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point on a line closest to another point Question:
Find the point on the line $2x+y+3=0$ that is closest to the point $(2,-6)$
This question really made me think for a while as to how to approach it.
I calculated distance as $\frac{1}{\sqrt{14}}$ using $D = \frac{ax_0+by_0+C}{\sqrt{a^2+b^2+c^2}}$
So would that mean that $x$ would be closest when the distance is $2- \frac{1}{\sqrt{14}} = 1.732738758$
Any suggestions?
|
The equation for the line perpendicular to $y = -2x - 3$ passing through $(2, -6)$ is $y = 0.5x - 7$.
($0.5$ is the negative reciprocal of $-2$ and the $y$ intercept is $-6$ minus $2\cdot 0.5$)
Subtracting one from the other, their intersection is where $0 = 2.5x - 4$, which is at $x = 1.6$ and therefore $y = -6.2$.
$d$ the distance between their intersection $(1.6, -6.2)$ and $(2, -6)$ is
$d = \sqrt{0.4^2 + 0.2^2} = \sqrt{0.2}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How does one solve $\sin x-\sqrt{3}\ \cos x=1$? I thought this one up, but I am not sure how to solve it. Here is my attempt:
$$\sin x-\sqrt{3}\ \cos x=1$$
$$(\sin x-\sqrt{3}\ \cos x)^2=1$$
$$\sin^2x-2\sqrt{3}\sin x\cos x\ +3\cos^2x=1$$
$$1-2\sqrt{3}\sin x\cos x\ +2\cos^2x=1$$
$$2\cos^2x-2\sqrt{3}\sin x\cos x=0$$
$$2\cos x(\cos x-\sqrt{3}\sin x)=0$$
$2\cos x=0\Rightarrow x\in \{\frac{\pi }2(2n-1):n\in\Bbb Z\}$
But how do I solve $$\cos x-\sqrt{3}\sin x=0$$
|
Avoid squaring whenever possible as it immediately introduces
extraneous root(s).
$$\sin x-\sqrt3\cos x=1$$
Method$\#1:$
Use Prosthaphaeresis Formulas
Method$\#2:$
Use Solving trigonometric equations of the form $a\sin x + b\cos x = c$
Method$\#3:$
Use Double Angle formula,
$\cos x=\cos^2\dfrac x2-\sin^2\dfrac x2$
and $1-\sin x=\left(\cos\dfrac x2-\sin\dfrac x2\right)^2$
We immediately have $\cos\dfrac x2-\sin\dfrac x2$ as common factor
|
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|
Evaluate $\lim\limits_{x \to 0+}\left[\frac{x^{\sin x}-(\sin x)^{x}}{x^3}+\frac{\ln x}{6}\right].$ Problem
Evaluate
$$\lim_{x \to 0+}\left[\frac{x^{\sin x}-(\sin x)^{x}}{x^3}+\frac{\ln x}{6}\right].$$
Attempt
First, we may obtain
$$\lim_{x \to 0+}\left[\frac{x^{\sin x}-(\sin x)^{x}}{x^3}+\frac{\ln x}{6}\right]=\lim_{x \to 0+}\frac{6e^{\sin x\ln x}-6e^{x\ln\sin x}+x^3\ln x}{6x^3}.$$
Here, you can apply L'Hôpital's rule, but it's too complicated. Moreover, you can also apply Taylor's formula, for example
$$e^{\sin x\ln x}=1+\sin x\ln x+\frac{1}{2}(\sin x\ln x)^2+\cdots,\\e^{x\ln\sin x}=1+x\ln\sin x+\frac{1}{2}(x\ln\sin x)^2+\cdots,$$
but you cannot cancel the terms, thus you cannot avoid differentiating, either. Is there any elegant solution?
P.S. Please don't suspect the existence of the limit. The result equals $\dfrac{1}{6}.$
|
The key here is obtaining the expansions for $x^{\sin x} $ and $\sin^xx$ but it is simpler to deal with their logarithms. Consider
\begin{align}
f(x) &=\sin x \log x-x\log\sin x\notag\\
&=(x\log x)\left(1-\frac{x^2}{6}+\frac{x^4}{120}-\dots \right)-x\log x-x\log\left(1-\frac{x^2}{6}+\frac{x^4}{120}-\dots\right)\notag\\
&=\notag\\
&=-\frac{x^3\log x} {6}+\frac{x^3}{6}+o(x^3)\notag
\end{align}
And then the expression under limit is $$\sin^xx\cdot\frac{e^{f(x)} - 1}{x^3}+\frac{\log x} {6}$$ which can be further written as $$\sin^xx \left(\frac{e^{f(x)} - 1}{x^3}+\frac{\log x} {6}\right) +\frac{\log x} {6}\cdot(1-\sin^xx) $$ From the expansion of $f(x) $ it is clear that $f(x) =o(x^2)$ and hence $$\frac{e^{f(x)} - 1}{x^3}=-\frac{\log x} {6}+\frac{1}{6}+o(1)$$ The desired limit is $1/6$ if we can prove that $\sin^xx \to 1$ and $(\log x) (1-\sin^xx) \to 0$ and this is not difficult to prove.
We just need to note that $$(\log x) \cdot(1-\sin^xx)=-\frac{\exp(x\log\sin x) - 1}{x\log\sin x} \cdot x\log \sin x\cdot\log x$$ and the fraction above tends to $1$ so that the limit of above expression is equal to the limit of $$-x\log x\left(\log x+\log\frac{\sin x} {x} \right) $$ which is same as $$-x(\log x) ^2-(x\log x) \log\frac{\sin x} {x} $$ and the above clearly tends to $0$.
|
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|
Prove $\tan\frac\pi{16}+2\tan\frac\pi8+4=\cot\frac{\pi}{16}$
Prove that $\tan\dfrac{\pi}{16}+2\tan\dfrac{\pi}{8}+4=\cot\dfrac{\pi}{16}$
My Attempt
\begin{align}
&\tan\dfrac{\pi}{16}+2\tan\dfrac{\pi}{8}+4=\dfrac{1}{\cot\dfrac{\pi}{16}}+\dfrac{2}{\cot\dfrac{\pi}{8}}+4\\
&=\dfrac{1}{\cot\dfrac{\pi}{16}}+2\dfrac{2\cot\dfrac{\pi}{16}}{\cot^2\dfrac{\pi}{16}-1}+4=\dfrac{\cot^2\dfrac{\pi}{16}-1+4\cot^2\dfrac{\pi}{16}+4\cot^3\dfrac{\pi}{16}-\cot\dfrac{\pi}{16}}{\cot\dfrac{\pi}{16}(\cot^2\dfrac{\pi}{16}-1)}\\
&=
\end{align}
I don't think its going anywhere with my attempt, Is there an easy way to prove this ?
I have checked a similar post Reducing $\tan\frac{\pi}{16} + 2\tan\frac{\pi}{8}+4$ to $\cot\frac{\pi}{16}$, but as it was a multiple choice question hope there \d be any direct way to solve this.
|
Given $$\tan\dfrac{\pi}{16}+2\tan\dfrac{\pi}{8}+4=\cot\dfrac{\pi}{16}$$
Now $$\cot \theta-\tan\theta=\dfrac{\cos\theta}{\sin\theta}-\dfrac{\sin\theta}{\cos\theta}=\dfrac{\cos^2\theta-\sin^2\theta}{\sin\theta\cos\theta}=\dfrac{\cos2\theta}{\dfrac12\sin2\theta}=2\cot2\theta$$
Since $\cot\theta-\tan\theta=2\cot2\theta\ \ $ we get
$$\cot\theta-2\cot2\theta=\tan\theta$$
Now
$\tan\dfrac{\pi}{16}+2\tan\dfrac{\pi}{8}+4=\cot\dfrac{\pi}{16}-2\cot\dfrac{\pi}{8}+2\left(\cot\dfrac{\pi}{8}-2\cot\dfrac{\pi}{4}\right)+4$
$$=\cot\dfrac{\pi}{16}-4\cot\dfrac{\pi}{4}+4$$$$=\cot\dfrac{\pi}{16}-4+4$$
$$=\cot\dfrac{\pi}{16}$$
Therefore,
$$\tan\dfrac{\pi}{16}+2\tan\dfrac{\pi}{8}+4=\cot\dfrac{\pi}{16}$$
|
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|
Computing $\;\lim\limits_{x\to 4}\left (\frac{\frac{\pi}{6} - \arcsin\left(\frac{\sqrt{x}}{4}\right)}{\sqrt[3]{2x-7}-1}\right) $ without L'Hôpital? $$
\lim_{x\to 4}\left(\frac{\frac{\pi}{6} - \arcsin\left(\frac{\sqrt{x}}{4}\right)}{\sqrt[3]{2x-7}-1}\right)
$$
Hello! I need to solve this limit. I had solved it with the rule of L'Hôpital, but i can't without it. I tried multiplication by conjugate expression and using Special Limits. Please help me, I must solve it using only Special Limits and simple transformations. I can't use derivatives.
|
HINT
We have that by definition of derivative by $f(x)=\arcsin(\frac{\sqrt{x}}{4})$ and $g(x)=\sqrt[3]{2x-7}$ we have
$$\lim_{x\to 4} \frac{\frac{\pi}{6} - \arcsin(\frac{\sqrt{x}}{4})}{\sqrt[3]{2x-7}-1}=\lim_{x\to 4} \frac{\frac{\pi}{6} - \arcsin(\frac{\sqrt{x}}{4})}{x-4}\frac{x-4}{\sqrt[3]{2x-7}-1}=-\frac{f'(4)}{g'(4)}$$
After editing
In order to use first order exapansion let $x=y+4$ with $y\to 0$ and we obtain
$$\lim_{x\to 4} \frac{\frac{\pi}{6} - \arcsin(\frac{\sqrt{x}}{4})}{\sqrt[3]{2x-7}-1}=\lim_{y\to 0} \frac{\frac{\pi}{6} - \arcsin(\frac{\sqrt{y+4}}{4})}{\sqrt[3]{2y+1}-1}$$
then we have
*
*$\sqrt{y+4}=2(1+y/4)^\frac12=2+\frac y4+o(y)$
*$\arcsin(\frac{\sqrt{y+4}}{4})=\arcsin\left(\frac12+\frac y{16}+o(y)\right)=\frac{\pi}6+\frac{\sqrt 3}{24}y+o(y)$
*$\sqrt[3]{2y+1}=(1+2y)^\frac13=1+\frac23 y+o(y)$
then we obtain
$$\frac{\frac{\pi}{6} - \arcsin(\frac{\sqrt{y+4}}{4})}{\sqrt[3]{2y+1}-1}=\frac{-\frac{\sqrt 3}{24}y+o(y)}{\frac23 y+o(y)}=\frac{-\frac{\sqrt 3}{24}+o(1)}{\frac23 +o(1)}\to -\frac{\sqrt 3}{16}$$
and you can check the result by the first hint.
|
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|
Evaluate $\lim_{x \to \infty}x^4\left(\arctan\frac{2x^2+5}{x^2+1}-\arctan\frac{2x^2+7}{x^2+2}\right)$
Evaluate $$\lim_{x \to \infty}x^4\left(\arctan\frac{2x^2+5}{x^2+1}-\arctan\frac{2x^2+7}{x^2+2}\right)$$
My Solution
Denote
$$f(t):=\arctan t.$$
By Lagrange's Mean Value Theorem，we have
$$f\left(\frac{2x^2+5}{x^2+1}\right)-f\left(\frac{2x^2+7}{x^2+2}\right)=f'(\xi)\left(\frac{2x^2+5}{x^2+1}-\frac{2x^2+7}{x^2+2}\right)=\frac{3}{(1+\xi^2)(x^2+1)(x^2+2)}$$
where $$\frac{2x^2+5}{x^2+1}\lessgtr \xi \lessgtr \frac{2x^2+7}{x^2+2}.$$
Here, applying the Squeeze Theorem, it's easy to see$$\lim_{x \to \infty}\xi=2.$$
It follows that
$$\lim_{x \to \infty}x^4\left(\arctan\frac{2x^2+5}{x^2+1}-\arctan\frac{2x^2+7}{x^2+2}\right)=\lim_{x \to \infty}\frac{3x^4}{(1+\xi^2)(x^2+1)(x^2+2)}=\frac{3}{5}.$$
Hope to see other solutions.THX.
|
Recall that by Arctangent addition formula
$$\arctan\dfrac{2x^2+5}{x^2+1}-\arctan\dfrac{2x^2+7}{x^2+2}=\arctan \left(\frac{\dfrac{2x^2+5}{x^2+1}-\dfrac{2x^2+7}{x^2+2}}{1+\dfrac{2x^2+5}{x^2+1}\dfrac{2x^2+7}{x^2+2}}\right)=$$
$$=\arctan \left(\frac{3}{5x^4+27x^2+37}\right)$$
therefore
$$x^4\left(\arctan\dfrac{2x^2+5}{x^2+1}-\arctan\dfrac{2x^2+7}{x^2+2}\right)=\frac{\arctan \left(\frac{3}{5x^4+27x^2+37}\right)}{\frac{3}{5x^4+27x^2+37}}\frac{3x^4}{5x^4+27x^2+37}\to \frac35$$
|
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|
Prove that $C_{3 \over 2}^n$ is bounded given $C_{a}^n = \frac{a(a-1)(a-2)\dots(a-n+1)}{n!}$
Let:
$$
\begin{cases}
C_{a}^n = \frac{a(a-1)(a-2)\dots(a-n+1)}{n!}\\
C_{a}^0 = 1
\end{cases}
$$
Prove $C_{3 \over 2}^n$ is bounded.
I've started with finding a reduced formula:
$$
C_{3\over 2}^n = \frac{{3\over 2}\left({3\over 2} - 1\right)\left({3\over 2} - 2\right)\left({3\over 2} - 3\right)\cdots\left({3\over 2} - n+1\right)}{n!} =\\
= \frac{3\cdot 1 \cdot(-1)\cdot(-3)\cdots(3-2n)}{2^n\cdot n!} = \\
\frac{1}{2^n \cdot n!}\prod_{k=1}^n\left(5-2k\right)
$$
This seems to converge to $0$ and therefore the sequence should be bounded, but how do i formally show that using inequalities? Moreover the elements change their sign depending on $n$ is even/odd (should i consider absolute values?).
Please note the precalculus tag. Thank you!
|
We should start from using the known expression of Gamma function:
$\Gamma(z)= \frac{\Gamma(z+n+1)}{z(z+1)....(z+n)}$ following that $z(z+1)....(z+n-1) =\frac{\Gamma(z+n+1)}{\Gamma(z) (n+z)}$
On the other hand:
$C_{a}^n = \frac{a(a-1)(a-2)\dots(a-n+1)}{n!}=(-1)^n\frac{(-a)(-a+1)(-a+2)\dots(-a+n-1)}{n!}$
If $z=-a=-\frac{3}{2}$ put it back to $\Gamma(z)$ we get:
$C_{\frac{3}{2}}^n =\frac{(-1)^n}{n!}\frac{\Gamma(n-\frac{1}{2})}{\Gamma(-\frac{3}{2}) (n-\frac{3}{2})}$
As $\Gamma(n-\frac{1}{2})=\frac{\Gamma(n+\frac{1}{2})}{(n-\frac{1}{2})}$ using that $\Gamma(n+\frac{1}{2})=\frac{(2n)! \sqrt{\pi}}{4^n n!}$ and $\Gamma(-\frac{3}{2})=\frac{4\sqrt{\pi}}{3}$
We have the following:
$C_{\frac{3}{2}}^n =3\frac{(-1)^n}{4^n}\binom{2n}{n}\frac{1}{2n-1}\frac{1}{2n-3}$
We handle the even and odd terms separately and perform the ratio test for both terms:
$n={2m}$, if n - even and ${2m+1}$, if n - odd.
Let's take $\frac{C^{2m+2}_{\frac{3}{2}}}{C^{2m}_{\frac{3}{2}}}$ and $\frac{C^{2m+3}_{\frac{3}{2}}}{C^{2m+1}_{\frac{3}{2}}}$
After perform the simplifications we have:
$\frac{C^{2m+2}_{\frac{3}{2}}}{C^{2m}_{\frac{3}{2}}}=\frac{(4m-1)(4m-3)}{(4m+2)(4m+4)}\lt 1$ so the even terms are decreasing and tends to zero, because $C_{\frac{3}{2}}^{2m} \gt 0$
$\frac{C^{2m+3}_{\frac{3}{2}}}{C^{2m+1}_{\frac{3}{2}}}=\frac{(4m+1)(4m-1)}{(4m+6)(4m+4)}\lt 1$ so the odd terms are increasing and tends to zero, because $C_{\frac{3}{2}}^{2m+1} \lt 0$
So $C_{\frac{3}{2}}^n \rightarrow{0}$ if $n\rightarrow {\infty}$.
|
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|
Characteristic polynomial modulo 12
Consider the vector space $V =\left\{a_0+a_1x+\cdots+a_{11}x^{11},\;a_i\in\mathbb{R}\right\}$. Define a linear operator $A$ on $V$ by $A(x^i) = x^{i+4}$ where $i + 4$ is taken modulo $12$.
Find $(a)$ the minimal polynomial of $A$ and $(b)$ the characteristic polynomial of $A$.
My try:
I coulnot find another way so I tried the brute force method. $\:$I found the matrix representation of the operator is $$A=\begin{bmatrix} 0&0&0&0&0&0&0&0&1&0&0&0\\0&0&0&0&0&0&0&0&0&1&0&0\\0&0&0&0&0&0&0&0&0&0&1&0\\0&0&0&0&0&0&0&0&0&0&0&1\\1&0&0&0&0&0&0&0&0&0&0&0\\0&1&0&0&0&0&0&0&0&0&0&0\\0&0&1&0&0&0&0&0&0&0&0&0\\0&0&0&1&0&0&0&0&0&0&0&0\\0&0&0&0&1&0&0&0&0&0&0&0\\0&0&0&0&0&1&0&0&0&0&0&0\\0&0&0&0&0&0&1&0&0&0&0&0\\0&0&0&0&0&0&0&1&0&0&0&0\\\end{bmatrix}.$$
The characteristic polynomial I found to be $\lambda^{12}-4\lambda^9+6\lambda^6-4\lambda^3+1$.$\:$(It took me almost 40 minutes. $\:$Is there another way to do this problem? Provide hints or suggestions please.
$\rule{17cm}{1pt}$
Taking forward the answer provided by $\textbf{Servaes}$ "The minimal polynomial of $A|_{U_i}$ is still $X^3−1$." Taking $U_1=span\{x_1,x_5,x_9\}$ we see $A(x)=x^5,\:A^2(x)=x^9,\:A^3(x)=x\implies (A^3-I)=0$ and since it factors into linear irredeucible factors, we have the minimal polynomial of $A|_{U_i}$ is $X^3−1$.
Completing the proof: We show that characteristic polynomial of $A$ is the product of characteristic polynomials of $A|_{U_i}$ where $V=\oplus U_i$. We have seen that minimal polynomial of $A|_{U_i}$ is $X^3−1$ which is precisely the characteristic polynomial. So let $p_i(\lambda)$ is characteristic polynomial corresponding to eigenvalue $\lambda_i$ and invariant subspace $U_i$ and $p(\lambda)$ is the characteristic polynomial of A. Then $\displaystyle p(\lambda_i)=0 \:\forall\;i \implies p_i(\lambda)|p(\lambda) \:\forall\;i \implies p(\lambda)=\prod_ip_i(\lambda)$.
$\Big($$p(\lambda)$ is atleast $\displaystyle\prod_ip_i(\lambda)$. If $\exists\lambda\neq\lambda_i \forall\: i$ such that $p(\lambda)=0$ then $V$ is not $\oplus U_i$ $\Big)$
$\rule{17cm}{1pt}$
Minimal polynomial : $X^3-1\qquad$ Characteristic polynomial : $(X^3-1)^4$.
|
It usually helps to find some nontrivial relation that the given operator satisfies. Clearly
$$A^3(x^i)=x^{i+12}=x^i,$$
for all $i$, so $A$ is a zero of $X^3-I$. This factors as
$$X^3-I=(X-I)(X^2+X+I),$$
which has no repeated factors, so this is the minimal polynomial of $A$. The characteristic polynomial has the same irreducible factors and has degree $12$, so the minimal polynomial equals
$$(X-I)^a(X^2+X+I)^b,$$
for some positive integers $a$ and $b$ satisfying $a+2b=12$.
Note that $a$ is the algebraic multiplicity of the eigenvalue $1$, which is at least $4$ because
$$1+x^4+x^8,\qquad x+x^5+x^9,\qquad x^2+x^6+x^{10},\qquad x^3+x^7+x^{11},$$
are four linearly independent eigenvectors with eigenvalue $1$. So $(a,b)$ is either $(4,4)$, $(6,3)$, $(8,2)$ or $(10,1)$.
The following is a bit contrived and implicitly assumes some slightly advanced ideas, but it is an (almost) computation-free way of determining the characteristic polynomial:
For $i\in\{1,2,3,4\}$ let $U_i:=\operatorname{span}(x^i,A(x^i),A^2(x^i))$. Then $U_i\cap U_j=0$ whenever $i\neq j$, and the $U_i$ together span $V$ and are invariant under $A$. This yields a decomposition
$$V=U_1\oplus U_2\oplus U_3\oplus U_4,$$
of $A$-invariant subspaces. The minimal polynomial of $A\vert_{U_i}$ is still $X^3-1$ (verify this!), hence it is the characteristic polynomial of the restriction. The characteristic polynomial of $A$ is the product of the characteristic polynomials of the $A\vert_{U_i}$, so it is $(X^3-1)^4$.
|
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|
Prove that $\binom{n}{1}^2+2\binom{n}{2}^2+\cdots +n\binom{n}{n}^2=n\binom{2n-1}{n-1}$
Prove that
$$
\binom{n}{1}^2+2\binom{n}{2}^2+\cdots + n\binom{n}{n}^2
= n \binom{2n-1}{n-1}.
$$
So
$$
\sum_{k=1}^n k \binom{n}{k}^2
= \sum_{k=1}^n k \binom{n}{k}\binom{n}{k}
= \sum_{k=1}^n n \binom{n-1}{k-1} \binom{n}{k}
= n \sum_{k=0}^{n-1} \frac{(n-1)!n!}{(n-k-1)!k!(n-k-1)!(k+1)!}
= n^2 \sum_{k=0}^{n-1} \frac{(n-1)!^2}{(n-k-1)!^2k!^2(k+1)}
=n^2 \sum_{k=0}^{n-1} \binom{n-1}{k}^2\frac{1}{k+1}.
$$
I do not know what to do with $\frac{1}{k+1}$, how to get rid of that.
|
$$\sum_{k=1}^{n}k\binom{n}{k}^{2}=n\sum_{k=1}^{n}\binom{n-1}{k-1}\binom{n}{n-k}=n\sum_{k=0}^{n-1}\binom{n-1}{k}\binom{n}{n-k-1}=n\binom{2n-1}{n-1}$$
Applying Vandermonde's identity in third equality.
|
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|
Determine whether the solution space of the system $Ax=0$ is a line through the origin, a plane through the origin, or the origin only Q:Determine whether the solution space of the system $Ax=0$ is a line through the origin, a plane through the origin, or the origin only. If it is a plane, find an equation for it; if it is a line, find parametric equations for it.$(i)
\left( \begin{array}{rrr}
1 & -2 & 7 \\
-4 & 8 & 5 \\
2 & -4 & 3 \\
\end{array}
\right)(ii)\left( \begin{array}{rrr}
1 & 2 & 3 \\
2 & 5 & 3 \\
1 & 0 & 8 \\
\end{array}
\right)(iii)\left( \begin{array}{rrr}
-1 & 1 & 1 \\
3 & -1 & 0 \\
2 & -4 & -5 \\
\end{array}
\right)$My Approach:First of all i transforming the matrices to reduced row echelon form:$(i)
\left( \begin{array}{rrr}
1 & -2 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
\end{array}
\right)(ii) \left( \begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)(iii) \left( \begin{array}{rrr}
1 & 0 & \frac{1}{2} \\
0 & 1 & \frac{3}{2} \\
0 & 0 & 0 \\
\end{array}
\right)$But now i get stuck because i don't know how to related them geometrically.Anyone can explain it elaborately that's make my intuition on it.Any hints or solution will be appreciated.
Thanks in advance.
|
Assuming that the reduction has been done correctly, (i) has a two-dimensional image space, so its null space is one-dimensional, hence a line.
To find a parametrisation, note that $y$ can be chosen freely, $z=0$ is forced by the second equation in reduced form, and the first one forces $x=2y$, so
$$\begin{pmatrix}2t\\ t\\0 \end{pmatrix}= t\begin{pmatrix}2\\ 1\\0 \end{pmatrix}$$ is a possible parametrisation of the solution of $Ax=0$.
(ii) has a three-dimensional image space, so the null space is just $\{(0,0,0)\}$.
In (iii) the third variable $z$ is free, $y - \frac{3}{2}z=0$ so $y=\frac{3}{2}z$, and $x - \frac{1}{2}z=0$, so $x=\frac{1}{2}z$, so a possible parametrisation is (taking $z=2t$ to get rid of the fractions):
$$\begin{pmatrix}t\\ 3t\\2t \end{pmatrix} = t\begin{pmatrix}1\\ 3\\2 \end{pmatrix} $$
among others.
|
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|
Proving trigonometric identities I’ve had a bit of difficulty of this question:
(1+sinA+cosA)/(1-sinA+cosA)=(1+sinA)/cosA
I tried to do:
(SinA)^2+(CosA)^2+sinA+cosA/(SinA)^2+(CosA)^2-sinA+cosA=(1+sinA)/cosA
But then I’m kind of lost. Any help will be appreciated! Additionally, I am not allowed to move one side to another (over the equal sign).
|
$$
\frac{1+\sin A +\cos A}{1-\sin A + \cos A} = \frac{1+\sin A}{\cos A}\\
\cos A + \sin A \cos A + \cos^2 A = (1- \sin A + \cos A) (1+\sin A)\\
\cos A + \sin A \cos A + \cos^2 A = 1- \sin A + \cos A + \sin A - \sin^2 A + \sin A \cos A\\
\cos A + \sin A \cos A + \cos^2 A + \sin^2 A = 1- \sin A + \cos A + \sin A + \sin A \cos A\\
\cos A + \sin A \cos A + 1 = 1 + \cos A + \sin A \cos A\\
$$
Read from bottom to top.
|
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|
$\frac{1}{15}<(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot \cdot \cdot \cdot\frac{99}{100})<\frac{1}{10}$. Show that
$$\frac{1}{15}<(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot \cdot \cdot\cdot\frac{99}{100})<\frac{1}{10}$$
My attempt: This problem is from a text book where is introduced as: https://en.wikipedia.org/wiki/Generalized_mean
This wouldn't be a problem if I knew the sum or a product of the given numbers. Then I will use AG inequality, but I don't know how.
|
Let $P_n:=\prod\limits_{k=1}^n\,\dfrac{2k-1}{2k}$ for each integer $n\geq 1$. We shall prove that $$\frac{2}{3\sqrt{2n}}<P_n<\frac{1}{\sqrt{2n}}\tag{*}$$
for every positive integer $n$, as suggested by Kemono Chen (I think there should be a factor $\dfrac{1}{\sqrt{2}}$ there so that (*) implies the OP's inequality for $n=50$). The asymptotic behavior of $P_n$ is given here.
Observe that
$$P_n^2\leq \frac{1}{2^2}\,\left(\prod_{k=2}^n\,\frac{2k-1}{2k}\right)\,\left(\prod_{k=2}^n\,\frac{2k}{2k+1}\right)=\frac{1}{4}\,\prod_{k=3}^{2n}\,\frac{k}{k+1}=\frac{3}{4(2n+1)}.$$
In addition,
$$P_n^2\geq \frac{1}{2^2}\,\left(\prod_{k=2}^n\,\frac{2k-1}{2k}\right)\,\left(\prod_{k=2}^n\,\frac{2k-2}{2k-1}\right)=\frac{1}{4}\,\prod_{k=2}^{2n-1}\,\frac{k}{k+1}=\frac{2}{4(2n)}\,.$$
This shows that
$$\frac{1}{2\,\sqrt{n}}\leq P_n\leq \frac{\sqrt{3}}{2\,\sqrt{2n+1}}\,.$$
Both the inequality on the right and the inequality on the left have a unique equality case: $n=1$. Note that this implies (*). In particular, for $n=50$, we have
$$\frac{1}{15}<0.0707<\frac{1}{2\cdot\sqrt{50}}<P_{50}<\frac{\sqrt{3}}{2\cdot \sqrt{101}}<0.0862<\frac{1}{10}\,.$$
|
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What am I doing wrong finding the derivative of $\frac{3-x}{2}\sqrt{1-2x-x^2}+2\arcsin{\frac{1+x}{\sqrt{2}}}$? $$y=\frac{3-x}{2}\sqrt{1-2x-x^2}+2\arcsin{\frac{1+x}{\sqrt{2}}}$$
For convenience, let
$$A=\frac{3-x}{2}\sqrt{1-2x-x^2},$$
$$B=2\arcsin{\frac{1+x}{\sqrt{2}}}.$$
$$y'=A'+B'$$
$$A'=(-\frac{1}{2})(\sqrt{1-2x-x^2})+(-2x)(\frac{1}{2}\cdot \frac{1}{{\sqrt{1-2x-x^2}}})(\frac{3-x}{2})$$
$$A'=(-\frac{1}{2})(\sqrt{1-2x-x^2})+(-x)(\frac{1}{{\sqrt{1-2x-x^2}}})(\frac{3-x}{2})$$
$$A'=(-\frac{1-2x-x^2}{2\sqrt{1-2x-x^2}})+(\frac{x^2-3x}{{2\sqrt{1-2x-x^2}}})$$
$$A'=(-\frac{1-2x-x^2}{2\sqrt{1-2x-x^2}})+(\frac{x^2-3x}{{2\sqrt{1-2x-x^2}}})$$
$$A'=\frac{2x^2-x-1}{{2\sqrt{1-2x-x^2}}}$$
$$B'=2\bigg(\arcsin{\frac{1+x}{\sqrt{2}}}\bigg)'$$
$$B'=2 \bigg( \frac{1+x}{\sqrt{2}} \bigg)' \bigg(\arcsin{\frac{1+x}{\sqrt{2}}}\bigg)'$$
$$B'=\sqrt{2} \bigg( \frac{1}{\sqrt{1- \frac{1+2x+x^2}{2}} } \bigg)$$
$$B'=\sqrt{\frac{4}{1-2x-x^2}}$$
$$B'=\frac{4}{2\sqrt{1-2x-x^2}}$$
$$y'=A'+B'=\frac{2x^2-x+3}{2\sqrt{1-2x-x^2}}$$
The answer in the book is
$$y'=\frac{x^2}{\sqrt{1-2x-x^2}}$$
|
In the first line of your derivation of $A'$: after the first summand, your second summand should be $$\color{blue}{(-2 -2x)}\left(\frac{1}{2}\cdot \frac{1}{{\sqrt{1-2x-x^2}}}\right)\left(\frac{3-x}{2}\right)$$
So in fact, $$A'=\left(-\frac{1}{2}\right)\left(\sqrt{1-2x-x^2}\right)+ \color{blue}{(-2 -2x)}\left(\frac{1}{2}\cdot \frac{1}{{\sqrt{1-2x-x^2}}}\right)\left(\frac{3-x}{2}\right)$$
|
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How to prove that $(-18+\sqrt{325})^{\frac{1}{3}}+(-18-\sqrt{325})^{\frac{1}{3}} = 3$ How to prove that $\left(-18+\sqrt{325}\right)^{\frac{1}{3}}+\left(-18-\sqrt{325}\right)^{\frac{1}{3}} = 3$ in a direct way ?
I have found one indirect way to do so: Define $t=\left(-18+\sqrt{325}\right)^{\frac{1}{3}}+\left(-18-\sqrt{325}\right)^{\frac{1}{3}}$, and then observe $t$ is a root of $f(x) = x^3 + 3x - 36 = 0$. Also observe that $f(3) = 0$. Now $f(x)$ has only one real root, since $f'(x) = x^2+3$ has no real solution. So $t =3$, as desired.
But I couldn't find any direct/algebraic way of fiddling it for a while. I guess morever this is true: $\left(-\frac{t(t^2+3)}{2}+\sqrt{\left[\frac{t(t^2+3)}{2}\right]^2 + 1}\right)^{\frac{1}{3}}+\left(-\frac{t(t^2+3)}{2}-\sqrt{\left[\frac{t(t^2+3)}{2}\right]^2 + 1}\right)^{\frac{1}{3}} = t$, which can possibly be proved in the same indirect way but I don't know how to prove this by direct manipulation.
One way maybe is to write $z_+ = \left(-\frac{t(t^2+3)}{2}+\sqrt{\left[\frac{t(t^2+3)}{2}\right]^2 + 1}\right)^{\frac{1}{3}}$, $z_- = \left(-\frac{t(t^2+3)}{2}-\sqrt{\left[\frac{t(t^2+3)}{2}\right]^2 + 1}\right)^{\frac{1}{3}}$, and writing $t = z_+ + z_-$, observe $z_+z_- = -1$, so $(x-z_+)(x-z_-) = x^2 - tx -1$, and then with Fundamental Theorem of Symmetric Polynomials, maybe calculating $(z_+^3 - z_-^3)^2$ and $(z_+^3 + z_-^3)$ in terms of $t$ would help ?
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Compute
$$\left(\frac{3+\sqrt{13}}2\right)^3=18+5\sqrt{13}.$$
Therefore
$$\sqrt[3]{18+5\sqrt{13}}=\frac{3+\sqrt{13}}2.$$
Similarly
$$\sqrt[3]{\pm 18\pm 5\sqrt{13}}=\frac{\pm 3\pm\sqrt{13}}2$$
where the signs on both sides correspond. Then
$$\sqrt[3]{-18+ 5\sqrt{13}}-\sqrt[3]{-18-5\sqrt{13}}
=\frac{-3+\sqrt{13}}2+\frac{-3-\sqrt{13}}2=-3.$$
|
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|
Let $3\sin x +\cos x =2 $ then $\frac{3\sin x}{4\sin x+3\cos x}=?$ Let $3\sin x +\cos x =2 $ then $\dfrac{3\sin x}{4\sin x+3\cos x}=\,?$
My try :
$$\frac{\frac{3\sin x}{\cos x}}{\frac{4\sin x+3\cos x}{\cos x}}=\frac{3\tan x}{4\tan x+3} =\;?$$
Now we have to find $\tan x$ from $3\sin x +\cos x =2 $ but how?
|
HINT
By tangent half-angle substitution with $t=\tan \frac x 2$ we have
$$3\sin x +\cos x =2 \iff 3\frac{2t}{1+t^2}+\frac{1-t^2}{1+t^2}=2 \iff 3t^2-6t+1=0$$
then
$$\dfrac{3\sin x}{4\sin x+3\cos x}=\frac{6t}{-3t^2+8t+3}=\frac{6t}{(-3t^2-1)+8t+4}=\frac{6t}{2t+4}=\frac{3t}{t+2}$$
As an alternative since $\cos x=\pm \frac1{\sqrt{1+\tan^2x}}$ we can use that
$$3\sin x +\cos x =2\iff \cos x\left(3\tan x+1\right)=2 \iff 3\tan x+1=\pm 2\sqrt{1+\tan^2x}$$
|
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|
The commutative law of multiplication. The commutative law of multiplication states that:
If '$a$' and '$b$' are any whole numbers,
then
'$a · b = b · a$'.
I've been looking for why multiplication has this property and how is saying that 3 times 4 the same thing as saying 4 times 3. I've found two interpretations:
*
*$3 \cdot 4 = 12$ and $4 \cdot 3 = 12$ so, $3 \cdot 4 = 4 \cdot 3$,
*Think of $3\times 4$ as the area of a rectangle with width $4$ and height $3$. If you rotate the rectangle by $90^\circ$, the area is unchanged, but now the width is $3$, and the height is $4$. So, $3\times4 =$ area of rectangle = area of rotated rectangle $= 4\times3$.
These explanations don't seem to show why multiplication has this property but are showing us a way to make it intuitive and, thus, accept it as common sense.
Is there a better way to show that multiplication is fundamentally commutative?
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a * b = $\sum_{n=1}^b a$ = (a / b) (b / a) $\sum_{n=1}^b a$ = (a / b) $\sum_{n=1}^b b$ = $\sum_{n=1}^a b$ = b * a
For example, 3 * 4 = 3 + 3 + 3 + 3 = (3 / 4) (4 / 3) (3 + 3 + 3 + 3) = (3 / 4) (4 + 4 + 4 + 4) = 4 + 4 + 4 = 4 * 3
In words, 3 * 4 is 3 added 4 times (3 + 3 + 3 + 3). If you take 4/3 of that, all the 4s become 3s (4 + 4 + 4 + 4). In order to keep the expression constant, you need to multiply by 3/4. Since there are 4 4s in (4 + 4 + 4 + 4), 3/4 of that is just 3 4s (4 + 4 + 4), which is 4 added 3 times or 4 * 3.
|
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|
Prove that $\int_0^{\frac{n\pi}{4}} \frac{1}{3\sin^4(x) + 3\cos^4(x) -1} dx = \frac{n\pi}{4} $
Prove that $$\int_0^\frac{n\pi}{4} \left(\frac{1}{3\sin^4(x) + 3\cos^4(x) -1}\right) dx = \frac{n \pi}{4} $$
is true for all integers $n$.
Here I've looked at the case where $n=1$ as $t=\tan(x)$ is injective in that domain. Then I could perhaps try to use some argument about periodicity and symmetry to extend it to all the integers.
$$
= \int_0^\frac{\pi}{4} \frac{\sec^4(x)}{3\tan^4(x)+3-\sec^4(x)} \text{d}x
= \frac12 \int_0^1 \frac{t^2+1}{t^4-t^2+1} \text{d}t
$$
Which looks a little tough but reminiscent of the well-known integral:
$$ \int_0^\infty \frac{1}{x^4 + 2x^2\cos(2a) + 1} \text{d}x = \frac12\int_0^\infty \frac{x^2+1}{x^4 + 2x^2\cos(2a) + 1} \text{d}x = \frac{\pi}{4\cos(a)} $$
Where we choose $a=\frac\pi3$. But I can't seem to adjust the bounds to get an answer. How would you evaluate this integral?
|
For $x \in (0,\pi/4)$,
the integral $$\frac12 \int_0^1 \frac{t^2+1}{t^4-t^2+1} \text{d}t$$ can also be written as $$\frac12 \int_0^1 \frac{1+1/t^2}{(t-1/t)^2+1} \text{d}t$$
Substituting $u=t-\frac1t$, the integral becomes $$\frac12 \int_{-\infty}^0 \frac{1}{u^2+1} \text{d}u$$ which is equal to $$\frac12[\arctan(u)]_{-\infty}^0=\pi/4$$
Similarly for $x \in (\pi/4,\pi/2)$, the integral becomes$$\frac12 \int_1^\infty \frac{1+1/t^2}{(t-1/t)^2+1} \text{d}t$$ which is equal to$$\frac12 \int_0^{\infty} \frac{1}{v^2+1} \text{d}v$$ or$$\frac12[\arctan(u)]_0^{\infty}=\pi/4$$
And since the integrand has a period of $\pi/2$ because $$\sin^4x+\cos^4x=1-\frac12\sin^22x$$
one can easily contemplate the proof by adding the areas of the curve on and on.
|
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|
Prove that $ \frac{1}{5}+ \frac{1}{9}+ \frac{1}{13}+\ldots+ \frac{ 1}{2005}< \frac{ 7}{4}$. Prove that
$ \dfrac{1}{5}+ \dfrac{1}{9}+ \dfrac{1}{13}+\ldots+ \dfrac{ 1}{2005}< \dfrac{ 7}{4}$.
Solution
$\sum_{k=1}^{501}\frac 1{4k+1}$ $=\frac 15+\sum_{k=2}^{501}\frac 1{4k+1}$ $<\frac 15+\int_1^{501}\frac{dx}{4x+1}$ $=\frac 15+\left[\frac{\ln(4x+1)}4\right]_1^{501}$
So $\sum_{k=1}^{501}\frac 1{4k+1}$ $<\frac 15+\frac{\ln 2005-\ln 5}4$ $=\frac 15+\frac{\ln 401}4$ $<\frac 74$
Q.E.D.
My questions:
Could you please post a simpler solution?
I just don't understand why $\frac 15+\frac{\ln 401}4<\frac 74$. Please explain why is that.
Thank you in advance!
|
It is enough to invoke the Hermite-Hadamard inequality. Since $f(x)=\frac{1}{4x+1}$ is convex over $\mathbb{R}^+$,
$$ \int_{1/2}^{501+1/2}f(x)\,dx \geq f(1)+f(2)+\ldots+f(501) $$
hence
$$ \sum_{k=1}^{501}\frac{1}{4k+1}\leq \frac{\log(669)}{4} $$
and it is enough to show that $e^7>669.$ On the other hand $e>2+\frac{2}{3}$ and
$$\left(2+\frac{2}{3}\right)^7=2^7\cdot\frac{4}{3}\cdot\left(2-\frac{2}{9}\right)^3=\frac{2^{12}}{3}\left(1-\frac{1}{9}\right)^3>\frac{2^{13}}{3^2}>\frac{8000}{9}>888.$$
|
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|
Evaluate the limit of $\lim_{n\to\infty}\frac{1}{n^2}\left(\frac{2}{1}+\frac{9}{2}+\frac{64}{9}+\cdots+\frac{(n+1)^{n}}{n^{n-1}}\right)$ $$\lim_{n\to\infty}\frac{1}{n^2}\left(\frac{2}{1}+\frac{9}{2}+\frac{64}{9}+\cdots+\frac{(n+1)^{n}}{n^{n-1}}\right)$$
My try:
The limit can be written as follows:
$$\lim_{n\to\infty}\left(\frac{1}{n^2}\cdot\sum_{k=1}^{n}\frac{(k+1)^{k}}{k^{k-1}}\right)$$
Evaluate the following series:
$\sum_{k=1}^{\infty}\frac{(k+1)^{k}}{k^{k-1}}$
$\frac{(k+1)^{k}}{k^{k-1}}=k\cdot\frac{(k+1)^{k}}{k^{k}}=k\cdot\left(1+\frac{k+1}{k}-1\right)^k=k\cdot\left(1+\frac{1}{k}\right)^k$
Then:
$\lim_{k\to\infty}\frac{(k+1)^{k}}{k^{k-1}}=\lim_{k\to\infty}k\cdot\left(1+\frac{1}{k}\right)^k=e\cdot\infty\neq0 \Longrightarrow \sum_{k=1}^{\infty}\frac{(k+1)^{k}}{k^{k-1}}$ diverges.
Therefore:
$$\lim_{n\to\infty}\frac{1}{n^2}\left(\frac{2}{1}+\frac{9}{2}+\frac{64}{9}+\cdots+\frac{(n+1)^{n}}{n^{n-1}}\right)=0\cdot\infty$$
What to do next?
|
Thanks to user xbh for the hint:
Using the Stolz-Cesàro theorem, we have:
$a_n=\frac{2}{1}+\frac{9}{2}+\frac{64}{9}+\cdots+\frac{(n+1)^{n}}{n^{n-1}}$
$b_n=n^2$
Two monotone and increasing sequences.
Apply the theorem to get:
$\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\frac{\frac{(n+2)^{n+1}}{(n+1)^n}}{2n+1}=\frac{(n+2)(n+2)^{n}}{(2n+1)(n+1)^n}=\frac{n+2}{2n+1}\cdot\left(1+\frac{n+2}{n+1}-1\right)^n=\frac{n+2}{2n+1}\cdot\left[\left(1+\frac{1}{n+1}\right)^{n+1}\right]^{\frac{1}{n+1}n}$
Then:
$\lim_{n\to\infty}\frac{1}{n^2}\left(\frac{2}{1}+\frac{9}{2}+\frac{64}{9}+\cdots+\frac{(n+1)^{n}}{n^{n-1}}\right)=\lim_{n\to\infty}\frac{n+2}{2n+1}\cdot\left[\left(1+\frac{1}{n+1}\right)^{n+1}\right]^{\frac{1}{n+1}n}=\frac{e}{2}$
|
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|
Find $A$ given $A^2$ and $B$ and $(A^2)^{-1} = A^{-1}B$ This excersice took place in class I had today. The exercise was the following:
Let the regular matrix $A$,$B$:
$$A^2 = \left[ \begin{matrix} 2 &-2 & 2\\ -2 & 2 & 2 \\ 4&4&-4 \end{matrix} \right] \hspace{2cm} B = \left[ \begin{matrix} 0 &-1 & 2\\ 0 & 2 & 0 \\ 1&3&-1 \end{matrix} \right]$$
Knowing that $(A^2)^{-1} = A^{-1}B$, find A.
My Attempt
During the exam I tried the following:
$(A^2)^{-1} = A^{-1}B \Leftrightarrow A^2(A^2)^{-1} = A^2A^{-1}B \Leftrightarrow I_n = AB \Leftrightarrow A = B^{-1}$
And procedeed to find the inverse of B, which is:
$A = B^{-1} = \left[ \begin{matrix} \frac{1}{2} &\frac{-5}{4} & 1\\ 0 & \frac{1}{2} & 0 \\ \frac{1}{2}& \frac{1}{4}&0 \end{matrix} \right]$
But, as I tend to make silly mistakes, I figured out that it would be a better way to do find $A$ just by using matrix products. We had $I_n = AB \Leftrightarrow A = A^2B$. Using this method I got that
$A = A^2B = \left[ \begin{matrix} 2 & 0 & 2\\ 2 & 12 & -6 \\ -4& -8&12 \end{matrix} \right]$
But as you can see the $2$ matrix are different. Where is the mistake in my logic?
Another thing I noticed is that $I_n = AB \Leftrightarrow A*I_n*B = A*(AB)*B \Leftrightarrow I_n = AB = A^2B^2$, but $A^2B^2$ isn't the identity matrix, so maybe the exercise is wrong, but I rather not rush into that thinking.
|
Use the properties of determinant to show that the given relationship does not hold.
$$\det ((A^2)^{-1})=-\frac1{64}=\det (A^{-1})\cdot(-4)$$
from the given relationship and properties of determinant. So $\det (A)=64\cdot 4$ and this cannot be true.
|
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|
Simplified form of $\cos^{-1}\big[\frac{3}{5}\cdot\cos x+\frac{4}{5}\cdot\sin x\big]$, where $x\in\big[\frac{-3\pi}{4},\frac{3\pi}{4}\big]$
Find the simplified form of $\cos^{-1}\bigg[\dfrac{3}{5}\cdot\cos x+\dfrac{4}{5}\cdot\sin x\bigg]$, where $x\in\Big[\dfrac{-3\pi}{4},\dfrac{3\pi}{4}\Big]$
My reference gives the solution $\tan^{-1}\frac43-x$, but is it a complete solution ?
My Attempt
Let $\alpha=\cos^{-1}\dfrac{3}{5}\implies \dfrac{3}{5}=\cos\alpha,\;\dfrac{4}{5}=\sin\alpha$
$$
\cos^{-1}\bigg[\dfrac{3}{5}\cdot\cos x+\dfrac{4}{5}\cdot\sin x\bigg]=\cos^{-1}\bigg[\cos\alpha\cdot\cos x+\sin\alpha\cdot\sin x\bigg]\\
=\cos^{-1}\bigg[\cos\Big(\alpha-x\Big)\bigg]=2n\pi\pm(\alpha-x)=2n\pi\pm\Big(\tan^{-1}\frac{4}{3}-x\Big)\\
=\tan^{-1}\frac{4}{3}-x\quad\text{iff }\tan^{-1}\frac{4}{3}-x\in[0,\pi]
$$
$$
-x\in\Big[\dfrac{-3\pi}{4},\dfrac{3\pi}{4}\Big]\quad\&\quad\alpha=\tan^{-1}\frac{4}{3}\in\Big(0,\frac{\pi}{2}\Big)\\
\implies\alpha-x\in\big[\frac{-3\pi}{4},\frac{5\pi}{4}\big]\not\subset[0,\pi]
$$
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Your book is wrong. $x\in\Big[-3\pi/4, 3\pi/4\Big]$, which is an interval of length $3\pi/2$. Whatever be the value of $\alpha, \alpha-x$ will belong to an interval of length $3\pi/2$, which means $\alpha-x$ is not confined to $[0, \pi].$
So the answer is $\begin{cases}2\pi-\alpha+x,&x\in[-3\pi/4,\alpha-\pi)\\\alpha-x,&x\in[\alpha-\pi,\alpha]\\x-\alpha,&x\in(\alpha, 3\pi/4]\end{cases}$
|
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Let $n$ be the product of all positive factors of $10^4$. Find $log_{10} n$ I can't see any relationship between the products of the positive factors of powers of 10.
The answer given here is 50, by the way, but I simply don't know where to start.
|
As @NL1992 pointed out, the key is write $10^4 = 2^4 \times 5^4$.
Consider the factors which have no contribution from the "2". These are $5^0, 5^1, 5^2, 5^3$, and $5^4$. Their product is $5^{10}$.
Now, consider the factors which exactly one "2". These are $2 \times 5^0$, $2 \times 5^1$, $2 \times 5^2$, $2 \times 5^3$, and $2 \times 5^4$. Their product is $2^5 \times 5^{10}$.
Generalizing, you can see that if you multiply all factors which have a $2^k$ term, you get $2^{5k} \times 5^{10}$.
So the final answer is $n = \prod_{k=0}^4 2^{5k}5^{10} = 2^{50} 5^{50}$, implying $\log_{10}(n) = 50$.
|
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|
Find all Pythagorean triples $x^2+y^2=z^2$ where $x=21$ Consider the following theorem:
If $(x,y,z)$ are the lengths of a Primitive Pythagorean triangle, then $$x = r^2-s^2$$ $$y = 2rs$$ $$z = r^2+z^2$$ where $\gcd(r,s) = 1$ and $r,s$ are of opposite
parity.
According to the previous theorem,My try is the following:
since $x = r^2-s^2$, $x$ is difference of two squares implying that $x \equiv 0 \pmod 4$. But $x=21 \not \equiv 0 \pmod 4$. Hence, there are no triangles having such $x$.
Is that right?
Added:
My argument is false here. Please refer to the appropriate answer.
|
Recall that
$$3^2+4^2=5^2 \implies (3\cdot 7)^2+(4\cdot 7)^2=(5\cdot 7)^2$$
and note that
$$(21, 220, 221)$$
is a primitive triple.
Your criterion doesn't works because the remainder of squares $\pmod 4$ are $0,1$ therefore we can't comclude that
$$z^2-y^2\equiv 0 \pmod 4$$
What we need to solve is
$$21^2=441=3^2\cdot 7^2=(z+y)(z-y)$$
that is we need to try with
*
*$z-y=1 \quad z+y=441\implies (x,y,z)=(21,200,221)$
*$z-y=3 \quad z+y=147\implies (x,y,z)=(21,72,75)$
*$z-y=7 \quad z+y=63\implies (x,y,z)=(21,28,35)$
*$z-y=9 \quad z+y=49\implies (x,y,z)=(21,20,29)$
|
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|
Prove $f(x) = x^2\sin\left(\frac{1}{x}\right)$ is Lipschitz (no use of derivative)
Prove that $f:\mathbb{R}\to \mathbb{R}$ such that
$$ f(x) = \left\{
\begin{array}{c l}
x^2\, \sin\left(\frac{1}{x}\right) & ,\quad x\neq 0\\
0 & ,\quad x=0
\end{array} \right.$$
is Lipschitz (without use of derivatives).
Attempt. I am aware (Lipschitz-continuous $f(x)=x^2\cdot \sin\left(\frac{1}{x}\right)$)
that: $$|f(x)-f(y)|\leq 3|x-y| ~~~\forall~x,~y\in \mathbb{R},$$
but I am looking for a proof, without use of derivatives.
I tried: for $x,y\neq 0$:
\begin{eqnarray}
x^2\sin\frac{1}{x} - y^2 \sin\frac{1}{y}
&=& (x^2-y^2)\sin\frac{1}{x} + y^2\left ( \sin\frac{1}{x} - \sin\frac{1}{y} \right ),\nonumber
\end{eqnarray}
so: $$\left | x^2\sin\frac{1}{x} - y^2 \sin\frac{1}{y} \right |
\leq |x^2 - y^2| + y^2 \left | \sin\frac{1}{x} - \sin\frac{1}{y} \right |.$$
Since:
$$\left | \sin\frac{1}{x} - \sin\frac{1}{y} \right | \leq \left| \frac{1}{x} - \frac{1}{y}\right |= \frac{|x - y|}{xy},$$
we get:
$$\left | x^2 \sin\frac{1}{x} - y^2 \sin\frac{1}{y} \right |
\leq \left(x+y+\frac{y}{x}\right)|x-y|.$$
Unfortunatelly , the quantity $x+y+\frac{y}{x}$ grows to
$+\infty$, either for large $x$, or for $x\approx 0.$
Thanks in advance for the help.
|
Writing
$$
f(x) - f(y) = (x^2-y^2)\sin\frac{1}{x} + y^2\left ( \sin\frac{1}{x} - \sin\frac{1}{y} \right )
$$
is a good start. The terms on the right-hand side can be estimated better if we assume that $0 < |y| \le |x|$:
$$
\left| (x^2- y^2) \sin \frac 1x \right| \le |x-y| |x+y|| \frac 1x| \le 2|x-y|
$$
and
$$
\left|y^2\left ( \sin\frac{1}{x} - \sin\frac{1}{y} \right )\right| \le |y^2|
\left| \frac 1x - \frac 1y \right| = \left|\frac yx\right||x-y| \le |x-y|
$$
and therefore
$$
|f(x)-f(y)|\leq 3|x-y| \, .
$$
For $0 < |x| \le |y|$ repeat the same calculation with $x$ and $y$ exchanged, or use the symmetry of $f$.
Finally, for $x \ne 0 = y$
$$
|f(x) - f(0)| = \left |x^2 \sin \frac 1x \right| \le |x| = |x-0| \, .
$$
|
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|
Compute $\sum\frac1{2-A_k}$ for $(A_k)$ the $n$th roots of unity
If $1,A_1,A_2,A_3....A_{n-1}$ are the $n^{th}$ roots of unity then prove that
$$\dfrac{1}{2-A_1} + \dfrac{1}{2-A_2}+\cdots+ \dfrac{1}{2-A_{n-1}} = \dfrac{2^{n-1}(n-2) + 1}{2^n-1}$$
What I did: I tried to use some of the following formulas:
$$1+ A_1 +A_2+A_3+\cdots+A_{n-1} = 0$$
$$\dfrac{2^n - 1}{2-1} = (2 -A_1)(2-A_2)\cdots(2-A_{n-1})$$
and the fact that $|A_i| = 1$ for $i =1,2,3,\cdots,n-1$.
|
Two ingredients for this proof:
*
*The numbers $A_k$ for $1\leqslant k<n$ and $A_n=1$ are the roots of the polynomial $X^n-1$.
*If $Y=\frac1{2-X}$ then $X=\frac{2Y-1}Y$.
Thus, the numbers $B_k=\frac1{2-A_k}$ for $1\leqslant k<n$ and $B_n=\frac1{2-A_n}=1$ are the solutions of the equation $$\left(\frac{2Y-1}Y\right)^n-1=0$$ that is, the roots of the polynomial $$(2Y-1)^n-Y^n$$
The highest degree terms of this polynomial are $$(2^n-1)Y^n-n2^{n-1}Y^{n-1}+\ldots$$ hence the sum of its roots is $$\sum_{k=1}^nB_k=1+\sum_{k=1}^{n-1}\frac1{2-A_k}=\frac{n2^{n-1}}{2^n-1}$$ from which the desired result follows.
|
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|
Isomorphism between two versions of $GF(2^3)$ I have $GF(2^3)$ generated by $\Pi_1(\alpha)=x^3+x+1$ and $GF(2^3)$ generated by $\Pi_2(\alpha)=x^3+x^2+1$.
$\bullet$ $\Pi_1(\alpha)=x^3+x+1$ $000=0, 100=1,010=\alpha,001=\alpha^2,110=\alpha^3, 011=\alpha^4,111=\alpha^5,101=\alpha^6$
$\bullet$ $\Pi_2(\alpha)=x^3+x^2+1$ $000=0, 100=1,010=\lambda,001=\lambda^2,101=\lambda^3, 111=\lambda^4,110=\lambda^5,011=\lambda^6$
The exampla say that $\alpha$ and $\lambda^3$ both have minimal polynomial $\Pi_1$ and thus $\alpha \iff \lambda^3$ form an isomorphism between the two version.
How can I see in a practical way this fact? Someone can explain me this concept?
|
$\Pi_1(\lambda^3) = (\lambda^3)^3 + \lambda^3 + 1 = \lambda^9 + \lambda^3 + 1$
But you know that $\lambda^7=1$. So $\lambda^9=\lambda^2$ and
$\Pi_1(\lambda^3) = \lambda^3 + \lambda^2 + 1 = \Pi_2(\lambda) = 0$
The complete isomorphism is as follows:
$0 \leftrightarrow 0 \\ 1 \leftrightarrow 1 \\ \alpha \leftrightarrow \lambda^3 \\ \alpha^2 \leftrightarrow \lambda^6 \\ \alpha^3 \leftrightarrow \lambda^2 \\ \alpha^4 \leftrightarrow \lambda^5 \\ \alpha^5 \leftrightarrow \lambda \\ \alpha^6 \leftrightarrow \lambda^4$
|
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|
Integer Solutions of the Equation $u^3 = r^2-s^2$ The question says the following:
Find all primitive Pythagorean Triangles $x^2+y^2 = z^2$ such that $x$ is a perfect cube.
The general solution for each variable are the following:
$$x=r^2-s^2$$
$$y=2rs$$
$$z=r^2+s^2$$
such that $\gcd(r,s) = 1$and $r+s \equiv 1 \pmod {2}$
In order to make $x$ a perfect cube, I shall have the equation $x=u^3=r^2-s^2$. However, I am stuck to find a general formula for such cubes.
I know that a subset of the solutions might be the difference between two consecutive squares. This difference is always an odd integer. I can collect some examples such $14^2-13^2 = 27$ but I cannot give a formula for such type either.
Any ideas?
|
$u^3=(r+s)(r-s)$ and $\gcd(r+s,r-s)=1$, so $r+s$ and $r-s$ are odd, coprime perfect cubes.
So let $r+s=a^3$, $r-s=b^3$. Then
$$r=\frac{a^3+b^3}2$$
$$s=\frac{a^3-b^3}2$$
where $a$ and $b$ are odd and coprime.
Conversely, if $a$ and $b$ are odd and coprime, let $r=(a^3+b^3)/2$ and $s=(a^3-b^3)/2$, which are coprime and have different parity. Indeed,
$$r+s=a^3$$
which is odd and coprime with
$$r-s=b^3$$
|
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|
Can't solve a quartic equation I'm trying to solve an algebraic question.The question wants me to solve $n^4+2n^3+6n^2+12n+25=m^2$.The question also states that n is a positive integer and the answer for $n^4+2n^3+6n^2+12n+25$ is a square number. Here's how I tried to solve it:
$$n^4+2n^3+6n^2+12n+25=\\n^4+6n^2+2n^3+12n+25=\\
n^2(n^2+6)+2n(n^2+6)+5^2=\\
(n\sqrt {n^2+6})^2+2n(n^2+6)+5^2.\\$$
Because $a^2+2ab+b^2=(a+b)^2$,so
$\sqrt {(n\sqrt {n^2+6})^2}\cdot\sqrt5^2=n(n^2+6)$
Then:
$$n\sqrt {n^2+6}\cdot 5=n(n^2+6)\\
\sqrt {n^2+6}\cdot 5=n^2+6\\
25(n^2+6)=n^4+12n^2+36\\
n^4+12n^2+36=25n^2+150\\
n^4-13n^2-114=0\\
(n^2+6)(n^2-19)=0\\
n^2=19\\
n=\sqrt 19$$
But $n$ is a positive integer.
Can anyone help?
|
$$ ( n^2 + n + 2 )^2 = n^4 + 2n^3 + 5n^2 + 4n + 4 $$
$$ ( n^2 + n + 3 )^2 = n^4 + 2n^3 + 7n^2 + 6n + 9$$
The second one is larger than yours, meaning yours cannot be square, when
$$ 7n^2 + 6n+9 - 6n^2 - 12 n - 25 > 0 \; , \; $$
$$ n^2 - 6n - 16 > 0 \; , \; $$
$$ (n-8)(n+2) > 0 \; . $$
You need check only $0 \leq n \leq 8.$
ADDED: the quartic in the question also lies strictly between consecutive squares when $n \leq -3.$ The squares are just $n=-2,0,8.$
|
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|
AMC 1834 - If $a, b $ and $c$ are nonzero numbers such that (a+b-c)/c=(a-b+c)/b=(-a+b+c)/a and x=((a+b)(b+c)(c+a))/abc and x<0, then x=? I am getting stuck on this question:
If a, b, and c, are nonzero numbers such that $\frac{a+b-c}{c}$=$\frac{a-b+c}{b}$=$\frac{-a+b+c}{a}$ and x=$\frac{(a+b)(b+c)(c+a)}{abc}$ and x<0, then x=?
What I have done so far: Consider, if $\frac{a}{b}$=$\frac{c}{d}$, and (b+d) is non-zero, then $\frac{a+c}{b+d}$=$\frac{a}{b}$=$\frac{c}{d}$. So with that fact, I did... $\frac{a+b-c}{c}$=$\frac{a-b+c}{b}$ $\implies$
$\frac{2a}{c+b}$ and $\frac{2a}{c+b}$=$\frac{-a+b+c}{a}$$\implies$ $\frac{(a+b+c)}{a+b+c}$ = 1. Then I took the original equations and did $\frac{a+b-c}{c}$=1, so $a+b=2c$ , and similarly, $(a+c)=2b$ and $b+c=2a$. So plugging that into x=$\frac{(a+b)(b+c)(c+a)}{abc}$ we have $x=\frac{(2a)(2b)(2c)}{abc}$, which is $x=\frac{8(abc)}{abc}=8$. This is wrong though, because it says in the problem x<0. So how do I do this?
|
$$\frac{a+b-c}{c}=\frac{a-b+c}{b}$$
$$ab+b^2-bc=ac-bc+c^2\implies a(b-c)=(b+c)(c-b)$$
$$\implies b=c \quad OR\quad a+b+c=0$$
If we take $b=c$, $$\frac{a}{c}=\frac{-a+2c}{a}\quad using (1)and (3).$$
$$\implies ({\frac{a}{c}})^2=1\quad OR\quad -2$$
Obviosuly $-2$ isn't possible, $a=c$ will give $x=1$ and $a=-c$ will give $x=0$.
So, $a+b+c=0\implies a=-(b+c), b=-(a+c), c=-(a+b)\implies x=-1.$
|
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|
Diagonal entries are zero, others are $1$. Find the determinant. $\det\begin{vmatrix}
0 & \cdots & 1& 1 & 1 \\
\vdots & \ddots & \vdots & \vdots & \vdots \\
1 & \cdots & 1 & 1 & 0
\end{vmatrix}=?$
Attempt:
First I tried to use linearity property of the determinants such that $$\det\binom{ v+ku }{ w
}=\det\binom{v }{ w
}+k\det \binom{ u }{ w
}$$
$v,u,w$ are vectors $k$ is scalar.
I have tried to divide it into $n$ parts and tried to compose with sense but didn't acomplish.
Second I tried to make use of "Row Reduction" i.e. adding scalar multiple of some row to another does not change the determinant, so I added the all different rows into other i.e. adding $2, 3,4,\dots,n$th row to first row and similarly doing for all rows we got
$$\det\begin{vmatrix}
0 & \cdots & 1& 1 & 1 \\
\vdots & \ddots & \vdots & \vdots & \vdots \\
1 & \cdots & 1 & 1 & 0
\end{vmatrix}=\det\begin{vmatrix}
n-1 & \cdots & n-1& n-1 & n-1 \\
\vdots & \ddots & \vdots & \vdots & \vdots \\
n-1 & \cdots & n-1 & n-1 & n-1
\end{vmatrix}=0$$
The last determinant is zero (I guess) so the given determinant is zero?
I don't have the answer this question, so I am not sure. How to calculate this determinant?
|
For $n = 1$, we have
$$ \det \left[ \begin{matrix} 0 \end{matrix} \right] = 0.$$
For $n=2$, we have
$$ \det \left[ \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right] = -1. $$
For $n = 3$, we have
$$ \det \left[ \begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix} \right] = 0 \det \left[ \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right] - 1 \det \left[ \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} \right] + 1 \det \left[ \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} \right] = 2. $$
For $n = 4$, we have
$$ \det \left[ \begin{matrix} 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \end{matrix} \right] = -3. $$
For $n = 5$, we have
$$ \det \left[ \begin{matrix} 0 & 1 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 & 1 \\ 1 & 1 & 0 & 1 & 1 \\ 1 & 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 1 & 0 \end{matrix} \right] = 4. $$
So, in general, if $A$ is your $n \times n$ matrix with all diagonal entries equal to $0$ and all off-diagonal entries equal to $1$, then we have
$$ \det A = (-1)^{n-1} (n-1). $$
Hope this helps.
For calculation of determinants, I've used this online tool.
|
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|
Inequality with $(x+y)(y+z)(z+w)(w+x)=1$ Let $x,y,z,w>0$ and such that
$$(x+y)(y+z)(z+w)(w+x)=1.$$
Show that
$$\sqrt[3]{xyz}+\sqrt[3]{yzw}+\sqrt[3]{zwx}+\sqrt[3]{wxy}\le 2.$$
I'm trying to use Holder's inequality
$$(\sqrt[3]{xyz}+\sqrt[3]{yzw})^3
\le (x+y)(y+z)(z+w)$$
$$(\sqrt[3]{zwx}+\sqrt[3]{wxy})^3\le (z+w)(w+x)(x+y)$$
so
$$\sqrt[3]{xyz}+\sqrt[3]{yzw}+\sqrt[3]{zwx}+\sqrt[3]{wxy}\le \sqrt[3]{(x+y)(y+z)(z+w)}+\sqrt[3]{(z+w)(w+x)(x+y)}.$$
|
Since $$\prod_{cyc}(x+y)-\sum_{cyc}x\sum_{cyc}xyz=(xz-yw)^2\geq0,$$ by Holder we obtain:
$$2=2\left(\prod_{cyc}(x+y)\right)^{\frac{1}{4}}\geq2\left(\sum_{cyc}x\sum_{cyc}xyz\right)^{\frac{1}{4}}=$$
$$=\sqrt[4]{\left(4^2\sum_{cyc}xyz\right)\sum_{cyc}x}\geq\sqrt[4]{\left(\sum_{cyc}\sqrt[3]{xyz}\right)^3\sum_{cyc}x}.$$
Thus, it's enough to prove that
$$\sum_{cyc}x\geq\sum_{cyc}\sqrt[3]{xyz},$$ which is true by AM-GM:
$$\sum_{cyc}x=\frac{1}{3}\sum_{cyc}(x+y+z)\geq\frac{1}{3}\sum_{cyc}3\sqrt[3]{xyz}=\sum_{cyc}\sqrt[3]{xyz}.$$
Done!
I used Holder for two sequences:
$$16\sum_{cyc}xyz=\left(\sum_{cyc}1\right)^2\sum_{cyc}xyz\geq\left(\sum_{cyc}\sqrt[3]{1^2xyz}\right)^3=\left(\sum_{cyc}\sqrt[3]{xyz}\right)^3$$
|
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|
How do I get from $x - x^2 = \frac{1}{4}$ to $x =\frac{1}{2}$? I'm working on a text book problem where I need to sketch the graph of $y = 4x^2 - 4x+1$ by finding where the curve meets the $x$ axis.
To start out I set $y = 0$ then tried to isolate $x$ then,
$4x - 4x^2 = 1$
$x - x^2 = \frac{1}{4}$
From here I want to continue algebraically to reach $x = \frac{1}{2}$ which I know is the solution from plotting the curve using an app and I can see that that makes sense.
However I am missing some concepts which allow me to turn $x - x^2 = \frac{1}{4}$ into $x = \frac{1}{2}$ and wanted some help to get unblocked.
|
Starting at $ x - x^2 = 1/4 $, subtract $ 1/4 $ to each side:
$$\mathrm{(1)} \qquad x - x^2 - 1/4 = 0 $$
Arrange the left hand side of the equation $ (1) $ in descending degree of $ x $:
$$\mathrm{(2)} \qquad -x^2 + x - 1/4 = 0 $$
I would recommend having the coefficient of the term with the highest degree of $ x $ not be negative to make the equation neater. So, after multiplying a $ -1 $ to each side:
$$\mathrm{(3)} \qquad x^2 - x + 1/4 = 0 $$
Now the left side of equation $ (3) $ is a perfect square of the form $ x^2 + 2ax + a^2 = (x + a)^2 $, where $ a = -1/2 $ since $ 2ax = -x $, or $ 2a = -1 $.
Thus, $ x^2 - x + 1/4 = (x - 1/2)^2 $, and equation $ (3) $ becomes:
$$\mathrm{(4)} \qquad (x - 1/2)^2 = 0 $$
Solving for $ x $,
$$ x = 1/2 $$
but with a multiplicity of 2. The significance of the multiplicity arises from $ x = 1/2 $ being the $ x $-coordinate of the vertex touching (but not crossing) the $x$-axis. In the following graph below, $f(x) = -x^2 + x - 1/4$ and $g(x) = 0$.
Zooming in closely at the graph clearly shows the parabola ($f(x)$) touching the $x$-axis ($g(x)$) at $x = 1/2$.
|
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|
Integrate $\int_{0}^{\infty}e^{-pt}\sin\left(\sqrt{t}\right)\mathrm dt$ I need the following Laplace transform to solve the Differential Equation
$$\int_{0}^{\infty}e^{-pt}\sin\sqrt{t}\, dt, \quad \text{where} \ \ \ p>0$$
I tried Integration by parts after substituting $t=x^2$, but didn't work.
\begin{align}
\int_{0}^{\infty}e^{-pt}\sin\sqrt{t}dt &\overset{t=x^2}= \int_{0}^{\infty}e^{-px^2}2x\sin xdx \ = \ \text{I}\\
& = \sin x \ \frac{e^{-px^2}}{-p} + \frac{1}{p}\int_{0}^{\infty}e^{-px^2}\cos xdx \\
& = \sin x \ \frac{e^{-px^2}}{-p} + \frac{1}{p}\left(-e^{-px^2}\sin x - \int_{0}^{\infty}e^{-px^2}(-2px)(-\sin x)dx\right) \\
& = -\sin x \ \frac{e^{-px^2}}{p} + \frac{1}{p}\left(-\sin xe^{-px^2} - p\int_{0}^{\infty}e^{-px^2}(2x)(\sin x)dx\right) \\
& = -\sin x \ \frac{e^{-px^2}}{p} + \frac{1}{p}\left(-\sin xe^{-px^2} - p\text{I}\right) \\
\end{align}
\begin{align}
& \text{I} = -\sin x \ \frac{e^{-px^2}}{p} + \frac{1}{p}\left(-\sin xe^{-px^2} - p\text{I}\right) \\
& 2\text{I} = -2\sin x \ \frac{e^{-px^2}}{p}\Big|_0^\infty \\
& 2\text{I} = 0
\end{align}
|
Hint:
Use the expansion of $\sin$
$$\sin\sqrt{t}=\sum_{n=0}^{\infty}(-1)^n\dfrac{t^{n+1/2}}{\Gamma(2n+2)}$$
Edit:
$${\cal L}(\sin\sqrt{t})=\sum_{n=0}^{\infty}(-1)^n\dfrac{\Gamma(n+3/2)}{\Gamma(2n+2)p^{n+3/2}}$$
with Legendre Duplication Formula we have
$${\cal L}(\sin\sqrt{t})=\dfrac{1}{p^{3/2}}\sum_{n=0}^{\infty}(-1)^n\dfrac{\Gamma(n+3/2)}{\sqrt{\pi}^{-1}2^{2n+1}\Gamma(n+1)\Gamma(n+3/2)p^n}$$
$$=\dfrac{\sqrt{\pi}}{p^{3/2}}\sum_{n=0}^{\infty}\dfrac{\left(\frac{-1}{4p}\right)^{n}}{n!}=\color{blue}{\dfrac{\sqrt{\pi}}{p^{3/2}}e^{\frac{-1}{4p}}}$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Integral $\int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^2} \frac{dx}{\sqrt x}$ I have stumbled upon the following integral:$$I=\int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^2} \frac{dx}{\sqrt x}=-\frac{\pi}{24}$$
Although I could solve it, I am not quite comfortable with the way I did it.
But first I will show the way. We can substitute $\ln x \rightarrow t\ $ which gives:
$$I=\int_{-\infty}^\infty \frac{t}{\pi^2+t^2}\frac{e^{\frac{t}{2}}}{(1+e^t)^2}dt\overset{t=-x}=\int_{-\infty}^\infty \frac{-x}{\pi^2+x^2}\frac{e^{-\frac{x}{2}}}{(1+e^{-x})^2}dx$$
Also adding the two integral from above and simplify some of it yields:
$$2I= \int_{-\infty}^\infty \frac{x}{\pi^2+x^2}\left(\frac{e^{\frac{x}{2}}}{(1+e^x)^2}-\frac{e^{-\frac{x}{2}}}{(1+e^{-x})^2}\right)dx$$
$$\Rightarrow I=-\frac{1}{4} \int_{-\infty}^\infty \frac{x}{\pi^2+x^2}\frac{\sinh \left(\frac{x}{2}\right)}{\cosh ^2\left(\frac{x}{2}\right)}dx$$
And now a round of IBP gives:
$$I=\frac12 \int_{-\infty}^\infty \left(\frac{x^2-\pi^2}{(x^2+\pi^2)^2}\right)\left(\frac{1}{\cosh \left(\frac{x}{2}\right)}\right)dx$$
Using the Plancherel theorem the integral simplifies to: $$I=\int_0^\infty \left(\sqrt{\frac{\pi}{2}}x\left(-e^{-\pi x}\right)\right)\left(\sqrt{2\pi}\frac{1}{\cosh(\pi x)}\right)dx\overset{\pi x\rightarrow x}=-\frac{1}{\pi}\int_0^\infty \frac{x}{\cosh( x)}e^{- x}dx$$
We also have the following Laplace tranform for:$$f(t)=\frac{t}{\cosh( t)}\rightarrow F(s)=\frac18\left(\psi_1\left(\frac{s+1}{4}\right)-\psi_1\left(\frac{s+3}{4}\right)\right)$$
Where $\displaystyle{\psi_1(z)=\sum_{n=0}^\infty \frac{1}{(z+n)^2}}\,$ is the trigamma function.
$$\Rightarrow I=-\frac{1}{\pi}F(s=1)=-\frac{1}{\pi}\cdot \frac18\left(\psi_1\left(\frac{1}{2}\right)-\psi_1 (1)\right)=-\frac{1}{\pi}\cdot \frac18\left(\frac{\pi^2}{2}-\frac{\pi^2}{6}\right)=-\frac{\pi}{24}$$
Have I done anything wrong, or can it be improved?
I have to admit that I mostly used wolfram when applying Plancherel theorem and Laplace transform which I'm not comfortable with, but I didn't find an alternative method myself.
For this question I would like to see a different proof that doesn't rely on that theorem.
Probably not needed, but I should mention that my contour integration knowledge is pretty low. Also maybe there is a conexion with this integral, but I didn't find any.
|
To show that $$(-1)^{n-1} \int_{0}^{\infty}\frac{ \mathrm dx}{ (\pi^{2}+\ln^{2} x)(1+x)^{n}} $$ is an integral representation of the Gregory coefficients, some textbooks integrate the function $\frac{1}{(\ln z - i\pi)(1+z)^{n}} $ around a keyhole contour. We can do something similar here.
Let's integrate the function $$f(z)= \frac{1}{(\ln z - i \pi) (1+z)^{2}} \frac{1}{\sqrt{z}}$$ around a keyhole contour where the branch cut is along the positive real axis.
Integrating around the contour, we get $$ \int_{0}^{\infty}\frac{1}{(\ln x - i \pi) (1+x)^{2}} \frac{\mathrm dx}{\sqrt{x}} + \int_{\infty}^{0} \frac{1}{(\ln x + 2 \pi i - i \pi) (1+x)^{2}} \frac{\mathrm dx}{\sqrt{e^{2 \pi i}x}} = 2 \pi i \operatorname{Res} \left[f(z), -1 \right] $$
The left side of the above equation is
$$ \int_{0}^{\infty}\frac{1}{(\ln x - i \pi) (1+x)^{2}} \frac{\mathrm dx}{\sqrt{x}} + \int_{0}^{\infty} \frac{1}{(\ln x + i \pi) (1+x)^{2}} \frac{\mathrm dx}{\sqrt{x}} $$
$$= \int_{0}^{\infty} \frac{2 \ln x}{\left(\ln^{2}(x)+\pi^{2}\right)(1+x)^{2}} \frac{\mathrm dx}{\sqrt{x}} $$
Since $f(z)$ has pole at $z=e^{i \pi}=-1$ of order three, calculating the residue of $f(z)$ at $z=-1$ is a bit tedious.
But at $z=-1$, $\ln(z) - i\pi$ has the Taylor series expansion $$\ln(z) - i \pi = - (z+1)-\frac{(z+1)^{2}}{2!}- \frac{(z+1)^{3}}{3!} + O \left((z+1)^{4} \right) $$
Using polynomial long division, one can then show that Laurent series expansion of $\frac{1}{\ln z - i \pi}$ about $z=-1 $ is $$\frac{1}{\ln z - i \pi} = - \frac{1}{z+1} + \frac{1}{2} + \frac{z+1}{12} + O \left((z+1)^{2} \right) $$
So near $z=-1$, $$f(z) = \frac{1}{\sqrt{z}} \left(- \frac{1}{(z+1)^{3}} + \frac{1}{2(z+1)^{2}} + \frac{1}{12(z+1)} +O(1)\right), $$
from which we get $$ \begin{align} \small \operatorname{Res}\left[f(z), -1 \right] &= \small -\operatorname{Res}\left[\frac{1}{\sqrt{z}} \frac{1}{(z+1)^{3}}, -1 \right] + \frac{1}{2} \, \operatorname{Res}\left[\frac{1}{\sqrt{z}} \frac{1}{(z+1)^{2}}, -1 \right] + \frac{1}{12} \, \operatorname{Res}\left[\frac{1}{\sqrt{z}} \frac{1}{z+1}, -1 \right] \\ &=- \frac{1}{2!} \frac{3/4}{(e^{i \pi})^{5/2}} - \frac{1}{2} \frac{1/2}{(e^{i \pi)^{3/2}}} + \frac{1}{12} \frac{1}{(e^{i \pi})^{1/2}} \\ &= - \frac{3}{8i} + \frac{1}{4i} + \frac{1}{12i} = -\frac{1}{24i} \end{align}$$
(Alternatively, we could have also expanded $\frac{1}{\sqrt{z}}$ at $z=-1$ to get the first few terms of the Laurent series expansion of $f(z)$ about $z=-1$.)
Therefore, $$\int_{0}^{\infty} \frac{\ln x}{(\pi^{2}+\ln^{2} x)(1+x)^{2}} \frac{\mathrm dx}{\sqrt{x}} = \frac{2 \pi i}{2} \left(-\frac{1}{24i} \right) = -\frac{\pi}{24}$$
If $a>0$ and $a \ne 1$, then the same approach shows that $$\int_{0}^{\infty} \frac{\ln x}{(\pi^{2}+ \ln^{2}x)(a+x)^{2}} \frac{\mathrm dx}{\sqrt{x}} = \pi \left(\frac{2 + \ln a}{2a^{3/2} \ln^{2}a}- \frac{1}{(a-1)^{2}} \right) $$
Applying L'Hôpital's rule 4 times shows that the limit of the right side of the above equation as $a \to 1$ is $-\frac{\pi}{24}$.
|
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"url": "https://math.stackexchange.com/questions/3058679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
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|
How do I find out that the following two matrices are similar? How do I find out that the following two matrices are similar?
$N =
\begin{pmatrix}
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix}$
and $M=
\begin{pmatrix}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0
\end{pmatrix}$
I initially tried to think of the left multiplication of a matrix $P$ as a row operation and tried
$P=
\begin{pmatrix}
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}$
such that $PN = \begin{pmatrix}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix}$ but then $PNP^{-1} \neq M$.
My linear algebra is a bit rusty. Is there a more elaborate way to do this?
|
The two matrices are made of Jordan blocks; in $2\times2$ block format, they are
$$
N=\begin{bmatrix} J & 0 \\ 0 & 0 \end{bmatrix}
\qquad
M=\begin{bmatrix} 0 & 0 \\ 0 & J \end{bmatrix}
$$
You get similar matrices if you perform a row switch together with the corresponding column switch; in this case there is only one possible switch:
$$
M=\begin{bmatrix} 0 & I_2 \\ I_2 & 0 \end{bmatrix} N
\begin{bmatrix} 0 & I_2 \\ I_2 & 0 \end{bmatrix}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3060424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Function Optimization with Non-linear Constraint, Lagrange Multipliers Fails I am trying to maximize the function $A(x,y)=\frac{1}{2}(x(12-x)+y(13-y))$ subject to the constraint $x^2+(12-x)^2-y^2-(13-y)^2=0$.
My attempt:
$\begin{align*} \nabla A=\frac{1}{2}\langle 12-2x,\,13-2y\rangle &= \lambda\langle4x-24,\, -4y+26\rangle\\ \implies&\begin{cases} -x+6=\lambda(4x-24)\\-y+\frac{13}{2}=\lambda(-4y+26)\\x^2+(12-x)^2-y^2-(13-y)^2=0\end{cases}\end{align*}$
But clearly there is no solution due to the first two equations.
Using Wolfram Alpha, however, yields a maximum at $\displaystyle \left(\frac{17}{2},\,\frac{13}{2}\right)$ being $A=36$ and shows a nice little graph.
|
For a constrained problem
\begin{align}
\max{} & f(x,y) \\
& g(x,y)\le 0
\end{align}
you must write the Lagrangian:
$$
L(x,y,\lambda)=f(x,y)+\lambda g(x,y)
$$
Then you must find stationary points of your Lagrangian (attention this is only necessary conditions, see wiki)
\begin{align}
\partial_x L &= 0 = \partial_x f+ \lambda \partial_x g \\
\partial_y L &= 0 = \partial_y f+ \lambda \partial_y g \\
\partial_\lambda L &= 0 = g
\end{align}
With your example, this is essentially computations:
$$
L(x,y,\lambda)=\frac{1}{2} ((12-x) x+(13-y) y)-\lambda
\left(x^2+(12-x)^2-y^2-(13-y)^2\right)
$$
your three equations are (after simplification):
\begin{align}
-(-6 + x) (1 + 4 \lambda) &= 0 \\
\frac{1}{2} (-13 + 2 y) (-1 + 4 \lambda) &= 0\\
-x^2-(12-x)^2+y^2+(13-y)^2 &=0 &\\
\end{align}
The (real) solutions are:
$$
(x,y,\lambda)=(\frac{7}{2},\frac{13}{2},-\frac{1}{4})
$$
and
$$
(x,y,\lambda)=(\frac{17}{2},\frac{13}{2},-\frac{1}{4})
$$
You can check that for these two solutions
$$
f(\frac{7}{2},\frac{13}{2})=f(\frac{17}{2},\frac{13}{2})=36
$$
Extra: to check that these points are maximizers you must check that the Hessian of $f(x,y)$ is symmetric definite negative, which is clearly the case as:
$$
\nabla^2f=\left(\begin{array}{cc}-1& 0 \\ 0 & -1\end{array}\right)
$$
|
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"url": "https://math.stackexchange.com/questions/3062247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Algebraic Closed Form for $\sum_{n=1}^{k}\left( n- 3 \lfloor \frac{n-1}{3} \rfloor\right)$ Let's look at the following sequence:
$a_n=\left\{1,2,3,1,2,3,1,2,3,1,2,3,...\right\}$
I'm trying to calculate:
$$\sum_{n=1}^{k} a_n$$
Attempts:
I have a Closed Form for this sequence.
$$a_n=n- 3 \bigg\lfloor \frac{n-1}{3} \bigg\rfloor$$
The problem is, I'm looking for a closed form for this summation:
$$\sum_{n=1}^{k}\left( n- 3 \bigg\lfloor \frac{n-1}{3} \bigg\rfloor\right)$$
Is it possible?
|
Sorry can't comment on @5xum's answer. Building on the quadratic function we can derive:
$S_k=\frac{1}{2}((k-3\lfloor\frac{k}{3}\rfloor)^2+k+9\lfloor\frac{k}{3}\rfloor)$, which is pretty neat.
To derive this, note that from @5xum's answer, $S_k=2k-1+\frac{1}{2}(x^2 -3x+2)$ where $x=k-3\lfloor\frac{k}{3}\rfloor$. We can write $k=x+3\lfloor\frac{k}{3}\rfloor$, so that:
$$S_k=\frac{1}{2}(x^2 -3x+2+4x+12\lfloor\frac{k}{3}\rfloor-2)=\frac{1}{2}(x^2+x+12\lfloor\frac{k}{3}\rfloor). $$
Substituting $x$ back yields the result.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Not getting the right answer with alternate completing the square method on $\int\frac{x^2}{\sqrt{3+4x-4x^2}^3}dx$ So I've looked up how to do this problem and when they complete the square it's using the $(\frac{-b}{a})^2$ method where, in the denominator, they factor out the $4$, and then make:
$$x^2-x=x^2-x+{1\over4}$$ and then
$$x^2-x+{1\over4}=(x-\frac{1}{2})^2$$
which I understand, but is not the first option that popped into my head.
What I don't understand is why all the tutorials I've looked up are not completing the square in the traditional method. Here's my work:
$\int\frac{x^2}{\sqrt{3+4x-4x^2}^3}dx$
denominator:
$$3+4x-4x^2$$
$$(-4x^2+4x-1)+4$$
$$-(4x^2-4x+1)+4$$
$$-(2x-1)^2+4$$
So the integral is now:
$$\int\frac{x^2}{(4-u^2)^\frac{3}{2}}$$
So I do the rest of the work by using trig substitution now:
$u=2x-1$
$x=\frac{u+1}{2}$
Now I subbed in the trig identities:
$u=2\sin\theta$
$du=2\cos\theta d\theta$
$$\int\frac{(\frac{u+1}{2})^22\cos\theta}{\sqrt{4-4\sin^2\theta}^3}d\theta$$
$$\int\frac{\frac{(2\sin\theta+1)^2}{4}2\cos\theta}{\sqrt{4\cos^2\theta}^3}d\theta$$
$$\int\frac{\frac{(2\sin\theta+1)^2}{4}2\cos\theta}{(2\cos\theta)^3}d\theta$$
$$\int\frac{\frac{(2\sin\theta+1)^2}{4}2\cos\theta}{8\cos^3\theta}d\theta$$
$$\int\frac{\frac{(4\sin^2\theta+4\sin\theta+1)}{4}2\cos\theta}{8\cos^3\theta}d\theta$$
$$\int\frac{(\frac{4\sin^2\theta}{4}+\frac{4\sin\theta}{4}+\frac{1}{4})2\cos\theta}{8\cos^3\theta}d\theta$$
$$\int\frac{(\sin^2\theta+\sin\theta+\frac{1}{4})2\cos\theta}{8\cos^3\theta}d\theta$$
$$\int\frac{(\sin^2\theta+\sin\theta+\frac{1}{4})}{8\cos^3\theta}*\frac{2\cos\theta}{1}d\theta$$
$$\int\frac{(\sin^2\theta+\sin\theta+\frac{1}{4})}{4\cos^2\theta}d\theta$$
$$\int\frac{\sin^2\theta}{4\cos^2\theta}d\theta+\int\frac{\sin\theta}{4\cos^2\theta}d\theta+\int\frac{\frac{1}{4}}{4\cos^2\theta}d\theta$$
$$\frac{1}{4}\int \tan^2\theta d\theta+\frac{1}{4}\int\frac{\sin\theta}{\cos^2\theta}d\theta+\int\frac{1}{16\cos^2\theta}d\theta$$
For the second integral, I subbed $u=\cos\theta$ and $-du=\sin\theta d\theta$
$$\frac{1}{4}\int \sec^2\theta-1d\theta-\frac{1}{4}\int\frac{du}{u^2}+\frac{1}{16}\int \sec^2\theta d\theta$$
$$[\frac{\tan\theta}{4}-\frac{\theta}{4}]+[\frac{1}{4\cos\theta}]+[\frac{\tan\theta}{16}]+C$$
On a right triangle where $\sin\theta=\frac{2x-1}{2}$ and $\cos\theta=\frac{\sqrt{4-(2x-1)^2}}{2}$, I subbed these values back in to make the integral in terms of $x$:
$$\frac{2x-1}{4\sqrt{4-(2x-1)^2}}-\frac{\arcsin(\frac{2x-1}{2})}{4}+\frac{1}{4(\frac{\sqrt{4-(2x-1)^2}}{2})}+\frac{2x-1}{16\sqrt{4-(2x-1)^2}}+C$$
This does not seem to be the right answer when I compare it to a calculator answer, so I'm guessing I either completed the square wrong or made an algebraic mistake? I've done this problem three or four times now and am getting a little frustrated over what I'm guessing is a easy fix. If anyone can help, I'd really appreciate it.
|
Except some minor mistakes, your solution is correct but you led to wrong answer.
Mistake 1
In the first integral after substituting $u=\sin \theta$ you used $dx=du$ while $dx={1\over 2}du$.
Mistake 2
From $2$nd to $3$rd integral, you used $|\cos \theta|=\cos \theta$ while you should have dealt with $|\cos \theta|$ till the result.
Fixing up
Fixing these two mistakes, the final integral to be solved would become$$\int\frac{\sin^2\theta}{8\cos\theta|\cos\theta|}d\theta+\int\frac{\sin\theta}{8\cos\theta|\cos\theta|}d\theta+\int\frac{1}{32\cos\theta|\cos\theta|}d\theta\\={4+5\sin\theta-4\theta\cos\theta\over 32|\cos\theta|}+C$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
The exponential of a skew-symmetric matrix in any dimension. The Rodrigues rotation formula gives the exponential of a skew-symmetric matrix in three dimensions, and the exponential of a skew-symmetric matrix in two dimensions is given by Euler's formula. Is there a general formula (or set of formulas) for the exponential of a skew-symmetric matrix in any dimension?
|
The spectral decomposition of any skew-symmetric matrix $A$ is given by $A=U Q U^\dagger$ where $U$ is unitary and
\begin{align*}
Q=\begin{bmatrix}
0 & \lambda_1 & \\
-\lambda_1 & 0 & \\
& & 0 & \lambda_2\\
& & -\lambda_2 &0\\
& & & & \ddots\\
& & & & & 0 & \lambda_r\\
& & & & & -\lambda_r & 0\\
& & & & & & & 0\\
& & & & & & & &\ddots\\
& & & & & & & & & 0
\end{bmatrix}
\end{align*}
where I don't put all the $0$s for visibility's sake.
The exponential of a matrix is defined as the extension of the tailor expansion (up to convergence matter you will need to take care of), hence
\begin{align*}
e^A = \sum_{n=0}^\infty \frac{1}{n!} A^n
\end{align*}
and since $U$ is unitary, $A^n = U Q U^\dagger \dots U Q U^\dagger=U Q^n U^\dagger$ so we aim to get an expression for $Q^n$, this is not trivial but the right way to go is to compute it for several $n$ and try to see a pattern and then prove the pattern is right. It is easy to prove by induction that for any $k\in\mathbb N$
\begin{align*}
Q^{2k} &= \begin{bmatrix}
(-1)^k\lambda_1^{2k} & & \\
& (-1)^k\lambda_1^{2k} & \\
& & (-1)^k\lambda_2^{2k} & \\
& & & (-1)^k\lambda_2^{2k}\\
& & & & \ddots\\
& & & & & (-1)^k\lambda_r^{2k}\\
& & & & & & (-1)^k\lambda_r^{2k}\\
& & & & & & & 0\\
& & & & & & & &\ddots\\
& & & & & & & & & 0
\end{bmatrix}\\
Q^{2k+1} &=\begin{bmatrix}
0 & (-1)^k\lambda_1^{2k+1} & \\
-(-1)^k\lambda_1^{2k+1} & 0 & \\
& & \ddots\\
& & & 0 & (-1)^k\lambda_r^{2k+1}\\
& & & -(-1)^k\lambda_r^{2k+1} & 0\\
& & & & & 0\\
& & & & & &\ddots\\
& & & & & & & 0
\end{bmatrix}
\end{align*}
For this proof the base case would be $Q^0=I$ and $Q$ then you do induction on $k$.
We had
\begin{align*}
e^A &= \sum_{n=0}^\infty \frac{1}{n!} A^n\\
&= \sum_{n=0}^\infty \frac{1}{n!} U Q^n U^\dagger\\
&= U \left( \sum_{n=0}^\infty \frac{1}{n!} Q^n \right) U^\dagger
\end{align*}
and by plugin in the $Q^n$ we had, we get that the diagonal terms will be of the form $\sum_{k=0}^\infty \frac{1}{(2k)!} (-1)^k \lambda_p^{2k}=\cos(\lambda_p)$ and the other elements are of the form $\pm \sum_{k=0}^\infty \frac{1}{(2k+1)!} (-1)^k \lambda_p^{2k+1}=\pm \sin(\lambda_p)$ hence
\begin{align*}
e^A = U e^Q U^\dagger
\end{align*}
with
\begin{align*}
e^Q = \begin{bmatrix}
\cos(\lambda_1) & \sin(\lambda_1) & \\
-\sin(\lambda_1) & \cos(\lambda_1) & \\
& & \cos(\lambda_2) & \sin(\lambda_2)\\
& & -\sin(\lambda_2) &\cos(\lambda_2)\\
& & & & \ddots\\
& & & & & \cos(\lambda_r) & \sin(\lambda_r)\\
& & & & & -\sin(\lambda_r) & \cos(\lambda_r)\\
& & & & & & & 1\\
& & & & & & & &\ddots\\
& & & & & & & & & 1
\end{bmatrix}
\end{align*}
So an algorithmic way to find the exponential of your matrix is finding the spectral decomposition and then applying the last formula.
|
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|
Minimum value of $\frac{4}{4-x^2}+\frac{9}{9-y^2}$ Given $x,y \in (-2,2)$ and $xy=-1$
Minimum value of $$f(x,y)=\frac{4}{4-x^2}+\frac{9}{9-y^2}$$
My try:
Converting the function into single variable we get:
$$g(x)=\frac{4}{4-x^2}+\frac{9x^2}{9x^2-1}$$
$$g(x)=\frac{4}{4-x^2}+1+\frac{1}{9x^2-1}$$
Using Differentiation we get:
$$g'(x)=\frac{8x}{(4-x^2)^2}-\frac{18x}{(9x^2-1)^2}$$
$$g'(x)=2x\left(\frac{4(9x^2-1)^2-9(4-x^2)^2}{(4-x^2)^2(9x^2-1)^2}\right)$$
$$g'(x)=70x\frac{9x^4-4}{(4-x^2)^2(9x^2-1)^2}=0$$
So the critical points are:
$x=0, x=\pm \sqrt{\frac{2}{3}}$
But $x \ne 0$ since $xy=-1$
$$g'(x)=70x \frac{(3x^2-2)(3x^2+2)}{(4-x^2)^2(9x^2-1)^2}$$
By using derivative test we get Minimum occurs when $x=\pm \sqrt{\frac{2}{3}}$
Hence $$x^2=\frac{2}{3}, y^2=\frac{3}{2}$$
Min value is $$\frac{4}{4-\frac{2}{3}}+\frac{9}{9-\frac{3}{2}}=\frac{12}{5}$$
Is there any other approach?
|
From “harmonic mean $\le$ arithmetic mean” one gets
$$
\frac{2}{f(x, y)} \le \frac12 \left(\frac{4-x^2}{4}+ \frac{9-y^2}{9} \right)
= \frac 12 \left(2- \frac{x^2}{4} - \frac{y^2}{9} \right)
$$
and from “arithmetic mean $\ge$ geometric mean”
$$
\frac{x^2}{4} + \frac{y^2}{9}\ge 2 \frac{|xy|}{6} = \frac 13
$$
so that
$$
f(x, y) \ge \frac{4}{2-\frac 13} = \frac{12}{5} \, .
$$
In both estimates equality holds if $\frac{x^2}{4} = \frac{y^2}{9} \iff x^2=\frac{2}{3}, y^2=\frac{3}{2},$ i.e. equality is attained the two points
$$
( \sqrt \frac 23, -\sqrt \frac 32) \quad \text{and} \quad
( -\sqrt \frac 23, \sqrt \frac 32)
$$
which are both in the given domain $(-2, 2) \times (-2, 2)$ of $f$.
|
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|
Expressing $1-\exp{(\lambda_1p+\lambda_3q)}$ as $x+y$, where $x$ is in terms of $\lambda_1$ and $y$ is in terms of $\lambda_3$ I have this simple equation
$$c = 1 - \exp\left(\lambda_1 R^2 \left(\frac{2\pi}{3}-\frac{\sqrt{3}}{2}\right) + \lambda_3 R^2 \left(\frac{2\pi}{3}-\frac{\sqrt{3}}{2}\right)\right)$$
I will like to express this in the form $c = x + y$, where $x$ is in terms of $\lambda_1$, and $y$ is in terms of $\lambda_3$. How can I go about this?
I have tried basic expansion. However, it seems that there might be a mathematical rule to get out of this.
|
Write $$c(\lambda_1,\lambda_3)=1 - e^{(\lambda_1 R^2 (\frac{2\pi}{3}-\frac{\sqrt{3}}{2}) + \lambda_3 R^2 (\frac{2\pi}{3}-\frac{\sqrt{3}}{2}))}$$
Note first that $c(0,0)=1-e^0=0$.
Suppose we had the desired expression $c(\lambda_1,\lambda_3)=x(\lambda_1)+y(\lambda_3)$. Then:
$$x(\lambda_1)+y(0)=c(\lambda_1,0)=1 - e^{\lambda_1 R^2 (\frac{2\pi}{3}-\frac{\sqrt{3}}{2}) }$$
$$x(0)+y(\lambda_3)=c(0,\lambda_3)=1 - e^{ \lambda_3 R^2 (\frac{2\pi}{3}-\frac{\sqrt{3}}{2})}$$
Adding the above, we get $$x(0)+y(0)+x(\lambda_1)+y(\lambda_3)=1 - e^{\lambda_1 R^2 (\frac{2\pi}{3}-\frac{\sqrt{3}}{2}) }+1 - e^{ \lambda_3 R^2 (\frac{2\pi}{3}-\frac{\sqrt{3}}{2})}$$
But also we have $$x(0)+y(0)+x(\lambda_1)+y(\lambda_3)=0+c(\lambda_1,\lambda_3)=1 - e^{(\lambda_1 R^2 (\frac{2\pi}{3}-\frac{\sqrt{3}}{2}) + \lambda_3 R^2 (\frac{2\pi}{3}-\frac{\sqrt{3}}{2}))}$$
Hence $$1+e^{(\lambda_1 R^2 (\frac{2\pi}{3}-\frac{\sqrt{3}}{2}) + \lambda_3 R^2 (\frac{2\pi}{3}-\frac{\sqrt{3}}{2}))}=e^{\lambda_1 R^2 (\frac{2\pi}{3}-\frac{\sqrt{3}}{2}) }+e^{ \lambda_3 R^2 (\frac{2\pi}{3}-\frac{\sqrt{3}}{2})}$$
Note that this can be written as $1+xy=x+y$, which can be rewritten as $0=xy-x-y+1=(x-1)(y-1)$ and therefore has solutions $x=1$ and $y=1$. Hence, IF we can decompose $c=x+y$ as desired, then either $1=e^{\lambda_1 R^2 (\frac{2\pi}{3}-\frac{\sqrt{3}}{2}) }$ or $1=e^{ \lambda_3 R^2 (\frac{2\pi}{3}-\frac{\sqrt{3}}{2})}$. In the first case, $\lambda_1=0$ (or $R=0$); in the second case, $\lambda_3=0$ (or $R=0$). These can be all thought of as trivial.
If $R\lambda_1\lambda_3\neq 0$, then it is impossible to have a decomposition as desired.
|
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|
Proof of $ x^2 + y = y^2 + x$ when $ x+ y =1$ and $x$ is larger than $y$ I know that whatever numbers you choose for x and y and their sum equals to 1 will satisfy the equation $x^2 + y = y^2 + x$
Algebraic proof:
Given: $x + y = 1$
$$LS = x^2+ y
= (1-y)^2 + y
= 1 - 2y+y^2 + y
= y^2 - y + 1$$
$$RS = y^2 + x
= y^2 + (1-y)
= y^2 - y + 1$$
Therefore,$$ LS = RS $$
How can this be proved geometrically? (Ex. in a diagram of rectangular areas)
I tried to add a square piece with side lengths y with a rectangle with side lengths x and x+y but I can't seem to prove it geometrically.
Can someone help?
|
Because $$0=x^2+y-(y^2+x)=(x-y)(x+y)-(x-y)=(x-y)(x+y-1).$$
|
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|
Counting, Probability and Binomial Coefficients If $$P_{2n+2}=\sum_{k=n+2}^{2n+2}{2n+2 \choose k}p^kq^{2n+2-k}$$
and,
$$P_{2n}=\sum_{k=n+1}^{2n}{2n \choose k}p^kq^{2n-k}$$
where $0<p<q<1$ and $q=1-p$
Prove that
$$P_{2n+2}=P_{2n}+{2n \choose n}p^{n+2}q^n-{2n \choose {n+1}}p^{n+1}q^{n+1}$$
$\mathbf {Inspiration:}$ A and B play a series of games where the probability of winning $\mathit p$ for A is kept less than 0.5. However A gets to choose in advance the total no. of plays. To win the game one must score more than half the games . If the total no. of games is to be even, How many plays should A choose?
$\mathbf {Here}$ $P_{2n}$ and $P_{2n+2}$ represents the probability of A winning the play in $2n$ and $2n+2$ games where $2n$ is considered the optimum number of games
|
We have for the first sum
$$S = \sum_{k=0}^n {2n+2\choose k+n+2} p^{k+n+2} q^{n-k}$$
and for the second one
$$T = \sum_{k=0}^{n-1} {2n\choose k+n+1} p^{k+n+1} q^{n-1-k}$$
and seek to show
$$S-T =
{2n\choose n} p^{n+2} q^n - {2n\choose n+1} p^{n+1} q^{n+1}$$
where $p+q=1.$ We see that with
$$Q_n = \sum_{k=0}^n {2n+2\choose k+n+2} p^{k+n+2} q^{n-k}$$
the claim becomes
$$Q_n - Q_{n-1} =
{2n\choose n} p^{n+2} q^n - {2n\choose n+1} p^{n+1} q^{n+1}.$$
Now
$$Q_n = p^{n+2} q^n
\sum_{k=0}^n {2n+2\choose k+n+2} p^{k} q^{-k}
= p^{n+2} q^n
\sum_{k=0}^n {2n+2\choose n-k} p^{k} q^{-k}
\\ = p^{n+2} q^n
\sum_{k=0}^n p^{k} q^{-k}
[z^{n-k}] (1+z)^{2n+2}
= p^{n+2} q^n [z^n] (1+z)^{2n+2}
\sum_{k=0}^n p^{k} q^{-k} z^k.$$
We may extend $k$ beyond $n$ owing to the coefficient extrator $[z^n]$
in front:
$$p^{n+2} q^n [z^n] (1+z)^{2n+2}
\sum_{k\ge 0} p^{k} q^{-k} z^k
= p^{n+2} q^n [z^n] (1+z)^{2n+2} \frac{1}{1-pz/q}
\\ = p^2 [z^n] (1+pqz)^{2n+2} \frac{1}{1-p^2z}.$$
We thus have
$$Q_{n-1} = p^2 [z^{n-1}] (1+pqz)^{2n} \frac{1}{1-p^2z}
\\ = p^2 [z^n] z (1+pqz)^{2n} \frac{1}{1-p^2z}.$$
Subtracting we find
$$Q_n - Q_{n-1}
= p^2 [z^n] ((1+pqz)^2-z) (1+pqz)^{2n} \frac{1}{1-p^2z}
\\ = p^2 [z^n] (1-p^2 z) (1-q^2 z)
(1+pqz)^{2n} \frac{1}{1-p^2z}
\\ = p^2 [z^n] (1-q^2 z) (1+pqz)^{2n}
\\ = p^2 [z^n] (1+pqz)^{2n}
- p^2 [z^n] q^2 z (1+pqz)^{2n}
\\ = p^2 p^n q^n {2n\choose n}
- p^2 q^2 [z^{n-1}] (1+pqz)^{2n}
\\ = p^{n+2} q^n {2n\choose n}
- p^2 q^2 p^{n-1} q^{n-1} {2n\choose n-1}.$$
This is indeed
$$\bbox[5px,border:2px solid #00A000]{
p^{n+2} q^n {2n\choose n}
- p^{n+1} q^{n+1} {2n\choose n+1}}$$
as claimed.
Remark. It might be simpler not to merge the $p^n q^n$ into the
coefficient extractor. Further commentary TBA.
|
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|
Solving for $x$ in $y=N\cdot\left(\frac{10}{x}\right)^{-2.6}$ I want to confirm my solution of $x$ from
$$y=N\cdot\left(\frac{10}{x}\right)^{-2.6}$$
My answer is:
$$x=\frac{N^{2.6}}{10\cdot y^{2.6}}$$
Is this right? How would you solve?
|
First, notice that
$$
y = N \cdot \left( \dfrac{10}{x} \right)^{-2.6} = N \cdot \left( \dfrac{x}{10} \right)^{2.6}
$$
Next, divide both sides by $N$:
$$
\dfrac{y}{N} = \left( \dfrac{x}{10} \right)^{2.6}
$$
Now exponentiate with $\frac{1}{2.6}$ to get
$$
\left( \dfrac{y}{N} \right)^{\frac{1}{2.6}} = \dfrac{x}{10}
$$
Finally, multiply by $10$:
$$
10 \cdot \left( \dfrac{y}{N} \right)^{\frac{1}{2.6}} = x
$$
|
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|
How to find the range of $y=\frac{x^2+1}{x+1}$ without using derivative? The only thing I know with this equation is $y=\frac{x^2+1}{x+1}=x+1-\frac{2x}{x+1}$.
Maybe it can be solved by using inequality.
|
Does not exist.
Try $x\rightarrow-1^-$.
For $x>-1$ by AM-GM we obtain:
$$\frac{x^2+1}{x+1}=\frac{x^2-1+2}{x+1}=x-1+\frac{2}{x+1}=$$
$$=x+1+\frac{2}{x+1}-2\geq2\sqrt{(x+1)\cdot\frac{2}{x+1}}-2=2\sqrt2-2.$$
The equality occurs for $x+1=\frac{2}{x+1},$ which says that we got a minimal value and the local minimal value.
Thus, the range for $x>0$ it's $[2\sqrt2-2,+\infty).$
For $x<-1$ we can get the range by the similar way.
|
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|
Determine all $a$ and $b$ natural numbers such that $\frac {a^2+2b} {b^2-2a}$ and $\frac {b^2+2a} {a^2-2b}$ are whole numbers. I proceeded in the following way:
It is clear that $a \ne 0$ and $b \ne 0$.
Let $\frac {a^2+2b} {b^2-2a} = k, k \in \mathbb{Z} \tag 1$
and $\frac {b^2+2a} {a^2-2b} = m, m \in \mathbb{Z} \tag 2$
1. If $a = b$:
Let $a = b = n$. Then $k = m = \frac {n^2+2n} {n^2-2n} = \frac {n(n+2)} {n(n-2)} = \frac {n+2} {n-2} = p, p \in \mathbb{Z}$. From here we get that $n \in \{1, 3, 4, 6\}$. I don't know if these are all so I can't prove that there are no others.
2. If $a \ne b$:
For $k$ and $m$ to be whole numbers the numerators have to be greater than or equal to the denominator in $(1)$ and $(2)$.
From $(1)$ we get that $a^2+2b \ge b^2-2a \implies a+2 \ge b \tag {3}$
From $(2)$ we get that $b^2+2a \ge a^2-2b \implies b+2 \ge a \tag {4}$
Here I am stuck again... WolframAlpha gives me these solutions to $(3)$ and $(4)$.
|
So we have $b\in \{a-2,a-1,a,a+1,a+2\}$
Case 1: $b= a-2$, then $$(a-2)^2-2a\mid a^2+2(a-2)$$
so $$a^2-6a+4\mid a^2+2a-4$$
and thus $$a^2-6a+4\mid (a^2+2a-4)-(a^2-6a+4)=8a-8$$ Now if $a=1$ then $b=-1$ which works. So $a>1$ and now we have $$a^2-6a+4\leq 8a-8\implies a^2-14a+12\leq 0$$
so $$ (a-7)^2\leq 37\implies |a-7|\leq 6...$$
We do similary for all other cases...
|
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|
Prove $a^{n} - 1 = (a-1)(a^{n-1}+a^{n-2}+a^{n-3} + \cdots + a + 1)$ for all $n \ge 1$ using induction Problem 1.1.3 in Burton's Elementary Number Theory (6th ed.) is stated as follows:
Use the Second Principle of Finite Induction to establish that for all $n \ge 1$,
\begin{align} a^{n} - 1 = (a-1)(a^{n-1}+a^{n-2}+a^{n-3} + \cdots + a + 1)
\end{align}
[Hint: $a^{n+1} - 1 = (a+1)(a^{n}-1) - a(a^{n-1}-1)$.]
I already came across a much simpler proof in terms of requiring fewer algebraic manipulations but it didn't use the hint given in the problem statement. Right below is my attempt at proving the statement and I wish someone could confirm what I've done.
Proof by induction.
Base case. When $n =1$, LHS = $a^{1} - 1 = a-1$ and RHS = $(a-1)(a^{1-1}) = a-1$. Thus LHS = RHS.
Inductive step. Let's assume that $ a^{k} - 1 = (a-1)(a^{k-1}+a^{k-2}+a^{k-3} + \cdots + a + 1) $ holds for $ 1, 2, \ldots k $. Then
\begin{align*}
a^{k+1} - 1 &= (a+1)(a^{k}-1) - a(a^{k-1}-1) \\
&= (a+1)(a^{k}-1) - (a^{k}-a) \\
&= (a+1)(a^{k}-1) - [(a^{k} -1) + (1 -a)] \\
&= (a+1)(a^{k}-1) - (a^{k} -1) - (1 -a) \\
&= [(a+1)(a^{k}-1) - (a^{k} -1)] + (a-1) \\
&= (a^{k}-1)[a+1-1] + (a-1) \\
&= (a^{k}-1)\cdot a + (a-1) \\
&= (a-1)(a^{k-1}+a^{k-2}+a^{k-3} + \cdots + a + 1)\cdot a + (a-1) \\
&= (a-1)[(a^{k-1}+a^{k-2}+a^{k-3} + \cdots + a + 1)\cdot a + 1] \\
&= (a-1)(a^{k}+a^{k-1}+a^{k-2} + \cdots + a^{2} + a + 1) \\
\end{align*}
which is precisely the right side of the formula in the problem statement when $n = (k+1)$. Therefore by the principle of mathematical induction the formula holds for all $n \ge 1$.
|
The only problem with your proof is that you failed to check and confirm that for the base case, when $n=1$, the given proposition holds. Add that to your proof, and you're good to go.
It may seem utterly obvious and not worth mentioning in a proof, but without proving the base case $n=1$, your proof is meaningless in terms of providing an inductive proof.
Note also that you are indeed using the hint when you write for first line in determining $a^{k+1}$.
|
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Integrals depending on a parameter: $\int_{0}^{\pi/2} \ln(a^2\sin^2{x} + \cos^2{x})dx $ I'm trying to calculate this integral:
$$ \int_{0}^{\pi/2} \ln(a^2\sin^2{x} + \cos^2{x})dx $$
For $a > 0$.
This is what I did:
$$ I(a) = \int_{0}^{\pi/2} \ln(a^2\sin^2{x} + \cos^2{x})dx \\\
I'(a) = \int_{0}^{\pi/2} \frac{2a\sin^2{x}}{(a^2\sin^2{x} + \cos^2{x})} dx \\\
I'(a) = \int_{0}^{\pi/2} \frac{2a}{(a^2 + \frac{\cos^2{x}}{\sin^2{x}})} dx \\\
I'(a) = 2a \int_{0}^{\pi/2} \frac{1}{a^2 + \cot^2{x}}dx \\\
I'(a) = \frac{2}{a} \int_{0}^{\pi/2} \frac{1}{1 + \frac{\cot^2{x}}{a^2}}dx $$
Here I tried to substitude $\frac{\cot^2{x}}{a^2} = t$. This should lead me to $\arctan(something)$ (I write $\arctan{x} = \tan^{-1}{x}$.)
But I got stuck.
After I get the derivative I should integrate it back to get $I(a)$. Also there are steps that need some prepositions to be checked. Could one help me with this integral and also to clarify the steps that need special attention; such as the second step, where I can get the derivative $I'(a)$ only if the function under the integral can be differentiated? It must be possible to find its derivative.
|
you can also use the series of $\ln x$ to find it
$$\begin{aligned}
\int_{0}^{\pi/2} \ln(a^2\sin^2{x} + \cos^2{x}) \mathrm{d}x
& = \int_{0}^{\pi/2} \ln(1+(a^2-1)\sin^2{x}) \mathrm{d}x\\
& = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \int_{0}^{\pi/2}(a^2-1)^n\sin^{2n}{x} \mathrm{d}x\\
& = \pi \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{2n} \frac{(2n-1)!!}{(2n)!!} (a^2-1)^n
\end{aligned}$$
where you need wallis' integral here, then using this identity
$$\operatorname{arsinh} x = \ln(2x) + \sum\limits_{n = 1}^\infty {\left( { - 1} \right)^{n - 1} \frac{{\left( {2n - 1} \right)!!}}{{2n\left( {2n} \right)!!}}} \frac{1}{{x^{2n} }}$$
let $x=1/\sqrt{a^2-1}$ and you will find the answer
$$\int_{0}^{\pi/2} \ln(a^2\sin^2{x} + \cos^2{x}) \mathrm{d}x = \pi\left( \operatorname{arsinh} \frac1{\sqrt{a^2-1}} - \ln\frac2{\sqrt{a^2-1}} \right) = \pi\ln\frac{a+1}{2}$$
|
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|
Finding the tenth derivative of $f(x) = e^x\sin x$ at $x=0$ I came across this Question where I have to find $$f^{(10)}$$ for the following function at $x = 0$
$$f(x) = e^x\sin x$$
I tried differentiating a few times to get a pattern but didn’t get one, can someone provide the solution.
|
Using power series: it is well-known that $e^x=\sum_{k=0}^\infty \frac{x^k}{k!}$ and $\sin(x)=\sum_{k=0}^\infty (-1)^k\frac{x^{2k+1}}{(2k+1)!}$ for any real number $x$, so
$$ e^x\sin(x)=(1+x+\frac{x^2}{2!}+\dots+\frac{x^{10}}{10!}+\dots)(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}+\dots) $$
By expanding, the coefficient of $x^{10}$ is $\frac{1}{9!1!}-\frac{1}{7!3!}+\frac{1}{5!5!}-\frac{1}{7!3!}+\frac{1}{9!1!}$
But this coefficient is also $\frac{f^{(10)}(0)}{10!}$, so
$$ f^{(10)}(0)=\frac{10!}{9!1!}-\frac{10!}{7!3!}+\frac{10!}{5!5!}-\frac{10!}{7!3!}+\frac{10!}{9!1!} = 10 -120 + 252 - 120 +10 = 32$$
|
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|
Integration by substitution to take out square root
Find $$\int (x+1)\sqrt{x^2+1}\,dx .$$
In order to not bother with the square root I thought of doing this:
$let$ $ x^2+1=(x+t)^2$ $\Rightarrow$ $x=\frac{1-t^2}{2t}$ $\Rightarrow x-t=\frac{1+t^2}{2t}$ $(1)$
$\Rightarrow$ $t=\sqrt{x^2+1}-x$ $(2)$
And
$dx=-\frac{t^2+1}{2t^2}dt $ $(3)$
and also $x+1=-\frac{t^2-2t-1}{2t}$ $(4)$
Then
$(1),(3),(4)\Rightarrow$$\int (x+1)\sqrt{x^2+1}dx=$ $\int\frac{t^2-2t-1}{2t} \frac{1-t^2}{2t} \frac{t^2+1}{2t}dt=$
$\frac{1}{8} \int \frac{-t^6+2t^5+t^4+t^2-2t-1}{t^3}dt=$
$\frac{1}{8} [-\int t^3 dt+2\int t^2 dt+\int t dt +\int \frac{1}{t} dt-2 \int \frac{1}{t^2} dt- \int \frac{1}{t^3}dt]=$
$\frac{1}{8}[-\frac{t^4}{4}+2\frac{t^3}{3}+\frac{t^2}{2}+lnt+\frac{2}{t}-\frac{1}{2t^2}]$
We do the substitution from $(2)$ and we get the result
You think that's correct?
|
Avoiding hyperbolic functions, I believe you can do a trigonometric substitution: consider a right triangle with angle $\theta$ for which the opposite side is $x$, the adjacent side is $1$, and (thus) the hypotenuse is $\sqrt{x^2 + 1}$.
Then $x = \tan{\theta}$, so $dx = \sec^2{\theta}d\theta$; note that $\sec{\theta} = \sqrt{x^2 + 1}$ and your integral becomes:
$$\int(x+1)\sqrt{x^2 + 1}dx = \int(\tan{\theta} + 1)\sec^3{\theta} d\theta = \int\tan{\theta}\sec^3{\theta} + \sec^3{\theta} d \theta$$
The first part of the integrand is not difficult, since $u = \sec{\theta}$ yields $du = \sec{\theta}\tan{\theta}d\theta$, whereby
$$\int \tan\theta \sec^3\theta d\theta = \int u^2 du$$
and finding the antiderivative of secant cubed is now the only task that remains. If you have seen this before (example link) then you have reduced the problem to something familiar. If not, then you can use the already provided answer(s) for an alternative derivation of this function's antiderivative!
|
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|
Find the coefficient of $x$ in the given polynomial We have polynomial
$$(x-2^0)(x-2^1)(x-2^2)···(x-2^n)$$
I know that the coefficient of $x$ equals to:
$$(-1)^{n-1}((2^0×2^1×···×2^{n-1})+(2^0×2^1×···×2^{n})+···+(2^1×2^2×···×2^{n}))$$
But it's hard to calculate this sum.
Any help is appreciated.
|
The coefficient of $x$ in $(x-a_1)\cdots(x-a_n)$ is
$$
(-1)^n \sum_{k=1}^n \frac{a_1 \cdots a_n}{a_k}
$$
If $a_k=2^{k-1}$, then $a_1 \cdots a_n = 2^0 2^1 \cdots 2^{n-1} = 2^{0+1+\cdots +(n-1)}=2^{n(n-1)/2}$.
Thus, the coefficient is
$$
(-1)^n \sum_{k=1}^n \frac{2^{n(n-1)/2}}{2^{k-1}}
=
(-1)^n 2^{n(n-1)/2} \sum_{k=1}^n \frac{1}{2^{k-1}}
= \cdots
$$
|
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|
Evaluate the limit: $\lim_{x\to 3} \frac{x^{2}+\sqrt{x+6}-12}{x^{2}-9}$ $$\lim_{x\to 3} \frac{x^{2}+\sqrt{x+6}-12}{x^{2}-9} $$
I want to know how to evaluate without using L'Hopital Rule. I'm unable to factorise or simplify it suitably.
|
$$\lim_{x\to 3} \frac{x^{2}+\sqrt{x+6}-12}{x^{2}-9} $$
$$ = \lim_{x\to 3} \frac{x^{2}+\sqrt{x+6}-9-3}{x^{2}-9}$$
$$ = \lim_{x\to 3} \frac{x^{2}-9+\sqrt{x+6}-3}{x^{2}-9}$$
$$ = \lim_{x\to 3} \Big(\frac{x^{2}-9}{x^{2}-9}+\frac{\sqrt{x+6}-3}{x^{2}-9}\Big) $$
$$ = \lim_{x\to 3} \frac{x^{2}-9}{x^{2}-9} + \lim_{x\to 3} \frac{\sqrt{x+6}-3}{x^{2}-9}$$
$$ = \lim_{x\to 3} 1 + \lim_{x\to 3} \frac{\sqrt{x+6}-3}{x^{2}-9}$$
$$ = 1 + \lim_{x\to 3} \frac{\sqrt{x+6}-3}{x^{2}-9}$$
After rationalizing the numerator to remove the radical:
$$\lim_{x\to 3}\frac{\sqrt{x+6}-3}{x^{2}-9} \cdot \frac{\sqrt{x+6}+3}{\sqrt{x+6}+3} = \lim_{x\to 3}\frac{x+6-9}{(x^{2}-9)(\sqrt{x+6}+3)} = \lim_{x\to 3}\frac{x-3}{(x^{2}-9)(\sqrt{x+6}+3)} $$
After factoring $ x^2 - 9 $, simplifying, and direct substitution:
$$ \lim_{x\to 3}\frac{x-3}{(x^{2}-9)(\sqrt{x+6}+3)} = \lim_{x\to 3}\frac{x-3}{(x-3)(x+3)(\sqrt{x+6}+3)} = \lim_{x\to 3}\frac{1}{(x+3)(\sqrt{x+6}+3)} = \frac{1}{6\cdot 6} = \frac{1}{36}$$
Thus $$\lim_{x\to 3} \frac{x^{2}+\sqrt{x+6}-12}{x^{2}-9} = 1 + \lim_{x\to 3} \frac{\sqrt{x+6}-3}{x^{2}-9} = 1 + \frac{1}{36} = \frac{36}{36} + \frac{1}{36} $$
$$= \frac{37}{36}$$
|
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|
Trouble showing $a_n=\frac{\sin \frac{n\pi}{3n+1}}{\sqrt{n+3}}$ is monotonically decreasing I've tried reducing the expression to something obviously true, but I couldn't get anywhere. Also tried induction but I got stuck at the inductive step and failed to obtain anything useful. I've also tried to prove it by contradiction, still to no avail.
Also tried to show that $a_n/a_{n+1} > 1$, again without success because I don't know how to deal with the sines that appear in the expression, and there I ran out of ideas.
EDIT: I could perhaps consider the continuos case of a function given with the same expression, by testing it's derivative, however that's not really what I'm trying to do here unless someone knows a more rigorous line of reasoning to use this idea somehow.
|
The fraction $\frac{n\pi}{3n+1}$ increases strictly with $n$. Therefore, by Aristarchus's inequality,
$$
\frac{\sin\left(\frac{(n+1)\pi}{3n+4}\right)}{\sin\left(\frac{n\pi}{3n+1}\right)} < \frac{(n+1)(3n+1)}{n(3n+4)} = 1 + \frac1{n(3n+4)}.
$$
Write $t = \frac1{n(3n+4)} < 1$. For $n \geqslant 2$,
$$
\left(1 + \frac1{n(3n+4)}\right)^2 = 1 + 2t + t^2 < 1 + 3t < 1 + \frac1{n(n+1)} < 1 + \frac1{n+3} = \frac{n+4}{n+3}.
$$
Hence:
$$
\frac{a_{n+1}}{a_n} = \frac{\sin\left(\frac{(n+1)\pi}{3n+4}\right)}{\sin\left(\frac{n\pi}{3n+1}\right)} \cdot \frac{\sqrt{n+3}}{\sqrt{n+4}} < 1 \quad (n \geqslant 2).
$$
We also have
$$
a_1 = \frac{1}{2\sqrt2} > 0.35 > \frac{\sin\left(\frac{2\pi}{7}\right)}{\sqrt5} = a_2,
$$
so the inequality does actually hold for all $n \geqslant 1$.
For a slightly tidier finish, one could argue thus:
$$
2\sin^2\left(\frac{2\pi}{7}\right) = 1 - \cos\left(\frac{4\pi}{7}\right) = 1 + \sin\left(\frac{\pi}{14}\right) < 1 + \frac{\pi}{14} < \frac{5}{4}.
$$
|
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|
What is the value of $\sqrt{49-x^2}+\sqrt{25-x^2}, x \in \mathbb{R}$ if $\sqrt{49-x^2}-\sqrt{25-x^2}=3$ Suppose that real number $x$ satisfies $$\sqrt{49-x^2}-\sqrt{25-x^2}=3$$What is the value of $\sqrt{49-x^2}+\sqrt{25-x^2}$?
This is what I did:
I try to multiply by the conjugate. Its value I believe is technically the solution. $(\sqrt {49-x^2} + \sqrt {25-x^2})(\sqrt {49-x^2} - \sqrt {25-x^2})=24$. Given that $(\sqrt {49-x^2} - \sqrt {25-x^2}) = 3$, $(\sqrt {49-x^2} + \sqrt {25-x^2})(\sqrt {49-x^2} - \sqrt {25-x^2})= 3(\sqrt {49-x^2} + \sqrt {25-x^2}) =24\implies \sqrt {49-x^2} + \sqrt {25-x^2}=8$
My question is that will this method work for similar problems, and is there a faster method?
Thanks!
|
Multiplying both sides by $\sqrt {49-x^2} + \sqrt {25-x^2}$, yields:
$$(\sqrt {49-x^2} + \sqrt {25-x^2})(\sqrt {49-x^2} - \sqrt {25-x^2})=3(\sqrt {49-x^2} + \sqrt {25-x^2})$$
$$(49-x^2 - (25-x^2)=3(\sqrt {49-x^2} + \sqrt {25-x^2})$$
Thus,
$$\sqrt {49-x^2} + \sqrt {25-x^2}=8.$$
|
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|
$ 0=1 $ ? Where is the mistake? I just found this formula although it can be easily derived.
Let $ n $ be any integer then,
$$n=\sqrt{n^2-n+\sqrt{n^2-n+.....}}$$
So if I plug in $ 0 $ in this equation I get,
$$0=\sqrt{0-0+\sqrt{0-0+....}}$$
$$0=\sqrt{0+\sqrt{0+\sqrt{0+....}}}$$—————->1
But if I plug in $1$ in the equation I get,
$$1=\sqrt{1^{2}-1+\sqrt{1^{2}-1+.....}}$$
$$1=\sqrt{0+\sqrt{0+\sqrt{0+....}}}$$,—————->2
Which gives me that 0=1.Where is the mistake?
Edit:
Equation 1 can also be derived by the following method
$$1=\sqrt{0+\sqrt{1}}$$
$$1=\sqrt{0+\sqrt{0+\sqrt{1}}}$$
.
.
.
$$1=\sqrt{0+\sqrt{0+\sqrt{0+....}}}$$
Ok so let me tell you how I derived it.
I came across this particular expression,
$$x=\sqrt{1+\sqrt{1+\sqrt{1+.....}}}$$ which actually gives me the golden ratio,
So I chose n to be an integer and let x be the value of the following expression,
$$x=\sqrt{n+\sqrt{n+\sqrt{n+...}}}$$.
After solving for x you get it’s value to be!
$$x=\frac{1+\sqrt{1+4n}}{2}$$
(I took a positive sign since x is greater than or equal to 0)
So x can be an integer whenever n=0,2,6,10,20.....
When n=2,x=2(since x>0) and we get,
$$2=\sqrt{2+\sqrt{2+......}}$$
Similarly for n=6,x=3,
$$3=\sqrt{6+\sqrt{6+......}}$$
So x is an integer whenever $$n=k^{2}-k$$ where k is a non-negative integer.Substituting $$n=k^{2}-k$$ you get x=k.
After all this I get,
$$n=\sqrt{n^2-n+\sqrt{n^2-n+.....}}$$
|
The problem arises because you didn't really defined the radical expression
$$\sqrt{n^2-n+\sqrt{n^2-n+\sqrt{n^2-n+\cdots}}}.$$
Let us rewrite it as the recurrence
$$a_{k+1}=\sqrt{n^2-n+a_k}.$$
Then with $n=0$ or $1$,
$$a_{k+1}=\sqrt{a_k}=\sqrt[4]{a_{k-1}}=\cdots\sqrt[2^{k+1}]{a_0}.$$
For $a_0=0$, we have the limit $a_\infty=0$, and for $a_0>0$, $a_\infty=1$, so you could indeed assign the expression the value $0$ or $1$, and the equation
$$n=\sqrt{n^2-n+\sqrt{n^2-n+\sqrt{n^2-n+\cdots}}}$$ may hold or not depending on the value of $a_0$.
In other words, the truth is that you should write
$$0=\sqrt{0^{2}-0+\sqrt{0^{2}-0+\sqrt{0^{2}-0+\cdots0}}}$$
$$1=\sqrt{1^{2}-1+\sqrt{1^{2}-1+\sqrt{1^{2}-1+\cdots1}}}$$
while you innocently assumed
$$1=\sqrt{1^{2}-1+\sqrt{1^{2}-1+\sqrt{1^{2}-1+\cdots\color{red}0}}}.$$
For other $n$, if there is convergence, we have
$$a(a-1)=n(n-1)$$ and
$$a=\frac{1\pm|1-2n|}2.$$
For $n>1$, the only positive possibility is $a=n$. Otherwise, there could be two solutions, presumably depending on the initial value. But you can't spare convergence analysis.
|
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|
Why does the CRT formula yield a solution of a congruence system? I understand there is a method for solving simultaneous modular equations. For example;
$$x = 2 \mod{3}$$
$$x = 3 \mod{5}$$
$$x = 2 \mod{7}$$
We find numbers equal to the product of every given modulo except one of them - giving $5 \cdot 7$, $3 \cdot 7$ and $3 \cdot 5$. We then find the multiplicative inverses of these numbers with modulo equal to the number missing from the product. The numbers found are then 2, 1 and 1 in this case. The value of x is then given by:
$$x = 2 \cdot (5 \cdot 7) \cdot 2 + 3 \cdot (3 \cdot 7) \cdot 1 + 2 \cdot (3 \cdot 5) \cdot 1 = 233 = 23 \mod{3\cdot5\cdot7}$$
But I do not understand how this method correctly gives the value of $x$. I understand that the Chinese remainder theorem proves that there is a unique value of $0\le x \lt 3\cdot5\cdot7 \mod{3\cdot5\cdot7}$ but can someone please explain why this method finds this value of x?
|
This is a generalisation of the formula for the solutions of a system of two congruences modulo two coprime numbers $a$ and $b$?. This formula uses a Bézout's relation: $\;ua+vb=1$ and it is:
$$\begin{cases}
x\equiv \alpha\mod a,\\
x\equiv \beta\mod b,
\end{cases}
\quad\text{which is }\qquad x\equiv \beta ua+\alpha vb\mod ab$$
Indeed we have $\;\beta ua+\alpha vb\equiv \alpha vb\equiv \alpha\mod a$ since $\;vb\equiv 1\mod a$. Similarly modulo $b$.
Now, as $v \equiv b^{-1}\bmod a\:$ and $\;u\equiv a^{-1}\bmod b$, this formula can be written as
$$x\equiv \beta\, a (a^{-1}\bmod b)+\alpha\, b(b^{-1}\bmod a)\mod ab.$$
Some details with the example in the question:
In each term of $x$:
$$ 2 \cdot (5 \cdot 7) \cdot 2 + 3 \cdot (3 \cdot 7) \cdot 1 + 2 \cdot (3 \cdot 5) \cdot 1 $$
the first factor is the r.h.s. of a congruence equation mod. $m_i$, the second (between parentheses) is the product of the other moduli and the last factor is the inverse of the former mod. $m_i$.
For instance, consider the first congruence: as $5\cdot 7\equiv 2\mod 3$, which is its own inverse, and $\equiv 0\mod 5,7$,we see that
$$(5\cdot7)\cdot 2\begin{cases}\equiv 1\mod3,\\[1ex]\equiv 0 \mod 5,7 \end{cases}\quad\text{hence }\quad\alpha\cdot(5\cdot7)\cdot 2\begin{cases}\equiv \alpha\mod3\\[1ex]\equiv 0 \mod 5,7 \end{cases}$$
So we obtain a formula analog to Lagrange's interpolation formula:
$$ \alpha \cdot (5 \cdot 7) \cdot 2 + \beta \cdot (3 \cdot 7) \cdot 1 + \gamma\cdot (3 \cdot 5) \cdot 1 \equiv\begin{cases}\alpha\mod 3, \\[1ex]\beta\mod 5,\\[1ex]\gamma\mod 7. \end{cases}$$
|
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|
Evaluate $\int \frac{x^2}{x-1} \,dx$ Evaluate $\int \frac{x^2}{x-1} \,dx$
(A) $2x^2+x+\ln|x-1|+C$
(B) $\frac{x^2}{2}+x+\ln|x+1|+C$
(C) $\frac{x^2}{2}+x+\ln|x-1|+C$
(D) $x^2+x+\ln|x-1|+C$
My attempt :
Let $u=x-1$, so : $du=dx$ and $u+1$
$\int \frac{(u+1)^2}{u}\,du\\
=\int u +2+\frac{1}{u}\,du\\
=\frac{u^2}{2}+2u+\ln|u|+C\\
=\frac{(x-1)^2}{2}+2(x-1)+\ln|x-1|+C$
Simplify :
$\frac{x^2-3}{2}+x+\ln|x-1|+C$
It's not on the option.
|
After some rather basic algebraic manipulations, all you're left with is a pretty simple u-substantiation problem:
$$
\begin{align}
\int \frac{x^2}{x-1} \,dx
&=\int \frac{x^2-1+1}{x-1} \,dx\\
&=\int \left(\frac{x^2-1}{x-1}+\frac{1}{x-1}\right) \,dx\\
&=\int \frac{(x-1)(x+1)}{x-1}\,dx+\int\frac{1}{x-1} \,dx\\
&=\int \left(x+1\right)\,dx+\int\frac{1}{x-1}\frac{d}{dx}\left(x-1\right) \,dx\\
&=\frac{x^2}{2}+x+\int\frac{1}{x-1}\,d\left(x-1\right)\\
&=\frac{x^2}{2}+x+\ln{\left|x-1\right|}+C.\\
\end{align}
$$
|
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|
In the triangle $\triangle ABC$, $|AB|^3 = |AC|^3 + |BC|^3$. Prove that $\angle ACB > 60^\circ$. The following was a question in the final of the Flanders Mathematics Olympiad 2018:
In the triangle $\triangle ABC$, $|AB|^3 = |AC|^3 + |BC|^3$. Prove that $\angle ACB > 60^\circ$.
In this competition, points are assigned for formulating a rigorous and mathematically sound proof.
I proved the above by contradiction. Let $\alpha = \angle BAC, \beta = \angle CBA, \gamma = \angle ACB$. Suppose $\gamma \le 60^\circ$:
$$\gamma \le 60^\circ \iff \sin(\gamma) \leq \frac{\sqrt{3}}{2}$$
Applying the sine rule:
$$\frac{\sin(\alpha)}{|BC|} = \frac{\sin(\beta)}{|AC|} = \frac{\sin(\gamma)}{|AB|}$$
$$\iff\frac{|AC|^3}{|AB|^3} + \frac{|BC|^3}{|AB|^3} = \frac{\sin^3(\alpha) + \sin^3(\beta)}{\sin^3(\gamma)} = 1$$
$$\iff \sin^3(\alpha) + \sin^3(\beta) = \sin^3(\gamma) \le \left( \frac{\sqrt{3}}{2} \right)^3$$
$$\iff\sin(\alpha) \le \frac{\sqrt(3)}{2}, \, \sin(\beta) \le \frac{\sqrt(3)}{2}\tag{1}$$
We also know that:
$$\alpha + \beta = 180^\circ - \gamma \ge 120^\circ\tag{2}$$
$$\alpha + \beta < 180^\circ\tag{3}$$
Without loss of generality, assume $\alpha \ge \beta$. From $(1)$, $(2)$ and $(3)$, it then follows that:
$$a \ge 120^\circ, \, \beta \le 60^\circ$$
$$\implies |BC| > |AB| \qquad \unicode{x21af}$$
Is this answer adequate enough? Can the notation be improved? Are there any alternative approaches to solve this problem?
|
In the standard notation we need to prove that
$$\frac{a^2+b^2-c^2}{2ab}<\cos60^{\circ}$$ or
$$c^2>a^2-ab+b^2$$ or
$$\sqrt[3]{(a^3+b^3)^2}>a^2-ab+b^2$$ or
$$(a+b)^2(a^2-ab+b^2)^2>(a^2-ab+b^2)^3$$ or
$$ab>0,$$ which is true.
Id est, $$\measuredangle ACB>60^{\circ}$$ and we are done!
|
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|
Convergence of $\sum_{n=1}^\infty\left(\frac{1}{a_n^2}-\frac{1}{a_{n+1}^2}\right)$ where $a_1=1$ and $a_n=2-\frac 1n$ for $n\geq 2$ Let $a_1=1\ $ and $\ a_n=2-\frac 1n$ for $n\geq 2$. Then $$\sum_{n=1}^\infty\bigg(\frac{1}{a_n^2}-\frac{1}{a_{n+1}^2}\bigg)$$
converges to?
I started by expanding the sum:
$\sum_{n=1}^\infty\bigg(\frac{1}{a_n^2}-\frac{1}{a_{n+1}^2}\bigg)=\bigg(\frac{1}{a_1^2}-\frac{1}{a_2^2}\bigg)+\bigg(\frac{1}{a_2^2}-\frac{1}{a_3^2}\bigg)+\bigg(\frac{1}{a_3^2}-\frac{1}{a_4^2}\bigg)+\ldots$
$=\frac{1}{a_1^2}+\bigg(-\frac{1}{a_2^2}+\frac{1}{a_2^2}\bigg)+\bigg(-\frac{1}{a_3^2}+\frac{1}{a_3^2}\bigg)+\bigg(-\frac{1}{a_4^2}+\frac{1}{a_4^2}\bigg)+\ldots$
$=\frac{1}{a_1^2}$
$=1$
Now, the strange part is that the answer to this question is known to me as $0.75$. Is there something wrong with what I did? How does one arrive at $0.75$ as a solution?
PS: I am aware that infinite sums can behave in strange ways, and sometimes more than one solution may follow logically. Is this one such case?
|
Let $$S_N=\sum_{n=1}^N\left(\frac{1}{a_n^2}-\frac{1}{a_{n+1}^2}\right)=\frac{1}{a_1^2}-\frac{1}{a_{N+1}^2}.$$
Since $a_1=1$ and $a_n\to 2$, we have
$$\lim_{N\to\infty}S_N=1-\frac{1}{2^2}=\frac{3}{4}.$$
|
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|
GCSE Probability question relating to trees, help please. There are only red counters, yellow counters and blue counters in a bag.
Kevin takes at random a counter from the bag.
He puts the counter back in the bag.
Lethna takes at random a counter from the bag.
She puts the counter back in the bag.
The probability that both counters are red or that both counters are yellow is 13/36.
The probability that the first counter is red and the second counter is not red is 1/4.
Seb takes at random a counter from the bag.
Work out the probability that Seb takes a yellow counter.
-
So far, from the text, I've formed $2$ equations but since the number of unknowns is greater than the number of equations, I cannot solve.
$[(R^2)/(R+Y+B)^2] + [(Y^2)/(R+Y+B)^2] = 13/36$
$[(R)/(R+Y+B)]*[(Y+B)/(R+Y+B)] = 1/4$
Any help would be appreciated. Thanks.
|
Let $R$, $Y$ and $B$ be the events in which a red, yellow and blue counter is drawn from the bag, respectively. We have:
$$P(R) \cdot (1 - P(R)) = \frac{1}{4} \iff P(R)^2 - P(R) + \frac{1}{4} = 0 \iff P(R) = \frac{1 \pm \sqrt{1 - 1}}{2} = \frac{1}{2}$$
$$P(R) \cdot P(R) + P(Y) \cdot P(Y) = \frac{13}{36} \iff P(Y)^2 = \frac{13 - 9}{36} = \frac{4}{36} = \frac{1}{9} \iff P(Y) = \frac{1}{3}$$
The probability that Seb draws a yellow counter, thus equals $\frac{1}{3}$.
|
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Can you solve $\frac{dy}{dx}=(1-x)y^2+(2x-1)y-x$? $$\frac{dy}{dx}=(1-x)y^2+(2x-1)y-x$$
This is a form of riccati differential equation, which can be reduced to Bernoulli's equation if one particular solution is given. Here $y=1$ is a given solution.
Hence the general solution must be in the form $y=1+z$ for some $z(x)$, but substituting this in main equation gives
$$\frac{dz}{dx}+z^2x=z^2+z$$
Which is not in the form of Bernoulli equation.
Please help !
Also
$$2x^2\frac{dy}{dx}=(x-1)(y^2-x^2)+2xy$$
This is asked in the problem section of bernoulli's differential equation but i have no idea how to solve this.
|
Bernoulli's equation has form,
$$
\frac{dy}{dx}+p(x)y=q(x)y^n
$$
Now, consider this,
$$
\frac{dz}{dx}+z^2x=z^2+z
$$
This easily simplifies to,
$$
\frac{dz}{dx}-z=(1-x^2)z^2
$$
where $p(x)=-1$ and $q(x)=1-x^2$ .
And similarly other one simplifies to,
$$
2x^2\frac{dy}{dx}=(x-1)(y^2-x^2)+2xy
$$
$$
2x^2\frac{dy}{dx}=xy^2-x^3-y^2+x^2+2xy
$$
$$
\frac{dy}{dx}=\frac{xy^2}{2x^2}-\frac{x}{2}-\frac{y^2}{2x^2}+\frac{1}{2}+\frac{y}{x}
$$
put $z=\frac{y}{x}$ and $\frac{dz}{dx}=\frac{-y}{x^2}+\frac{1}{x}\frac{dy}{dx}$ which is same as $\frac{dz}{dx}=\frac{-z}{x}+\frac{1}{x}\frac{dy}{dx}$ implies $\frac{dy}{dx}=x\frac{dz}{dx}+z$. Hence, the equation will be,
$$
x\frac{dz}{dx}+z=\frac{xz^2}{2}-\frac{x}{2}-\frac{z^2}{2}+\frac{1}{2}+z
$$
$$
2x\frac{dz}{dx}=xz^2-x-z^2+1
$$
$$
2x\frac{dz}{dx}=z^2(x-1)-(x-1)
$$
$$
2x\frac{dz}{dx}=(z^2-1)(x-1)
$$
Finally,
$$
\frac{2}{z^2-1}\frac{dz}{dx}=\frac{x-1}{x}
$$
or
$$
\frac{2}{z^2-1}dz=\frac{x-1}{x}dx
$$
which can be solved by taking integration on both sides...
|
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Equality in triangle obtuse
$m_1, m_2, m_3 $ are sides-lengths of a triangle such that $m_1\sqrt{m_1}+m_2\sqrt{m_2}=m_3\sqrt{m_3}$.
Prove that this triangle is an obtuse-angled triangle.
I don't have idea make run this example
when I raised to square argument it makes a little bit complicated like this
$m_1^3+2m_1\sqrt{m_1}m_2\sqrt{m_2}+m_2^3=m_3^3$
|
The hint.
Let $m_1=a$, $m_2=b$ and $m_3=c$.
Thus, $$c^2=\left(\sqrt{c^3}\right)^{\frac{4}{3}}=\left(\sqrt{a^3}+\sqrt{b^3}\right)^{\frac{4}{3}}>a^2+b^2.$$
The last inequality it's
$$(\sqrt{a^3}+\sqrt{b^3})^4>(a^2+b^2)^3,$$ which is true because
$$(\sqrt{a^3}+\sqrt{b^3})^4=a^6+4\sqrt{a^9b^3}+6a^3b^3+4\sqrt{a^3b^9}+b^6>$$
$$>a^6+3\sqrt{a^9b^3}+3\sqrt{a^3b^9}+b^6\geq a^6+3a^4b^2+3a^2b^4+b^6=(a^2+b^2)^3.$$
$$3\sqrt{a^9b^3}+3\sqrt{a^3b^9}\geq3a^4b^2+3ab^4$$ it's
$$\sqrt{a^3b^3}(a^3+b^3)\geq a^2b^2(a^2+b^2)$$ or
$$a^3+b^3\geq\sqrt{ab}(a^2+b^2)$$ or
$$a^3-a^2\sqrt{ab}-b^2\sqrt{ab}+b^3\geq0$$ or
$$a^2\sqrt{a}(\sqrt{a}-\sqrt{b})-b^2\sqrt{b}(\sqrt{a}-\sqrt{b})\geq0$$ or
$$(\sqrt{a}-\sqrt{b})(\sqrt{a^5}-\sqrt{b^5})\geq0,$$
which is true because if $\sqrt{a}\geq\sqrt{b}$ so $\sqrt{a^5}\geq\sqrt{b^5}$ and if $\sqrt{a}\leq\sqrt{b}$ so $\sqrt{a^5}\leq\sqrt{b^5}$.
|
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|
Prove by induction: $C(2n, 2) = 2C(n, 2) + n^2$
Show that if $n$ is a positive integer, then $C(2n, 2) = 2C(n, 2) + n^2$. Here, $C(a, b)$ means the binomial coefficient $\dbinom{a}{b}$.
Prove this by induction.
Here is my calculation:
$n$ cannot be $1$ because $n$ should be equal or larger than $2$ for $C(n, 2)$.
If $n=2$: $C(4, 2) = 2C(2, 2) + 2^2 = 6$
If $n=m$: $C(2m, 2) = 2C(m, 2) + m^2 = 2m^2 - m$
If $n=m+1$: $C(2m+2, 2) = 2C(m+1, 2) + (m+1)^2 = 2m^2 + 3m + 1$
And I think I have to prove that:
\begin{align*}
C(2m+2, 2) & = [\dots]\\
& = [2C(m, 2) + m^2] + [2C(m, 1) + (2m + 1)]\\
& = 2C(m+1, 2) + (m+1)^2 && \text{(by Pascal's Identity).}
\end{align*}
But I have no idea what to do on $[\dots]$ part.
Thanks.
|
Hint:
Show and then use that
$$\binom nk=\binom{n-1}{k-1}+\binom{n-1}k$$
|
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|
Find a generating function for which $A(n)={n \choose 2}$ In the book I'm using, $A(x)$ denotes the formal power series (generating function), $A(x) = \sum a_ix^i$.
I'm really stuck on this problem. Thanks for any help.
My attempt after the given hint:
$$\begin{align}
A(x)&=\sum_{n\geq 0} \binom{n}{2}x^n\\
&=\sum_{n\geq 0} \frac{x^2}{2}\frac{d^2}{dx^2}(x^{n})\\
&=\frac{x^2}{2} \frac{d^2}{dx^2}\sum_{n\geq 0} x^{n}\\
&=\frac{x^2}{2} \frac{d^2}{dx^2}\frac{1}{1-x}\\
&=\frac{x^2}{2} \frac{2}{(1-x)^3}
\end{align}$$
Sorry, I'm new to $\rm \LaTeX$.
|
Do you mean $a_n=\binom{n}{2}$ and $A(x) =\sum_{n\geq 0}a_nx^n= \sum_{n\geq 0} \binom{n}{2}x^n$?
Then note that
$$\binom{n}{2}x^n=\frac{x^2}{2}\cdot n(n-1) x^{n-2}=\frac{x^2}{2}\cdot\frac{d^2}{dx^2}(x^{n})$$
and recall the basic generating function $\sum_{n\geq 0}x^n=\frac{1}{1-x}$.
What is $A(x)$?
|
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|
Evaluate the values of $x$ in $\sqrt{2x-5} = \sqrt[3]{6x-15}$
$$\sqrt{2x-5} = \sqrt[3]{6x-15}$$
*
*Evaluate the values of $x$
I believe that there would be an easier approach to this problem because I will have to expand 3th degree binomial as shown below
$$(2x-5)^3 = (6x-15)^2$$
What am I missing?
Regards
|
because I will have to expand 3th degree binomial as shown below: $(2x-5)^3 = (6x-15)^2$
Which is not a difficult thing to do:
$8x^3 - 60x^2 + 150x - 125 = 36x^2 - 180x +225$ so
$8x^3 -96x^2 + 330x - 350=0$
which is not an unreasonable thing to expect a student to be able to factor and solve (although I pity the student who tries).
However the student should at this point see:
$(2x -5)^3 = (6x-15)^2 = (3(2x-5))^2 = 9(2x-5)^2$
So either $2x-5 =0$ or $2x -5 = 9$.
=====
.... or one can do it the sweaty he-man way:
$\sqrt{2x - 5} = \sqrt[3]{6x - 15} = \sqrt[3]3\sqrt[3]{2x - 5}$.
If $x - 5 = 0$ we have $x = 2 \frac 12$ as a potential answer.
If we assume $2x -5\ne 0$ we have
$\frac {\sqrt{2x - 5}}{\sqrt[3]{2x - 5}} = \sqrt[3] 3$.
Assume $2x - 5 > 0$ (which we must for $2x - 5 \ne 0$ and $\sqrt{2x - 5}$ to be defined) we have:
$\frac {(2x-5)^{\frac 12}}{(2x-5)^{\frac 13}}=(2x-5)^{\frac 12 - \frac 13} = (2x-5)^{\frac 16} = 3^{\frac 13}$
So $2x-5 = (3^{\frac 13})^6 = 3^2 = 9$.
So $x = 7$.
So solutions are
$x = 2\frac 12, 7$.
|
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|
Let $\alpha$ denote the image of $x$ in $\mathbb Q[x]/(x^3 + 3x + 3)$
Let $\alpha$ denote the image of $x$ in $\mathbb Q[x]/(x^3 + 3x + 3)$. Express each of $1/α,1/(1+ \alpha),1/(1+ \alpha^2)$ in the form $c_2\alpha^2+c_1\alpha+c_0$ with $c_0, c_1,c_2 \in \mathbb Q$.
I can prove $x^3 + 3x + 3$ is irreducible in $\mathbb Q[x]$ and $\mathbb Q[x]/(x^3 + 3x + 3)$ is a field. However, I don't know how to proceed from here.
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Let $P(x)=x^3+3x+3$, $Q_1(x)=x$, $Q_2(x)=1+x$ and $Q_3(x)=1+x^2.$
By Euclid's algorithm, we have that $$P(x)=(x^2+3)Q_1(x) + 3.$$
If we view this relation in $\mathbb{Q}[x]/(P(x))$, we get that $$0=P(\alpha)=(\alpha^2+3)Q_1(\alpha)+3,$$ hence $$3 = -(\alpha^2+3)\alpha,$$ and is $$\frac{1}{\alpha} = \frac{-(\alpha^2+3)}{3}=-\frac{1}{3}\alpha^2-1.$$
For $\frac{1}{1+\alpha}$, we solve in a similar way. We have that $$P(x)=(x^2-x+4)Q_2(x)-1,$$ then $$0=P(\alpha)=(\alpha^2-\alpha+4)Q_2(\alpha)-1.$$ So $$\frac{1}{1+\alpha} = \frac{1}{Q_2(\alpha)} = \alpha^2-\alpha+4.$$
For $\frac{1}{1+\alpha^2}$ we found it in two steps (I'm avoiding the use of the mentioned Extended Euclidean Algorithm). First, we know as above that $$P(x)=xQ_3(x)+2x+3,$$ and in $\mathbb{Q}[x]/(P(x))$ we have $$0 = P(\alpha)=\alpha Q_3(\alpha)+2\alpha+3,$$ so $$\frac{1}{1+\alpha^2}=\frac{1}{Q_3(\alpha)} = \frac{-\alpha}{2\alpha+3}.$$
Now, we compute $\frac{1}{2\alpha+3}$. Since $$P(x)=(\frac{1}{2}x^2-\frac{3}{4}x+\frac{21}{8})(2x+3)-\frac{39}{8},$$ is $$\frac{1}{2\alpha+3} = \frac{8}{39}(\frac{1}{2}\alpha^2-\frac{3}{4}\alpha+\frac{21}{8}).$$ You now only have to multiply $\frac{1}{2\alpha+3}$ by $-\alpha$ and simplify the expresion using the relation $\alpha^3+3\alpha+3=0.$
|
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Provide a counterexample: If $n^2-1$ is divisible by $5$, then $n$ is divisible by $2$ or $3$ Provide a counterexample: If $n^2-1$ is divisible by $5$, then $n$ is divisible by $2$ or $3$
My book doesn't have an answer to this question, but I think it's $n=6$.
Since $6^2-1=35$, which is divisible by $5$ but not by $2$ or $3$. Is this the right way to solve these counterexample problems? Are these problems solved by just plugging in numbers until you get to the right one?
|
$n^2 - 1= (n-1)(n+1)$ so if $5|n^2 -1$, as $5$ is prime we know either $5|n-1$ or $5|n+1$.
Now if we're clever we can try to take counter examples $n=6$ or $4$. nope. $n = 11$ or $9$. Oh, wait there's a counter example. $n=11$. (Also $n=1$ and $5|0$ is a counter example.)
But plugging in numbers in the hope of a counterexample is ineffiecient. What if the first counter example is really huge? Or what if the statement is true and we never find a counter example?
we should evaluate further.
So $n\equiv \pm 1 \pmod 5$. And we are asked to show or find a counter example that $n \equiv 0 \pmod 3$ or $n \equiv 0\pmod 2$.
Now By Chinese remainder theorem $\gcd(5,2); \gcd(5,3); \gcd(2,3)=1$ so ther is a unique solution to
$n \equiv a \pmod 5$ and $n \equiv b \pmod 3$ and $n \equiv c \pmod 2$ will have a unique solution $\mod 30$ and we just need to choose $a = \pm 1$ and $b \ne 0$ and $c \ne 0$.
Example $n \equiv 1 \pmod 5$ and $n\equiv 1 \pmod 3$ and $n\equiv 1 \mod 2$ will have solutions and that solution would be a counter example.
We don't even have to find it! We know it exists!
.... but it's $1 \pmod {30}$ so $n =1$ is counter example. $n = 31$ is a counterexample and $n = 30k + 1$ is a counter example. $(30k + 1)^2- 1 = 900k^2 + 60k + 1 - 1= 5(180k^2 + 12k)$ but $2|30k$ and $3|30k$ so $2,3\not \mid 30k + 1$).
We can find other counter examples.
$n \equiv -1 \pmod 5$ and $n \equiv 1 \pmod 3$ $n\equiv 1 \pmod 2$ would give us $n \equiv 19\pmod 30$ so $19, 49, 79$ etc are counter examples.
And $n \equiv 1 \pmod 5$ and $n \equiv -1 \pmod 3$ and $n \equiv 1 \pmod 2$ will have solution $n \equiv 11 \pmod 30$. So $11, 41, 71$ etc are counter examples.
And $n \equiv -1\pmod 5$ and $n \equiv -1 \pmod 3$ and $n \equiv 1 \pmod 2$ will have soultion $n \equiv 29 \pmod 30$. So $29, 59, 89$ etc are counter examples.
|
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|
Showing an inequality using Cauchy-Schwarz I managed to solve the following inequality using AM-GM:
$$
\frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+\frac{c}{(c+1)(a+1)} \geq \frac{3}{4}
$$
provided that $a,b,c >0$ and $abc=1$.
However it was hinted to me that this could also be solved with Cauchy-Schwarz inequality but I have not been able to find a solution using it and I'm really out of ideas.
|
Let $a=\frac{y}{x}$ and $b=\frac{z}{y},$ where $x$, $y$ and $z$ are positives.
Thus, $c=\frac{x}{z}$ and by C-S we obtain:
$$\sum_{cyc}\frac{a}{(a+1)(b+1)}=\sum_{cyc}\frac{\frac{y}{x}}{\left(\frac{y}{x}+1\right)\left(\frac{z}{y}+1\right)}=\sum_{cyc}\frac{y^2}{(x+y)(y+z)}\geq\frac{(x+y+z)^2}{\sum\limits_{cyc}(x+y)(y+z)}.$$
Id est, it's enough to prove that:
$$\frac{(x+y+z)^2}{\sum\limits_{cyc}(x+y)(y+z)}\geq\frac{3}{4}$$ or
$$4(x+y+z)^2\geq3\sum_{cyc}(x+y)(y+z)$$ or
$$4\sum_{cyc}(x^2+2xy)\geq3\sum_{cyc}(x^2+3xy)$$ or$$\sum_{cyc}(x^2-xy)\geq0$$ or $$\sum_{cyc}(x-y)^2\geq0.$$
|
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|
Stuck at proving whether the sequence is convergent or not I have been trying to determine whether the following sequence is convergent or not. This is what I got:
Exercise 1: Find the $\min,\max,\sup,\inf, \liminf,\limsup$ and determine whether the sequence is convergent or not:
$X_n=\sin\frac{n\pi}{3}-4\cos\frac{n\pi}{3}$
I wrote down a few cases:
$X_1 = \sin\frac{\pi}{3}-4\cos\frac{\pi}{3}= \frac{\sqrt3-4}{2} $
$X_2 = \sin\frac{2\pi}{3}-4\cos\frac{2\pi}{3}= \frac{\sqrt3+4}{2} $
$X_3 = \sin\frac{3\pi}{3}-4\cos\frac{3\pi}{3}=4 $
$X_4 = \sin\frac{4\pi}{3}-4\cos\frac{4\pi}{3}= \frac{4-\sqrt3}{2} $
$X_5 = \sin\frac{5\pi}{3}-4\cos\frac{5\pi}{3}= \frac{4-\sqrt3}{2} $
$X_6 = \sin\frac{6\pi}{3}-4\cos\frac{6\pi}{3}= -4 $
So as found above, $\min = -4$, $\max = 4$, $\inf = -4$, $\sup = 4$, $\liminf = -4$, $\limsup = 4$.
Let's check whether its convergent or not:
The sequence is bounded as stated above so lets check if its decreasing or increasing.
$X_n \geq X_{n+1}$
$\sin\frac{n\pi}{3}-4\cos\frac{n\pi}{3} \geq\sin\frac{(n+1)\pi}{3}-4\cos\frac{(n+1)\pi}{3}$
$\sin\frac{n\pi}{3} - \sin\frac{(n+1)\pi}{3} \geq -4\cos\frac{(n+1)\pi}{3} + 4\cos\frac{n\pi}{3}$
I used trigonometrical identity for $\sin\alpha+\sin\beta$ and $\cos\alpha-\cos\beta$ :
$-\cos\frac{\pi(2n+1)}{6} \geq 4\sin\frac{\pi(2n+1)}{6}$
What should I do next? I am stuck here.
Thanks, and sorry if I made mistakes.
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The limit of a sequence, if it exists, is equal to its lim inf and lim sup. Accordingly, if the lim inf and lim sup of a sequence are different, then its limit cannot exist.
|
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|
Computing the product $(\frac{d}{dx}+x)^n(-\frac{d}{dx}+x)^n$ I want to compute the product
$$
(\frac{d}{dx}+x)^n(-\frac{d}{dx}+x)^n,
$$
for a natural number $n$. For $n$ equal to 0 or 1, the computation is very simple but for such a low number as 2 the brute force calculation begins to be rather cumbersome and I cannot see any pattern emerging. I tried to find some connection with the Rodrigues' formula for the Hermite polynomials but I could not.
These operators come up in the algebraic approach to the quantum harmonic oscillator.
Explicit Example
To avoid any misunderstanding, I am going to show explicitly the computation for the case $n=1$:
$$
(\frac{d}{dx}+x)(-\frac{d}{dx}+x)=-\frac{d^2}{dx^2}+1+x\frac{d}{dx}-x\frac{d}{dx}+x^2=-\frac{d^2}{dx^2}+x^2+1.
$$
One can think of a function $f$ the operators are acting on. For example,
$$
(\frac{d}{dx}\circ x) f= (\frac{d}{dx}x)f+x\frac{d}{dx}f=(1+\frac{d}{dx})f,
$$
then
$$
\frac{d}{dx}\circ x=1+x\frac{d}{dx}
$$
|
For convenience let us rewrite $x,\partial_x$ with $a,b$ so $[a,b]=x\partial_x-\partial_x x=-1$ (in the operator sense on the Schwartz space).
Lemma. For all $n\in\mathbb N$
$$
[(a+b)^n,a-b]=2n(a+b)^{n-1}
$$
Proof. As darij pointed out, one has $[a+b,a-b]=2$ (i.e. the case $n=1$). The trick then is
$$
\begin{split}
[(a+b)^{n+1},a-b]&=(a+b)[(a+b)^n,a-b]+[a+b,a-b](a+b)^n\\
&=(a+b)2n(a+b)^{n-1}+2(a+b)^{n}=2(n+1)(a+b)^{(n+1)-1}
\end{split}
$$
which concludes the proof via induction. $\square$
Proposition. For all $n\in\mathbb N_0$
$$
(a+b)^n(a-b)^n=\prod_{j=1}^n (a^2-b^2+(2j-1))
$$
Proof. ($n=0$ is obvious). Note that $(a-b)(a+b)=a^2-b^2-1$. Using the previous lemma
$$
\begin{split}
(a+b)^{n+1}(a-b)^{n+1}&=[(a+b)^{n+1},a-b](a-b)^n+(a-b)(a+b)(a+b)^n(a-b)^n\\
&=2(n+1)(a+b)^n(a-b)^n +(a^2-b^2-1)(a+b)^n(a-b)^n\\
&=\big( a^2-b^2+2n+1)(a+b)^n(a-b)^n=\prod_{j=1}^{n+1} (a^2-b^2+(2j-1))
\end{split}
$$
which again concludes the proof via induction. $\square$
This result reproduces the cases (aside from $n=0$, obvious)
*
*$(a+b)(a-b)=a^2-b^2+1$
*Making use of $[a^2,b^2]=-4ab-2$ (similar techniques) one gets
$$
\begin{split}
(a+b)^2(a-b)^2=(a^2-b^2+1)(a^2-b^2+3)&=a^4-a^2b^2-b^2a^2+4a^2-4b^2+b^4+3\\
&=a^4-2a^2b^2-4ab+4a^2-4b^2+b^4+1
\end{split}
$$
etc... I feel like this formula is the best thing one can hope for in terms of structure.
Edit: Thanks darij for the +200 rep!
|
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Radical equation - can I square both sides with more than 1 radical on one side? I'm familiar with equations like:
$\sqrt{x+1} - \sqrt{x+2} = 0 $
Has no solutions, it's just an example off the top of my head
Just move the negative square root to the other side, square both sides and solve.
$\sqrt{x+1} = \sqrt{x+2}$
$x+1 = x+2$
0 = 1
My question is, if there are two square roots on one side, then can I still square both sides in this way:
$\sqrt{x+1} - \sqrt{x+2} = \sqrt{x+3}$
$x+1 - (x+2) = x+3$
$x+1 - (x-2) = x+3$
$x = -4$
Or does squaring both sides cause something strange to happen on the left hand side?
|
No, what you did was wrong.
this is because :
$\sqrt{x+1} - \sqrt{x+2} = \sqrt{x+3}$ does not imply $(x+1)-(x+2)=x+3$.
You need to square both sides, which means:
$(\sqrt{x+1} - \sqrt{x+2})^2=x+3$
$(x+1)+(x+2)-2\sqrt{(x+1)(x+2)}=x+3$.
As a sidenote, you can plug x=-4 into the original equation to check whether x=-4 is not a solution.
|
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|
Find all primes $p$ for which there are positive integers $x, y$ such that $p+1=2x^2$ and $p^2+1=2y^2$ Find all primes $p$ for which there exist positive integers $x, y$ such that $p+1=2x^2$ and $p^2+1=2y^2$.
I have tried coming up with an equation for $p$ or $p^2$ and this is what I've got
$p=2x^2-1$;
$p^2=(2x^2-1)^2$
$p^2=2y^2-1$
which means $(2x^2-1)^2=2y^2-1$ but this doesn't do much...
|
So $$p(p-1)=p^2-p=2y^2-2x^2=2(y+x)(y-x)$$
One of the three factors on the right must be $p$ or a multiple thereof, and it must be the largest among the three. We conclude $p\mid x+y$. But clearly $x<p$ and $y<p$, hence we can only have $p=x+y$. Then $p-1=2(y-x)$, hence $y=3x-1$.
Now by simple algebraic manipulations,$$(2x^2-1)^2=p^2=2(3x-1)^2-1 $$
$$4x^4-4x^2+1=18x^2-12x+1 $$
$$4x^4-22x^2+12x=0 $$
$$x(x-2)(x^2+x-\tfrac32)=0 $$
showing that $x=2$ is the only possible positive integer, and it leads to $y=5$ and the only solution $$p=7.$$
|
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|
Is this a Riemann sum (if so, I can't figure out which one)? This was supposedly an easy limit, and it is suspiciously similar to a Riemann sum, but I can't quite figure out for what function.
$$\lim_{n\to\infty}{\frac{1}{n} {\sum_{k=3}^{n}{\frac{3}{k^2-k-2}}}}$$
Well, even the fact that $\frac{3}{k^2-k-2} = \frac{1}{k-1}-\frac{1}{k+2}$ doesn't seem to simplify the problem. I thought this would be a telescoping sum, but it's clearly not.
Is that a Riemann sum at all?
|
$$\sum_{k=3}^n \frac{3}{k^2-k-2} = \sum_{k=3}^n \frac{1}{k-2}- \sum_{k=3}^n \frac{1}{k+1}$$
You (and I) were mistaken before, see @Romeo 's answer.
Notice that $$\sum_{k=3}^n \frac{1}{k-2}=\sum_{k=0}^{n-3} \frac{1}{k+1}$$
Insert above you get
$$\sum_{k=3}^n \frac{3}{k^2-k-2} = \sum_{k=0}^{n-3} \frac{1}{k+1} - \sum_{k=3}^n \frac{1}{k+1} = \sum_{k=0}^2 \frac{1}{k+1} - \sum_{k=n-2}^{n} \frac{1}{k+1}= $$$$=1+\frac{1}{2}+\frac{1}{3} - \frac{1}{n-1}-\frac{1}{n}-\frac{1}{n+1}$$
Of course this argument requires $n\geq 3$.
|
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"url": "https://math.stackexchange.com/questions/3115090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Evaluate : $\lim_{x\to -\infty} \frac{\sqrt{4x^2-2x}}{\sqrt[3]{x^3+1}}$
Evaluate: $\lim_{x\to -\infty} \frac{\sqrt{4x^2-2x}}{\sqrt[3]{x^3+1}}$
My solution:
\begin{align}
\lim_{x\to -\infty} \frac{\sqrt{4x^2-2x}}{\sqrt[3]{x^3+1}} & = \lim_{x\to -\infty} \frac{\sqrt{4-\frac{2}{x}}}{\sqrt[3]{1+\frac{1}{x^3}}}\\
& = \frac{\sqrt{4-0}}{\sqrt[3]{1+0}} \\
& = 2
\end{align}
Despite the steps I've taken seems plausible to me, the answer is given as $-2$.
Is dividing both the numerator and the denominator by $x$ allowed here? Where am I making a mistake?
|
Note that
$$\frac{\sqrt{x^2}}{\sqrt[3]{x^3}}= \frac{|x|}{x}\stackrel{x<0}{=}-1$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3116034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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|
If $\sec\theta=-\frac{13}{12}$, then find $\cos{\frac{\theta}{2}}$, where $\frac\pi2<\theta<\pi$. The official answer differs from mine.
Given $\sec\theta=-\frac{13}{12}$ find $\cos{\frac{\theta}{2}}$, where $\frac\pi2<\theta<\pi$.
If the $\sec\theta$ is $-\frac{13}{12}$ then, the $\cos \theta$ is $-\frac{12}{13}$, and the half angle formula tells us that $\cos{\frac{\theta}{2}}$ should be
$$\sqrt{\frac{1+\left(-\frac{12}{13}\right)}{2}}$$
which gives me $\sqrt{\dfrac{1}{26}}$ which rationalizes to $\dfrac{\sqrt{26}}{26}$.
The worksheet off which I'm working lists the answer as $\dfrac{5\sqrt{26}}{26}$.
Can someone explain what I've done wrong here?
|
The answer they gave $\left(\frac {5 \sqrt{26}}{26}\right)$ is the value for $$\sin \dfrac {\theta}{2} = \pm \sqrt {\dfrac {1-\cos \theta}{2}}$$ however they're looking for $$\cos \dfrac {\theta}{2} = \pm \sqrt {\dfrac {1+\cos \theta}{2}}$$
EDIT (thanks, DanielWainfleet!): For the range $\pi/2 \lt \theta \lt \pi$, $\pi/4 \lt \theta/2 \lt \pi/2$, so $\cos \dfrac {\theta}{2}$ will be positive. Thus, your answer will be $\left(\frac {\sqrt{26}}{26}\right).$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3117696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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|
Limit involving inverse functions When I am given the limit
$$\lim\limits_{x \rightarrow \infty}\frac{x\arctan\sqrt{x^2 +1}}{\sqrt{x^2+1}}$$
would it be possible to evaluate it giving some substitution?
L'Hospital's rule seemed an option but I ended up going in circles.
|
When $x>0$, $|x|=x$ and obviously if $x\rightarrow\infty$, then $\sqrt{x^2+1}\rightarrow\infty$. And the last thing that you will need is the fact that the inverse tangent function approaches a value of $\pi/2$ as its argument goes to infinity:
$$
\lim\limits_{x \rightarrow \infty}\frac{x\arctan\sqrt{x^2+1}}{\sqrt{x^2+1}}=
\lim\limits_{x \rightarrow \infty}\frac{x\arctan\sqrt{x^2 +1}}{\sqrt{x^2}\sqrt{1+\frac{1}{x^2}}}=
\lim\limits_{x \rightarrow \infty}\frac{x\arctan\sqrt{x^2 +1}}{|x|\sqrt{1+\frac{1}{x^2}}}=\\
\lim\limits_{x \rightarrow \infty}\frac{x\arctan\sqrt{x^2 +1}}{x\sqrt{1+\frac{1}{x^2}}}=
\lim\limits_{x \rightarrow \infty}\frac{\arctan\sqrt{x^2 +1}}{\sqrt{1+\frac{1}{x^2}}}=
\frac{\pi/2}{\sqrt{1+0}}=\frac{\pi}{2}.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3118910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Minimum number of steps in water measure problem I've been struggling with this problem for a while and gone through questions about the "Water Jug Problem/Puzzle".
A person wants to have $2$ separate $1$ L measures of water at the same time. However the only measures she has are for $6, 10$ and $15$ L. Show how this could be done with the minimum number of steps without marking the measures or using any container other than the original large beaker of water. The only steps allowed are filling or emptying a measure or transferring water from one measure to another.
I've read a question here which is pretty similar (also involving minimum operations) to this but only used $2$ measures/jugs instead of $3$. I didn't understand how I could adapt it into my problem. It would be a great help if someone could explain in simpler terms...
|
Solutions in $9$ steps (without proof of optimality yet):
denote jugs as $a$($6$L), $b$($10$L), $c$($15$L); then:
\begin{array}{|c|c|c|c|c|}
\hline
\# & move & a & b & c \\
\hline
1) & \bullet \rightarrow a & 6 & - & - \\
2) & \bullet \rightarrow c & 6 & - & 15 \\
3) & c \rightarrow b & 6 & 10 & 5 \\
4) & b \rightarrow \bullet & 6 & - & 5 \\
5) & c \rightarrow b & 6 & 5 & - \\
6) & a \rightarrow c & - & 5 & 6 \\
7) & \bullet \rightarrow a & 6 & 5 & 6 \\
8) & a \rightarrow b & 1 & 10 & 6 \\
9) & b \rightarrow c & 1 & 1 & 15 \\
\hline
\end{array}
\begin{array}{|c|c|c|c|c|}
\hline
\# & move & a & b & c \\
\hline
1) & \bullet \rightarrow a & 6 & - & - \\
2) & \bullet\rightarrow b & 6 & 10 & - \\
3) & b \rightarrow c & 6 & - & 10 \\
4) & a \rightarrow b & - & 6 & 10 \\
5) & \bullet \rightarrow a & 6 & 6 & 10 \\
6) & a \rightarrow c & 1 & 6 & 15 \\
7) & c \rightarrow b & 1 & 10 & 11 \\
8) & b \rightarrow \bullet & 1 & 0 & 11 \\
9) & c \rightarrow b & 1 & 10 & 1 \\
\hline
\end{array}
Actually there are up to $12$ valid moves at each stage:
$\bullet \rightarrow a$, $\;\bullet \rightarrow b$, $\;\bullet \rightarrow c$;
$a \rightarrow b$, $\;a \rightarrow c$, $\;a \rightarrow \bullet$;
$b \rightarrow a$, $\;b \rightarrow c$, $\;b \rightarrow \bullet$;
$c \rightarrow a$, $\;c \rightarrow b$, $\;c \rightarrow \bullet$;
so there are not more than $12^{8}\approx 430\:000\:000$ chains of $8$-step moves (not too much as for computer check).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3119429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Finding the ratio of a side of $\triangle ABC$ and its segment where one cevian line from the opposite vertex intersect the side in any point
In $\triangle ABC$, $L$ and $M$ are two points on $AB$ and $AC$ such that $AL = \frac{2AB}{5}$ and $AM = \frac{3AC}{4}$. $BM$ and $CL$ intersect at the point $P$ and the extension line of $AP$ and the side $BC$ intersect $BC$ at the point $N$. What is the value of $\frac{BN}{BC}$?
SOURCE: BANGLADESH MATH OLYMPIAD
My Attempt:
Let denote the area of $\triangle ALP$, $[ALP] = x$ and $[APM] = y$.
So, from $\triangle APB$, showing the relation of the base of both the triangle $\triangle APL$ and $\triangle LPB$, I got $[LPB] = \frac{3x}{2}$.
Similarly, from $\triangle APC$, doing the above likewise approach, I got $[MPC] = \frac{y}{3}$. After that, getting two triangle $\triangle ACL$ and $\triangle BCL$ and showing their relation of area with their particular base, I got
$[BPC] = \frac{5x}{6}$....(given that $AL:AB =2:5$)
Again from $\triangle ABM$ and $\triangle CBM$,
$[BPC] = 2y$.....($AM:AC = 3:4$)
So, $2y = \frac{5x}{6}$ $\implies$ $x =\frac{12y}{5}$
And then expressing the area of all the triangle by $y$, I got $[ABC] = \frac{28y}{3}$
But, I can't anyhow relate the area of $\triangle BPN$ with $y$. Here, I got stuck. What should I do to find out the area of $\triangle BPN$. I need really some help Thank you.
|
$$\frac{BN}{NC}\cdot\frac{CM}{MA}\cdot\frac{AL}{LB}=\frac{S_{\Delta PBN}}{S_{\Delta PCN}}\cdot\frac{S_{\Delta PCM}}{S_{\Delta PAM}}\cdot\frac{S_{\Delta PAL}}{S_{\Delta PBL}}=$$
$$=\frac{\frac{1}{2}PB\cdot PN\sin\measuredangle BPN}{\frac{1}{2}PC\cdot PN\sin\measuredangle CPN}\cdot\frac{\frac{1}{2}PC\cdot PM\sin\measuredangle CPM}{\frac{1}{2}PA\cdot PM\sin\measuredangle APM}\cdot\frac{\frac{1}{2}PA\cdot PL\sin\measuredangle APL}{\frac{1}{2}PB\cdot PL\sin\measuredangle BPL}=1.$$
Thus, $$\frac{BN}{NC}\cdot\frac{1}{3}\cdot\frac{2}{3}=1.$$
Can you end it now?
I got $$\frac{BN}{BC}=\frac{9}{11}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3120343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Finding $\int\frac{\sin^4 x+\cos^4 x}{\sin^3 x+\cos^3 x}dx$
Find
$$\int\frac{\sin^4 x+\cos^4 x}{\sin^3 x+\cos^3 x}dx$$
What I tried:
$$\sin^4(x)+\cos^4(x)=(\sin^2 x+\cos^2 x)^2-2\sin^2 x\cos^2 x=1-2\sin^2 x\cos^2 x$$
and $$\sin^3 x+\cos^3 x=(\sin x+\cos x)(1-\sin x\cos x)$$
so
$$\int\frac{1-\sin^2 x\cos^2 x}{(\sin x+\cos x)(1-\sin x\cos x)}dx-\int\frac{\sin^2 x\cos^2 x}{(\sin x+\cos x)(1-\sin x\cos x)}dx$$
How do I solve it? Help me, please.
|
A brutal way is just to enforce the substitution $x=2\arctan t$, leading to
$$ \int\frac{\sin x\cos x}{\sin^3 x+\cos^3 x}\,dx=2\int\frac{\frac{2t(1-t^2)}{(1+t^2)^3}}{\frac{(2t)^3}{(1+t^2)^3}+\frac{(1-t^2)^3}{(1+t^2)^3}}\,dt=\int\frac{4t(1-t^2)}{(2t)^3+(1-t^2)^3}\,dt $$
an ugly integral, but perfectly solvable by partial fraction decomposition. The roots of $(2t)^3+(1-t^2)^3$ can be found by solving the quadratic equations given by
$$\frac{1-t^2}{2t}\in\left\{-1,\frac{1+i\sqrt{3}}{2},\frac{1-i\sqrt{3}}{2}\right\}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3121413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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|
Integral problem. Unsure of the approach. I have this integral:
$$\int_0^1 \frac{1 + 12t}{1 + 3t}dt$$
I can split this up into:
$$\int_0^1 \frac{1}{1+3t} dt + \int_0^1\frac{12t}{1+3t}dt$$
The left side:
$u = 1+3t$ and $du = 3dt$ and $\frac{du}{3} = dt$
so $$\frac{1}{3} \int_0^1 \frac{1}{u} du = \frac{1}{3} \ln |u| + C$$
But what about the right?
|
$u=1+3t$:
$$
\begin{align}
\int\frac{1+12t}{1+3t}\,dt
&=\int\frac{1}{1+3t}\,dt+12\int\frac{t}{1+3t}\,dt\\
&=\frac{1}{3}\int\frac{1}{1+3t}\frac{d}{dt}(1+3t)\,dt+\frac{12}{3}\int\frac{3t}{1+3t}\,dt\\
&=\frac{1}{3}\int\frac{1}{u}\,du+4\int\frac{1+3t-1}{1+3t}\,dt\\
&=\frac{1}{3}\ln{|u|}+4\left(\int\frac{1+3t}{1+3t}\,dt-\int\frac{1}{1+3t}\,dt\right)\\
&=\frac{1}{3}\ln{|u|}+4\int\,dt-\frac{4}{3}\int\frac{1}{1+3t}\frac{d}{dt}(1+3t)\,dt\\
&=\frac{1}{3}\ln{|u|}+4t-\frac{4}{3}\int\frac{1}{u}\,du\\
&=\frac{1}{3}\ln{|u|}+4t-\frac{4}{3}\ln{|u|}\\
&=\ln{\left(\frac{|u|^{1/3}}{|u|^{4/3}}\right)}+4t\\
&=\ln{|u|^{-1}}+4t\\
&=4t-\ln{|1+3t|}+C.
\end{align}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3122196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
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|
Solve $\log_2(3^x-1)=\log_3(2^x+1)$
Solve the following equation over the real number(preferably without calculus):
$$\log_2(3^x-1)=\log_3(2^x+1).$$
This problem is from a math contest held where I learn; I was unable to do much at all tinkering with it; I have observed the solution $x=1$ but haven't been able to prove there are no others or determine them if there are.
|
We'll use the following: $\log_a(b) = \frac{\log_c(b)}{\log_c(a)}$
So: $ \log_2(3^x - 1) = \log_3(3^x-1) \iff \frac{\ln(3^x-1)}{\ln(2)} = \frac{\ln(2^x+1)}{\ln(3)}$
That is:
$ \ln(3)\ln(3^x-1) = \ln(2)\ln(2^x+1) $
So we're left with a function $f:(0,+\infty) -> \Bbb R$ , $f(x) = \ln(3)\ln(3^x-1) -\ln(2)\ln(2^x+1) $
Looking at its derivative:
$f'(x) = \ln(3) \frac{3^x\ln(3)}{3^x-1} - \ln(2) \frac{2^x\ln(2)}{2^x+1} = \frac{6^x(\ln^2(3) - \ln^2(2)) + 3^x\ln^2(3) - 2^x\ln^2(2)}{(3^x-1)(2^x+1)} $
We see, that the sign of it depends on the sign of the numerator.
Let $g(x) = 6^x(\ln^2(3) - \ln^2(2)) + 3^x\ln^2(3) - 2^x\ln^2(2) $
Which is clearly positive ( cause $3^x\ln^2(3) > 2^x\ln^2(2) $ and $6^x(\ln^2(3) - \ln^2(2)) > 0 $ )
So our function $f$ is increasing, and that means (because $\lim_{x \to 0^+} f(x) = -\infty$, $\lim_{x \to +\infty} f(x) = +\infty$),
that our function has only one root. However, it is a little bit of a guess to tell it's $x=1$.
EDIT:
In fact, it isn't that hard to find that solution. We have an equation involving $\ln$.
Let $a(x) = 3^x-1$, $b(x) = 2^x+1$
Then, we arrive with: $\ln(3)\ln(a(x)) = \ln(2)\ln(b(x))$
Which clearly has a solution when $a(x) = 2$ and $b(x) = 3$, and fortunately $3^x = 3$ and $2^x=2$ have a solution $x=1$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3122795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
Find matrix $B$ from kernel of $A$ Here's the question:
For the matrix
$$
A=\begin{bmatrix}
1 & 1 & 0 & 1 & 1\\
1 & 0 & 1 & 1 & 0\\
0 & 1 & 1 & 1 & 1\\
\end{bmatrix}
$$
find a matrix B such that $\text{Im}(\phi_B)=\text{Ker}(\phi_A)$ and $\text{Ker}(\phi_B)=0$
I have found the null space of $A$ which must be the image of B but I have no idea how to proceed from here. Any tips?
|
I like using the language of column spaces and null spaces when discussing matrices.
Presumably, the notation $\phi_M$ refers to the linear map $\phi_M:\Bbb R^n\to \Bbb R^m$ given by $\phi_M(\vec{x})=M\vec{x}$ where $M$ is an $m\times n$ matrix. Then, in our notation, for any matrix $M$ we have
\begin{align*}
\operatorname{Ker}(\phi_M) &= \operatorname{Null}(M) & \operatorname{Im}(\phi_M) &= \operatorname{Col}(M)
\end{align*}
I think this language is useful because it hints at how we might construct our desired matrix $B$.
We are given the matrix
$$
A=\left[\begin{array}{rrrrr}
1 & 1 & 0 & 1 & 1 \\
1 & 0 & 1 & 1 & 0 \\
0 & 1 & 1 & 1 & 1
\end{array}\right]
$$
and wish to find $B$ so that $\operatorname{Null}(A)=\operatorname{Col}(B)$. Since $\operatorname{Col}(B)$ is the span of the columns of $B$, we can start by finding a basis of $\operatorname{Null}(A)$ and then inserting these basis vectors into the columns of $B$.
To find a basis of $\operatorname{Null}(A)$, note that
$$
\operatorname{rref}(A)
=
\left[\begin{array}{rrrrr}
1 & 0 & 0 & \frac{1}{2} & 0 \\
0 & 1 & 0 & \frac{1}{2} & 1 \\
0 & 0 & 1 & \frac{1}{2} & 0
\end{array}\right]
$$
This means that every solution $\vec{x}$ to $A\vec{x}=\vec{O}$ is of the form
$$
\vec{x}
=\left[\begin{array}{r}
x_{1} \\
x_{2} \\
x_{3} \\
x_{4} \\
x_{5}
\end{array}\right]
= \left[\begin{array}{r}
-\frac{1}{2} \, x_{4} \\
-\frac{1}{2} \, x_{4} - x_{5} \\
-\frac{1}{2} \, x_{4} \\
x_{4} \\
x_{5}
\end{array}\right]
= x_4\left[\begin{array}{r}
-\frac{1}{2} \\
-\frac{1}{2} \\
-\frac{1}{2} \\
1 \\
0
\end{array}\right]+x_5\left[\begin{array}{r}
0 \\
-1 \\
0 \\
0 \\
1
\end{array}\right]
$$
This gives the basis
$$
\operatorname{Null}(A)=\operatorname{Span}\left\{\left[\begin{array}{r}
-\frac{1}{2} \\
-\frac{1}{2} \\
-\frac{1}{2} \\
1 \\
0
\end{array}\right], \left[\begin{array}{r}
0 \\
-1 \\
0 \\
0 \\
1
\end{array}\right]\right\}
$$
Inserting these two vectors into the columns of a matrix $B$ gives
$$
B=\left[\begin{array}{rr}
-\frac{1}{2} & 0 \\
-\frac{1}{2} & -1 \\
-\frac{1}{2} & 0 \\
1 & 0 \\
0 & 1
\end{array}\right]
$$
How could we verify that this $B$ has our desired properties?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3122903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Question from the 2011 IMC (International Mathematics Competition) Key Stage III paper, about the evaluation of a quadratic equation
When $a=1, 2, 3, ..., 2010, 2011$, the roots of the equation $x^2-2x-a^2-a=0$ are $(a_1, b_1), (a_2, b_2), (a_3, b_3),\cdots, (a_{2010}, b_{2010}), (a_{2011}, b_{2011})$ respectively. Evaluate:
$$
\frac{1}{a_1} + \frac{1}{b_1} + \frac{1}{a_1} + \frac{1}{b_2} + \frac{1}{a_3} + \frac{1}{b_3} +\cdots + \frac{1}{a_{2010}} + \frac{1}{b_{2010}} + \frac{1}{a_{2011}} + \frac{1}{b_{2011}}
$$
I tried solving this question, with the use of the quadratic equation.
Using the quadratic equation, I concluded that $a_1=\frac{2+\sqrt{12}}{2}=1+\sqrt{3}$ and that $b_1=\frac{2-\sqrt{12}}{2}=1-\sqrt{3}$. The reason that I concluded to this is because the quadratic equation states:
$$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
So I substituted $a$ with $1$ ($1$ is multiplying $x^2$, in the original equation), $b$ with $-2$ ($x$ is getting multiplied by $-2$ in the original equation) and $c$ with $(-a^2-a)$, as they are the only ones which are not getting directly multiplied by $x$ in the original equation.
Hence, I subsequently worked out that $\frac{1}{a_1} + \frac{1}{b_1} = \frac{2}{-2}=-1$
Continuing to do the same thing I worked out $\frac{1}{a_2} + \frac{1}{b_2}= \frac{2}{-6} = -\frac{1}{3}$ and $\frac{1}{a3} + \frac{1}{b_3}= \frac{2}{-12}=-\frac{1}{6}$ and $\frac{1}{a_4} + \frac{1}{b_4} = \frac{2}{-20}=-\frac{1}{10}$ and $\frac{1}{a_5} + \frac{1}{b5} = \frac{2}{-30} =-\frac{1}{15}$.
I subsequently realised that a pattern was emerging, the denominator, each time is getting increased by the degree of $n$ at which $a$ and $b$ are (for instance $\frac{1}{a_4} + \frac{1}{b_4}= -\frac{1}{6+4}$)
Had I not been dealing with fractions, I would have solved it using arithmetic progressions, but unfortunately that is not possible.
I can think of no other way of finishing off my thoughts, nor any other way to solve this problem. Can you please help me? Can you please tell me if there is any method of finishing off my thoughts and if there isn't, can you please suggest a method of solving the problem
Thank you in advance
|
$$\sum_{k=1}^{2011}\left(\frac{1}{a_k}+\frac{1}{b_k}\right)=\sum_{k=1}^{2011}\frac{a_k+b_k}{a_kb_k}=\sum_{k=1}^{2011}\frac{2}{-k(k+1)}=$$
$$=-2\sum_{k=1}^{2011}\left(\frac{1}{k}-\frac{1}{k+1}\right)=-2\left(1-\frac{1}{2012}\right)=-\frac{2011}{1006}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
prove that the triangle is isosceles
In a $\triangle ABC$, If
$\begin{vmatrix}
1 & \;\;1\;\;&\;\; 1\;\;\\\\
\displaystyle \cot \frac{A}{2} & \displaystyle \cot \frac{B}{2} & \displaystyle \cot \frac{C}{2} \\\\
\displaystyle \tan\frac{B}{2}+\tan \frac{C}{2} &\;\;\displaystyle \tan \frac{C}{2}+\tan\frac{A}{2} & \;\;\displaystyle\tan \frac{A}{2}+\tan \frac{B}{2}\end{vmatrix}=0$
Then prove that the triangle is isosceles.
Try: Let $\displaystyle \tan \frac{A}{2}=p\;\;,\tan \frac{B}{2}=q\;\;,\tan \frac{C}{2}=r$
Using $$\tan\bigg(\frac{A}{2}+\frac{B}{2}\bigg)=\tan\bigg(\frac{\pi}{2}-C\bigg)$$
So $$\sum \tan\frac{A}{2}\tan\frac{B}{2}=1\Rightarrow pq+qr+rp=1$$
So $$\begin{vmatrix}1& 1& 1\\
\displaystyle \frac{1}{p}& \displaystyle \frac{1}{q}& \displaystyle \frac{1}{r}\\
q+r & r+p& p+q\end{vmatrix}=0$$
So $$\frac{1}{pqr}\begin{vmatrix}p& q& r\\
\displaystyle 1 & \displaystyle 1 & \displaystyle 1 \\
p(q+r) & q(r+p)& r(p+q)\end{vmatrix}=0$$
So $$\frac{1}{pqr}\begin{vmatrix}p& q& r\\
\displaystyle 1 & \displaystyle 1 & \displaystyle 1 \\
1-qr & 1-rp& 1-pq\end{vmatrix}=0$$
So $$-\begin{vmatrix}p^2& q^2& r^2\\
\displaystyle p & \displaystyle q & \displaystyle r \\
1 & 1& 1\end{vmatrix}=0$$
So we have $(p-q)(q-r)(r-p)=0.$
So either $p=q$ or $q=r$ and $r=p.$
Could some help me some short way to solve it? Thanks.
|
Hint:
Applying $C_2'=C_2-C_1,C_3=C_3-C_1$
$$\begin{vmatrix}1& 1& 1\\
\displaystyle \frac{1}{p}& \displaystyle \frac{1}{q}& \displaystyle \frac{1}{r}\\
q+r & r+p& p+q\end{vmatrix} =\begin{vmatrix}1& 1-1& 1-1\\
\displaystyle \dfrac1p & \displaystyle \dfrac1q-\dfrac1p & \displaystyle \dfrac1r-\dfrac1p\\
q+r & r+p-(q+r)& p+q-(q+r)\end{vmatrix}$$
$$=1\cdot\begin{vmatrix}
\displaystyle \displaystyle \dfrac1q-\dfrac1p & \displaystyle \dfrac1r-\dfrac1p\\r+p-(q+r)& p+q-(q+r)\end{vmatrix} $$
$$=\dfrac{(p-q)(p-r)}{pq}-\dfrac{(p-r)(p-q)}{pr}=?$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3126173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Find $\mathbf a$ for which the following equation has triple roots Find $\mathbf a \in\mathbb{R}$ for which the following equation has triple roots:
$$ x^4-5x^3+\mathbf ax^2-7x+2=0$$
|
Suppose $b$ is the triple root. Then the polynomial can be factored as $(x-b)^3(x-c)$ and we get
$$
x^4 - (3b+c)x^3 + (3b^2 + 3bc)x^2 - (b^3+3b^2c)x + b^3c
$$
so we need
\begin{cases}
3b+c=5 \\
3b^2+3bc=a\\
b^3+3b^2c=7\\
b^3c=2
\end{cases}
Multiplying the third equation by $b$ and taking into account the fourth equation we get
$$
b^4-7b+6=0
$$
The fourth equation, together with the first, becomes $5b^3-3b^4=2$
$$
3b^4-5b^3+2=0
$$
Obviously $b=1$ is a solution for both. If $b\ne1$ we get
\begin{cases}
b^3+b^2+b-6=0 \\
3b^3-2b^2-2b-2=0
\end{cases}
and, eliminating $b^3$,
$$
5b^2+5b-16=0
$$
Multiplying the top equation by $2$ and summing up,
$$
5b^3-14=0
$$
Since $5b^3+5b^2-16b=0$, we get $5b^2-16b+14=0$, that finally yields $5b-16=-16b+14$, or $b=10/7$, that doesn't satisfy $5b^3=14$.
Therefore $b=1$ and $c=2$, so $a=9$.
|
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|
How to solve $\ x^{12}\equiv 87\pmod{101}$ Can someone help me to solve the following congruent equation. I have tried to use Fermat's little theorem but failed to solve this:
$x^{12}=87(\mod 101)$
|
Numbers congruent to $87 \pmod {101}$ are $87,\; 87+101=188,\; 87+2\times 101=\color{red}{289}, ...$
and $87-101=-14,\;87-2\times 101=-115,\;87-3\times 101=\color{blue}{-216},...$
Note that therefore $17^2=289 \equiv87\pmod{101}$
and $(-6)^3=-216 \equiv 87 \pmod{101}.$
Furthermore $17\times6=102 \equiv 1 \pmod{101}.$
Therefore $17^5\equiv17^2/6^3\equiv -1 \pmod{101}.$
Thus $17^{12}=17^{10}\times17^2=(17^5)^2\times17^2\equiv87 \pmod {101},$
so $x=17$ is a solution to $x^{12}\equiv87\pmod{101}$.
|
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|
Determinant of block matrix with equal diagonals
Let $$Q=\begin{bmatrix}A&B\\-B&A\end{bmatrix}$$ where $A,B\in \mathbb{R}^{n\times n}$. Prove that $$\det(Q)=\det(A^2+B^2)$$
Since $A$ and $B$ do not commute, I cannot use Schur's formula. Also, $A$ and $B$ may not be invertible. Any comment or response is appreciated.
|
You need to require that the matrices $A$ and $B$ commute (i.e., that
$AB=BA$). Otherwise, for example, $A=\begin{pmatrix}
1 & 1\\
0 & 1
\end{pmatrix}$ and $B=\begin{pmatrix}
1 & 0\\
1 & 1
\end{pmatrix}$ yield a counterexample (since $\det\begin{pmatrix}
A & B\\
-B & A
\end{pmatrix}=\det\begin{pmatrix}
1 & 1 & 1 & 0\\
0 & 1 & 1 & 1\\
-1 & 0 & 1 & 1\\
-1 & -1 & 0 & 1
\end{pmatrix}=1\neq0=\det\left( A^2 +B^2 \right) $ in this case).
But if $A$ and $B$ do commute, then your claim holds:
Theorem 1. Let $n\in\mathbb{N}$. Let $A$ and $B$ be two $n\times
n$-matrices over a commutative ring $\mathbb{K}$ such that $AB=BA$. Then, the
block matrix $\begin{pmatrix}
A & B\\
-B & A
\end{pmatrix}$ satisfies
\begin{align}
\det\begin{pmatrix}
A & B\\
-B & A
\end{pmatrix}=\det\left( A^2 +B^2 \right) .
\end{align}
First proof of Theorem 1 (sketched). One fact about block matrices is the
following: If $A$, $B$, $C$ and $D$ are four $n\times n$-matrices over
$\mathbb{K}$ such that $AB=BA$, then
\begin{align}
\det\begin{pmatrix}
A & B\\
C & D
\end{pmatrix}=\det\left( DA-CB\right) .
\label{darij1.pf.t1.1st.1}
\tag{1}
\end{align}
(This is mentioned in
https://math.stackexchange.com/a/548487/ , and can be proven using the Schur
complement in the case when $A$ is invertible. When $A$ is not invertible,
replace $A$ by $A+xI_{n}$, where $x$ is a polynomial indeterminate. This
argument is probably all over math.stackexchange. For a specific reference, see (16) in John R. Silvester, Determinants of Block Matrices, The Mathematical
Gazette, Vol. 84, No. 501 (Nov., 2000), pp. 460--467.)
Applying \eqref{darij1.pf.t1.1st.1} to $C=-B$ and $D=A$, we find
\begin{align}
\det\begin{pmatrix}
A & B\\
-B & A
\end{pmatrix}=\det\underbrace{\left( AA-\left( -B\right) B\right) }
_{=A^2 +B^2 }=\det\left( A^2 +B^2 \right) .
\end{align}
This proves Theorem 1. $\blacksquare$
A second proof of Theorem 1 will result from proving a somewhat more
general result, which however relies on the existence of an "imaginary
unit" in our ring $\mathbb{K}$ (that is, an element $i$ such that
$i^2 = -1$):
Theorem 2. Let $n\in\mathbb{N}$. Let $A$ and $B$ be two $n\times
n$-matrices over a commutative ring $\mathbb{K}$.
Let $i \in \mathbb{K}$ be such that $i^2 = -1$.
Then, the
block matrix $\begin{pmatrix}
A & B\\
-B & A
\end{pmatrix}$ satisfies
\begin{align}
\det\begin{pmatrix}
A & B\\
-B & A
\end{pmatrix}=\det\left( A-iB\right) \det \left( A+iB\right).
\end{align}
Proof of Theorem 2. It is straightforward to see
that the block matrix $\begin{pmatrix}
I_{n} & iI_{n}\\
0_{n\times n} & I_{n}
\end{pmatrix}$
(where $0_{n\times n}$ denotes the $n\times n$ zero matrix)
is invertible (with inverse $\begin{pmatrix}
I_{n} & -iI_{n}\\
0_{n\times n} & I_{n}
\end{pmatrix} $) and satisfies
\begin{align}
\begin{pmatrix}
I_{n} & iI_{n}\\
0_{n\times n} & I_{n}
\end{pmatrix}
\begin{pmatrix}
A & B \\
-B & A
\end{pmatrix}
=
\begin{pmatrix}
A-iB & 0\\
-B & A+iB
\end{pmatrix}
\begin{pmatrix}
I_{n} & iI_{n}\\
0_{n\times n} & I_{n}
\end{pmatrix} .
\end{align}
Hence,
\begin{equation}
\begin{pmatrix}
A & B\\
-B & A
\end{pmatrix}
=
\begin{pmatrix}
I_{n} & iI_{n}\\
0_{n\times n} & I_{n}
\end{pmatrix} ^{-1}
\begin{pmatrix}
A-iB & 0\\
-B & A+iB
\end{pmatrix}
\begin{pmatrix}
I_{n} & iI_{n}\\
0_{n\times n} & I_{n}
\end{pmatrix} .
\end{equation}
Thus, the matrices $\begin{pmatrix}
A & B\\
-B & A
\end{pmatrix}$ and $\begin{pmatrix}
A-iB & 0\\
-B & A+iB
\end{pmatrix}$ are similar, and therefore have the same determinant. Hence,
\begin{align*}
\det\begin{pmatrix}
A & B\\
-B & A
\end{pmatrix} & =\det\begin{pmatrix}
A-iB & 0\\
-B & A+iB
\end{pmatrix}\\
& =\det\left( A-iB\right) \cdot\det\left( A+iB\right)
\end{align*}
(because the determinant of any block-triangular matrix whose diagonal blocks
are square matrices always equals the product of the determinants of these
diagonal blocks).
This proves Theorem 2. $\blacksquare$
Second proof of Theorem 1 (sketched).
We can find a commutative ring $\mathbb{L}$ such that $\mathbb{K}$ is a
subring of $\mathbb{L}$ and such that there exists some $i\in\mathbb{L}$
satisfying $i^2 =-1$. (For example, if $\mathbb{K}=\mathbb{R}$ or
$\mathbb{K}=\mathbb{C}$, then we can take $\mathbb{L}=\mathbb{C}$. In the
general case, we can let $\mathbb{L}$ be the quotient ring $\mathbb{K}\left[
x\right] /\left( x^2 +1\right) $, which is a free $\mathbb{K}$-module with
basis $\left( \overline{1},\overline{x}\right) $ because $x^2 +1$ is a
monic polynomial; then, $i$ should be taken to be the residue class
$\overline{x}$ of the indeterminate $x$.)
Anyway, having picked our ring $\mathbb{L}$ and element $i$, let us now regard
our matrices as matrices over $\mathbb{L}$. Now, Theorem 2 (applied to
$\mathbb{L}$ instead of $\mathbb{K}$) yields
\begin{align*}
\det\begin{pmatrix}
A & B\\
-B & A
\end{pmatrix} & =\det\left( A-iB\right) \cdot\det\left( A+iB\right) \\
& =\det\left( \underbrace{\left( A-iB\right) \left( A+iB\right)
}_{=AA+iAB-iBA-i^2 BB}\right) \\
& =\det\left( \underbrace{AA}_{=A^2 }+i\underbrace{AB}_{=BA}
-iBA-\underbrace{i^2 }_{=-1}\underbrace{BB}_{=B^2 }\right) \\
& =\det\underbrace{\left( A^2 +iBA-iBA-\left( -1\right) B^2 \right)
}_{=A^2 +B^2 }=\det\left( A^2 +B^2 \right) .
\end{align*}
This proves Theorem 1. $\blacksquare$
|
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|
If nine coins are tossed, what is the probability that the number of heads is even? If nine coins are tossed, what is the probability that the number of heads is even?
So there can either be 0 heads, 2 heads, 4 heads, 6 heads, or 8 heads.
We have $n = 9$ trials, find the probability of each $k$ for $k = 0, 2, 4, 6, 8$
$n = 9, k = 0$
$$\binom{9}{0}\bigg(\frac{1}{2}\bigg)^0\bigg(\frac{1}{2}\bigg)^{9}$$
$n = 9, k = 2$
$$\binom{9}{2}\bigg(\frac{1}{2}\bigg)^2\bigg(\frac{1}{2}\bigg)^{7}$$
$n = 9, k = 4$
$$\binom{9}{4}\bigg(\frac{1}{2}\bigg)^4\bigg(\frac{1}{2}\bigg)^{5}$$
$n = 9, k = 6$
$$\binom{9}{6}\bigg(\frac{1}{2}\bigg)^6\bigg(\frac{1}{2}\bigg)^{3}$$
$n = 9, k = 8$
$$\binom{9}{8}\bigg(\frac{1}{2}\bigg)^8\bigg(\frac{1}{2}\bigg)^{1}$$
Add all of these up:
$$=.64$$ so there's a 64% chance of probability?
|
If $h$ denotes getting a heads and $p$ denotes getting tails, let's write $\frac{1}{2}h+\frac{1}{2}p$ for the notion of a fair coin: half the time it turns up heads, and half the time tails.
If we expand out the following product while keeping track of multiplication order,
$$\left(\frac{1}{2}h+\frac{1}{2}p\right)\left(\frac{1}{2}h+\frac{1}{2}p\right)=\frac{1}{4}hh+\frac{1}{4}hp+\frac{1}{4}ph+\frac{1}{4}pp,$$
we see that the sequences $hh$, $hp$, $ph$, and $pp$ are equally likely. Forgetting about multiplication order, which we want to do because we only care about how many times heads or tails showed up, corresponds to just treating this like a polynomial:
$$=\frac{1}{4}h^2+\frac{1}{2}hp+\frac{1}{4}p^2.$$
We can keep multiplying copies of a fair coin together to find the probability that a certain number of heads or tails happened, where the coefficient in front of $h^kp^\ell$ is the probability of $k$ heads and $\ell$ tails.
Nine coins is the expansion
$$\left(\frac{1}{2}h+\frac{1}{2}p\right)^9=\sum_{k=0}^9\binom{9}{k}\left(\frac{1}{2}h\right)^k\left(\frac{1}{2}p\right)^{9-k}=\sum_{k=0}^9\frac{1}{2^9}\binom{9}{k}h^kp^{9-k}.$$
So far, all this has done is explain why you were adding up $2^{-9}\binom{9}{k}$ for $k=0,2,4,\dots,8$. Here, now, is a nice trick. If we formally set $h=1$ and $p=1$, then we get
$$1=\sum_{k=0}^9\frac{1}{2^9}\binom{9}{k},$$
and if we formally set $h=-1$ and $p=1$, then we get
$$0=\sum_{k=0}^9\frac{1}{2^9}\binom{9}{k}(-1)^k.$$
The average of these two equations is
$$\frac{1}{2}=\sum_{k=0,k\text{ even}}^9\frac{1}{2^9}\binom{9}{k},$$
since $\frac{1}{2}(1+(-1)^k)$ is $1$ or $0$ depending on whether $k$ is even or odd. Thus the probability of an even number of heads is $\frac{1}{2}$.
Notice that this did not use the fact nine coins were tossed at any point! (Other than the fact that at least one coin was tossed. In the case of tossing no coins, an even number of heads happens with probability $1$. What part of my argument goes wrong for the case of zero coins?)
|
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|
Calculate the sum $S_n = \sum\limits_{k=1}^{\infty}\left\lfloor \frac{n}{2^k} + \frac{1}{2}\right\rfloor $ I am doing tasks from Concrete Mathematics by Knuth, Graham, Patashnik for trainning, but there are a lot of really tricky sums like that:
Calculate sum $$S_n = \sum_{k=1}^{\infty} \left\lfloor \frac{n}{2^k} + \frac{1}{2} \right\rfloor $$
My idea
I had the idea to check when $$\frac{n}{2^k} < \frac{1}{2}$$
because then $$ \forall_{k_0 \le k} \left\lfloor \frac{n}{2^k} + \frac{1}{2} \right\rfloor=0$$
It should be $$ k_0 = \log_2(2n) $$ but I don't know how it helps me with this task (because I need not only "stop moment" but also sum of considered elements
Book idea
Let $$S_n = \sum_{k=1}^{\infty} \left\lfloor \frac{n}{2^k} + \frac{1}{2} \right\rfloor $$
then $$ S_n-S_{n-1} = 1$$
and then solve this recursion. But I write $S_n - S_{n-1}$ and I don't see how it can be $1$ , especially that is an infinite sum.
|
Let
$$2^{b}\le n<2^{b+1}$$
For all $1\le k\le b$,
$$\left\lfloor\frac{n}{2^k}+\frac12\right\rfloor=\frac{2^b}{2^k}+\left\lfloor\frac{n-2^b}{2^k}+\frac12\right\rfloor.$$
For $k=b+1$,
$$\left\lfloor\frac{n}{2^k}+\frac12\right\rfloor=1.$$
And for $k>b+1$,
$$\left\lfloor\frac{n}{2^k}+\frac12\right\rfloor=0.$$
This allows us to write, by summing,
$$S_n=2^{b-1}+2^{b-2}+\cdots 2^0+1+S_{n-2^b}=2^b+S_{n-2^b},$$
which shows that $n$ and $S_n$ share the same binary representation.
E.g., inductively,
$$S_{123}=64+S_{59}=64+32+S_{27}=64+32+16+S_{11}\\\cdots\\=64+32+16+8+2+1.$$
|
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|
Prove that $1+\frac{1}{8n}\sqrt{\pi_1n}<\frac{2.4.6\ldots(2n-2)(2n)}{1.3.5\ldots(2n-3)(2n-1)}<(1+\frac{1}{8n}+\frac{1}{128n^{2}})\sqrt{\pi_2n}$ Within the confines of the O-level syllabus, prove that:
$$1+\dfrac{1}{8n}\sqrt{\pi_1n}<\dfrac{2.4.6\ldots(2n-2)(2n)}{1.3.5\ldots(2n-3)(2n-1)}<\left(1+\dfrac{1}{8n}+\dfrac{1}{128n^{2}}\right)\sqrt{\pi_2n}$$
for all positive integer $n$, where $\pi_1=3.141$ and $\pi_2=3.142$
Using A-level mathematics, show that the result remains true with $\pi_1=\pi_2=\pi$ where $\pi$ is the familiar constant associated with a circle.
This is a question found in Hammersley's "On the enfeeblement of mathematical skills by "Modern Mathematics" and by similar soft intellectual trash in schools and universities"
When I first look at this problem, I wonder if I can prove it via Wallis'product?
$$\prod_{i=1}^{\infty}\dfrac{2n}{2n-1}.\dfrac{2n}{2n+1}=\dfrac{\pi}{2}$$
|
$$
\dfrac{2.4.6\ldots(2n-2)(2n)}{1.3.5\ldots(2n-3)(2n-1)} = \dfrac{2n!}{(2n-1)!} = 2n \\ \therefore \;\; 1+\dfrac{1}{8n}\sqrt{\pi_1n}<\dfrac{2.4.6\ldots(2n-2)(2n)}{1.3.5\ldots(2n-3)(2n-1)}<\left(1+\dfrac{1}{8n}+\dfrac{1}{128n^{2}}\right)\sqrt{\pi_2n}
\\ = 1+\dfrac{1}{8n}\sqrt{\pi_1n}<2n<\left(1+\dfrac{1}{8n}+\dfrac{1}{128n^{2}}\right)\sqrt{\pi_2n}
$$
Considering the case of $$\pi_1 = \pi_2 = \pi$$ since this generality can be used for the other case too.
Splitting the inequalities into two,
\begin{align}
I_1:
1+\dfrac{1}{8n}\sqrt{\pi n}<2n \\
1+\dfrac{1}{8\sqrt n}\sqrt{\pi}<2n \\
\dfrac{1}{8\sqrt n}\sqrt{\pi}<2n - 1 \\
\dfrac{1}{8}\sqrt{\pi}<(2n - 1)\cdot\sqrt n \tag{1}\label{1}
\end{align}
Since n is a positive integer, minimum value of n is 1, which when substituted in \eqref{1} gives $$ 0.2215 < 1 $$ which is true.
\begin{align}
I_2:
2n<\left(1+\dfrac{1}{8n}+\dfrac{1}{128n^{2}}\right)\sqrt{\pi n}\\
\dfrac{2n}{\sqrt{\pi n}}<\left(1+\dfrac{1}{8n}+\dfrac{1}{128n^{2}}\right) \\
\dfrac{2n}{\sqrt{\pi n}} - 1<\left(\dfrac{1}{8n}+\dfrac{1}{128n^{2}}\right) \\
\dfrac{2n}{\sqrt{\pi n}} - 1<\dfrac{1}{8n}\left(1+\dfrac{1}{16n}\right)\\
\dfrac{2n}{\sqrt{\pi n}} - 1<\dfrac{1}{8n}\left(\dfrac{16n+1}{16n}\right)\\
(128n^{2})\cdot\dfrac{2\sqrt n}{\sqrt{\pi}} - 1< 16n+1\\
\dfrac{256n^{\frac{5}{2}}}{\sqrt \pi} - 128n^{2}<16n+1\\
\dfrac{256}{\sqrt \pi} < ( 128n^2 + 16n + 1 )\cdot n^{\frac{2}{5}} \tag{2}\label{2}
\end{align}
Since n is a positive integer, minimum value of n is 1, which when substituted in \eqref{2} gives $$ 144.43 < 145 $$ which is true.
|
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|
Finding the least possible value of perimeter of $\triangle ABC$ with given ranges of angle
In $\triangle ABC$,$\angle A >2\angle B$ and $\angle C > 90^\circ$. If the length of all side of triangle $\triangle ABC$ are positive integers, then what is the least possible value of perimeter of $\triangle ABC$?
However, I can't think even of the length of the sides related with the possible values for all angle $\angle A, \angle B$ and $\angle C$. How can I construct the triangle and then get all the side having a length belonging to the positive integers? The problem was very weird for me and all of my effort can be hardly shown or described. And how can I get the minimum possible perimeter?
Thanks in advance.
|
So, how about a really simple approach: run through positive integer triples $b<a<c$. If they're sides of a triangle, the angles will be in order $B<A<C$.
*
*$(1,2,3)$: Not a triangle. In fact, we can't have a triangle if the small side is $1$.
*$(2,3,4)$: Since $4^2=16>13=2^2+3^2$, it's obtuse. Now, the smaller angles: $\cos B = \frac{4^2+3^2-2^2}{2\cdot 4\cdot 3}=\frac{7}{8}$ and $\cos A = \frac{4^2+2^2-3^2}{2\cdot 4\cdot 2}=\frac{11}{16}$. By the double-angle formula, $\cos 2B = 2\cdot\frac{7^2}{8^2}-1=\frac{98-64}{64}=\frac{17}{32}<\frac{11}{16}$. Not this one.
*$(2,3,5)$: Not a triangle.
*$(2,4,5)$: $25>20=16+4$, so it's obtuse. $\sin A=2\sin B>\sin 2B$, so it satisfies the second condition as well.
There it is - the second smallest triangle with three different integer sides. Sometimes, the simple approach pays off.
|
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|
Solve a system equation in $\mathbb{R}$ - $\sqrt{x+y}+\sqrt{x+3}=\frac{1}{x}\left(y-3\right)$ how to solve a system equation with radical
$$\sqrt{x+y}+\sqrt{x+3}=\frac{1}{x}\left(y-3\right)$$
And $$\sqrt{x+y}+\sqrt{x}=x+3$$
This system has $1$ root is $x=1;y=8$,but i have no idea which is more clearly to solve it. I tried substituting and squaring to find the factor but failed.
|
From the second equation we get
$$\sqrt{x+y}=x+3-\sqrt{x}$$ y squaring this equation we obtain
$$y=(x+3-\sqrt{x})^2-x$$ plugging this in the first equation we get
$$\sqrt{(x+3-\sqrt{x})^2}+\sqrt{x+3}=\frac{1}{x}((x+3-\sqrt{x})^2-x-3)$$
Can you proceed?
Hint: $$x=1,y=8$$
Ok, we can also eliminate the square root:
$$\sqrt{x+y}=x+3-\sqrt{x}$$ so
$$x+3-\sqrt{x}+\sqrt{x+3}=\frac{1}{x}(y-3)$$
this is only a littlebit better.
|
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|
Number of attempts to see all faces $1, 2, 3, ..., 6$ when rolling a dice? Recently I have been thinking about the following random experiment: we repeatedly roll a dice until we see all the faces $1, 2, 3, 4, 5, 6$ of the dice at least once.
Let $X$ = number of attempts necessary to see all the faces.
Obviously $X(\Omega) = \{6, 7, 8, ...\}$
Can we describe precisely the law (and maybe also the expected value) of $X$?
(It did not look as simple as it seems, thus this question).
Note: linked to Expected time to roll all 1 through 6 on a die but in my question here, the law of $X$ is also discussed ($P(X=k)$ for $k \geq 6$).
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Set up the problem as a Markov chain with states being the number of faces that have shown up to present, thus from 1 to 6. The transition matrix (with columns adding to 1) is $$P=\frac{1}{6}\begin{pmatrix}1&0&0&0&0&0\\5&2&0&0&0&0\\0&4&3&0&0&0\\0&0&3&4&0&0\\0&0&0&2&5&0\\0&0&0&0&1&6\end{pmatrix}$$ The 'absorbing' state is 6, so let $P=\begin{pmatrix}Q&0\\x&1\end{pmatrix}$ where $Q$ is a $5\times5$ sub-matrix of $P$ that represents the chain for the first five states. The required expectation is given by $$(I+Q+Q^2+\cdots)\mathbf{i}=(I-Q)^{-1}\mathbf{i}=6\begin{pmatrix} \frac{1}{5} & 0 & 0 & 0 & 0 \\ \frac{1}{4} & \frac{1}{4} & 0 & 0 & 0 \\
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} & 0 & 0 \\
\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & 0 \\
1 & 1 & 1 & 1 &1 \end{pmatrix}\begin{pmatrix}1\\0\\0\\0\\0\end{pmatrix}=\begin{pmatrix}\frac{6}{5}\\ \frac{3}{2}\\ 2\\ 3\\ 6\end{pmatrix}$$ This vector signifies the expected number of steps each state has been visited before reaching the sixth state. Hence their sum is what is needed: $137/10$. Since this model assumes it starts with state 1 already, you need to add 1 for the first throw. Hence the expected number of times before all six faces appear is $$147/10.$$
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Prime numbers $pFind the prime numbers $p<q<r$ such that $r^2-q^2-p^2$ is a perfect square.
I think the only solution is (2,3,7) but i cannot prove it.
The equation would be $r^2-q^2-p^2=k^2$ equivalently
$q^2+p^2+k^2=r^2$ which is somehow a classical diophantine equation (of the pythagorean quadruples) which have a parametrization but how do i use it? The problem is from a magazine at 8 th grade and i don't think the solution is supposed to use the general solution.
Actually the general solution of $q^2+p^2+k^2=r^2$ is given by:
$ q=a^2+b^2-c^2-d^2$
$p=2(mq+np)$
$k=2(nq-mp)$
$r=a^2+b^2+c^2+d^2$
IT would follow that one of the numbers q or p is even and prime, so it is 2.
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Hint
All prime numbers $p\geq5$ satisfy $p\equiv \pm 1\mod 6$
Therefore, if $p,q,r\geq5$
$$r^2-q^2-p^2\equiv (\pm1)^2-(\pm1)^2-(\pm1)^2\equiv 1-1-1\equiv5\mod 6$$
Observe now, that there's no perfect square with the residue $5$ modulo 6.
This follows from the simple fact that (taking the equations modulo $6$)
$$\begin{array}l 1^2\equiv\color{red}1\\ 2^2\equiv \color{red}4\\ 3^2\equiv9\equiv\color{red}3\\4^2\equiv16\equiv \color{red}4\\5^2\equiv25\equiv \color{red}1 \\6^2\equiv\color{red}0\end{array}$$
Can you end it now?
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How do I find ordered pair, given slope of the tangent line? The function is $f(x) = x^3 + 9x^2 + 36x + 10$ and the slope given is $9$.
I found the derivative and set it equal to $9$, but I ended up with $x = (-9,-33)$ and the answer is $(-3,-44)$.
I've asked two Math majors and neither knows how to find it.
Where did I go wrong and how can I answer the next one correctly?
Work:
\begin{align*}
& f(x) = x^3 + 9x^2 + 36x + 10 \Rightarrow f^{\prime}(x) = 3x^2 + 18x + 36 \Rightarrow 3x^2 + 18x + 36 = 9 \Rightarrow\\\\
& 3x^2 + 18x = -27 \Rightarrow 3x ( x + 6 ) = -27 \Rightarrow 3x = -27 x + 6 = -27 \Rightarrow
x = -3 x = -33
\end{align*}
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From your second line at start you get by simplification for given slope $9$
$$ (x+3)^2 =0$$
Plug $x=-3$ into $y=f(x)$ expression and you directly get $ y=-44$ so that this point ( called inflection point ) is described as
$$ y= f(x)= x^3+9x^2+36x+10 $$
$$\frac{dy}{dx}= 3x^2+18x +36 $$
$$ \frac{d^2y}{dx^2}=6x+18 $$
$$ \frac{d^3y}{dx^3}=6 $$
$$ x=-3,y=-44,\frac{dy}{dx}=9, \frac{d^2y}{dx^2}=0 $$
all of which is seen in the WA plot
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Is $n^2+3n+6$ divisible by 25, where $n$ is a integer? If we want $n^2+3n+6$ to be divisible by $25$, it firstly has to be divisible by $5$.
So, let's take a look at couple of cases:
$n=5k: 25k^2+15k+6$. The remainder is $6$, so it's not divisible by $25$.
$n=5k+1: 25k^2+25k+10$. The remainder is $10$, so it is not divisible by $25$.
And so on for $n=5k+2, n=5k+3, n=5k+4.$
*
*Is this a good way to do this?
The other way that came to my mind would be following:
Let's say that $n^2+3n+6$ is divisible by $25$: $$n^2+3n+6=25k$$
$$n^2+3n+(6-25k)=0$$
If we solve this equation, we get $$n_{1,2}=\frac{-3 \pm \sqrt{5} \sqrt{20k-3}}{2}.$$ But $\sqrt{5}$ is a irrational number, so is $n$ irrational number too. This is contradiction, so it's not divisible by $25$.
*Is this a good way to solve this problem?
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Hint: We have
$$
n^2+3n+6=(n+4)^2 \bmod 5
$$
Now look at Bill's answer at this duplicate:
$n^2 + 3n +5$ is not divisible by $121$
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Last digit of large powers Define the sequence $a_1, a_2,...$ by $a_1=7$ and where $a_{n+1} = a_n^7$.
1) Find the last digit of $a_{1876}$, ie $7(^7)(^7)(^7)(^7)(^...)$
Now, I understand that the last digit of powers of 7 repeat every 4 powers (i think), so the answer has something to do with integers modulo 4. While I was working on this, I found than $a_n$ can be written as $a_n = 7^{(7^{n-1})}$, not sure if this is correct or if it helps.
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The order of $7$ mod $100$ is $4$, i.e. $7^4 = 2401 \equiv 1 \mod 100$, so $7^{4i+j} \equiv 7^j (\bmod 100)$ for any integers $i$ and $j$.
Since $4$ divides $100$, this is also true mod $4$.
Now $a_n = 7^{7^{n-1}}$, and $2008$ is divisible by $4$ so $7^{2008} \equiv 7^0 = 1 (\bmod 4)$, and $a_{2009} \equiv 7^{1} = 7 (\bmod 100)$.
As for $b_{2009}$, we have $b_1 = 7 \equiv 3 (\bmod 4)$, and since $3^3 = 27 \equiv 3 (\bmod 4)$ we get $b_n \equiv 3 (\bmod 4)$ for all $n \ge 1$,
and so for $n \ge 2$, $b_n \equiv 7^{b_{n-1}} \equiv 7^3 \equiv 43 (\bmod 100)$.
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If rank of a given matrix of order $3 \times 4$ is $2$ then the value of $b$ is Q) Suppose the rank of the matrix
$\begin{pmatrix} 1 &1 &2 &2 \\ 1&1 &1 &3 \\ a&b &b &1 \end{pmatrix}$
is $2$ for some real numbers $a$ and $b$. Then $b$ equals
$(A)$ $1\;\;\;$ $(B)$ $3\;\;\;$ $(C)$ $1/2\;\;\;$ $(D)$ $1/3\;\;\;$
My Approach :- Since rank is $2$ , So, the determinant of all the submatrices of order $3 \times 3$ must be zero.
So, $\begin{vmatrix} 1&1 &2 \\ 1&1 &1 \\ a&b &b \end{vmatrix}$ = $0$. After solving it, I am getting $a=b$
and $\begin{vmatrix} 1&2 &2 \\ 1&1 &3 \\ b&b &1 \end{vmatrix}$ = $0$. After Solving it , $b=\frac{1}{3}$
Now, Rank of a matrix is also defined as no. of non-zero rows in row echelon form of that matrix. So, If I convert it into Row Echelon form then
$\begin{bmatrix} 1 &1 &2 &2 \\ 1&1 &1 &3 \\ a&b &b &1 \end{bmatrix}$
Applying $R_{2} \leftarrow R_{2}-R_{1}$
$\begin{bmatrix} 1 &1 &2 &2 \\ 0&0 &-1 &1 \\ a& b &b &1 \end{bmatrix}$
Now, Applying $R_{3}\leftarrow R_{3}-aR_{1}$
$\begin{bmatrix} 1 &1 &2 &2 \\ 0&0 &-1 &1 \\ 0&(b-a) &(b-2a) &(1-2a) \end{bmatrix}$
Now, to make rank of this matrix = $2$ means I should have 2 non-zero rows or all the elements of the last row must be zero simultaneously.
So, $(b-a) =0$ and $(b-2a) =0$ and $(1-2a) =0$
So, $b=a$ and $b=2a$ and $a=\frac{1}{2}$
Now, My doubt is how $b=a$ and $b=2a$ is possible here simultaneously and why this method is giving wrong result. I must be doing some mistake here but I am not getting what mistake I am doing. Please help.
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Consider the submatrix $M$ obtained from the first two rows
$$
M = \left[\begin{array}{rrrr}
1 & 1 & 2 & 2 \\
1 & 1 & 1 & 3
\end{array}\right]
$$
The two vectors
\begin{align*}
\vec{n}_1 &= \left\langle1,\,3,\,-1,\,-1\right\rangle & \vec{n}_2 &= \left\langle0,\,4,\,-1,\,-1\right\rangle
\end{align*}
form a basis of $\operatorname{Null}(M)$. Since we want our $3\times 4$ matrix $A$ to have the same rank as $M$, and hence the same nullity as $M$, we must have $A\vec{n}_1=A\vec{n}_2=\vec{O}$. This gives
\begin{align*}
\overset{A\vec{n}_1}{\left\langle0,\,0,\,a + 2 \, b - 1\right\rangle} &= \left\langle0,\,0,\,0\right\rangle & \overset{A\vec{n}_2}{\left\langle0,\,0,\,3 \, b - 1\right\rangle} &= \left\langle0,\,0,\,0\right\rangle
\end{align*}
These two equations give $a=b=\frac{1}{3}$.
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Weierstrass Zeta function Series Let $\Gamma := \mathbb{Z} + \mathbb{Z} \tau$ be a lattice.
Then \begin{align*} \zeta_\Gamma(z+1) - \zeta_\Gamma(z) \end{align*} is a function independent from $z$. Why is that so?
The function is defined as follows
\begin{align}
\zeta_\Gamma(z) = \frac{1}{z} + \sum_{(m,n)\neq (0,0)} \frac{1}{z - (m\tau+n)} + \frac{1}{m\tau+n} + \frac{z}{(m\tau+n)^2}
\end{align}
I see how the middle term of the sum cancels in this substraction but i dont really see how the $z$ vanish, especially the $\frac{1}{z+1} - \frac{1}{z}$ in front of the sum.
\begin{align*}
\zeta_\Gamma(z+1)- \zeta_\Gamma(z+1) &= \frac{1}{z+1} + \sum_{(m,n+1)\neq (0,0)} \frac{1}{(z+1) - (m\tau+n)} + \frac{1}{m\tau+n} + \frac{(z+1)}{(m\tau+n)^2} \\ &-\left( \frac{1}{z} + \sum_{(m,n)\neq (0,0)} \frac{1}{z - (m\tau+n)} + \frac{1}{m\tau+n} + \frac{z}{(m\tau+n)^2} \right) \\ &= \left(\frac{1}{z+1} - \frac{1}{z}\right) \\&+ \left( \sum_{(m,n+1)\neq (0,0)}\frac{1}{(z+1) - (m\tau+n)} + \frac{(z+1)}{(m\tau+n)^2} - \sum_{(m,n)\neq 0}\frac{1}{z - (m\tau+n)} + \frac{z}{(m\tau+n)^2} \right)
\end{align*}
Do we need some index shifting here? I am totally clueless so any help is appreciated.
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Its derivative is $-\wp_\Gamma(z+1)+\wp_\Gamma(z)=0$, as $1\in\Gamma$
and $\Gamma$ is the
period lattice of $\wp_\Gamma$.
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Proving an inequality without an integral: $\frac {1}{x+1}\leq \ln (1+x)- \ln (x) \leq \frac {1}{x}$ I would like to prove the following inequality without integration; could you help?
$$\frac {1}{x+1}\leq \ln (1+x)- \ln (x) \leq \frac {1}{x}, \quad x > 0. $$
I can however differentiate this.
Thanks in advance.
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Lemma: For all real $x, e^x \ge 1 + x$.
Proof of lemma: Consider $f(x)=e^x-x-1.$ Its first derivative ($e^x-1$) is $0$ when and only when $x=0$, and its second derivative ($e^x$) is positive for all real $x, $ so $e^x-x-1$ has a global minimum when $x=0\; (f(0)=0$) and $e^x-x-1\ge 0$ for all real $x.$
From the lemma, $\exp\left({\frac 1x}\right)\ge 1+ \frac 1 x,$
which implies (take logarithm of both sides, noting that preserves order) $ \frac 1 x \ge \ln\left(1 + \frac 1 x\right)=\ln\left(\frac{x+1}x\right)=\ln(x+1)-\ln(x),$
which is one part of the desired inequality.
Also from the lemma, $\exp\left(\frac {-1}{1+x}\right)\ge 1-\frac 1 {1+x}=\frac x {1+x}$ implies (taking reciprocal of both sides, reversing the order) $\exp\left(\frac {1}{1+x}\right)\le \frac {1+x} {x}$ so (again, taking logarithm of both sides) $ \frac 1 {1+x} \le \ln \frac{1+x}{x}=\ln (1+x) - \ln(x),$ the other part of the desired inequality.
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.