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Dimension of the spanned subspace Let $a,b\in \mathbb{R}.$ Determine the dimension of the subspace of $\mathbb{R}^5$ generated by the following vectors.
$$\begin{pmatrix} 1\\2\\1\\2\\0 \end{pmatrix}, \begin{pmatrix} 2\\5\\1\\3\\1 \end{pmatrix}, \begin{pmatrix} 5\\8\\3\\6\\2 \end{pmatrix}, \begin{pmatrix} 2\\1\\1\\1\\1 \end{pmatrix}, \begin{pmatrix} 0\\2a-b+2\\-a+b-1\\-a+2b-1\\3a-2b \end{pmatrix}.$$
My stategy is to construct the matrix
\begin{pmatrix}
1 & 2 & 5 &2 &0\\
2 & 5 & 8 &1 &2a-b+2\\
1& 1 & 3 &1 &-a+b-1\\
2 & 3 & 6 &1 &-a+2b-1\\
0 &1 &2 &1 &3a-2b
\end{pmatrix}
and compute its rank by row reduction. Since the rank of the matrix above should be equal to the dimension of its column space, then in theory this computation should give the dimension of the subspace being asked. Is my strategy correct?
|
Yes, it is, that is the quickest/most common method to compute the dimension of the subspace generated by those vectors :)
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Prove that numerator of fraction $\sum \limits_{k=1}^{p-2}\frac{k}{(k+1)^2}$ is divisible by $p$ for all prime $p>3$ Let $p>3$ be a prime number. Let $\sum \limits_{k=1}^{p-2}\frac{k}{(k+1)^2}=\frac{r}{s}$, where $r,s$ are some integers numbers. Prove that $r$ is divisible by $p$.
My work. Let $\sum \limits_{k=1}^{p-1}\frac{1}{k}=\frac{a}{(p-1)!}$. Then $\sum \limits_{k=1}^{p-1}\frac{1}{k}=\sum \limits_{k=1}^{\frac{p-1}{2}}\left(\frac{1}{k}+\frac{1}{p-k} \right)=\sum \limits_{k=1}^{\frac{p-1}{2}}\frac{p}{k(p-k)}$. Then $a$ is divisible by $p$. Let $\sum \limits_{k=1}^{p-1}\frac{1}{k^2}=\frac{b}{((p-1)!)^2}$. Then $\frac{r}{s}=\sum \limits_{k=1}^{p-2}\frac{k}{(k+1)^2}=\sum \limits_{k=1}^{p-2} \left( \frac{1}{k+1}-\frac{1}{(k+1)^2}\right)=\sum \limits_{k=1}^{p-1}\frac{1}{k}-\sum \limits_{k=1}^{p-1}\frac{1}{k^2}=\frac{a}{(p-1)!}-\frac{b}{((p-1)!)^2}=\frac{a(p-1)!-b}{((p-1)!)^2}$. Since $a$ is divisible by $p$, we need to prove that $b$ is divisible by $p$.
|
Note that $$\sum_{k=1}^{p-2} \frac{k}{(k+1)^2} = \sum_{i=2}^{p-1} \frac{1}{i} - \frac{1}{i^2} = \sum_{i=1}^{p-1} \frac{1}{i} - \frac{ 1}{i^2}. $$
We work mod $p$.
*
*$ \{ \frac{1}{i} | i = 1\text{ to }p-1\} = \{ j | j = 1\text{ to }p-1\}$.
Hence $ \sum \frac{1}{i} = \sum j = \frac{p(p-1)}{2} \equiv 0\pmod{p}$ for $ p > 2$.
*$ \{ \frac{1}{i^2} | i = 1\text{ to }p-1\} = \{ j^2 | j =1\text{ to }p-1\}$.
Hence $\sum \frac{1}{i^2} = \sum j^2 = \frac{ p(p-1)(2p-1)}{6} \equiv 0 \pmod{p}$ for $p>3$.
Hence, $\sum_{k=1}^{p-2} \frac{k}{(k+1)^2} \equiv 0 \pmod{p}$, which means that the numerator is a multiple of $p$ for $p > 3$.
And yes, the equations generalize to higher powers. You just have to be careful with the denominator that appears in $\sum j^n$. By the method of differences, we know the denominator is factor of $(n+1)!$, hence the equation holds for $p > n+1$ (and possibly some other cases).
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If $x,y,z>0.$Prove: $(x+y+z) \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right) \geq9\sqrt[]\frac{x^2+y^2+z^2}{xy+yz+zx}$ If $x,y,z>0.$Prove:
$$(x+y+z) \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right)\geq9\sqrt[]\frac{x^2+y^2+z^2}{xy+yz+zx}$$
I was not able to solve this problem instead I could solve similar inequality when we have two variable.I assumed $y=tx$ and uesd derivative.
Can this be generalized as:
If ${a_i>0}\quad(i=1,2,...,n)$
$$\sum_{i=1}^n a_{i} \sum_{i=1}^n \frac{1}{a_{i}}\geq n^2\sqrt[]\frac{\sum_{i=1}^n a^2_{i} }{\sum_{i=1}^n a_{i}a_{i+1} }$$
$a_{n+1}=a_{1}$
Question from Jalil Hajimir
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Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, we need to prove that $f(w^3)\geq0,$ where
$$f(w^3)=\frac{uv^2}{w^3}-\sqrt{\frac{3u^2-2v^2}{v^2}}.$$
We see that $f$ decreases, which says that it's enough to prove our inequality for a maximal value of $w^3$, which by $uvw$ ( https://artofproblemsolving.com/community/c6h278791 )
happens for equality case of two variables.
Since our inequality is homogeneous, we can assume $y=z=1,$ which gives
$$(x+2)^2(2x+1)^3\geq81x^2(x^2+2)$$ or
$$(x-1)^2(8x^3-21x^2+36x+4)\geq0,$$ which is obvious.
|
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|
Roots of equation form infinite sequence
The sequence $a_n$ has the property that $a_n$ and $a_{n+1}$ are the roots of the equation
$$x^2-c_nx+\frac{1}{3^n}=0$$ and $a_1=2$. What is $\sum_{n=1}^{\infty}c_n?$
By Vieta's,
$a_{n+1}=\frac{1}{3^na_n}$ and $c_n=a_n+a_{n+1}$. Additionally listing the first numbers in $a_n$
$$2,\frac{1}{6},\frac{2}{3}, \frac{1}{18}, \frac{2}{9},\cdots$$
doesn't reveal anything (even though $c_n$ seems like a geometric sequence). Thanks!
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Instead of simplifying all the terms, try to find a relation between every term in terms of $a_1$.
Now, your sequence of $a$ shall look like this $a_1 ,\frac{1}{3a_1},\frac{a_1}{3},\frac{1}{9a_1}\cdots$
Notice that, every even terms of the sequence are in a $g.p$ with common ratio $\frac{1}{3}$
And every odd terms are also in $g.p$ with common ratio $\frac{1}{3}$ again !
We need to find $\Sigma_{n=1}^{\infty}c_{n} = c_1 + c_2 + c_3 \cdots = (a_1+a_2)+(a_2+a_3) + (a_3+a_4) \cdots$
Notice that, except the first term i.e $a_1$ every other term occurs 2 times, so our sum is simply,
$a_1 + 2(a_3+a_5+a_7+\cdots)+2(a_2+a_4+a_6+\cdots)$
We know that the 2 sequences in the above expression have a common ratio $\lt 1$, So their infinite sum will converge,
and hence our answer is $a_1 + \frac{2a_3}{1-\frac{1}{3}}+\frac{2a_2}{1-\frac{1}{3}} = 2 + \frac{2\cdot\frac{2}{3}}{1-\frac{1}{3}} + \frac{2\cdot\frac{1}{6}}{1-\frac{1}{3}} = 2 + 2 + \frac{1}{2} = \frac{9}{2}$
So our answer is $\frac{9}{2}$. Hope this helps !
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|
Finding the Maclaurin series for $f(x)=x\ln(x+1)$
Find the Maclaurin series for the function
$$f(x)=x\ln(x+1)$$
So finding the derivatives is the first step. How many derivatives I need to find is explicitly said so I'll just go till the $4^{th}$ derivative.
$$\begin{align}
f'(x)&=\frac{x}{x+1}+\ln(x+1) & f'(0) &=0 \\
f''(x)&=\frac{1}{(x+1)^2}+\frac{1}{x+1} & f''(0)&=2 \\
f^{(3)}(x)&=\frac{-x-3}{(x+1)^3} & f^{(3)}(0)&=0 \\
f^{(4)}(x)&=\frac{2x+8}{(x+1)^4} & f^{(4)}(0)&=8
\end{align}$$
So now plugging in for the Maclaurin form
$$P_n(x)=0+x^2-\frac{1}{2}x^3+\frac{1}{3}x^4+....$$
Is this correct?
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Note that $$\ln(1+x)=\sum_{k=1}^{\infty} \frac{(-1)^{k+1}x ^k}{k}$$
So the Mclaurin series for $x\ln(1+x)$
is $$ x\ln(1+x)= \sum_{k=1}^{n} \frac{(-1)^{k+1} x^{k+1}}{k}=x^2-\frac{x^3}{2}+\frac{x^4}{3}-\frac{x^5}{4}+....$$
|
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Show that there are infinitely many natural numbers such that $a^2+b^2=c^2+3 .$ This question was asked in the Crux Mathematicorum , October edition , Pg -$5$ , which can found be Here.
The question states that :
Both $4$ and $52$ can be expressed as the sum of two squares as well as
exceeding another square by $3$ :
$$4 = 0^2+2^2 \quad\,, \, 4-3=1^2 $$
$$52 = 4^2+6^2 \quad\,, \, 52-3=7^2 $$
Show that there are an infinite number of such numbers that have these two
characteristics.
My attempt :
I Found $4,52$ and $292$ to have this characteristics. An interesting feature I noticed was $$4 = \color{red}{0^2}+2^2 \quad\,, \, 4-3=\color{green}{1^2}$$
$$52 =4^2 + \color{red}{6^2} \quad\,, \, 52-3=\color{green}{7^2}$$
$$292 = 6^2 + \color{red}{16^2} \quad\,, \, 292-3=\color{green}{17^2}$$
If a number $y$ satisfies this property , then it might be written as :
$$y= a^2+b^2 = c^2+3$$
And based on the above examples , I conjectured that :
$$(k)^2 + b ^2 = (k+1)^2 + 3$$
where $a=k$ and $c=k+1$.
This expression on simplifying gives us :
$b^2 = 2(k+2)$ . For R.H.S to be a perfect square , $(k+2)$ must be of the form $2x^2$ , which on solving , gives $k = 2x^2-2$ and $b$ comes out to be $2x$.
So a solution is given by $\color{blue}{(a,b,c) = (2x^2-2,2x,2x^2-1)}$
And our number becomes $y = 4(x^4 - x^2 + 1)$ for all $x\in\mathbb{N}$ and hence there are infinite numbers with this characteristics.
Although ' maybe ' this proves the question , it is not quite rigorous method to do this . Also , it does not provide all the possible numbers as we have only taken the special case when $a=k \,\,, c= k+1$.
What is the better way to solve this problem and the general formula for the number?
Bonus Question : Prove that the highest power of two dividing the number is $2.$ Or more generally ,show that :
$$2^c\nmid y \quad \quad \text { For any } c \ge 3.$$
Verified it for all $y\le1.5\times10^5$ and it seems quite likely to be true . For my case , where $y = 4(x^4 - x^2 +1)$ , this is obvious as $x^4-x^2+1$ is always odd and hence the number is only divisible by $4$. But what about the other numbers ?
Edit :
The first $5$ numbers with this characteristics are :
$$4 = 0^2+2^2 \quad\,, \, 4-3={1^2}$$
$$52 =4^2 + {6^2} \quad\,, \, 52-3=7^2$$
$$292 = 6^2 + {16^2} \quad\,, \, 292-3={17^2}$$
$$628 = 12^2 + {22^2} \quad\,, \, 628-3={25^2}$$
$$964 = 8^2 + {30^2} \quad\,, \, 964-3={31^2}$$
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There are infinitely many even numbers. Let us assume that $b$ is even, as doing so does not limit the number of possible solutions. We can rearrange the original equation to $$c^2-a^2=b^2-3$$ where $b^2-3$ must be odd.
It is well know that the sum of the first $n$ odd numbers equals $n^2$. Consequently, every odd number is the difference of two consecutive squares. It may also be the difference of other pairs of squares, but it is at least the difference of two consecutive squares.
Therefore, every odd number $b^2-3$ can be represented in at least one way as the difference of two squares.
For example, choose $b=6$ (an even number), so $b^2-3=33$.
$33$ is the $17$th odd number, and the sum of the first $17$ odd numbers is $17^2=289$. The $17$th odd number succeeds the $16$th odd number, and the sum of the first $16$ odd numbers is $16^2=256$. Plainly, $289-256=33=b^2-3$. It is also the case that $49-16=33$ (which is one of OP's original instances), but that is not necessary to establish that the number of solutions is infinite.
Added by edit: In my answer as originally posted, I addressed only the question as to whether there are infinite solutions. That garnered a query by OP as to whether all solutions can be found. The answer to that is yes, but the formulation is somewhat involved.
Based on the fact that $\sum_{i=1}^k(2i-1)=k^2$ we can show that any odd number is the difference of two squares: $n=r^2-s^2$
The formula $(\sum_{i=1}^{\frac{n+1}{2}}(2i-1))-(\sum_{i=1}^{\frac{n-1}{2}}(2i-1))$ represents the difference of two consecutive squares. Each sum contains the same terms, which by subtraction cancel, except the first sum contains an extra term corresponding to $2(\frac{n+1}{2})-1$ which simply equals $n$. In order that the indices on the sums be integers, it is required that $n$ be odd, but beyond that, any odd number affords a solution. This more fully explains the conclusion presented in my first answer.
If $n$ has integer factors, a larger set of solutions is available. Let $t$ be a factor of $n$.
Then $$n=\Biggl(\sum_{i=1}^{\frac{\frac{n}{t}+t}{2}}(2i-1)\Biggr)-\Biggl(\sum_{i=1}^{\frac{\frac{n}{t}-t}{2}}(2i-1)\Biggr)$$
with the following constraints: $\frac{n}{t}>t$ so that the indices are greater than $0$ and $\frac{n}{t}\equiv t \bmod 2$ so that the indices are integers. The first sum is the square of the superior index $$\Bigl(\frac{\frac{n}{t}+t}{2}\Bigr)^2=\frac{(\frac{n}{t})^2+2n+t^2}{4}$$
The second sum is the square of the superior index $$\Bigl(\frac{\frac{n}{t}-t}{2}\Bigr)^2=\frac{(\frac{n}{t})^2-2n+t^2}{4}$$
Subtracting the second from the first yields $\frac{4n}{4}=n$
So when $n$ has appropriate factors, it can be represented in additional ways as the difference of two squares.
In the context of the question, we can set $n=b^2-3$. It is not necessary to write out all of the terms and indices explicitly here; the principles have been demonstrated.
In the example I used, $b=6$, the resulting $b^2-3=33$ is divisible by $3$. this affords indices $\frac{\frac{33}{3}\pm 3}{2}=7,4$ yielding $7^2-4^2=33$. Since the other factor of $33$ is $11$ and $\frac{33}{11}\not > 11$, this factor does not satisfy the constraints, and we have found all solutions corresponding to $b=6$.
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What fraction is shaded? Conrguent $\frac{5}{6}$ circles in a circle. What fraction is shaded?
]1
Solution:
Let $r$ be the radius of the small circles and $R$ the radius of the big one.
The colored section is three times five sixths of the area of one of the small circles.
Colored Section area= $3\times\dfrac{5}{6}\times\pi r^2=\dfrac{5\pi r^2}{2}$
The radius $R$ of the big circle is equal to $r$ plus the radius of the circumscribed circle of equailateral triangle ABC, whose side is $2r$.
The radius of the circumscribed circle of an equilateral triangle is the length of the sides divided by $\sqrt{3}$. Since the side here measures $2r$, the radius of the circumscribed circle is $\dfrac{2r}{\sqrt{3}}$.
So we have $R = r+\dfrac{2r}{\sqrt{3}}$
The area of the big circle is $\pi \times R^2$, which here is equal to $(r+\dfrac{2r}{\sqrt{3}})^2$
which, when expanded, gives
Big Circle area = $\dfrac{\pi r^2(7+4\sqrt{3})}{3}$
To obtain the shaded fraction, we need to divide the area of the colored region by the area of the big circle:
Shaded fraction = $\dfrac{\dfrac{5\pi r^2}{2}}{\dfrac{\pi r^2(7+4\sqrt{3})}{3}}$
Shaded fraction = $\dfrac{5\pi r^2}{2} \times \dfrac{3}{\pi r^2(7+4\sqrt{3})}$
Shaded fraction = $\dfrac{15}{2(7+4\sqrt{3})} \simeq 53.847 \% $
I think it's wrong. In the drawing the smaller circles are not tangent to the largest
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We need to find the radius of the large circle and the rest is straight forward:
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particular solution of $(D^2+4)y=4x^2\cos 2x$
Find a particular solution to the equation $$(D^2+4)y=4x^2\cos 2x$$
\begin{align}
y_p&=\frac{1}{D^2+4}(4x^2\cos 2x)\\
&=\frac{1}{D^2+4}\left[2x^2(e^{2ix}+e^{-2ix})\right]\\
&=2\left[e^{2ix}\frac{1}{(D+2i)^2+4}x^2+e^{-2ix}\frac{1}{(D-2i)^2+4}x^2\right]\\
&=2\left[e^{2ix}\frac{1}{D(D+4i)}x^2+e^{-2ix}\frac{1}{D(D-4i)}x^2\right]\\
\end{align}
But I stuck at this point because how to solve $\frac{1}{D(D+4i)}x^2?$ The solution provided this
\begin{align}
&=2\left[e^{2ix}\frac{1}{D(D+4i)}x^2+e^{-2ix}\frac{1}{D(D-4i)}x^2\right]\\&=\frac{1}{4i}\left[e^{2ix}D^{-1}\left(2x^2+ix-\frac{1}{4}\right)-e^{-2ix}D^{-1}\left(2x^2-ix-\frac{1}{4}\right)\right]\\
&=\frac{1}{24}[6x^2\cos 2x+x(8x^2-3)\sin 2x]
\end{align}
I still can't figure out how they do this. Any help will be appreciated.
Update: Using @Isham answer I tired
\begin{align}
&=\frac{1}{4i}\left[e^{2ix}D^{-1}\left(2x^2+ix-\frac{1}{4}\right)-e^{-2ix}D^{-1}\left(2x^2-ix-\frac{1}{4}\right)\right]\\
&=\frac{1}{4i}\left[e^{2ix}\left(2\frac{x^3}{3}+i\frac{x^2}{2}-\frac{x}{4}\right)-e^{-2ix}\left(2\frac{x^3}{3}-i\frac{x^2}{2}-\frac{x}{4}\right)\right]\\
&=\frac{1}{4i}\left[\frac{2x^3}{3}(e^{2ix}-e^{-2ix})+\frac{x^2i}{2}(e^{2ix}-e^{-2ix})-\frac{x}{4}(e^{2ix}-e^{-2ix})\right]\\
&=\frac{1}{4i}\left[\frac{2x^3}{3}2\cos 2x+\frac{x^2i}{2}2\cos 2x-\frac{x}{4}2\cos 2x\right]
\end{align}
But I think I am again Lost.
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Given differential equation is$$(D^2+4)y=4x^2\cos 2x$$
We have to find the particular integral (P.I.).
\begin{equation}
\text{P.I.}~=4\dfrac{1}{D^2+4}~x^2\cos 2x\\
=\text{R.P. of }~\left(4\dfrac{1}{D^2+4}~x^2~e^{2ix}\right)\\
=4~\cdot~\left[\text{R.P. of }~\left\{e^{2ix}~\dfrac{1}{(D+2i)^2+4}~x^2\right\}\right]\\
=4~\cdot~\left[\text{R.P. of }~\left\{e^{2ix}~\dfrac{1}{D^2+4iD}~x^2\right\}\right]\\
=4~\cdot~\left[\text{R.P. of }~\left\{e^{2ix}~\dfrac{1}{4iD}\left(1+\dfrac{D}{4i}\right)^{-1}~x^2\right\}\right]\\
=4~\cdot~\left[\text{R.P. of }~\left\{e^{2ix}~\dfrac{1}{4iD}\left(1+i\dfrac{D}{4}-\dfrac{D^2}{16}+\cdots\right)~x^2\right\}\right]\\
=4~\cdot~\left[\text{R.P. of }~\left\{e^{2ix}~\dfrac{1}{4iD}\left(x^2+i~\dfrac{x}{2}-\dfrac{1}{8}\right)\right\}\right]\\
=-~\left[\text{R.P. of }~\left\{i~e^{2ix}~\dfrac{1}{D}\left(x^2+i~\dfrac{x}{2}-\dfrac{1}{8}\right)\right\}\right]\\
=-~\left[\text{R.P. of }~\left\{i~e^{2ix}~\left(\dfrac{x^3}{3}+i~\dfrac{x^2}{4}-\dfrac{x}{8}\right)\right\}\right]\\
=-~\left[\text{R.P. of }~\left\{i~\left(\cos 2x+i~\sin 2x\right)~\left(\dfrac{x^3}{3}+i~\dfrac{x^2}{4}-\dfrac{x}{8}\right)\right\}\right]\\
=-~\left[\text{R.P. of }~\left\{\left(i~\cos 2x-~\sin 2x\right)~\left(\dfrac{x^3}{3}+i~\dfrac{x^2}{4}-\dfrac{x}{8}\right)\right\}\right]\\
=-~\left[\text{R.P. of }~\left\{\left(-~\dfrac{x^2}{4}~\cos 2x-\left(\dfrac{x^3}{3}-\dfrac{x}{8}\right)~\sin 2x\right)~+~i~\left(-\dfrac{x^2}{4}\sin 2x+\left(\dfrac{x^3}{3}-\dfrac{x}{8}\right)\cos 2x\right)\right\}\right]\\
=-~\left[-~\dfrac{x^2}{4}~\cos 2x-\left(\dfrac{x^3}{3}-\dfrac{x}{8}\right)~\sin 2x\right]\\
=\dfrac{x^2}{4}~\cos 2x+\left(\dfrac{x^3}{3}-\dfrac{x}{8}\right)~\sin 2x
\end{equation}
Here R.P. stands for real part.
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|
What is the expected length of the hypotenuse formed by bending a unit length randomly at a right angle? This is easy enough to simulate and find the answer is somewhere around .812. However I am not finding it so easy to solve the integral involved which I believe is...
$$\int_0^1 \sqrt{x^2+(1-x)^2} dx$$
I don't seem to get the right answer if I use this derivation. Bonus points for finding the expected area which is a much easier problem!
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Use the substitution $u=x-\frac{1}{2}$
$$\int_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{\left(u+\frac{1}{2}\right)^2+\left(u-\frac{1}{2}\right)^2}du = \int_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{2u^2 + \frac{1}{2}}du = 2\sqrt{2}\int_0^{\frac{1}{2}} \sqrt{u^2+\frac{1}{4}}du$$
Then there are two ways to approach this problem, either let $u=\frac{1}{2}\tan\theta$ or $u=\frac{1}{2}\sinh t$. Personally I find the second one easier as it doesn't require an integration by parts
$$ \frac{1}{\sqrt{2}} \int_0^{\sinh^{-1}\left(1\right)} \cosh^2 t dt = \frac{1}{2\sqrt{2}} \int_0^{\sinh^{-1}\left(1\right)} 1 + \cosh(2t) dt = \frac{1}{2\sqrt{2}}\Biggr[t + \frac{1}{2}\sinh(2t)\Biggr]_0^{\sinh^{-1}\left(1\right)}$$
$$=\frac{1}{2\sqrt{2}}\Biggr[t + \sinh(t)\cosh(t)\Biggr]_0^{\sinh^{-1}\left(1\right)}=\frac{1}{2\sqrt{2}}\Biggr[t + \sinh(t)\sqrt{1+\sinh^2(t)}\Biggr]_0^{\sinh^{-1}\left(1\right)}$$
$$ = \frac{\sinh^{-1}\left(1\right)}{2\sqrt{2}} + \frac{1}{2} \approx 0.8116$$
by using the hyperbolic identities $\cosh(2t) = \cosh^2(t) + \sinh^2(t)$ and $\cosh^2(t)-\sinh^2(t) = 1$.
|
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If point $P$ varies on a circle about the center of rectangle $\square ABCD$, then $PA^2+PB^2+PC^2+PD^2$ remains constant
Let $\square ABCD$ be a rectangle with center $O$. Prove that if a $P$ is a point that varies over a circle about $O$, then $PA^2+PB^2+PC^2+PD^2$ remains constant
Attempt:
|
Place $O$ at $(0,0)$ on the Cartesian Plane, and place $P$ at a point $(x,y)$ at a distance $r$ from $O$. Furthermore, place the vertices of the rectangle at the points of the form $(\pm a,\pm b)$, for the corresponding $a$, $b$. Then, $$PA^2+PB^2+PC^2+PD^2=$$ $$2\left(\left(x-a\right)^2+\left(x+a\right)^2+\left(y-b\right)^2+\left(y+b\right)^2\right)=$$ $$4\left(x^2+y^2+a^2+b^2\right)=$$ $$4\left(r^2+a^2+b^2\right),$$ which is clearly a constant. $\blacksquare$
|
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evaluation of Trigonometric limit
Evaluation of $$\lim_{n\rightarrow \infty}\frac{1}{n\bigg(\cos^2\frac{n\pi}{2}+n\sin^2\frac{n\pi}{2}\bigg)}$$
What i try
Put $\displaystyle \frac{n\pi}{2}=x,$ when $n\rightarrow\infty,$ Then $x\rightarrow \infty$
$$\frac{\pi^2}{2}\lim_{x\rightarrow \infty}\bigg(\frac{ x^{-1}}{\pi\cos^2(x)+2x\sin^2(x)}\bigg)$$
How do i solve it help me please
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$$\lim_{n\rightarrow \infty}\dfrac{1}{n\bigg(\cos^2\frac{n\pi}{2}+n\sin^2\frac{n\pi}{2}\bigg)}=\lim_{n\rightarrow \infty}\dfrac{1}{n\cos^2\frac{n\pi}{2}+n^2\sin^2\frac{n\pi}{2}}$$
As $n\to\infty$, since $\sin^2u$ and $\cos^2u$ both oscillate in $[0,1]$ and $n,n^2\to\infty$, So, $$(n\cos^2\frac{n\pi}{2}+n^2\sin^2\frac{n\pi}{2})\to\infty\implies\lim_{n\rightarrow \infty}\dfrac{1}{n(\cos^2\frac{n\pi}{2}+n\sin^2\frac{n\pi}{2})}=0$$
|
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|
prove that : $a^{2}+b^{2}+c^{2}+4abc<\frac12$ Let $a,b,c$ be a sides of triangle
Such that : $a+b+c=1$ Then
prove that : $a^{2}+b^{2}+c^{2}+4abc<\frac{1}{2}$
My effort :
Since $a+b+c=1$ $\implies$ $2S=sr=bc\sin A=\frac{abc}{2R}$
Also : $S=\sqrt{s(s-a)(s-b)(s-c)}$
Also :
$a^{2}+b^{2}+c^{2}+2(ab+ac+bc)=1$
But I don't know how to complete it,any help is appreciated !
|
A standard trick to solve such problems is to use the substitution $a= y + z, b =x+z, c = x+y$, where $x, y, z>0$ (to see why this is so, draw the inscribed circle and pairs of equal tangents: $x, y, z$ are the lengths of the tangents)
So, we have $$x+y+z = \frac{1}{2}$$
and the inequality is $$ (x+y)^2 + (x+z)^2 + (y+z)^2 + 4 (x+y)(x+z)(y+z) < x+y+z$$
$$2x^2 + 2y^2 + 2z^2 + 2(xy+yz+xz) + 4 (x+y)(x+z)(y+z) < x+y+z $$
$$x^2 + y^2 +z^2 + (xy+yz+xz) + 2(x+y)(x+z)(y+z) < (x+y+z)^2 $$
$$ 2(x+y)(x+z)(y+z) < xy + xz +yz$$
$$(x+y)(x+z)(y+z) < (x+y+z) (xy+xz +yz)$$
$$ x^2y + xzy + x^2z + xz^2 + y^2x + y^2z + yxz + yz^2 < x^2y + x^2z + xyz + xy^2 +xyz +y^2z + xyz +xz^2 + yz^2 $$
$$2 xyz < 3xyz $$
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|
Compute $\cos(\frac{2\pi}{7})\cos(\frac{4\pi}{7})+\cos(\frac{2\pi}{7})\cos(\frac{6\pi}{7})+\cos(\frac{4\pi}{7})\cos(\frac{6\pi}{7})$
$\cos(\frac{2\pi}{7})\cos(\frac{4\pi}{7})+\cos(\frac{2\pi}{7})\cos(\frac{6\pi}{7})+\cos(\frac{4\pi}{7})\cos(\frac{6\pi}{7}) = -\frac12$
I tried showing the equation, but my attempts did not get the result.
I already showed that
$$\cos(\frac{2\pi}{7})\cos(\frac{4\pi}{7})\cos(\frac{6\pi}{7})=\frac18$$ $$\cos(\frac{2\pi}{7})+\cos(\frac{4\pi}{7})+\cos(\frac{6\pi}{7})=-\frac12$$
Using in particular the trigonometric formulas:
$\sin(2\alpha)=2\sin(\alpha)\cos(\alpha)$ , $\sin(\pi+\alpha)=-\sin(\alpha)$ , $\sin(\pi-\alpha)=\sin(\alpha)$ , $\cos(\pi-\alpha)=-\cos(\alpha)$ , $\sin(-\alpha)=-\sin(\alpha)$ , and $2\sin(\alpha)\cos(\beta)=\sin(\alpha+\beta)+\sin(\alpha-\beta)$
Any hint would be helpful. Thanks in advance.
|
$$ 2 \cos \left( \frac{2 \pi}{7} \right) \; , \; \; 2 \cos \left( \frac{4 \pi}{7} \right) \; , \; \; 2 \cos \left( \frac{8 \pi}{7} \right) \; , \; \; $$
are the roots of
$$ x^3 + x^2 - 2x - 1 $$
so
$$ \cos \left( \frac{2 \pi}{7} \right) \; , \; \; \cos \left( \frac{4 \pi}{7} \right) \; , \; \; \cos \left( \frac{8 \pi}{7} \right) \; , \; \; $$
are the roots
$$ 8 x^3 + 4 x^2 - 4x - 1 $$
or
$$ x^3 + \frac{1}{2}x^2 - \frac{1}{2} x - \frac{1}{8} $$
The sum of products of two roots ate a time is $-\frac{1}{2}$
The original claim is from a method due to Gauss, this time we take
$$ \omega = e^{2 \pi i / 7} , $$
next take
$$ \eta = \omega + \frac{1}{\omega}, $$
then look for a cubic (in $\eta$) that vanishes because of the relation
$$ \omega^6 + \omega^5 + \omega^4 + \omega^3 + \omega^2 + \omega + 1=0$$
Here we find
$$
\eta^3 + \eta^2 - 2\eta - 1 =
\frac{ \omega^6 + \omega^5 + \omega^4 + \omega^3 + \omega^2 + \omega + 1}{\omega^3}
$$
|
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|
Without calculating the square roots, determine which of the numbers:$a=\sqrt{7}+\sqrt{10}\;\;,\;\; b=\sqrt{3}+\sqrt{19}$ is greater.
Without calculating the square roots, determine which of the numbers:
$$a=\sqrt{7}+\sqrt{10}\;\;,\;\; b=\sqrt{3}+\sqrt{19}$$
is greater.
My work (I was wondering if there are other ways to prove this):
$$a^2=17+2\sqrt{70}, \;\;b^2=22+2\sqrt{57}$$
$$\sqrt{64}<\sqrt{70}<\sqrt{81}\implies 8<\sqrt{70}<9\implies a^2<35$$
$$\sqrt{49}<\sqrt{57}<\sqrt{64}\implies 7<\sqrt{57}<8\implies b^2>36$$
$$a^2<35<36<b^2\implies a^2<b^2\implies |a|<|b|$$
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$$a<b\iff a^2<b^2\iff 7+10+2\sqrt {70}<3+19+2\sqrt {57}$$ $$\iff 2\sqrt {70}<5+2\sqrt {57}$$ $$\iff (2\sqrt {70})^2<(5+2\sqrt {57})^2$$ $$\iff 280<25+228+20\sqrt {57}$$
$$\iff 27<20\sqrt {57}$$ and we have $20\sqrt {57}>20\sqrt 4=40>27.$
Another way, from one of the intermediate steps above, is $$a<b\iff 2\sqrt {70}<5+2\sqrt {57}$$ $$\iff 2(\sqrt {70}-\sqrt {57})<5$$ $$\iff 2(\sqrt {70}-\sqrt {57})(\sqrt {70}+\sqrt {57})<5(\sqrt {70}+\sqrt {57})$$ $$\iff 2(70-57)<5(\sqrt {70}+\sqrt {57})$$ and we have $5(\sqrt {70}+\sqrt {57})>5\sqrt {70}>5\sqrt {64}=40>26=2(70-57).$
Or we might notice that $2(\sqrt {70}-\sqrt {57})<2(\sqrt {81}-\sqrt {49})=4<5.$
|
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|
Fibonacci sum for $\pi$: $\sum_{n=1}^\infty\frac{F_{2n}}{n^2\binom{2n}{n}}=\frac{4\pi^2}{25\sqrt5}$ Wikipedia's "List of formulae involving $\pi$" entry states that
$$\sum_{n=1}^\infty\frac{F_{2n}}{n^2\binom{2n}{n}}=\frac{4\pi^2}{25\sqrt5}.$$
Why is this true?
If $\varphi=(1+\sqrt5)/2$ and $\psi=(1-\sqrt5)/2$, then we can rewrite this identity as
$$F(\varphi)-F(\psi)=\frac{4\pi^2}{25}\quad\text{ where }F(x)=\sum_{n=1}^\infty\frac{x^{2n}}{n^2\binom{2n}{n}}.$$
$F(x)$ looks similar to the power series for arcsin.
|
$$F(x)=\sum_{n=1}^\infty\frac{x^{2n}}{n^2\binom{2n}{n}}=2 \left[\sin ^{-1}\left(\frac{x}{2}\right)\right]^2$$ makes
$$F(\varphi)=\frac{9 \pi ^2}{50}\qquad \text{and}\qquad F(\psi)=\frac{\pi ^2}{50}$$
since
$$\varphi=1-2 \cos \left(\frac{3 \pi }{5}\right)$$
|
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|
If $m$ and $n$ are integers, show that $\left|\sqrt{3}-\frac{m}{n}\right| \ge \frac{1}{5n^{2}}$ If $m$ and $n$ are integers, show that $\biggl|\sqrt{3}-\dfrac{m}{n}\biggr| \ge \dfrac{1}{5n^{2}}$.
Since $\biggl|\sqrt{3}-\dfrac{m}{n}\biggr|$ is equivalent to $\biggl|\dfrac{ \sqrt{3}n-m}{n}\biggr|$
So I performed the following operation $\biggl|\dfrac{\sqrt{3}n-m}{n}\biggr|\cdot \biggl|\dfrac{\sqrt{3}n+m}{\sqrt{3}n+m}\biggr|$ to get $$\biggl|\dfrac{3n^{2}-m^{2}}{\sqrt{3}n^{2}+mn}\biggr|$$
Since $n,m \ne 0$, we have that $|3n^{2}-m^{2}| \ge 1$. Now for the denominator, we have $$ |\sqrt{3}n^{2}+mn| \le |\sqrt{3n^{2}}| + |mn| $$
Thus it follows that $$\dfrac{1}{|\sqrt{3}n^{2}+mn|} \ge \dfrac{1}{|\sqrt{3}n^{2}| + |mn|}$$
Would I have to work in cases where $m<n$, for example? Then we have $$|\sqrt{3}n^{2}| + |mn| < |\sqrt{3}n^{2}| + n^{2} < 3n^{2} + n^{2} < 5n^{2}$$ which gives us the desired result. Although, the same method doesn't work when $n >m$.
|
The original inequality is equivalent to
$$
(5 m n+1)^2\leq 75 n^4
$$
or
$$
75 n^4\leq (5 m n-1)^2
$$
This reduces to
$$
\left(n\leq -1\land \left(m\leq \frac{1}{5 n}-\sqrt{3} \sqrt{n^2}\lor m\geq -\sqrt{3} \sqrt{n^2}-\frac{1}{5 n}\right)\right)
$$
or
$$
\left(n\geq 1\land \left(m\leq \sqrt{3} \sqrt{n^2}-\frac{1}{5 n}\lor m\geq \sqrt{3} \sqrt{n^2}+\frac{1}{5 n}\right)\right)
$$
Since $|n| \ge 1$ implies $2/(5n) \le 1$, we are done.
|
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|
Find the value of the double integral: $\int\limits_0^{2a}\int\limits_0^{\sqrt{2ax-x^2}}\frac{\phi'(y)(x^2+y^2)xdxdy}{\sqrt{4a^2x^2-(x^2+y^2)^2}}$ Evaluate $$\int_0^{2a}\int_0^{\sqrt{2ax-x^2}}\frac{\phi'(y)(x^2+y^2)xdxdy}{\sqrt{4a^2x^2-(x^2+y^2)^2}}$$
Here I changed the order of integration but after that I am not able to think of the relevant substitution to proceed with the integration.
I have tried to substitute it with polar co-ordinates but didn't succeed in solving that either.
I have checked the solution provided here Evaluate the integral $\int_0^{2a}\int_0^\sqrt{2ax-x^2}\frac{\phi'(y)(x^2+y^2)x dxdy}{\sqrt{4a^2x^2-(x^2+y^2)^2}}$ , but this solution is very lengthy.
I am looking for a shorter way of solving this question which is more obvious to hit during exam and also considers the space constraint.
If someone could suggest what substitutions to make to solve it in lesser steps.
|
Changing the order and taking $x^2+y^2=t,dt=2xdx$ gives,$$\frac12\int_0^a\phi'(y)\int_{2a^2-2a\sqrt{a^2-y^2}}^{2a^2+2a\sqrt{a^2-y^2}}\frac{t~dt}{\sqrt{4a^2(t-y^2)-t^2}}~dy$$Now$$\frac t{\sqrt{4a^2(t-y^2)-t^2}}=\frac{-1}2\left[\frac{4a^2-2t}{\sqrt{4a^2(t-y^2)-t^2}}\right]+\frac{2a^2}{\sqrt{4a^2(a^2-y^2)-(t-2a^2)^2}}$$and so$$\int\frac{t~dt}{\sqrt{4a^2(t-y^2)-t^2}}=-\sqrt{(4a^2-t)t-4a^2y^2}+2a^2\sin^{-1}\left(\frac{t-2a^2}{2a\sqrt{a^2-y^2}}\right)$$Applying the limits should give$$\frac12\int_0^a\phi'(y)[0+2a^2\pi]dy=\pi a^2[\phi(a)-\phi(0)]$$
|
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Solve system of equations $\frac{x}{y}+\frac{y}{x}=\frac{10}{3}$, $x^2-y^2=8$
Solve the system:
$$\begin{array}{l}\dfrac{x}{y}+\dfrac{y}{x}=\dfrac{10}{3} \\
x^2-y^2=8\end{array}$$
First, we have $x \ne 0$ and $y \ne 0$. We can rewrite the first equation as $$\dfrac{x^2+y^2}{xy}=\dfrac{10}{3}$$
What should I do next?
|
Multiplying the first equation by $xy$ gives us
$$3x^2-10xy+3y^2=0$$
Let $a=x+y$ and $b=x-y$ to get
$$4b^2-a^2=3x^2-10xy+3y^2=0$$
$$ab=8$$
Solving for $a$ in the second equation we get $a=8/b$, and substituting into the first equation gives
$$b^2-\frac{64}{b^2}=0$$
Solving for $b$ we get $b^4=16$, which has the four solutions given by $b=\pm2,\pm2i$. Correspondingly, the solutions for $a$ are given by $\pm4,\mp4i$. Solving back for $x$ and $y$ we then get the final solutions of $(x,y)=(\pm3,\pm1)$ and $(x,y)=(\mp i,\mp3i)$.
Alternatively, it is easy to see we have
$$4b^2-a^2=(2b-a)(2b+a)=0$$
and hence
$$\begin{cases}2b-a=0\\ab=8\end{cases}\text{ or }\begin{cases}2b+a=0\\ab=8\end{cases}$$
which can be solved just as easily.
|
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|
If $a^2>b^2$ prove that $\int\limits_0^{\pi} \frac{dx}{(a+b\cos x)^3}=\frac{\pi (2a^2+b^2)}{2(a^2-b^2)^{5/2}}$. Problem: If $a^2>b^2$ prove that $\int\limits_0^\pi \dfrac{dx}{(a+b\cos x)^3} = \dfrac{\pi (2a^2+b^2)}{2(a^2-b^2)^{5/2}}$.
My effort:
If we choose $$x=\tan\frac{\theta}{2}\Longrightarrow d\theta=\frac{2}{x^2+1} \, dx\;\;,\;\;\cos\theta=\frac{1-x^2}{1+x^2}$$ then the integral becomes critical. What is the simplest way to solve?
|
As question is closed let me bring short answer here:
Having $I_n=\int\frac{dx}{(a+b\cos x)^n}$ partial integration gives
$$I_2(a^2-b^2) = aI_1-\frac{b\sin x}{a+b\cos x}$$
$$I_3 2(a^2-b^2) = 3aI_2 - I_1 -\frac{b\sin x}{a+b\cos x}$$
So $I_3$ can be represented as sum of constant member with $I_1$ plus function members.
Knowing, for $0\leqslant \varepsilon < 1$
$$\int\limits_{0}^{2\pi}\frac{dx}{1+\varepsilon \cos x} = \frac{2\pi}{\sqrt{1-\varepsilon^2}}$$
we can obtain
$$\int\limits_{0}^{2\pi}\frac{dx}{(a+b \cos x)^3}=\frac{\pi (2a^2+b^2)}{2(a^2-b^2)^{5/2}}$$
|
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|
Calculate $ f [f (x)] $ Where $ f (x) = \frac {1} {1 + x ^ 2} $, $ f [f (x)] $ is equal to:
a
I was factoring and found $f(f(x))= \frac{1}{1+(\frac{1}{1+x^2})^2} = \frac{1}{\frac{x^4+2x^2+2}{x^4+2x^2+1}}= \frac{x^4+2x^2+1}{x^4+2x^2+2}$, but the answer is x. I am wrong?
|
The answer is not $x$, as Desmos (and the rest of math!) suggest(s):
|
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Find the total number of 7-digit Find the total number of 7-digit natural numbers whose digits sum is 20.
Solution: This is not a exactly answer, as well we can use a way to subtracts the cases when exists a digit greater than $9$.
Let $x_1,\cdots, x_7$ the digits.
As well there are ${25\choose 6}$ ways to give us the sums of numbers, but this permits $x_i>9$.
We need to analyze a few cases (or you can abbreviate this if you're awesomely intellectual, I'm not).
$Case~1)$ All the cases (ways) where $x_1>9$ and the others are less than or equal to $9$.
Then you need to write $x_1=9+x_1'$. And why is this a particular case? Because Stars and Bars here admits numbers equal to zero, and in this ways $0$ is permitted.
But see that this case counts ways that is not permitted by our hypothesis, is when$ (x_1,x_i,x_j,\cdots, x_{j+3})=(10,10,0,\cdots,0) $
To calculate the ways it's easy only have that $x_1'+x_2+\cdots +x_7= 11$
Then clearly all the ways are ${16\choose 6} - 6$
$Case~2)$ Some $x_i\neq x_1$ is greater than $9$.
This is only ${15\choose 6} $ but for individual digits, then all the ways are $6 {15\choose 6} $
This give us that the quantity of numbers such that the sum of digits is $20$ are:
$${25\choose 6}-\left({16\choose 6}+6 {15\choose 6}-6\right) $$
Correct
|
We count the number of integers $x$ with $1\,000\,000\leq x\leq 9\,999\,999$ which have a digit sum equal to $20$ with the help of generating functions. It is convenient to use the coefficient of operator $[x^{n}]$ to denote the coefficient of $x^n$ in a series.
We obtain
\begin{align*}
\color{blue}{[x^{20}]}&\color{blue}{\left(1+x+x^2+\cdots+x^9\right)^7-[x^{20}]\left(1+x+x^2+\cdots+x^9\right)^6}\tag{1}\\
&=[x^{20}]\left(\frac{1-x^{10}}{1-x}\right)^7-[x^{20}]\left(\frac{1-x^{10}}{1-x}\right)^6\tag{2}\\
&=[x^{20}]\left(1-7x^{10}+\binom{7}{2}x^{20}\right)\sum_{j=0}^\infty\binom{-7}{j}(-x)^j\\
&\qquad-[x^{20}]\left(1-6x^{10}+\binom{6}{2}x^{20}\right)\sum_{j=0}^\infty\binom{-6}{j}(-x)^j\tag{3}\\
&=\left([x^{20}]-7[x^{10}]+21[x^0]\right)\sum_{j=0}^\infty\binom{j+6}{6}x^j\\
&\qquad-\left([x^{20}]-6[x^{10}]+15[x^0]\right)\sum_{j=0}^\infty\binom{j+5}{5}x^j\tag{4}\\
&=\left(\binom{26}{6}-7\binom{16}{6}+21\binom{6}{6}\right)-\left(\binom{25}{5}-6\binom{15}{5}+15\binom{5}{5}\right)\tag{5}\\
&=\left(230\,230-56\,056+21\right)-\left(53\,130-18\,018+15\right)\\
&=174\,195-35\,127\\
&\,\,\color{blue}{=139\,068}
\end{align*}
Comment:
*
*In (1) we count the strings with length $7$ consisting of digits from $0$ to $9$ with digit-sum $20$. Since numbers do not start with a leading zero, we subtract all strings of length $6$ with digit-sum $20$.
*In (2) we use the finite geometric series formula.
*In (3) we expand the numerator omitting terms which do not contribute to $[x^{20}]$ and we use the binomial series expansion.
*In (4) we apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$ and we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{p-1}(-1)^q$.
*In (5) we select the coefficients accordingly.
|
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Proof by mathematical induction proof verification If $n\in\Bbb N$ then $\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{n}{(n+1)!}=1-\frac{1}{(n+1)!}$.
The base case where $n=1$ is clearly true: $\frac{1}{2!}=\frac{1}{2}$ and $1-\frac{1}{(1+1)!}=\frac{1}{2}$.
Now to show that $S_k\Rightarrow S_{k+1}$, we assume that $\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{k}{(k+1)!}=1-\frac{1}{(k+1)!}$ and observe the following equalitites:
$\begin{align*}\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{k+1}{((k+1)+1)!}&=\\
(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{k}{(k+1)!})+\Big(\frac{(k+1)}{((k+1)+1)!}\Big)&=\\
1-\frac{1}{(k+1)!}+\frac{k+1}{((k+1)+1)!}&=\frac{(k+2)(k+1)!-(k+2)+(k+1)}{(k+2)(k+1)!}\\
&=\frac{(k+2)(k+1)!-1}{(k+2)(k+1)!}\\
&=1-\frac{1}{(k+2)!}\\
&=1-\frac{1}{((k+1)+1)!}\end{align*}$
Therefore we conclude that if $n\in\Bbb N$ then $\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{n}{(n+1)!}=1-\frac{1}{(n+1)!}$.
I know this is a relatively straightforward example but I'm very new to the concept of induction and I'm wondering if I'm on the right track with this proof. It seems correct to me but I'm worried I may have overlooked something. Any advice is helpful, thanks!
|
A couple of your steps could possibly be arranged in a more readable way. However, I do believe your proof is core.
|
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|
value of $n$ in limits
If $\displaystyle \lim_{x\rightarrow 0}\frac{x^n\sin^{n}(x)}{x^{n}-(\sin x)^{n}}$ is a finite non zero number.
Then value of $n$ is equals
What I try:
$$\sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots $$
$$\lim_{x\rightarrow 0}\frac{x^n\bigg(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots \bigg)^n}{x^{n}-\bigg(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots \bigg)^n}$$
$$\lim_{x\rightarrow 0}\frac{x^{n+1}\bigg(1-\frac{x^2}{3!}+\frac{x^4}{5!}+\cdots \bigg)^n}{x^{n}-\bigg(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots \bigg)^n}$$
How do I solve it? Help me please.
|
$$L=\lim_{x \rightarrow 0} \frac{x^n \sin^{n} x}{x^n-\sin^n x}$$ $$= \lim_{x \rightarrow 0} \frac{x^{2n}(1-x^2/6)^n}{x^{n}-x^n(1-x^2/6)^n}$$
$$=\lim_{x \rightarrow 0} \frac{x^n (1-nx^2/6)}{nx^2/6}=\lim_{x \rightarrow 0}\frac {6x^{n-2}}{n}= 3,~ if~ n =2,$$
$$= 0, ~if~ n > 2, $$ $$=\infty ~~if~~n= n <2 $$
|
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|
Integer triangle $ABC$ such that $IHO$ is also an integer triangle. An infinite number of such non-similar triangles $ABC$.
Let $I, H, O$ be the incenter, orthocenter and circumcenter of non-isosceles triangle $ABC$ respectively. Prove that there are infinitely many integer triangles $ABC$, none of which are similar, suchs that for each of them the triangle $IHO$ is also an integer triangle. (An integer triangle is a triangle all of whose sides have lengths that are integers)
My work. $$IO=\sqrt{R^2-2rR}$$ $$OH=\sqrt{9R^2-(a^2+b^2+c^2)}$$ $$IH=\sqrt{2r^2+4R^2-\frac{1}{2}(a^2+b^2+c^2)}$$
where $R$, $r$ are circumradius and inradius of triangle $ABC$ respectively; $a$, $b$, $c$ are the lengths of the sides of triangle $ABC$. Unfortunately, these formulas are very inconvenient for analysis.
|
(Some small examples that I found satisfy $IO=IH$. So...)
Let us consider $\triangle{ABC}$ satisfying $IO=IH$. We have
$$IO=IH\iff (a^2 - a b + b^2 - c^2) (a^2 - b^2 + b c - c^2) (a^2 - a c - b^2 + c^2)=0$$
So, in the following, let us consider $\triangle{ABC}$ such that $$a^2=b^2-bc+c^2\tag1$$
(which means that $\angle A=60^\circ$.)
Here, it is known that
$$a=m^2+mn+n^2,\quad b=m^2+2mn,\quad c=m^2-n^2$$
satisfy $(1)$. Then, we get
$$IO=IH=n\sqrt{m^2+mn+n^2},\qquad HO=2mn+n^2\in\mathbb Z$$
Also, it is known that
$$m=s^2-1,\quad n=2s+1,\quad z=s^2+s+1$$
satisfy $z^2=m^2+mn+n^2$, so if
$$\begin{cases}a=(s^2+s+1)^2
\\\\b=(s - 1) (s + 1) (s^2 + 4 s + 1)
\\\\c=s (s + 2) (s^2 - 2 s - 2)\tag2\end{cases}$$
where $s\ (\ge 3)$ is an integer, then we get $$IO=IH=(2s+1)(s^2+s+1)\in\mathbb Z$$
Let us prove that if $(2)$, then $\triangle{ABC}$ are not isosceles triangles.
Proof :
We have
$$b-a=(2 s + 1) (s^2 - 2 s - 2)\gt 0$$
$$a-c=(2 s + 1) (s^2 + 4 s + 1)\gt 0$$
$$a+c-b=(s - 1) (s + 1) (s^2 - 2 s - 2)\gt 0\qquad\square$$
Next, let us prove that if $(2)$, then no two are similar.
Proof :
Suppose that a triangle with $(2)$ and a triangle with
$$\begin{cases}a=(t^2+t+1)^2
\\\\b=(t - 1) (t + 1) (t^2 + 4 t + 1)
\\\\c=t (t + 2) (t^2 - 2 t - 2)\end{cases}$$
where $t\not=s\ (t\ge 3)$ are similar. Then, we have
$$\frac{(s^2+s+1)^2}{(s - 1) (s + 1) (s^2 + 4 s + 1)}=\frac{(t^2+t+1)^2}{(t - 1) (t + 1) (t^2 + 4 t + 1)}$$
$$\iff (s - t) (2 s t + s + t + 2) (s^2 t^2 - 2 s^2 t - 2 s^2 - 2 s t^2 - 8 s t - 2 s - 2 t^2 - 2 t + 1) = 0$$
$$\iff s^2 t^2 - 2 s^2 t - 2 s^2 - 2 s t^2 - 8 s t - 2 s - 2 t^2 - 2 t + 1 = 0$$
$$\iff s = \frac{2 (t^2 + 4 t + 1) ± 2(t^2+t+1)\sqrt{3}}{2 (t^2 - 2 t - 2) }$$
which is impossible since the RHS is irrational.$\qquad\square$
Conclusion :
For $$\begin{cases}a=(s^2+s+1)^2
\\\\b=(s - 1) (s + 1) (s^2 + 4 s + 1)
\\\\c=s (s + 2) (s^2 - 2 s - 2)\end{cases}$$
where $s\ (\ge 3)$ is an integer, $\triangle{ABC}$ are not isosceles triangles and no two are similar, with $$IO=IH=(2s+1)(s^2+s+1),\qquad HO=(2 s + 1) (2 s^2 + 2 s - 1)$$
where $IO+IH-HO=6s+3\gt 0$.
|
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|
Evaluate $\int \frac1{\left(1+x^4\right)\left(\sqrt{\sqrt{1+x^4}}-x^2\right)}dx$ Evaluate $\int \dfrac{1}{\left(1+x^4\right)\left(\sqrt{\sqrt{1+x^4}}-x^2\right)}dx$
My attempt is as follows:-
$$x^2=\tan\theta$$
$$2xdx=\sec^2\theta d\theta$$
$$dx=\dfrac{\sec^2\theta d\theta}{2\sqrt{\tan\theta}}$$
$$\int \dfrac{1}{\left(1+\tan^2\theta\right)\left(\sqrt{\sec\theta}-\tan\theta\right)}\cdot\dfrac{\sec^2\theta}{2\sqrt{\tan\theta}}d\theta$$
$$\dfrac{1}{2}\cdot\int\dfrac{d\theta}{\sqrt{\tan\theta}\left(\sqrt{\sec\theta}-\tan\theta\right)}$$
$$\dfrac{1}{2}\cdot\int\dfrac{(\cos\theta)^\frac{3}{2} d\theta}{\sqrt{\sin\theta}\left(\sqrt{\cos\theta}-\sin\theta\right)}$$
$$\sqrt{\sin\theta}=y$$
$$\dfrac{\cos\theta}{2\sqrt{\sin\theta}}d\theta=dy$$
$$\int \dfrac{\sqrt{1-y^4}}{\sqrt{1-y^4}-y^2}dy$$
$$\int \dfrac{\sqrt{\dfrac{1}{y^4}-1}}{\sqrt{\dfrac{1}{y^4}-1}-1}dy$$
How to proceed from here or feel free to suggest shorter and clean approach
|
In the same spirit as Mariusz Iwaniuk's comment, a CAS gives
$$\int \dfrac{\sqrt{1-y^4}}{\sqrt{1-y^4}-y^2}dy=\frac{1}{3} y^3 F_1\left(\frac{3}{4};-\frac{1}{2},1;\frac{7}{4};y^4,2
y^4\right)+\frac{1}{8} \left(4 y+2^{3/4} \left(\tan ^{-1}\left(\sqrt[4]{2}
y\right)+\tanh ^{-1}\left(\sqrt[4]{2} y\right)\right)\right)$$ where appears the Appell hypergeometric function of two variables.
|
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|
Prove that $R$ is reflexive, symmetric, and transitive. Define a relation $R$ on $\Bbb Z$ by declaring that $xRy$ if and only if $x^2\equiv y^2\pmod{4}$. Prove that $R$ is reflexive, symmetric, and transitive.
Suppose $x\in\Bbb Z$. Then $x^2\equiv x^2\pmod {4}$ means that $4\mid (x^2-x^2)$, so $x^2-x^2=4a$ where $a=0\in\Bbb Z$. Therefore $R$ is reflexive.
Now suppose $x^2\equiv y^2\pmod {4}$. This means $4\mid (x^2-y^2)$ and so $x^2-y^2=4a$, for some $a\in\Bbb Z$. Multiplying by $-1$ we have $-1(x^2-y^2=4a)\\\rightarrow -x^2+y^2=-4a\\\rightarrow y^2-x^2=4(-a)$
so $4\mid(y^2-x^2)$ and $y^2\equiv x^2\pmod{4}$. This shows that $R$ is symmetric.
Now we assume that $x^2\equiv y^2\pmod{4}$ and $y^2\equiv z^2\pmod{4}$. This means $4\mid(x^2-y^2)$ and $4\mid(y^2-z^2)$. Then we have $x^2-y^2=4a$ and $y^2-z^2=4b$ for some $a,b\in\Bbb Z$. Rearranging we get $x^2=4a+y^2$ and $z^2=y^2-4b$.
Then
$\begin{align*}x^2-z^2&=(4a+y^2)-(y^2-4b)\\&=4a+4b\\&=4(a+b)\end{align*}$
This shows that $4\mid(x^2-z^2)$ and $x^2\equiv z^2\pmod{4}$, therefore $R$ is transitive.
$\blacksquare$
Please forgive my rushed formatting, just wondering if my arguments here work and if this is a valid proof. Any feedback is appreciated, thanks!
|
I agree with J. W. Tanner's question comment that what you've done looks all right.
I have just one small suggestion. With your $x^2−y^2=4a$ and $y^2−z^2=4b$ equations, you don't need to do any rearranging. Instead, you can just add these $2$ equations, as the $y^2$ terms cancel, to more directly get your result of $x^2−z^2=4a+4b=4(a+b)$. This will make your proof a bit shorter & more succinct.
|
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|
Finding the determinant of a matrix given by three parameters.
Show that for $a,b,c\in\mathbb R$ $$\begin{vmatrix}b^2+c^2&ab&ac\\ba&c^2+a^2&bc\\ca&cb&a^2+b^2\end{vmatrix} = \begin{vmatrix}0&c&b\\c&0&a\\b&a&0\end{vmatrix}^2 = 4a^2b^2c^2. $$
There must be some trick, like using elementary row operations, to get the determinant into that form, but I am not seeing it. And directly computing the determinant by the cofactor expansion looks very nasty. So is there a simpler way to compute this determinant?
|
Hint:
$$\triangle=\begin{vmatrix}b^2+c^2&ab&ac\\ba&c^2+a^2&bc\\ca&cb&a^2+b^2\end{vmatrix} = \dfrac1a\begin{vmatrix}a(b^2+c^2)&ab&ac\\a(ba)&c^2+a^2&bc\\a(ca)&cb&a^2+b^2\end{vmatrix} $$
$$C_1'=C_1-bC_2-cC_3$$
$$\triangle=\dfrac1a\begin{vmatrix}0&ab&ac\\-2bc^2&c^2+a^2&bc\\-2bc^2&cb&a^2+b^2\end{vmatrix}=-2bc\begin{vmatrix}0&b&c\\c&c^2+a^2&bc\\b&cb&a^2+b^2\end{vmatrix}=2bc\begin{vmatrix}0&c&b\\c&bc&c^2+a^2\\b&a^2+b^2&cb\end{vmatrix}$$
$$=2\begin{vmatrix}0&bc&bc\\c&b^2c&(c^2+a^2)c\\b&b(a^2+b^2)&(bc)c\end{vmatrix}$$
$$=2bc(c)(b)\begin{vmatrix}0&1&1\\1&b^2&c^2+a^2\\1&a^2+b^2&(c)c\end{vmatrix}$$
|
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|
Matrix with 2 unknown variables, how to solve? Given that
$$Ax=b$$
where $$x = \begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix},$$
$$A= \begin{bmatrix}
1 & 1 & 3 \\
1 & 2 & 4 \\
k_1 & 3 & 5
\end{bmatrix},$$
and
$$ b= \begin{bmatrix}
2\\
3\\
k_2
\end{bmatrix}$$
and $k_1 ,k_2\in\mathbb R$, I need to find the value of $k_1$ and $k_2$, so that $Ax=b$ works out.
when its
*
*inconsistent
*1 solution
*inf many solutions
and solve the system when its inf many solutions
I don't understand how to solve the system for both unknown variables. I tried to solve the matrix as I would if there was only one variable and I got
$$\begin{bmatrix}-k_1 & -2 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix}0 \\ 1 \\ k_2 - 5\end{bmatrix}$$
but this doesn't seem to make any sense to me because the R1 still has another unknown variable
if $k_2=5$ then the matrix would have inf many solutions, but still, there is $k_1$.
I would be grateful for any tip or solution.
|
I'll call $k_1=x$, $k_2=y$ for simplicity. I don't think you did your row reduction right. you have
$
\begin{bmatrix}
1&1&3&2\\
1&2&4&3\\
x&3&5&y
\end{bmatrix}
\xrightarrow{R_2-R_1}
\begin{bmatrix}
1&1&3&2\\
0&1&1&1\\
x&3&5&y
\end{bmatrix}
\xrightarrow{R_3-xR_1}
\begin{bmatrix}
1&1&3&2\\
0&1&1&1\\
0&3-x&5-3x&y-2x
\end{bmatrix}
\xrightarrow{R_1-R_2}
\begin{bmatrix}
1&0&2&1\\
0&1&1&1\\
0&3-x&5-3x&y-2x
\end{bmatrix}
\xrightarrow{R_3-(3-x)R_2}
\begin{bmatrix}
1&0&2&1\\
0&1&1&1\\
0&0&2-2x&y-x-3
\end{bmatrix}
$
For the system to inconsistent, you need $2-2x=0$ and $y-x-3\ne0$. That is, $x=1$ and $y\ne4$. When $x\ne1$ the system is consistent, and it will have a unique solution. When $x=1$ and $y-x-3=0$, the last row will be zero and so there will be infinitely many solutions. In summary,
*
*$x\ne1$: consistent, unique solution.
*$x=1$, $y=4$: consistent, infinitely many solutions.
*$x=1$, $y\ne4$: inconsistent.
In the case where $x=1$, $y=4$, the first two rows spell
$$
x_1+2x_3=1,\ \ \ x_2+x_3=1.
$$
So you can take $x_3=t$ as a parameter, and then the solution is
$$
x_1=1-2t,\ \ \ x_2=1-t,\ \ \ x_3=t.
$$
|
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|
How to solve this quadratic equation (with $x$ represented by a fraction containing a square root)? If $x-\frac{4}{5} = \pm \frac{\sqrt{31}}{5}$, how can I find the values of a and b in the following equation?
$$5x^2+ax+b=0?$$
I've tried substituting $(\frac{4}{5} \pm \frac{\sqrt{31}}{5})$ for $x$ and got totally confused there.
|
Square both sides, multiply both sides by $5$, then move all terms to the left side. Note that your original equation is really two equations, but after squaring they are the same:
\begin{align}x-\tfrac45=\pm\tfrac{\sqrt{31}}{5}
&\implies \left(x-\tfrac45\right)^2=\left(\pm\tfrac{\sqrt{31}}{5}\right)^2 \\
&\implies x^2-\tfrac85 x + \tfrac{16}{25}=\tfrac{31}{25} \\
&\implies 5x^2-8 x + \tfrac{16}{5}=\tfrac{31}{5} \\
&\implies 5x^2-8 x - \tfrac{15}{5} = 0 \\
&\implies 5x^2-8 x - 3 = 0.
\end{align}
So $a=-8$ and $b=-3$.
|
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|
Turning $\frac{2x-y}{2x+y}\frac1{\left(1-\frac{y}{2x}\right)^2}$ into $\frac1{\frac{y}{x}(1-\frac{y}{2x})+(1-\frac{y}{2x})^2}$ I stopped at this equality in the solution of a question in my book
$$\left(\frac{2x-y}{2x+y}\right)\left(\frac{1}{\left(1-\frac{y}{2x}\right)^2}\right) = \frac{1}{\big( \frac{y}{x}\big) \Big(1-\frac{y}{2x} \Big) +\Big( 1-\frac{y}{2x}\Big)^2}$$
I already verified that starting in R.H.S will give the L.H.S but I am wondering how can we simplify the "L.H.S" to arrive to the R.H.S ?
|
Divide top and bottom by $2x-y$. The numerator becomes 1 and the denominator becomes
$$\frac{2x+y}{2x-y}\left(1-\frac{y}{2x}\right)^2$$
$$=\frac{1-\frac{y}{2x}+\frac{y}{x}}{1-\frac{y}{2x}}\left(1-\frac{y}{2x}\right)^2$$
$$=\frac{y}{x}\left(1-\frac{y}{2x}\right)+\left(1-\frac{y}{2x}\right)^2$$
|
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|
Given sequence $(a_n)$ such that $a_{n + 2} = 4a_{n + 1} - a_n$. Prove that $\exists \frac{a_i^2 + 2}{a_j}, \frac{a_j^2 + 2}{a_i} \in \mathbb N$.
Given sequence $(a_n)$ such that $a_1 = a_2 = 1$ and $$\large a_{n + 2} = 4a_{n + 1} - a_n, \forall n \ge 2$$
Prove that $$\large \exists i,j \in \mathbb N \colon \frac{a_i^2 + 2}{a_j}, \frac{a_j^2 + 2}{a_i} \in \mathbb N$$
A fundamental observation is that all numbers in the sequence are odd positive integers.
Let $\gcd(a_i, a_j) = d \ (d \ne 2)$
$$\implies \gcd(a_i, a_j - a_i^2) = d \implies \gcd(a_i, 2) = d \implies \gcd(a_i, a_j) = d = 1$$
We have that $$\left\{ \begin{align} a_j &\mid a_i^2 + 2\\ a_i &\mid a_j^2 + 2\end{align} \right. \implies \text{lcm}(a_i, a_j) \mid a_i^2 + a_j^2 + 2 \implies a_ia_j \mid a_i^2 + a_j^2 + 2$$
$$\implies a_i^2 + a_j^2 + 2 = ka_ia_j \ (k \in \mathbb N)$$
If $a_i = a_j$ then we have $a_i^2 \mid 2(a_i^2 + 1)$ and thus $a_i = a_j = 1$.
Otherwise, suppose that $1 \le a_i < a_j$ is a solution.
We know that $x_1 = a_j$ would be a solution to the quadratic equation $x^2 - ka_ix + (a_i^2 + 2) = 0$ and the other solution would be $x_2 = ka_i - a_j \ (\in \mathbb N) = \dfrac{a_i^2 + 2}{a_j} \ (< a_i)$, implying that $x_2$ is an integer and we get a smaller solution $(x_2, a_j)$.
Hence, starting with any solution $(a_i, a_j)$, we can always backtrack along with smaller solutions until $a_i = 1$.
This gives us $a_j \mid a_j^2 + 3$ or that $a_j \mid 3$. Hence $a_j = 1$, $a_j = 2$ or $a_j = 3$.
However, $a_j$ is an odd natural number $\implies a_j = 1$ or $a_j = 3$.
It can calculated that if $(a_i, a_j) = (1, 1)$ and $(a_i, a_j) = (1, 3)$, we have that $$\dfrac{a_i^2 + a_j^2 + 2}{a_ia_j} = \dfrac{1^2 + 1^2 + 2}{1 \cdot 1} = \dfrac{1^2 + 3^2 + 2}{1 \cdot 3} = 4$$
Hence, the value of $k = 4$ stays the same throughout.
Hence, we must always have $k = 4$, or that $a_i^2 + a_j^2 + 2 = 4a_ia_j \implies 4a_ia_j - (a_i^2 + a_j^2) = 2$
$$\implies \dfrac{1}{2}(a_i + a_j)^2 - \dfrac{3}{2}(a_j - a_i)^2 = 2 \implies \left(\dfrac{a_i + a_j}{2}\right)^2 - 3 \cdot \left(\dfrac{a_j - a_i}{2}\right)^2 = 1$$
Since $a_i$ and $a_j$ are odd positive integers, $x = \dfrac{a_i + a_j}{2}$ and $y = \left|\dfrac{a_j - a_i}{2}\right|$ are natural numbers.
The Pell equation $x^2 - 3y^2 = 1$ has the two nonnegative integer solutions $(x_1, y_1)$ and $(x_2, y_2)$ where $x_1 + y_1$ and $x_2 + y_2$ are respectively at their smallest and second smallest values are $(1, 0)$ and $(2, 1)$.
Consider the sequences $(x_n)$ and $(y_n)$ satisfying $$\left\{ \begin{align} x_1 = 1, x_2 = 2, x_{n + 2} &= 4x_{n + 1} - x_n\\ y_1 = 0, y_2 = 1, y_{n + 2} &= 4y_{n + 1} - y_n \end{align} \right. \ (n \in \mathbb Z^+, n \ge 2)$$
There exists $m \in \mathbb Z^+$ such that $(x, y) = (x_m, y_m)$
$\implies (a_i, a_j)$ is a permutation of $(x_m + y_m, x_m - y_m)$. Let $\left\{ \begin{align} u_n = x_n + y_n\\ v_n = x_n - y_n \end{align} \right. \ (n \in \mathbb Z^+)$.
We have that
$$\left\{ \begin{align} u_1 = 1, u_2 = 3, u_{n + 2} &= 4u_{n + 1} - u_{n}\\ v_1 = 1, v_2 = 1, v_{n + 2} &= 4v_{n + 1} - v_n \end{align} \right. \implies u_{n - 1} = v_n = a_n \ (n \in \mathbb Z^+, n \ge 2)$$
Then I don't know what to do next. Great.
|
The condition holds true for all pairs $a_i, a_{i+1}$.
In fact, the relation $a_i^2+2=a_{i-1}a_{i+1}$ holds for all $i$, from which the above result follows.
This can be proven by induction. Assume $a_i^2+2=a_{i-1}a_{i+1}$ is true for $1<i<n$. That makes
$$
\begin{split}
a_{n-1}a_{n+1}
& = a_{n-1}\cdot(4a_n-a_{n-1})
= 4a_{n-1}a_n-a_{n-1}^2 \\
& = 4a_{n-1}a_n-(a_{n-2}a_n-2)
= (4a_{n-1}-a_{n-2})\cdot a_n + 2 \\
& = a_n^2 + 2
\end{split}
$$
for $n>2$, and you just have to verify that the relation holds for $i=2$ which is the first case not covered by the induction step.
|
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|
Factoring $1 - y-x^2-y^2-yx^2+y^3$ I was struggling to factorize this expression, I need it cleaned to bracket form but I was struggling to understand the process to get the answer? Is there a special formula or technique for factorizing these sorts of expressions?
$$1 - y-x^2-y^2-yx^2+y^3$$
Answer: $$ \ ( 1+y)[(1-y)^2-x^2] $$
|
Factoring multivariable polynomials usually needs a bit of creativity. However, when looking for an approach, you can always think of multivariable polynomials as polynomials over a single variable whose coefficients are another polynomials!
With a bit of creativity: $$\begin{align}
1 - y-x^2-y^2-yx^2+y^3 &= \underbrace{(1-y-y^2 +y ^3)}_{\text{only y dependent}}-\overbrace{(x^2 +yx^2)}^\text{also x dependent}\\ \\
&= (1-y-y^2 +y ^3)-x^2(1+y). \\\\ \end{align}$$
If we plug $y = -1$, we have $(1+1-1-1)+x^2(1-1) \equiv 0.$ So we can divide this expression by $(1+y)$, and obtain the desired result or factorize even further to get
$$\begin{align}(1+y)( (y^2-2y+1) - x^2) &= (1+y)((y-1)^2-x^2)\\\\
&= \boxed{(1+y)(y-1-x)(y-1+x).}\end{align}$$
A more deep argument of why I tried this method: note that this expression is cubic over $y$. So if you can factorize this polynomial $p(x,y)$ into $q(x,y)r(x,y)$, either $q$ has degree $1$ over $y$ or $q$ has degree $2 \implies r$ has degree $1$. So there's a rational function $g(x)$ such that $p(x,g(x)) = 0$. So we need a factor of the form $k(x)y - l(x)$.
Note that you could use this last argument to deduce the same thing for $x$. So getting $x^2 = k(y)/l(y)$ is also a good guess.
|
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|
Prove that $a^{4b}+b^{4a}\geq \frac{1}{2}$ Inspired by a problem of Vasile Cirtoaje I propose this :
Let $a,b>0$ such that $a+b=1$ then we have :
$$a^{4b}+b^{4a}\geq \frac{1}{2}$$
I compute the derivative of $f(x)=x^{4(1-x)}+(1-x)^{4x}$ on $]0,1]$ we get :
$$ f'(x)=x^{4 (1 - x)} (\frac{4 (1 - x)}{x} - 4 \log(x)) + (1 - x)^{(4 x)} (4 \log(1 - x) - \frac{4 x}{1 - x})$$
If we denote by $g(x)$ the function :
$$g(x)=x^{4 (1 - x)} (\frac{4 (1 - x)}{x} - 4 \log(x))$$
We can rewrite the derivative as :
$$f'(x)=g(x)-g(1-x)$$
So it's remains to show that $g(x)\geq g(1-x)$ or $g(x)\leq g(1-x)$
So it remains to show that $g(x)$ is increasing or decreasing .
After that I'm stuck...
Any helps are very appreciated !
Thanks a lot for your time .
|
Alternative proof:
I give a proof following @Malper's nice idea.
Fact 1: Let $x\in (0, 1/2]$ and $t \in (0, 1/2]$. Then $x^{4 - 4t} + (1 - x)^{4t} \ge \frac12$.
The proof is given at the end.
By Fact 1, we have
$x^{4 - 4x} + (1 - x)^{4x} \ge \frac12$
for all $x$ in $(0, 1/2]$.
We are done.
Proof of Fact 1: Let $f(x, t) = x^{4 - 4t} + (1 - x)^{4t}$. We have
\begin{align}
\frac{\partial f}{\partial t}
&= 4(1 - x)^{4t}\ln (1 - x) - 4x^{4 - 4t}\ln x\\
&= 4x^{-4t}\Big([x(1 - x)]^{4t}\ln(1 - x) - x^4\ln x\Big)\\
&\le 4x^{-4t}\Big([x(1 - x)]^{4\cdot \frac12}\ln(1 - x) - x^4\ln x\Big)\\
&= 4x^{2 - 4t}\Big((1 - x)^2\ln(1 - x) - x^2\ln x\Big)\\
&\le 0
\end{align}
where we have used $(1 - x)^2\ln(1 - x) - x^2\ln x \le 0$.
Note: Let $g(x) = (1 - x)^2\ln(1 - x) - x^2\ln x$. We have $g''(x) = 2\ln(1 - x) - 2\ln x \ge 0$ on $(0, 1/2]$. Thus, $g(x)$ is convex on $(0, 1/2]$. Also, $g(0^{+}) = 0$ and $g(1/2) = 0$. Thus, $g(x) \le 0$ on $(0, 1/2]$.
Thus, we have $f(x, t) \ge f(x, 1/2) = x^2 + (1 - x)^2 \ge 1/2$.
We are done.
|
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|
Sum of matrices value Let $\mspace{10mu}A,B\in M^{4\times 4}(\mathbb{R})\mspace{10mu}, $
$A=\begin{pmatrix}
1 &0 &2 &-1 \\
0 &1 &-1 &1 \\
1 &1 &1 &0 \\
1 &-1 &3 &-2
\end{pmatrix} \mspace{10mu} \mathrm{and} \mspace{10mu} \mathrm{rk}(B)=1$
What values may have $\mspace{10mu}\mathrm{rk}(A+B)\mspace{10mu}?$
I only figured out that $\mathrm{rk}(A)=2$.
|
It cannot be $A = -B$ since $\operatorname{rank}(A) = 2$ and $\operatorname{rank}(B) = 1$ so $A+B \ne 0$. Therefore by the subadditivity of rank we have
$$1 \le \operatorname{rank}(A+B) \le \operatorname{rank}(A) + \operatorname{rank}(B) = 3$$
so $\operatorname{rank}(A+B) \in \{1,2,3\}$.
All possibilities are indeed attainable, consider $B$ equal to
$$\begin{pmatrix}
0 &0 &0 &0 \\
0 &0 &0 &0 \\
0 &0 &1 &0 \\
0 &0 &0 &0
\end{pmatrix}, \quad\begin{pmatrix}
0 &0 &0 &0 \\
1 &1 &1 &1 \\
1 &1 &1 &1 \\
-1 &-1 &-1 &-1
\end{pmatrix}, \quad\begin{pmatrix}
0 &0 &0 &0 \\
0 &-1 &1 &-1 \\
0 &-1 &1 &-1 \\
0 &1 &-1 &1
\end{pmatrix}$$
respectively.
|
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|
Find $\left( \frac{2}{7} \right) \left(\log_2 \frac{y}{x} \right)$ if- If $$x= (\cos1°) (\cos2°) (\cos3°) .............(\cos89°)$$
and
$$y=(\cos 2°)(\cos 6°)(\cos 10°).............(\cos 86°)$$
Then what is the integer nearest to $$\left( \frac{2}{7} \right) \left( \log_2 \frac{y}{x} \right) ?$$
I tried putting it in $\Pi$ form but had no luck.
|
$$
x = \Bigl(\cos(1^{\circ})\cos(2^{\circ})\cdots\cos(44^{\circ})\Bigr) \cdot \cos(45^{\circ}) \cdot \Bigl(\sin(44^{\circ})\sin(43^{\circ})\cdots\sin(1^{\circ})\Bigr)\tag{1}
$$
$$
=\dfrac{1}{\sqrt{2}} \cdot 2^{-44} \sin(2^{\circ})\sin(4^{\circ})\cdots\sin(88^{\circ}).
$$
Next step:
denote $$z=\sin(2^{\circ})\sin(4^{\circ})\cdots\sin(88^{\circ}).\tag{2}$$
Then
$$
z = \Bigl(\sin(2^{\circ})\sin(4^{\circ})\cdots\sin(44^{\circ})\Bigr) \cdot \Bigl(\cos(44^{\circ})\cos(42^{\circ})\cdots\cos(2^{\circ})\Bigr)\tag{3}
$$
$$
=2^{-22}\sin(4^{\circ})\sin(8^{\circ})\cdots\sin(88^{\circ})
$$
$$
=2^{-22}\cos(86^{\circ})\cos(82^{\circ})\cdots\cos(2^{\circ})
$$
$$
=2^{-22}y.
$$
Therefore:
$$
x = \dfrac{1}{\sqrt{2}}\cdot 2^{-44}\cdot 2^{-22}y = 2^{-66.5}y.\tag{4}
$$
Hence
$$
\log_2(y/x) = 66.5;
$$
$$
\dfrac{2}{7}\log_2(y/x) = \dfrac{2}{7}\cdot \dfrac{133}{2} = 19.\tag{5}
$$
|
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|
Minimum value of the given function- Find the minimum value of-
$$\frac{(x+\frac{1}{x})^6-(x^6+\frac{1}{x^6})-2}{(x+\frac{1}{x})^3+(x^3+\frac{1}{x^3})}$$
I tried opening the brackets and trying to cancel the terms and using AM-GM.
|
Write $z=x+\frac1x$. Then the function is equal to
$$\frac{z^6-(z^6-6z^4+9z^2-2)-2}{z^3+z^3-3z}=\frac{6z^4-9z^2}{2z^3-3z}=\frac{3z^2(2z^2-3)}{z(2z^2-3)}$$
Now $z$ can never assume a value in $(-2,2)$ and $2z^2-3=0\implies z=\sqrt{3/2}\in(-2,2)$, so we can cancel and get $3z$. Since $z$ has no minimum over the reals, so does the original function, but over the positive reals it does have a minimum of $6$.
|
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|
Extrema of $f(x,y) = xy\sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}}$ From Demidovich
Find the extrema of the function of $z = xy \sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}}$
What I've done so far:
I've make $z_{x}'= 0$ and $z_{y}' = 0$, wich gives me:
$$\frac{\partial z}{\partial x} = y \sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}} - \frac{yx}{a^2 \sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}}} = 0$$
$$\frac{\partial z}{\partial y} = x \sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}} - \frac{yx}{b^2 \sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}}} = 0$$
And I've tried to simplify, and I got:
$$\frac{y a^2 (1 - \frac{x^2}{a^2} - \frac{y^2}{b^2} ) - xy}{a^2 \sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}}} = 0$$
$$\frac{x b^2 (1 - \frac{x^2}{a^2} - \frac{y^2}{b^2} ) - xy}{b^2 \sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}}} = 0$$
Now the problem is, some cubic terms appear and I don't know how to solve this and find the critical points to proceed with the exercise. Of course the denominator cannot be zero, so my system is composed only by the numerators.
I'm pretty sure I'm doing something wrong, 'cause this seems too much trouble for an introductory exercise (it's one of the firsts in the chapter)
|
With the polar coordinates $\frac xa = r\cos t$ and $\frac yb =r \sin t$,
$$z=\frac12ab r^2\sin 2t \sqrt{1-r^2}$$
For any given $r\le 1$,
$$|z(r)|\le \frac12abr^2 \sqrt{1-r^2}$$
Then, set $(r^2 \sqrt{1-r^2})'=0$ to locate the extreme at $r^2=\frac23$. Thus,
$$-\frac{ab}{3\sqrt3}\le z \le \frac{ab}{3\sqrt3}$$
|
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|
Summation to n terms of series \begin{gather}
In\ a\ problem\ given\ to\ us\ by\ our\ teacher,\ one\ part\ of\ it\ requires\ the\ summation\ of\ the\ \notag\\
following\ sequence\ to\ a\ finite\ number\ of\ terms\ ( n) , \notag\\
t_{r\ } \ =\ \frac{1}{r\times 2^{r}} \notag\\
\notag\\
S\ =\ \frac{1}{1\times 2} +\frac{1}{2\times 2^{2}} +\frac{1}{3\times 2^{3}} +....\ \frac{1}{r\times 2^{r}} \notag\\
\notag\\
I\ tried\ the\ following\ approach\ :\ \notag\\
\notag\\
\ S\ =\ \frac{1}{1\times 2} +\frac{1}{2\times 2^{2}} +\frac{1}{3\times 2^{3}} +....\ \frac{1}{r\times 2^{r}} \ \ \ ( i) \notag\\
\ \frac{S}{2} \ =\ \frac{1}{1\times 2^{2}} +\frac{1}{2\times 2^{3}} +\frac{1}{3\times 2^{4}} +....\ \frac{1}{( r-1) \times 2^{r}} \ +\ \frac{1}{r\times 2^{r+1}} \ \ \ \ ( ii) \notag\\
( i) \ -\ ( ii) \ \notag\\
\frac{S}{2} \ =\ \frac{1}{1\times 2} \ -\left( \ \left( 1-\frac{1}{2}\right) 2^{-1} \ +\ \left(\frac{1}{2} \ -\ \frac{1}{3}\right) 2^{-2} \ +...\left(\frac{1}{r-1} -\frac{1}{r}\right) 2^{-r}\right) \ -\ \frac{2^{-r}}{r}\\
\notag\\
\frac{S}{2} \ =\ \frac{1}{1\times 2} \ -\left( \ \left(\frac{1}{2}\right) 2^{-1} \ +\ \left(\frac{1}{6}\right) 2^{-2} \ +...\left(\frac{1}{r( r-1)}\right) 2^{-r}\right) \ -\ \frac{2^{-r}}{r} \notag\\
Here\ I\ got\ stuck\ ,\ as\ the\ series\ generated\ above\ again\ generates\ another\ similar\ sequence.\ \notag\\
\notag\\
\notag\\
\notag
\end{gather}
I don't think the sum telescopes as well. Can calculus be used in some way to calculate the sum?
I'll just add the actual problem, in case it offers some help :
\begin{gather*}
t_{r} \ =\ \frac{r+2}{r( r+1)} \ 2^{-r}\\
\\
\\
\end{gather*}
|
Let $$f(x):=\sum_{r=1}^n\frac{x^r}r$$ and
$$f'(x)=\sum_{r=1}^n x^r=\frac{1-x^r}{1-x}.$$
Then by integration,
$$f(\tfrac12)=-\log(\tfrac12)-\int_0^{1/2}\frac{x^r}{1-x}dx.$$
The last integral is known as an incomplete Beta, and has no closed-form expression (other than the explicit summation for integer $r$).
|
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|
Solve in $\mathbb{R^{3}}$ : $\begin{cases}(1+4x^{2})y=4z^{2}\\(1+4y^{2})z=4x^{2}\\(1+4z^{2})x=4y^{2}\end{cases}$ Solve the system in $\mathbb{R^{3}}$ :
$$\begin{cases}(1+4x^{2})y=4z^{2}\\(1+4y^{2})z=4x^{2}\\(1+4z^{2})x=4y^{2}\end{cases}$$
My try :
By imaging I see $(\frac{1}{2},\frac{1}{2},\frac{1}{2})$ is a solution!
From a first equation :
$$x=\frac{1}{2}\sqrt{\frac{4z^{2}}{y}-1}$$
So by second equation :
$$y=\frac{1}{2}\sqrt{\frac{\frac{4z^2}{y}-1}{z}-1}$$
But after applied I get difficult equation for $z$ ?
|
We see that all are nonnegative.
If one of them is $0$ then all are $x=y=z=0$. Now suppose all are $>0$.
Now $1+4x^2\geq 4x$ so $$4z^2=y(1+4x^2)\geq 4xy\implies z^2\geq xy$$
In the same manner we get $y^2\geq xz$ and $x^2\geq yz$ so, if say $x^2>yz$ we get $$x^2y^2z^2>xyxzzy$$
a contradiction. So $x^2=yz$ and similary for other, so we have equality sign every where and thus $$1+4x^2 = 4x\implies x={1\over 2} = y=z$$
|
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|
Every natural number is covered by consecutive numbers that sum to a prime power. Conjecture. For every natural number $n \in \Bbb{N}$, there exists a finite set of consecutive numbers $C\subset \Bbb{N}$ containing $n$ such that $\sum\limits_{c\in C} c$ is a prime power.
A list of the first few numbers in $\Bbb{N}$ has several different covers by such consecutive number sets.
One such is:
3 7 5 13 8 19 11 25 29 16 37 41 49 53
___ ___ _ ___ _ ____ __ _____ _____ __ __ _____ _____ _____ _____ __
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29
___ ___ _____ _____ __ ________
11 17 31 43 81
59 71 3^5
___ __ _____ _________________
30 31 32 33 34 35 36 37 38 39 40 41 42 43 .....
_____ _____ _____
61 67 73
Has this been proved already?
|
While nickgard's answer shows how to solve the problem using sums being squares of increasing primes, this answer shows how to do it using the sums being just odd powers of $3$.
As suggested in joriki's question comment, for any integers $1 \le j \le k$, you have
$$\begin{equation}\begin{aligned}
\sum_{i=j}^{k}i & = \sum_{i=1}^{k}i - \sum_{i=1}^{j-1}i \\
& = \frac{k(k+1)}{2} - \frac{(j-1)(j)}{2} \\
& = \frac{k^2 + k - j^2 + j}{2} \\
& = \frac{(k-j)(k+j) + k + j}{2} \\
& = \frac{(k+j)(k-j+1)}{2}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Consider the ranges $\left[\frac{3^m + 1}{2},\frac{3^{m+1} - 1}{2}\right]$ for $m = 0, 1, 2, \ldots$. The union of these disjoint subsets cover all positive integers. Thus, for any $n \ge 1$, there's a unique $m$ where $n \in \left[\frac{3^m + 1}{2},\frac{3^{m+1} - 1}{2}\right]$. For that $m$, since $\frac{5\left(3^{m}\right)-1}{2} \gt \frac{3^{m+1} - 1}{2}$, you can have $j = \frac{3^m + 1}{2}$ and $k = \frac{5\left(3^{m}\right)-1}{2}$ with $n \in [j,k]$. Using this in \eqref{eq1A} gives
$$\begin{equation}\begin{aligned}
\sum_{i=j}^{k}i & = \frac{(k+j)(k-j+1)}{2} \\
& = \frac{\left(\frac{5\left(3^{m}\right)-1}{2}+\frac{3^m + 1}{2}\right)\left(\frac{5\left(3^{m}\right)-1}{2}-\frac{3^m + 1}{2}+1\right)}{2} \\
& = \frac{\left(\frac{6\left(3^{m}\right)}{2}\right)\left(\frac{4\left(3^{m}\right)}{2}-\frac{2}{2}+1\right)}{2} \\
& = \frac{\left(3\left(3^{m}\right)\right)\left(2\left(3^{m}\right)\right)}{2} \\
& = \left(3^{m+1}\right)\left(3^{m}\right) \\
& = 3^{2m+1}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
The first few examples for $m = 0, 1$ and $2$ are
$$1 + 2 = 3 = 3^{1} \tag{3}\label{eq3A}$$
$$2 + 3 + 4 + 5 + 6 + 7 = 27 = 3^{3} \tag{4}\label{eq4A}$$
$$5 + 6 + \ldots + 21 + 22 = 243 = 3^{5} \tag{5}\label{eq5A}$$
|
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|
Prove that : $\frac{10^{18n+12}-7}{3}\equiv 0\pmod{19}$ Prove that : ($n\in\mathbb{N}$)
$$\frac{10^{18n+12}-7}{3}\equiv 0\pmod{19}$$
I know that from little theorem :
$$10^{18}\equiv 1\pmod{19}$$
So :
$$10^{18k}\equiv 1\pmod{19}$$
Now i will go to this step if $\operatorname{correct}$
$$10^{12}=100^{6}\equiv 5^{6}\pmod{19}$$
$$5^{6}=(5^{2})^{3}\equiv 6^{3}\pmod{19}$$
$$6^{3}\equiv 7\pmod{19}$$
So :
$$10^{18k+12}-7\equiv 0\pmod{19}$$
But my be :
$$\frac{10^{18n+12}-7}{3}\not\equiv 0\pmod{19}$$
|
Let $a=10^{18k+12}-7$.
If $19|a$ and $3|a$, then $19|(\frac a3)3$, and $19\nmid3$,
so, by Euclid's lemma, $19|\frac a3$.
|
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|
How to prove the inequality $\sqrt{\frac{a^2+b^2}2}+\frac{a+b}2\geq\frac{a^2+b^2}{a+b}+\sqrt{ab}$? I want to prove that for all $a,b>0$: $$\sqrt{\frac{a^2+b^2}2}+\frac{a+b}2\geq\frac{a^2+b^2}{a+b}+\sqrt{ab}.$$
My attempts:
Failed attempt. We know by QM-AM inequality that $\sqrt{\frac{a^2+b^2}2}\geq\frac{a+b}2$ so it would be enough to prove $$a+b\geq\frac{a^2+b^2}{a+b}+\sqrt{ab}$$ but this is wrong, try $a=1$ and $b=4$.
|
Let
$$\sqrt{\frac{a^2+b^2}{2}}+\frac{a+b}{2} \ge \frac{a^2+b^2}{a+b}+\sqrt{ab}$$
$$\implies \sqrt{\frac{a^2+b^2}{2}}\ge \frac{a^2+b^2}{a+b}-\frac{a+b}{2}+\sqrt{ab}$$
Let $a=1, b=t^2$, then
$$\implies \sqrt{\frac{1+t^4}{2}} \ge \frac{1+t^4}{1+t^2}-\frac{1+t^2}{2}+t=\frac{(1-t^2)^2+2t(1+t^2)}{2(1+t^2)}$$ So now we need to prove that
$$F=2(1+t^4)(1+t^2)^2-[(1-t^2)^2+2t(1+t^2)]^2 \ge 0$$
By expanding it we get
$$F=t^8-4t^7+4t^6+4t^5-10t^4+4t^3+4t^2-4t+1=(t-1)^2(t+1)^6 \ge 0.$$
Hence proved and the equality hold when $t=1$ $(a=b).$
|
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|
Given positives $a, b, c$, prove that $\frac{a}{(b + c)^2} + \frac{b}{(c + a)^2} + \frac{c}{(a + b)^2} \ge \frac{9}{4(a + b + c)}$.
Given positives $a, b, c$, prove that $$\large \frac{a}{(b + c)^2} + \frac{b}{(c + a)^2} + \frac{c}{(a + b)^2} \ge \frac{9}{4(a + b + c)}$$
Let $x = \dfrac{b + c}{2}, y = \dfrac{c + a}{2}, z = \dfrac{a + b}{2}$
It needs to be sufficient to prove that $$\sum_{cyc}\frac{\dfrac{a + b}{2} + \dfrac{b + c}{2} - \dfrac{c + a}{2}}{\left(2 \cdot \dfrac{c + a}{2}\right)^2} \ge \frac{9}{\displaystyle 4 \cdot \sum_{cyc}\dfrac{c + a}{2}} \implies \sum_{cyc}\frac{z + x - y}{y^2} \ge \frac{9}{y + z + x}$$
According to the Cauchy-Schwarz inequality, we have that
$$(y + z + x) \cdot \sum_{cyc}\frac{z + x - y}{y^2} \ge \left(\sum_{cyc}\sqrt{\frac{z + x - y}{y}}\right)^2$$
We need to prove that $$\sum_{cyc}\sqrt{\frac{z + x - y}{y}} \ge 3$$
but I don't know how to.
Thanks to Isaac YIU Math Studio, we additionally have that $$(y + z + x) \cdot \sum_{cyc}\frac{z + x - y}{y^2} = \sum_{cyc}(z + x - y) \cdot \sum_{cyc}\frac{z + x - y}{y^2} \ge \left(\sum_{cyc}\frac{z + x - y}{y}\right)^2$$
We now need to prove that $$\sum_{cyc}\frac{z + x - y}{y} \ge 3$$, which could be followed from Nesbitt's inequality.
I would be greatly appreciated if there are any other solutions than this one.
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By Hölder's inequality:
$$ ..... \geq \frac{(a+b+c)^3}{4(ab+bc+ac)^2} \geq \frac{9(a+b+c)^3}{4(a+b+c)^4}$$
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|
Find the coordinates of $3x^3-2x^2+x+4$ in the base $(3,-1+x,-2+x^2,-3+x^3)$ I did:
$$\alpha_1(3)+\alpha_2(-1+x)+\alpha_3(-2+x^2)+\alpha_4(-3+x^3) = \\
x^3(\alpha_4)+x^2(\alpha_3)+x(\alpha_2)+(3\alpha_1-\alpha_2-2\alpha_3-3\alpha_4)$$
$$\left\{
\begin{array}{c}
\alpha_4 = 3 \\
\alpha_3=-2\\
\alpha_2=1 \\
3\alpha_1-\alpha_2-2\alpha_3-3\alpha_4=4
\end{array}
\right. \\
\Leftrightarrow \\
\left\{
\begin{array}{c}
\alpha_4 = 3 \\
\alpha_3=-2\\
\alpha_2=1 \\
\alpha_1=2
\end{array}
\right. \\$$
The coordinates are $(2,1,-2,3)$. Is this correct?
Bonus question: Determine the vector $(1,-2,3,-3)$ in the given base.
I did:
$$(1)*(3)+(-2)*(-1+x)+(3)(-2+x^2)+(-3)*(-3+x^3) = \\
3+2-2x-6+3x^2+9-3x^3=\\
8-2x+3x^2-3x^3$$
Is this correct?
|
Looks like you did the first part right, barring a careless mistake.
For the second part, $x^3-2x^2+3x-3=3a+(-1+x)b+(-2+x^2)c+(-3+x^3)d$. So solve as above.
$d=1, c=-2, b=3$ and $a=-1/3$.
So $(-1/3, 3, -2, 1)$.
This problem can also be done by inverting the change of basis matrix: $\begin{pmatrix}3&-1&-2&-3\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}$, and multiplying by $(1,-2,3,-3)^t$.
|
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|
How to deal with negative area when evaluating a definite integral
Find the area bounded by the curves $y=2-x^2$ and $x+y=0$
$$
-x=2-x^2\implies x^2-x-2=(x-2)(x+1)=0
$$
My Attempt
$A_1:$ Area above the x-axis and $A_2:$ Area below the x-axis
$$
A_1=\int_{-1}^\sqrt{2}(2-x^2)dx-\int_{-1}^0(-x)dx=\Big[2x-\frac{x^3}{3}\Big]_{-1}^{\sqrt{2}}-\Big[-\frac{x^2}{2}\Big]_{-1}^0\\
=2\sqrt{2}-\frac{2\sqrt{2}}{3}+2-\frac{1}{3}-(\frac{1}{2})=\frac{4\sqrt{2}}{3}+\frac{7}{6}=\frac{8\sqrt{2}+7}{6}\\
A_2=\Big|\int_0^{{2}}(-x)dx\Big|-\Big|\int_\sqrt{2}^2(2-x^2)\Big|=\Big|\Big[-\frac{x^2}{2}\Big]_{0}^{2}\Big|-\Big|\Big[2x-\frac{x^3}{3}\Big]_{\sqrt{2}}^{2}\Big|\\
=|-2|-|4-\frac{8}{3}-2\sqrt{2}+\frac{2\sqrt{2}}{3}|=2-\Big|(\frac{4-4\sqrt{2}}{3})\Big|=2-(\frac{4\sqrt{2}-4}{3})=\bigg|\frac{10-4\sqrt{2}}{3}\bigg|\\
=\frac{20-8\sqrt{2}}{6}\\
A=A_1+A_2=\frac{8\sqrt{2}+7+20-8\sqrt{2}}{6}=\frac{27}{6}=\frac{9}{2}
$$
Reference
$$
A=\int_{-1}^2(2-x^2+x)dx=\bigg[2x-\frac{x^3}{3}+\frac{x^2}{2}\bigg]_{-1}^2=4-\frac{8}{3}+2+2-\frac{1}{3}-\frac{1}{2}\\
=5-\frac{1}{2}=9/2
$$
Isn't the attempt in my reference factually incorrect ?
yet why am I getting a same solutions in my attempt, ie. after splitting the areas and subtracting absolute values ?
$\color{red}{\text{Another Example}}$
Area bounded by the curve $y=x^3$, x-axis at $x=-2$ and $x=1$
Method 1
$$
A=|\int_{-2}^{0}(x^3)dx|+|\int_0^1x^3dx|=|\Big[\frac{x^4}{4}\Big]_{-2}^0|+|\Big[\frac{x^4}{4}\Big]_{0}^1|=|-4|+|\frac{1}{4}|=4+\frac{1}{4}=17/4
$$
Method 2
$$
A=|\int_{-2}^{1}(x^3)dx|=|\bigg[\frac{x^4}{4}\bigg]_{-2}^1|=|\frac{1}{4}-4|=|\frac{-15}{4}|=\frac{15}{4}
$$
Here I think we are not getting the correct answer in method 2 because the area is counted negative, right ?
|
Example 1:
When trying to find the area between curves $f(x)$ and $g(x)$ you can achieve this by integrating the function $H(x) = |f(x) - g(x)|$. However integration of an absolute value function is piecewise. In this case though $f(x) \ge g(x)$ for $-1\le x \le 2$ so integrating $\int_{-1}^2 (f(x) - g(x)) dx$ is valid.
Now we can go into greater (but unnecessary) analysis and do what you did and note that for $-1\le x \le 0$ we have $f(x)\ge g(x) >0$. While for $0< x\le \sqrt 2$ we have $f(x)\ge 0 > g(x)$ and for $\sqrt 2 < x \le 2$ we have $0 > f(x)\ge g(x)$. and we can break it into sections of $\int_{-1}^{\sqrt 2} f(x) d(x)$ (where $f(x)\ge 0$) and subtract $-\int_{-1}^0 g(x)dx$ (where $g(x) \ge 0$) . The we can add then absolute value of $g(x)$ where $g(x) < 0$ so $+\int_{0}^2 [-g(x)]dx$ and the subtract the absolute value of $f(x)$ where $f(x) < 0$ so $-\int_{\sqrt 2}^2 [-f(x)]dx$.
This is exactly the same result.
Example 2:
Here $f(x) = x^3$ and $g(x) = 0$ and we are trying to integrate $H(x)=|f(x)-g(x)|= |x^3|$.
Here it is NOT the case that $f(x) \ge g(x)$ and it is not the case that $|f(x)-g(x)| = f(x)-g(x)$.
So we can solve this by
Method 1: $\int_{-2}^1 |x^3|dx = \int_{-2}^1 \begin{cases}-x^3&x< 0\\x^3&x\ge 0\end{cases} dx=[\begin{cases}-\frac {x^4}4&x< 0\\\frac {x^4}4&x\ge 0\end{cases}]_{-2}^1= (\frac 14 -(-\frac {16}4))$
Method 2: for $-2 \le x < 0$ we have $x^3 < 0$ and for $0\le x \le 1$ we have $x^3 \ge 0$ so $|\int_{-2}^0 x^3 dx| = |0 - \frac {16}{4}|$ and $|\int_{0}^1 x^3 dx| = |\frac 14 - 0|$. And the result follows.
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|
Divide the rational expressions below and simplify as much as possible: The question is :
$$\frac{x^2 + 8}{2x^2 + x - 3}\div\frac{x + 2}{x - 1}$$
the answer I get is
$$\frac{x^2 + 8}{(2x + 3)(x + 2)}$$
but a further factor of $x+2$ is factored out somehow from the $x^2+8$.
I'm having doubts about this. please help, thanks
ANSWER GIVEN BY PROF:
$$\frac{(x+2)\left(x^{2}-2 x+4\right)}{(x-1)(2 x+3)} \cdot \frac{x-1}{x+2}=\frac{x^{2}-2 x+4}{2 x+3}$$
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In the answer given by the prof $(x+2)(x^2-2x+4)=x^3+8$. It seems the problem statement has the wrong exponent on the $x$ at the start. Your answer is correct for the problem as given.
|
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|
Solve the differential equation $(1-x^2)y' - 2xy^2 = xy$ I solved this equation as a Bernoulli equation:
$y' - \frac{x}{1-x^2}y = \frac{2x}{1-x^2}y^2$
I get expression for $y$:
$y = -C(x)\sqrt{|1-x^2|}$
And then I don’t know how to differentiate the module
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$$y' = \frac{x}{1-x^2}y + \frac{2x}{1-x^2}y^2=\frac{x}{1-x^2}(y+2y^2)$$
This is a separable ODE.
$$\frac{2dy}{y+2y^2}=\frac{2x\:dx}{1-x^2}$$
Integrate :
$$2\ln\left|\frac{y}{2y+1}\right|=\ln\left|\frac{1}{1-x^2}\right|+\text{constant}$$
$$\left(\frac{y}{2y+1}\right)^2=C\frac{1}{1-x^2}\quad\begin{cases}
C>0 \quad \text{if}\quad 1-x^2>0\\
C<0 \quad \text{if}\quad 1-x^2<0\end{cases}\quad\text{for real solution}$$
Equivalent to :
$$\left(\frac{y}{2y+1}\right)^2=\frac{1}{c^2|1-x^2|}\qquad c\neq 0$$
Solving for $y$ leads to :
$$\boxed{y=\frac{1}{-2+c\:\sqrt{|1-x^2|}}}$$
The equation remains valid for $c=0$ according to the particular solution $y(x)=-\frac12$ which can be checked directly in the original ODE.
Also the equation remains valid for $|c|=\infty$ according to the particular solution $y(x)=0$ which is an obvious trivial solution of the original ODE.
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|
Is there an easy way to see that ${1\over5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12} > 1$? The sum $$\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12}$$ is just a bit larger than $1$. Is there some clever way to show this other than to add the fractions together by brute-force? For example, is there some way to group terms together and say something like "These terms sum to more than $\frac{1}{3}$, these terms sum to more than $\frac{1}{2}$, and these terms sum to larger than $\frac{1}{6}$, so the whole thing sums to more than $1$"?
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$\dfrac{1}{5}+\dfrac{1}{10}=\dfrac{3}{10}=\dfrac{6}{20}$
$\dfrac{1}{6}+\dfrac{1}{12}=\dfrac{3}{12}=\dfrac{5}{20}$
$\dfrac{1}{7}+\dfrac{1}{8}>\dfrac{2}{8}=\dfrac{5}{20}$
$\dfrac{1}{9}+\dfrac{1}{11}=\dfrac{20}{99}>\dfrac{4}{20}$
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|
$a^3 + b^3 + c^3 = 3abc$ , can this be true only when $a+b+c = 0$ or $a=b=c$, or can it be true in any other case? Since
$$
a^3 + b^3 + c ^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc ),
$$ from this it is clear that if $(a+b+c) = 0$ , then $a^3 + b^3 + c ^3 - 3abc = 0$ and
$a^3 + b^3 + c ^3 = 3abc$ . Also if $a=b=c$ , $a^3 + b^3 + c^3 = a^3 + a^3 +a^3 = 3a^3 = 3aaa = 3abc$ , hence $a^3 + b^3 + c^3 = 3abc$.
But I was wondering if $a^3 + b^3 + c^3 = 3abc$ can be true even when none of the above two relations are true. Please guide me toward a solution.
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The polynomial factors fully over the complex numbers,
$$ (a+b+c)(a+b \omega + c \omega^2)(a + b \omega^2 + c \omega) \; , \; $$
where $\omega$ is a primitive cube root of unity, either solution of $x^2 + x + 1 =0$
There is actually a concrete calculation that tells us whether a homogeneous cubic factors completely over the complexes. Here is an excerpt from an article by Brookfield:
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|
JBMO TST macedonia Number Theory
Find all $x,y$ positive integers $$x + y^2 + (\gcd(x, y))^2 = xy \cdot
\gcd(x, y)$$
I tried supposing $\gcd(x,y)=d$ and letting $x=ad$, $y=bd$ but didn't get anything the closest thing I think is usefull that $(b+1)^2+4a^2b$ is a perfect square.
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Your start is very good. Now you have $$ad+b^2d^2+d^2=abd^3$$ thus $$a+db^2+d=abd^2\implies d\mid a$$ so $a=dc$ for some integer $c$ and now we have: $$c+b^2+1=cbd^2$$
From here we have $$c={b^2+1\over bd^2-1}\implies \boxed{bd^2-1\mid b^2+1}$$
Now we have $$\color{red}{bd^2-1}\mid (b^2+1)d^2-(bd^2-1)b =\color{red}{ d^2+b }$$
From here we deduce $$bd^2-1\leq d^2+b\implies d^2\leq {b+1\over b-1}$$
if $b\ne 1$. Now if $b\geq 2 $ we get $d^2\leq 3$ so $d=1$ and so $$b-1\mid b^2+1 \implies b-1 \mid (b^2+1)-(b^2-1)=2$$
This means $b = 2=y$ so $a=5=x$ or $b=3=y$ and $a=5=x$.
If $b=1$ we get from boxed relation $d-1\mid 2$ so $d=2$ or $d=3$...
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Prove $\sqrt{a^2+4}+\sqrt{b^2+4}\leq\frac{\sqrt{2}}{4}(a+b)^2$ with $\frac1a+\frac1b=1$. Let $ a, b> 0 $ and $\frac{1}{a}+\frac{1}{b}=1.$ Prove that$$\sqrt{a^2+4}+\sqrt{b^2+4}\leq\frac{\sqrt{2}}{4}(a+b)^2.$$
Obviously $a+b=ab\ge4$. Other than that, I do not know what trick to use to deal with the constraint. The Lagrange multiplier method has not gotten me further either.
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To get rid of some square roots and simplify, substitute $(a,b)\to (2x, 2y)$. The inequality can be written as:
Let $x,y >0$ and $\frac{1}{x}+\frac{1}{y} = 2$. Prove that:
$$\sqrt{2(x^2+1)}+\sqrt{2(y^2+1)} \leq (x+y)^2$$
In this case, from AM-GM we can see that:
$$\sqrt{2(x^2+1)}+\sqrt{2(y^2+1)} \leq \frac{x^2+1+2}{2}+\frac{y^2+1+2}{2} = \frac{x^2+y^2}{2}+3$$
It will be enough to prove:
$$\frac{x^2+y^2}{2}+3 \leq (x+y)^2$$
or
$$x^2+y^2+4xy \geq 6$$
From AM-GM on the initial condition, we can see that $xy \geq 1$. Therefore
$$x^2+y^2+4xy \geq 6xy \geq 6$$
Equality occurs when $(x,y) = (1,1)$, so when $(a,b)=(2,2)$.
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How do we integrate the expression $\frac{b^{3}}{1 - \theta(1-b)^{2}}$ where $0\leq b\leq 1$ and $\theta\in(0,1)$? MY ATTEMPT
In order to solve this integral, I have tried using the integration by parts method, as suggested by the expression
\begin{align}
\int_{0}^{1}\frac{b^{3}}{1-\theta(1-b)^{2}}\mathrm{d}b = \int_{0}^{1}\frac{b^{2}}{1-\theta(1-b)^{2}}\mathrm{d}b - \int_{0}^{1}\left[\int\frac{b^{2}}{1-\theta(1-b)^{2}}\mathrm{d}b\right]\mathrm{d}b
\end{align}
Based on it, the first summand is given by
\begin{align}\label{eq8}
\int_{0}^{1}\frac{b^{2}}{1 - \theta(1-b)^{2}}\mathrm{d}b & = \int_{0}^{1}\frac{(1 - 2b + b^{2}) - (1 - 2b)}{1 - \theta(1-b)^{2}}\mathrm{d}b\\
& = \int_{0}^{1}\frac{(1 - b)^{2}}{1 - \theta(1-b)^{2}}\mathrm{d}b - \int_{0}^{1}\frac{(2-2b) - 1}{1 - \theta(1-b)^{2}}\mathrm{d}b\\
& = \int_{0}^{1}\frac{(1 - b)^{2}}{1 - \theta(1-b)^{2}}\mathrm{d}b - 2\int_{0}^{1}\frac{1-b}{1 - \theta(1-b)^{2}}\mathrm{d}b + \int_{0}^{1}\frac{1}{1 - \theta(1-b)^{2}}\mathrm{d}b
\end{align}
The first of the last three integrals is given by
\begin{align}
\int_{0}^{1}\frac{(1 - b)^{2}}{1 - \theta(1-b)^{2}}\mathrm{d}b & = \frac{1}{\theta}\int_{0}^{1}\frac{\theta(1 - b)^{2}}{1 - \theta(1-b)^{2}}\mathrm{d}b\\
& = -\frac{1}{\theta}\int_{0}^{1}\frac{1 - \theta(1 - b)^{2}}{1 - \theta(1-b)^{2}}\mathrm{d}b + \frac{1}{\theta}\int_{0}^{1}\frac{1}{1 - \theta(1-b)^{2}}\mathrm{d}b\\
& = -\frac{1}{\theta} + \frac{1}{\theta}\int_{0}^{1}\frac{1}{1 - \theta(1-b)^{2}}\mathrm{d}b
\end{align}
According to the substitution $u = \sqrt{\theta}(1-b)$, it results that
\begin{align}\label{eq10}
\int_{0}^{1}\frac{1}{1 - \theta(1-b)^{2}}\mathrm{d}b = \frac{1}{\sqrt{\theta}}\int_{0}^{\sqrt{\theta}}\frac{1}{1 - u^{2}}\mathrm{d}u = \frac{1}{2\sqrt{\theta}}\ln\left|\frac{1+\sqrt{\theta}}{1-\sqrt{\theta}}\right|
\end{align}
Finally, based on the same substitution, we have
\begin{align}\label{eq11}
\int_{0}^{1}\frac{1-b}{1 - \theta(1-b)^{2}}\mathrm{d}b = \frac{1}{\theta}\int_{0}^{\sqrt{\theta}}\frac{u}{1-u^{2}}\mathrm{d}u = -\frac{1}{\theta}\ln|1-\theta^{2}|
\end{align}
Combining all these results, we get the first summand from the integration by parts method. The problem arises when I try to determine the second part.
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$\int_{0}^{1}\dfrac{b^3}{1-\theta(1-b)^2}db$. Integrating by parts, taking u=b and dv as the integral of the rest we have:
$(b\int\dfrac{b^2}{1-\theta(1-b)^2}db)\Big|_0^1 -\int_{0}^{1}(\int\dfrac{b^2}{1-\theta(1-b)^2}db)db$.
Lets solve the first integral.
$\int\dfrac{b^2}{1-\theta(1-b)^2}db=\dfrac{-1}{\theta}\int\dfrac{1-\theta+2\theta b-\theta b^2-1+\theta-2\theta b}{1-\theta(1-b)^2}db$
$=\dfrac{-1}{\theta}\int(1+\dfrac{-1+\theta -2\theta b}{1-\theta(1-b)^2})db$
$=\dfrac{-1}{\theta}\int(1+\dfrac{2\theta(1 - b)}{1-\theta(1-b)^2}+\dfrac{-1-\theta}{1-\theta(1-b)^2})db$
Taking $\sqrt\theta(1-b)=sen(t)$ in the last term
$=\dfrac{-1}{\theta}(\int(1+\dfrac{2\theta(1 - b)}{1-\theta(1-b)^2})db -\dfrac{1}{\sqrt{\theta}}(-1-\theta)\int \frac{cos(t)}{1-sen^2(t)}dt)$
We get $-\dfrac{1}{\theta}(b+\ln|1-\theta(1-b)|+\dfrac{(1+\theta)}{\sqrt{\theta}}(\ln|1+\sqrt \theta(1-b)|-\ln|\sqrt{1-\theta(1-b)^2}|)$
For the second integral we have to integrate the last result, and are only logarithms.
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|
Number of Divisors of $N = 3^55^77^9$ of the form $4n+1$ I need to find the number of divisors of $N = 3^55^77^9$ that are of the form $4n+1$, $n\geq 0$.
My try:
I noticed that $5$ itself is a number of the form $4n+1$ so all of its power satisfy the required condition, so number for the exponent of $5$ will be $7+1 = 8$.
Also, even powers of $3$ and $7$ satisfy the required condition.
But, I also noticed that odd powers of $3$ and $7$ multiplied together also satisfy the given condition as $(4m+3)(4k+3) = 4q+1$. How would I take that into account?
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$N = 3^55^77^9$ gives $(7+1)((3)(5)+(3)(5)) =240$ the first $(3)(5)$ deals with $\{0,2,4\}$ and $\{0,2,4,6,8\}$ the second with $\{1,3,5\}$and $\{1,3,5,7,9\}$
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|
Representing $f(x)=\frac{1}{x^2}$ as power series, with powers of $(x+2)$ As stated in the title, I want $f(x)=\frac{1}{x^2}$ to be expanded as a series with powers of $(x+2)$.
Let $u=x+2$. Then $f(x)=\frac{1}{x^2}=\frac{1}{(u-2)^2}$
Note that $$\int \frac{1}{(u-2)^2}du=\int (u-2)^{-2}du=-\frac{1}{u-2} + C$$
Therefore, $\frac{d}{du} (-\frac{1}{u-2})= \frac{1}{x^2}$ and
$$\frac{d}{du} (-\frac{1}{u-2})= \frac{d}{du} (-\frac{1}{-2(1-\frac{u}{2})})=\frac{d}{du}(\frac{1}{2} \frac{1}{1-\frac{u}{2}})=\frac{d}{du} \Bigg( \frac{1}{2} \sum_{n=0}^\infty \bigg(\frac{u}{2}\bigg)^n\Bigg)$$
$$= \frac{d}{du} \Bigg(\sum_{n=0}^\infty \frac{u^n}{2^{n+1}}\Bigg)= \frac{d}{dx} \Bigg(\sum_{n=0}^\infty \frac{(x+2)^n}{2^{n+1}}\Bigg)= \sum_{n=0}^\infty \frac{d}{dx} \bigg(\frac{(x+2)^n}{2^{n+1}}\bigg)=$$
$$\sum_{n=0}^\infty \frac{n}{2^{n+1}} (x+2)^{n-1}$$
From this we can conclude that
$$f(x)=\frac{1}{x^2}=\sum_{n=0}^\infty \frac{n}{2^{n+1}} (x+2)^{n-1}$$
Is this solution correct?
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Why not a Taylor series expansion at $x=-2$? That would be
$$f(x) = \sum_{n\geq 0} \frac{f^{(n)}(-2)}{n!}(x+2)^n, $$
with radius of convergence of $2$. You may calculate the $n$-th derivative of $f(x)=1/x^2$ to find
$$\frac{d^nf}{dx^n}(x)= \frac{(-2)(-3) \cdots (-2-n+1)}{x^{n+2}} = (-1)^n \frac{(n+1)!}{x^{n+2}} $$
and so
$$\frac{1}{x^2} = \sum_{n\geq 0} \frac{(n+1)}{2^{n+2}}(x+2)^n, \qquad x\in(-4,0) .$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3527004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
}
|
Integral of $\left(\frac{x^4}{1-x^4}\right)^k$ I'm stumped. How does one solve the following expression:
$$\int\left(\frac{x^4}{1-x^4}\right)^kdx, k\in\mathbb{N}$$
I thought it would be a fun challenge, but it beats me.
|
You can obtain a recurrence relation and then find a formula using sums and products for a closed form solution. $k\geq 1$, let
$$ I_k(x) = \int \left( \frac{x^4}{1-x^4} \right) ^k {\rm d}x$$
Using by parts with $u = (1-x^4)^{-k}$ and ${\rm d}v = x^{4k} {\rm d}x$, we will get
$$
\begin {align}
I_k(x) &= \frac{x^{4k+1}}{(4k+1)(1-x^4)^k} - \frac{4k}{4k+1} \int \frac{x^{4(k+1)}}{(1-x^4)^{k+1}} {
\rm d}x
\\ & = \frac{x^{4k+1}}{(4k+1)(1-x^4)^k} - \frac{4k}{4k+1} I_{k+1}(x)
\end {align}
$$
Rewriting, we get
$$
I_{k+1}(x) = \frac{x^{4k+1}}{4k(1-x^4)^k} - \frac{4k+1}{4k} I_k(x)
$$
Now, getting a formula for $I_k$ in terms of sums might be a little tedious but shouldn't be too difficult. First solve it for $k=1$ using partial fractions,
$$ I_1(x) = \frac12 \arctan(x) +\frac14 \ln \left| \frac{1+x}{1-x}
\right| -x + C$$
Then, for $k \geq 2$,
$$
I_k(x) = I_1(x) \prod_{n=1}^{k-1} \left(-\frac{4n+1}{4n} \right) + \sum_{n=1}^{k-1} \left( \prod_{m=n+1}^{k-1} -\frac{4m+1}{4m} \right)\frac{x^{4n+1}}{4n(1-x^4)^n} +C
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3527301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Numbers $x,y,z$ are such that $x+y+z=0$ and $x^2+y^2+z^2=1$. Show that at least one of $xy,xz,yz$ is not bigger than $-\frac{1}{3}$. So i had to prove:
Numbers $x,y,z$ are such that $x+y+z=0$ and $x^2+y^2+z^2=1$. Show that at least one of $xy,xz,yz$ is not bigger than $-\frac{1}{3}$.
Here is my proof:
So we have
$(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx=1+2(xy+yz+zx)=0\Rightarrow2(xy+yz+xz)=-1$
Let's assume that all of $xy,zx,yz$ are bigger than $-\frac{1}{3}$
$xy>-\frac{1}{3}$
$yz>-\frac{1}{3}$
$zx>-\frac{1}{3}$
Adding that we have $xy+yz+zx>-1$
$2(xy+yz+zx)>xy+yz+zx>-1\Rightarrow 2(xy+yz+zx)>-1$
Which is contradiction because $2(xy+yz+xz)=-1$, so one of $xy,yz,zx$ must be less or equal $-\frac{1}{3}$
Is my proof valid?
|
Regarding my notations here, I refer to my comments under the question. The idea of the parametrization in my comments arises from the following observation.
For $(x,y,z)\in S$, the maximum value of $|x|$ is $\sqrt{2/3}$. To show this, we note that $x=-y-z$ and
$$1=x^2+y^2+z^2=(-y-z)^2+y^2+z^2=2(y^2+yz+z^2) \geq \frac{3}{2}\left(y+z\right)^2=\frac{3}{2}x^2,$$
since $$4(y^2+yz+z^2)=(y^2+z^2)+(3y^2+4yz+3z^2)\geq 2yz+(3y^2+4yz+3z^2)=3(y+z)^2.$$
The maximum value $|x|=\sqrt{\frac23}=\frac2{\sqrt6}$ is attained when $(x,y,z)=\pm\left(\frac{2}{\sqrt6},-\frac{1}{\sqrt6},-\frac{1}{\sqrt6}\right)$. Hence we may write $x=\sqrt{\frac{2}{3}}\cos t$ for some $t\in[0,2\pi)$. Due to symmetry, we expect $y=\sqrt{\frac{2}{3}}\cos\left(t-\frac{2\pi}{3}\right)$ and $z=\sqrt{\frac{2}{3}}\cos\left(t+\frac{2\pi}{3}\right)$. We now need to check whether $x+y+z=0$ and $x^2+y^2+z^2=1$ hold, but this is an easy exercise. (Note that for a given $x$, $(y,z)$ is unique up to swapping because $q=y$ and $q=z$ are roots of $q^2+xq+x^2-\frac12$. Furthermore, the assignment $t\mapsto 2\pi-t$ swaps the values of $y$ and $z$ while keeping $x$ fixed. Thus, this parametrization gives all points on $S$.)
Now, as mentioned by YiFan, the function $f:S\to\Bbb R$ defined by $$f(p)=\min\{xy,yz,zx\}$$ for $p=(x,y,z)\in S$ is invariant under permuting entries of $p$ and also under the antipodal map $p\mapsto -p$. Therefore we can assume that $x>0\geq y\geq z$, which ultimately implies that $t\in[0,\pi/6]$. For such points $p$, $$f(p)=xz=\frac23\cos t\cos\left(t+\frac{2\pi}{3}\right)=\frac{1}{3}\Biggl(\cos\frac{2\pi}{3}+\cos\left(2t+\frac{2\pi}{3}\right)\Biggr).$$
For $t\in[0,\pi/6]$, the maximum value of $\cos\left(2t+\frac{2\pi}{3}\right)$ is attained at $t=0$ since $$\frac{2\pi}{3}\leq 2t+\frac{2\pi}{3}\leq \pi.$$ Therefore the maximum value of $f(p)$ is
$$\frac{1}{3}\Biggl(\cos\frac{2\pi}{3}+\cos\left(2\cdot 0+\frac{2\pi}{3}\right)\Biggr)=\frac{1}{3}\left(-\frac12-\frac12\right)=-\frac13.$$
(It is also not difficult to see that the minimum value of $f(p)$ is $-1/2$. This is attained when $p$ is a permutation of $\left(-\frac1{\sqrt2},0,\frac1{\sqrt2}\right)$; that is $-\frac12\le\min\{xy,yz,zx\}\le-\frac13$. We can also show that $0\le \max\{xy,yz,zx\}\le\frac16$ and $-\frac13\le\operatorname{median}\{xy,yz,zx\}\le0$.)
Alternatively if $\mu=\min\{xy,yz,zx\}$, then show that
$$\operatorname{median}\{xy,yz,zx\}+\max\{xy,yz,zx\}=-\frac{1+2\mu}{2}$$
and
$$\operatorname{median}\{xy,yz,zx\}\cdot\max\{xy,yz,zx\}=\frac{\mu(1+2\mu)}{2}.$$ This implies
$$\operatorname{median}\{xy,yz,zx\}=\frac{-(1+2\mu)-\sqrt{(1-6\mu)(1+2\mu)}}{4}$$
and
$$\max\{xy,yz,zx\}=\frac{-(1+2\mu)+\sqrt{(1-6\mu)(1+2\mu)}}{4}.$$
This means $\mu\geq -1/2$. We must have
$$\mu \leq \frac{-(1+2\mu)-\sqrt{(1-6\mu)(1+2\mu)}}{4},$$
or
$$\sqrt{(1-6\mu)(1+2\mu)}\leq -(1+6\mu).$$
That is,
$$1-4\mu-12\mu^2\leq 1+12\mu+36\mu^2,$$
or
$$16\mu(1+3\mu)\ge0.$$
It is clear that $\mu<0$, so $1+3\mu \leq 0$, and hence
$$\mu\le -\frac13.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3527542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
If $S_{100} - k + k² = 7500$ and $S_{100}$ it is the sum of $100$ consecutive positive integers, then what it is the value o k? $k$ must be one of the 100 consecutive integers.
I know the answers are $50$ and $26$.
But I got stuck into getting these numbers.
Here is what I tried:
$S_{100} = \frac{(n+n+99)\times100}{2} \implies S_{100} - k + k² = 7500 \implies \frac{(n+n+99)\times100}{2} + k(k-1) = 7500 \implies k(k-1) = 7500 - 50(2n+99)\implies k(k-1)= 50(51-2n) $.
It's easy to see that $n = 1 \implies k = 50$ is an answer. But how can I discover that $n = 19 \implies k = 26$?
|
Let's start with $$k^2-k = 50(51-2n)$$
Then we must have $k^2-k \equiv 0 \pmod{50}$.
We split this up into $k^2-k \equiv 0 \pmod 2$ and $k^2-k \equiv 0 \pmod{25}$.
\begin{align}
k^2-k &\equiv 0 \pmod{25} \\
k(k-1) &\equiv 0 \pmod{25} \\
k &\equiv 0, 1 \pmod {25}
\end{align}
Similarly, $k \equiv 0,1 \pmod 2$.
\begin{array}{c|ccc}
& \pmod{2} & \pmod{25} \\
\hline
25 & 1 & 0 \\
2 & 0 & 2 \\
\hline
25 & 1 & 0 \\
26 & 0 & 1 \\
\hline
\end{array}
Hence, if $k \equiv x \pmod 2$ and $k \equiv y \pmod{25}$, Then
$k \equiv 25x + 26y \pmod {50}$.
\begin{array}{cc|ccc}
x \pmod 2 & y \pmod{25} & k \pmod{50} & \dfrac{k(k-1)}{50} & n \\
\hline
0 & 0 & 0 & 0 & *\\
\color{red}0 & \color{red}1 & \color{red}{26} & \color{red}{13} & \color{red}{19}\\
1 & 0 & 25 & 12 & *\\
1 & 1 & 1 & 0 & *\\
\hline
\end{array}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3528435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Proving that the limit is bounded. I am trying the following exercise that's actually stated as a lemma on an Econometrics textbook:
Let $\{a_t\}_{t = 1,2.,..}$ be a sequence of nonnegative scalars such that $\sum\limits_{t = 1}^{T}\frac{a_t}{T} < M < \infty$ for every $T$ for some finite $M$. Prove that $\lim\limits_{T\rightarrow \infty}\sum\limits_{t = 1}^{T}\frac{a_t}{t^2} < \infty$.
I tried using the Cauchy-Schwarz inequality but I didn't take me very far. So I did this trick, although I'm not totally sure I'm taking a safe path:
For any given $T$:
$\sum\limits_{t = 1}^{T}\frac{a_t}{t^2} = \sum\limits_{t \leq \sqrt{T}}\frac{a_t}{t^2} + \sum\limits_{t > \sqrt{T}}\frac{a_t}{t^2} \leq \sum\limits_{t \leq \sqrt{T}} a_t + \sum\limits_{t > \sqrt{T}}\frac{a_t}{T}$
The second term cannot be bigger than $M - M\sqrt{T} + 1$. The first term is smaller than $M\sqrt{T}$. Then this sum is bounded for any $T$ and we would have the desired result.
I would appreciate some comments on if this is true or if I am missing something! Thanks a lot in advance!
|
We have that
$$
\forall T \in \mathbb{N}^+ \qquad \sum_{i=1}^{T}a_i < MT.
$$
We can define $s_i$ as the cumulative sum, i.e.,
$$
s_i = \sum_{j=1}^{i} a_j
$$
and we notice that
$$
A
\left(
\begin{array}{c}
s_1\\
s_2\\
\vdots\\
s_T
\end{array}
\right )
=
\left(
\begin{array}{c}
a_1\\
a_2\\
\vdots\\
a_T
\end{array}
\right ).
$$
Where
$$
A=\left (
\begin{array}{cccc}
1 & 0 & & 0\\
-1 & \ddots & \ddots & \\
& \ddots & \ddots & 0 \\
0 & & -1 & 1
\end{array}
\right )$$
Therefore
$$
\sum_{i=1}^{T}\frac{a_i}{i^2} =
\left(
\begin{array}{cccc}
1 & \frac{1}{2^2} & \cdots & \frac{1}{T^2}
\end{array}
\right)
\left(
\begin{array}{c}
a_1\\
a_2\\
\vdots\\
a_T
\end{array}
\right )
=
\left(
\begin{array}{cccc}
1 & \frac{1}{2^2} & \cdots & \frac{1}{T^2}
\end{array}
\right)
A
\left(
\begin{array}{c}
s_1\\
s_2\\
\vdots\\
s_T
\end{array}
\right ).
$$
By associativity we rewrite the expression as
$$
\sum_{i=1}^{T}\frac{a_i}{i^2} =
\left(
\begin{array}{cccc}
1 - \frac{1}{2^2} & \frac{1}{2^2} -\frac{1}{3^2} & \cdots & \frac{1}{T^2}
\end{array}
\right)
\left(
\begin{array}{c}
s_1\\
s_2\\
\vdots\\
s_T
\end{array}
\right )
<
\left(
\begin{array}{cccc}
1 - \frac{1}{2^2} & \frac{1}{2^2} -\frac{1}{3^2} & \cdots & \frac{1}{T^2}
\end{array}
\right)
\left(
\begin{array}{c}
M\\
2 \cdot M\\
\vdots\\
T \cdot M
\end{array}
\right )
.
$$
Where the inequality follows from the fact that all the components of the horizontal vector are positive.
The RHS can be rewritten again as
$$
\left(
\begin{array}{cccc}
1 & \frac{1}{2^2} & \cdots & \frac{1}{T^2}
\end{array}
\right)
A
\left(
\begin{array}{c}
M\\
2 \cdot M\\
\vdots\\
T \cdot M
\end{array}
\right )
=
M \cdot \left(
\begin{array}{cccc}
1 & \frac{1}{2^2} & \cdots & \frac{1}{T^2}
\end{array}
\right)
\left(
\begin{array}{c}
1\\
1\\
\vdots\\
1
\end{array}
\right )
= M \sum_{i=1}^{T}\frac{1}{i^2}<M\frac{\pi^2}{6}
$$
where the last inequality is a well known result (https://en.wikipedia.org/wiki/Basel_problem).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3530793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Solve $\sin x + \cos x = \sin x \cos x.$ I have to solve the equation:
$$\sin x + \cos x = \sin x \cos x$$
This is what I tried:
$$\hspace{1cm} \sin x + \cos x = \sin x \cos x \hspace{1cm} ()^2$$
$$\sin^2 x + 2\sin x \cos x + \cos^2 x = \sin^2 x \cos^2x$$
$$1 + \sin(2x) = \dfrac{4 \sin^2 x \cos^2x}{4}$$
$$1 + \sin(2x) = \dfrac{\sin^2(2x)}{4}$$
$$\sin^2(2x) - 4 \sin(2x) -4 = 0$$
Here we can use the notation $t = \sin(2x)$ with the condition that $t \in [-1,1]$.
$$t^2-4t-4=0$$
Solving this quadratic equation we get the solutions:
$$t_1 = 2+ 2\sqrt{2} \hspace{3cm} t_2 = 2 - 2\sqrt{2}$$
I managed to prove that $t_1 \notin [-1, 1]$ and that $t_2 \in [-1, 1]$. So the only solution is $t_2 = 2 - \sqrt{2}$. So we have:
$$\sin(2x) = 2 - 2\sqrt{2}$$
From this, we get:
$$2x = \arcsin(2-2\sqrt{2}) + 2 k \pi \hspace{3cm} 2x = \pi - \arcsin(2-2\sqrt{2}) + 2 k \pi$$
$$x = \dfrac{1}{2} \arcsin(2-2\sqrt{2}) + k \pi \hspace{3cm} x = \dfrac{\pi}{2} - \dfrac{1}{2}\arcsin(2 - 2\sqrt{2}) + k \pi$$
Is this solution correct? It's such an ungly answer, that I kind of feel like it can't be right. Did I do something wrong?
|
There is a subtle way to get rid of the extra solutions that uses the work you've done.
When you render
$\sin 2x=2\sin x\cos x =2-2\sqrt{2}$
you then have with $u=\sin x, v=\cos x$:
$uv=1-\sqrt{2}$
$\color{blue}{u+v=uv=1-\sqrt{2}}$
where the blue equation reimposes the original requirement and it's goodbye extraneous roots. Using the Vieta formulas for a quadratic polynomial this is solved by rendering $u$ and $v$ as the two roots of the quadratic equation
$w^2-(1-\sqrt2)w+(1-\sqrt2)=0$
The roots of this equation are obtained by the usual methods, and because of the symmetry of the original equation between $\sin x$ and $\cos x$ you may take either root as the sine and the other as the cosine. Note that with the negative product the values must be oppositely signed which informs us of quadrant location.
$\sin x=\dfrac{1-\sqrt2+\sqrt{2\sqrt2-1}}{2},\cos x=\dfrac{1-\sqrt2-\sqrt{2\sqrt2-1}}{2}$ ($x$ in 2nd quadrant)
$\sin x=\dfrac{1-\sqrt2-\sqrt{2\sqrt2-1}}{2},\cos x=\dfrac{1-\sqrt2+\sqrt{2\sqrt2-1}}{2}$ ($x$ in 4th quadrant)
Then with the quadrant information above we may render the correct roots for $x$:
$x=\pi-\arcsin(\dfrac{1-\sqrt2+\sqrt{2\sqrt2-1}}{2})+2n\pi$
$x=\arcsin(\dfrac{1-\sqrt2-\sqrt{2\sqrt2-1}}{2})+2n\pi$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3530949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 2
}
|
Solve for the real scalar value of $k$, if $z =\frac{2}{1+ki} -\frac{i}{k-i}$ and $z$ lies on the line, $y= 2x$ The complex number $z$ is given by $\frac{2}{1+ki} - \frac{i}{k-i}$. If it is given that $z$ lies on the line, $y = 2x$, find the value of the real scalar $k$.
So far this is what I have come up with : (See attached photo)
I am stuck at this point. I think that, because the question asks for the value of the real scalar $k$, I should move forward with only the "$x$" part of the coordinate system, but I do not know what to do with it.
Also, because it asks for the scalar value, I am wondering if I need to find the modulus of anything.
Any help would be appreciated.
|
The equations derived in the picture for $(x,y)$ are wrong from line 4 onward. The corrected ones are
$$z=\frac{3k-3i}{2k-i(1-k^2)}\times\frac{2k+i(1-k^2)}{2k+i(1-k^2)}=\frac{3k^2+3+i(-3k^3-3k)}{4k^2+(1-k^2)^2}$$
We have $y=2x$ and $x = \frac{3k^2+3}{4k^2+(1-k^2)^2}$ and $y = \frac{-3k^3-3k}{4k^2+(1-k^2)^2}$ therefore
$$\frac{-3k^3-3k}{4k^2+(1-k^2)^2}=2\times \frac{3k^2+3}{4k^2+(1-k^2)^2} \Rightarrow -3k^3-3k=6k^2+6 \Rightarrow k^3+2k^2+k+2=0\\ \Rightarrow k^2(k+2)+(k+2)=0
\Rightarrow(k+2)(k^2+1)=0 \Rightarrow k=-2,i,-i
$$
That's all the possible answers, of course only $k=-2$ is real and obviously $k=i,-i$ are not in the domain of $z=\frac{1}{1+ki}-\frac{i}{k-i}$ the original problem.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3532332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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|
What's wrong with my calculation of a limit here? So, I have to find the following limit:
$\lim _{x\to \:0+}\left(\frac{\left(1-\cos \left(2x\right)\right)^{14}\left(1-\cos \left(7x\right)\right)^2\sin ^{14}\left(9x\right)}{\tan ^{14}\left(x\right)\left(\ln \left(8x+1\right)\right)^{30}}\right)$
I solved it by splitting it into three limits as follows:
$$\begin{align}
&\lim _{x\to \:0+}\left(\frac{\left(1-\cos \left(2x\right)\right)^{14}\left(1-\cos \left(7x\right)\right)^2\sin ^{14}\left(9x\right)}{\tan ^{14}\left(x\right)\left(\ln \left(8x+1\right)\right)^{30}}\right)
\\
&=\lim _{x\to 0+}\frac{\left(1-cos\left(2x\right)\right)^{14}}{\tan ^{14}\left(x\right)}\cdot \frac{\left(1-cos\left(7x\right)\right)^2}{\left(\ln \left(8x+1\right)\right)^2}\cdot \frac{sin\left(9x\right)^{14}}{\left(\ln \left(8x+1\right)\right)^{28}}
\\
&=\lim_{x\to 0+}\frac{\left(1-cos\left(2x\right)\right)^{14}}{\tan ^{14}\left(x\right)}\cdot \lim_{x\to\:0+}\frac{\left(1-cos\left(7x\right)\right)^2}{\left(\ln\left(8x+1\right)\right)^2}\cdot\lim_{x\to 0+}\frac{sin\left(9x\right)^{14}}{\left(\ln\left(8x+1\right)\right)^{28}}
\\
&=\left(\lim _{x\to \:\:0+}\frac{\left(1-cos\left(2x\right)\right)}{\tan \left(x\right)}\right)^{14}\cdot \left(\lim _{x\to 0+}\frac{\left(1-cos\left(7x\right)\right)}{\:\ln \left(8x+1\right)}\right)^2\cdot \left(\lim _{x\to 0+}\frac{sin\left(9x\right)}{\left(\ln\left(8x+1\right)\right)^2}\right)^{14}
\end{align}$$
Using L'Hospital's rule to solve these separate limits, you get
$$\left(\lim _{x\to \:\:0+}\frac{\left(1-cos\left(2x\right)\right)}{\tan \left(x\right)}\right)^{14}\cdot \left(\lim _{x\to 0+}\frac{\left(1-cos\left(7x\right)\right)}{\:\ln \left(8x+1\right)}\right)^2\cdot \left(\lim _{x\to 0+}\frac{sin\left(9x\right)}{\left(\ln\left(8x+1\right)\right)^2}\right)^{14}
\\
=0^{14}\cdot 0^2\cdot \left(-\frac{9}{16}\right)^{14}
\\
=0$$
However, the automated homework system did not accept 0 as the correct answer. What did I do wrong?
|
Sanity check:
For small $x$,
$$1-\cos x=\Theta(x^2),\sin x=\Theta(x),\tan x=\Theta(x),\log(x+1)=\Theta(x).$$
Then the expression is of order
$$\frac{x^{28}x^4x^{14}}{x^{14}x^{30}}=x^2.$$
(Which tends to $0$, of course.)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3532764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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|
Find $a$ such that the minimum and maximum distance from a point to the curve $x^2+y^2=a$ are $\sqrt{5}$ and $3\sqrt{5}$ Context: I have to solve the following problem: find $a\in \mathbb{R},\, a>0\,\, /$ the minimum and maximum distance from $(4,2)$ to the curve $x^2+y^2=a$ are $\sqrt{5}$ and $3\sqrt{5}$ respectively.
I can't use polar coordinates. I decided to used Lagrange multipliers.
Problem: after using Lagrange multipliers I got this system:
$$
\left\{
\begin{array}{c}
\frac{x-4}{x}=\frac{y-2}{y} \\
x^2+y^2=a
\end{array}
\right.
$$
And then I add one more equation to the system: $5=(x-4)^2+(y-2)^2$ and $45=(x-4)^2+(y-2)^2$, so I have to find the $a$ that satisfies both system of equations:
$$
\left\{
\begin{array}{c}
\frac{x-4}{x}=\frac{y-2}{y} \\
x^2+y^2=a \\
5=(x-4)^2+(y-2)^2
\end{array}
\right.
$$
and
$$
\left\{
\begin{array}{c}
\frac{x-4}{x}=\frac{y-2}{y} \\
x^2+y^2=a \\
45=(x-4)^2+(y-2)^2
\end{array}
\right.
$$
The thing is that I can't resolve those systems, I already tried a lot of ways but I never get a result. I would really appreciate some help.
|
It can be geometrically seen that once you draw the diameter of the circle that passes through $(4,2)$, contributes the maximum and minimum distances. That diameter must have an equation like $y={x\over2}$ which together with $x^2+y^2=a$, leads to two points $(2\sqrt{a\over 5},\sqrt{a\over 5})$ and $(-2\sqrt{a\over 5},-\sqrt{a\over 5})$. The right choice for $a$ is then straightforward, but the problem may or may not have a solution.
|
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|
Integrate $\int \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}dx$
Evaluate
$$\int \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$$
I have tried substitution of $\sin x$ as well as $\cos x$ but it is not giving an answer.
Do not understand if there is a formula for this or not.
|
\begin{align}
I&=\int \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx\\
&=\int \frac{1}{\sqrt{\cot x}+1}dx
=\frac12x -\frac12 \int \frac{\sqrt{\cot x}-1}{\sqrt{\cot x}+1}dx\tag1 \\
\end{align}
where, with $t=\sqrt{\cot x}$
\begin{align}
\int \frac{\sqrt{\cot x}-1}{\sqrt{\cot x}+1}dx
&=\int \frac{2t(1-t)}{(1+t)(1 + t^4)}dt\\
&= 2 \int \frac{t^3}{1+t^4}-\frac1{1+t}+\frac{1-t^2}{1+t^4}\ dt\\
&= \frac12\ln(1+t^4)-2\ln(1+t)+2\int \frac{d(t+\frac1t)}{2-(t+\frac1t)^2}dt\\
&= 2\ln \frac{\sqrt[4]{1+t^4}}{1+t}+\sqrt2\coth^{-1}\frac{1+t^2}{\sqrt2t}
\end{align}
Substitute into (1) to obtain
\begin{align}
I= \frac12x+\ln (\sqrt{\sin x}+\sqrt{\cos x})
-\frac1{\sqrt2}\coth^{-1}\frac{\sqrt{\tan x}+\sqrt{\cot x}}{\sqrt{2}}
\end{align}
|
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|
Is $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\dots$ convergent?
Let $a_n= (-1)^k$ ,when $2^k \leq n < 2^{k+1} $.
Then determine if $\sum_{n=1}^{\infty}\frac{a_n}{n}$ converge or not.
[my attempt]
I try to apply Dirichlet tset but I fail.
So I tried following method
$$1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\dots \geq 1-\left(\frac{1}{2}+\frac{1}{2}\right)+\left(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}\right)-\frac{1}{8}-\dots = 1 - 1 +1 + \dots $$ However this has no information so I fail,too
May I ask you how to solve this problem?
|
Bunching terms with the same sign together already gives you enough information. Each such bunched term has a magnitude of at least $\frac12$, which means that the series diverges by Cauchy's test (all sums of sufficiently far sequences of arbitrary length must be less than any $\varepsilon$ for the series to converge).
|
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|
Finding $\sqrt{625}$ without a calculator We were factoring $25n^2+15n-4$ in class and my professor used the quadratic formula. This root came up and he wrote the answer directly. He doesn't want us to use calculators.
Is there any trick for this?
|
I guess your professor has been using the following trick: the square of a number of the form $10n + 5$ is $100n(n+1) + 25$. Indeed $(10n + 5)^2 = 100n^2 + 100n + 25 = 100n(n+1) + 25$.
For instance:
\begin{align}
n &= 0: && 5^2 = 25 &&\text{since $0 \times 1 = 0$} \\
n &= 1: && 15^2 = 225 &&\text{since $1 \times 2 = 2$} \\
n &= 2: && 25^2 = 625 &&\text{since $2 \times 3 = 6$} \\
n &= 3: && 35^2 = 1225 &&\text{since $3 \times 4 = 12$} \\
n &= 4: && 45^2 = 2025 &&\text{since $4 \times 5 = 20$} \\
n &= 5: && 55^2 = 3025 &&\text{since $5 \times 6 = 30$} \\
&\ \vdots
\end{align}
If you know this trick, it is not difficult to remember that $25^2 = 625$...
|
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|
About the sequence: $1, 3, 8, 24, 29, 87, 92, ?...$ find $a_n$ This is a classic and curious recurrence sequence used in logic tests. Your rule can be determined as follows:
for $n$ even: $a_n = 3a_{n-1}$
for $n$ odd: $a_n = a_{n-1} + 5$
That is, the ratio alternates with each term.
Will there be a single formula for $a_n$ in terms of $n$?
|
A bit of a sneaky answer. The subsequence of only odd terms, $1,8,29,92,\ldots$, is listed in the OEIS as sequence A116952, $b_n=3b_{n-1} + 5$ with $b_0=1$ and has the explicit formula $b_n=\frac723^n-\frac52$. So then, the even terms satisfy $c_n=3b_n=3\left(\frac723^n-\frac52\right)$. Simple indicator functions for odd and even numbers outputting $1$ and $0$ can be constructed as $\frac{\pm(-1)^n+1}{2}$ so we have the explicit formula
$$\begin{align}a_n=\left(\frac{7}{2}3^{\frac{n-1}{2}}-\frac{5}{2}\right)\left(\frac{-\left(-1\right)^{n}+1}{2}\right)+3\left(\frac{7}{2}3^{\frac{n-1-1}{2}}-\frac{5}{2}\right)\left(\frac{\left(-1\right)^{n}+1}{2}\right)
\\
\\
=\frac14\left[{7\cdot3^{{n}/{2}}\left(1+\frac{1}{\sqrt{3}}\right)-20+\left(-1\right)^{n}\left(7\cdot3^{{n}/{2}}\left(1-\frac{1}{\sqrt{3}}\right)-10\right)}\right]
\end{align}
$$
|
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|
Deriving $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$ using Fourier Series for $f(x)=|x|$ One can find that $$f(x) = \frac{\pi}{2} + \sum_{n=1}^{\infty}\frac{2(-1+(-1)^n)}{\pi n^2}\cos(nx)$$
Now look at the case for $x=0$. We can find that $\sum_{n \text{ odd}, n \geq 1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{8}$. However, I am having a hard time proving the original equality since all the even terms would vanish. Thus it seems like the total sum should be equivalent to the sum of odd terms, but I know this is not the case. What nuance am I missing?
|
\begin{align}
& \text{sum of even terms} = \sum_{n=1}^\infty \frac 1 {(2n)^2} \\[8pt]
= {} & \frac 1 4 \sum_{n=1}^\infty \frac 1 {n^2} = \left( \frac 1 4 \times\big(\text{sum of all terms} \big)\right).
\end{align}
If the sum of the even terms is $1/4$ times the sum of all of the terms, then the sum of the odd terms is $3/4$ times the sum of all of the terms.
So multiply the sum of the odd terms by $4/3$ to get the sum of all of the terms.
|
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|
Consider the linear function$ f : \mathbb{R}^{2\times2}\to \mathbb{R}^{3\times2}$ defined as follows: Consider the linear function $f : \mathbb{R}^{2\times2} → \mathbb{R}^{3\times2}$ defined as follows:
$$\begin{pmatrix} r_1 & r_2 \\ r_3 & r_4 \\ \end{pmatrix} \in \mathbb{R}^{2\times2}\mapsto f\begin{pmatrix} r_1 & r_2 \\ r_3 & r_4 \\ \end{pmatrix} :=\begin{pmatrix} 3 & 2 \\ -2 & 1 \\ 0 & 4 \end{pmatrix} \begin{pmatrix} r_1 & r_2 \\ r_3 & r_4 \\ \end{pmatrix}$$
Check whether $f$ is injective and/or surjective. If it is bijective, find its inverse function. Finally, find bases for its Kernel and its Range.
Attempted solution:
This matrix multiplication evaluates to:
$$\begin{pmatrix} 3r_1+2r_3 & 3r_2+2r_4 \\ -2r_1+r_3 & -2r_2+r_4 \\ 4r_3 & 4r_4 \end{pmatrix}$$
$f$ cannot be surjective since $m>n$. However, it can be injective if $n=\text{rank}$. Therefore, it is not bijective.
Now I get stuck. I don't understand the technique to find the bases for its kernel or range nor can I calculate its rank. I know we can row reduce this but it isn't in the proper form so I'm unsure of how to proceed.
|
The rank of a matrix is the number of non empty rows once it is une the reduced echelon form. Your matrix could be reduced by doing the following operations $3L_2+2L_1$ then $7L_3-4L_2$.
$$\begin{pmatrix}3&2\\ -2&1\\ 0&4\end{pmatrix}\sim\begin{pmatrix}3&2\\ 0&7\\ 0&4\end{pmatrix}\sim\begin{pmatrix}3&2\\ 0&7\\ 0&0\end{pmatrix}$$
The matrix has two non empty lines, so $\text{rank}{A}=2$. The function is injective.
There is an other way to know the function is injective: if the kernel has only one element.
The kernel is the set of all matrix from the domain that are maps to the $0$ of the image. We need to find all value of $r_1$, $r_2$, $r_3$ and $r_4$ such that
$$\begin{pmatrix}3r_1+2r_3&3r_2+2r_4\\ -2r_1+r_3&-2r_2+r_4\\ 4r_3&4r_4\end{pmatrix}=\begin{pmatrix}0&0\\ 0&0\\ 0&0\end{pmatrix}$$
This gives us six equations with four unknows. Starting with the bottom line, $r_3=r_4=0$ once those are known, the other equations give $r_1=r_2=0$. Only the null matrix is part of the kernel.
To find a basis for the image, we take a basis of the domain and apply the function. It will give us four linearly independant matrix that generate the image of $f$. A basis of $\Bbb R^{2\times2}$ could be
$$e_1=\begin{pmatrix}1&0\\ 0&0\end{pmatrix}, e_2=\begin{pmatrix}0&1\\ 0&0\end{pmatrix}, e_3=\begin{pmatrix}0&0\\ 1&0\end{pmatrix}, e_4=\begin{pmatrix}0&0\\ 0&1\end{pmatrix}$$
We apply the function to each matrix.
$$f(e_1)=\begin{pmatrix}3&0\\ -2&0\\ 0&0\end{pmatrix}, f(e_2)=\begin{pmatrix}0&3\\ 0&-2\\ 0&0\end{pmatrix}, f(e_3)=\begin{pmatrix}2&0\\ 1&0\\ 4&0\end{pmatrix}, f(e_4)=\begin{pmatrix}0&2\\ 0&1\\ 0&4\end{pmatrix}$$
These four matrix are the basis of the image.
|
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|
Proof that both eigenvalues lie inside the unit circle iff $\det < 1$ and $\operatorname{tr }< 2$ Let $A = \begin{pmatrix}a&b \\c &d \end{pmatrix}$, prove that both eigenvalues lie within the unit circle if and only if $\det(A) < 1$ and $|\operatorname{tr}(A)| < 1 +\det(A)$.
This is what I have done so far:
$|\operatorname{tr}(A)| < 1 +\det(A) \Rightarrow |tr(A)| < 2$.
The characteristic polynomial of A is
$$\lambda^2 - (a+b)\lambda +ab+bc = 0$$
and the solutions are
$$\lambda_{1,2} = \frac{a+d}{2} \pm \sqrt{\frac{(a+d)^2}{4}-(ad-bc)} \qquad (\star)$$
If the eigenvalues $x$ and $y$ are complex then they are also conjugated. So $\det(A) = xy < 1$ is true only if the real part is less then 1 and $|\operatorname{tr}(A)| = |x+y| < 2$ is obviously true then as well (the two complex parts take out each other since conjugated).
This is the part that I'm a bit uncertain about. If $x$ and $y$ are real then we could express $(\star)$ as $k \pm t$ where both $k$ and $t$ are constants.
$\det(A) =xy < 1$ gives us two possibilities, either $x$ and $y$ are both less then 1 or if say $x > 1$ then $y < \frac{1}{x}$.
So if $x = k+t > 1$ and $y = k-t < \frac{1}{k+t} \Leftrightarrow y = \frac{1}{k-t} > k+t \Rightarrow y > 1$ which is a contradiction. Thus both $x$ and $y$ must be less then 1.
Thanks in advance.
|
Hint:
Assume $x,y \in \mathbb{R}$.
We have $0 \le |x+y| < 1+xy$ which implies that actually $ -1 \le xy < 1$, or $x^2y^2 \le 1$.
Squaring the relation $|x+y| < 1+xy$ gives
$$x^2+y^2 < 1+x^2y^2 \implies (1-x^2)(1-y^2) > 0$$
so $x^2, y^2 > 1$ or $x^2, y^2 < 1$. However, the first possibility contradicts $x^2y^2 \le 1$.
|
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|
Proof of $n^3/3 + n^2/2 + n/6 = [1^2 + 2^2 + ... + n^2]$? I am having a hard time following this proof. Here is how it goes.
$$
(k+1)^3 = k^3+3k^2+3k+1\\
3k^2+3k+1 = (k+1)^3-k^3\\
$$
if $ k = 1, 2, 3, ... , n-1$ we add all the 5 formulas like this
$$
3(1)^2+3(1)+1 = ((1)+1)^3-(1)^3\\
3(2)^2+3(2)+1 = ((2)+1)^3-(2)^3\\
3(3)^2+3(3)+1 = ((3)+1)^3-(3)^3\\
3(4)^2+3(4)+1 = ((4)+1)^3-(4)^3\\
\vdots \\
3(n-1)^2+3(n-1)+1 = ((n-1)+1)^3-(n-1)^3\\
3n^2+3n+1 = (n+1)^3-(n-1)^3\\
$$
The result of adding these formulas is
$$
3[1^2+2^2 + ... + (n-1)^2+n^2] + 3[1 + 2 + ... + (n-1)+n] + n = (n+1)^3-1^3
$$
I am able to follow up-to this point easily. I don't understand how this last equation goes from that to this $ n^3/3 + n^2/2 + n/6 = [1^2 + 2^2 + ... + n^2]$ I know thee sum of arithmetic series to n is $n(n-1)/2$ and that replaces the second expression on the left side.
Can some one please show me the algebra step by step?
|
what you need to show after applying the inductive method is that:
$\frac{k^3}{3}+\frac{k^2}{2}+\frac{k}{6}+(k+1)^2=\frac{(k+1)^3}{3}+\frac{(k+1)^2}{2}+\frac{k+1}{6}$
this expands to:
$2k^3+3k^2+k+6k^2+12k+6=2k^3+6k^2+6k+2+3k^2+6k+3+k+1$
after simplifying:
$2k^3+9k^2+13k+6=2k^3+9k^2+13k+6$
left side = right side.
|
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|
minimum value of $f(x)=\frac{(x^2-x+1)^3}{x^6-x^3+1}$
If $f:\mathbb{R}\rightarrow \mathbb{R}.$ Then minimum value of $$\displaystyle f(x)=\frac{(x^2-x+1)^3}{x^6-x^3+1}$$
what i try
If $x\neq 0,$ Then divide numerator and denominator by $x^3$
$$f(x)=\frac{\bigg(x+\frac{1}{x}-1\bigg)^3}{x^3+\frac{1}{x^3}-1}$$
put $\displaystyle x+\frac{1}{x}=t,$ then $\displaystyle x^3+\frac{1}{x^3}=t^3-3t$
$$f(t)=\frac{(t-1)^3}{t^3-3t}$$
How do i solve it without derivatives
Help me please
|
You have some mistakes in $f(t)$. Correct is, it should be:
$$f(t) = \frac{(t-1)^3}{t^3-3t\color{red}{-1}}$$
and it's important to notice that $|t| \geq 2$ for any real $x$. So, we have to minimize $f:(-\infty,-2] \cup [2,\infty) \to \mathbb{R}$ with
$$f(t) = \frac{(t-1)^3}{t^3-3t-1}$$
We have:
$$f'(t) = \frac{3 (t-1)^2 (t^2-2t-2)}{(t^3-3t-1)^2}$$
$f'$ has only one real root in the definition domain of $f$ and that is $t_0=1+\sqrt{3}$. Also $f'(t) < 0$ for $t < t_0$ and $f'(t) > 0$ for $t > t_0$, so $t$ is a local minima. Therefore:
$$\min f(t)=f(1+\sqrt{3})=2\sqrt{3}-3$$
This is attained when $x+\frac{1}{x} = 1+\sqrt{3}$, which gives two values for $x = \frac{1}{2}(1 \pm \sqrt[4]{12} + \sqrt{3})$.
|
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|
Solving an integral: $\int \frac{x}{x^3-1}\,\mathrm dx$ I'm trying to solve this integral:
$$\int \frac{x}{x^3-1}\,\mathrm dx$$
What I did was:
$$\int \frac{x}{(x-1)(x^2+x+1)}\,\mathrm dx.$$
$$\frac{x}{(x-1)(x^2+x+1)} = \frac{a}{x-1}+ \frac{bx+c}{x^2+x+1}$$
Then I got this in the numerator:
$$ax^2+ax+a+bx^2-bx+cx-c $$
$$a+b=0;a-b+c=1; a-c=0 $$
$$a=c=\frac{1}3 \qquad b=-\frac{1}3$$
Then I wrote:
$$\frac{1}3\int \frac{1}{x-1}\,\mathrm dx-\frac{1}3\int\frac{x-1}{x^2+x+1} \, \mathrm dx$$
so the first one is just $\frac{1}{3}\ln|x-1|$. Which makes my calculations already wrong, most likely.
With the second one I tried a few different things ( involving u-substitution mostly) and got stuck.
I know I'm supposed to get this:
$$\frac{1}6\ln \frac{(x-1)^3}{x^2+x+1}+\frac{1}{\sqrt{3}} \arctan\frac{2x+1}{\sqrt{3}}$$
What have I already done wrong? What am I supposed to do?
|
You did right, but gave up soon. For
$$\int\frac{x-1}{x^2+x+1}$$
write
$$\frac{x-1}{x^2+x+1}=\frac{1}{2}\frac{2x-2}{x^2+x+1}=\frac{1}{2}\frac{2x+1}{x^2+x+1}
+\frac{1}{2}\frac{-3}{x^2+x+1}$$
then the integral becomes
$$\int\frac{x-1}{x^2+x+1}dx=\frac{1}{2}\ln(x^2+x+1)+\frac{3}{2}\int\frac{dx}{x^2+x+1}$$
and then...
|
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|
Prove that $6 | (a+b+c)$ if and only if $6 | (a^3 + b^3 +c^3).$ I have tried the question but not sure if my solution is correct or not...
My try..
\begin{align}a^3 + b^3 + c^3 = (a+b+c)^3 - 3(a+b)(b+c)(c+a)\end{align}
So , if
\begin{align}6 |(a^3 + b^3 + c^3)\end{align}
Then,
\begin{align}6 | [(a+b+c)^3 - 3(a+b)(b+c)(c+a)]\end{align}
So, also
\begin{align}6 | (a+b+c)^3\end{align}
\begin{align}6 | (a+b+c)(a+b+c)(a+b+c)\end{align}
So,
\begin{align}6|(a+b+c)\end{align}
Is my solution correct ?
|
I'm not sure how you got
Then,
6 | [(a+b+c)^3 - 3(a+b)(b+c)(c+a)]
So, also
6 | (a+b+c)^3
6 | (a+b+c)(a+b+c)(a+b+c)
As Iris's question comment says, because $6$ divides the first expression of $(a+b+c)^3 - 3(a+b)(b+c)(c+a)$ doesn't mean it divides the first term, i.e., $(a+b+c)^3$ and then a different expression of $(a+b+c)(a+b+c)(a+b+c)$.
Instead, here is a simpler way. Note that odd integers cubed are odd & even integers cubed are even, so for all integers $n$ you have
$$n^3 \equiv n \pmod 2 \tag{1}\label{eq1A}$$
Also, as $3$ is prime, then by Fermat's little theorem, for all integers $n$,
$$n^3 \equiv n \pmod 3 \tag{2}\label{eq2A}$$
Since $2$ and $3$ are relatively prime, you can put these together to get
$$n^3 \equiv n \pmod 6 \tag{3}\label{eq3A}$$
Thus, you get
$$a + b + c \equiv a^3 + b^3 + c^3 \pmod 6 \tag{4}\label{eq4A}$$
As such,
$$a + b + c \equiv 0 \pmod 6 \iff a^3 + b^3 + c^3 \equiv 0 \pmod 6 \tag{5}\label{eq5A}$$
|
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|
Direct formula for sequence I want to find out the function to generate the sequence with the following pattern.
1 2 3 1 1 2 2 3 3 1 1 1 1 2 2 2 2 3 3 3 3 ....
Where the lower bound number is 1 upper number bound number is 3. Each time numbers start from 1 and each number repeats 2 ^ n times, with n starting with 0.
|
The following PARI/GP code will do what you want
a(n) = (n-3*2^(k=exponent((n-1)\3+1))+2)\2^k+1;
Note that exponent(n) is the binary exponent of $n$. In other words, the floor of the base $2$ logarithm of $n$. Also, n\m is the floor of $n/m$. Here is an example of use of the code:
? vector(45,n,a(n))
[1,2,3,1,1,2,2,3,3,1,1,1,1,2,2,2,2,3,3,3,3,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3]
The logic is not hard. We need $\lfloor\log_2(n)\rfloor+1$ for something like $[1,2,2,3,3,3,3,4,4,4,4,4,4,4,4,5,\dots]$. Because we repeat 3 times in each cycle we need to replace $n$ with something like $\lfloor n/3\rfloor$. The actual number we need is
$$k = \lfloor\log_2(\lfloor (n-1)/3\rfloor +1)\rfloor. $$ This ensures that
$\, 0 \le n - 3\cdot 2^k +2 < 3\cdot 2^k .\,$ Note that
$\, 1 \le (n - 3\cdot 2^k + 2)/2^k + 1 < 4.$
which is what we want.
|
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|
Find : $\lim\limits_{n\to +\infty}\frac{\left(1+\frac{1}{n^3}\right)^{n^4}}{\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^4}}$ Find :
$$\lim\limits_{n\to +\infty}\frac{\left(1+\frac{1}{n^3}\right)^{n^4}}{\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^4}}$$
My attempt : i don't know is correct or no!
I use this rule :
$$\lim\limits_{n\to +\infty}(f(x))^{g(x)}=1^{\infty}$$
Then :
$$\lim\limits_{n\to +\infty}(f(x))^{g(x)}=\lim\limits_{n\to +\infty}e^{g(x)(f(x)-1)}$$
So :
$$\lim\limits_{n\to +\infty}\frac{\left(1+\frac{1}{n^3}\right)^{n^4}}{\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^4}}$$
$$=\lim\limits_{n\to +\infty}\frac{e^{\frac{n^{4}}{n^{3}}}}{e^{\frac{(n+1)^{4}}{(n+1)^{3}}}}$$
$$=\lim\limits_{n\to +\infty}\frac{e^{n}}{e^{n+1}}$$
$$=\frac{1}{e}$$
Is my approach wrong ?
is this called partial limit calculation ?
|
HINT:
Use the inequality
$$\left(1+\frac{1}{n}\right)^n < e < \left(1+\frac{1}{n}\right)^{n+1}$$
so
$$e^{\frac{1}{n+1}}< \left(1+\frac{1}{n}\right)< e^{\frac{1}{n}}$$
We get lower and upper bounds:
$$e^{\frac{n^4}{n^3+1}}< \left(1+\frac{1}{n^3}\right)^{n^4}< e^n $$
$$e^{\frac{(n+1)^4}{(n+1)^3 +1}}< \left(1 + \frac{1}{(n+1)^3}\right)^{(n+1)^4} < e^{n+1}$$
Conclude:
$$e^{\frac{n^4}{n^3+1}}/e^{n+1}< \ldots < e^n/e^{\frac{(n+1)^4}{(n+1)^3 +1}}$$
so the limit is $\frac{1}{e}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that if $\gcd(a,3)=1$ then $a^7 \equiv a\pmod{63}$. Why is this assumption necessary? Question:
Show that if $\gcd(a,3)=1$ then $a^7 \equiv a\pmod{63} $. Why is this assumption necessary?
Proof:
Since $\gcd(a,3)=1$ $\Leftrightarrow a\equiv 1\pmod 3$ $\Leftrightarrow a^7\equiv 1\pmod3\equiv a\pmod3$
Then using Fermat's Little Theorem:
If $a,p\in\mathbb N$ and $p$ is prime then $a^7\equiv a\pmod7$
$\Rightarrow 3 |a^7-a$ and $7 |a^7-a$
$\Leftrightarrow a^7-a=3k_1$ and $a^7-a=7k_2$
$\Rightarrow (a^7-a)^3=63(k_1)^2k_2$ $\Rightarrow (a^7-a)^3\pmod{63}\equiv 0$
Since $x^m\pmod n\equiv x\pmod n$
$\Rightarrow (a^7-a)^3\pmod{63}\equiv a^7-a\pmod{63}\equiv 0$
$\Leftrightarrow a^7\equiv a\pmod{63}$
The thing I am struggling with is that the question says that the only assumption necessary is that $\gcd(a,3)=1$. Surely there are two assumptions neccessary, since to use Fermat's Little Theorem (in this situation) we need $a\neq 0 \pmod7 \space\space\space(\gcd(a,7)=1)$. I am sure there is something obvious I'm missing -would be great if someone could check over what I have done and point out any mistakes :)
|
$a^7-a=a\left(a^6-1\right)=a(a^2-1)(a^4+a^2+1)=\underbrace{(a-1)}_{\equiv 0\pmod{3}}a(a+1)(a^4+a^2+1)$ is clearly always divisible by $3$ because it's a product of $3$ consecutive integers.
$(\forall a\in\mathbb Z)$
$a^7\equiv a\pmod{7}$
$$a^7-a\equiv 0\pmod{3}\;\land\;a^7-a\equiv 0\pmod{7}\implies a^7-a\equiv 0\pmod{63}$$
|
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|
Simplify $\frac{1}{(p+q)^3}\left(\frac{1}{p^3}+\frac{1}{q^3}\right)+...$
Simplify $\dfrac{1}{(p+q)^3}\big(\dfrac{1}{p^3}+\dfrac{1}{q^3}\big)+\dfrac{3}{(p+q)^4}\big(\dfrac{1}{p^2}+\dfrac{1}{q^2}\big)+\dfrac{6}{(p+q)^5}\big(\dfrac{1}{p}+\dfrac{1}{q}\big)$ if $p\ne -q, p\ne0$ and $q\ne 0$.
$\dfrac{1}{(p+q)^3}\big(\dfrac{1}{p^3}+\dfrac{1}{q^3}\big)+\dfrac{3}{(p+q)^4}\big(\dfrac{1}{p^2}+\dfrac{1}{q^2}\big)+\dfrac{6}{(p+q)^5}\big(\dfrac{1}{p}+\dfrac{1}{q}\big)=\dfrac{p^3+q^3}{p^3q^3(p+q)^3}+\dfrac{3(p^2+q^2)}{p^2q^2(p+q)^4}+\dfrac{6(p+q)}{pq(p+q)^5}$
I guess now it is when I am supposed to calculate the LCD (least common denominator). I have forgotten the algorithm. Can you give me a hint?
|
$$
=\dfrac{(p+q)^2}{(p+q)^5} \times \dfrac{p^3 + q^3}{p^3 q^3} + \dfrac{3(p+q)}{(p+q)^5} \times \dfrac{(p^2+q^2)pq}{p^3q^3} + \dfrac{6}{(p+q)^5} \times \dfrac{(p+q)p^2q^2}{p^3q^3}
$$
$$
=\dfrac{(p+q)^2(p^3+q^3)+3(p+q)(p^2+q^2)pq+6(p+q)p^2q^2}{(p+q)^5p^3q^3}
$$
$$
=\dfrac{(p+q)(p^3+q^3)+3(p^2+q^2)pq+6p^2q^2}{(p+q)^4p^3q^3}
$$
numerator $=p^4+p^3q+pq^3+q^4+3q^3q+3pq^3+6p^2q^2=p^4+4p^3q+6p^2q^2+4pq^3+q^4 = (p+q)^4$
back to the original:
$=\dfrac{(p+q)^4}{(p+q)^4p^3q^3}$
so, the answer is:
$$
\dfrac{1}{p^3q^3}
$$
|
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|
How to know the most efficient way of finding the rank of matrix and how do I apply it to this example For the matrix $
\begin{bmatrix} 1 & 2 & -3 \\ 1 & -2 & 3 \\ 4 & 8 & -12 \\ 1 & -1 & 5 \end{bmatrix}
$
The reduced row echelon I obtained is $
\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}
$
However, the answer is that the rank of this matrix is 2. How do I know if the rank is indeed 2?
|
Doing Gaussian Elimination, you get
$$
\begin{split}
\begin{bmatrix}
1 & 2 & -3 \\
1 & -2 & 3 \\
4 & 8 & -12 \\
1 & -1 & 5
\end{bmatrix}
&\to
\begin{bmatrix}
1 & 2 & -3 \\
0 & 4 & -6 \\
0 & 0 & 0 \\
0 & -3 & 8
\end{bmatrix}
\to
\begin{bmatrix}
1 & 2 & -3 \\
0 & 1 & -3/2 \\
0 & 0 & 0 \\
0 & -3 & 8
\end{bmatrix}
\\ &\to
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 0 \\
0 & 0 & 1
\end{bmatrix}
\to
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{bmatrix}
\end{split}
$$
so you are right, and the rank is 3. Must be a typo in the book.
|
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|
finding a formula for the generating function for a recurring sequence I have the sequence
$a_0=0$, $a_1=3$, $a_2=0$, $a_3=23$,
and
$a_n=6a_{n-2} + 8a_{n-3} + 3a_{n-4}$ for $n\ge 4$
and I have to find the formula for the generating function $A(t)=\sum_{n=0}^\infty a_n t^n$ and find a formula for $a_n$.
So far I have found
$A(t)=(26-18t^3) / (1-6t^2-8t^3-3t^4)$
but when I use this I get a formula for $a_n$ that does not work.
|
The recurrence relation and initial conditions imply
$$A(t)-0 t^0-3 t^1-0 t^2-23 t^3=6t^2 (A(t)-3t^1) + 8t^3 A(t) + 3t^4 A(t).$$
Solving for $A(t)$ yields
$$A(t)=\frac{3t+5t^3}{1-6t^2-8t^3-3t^4}=\frac{1/2}{1-3 t} - \frac{3/2}{1+t} + \frac{3}{(1+t)^2} - \frac{2}{(1+t)^3},$$
which immediately implies that
$$a_n=\frac{1}{2}\cdot 3^n - \frac{3}{2}(-1)^n + 3\binom{n+1}{1}(-1)^n - 2\binom{n+2}{2}(-1)^n=\frac{3^n-(-1)^n (2 n^2 + 1)}{2}.$$
|
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|
Taylor's Polynomial Evaluation Hi guys now started off doing some taylor expansion wanted some help with the following question:
$$ f:[0,\infty)\to \mathbb{R}$$
$$ f(x) = 2e^{-x/2} + e^{-x} $$
Determine the nth degree taylor's polynomila $T_n$ of f about c = 0
My attempt to this question can be seen below;
$$P(x) = f(x) + f'(x)(x-c) + \frac{f''(x)(x-c)^2}{2!} + \frac{f'''(x)(x-c)^3}{3!} + \frac{f''''(x)(x-c)^4}{4!}$$
Evaluation of the function up to the fourth derivative
$$f'(x) = -e^{-x/2} - e^{-x}$$
$$f''(x) = \frac{1}{2}e^{-x/2} + e^{-x}$$
$$f'''(x) = -\frac{1}{4}e^{-x/2} - e^{-x}$$
$$f''''(x) = \frac{1}{8}e^{-x/2} + e^{-x}$$
substituting x = 0 and we obtain the following
$$f'(x) = -2 $$
$$f''(x) = \frac{3}{2} $$
$$f'''(x) = \frac{-5}{4}$$
$$f''''(x) = \frac{9}{8} $$
substituting c = 0 and the derivatives into the taylor's polynomial we obtain the following:-
$$P(x) = 3 - 2x + \frac{\frac{3}{2}x^2}{2!} - \frac{\frac{5}{4}x^3}{3!} + \frac{\frac{9}{8}x^3}{4!} $$
I am not sure if my evaluation is correct and if so I am having difficulty putting into the form as required below hoping someone can help.
$$T_n (x) = \sum_{k=o}^\infty \frac{f^{k}(c)(x-c)^k}{k!}$$
|
You got all the derivatives right, with $f(0) = 3, $$f'(0) = -2$, $f''(0) = \dfrac{3}{2}$, $f'''(0) = -\dfrac{5}{4}$, and so on. As a side note, keep in mind you're evaluating $f^{(n)}$ at $x = 0$, so you saying $f'(x) = -2$, $f''(x) = \dfrac{3}{2}$, and so on isn't correct. (Note that by "derivatives," I'm referring to their values at $x = 0$.)
So the $n^{\text{th}}$-degree Taylor polynomial at $x = 0$ is given by $T_n(x) = \displaystyle\sum_{k = 0}^{n} \dfrac{f^{(k)}(0)}{k!}x^k$. Now, you need to find some way to represent $f^{(n)}(0)$:
*
*First, notice that the signs are alternating. Since the even-order derivatives are positive, you can introduce a factor of $(-1)^k$. This means even $k$'s give $+1$ while odd $k$'s give $-1$.
*Notice how the numerator is always one more than the denominator. Also, notice how the denominators are just powers of two, with the power always being one less than the order. For instance, you have $4 = 2^{3-1}$ for the third derivative. This would indicate a factor of $\dfrac{2^{k-1}+1}{2^{k-1}}$. Does this work for $f(0)$ (which you can think of as the zero-order derivative)? Testing it shows that it does:
$$\frac{2^{0-1}+1}{2^{0-1}} = \frac{\dfrac{1}{2}+1}{\dfrac{1}{2}} = 3 = f(0)$$
Combining the two points gives $f^{(k)} (0) = (-1)^k \dfrac{2^{k-1}+1}{2^{k-1}}$. This means
$$\boxed{T_n(x) = \displaystyle\sum_{k = 0}^{n} (-1)^k \dfrac{2^{k-1}+1}{\left(2^{k-1}\right)k!}x^k}$$
|
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|
can the number $x^2 +y^2$ when with x and y positive integers
can the number $x^2 +y^2$ when with $x$ and $y$ positive integers,
end in $03$?
I know that $x^2+y^2$ can never end with unit digit 03 but am not sure how would I show the proof of that.
|
Let $x = 10a \pm b$ and $y = 10c \pm d$ where $b,d = 0,1,2,3,4$ or $5$.
So $x^2 + y^2 = 100(a^2 + c^2) + 20(\pm ab \pm cd) + (b^2 + d^2)$.
The only way for $b^2 +d^2$ to end in $3$ is if $b=2$ and $d=3$ (or vice versa)
So $x^2 + y^2 =100(a^2 + c^2) + 20(\pm 2a \pm 3c) + 13$
And the only we for that to end in $03$ is for $20(\pm 2a \pm 3c)+10$ to end in $00$ or for $2(\pm 2a \pm 3c)+1$ to end in $0$.
But $2(\pm 2a\pm 3c) +1$ is odd so that can never happen.
|
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|
Find minimum of $a^2+b^2+\frac{a^2b^2}{(a+b)^2}-\frac{2(a^2+ab+b^2)}{a+b}$ If $a,b$ are real numbers, find the minimum value of:
$$a^2+b^2+\frac{a^2b^2}{(a+b)^2}-\frac{2(a^2+ab+b^2)}{a+b}$$
This is what I did: I tried some values and I set $a=0$. Then, it becomes a quadratic of $b$:
$$b^2-2b$$
Here, the minimum is $-1$. So, I tried to prove that:
$$a^2+b^2+\frac{a^2b^2}{(a+b)^2}-\frac{2(a^2+ab+b^2)}{a+b}\ge -1$$
Using Wolfram, I found this is a square:
$$a^2+b^2+\frac{a^2b^2}{(a+b)^2}-\frac{2(a^2+ab+b^2)}{a+b} + 1 = \frac{(a^2+b^2+ab-a-b)^2}{(a+b)^2} $$
so it is positive.
My question is, can we prove this with more traditional and natural solution, maybe with Cauchy-Schwarz?
|
To prove that:
$$a^2+b^2+\frac{a^2b^2}{(a+b)^2}+1 \ge \frac{2(a^2+ab+b^2)}{a+b}$$
we can use AM-GM:
$$
\begin{aligned}
a^2+b^2+\frac{a^2b^2}{(a+b)^2}+1 &\geq 2\sqrt{a^2+b^2+\frac{a^2b^2}{(a+b)^2}}\\
&= 2\sqrt{\frac{(a^2+b^2)^2+2(a^2+b^2)ab+a^2b^2}{(a+b)^2}}\\
&= 2\sqrt{\frac{(a^2+ab+b^2)^2}{(a+b)^2}}\\
&=2\left|\frac{a^2+ab+b^2}{a+b}\right|\\
&\geq \frac{2(a^2+ab+b^2)}{a+b}
\end{aligned}
$$
Equality occurs when $a^2+ab+b^2=a+b$.
Later edit: As a matter of fact, we have
$$a^2+b^2+\frac{a^2b^2}{(a+b)^2}=\frac{(a^2+ab+b^2)^2}{(a+b)^2}$$
therefore, if you substitute $x=\dfrac{a^2+ab+b^2}{a+b}$, the question is rephrased as minimize the function $f(x)=x^2-2x$.
|
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|
Taylor approximation is different from the actual value The problem is to use the $n$th Taylor polynomial to approximate the value of $F(x)=x^{2}\cos{x}$ with $n=3,x_{0}=-2,x=-1.09$.
I found that the $3$rd Taylor polynomial with remainder term is \begin{align*}
x^{2}\cos{x} =& 4\cos(2)+[-4\cos(2)+4\sin(2)][x+2]+[-\cos(2)-4\sin(2)][x+2]^{2}+ \\
& [2\cos(2)+ \dfrac{1}{3}\sin(2)][x+2]^{3} + [\dfrac{(x+2)^{4}}{24}][-8\cos(\xi(x))+16\sin(\xi(x))]
\end{align*}
approximating $x=-1.09$ using the Taylor polynomial,
\begin{align*}
(-1.09)^{2}\cos(-1.09) =& 4\cos(2)+[-4\cos(2)+4\sin(2)][(-1.09)+2]+[-\cos(2)-4\sin(2)][(-1.09)+2]^{2}+ \\
& [2\cos(2)+ \dfrac{1}{3}\sin(2)][(-1.09)+2]^{3} + [\dfrac{((-1.09)+2)^{4}}{24}][-8\cos(\xi(x))+16\sin(\xi(x))] \\
=& 0.0938985 + [\dfrac{((0.91)^{4}}{24}][-8\cos(\xi(x))+16\sin(\xi(x))]
\end{align*}
However, $(-1.09)^{2}\cos(-1.09)=0.549479$ which is a lot of decimals different from Taylor approximation of $0.0938985$.
Is there something that I miss?
|
This just means that teh third order is not sufficient in particular because $-1.09$ is very far away from $-2$.
To make the story short, what you have is
$$x^2 \cos(x)=\sum_{n=0}^\infty \frac{\left(-n^2+n+4\right) \cos \left(2-\frac{\pi n}{2}\right)+4 n \sin \left(2-\frac{\pi n}{2}\right)}{n!}(x+2)^n$$ So, if you truncate to $O((x+2)^{p+1})$ you will have the following results
$$\left(
\begin{array}{cc}
2 & 0.4926841712 \\
3 & 0.0938985194 \\
4 & 0.6047229336 \\
5 & 0.5723469909 \\
6 & 0.5466012534 \\
7 & 0.5489493571 \\
8 & 0.5495411050 \\
9 & 0.5494858556 \\
10 & 0.5494781119 \\
11 & 0.5494788050 \\
12 & 0.5494788702 \\
13 & 0.5494788647 \\
14 & 0.5494788644
\end{array}
\right)$$
|
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|
Find the value of C for which the following integral converges $$I=\int_0^\infty \frac {x} {x^2+1} - \frac {C} {3x+1} dx$$
Find the value of C for which the integral converges.
I've gotten as far as solving the integral
$\frac 1 2 \ln|x^2+1| - \frac 1 3 c\ln|3x+1|$ but not sure where to go from here as evaluating it gives $\infty - 0 - c \infty + 0$. Any help would be appreciated!
|
You can do this without finding the actual integral. The basic question is what is the behaviour of the integrand as $x \to \infty$. We have $$ \frac{x}{x^2+1} = \frac{1}{x (1+1/x^2)} = \frac{1}{x} \left(1 - \frac{1}{x^2} + \ldots \right) = \frac{1}{x} + O\left(\frac{1}{x^3}\right)\ \text{as}\ x \to \infty$$
while
$$\frac{C}{3x+1} = \frac{C}{3 x (1 + 1/(3x))} = \frac{C}{3x} + O\left(\frac{1}{x^2}\right) $$
So if $C = 3$,
$$ \frac{x}{x^2+1} - \frac{3}{3x+1} = O\left(\frac{1}{x^2}\right) $$
which is integrable as $x \to \infty$. If $C \ne 3$, you have a term in $1/x$ which
is not integrable.
|
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|
Prove that if $a,b,c > 0$ and $a + b + c = 1$, we have: $\frac{a^2}{a^3 + 5} + \frac{b^2}{b^3 + 5} + \frac{c^2}{c^3 + 5} \leq \frac{1}{4}$ Prove that if $a,b,c > 0$ such that $a + b + c = 1$, then the following inequality holds:
$$S = \frac{a^2}{a^3 + 5} + \frac{b^2}{b^3 + 5} + \frac{c^2}{c^3 + 5} \leq \frac{1}{4}$$
What I have tried so far is the following:
Firstly I rewrote $S$ as:
$$ S = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} - 5\left[ \frac{1}{a(a^3 + 5)} + \frac{1}{b(b^3 + 5)} + \frac{1}{c(c^3 + 5)}\right]$$
Then, in order to upper bound $S$, I used the inequality: $ \frac{x^2}{u} + \frac{y^2}{v} + \frac{z^2}{w} \geq \frac{(x + y + z)^2}{u + v + w} $ for any $u,v,w > 0$.
Therefore, I got:
$$ S \leq \frac{1}{a} + \frac{1}{b} + \frac{1}{c} - 5 \cdot \frac{(1 + 1 + 1)^2}{a^4 + b^4 + c^4 + 5(a + b + c)} = \frac{45}{a^4 + b^4 + c^4 + 5}$$
Then, given what we want to show about $S$, this would reduce to proving that:
$$ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \leq \frac{185 + a^4 + b^4 + c^4}{4(a^4 + b^4 + c^4 + 5)} $$
At which point I got stuck and I am not sure whether I started the right way.
I would be grateful for any suggestions. Many thanks!
|
Based on Michael Rozenberg's idea, we also have:
$$\frac{a^2}{a^3+5}=\frac{a}{6}-\frac{a(1-a)(5-a^2-a)}{a^3+5}\leq \frac{a}{6}$$
so that:
$$\frac{a^2}{a^3+5}+\frac{b^2}{b^3+5}+\frac{c^2}{c^3+5}\leq \frac{1}{6}(a+b+c)=\frac{1}{6}$$
Equality is attained when $(a,b,c)=(1,0,0)$ up to any cyclic permutation.
We can use the tangent line method to find the minimum:
$$
\begin{aligned}
\frac{a^2}{a^3+5}&=\frac{3}{136}+\frac{2421}{18496}\left(a-\frac{1}{3}\right)+\frac{(3a-1)^2(1995-135a-269a^2)}{18496(a^3+5)}\\
&\geq \frac{3}{136}+\frac{2421}{18496}\left(a-\frac{1}{3}\right)
\end{aligned}
$$
so that the minimum value is $\frac{9}{136}$ when $a=b=c=\frac{1}{3}$. Therefore the entire range of the left hand side is $\left[\frac{9}{136},\frac{1}{6}\right]$.
|
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|
Prove $\sqrt[3]{\frac{(a^4+b^4)(a^4+c^4)(b^4+c^4)}{abc}}\ge \sqrt{\frac{(a^3+bc)(b^3+ca)(c^3+ab)}{1+abc}}$ If $a,b,c>0$, prove that:
$$\sqrt[3]{\frac{(a^4+b^4)(a^4+c^4)(b^4+c^4)}{abc}}\ge \sqrt{\frac{(a^3+bc)(b^3+ca)(c^3+ab)}{1+abc}}$$
My try: I used $a^4+b^4 \geq ab(a^2+b^2)$ to get:
$$(a^4+b^4)(a^4+c^4)(b^4+c^4) \geq a^2b^2c^2(a^2+b^2)(a^2+c^2)(b^2+c^2)$$
I am stuck with this inequality:
$$\sqrt[3]{(a^2+b^2)(a^2+c^2)(b^2+c^2)}\ge \sqrt{\frac{(a^3+bc)(b^3+ca)(c^3+ab)}{1+abc}}$$
I am not sure if it is true or not.
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It's wrong.
Let $a=b=c.$
Thus, we have $$2a^2\geq\sqrt{\frac{(a^3+a^2)^3}{1+a^3}}$$ or
$$2\geq a\sqrt{\frac{(a+1)^2}{a^2-a+1}},$$ which is wrong for $a\rightarrow+\infty$.
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"url": "https://math.stackexchange.com/questions/3569404",
"timestamp": "2023-03-29T00:00:00",
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|
$\log_{yz} \left(\frac{x^2+4}{4\sqrt{yz}}\right)+\log_{zx}\left(\frac{y^2+4}{4\sqrt{zx}}\right)+\log_{xy}\left(\frac{z^2+4}{4\sqrt{xy}}\right)=0$ Can I ask how to solve this type of equation:
$$\log_{yz} \left(\frac{x^2+4}{4\sqrt{yz}}\right)+\log_{zx}\left(\frac{y^2+4}{4\sqrt{zx}}\right)+\log_{xy}\left(\frac{z^2+4}{4\sqrt{xy}}\right)=0$$
It is given that $x,y,z>1$. Which properties of the logarithm have to be used?
I know that $\log_a b=\log a/\log b\to \log_{yz}((x^2+4)/(4\sqrt{yz}))=\log ((x^2+4)/(4\sqrt{yz}))/\log xy$
and $\log_a b/c=\log_a b/\log_a c\to \log_{yz}((x^2+4)/(4\sqrt{yz}))=\log_{yz} (x^2+4)/\log_{yz} (4\sqrt{yz})$
And how to solve this type of equation in general?
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By AM-GM
$$0=\sum_{cyc}\log_{yz}\frac{x^2+4}{4\sqrt{yz}}\geq\sum_{cyc}\log_{yz}\frac{2\sqrt{x^2\cdot4}}{4\sqrt{yz}}=\sum_{cyc}\left(\log_{yz}x-\frac{1}{2}\right)=$$
$$=\sum_{cyc}\frac{2\ln{x}-\ln{y}-\ln{z}}{2(\ln{y}+\ln{z})}=\frac{1}{2}\sum_{cyc}\frac{\ln{x}-\ln{y}-(\ln{z}-\ln{x})}{\ln{y}+\ln{z}}=$$
$$=\frac{1}{2}\sum_{cyc}(\ln{x}-\ln{y})\left(\frac{1}{\ln{y}+\ln{z}}-\frac{1}{\ln{z}+\ln{x}}\right)=$$
$$=\sum_{cyc}\frac{(\ln{x}-\ln{y})^2}{2(\ln{y}+\ln{z})(\ln{x}+\ln{z})}\geq0.$$
The equality occurs only for $x=y=z=2$ and $\ln{x}=\ln{y}=\ln{z},$
which gives that $\{(2,2,2)\}$ is an answer.
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"timestamp": "2023-03-29T00:00:00",
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|
Find the relationship between and (complex numbers) $ z = x + \text{j} \cdot y $ and $ \arg(z) = \frac{\pi}{4} $. Find the relationship between $x$ any $y$, then find $z$ for
$|z - 3 + 2 \cdot \text{j}| = |z +3 \cdot \text{j}|$, where $\text{j}^2= -1$,
In the above question do I have to compare like this:
$Z= x+\text{j}\cdot y$, $Z=\tan^{-1}\left(\dfrac{\text{j} \cdot y}{x}\right)=arg(z)= \dfrac{\pi}{4}$? to find $x$, $y$ and $z$?
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Well, when we know that the argument of a complex number is $\frac{\pi}{4}$ we can write:
$$\arg\left(\text{a}+\text{b}i\right)=\arctan\left(\frac{\text{b}}{\text{a}}\right)=\frac{\pi}{4}\space\Longleftrightarrow\space\frac{\text{b}}{\text{a}}=\tan\left(\frac{\pi}{4}\right)=1\space\Longleftrightarrow\space\text{a}=\text{b}>0\tag1$$
So, we get:
$$\left|\text{z}-3+2i\right|=\left|\text{z}+3i\right|\space\Longleftrightarrow\space\sqrt{\left(\text{a}-3\right)^2+\left(\text{a}+2\right)^2}=\sqrt{\text{a}^2+\left(\text{a}+3\right)^2}\tag2$$
Which give:
$$\text{a}=\frac{1}{2}\tag3$$
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"timestamp": "2023-03-29T00:00:00",
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|
For how many integers $n$ is $n^6+n^4+1$ a perfect square? QUESTION
For how many integers $n$ is $n^6+n^4+1$ a perfect square?
I am completely blank on how to start. Could anyone please provide tricks on how to get a start on such questions?
Thanks for any answers!
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Let $m$ and $n$ be integers such that $m^2=n^6+n^4+1$. Without loss of generality, we may assume that $m$ and $n$ are nonnegative. Clearly, $(m,n)=(1,0)$ is the only solution when $n\in\{0,1\}$.
If $n\ge 2$, then $n^2\geq 4$, so that $$(2n^3+n)^2=4n^6+4n^4+n^2\geq 4n^6+4n^4+4=4m^2\,.$$ On the other hand, $$\begin{align}4n^3-n^2+2n+3&\geq 8n^2-n^2+2n+3=7n^2+2n+3\\&\geq 7\left(n+\frac{1}{7}\right)^2+\frac{20}{7}\geq \frac{20}7>0\,,\end{align}$$
whence
$$\begin{align}(2n^3+n-1)^2&=4n^6+4n^4-4n^3+n^2-2n+1\\&<4n^6+4n^4+4=4m^2\,.\end{align}$$
Therefore, $$(2n^3+n-1)^2<(2m)^2\leq (2n^3+n)^2\,.$$
Ergo, $4n^6+4n^4+4=(2m)^2=(2n^3+n)^2$, which means $(m,n)=(9,2)$.
With signs involved, there are in total $3$ possible values of integers $n$: $-2$, $0$, and $+2$. The equation $m^2=n^6+n^4+1$ has $6$ solutions $(m,n)\in\mathbb{Z}\times\mathbb{Z}$: $(\pm 1,0)$ and $(\pm 9,\pm 2)$.
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|
Prove $\sum_{n=1}^{\infty}\frac{n-1}{2^{n+1}}=\frac{1}{2}$, without using any $\sum\frac{n}{2^{n}}$ Problem: Prove $\sum_{n=1}^{\infty}\frac{n-1}{2^{n+1}}=\frac{1}{2}$, without using any $\sum\frac{n}{2^{n}}$.
I manage to make progress, but then end up always getting a $\sum\frac{n}{2^{n}}$. For example, $\sum_{n=1}^{\infty}\frac{n-1}{2^{n+1}}$ = $\sum_{n=0}^{\infty}\frac{n}{2^{n+2}}$ = $\frac{1}{4}\sum_{n=0}^{\infty}\frac{n}{2^{n}}$, which is the sum we're not allowed to use.
Any help on how to proceed would be appreciated!
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Here is a fairly general approach. Note that $n-1=\sum_{m=1}^{n-1}(1)$ so that
$$\begin{align}
\sum_{n=1}^\infty (n-1)a_n &=\sum_{n=2}^\infty (n-1)a_n\\\\
&=\sum_{n=2}^\infty \sum_{m=1}^{n-1} a_n\\\\
&=\sum_{m=1}^\infty \sum_{n=m+1}^\infty a_n\tag1
\end{align}$$
Now, take $a_n=x^n$, $|x|<1$ in $(1)$ and using $\sum_{n=m+1}^\infty x^n=\frac{x^{m+1}}{1-x}$ we have
$$\begin{align}
\sum_{n=1}^\infty (n-1)x^n&=\sum_{m=1}^\infty \frac{x^{m+1}}{1-x}\\\\
&=\frac{x^2}{(1-x)^2}\tag2
\end{align}$$
Using $x=1/2$ in $(2)$ yields
$$\sum_{n=1}^\infty \frac{n-1}{2^n}=1$$
from which we see that
$$\begin{align}
\sum_{n=1}^\infty \frac{n-1}{2^{n+1}}&=\frac12
\end{align}$$
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|
For non-negative reals $a$, $b$, $c$, show that $3(1-a+a^2)(1-b+b^2)(1-c+c^2)\ge(1+abc+a^2b^2c^2)$ A 11th grade inequality problem:
Let $a,b,c$ be non-negative real numbers. Prove that
$$3(1-a+a^2)(1-b+b^2)(1-c+c^2)\ge(1+abc+a^2b^2c^2)$$
Do you have any hints to solve this inequality? Any hints would be fine.
I tried this:
$$3(1-a+a^2)(1-b+b^2)(1-c+c^2)\ge(1+abc+a^2b^2c^2)$$
By Cauchy's inequality, $$a^2+1\ge(2a)$$ and I did the same to $b$ and $c$ and applied it to the problem but the results are $$2abc\ge1+a^2b^2c^2$$ and this is wrong.
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Another way.
Since $$3(1-a+a^2)^3-1-a^3-a^6=(a-1)^4(2a^2-a+2),$$ by Holder we obtain:
$$3\prod_{cyc}(1-a+a^2)\geq\sqrt[3]{\prod_{cyc}(1+a^3+a^6)}\geq1+abc+a^2b^2c^2.$$
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|
Looking for a short proof of a harmless looking binomial identity I managed to prove for this MSE post the rather harmless looking binomial identity for natural $1\leq k\leq n$:
\begin{align*}
\color{blue}{\sum_{j=0}^k\binom{2n}{2j}\binom{n-j}{k-j}=\binom{n+k}{n-k}\frac{4^kn}{n+k}}\tag{1}
\end{align*}
using the coefficient of operator method. Admittedly, there are a lot of intermediate steps used to show the validity of (1).
Question: I'm wondering if there is a more direct, less lengthy derivation than the one I've provided below.
We obtain for $1\leq k\leq n$:
\begin{align*}
\color{blue}{\sum_{j=0}^k}&\color{blue}{\binom{2n}{2j}\binom{n-j}{k-j}}\\
&=\sum_{j=0}^n\binom{2n}{2j}\binom{n-j}{n-k}\tag{2}\\
&=\sum_{j=0}^n\binom{2n}{2j}[z^{n-k}](1+z)^{n-j}\tag{3}\\
&=[z^{n-k}](1+z)^n\sum_{j=0}^n\binom{2n}{2j}\frac{1}{(1+z)^j}\\
&=\frac{1}{2}[z^{n-k}](1+z)^n\left(\left(1+\frac{1}{\sqrt{1+z}}\right)^{2n}+\left(1-\frac{1}{\sqrt{1+z}}\right)^{2n}\right)\\
&=\frac{1}{2}[z^{n-k}]\left(\left(1+\sqrt{1+z}\right)^{2n}+\left(1-\sqrt{1+z}\right)^{2n}\right)\\
&=\frac{1}{2}[z^{n-k}]\left(1+\sqrt{1+z}\right)^{2n}\tag{4}\\
&=\frac{1}{2}[z^{-1}]z^{-n+k-1}\left(1+\sqrt{1+z}\right)^{2n}\tag{5}\\
&=\frac{1}{2}[w^{-1}]\left(w^2-1\right)^{n-k-1}(1+w)^{2n}2w\tag{6}\\
&=[w^{-1}]w(w-1)^{-n+k-1}(w+1)^{n+k-1}\\
&=[u^{-1}](u+1)u^{-n+k-1}(u+2)^{n+k-1}\tag{7}\\
&=\left([u^{n-k}]+[u^{n-k-1}]\right)\sum_{j=0}^{n+k-1}\binom{n+k-1}{j}u^j2^{n+k+1-j}\\
&=\binom{n+k-1}{n-k}2^{2k-1}+\binom{n+k-1}{n-k-1}2^{2k}\tag{8}\\
&=\binom{n+k}{n-k}\frac{2k}{n+k}2^{2k-1}+\binom{n+k}{n-k}\frac{n-k}{n+k}2^{2k}\tag{9}\\
&\,\,\color{blue}{=\binom{n+k}{n-k}\frac{4^kn}{n+k}}
\end{align*}
and the claim follows.
Comment:
*
*In (2) we use the binomial identity $\binom{p}{q}=\binom{p}{p-q}$. We also set the upper index to $n$ without changing anything, since we are adding zeros only.
*In (3) we use the coefficient of operator method.
*In (4) we skip $\left(1-\sqrt{1+z}\right)^{2n}=cz^{2n}+\cdots$ since it has only powers of $z$ greater than $n$ and does not contribute to $[z^{n-k}]$.
*In (5) we apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.
*In (6) we use the transformation of variable formula $[z^{-1}]f(z)=[w^{-1}]f(g(w))g^\prime(w)$ with $1+z=w^2, \frac{dz}{dw}=2w$.
*In (7) we use the transformation of variable formula again, with $w-1=u, \frac{dw}{du}=1$.
*In (8) we select the coefficients accordingly.
*In (9) we use the binomial identities $\binom{p-1}{q}=\binom{p}{q}\frac{p-q}{p}$ and $\binom{p}{q}=\binom{p-1}{q-1}\frac{p}{q}$.
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Here is an alternate solution, where the number of steps is about
the same as what OP provided. Could use additional streamlining by
removing some of the simpler proceedings. Start as follows:
$$\sum_{j=0}^k {2n\choose 2j} {n-j\choose k-j}
= \sum_{j=0}^k {2n\choose 2k-2j} {n-k+j\choose j}
\\ = [z^{2k}] (1+z)^{2n} \sum_{j=0}^k z^{2j} {n-k+j\choose j}.$$
Here the coefficient extractor enforces the range:
$$[z^{2k}] (1+z)^{2n} \sum_{j\ge 0} z^{2j} {n-k+j\choose j}
\\ = [z^{2k}] (1+z)^{2n} \frac{1}{(1-z^2)^{n-k+1}}
= [z^{2k}] (1+z)^{n+k-1} \frac{1}{(1-z)^{n-k+1}}.$$
This is
$$\mathrm{Res}_{z=0} \frac{1}{z^{2k+1}}
(1+z)^{n+k-1} \frac{1}{(1-z)^{n-k+1}}
\\ = (-1)^{n-k+1} \mathrm{Res}_{z=0} \frac{1}{z^{2k+1}}
(1+z)^{n+k-1} \frac{1}{(z-1)^{n-k+1}}.$$
Now the residue at infinity is zero so this is minus the residue
at one:
$$(-1)^{n-k} \mathrm{Res}_{z=1} \frac{1}{(1+(z-1))^{2k+1}}
(2+(z-1))^{n+k-1} \frac{1}{(z-1)^{n-k+1}}
\\ = (-1)^{n-k} \sum_{j=0}^{n-k} {n+k-1\choose j} 2^{n+k-1-j}
(-1)^{n-k-j} {n-k-j+2k\choose 2k}
\\ = 2^{n+k-1} \sum_{j=0}^{n-k} {n+k-1\choose j} 2^{-j}
(-1)^{j} {n+k-j\choose n-k-j}.$$
Coefficient extractor enforces range:
$$2^{n+k-1} [z^{n-k}] (1+z)^{n+k}
\sum_{j\ge 0} {n+k-1\choose j} 2^{-j}
(-1)^{j} \frac{z^j}{(1+z)^j}
\\ = 2^{n+k-1} [z^{n-k}] (1+z)^{n+k}
\left(1-\frac{z}{2(1+z)}\right)^{n+k-1}
\\ = [z^{n-k}] (1+z) (2+z)^{n+k-1}
\\ = [z^{n-k}] (2+z)^{n+k-1} + [z^{n-k-1}] (2+z)^{n+k-1}
\\ = {n+k-1\choose n-k} 2^{n+k-1-(n-k)}
+ {n+k-1\choose n-k-1} 2^{n+k-1-(n-k-1)}
\\ = \frac{1}{2} 4^k \frac{2k}{n+k} {n+k\choose n-k}
+ \frac{n-k}{n+k} 4^k {n+k\choose n-k}
\\ = \frac{4^k n}{n+k} {n+k\choose n-k}.$$
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"url": "https://math.stackexchange.com/questions/3583191",
"timestamp": "2023-03-29T00:00:00",
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|
The integral: $\int_{0}^{\pi/2} \frac{\sin x \cos^5 x}{(1-2\sin^2x\cos^2x)^2}dx$ The integral: $$\int_{0}^{\pi/2} \frac{\sin x \cos^5 x}{(1-2\sin^2x\cos^2x)^2}dx$$
has been encountered today while solving a longlish problem at MSE. The question here is: How would one evaluate it?
Addendum
For an interesting use of this integral see my Answer to: Evaluate: $S=\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2k}{j+k+1}$
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Amazing : the tangent half angle substitution works. Let $x=2\tan^{-1}(t)$ to get
$$I=-4\int\frac{ t \left(t^2-1\right)^5 \left(t^2+1\right)}{\left(t^8-4 t^6+22 t^4-4
t^2+1\right)^2}\,dt$$ Now, a non-trivial substitution
$$t=\sqrt{1+\frac{2 \left(\sqrt{z+1}\right)}{z}}\implies dt=-\frac{1}{2} \sqrt{\frac{z+ 2 \left(\sqrt{z+1}\right)}{z^3(z+1)}}\,dz$$
$$I=\frac{1}{2} \int \frac{dz}{ \left(z^2+1\right)^2}=\frac{1}{4} \left(\frac{z}{z^2+1}+\tan ^{-1}(z)\right)$$
Back to $t$
$$I=\frac{t^2 \left(t^2-1\right)^2}{t^8-4 t^6+22 t^4-4 t^2+1}+\frac{1}{4} \tan ^{-1}\left(\frac{4 t^2}{\left(1-t^2\right)^2}\right)$$
Expanded as a series around $t=1$ gives for the definite integral
$$\frac{\pi }{8}-\frac{1}{6} (1-t)^6+O\left((1-t)^7\right)$$
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|
The value of parameter $a$ for which $\dfrac{ax^2+3x-4}{a+3x-4x^2}$ takes all real values for $x\in R$ are: => The value of parameter $a$ for which $\dfrac{ax^2+3x-4}{a+3x-4x^2}$
takes all real values for $x\in R$ are:
My question is why we need to validate end points i.e. $1,7$ (Refer the last part of my attempt)
My attempt is as follows:-
$$y=\dfrac{ax^2+3x-4}{a+3x-4x^2}$$
$$ya+3yx-4yx^2=ax^2+3x-4$$
$$x^2(-4y-a)+x(3y-3)+ya+4=0$$
As $x$ can be any real, so $D\ge0$
$$9y^2+9-18y-4(ya+4)(-4y-a)\ge0$$
$$9y^2+9-18y+4(4y^2a+ya^2+16y+4a)\ge0$$
$$y^2(9+16a)+y(4a^2+64-18)+9+16a\ge0$$
$$y^2(9+16a)+y(4a^2+46)+9+16a\ge0$$
As range is $R$, so discriminant of quadratic in $y$ should be less than equal to zero
$$4(2a^2+23)^2-4(9+16a)^2\le0$$
$$(2a^2+23-9-16a)(2a^2+23+9+16a)\le0$$
$$(2a^2-16a+14)(2a^2+16a+32)\le0$$
$$(a^2-8a+7)(a^2+8a+16)\le0$$
$$a\in[1,7]$$
But in such type of questions, we always check at endpoints like here we need to check at $a=1$ and $a=7$. But I don't understand what is so special about endpoints.
From the above calculation I can only say at $a=1,7$ discriminant of quadratic in $y$ is zero, but what is so special about this. Please help me in this.
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Let's denote $f(x)=ax^2+3x−4$, $g(x)=-4x^2+3x+a$, and $h(x)=f(x)/g(x)$.
Since $a>0>-9/16>-4$, $f(x)$ and $g(x)$ cannot have the same roots, and each of them has two distinct roots. However, they may have a common root. If $a=1,7$ then there exists $y_0\in \mathbb{\mathbb{R}}$ for which only one $x$ may give $h(x)=y_0$. This $x$ may be the root of the denominator, so we must check. Else, when $1<a<7$, there exists two distinct candidates $x_1,x_2$ which may give $h(x)=y$. In this case at least one of $x_1,x_2$ is not a root for $g$, since otherwise they will be the roots of $f$.
|
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|
Prove $\tan^{-1}m+\tan^{-1}n=\cos^{-1}\frac{1-mn}{\sqrt{1+m^2}\sqrt{1+n^2}}$ (part of sum) I'm looking for a specific clarification for a part of the solution in proving the following identity.
$\tan^{-1}m+\tan^{-1}n=\cos^{-1}\frac{1-mn}{\sqrt{1+m^2}\sqrt{1+n^2}}$
Here I'm taking, $\theta=\tan^{-1}m;$ $-\pi/2<\theta<\pi/2$
so I get $\tan\theta=m$---(1)
I need to find $\sin \theta $and $\cos \theta$ in terms of m
By Trigonometric identity I can easily derive, $\cos \theta$
$\tan^2\theta+1=\sec^2\theta$
$\cos^2\theta=\frac{1}{m^2+1}$
$\cos\theta=+\sqrt\frac{1}{m^2+1}$ ( here only plus due to range of $\theta$)
Now if I deduce $\sin\theta$ from,
$\sin^2\theta+\cos^2\theta=1$
I get, $\sin\theta=\pm\sqrt\frac{m^2}{m^2+1}$ ( I have to take $\pm$ because of range of $\theta$)
But if I deduce $\sin\theta$ from (1)
I get, $\sin\theta=\frac{m}{\sqrt {m^2+1}}$
Which of the following method is correct to find $\sin\theta$? Please help me. Thank you!
P.S. I'm not interested in the solution. What I need to know is how to find $\sin\theta$
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Using this
We need $\tan^{-1}m+\tan^{-1}n\ge0$ to admit the equality
$$\tan^{-1}m+\tan^{-1}n\ge0\iff\tan^{-1}m>-\tan^{-1}n=\tan^{-1}(-n)$$
$$\iff m\ge-n\iff m+n\ge0$$
Now use Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$
Then if $\tan^{-1}x=y,-\dfrac\pi2<y<\dfrac\pi2,\cos y>0$
$x=\tan y\implies\cos y=\dfrac1{\sqrt{1+x^2}}$
$\tan^{-1}x=\text{sign of}(x)\cdot\cos^{-1}\dfrac1{\sqrt{1+x^2}}$
See also : Proving that $\cos^{-1}\frac{4}{5}+\cos^{-1}\frac{12}{13}=\cos^{-1}\frac{33}{65}$
|
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|
Solve the following system of equations by Cross Multiplication Method Solve the following System of equations by Cross Multiplication Method
$$2x-2y+3z=20$$
$$x+3y-z=12$$
$$3x-y+4z=22$$
My Attempt:
Let us solve first two equations:
$$\dfrac {x}{2-9}=\dfrac {y}{3-(-2)}=\dfrac {z}{6-(-2)} = k(let)$$
$$\dfrac {x}{-7}=\dfrac {y}{5}=\dfrac {z}{8}=k$$
So we have: $x=-7k$, $y=5k$ and $z=8k$
Now, putting these values of $x$,$y$ and $z$ in third equation,
$$3(-7k)-5k+4(8k)=22$$
$$k=\dfrac {11}{3}$$
Thus,
$x=\dfrac {-77}{3}$,
$y=\dfrac {55}{3}$ and
$z=\dfrac {88}{3}$.
However the actual solution is $(x,y,z)=(21,-7,-12)$. Where has the error occured in this solution?
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We want to use the Cross Multiplication Method (CMM) to solve
$$\begin{align} 2x-2y+3z&=20 \tag 1 \\ x+3y-z&=12 \tag 2 \\ 3x-y+4z&=22 \tag 3 \end{align}$$
We can write $(1)$ and $(2)$ as
$$\begin{align} 2x-2y+ (3z - 20)&=0 \\ x+3y-(z+12)&=0 \end{align}$$
By the CMM
$$\dfrac{x}{-2(-z-12)-3(3z - 20)} = \dfrac{y}{(3z-20)-2(-z-12)} = \dfrac{1}{2(3) - 1(-2)}$$
Simplifying
$$\dfrac{x}{84-7z} = \dfrac{y}{4 + 5z} = \dfrac{1}{8}$$
Solving
$$x = \dfrac{84-7z}{8}, y = \dfrac{4+5z}{8}$$
We now substitute these into $(3)$
$$3\left(\dfrac{84-7z}{8}\right) - \left(\dfrac{4+5z}{8}\right) + 4 z = 22$$
This gives $z = -12$, hence
$$(x, y, z) = (21, -7, -12)$$
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int\sqrt{4x^2+8x}dx$ using $\int\sqrt{u^2-a^2}dx=\frac u2\sqrt{u^2-a^2}-\frac{a^2}{2}\ln|u+\sqrt{u^2-a^2}|+C.$ I completed the square to get $$2 \sqrt{(x+1)^2 - 1}$$
set $u=x+1$ and $a=1$, and you end up with
$$(x+1) \sqrt{(x+1)^2 - 1} - \ln\left\lvert x+1 + \sqrt{(x+1)^2-1}\right\rvert+C$$
This is not the right answer. I don't see how you could simplify it more, but even if you could it would still be wrong (I put this and the answer given by Symbolab.com into Desmos.com and they didn't line up).
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Your answer is equivalent to the answer provided by Symoblab
$$x\sqrt{x^2+2x}-\ln\left\lvert x + \sqrt{x^2+2x}+1\right\rvert+\sqrt{x^2+2x}+C$$
since starting from
$$(x+1) \sqrt{(x+1)^2 - 1} - \ln\left\lvert x+1 + \sqrt{(x+1)^2-1}\right\rvert+C$$
we have that
$$\sqrt{(x+1)^2 - 1}=\sqrt{x^2+2x}$$
therefore
$$(x+1) \sqrt{(x+1)^2 - 1}=x\sqrt{x^2+2x}+\sqrt{x^2+2x}$$
and
$$\ln\left\lvert x+1 + \sqrt{(x+1)^2-1}\right\rvert=\ln\left\lvert x + \sqrt{x^2+2x}+1\right\rvert$$
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"url": "https://math.stackexchange.com/questions/3596609",
"timestamp": "2023-03-29T00:00:00",
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Show that the line $x+y=q$ intersects the ellipse $x^2-2x+2y^2=3$ at two different points if $q^2<2q+5$ What should be the proper way to answer this question?
My working is as follows, but is it the right way to answer the question, since it seems like I am only showing that $q^2<2q+5$?
$x+y=q \tag{1}$
$x^2-2x+2y^2=3 \tag{2}$
From $(1)$, $y=q-x\tag{3}$
Substitute $(3)$ into $(2)$,
$\begin{align}x^2-2x+2(q-x)^2&=3\\x^2-2x+2(q^2-2qx+x^2)&=3\\3x^2-(2+4q)x+2q^2-3&=0\end{align}$
Since the line intersects the ellipse at two different points, therefore $b^2-4ac>0$
$(-2-4q)^2-4(3)(2q^2-3)>0$
$4+16q+16q^2-24q^2+36>0$
$8q^2-16q-40<0$
$q^2-2q-5<0$
$q^2<2q+5$
Does my working answer the quesiton, or are there more accurate approaches to solve this question?
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Your calculations are correct. Here you can find also for what values of $q$ the line $x+y=q$ and the ellpise $x^2-2x+2y^2=3$ have two points in common.
First, you can rewrite the equation of the ellipse in the form:
$$(x-1)^2+2y^2=4$$
Now, substituing $y=q-x$, you have:
$$(x-1)^2+2(q-x)^2=4\leftrightarrow 3x^2-2x(2q+1)+2q^2-3=0\leftrightarrow x=\frac{2(2q+1)\pm\sqrt{4(2q+1)^2-12(2q^2-3)}}{6}$$
The $|Delta$ must be greater then $0$ and so:
$$q^2-2q-5<0 \leftrightarrow q<1-\sqrt{6} \lor q>1+\sqrt{6}$$
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Solve for $x$: $\log_{\frac {3}{4}}(\log_{8} (x^2+7))+\log_{\frac {1}{2}} (\log_{\frac {1}{4}} (x^2+7)^{-1})=-2$ Solve for $x$: $\log_{\frac {3}{4}}(\log_{8} (x^2+7))+\log_{\frac {1}{2}} (\log_{\frac {1}{4}} (x^2+7)^{-1})=-2$
My Attempt:
$$\log_{\frac {3}{4}}(\frac {1}{3} \log_2 (x^2+7))+\log_{\frac {1}{2}} (\frac {1}{2}\log_2 (x^2+7))=-2$$
$$\log_{\frac {3}{4}}(\frac {1}{3} \log_2 (x^2+7))+1+\log_{1}{2}(\frac {1}{2} \log_2(x^2+7))=-1$$
$$\log_\frac {3}{4} (\frac {1}{3} \log_2 (x^2+7))+\log_{\frac {3}{4}} (\dfrac {3}{4})=-(1+\log_{\frac {1}{2}} (\frac {1}{2} \log_2 (x^2+7)))$$
$$\log_{\frac {3}{4}} (\frac {1}{4} \log_2 (x^2+7))=-(\log_{\frac {1}{2}} (\frac {1}{2})+\log_{\frac {1}{2}} (\frac {1}{2} \log_2 (x^2+7)))$$
$$\log_{\frac {3}{4}} (\frac {1}{4} \log_2 (x^2+7))=-\log_{\frac {1}{2}} (\frac {1}{4} \log_2 (x^2+7))$$
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Let's use
\begin{eqnarray*}
\log_b(a) = \frac{\ln(a)}{\ln(b)}
\end{eqnarray*}
to change everything into natural logarithms.
Your equation becomes
\begin{eqnarray*}
\frac{\ln \left( \frac{\ln(x^2+7)}{\ln(8)} \right) }{\ln(3/4)} + \frac{\ln \left( - \frac{\ln(x^2+7)}{\ln(1/4)} \right) }{\ln(1/2)} =-2 \\
\end{eqnarray*}
\begin{eqnarray*}
\frac{\ln ( \ln(x^2+7)) -\ln(\ln(8)) }{\ln(3/4)} + \frac{ \ln(\ln(x^2+7)) -\ln (\ln(4)) }{\ln(1/2)} =-2 \\
\end{eqnarray*}
It is now just linear algebra to isolate $\ln ( \ln(x^2+7))$.
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|
Find $\lim_{(x,y)\rightarrow(0,0)} \cos(\frac{x^2-y^2}{\sqrt{x^2+y^2}})$ $\lim_{(x,y)\rightarrow(0,0)} \cos(\frac{x^2-y^2}{\sqrt{x^2+y^2}})$
I tried replacing x and y with several values and kept getting 1 so I tried:
$$0 \le |\cos(\frac{x^2-y^2}{\sqrt{x^2+y^2}})| \le |\cos(|(\frac{x^2-y^2}{\sqrt{x^2+y^2}}|)|\le |\cos(\frac{|x|^2+|y|^2}{\sqrt{x^2+y^2}})|\le |\cos(\frac{2\sqrt{x^2+y^2}^2}{\sqrt{x^2+y^2}})| \le |\cos(2\sqrt{x^2+y^2})|$$
Is this correct? I couldn't get Wolfram to compute this properly for some reason.
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The hint:
Prove that $$-(|x|+|y|)\leq\frac{x^2-y^2}{\sqrt{x^2+y^2}}\leq|x|+|y|.$$
The right inequality.
We need to prove that:
$$\frac{(|x|-|y|)(|x|+|y|)}{\sqrt{x^2+y^2}}\leq|x|+|y|$$ or $$ |x|-|y|\leq\sqrt{x^2+y^2}.$$
If $|x|-|y|\leq0$ it's obvious.
But for $|x|-|y|\geq0$ it's enough to prove that
$$(|x|-|y|)^2\leq x^2+y^2$$ or
$$-2|xy|\leq0,$$ which is obvious.
By the same way we can prove a left inequality.
Thus, since $$\lim_{(x,y)\rightarrow(0,0)}(|x|+|y|)=\lim_{(x,y)\rightarrow(0,0)}(-(|x|+|y|))=0,$$
we obtain: $$\lim_{(x,y)\rightarrow(0,0)}\frac{x^2-y^2}{\sqrt{x^2+y^2}}=0.$$
Id est, since $\cos$ is a continuous function, we obtain:
$$\lim_{(x,y)\rightarrow(0,0)}\cos\frac{x^2-y^2}{\sqrt{x^2+y^2}}=\cos\left(\lim_{(x,y)\rightarrow(0,0)}\frac{x^2-y^2}{\sqrt{x^2+y^2}}\right)=\cos0=1.$$
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|
Subsets and Splits
Fractions in Questions and Answers
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