_1=1$, $a_n\ge\frac34$.
|
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|
A circle is drawn with its centre on the line $x + y = 2$ to touch the line $4x – 3y + 4 = 0$ and pass through the point $(0, 1)$. Find its equation. A circle is drawn with its centre on the line $x + y = 2$ to touch the line $4x – 3y + 4 = 0$ and pass through the point $(0, 1)$. Find its equation.
My attempt is as follows:-
Let the equation of circle be $x^2+y^2+2gx+2fy+c=0$
As center lies on the line $x+y=2$
$$-g-f=2$$
$$g+f=-2\tag{1}$$
As circle passes through the point $(0,1)$
$$1+2f+c=0$$
$$2f+c=-1\tag{2}$$
As $4x – 3y + 4 = 0$ is tangent to the circle
$$\dfrac{\left|-4g+3f+4\right|}{5}=\sqrt{g^2+f^2-c}$$
Squaring both sides
$$16g^2+9f^2-24gf+16+8(-4g+3f)=25g^2+25f^2-25c$$
$$9g^2+16f^2+24gf+32g-24f-25c-16=0$$
Eliminating $g$ with the help of equation $(1)$
$$9(-2-f)^2+16f^2+24(-2-f)f+32(-2-f)-24f-25c-16=0$$
$$9(4+f^2+4f)+16f^2-48f-24f^2-64-32f-24f-25c-16=0$$
$$f^2-68f-44-25c=0$$
Eliminating $c$ with the help of equation $(2)$
$$f^2-68f-44-25(-1-2f)=0$$
$$f^2-18f-19=0$$
$$f^2-19f+f-19=0$$
$$f=19,-1$$
$$(g,f,c)\equiv (-1,-1,1),(-21,19,-39)$$
So equations are $x^2+y^2-2x-2y+1=0$, $x^2+y^2-42x+38y-39=0$
But this got too long, any shorter method?
|
A different method (but not much shorter):
The centre of the circle also lies on the line $3x+4y=k$ for some $k\in\mathbb{R}$. Solving with $x+y=2$, the centre is $(8-k,k-6)$. The radius is $\dfrac{|4(8-k)-3(k-6)+4|}5=\dfrac{|54-7k|}5$.
Therefore, $\displaystyle (8-k)^2+(k-6-1)^2=\frac{(54-7k)^2}{25}$.
$25(2k^2-30k+113)=49k^2-756k+2916$
$k^2+6k-91=0$
$k=-13$ or $7$
The equation of the circle is $(x-21)^2+(y+19)^2=21^2+20^2$ or $(x-1)^2+(y-1)^2=1^2+0^2$.
|
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|
Finding the minimum of a discrete function from a continuous version In another post is asked to minimize
$$n+\left\lceil\frac dn\right\rceil$$ where $n$ is a positive integer ($d$ is also an integer, but we can relax this condition).
We obviously have
$$n+\dfrac dn\le n+\left\lceil\frac dn\right\rceil< n+\dfrac dn+1$$ and there is some relation between the requested minimum and that of the function
$$x+\frac dx$$ over $\mathbb R$ (at $x=d$).
But I cannot pinpoint the exact relation and deduce a procedure to solve the initial problem from the solution of the latter.
Update:
I am looking for some generality in the approach. I am not really after the solution of this particular problem.
|
Let $f(n,d)=n+\lceil\frac{d}{n}\rceil$.
Let $m(d)$ denote the minimum value of $f(n,d)$ for positive integers $n$. The function $x+\frac{k^2}{x}$ has minimum $2k$ over the reals and $x+\lceil\frac{k^2}{x}\rceil\ge x+\frac{k^2}{x}$, so $m(k^2)=2k$.
Note that $f(n,d+1)=f(n,d)$ or $f(n,d)+1$, so $m(d+1)=m(d)$ or $m(d)+1\quad(*)$.
We have $f(k,k^2+1)=k+\lceil k+\frac{1}{k}\rceil=2k+1$. Also $k^2+1>k^2-a^2$, so $\frac{k^2+1}{k\pm a}>k\mp a$ and hence $f(k\pm a,k^2+1)\ge 2k+1$. So $m(k^2+1)=2k+1$.
We have $f(k,k^2+k)=2k+1$, so by $(*)$ we have $m(k^2+a)=2k+1$ for $0\le a\le k$.
Similarly, $(k-a+1)(k+a)=k^2+k-a^2+a<k^2+k+1$, so $\frac{k^2+k+1}{k+a}>k-a+1$ and hence $f(k+a,k^2+k+1)\ge 2k+2$. But $f(k+1,k^2+k+1)=k+1+\lceil k+\frac{1}{k+1}\rceil=2k+2$. Hence $m(k^2+k+1)=2k+2$.
But $f(k^2+2k,k)=2k+2$, so by $(*)$ we have $m(k^2+k+i)=2k+2$ for $i=1,2,\dots,k$.
Summary: $m(d)=2k+1$ for $k^2<d\le k^2+k$ and $2k+2$ for $k^2+k+1\le d\le(k+1)^2$.
Let $M(d)=\lceil2\sqrt{d}\rceil$. Then $M(k^2)=m(k)$. Also $$k^2<k^2+1<k^2+k<(k+\frac{1}{2})^2$$ so $M(d)=m(d)$ for $k^2<d\le k^2+k$. And $$(k+\frac{1}{2})^2<k^2+k+1<(k+1)^2-1<(k+1)^2$$ so $M(d)=m(d)$ for $k^2+k<d\le(k+1)^2$. So we have $$m(d)=M(d)\quad\text{ for all }d$$ But $M(d)$ is just the minimum for the real valued function $x+\frac{d}{x}$ rounded up to the nearest integer. So there is apparently a close and suggestive relationship between the solutions in the discrete and continuous cases.
|
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"url": "https://math.stackexchange.com/questions/3511096",
"timestamp": "2023-03-29T00:00:00",
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|
How to solve $\sin 2x \sin x+(\cos x)^2 = \sin 5x \sin 4x+(\cos 4x)^2$? How to solve $\sin 2x \sin x+(\cos x)^2 = \sin 5x \sin 4x+(\cos 4x)^2$?
\begin{align*}
2\cos x(\sin x)^2+(\cos x)^2 & = \frac{1}{2}(\cos x- \cos 9x) +(\cos 4x)^2\\
4\cos x(1-(\cos x)^2)+2(\cos x)^2 & = \cos x + \cos 9x +(2(\cos 4x)^2-1)+1\\
\cos x(4(1-(\cos x)^2 + 2\cos x -1) & = \cos 9x + \cos 8x + 1
\end{align*}
I don't see any way to get rid of $8x$ and $9x$ as arguments to have same angles from there
|
I put this to Wolfram Alpha with $x=\cos\theta$ and got an output of
$$(2 x - 1) (2 x + 1) (x - 1) (x + 1) (32 x^5 - 16 x^4 - 32 x^3 + 12 x^2 + 6 x - 1)$$
It it returned roots of
$$x\in\{-0.841254,-0.415415,0.142315,0.654861,0.959493\}$$
for the last factor, so you need to solve a quintic to get all the roots it seems.
EDIT: Oops, spoke too soon. Cheating from @Math1000's answer, we can see now that
$$\begin{align}\frac{\sin\frac{11y}2}{\sin\frac y2}&=-\left(32\cos^5(\pi-y)-16\cos^4(\pi-y)-32\cos^3(\pi-y)\right.\\
&\quad\left.+12\cos^2(\pi-y)+6\cos(\pi-y)-1\right)\end{align}$$
And then $11y/2=n\pi$ so the roots are
$$\theta\in\{\frac{\pi}3,-\frac{\pi}3,\frac{2\pi}3,-\frac{2\pi}3,0,\pi,\frac{9\pi}{11},-\frac{9\pi}{11},\frac{7\pi}{11},-\frac{7\pi}{11},\frac{5\pi}{11},-\frac{5\pi}{11},\frac{3\pi}{11},-\frac{3\pi}{11},\frac{\pi}{11},-\frac{\pi}{11}\}$$
EDIT: After a nice walk in the chill air, the most direct path becomes clear: starting from
$$\sin2x\sin x+\cos^2x=\sin5x\sin4x+\cos^24x$$
We can apply $\sin A\sin B=\frac12\left(\cos(A-B)-\cos(A+B)\right)$ and $\cos^2A=\frac12(\cos2A+1)$ to obtain
$$\frac12(\cos x-\cos3x)+\frac12(\cos2x+1)=\frac12(\cos x-\cos9x)+\frac12(\cos8x+1)$$
Rearrange to
$$\cos9x+\cos2x-\cos8x-\cos3x=0$$
Then apply
$$\cos A+\cos B=2\cos\left(\frac{A+B}2\right)\cos\left(\frac{A-B}2\right)$$
Arriving at
$$\begin{align}0&=2\cos\left(\frac{11x}2\right)\cos\left(\frac{7x}2\right)-2\cos\left(\frac{11x}2\right)\cos\left(\frac{5x}2\right)\\
&=2\cos\left(\frac{11x}2\right)\left(\cos\left(\frac{7x}2\right)-\cos\left(\frac{5x}2\right)\right)\end{align}$$
And finally we may avail ourselves of
$$\cos A-\cos B=-2\sin\left(\frac{A+B}2\right)\sin\left(\frac{A-B}2\right)$$
So that
$$-4\cos\left(\frac{11x}2\right)\sin(3x)\sin\left(\frac x2\right)=0$$
Our solutions are therefore
$$\frac{11x}2=\left(n+\frac12\right)\pi$$
for $-5\le n\le4$ and
$$3x=n\pi$$
for $-2\le n\le3$, $n\in\mathbb{Z}$. This accounts for all $16$ solutions $\pmod{2\pi}$ with much less effort. If someone were to have whispered "$11$" in your ear this solution would have become immediately apparent.
|
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|
Is there an easy way to see that ${1\over5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12} > 1$? The sum $$\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12}$$ is just a bit larger than $1$. Is there some clever way to show this other than to add the fractions together by brute-force? For example, is there some way to group terms together and say something like "These terms sum to more than $\frac{1}{3}$, these terms sum to more than $\frac{1}{2}$, and these terms sum to larger than $\frac{1}{6}$, so the whole thing sums to more than $1$"?
|
Since $y=\frac{1}{x}$ is convex we have:-
$\dfrac15+\dfrac16+\dfrac17>\dfrac36=\dfrac12$
$\dfrac18+\dfrac19+\dfrac1{10}+\dfrac1{11}+\dfrac1{12}>\dfrac5{10}=\dfrac12$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3514103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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|
Rewriting inequality given by a fraction so I got the following solution for a problem:
$u(x,)=\begin{cases}
1 & x\leq y \\
\sqrt{\frac{x-1}{y-1}} & 1< x < y\\
0 & x\geq 1
\end{cases} \tag{1}$
and the master solution is
$u(x,)=\begin{cases}
1 & x\leq y \\
\sqrt{\frac{1-x}{1-y}} & y< x<1\\
0 & x\geq 1
\end{cases} \tag{2}$
Look at the second case of both solutions. To get there, I started at $s=\frac{y-x}{y-1}\tag{3}$ whereas the solution started at $s=\frac{x-y}{1-y} \tag{4}$ which is the same since we can multiply it with $1=\frac{-1}{-1}$
Further we know that $0<s<1\tag{5}$
Now we plug in (3) resp. (4) into (5). We multiply with the denominator, resulting in implicitly assuming $y-1>0$ for (3) and $1-y>0$ in (4). We then get (1) resp. (2).
So far, so good. But isn't $\sqrt{\frac{1-x}{1-y}}=\sqrt{\frac{x-1}{y-1}} \tag{6}$? And wouldn't that lead to a contradiction because we could plug that into (1) and (2) and get:
$\sqrt{\frac{1-x}{1-y}}=\sqrt{\frac{x-1}{y-1}}$
for
$1< x < y$ resp. $y < x < 1$? Which contradicts itself, no?
Where is my mistake?
|
No, it's true:$$\sqrt{\frac{1-x}{1-y}}=\sqrt{\frac{-(1-x)}{-(1-y)}}=\sqrt{\frac{x-1}{y-1}}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
The value of the series: $\sum_{p=1}^{n} \sin(\frac{p}{n^2})$
Find the limit as $n\rightarrow \infty$ of:
$$\sum_{p=1}^{n} \sin\left(\frac{p}{n^2}\right)$$
This question seems odd to me because it was included in my differentiability problems set.
My Attempt:
We have: $\sin(x) \le x $, Thus clearly:
$$\sum_{p=1}^{n} \sin\left(\frac{p}{n^2}\right) \le \sum_{p=1}^{n} \frac{p}{n^2}$$
I have to prove that:
$$\sum_{p=1}^{n} \frac{p}{n^2} \rightarrow \frac12$$
And:
$$\frac{1}{2}\le\sum_{p=1}^{n} \sin\left(\frac{p}{n^2}\right)$$
To conclude that the series converges to $\frac12$.
Though, I think there exists another solution that uses derivative or something similar.
Update:
I proved by double summation that:
$$\sum_{p=1}^{n} \frac{p}{n^2} = \frac{1+\frac1n}{2}$$
So it clearly goes to $\frac12$
|
Maple says
$$
\sum_{p=1}^n\sin\left(\frac{p}{n^2}\right) =
{\frac {\sin \left( \frac{1}{n} \right) \cos \left( \frac{1}{n^2} \right) +
\left( \cos \left( \frac{1}{n} \right) -1 \right) \sin \left( \frac{1}{n^2}
\right) -\sin \left( \frac{1}{n} \right) }{2\,\cos \left( \frac{1}{n^2}
\right) -2}}
$$
|
{
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"url": "https://math.stackexchange.com/questions/3516525",
"timestamp": "2023-03-29T00:00:00",
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|
Compute $\sigma$, given $\sigma^{11}$ in $S_{10}.$
Let $\sigma \in S_{10}$ with $\sigma^{11}=(3,1,4)(5,2,9,6,8,7,10)$. How to compute $\sigma$?
I tried:
$\sigma^{11}=(3,1,4)(5,2,9,6,8,7,10)=\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 4 & 9 & 1 & 3 & 2 & 8 & 10& 7 & 6 & 5\end{pmatrix}$
Now I tried to find two permutations such that
$\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ & & & & & & & & & \end{pmatrix}\cdot \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ & & & & & & & & & \end{pmatrix}=\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 4 & 9 & 1 & 3 & 2 & 8 & 10& 7 & 6 & 5\end{pmatrix}$
But it's not working.
How to find $\sigma$ here?
|
The first thing to note is that $11$ is $-1$ modulo $3$ and $4$ modulo $7$. In particular, we note that $(1,3,4)^{11} = (3,1,4)$, so if we can find an element $\tau$ of the symmetric group on $\{2,5,6,7,8,9,10\}$ whose eleventh power is $(5,2,9,6,8,7,10)$, we're done. Now, we note that the $n$th power of a $k$ cycle is a $k$ cycle whenever $k$ and $n$ are coprime, as in this case, so it's reasonable to guess that we're looking for a $7$-cycle, and we note that the eleventh power of a $7$-cycle is equal to its $4$th power, so we're looking for $(a,b,c,d,e,f,g)$ such that $(a,b,c,d,e,f,g)^4 = (5,2,9,6,8,7,10)$. But $(a,b,c,d,e,f,g)^4 = (a,e,b,f,c,g,d)$, so we have $(a,b,c,d,e,f,g) = (5,9,8,10,2,6,7)$. Putting these together (since disjoint cycles commute), we have $$((1,3,4)(5,9,8,10,2,6,7))^{11} = (3,1,4)(5,2,9,6,8,7,10).$$
|
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|
Representing $f(x)=\frac{1}{x^2}$ as power series, with powers of $(x+2)$ As stated in the title, I want $f(x)=\frac{1}{x^2}$ to be expanded as a series with powers of $(x+2)$.
Let $u=x+2$. Then $f(x)=\frac{1}{x^2}=\frac{1}{(u-2)^2}$
Note that $$\int \frac{1}{(u-2)^2}du=\int (u-2)^{-2}du=-\frac{1}{u-2} + C$$
Therefore, $\frac{d}{du} (-\frac{1}{u-2})= \frac{1}{x^2}$ and
$$\frac{d}{du} (-\frac{1}{u-2})= \frac{d}{du} (-\frac{1}{-2(1-\frac{u}{2})})=\frac{d}{du}(\frac{1}{2} \frac{1}{1-\frac{u}{2}})=\frac{d}{du} \Bigg( \frac{1}{2} \sum_{n=0}^\infty \bigg(\frac{u}{2}\bigg)^n\Bigg)$$
$$= \frac{d}{du} \Bigg(\sum_{n=0}^\infty \frac{u^n}{2^{n+1}}\Bigg)= \frac{d}{dx} \Bigg(\sum_{n=0}^\infty \frac{(x+2)^n}{2^{n+1}}\Bigg)= \sum_{n=0}^\infty \frac{d}{dx} \bigg(\frac{(x+2)^n}{2^{n+1}}\bigg)=$$
$$\sum_{n=0}^\infty \frac{n}{2^{n+1}} (x+2)^{n-1}$$
From this we can conclude that
$$f(x)=\frac{1}{x^2}=\sum_{n=0}^\infty \frac{n}{2^{n+1}} (x+2)^{n-1}$$
Is this solution correct?
|
Binomial theorem:
$\begin{align*}
\frac{1}{x^2}
&= (2 + (x - 2))^{-2} \\
&= \frac{1}{4}
\cdot \left(
1 + \frac{1}{2}(x - 2)
\right)^{-2} \\
&= \frac{1}{4}
\cdot \sum_{k \ge 0}
\binom{-2}{k}
\left(
\frac{x - 2}{2}
\right)^k \\
&= \sum_{k \ge 0} \frac{(-1)^k}{2^{k + 2}}
\binom{k + 1}{2} (x - 2)^k \\
&= \sum_{k \ge 0}
\frac{(-1)^k (k + 1) k}{2^{k + 3}}(x - 2)^k
\end{align*}$
Any way you get a power series that converges to your function will give the same series. What way you select depends on taste/ease/familiarity.
|
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|
How to solve the ordinary differential equation $ x^2 y'' - 2 x y' + 2y = x^4 \mathrm{e}^x $ Please tell me how to solve this differential equation.
$$ x^2 y'' - 2 x y' + 2y = x^4 \mathrm{e}^x $$
I tried to solve it but finally I stuck tell me how to go further more if anyone gives me a hint(I want to know how to find paticular integral?).I will appreciate about it.i posted my way below please tell me any hint. Here's my work:
|
Operators and first and second integrals are two methods to find a solution. This solution uses the series solution.
Let
$$y(x) = \sum_{n=0}^{\infty} a_{n} \, x^n$$
which easily leads to
\begin{align}
x^2 \, y'' - 2 x y' + 2 y &= x^4 \, e^{x} \\
\sum_{n=0}^{\infty} a_{n} \, (n(n-1) - 2n + 2) \, x^n &= \sum_{n=0}^{\infty} \frac{x^{n+4}}{n!} \\
\sum_{n=0}^{\infty} (n-1)(n-2) \, a_{n} \, x^n &= \sum_{n=4}^{\infty} \frac{x^n}{(n-4)!} \\
a_{1} \, x + a_{2} \, x^2 + \sum_{n=4}^{\infty} (n-1)(n-2) \, a_{n} \, x^n &= \sum_{n=4}^{\infty} \frac{x^n}{(n-4)!}
\end{align}
This leads to $a_{1}$ and $a_{2}$ along with $$a_{n} = \frac{n(n-3)}{n!} \hspace{5mm} \text{for} \hspace{5mm} n \geq 4.$$
Now,
\begin{align}
y(x) &= a_{1} \, x + a_{2} \, x^2 + \sum_{n=0}^{\infty} \frac{n(n-3) \, x^n}{n!} \\
&= a_{1} \, x + a_{2} \, x^2 + x \, \frac{d}{dx} \, \sum_{n=0}^{\infty} (n-3) \, \frac{x^n}{n!} \\
&= a_{1} \, x + a_{2} \, x^2 + x \, \frac{d}{dx} \left[ x \, \frac{d}{dx} e^x - 3 \, e^x \right] \\
&= a_{1} \, x + a_{2} \, x^2 + x \, \frac{d}{dx} \left( (x-3) \, e^x \right) \\
&= a_{1} \, x + a_{2} \, x^2 + x(x-2) \, e^x.
\end{align}
|
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|
Combination of problem having identical objects without using generating function approach
A person goes in for an examination in which there are $4$ papers with a maximum of $m$ marks from each paper. The number of ways in which one can get $2m$ marks is
My Attempt
Number of ways of partitioning $n$ identical objects into $r$ distinct groups, if empty groups are allowed is:${}^{n+r-1}C_{r-1}$.
Here, $n=2m$ and $r=4$
Req. number of ways = ${}^{n+r-1}C_{r-1}-$[#{more than m marks for paper 1}+#{more than m marks for paper 2}+#{more than m marks for paper 3}+#{more than m marks for paper 4}]=$${}^{2m+4-1}C_{4-1}-[{}^{m+4-1}C_{4-1}+{}^{m+4-1}C_{4-1}+{}^{m+4-1}C_{4-1}+{}^{m+4-1}C_{4-1}]={}^{2m+4-1}C_{4-1}-4.{}^{m+4-1}C_{4-1}\\
={}^{2m+3}C_{3}-4.{}^{m+3}C_{3}=\frac{(2m+3)(2m+2)(2m+1)}{3.2}-4\frac{(m+3)(m+2)(m+1)}{3.2}\\
=\frac{m+1}{3}.[4m^2+8m+3-2m^2-10m-12]=\frac{(m+1)(2m^2-2m-9)}{3}
$$
But my reference gives the solution $\dfrac{(m+1)(2m^2+4m+3)}{3}$, so what is going wrong here ?
Note: I am trying to solve this problem without using the generating function approach
|
$\#$ of ways $n$ identical objects can be divided into $r$ distinct groups if empty groups are allowed is:${}^{n+r-1}C_{r-1}$
Req. # of ways=$${}^{2m+4-1}C_{4-1}-\Big[\text{#{> m for paper 1}+#{> m for paper 2}+#{> m for paper 3}+#{> m for paper 4}}\Big]\\+\Big[\text{#{strictly m for paper 1}+#{strictly m for paper 2}+#{strictly m for paper 3}+#{strictly m for paper 4}}\Big]\\
={}^{2m+4-1}C_{4-1}-4\Big[{}^{m+4-1}C_{4-1}\Big]+4\Big[{}^{m+3-1}C_{3-1}\Big]\\
={}^{2m+3}C_{3}-4\Big[{}^{m+3}C_{3}\Big]+4\Big[{}^{m+2}C_{2}\Big]\\
=\frac{(2m+3)(2m+2)(2m+1)}{3.2}-4.\frac{(m+3)(m+2)(m+1)}{3.2}+4.\frac{(m+2)(m+1)}{2}\\
=\frac{m+1}{3}.[4m^2+8m+3-2m^2-10m-12+6m+12]=\frac{m+1}{3}.[2m^2+4m+3]
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding limit for given function
If $f(x)=\lim\limits_{n \to \infty} \frac{x}{n} \left(\frac{1}{1+e^{-x/n}}+\frac{1}{1+e^{-2x/n}}+\dots+\frac{1}{1+e^{-x}} \right)$
then what is $\lim\limits_{x\to 0} \left(\frac{2f(x)-x}{x^2}\right)$?
My solution is :
$f(x)=0$ since $\lim\limits_{n \to \infty} \frac{x}{n}$ part will be zero for $n=\infty$, making entire term $0$.
then $\lim\limits_{x\to 0} \left(\frac{2f(x)-x}{x^2}\right)$ will become $\lim\limits_{x\to 0} \left(\frac{-x}{x^2}\right)$ or $\lim\limits_{x\to 0} \left(\frac{-1}{x}\right)$ = $\infty$.
This answer is not avaiable in the given options. What should be the correct answer for this questions?
|
Your answer is not correct because $f$ is not identically zero.
Note that the definition of $f$ is the limit of a Riemann sum:
$$\begin{align}f(x)&=\lim\limits_{n \to \infty} \frac{x}{n} \left(\frac{1}{1+e^{-x/n}}+\frac{1}{1+e^{-2x/n}}+\dots+\frac{1}{1+e^{-x}} \right)\\&=x\int_0^1\frac{dt}{1+e^{-tx}}=\ln\left(\frac{e^x+1}{2}\right).\end{align}$$
Therefore, as $x\to 0$,
$$\begin{align}
2f(x)-x&=2\ln\left(\frac{1+x+\frac{x^2}{2}+o(x^2)+1}{2}\right)-x\\
&=2\ln\left(1+\frac{x}{2}+\frac{x^2}{4}+o(x^2)\right)-x\\
&=2\left(\frac{x}{2}+\frac{x^2}{4}-\frac{x^2}{8}+o(x^2)\right)-x
=\frac{x^2}{4}+o(x^2)
\end{align}$$
and we may conclude that the given limit is $1/4$.
|
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|
Let $X,Y$ be independent normally distributed random variables. Find the density of $\frac{X^2}{Y^2+X^2}$ Let $X,Y$ be independent standard normally distributed random variables and $X,Y\neq 0$. Find the density of $\frac{X^2}{Y^2+X^2}$
I was given the tip of first calculating the density of $(X^2,Y^2)$ and then calculate the density of $(\frac{X^2}{Y^2+X^2},Y^2+X^2)$
When I follow the tips: I know that $X^2\sim\Gamma(\frac{1}{2},\frac{1}{2})$ and $Y$ too. Furthermore, $X^2$ and $Y^2$ are still independent. Therefore the density $f_{(X^2,Y^2)}(x,y)$ can be written as $f_{X^2}(x)f_{Y^2}(y)$ where $f_{X^2}$ and $f_{Y^2}$ are the density functions of $X^2$ and $Y^2$
My next idea, with the tip above in mind is to consider a map $\varphi: (x,y) \mapsto (\frac{x}{x+y},x+y)$
It then follows that $(\frac{X^2}{Y^2+X^2},Y^2+X^2)=\varphi(X^2,Y^2)$
and $f_{\frac{X^2}{Y^2+X^2},Y^2+X^2}(a,b) = f_{\varphi(X^2,Y^2)}(a,b)
$
Note that $\varphi^{-1}: (a,b)\mapsto (ba,b-ba)$ and thus $\det D \varphi^{-1}(a,b)=\det\begin{pmatrix} b & a \\
-b & 1-a
\end{pmatrix}=b(1-a)+ab\implies \det D \varphi^{-1}(\frac{X^2}{Y^2+X^2},Y^2+X^2)=(Y^2+X^2)(1-\frac{X^2}{Y^2+X^2})+(\frac{X^2}{Y^2+X^2})(Y^2+X^2)=Y^2+X^2$
And hence $P_{(\frac{X^2}{Y^2+X^2},Y^2+X^2)}(A)=\int_A f_{(\frac{X^2}{Y^2+X^2},Y^2+X^2)}(x,y)\,dx\,dy = \int_{\varphi^{-1}(A)}f_{(X,Y)}(x,y)\times (X^2+Y^2)\,dx\,dy = \int_{\varphi^{-1}(A)} f_X(x)\times f_Y(y)\times (X^2+Y^2)\,dx\,dy$
Where do I go from here?
|
Observe that the distribution of the vector $(X,Y)$ is rotationally invariant. That is, if we rotate it at any angle, the resulting vector has the same distribution as $(X,Y)$.
Why? Rotation $\alpha$ radians counterclockwise gives the vector $(X\cos \alpha-Y\sin \alpha,X\sin \alpha+Y \cos \alpha)$, which is centered gaussian whose covariance matrix is the identity, hence the components are independent standard normal.
Since the distribution of $(X,Y)$ is invariant under rotation, the angle it makes with the positive $x$-axis, $\Theta$ is uniformly distributed on $[0,2\pi)$.
Let $Z= X^2/(X^2+Y^2)$. Then from the definition of $\Theta$,
$Z = \cos^2 \Theta$, so this is nothing but a transformation of a uniform RV, which after a routine calculation gives us the density
$$f_Z (z) = \frac{1}{\pi\sqrt{(1-z)z}},~z \in (0,1).$$
|
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Help with proof by induction, please! I may not be understanding how to work with factorials.
Prove that for all integers $n \ge 1, 1\cdot1! + 2\cdot2! + 3\cdot3! + \cdots + n\cdot n! = (n+1)! - 1$.
OK, so I verified base case $n = 1: 2! - 1 = 1$
I assumed that for all $k \ge n, P(k) = (k+1)! - 1$
I am unsure how to prove true for $P(k+1)$, though I understand that the first step is to replace all $k$ values with $k+1$ which yields $$1\cdot1! + 2\cdot2! + 3\cdot3! + \cdots + k!\cdot k! + (k+1\cdot k+1)! = (k+2)! - 1$$
What next?
|
Hint: If $f(n) = 1*1! + 2*2! + ..... + n*n!$ then $f(n+1) = 1*1! + 2*2! + ..... + n*n! + (n+1)(n+1)! = f(n) + (n+1)(n+1)!$.
Hint 2: $m!(m+1) = (m+1)!$ and $m!*m + m! {=\over{\text{factor}}} m!(m+1) = m!(m+1)= m!$.
So if we assume $f(k) = 1*1! + 2*2! + ..... + k*k!= (k+1)! - 1$.
.
Then $f(k+1) = f(k) + (k+1)(k+1)! =$
.
$(k +1)! -1 + (k+1)(k+1)!=(k+1)! + (k+1)(k+1)! - 1 = (k+1)![1+(k+1)]-1=$
.
$(k+1)![k+2]-1 = (k+2)!-1$
|
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Solving the inequality $1+1.2 + 1.2^2 + 1.2 ^3 + \cdots + 1.2^n < N/16?$ Given the inequality $$1+1.2 + 1.2^2 + 1.2 ^3 +\cdots + 1.2^n < \frac{N}{16}?$$
I need the value of $n$, or just an approximation. $N$ is known.
|
$1+1.2+1.2^2+\ldots +1.2^n = \frac{1.2^{n+1}-1}{1.2-1}$ $= 5(1.2^{n+1}-1)$.
So you want to find the values of $n$ such that the inequality
$$5(1.2^{n+1}-1) < \frac{N}{16}$$
This gives
$$1.2^{n+1} < \frac{N}{80} - 1$$
Taking logs this gives
$$n \log 1.2 < \log \left(\frac{N}{80}-1\right)$$
which as $\log 1.2 < .2$ and $\log 80 \approx .4.4$ , will hold for large $N$ if the following is satisfied:
$$.2n < \log N - 4.5$$
or $$n < 5 \log N - 18$$
|
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|
Integral $\int{\frac{1}{(x^{3} \pm 1)^2}}$ I need to solve the following integrals:
$$\int{\frac{1}{(x^3+1)^2}}dx$$
and
$$\int{\frac{1}{(x^3-1)^2}}dx$$
My first thought was to use a trigonometric substitution but the $x^3$ messed it up. Can you guys suggest another method?
|
The standard method uses partial fractions decomposition.
Here is how it begins for the first integral: factorout the denominator into irreducible factors with high-school identities:
$$x^3+1=(x+1)(x^2-x+1)$$
Newt proceed to decompose into partial fractions:
$$\frac{1}{(x^3+1)^2}=\frac{1}{(x+1)^2(x^2-x+1)^2}=\frac A{x+1}+\frac B{(x+1)^2}+\frac {Cx+D}{x^2-x+1}+\frac{Ex+F}{(x^2-x+1)^2}.$$
|
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|
Find all integer values of $m$ such that the equation $\sqrt{9 - x} = \sqrt{3m - x^2 + 9x} - \sqrt{x}$ has exactly four distinct real roots.
Find all integer values of $m$ such that the equation $$\large \sqrt{9 - x} = \sqrt{3m - x^2 + 9x} - \sqrt{x}$$ has exactly four distinct real roots.
$\left(x \in [0, 9], m \in \left[0, \dfrac{27}{4}\right]\right)$
Let $9 - x = y \ (\iff x + y = 9)$, we have that $$\sqrt{9 - x} = \sqrt{3m - x^2 + 9x} - \sqrt{x} \implies \sqrt{x} + \sqrt{y} = \sqrt{3m + xy}$$
$$\implies x + y + 2\sqrt{xy} = 3m + xy \iff (x - 1)y - 2\sqrt{xy} + (3m - x) = 0$$
For the equation $$\sqrt{9 - x} = \sqrt{3m - x^2 + 9x} - \sqrt{x}$$ to have exactly four distinct real roots, the equation $$(x - 1)y - 2\sqrt{xy} + (3m - x) = 0$$ must have two distinct real roots $x$.
(note that $x_0$ and $9 - x_0$ are both solutions to $\sqrt{9 - x} = \sqrt{3m - x^2 + 9x} - \sqrt{x}$)
$$\implies \Delta' = x - (x - 1)(3m - x) > 0$$, in which case, the solutions are $\begin{cases} m > \dfrac{y^2}{3(y - 1)} &\text{if $x \in [0, 1)$}\\ m \in \mathbb R &\text{if $x = 1$}\\ m < \dfrac{y^2}{3(y - 1)} &\text{if $x \in (1, 9]$} \end{cases}$, (according to WolframAlpha, of course), which lead to suspect that $m \in \{0, 1\}$ are the integer solutions. But I don't really know.
|
$\sqrt{9-x}=\sqrt{3m-x^2+9x}-\sqrt x$
Equations $\sqrt{9-x}=\sqrt x$ has no solution. For the moment we assume that equation $\sqrt{3m-x^2+9x}=\sqrt x$ has also no solution. Therefore the roots of equation $\sqrt{9-x}=\sqrt{3m-x^2+9x}$ must be satisfied by $\sqrt x$; considering $x∈[0, 9]$, we may write for integer x:
$x=0. 1, 4, 9$
$\sqrt{9-x}=\sqrt{3m-x^2+9x}$
$m= \frac{x^2-10x+9}{3}$
Which gives :
$(x, m)=(0, 3), (1, 0), (4, -5), (9, 0)$
Among these only $(x, m)=(0, 3)$ satisfies the original equation.Moreover for $m=3$ we also get $x=9$ in equation.
If we assume that equation $\sqrt{3m-x^2+9x}=\sqrt x$ has solution then it must satisfy the original equation, I checked; it does not.
|
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How can we get all positive integer solution for these two equations While solving a problem i was able to reduce the question to finding positive integer solutions that satisfy the equations:-
$x+y+z = 2020$ and $x^2 + y^2 + z^2 = xyz+4$.
I found
$$(x,y,z)= (2,1009,1009),\ (1009,2,1009) \text{ and }(1009,1009,2)$$
as few solutions by guessing and hit and trial. How can I find other solutions (if they exist) to this equations
|
You can write second equation as $(x+y+z)^2-2(xy+yz+xz)=xyz+4$ that is $2020^2-4= xyz+ 2(xy+yz+xz)$. Now it could be useful to write RHS as a product. You can calculate $(2+x)(2+y)(2+z)=8+4(x+y+z)+ 2(xy+yz+xz)+xyz=8088+ 2(xy+yz+xz)+xyz$.
So we can write $(2+x)(2+y)(2+z)=8088+2020^2-4=2020^2+4\cdot 2020+4=2022^2$. Are you able to continue now?
|
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|
How many integer solutions are there to $x+y+z=8$
How many integer solutions are there to $x+y+z=8$ When $x,y,z>0$? When $x,y,z\geq -3$?
So I know there is a formula for computing the number of nonnegative solutions
${8+3-1 \choose 3-1}={10\choose 2}$
So I then just subtracted cases where one or two integers are $0$.
If just $x=0$ then there are $6$ solutions where neither $y,z=0$.
So I multiplied this by $3$, then added the cases where two integers are $0$
$3\cdot 6+3=21$. So I get ${10 \choose 2}+21=66$
For the last problem where $x,y,z\geq -3$ I'm not sure how to deal with it.
|
Start by looking at the number of solutions of
\begin{eqnarray*}
X+Y+Z=17
\end{eqnarray*}
which has $\binom{17+3-1}{3-1}$ solutions.
This is the same as the number of solutions of
\begin{eqnarray*}
x+y+z=8
\end{eqnarray*}
where $x=X-3$,$y=Y-3$ and $z=Z-3$.
|
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|
find $\alpha,\beta$ such that $A^{-1}=\alpha A+\beta I$. Let $A$be a $5 \times 5$ matrix whose characteristic polynomial is given by $(\lambda -2)^3(\lambda+2)^2.$ If $A$ is diagonalizable,find $\alpha,\beta$ such that $A^{-1}=\alpha A+\beta I$
Solution:
Characteristic polynomial of $A$,$Ch_{A}(\lambda)=(\lambda -2)^3(\lambda+2)^2$.
By Cayley-Hamilton Theorem, we have $Ch_{A}(A)=0\implies (A -2)^3(A+2)^2=0\implies$ Either $(A -2)^3=0$ or $(A+2)^2=0$
Considering
$$(A+2)^2=0\implies A^2+4A+4I=0$$
$$\implies I=\left(-\frac{1}{4}\right)A^2+(-A)$$
$$\implies A^{-1}=\left(-\frac{1}{4}\right)A+(-I).$$
On comparing just recent equation with $A^{-1}=\alpha A+\beta I$,we get $\alpha =-\frac{1}{4}$ and $\beta =-1$
But the correct answer is $\alpha=\frac{1}{4}$ and $\beta =0$..
I wanted to know where I'm wrong?
|
Here's another way of solving your problem:
We're given that the characteristic polynomial of $A$ is $c_A(x)=(x-2)^3(x+2)^2$. Since $A$ is diagonalizable, that implies that the minimal polynomial must be $m_A(x)=(x-2)(x+2)$. Hence we have that
$$\begin{align*}
0 &= m_A(A)\\
&= (A-2I)(A+2I)\\
&= A^2-4I
\end{align*}$$
So $A^2-4I=0$, or equivalently $A^2=4I$.
Since $0$ is not an eigenvalue of $A$, we also know that $A$ is invertible. Multiplying by $A^{-1}$, we get that $A=4A^{-1}$. Thus $A^{-1}=\frac{1}{4}A$, $\alpha=\frac{1}{4}$, and $\beta=0$.
|
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|
Tough multivariable inequality: Minimize $a^2 + b^3 + c^4$ given $a + b^2 + c^3 = \frac{325}{9}$ Let $a,$ $b,$ and $c$ be positive real numbers such that $a + b^2 + c^3 = \frac{325}{9}.$ Find the minimum value of
$a^2 + b^3 + c^4.$
|
Note that for the surface $a^2 + b^3 + c^4=k$ to have the minimum value $k$, it is tangential to the surface $a + b^2 + c^3 = \frac{325}{9}$. Calculate their normal vectors
$(1,2b,3c^2)$ and $(2a,3b^2,4c^3)$, and match them to get
$$\frac{2a}1=\frac{3b}2=\frac{4c}3$$
Then, plug $b= \frac43a$ and $c=\frac32a$ into $a + b^2 + c^3 = \frac{325}{9}$ to obtain the tangential point at
$$a=2,\>\>\>b = \frac83,\>\>\>c=3$$
As a result, the minimum value is
$$a^2 + b^3 + c^4=2^2+\left(\frac83\right)^3+3^4=\frac{2807}{27}$$
|
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|
Solving $15^{\log_5(3)}\cdot x^{\log_5(9x)+1}=1$ I need help solving this logarithm exercise:
$$15^{\log_5(3)}\cdot x^{\log_5(9x)+1}=1$$
What I've done is re-writing the equation
$$\Rightarrow \qquad 5^{\log_5(3)}\cdot 3^{\log_5(3)}\cdot x^{\log_5(9x)+1}=1 \tag{1}$$
Then applying logarithms on both sides
$$\Rightarrow \qquad \log_5(3^{\log_5(3)}\cdot5^{\log_5(3)}\cdot x^{log_5(9x)+1})=\log_5(1) \tag{2}$$
re-writing the equation a little bit
$$\Rightarrow \qquad \log_53+\log_53^{\log_5(3)}+\log_5x^{\log_5(9x)+1}=\log_55 \tag{3}$$
But then I'm not entirely sure how to proceed
|
$15^{\log_5(3)}\cdot x^{\log_5(9x)+1}=1\\
x^{\log_5(9x)+1}=15^{-\log_5(3)}$
Take the $\log_5$ of both sides
$(\log_5(9x)+1)\log_5 x =(-\log_5(3))(\log_5 15)\\
(\log_5 x + \log_5 9+1)\log_5 x =(-\log_5(3))(\log_5 15)\\
(\log_5 x)^2 + (\log_5 9+1)\log_5 x + \log_5(3)(\log_5 3 + 1) = 0\\
$
let $u = \log_5 x, b = 2\log_5 3 + 1, c = (\log_5 3)^2 + \log_5 3$
$u^2 + bu + c = 0$
$u = \frac {-b \pm \sqrt {b^2 - 4c}}{2}$
$b^2 - 4c = (2\log 3 + 1)^2 - 4(\log^2 3 + \log 3)\\
4\log^2 3 + 4\log 3 + 1 - 4\log^2 3 - 4\log 3 = 1$
$u = -\log_5 3, -\log_5 3 - 1\\
x = 5^u$
$x = \frac {1}{3}$ or $\frac {1}{15}$
|
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|
Let $x∈R$ and suppose that$ |x−2| < 1/3$. Prove that $|x^2+x−6| < 2$. I need help studying for a midterm coming up and am doing this problem for practice:
Let $x∈R$ and suppose that $|x−2| < 1/3$. Prove that $|x^2+x−6| < 2$.
If $|x-2| < 1/3, x < 7/3$ and $x > 5/3$.
$|x^2 + x - 6| = |x-2||x+3| < 2$.
I'm really not sure where to go from here. What I tried was the following:
$(1/3)|x+3| < 2$, so $|x+3|<6$
Since the maximum of $x$ is $7/3$, the most $|x+3|$ can be is $|(7/3)+3|= 16/3 < 6$.
Then, the minimum of $x$ is $5/3$, so the least $|x+3|$ can be is $|(5/3)+3| = 14/3 < 6$.
Thus, $|x-2||x+3| < 2$.
Anyone know if this is correct? If not, can you point me in the right direction? Thanks in advance.
|
Let $x \in \mathbb R$ and suppose that $|x−2| < \dfrac 13$. Prove that $|x^2+x−6| < 2$.
\begin{align}
|x-2| < \dfrac 13
&\implies \dfrac 53 < x < \dfrac 73 \\
&\implies 3+\dfrac 53 < x+3 < 3 + \dfrac 73 \\
&\implies -\dfrac{16}{3} < x+3 < \dfrac{16}{3} \\
&\implies |x+3| < \dfrac{16}{3} \\
\end{align}
Hence $|x^2 + x - 6| = |x-2||x+3| < \dfrac 13 \cdot \dfrac{16}{3} < \dfrac{16}{9} < 2$.
|
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|
Find : $\lim\limits_{n\to +\infty}\frac{\left(1+\frac{1}{n^3}\right)^{n^4}}{\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^4}}$ Find :
$$\lim\limits_{n\to +\infty}\frac{\left(1+\frac{1}{n^3}\right)^{n^4}}{\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^4}}$$
My attempt : i don't know is correct or no!
I use this rule :
$$\lim\limits_{n\to +\infty}(f(x))^{g(x)}=1^{\infty}$$
Then :
$$\lim\limits_{n\to +\infty}(f(x))^{g(x)}=\lim\limits_{n\to +\infty}e^{g(x)(f(x)-1)}$$
So :
$$\lim\limits_{n\to +\infty}\frac{\left(1+\frac{1}{n^3}\right)^{n^4}}{\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^4}}$$
$$=\lim\limits_{n\to +\infty}\frac{e^{\frac{n^{4}}{n^{3}}}}{e^{\frac{(n+1)^{4}}{(n+1)^{3}}}}$$
$$=\lim\limits_{n\to +\infty}\frac{e^{n}}{e^{n+1}}$$
$$=\frac{1}{e}$$
Is my approach wrong ?
is this called partial limit calculation ?
|
Assuming that you would like to get more than the limit itself, starting with
$$a_n=\frac{\left(1+\frac{1}{n^3}\right)^{n^4}}{\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^4}}$$ take logarithms
$$\log(a_n)=n^4\log\left(1+\frac{1}{n^3}\right)-(n+1)^4\log\left(1+\frac{1}{(n+1)^3}\right)$$ Use twice the expansion
$$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+O\left(x^5\right)$$ and expand to get
$$\log(a_n)=-1-\frac{1}{n^3}+\frac{3}{2n^4}+O\left(\frac{1}{n^5}\right)$$ and continue with Taylor series
$$a_n=e^{\log(a_n)}=e\left(1-\frac{1}{n^3}+\frac{3}{2n^4}\right)+O\left(\frac{1}{n^5}\right)$$ which shows the limit and how it is approached. Moreover, this gives a shortcut estimation of $a_n$.
Suppose that, using your pocket calculator, you compute $a_3$. You should get
$a_3=0.3594$ while the above truncated expansion gives $\frac{53}{54 e}=0.3611$
|
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Find $\lim\limits_{x\rightarrow 0^+}\frac{1}{\ln x}\sum\limits_{n=1}^{\infty}\frac{x}{(1+x)^n-(1-x)^n}$
Find $$\lim\limits_{x\rightarrow 0^+}\dfrac{1}{\ln x}\sum_{n=1}^{\infty}\dfrac{x}{(1+x)^n-(1-x)^n}.$$
Consider $$f(x):=(1+x)^n,$$
By Lagrange MVT, we can obtain
$$\frac{2x}{f(x)-f(-x)}=\frac{1}{f'(\xi)}, -x\gtrless \xi\gtrless x$$
Thus $$\frac{x}{(1+x)^n-(1-x)^n}=\frac{1}{2f'(\xi)}=\frac{1}{2n(1+\xi)^{n-1}}$$
Can we go on from here?
|
If $x>0$, then by the mean value argument
$$
\frac{x}{{(1 + x)^n - (1 - x)^n }} > \frac{1}{{2n(1 + x)^{n - 1} }}.
$$
Assume now that $0<x<1$. Then
$$
\frac{1}{{\log x}}\sum\limits_{n = 1}^\infty {\frac{x}{{(1 + x)^n - (1 - x)^n }}} < \frac{1}{{\log x}}\sum\limits_{n = 1}^\infty {\frac{1}{{2n(1 + x)^{n - 1} }}}
\\
= - \frac{{1 + x}}{{2\log x}}\log \left( {1 - \frac{1}{{1 + x}}} \right) = - \frac{{1 + x}}{2} + \frac{{1 + x}}{{2\log x}}\log (1 + x).
$$
Consequently,
$$
\mathop {\lim }\limits_{x \to 0 + } \frac{1}{{\log x}}\sum\limits_{n = 1}^\infty {\frac{x}{{(1 + x)^n - (1 - x)^n }}} \le - \frac{1}{2}.
$$
This, together with Robert Z's answer, imply that the limit is $-1/2$.
|
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proof that there doesn't exist integers $a, b, c>0$, that satisfy $a^2bc+2$, $ab^2c+2$ and $abc^2+2$ are all square numbers. please proof that there doesn't exist a,b,c which are all positive integers, that satisfy a^2bc+2, ab^2c+2 and abc^2+2 are all square numbers.
I know squares remain 1 or 0 when divided by 4, so a^2bc, ab^2c and abc^2 must remain 2 or 3 when divided by 4. Since a^2,b^2,c^2 must remain 1 ,we can proof ab,bc,ac can’t all remain 2 or 3 at the same time.If any two of a,b,c remain the same, one of ab, bc or ac will remain 1.Hence a,b,c must remain 1,2 and 3.
But this is not a contradiction to my suppose. Which part of my proof is wrong, or should I use other way to solve this problem?
I’ve tried mod 3 and it doesn’t work either.
Any help is appreciated.
|
You idea will work. We just need to be careful.
Suppose all $3$ $a^2bc + 2$ and $ab^2c + 2$ and $abc^2+2$ are square numbers.
Then all $a^2bc, ab^2c, abc^2$ will be $\equiv 2,3 \pmod 4$.
Now suppose $a$ is even. Then $a^2 \equiv 0\pmod 4$. And $a^2bc \equiv 0 \pmod 4$ . That is a contradiction. So $a$ is odd.
The same argument will prove that $b$ and $c$ are odd and thus either $1,3 \pmod 4$.
So $a^2bc, ab^2c, abc^2$ are all odd and $\equiv 3\pmod 4$.
$a^2,b^2,c^2$ are all $\equiv 1\pmod 4$, so $bc,ac, ab$ are all $\equiv 3\pmod 4$.
If $b,c$ are both $\equiv 1\pmod 4$ or both $\equiv 3 \pmod 4$ then $bc\equiv 1\pmod 4$. So $b,c$ must have different equivalencies.
And the same argument will show that $a$ and $c$, and $a$ and $b$ will have different equivalencies from each other.
So $b$ and $c$ both have a different equivalency than $a$. But as there are only two options, $1$ or $2\pmod 4$ that would mean $b$ and $c$ must have the same as each other.
Which is a contradiction.
It's possible for two to be perfect squares but not all three.
|
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|
Find minimum of $a^2+b^2+\frac{a^2b^2}{(a+b)^2}-\frac{2(a^2+ab+b^2)}{a+b}$ If $a,b$ are real numbers, find the minimum value of:
$$a^2+b^2+\frac{a^2b^2}{(a+b)^2}-\frac{2(a^2+ab+b^2)}{a+b}$$
This is what I did: I tried some values and I set $a=0$. Then, it becomes a quadratic of $b$:
$$b^2-2b$$
Here, the minimum is $-1$. So, I tried to prove that:
$$a^2+b^2+\frac{a^2b^2}{(a+b)^2}-\frac{2(a^2+ab+b^2)}{a+b}\ge -1$$
Using Wolfram, I found this is a square:
$$a^2+b^2+\frac{a^2b^2}{(a+b)^2}-\frac{2(a^2+ab+b^2)}{a+b} + 1 = \frac{(a^2+b^2+ab-a-b)^2}{(a+b)^2} $$
so it is positive.
My question is, can we prove this with more traditional and natural solution, maybe with Cauchy-Schwarz?
|
Call the function you're minimizing $f(a,b)$. Note that the first three terms are homogeneous of order $2$ in $a$ and $b$, while the fourth is homogeneous of order $1$.
Thus $$f(ta, tb) = t^2 \left(a^2 + b^2 + \frac{a^2 b^2}{(a+b)^2}\right) - 2 t \frac{a^2+ab+b^2}{a+b} = t^2 g(a,b) + t h(a,b)$$
where $g \ge 0$, and this is minimized with respect to $t$ when $t = -h(a,b)/(2 g(a,b))$,
the minimum value being $-h(a,b)^2/(4 g(a,b))$, which simplifies to $-1$.
|
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|
Limit of $\lim_{x\to 3} \frac{3-x}{6\sin \frac{\pi x}{18}-x}$ I want to solve a limit without l'Hospital, just with algebraic manipulation:
$$\lim_{x\to 3} \frac{3-\sqrt{6+x}}{6\sin \frac{\pi x}{18}-x}$$
I started with:
$$\lim_{x\to 3} \frac{3-\sqrt{6+x}}{6\sin \frac{\pi x}{18}-x}=\lim_{x\to 3} \frac{(3-\sqrt{6+x})(3+\sqrt{x+6})}{(6\sin \frac{\pi x}{18}-x)(3+\sqrt{x+6})}=\lim_{x\to 3} \frac{3-x}{(6\sin \frac{\pi x}{18}-x)(3+\sqrt{x+6})}$$
$3+\sqrt{x+6} \to 6$ is determinate, so I only need to calculate:
$$\lim_{x\to 3} \frac{3-x}{6\sin \frac{\pi x}{18}-x}$$
but I could only do it with l'Hospital:
$$\lim_{x\to 3} \frac{3-x}{6\sin \frac{\pi x}{18}-x}=\lim_{x\to 3} \frac{-1}{6\cdot \frac{\pi}{18}\cos \frac{\pi x}{18}-1}=\frac{-1}{\frac{\pi\sqrt{3}}{6}-1}$$
Can I get some help without l'Hospital?
|
Clearly, it is enough to compute$$\lim_{x\to3}\frac{\sin\left(\frac{\pi x}{18}\right)-\frac x6}{x-3}\tag1$$and, in order to compute that, it will be enough to compute$$\lim_{x\to3}\frac{\sin\left(\frac{\pi x}{18}\right)-\frac 12}{x-3}\text{ and }\lim_{x\to3}\frac{\frac x6-\frac12}{x-3},\tag2$$since $(1)$ is equal to$$\lim_{x\to3}\frac{\sin\left(\frac{\pi x}{18}\right)-\frac 12}{x-3}-\lim_{x\to3}\frac{\frac x6-\frac12}{x-3}$$(assuming that both limits from $(2)$ exist). But, clearly,$$\lim_{x\to3}\frac{\frac x6-\frac12}{x-3}=\lim_{x\to3}\frac16\frac{x-3}{x-3}=\frac16,$$whereas$$\lim_{x\to3}\frac{\sin\left(\frac{\pi x}{18}\right)-\frac 12}{x-3}$$is the derivative at $3$ of $x\mapsto\sin\left(\frac{\pi x}{18}\right)$, which is equal to $\frac{\pi\sqrt3}{36}$.
|
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|
Stability Function system Hi I have this question
For the system
$$\begin{align}
\dot p_1 &= -p_1 + 6p_2 \\
\dot p_2 &= -7p_2+(p_1+p_2)\cos p_1
\end{align}
$$
Use the function $L(p)=\frac{1}{2}(p_1^2 + p_2^2)$ ,and show that the origin is locally asymptotically stable.
What I have done so far:
*
*Differentiated $L(p)$ to get $\dot L(p)$
*Substituted in the values of $\dot p_1$ and $\dot p_2$
|
Since
$$x_1^2 + x_2^2 \ge 2x_1x_2 \implies x_1x_2 \le \frac{1}{2}x_1^2+\frac{1}{2}x_2^2$$
for all real numbers $x_1,x_2$, we have
\begin{align}\dot V(x)&=-x_1^2+x_1x_2+(x_1x_2+x_2^2)\sin x_1-3x_2^2\\&\le
-x_1^2+\frac{1}{2}x_1^2+\frac{1}{2}x_2^2+\left(\frac{1}{2}x_1^2+\frac{1}{2}x_2^2+x_2^2\right)|\sin x_1|-3x_2^2\\&=
-\frac{1}{2}x_1^2-\frac{5}{2}x_2^2+\left(\frac{1}{2}x_1^2+\frac{3}{2}x_2^2\right)|\sin x_1|\\&\le-x_2^2\\&
\le 0
\end{align}
as $|\sin x_1|\le 1$. Global asymptotic stability follows from $\dot V(x) < 0$ when $(x_1,x_2)\neq 0$.
|
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|
Interesting thing about sum of squares of prime factors of $27$ and $16$. Let
$$n=p_1×p_2×p_3×\dots×p_r$$
where $p_i$ are prime factors and
$f$ is the functions
$$f(n)=p_1^2+p_2^2+\dots+p_r^2$$
If we put $n=27,16$ and $27=3×3×3$, $16=2×2×2×2$ then
$$\begin{split}f(27)&=3^2+3^2+3^2=27\\f(16)&=2^2+2^2+2^2+2^2=16.\end{split}$$
I checked it upto $n=10000$, I did not find another number with this property $f(n)=n$.
Can we prove that other such numbers do not exist?
Some approaching values
$f(992)=981\\f(1058)=1062\\f(1922)=1926\\f(5396)=5410\\f(7198)=7206\\f(9506)=9511$
Sequence: A067666, Sum of squares of prime factors of n (counted with multiplicity).
Edit
We can show there are infinitely many $n$ s.t. $f(n)=n+4$
Proof: put $n=2\cdot p^2$ where $p$ prime number
gives $f(2\cdot p^2)=2^2+p^2+p^2=4+2\cdot p^2$.
|
For two factors $$f(pq)=p^2+q^2\gt pq$$ so $pq$ is not a solution.
For three factors: If $3$ is a factor then $3^2+p^2+q^2$ is only a multiple of $3$ if $p=q=3$ as well. If $3$ is not a factor then $p^2=q^2=r^2=1\pmod3$, so the sum is a multiple of $3$, and $pqr$ is not a solution. So $27$ is the only solution with three factors.
For four factors, they can't all be odd as the sum would be even. Then there must be an even number of odd factors. So it is a multiple of $4$, and looking $\pmod4$, the factors are either all odd or all even. So $16$ is the only solution with exactly four factors.
For five factors, I think they must all be odd; so $n=5\pmod8$.
For six factors, two of them must be 2, three must be 3, leaving $35+p^2=108p$ which has no solution.
For eight factors, all of them must be even, but $256$ doesn't work so there is no solution.
Edit:
Good news, bad news.
Good news: $$3^2+3^2+5^2+1979^2+89011^2\\=3×3×5×1979×89011$$
Bad news: $89011$ is not prime.
My idea was that the equation is a quadratic in the final prime. The quadratic's discriminant must be a perfect square, and that is a Pellian equation in the second-last prime. If the other primes are $3,3,5$, this Pellian has solutions $$1,44,1979,89011,...$$ with $$a_{n+1}=45a_n-a_{n-1}$$
If two consecutive terms are prime, then I think $3×3×5×a_n×a_{n+1}$ is a solution to the current problem
EDIT: Let
$$\alpha=\frac12(\sqrt{47}+\sqrt{43}),\beta=\frac12(\sqrt{47}-\sqrt{43})\\
A = \frac1{\sqrt{47}}(\alpha^{107}+\beta^{107}),B=\frac1{\sqrt{47}}(\alpha^{109}+\beta^{109})$$
$A$ and $B$ are consecutive terms from the sequence in the previous edit. Maple confirms that $A$ and $B$ are prime, and $$3\times3\times5\times A
\times B=3^2+3^2+5^2+A^2+B^2$$
|
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|
Complex numbers is an algebraic extension of Reals so what real Polynomial has root x+iy If I have an arbitrary complex number $x+iy$ then how do I work out a polynomial in the ring of polynomials over $\mathbb{R}$ for which it is a root?
|
$z= (x+yi)$
$z-x = yi$
$(z-x)^2 = -y^2$
$z^2 -2xz + (x^2 + y^2)$ is a polynomial that has $x+yi$ as a root.
..... or .....
$P(z) = az^2 + bz + c$ then the root of $P$ is $\frac {-b\pm\sqrt{b^2-4ac}}{2a}$ so we let $2a=1$ need $x=-b$ and $b^2-4ac=b^2+2c= -y^2$
So let $b=-x$, $a=\frac 12$ then $c=\frac {y^2 + x^2}2$.
So $x+yi$ is root to $\frac 12z^2 -xz +\frac {y^2 + x^2}2$ or to $z^2 -2xz + (y^2 + x^2)$
|
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|
Integrate $\int \frac{1}{1+x+x^4}dx$
Find $$\int \frac{{\rm d}x}{1+x+x^4}.$$
WA gives a result, which introduces complex numbers. But according to the algebraic fundamental theorem, any rational fraction can be integrated over the real number field.
How to do this?
|
Why did you have to choose this rational function... ? Well there is a factorization like this:
$$1+x+x^4=\left(x^2-\sqrt{\frac{\sqrt[3]{\frac{1}{2} \left(9+i
\sqrt{687}\right)}}{3^{2/3}}+\frac{4}{\sqrt[3]{\frac{3}{2} \left(9+i \sqrt{687}\right)}}}
x+\frac{1}{2} \left(\frac{\sqrt[3]{\frac{1}{2} \left(9+i
\sqrt{687}\right)}}{3^{2/3}}+\frac{4}{\sqrt[3]{\frac{3}{2} \left(9+i \sqrt{687}\right)}}-4
\sqrt{\frac{\sqrt[3]{\frac{1}{2} \left(9+i \sqrt{687}\right)}}{3^{2/3}}+\frac{4}{\sqrt[3]{\frac{3}{2}
\left(9+i \sqrt{687}\right)}}}+\left(\frac{\sqrt[3]{\frac{1}{2} \left(9+i
\sqrt{687}\right)}}{3^{2/3}}+\frac{4}{\sqrt[3]{\frac{3}{2} \left(9+i
\sqrt{687}\right)}}\right)^{5/2}\right)\right) \left(x^2+\sqrt{\frac{\sqrt[3]{\frac{1}{2} \left(9+i
\sqrt{687}\right)}}{3^{2/3}}+\frac{4}{\sqrt[3]{\frac{3}{2} \left(9+i \sqrt{687}\right)}}}
x+\frac{1}{2} \left(\frac{\sqrt[3]{\frac{1}{2} \left(9+i
\sqrt{687}\right)}}{3^{2/3}}+\frac{4}{\sqrt[3]{\frac{3}{2} \left(9+i \sqrt{687}\right)}}+4
\sqrt{\frac{\sqrt[3]{\frac{1}{2} \left(9+i \sqrt{687}\right)}}{3^{2/3}}+\frac{4}{\sqrt[3]{\frac{3}{2}
\left(9+i \sqrt{687}\right)}}}-\left(\frac{\sqrt[3]{\frac{1}{2} \left(9+i
\sqrt{687}\right)}}{3^{2/3}}+\frac{4}{\sqrt[3]{\frac{3}{2} \left(9+i
\sqrt{687}\right)}}\right)^{5/2}\right)\right)$$
(you can check that, even though $i$ is used in the construction of the coefficients, they are actually all real.)
|
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|
Find $\lim _{x\to \infty }\left(x\left(\arctan2x\:-\arccos\left(\frac{1}{x}\right)\right)\right)$ Find $$\lim _{x\to \infty }\left(x\left(\arctan(2x)-\arccos\left(\frac{1}{x}\right)\right)\right)$$
My idea was to let $t=\frac{1}{x}$ then I get
$$\lim _{t\to 0 }...$$
but i did not know what to do with $\arctan(2x)$
any hint how to solve this ?
thanks
|
The parantheses goes to $0$, so using $\lim\limits_{x\to 0} \dfrac{\sin x}{x} = 1$, the limit equals:
$$\lim _{x\to \infty }\left(x\left(\arctan 2x-\arccos \frac{1}{x}\right)\right)=\lim_{x\to \infty}x\left(\sin\left(\arctan 2x-\arccos\frac{1}{x}\right)\right)$$
and using the trigonometric identities:
$$\sin(a-b)=\sin a\cos b-\sin b\cos a,\\ \ \arctan x=\arcsin \frac{x}{\sqrt{1+x^2}},\ \arccos x=\arcsin \sqrt{1-x^2}$$
the limit is:
$$
\begin{aligned}
\lim_{x\to \infty}x\sin\left(\arctan 2x -\arccos \frac{1}{x}\right)
&= \lim_{x\to \infty} x\left(\frac{2x}{\sqrt{1+x^2}}\cdot \frac{1}{x}-\frac{1}{\sqrt{1+4x^2}}\cdot \sqrt{1-\frac{1}{x^2}}\right)\\
&=\lim_{x\to \infty} \frac{2-\sqrt{1-\frac{1}{x^2}}}{\sqrt{\frac{1}{x^2}+4}}\\
&=\frac{1}{2}
\end{aligned}
$$
|
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|
Show convergence of series $\sum_{k=1}^{\infty}\frac{k!}{k^k} $ I want to prove that $\sum_{k=1}^{\infty}\frac{k!}{k^k} $ converges. My idea is that, for any integer $k \ge 1 $, we have
\begin{align*}
\frac{k!}{k^k}
&= \frac{1}{k}\cdot\frac{2}{k}\cdots\frac{k-2}{k}\cdot\frac{k-1}{k}\cdot\frac{k}{k} \\
&= \frac{1}{k}\cdot\frac{2}{k}\cdots\frac{k-2}{k}\cdot\frac{k-1}{k}\\
&\leq \frac{1}{k}\cdot\frac{2}{k}\cdots\cdot\frac{k-2}{k} \\
&\qquad \vdots \\
&\leq \frac{1}{k}\cdot\frac{2}{k}
= \frac{2}{k^2}
\end{align*}
That is
$$\sum_{k=1}^{\infty}\frac{k!}{k^k} \leq \sum_{k=1}^{\infty}\frac{2}{k^2} $$
And since right-hand side of the inequality is finite, so is left-hand side and therefore the series is convergent.
However, I dont find this way of solving the assignment elegant and I believe there is a cleaner way. Appreciates all help I can get.
|
You can apply the Ratio Test and use $\displaystyle{\lim_{k\to\infty}\left(\frac{k+1}{k}\right)^k=e}$.
|
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|
Value of $\arg(\sin \theta +i\cos \theta)$
If $\displaystyle \frac{3+i\sin \theta}{4-i\cos \theta}$ is purely real number , where $\theta \in [0,2\pi].$ Then what is $\arg(\sin \theta +i\cos \theta)$?
What I tried:
\begin{align*}
\frac{3+i\sin \theta}{4-i\cos \theta} & =\frac{(3+i\sin \theta)(4+i\cos \theta)}{(4-i\cos \theta)(4+i\cos \theta)}\\
&=\frac{12-(\sin \theta\cos \theta)+i(4\sin \theta+3\cos \theta)}{16+\cos^2 \theta}\in \mathbb{R}
\end{align*}
means $(4\sin \theta+3\cos \theta)=0$, namely $\displaystyle \tan \theta = -3/4$. So either $\theta\in (\pi/2,\pi)$ or $\theta\in(3\pi/2,2\pi)$.
Now $\arg(\sin \theta+i\cos\theta)=\arctan\left(\frac{\cos \theta}{\sin \theta}\right)=\arctan(\cot\theta)=-4/3$, but the answer given as $\displaystyle \pi-\tan^{-1}(4/3)$.
How do I solve this? Help me please.
|
So your work is good until the third from last line. You forgot the arctan.
So your solution should be $\arctan(\frac{-4}{3})$ now remember arctan is an odd function so $\arctan(\frac{-4}{3})=-\arctan(\frac{4}{3})$ which will be a negative number and you want $\theta \in [0,2 \pi]$ So we add $\pi$ to get what we want. We can do this because tangent is periodic of period $\pi$
Hope this helps.
|
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|
A geometric inequality for acute triangle $\sum \sqrt \frac{b^2+c^2-a^2}{a^2+2bc} \leq \sqrt3$ I am trying to prove $\sum \sqrt \frac{b^2+c^2-a^2}{a^2+2bc} \leq \sqrt3$ for an acute triangle. There is:
$$\sum \sqrt \frac{b^2+c^2-a^2}{a^2+2bc} \leq \sqrt{3\sum\frac{b^2+c^2-a^2}{a^2+2bc}}$$
So we have to prove: $\sum\frac{b^2+c^2-a^2}{a^2+2bc}\leq 1 $
We observe that $\frac{b^2+c^2-a^2}{a^2+2bc}=\frac{(b+c)^2}{a^2+2bc}-1$ and I can't go further
|
We can end the beautiful Atticus's idea also by the following C-S:
$$\sum_{cyc}\frac{x}{\sqrt{\frac{y^2+z^2}{2}+x^2+yz}}=\sum_{cyc}\frac{x\sqrt6}{\sqrt{(1+2)(2x^2+(y+z)^2}}\leq\sum_{cyc}\frac{x\sqrt6}{\sqrt2x+\sqrt2(y+z)}=\sqrt3.$$
|
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|
Evaluate $\sum_{r=1}^{\infty} \frac{1 \cdot 3 \cdot \ldots (2r-1)}{r!}\left(\frac{2}{5} \right)^{r}$
Evaluate $$\sum_{r=1}^{\infty} \frac{1 \cdot 3 \cdots (2r-1)}{r!}\left(\frac{2}{5} \right)^{r}$$
Let $$y=x + \frac{1 \cdot 3 \cdot}{2!} x^2 + \frac{1 \cdot 3 \cdot 5}{3!} x^3+\ldots$$ be the given expression.(replacing $2/5$ with $x$)
After some manipulations,
$$y+1=(1-2x)\frac{dy}{dx}$$
Integrating and substituting $x=\dfrac{2}{5}$, we get $y=\sqrt{5}-1$.
Is there any other way to solve this question?
|
$$S=\sum_{r=1}^{\infty} \frac{1.3.5....(2r-1)}{r!}\left(\frac{2}{5}\right)^r$$
$$\implies S=\sum_{r=1}^{\infty} \frac{(2r)!}{r!~ r!} 5^{-r} =\sum_{r=1}^{\infty} {2r \choose r} 5^{-r}=\frac{1}{\sqrt{1-4/5}}-1=\sqrt{5}-1.$$
Here we have used $$\sum_{k=0}^{\infty} {2k \choose k} x^k=\frac{1}{\sqrt{1-4x}}$$
|
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|
Range of $f(x) = \frac{\sin^2 x + \sin x - 1}{\sin^2 x - \sin x + 2} $ Here's what I did :
$$\text{Let }\quad y = \frac{\sin^2 x + \sin x - 1}{\sin^2 x - \sin x + 2} $$
$$\text{Let } \sin^2x=t$$
$$\Rightarrow\ (y-1)t^2-(y+1)t+2y+1=0 $$
$$\text{Since } t= \sin x\text{ is real,} $$
$$\text{Discriminant} \geqslant 0 $$
$$ \Rightarrow (y+1)^2-4(y-1)(2y+1) \geqslant 0$$
$$ \Rightarrow\ -7y^2+6y+5 \geqslant 0$$
$$\Rightarrow\ 7y^2-6y-5 \leqslant 0$$
$$\Rightarrow \ y\in\left[\frac{3-2\sqrt{11}}{7},\frac{3+2\sqrt{11}}{7}\right] $$
But the correct answer (according to wolfram) is
$$y \in \left[\frac{3-2\sqrt{11}}{7},\frac{1}{2}\right] $$
Please guide me how I should proceed further .
|
Note that to find the maximum value of the range you can just simply reason as follows:
Max($\sin^2(x)) = 1$ = Max($\sin(x)$). Hence the max is
$$y_{\text{max}} = \frac{1+1-1}{1-1+2} = \frac{1}{2}$$
|
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|
Hopf map $S^3 \to S^2$ stereographic projection I'm reading up on this topic and I've seen lots of literature suggest that the complex number representation of the stereographic projection of $S^2$ can be expressed as
$\frac{\zeta^1 + i\zeta^2}{1-\zeta^3} = \frac{x^1 + ix^2}{x^3 +ix^4}$
where
$\zeta^1 = 2(x^1x^3 + x^2x^4)\\
\zeta^2 = 2(x^2x^3 - x^1x^4)\\
\zeta^3 = (x^1)^2 + (x^2)^2 - (x^3)^2 - (x^4)^2$
and
$S^3: (x^1)^2 + (x^2)^2 + (x^3)^2 + (x^4)^2 = 1\\
S^2: (\zeta^1)^2 + (\zeta^2)^2 + (\zeta^3)^2 = 1.$
I understand how to get to the LHS, but the RHS doesn't seem clear to me. Any help would be much appreciated.
|
Write points of $S^3$ as things in $\Bbb C^2$, i.e., pairs $(z, w)$ where $|z|^2 + |w|^2 = 1$. Then the Hopf-map (to $\Bbb C \cup \{ \infty \}$) is defined by
$$
(z, w) \mapsto z/w
$$
Now the codomain (the extended complex plane) isn't exactly $S^2$, so you need to stereographically project a point $p + q i$ up to the unit 2-sphere. There are multiple stereographic projections from the extended complex plane to the 2-sphere, so things are a little ambiguous. But the first step isn't so bad: the point $(z, w) \in S^3$, where $z = a + bi, w = c + di$, gets sent to
\begin{align}
u + iv
&= \frac{z}{w} \\
&= \frac{a + bi}{c + di} \\
&= \frac{a + bi}{c + di}\frac{c - di}{c - di} \\
&= \frac{(a + bi)(c-di)}{c^2 + d^2}\\
&= \frac{(ac + bd) + (bc -ad)i}{c^2 + d^2}\\
&= \frac{(ac + bd)}{c^2 + d^2} + \frac{(bc -ad)i)}{c^2 + d^2}
\end{align}
so that $u = \frac{(ac + bd)}{c^2 + d^2}$ and $v = \frac{(bc -ad)}{c^2 + d^2}$. (You can see that the numerators of these things look a lot like $\zeta^1$ and $\zeta^2$!)
The point $u + iv$ is sent, by stereographic projection from the south pole (say) to the point on the line from
$$
(0,0,-1) \to (u,v,0)
$$
whose squared coordinates sum to $1$, i.e., the point
$$(p,q,r) = (1-t)(0,0,-1) + t(u, v, 0).$$
Here I'm parameterizing the line through those two points as
$$
L(t) = (1-t) (0, 0, -1) + t (u, v, 0) = (1-t)A + t B
$$
which is an expression linear in $t$, so if we look at
$$
H = \{ L(t) \mid t \in \Bbb R \}
$$
we get a straight line in $3$-space. When $t = 0$, we're at $A$ (i.e., the south pole); when $t = 1$, we get $B$, i.e., the point $(u,v, 0)$, so $H$ really is the line we want.
For other values of $t$, we get points on the segment between $A$ and $B$ (if $0 < t < 1$) or outside of the segment (for other $t$ values).
When we ask "For what value of $t$ is $L(t)$ on the unit sphere?", we get a quadratic in $t$ (as you'll see below); one solution to this quadratic is $t = 0$, corresponding to the point $(0,0,-1)$, the south pole. The nonzero root of this quadratic will be the one that interests us in general. When $u$ and $v$ are both $\infty$, the quadratic degenerates into a linear expression with its only root at $t = 0$, i.e., the "point at infinity" maps to the south pole of $S^2$.
with the property that $p^2 + q^2 + r^2 = 1$, i.e.,
\begin{align}
1 &= t^2 u^2 + t^2 v^2 + (1-t)^2\\
1 &= t^2 u^2 + t^2 v^2 + 1-2t + t^2\\
0 &= t^2 u^2 + t^2 v^2 + -2t + t^2\\
0 &= t^2 (u^2 + v^2 + 1) + -2t\\
2t &= t^2 (u^2 + v^2 + 1) \\
2 &= t (u^2 + v^2 + 1) \\
t &= \frac{u^2 + v^2 + 1}{2}
\end{align}
From this, you can compute $(p,q,r)$, and I'll bet that they look a whole lot like $\zeta^1, \zeta^2, \zeta^3$.
|
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|
Use Taylor series to compute $\lim\limits_{x \to 0} \left( \frac{1}{\sin(x)}- \frac{1}{x} \right) $ Use Taylor series to solve $$\lim\limits_{x \to 0} \left( \frac{1}{\sin(x)}- \frac{1}{x} \right) $$
This equals $$\lim\limits_{x \to 0} \left( \sum_{n=0}^{\infty}\frac{(2n+1)!}{x^{2n+1}(-1)^{n+1}} - \sum_{n=0}^{\infty}\frac{1}{(-x+1)^n} \right)$$
I am unsure how to proceed. I tried writing out the first few terms, but nothing seemed to cancel.
|
I'm not quite sure how you got your formula. Using just the first $2$ terms of the Taylor series for $\sin(x)$ (you actually only need just $1$ term, as shown in some of the other answers, but I thought using $2$ may help to better see what is going on), such as given in Trigonometric functions, and as suggested in copper.hat's question comment, you have
$$\begin{equation}\begin{aligned}
\lim_{x\to 0}\left(\frac{1}{\sin(x)} - \frac{1}{x}\right) & = \lim_{x\to 0}\left(\frac{x - \sin(x)}{x\sin(x)}\right) \\
& = \lim_{x\to 0}\left(\frac{x - \left(x - \frac{x^3}{6} + O(x^5)\right)}{x\left(x - \frac{x^3}{6} + O(x^5)\right)}\right) \\
& = \lim_{x\to 0}\left(\frac{\frac{x^3}{6} - O(x^5)}{x^2 - \frac{x^4}{6} + O(x^6)}\right) \\
& = \lim_{x\to 0}\left(\frac{\frac{x}{6} - O(x^3)}{1 - \frac{x^2}{6} + O(x^4)}\right) \\
& = \frac{0}{1} \\
& = 0
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
|
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|
Solving for $x$ in a (modified) quadratic I've come across this question which can be simplified (with a generous amount of manipulation) to a quadratic. Here it is:
$$\left(\sqrt{49+20\sqrt6}\right)^{\sqrt {a \sqrt{a\sqrt{a\cdots \infty}}}}+(5-2\sqrt{6})^{x^2+x-3-\sqrt{x\sqrt{x\sqrt{x}\cdots \infty}}}=10 $$
$ \text{ where } a=x^2-3,(a\neq 0), \text{ Solve for }$ $x$.
Here's what I tried:
I first tried to simplify the powers in the given expression
$$y={\sqrt {a \sqrt{a\sqrt{a\cdots \infty}}}} \\
\Rightarrow y=\sqrt{ay} \\
\Rightarrow y=a$$
Similarly, the second power reduces to $x^2-3$ (equal to $a$). I've also condensed the first term in the expression. So our equation simplifies to:
$$(5+2\sqrt{6})^{x^2-3} +(5-2\sqrt{6})^{x^2-3}=10$$
Since $5+2\sqrt{6} =\frac{1}{5-2\sqrt{6}}$, $(5+2\sqrt{6})^{x^2-3}$ can be taken as $t$ to furthur simplify the equation into:
$$t^2-10t+1=0$$
Solving for $t$, we get:
$$ t=5 \pm2\sqrt{6}$$
plugging back the value of $t$, we get:
$$(5+2\sqrt{6})^{x^2-3}=(5+2\sqrt{6}) \text{ or } (5+2\sqrt{6})^{x^2-3}=(5+2\sqrt{6})^{-1}$$
$\Rightarrow$ $x^2-3=\pm 1$ . Solving this gives the values of $x$ to be: $\pm 2 , \pm \sqrt{2}$. $x$ cannot take the negative values as it will not satisfy the domain of the original question.
Hence the possible values of $x$ can be either $2$ or $\sqrt{2}$. But the answer tells me $x$ can only be $2$.
Where have I gone wrong or what have I overlooked?
|
There is nothing wrong but if $x=\pm\sqrt 2$ then $a=-1<0$ so $\sqrt{a\sqrt{a\dots}}$ is not well-defined. Hence, $x=\pm\sqrt 2$ must be "disqualified" as a solution.
Usually it is always good to check all solutions you got with the original equation since we usually only prove the direction that if $x$ satisfies then equation then $x$ is equal to this and this, but we actually want to prove the other direction as well.
|
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|
Evaluate $\int \frac{x^2}{(7x^3+2)^2} dx$ using u substitution Section 5.2
Can somebody verify my solution? Thanks!!
Evaluate $\int \frac{x^2}{(7x^3+2)^2} dx$ using u substitution
Let $u=7x^3+2$. Then $\frac{du}{dx}=21x^2$ and so $\frac{du}{21x^2}=dx$.
Thus we have:
\begin{align}
& \int \frac{x^2}{(7x^3+2)^2} \, dx \\[8pt]
= {} & \int \frac{x^2}{u^2} \frac{du}{21x^2}\,dx \\[8pt]
= {} & \frac{1}{21} \int \frac{1}{u^2} \, du \\[8pt]
= {} & \frac{1}{21} \int u^{-2} \, du \\[8pt]
= {} & \frac{1}{21} \frac{u^{-1}}{-1}+C \\[8pt]
= {} & \frac{-1}{21} u^{-1}+C \\[8pt]
= {} & \frac{-1}{21} (7x^3+2)^{-1}+C \\[8pt]
= {} & \frac{-1}{21(7x^3+2)}+C
\end{align}
|
I worked it out and I got the same answer you did.
One thing to point out: When you have $$\dfrac {du}{dx} = 21 x^2$$ you can write the subsitution as either $$\dfrac {du}{21 dx} = x^2$$ if you prefer to use $\dfrac {du}{dx}$ notation or $$\dfrac {1}{21}du = x^2 dx$$ if you prefer differential notation.
|
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|
Line Integral with Change of Variables I'm a bit rusty on my computational math, and genuinely can't solve this question which is frustrating me.
QUESTION:
$$\int_Csin(y)dx+xcos(y)dy$$ where C is the ellipse defined as follows: $x^2+xy+y^2=1$
MY ATTEMPT:
Define variable $u= \sqrt{\frac{3}{4}}x+\sqrt{\frac{3}{4}y}$
Define variable $v= \sqrt{\frac{1}{4}}x-\sqrt{\frac{1}{4}}y$
Hence:
$u^2+v^2=(\sqrt{\frac{3}{4}}x+\sqrt{\frac{3}{4}y})^2+(\sqrt{\frac{1}{4}}x-\sqrt{\frac{1}{4}}y)^2$
$=(\frac{3}{4}x^2+\frac{(2)(3)}{4}xy+\frac{3}{4}y^2)+(\frac{1}{4}x^2-\frac{2}{4}xy+\frac{1}{4}y^2)$
$=(\frac{3}{4}+\frac{3}{4})x^2+(\frac{3}{2}-\frac{1}{2})xy+(\frac{3}{4}+\frac{1}{4})y^2$
$=(x^2+xy+y^2) = C$
Therefore, we have Jacobian:
$$
\begin{bmatrix}
\frac{\partial{u}}{\partial{x}} & \frac{\partial{u}}{\partial{y}} \\
\frac{\partial{v}}{\partial{x}} & \frac{\partial{v}}{\partial{y}} \\
\end{bmatrix}
$$
$$
\begin{bmatrix}
\sqrt\frac{3}{4} & \sqrt\frac{3}{4} \\
\sqrt\frac{1}{4} & -\sqrt\frac{1}{4} \\
\end{bmatrix}
$$
$$ = \frac{\sqrt{3}}{2} $$
Beyond this point, I must incorporate the original dot product with the differential $(dx,dy)$ but don't know how
|
Hint: Use Green's theorem instead to ease out calculations:
$\int_C Pdx+Qdy=\int\int_R (\partial Q/\partial x-\partial P/\partial y) dx dy$
where $R$ is the region bounded by closed curve $C$ on $xy-$ plane oriented counterclockwise.
|
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|
Roots of the equation $(1-4x)^4+32x^4=\frac{1}{27}$ find all real roots of the equation $(1-4x)^4+32x^4=\dfrac{1}{27}$
i try to use binomial expansion and am-gm inequality but i don't know how to do next.
|
Clearly there are no solutions if $ x > \frac{1}{4}$ or $ x < 0$. So, we may apply the power mean inequality to $ 1-4x, 2x, 2x$.
$$\sqrt[4]{\frac{( 1- 4x)^ 4 + (2x)^4 + (2x)^4 }{3}}\geq \frac{ (1-4x) + 2x + 2x } { 3} = \frac{1}{3} $$
This tells us that $( 1- 4x)^ 4 + 32x^4 \geq \frac{1}{27}$.
Equality holds iff $ 1 -4x = 2x = 2x$, or that $ x = \frac{1}{6}$.
|
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|
Compute: $\lim_{n\to\infty}\sum_{i=1}^{2n^4+n^2}\frac{5n^2+1}{n^4+i}$ The question is to find $\lim_{n\to\infty} \left ( \frac{5n^2+1}{n^4+1}+\frac{5n^2+1}{n^4+2}+\frac{5n^2+1}{n^4+3}+...\frac{5n^2+1}{3n^4+n^2} \right )$
The answer is 5, according to the booklet.
I tried to bring the sum to the form $ \sum \frac{1}{n}f(\frac{k}{n})$ in order to use Rieman sum. Also I tried to apply Cauchy's first theorem on limits. Didn't get me closer to something helpful.
|
Let $a_n=\frac{5n^2+1}{n^4+1}+\frac{5n^2+1}{n^4+2}+\frac{5n^2+1}{n^4+3}+...+\frac{5n^2+1}{n^4+n^2}$, then notice
$$
a_n \geq n^2\cdot \frac{5n^2+1}{n^4+n^2}=\frac{5+\frac{1}{n^2}}{1+\frac{1}{n^2}}
$$
and similarly
$$
a_n \leq n^2 \cdot \frac{5n^2+1}{n^4+1} = \frac{5+\frac{1}{n^2}}{1+\frac{1}{n^4}}.
$$
Now just use the squeeze theorem.
|
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Diophantine equation from "Solving mathematical problems" by Terence Tao
Find all integers $n$ such that the equation $\frac{1}{a} + \frac{1}{b} = \frac{n}{a+b}$ is satisfied for some non-zero values of $a$ and $b$ (with $a + b \neq 0$).
I'm reading "Solving mathematical problems" by Terence Tao and I'm a bit stuck on this particular example. It starts as follows:
First multiplying out the denominators we get
$$\frac{a+b}{ab} = \frac{n}{a+b}$$
and from here follows $$(a+b)^2 = nab.$$
Now expanding this we get
$$a^2+2ab+b^2-nab=0 \Leftrightarrow a²+ab(2-n)+b^2.$$
From here on he suggests to use the quadratic formula to get
$$a= \frac{b}{2}[(n-2) \pm \sqrt{(n-2)^2-4}]$$
which I don't quite see how he came up with...
Also he then notes that "This looks very messy, but actually we can turn this messiness to our advantage. We know that $a, b$, and $n$ are integers, but there is a square root in the formula. Now this can only work if the term inside the square root, $(n-2)^2-4$ is a perfect square."
Could someone enlighten me on the part "Now this can only work if the term inside the square root, $(n-2)^2-4$ is a perfect square." what is he stating right here?
|
Regarding your first equation, with the formula
$$a^2+ab(2-n)+b^2 \tag{1}\label{eq1A}$$
consider that $n$ and $b$ are constants, with only $a$ being a variable. In that case, it's a quadratic polynomial in $a$, of the form
$$a^2 + ca + d \tag{2}\label{eq2A}$$
where $c = b(2-n)$ and $d = b^2$. Thus, using the quadratic formula gives
$$\begin{equation}\begin{aligned}
a & = \frac{-c \pm \sqrt{c^2 - 4d}}{2} \\
& = \frac{-b(2-n) \pm \sqrt{(-b(2-n))^2 - 4(b^2)}}{2} \\
& = \frac{b}{2}\left((n - 2) \pm \sqrt{(n-2)^2 - 4}\right)
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Next, note that since $n$ is an integer, if $(n-2)^2 - 4$ is not a perfect square, then $\sqrt{(n-2)^2 - 4}$ would be irrational, so that $a$ determined from \eqref{eq3A} would also be irrational and, thus, not an integer as required. This is why it's required for $(n-2)^2 - 4$ to be a perfect square.
|
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|
How do I get from $x^4+2x^3y+3x^2y^2+2xy^3+y^4$ to $(x^2+xy+y^2)^2$? I was doing an example $$(x+y)^4+x^4+y^4$$ and I need to factor it. I've tried and couldn't really do much, so I checked if there was anything to help, and I came across a post asking about the same thing. But my question is how do I know that $$x^4+2x^3y+3x^2y^2+2xy^3+y^4=(x^2+xy+y^2)^2$$ without knowing the answer.
I know about $$(a+b+c)^2$$ but then how would i chose my $a, b$ and $c$?
There are many $xy$ combinations here, with different powers, so would I chose the lowest as my $b$, and $x^2$ as my $a$ and $y^2$ as my $c$, or perhaps is there another formula that can help me.
|
The given quartic form can be written as follows
$$q (x,y) := x^4+2x^3y+3x^2y^2+2xy^3+y^4 = \begin{bmatrix} x^2\\ x y\\ y^2\end{bmatrix}^\top \begin{bmatrix} 1 & 1 & t\\ 1 & 3-2t & 1\\ t & 1 & 1\end{bmatrix} \begin{bmatrix} x^2\\ x y\\ y^2\end{bmatrix}$$
Choosing $t = 1$, we obtain a symmetric, positive semidefinite, rank-$1$ matrix, namely, ${\bf 1}_3 {\bf 1}_3^\top$. Hence, we obtain the following (single term) sum of squares (SOS) decomposition
$$q (x,y) = \begin{bmatrix} x^2\\ x y\\ y^2\end{bmatrix}^\top \begin{bmatrix} 1\\ 1\\ 1\end{bmatrix} \begin{bmatrix} 1\\ 1\\ 1\end{bmatrix}^\top \begin{bmatrix} x^2\\ x y\\ y^2\end{bmatrix} = \color{blue}{\left( x^2 + x y + y^2 \right)^2}$$
factoring polynomials matrices rank-1-matrices
|
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|
How to find the common factor I am currently stomped at this problem $(2 x^2-(y+z) (y+z-x))/(2 y^2-(z+x) (z+x-y))$ I am supposed to factor it and I can't find a way how to get to the answer.
It says that
$(2 x^2+x y-y^2+x z-2 y z-z^2)/(2 y^2-x^2+x y-2 x z+y z-z^2)$ has a common factor of x+y+z. Can anyone elaborate on how that is possible? What am I not seeing...
|
From the numerator, you can pick up a $2x-y-z$, obtaining:
$$ x^2+x y-y^2+x z-2 y z-z^2=(2x-y-z)\cdot(x+y+z)$$
From the denominator, you can pick up $2y-x-z$, arriving at:
$$2 y^2-x^2+x y-2 x z+y z-z^2=(2y-x-z)\cdot(x+y+z)$$
Now, you can rewrite your fraction as:
$$\frac{(2 x^2+x y-y^2+x z-2 y z-z^2)}{(2 y^2-x^2+x y-2 x z+y z-z^2)}=\frac{(2x-y-z)\cdot(x+y+z)}{(2y-x-z)\cdot(x+y+z)}$$
If $x+y+z\neq 0$, you can cancel out the common factor $x+y+z$ and arrive at:
$$\frac{(2x-y-z)}{(2y-x-z)}$$
This fraction can't further be simplified.
|
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|
Integral involving elliptic integral functions Recently I came across this identity:
$$\int _0^1\:\frac{K\left(x\right)}{1+x}dx=\frac{\pi ^2}{8}$$
Where:
$$K\left(x\right)=\int _0^{\frac{\pi }{2}}\:\frac{1}{\sqrt{1-\left(x\sin \left(\theta \right)\right)^2}}d\theta $$
Any hints to how to prove this identity?
|
Well, we are trying to find the following integral:
$$\mathcal{I}:=\int_0^1\frac{1}{1+x}\cdot\left(\int_0^\frac{\pi}{2}\frac{1}{\sqrt{1-\left(x\sin\left(\theta\right)\right)^2}}\space\text{d}\theta\right)\space\text{d}x\tag1$$
Now, the inner integral is known as the complete elliptic integral of the first kind. And can be written as:
$$\int_0^\frac{\pi}{2}\frac{1}{\sqrt{1-\left(x\sin\left(\theta\right)\right)^2}}\space\text{d}\theta=\frac{\pi}{2}\sum_{\text{n}=0}^\infty\left(\frac{\left(2\text{n}\right)!}{2^{2\text{n}}\cdot\left(\text{n}!\right)^2}\right)^2\cdot x^{2\text{n}}\tag2$$
So, we can write:
$$\mathcal{I}=\int_0^1\frac{1}{1+x}\cdot\left(\frac{\pi}{2}\sum_{\text{n}=0}^\infty\left(\frac{\left(2\text{n}\right)!}{2^{2\text{n}}\cdot\left(\text{n}!\right)^2}\right)^2\cdot x^{2\text{n}}\right)\space\text{d}x=$$
$$\frac{\pi}{2}\sum_{\text{n}=0}^\infty\left(\frac{\left(2\text{n}\right)!}{2^{2\text{n}}\cdot\left(\text{n}!\right)^2}\right)^2\int_0^1\frac{x^{2\text{n}}}{1+x}\space\text{d}x\tag3$$
Now, we need to look at using long division on the integrand. Which gives (when $\text{n}\in\mathbb{N}^+$):
$$\frac{x^{2\text{n}}}{1+x}=\frac{1}{1+x}-1+\sum_{\text{k}=1}^\text{n}x^{2\text{k}-1}-\sum_{\text{p}=1}^{\text{n}-1}x^{2\text{p}}\tag4$$
When $\text{n}=0$, we get for $(3)$:
$$\left(\frac{\left(2\cdot0\right)!}{2^{2\cdot0}\cdot\left(0!\right)^2}\right)^2\int_0^1\frac{x^{2\cdot0}}{1+x}\space\text{d}x=\ln\left(2\right)\tag5$$
And it is not hard to prove that:
$$\text{P}_\text{n}:=\int_0^1\frac{x^{2\text{n}}}{1+x}\space\text{d}x=$$
$$\int_0^1\left(\frac{1}{1+x}-1+\sum_{\text{k}=1}^\text{n}x^{2\text{k}-1}-\sum_{\text{p}=1}^{\text{n}-1}x^{2\text{p}}\right)\space\text{d}x=$$
$$\ln\left(2\right)-1+\sum_{\text{k}=1}^\text{n}\frac{1}{2\text{k}}-\sum_{\text{p}=1}^{\text{n}-1}\frac{1}{1+2\text{p}}\tag6$$
So, we can write:
$$\mathcal{I}=\frac{\pi}{2}\left\{\ln\left(2\right)+\sum_{\text{n}=1}^\infty\left(\frac{\left(2\text{n}\right)!}{2^{2\text{n}}\cdot\left(\text{n}!\right)^2}\right)^2\text{P}_\text{n}\right\}\tag7$$
|
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|
If $x$ and $y$ are rational numbers such that : $(x+y)+(x-2y)\sqrt {2}=2x-y+(x-y-1)\sqrt {6}$
If $x$ and $y$ are rational numbers such that : $$(x+y)+(x-2y)\sqrt {2}=2x-y+(x-y-1)\sqrt {6}$$ then which of the following is true?
A) $\;x=1$, $y=1$
B) $\;x=2$, $y=1$
C) $\;x=5$, $y=1$
D) $\;x$ and $y$ can take infinitely many values.
Given:
$$(x+y)+(x-2y)\sqrt {2} = (2x-y)+(x-y-1)\sqrt {6}$$
$$x+y-2x+y+(x-2y) \sqrt {2}=(x-y-1)\sqrt {6}$$
$$(x-2y)\sqrt {2} - (x-2y)=(x-y-1)\sqrt {6}$$
$$(x-2y)(\sqrt {2} -1)=(x-y-1)\sqrt {6}$$
Only the values at option B satisfies the equation. I just did that by hit and trial. Is there any elaboration on that?
|
Since
$\sqrt{2}$
and
$\sqrt{6}$
are linearly independent, their coefficients must be zero.
Therefore
$0 = x-2y = x-y-1$
so $x=2, y=1$.
They threw in
$x+y=2x-y$
for the fun of it.
|
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|
if $x_{m}$be prime number, show that $2m+1$ must prime number! give the integer $a>1$, and define
$$x_{0}=1,x_{1}=4a+1,x_{n+1}=(4a+2)x_{n}-x_{n-1},n\ge 1$$
if $x_{m}$be prime number, show that $2m+1$ must prime number!
this problem seem interesting,Now I post follow my try :
since Characteristic equation
$$r^2-2(2a+1)r=1\Longrightarrow r_{1,2}=2a+1+\sqrt{4a^2+4a},2a+1-\sqrt{4a^2+4a}$$
so we have
$$x_{n}=A(r_{1})^n+B(r_{2})^n$$
where
$$\begin{cases}
A+B=1\\
(2a+1)(A+B)+\sqrt{4a^2+4a}(A-B)=4a+1
\end{cases}$$
so we have
$$A=\dfrac{\sqrt{a+1}+\sqrt{a}}{2\sqrt{a+1}},B=\dfrac{\sqrt{a+1}-\sqrt{a}}{2\sqrt{a+1}}$$
Note $$r_{1}=(\sqrt{a}+\sqrt{a+1})^2,r_{2}=(\sqrt{a+1}-\sqrt{a})^2$$
Let $$\alpha=\sqrt{a}+\sqrt{a+1},\beta=\sqrt{a+1}-\sqrt{a},\alpha\beta=1$$
so we have
$$x_{n}=\dfrac{\alpha^{2n+1}+\beta^{2n+1}}{\alpha+\beta}$$
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As far as I can tell, everything you've shown is correct. As such, as you stated near the end of your question text, with
$$\alpha = \sqrt{a + 1} + \sqrt{a}, \; \; \beta = \sqrt{a + 1} - \sqrt{a}, \; \; \alpha\beta = 1 \tag{1}\label{eq1A}$$
you get
$$x_{n} = \frac{\alpha^{2n + 1} + \beta^{2n + 1}}{\alpha + \beta} \tag{2}\label{eq2A}$$
First, note for any integer $k \ge 0$ that
$$\begin{equation}\begin{aligned}
\alpha^{2k} + \beta^{2k} & = (\sqrt{a + 1} + \sqrt{a})^{2k} + (\sqrt{a + 1} - \sqrt{a})^{2k} \\
& = \sum_{i=0}^{2k}\binom{2k}{i}(\sqrt{a + 1})^{2k-i}(\sqrt{a})^{i} + \sum_{i=0}^{2k}\binom{2k}{i}(\sqrt{a + 1})^{2k-i}(-\sqrt{a})^{i} \\
& = \sum_{i=0}^{2k}\binom{2k}{i}\left((\sqrt{a + 1})^{2k-i}(\sqrt{a})^{i} + (-1)^i(\sqrt{a + 1})^{2k-i}(\sqrt{a})^{i}\right) \\
& = \sum_{i=0}^{2k}\binom{2k}{i}(\sqrt{a + 1})^{2k-i}(\sqrt{a})^{i}(1 + (-1)^i) \\
& = 2\sum_{i=0}^{k}\binom{2k}{2i}(\sqrt{a + 1})^{2k-2i}(\sqrt{a})^{2i} \\
& = 2\sum_{i=0}^{k}\binom{2k}{2i}(a + 1)^{k-i}(a)^{i}
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
The second last line comes from all of the terms with odd $i$ canceling as $1 + (-1)^{i} = 1 - 1 = 0$ and the even terms having a factor of $1 + (-1)^{i} = 1 + 1 = 2$. Also, since $a$ is an integer, this shows $\alpha^{2k} + \beta^{2k}$ is an integer.
Next, consider that $x_n$ in \eqref{eq2A} is a prime, but $2n + 1$ is not a prime. Since $x_0 = 1$ is not a prime, this means $n \ge 1$, so $2n + 1 \gt 1$ and, thus, must be composite. This means there are integers $q \gt 1$ and $r \gt 1$ where
$$2n + 1 = qr \tag{4}\label{eq4A}$$
Also, since $2n + 1$ is odd, this means $q$ and $r$ are also odd, so $q \ge 3$ and $r \ge 3$.
Note for all odd positive integers $s$, we have
$$\begin{equation}\begin{aligned}
x^s + y^s & = (x + y)(x^{s-1} - x^{s-2}y + \ldots - x(y^{s-2}) + y^{s-1}) \\
& = (x + y)\left(\sum_{i=0}^{s-1}(-1)^{i}x^{s-1-i}y^{i}\right)
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
Using \eqref{eq4A} and \eqref{eq5A}, then \eqref{eq2A} becomes
$$\begin{equation}\begin{aligned}
x_n & = \frac{\alpha^{qr} + \beta^{qr}}{\alpha + \beta} \\
& = \frac{(\alpha^{q})^{r} + (\beta^{q})^{r}}{\alpha + \beta} \\
& = \frac{(\alpha^{q} + \beta^{q})\left(\sum_{i=0}^{r-1}(-1)^{i}(\alpha^{q})^{r-1-i}(\beta^{q})^{i}\right)}{\alpha + \beta} \\
& = \left(\frac{\alpha^{q} + \beta^{q}}{\alpha + \beta}\right)\left(\sum_{i=0}^{r-1}(-1)^{i}(\alpha^{q})^{r-1-i}(\beta^{q})^{i}\right)
\end{aligned}\end{equation}\tag{6}\label{eq6A}$$
With the first factor, letting $q = 2m + 1$, we can see from \eqref{eq2A} it's $x_m$ and, thus, an integer. With the second factor, note the first & last terms are
$$(\alpha^{q})^{r-1} + (\beta^{q})^{r-1} \tag{7}\label{eq7A}$$
Since $r$ is odd, then $r - 1$ is even so by \eqref{eq3A} the value in \eqref{eq7A} is an integer. Next, consider the second term & the second last term of that factor,
$$\begin{equation}\begin{aligned}
-(\alpha^{q})^{r-2}(\beta^{q}) - (\alpha^{q})(\beta^{q})^{r-2} & = -(\alpha\beta)^{q}((\alpha^{q})^{r-3} + (\beta^{q})^{r-3}) \\
& = -((\alpha^{q})^{r-3} + (\beta^{q})^{r-3})
\end{aligned}\end{equation}\tag{8}\label{eq8A}$$
where $\alpha\beta = 1$ from \eqref{eq1A} was used. Once again, $r - 3$ is even so the value in \eqref{eq8A} is an integer.
You can repeat this pairing of terms from the start and end of the second factor in \eqref{eq6A} until you get to the middle term of
$$\begin{equation}\begin{aligned}
(-1)^{\frac{r-1}{2}}(\alpha^{q})^{\frac{r-1}{2}}(\beta^{q})^{\frac{r-1}{2}} & = (-1)^{\frac{r-1}{2}}\left(\alpha\beta\right)^{\frac{q(r-1)}{2}} \\
& = (-1)^{\frac{q(r-1)}{2}}
\end{aligned}\end{equation}\tag{10}\label{eq10A}$$
This is also, of course, an integer. Thus, the second factor of \eqref{eq6A} is a sum of integers so it, too, is also an integer. In addition, you can easily show both these factors in \eqref{eq6A} are $\gt 1$. As such, this shows that $x_n$ is a product of $2$ integer factors each $\gt 1$ and, thus, it cannot be prime. However, since it was stated it was a prime, this is a contradiction. This means the assumption of $2n + 1$ being composite must be false, thus proving $2n + 1$ is actually a prime if $x_n$ is a prime.
|
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"timestamp": "2023-03-29T00:00:00",
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|
What are the asymptotics of the solutions to $a_n\ ^{2} = (1 - a_n)a_{n-1} \ ^2$ as $n \to \infty$? In
Prove $a_t \leq \frac{c}{t}$
I showed that
the solutions to
$a_0 \in (0, 1]$,
$a_n^2 = (1 - a_n)a_{n-1}^2$
satisfy
$a_n \le \dfrac{2}{n}$.
Numerical experiments
seem to show that
$na_n \to 2$
for any initial
$a_0 \in (0, 1]$.
However,
I have not been able to prove this.
These experiments also seem to show that
$n(2-na_n)$
does not seem to approach a limit,
although it does seem to increase slowly,
and this may be due to
the limited precision
(double precision)
used.
So,
my questions are
(1) does $na_n \to 2$?
(I feel that the
answer to this is
certainly yes)
(2) Does
$\lim_{n \to \infty} n(2-na_n)$
exist?
(My feeling is probably)
(3) is there an expansion
$a_n
=\sum_{k=1}^m \dfrac{c_k}{n^k}
$
where
$c_1 = 2$?
This may help:
My proof that
$a_n \le \dfrac{2}{n}
$
was by induction
based on
rewriting the iteration as
$a_n
=\dfrac{2}{1+\sqrt{1+4/a_{n-1}^2}}
$.
|
I will answer (1) and (2) and try to explain a potential strategy to get the finer results of (3).
First of all let $$f(x) = \frac{2}{1 + \sqrt{1 + \frac{4}{x^2}}},$$ then the sequence $(a_n)$ is defined by $$a_0 > 0, \forall n \geq 0, a_{n+1} = f(a_n).$$
Now you have already shown that $\lim_{n \rightarrow \infty} a_n = 0$, to get finer results we study $f$ at $0$.
Basic computations give :
$$f(x) = x - \frac{1}{2}x^2 - \frac{1}{8}x^3 + o_0(x^3)$$
Then set $b_n = \frac{1}{a_{n+1}} - \frac{1}{a_n}$, using our approximation we have :
\begin{align}
\frac{1}{a_{n+1}} &= \frac{1}{a_n - \frac{1}{2}a_n^2 + o(a_n^2)} \\
&= \frac{1}{a_n}(1 + \frac{1}{2}a_n + o(a_n)) \\
&= \frac{1}{a_n} + \frac{1}{2} + o(1).
\end{align}
Thus $b_n \rightarrow 2$, and the series $\sum_n b_n$ diverges, using some comparison theorems for divergent series we have :
\begin{align}
\sum_{k = 1}^{n} b_k &\sim \frac{n}{2} \\
\frac{1}{a_{n+1}} - \frac{1}{a_0} &\sim \frac{n}{2}.
\end{align}
And finally $a_n \sim \frac{2}{n}$ answering (1). Now question (2) is about a second order term, to do this we redo the same computation using the third order approximation of $f$ at $0$. We have :
\begin{align}
\frac{1}{a_{n+1}} &= \frac{1}{a_n}\frac{1}{1 - \frac{1}{2}a_n - \frac{1}{8}a_n^2 + o(a_n^2)} \\
&= \frac{1}{a_n} + \frac{1}{2} + \frac{3}{8}a_n + o(a_n).
\end{align}
Now the estimate $a_n \sim \frac{2}{n}$ yields $b_n - \frac{1}{2} \sim \frac{3}{8}a_n \sim \frac{3}{4n}$. Using some comparison theorem for divergent series (as before) we get
$$ \frac{1}{a_n} - \frac{n}{2} \sim \frac{3}{4}\ln(n). $$
Finally we need only to inverse this relation, it translates as :
\begin{align}
\frac{1}{a_n} - \frac{n}{2} &= \frac{3}{4}\ln(n) + o(\ln(n)) \\
a_n &= \frac{1}{\frac{n}{2} + \frac{3}{4}\ln(n) + o(\ln(n))} \\
&= \frac{2}{n} - 3\frac{\ln(n)}{n} + o(\frac{\ln(n)}{n}).
\end{align}
So $n(2 - na_n) \sim 3\ln(n)$ and the limit does not exists. If you want to compute the higher order terms you need to compute the higher order terms of $f$ at $0$ and plug them into the equation, but instead of a theorem for divergent series you will probably need to use the version for convergent series.
The general strategy to determine an approximation of a sequence defined as $u_{n+1} = f(u_n)$, with $f(x) = x - cx^{\alpha} + o_0(x^{\alpha})$ for some $c, \alpha > 0$ is to find a suitable $\gamma$ such that $u_{n+1}^{\gamma} - u_n^{\gamma}$ converges to some nonzero real and then use comparison theorem (or Cesaro's theorem also works in this situation) for convergent/divergent series.
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"timestamp": "2023-03-29T00:00:00",
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|
For $P(x)$ a degree-$n$ polynomial such that $P(k)=\frac{k}{k+1}$ for $k=0,1,2,\ldots,n$, find $P(n+1)$. The problem, from USAMO $1975$, reads:
If $P(x)$ denotes a polynomial of degree $n$ such that $P(k) = \frac{k}{k+1}$ for $k = 0,1,2, \ldots, n$. Find $P(n+1)$.
My attempt:
We can represent $p(x)$ as another polynomial which would satisfy all the given conditions. This polynomial would be:
$P(x)$ = $x(x-1)(x-2).....(x-n) + \frac{x}{x+1}$
In other words,
$\frac{P(x)(x+1)-x}{(x+1)}$ = $x(x-1)(x-2)....(x-n)$, or:
${P(x)(x + 1)-(x)}$ = $x(x^2$ $- 1)(x - 2)......(x-n)$
now, simply putting $x = n + 1$, we have
${P(n+1)(n+2)-(n+1)} = \frac{(n + 1)(n^2+ 2n)(n - 1)!}{2}$ (after some simplification)
thus, our final expression would look something like:
$P(n + 1) = \frac{(n + 1)(n^2+ 2n)(n - 1)!+2(n+1)}{2(n + 2)}$
which is not the final answer.
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So $$Q(x) = (x+1)P(x)-x$$ is of degree $n+1$ and has zeroes $0,1,2,...,n$, thus we can write it $$Q(x) = ax(x-1)(x-2)\cdots (x-n)$$ so $$(x+1)P(x) =ax(x-1)(x-2)\cdots (x-n)+x$$ which is valid for all $x$ and in particulary for $x=-1$: $$0 =a(-1)^{n+1}(n+1)!-1$$ and for $x=n+1$ we get: $$(n+2)P(n+1) =a(n+1)!+n+1$$ so $$P(n+1)= {(-1)^{n+1} +n+1\over n+2}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the triangle with the maximum area with a given perimeter
Question: Of all triangles with a given perimeter, find the triangle with the maximum area. Justify your answer.
My approach: Let us have any $\Delta ABC$ such that the sides opposite to $A$ is of length $a$, the side opposite to $B$ is of length b and the side opposite to $C$ is of length $c$. Now since the perimeter to $\Delta ABC$ is fixed, thus we must have $P=a+b+c$ to be constant.
Now fix any side of the triangle, let us fix $BC$. Therefore $b+c=P-a$, which implies that $b+c$ is constant. Thus the locus of the point $A$ must be an ellipse having one of it's axis as the side $BC$. Now let us select any point $A$ on the ellipse and drop a perpendicular to the axis $BC$. Let it meet the major axis at $P$. Now let $AP=h$. Therefore, the area of the $\Delta ABC=\frac{1}{2}.h.BC=\frac{1}{2}.h.a.$ Now since $\frac{1}{2}a$ is constant, implies the area of $\Delta ABC$ can be maximized by maximizing $h$. Now clearly $h$ attains it's maximum value when it coincides with the other axis of the ellipse under consideration, that is when $AB=AC$. Thus $\Delta ABC$ must be isosceles with $AB=AC$ to get a maximum value of the area of $\Delta ABC$.
Thus it is clear that the triangle which will have the maximum area must be one of the isosceles triangles $ABC$ having $BC$ as the base.
Thus in any such $\Delta ABC$, we must have $a+b+c=a+2b \hspace{0.2 cm}(\because b=c)=P\implies b=\frac{1}{2}(P-a).$ Thus by Heron's formula we have $$|\Delta ABC|=\sqrt{\frac{P}{2}\left(\frac{P}{2}-a\right)\left(\frac{a}{2}\right)\left(\frac{a}{2}\right)}=\frac{a\sqrt{P}}{4}\sqrt{P-2a}.$$
Now to obtain the condition for maximizing $|\Delta ABC|$ we have to check when $\frac{d}{da}\frac{a\sqrt{P}}{4}\sqrt{P-2a}=0.$ Now $$\frac{d}{da}\frac{a\sqrt{P}}{4}\sqrt{P-2a}=0\\\iff (P-2a)^{1/2}-a(P-2a)^{-1/2}=0\\\iff P-2a=a\\\iff 3a=P\\\iff a=\frac{P}{3}.$$
Now note that $\frac{d^2}{da^2}|\Delta ABC|<0$, which implies that $|\Delta ABC|$ attain it's maximum value when $a=\frac{P}{3}.$ This implies that $b=c=\frac{P}{3}$. Thus we have $a=b=c=\frac{P}{3}$. Therefore, $|\Delta ABC|$ is maximized if and only if $a=b=c$, i.e, the triangle is equilateral.
Can someone check if this solution is correct or not? And other solutions are welcomed. Please ensure that the solutions are based on geometry and one-variable calculus. This problem can be solved using Lagrange multipliers or multi-variable calculus, but I do not want a solution using the same.
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Use the Heron's formula
$$A =\sqrt{s(s-a)(s-b)(s-c)}, \>\>\>\>\> s=\frac p2$$
and apply the AM-GM inequality to get
$$A =\sqrt{s(s-a)(s-b)(s-c)} \le \left[ s \left( \frac{3s-(a+b+c)}{3} \right) ^3\right]^{1/2}
= \frac{s^2}{3\sqrt3}=\frac{p^2}{12\sqrt3}$$
where the equality, or the maximum area, occurs at $a=b=c=\frac p3$.
|
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|
Find $ \lim\limits_{x\to 0}{\frac{1-\sqrt{1+x}\sqrt[3]{1-x}\cdots\sqrt[2n+1]{1-x}}{x}} $ without using L'Hopital's rule. Let $ n $ be a positive integer greater than $ 1 $.
Find : $$ \lim\limits_{x\to 0}{\frac{1-\sqrt{1+x}\sqrt[3]{1-x}\cdots\sqrt[2n+1]{1-x}}{x}} $$
Without L'Hopital's rule or series expansion.
Here is What I did to solve the problem.
\begin{aligned}\displaystyle\lim_{x\to 0}{\displaystyle\frac{1-\prod\limits_{k=2}^{2n+1}{\sqrt[k]{1+\left(-1\right)^{k}x}}}{x}}&=\displaystyle\lim_{x\to 0}{\displaystyle\sum_{k=2}^{2n+1}{\displaystyle\frac{\prod\limits_{i=2}^{k-1}{\sqrt[i]{1+\left(-1\right)^{i}x}}-\prod\limits_{i=2}^{k}{\sqrt[i]{1+\left(-1\right)^{i}x}}}{x}}}\\&=\displaystyle\lim_{x\to 0}{\displaystyle\sum_{k=2}^{2n+1}{\displaystyle\frac{1-\sqrt[k]{1+\left(-1\right)^{k}x}}{x}\displaystyle\prod\limits_{i=2}^{k-1}{\sqrt[i]{1+\left(-1\right)^{i}x}}}}\\ &=\displaystyle\lim_{x\to 0}{\displaystyle\sum_{k=2}^{2n+1}{\displaystyle\frac{\left(-1\right)^{k+1}}{\sum\limits_{j=0}^{k-1}{\sqrt[k]{1+\left(-1\right)^{k}x}^{j}}}\displaystyle\prod\limits_{i=2}^{k-1}{\sqrt[i]{1+\left(-1\right)^{i}x}}}}\\ &=\displaystyle\sum_{k=2}^{2n+1}{\displaystyle\frac{\left(-1\right)^{k+1}}{k}} \\ &=\displaystyle\sum_{k=1}^{n}{\displaystyle\frac{1}{2k+1}}-\displaystyle\sum_{k=1}^{n}{\displaystyle\frac{1}{2k}}\\ &=-1+\displaystyle\sum_{k=0}^{n}{\displaystyle\frac{1}{2k+1}}+\displaystyle\sum_{k=1}^{n}{\displaystyle\frac{1}{2k}}-\displaystyle\sum_{k=1}^{n}{\displaystyle\frac{1}{k}}\\ &=-1+\displaystyle\sum_{k=1}^{2n+1}{\displaystyle\frac{1}{k}}-\displaystyle\sum_{k=1}^{n}{\displaystyle\frac{1}{k}}\\ \displaystyle\lim_{x\to 0}{\displaystyle\frac{1-\prod\limits_{k=2}^{2n+1}{\sqrt[k]{1+\left(-1\right)^{k}x}}}{x}}&=H_{2n+1}-H_{n}-1 \end{aligned}
What's your approach to solve the problem ?
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By definition of the derivative, the desired limit is equal to $-f'(0)$ where
$$f(x) = \sqrt{1+x} \sqrt[3]{1-x} \sqrt[4]{1+x} \cdots \sqrt[2n+1]{1-x}.$$
We can now use logarithmic differentiation to find
$$f'(x) = f(x) \sum_{k=2}^{2n+1} \frac{(-1)^k}{k} (1 + (-1)^k x)^{-1}.$$
Plugging in $x = 0$ and $f(0) = 1$ will give the desired result.
|
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|
Evaluate $\sum_{n = 1}^{99} \frac{2}{\sqrt{n} + \sqrt{n + 2}}.$ Evaluate $\sum_{n = 1}^{99} \frac{2}{\sqrt{n} + \sqrt{n + 2}}.$
Listing out a few terms and simplifying the denominators gets me $$-1+\sqrt{3}+2-\sqrt{2}-\sqrt{3}+\sqrt{5}\cdots-\sqrt{97}+3\sqrt{11}-7\sqrt{2}+10-3\sqrt{11}+\sqrt{101}.$$
A lot of these terms cancel, and I think that the first and last terms will be left.
I feel like I'm close, but missing something.
|
We can rewrite the sum as:
$$\sum_{n=1}^{99} \frac{2}{\sqrt{n} + \sqrt{n+2}}\frac{\sqrt{n+2} - \sqrt{n}}{\sqrt{n+2} - \sqrt{n}}=
\sum_{n=1}^{99} \sqrt{n+2} - \sqrt{n}$$
It's a telescopic sum, where most of the terms are cancelled.
We can expand it:
$$\sum_{n=1}^{99} \sqrt{n+2} - \sqrt{n}=
(\sqrt{3}-\sqrt{1})+(\sqrt{4}-\sqrt{2})+(\sqrt{5}-\sqrt{3})+\dots+(\sqrt{101}-\sqrt{99})=\sqrt{100}+\sqrt{101}-\sqrt{1}-\sqrt{2}$$
|
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|
Evaluating $\lim_{x\to1^+}\frac{\sqrt{x+1}+\sqrt{x^2 -1}-\sqrt{x^3+1}}{\sqrt{x-1}+\sqrt{x^2 +1}-\sqrt{x^4+1}}.$ I don't know how to solve these kind of limit problems. I was wondering if someone could help me about it to learn them. $$\lim_{x\to 1^+}\frac{\sqrt{x+1}+\sqrt{x^2 -1}-\sqrt{x^3+1}}{\sqrt{x-1}+\sqrt{x^2 +1}-\sqrt{x^4+1}}.$$ Thanks in advance.
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We can rewrite the numerator as $$N=\sqrt{x-1}\sqrt{x+1}+\frac{x(1-x)(1+x)}{\sqrt{1+x}+\sqrt{1+x^3}}=\sqrt{x-1}\left(\sqrt{1+x}-\frac{x(1+x)\sqrt{x-1}}{\sqrt{1+x}+\sqrt {1+x^3}}\right)$$ and hence $N/\sqrt {x-1}\to\sqrt{2}$. Similarly one can show that if $D$ is the denominator then $D/\sqrt{x-1}\to 1$. And therefore $N/D\to\sqrt {2}$.
Most algebraic limits can be handled using simple algebraic manipulation combined with the standard algebraic limit $\lim\limits _{x\to a} \dfrac{x^n-a^n} {x-a} =na^{n-1}$.
|
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|
How to solve the following radical equation?$\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-x}}}} =x$
How to solve the following radical equation?
$$\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-x}}}} =x$$
I'm looking for an easy way to solve it.
|
As hinted
For real $x,\sqrt{2-x}\ge0\implies x\le2$
But $x=2$ doesn't satisfy the given equation
and $x=\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-x}}}}>0$
So, $0<x,<2$
WLOG let $x=2\cos32t, 0<32t<\dfrac\pi2$
Use $\cos2x=2\cos^2x-1=2-2\sin^2x$
$$\sqrt{2-x}=+2\sin16t$$
Now $\sqrt{2+\sqrt{2-x}}=\sqrt{2(1+\sin16t)}=+\sqrt2(\sin8t+\cos8t)$ as $\sin8t+\cos8t>0$
$\implies\sqrt{2+\sqrt{2-x}}=2\cos\left(\dfrac\pi4-8t\right)$
$\sqrt{2-\sqrt{2+\sqrt{2-x}}}=\sqrt{2-2\cos\left(\dfrac\pi4-8t\right)}=2\sin\left(\dfrac\pi8-4t\right)$ as $\dfrac\pi8-4t>0$
$\implies\sqrt{2-\sqrt{2+\sqrt{2-x}}}=2\cos\left(4t+\dfrac{3\pi}8\right)$
Finally, $\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-x}}}}=\sqrt{2+2\cos\left(4t+\dfrac{3\pi}8\right)}=2\cos\left(2t+\dfrac{3\pi}{16}\right)$
So, we have $$2\cos32t=2\cos\left(2t+\dfrac{3\pi}{16}\right)$$
$$\implies32t=2m\pi\pm\left(2t+\dfrac{3\pi}{16}\right)$$
Case $\#1:$ Taking the '+' sign, $$30t=2m\pi+\dfrac{3\pi}{16}=\dfrac{\pi(32m+3)}{16}$$
$$\implies32t=\dfrac{\pi(32m+3)}{16}\cdot\dfrac{32}{30}=\dfrac{\pi(32m+3)}{15}$$
We need $0<\dfrac{\pi(32m+3)}{15}<\dfrac\pi2$
$\iff0<2(32m+3)<15\implies m=0\implies x=2\cos\dfrac{\pi(32\cdot0+3)}{15}$
Case $\#2:$ Take the '-' sign
|
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For positive a,b,c,d, if $a^2+b^2+c^2+d^2+abcd=5$, show $a+b+c+d\leq 4$. One can use Lagrange multiplier, but I am looking for a more elementary proof.
I try to find the maximum of $a+b+c+d$ and follow the standard approach.
Construct $a+b+c+d+\lambda (a^2+b^2+c^2+d^2+abcd-5)$.
One can obtain $1+2a\lambda +bcd =0$ and so on, and eventually have $a=b=c=d$.
Then you are done.
|
Alternative solution:
Fact 1: For $a, b, c, d \ge 0$, it holds that
$$a^2 + b^2 + c^2 + d^2 + abcd - 5 - 3(a+b+c+d - 4) \ge 0.\tag{1}$$
From Fact 1, the desired result follows.
Edit 2022/02/14: A nice proof of Fact 1 is given by mudok@AoPS.
See: #4 in https://artofproblemsolving.com/community/c6t243f6h2780412_hard_inequality
$\phantom{2}$
Proof of Fact 1:
From Vasc's Equal Variable Theorem [1, Corollary 1.9], we only need to prove the case
when $a = b = c$.
It suffices to prove that
$d^2+(a^3-3)d+3a^2-9a+7 \ge 0$.
Since $3a^2 - 9a + 7 > 0$, by AM-GM, we have
$d^2+(a^3-3)d+3a^2-9a+7 \ge (a^3-3)d + 2d\sqrt{3a^2-9a+7}$.
It suffices to prove that $a^3 - 3 + 2\sqrt{3a^2-9a+7} \ge 0$.
If $a > \frac{3}{2}$, then $a^3 > 3$ and the inequality is true.
If $a \le \frac{3}{2}$, then it suffices to prove that
$4(3a^2-9a+7) - (a^3 - 3)^2 \ge 0$ or $(-a^4 - 2a^3 - 3a^2 + 2a + 19)(a-1)^2 \ge 0$.
We have $-a^4 - 2a^3 - 3a^2 + 2a + 19 \ge - (3/2)^4 - 2\cdot (3/2)^3 - 3\cdot (3/2)^2 + 19 > 0$.
We are done.
Reference
[1] Vasile Cirtoaje, “The Equal Variable Method”, J. Inequal. Pure and Appl. Math., 8(1), 2007.
https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf
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Prove: $\frac{\sin{nx}}{\sin{x}}\geqslant{n-\frac{n(n^2-1)x^2}{6}},n\in{\mathbb{N}\setminus\{0\}, x\in{\mathbb{R}, x\neq k\pi}} $
Prove :
$\frac{\sin{nx}}{\sin{x}}\geqslant{n-\frac{n(n^2-1)x^2}{6}},n\in{\mathbb{N\setminus\{0\}}}, x\in{\mathbb{R}} $
I proved this relationship by incident. I tried to directly prove this afterwards, but failed. I would love to see another proof to this Problem.
-A proof-: we know that $\displaystyle\sum_{i=1}^{n}\cos{a_i}= \sum_{k=1}^{n}\cos{(a+(k-1)x)}=\frac{\cos{\frac{a+a_n}{2}}\sin{\frac{nx}{2}}}{\sin{\frac{x}{2}}}=T(x)\quad(1)$
(Here's the source for $(1)$).
Also, $\displaystyle\sum_{i=1}^{n}a^2_i=\sum_{k=1}^{n}(a+(k-1)x)^2\\=\displaystyle\sum_{k=1}^{n}(a^2+2ax(k-1)+x^2(k-1)^2)\\=\displaystyle\sum_{k=1}^{n}a^2+2ax\sum_{k=1}^{n}(k-1)+x^2\sum_{k=1}^{n}(k-1)^2\\=na^2+2ax\frac{n(n-1)}{2}+x^2\frac{n(n-1)(2n-1)}{6}\quad(2)$
We consider the function $f(x)=\frac{x^2}{2}+\cos{x}\implies f''(x)=1-\cos{x}\geq{0}$ therefore $f(x)$ is a concave function. From Jensens inequality for concave functions:$ \displaystyle\sum_{i=1}^{n}f(a_i)\geq{nf\left(\frac{\sum_{i=1}^{n}a_i}{n}\right)}\iff\displaystyle\sum_{i=1}^{n}\left(\frac{a^2_i}{2}+\cos{a_i}\right)\geq n\Big(\frac{1}{2}\left(\frac{\frac{n}{2}(a+a_n)}{n}\right)^2+\cos{\frac{\frac{n}{2}(a+a_n)}{n}}\Big)\iff\frac{1}{2}\sum_{i=1}^{n}a^2_i+\sum_{i=1}^{n}\cos{a_i}\geq \frac{n}{2}\left(\frac{a+a_n}{2}\right)^2+n\cos{\frac{a+a_n}{2}}\overset{(1),(2)}{\iff}\frac{1}{2}\left(na^2+2ax\frac{n(n-1)}{2}+x^2\frac{n(n-1)(2n-1)}{6}\right)+T(x)\geqslant \frac{n\left(2a+(n-1)x\right)^2}{8}+n\cos{\frac{a+a_n}{2}}\overset{\ldots}{\iff} \frac{x^2n(n^2-1)}{3}\geqslant n\cos{\frac{a+a_n}{2}}-T(x)\overset{(1)}{\iff} \frac{x^2n(n^2-1)}{3}\geq \cos{\frac{a+a_n}{2}}\left(n-\frac{\sin{\frac{nx}{2}}}{\sin{\frac{x}{2}}}\right)\iff \frac{x^2n(n^2-1)}{3}\geqslant \cos{\left(a+\frac{(n-1)x}{2}\right)}\left(n-\frac{\sin{\frac{nx}{2}}}{\sin{\frac{x}{2}}}\right)\quad(3) $
$Lemma.$ For every dinstict $x,x\in\mathbb{R}_{\neq kπ}$ there exists at least one value of $a$ such that $\cos{(a+\frac{(n-1)x}{2}})=1\quad(5)$
Proof of the lemma: $(5)\iff (n-1)x=2-2a+4kπ\iff a=2kπ+1-\frac{(n-1)x}{2}$.
Hence, for every $x,x\in\mathbb{R}_{\neq k\pi}$
we have $\frac{x^2n(n^2-1)}{3}\geqslant \cos{\left(a+\frac{(n-1)x}{2}\right)}\left(n-\frac{\sin{\frac{nx}{2}}}{\sin{\frac{x}{2}}}\right)\overset{(4,x\to 2x)}{\iff}\frac{\sin{nx}}{\sin{x}}\geqslant{n-\frac{n(n^2-1)x^2}{6}} \square$
*
*To me it looks like $n-\frac{x^2n(n^2-1)}{6}$ is a good approximation function of $\frac{\sin{nx}}{\sin{x}}$ for small values of $x$ (or for small values of $x+ z\pi$, $z\in\mathbb{Z}$, it depends on $n$, because the second one is periodic with period multiple of $\pi$).
|
Remark: My previous solution is ugly. I give another solution.
Clearly, we only need to prove the case when $x > 0$.
For $n=1$, clearly the inequality is true.
For $n=2$, the inequality is equivalent to $2\cos x - 2 + x^2\ge 0$ which is true.
For $n\ge 3$ and $x\in [\frac{\pi}{2}, \infty)$:
It is easy to prove that
$-n \le \frac{\sin nx}{\sin x} \le n$ (by math induction) and $-n \ge n - \frac{n(n^2-1)x^2}{6}$. The inequality is true.
For $n \ge 3$ and $x\in (0, \frac{\pi}{2})$: Let $f(n) \triangleq \frac{\sin nx}{\sin x} - n + \frac{n(n^2-1)x^2}{6}$.
We have
\begin{align}
f(n+2) - 2f(n+1) + f(n) &= \frac{-4(\sin\frac{x}{2})^2\sin (n+1) x}{\sin x} + (n+1)x^2\\
&\ge -4(n+1)(\sin\frac{x}{2})^2 + (n+1)x^2\\
& \ge 0
\end{align}
where we have used
$\sin (n+1) x \le (n+1)\sin x$.
Thus, we have $f(n+2) - f(n+1) \ge f(n+1) - f(n)$ for all $n\ge 3$.
Thus, we have
\begin{align}
f(n+1) - f(n) &\ge f(4) - f(3) \\
&= 8(\cos x)^3- 4(\cos x)^2+6x^2-4\cos x\\
&= 2(4\cos x + 3)(1-\cos x)^2 + 6x^2 - 6(\sin x)^2\\
& \ge 0.
\end{align}
Thus, we have $f(n) \ge f(3) = -4(\sin x)^2+4x^2\ge 0$ for all $n\ge 3$.
We are done.
|
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Intersection of a circle $x^2+(y-1)^2=1$ and a parabola $ax^2=y$, bounds for a I have a circle $x^2+(y-1)^2=1$ and a parabola $x^2=\frac{y}{a}$ and it is required to find the bounds for $a$ if they intersect at point other than origin.
So I proceeded as follow, substituted $x^2=y/a$ into $x^2+(y-1)^2=1$
which led me to get,
$$\implies y^2+y\bigg(\frac{1}{a}-2\bigg)= 0$$
So for more than one point of intersection, I would necessarily have $D>0$, $D$ being the discriminant of the quadratic in $y$
From there I got the bounds as $a \in \bigg(R-\{0, \frac{1}{2}\}\bigg)$
But if I substitute the value of $y$ other than $0$, ie, $y = 2-\frac{1}{a}$ into $x^2=\frac{y}{a}$, I get $x^2 = \frac{2a-1}{a^2}$, which is always greater than $0$, and as a result from here I get the bound that $a \in \bigg(\frac{1}{2}, \infty\bigg)$
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We can compute very easily the intersections of the two conics, or:
$$\frac{y}{a}+(y-1)^2=1 \leftrightarrow ay^2+y(1-2a)=0 \leftrightarrow y(ay+1-2a)=0$$
Now, a solution is $y=0$ that holds for evry $a$. The other solution is $y=\frac{2a-1}{a}$ and so $x=\frac{1}{a}\sqrt{2a-1}$. Now, taking from here, it's very simple. In fact we want the square root to be defined, so:
$$2a-1>0 \leftrightarrow a>\frac{1}{2}$$
Note that your error has been committed when stating tehe range of $a$ without considering the fact that $x=\pm\sqrt{\frac{y}{a}}$. In fact, as you stated, if $a\in\left(R-\left\{0,\frac{1}{2}\right)\right)$, then for example $a=-1$ would be OK, but if we plug into the equation for $x$ we have a contraddiction. In fact:
$$y(-y+3)=0 \leftrightarrow y=0\vee y=3$$
But now:
$$x=\pm\sqrt{\frac{y}{a}}=\pm\sqrt{\frac{3}{-1}}=\text{IMPOSSIBLE}$$
|
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Conditional probability of continuous throwing a dice A person will keep throwing a dice until he gets $6 .$ If he gets 6 then the experiment will be stopped.
The probability that experiment ends at 6 th trial, given that prime number does not appear is
(A) $\left(\frac{2}{5}\right)^{5}\left(\frac{1}{6}\right)$
(B) $\left(\frac{2}{3}\right)^{5}\left(\frac{1}{3}\right)$
(C) $^{6} \mathrm{C}_{1}\left(\frac{2}{3}\right)^{5}\left(\frac{1}{3}\right)$
(D) $^{6} \mathrm{C}_{2}\left(\frac{2}{3}\right)^{5}\left(\frac{1}{3}\right)$
My approach:
$\begin{aligned} A=& \text { throwing a dice until } 6 \text { occurd and } \\ \text { prime } & \text { no. does not appear. } \end{aligned}$
$B=$ exp. ends at $6^{\text {th }}$ trial.
$P(B / A)=\frac{P(B \cap A)}{P(A)}$
$=\frac{\left(\frac{2}{6}\right)^{5} \frac{1}{6}}{\frac{1}{6}+\frac{2}{6} \cdot \frac{1}{6}+\left(\frac{2}{6}\right)^{2} \frac{1}{6}+\cdots}$
$=\frac{\left(\frac{1}{3}\right)^{5} \frac{1}{6}}{\frac{116}{1-1 / 3}}=\left(\frac{1}{3}\right)^{5} \cdot \frac{1}{6} \cdot \frac{2 / 3}{16}$
No correct ans in option. Where did I go wrong.
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Context is a bit confusingly worded to me... isn't it just simply $(B)$ because if no prime numbers appear then $2,3$ and $5$ won't appear, reducing the dice to only 3 possibilities of which only 1 of those 3 is a 6 so isn't it simply just $({2 \over 3})^5*(1/3)$? I could be wrong however...
|
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Use mathematical induction to prove that for all integers $n \geq 3,\, 2n + 1 < 2^{n}$ This is what I've got so far.
Let $P(n)$ be the statement that $2n + 1 < 2^n$
Basis:
Let $n = 3$. Show that $P(3)$ is true.
$2(3) + 1 = 7$ and $2^3 = 8$.
Since $7 < 8$, $P(3)$ is true.
Inductive Hypothesis:
Suppose that $P(k)$ is true for an arbitrary integer $k\ge 3$.
This implies that $2k + 1 < 2^k$.
Show that $2(k + 1) + 1 < 2^{k + 1}$.
\begin{align*}
2(k + 1) + 1 &= 2k + 2 + 1\\
&= (2k + 1) + 2
\end{align*}
From the IH,
\begin{align*}
2k + 1 < 2^k &\Rightarrow (2k + 1) + 2 < 2^k + 2\\
&\Rightarrow (2k + 1) + 2 < 2^k + 2 \\
&\Rightarrow (2k + 2) + 1 < 2^k + 2 \\
&\Rightarrow 2(k + 1) + 1 < 2^k + 2
\end{align*}
Now we need to show that $2^k + 2 < 2^{k + 1}$.
$2^k < 2^{k + 1}\Rightarrow 2^k + 2 < 2^{k + 1} + 2$
This is where I'm stuck. Any help on how to proceed is much appreciated.
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Continuing from your step $2(k+1)+1<2^{k}+2$.
Since $k\geq 3$, then $2<2^{k}$.
So, $2(k+1)+1<2^{k}+2<2^{k}+2^{k}=2^{k+1}$
$\therefore 2(k+1)+1<2^{k+1}$.
|
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How to prove such $|\mu_{1}a_{1}+\mu_{2}a_{2}+\cdots+\mu_{n}a_{n}|\le\frac{1}{a}$ let $a$ is give postive real number,and $a_{1},a_{2},\cdots,a_{n}$ be postive real numbers,and such
$$a^2_{1}+a^2_{2}+\cdots+a^2_{n}=1,a_{1}+a_{2}+\cdots+a_{n}=a$$
show that: there exist $\mu_{i}\in\{-1,1\},i=1,2,\cdots,n$ such
$$|\mu_{1}a_{1}+\mu_{2}a_{2}+\cdots+\mu_{n}a_{n}|\le\dfrac{1}{a}\tag{1}$$
I think use this identity:
$$\sum_{\mu_{i}\in\{-1,1\}}|\mu_{1}a_{1}+\mu_{2}a_{2}+\cdots+\mu_{n}a_{n}|^2=2^n\sum_{k=1}^{n}a^2_{i}=2^n$$
By the Pigeonhole Principle we obatain:
$$|\mu_{1}a_{1}+\mu_{2}a_{2}+\cdots+\mu_{n}a_{n}|^2\le\dfrac{2^n}{2^n}= 1$$
or
$$|\mu_{1}a_{1}+\mu_{2}a_{2}+\cdots+\mu_{n}a_{n}|\le 1$$
or
but I can't prove the stronger inequality $(1)$,thanks
|
Let us show the following more general proposition.
Let $a_{1},a_{2},\cdots,a_{n}$ be positive real numbers. Then there exist $\mu_{i}\in\{-1,1\},i=1,2,\cdots,n$ such
$$0\le(\mu_{1}a_{1}+\mu_{2}a_{2}+\cdots+\mu_{n}a_{n})(a_{1}+a_{2}+\cdots+a_{n})\le a_{1}^2+a_{2}^2+\cdots+a_{n}^2\tag{1}$$
Proof by induction on $n$.
The proposition is trivially true for the base case, $n=1$.
Assume it is true for $n$. Consider the case of $n+1$. WLOG, suppose $a_1\ge a_2\ge\cdots\ge a_n\ge a_{n+1}\gt0$. By assumption, there exist $\mu_{i}\in\{-1,1\},i=1,2,\cdots,n$ such that inequality $(1)$ holds.
There are two cases.
*
*$\mu_{1}a_{1}+\mu_{2}a_{2}+\cdots+\mu_{n}a_{n} \ge a_{n+1}$. Let $\mu_{n+1}=-1$. Then,
$$0\le\mu_{1}a_{1}+\mu_{2}a_{2}+\cdots+\mu_{n}a_{n}+\mu_{n+1}a_{n+1}.$$
Moreover, $$\begin{aligned}
&\qquad(\mu_{1}a_{1}+\mu_{2}a_{2}+\cdots+\mu_{n}a_{n}+\mu_{n+1}a_{n+1}) (a_{1}+a_{2}+\cdots+a_{n}+a_{n+1}) \\
&=(\mu_{1}a_{1}+\mu_{2}a_{2}+\cdots+\mu_{n}a_{n}- a_{n+1})(a_{1}+a_{2}+\cdots+a_{n}+a_{n+1}) \\
&=(\mu_{1}a_{1}+\mu_{2}a_{2}+\cdots+\mu_{n}a_{n})(a_{1}+a_{2}+\cdots+a_{n})\\ &\qquad-a_{n+1}(a_{1}(1-\mu_1)+a_{2}(1-\mu_2)+\cdots+a_{n}(1-\mu_{n}))-a_{n+1}^2\\
&\lt (\mu_{1}a_{1}+\mu_{2}a_{2}+\cdots+\mu_{n}a_{n})(a_{1}+a_{2}+\cdots+a_{n})\\
&\le a_{1}^2+a_{2}^2+\cdots+a_{n}^2\\
&\lt a_{1}^2+a_{2}^2+\cdots+a_{n}^2+a_{n+1}^2.\\
\end{aligned}$$
*$\mu_{1}a_{1}+\mu_{2}a_{2}+\cdots+\mu_{n}a_{n} \lt a_{n+1}$. Let $\nu_i=-\mu_i$ for $1\le i\le n$ and $\nu_{n+1}=1$. Then,
$$0\le \nu_{1}a_{1}+\nu_{2}a_{2}+\cdots+\nu_{n}a_{n}+\nu_{n+1}a_{n+1}.$$
Moreover, $$\begin{aligned}
&\qquad(\nu_{1}a_{1}+\nu_{2}a_{2}+\cdots+\nu_{n}a_{n}+\nu_{n+1}a_{n+1}) (a_{1}+a_{2}+\cdots+a_{n}+a_{n+1}) \\
&=(a_{n+1}-(\mu_{1}a_{1}+\mu_{2}a_{2}+\cdots+\mu_{n}a_{n}))(a_{1}+a_{2}+\cdots+a_{n}+a_{n+1}) \\
&\le a_{n+1}(a_{1}+a_{2}+\cdots+a_{n}+a_{n+1}) \\
&\le a_{1}^2+a_{2}^2+\cdots+a_{n}^2+a_{n+1}^2.\\
\end{aligned}$$
|
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Prove this has real roots $(a^2-bc)x^2+(a+b)(a-c)x+a(b-c)=0$ Prove this has real roots
$(a^2-bc)x^2+(a+b)(a-c)x+a(b-c)=0$
My Work
\begin{align*}
\Delta&=(a+b)^2(a-c)^2-4a(b-c)(a^2-bc) \\
&=a^4+2a^3c+a^2c^2-2a^3b+b^2a^2-4a^2bc-2abc^2+2ab^2c+b^2c^2.
\end{align*}
How do I show that this is positive? Simplification doesn't help either...How to factor them? please help!
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Hint: remarkably, that expression for $\Delta$ is a perfect square...! Can you factor the expression knowing that?
|
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Solving $ \sqrt{3-x} - \sqrt{x-1} > \sqrt{4-x} - \sqrt{x} $ I am reviewing my high-school math (year 10-ish) but I am hitting a wall with this nasty inequality:
$$
\sqrt{3-x} - \sqrt{x-1} > \sqrt{4-x} - \sqrt{x}
$$
I find (steps below) the solution to be
$$
1 \le x \le 3 \text{ with } x \ne 2
$$
but the text-book and WolframAlpha both agree that the solution is
$$
1 \le x < 2
$$
I would really appreciate if you could point out my mistake.
I also tried using the patterns that the text-book offers in the theory section, $$\sqrt{A(x)}<B(x)$$ and $$\sqrt{A(x)}>B(x)$$ but the terms increased in number instead of decreasing. I can provide that derivation, but I suspect it goes in the wrong direction.
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$$\begin{cases}\sqrt{3-x}-\sqrt{x-1}>\sqrt{4-x}-\sqrt x\\ 1\le x\le 3\end{cases}$$
If you want to square both current sides, then you must first evaluate their signs, namely using $A>B\iff \begin{cases}A\ge 0\\ B\ge 0\ \\ A^2>B^2\end{cases}\lor \begin{cases}A\ge 0\\ B<0\end{cases}\lor\begin{cases}A< 0\\ B< 0\ \\ A^2<B^2\end{cases}$. I think that delaying such considerations by making both sides non-negative is better (recall that, in the reals, $\sqrt A\ge 0$ when defined):
\begin{align}&\begin{cases}\sqrt{3-x}+\sqrt x>\sqrt{4-x}+\sqrt{x-1}\\ 1\le x\le 3\end{cases}\\&\begin{cases}3+2\sqrt{x(3-x)}>3+2\sqrt{(4-x)(x-1)}\\ 1\le x\le 3\end{cases}\\ &\begin{cases}\sqrt{x(3-x)}>\sqrt{(4-x)(x-1)}\\ 1\le x\le 3\end{cases}\\&\begin{cases}x(3-x)>(4-x)(x-1)\\ 1\le x\le 3\end{cases}\end{align}
|
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how to find positive integer $m,n , (m^2 + n^2 )(n^2-m)= 2mn(m+n^2) $ how to find positive integer $m,n $
$(m^2 + n^2 )(n^2-m)= 2mn(m+n^2) $
answer is $(36,18), (36, 12) $
$n$-th try........
$ gcd (m,n)= p$, then $m=ap, n= bp$
$(a^2 p^2 + b^2 p^2 )(b^2p^2 -ap)= abcp^2(ap+ b^2p^2)$
$(a^2+ b^2 )(b^2p-a) = 2ab (a+ b^2p ) $
$ gcd(a^2+ b^2,b)=1$, $ gcd(b^2p-a,b)=1$
so, $b=1$
$(a^2+1)(p-a)= 2a (a+p)$
$ p= a (\frac{a+1}{a-1} )^2 $
$p$ is integer, so, $a=2,3$,
$p=n=18,12$
is this possible? and is there some relationship with pythagorean triple?
|
$(36, 18)$ and $(36, 12)$ are the only positive integer solutions.
Expand the equation and move all the terms to one side:
$$m^3+mn^2+2m^2n-m^2n^2-n^4+2mn^3=0\tag{1}$$
Let $p=gcd(m,n)$ and $m=ap, n=bp$. Then by rewriting as a cubic equation in $a$, we get
$$a^3-(b^2p-2b)a^2+(2b^3p+b^2)a-b^4p=0$$
By the Rational Root Theorem, if $a$ is an integer, $a|b^4p$. Since $gcd(a,b)=1$, then $a|p$.
Now go back to equation $(1)$, and notice that it has some nice patterns that look like the $(x+y)^2=x^2+2xy+y^2$ formula. It can be rewritten as
$$m(m^2+n^2+2mn)-n^2(m^2+n^2-2mn)=0$$
$$m(m+n)^2-n^2(m-n)^2=0$$
$$m(m+n)^2=n^2(m-n)^2$$
Now sub in $ap$ and $bp$ again: (The reason I didn't sub them in earlier is because it makes the equation more cluttered and makes it a lot harder to see patterns in the equation, like this square-of-sum formula.)
$$a(a+b)^2=pb^2(a-b)^2\tag{2}$$
We know $a|p$, so
$$\frac{p}{a}=\frac{(a+b)^2}{b^2(a-b)^2}=\left(\frac{a+b}{b(a-b)}\right)^2\in\mathbb{N}$$
so
$$b(a-b)|(a+b)$$
Since $gcd(b,(a+b))=1$, then $b$ must equal 1. That leaves us with
$(a-1)|(a+1)$, so $a$ can only be 2 or 3. Substituting these two possibilities back into $(2)$ quickly yields $p=18$ or $p=12$ respectively, so $(m,n)=(36,18)$ or $(36,12)$.
As for the Pythagorean triples, I'm not sure if there is any connection, since the formula is
$$a=m^2-n^2$$
$$b=2mn$$
$$c=m^2+n^2$$
so $b$ and $c$ appear, but $a$ does not.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find volume of region bounded by plane, bounded by $y^2 = 1 - x$ and $x = -1$, and parabolic cylinder, $z = 1 - x^2$ Question:
Let $R$ denote the finite plane region on the $xy$-plane bounded by the line $x = -1$ and the parabola $y^2 = 1 - x$. Let $D$ denote the solid region under the surface of the parabolic cylinder $z = 1 - x^2$ and over the plane region $R$. Find the volume of $D$.
My steps:
First of all, I went to find the area of region R.
Substitute in $x = -1$, $y^2 = 2 - x$
Define in terms of $y$, $x = 2 - y^2$
When $x = 0$, $y = \pm \sqrt{2}$
Integrate:
$\int^\sqrt{2} _{-\sqrt{2}} \sqrt {2 - y^2} dx$
= $[2y - \cfrac{(y)^{3}}{3} + C]^\sqrt{2} _{-\sqrt{2}}$
= $[2\sqrt{2} - \cfrac{2^{1.5}}{3}] - [-2\sqrt{2} - \cfrac{-2^{1.5}}{3}]$
= $4\sqrt{2} - \cfrac{2^{2.5}}{3}$
After this, I'm not sure how I should proceed with finding the region bounded by this area of the plane the parabolic cylinder..
|
Note: you don't have to find the area of region $R$ first. As $D$ is the volume under the surface of the form $z=f(x,y)$, we first set the limits of $z$ and then find the projection of the given surface on $xy$-plane that is our region $R$ and then we set the limits of $x, y$.
HINT:
The volume of the solid region $D$ can be given by the triple integration as $$V= \int_{y=-\sqrt{2}}^{\sqrt{2}} \int_{x=-1}^{1-y^2} \int_{z=0}^{1-x^2} dxdydz$$
i.e. $$V=\int_{y=-\sqrt{2}}^{\sqrt{2}} \int_{x=-1}^{1-y^2} (1-x^2) dx dy$$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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|
Prove $(\frac{n+1}{n})^{n+1}$ is decreasing I managed to prove a similar fact: the following sequence is increasing: $\left( \dfrac{n+1}{n}\right)^n$, which means $\left( \dfrac{n+1}{n}\right)^n < \left( \dfrac{(n+1)+1}{(n+1)}\right)^{n+1}$.
All this took was some simple algebra and bernoulli's inequality. For the one in the title, I am not sure how.
My attempt was:
$$\left(\frac{n+1}{n}\right)^n \left(\frac{n+1}{n}\right)\stackrel{?}{>}\left(\frac{(n+1)+1}{(n+1)}\right)^{n+1} \left(\frac{(n+1)+1}{(n+1)}\right)$$
But this doesn't help me. Do you have other ideas?
|
Easy approach with A.M$\gt $G.M
Let $u_n=(\frac{n+1}{n})^{n+1}=\left(1+\frac{1}{n}\right)^{n+1}$
Let us consider n+2 positive numbers $(1-\frac{1}{n+1}),(1-\frac{1}{n+1}),(1-\frac{1}{n+1}),••••,(1-\frac{1}{n+1}),[(n+1)times] $and $1$.
Applying A.M$\gt $G.M.,we have,
$$\frac{(n+1)(1-\frac{1}{n+1})+1}{n+2}\gt (1-\frac{1}{n+1})^{\frac{n+1}{n+2}}$$
Or,$(\frac{n+1}{n+2})^{n+2}\gt (\frac{n}{n+1})^{n+1}$
Or,$(\frac{n+1}{n})^{n+1}\gt (\frac{n+2}{n+1})^{n+2}$
That is $ u_n\gt u_{n+1}$ for all $n$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3643208",
"timestamp": "2023-03-29T00:00:00",
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|
How to calculate the sum of the first n natural numbers? I already know one way to prove the sum of the n first natural number is equal to $\frac{n(n+1)}{2}$, but I found another way which involves calculating $(k+1)^2 - k^2, k\in \mathbb{N},\ k\geqslant 1$ . Here is the whole demonstration:
$
\begin{gather}
(k+1)^2 -k^2 = 2k +1\\
\sum_{k=1}^n{((k+1)^2 -k^2)} = 2\sum_{k=1}^n{k} +\sum_{n=1}^n{1} \\
(n+1)^2-1=2\sum_{k=1}^n{k} +n \\
2\sum_{k=1}^n{k} = n^2 +n\\
\sum_{k=1}^n{k} = \frac{n(n+1)}{2}
\\
\end{gather}$
My question is: How do we get from $\sum_{k=1}^n{((k+1)^2 -k^2)}$ to $(n+1)^2-1$
|
Let's try writing some terms out:
\begin{align*}
\sum\limits_{k=1}^n ((k+1)^2 - k^2) & = ((1+1)^2 - 1^2) + ((2+1)^2 - 2^2) + ((3+1)^2 - 3^2) + \ldots + ((n+1)^2 - n^2) \\
& = (2^2 - 1^2) + (3^2 - 2^2) + (4^2 - 3^2) + \ldots + ((n+1)^2 - n^2) \\
& = (2^2 - 2^2) + (3^2 - 3^2) + \ldots + (n^2 - n^2) + (n+1)^2 - 1^2 \\
& = (n+1)^2 - 1
\end{align*}
It telescopes!
|
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"timestamp": "2023-03-29T00:00:00",
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|
MOP 2011 inequality If $a,b,c$ are positive integers prove that
$\sqrt{(a^2-ab+b^2)} +\sqrt{(b^2+c^2-bc)} +\sqrt{(a^2+c^2-ac)} +9(abc)^{1/3} \le 4(a+b+c)$
My attempt:
I tried to split inequality and prove it bit by bit
$9(abc)^{1/3} \le 3(a+b+c)$
It suffices to prove that
$\sqrt{(a^2-ab+b^2)} +\sqrt{(b^2+c^2-bc)} +\sqrt{(a^2+c^2-ac)} \le a+b+c$
After trying out few values I realized it is wrong.So I decided to tackle it all at your once. I tried few well known inequalities and played around but nothing came useful out of it. Any help??
|
Another way.
By AM-GM $$\sum_{cyc}\sqrt{a^2-ab+b^2}\leq\sum_{cyc}\left(\frac{a^2-ab+b^2}{a+b}+\frac{a+b}{4}\right)=$$
$$=\sum_{cyc}\left(\frac{a^2+2ab+b^2-3ab}{a+b}+\frac{a+b}{4}\right)=\frac{5}{2}(a+b+c)-3\sum_{cyc}\frac{ab}{a+b}=$$
$$=4(a+b+c)-3\sum_{cyc}\left(\frac{ab}{a+b}+\frac{a+b}{4}\right)\leq$$
$$\leq 4(a+b+c)-3\sum_{cyc}\sqrt{ab}\leq4(a+b+c)-9\sqrt[3]{abc}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Dividing polynomial $f(x^3)$ by $x^2+x+1$ Given two polynomials with real coefficients $g(x)$ and $h(x)$.
Additionally, $x^2+x+1$ is a factor of $f(x^3)= h(x^3)+xg(x^3)$.
Prove that $x-1|h(x)$ and $x-1|g(x)$
I have tried to solve it by doing this;
$(x-1)(x^2+x+1)|(x-1)(h(x^3)+xg(x^3))$
then;
$(x^3-1)|(x)(h(x^3)-g(x^3)) + (x^2)g(x^3) - h(x^3)$
consider degree of $3k, 3k+1, 3k+2$ separately, I got;
$(x^3-1)|(x^2)g(x^3)$ and $h(x^3)$
Lastly, $(x^3-1)|g(x^3), h(x^3)$ and so on finish proving.
I wonder, is it correct to separate three cases of degrees in this problem?
|
Suppose
$$
\left.x^2+x+1\,\,\middle|\,\,h\!\left(x^3\right)+x\,g\!\left(x^3\right)\right.\tag1
$$
Let ${\bar h}(x)=h(x+1)$ and ${\bar g}(x)=g(x+1)$, then $(1)$ becomes
$$
\left.x^2+x+1\,\,\middle|\,\,{\bar h}\!\left(x^3-1\right)+x\,{\bar g}\!\left(x^3-1\right)\right.\tag2
$$
Write $\bar h$ and $\bar g$ as
$$
\bar h(x)=\sum_{k=0}^n\bar h_kx^k\tag3
$$
and
$$
\bar g(x)=\sum_{k=0}^n\bar g_kx^k\tag4
$$
Since $\left.x^2+x+1\,\middle|\,x^3-1\right.$, we immediately have that $x^2+x+1$ divides each term in the sum for ${\bar h}\!\left(x^3-1\right)$ except for $\bar h_0$. Similarly, $x^2+x+1$ divides each term in the sum for ${\bar g}\!\left(x^3-1\right)$ except for $\bar g_0$. Therefore, $(2)$ implies
$$
\left.x^2+x+1\,\,\middle|\,\,{\bar h}_0+x\,{\bar g}_0\right.\tag5
$$
which means that ${\bar h}_0={\bar g}_0=0$; that is, the constant coefficients of $\bar h$ and $\bar g$ are $0$. Thus,
$$
\left.x-1\,\middle|\,\bar h(x-1)=h(x)\right.\tag6
$$
and
$$
\left.x-1\,\middle|\,\bar g(x-1)=g(x)\right.\tag7
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Which integration is correct? Or are both correct? What's the correct method to integrate the function $\frac{1}{2(3x+1)}$?
\begin{align}
\int\frac{1}{2(3x+1)} dx &= \int\frac{1}{6x+2}dx\\
&= \frac{1}{6}ln(6x+2)+c
\end{align}
$$or$$
\begin{align}
\int\frac{1}{2(3x+1)} dx &= \frac{1}{2}\int\frac{1}{3x+1}dx \\
&= (\frac{1}{2})(\frac{1}{3})ln(3x+1)+c\\
&= \frac{1}{6}ln(3x+1)+c
\end{align}
Both methods seem feasible, however the answers given are different. Is it because the constant c is of a different value?
|
Yes, you are right
Both the feasible answers are correct & the only difference between two is constant of integration
|
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|
Rewriting elliptic curve with point of order 3 I have a question regarding a change of coordinates with elliptic curves. I tried to show that an elliptic curve $E: y^2 = x^3 + ax^2 + bx +c$ with coefficients in $\mathbb{F}_q[t]$ and where $P$ is a rational point of order 3 can be rewritten as:
\begin{equation*}\label{}
E\colon y^2 = x^3 + A(x-B)^2,
\end{equation*}
where the coefficients $A$ and $B$ are from $\mathbb{F}_q[t]$. So I let $P=(\alpha,\beta)$ be a $\mathbb{F}_q(t)$-rational point of order $3$ on an elliptic curve given by $E: y^2 = x^3 + ax^2 + bx + c$. Shifting the $x$-coordinate of $P$ to $0$, i.e. replacing $x$ by $x+\alpha$ gives:
\begin{equation*}\label{}
\begin{split}
y^2&= (x+\alpha)^3 + a(x+\alpha)^2 + b(x+\alpha) + c \\
&= x^3 + x^2(3\alpha + a) +x(3\alpha^2 + 2a\alpha+b)+\alpha^3 + a\alpha^2 + \alpha b +c \\
&= x^3 + x^2(3\alpha + a) +x(3\alpha^2 + 2a\alpha+b)+\beta^2.
\end{split}
\end{equation*}
Renaming $3\alpha + a$ to $a$ and $3\alpha^2 + 2a\alpha+b$ to $b$ gives an equation of the form:
\begin{equation*}\label{}
y^2 = x^3 + ax^2 + bx + \beta^2.
\end{equation*}
The point $(0,\beta)$ is of order $3$, hence we need $2P = -P$. Using the duplication formulas we obtain that $\left(\frac{b^2-4a\beta^2}{4\beta^2},\frac{4ab\beta^2 - b^3 - 8\beta^4}{-8\beta^3}\right) = (0,-\beta)$ and hence $b^2 = 4a\beta^2$. Suppose that $b \neq 0$, then $a\neq 0$ and $a$ must be a perfect square. All of this means that we can take an element $z \in \mathbb{F}_q(t)$ such that $z^2 = \frac{\beta^2}{a}$. Combining everything yields the following equation:
\begin{equation*}\label{}
\begin{split}
y^2 &= x^3 + ax^2 + \sqrt{4a^2z^2}x + z^2a \\
&= x^3 + ax^2 + 2azx + z^2a \\
&= x^3 + a(x+z)^2.
\end{split}
\end{equation*}
Here is where I don't know how to continue. I can't seem to find a way to force $z$ to be an element of $\mathbb{F}_q[t]$. Any help?
|
Let us write $z=f/g$ with polynomials $f,g\in\Bbb F_p[t]$. We multiply with $g^6$ the last equation $y^2=x^3+a(x+z)^2$ getting:
$$
(yg^3)^2 = (xg^2)^3 + ag^2(xg^2+fg)^2\ ,
$$
and use the new "coordinates" $X = xg^2$, $Y=yg^3$.
Note: $y^2=x^3+C^2$ is not of the wanted form. It has the torsion points $T_\pm=(0,\pm C)$ of order $3$.
Note: If the constant $a$ in $y^2=x^3+a(x+z)^2$ has a denominator $d$, we multiply with $d^6$, group $y^2d^6=(yd^3)^2$, and $x^3d^6=(xd^2)^3$, then push from $ad^6=ad^2\cdot d^4$ the factor $d^4$ inside the square, so that $(xd^2)$ also appears inside the square.
|
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|
How many integers solutions does $x^2 + xy - y^2 = 1$ have? Also, if we know if a binary form represents an integer $n$, is there an algorithm to find all the solutions?
|
Multiply equation with 4: $$4x^2+4xy-4y^2=4$$ so
$$(2x+y)^2-5y^2=4$$
$$(2x+y-2)(2x+y+2)=5y^2$$ Clearly $\gcd(x,y)=1$. Let prime $p\mid \gcd(2x+y-2,2x+y+2)$
then $p\mid 4$ (so $p\mid y$). Let $y=2z$, so $$(x+z-1)(x+z+1)=5z^2$$
Let prime $q\mid \gcd(x+z-1,x+z+1)$, then $q\mid 2$ and thus $2\mid z$ and $2\mid x-1$. Now we have $z=2t$ and $x=2s+1$ so
$$(s+t)(s+t+1)=5t^2$$
Since $s+t$ and $s+t+1$ are consecutive we have $$s+t = a^2\;\;\wedge \;\;s+t+1=5b^2$$ or $$s+t = 5a^2\;\;\wedge \;\; s+t+1=b^2$$ where $a,b$ are relatively prime such that $t=ab$.
I believe you can finish from here...
|
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|
How to solve $3\lfloor x \rfloor - \lfloor x^{2} \rfloor = 2\{x\}$? I am trying to solve the following question involving floor/greatest integer functions.
$3\lfloor x \rfloor - \lfloor x^{2} \rfloor = 2\{x\}$ with the notations $\lfloor x \rfloor$ denoting the greatest integer less than or equal to $x$ and $\{x\}$ to mean the fractional part of $x$.
I used the following property for floor functions.
$n\leq x$ if and only if $n \leq \lfloor x \rfloor$ where $n\in \mathbb{Z}$
Let $p=\lfloor x^{2} \rfloor$, then
$p\leq \lfloor x^{2} \rfloor < p+1$
$\rightarrow p \leq x^{2} < p+1$
$\rightarrow \sqrt{p} \leq x < \sqrt{p+1}$ , since $\sqrt{p} \in \mathbb{Z}$
$\rightarrow \sqrt{p} \leq \lfloor x \rfloor < \sqrt{p+1}$ We then have $\sqrt{p} = \lfloor x \rfloor$
Since $\{x\}=x-\lfloor x \rfloor,$
$3\lfloor x \rfloor - \lfloor x^{2} \rfloor - 2\{x\}= 3\lfloor x \rfloor - \lfloor x^{2} \rfloor - 2(x-\lfloor x \rfloor)= 5\lfloor x \rfloor - \lfloor x^{2} \rfloor - 2x=0$
Substituting $p$, $\sqrt{p}$ for $\lfloor x^{2} \rfloor$ and $\lfloor x \rfloor$ respectively, and also letting $x= \sqrt{p}, $ we get $p = 3\sqrt{p}$ solving for $p$ gives $p=0, 9$, and hence $x=0, 3$
The problem is that according to the solution for the problem, $x$ also equals to $\frac{3}{2}$ for $\{x\}=\frac{1}{2}$ since $2\{x\}\in \mathbb{Z}$. However, by definition for $\{x\}$, $0 \leq \{x\} < 1$, then $0 \leq 2\{x\} < 2$. How can $\{x\}=\frac{1}{2}$ and how do I use this to obtain $x=\frac{3}{2}$. I am not sure what I am missing. IF I made any mistakes in my reasoning. Can someone point it out to me please. Thank you in advance.
|
Write $\{x\}=x-\lfloor x\rfloor$. Then we have
$$
5\lfloor x\rfloor - \lfloor x^2\rfloor = 2x
$$Since the LHS is an integer, the RHS must be as well. There are two cases: $x$ is an integer, or $x$ is a half-integer.
*
*$x$ an integer. Drop the brackets:
$$
5x-x^2=2x;\qquad x=0,3
$$
*$x$ is a half-integer. Write $x=y+1/2$. Then $x^2 = y^2+y+1/4$, and again we can drop the brackets:
$$
5y-(y^2+y)=2y+1; \qquad y=1, x=3/2
$$
|
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|
How to find k'th integer not divisible by n? Although this was a programming question I want the mathematical intuition behind it. So we were given two numbers $n$ and $k$. We were asked to find out $k{-th}$ number **not divisible** by $n$. For example, $n=3$ and
$k=7$.
If we see numbers $1$ to $12$. $$1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12$$ Then $3,6,9$ are divisible by $3$ so we remove it. So we are left with $1, 2, 4, 5, 7, 8, 10, 11$. Clearly the $7th$ number not divisible by $3$ is $10$. So answer is $10$. It turns out the answer is:
$$k+⌊\tfrac{k-1}{n-1}⌋$$
I want the mathematical intuition behind this formula.
|
We need the $k^{th}$ integer not divisible by $n$.
At first, we list all the integers not divisible by $n$. We don't know the values of these integers yet.
$$1^{st}, 2^{nd}, 3^{rd}, ..., k^{th}$$
This list omits the integers divisible by $n$. We need to know exactly how many such integers are omitted to find the value of the $k^{th}$ integer.
Exactly one integer (which is divisible by n) is omitted after $(n-1)$ integers. So, we need to know how many groups of $(n-1)$ integers do we encounter before reaching the $k^{th}$ integer. In other words, we will try to count the number of blocks of $(n-1)$ integers up till the $(k-1)^{th}$ non-divisible integer. And, that number is $\lfloor \frac{k-1}{n-1} \rfloor$.
The point of confusion might be the appearance of $(k-1)$. Remember, we need to find the number of $(n-1)$ integer blocks before we reach the $k^{th}$ integer. Examples might shed some light upon this.
Take $k=6$ and $n=3$ first. (I'll be omitting the positional identifiers, such as 'st, 'nd, 'th, etc.)
$$1,2,|,3,4,|,5,6,|$$
The pipe symbols express the position of the omitted integers. If we take $\lfloor \frac{k}{n-1} \rfloor$, we count the integer after the $6^{th}$ as well, which we don't want. Hence, we take $\lfloor \frac{k-1}{n-1} \rfloor$, which counts the missing integers before the $6^{th}$.
For the next example, consider $k=6$ and $n=2$. The missing integers will be placed like this.
$$1,|,2,|,3,|,4,|,5,|,6,|$$
Again, to count the missing integers before the $6^{th}$, we are gonna check how many groups of $(2-1)$ we can make up till the $5^{th}$. Hence $\lfloor \frac{6-1}{2-1} \rfloor$.
|
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|
Why $c$ closed to $-2\times10^n$ in $(1-c^2)^3+(c^3+10^nc^2-1)^3+(10^n c^2-1)^3=n$ for $n >1$? I have tried many times to evaluate $(1-c^2)^3+(c^3+10^nc^2-1)^3+(10^n c^2-1)^3=n$ for $n >1$ as polynomial for some values of integer $n$ which are greater than $1$ for the solution of the titled equation for looking why exactly $c$ always closed to integer which is $-2*10^n$ looking to the behavior of polynomial coefficients but I can't expand that, For numerical evidence I have used such that an example for $n=2$ is montioned here, Now my question here is :
Why $c$ closed to $-2\times10^n$ in $(1-c^2)^3+(c^3+10^nc^2-1)^3+(10^n c^2-1)^3=n$ for $n >1$ ? And does this gives any evidence about unknown representation numbers as sum of cubic like $n=390,732,\cdots $, because What I have tried the decimal expansion of $c$ increases with the digits $9$ ?
Addendum:This part is the motivation of this question, Now I should deduce this question for Unknown representations numbers as $390$, Could we have an integer $c$ for $n=390$ in the titled equation since we have increasing decimal expansion of $c$ with $999999999\cdots$?
|
Your expression is
$$(1 - c^2)^3 + (c^3 + 10^n c^2 - 1)^3 + (10^n c^2 - 1)^3 = n \tag{1}\label{eq1A}$$
for $n \gt 1$. For some real $d$, let
$$c = d \times 10^n \tag{2}\label{eq2A}$$
Note that within each of the $3$ terms on the left side of \eqref{eq1A}, there's a term of $1$ or $-1$, but the other terms are all at least $c^2$. If $d$ in \eqref{eq2A} is near $0$, this gives you other solutions, such as the $2$ that Wolfram Alpha gives of $c \approx \pm 0.134$ in your link.
Assume $d$ is not very close to $0$ from this point on (such as $|d| \gt 1$), so you have $c^2 \gg 1$. To make the analysis easier, for now ignore those terms of $\pm 1$. As such, using this and \eqref{eq2A} in the LHS of \eqref{eq1A}, gives
$$\begin{equation}\begin{aligned}
& (1 - c^2)^3 + (c^3 + 10^n c^2 - 1)^3 + (10^n c^2 - 1)^3 \\
& \approx (c^2)^3 + ((c^2)(c + 10^n))^3 + (10^n c^2)^3 \\
& = ((d\times10^n)^2)^3 + (((d\times10^n)^2)(d\times10^n + 10^n))^3 + (10^n (d\times10^n)^2)^3 \\
& = d^6\times 10^{6n} + d^6(d+1)^3\times 10^{9n} + d^6\times10^{9n} \\
& = d^6\times 10^{6n}(1 + 10^{3n}((d+1)^3 + 1))
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Next, you have $d^6\times 10^{6n} \gg n$, so the value within the outside brackets in \eqref{eq3A} should be close to $0$. Also, for $n \gt 1$, you have $10^{3n} \gg 1$, so this means you will then need to have
$$\begin{equation}\begin{aligned}
10^{3n}((d+1)^3 + 1) & \approx 0 \\
(d + 1)^3 + 1 & \approx 0 \\
d + 1 & \approx -1 \\
d & \approx -2
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
With $d = -2$, so $c = -2 \times 10^n$, note in the LHS of \eqref{eq1A} that the first term is negative, but much less in magnitude than either of the next two terms. Also, the second term is negative, with it being just slightly less than $-64 \times 10^{9n}$. Finally, the third terms is positive, just about equal in magnitude to the second term, but just slightly less than $64 \times 10^{9n}$. Thus, the overall result is negative. However, as indicated in \eqref{eq3A} and \eqref{eq4A}, as $d$ increases, the overall result increases, so you would expect a value just a bit greater than $c = -2 \times 10^n$ to be a root of \eqref{eq1A}. In fact, in your Wolfram Alpha result, if you click on the "More Digits" button for the $c = -200$ result, you get that $c \approx -199.99991666909716587$.
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"language": "en",
"url": "https://math.stackexchange.com/questions/3670521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Prove the Condition for Construction of a Triangle Let $a,b,c>0$ be real numbers. Show that you can construct the triangle with length sides $a,b,c$ if and only if $pa^2+qb^2>pqc^2$, for any $p,q$ such that $p+q=1$.
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As Exodd said, the statement is false. Let $f_{a,b,c}(p)=pa^2+2b^2-p(1-p)c^2=c^2p^2+(a^2-c^2)p+2b^2$. Then your statement says: $a,b,c>0$ are the sides of a triangle if, and only if, $f_{a,b,c}>0$.
Take $a=c=1$, $b=3$. Then $f_{a,b,c}(p)=p^2+6$, which is always positive, but $(1,1,3)$ cannot be the three sides of a triangle.
If we replace $2b^2$ by $qb^2$, then we get a much more interesting problem that we can prove. Once again, define
$$\begin{aligned}
f_{a,b,c}(p)&=pa^2+(1-p)b^2-p(1-p)c^2 \\
&=c^2p^2+(a^2-b^2-c^2)p+b^2,
\end{aligned}$$
and the problem becomes: $a,b,c>0$ are the sides of a triangle if, and only if, $f_{a,b,c}>0$. Moreover, we can suppose $a<b$, since their roles are symmetrical.
Since the leading coefficient $c^2$ of the quadratic function $f_{a,b,c}$ is positive, we have
$$\begin{aligned}
f_{a,b,c}>0&\Leftrightarrow \Delta=(a^2-b^2-c^2)^2-4b^2c^2<0\\
&\Leftrightarrow (a^2-b^2-c^2)^2<4b^2c^2 \\
&\Leftrightarrow 0<b^2+c^2-a^2<2bc \\
&\Leftrightarrow b^2-2bc+c^2=(b-c)^2<a^2 \\
&\Leftrightarrow |b-c|<a.
\end{aligned}$$
And by hypothesis, $a<b<b+c$, so we have
$$f_{a,b,c}>0\Leftrightarrow |b-c|<a<b+c,$$
and the right hand side is the triangle inequality.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/3670964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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|
Find $4\sin (x)\,\big(\sin(3x)+\sin (7x)+\sin(11x)+\sin(15x)\big)$ when $x=\frac{\pi}{34}$.
If $x=\frac{\pi}{34}$, then find the value of
$$S=4\sin (x)\,\big(\sin(3x)+\sin (7x)+\sin(11x)+\sin(15x)\big).$$
Source: Joszef Wildt International Math Competition Problem
My attempt:
$$\sin(3x)+\sin(15x)=2\sin(9x)\cos(6x)$$
$$\sin(7x)+\sin(11x)=2\sin(9x)\cos(2x)$$
So $$S=8\sin(x)\sin(9x)\big(\cos(6x)+\cos(2x)\big)$$
$$S=16\sin(x)\sin(9x)\cos(4x)\cos(2x)$$
$$S=16\sin(x)\sin(9x)\sin(13x)\sin(15x)$$
any clue here?
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as You simplified and as @WE Tutorial School pointed out,
$$S=16\cos\frac{\pi}{17}\cos\frac{2\pi}{17}\cos\frac{4\pi}{17}\cos\frac{8\pi}{17}.$$
Applying $\sin 2x = 2 \sin x \cos x$ 4 times,
$$ S \cdot \sin \frac \pi{17} = \sin \frac {16\pi}{17}$$
$$ S = \frac {\sin \left( \pi - \frac \pi{17} \right)}{\sin \frac \pi{17}}$$
$$S=1$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/3671242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
contour integration of $\int_0^\infty \frac{\ln(x)}{x^2-1}dx$ I asked for a problem for few days ago, regarding integration of
$$I=\int_0^\infty \frac{\ln(x)}{x^2-1}dx$$
I know that I could do the substitution $x=it$ and do the integral where the denominater is $x^2+1$ and so on... But I tried anyway to create a contour and compute my integral. Down below my contour $C$ can be seen.
Let $f(z)=\ln(z)/(z^2+1)$. I compute my residue and obtain 0. I.e. $\oint_C f(z)dz=0$. The following are also giving me zero, i.e. $$\begin{align}
\int_{\gamma_R}f(z)=0\\
\int_{\gamma_r}f(z)=0
\end{align}$$
by the estimation lemma. So now my integral is,
$$0=e^{\pi/3i}\int_r^R \frac{\ln|z|+\pi/3i}{z^2-1}dz+e^{2\pi/3i}\int_r^R \frac{\ln|z|+2\pi/3i}{z^2-1}dz$$
I rewrite the above to
$$0=(e^{\pi/3i}+e^{2\pi/3i})\int_r^R \frac{\ln(x)}{z^2-1}dz+(\frac{\pi}{3}ie^{\pi i/3}+\frac{2\pi}{3}ie^{2\pi i/3})\int_r^R \frac{1}{z^2+1}dz$$
And calculating the above gives me,
$$\int_r^R \frac{\ln(x)}{z^2-1}dz=\frac{\pi^2}{4}+\frac{i\sqrt{3}\pi^2}{36}$$
But if I compute the integral $I$ in Maple, I obtain $I=\frac{\pi^2}{4}$. My questions:
*
*Are my contour plot correct?
*Are my integrals correct, i.e. arc and residue?
*If yes to 1. and 2. Can I just drop the imaginary part and just use the real part?
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\begin{eqnarray*}
\int_0^\infty\int_0^\infty\frac{dxdy}{(1+y)(1+x^2y)}
&=&\int_0^\infty\frac{dy}{1+y}\int_0^\infty\frac{dx}{1+x^2y}\\
&=&\int_0^\infty\frac{dy}{1+y}\int_0^\infty\frac{1}{y}\frac{dx}{\frac{1}{y}+x^2}\\
&=&\int_0^\infty\frac{dy}{1+y}\int_0^\infty\frac{1}{y}\frac{1}{\frac{1}{\sqrt{y}}}arctan \frac{x}{\frac{1}{\sqrt{y}}}|_0^\infty\\
&=&\frac{\pi}{2}\int_0^\infty\frac{dy}{\sqrt{y}(1+y)}\\
&=&\frac{\pi}{2}2\int_0^\infty\frac{dt}{1+t^2}\\
&=&\frac{\pi^2}{2}
\end{eqnarray*}
\begin{eqnarray*}
\int_0^\infty\int_0^\infty\frac{dxdy}{(1+y)(1+x^2y)}
&=&\int_0^\infty\frac{dx}{1-x^2}\int_0^\infty\left(\frac{x^2}{1+x^2y}-\frac{1}{1+y}\right)dy\\
&=&\int_0^\infty\frac{1}{1-x^2}\left(\ln\frac{1+x^2y}{1+y}\big|_0^\infty\right)dx\\
&=&2\int_0^\infty\frac{\ln x dx}{1-x^2}
\end{eqnarray*}
Thus
\begin{eqnarray*}
\int_0^\infty\frac{\ln x dx}{1-x^2}=\frac{\pi^2}{4}.
\end{eqnarray*}
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"language": "en",
"url": "https://math.stackexchange.com/questions/3673393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
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Fractions in Questions and Answers
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