Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Motivate why $a^{-n}$ equals to $\frac{1}{a^n}$ I have to prove that $a^{-n} = \frac{1}{a^n}$ with $\frac{3^4}{3^7}$, but before I can do that I have to understand the background.
The background says:
we know that $\frac{3^4}{3^7} = \frac{1}{3^3}$ and that $3^{4-7} = 3^{-3} = \frac{1}{3^3}$
I don't understand why $\frac{3^4}{3^7} = \frac{1}{3^3}$, before I can move on and solve why $a^{-n}$ equals to $\frac{1}{a^n}$.
| If $a^b \cdot a^c = a^{b + c}$ generalizes from positive $b, c$ to all integer values of $b, c$, then we have
$$1 = a^0 = a^{1 - 1} = a^1 \cdot a^{-1} = a \cdot a^{-1}.$$
That is, $$a \cdot a^{-1} = 1$$ Hence we are motivated to now define
$$a^{-1} = \frac{1}{a}$$
Similarly,
$$a^n \cdot a^{-n} = 1$$
and hence we also define
$$a^{-n} = \frac{1}{a^n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Trigonometry problem Okay..this one simple problem but I am really stuck and have no idea how to start..
$\cos(a-b)+\cos(b-c)+\cos(c-a)=-\frac32$ we need to prove $\cos(a)+\cos(b)+\cos(c)=\sin(a)+\sin(b)+\sin(c)=0 $
| This is a detour to the solution compared to my original solution
Like Prove that in any triangle $ABC$, $\cos^2A+\cos^2B+\cos^2C\geq\frac{3}{4}$,
or In a triangle, find the minimum and maximum of $\cos(A-B)\cos(B-C)\cos(C-A)$
let $2x=A-B$ etc. $\implies x+y+z=0$
Now $$-\dfrac32=\cos2x+\cos2y+\cos2z=2\cos(x-y)\cos(x+y)+2\cos^2z-1$$
As $x+y=-z,$
$$-\dfrac32=2\cos(x-y)\cos z+2\cos^2z-1\iff2\cos^2z+2\cos(x-y)\cos z+\dfrac12=0$$
which is a quadratic equation $\cos z$
As $\cos z$ is real, the discriminant must be $\ge0$
i.e., $(2\cos(x-y))^2-4\cdot2\cdot\dfrac12\ge0\iff\sin^2(x-y)\le0\implies\sin(x-y)=0$
Consequently, $\cos z=\dfrac{-2\cos(x-y)}4=\mp\dfrac12\implies\cos2z=2\cos^2z-1=-\dfrac12$
Using Clarification regarding a question,
we can say the angles namely, $A,B,C$ have to differ by $\dfrac{2\pi}3\pmod{2\pi}$
WLOG $A-B=\dfrac{2\pi}3, B-C=\dfrac{2\pi}3, A-C=\dfrac{4\pi}3$
Now $\sin A+\sin B=\sin\left(C+\dfrac{4\pi}3\right)+\sin\left(C+\dfrac{2\pi}3\right)=2\sin(\pi+C)\cos\dfrac\pi3=-\sin C$
$\implies\sin A+\sin B+\sin C=0$
Similarly, for cosines.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Pair of tangents to a hyperbola How do I find the joint equation of the pair of tangents drawn to the hyperbola $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ from an external point $(x_1, y_1)$.
My book says that the answer is $SS_1 = T^2$, where
$$
S = \frac{x^2}{a^2} - \frac{y^2}{b^2} - 1\\
S_1 = \frac{x_1^2}{a^2} - \frac{y_1^2}{b^2} - 1\\
T = \frac{x x_1}{a^2} - \frac{y y_1}{b^2} - 1
$$
But I can't get this result. Please help me go about this problem.
| Let $P(x_1,y_1)$, and let $P'(\alpha,\beta)$ be any point in the plane. Let $V(x',y')$ lie on the line $PP'$ so that the ratio $PV:VP'=m:n$.
Then $$V(x',y')=\left(\frac{m\alpha+nx_1}{m+n},\frac{m\beta+ny_1}{m+n}\right)$$
Now let $V$ lie on the hyperbola, so that $$\left(\frac{m\alpha+nx_1}{a(m+n)}\right)^2-\left(\frac{m\beta+ny_1}{b(m+n)}\right)^2=1$$
This can be rearranged, after a couple of lines, to form a quadratic in $\frac mn$, namely, $$\left(\frac mn\right)^2\left(b^2\alpha^2-a^2\beta^2-a^2b^2\right)+2\left(\frac mn\right)\left(b^2\alpha x_1-a^2\beta y_1-a^2b^2\right)+\left(b^2x_1^2-a^2y_1^2-a^2b^2\right)=0$$
Now if $P'$ lies on either of the tangents to the hyperbola, this quadratic has double roots, which gives us $$\left(b^2\alpha^2-a^2\beta^2-a^2b^2\right)^2=\left(b^2\alpha x_1-a^2\beta y_1-a^2b^2\right)\left(b^2x_1^2-a^2y_1^2-a^2b^2\right)$$
Now replace $(\alpha, \beta)$ with $(x,y)$ and divide throughout by $\left(a^2b^2\right)^2$ and you get the stated result $SS_1=T^2$
| {
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"timestamp": "2023-03-29T00:00:00",
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Determine if matrices A and B are similar Determine if matrices $A$ and $B$ are similar.
$A=\begin{pmatrix}1 & 1 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & 3 & 0 \\0 & 0 & 0 & 3\end{pmatrix}\: $
and
$B=\begin{pmatrix}1 & 1 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & 3 & 1 \\0 & 0 & 0 & 3\end{pmatrix}$
I found the characteristic polynomials to be the same for both matrices
$(1-λ)(3-λ)(3-λ)(1-λ)$
so I went to plug in the values to find the eigenvectors for matrix $A$ and I think I may have screwed up because
$\begin{pmatrix}0 & 1 & 0 & 0 \\0 & 0 & 0 & 0 \\0 & 0 & 2 & 0 \\0 & 0 & 0 & 2\end{pmatrix}$
I got an eigenvector for $λ=1$ to be $\begin{pmatrix}0 \\0 \\0 \\0\end{pmatrix}$ which I'm not even sure if I did it correctly or not because I don't think this is possible.
for $λ=3$ I got $\:\begin{pmatrix}1 \\\frac{1}{2} \\0 \\0\end{pmatrix}$
If I did do this correctly then is the geometric multiplicity for both = 1?
| Both matrices are in Jordan normal form. This normal form is unique for all similar matrices. Henece $A$ and $B$ are not similar.
Besides, these Jordan normal forms tell us the eigenspace $E_1$ (for the eigenvalue $1$ has dimension $1$ for both matrices, while the eigenspace $E_3$ has dimension $2$ for $A$ and $1$ for $B$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Multiply inverse of matricies Let ${A}$ be a $2 \times 2$
matrix such that ${A}$ * $\begin{pmatrix} 3 \\ -8 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \quad \text{and} \quad {A} * \begin{pmatrix} 5 \\ 7 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}.$
Find
${A}^{-1} *\begin{pmatrix} -2 \\ -1 \end{pmatrix}.$
What is the easiest way to start this problem?
Thanks
| Given
$$
\mathbf{A} \begin{pmatrix} 3 \\ -8 \end{pmatrix}
= \begin{pmatrix} 1 \\ 0 \end{pmatrix} \quad \textrm{and} \quad
\mathbf{A} \begin{pmatrix} 5 \\ 7 \end{pmatrix}
= \begin{pmatrix} 0 \\ 1 \end{pmatrix}
$$
IF $\mathbf{A}^{-1}$ exists, we can write
$$
\mathbf{A}^{-1} \mathbf{A} \begin{pmatrix} 3 \\ -8 \end{pmatrix}
= \mathbf{A}^{-1} \begin{pmatrix} 1 \\ 0 \end{pmatrix} \quad \textrm{and} \quad
\mathbf{A}^{-1} \mathbf{A} \begin{pmatrix} 5 \\ 7 \end{pmatrix}
= \mathbf{A}^{-1} \begin{pmatrix} 0 \\ 1 \end{pmatrix}
$$
Note that
$$
\mathbf{A}^{-1} \begin{pmatrix} -2 \\ -1 \end{pmatrix}
= -2 \mathbf{A}^{-1} \mathbf{A} \begin{pmatrix} 3 \\ -8 \end{pmatrix}
- \mathbf{A}^{-1} \mathbf{A} \begin{pmatrix} 5 \\ 7 \end{pmatrix}
$$
So
$$
\mathbf{A}^{-1} \begin{pmatrix} -2 \\ -1 \end{pmatrix}
= -2 \begin{pmatrix} 3 \\ -8 \end{pmatrix}
- \begin{pmatrix} 5 \\ 7 \end{pmatrix}
= \begin{pmatrix} -11 \\ 9 \end{pmatrix}
$$
Thus
$$
\bbox[16px,border:2px solid #800000] { \mathbf{A}^{-1} \begin{pmatrix} -2 \\ -1 \end{pmatrix}
= \begin{pmatrix} -11 \\ 9 \end{pmatrix} }
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Getting the quadratic function given the vertex and one point.
Find out the quadratic function for the parable that contains the
point $(1,1)$ and the vertex $(-2,3)$.
The notes I got are pretty vague:
$$b = 4\frac{-2}{9} = \frac{-8}{9}\\ c = -\frac{19}{9}\\ a =
-\frac{2}{9}$$
So
$$f(x) = -\frac{2}{9}x^2-\frac{8}{9}+\frac{19}{9}$$
Didn't really help me. Why did they multiply $4$ by $\frac{-2}{9}$ to get $b$? And of course, how were $c$ and $a$ calculated?
| If a quadratic function has vertex $(h,k)$, then it can be written as
$$
f(x) = a (x - h)^2 + k
$$
where $a$ is some other real number. (This is one of the things they usually expect you to just memorize.)
So since the vertex is $(-2, 3)$, you get
$$
f(x) = a ( x - (-2))^2 + (3)
= a ( x + 2)^2 + 3.
$$
Now, we use $f(1) = 1$ to solve for $a$.
Plugging in $x = 1$, we get
\begin{align*}
(1) &= a((1) + 2)^2 + 3 \\
1 &= 9a + 3 \\
-2 &= 9a \\
a &= -\frac{2}{9}
\end{align*}
Therefore, we have
$$
f(x) = - \frac{2}{9} (x + 2)^2 + 3.
$$
We can also expand this out to get the usual form $ax^2 + bx + c$:
\begin{align*}
- \frac{2}{9} (x + 2)^2 + 3
&= -\frac{2}{9} (x^2 + 4x + 4) + 3 \\
&= -\frac{2}{9} x^2 - \frac{8}{9} x - \frac{8}{9} + 3 \\
&= -\frac{2}{9} x^2 - \frac{8}{9} x - \frac{8}{9} + \frac{27}{9} \\
&= \boxed{-\frac{2}{9} x^2 - \frac{8}{9} x + \frac{19}{9}}.
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding real roots of a Polynomial Equation without graphs. I am interested in finding the number of real roots of the polynomial equation
$$
x^9 + \frac{9}{8}x^6 + \frac{27}{64}x^3 - x + \frac{219}{512} = 0.
$$
I know that graphing it would tell me how many real roots it has: the graph cuts the x-axis three times. But the coefficients are telling me some factorization is possible. I tried to write it like
$$
\left(x^3 + \frac{3}{8}\right)^3 = x - \frac{3}{8},
$$
but what next? Or is a graphical solution is the only possibility?
| We have
$$
\begin{align}
&x^9 + \frac{9}{8}x^6 + \frac{27}{64}x^3 - x + \frac{219}{512} \\
&\qquad = \frac{1}{512} (2 x-1) \left(4 x^2+2 x-3\right) \left(64 x^6+64 x^4+48 x^3+64 x^2+24 x+73\right).
\end{align}
$$
Since
$$
\begin{align}
&64 x^6+64 x^4+48 x^3+64 x^2+24 x+73 \\
&\qquad \geq 64 x^6+48 x^3+24 x+73 \\
&\qquad > 28 x^6+48 x^3+24 x+44 \\
&\qquad = 4 (x+1)^2 \left(7 x^4-14 x^3+21 x^2-16 x+11\right) \\
&\qquad \geq 4 (x+1)^2 \left(7 x^4-14 x^3+15 x^2-16 x+8\right) \\
&\qquad = 4 (x+1)^2 (x-1)^2 \left(7 x^2+8\right) \\
&\qquad \geq 0,
\end{align}
$$
the only real roots are $x=1/2$ and
$$
x = \frac{-1 \pm \sqrt{13}}{4}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate $S=3\sqrt{\sqrt[3]{5}-\sqrt[3]{4}}-\sqrt[3]{2}-\sqrt[3]{20}+\sqrt[3]{25}$ Calculate $$S=3\sqrt{\sqrt[3]{5}-\sqrt[3]{4}}-\sqrt[3]{2}-\sqrt[3]{20}+\sqrt[3]{25}$$ $\color{red}{\text{without using calculator}.}$
Please help me, I can't find any solution to sovle it.
| Let
$$s=\sqrt[3]2+\sqrt[3]{20}-\sqrt[3]{25}$$
If you carefully square $s$, you find
$$s^2=9(\sqrt[3]5-\sqrt[3]4)$$
Note that $\sqrt[3]2\gt1$, $\sqrt[3]{20}\gt2$, and $\sqrt[3]{25}\lt3$, so $s\gt0$. Thus $s=3\sqrt{\sqrt[3]5-\sqrt[3]4}$, hence $S=0$.
| {
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Why does this "miracle method" for matrix inversion work? Recently, I answered this question about matrix invertibility using a solution technique I called a "miracle method." The question and answer are reproduced below:
Problem: Let $A$ be a matrix satisfying $A^3 = 2I$. Show that $B = A^2 - 2A + 2I$ is invertible.
Solution: Suspend your disbelief for a moment and suppose $A$ and $B$ were scalars, not matrices. Then, by power series expansion, we would simply be looking for
$$ \frac{1}{B} = \frac{1}{A^2 - 2A + 2} = \frac{1}{2}+\frac{A}{2}+\frac{A^2}{4}-\frac{A^4}{8}-\frac{A^5}{8} + \cdots$$
where the coefficient of $A^n$ is
$$ c_n = \frac{1+i}{2^{n+2}} \left((1-i)^n-i (1+i)^n\right). $$
But we know that $A^3 = 2$, so
$$ \frac{1}{2}+\frac{A}{2}+\frac{A^2}{4}-\frac{A^4}{8}-\frac{A^5}{8} + \cdots = \frac{1}{2}+\frac{A}{2}+\frac{A^2}{4}-\frac{A}{4}-\frac{A^2}{4} + \cdots $$
and by summing the resulting coefficients on $1$, $A$, and $A^2$, we find that
$$ \frac{1}{B} = \frac{2}{5} + \frac{3}{10}A + \frac{1}{10}A^2. $$
Now, what we've just done should be total nonsense if $A$ and $B$ are really matrices, not scalars. But try setting $B^{-1} = \frac{2}{5}I + \frac{3}{10}A + \frac{1}{10}A^2$, compute the product $BB^{-1}$, and you'll find that, miraculously, this answer works!
I discovered this solution technique some time ago while exploring a similar problem in Wolfram Mathematica. However, I have no idea why any of these manipulations should produce a meaningful answer when scalar and matrix inversion are such different operations. Why does this method work? Is there something deeper going on here than a serendipitous coincidence in series expansion coefficients?
| The basic rule of thumb for such problems is that as long as all of the components of your equations are invertible and any multiplications used will commute, things will work just fine (this is the naive way of making the Banach Algebra comment). Since all along we knew we were going to find out that $B$ was invertible (and since powers of $A$ are clearly invertible), you're safe to go ahead.
However, these sorts of problems often have easier solutions. Hint: $B = A^2 - 2A + 2I = A^3 + A^2 - 2A = A(A^2 + A - 2I) = A(A+2I)(A-I)$. Now finish with applications of the Neumann Lemma.
When these types of problems also give you the information that $A$ is, say, symmetric or normal, or has such-and-such eigenvalues, you should look to abuse the spectral theorem to get very quick proofs by proving for diagonal, then generally.
Exercise: If $A$ is real normal s.t. $A^3 = A^2 +A - I$, prove that $A$ is invertible - in fact, $A^2 = I$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\int \frac{\sqrt{1-x^2}}{1+x^2}\hspace{1mm}dx$ Find $$\int \dfrac{\sqrt{1-x^2}}{1+x^2}\hspace{1mm}dx$$
Any hints! I will do the work, just give me a clue
| Notice, $$\int\frac{\sqrt{1-x^2}}{1+x^2}dx$$ Let, $x=\sin\alpha\implies dx=\cos\alpha d\alpha$ $$\int \frac{\sqrt{1-\sin^2\alpha}}{1+\sin^2\alpha}\cos\alpha d\alpha $$ $$=\int\frac{\cos^2\alpha}{1+\sin^2\alpha}d\alpha$$ $$=\int\frac{\sec^4\alpha\cos^2\alpha}{\sec^4\alpha(1+\sin^2\alpha)}d\alpha$$
$$=\int\frac{\sec^2\alpha}{(1+\tan^2\alpha)(\sec^2\alpha+\sec^2\alpha\sin^2\alpha)}d\alpha$$
$$=\int\frac{\sec^2\alpha}{(1+\tan^2\alpha)(1+\tan^2\alpha+\tan^2\alpha)}d\alpha$$
$$=\int\frac{\sec^2\alpha}{(1+\tan^2\alpha)(1+2\tan^2\alpha)}d\alpha$$ Let, $\tan\alpha=t\implies \sec^2\alpha d\alpha=dt$
$$=\int\frac{dt}{(1+t^2)(1+2t^2)}dt$$ $$=\int\left(\frac{2}{1+2t^2}-\frac{1}{1+t^2}\right)dt$$ $$=2\int\frac{dt}{1+(t\sqrt 2)^2}-\int\frac{dt}{1+t^2}dt$$
$$=\sqrt 2\tan^{-1}\left(t\sqrt 2\right)-\tan^{-1}\left(t\right)+c$$
Setting $t=\tan \alpha$
$$=\sqrt 2\tan^{-1}\left(\sqrt 2\tan\alpha\right)-\tan^{-1}\left(\tan\alpha\right)+c$$
Setting $\tan \alpha=\frac{\sin\alpha}{\cos \alpha}=\frac{x}{\sqrt{1-x^2}}$
$$=\sqrt 2\tan^{-1}\left(\frac{x\sqrt 2}{\sqrt{1-x^2}}\right)-\tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)+c$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Completing the square help The textbook gives this equation:
${12x^2 + 24x -8x = 0}$ with an answer of ${x = 0}$ or ${x = -{4\over3}}$
But I suspect it should be ${12x^2 + 24x -8 = 0}$
So in order to solve this, I would first isolate the x terms on one side of the equation by adding 8 to both sides:
${12x^2 + 24x = 8}$
I would then divide both sides by the coefficient of the ${x^2}$ or 12 in this case which gives:
${x^2 + 2x = {8\over12}}$
I then divide the coefficient of x by 2 and square the result and add it to both sides
${x^2 + 2x + 1 = {8\over12}}$
=> ${(x + 1)^2 = {8\over12}}$
=> ${x + 1 = \pm \sqrt{8\over12}}$
=> ${x + 1 = \pm \sqrt{2\over3}}$
=> ${x = - 1\pm \sqrt{2\over3}}$
I've taken a wrong turn somewhere, I'm not sure how to get to ${x = 0}$ or ${x = -{4\over3}}$.
| No, if you consider $12x^2 + 24x = 8$, then considering $x=0$ you get $0=8$, which is clearly wrong.
Considering $12x^2 + 24x = 8x$ this is the same as $12x^2 + 16x = 0$, and from here:
$$ x(12x + 16) = 0$$
Which is fulfilled if $x=0$ or $12 x = -16 \implies x = - \frac{4}{3}$
| {
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"timestamp": "2023-03-29T00:00:00",
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How can I factor the equation $a^3 - 5a^2 + 8a -4 = 0$? Consider the equality
$$
a^3 - 5a^2 + 8a -4 = 0
$$
What are the steps to factor this into the following?
$$
(a − 1)(a − 2)^2 = 0
$$
Is there some technique to do this?
| \begin{align}
a^3 - 5a^2 + 8a -4 &= a^3 - a^2 -4a^2 + 8a -4 \\
&= a^2(a-1) - 4(a^2-2a+1) \\
&= a^2(a-1) - 4(a-1)^2\\
&= (a-1) \left[ a^2 - 4(a-1)\right] \\
&= (a-1)(a-2)^2
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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The pattern in mathematical induction proofs When given a statement to be proven by mathmatical induction
the statement tends to look like this
$1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}$
so going about the proof.
1) Prove the base case
$\frac{1\times(1+1)}{2} = 1$
2) Prove the inductive case
Assume
$1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}$
Now add $n+1$ to both sides of the equation.
\begin{align}
1 + 2 + 3 + \dots + n + (n + 1) & = \frac{n(n+1)}{2} + (n + 1) \\
& = \frac{n(n+1)}{2} + \frac{2(n+1)}{2} \\
& = \frac{(n + 1)(n + 2)}{2} \\
& = \frac{(n + 1)((n + 1) + 1)}{2}
\end{align}
Thus the proof is complete. Now my question is, can we just remove the $1 + 2 + 3 + \dots + n$ side of the equation, and put anything there? for example, if we used $f(n)$, Then the proof would look like this.
1) Prove the base case
$f(1) = \frac{1\times(1+1)}{2} = 1$
2) Prove the inductive case
Assume
$f(n) = \frac{n(n+1)}{2}$
Now add $n+1$ to both sides of the equation.
\begin{align}
f(n) + (n + 1) & = \frac{n(n+1)}{2} + (n + 1) \\
& = \frac{n(n+1)}{2} + \frac{2(n+1)}{2} \\
& = \frac{(n + 1)(n + 2)}{2} \\
& = \frac{(n + 1)((n + 1) + 1)}{2}
\end{align}
Thus we now know $f(n) = \frac{n(n+1)}{2}$. But what pattern represents $f(n)$?
it's not immediately obvious that $f(n)$ is the sum of the first $n$ positive integers.
| The general theorem of mathematical induction goes like this:
Suppose that "$\Phi$" is a proposition involving the variable "n". Actually let's call the proposition $\Phi(n)$, and agree that whenever we write "$\Phi(\text{something})$" the reader is meant to substitute the "something" in for $n$. If $\Phi(0)$ holds, and if $\Phi(n+1)$ holds whenever $\Phi(n)$ holds, then we may conclude that $\Phi(n)$ holds for all $n$.
We can express this more precisely using quantifiers. With the same setup, here is what induction says:
$$
\Phi(0)\text{ and }(\forall k\in\mathbb{N}\ |\ \Phi(k)\Rightarrow \Phi(k+1))\ \ \ \ \Longrightarrow\ \ \ \ \ (\forall n\in\mathbb{N}\ |\ \Phi(n))\ .
$$
(The $k$ could just as well be an $n$, but sometimes changing variable naming inside quantifiers helps to avoid confusion).
In the example with which you began your post, the proposition $\Phi(n)$ is
$$
1+2+3+\cdots+n=n(n+1)/2\ .
$$
The proposition you do your induction on can really be any proposition! It doesn't need to be an equality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1404480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determine a positive integer $n\leq5$,such that $\int_{0}^{1}e^x(x-1)^ndx=16-6e$ Determine a positive integer $n\leq5$,such that $\int_{0}^{1}e^x(x-1)^ndx=16-6e$.
I tried to solve it.But since $n$ is given to be $\leq$ 5,my calculations went lengthy.
Applying integration by parts repeatedly,we get
\begin{align}
\int e^x(x-1)^n \, dx &= \left[(x-1)^ne^x-n(x-1)^{n-1}e^x+n(n-1)(x-1)^{n-2}e^x \right. \\
& \hspace{5mm} \left.-n(n-1)(n-2)(x-1)^{n-3}e^x+n(n-1)(n-2)(n-3)(x-1)^{n-4}e^x \right. \\
& \hspace{5mm} \left.-n(n-1)(n-2)(n-3)(n-4)e^x\right]
\end{align}
\begin{align}
\int_{0}^{1}e^x(x-1)^n \, dx &= -n(n-1)(n-2)(n-3)(n-4)e-(-1)^n+n(-1)^{n-1}-n(n-1)(-1)^{n-2} \\
& \hspace{5mm} +n(n-1)(n-2)(-1)^{n-3}-n(n-1)(n-2)(n-3)(-1)^{n-4} \\
& \hspace{5mm} +n(n-1)(n-2)(n-3)(n-4) \\
&=16-6e
\end{align}
Now solving this is very difficult,is there another simple and elegant method to find $n=3.$
| If we set
$$ I_n = \int_{0}^{1}e^x(1-x)^n\,dx = \int_{0}^{1}x^n e^{1-x}\,dx\tag{1}$$
integration by parts gives:
$$ I_{n+1} = -1 + (n+1)\, I_n\tag{2}$$
Since $I_0 = -1+e$, by using the previous formula and induction we have:
$$ \forall n\in\mathbb{N},\qquad I_n = A_n + B_n e,\quad A_n,B_n\in\mathbb{Z}\tag{3}$$
as well as
$$ B_n = n! \tag{4} $$
so the solution to the original problem is clearly $n=\color{red}{3}$.
| {
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"url": "https://math.stackexchange.com/questions/1405064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to Prove : $\frac{2}{(n+2)!}\sum_{k=0}^n(-1)^k\binom{n}{k}(n-k)^{n+2}=\frac{n(3n+1)}{12}$ While I calculate an integral
$$
\int\limits_{[0,1]^n}\cdots\int(x_1+\cdots+x_n)^2\mathrm dx_1\cdots\mathrm dx_n
$$
I used two different methods and got two answers. I am sure it's equivalent, but how can I prove it?
$$\displaystyle\dfrac{2}{(n+2)!}\sum_{k=0}^n(-1)^k\binom{n}{k}(n-k)^{n+2}=\dfrac{n(3n+1)}{12}$$
Sincerely thanks!
| Fix $n$ and define $f(x)$ by
$$ f(x) = (1 - e^{-x})^n = \sum_{k=0}^{n} \binom{n}{k} (-1)^k e^{-kx}. $$
Then the left-hand side is equal to
$$ \frac{2}{(n+2)!} f^{(n+2)}(0). $$
So it suffices to find the Taylor polynomial of degree $n+2$ for $f(x)$ at $x = 0$. This can be done as follows:
\begin{align*}
f(x)
&= x^n \left( \frac{1 - e^{-x}}{x} \right)^n \\
&= x^n \exp\left[ n \log\left( \frac{1 - e^{-x}}{x} \right)\right] \\
&= x^n \exp\left( -\frac{n}{2} x + \frac{n}{24} x^2 + \cdots \right) \\
&= x^n \left( 1 - \frac{n}{2} x + \frac{(3n+1)n}{24} x^2 + \cdots \right) \\
&= x^n - \frac{n}{2} x^{n+1} + \frac{(3n+1)n}{24} x^{n+2} + \cdots.
\end{align*}
Therefore we have
$$ \frac{2}{(n+2)!} f^{(n+2)}(0) = \frac{2}{(n+2)!} \cdot \frac{(3n+1)n}{24} (n+2)! = \frac{(3n+1)n}{12}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1406385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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} |
Integrate $ \int \sqrt{\frac{x^3-3}{x^{11}}}\ dx$ $$ \int \sqrt{\frac{x^3-3}{x^{11}}}\ dx $$
It seems like substitution could not do it. Is there another way?
| Notice, $$\int\sqrt{\frac{x^3-3}{x^{11}}}dx$$
$$=\int\frac{\sqrt{x^3-3}}{x^{11/2}}dx$$
Let, $x^3=3\sec^2\theta\implies 3x^2dx=6\sec^2\theta\tan \theta d\theta$
$$dx=\frac{2\sec^2\theta\tan \theta d\theta}{(3\sec^2\theta)^{2/3}}$$
$$\int\frac{\sqrt{3\sec^2\theta-3}\frac{2\sec^2\theta\tan \theta d\theta}{(3\sec^2\theta)^{2/3}}}{(3\sec^2\theta)^{11/6}}$$
$$=\int\frac{2\sqrt 3\sec^2\theta\tan^2 \theta d\theta}{(3\sec^2\theta)^{11/6}(3\sec^2\theta)^{2/3}}$$
$$=\int\frac{2\sqrt 3\sec^2\theta\tan^2 \theta d\theta}{(3)^{5/2}\sec^5\theta}$$
$$=\frac{2}{9}\int\frac{\tan^2 \theta d\theta}{\sec^3\theta}$$
$$=\frac{2}{9}\int\cos^3\theta \frac{\sin^2 \theta d\theta}{\cos^2\theta}$$
$$=\frac{2}{9}\int\sin^2 \theta\cos\theta d\theta$$
Let $\sin \theta=t\implies \cos \theta d\theta=dt$
$$\frac{2}{9}\int t^2dt=\frac{2}{9}\frac{t^3}{3}+C$$
$$=\frac{2t^3}{27}+C$$
$$=\frac{2\sin^3 \theta}{27}+C$$
Now, substitute $$\sin\theta=\sqrt{1-\cos^2\theta}=\sqrt{1-\frac{1}{\sec^2\theta}}=\sqrt{1-\frac{3}{x^3}}=\sqrt{\frac{x^3-3}{x^3}}$$
$$\iff\sin^3\theta=\left(\frac{x^3-3}{x^3}\right)^{3/2}$$
$$=\frac{2}{27}\left(\frac{x^3-3}{x^3}\right)^{3/2}+C$$
Hence, we get
$$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{\int\sqrt{\frac{x^3-3}{x^{11}}}dx=\frac{2}{27}\left(\frac{x^3-3}{x^3}\right)^{3/2}+C}}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove $((a+b)/2)^n\leq (a^n+b^n)/2$ Struggling with this proof.
Prove that $$\left(\frac{a+b}{2}\right)^n≤\frac{a^n+b^n}{2},$$ where $a$ and $b$ are real numbers such that $a+b≥0$ and $n$ is a positive integer.
What technique would you use to prove this (e.g. induction, direct, counter example). How would you go about proving it?
Thanks in advance.
| When $n$ is a positive integer, the function $f(x) = x^n$ is convex on $(0, +\infty)$, thus by Jensen's inequality:
$$f\left(\frac{a}{2} + \frac{b}{2}\right) \leq \frac{1}{2}f(a) + \frac{1}{2}f(b),$$
gives the desired inequality.
As pointed out by Andrews, the $f$ defined above is convex on the whole line only when $n$ is even for which case Jensen's inequality can be applied directly. The inequality also holds if both $a$ and $b$ are nonnegative.
To the case that $n$ is an odd positive integer and $a \leq 0 < b$ (without losing of generality), $a + b \geq 0$, write $n = 2k + 1$, then the right hand side of the inequality is
$$\frac{a^{2k + 1} + b^{2k + 1}}{2} = \frac{1}{2}(a + b)(a^{2k} - a^{2k - 1}b + \cdots + b^{2k}) \geq \frac{1}{2}(a + b)(a^{2k} + b^{2k}) > \frac{a + b}{2}\frac{a^{2k} + b^{2k}}{2}$$
So if we can show
$$\left(\frac{a + b}{2}\right)^{2k} \leq \frac{a^{2k} + b^{2k}}{2}$$
the inequality holds, but this is the case which we can use Jensen's inequality.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving this Recurrence Relation in terms of previous values. What will be the value of $X(n)$ and $Y(n)$ in terms of given $n,X(0),Y(0)$.
$$
X(n) = X(n-1) + Y(n-1) \\
Y(n) = 2X(n-1) + Y(n-1)
$$
| This relation can be written as
$$\begin{pmatrix}X(n) \\ Y(n)\end{pmatrix} = \begin{pmatrix}1 & 1 \\ 2 & 1\end{pmatrix} \begin{pmatrix}X(n - 1) \\ Y(n - 1)\end{pmatrix}. $$
If you define $A = \begin{pmatrix}1 & 1 \\ 2 & 1\end{pmatrix}$, this yields
$$\begin{pmatrix}X(n) \\ Y(n)\end{pmatrix} = A^n \begin{pmatrix}X(0) \\ Y(0)\end{pmatrix}. $$
Now you only need to calculate the powers of $A$, this works straightforward via diagonalzation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1411307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
cos(4v) + cos(v) = 0 I am given the following equation:
$$\cos 4v + \cos v = 0$$
My attempt:
$$\cos4v = -\cos v$$
$$\cos4v = \cos(\pi \pm v)$$
$$4v = \pm \pi \pm v + 2\pi n$$
$$4v_1 = \pi + v_1 + 2\pi n$$
$$4v_2 = -\pi - v_2 + 2\pi n$$
$$4v_3 = +\pi - v_3 + 2\pi n$$
$$4v_4 = -\pi + v_4 + 2\pi n$$
$$v_1 = \frac{\pi}{3} + \frac{2\pi n}{3}$$
$$v_2 = -\frac{\pi}{5} + \frac{2\pi n}{5}$$
$$v_3 = \frac{\pi}{5} + \frac{2\pi n}{5}$$
$$v_4 = -\frac{\pi}{3} + \frac{2\pi n}{3}$$
However, the answer is simply the positive solutions i.e:
$$v_1 = \frac{\pi}{3} + \frac{2\pi n}{3}$$
$$v_3 = \frac{\pi}{5} + \frac{2\pi n}{5}$$
Why? What am I missing?
| $n$ is not necessarily positive.
$$ -\dfrac{\pi}{3} = \dfrac{\pi}{3} + \dfrac{2\pi(-1)}{3}$$
etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1414895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A triple integral dancing in the unit cube Straight integration seems pretty tedious and difficult, and I guess the symmetry may open some new ways of which I'm not aware. What would your idea be?
Calculate
$$\int_0^1 \int_0^1 \int_0^1 \frac{x^2}{\sqrt{x^2+1} \left(x^2-y^2\right) \left(x^2-z^2\right)}+\frac{y^2}{\sqrt{y^2+1} \left(y^2-x^2\right) \left(y^2-z^2\right)}+\frac{z^2}{\sqrt{z^2+1} \left(z^2-x^2\right) \left(z^2-y^2\right)} \, dx \ dy \ dz.$$
A 300 points bounty moment: After 2 years and 10 months since the problem was posed no full solution was provided yet. Is it possible to find a slick solution (like a bolt of lightning)? Good luck!
| Let us set $A=x^2,B=y^2,C=z^2$ and:
$$ S_n(A,B,C) = \frac{A^n}{(A-B)(A-C)}+\frac{B^n}{(B-A)(B-C)}+\frac{C^n}{(C-A)(C-B)} .\tag{1}$$
By partial fraction decomposition, it is straightforward to check that $S_0=S_1=0$ and $S_2=1$.
For every $n>2$ induction gives:
$$ S_n(A,B,C) = h_{n-2}(A,B,C)\tag{2} $$
where $h_k$ is a complete homogeneous symmetric polynomial, satisfying:
$$ \sum_{k\geq 0}h_k(A,B,C)t^k=\frac{1}{(1-A t)(1-B t)(1- Ct)}.\tag{3} $$
Assume now that the Taylor series of $g(u)=\frac{u}{\sqrt{1+u}}$ around $u=0$ is given by $\sum_{n\geq 0}g_n u^n$.
Our integral is given by:
$$ \iiint_{(0,1)^3}\sum_{n\geq 0}g_n\,S_n(x^2,y^2,z^2)\,d\mu=g_2+\iiint_{(0,1)^3}\sum_{n\geq 3}g_n\cdot h_{n-2}(x^2,y^2,z^2)\,d\mu\tag{4} $$
Now there are two possible approaches. The first approach is to look for a linear operator that maps $t^k$ to $g_k$, then exploit $(3)$ in order to write the RHS of $(4)$ as a simpler integral. The second approach is to apply a double-counting argument on the monomials appearing in the RHS of $(4)$. Since $\iiint_{(0,1)^3}x^{2a}y^{2b}z^{2c}\,d\mu=\frac{1}{(2a+1)(2b+1)(2c+1)}$, we get:
$$ \iiint_{(0,1)^3}\sum_{n\geq 3}g_n\cdot h_{n-2}(x^2,y^2,z^2)\,d\mu =\\= \sum_{\substack{a,b,c\geq 0\\(a,b,c)\neq (0,0,0)}}\frac{g_{a+b+c+2}}{(2a+1)(2b+1)(2c+1)}\tag{5} $$
and the last series can be further simplified by computing:
$$ l_n=\sum_{\substack{a,b,c\geq 0\\a+b+c=n}}\frac{1}{(2a+1)(2b+1)(2c+1)}\tag{6} $$
then computing $\sum_{n\geq 1}l_n g_n$, that is obviously related with the integral: $$\iiint_{(0,1)^3}g(x^2 y^2 z^2)\,d\mu = \iiint_{(0,1)^3}\frac{x^2 y^2 z^2}{\sqrt{1+x^2 y^2 z^2}}\,d\mu.\tag{7}$$
2018 update. Following the first approach, the given integral equals the opposite of
$$\iiint_{(0,1)^3}\frac{2}{\pi}\int_{0}^{\pi/2}\frac{\sin^2(\theta)\,d\theta}{(1+x^2\sin^2\theta)(1+y^2\sin^2\theta)(1+z^2\sin^2\theta)}\,dx\,dy\,dz$$
which by Fubini's theorem boils down to
$$ \frac{2}{\pi}\int_{0}^{\pi/2}\frac{\arctan^3(\sin\theta)}{\sin\theta}\,d\theta = \frac{2}{\pi}\int_{0}^{\pi/2}\frac{\arctan^3(\cos\theta)}{\cos\theta}\,d\theta=\frac{4}{\pi}\int_{0}^{1}\frac{\arctan^3\left(\frac{1-t^2}{1+t^2}\right)}{1-t^2}\,dt $$
then to
$\frac{2}{\pi}\int_{0}^{1}\frac{\arctan^3\left(\frac{1-t}{1+t}\right)}{\sqrt{t}(1-t)}\,dt = \frac{2}{\pi}\int_{0}^{1}\frac{\left(\frac{\pi}{4}-\arctan t\right)^3}{\sqrt{t}(1-t)}\,dt = \frac{2}{\pi}\int_{0}^{+\infty}\left(\frac{\pi}{4}-\arctan\tanh^2\frac{u}{2}\right)^3\,du$ which is related to the Gudermannian function. Probably it can be expressed in a closed form through high order polylogarithms / Euler sums with high weight. Accurate numerical approximation are pretty simple since the last integrand function essentially behaves like $\frac{\pi^3}{64}\exp\left(-\frac{3}{\pi}u^2\right)$ on $\mathbb{R}^+$.
We may also notice that the two approaches are pretty obviously equivalent, since $l_n$ appearing in $(6)$ is clearly given by a coefficient of the MacLaurin series of $\text{arctanh}^3(z)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Determine polynomial whose roots are a linear combination of roots of another polynomial Let $\alpha_1, \alpha_2, \alpha_3$ be the roots of the polynomial $p(x)=x^3+5x^2+7x+11$. Find a polynomial whose roots are $\frac{\alpha_1+\alpha_2}{2}, \frac{\alpha_2+\alpha_3}{2}, \frac{\alpha_1+\alpha_3}{2}.$ Calculating the exact roots doesn't really help because only one root is real. I've tried expanding
$$
\left(x-\frac{\alpha_1+\alpha_2}{2}\right)\left(x-\frac{\alpha_2+\alpha_3}{2}\right)\left(x-\frac{\alpha_1+\alpha_3}{2}\right)
$$
but it gets kind of messy. Any "nicer" ideas?
| Let $\alpha,\beta,\gamma$ are the roots of the equation $x^3+5x^2+7x+11=0\;,$ Then $\alpha+\beta +\gamma = -5$
and $\alpha\cdot \beta+\beta\cdot \gamma+\gamma\cdot \alpha = 7$ and $\alpha\cdot \beta \cdot \gamma = -11$
Now we have to calculate a polynomial equation
whose roots are $\displaystyle \left(\frac{\alpha+\beta}{2}\right)\;,\left(\frac{\beta+\gamma}{2}\right)$ and $\displaystyle \left(\frac{\gamma+\alpha}{2}\right).$
So Let $$\displaystyle P = \left(\frac{\alpha+\beta}{2}\right)+\left(\frac{\beta+\gamma}{2}\right)+\left(\frac{\gamma+\alpha}{2}\right)=\alpha+\beta+\gamma = -5$$
and Let $$\displaystyle Q = \left(\frac{\alpha+\beta}{2}\right)\cdot \left(\frac{\beta+\gamma}{2}\right)+\left(\frac{\beta+\gamma}{2}\right)\cdot \left(\frac{\gamma+\alpha}{2}\right)+\left(\frac{\gamma+\alpha}{2}\right)\cdot \left(\frac{\alpha+\beta}{2}\right)$$
So we get $$\displaystyle Q = \frac{1}{4}\left[\alpha^2+\beta^2+\gamma^2+3\alpha\beta+3\beta\gamma+3\gamma\alpha\right] = \frac{1}{4}\left[(\alpha+\beta+\gamma)^2+(\alpha\beta+\beta\gamma+\gamma\alpha)\right]$$
So we get $\displaystyle Q =\frac{1}{4}\left[25+7\right] = 8$
Now Let $$\displaystyle R = \left(\frac{\alpha+\beta}{2}\right)\cdot \left(\frac{\beta+\gamma}{2}\right)\cdot \left(\frac{\gamma+\alpha}{2}\right) = \frac{1}{8}\left[(5+\alpha)\cdot (5+\beta)\cdot (5+\gamma)\right]$$
So we get $$\displaystyle R = -\frac{1}{8}\left[125+5(\alpha\beta+\beta\gamma+\gamma\alpha)+\alpha\beta\gamma+25(\alpha+\beta+\gamma)\right] = -\frac{24}{8} = -3$$
above we used the relation $$\displaystyle \alpha+\beta+\gamma = -5\Rightarrow \alpha+\beta = -5-\alpha$$ etc...
So our polynomial equation is $$\displaystyle x^3-Px^2+Qx+R = 0\;,$$ Now put $P,Q$ and R
So we get a polynomial equation $$\displaystyle x^3+5x^2+8x+3 = 0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Given $a,b,c\ge1;abc\ge8$. Proving that $\sqrt{a^2-1}+\sqrt{b^2-1}+\sqrt{c^2-1}\ge 3\sqrt3$
Given $a,b,c\ge1;abc\ge8$. Proving that $$\sqrt{a^2-1}+\sqrt{b^2-1}+\sqrt{c^2-1}\ge 3\sqrt3$$
I have tried by using Jensen's inequality:
We consider the inequality: $\displaystyle\sqrt{x^2-1}\ge\sqrt3+\frac4{\sqrt3}\ln\frac x2\tag{i}$
When $x>\frac85$, I can prove that $(\text i)$ true. But it is clear that $(\text i)$ is not true for all $x<\frac85$, and I can't prove that $\displaystyle\sqrt{a^2-1}+\sqrt{b^2-1}+\sqrt{c^2-1}\ge 3\sqrt3$ in this case.
| $x^2=a^2-1,y^2=b^2-1,z^2=c^2-1 \implies (x^2+1)(y^2+1)(z^2+1)\ge 64 \to x+y+z \ge 3\sqrt{3}$
let $3u=x+y+z,3v^2=xy+yz+xz,w^3=xyz \implies u \ge v \ge w,(x^2+1)(y^2+1)(z^2+1)\ge 64 \iff w^6-6uw^3+9u^2-6v^2+9v^4-63=f(w^3) \ge 0 \implies \Delta\le 0 \implies 36u^2-4(9u^2-6v^2+9v^4-63) \le 0 \iff 3v^4-2v^2-63 \ge 0 \iff (v^2-3)(3v^2+7) \ge 0 \implies v^2 \ge 3 \implies u^2 \ge v^2 \ge 3 \implies x+y+z \ge 3\sqrt{3}$
| {
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"url": "https://math.stackexchange.com/questions/1418174",
"timestamp": "2023-03-29T00:00:00",
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How to solve $2 \tan x / (1 - (\tan x)^2) = (\sin 2x)^2$? $$\frac {2\tan {x}}{1-(\tan {x})^2} = (\sin {2x})^2$$
I tried a lot but I get nowhere
| $$\frac {2\tan {x}}{1-(\tan {x})^2} = \frac{2\times \dfrac{\sin(x)}{\cos(x)}}{1-\dfrac{\sin^2(x)}{\cos^2(x)}}=\frac{2\times \dfrac{\sin(x)}{\cos(x)}\times \cos^2(x)}{\cos^2(x)-\sin^2(x)}$$
$$=\frac{\sin(x)\cos(x) + \sin(x)\cos(x)}{\cos(x)\cos(x)-\sin(x)\sin(x)}=\frac{\sin(2x)}{\cos(2x)}=\tan(2x)=\sin^2(2x)$$
$$\sin(2x)(1-\sin(2x)\cos(2x))=0$$
So : $ 1-\sin(2x)\cos(2x)=1-\frac 12 \sin(4x)=0 \implies \sin(4x)=2$ which is impossible.
$\color{red}{\text{or}}$ $\sin(2x)=0\implies x=\dfrac 12 n\pi, n\in \mathbb{Z}$
| {
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How many ways to write $2010$?
Let $ N$ be the number of ways to write $ 2010$ in the form $ 2010 = a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10 + a_0$, where the $ a_i$'s are integers, and $ 0 \le a_i \le 99$. An example of such a representation is $ 1\cdot10^3 + 3\cdot10^2 + 67\cdot10^1 + 40\cdot10^0$. Find $ N$.
I picked the biggest $a_1$ so: $a_1 = 2$, there are only two ways to form $2010$.
Take $a_1 = 1$ now. This opens up to a lot of possibilities.
Specific Casework should work:
Cases 1-1: $a_2 = 10$, then possibilities are: $a_1 = 1, a_0 = 0$ or $a_1 = 0, a_0 = 10$
Actually, I think a number-theoretic way is easier.
But still.
Case 1: $a_1 = 1$ then we must solve:
$100x + 10y + z = 1010$.
Since $0 \le x \le 10$, we can casework $x$ so that:
Case 1-1:$x = 0$. So that:
$10y + z = 1010 \implies z \equiv 0 \pmod{10}, z = 10k$ and $y = 101 - k$.
Hence, $(0, 101 - k, 10k)$. $\min{k} = 0 $ and we need to find the max of $k$. We must have, $101 - k \le 99$ and $10k \le 99$. This suggests, $k \le 9$.
Cases 1-2: $x=1$. So that:
$10y + z = 910 \implies z \equiv 0 \pmod{10}$ Again, $z = 10k$ and $y = 91 - k$. Giving a set of $(1, 91 - k, 10k)$.Here again, $\min{k} = 0$ and $10k \le 99$ so $k \le 9$.
I am conjecturing that since we are always increasing $x$ values, the value on the RHS will always be divisible by $10$.
$x = 9$ so that:
$10y + z = 110 \implies z = 10k$ and $y = 11 - k$, which again there are $9$ values.
Except if $x=10$ then there is: $10y + z = 10$ then $z = 10k$ and $y = 1 - k$. Then $k$ must be $1$.
So there are: $10(9) + 1 + 2 = 93$ solutions total.
This is just an attempt!
Bump: anybody have anything?
| Here's a solution that's not particularly elegant but gives the right answer (checked by brute force).
First express each coefficient uniquely as $b_i\cdot 10 + a_i$.
$2010 = b_3 \cdot 10^4 + a_3 \cdot 10^3 + b_2 \cdot 10^3 + a_2 \cdot 10^2 +
b_1 \cdot 10^2 + a_1 \cdot 10 + b_0 \cdot 10 + a_0$.
We must have $b_3 = 0, a_0 = 0$, so
$201 = (a_3 + b_2) \cdot 10^2 + (a_2 + b_1) \cdot 10^1 + (a_1 + b_0)$.
Each coefficient is in $0, \ldots, 18$.
If $a_3 + b_2 = 1$, then $101 = (a_2 + b_1) \cdot 10^1 + (a_1 + b_0)$. We have
$a_2 + b_1 = 9, a_1 + b_0 = 11$ or $a_2 + b_1 = 10, a_1 + b_0 = 1$, this gives
$2 \cdot (10 \cdot 8 + 9 \cdot 2) = 196$ solutions.
If $a_3 + b_2 = 2$, then $1 = (a_2 + b_1) \cdot 10^1 + (a_1 + b_0)$. We have
$a_2 + b_1 = 0, a_1 + b_0 = 1$, this gives $3 \cdot 2 = 6$ solutions.
So together we have 202 solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If the circumcenter of the triangle $ABC$ is on the incircle of the triangle,then prove that $\cos A+\cos B+\cos C=\sqrt2$ If the circumcenter of the triangle $ABC$ is on the incircle of the triangle,then prove that $\cos A+\cos B+\cos C=\sqrt2$
How should i attempt this question?I thought over it hard but could not crack through.Some hints will help me.Please guide me.
| Let me try.
Let $d$ be a distance between the circumcentre and incentre.
From Euler's theorem, you have $$d^2= R(R-2r).$$
When the circumcentre is on the incircle, we have $d=r$. Then, $$r^2 = R^2 - 2Rr.$$
So, you get $$\left(\frac{r}{R}\right)^2 + 2\frac{r}{R} - 1 = 0.$$
Then, you get $$\frac{r}{R} = -1 + \sqrt{2}$$ (Do you see why not $-1-\sqrt{2}$?).
Recall that we have $$rR = \frac{abc}{2(a+b+c)}$$ (For example, see link)
Note that $a = 2R\sin A$, $b = 2R\sin B$ and $c = 2R\sin C$. We have
$$\frac{r}{R} = 2 \frac{\sin A\sin B\sin C}{\sin A + \sin B + \sin C}.$$
We have $$\sin A + \sin B +\sin C = 4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}.$$
Now, we have $$\frac{r}{R} = 4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}.$$
Note that $$\cos A + \cos B +\cos C = 4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2} + 1.$$
So, we have $$-1+\sqrt{2} = 4 \sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}.$$
Thus, we get $$\cos A +\cos B + \cos C = \sqrt{2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the expansion in power series of ${x \over \sin x}$ How can I expand in power series the following function:
$$
{x \over \sin x}
$$
? I know that:
$$
\sin x = x - {x^3 \over 3!} + {x^5 \over 5!} - \ldots,
$$
but a direct substitution does not give me a hint about how to continue.
| You can have the expansion to any order you please using polynomial division along increasing powers. Finding a general formula for the coefficients is another problem.
I'll show how to obtain the development at order $6$, for instance: start from the development of $\sin x$ at order $7$ (this is because there will be a simplifacation by $x$):
$$\frac x{\sin x}=\frac x{x-\dfrac{x^3}6+\dfrac{x^5}{120}-\dfrac{x^7}{5040}+o(x^8)}=\frac 1{1-\dfrac{x^2}6+\dfrac{x^4}{120}-\dfrac{x^6}{5040}+o(x^7)}$$
Division along increasing powers up to order $6$:
$$\begin{array}{rcr@{}r@{}r@{}l@{\,}}
& & 1 + &\!\!\!\! \dfrac{x^2}{6} + & \hskip-12mu\dfrac{7x^4}{360}+ & \hskip-12mu\dfrac{31x^6}{15120}\\[-18mu]
1-\dfrac{x^2}6+\dfrac{x^4}{120}-\dfrac{x^6}{5040} & \biggl( & 1\hphantom{{}+{}}\\[-18mu]
& & -1 + &\hskip-12mu\dfrac{x^2}6-&\hskip-12mu \dfrac{x^4}{120} + & \hskip-12mu\dfrac{x^6}{5040} \\
& & &\hskip-12mu \dfrac{x^2}6-&\hskip-12mu \dfrac{x^4}{120} + &\hskip-12mu \dfrac{x^6}{5040} \\
& &- &\hskip-12mu\dfrac{x^2}6 + &\hskip-12mu \dfrac{x^4}{36} - &\hskip-12mu \dfrac{x^6}{720} \\
& & & & \hskip-12mu\dfrac{7x^4}{360} -&\hskip-12mu \dfrac{x^6}{840} \\
& & &- & \hskip-12mu\dfrac{7x^4}{360} + & \hskip-12mu\dfrac{7x^6}{6\cdot360} \\
& & & & &\hskip-12mu \dfrac{31x^6}{15120}
\end{array}$$
Thus we've found that
$$\frac x{\sin x}=1 + \dfrac{x^2}{6} + \dfrac{7x^4}{360}+\dfrac{31x^6}{15120}+o(x^7)$$
| {
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Find a formula for the inverse of the function. $f(x) = \frac{4x − 1}{2x + 3}$ Please tell me know if my answer is right and whether the steps are correct? Thanks.
$f(x) = \frac{4x − 1}{2x + 3}$
Step 1: Write $y=f(x)$
$y=\frac{4x-1}{2x+3}$
Step 2: Solve this equation for $x$ in terms of $y$ (if possible)
2(a) Multiply both sides by $2x+3$
$(2x+3)\cdot(y)\ =\frac{4x-1}{2x+3}\cdot(2x+3)$
2(b) Distribute y term
$2xy+3y = 4x-1$
2(c) Isolate $x$ and $y$ terms
$2xy + 3y = 4x - 1 $
$-2xy+1 = -2xy$
$3y+1 = 4x-2xy$
$3y+1 = x (4-2y)$
2(d) Divide both sides by $(4-2y)$
$x=\frac{3y+1}{4-2y}$
$f^{-1}=\frac{3x+1}{4-2x} $
| Your process is correct and your result is correct. You could however stand to delete the line $-2xy+1=-2xy$ from the proof. I get that you were demonstrating what you did to both sides, but it doesn't make sense with the equals sign, and it's possible to go directly from $2xy+3y=4x-1$ to $3y+1=4x-2xy$ anyway.
You can always double check you are correct by plugging in a few value of $x$ and $y$. For example, we can try $x=0$ in the first equation, which gives $y=-\frac{1}{3}$, and plugging in $x=-\frac13$ into the $f^{-1}$ equation gives you $0$ as desired.
| {
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When the $a^2+b+c+d,b^2+a+c+d,c^2+a+b+d,d^2+a+b+c$ are all perfect squares? Find all $a,b,c,d\in \mathbb{Z}^+$, which $a^2+b+c+d,b^2+a+c+d,c^2+a+b+d,d^2+a+b+c$ are all perfect squares.
I found $(1,1,1,1)$, but I can't find more.
Is $a=b=c=d$ true?
| The solutions are $(11, 11, 6, 6)$, $(96, 96, 57, 40)$, and $(x, x, x, 1)$, where $3x + 1$ is a square.
Without loss of generality, $a \ge b \ge c \ge d$, since it is symmetric (not just cyclic). Then$$a^2 < a^2 + b + c + d \le a^2 + 4a < (a+2)^2 \implies a^2 + b + c + d = (a+1)^2.$$Thus, $b + c + d = 2a + 1$. Then$$b \ge {1\over3}(2a + 1) \implies a \le {3\over2}b,$$so$$b^2 < b^2 + c + d + a \le b^2 + {7\over2}b < (b+2)^2 \implies b^2 + c + d + a = (b+1)^2.$$Thus, $b + c + d = 2a + 1$ and $c + d + a = 2b + 1$, which implies $a = b$, thus $c + d = a + 1$.
Consequently,$$c \ge{1\over2}(2a+1) \implies a < 2c,$$so$$(c+1)^2 \le c^2 + 2a + d \le c^2 + 5c < (c+3)^2,$$and hence we have two cases.
Case 1. If $c^2 + 2a + d = (c+1)^2$, we must have $a = b = c$ and $d = 1$. Thus, we find that $(a, b, c, d) = (x, x, x, 1)$ where $x$ is such that $3x + 1$ is a square. This is one set of solutions.
Case 2. If $c^2 + 2a + d = (c + 2)^2$, so $2a + d = 3c + 4$. Combined with $c + d = 2a + 1$, we derive that $a = (5/2)d - 4$ and $c = (3/2)d - 3$. So$$(d+1)^2 \le d^2 + 2a + c = d^2 + {{13}\over2}d - 11 < (d+4)^2.$$So setting this equal to $(d+1)^2$, $(d+2)^2$, $(d+3)^2$, we find the integer solutions $d = 6$ and $d = 40$. This gives the solutions $(11, 11, 6, 6)$ and $(96, 96, 57, 40)$.
Hence, the solutions are $(11, 11, 6, 6)$, $(96, 96, 57, 40)$, and $(x, x, x, 1)$, where $3x + 1$ is a square.
| {
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"timestamp": "2023-03-29T00:00:00",
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Augmented Matrix For which real numbers $ a$ and $b$ does the following linear system have
(a) no solution,
(b) exactly one solution, or
(c) infinitely many solutions? Justify your answers.
$$ (a-1)x + (a+3)y + z = 1 $$
$$ax + a^2y + z = 1$$
$$x + ay + z = b$$
i start by putting it as an augmented matrix, but then i dont know how to reduce it into row echelon form or reduced ref. because of the algebraic terms. is there anyone that can help? thanks a lot...
| $$\begin{cases}(a-1)x + (a+3)y + z = 1 \\
ax + a^2y + z = 1 \\
x + ay + z = b\end{cases} \iff \left[\begin{array}{ccc|c} a-1 & a+3 & 1 & 1 \\ a & a^2 & 1 & 1 \\ 1 & a & 1 & b \end{array}\right]$$
OK, now we just do Gauss-Jordan:
$$\left[\begin{array}{ccc|c} a-1 & a+3 & 1 & 1 \\ a & a^2 & 1 & 1 \\ 1 & a & 1 & b \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 & a & 1 & b \\ a & a^2 & 1 & 1 \\ a-1 & a+3 & 1 & 1 \end{array}\right]\\ \sim \left[\begin{array}{ccc|c} 1 & a & 1 & b \\ 0 & 0 & 1-a & 1-ab \\ 0 & a+3-(a^2-a) & 1-(a-1) & 1-b(a-1) \end{array}\right] = \left[\begin{array}{ccc|c} 1 & a & 1 & b \\ 0 & 0 & -a+1 & -ab+1 \\ 0 & -a^2 + 2a+3 & -a+2 & -ab+b+1 \end{array}\right] \\ \sim \left[\begin{array}{ccc|c} 1 & a & 1 & b \\ 0 & -a^2 + 2a+3 & -a+2 & -ab+b+1 \\ 0 & 0 & -a+1 & -ab+1 \end{array}\right] \\ \sim \left[\begin{array}{ccc|c} 1 & a & 1 & b \\ 0 & 1 & \frac{a-2}{(a+1)(a-3)} & \frac{ab-b-1}{(a+1)(a-3)} \\ 0 & 0 & 1 & \frac{ab-1}{a-1} \end{array}\right]$$
Be sure to keep track of anything you divide by -- for instance I divided by $-a^2+2a+3=-(a+1)(a-3)$ in the second row of the last step and $-a+1$ in the last row of the last step. So you'll need to try $a=-1$, $a=3$, and $a=1$ individually.
Does that help?
| {
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How to determine Coercive functions A continuous function $f(x)$ that is defined on $R^n$ is called coercive if $\lim\limits_{\Vert x \Vert \rightarrow \infty} f(x)=+ \infty$.
I am finding it difficult to understand how the norm of these functions are computed in order to show that they are coercive.
$a) f(x,y)=x^2+y^2
\\b)f(x,y)=x^4+y^4-3xy\\c)f(x,y,z)=e^{x^2}+e^{y^2}+e^{z^2}$
To show that they are coercive I have to show that as norm goes to infinity the function too should go to infinity right?
| To prove that the function is coercive, we need to show that its value goes to $\infty$, as the norm becomes $\infty$.
1)$$ f(x,y)=x^2+y^2= \infty \\ as \left \| x \right \|\rightarrow \infty $$
i.e. $||x||=\sqrt(x^2+y^2)$
Hence , $f(x)$ is coercive.
2)$$ f(x,y)=x^4+y^4- 3xy
\\ \because ((x+y)^2-(x^2+y^2))=3xy (\frac{2}{3})
\\f(x,y)=x^4+y^4-(\frac{3}{2})( (x+y)^2)-(x^2+y^2))
\\ \leq x^4+y^4 + (\frac{3}{2})(x^2+y^2)\\
\leq (x^2+y^2)^2 + (\frac{3}{2})(x^2+y^2)
\\
\therefore f(x,y)=\infty
\\ as \left \| x \right \|\rightarrow \infty $$
i.e. $||x||=\sqrt(x^2+y^2)$
Hence , $f(x)$ is coercive.
3)$$ f(x,y,z)=e^{x^{2}} + e^{y^{2}}+ e^{z^{2}}
\\
\approx (1+x^{2})+(1+y^{2})+(1+z^{2})
= \infty $$
$$\\ as \left \| x \right \|\rightarrow \infty $$
i.e. $||x||=\sqrt(x^2+y^2+z^2)$
Hence , $f(x)$ is coercive.
| {
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"timestamp": "2023-03-29T00:00:00",
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I got a negative delta when computing deltas algebraically I am trying to find $\delta$ of $f(x)=4-x^2$ for $c=-1, L=3, \epsilon=\frac{1}{4}$. Here is what I did:
$$|f(x)-L|<\epsilon$$
$$|4-x^2-3|<\frac{1}{4}$$
$$-\frac{1}{4}<1-x^2<\frac{1}{4}$$
$$-\frac{5}{4}<-x^2<-\frac{3}{4}$$
$$\frac{5}{4}>x^2>\frac{3}{4}$$
$$\frac{\sqrt{5}}{2}>x>\frac{\sqrt{3}}{2}$$
$$\frac{\sqrt{3}}{2}<x<\frac{\sqrt{5}}{2}$$
Then the second step:
$$|x-c|<\delta$$
$$|x+1|<\delta$$
$$-\delta<x+1<\delta$$
$$-\delta-1<x<\delta-1$$
Left substitution:
$$-\delta-1=\frac{\sqrt{3}}{2}$$
$$\delta+1=-\frac{\sqrt{3}}{2}$$
$$\delta=-1-\frac{\sqrt{3}}{2}$$
Right substitution:
$$\delta-1=\frac{\sqrt{5}}{2}$$
$$\delta=\frac{\sqrt{5}}{2}+1$$
As you can see, I got a negative delta when doing the left substitution of the inequality ($\delta=-1-\frac{\sqrt{3}}{2}$). I heard it said somewhere that if you get a negative delta, you have an error in your algebra. Is this true? If so, what did I do wrong?
| Hint:
Your passage : $\dfrac{5}{4}<x^2<\dfrac{3}{4} \rightarrow \dfrac{\sqrt{5}}{2}<x<\dfrac{\sqrt{3}}{2}$ is wrong. You have two second degree inequalities:
$$
{5}{4}<x^2 \qquad \land \qquad x^2<\dfrac{3}{4}
$$
and you have to solve separetely.
The correct solution is:
$$
-\dfrac{\sqrt{5}}{2}<x<-\dfrac{\sqrt{3}}{2} \quad \lor \quad \dfrac{\sqrt{3}}{2}<x<\dfrac{\sqrt{5}}{2}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\mathop {\lim }\limits_{x \to \infty } (\sqrt[3]{{{x^3} + 5{x^2}}} - \sqrt {{x^2} - 2x} ) $ $$\mathop {\lim }\limits_{x \to \infty } (\sqrt[3]{{{x^3} + 5{x^2}}} - \sqrt {{x^2} - 2x} )
$$
My try:
$${a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})
$$
$$\mathop {\lim }\limits_{x \to \infty } \frac{{(\sqrt[3]{{{x^3} + 5{x^2}}} - \sqrt {{x^2} - 2x} )(\sqrt[{\frac{3}{2}}]{{{x^3} + 5{x^2}}} + \sqrt[3]{{{x^3} + 5{x^2}}}\sqrt {{x^2} - 2x} + {x^2} - 2x)}}{{(\sqrt[{\frac{3}{2}}]{{{x^3} + 5{x^2}}} + \sqrt[3]{{{x^3} + 5{x^2}}}\sqrt {{x^2} - 2x} + {x^2} - 2x)}} =
$$
$$\mathop {\lim }\limits_{x \to \infty } \frac{{{x^3} + 5{x^2} - {x^2} - 2x}}{{(\sqrt[{\frac{3}{2}}]{{{x^3} + 5{x^2}}} + \sqrt[3]{{{x^3} + 5{x^2}}}\sqrt {{x^2} - 2x} + {x^2} - 2x)}}
$$
And what's next...?
This task in first and second remarkable limits. I think i can replace variable, but how i will calculate it...
| If you can use the Taylor expansion $(1+t)^\alpha=1+\alpha t + \hbox{higher order terms}$ as $t\to0$, then it is easy:
$$
\sqrt[3]{{{x^3} + 5{x^2}}} - \sqrt {{x^2} - 2x} =x\left(\sqrt[3]{{1 + 5/x}} - \sqrt {1 - 2/x}\right)=x\left(1 + 5/(3x) - 1 + 1/x+ \hbox{h. o. t.}\right)=8/3+ \hbox{h. o. t.}\to8/3
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Given that $\ln (2x^2+9x-5)=1+\ln (x^2+2x-15)$ find $x$ in terms of $e$ Given that $\ln (2x^2+9x-5)=1+\ln (x^2+2x-15)$ find $x$ in terms of $e$. Then the left side is $2x^2+9x-5=\ln (1)+x^2+2x-15$, is that true? How should I proceed?
| Notice, $$\ln(2x^2+9x-5)=1+\ln(x^2+2x-15)$$
$$\ln(2x^2+9x-5)=\ln e+\ln(x^2+2x-15)$$
$$\ln(2x^2+9x-5)=\ln(e(x^2+2x-15))$$
$$2x^2+9x-5=e(x^2+2x-15)$$
$$(e-2)x^2-(9-2e)x-5(3e-1)=0$$
$$\iff x=\frac{(9-2e)\pm\sqrt{(9-2e)^2+20(e-2)(3e-1)}}{2(e-2)}$$
$$=\frac{(9-2e)\pm(8e-11)}{2(e-2)}$$
$$x=\frac{3e-1}{e-2}$$
or $$x=\frac{10-5e}{e-2}=-5$$ but for $x=-5$ log is undefined thus the above value is unacceptable
Hence, the correct value of $x$
$$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{x=\frac{3e-1}{e-2}}}$$
| {
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Integral $\int\frac{\sin2x+2tanx}{\cos^{6}x+6\cos^{2}x+4}dx$ Q) $\int\frac{\sin2x+2tanx}{\cos^{6}x+6\cos^{2}x+4}dx$
Tried to split it and integrate individual but , but getting an irreducible cubic equation in denominator .
Any magic substitution ?
| The substitution $u=\tan(x/2)$ will give a rational function, which in principle could be integrated. Here is another solution, that we are lucky to do. One could also in this case do $u=\tan x$ directly...
Using that
$$
\sin 2x=2\sin x\cos x=2\tan x\cos^2 x=\frac{2\tan x}{1+\tan^2x}
$$
and
$$
\cos^6x+6\cos^2x+4=\frac{1+6(1+\tan^2x)+4(1+\tan^2x)^3}{(1+\tan^2x)^3},
$$
we get, after some expanding,
$$
\frac{1}{12}\int \frac{48\tan x+72\tan^3x+24\tan^5x}{11+24\tan^2x+18\tan^4x+4\tan^6x}(1+\tan^2x)\,dx.
$$
Here, we are really lucky(?). Since
$$
D(11+24u^2+18u^4+4u^6)=48u+72u^3+24u^5,
$$
and since $D\tan x=1+\tan^2x$, we find that
$$
\begin{gathered}
\frac{1}{12}\int \frac{48\tan x+72\tan^3x+24\tan^5x}{11+24\tan^2x+18\tan^4x+4\tan^6x}(1+\tan^2x)\,dx\\
=\frac{1}{12}\log(11+24\tan^2x+18\tan^4x+4\tan^6x)+C.
\end{gathered}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Condition for a polynomial to have root of modulus 1
Prove that the polynomial$$P(X) = X^{n+1} - X^{n} - 1,\text{ }P \in \mathbb {C}[X]$$has a root $z$ with $\left|z\right|=1$ if and only if $6\,|\,(n+2)$.
One implication, from left to right, is quite simple because$$z=\cos \alpha + i\sin \alpha$$and$$z^n=\cos n\alpha + i\sin n\alpha$$allows me to find out the $\alpha$ value. It is the other implication that I wasn't able to prove.
| Note that $X^{n+1} - X^{n} - 1 = (X^2-X+1)X^{n-1}-(X^{n-1}+1)$.
Let $\omega$ be a root of $X^2-X+1$. Then $\omega$ is a primitive $6$-root of unity and so $\omega^3=-1$.
If $6 \mid (n+2)$, then $n=6k-2$ and $n-1=6k-3$.
Therefore, $\omega^{n-1}=\omega^{6k-3}=\omega^{-3}=-1$, and so $\omega$ is a root of $X^{n-1}+1$.
Thus, $\omega$ is a root of $X^{n+1} - X^{n} - 1 = (X^2-X+1)X^{n-1}-(X^{n-1}+1)$.
| {
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Integrate $\int\frac{x^{6}-2x^{3}}{\left(x^{3}+1\right)^{3}}dx$ . $$\int\frac{x^{6}-2x^{3}}{\left(x^{3}+1\right)^{3}}dx$$
I tried adding and subtracting 1 to bring a square expression with numerator as $(x^3-1)^2 -1$ but always going to partial fraction which obviously is very lengthy considering the cubic factor in denominator.
| Let $\displaystyle I = \int\frac{x^6-2x^3}{(x^3+1)^3}dx = \int\frac{x^6-2x^3}{x^6\left(x+\frac{1}{x^2}\right)^3}dx = \int\frac{1-2x^{-3}}{\left(x+\frac{1}{x^2}\right)^3}dx$
Now Put $\displaystyle x+\frac{1}{x^2} = t\;,$ Then $\displaystyle \left(1-\frac{2}{x^3}\right)dx = dt$
So Integral $\displaystyle I = \int\frac{1}{t^3}dt = -\frac{1}{2t^2}+\mathcal{C} = -\frac{1}{2}\cdot \left(\frac{x^2}{x^3+1}\right)^2+\mathcal{C}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1433339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
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Proof of $n^{1/n} - 1 \le \sqrt{\frac 2n}$ by induction using binomial formula Using $$(a+b)^{n} = \sum_{i=0}^n {n \choose k} a^{n-k} b^{K}$$
prove that
$$n^{1/n} - 1 \le \sqrt{\frac 2n}$$ for n= 2,3,4....
I know the first step is to set $$ n^{\frac 1n} = 1 + x $$ for some x>0 and then raise both sides to the n, but I'm lost after that.
| If $x=n^{1/n}-1$, then $n=(1+x)^n=1+nx+\frac{n(n-1)}{2}x^2+\cdots\ge 1+\frac{n(n-1)}{2} x^2$ for $n\ge 2$.
Then $n-1\ge \frac{n(n-1)}{2} x^2\implies 1\ge \frac{n}{2}x^2\implies x^2\le\frac{2}{n}\implies x\le\sqrt{\frac{2}{n}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1436663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Linear system - number of solutions depending on the parameter k Determine for what value of $k$ the following system has
*
*unique solution,
*no solution and
*infinitely many solutions.
\begin{cases}
x+2y+z=3\\
2x-y-3x=5\\
4x+3y-z=k
\end{cases}
I did till
$$
\left[\begin{array}{ccc|c}
1 & 2 & 1 & 3\\
1 & 1 & 1 & 1/5\\
0 & 0 & 1 & -k+42/5
\end{array}\right]
$$
I am not sure how to continue.
| You're not arrived at the end of the elimination:
\begin{align}
\left[\begin{array}{ccc|c}
1 & 2 & 1 & 3\\
2 & -1 & -3 & 5\\
4 & 3 & -1 & k
\end{array}\right]
&\to
\left[\begin{array}{ccc|c}
1 & 2 & 1 & 3\\
0 & -5 & -5 & -1\\
4 & 3 & -1 & k
\end{array}\right]
&&R_2\gets R_2-2R_1
\\&\to
\left[\begin{array}{ccc|c}
1 & 2 & 1 & 3\\
0 & -5 & -5 & -1\\
0 & -5 & -5 & k-12
\end{array}\right]
&&R_3\gets R_3-4R_1
\\&\to
\left[\begin{array}{ccc|c}
1 & 2 & 1 & 3\\
0 & 1 & 1 & 1/5\\
0 & -5 & -5 & k-12
\end{array}\right]
&&R_2\gets -\frac{1}{5}R_2
\\&\to
\left[\begin{array}{ccc|c}
1 & 2 & 1 & 3\\
0 & 1 & 1 & 1/5\\
0 & 0 & 0 & k-11
\end{array}\right]
&&R_3\gets R_3+5R_2
\end{align}
Now you should be able to end.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit of $(\frac{2^n+3^n+4^n}{5^n+6^n})^{\frac{1}{n}}$ I assume it is $1$.
Though $1$ is a limit of the function $(\frac{2^x+3^x+4^x}{5^x+6^x})^{\frac{1}{x}}$. Besides, even if it is $1$, I still need to solve the following inequality
$|(\frac{2^n+3^n+4^n}{5^n+6^n})^{\frac{1}{n}}-1|< \epsilon$
for arbitrary $\epsilon > 0$. That I don't know how to do as well.
| Here is a more classical approach. Take the $ln$ on both sides to get: $\frac{1}{n}*ln\frac{2^n+3^n+4^n}{5^n+6^n}$=$\frac{ln(2^n+3^n+4^n)-ln(5^n+6^n)}{n}$=$\frac{ln(2^n+3^n+4^n)}{n}-\frac{ln(5^n+6^n)}{n}$ With $n$ to infinity, we can apply L'Hospital's Rule: $\frac{2^nln2+3^nln3+4^nln4}{2^n+3^n+4^n}-\frac{5^nln5+6^nln6}{5^n+6^n}$ Dividing all terms in the first fraction by $4^n$ and the second fraction by $6^n$ and let $n$ go to infinity, we arrive at $ln4-ln6$ which is $ln\frac{2}{3}$ (please verify!) Since this is really the $ln$ of our limit, the actual limit is $2/3$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving that $y=x$ is an asymptote. I'm currently investigating the curve implicitly defined by $x^2-y^2= \ln (xy)$. Now I can see that by considering the region $xy>1$ and $xy<1$, we can determine where the curve is above or below the line $y=x$.
However, I do not feel that this is not enough to prove that the line $y=x$ is an asymptote in itself, just whether it approaches it from above or below.
For an implicitly defined curve like this, what kind of approach would I take to show that $y=x$ is an asymptote?
| You could separate the variables:
$$x^2 - \ln x = y^2 + \ln y.$$
It's easy then to show that $y\to\infty$ as $x\to\infty$.
Take the ratio of the two sides:
$$\frac{y^2 + \ln y}{x^2 - \ln x} = 1.$$
For large positive $x$, you have large positive $y$ and $y^2/x^2 \approx 1$,
so $y/x \approx 1$.
You can even write
\begin{align}
\frac{y^2}{x^2}
& = \left( \frac{y^2}{y^2 + \ln y} \right) \left( \frac{x^2 - \ln x}{x^2}\right)
\frac{y^2 + \ln y}{x^2 - \ln x} \\
& = \left( \frac{y^2}{y^2 + \ln y} \right) \left( \frac{x^2 - \ln x}{x^2}\right)
\end{align}
Each of the quantities in parentheses goes to $1$ as $x\to\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1443310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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prove $2\sin^4(x)+2\cos^4(x)+\sin^2(2x)=2$ \begin{align}2\sin^4(x) + 2\sin^2(x)\cos^2(x) + 2\cos^4(x)&=2(\sin^4(x) + \sin^2(x)\cos^2(x) + \cos^4(x))\\
&=2(\sin^2(x) + \cos^2(x))^2 -\sin^2(x)\cos^2(x)\\
&=2(1) -\sin^2(x)\cos^2(x) \\
&=2-\sin^2(x)\cos^2(x)
\end{align}
And I'm stuck
| $$ \sin^2 2x = 4\sin^2 x \cos^2 x$$
Let $\sin x = a$ and $\cos x = b$
RHS of your equation becomes:
$$ 2a^4 + 4a^2b^2+2b^4 = 2(a^2+b^2)^2= 2(\sin^2 x + \cos^2 x)^2 = 2 =RHS$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Evaluate $a+2b+3c$ If $$\lim_{h \to 0}\frac { \int _{ 0 }^{ h }{ f(x)dx-h(af(0)+bf(h/3)+cf(h)) } }{ { h }^{ 4 } } $$ is a finite non-zero number,how to evaluate $a+2b+3c$?
Ok only hints at first.I want to solve this myself.But can't understand how to start.Any suggestion appreciated.Thanks.
| Set $f(x) = 1, x, x^2, x^3$.
You will get a series of equations
for $a, b, c$.
Solve them and here is what you get.
$\frac { \int _{ 0 }^{ h } f(x)dx-h(af(0)+bf(h/3)+cf(h)) }{ { h }^{ 4 } }
$
For $f(x) = 1$:
$ \int _{ 0 }^{ h } dx-h(a+b+c)
=h-h(a+b+c)
$,
so
$a+b+c = 1$.
For $f(x) = x$:
$ \int _{ 0 }^{ h } xdx-h(a\cdot 0+b(h/3)+ch)
=h^2/2-h^2(b/3+c)
$,
so
$b/3+c = 1/2$
or
$b+3c = 3/2$.
For $f(x) = x^2$:
$ \int _{ 0 }^{ h } x^2dx-h(a\cdot 0+b(h/3)^2+ch^2)
=h^3/3-h^3(b/9+c)
$,
so
$b/9+c = 1/3$
or
$b+9c = 3$.
We get
$a+b+c = 1$,
$b+3c = 3/2$,
$b+9c = 3$.
Subtracting the 2nd and 3rd,
$9c = 3/2$
or
$c = 1/6$.
From the last,
$b = 3-9/6 = 3/2$.
From the first,
$a = 1-3/2-1/6
=-2/3
$.
So
$a+2b+3c
=-2/3+3+1/2
=17/6
$.
Note that,
for $f(x) = x^3$:
$ \int _{ 0 }^{ h } x^3dx-h(a\cdot 0+b(h/3)^3+ch^3)
=h^4/4-h^4(b/27+c)
=h^4(1/4-3/(2\cdot 27)+1/6)
=(h^4/(4\cdot 27)(27-4-18)
=5h^4/(4\cdot 27)
$,
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1446598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Normalized form of Triangule Inequality Extension
How I can prove $$\mathit{\frac{\left\vert a-b\right\vert}{a+b}\leq\frac{\left\vert a-c\right\vert}{a+c}+\frac{\left\vert c-b\right\vert}{c+b}}$$ where $ a,b,c \in \mathbb N?$
| This nice inequality holds for non-negative REAL numbers $a,b,c$.
Suppose $a\ge b$.
Consider three cases:
*
*$a\ge c \ge b$: we have $$RHS = \frac{a-c}{a+c}+\frac{c-b}{c+b}=\frac{2c(a-b)}{(a+c)(c+b)}.$$ It suffices to show that $$\frac{2c}{(a+c)(c+b)} \ge \frac{1}{a+b},$$ which is equivalent to $2c(a+b) \ge (a+c)(c+b) \Leftrightarrow (a-c)(c-b)\ge 0$, clearly true.
*$c\ge a \ge b$: in this case we have $$\frac{a-b}{a+b} \le \frac{c-b}{c+b}$$
(equivalent to $(a-b)(c+b)\le (a+b)(c-b) \Leftrightarrow 2b(c-a)\ge 0)$, and thus the original inequality is true.
*$a \ge b \ge c$: in this case, similarly we have $$\frac{a-b}{a+b} \le \frac{a-c}{a+c}.$$
We are done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Compute $\lim_{n\to +\infty}n\left(\tan\left(\frac{\pi}{3}+\frac{1}{n} \right)-\sqrt{3}\right)$ without using L' Hôpital
Compute $$\lim_{n\to +\infty}n\left(\tan\left(\frac{\pi}{3}+\frac{1}{n} \right)-\sqrt{3}\right)$$ without using L'Hospital's rule.
By using L'Hospital's rule and
$$\tan'( \Diamond )=( \Diamond )'(1+\tan^{2}( \Diamond ))$$
I mean by $\Diamond $ a function
so I got
\begin{align}
\lim_{n\to +\infty}n\left(\tan\left(\dfrac{\pi}{3}+\dfrac{1}{n} \right)-\sqrt{3}\right)
&=\lim_{n\to +\infty}\dfrac{\left(\tan\left(\dfrac{\pi}{3}+\dfrac{1}{n} \right)-\sqrt{3}\right)}{\dfrac{1}{n}}\\
&=1+\tan^{2}\left(\dfrac{\pi}{3}\right)=1+\sqrt{3}^{2}=1+3=4
\end{align}
I'm interested in more ways of computing limit for this sequence.
| The expression is a derivative. Since $\sqrt{3} = \tan\left(\frac{\pi}{3}\right)$, the limit is
$$\lim_{n \to \infty} n \left( \tan\left(\frac{\pi}{3} + \frac{1}{n} \right) - \tan\left(\frac{\pi}{3}\right)\right)$$
That is by the substitution $h = \frac{1}{n}$, $$\lim_{h \to 0} \frac{1}{h} \left(\tan\left(\frac{\pi}{3} + h \right) - \tan \left(\frac{\pi}{3} \right)\right)$$
That is, $$\tan'(\pi/3)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Find intersection of the two surfaces $x^2-y^2-z^2=1$ and $x+y=1$ Find intersection of this two surfaces
$$x^2-y^2-z^2=1$$
and
$$x+y=1.$$
I know that the first is hyperboloid of two sheet and the second is plane,
but how can i find the intersection? Is it possible do calculus in one variable? How?
| On the surface of the hyperboloid, given any $z$ the quantities $x$ and $y$ are related by $x^2 - y^2 = 1 + z^2$. On the plane $x$ and $y$ are related by $x+y=1$, so on the intersection of the surfaces they satisfy $$x-y = (x-y)(x+y) = 1 + z^2.$$
Add the equations $x-y = 1+z^2$ and $x+y=1$ to get $$x = 1 + \frac{z^2}{2}$$ and thus $$y = - \frac{z^2}{2}.$$ Since $z$ is arbitrary you can use it to parameterize the curve: $$ \left( 1 + \frac{t^2}{2}, - \frac{t^2} 2, t \right), \quad -\infty < t < \infty.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1448568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How many nonnegative integer solutions does the following equation have: $x + y + z = 15$
How many nonnegative integer solutions does the following equation
have: $x + y + z = 15$
(1) if $1 \le x \le 6$
(2) if $x \ge 2$ and $y \le 3$
(3) if $x \ge 3$, $y \ge 2$, and $1 \le z \le 3$
I solved the first question (which is C(3+11-1,11)). Still very stuck on the next two
| There are $75$ ways to do it. Using your procedure we get $\binom{17}{15}-\binom{10}{8}-\binom{16}{15}$. The last subtraction is to remove the cases where $x=0$. (I also did it another way, again getting $75$. Makes $75$ twice as true.)
Added: For the second question, we want the number of solutions of $x+y+z=13$, with the restriction that $y\le 3$.
Count the number of solutions of $x+y+z=13$. Subtract the number of solutions of $x+y+z=9$. That's because to count the number of "bad" solutions of $x+y+z=13$, bad because $y\ge 4$, we give $4$ to $y$ and distribute the remaining $9$ between $x$, $y$, and $z$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to find the maximum and minimum value of $2^{\sin x}+2^{\cos x}$ My try: Let $y$ = $2^{\sin x}+2^{\cos x}$
*
*Applying AM GM inequality I get
$y$ $> 2.2^{(\sin x+\cos x)/2}$.
Now, the highest value of R.H.S is $2^{\frac{\left(2+\sqrt{2}\right)}{2}}$.
Should this mean that $y$ is always greater than $2^{\frac{ \left ( 2+\sqrt{2}\right ) }{2}}$?
But this is not true (we can see in the graph).
*Calculus method:
$dy/dx$ = $\ln\left(2\right){\cdot}{2}^{\sin\left(x\right)}\cos\left(x\right){-\ln\left(2\right){\cdot}{2}^{\cos\left(x\right)}\sin\left(x\right)}$
When $dy/dx$ =0,
$\tan x = 2^{\sin x- \cos x}$ and I am stuck here.
https://www.desmos.com/calculator/p3zfvkq2mn
| Let $f(u,v)=2^u+2^v;\;$ we want to find the extrema of $f$ on the circle $u^2+v^2=1$.
Using Lagrange multipliers, $\;\;2^u\ln2=\lambda(2u)$ and $2^v\ln 2=\lambda(2v)\;$ so $\;\displaystyle\frac{2^u\ln2}{2u}=\frac{2^v\ln2}{2v}$.
Then $\displaystyle\frac{2^u}{u}=\frac{2^v}{v}$ and therefore $u=v$,
$\;\;\;$since the function $\displaystyle g(t)=\frac{2^t}{t}$ is 1-1 on $[-1,0)$ and $(0,1]$
$\;\;\;$because $\displaystyle g^{\prime}(t)=\frac{2^t(t\ln 2-1)}{t^2}<0$ for $t\in(-1,0)\cup(0,1)$.
Then $u^2+v^2=1\implies 2u^2=1\implies u=v=\pm\frac{\sqrt{2}}{2}$.
Thus $\displaystyle f\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)=2^{1+\frac{\sqrt{2}}{2}}$ is the maximum, and $\displaystyle f\left(-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)=2^{1-\frac{\sqrt{2}}{2}}$ is the minimum.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "9",
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Shortest method for $\int_{0}^{1}\frac{x^{4}\left(1-x\right)^{4}}{1+x^{2}}$ I don't want to solve by expanding it and all, I tried corollary but denominator becomes messy, also in the options there is $\pi$ so tried several trigonometric substitutions too
| Let $$\displaystyle I = \int_{0}^{1}\frac{x^4(1-x)^4}{1+x^2}dx\;,$$ Let $\displaystyle x= \tan \phi\;,$ Then $dx = \sec^2 \phi.$
And Changing Limit, We Get $$\displaystyle I = \int_{0}^{\frac{\pi}{4}}\tan^4 \phi\cdot \left(1-\tan \phi\right)^4d\phi.$$
$$\displaystyle =\int_{0}^{\frac{\pi}{4}}\tan^4 \phi \cdot \left(1-4\tan \phi+6\tan^2 \phi-4\tan^3 \phi+\tan^4 \phi\right).$$
$$\displaystyle = \int_{0}^{\frac{\pi}{4}}\left\{\tan^4 \phi-4\tan^5 \phi+6\tan^6 \phi-4\tan^7 \phi+\tan^8 \phi \right\}d\phi.$$
$$\displaystyle = \int_{0}^{\frac{\pi}{4}}\left\{\left[\tan^4 \phi+\tan^6 \phi \right]-4\left[\tan^5 \phi \cdot \sec^2 \phi \right]+\left[\tan^6 \phi+\tan^8 \phi \right]+4\tan^6 \phi \right\}d\phi.................(\star)\color{\red}\checkmark.$$
Now Let $$\displaystyle J_{n} = \int_{0}^{\frac{\pi}{4}}\tan^{n}\phi d\phi\;,$$ Then $$\displaystyle J_{n+2} = \int_{0}^{\frac{\pi}{4}}\tan^{n+2}\phi d\phi$$
Now $$\displaystyle J_{n+2}+J_{n} = \int_{0}^{\frac{\pi}{4}}\tan^{n} \phi \cdot \sec^2 \phi d\phi = \frac{1}{n+1}\Rightarrow J_{n+2}+J_{n} = \frac{1}{n+1}............(\star \star)\color{\red}\checkmark.$$
Now at $n=0\;,$ we get $\displaystyle J_{0} = \frac{\pi}{4}\;,$ Then $\displaystyle J_{2} = \left(1-\frac{\pi}{4}\right)$ and $\displaystyle $ When $ n= 2\;,$ We get $$\displaystyle J_{4} = \left(\frac{\pi}{4}-\frac{2}{3}\right).$$
and When $n=4\;,$ We get $\displaystyle J_{6} = \left(\frac{13}{15}-\frac{\pi}{4}\right).$ and We get $\displaystyle J_{4}+J_{6}=\frac{1}{5}$ And $\displaystyle J_{6}+J_{8} = \frac{1}{7}$
Now Put these values in $............(\star)\color{\red}\checkmark.\;,$ We get
$$\displaystyle I = \frac{1}{5}-\frac{2}{3}+\frac{1}{7}+\frac{52}{15}-\pi = \left(\frac{22}{7}-\pi\right).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Please help me for Partial fractions decomposition I need a partial fractions decomposition this function $$\frac{1}{(x^3+x+1)^3}$$
I write in form: $$\frac{1}{(x^3+x+1)^3}=\frac{Ax^2+Bx+C}{x^3+x+1}+\frac{Dx^2+Ex+F}{(x^3+x+1)^2}+\frac{Gx^2+Hx+I}{(x^3+x+1)^3}$$
but I do not know whether I have done well. Please help me. Thanks.
| Assume that $\zeta_1,\zeta_2,\zeta_3$ are the roots of $x^3+x+1$ - they are a negative real number and two conjugated complex numbers. Since:
$$\text{Res}\left(\frac{1}{x^3+x+1},x=\zeta_i\right)=\lim_{z\to \zeta_i}\frac{(z-\zeta_i)}{z^3+z+1}=\lim_{z\to \zeta_i}\frac{1}{3z^2+1}\tag{1}$$
we have:
$$ \frac{1}{(x^3+x+1)}=\sum_{i=1}^{3}\frac{1}{(3\zeta_i^2+1)(x-\zeta_i)}.\tag{2}$$
Are you able to compute the wanted partial fraction decomposition by "cubing" both terms of $(2)$?
Recall that by Viète's formulas, $\zeta_1+\zeta_2+\zeta_3=0,\zeta_1^2+\zeta_2^2+\zeta_3^2=-2$ and $\zeta_1 \zeta_2 \zeta_3=-1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $x_0=2\cos\frac{\pi}{6}$ and $x_n=\sqrt{2+x_{n-1}},n=1,2,3...,$ prove that $\lim_{n\to \infty}2^{n+1}\sqrt{2-x_n}=\frac{\pi}{3}$ Let $x_0=2\cos\frac{\pi}{6}$ and $x_n=\sqrt{2+x_{n-1}},n=1,2,3...,$ prove that $\lim_{n\to \infty}2^{n+1}\sqrt{2-x_n}=\frac{\pi}{3}$
I am not able to correctly solve it,made some attempts,but no luck.This $2^{n+1}$ is creating problem.How should i prove this limit to be $\frac{\pi}{3}$.
| As Suggested by @Did in comment,here is a solution:
First using induction on $n$ prove that $x_n=2 cos(\frac{\pi}{6 \cdot 2^n})$,then
From identity $\sin^2 \frac{t}{2}=\frac{1}{2}(1-\cos t)$ we have
\begin{align}
\lim_{n\rightarrow\infty}{2^{n+1}\sqrt{2-2\cos\left(\frac{\pi}{6 \cdot2^n}\right)}} & = \lim_{n\rightarrow \infty}{2^{n+2}\sin\left(\frac{\pi}{6\cdot 2^{n+1}}\right)} \\
& = \lim_{n\rightarrow \infty}\frac{\sin\left(\frac{\pi}{3 \cdot2^{n+2}}\right)}{\frac{1}{2^{n+2}}} \\
& = \lim_{t\rightarrow 0^+}\frac{\sin\left(\pi t/3\right)}{t} \\
& = \pi/3 \\
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculating $\lim_{x\to 0}\frac{\sec (x)-1}{x^2\sec(x)}$ $$\lim_{x\to 0}\frac{\sec (x)-1}{x^2\sec(x)}$$
The first thing to do, as I was taught, was to rewrite this in terms of sine and cosine. Since $\sec(x) = \frac{1}{\cos(x)}$ we have
$$\frac{\frac{1}{\cos(x)}-1}{x^2\frac{1}{\cos(x)}}$$
And that is
$$\frac{\frac{1-\cos(x)}{\cos(x)}}{\frac{x^2}{\cos(x)}}$$
Then
$$\frac{(1-\cos(x))\cdot\cos(x)}{\cos(x) \cdot x^2}$$
And
$$\frac{(1-\cos(x))}{x^2}$$
Alright. It is my understanding that there is a formula or something that states $\frac{1-\cos(x)}{x} = 0$, so I can apply it here because we actually have
$$\frac{\color{green}{(1-\cos(x))}}{\color{green}{x}\cdot x}$$
Leaving us with
$$\color{green}0\cdot\frac{1}{x}$$
But if we evaluate, we get
$$0\cdot\frac{1}{0}$$
Which I guess is indeterminate because of $\frac{1}{0}$.
Also, the correct answer should be $\frac{1}{2}$.
What was wrong with my procedure?
| Note that you get an undefined form :
$$
\lim_{x\to 0} \frac{1-\cos{(x)}}{x^2}=\frac{0}{0}
$$
So you can use l'Hopital rule to get:
$$
\lim_{x\to 0} \frac{1-\cos{(x)}}{x^2}=\lim_{x\to 0} \frac{\sin{(x)}}{2x}=\frac{1}{2}
$$
Because as you probably know :
$$
\lim_{x\to 0} \frac{\sin{(x)}}{x}={(\frac{d}{dx})}_{x=0} {\sin{(x)}}=cos{(0)}=1
$$
You can also use series expansion of $\cos{x}$ near $0$:
$$
\cos{(x)}=1-\frac{x^2}{2}+o(x^4)
$$
To get:
$$
\lim_{x\to 0}\frac{1-\cos{(x)}}{x^2}=\lim_{x\to 0}\frac{1-(1-\frac{x^2}{2}+o(x^4)))}{x^2}\\
=\lim_{x\to 0} \frac{\frac{x^2}{2}}{x^2}=\frac{1}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1452853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Tricky trig question from GRE
Please evaluate
\begin{align}
1-\sin^2\left(\arccos \frac{\pi}{12}\right)
\end{align}
What I've tried so far is to use Pythagorean identity and I got
\begin{align}
\cos^2\left(\arccos \frac{\pi}{12}\right)
\end{align}
If \begin{align}
\arccos \frac{\pi}{12}=y
\end{align}
then
\begin{align}
\cos y =\frac{\pi}{12}
\end{align}
and here I can't continue because the answers are in such form:
*
*(A) $\sqrt{\frac{1-\cos\frac\pi{24}}{2}}$
*(B) $\sqrt{\frac{1-\cos\frac\pi{6}}{2}}$
*(C) $\sqrt{\frac{1+\cos\frac\pi{24}}{2}}$
*(D) $\frac\pi6$
and one more is missing, but that's pdf's issue
| You have that $\sin(\arccos(x))=\sqrt{1-x^2}$, therefore
$$\sin^2\left(\arccos\left(\frac{\pi}{12}\right)\right)=1-\frac{\pi^2}{144}\implies 1-\sin^2\left(\arccos\left(\frac{\pi}{12}\right)\right)=\frac{\pi^2}{144}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1453443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Compute $\lim_{x \to 0} x \lfloor x - \frac{1}{x} \rfloor$ Let $f: \mathbb{R} \to \mathbb{R}$ be the following function:
$f(x) = x \lfloor x - \frac{1}{x} \rfloor$
Show that $f(x)$ admits a limit at zero, and compute its value.
Using epsilon delta, I can prove that $\lim_{x \to 0} x \lfloor x - \frac{1}{x} \rfloor = - 1$:
$|x \lfloor x - \frac{1}{x} \rfloor + 1|$
$\leq |x \lfloor x - \frac{1}{x} \rfloor| + |1|$
$\leq |x| |\lfloor x - \frac{1}{x} \rfloor| + |1|$
$\leq |x| |x| + |1|$
$\leq |x|^2 + |1| < \epsilon$ if $|x| < \delta = \sqrt{\epsilon - 1}$
Is that right? If not, how can I do prove it?
But how do I show that $\lim_{x \to 0} x \lfloor x - \frac{1}{x} \rfloor$ exists?
| Suppose that $\frac1{n+1}\le x<\frac1n$; then $n<\frac1x\le n+1$, so
$$-n-\frac{n}{n+1}=\frac1{n+1}-(n+1)\le x-\frac1x<\frac1n-n=-n+\frac1n\;,$$
and $\left\lfloor x-\frac1x\right\rfloor$ is either $-n-1$ or $-n$. Thus, $x\left\lfloor x-\frac1x\right\rfloor$ is either $-nx$ or $-(n+1)x$. Now use the limits on $x$ to bound $-nx$ and $-(n+1)x$, and let $n$ increase without bound.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1456621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Prove that a number is divisible by 3 iff the sum of its digits is divisible by 3 I'm trying to teach myself some number theory. In my textbook, this proof is given:
Example (2.3.1) Show that an integer is divisible by 3 if and only if the sum of its digits is a multiple of 3.
Let $n=a_0a_1\ldots a_k$ be the decimal representation of an integer $n$. Thus $n=a_k+a_{k-1}10+a_{k-2}10^2+\cdots+a_010^k$ where $0\le a_i<10$ The key observation is that $10\equiv1\pmod3$, i.e., $[10]=[1]$. Hence $[10^i]=[10]^i=[1]$, i.e., $10^i\equiv1\pmod 3$. The assertion is an immediate consequence of this congruence.
I don't understand the last statement. Why does it follow from that congruence?
| A simple way to see this (that actually generalizes nicely to Fermat's little theorem):
$$10 - 1 = 9 = 9 \times 1$$
$$100 - 1 = 99 = 9 \times 11$$
$$1000 - 1 = 999 = 9 \times 111$$
In general $$10^n - 1 = 9 \times \underbrace{111...111}_\mbox{$n$ times}.$$ This is just the algebraic identity
$$x^n - 1 = (x - 1)(x^{n-1} + x^{n - 2} + ... + x + 1)$$
when $x = 10$. The identity is easy to prove - just multiply it out term by term. All but the first and last terms cancel. Thus any power of $10$ less $1$ is divisible by $9$, and therefore also by $3$.
Now consider a multi-digit natural number, $43617$ for example. $$\begin{align} 43617 &= 4\times 10^4 + 3 \times 10^3 + 6 \times 10^2 + 1 \times 10 + 7 \\ &= 4\times 9999 + 3 \times 999 + 6 \times 99 + 1 \times 9 + (4 + 3 + 6 + 1 + 7) \end{align}$$
Every term on the right other than the sum of the digits is divisible by $3$. So the remainder when dividing the original number by $3$ and the sum of the digits by $3$ must be the same.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1457478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Finding the equation whose roots are Okay I found this one on a test and it must be something really silly I am missing $ \alpha$ is not equal to $\beta$ and $\alpha^2=5\alpha-3,5\beta-3=\beta^2$ then the equation whose roots are $\alpha/\beta$ and $ \beta/\alpha$ is $ (a)3x^2-25+3=0,(b)x^2-5x+3=0,(c)x^2+5x-3=0,(d)3x^2-19x+3=0$ Of course the answer is either a or d I tried solving for $\alpha$ and $\beta$ but the equation became too complicated
| HINT:
Method$\#1:$
So, $\alpha, \beta$ are the unequal roots of $$t^2-5t+3=0$$
$$\alpha+\beta=?, \alpha\cdot\beta=?$$
Method$\#2:$
If $a^2=5a-3, b^2=5b-3$
$a^2-b^2=5(a-b)\iff a+b=5$ as $a-b\ne0$
$a^2+b^2=5(a+b)-6,5^2-2ab=5\cdot5-6\iff ab=3$
Now from any one of the above methods,
$$\dfrac\alpha\beta+\dfrac\beta\alpha=\dfrac{\alpha^2+\beta^2}{\alpha\beta}=\dfrac{(\alpha+\beta)^2}{\alpha\beta}-2$$
$$\dfrac\alpha\beta\cdot\dfrac\beta\alpha=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1459332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the derivative of the function $f(x)=x/(x^2+1)$ at a point a $$\frac{x}{x^2+1}$$
I have to find the derivative and set a point by myself.
So I just set $x=a$ ( is it correct?)
and using the limit,
$$ \lim_{h\to 0}\frac{f(a+h)-f(a)}{h} = \lim_{h \to 0} \frac{\frac{a+h}{(a+h)^2 + 1} - \frac{a}{a^2+1}}{h}$$
therefore, the answer is $$\frac{ 1-a^2}{(a^2+1)^2}$$
is it correct?
| $\implies f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{\frac{x+h}{(x+h)^2+1}-\frac{x}{x^2+1}}{h}$
= $ \lim_{h\to 0}\frac{\frac{x+h}{(x^2 + 1 + 2h +h^2}-\frac{x}{x^2+1}}{h}$
$f(x + h) - f(x) = \frac{x+h}{(x+h)^2+1}-\frac{x}{x^2+1} = \frac{(x+h)(x^2 -1)}{((x+h)^2+1)(x^2 + 1)}-\frac{x((x+h)^2+1)}{((x+h)^2+1)(x^2 + 1)}$
$= \frac{h(1-x^2 -hx)}{(x^2 + 1)^2 + (2h + h^2)(x+1) } $
so $f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}= \lim_{h\to 0} \frac{\frac{h(1-x^2 -hx)}{(x^2 + 1)^2 + (2h + h^2)(x+1) }}{h} = \lim_{h\to 0}\frac{(1-x^2 -hx)}{(x^2 + 1)^2 + (2h + h^2)(x+1) } = \frac{1-x^2}{(x^2 + 1)^2}$
I don't know what "set a point" means unless it means try it for x = 0 or something. [In which case we'd get
$f'(0)=\lim_{h\to 0}\frac{f(h)-f(0)}{h}= \lim_{h\to 0} \frac{\frac{h}{h^2 + 1} -0}{h} = \lim_{h\to 0}\frac{1}{h^2 + 1} = 1$
]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1459681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Each of the two persons makes a single throw with a pair of unbiased dice.What is the probability that the throws are equal? Each of the two persons makes a single throw with a pair of unbiased dice.What is the probability that the throws are equal?
Since the same throws can result if either both of them get 1 or both of them get 2 or both of them get 3 or both of them get 4 or both of them 5 or both of them get 6.
So I calculated probability as $\frac{1}{6}\times\frac{1}{6}+\frac{1}{6}\times\frac{1}{6}+\frac{1}{6}\times\frac{1}{6}+\frac{1}{6}\times\frac{1}{6}+\frac{1}{6}\times\frac{1}{6}+\frac{1}{6}\times\frac{1}{6}=\frac{1}{6}$
But my answer is wrong and the correct answer is $\frac{73}{648}$.Please help me with the correct approach to solve it.Thanks.
| Deducing, like Sherlock Holmes, that the question is:
Two people throw two dice each. What is the probability that both get the same sum ?
Ways of throwing sums of $2 - 12$ with $2$ dice follows the pattern
$1-2-3-4-5-6-5-4-3-2-1$
So P(both get the same sum) = $\dfrac{(1^2+2^2+3^2+ ....+6^2+5^2+4^2... +1^2)}{36^2}= \dfrac{146}{1296}=\dfrac{73}{648}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1460864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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How to solve $\int_{0}^{\frac{π}{2}} \frac{dx}{4\sin^2(x) +5\cos^2(x)} $ $?$ I apply the substitutions:
$$t=\tan(x), \sin(x)=\frac{t}{\sqrt{1+t^2}}, \cos(x)=\frac{1}{\sqrt{1+t^2}}\ \&\ dx=\frac{dt}{1+t^2}$$
(using $t=\tan(x)$ you can draw a right angled triangle to find the other substitutions)
So we get:
$$\int_{0}^{\frac{π}{2}} \frac{1}{\frac{4t^2}{1+t^2}+\frac{5}{1+t^2}}\frac{dt}{1+t^2}=\frac{1}{4}\int_{0}^{\frac{π}{2}} \frac{1}{t^2+\Big(\frac{\sqrt{5}}{2}\Big)^2} dt$$
This is in a standard integral form, thus:
$$\frac{1}{4}\cdot\frac{2}{\sqrt{5}}\tan^{-1}\Bigg(\frac{2t}{\sqrt{5}}\Bigg)=\frac{1}{2\sqrt{5}}\tan^{-1}\Bigg(\frac{2\tan(x)}{\sqrt{5}}\Bigg)$$
this is from $0$ to $π/2$.
But I can't substitute for $π/2$, because $\tan(x)$ is undefined for $π/2$. If I input this integral into Mathcad I get $\frac{π\sqrt{5}}{20}$. How can I get the right answer out of this? Thanks in advance!
| Suppose we seek to evaluate
$$\int_0^{\pi/2} \frac{1}{4\sin^2(x)+5\cos^2(x)} dx
= \frac{1}{4} \int_0^{2\pi} \frac{1}{4\sin^2(x)+5\cos^2(x)} dx.$$
Introduce $z=\exp(ix)$ so that $dz=iz \; dx$ to get
$$\frac{1}{4}\int_{|z|=1}
\frac{1}{4(z-1/z)^2/(-4)+5(z+1/z)^2/4} \frac{dz}{iz}
\\ = \int_{|z|=1}
\frac{1}{-4(z-1/z)^2+5(z+1/z)^2} \frac{dz}{iz}
\\ = \int_{|z|=1}
\frac{z^2}{-4(z^2-1)^2+5(z^2+1)^2} \frac{dz}{iz}
\\ = \frac{1}{i} \int_{|z|=1}
\frac{z}{-4(z^2-1)^2+5(z^2+1)^2} \; dz
\\ = \frac{1}{i} \int_{|z|=1}
\frac{z}{z^4 + 18z^2 + 1} \; dz.$$
We have $$z^4 + 18z^2 + 1 = (z^2+9)^2 - 80$$
so we get for the poles
$$\rho_{1,2,3,4} = \pm\sqrt{\pm\sqrt{80} -9}.$$
Now have by inspection that only the two poles
$$\rho_{1,2} = \pm\sqrt{\sqrt{80} -9}$$
contribute being inside the unit circle and we get for the integral
$$\frac{1}{i} 2\pi i
(\mathrm{Res}(f(z); z=\rho_1)
+ \mathrm{Res}(f(z); z=\rho_2))$$
where we have set
$$f(z) = \frac{z}{z^4 + 18z^2 + 1}.$$
These residues are
$$\left.\frac{z}{4z^3+36z}\right|_{z=\rho_{1,2}}
= \left.\frac{1}{4z^2+36}\right|_{z=\rho_{1,2}}
= \frac{1}{4}\left.\frac{1}{z^2+9}\right|_{z=\rho_{1,2}}.$$
By definition of $\rho_{1,2}$ we have $\rho_{1,2}^2 + 9
= \sqrt{80}$ so we finally obtain
$$\frac{1}{i} 2\pi i \times 2\times\frac{1}{4} \frac{1}{\sqrt{80}}
= \pi \frac{1}{\sqrt{80}}
= \frac{\pi}{4\sqrt{5}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1462259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Prove that $\sin^{6}{\frac{\theta}{2}}+\cos^{6}{\frac{\theta}{2}}=\frac{1}{4}(1+3\cos^2\theta)$ A question on submultiple angles states...
Prove that:$$\sin^{6}{\frac{\theta}{2}}+\cos^{6}{\frac{\theta}{2}}=\frac{1}{4}(1+3\cos^2\theta)$$
My efforts
I tried using the formula
$$a^3+b^3=(a+b)^3-3ab(a+b)$$
and
$$\cos^2{\frac{\theta}{2}=\frac{\cos\theta + 1}{2}}$$
Then I tried simplifying it:
$$\require{cancel} \begin{align} \sin^{6}{\frac{\theta}{2}}+\cos^{6}{\frac{\theta}{2}} & = (\sin^{2}{\frac{\theta}{2}}+\cos^{2}{\frac{\theta}{2}})^3 - 3\sin^2{\frac{\theta}{2}}\cos^2{\frac{\theta}{2}}(\sin^{2}{\frac{\theta}{2}}+\cos^{2}{\frac{\theta}{2}})\\
& = 1 - 3\sin^2{\frac{\theta}{2}}\cos^2{\frac{\theta}{2}}\\
& = 1 - 3(1 - \cos^2{\frac{\theta}{2}})\cos^2{\frac{\theta}{2}}\\
& = 1 - \frac{3}{2}(\cos\theta+1) + \frac{3}{4}(\cos\theta+1)^2\\
& = \frac{4\cancel{-6\cos\theta}-2+3\cos^2+3+\cancel{6\cos\theta}}{4}\\
& = \frac{1}{4}(5+3\cos^2\theta)\end{align} $$
I suspect I must have messed up with some sign somewhere. The trouble is, I can't seem to find where. Please help me in this regard.
Update: I am not accepting an answer because all the answers are equally good. It would be an injustice to the other answers.
Note to the editors: I also suspect that my post is a little short on appropriate tags. If required, please do the needful.
| You may use
$$
2\sin x \cos x= \sin (2x) \tag1
$$
giving
$$
\begin{align} \sin^{6}{\frac{\theta}{2}}+\cos^{6}{\frac{\theta}{2}} & = (\sin^{2}{\frac{\theta}{2}}+\cos^{2}{\frac{\theta}{2}})^3 - 3\sin^2{\frac{\theta}{2}}\cos^2{\frac{\theta}{2}}(\sin^{2}{\frac{\theta}{2}}+\cos^{2}{\frac{\theta}{2}})\\
& = 1 - 3\sin^2{\frac{\theta}{2}}\cos^2{\frac{\theta}{2}} \\
& = 1 - \frac34\sin^2{\theta}\quad (\text{using} \,(1))\\
& = 1 - \frac34(1-\cos^2{\theta})\\
& = \frac14(1+3\cos^2{\theta})
\end{align}
$$ as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1467556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
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Prove that $2^{n-1}$ divides $\binom{n}{1} + \binom{n}{3}5 + \binom{n}{5}25 + \binom{n}{7}125 + \cdots$ for $n \geqslant 1$. Prove that $2^{n-1}$ divides $\binom{n}{1} + \binom{n}{3}5 + \binom{n}{5}25 + \binom{n}{7}125 + \cdots$ for $n \geqslant 1$.
Assume $\binom{n}{k} = 0$ if $k>n$.
Does anyone know an elementary proof without using the Binet formula?
| This proof may not qualify as elementary, as it uses some algebraic number theory.
Let $R=\mathbb{Z}[(1+\sqrt{5})/2]$, the ring of integers of $\mathbb{Q}(\sqrt{5})$, and define $\alpha:=(1+\sqrt{5})/2$ and $\beta:=(1-\sqrt{5})/2\in R$. The binomial theorem implies
$$
\begin{align*}
S_n:=
{n\choose 1}+5{n\choose 3}+25{n\choose 5}\ldots&=\frac{(1+\sqrt{5})^n-(1-\sqrt{5})^n}{2\sqrt{5}}\\
&=2^{n-1}\frac{\alpha^n-\beta^{n}}{\sqrt{5}}.
\end{align*}
$$
Now $2$ and $\sqrt{5}$ are coprime, so $2^{n-1}$ must divide $S_n$ in $R$. But the only rational numbers in $R$ are integers, so $2^{n-1}$ divides $S_n$ as integers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1469026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
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A golden ratio series from a comic book The eighth installment of the Filipino comic series Kikomachine Komix features a peculiar series for the golden ratio in its cover:
That is,
$$\phi=\frac{13}{8}+\sum_{n=0}^\infty \frac{(-1)^{n+1}(2n+1)!}{(n+2)!n!4^{2n+3}}$$
How might this be proven?
| Using the series
$$
(1-4x)^{-1/2}=\sum_{n=0}^\infty\binom{2n}{n}x^n\tag{1}
$$
we get
$$
\begin{align}
f(x)
&=\sum_{n=0}^\infty\frac{(-1)^{n+1}(2n+1)!}{(n+2)!\,n!}x^{n+2}\\
&=\frac12\sum_{n=1}^\infty(-1)^n\binom{2n}{n}\frac{x^{n+1}}{n+1}\\
&=\frac12\int_0^x\left[(1+4t)^{-1/2}-1\right]\,\mathrm{d}t\\
&=\frac14(1+4x)^{1/2}-\frac x2-\frac14\tag{2}
\end{align}
$$
Therefore,
$$
\begin{align}
\frac{13}8+\sum_{n=0}^\infty\frac{(-1)^{n+1}(2n+1)!}{(n+2)!\,n!4^{2n+3}}
&=\frac{13}8+4f\left(\frac1{16}\right)\\
&=\frac{13}8+4\left(\frac14\left(1+\frac14\right)^{1/2}-\frac1{32}-\frac14\right)\\
&=\frac{1+\sqrt5}2\\[8pt]
&=\phi\tag{3}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1470099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "45",
"answer_count": 5,
"answer_id": 0
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Show $\sum \frac{n}{2^n + n} <\frac 32$ Let $n\in \mathbb N^{+}$,show that
$$\dfrac{1}{2+1}+\dfrac{2}{2^2+2}+\dfrac{3}{2^3+3}+\cdots+\dfrac{n}{2^n+n}<\dfrac{3}{2}$$
| $$\sum_{n\geq 1}\frac{n}{2^n+n} = \frac{31}{33}+\sum_{n\geq 4}\sum_{k\geq 1}(-1)^{k+1} n^k 2^{-nk}\leq \frac{31}{33}+\frac{5}{8}-\frac{173}{1728}+\frac{25157}{1229312}<\frac{3}{2}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1470205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Question related to tan in a ratio in a triangle If in a triangle $\tan A:\tan B:\tan C = 1:2:3$ then, what are the ratio of the sides $a,b,c $?
| Let $AD, BE, CF$ be altitudes of triangle $ABC$. Denote $BC=a, CA=b, AB=c$.
Observe that $$\frac 23=\frac{\tan B}{\tan C} = \frac{\frac{AD}{BD}}{\frac{AD}{DC}} = \frac{DC}{BD}$$ so $BD=\frac 35 a$ and $DC=\frac 25 a$.
On the other hand Pythagorean theorem says that $$c^2-b^2=AB^2-AC^2=(BD^2+AD^2)-(CD^2+AD^2)=BD^2-CD^2=\left(\frac 35 a\right)^2 - \left(\frac 25 a\right)^2 = \frac 15 a^2$$
Analogously we find $CE=\frac 14 b$, $AE=\frac 34 b$ and $$c^2-a^2=\frac 12 b^2$$
The above results imply that $b^2=\frac 85 a^2$ and $c^2=\frac 95 a^2$. Thus $a^2:b^2:c^2=5:8:9$ which leads to the answer: $$a:b:c=\sqrt 5 : 2\sqrt 2 : 3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1473159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
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Numerical Series nature I need someone to approve if my way to prove that the following numerical series is divergent.
*
*Consider the Series $\sum_{n=1} \frac{1}{ln(n+n^2+n^3)}$. following the comparison test i found that the series diverge.
*My approach:
since $e^n +\frac{n^2}{2!} +\frac{5n^3}{3!} \ge n+n^2+n^3$ then $ \ln(e^n +\frac{n^2}{2!} +\frac{5n^3}{3!}) \ge ln(n+n^2+n^3) $
i.e $\frac{1}{ln(e^n +\frac{n^2}{2!} +\frac{5n^3}{3!})} \le \frac{1}{ln(n+n^2+n^3)}$ but since $e^n +\frac{n^2}{2!} +\frac{5n^3}{3!} \sim e^n$ as $n\rightarrow \infty$
hence$\frac{1}{ln(e^n +\frac{n^2}{2!} +\frac{5n^3}{3!})} \sim \frac{1}{n}$ where $\sum\frac{1}{n}$ is divergent according to Reimann and so we get that the sum $\sum_{n=1} \frac{1}{ln(n+n^2+n^3)}$ is also divergent.
Am i right doing so ?
| $$\log{(n+n^2+n^3)} = \log{n^3} + \log{\left (1+\frac1{n}+\frac1{n^2} \right )} = 3 \log{n}+O\left ( \frac1{n} \right )$$
Thus
$$\frac1{\log{(n+n^2+n^3)}} = \frac1{3 \log{n}} + O\left ( \frac1{n \log{n}} \right )$$
Thus the sum diverges by comparison with, e.g., then sum over $1/n$ which diverges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1473247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Can someone prove this inequality? I am trying to prove the following inequality, but I cannot.
Can someone help me?
I'll appreciate any small hints.
$$
\sum_{k=2}^{n-1}k\log_2k \le \frac12 n^2 \log_2n - \frac18 n^2
$$
| I am really appreciate many hints and advice.
Thanks to them, I could success proving this proposition state.
Prove that
$$
\sum_{k=2}^{n-1}k\log_2k \le \frac12n^2 \log_2n-\frac18n^2
$$
Proof:
\begin{align*}
\sum_{k=2}^{n-1} k \log_{2}{k} &\le \int_{2}^{n} x \log_{2}{x} {dx}\\
&=\left[\frac{x^2}{2}{\log_2x}\right]_{2}^{n} - \int_{2}^{n} \frac{x^2}{2}\frac{1}{x\ln2}dx\\
&=\left[\frac{x^2}{2}{\log_2x}\right]_{2}^{n} - \int_{2}^{n} \frac{x}{2\ln2}dx\\
&=\left[\frac{x^2}{2}{\log_2x}\right]_{2}^{n} - \left[\frac{x^2}{4\ln2}\right]_{2}^{n}\\
&=\left[\frac{x^2}{2}{\log_2x} - \frac{x^2}{4\ln2}\right]_{2}^{n}\\
&=\left[\frac{n^2}{2}{\log_2n} - \frac{n^2}{4\ln2}\right] - \left[2 - \frac{1}{\ln2}\right]\\
&=\frac{1}{2}{n^2}{\log_2n} - \frac{n^2}{4\ln2} - 2 + \frac{1}{\ln2}\\
&=\frac{1}{2}{n^2}{\log_2n} - \frac{1}{8}n^2\left(\frac{2}{\ln2}+\frac{16}{n^2}-\frac{8}{n^2\ln2}\right)
\end{align*}
\begin{align*}
\frac{2}{\ln2}+\frac{16}{n^2}-\frac{8}{n^2\ln2} &\ge 1\\
\frac{2n^2+16\ln2-8}{n^2\ln2} &\ge 1\\
2n^2+16\ln2-8 &\ge n^2\ln2\\
(2-\ln2)n^2 &\ge 8(1-2\ln2) = 8\left(\ln\frac{e}{4}\right)\\\\\\
\therefore \frac{2}{\ln2}+\frac{16}{n^2}-\frac{8}{n^2\ln2} &\ge 1\\
\big(\because (2-\ln2)n^2 &\gt 0 , \ \ \ 8\left(\ln\frac{e}{4}\right) \lt 0\big)
\end{align*}
\begin{align*}
\sum_{k=2}^{n-1}k\log_2k &\le \int_{2}^{n} x \log_{2}{x} {dx}\\
&= \frac{1}{2}{n^2}{\log_2n} - \frac{1}{8}n^2\left(\frac{2}{\ln2}+\frac{16}{n^2}-\frac{8}{n^2\ln2}\right)\\
&\le \frac12n^2 \log_2n-\frac18n^2
\end{align*}
| {
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"source": "stackexchange",
"question_score": "1",
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Show that this sum is an integer. I have to show that
$$g\left(\frac{1}{2015}\right) + g\left(\frac{2}{2015}\right) +\cdots + g\left(\frac{2014}{2015}\right) $$
is an integer. Here $g(t)=\dfrac{3^t}{3^t+3^{1/2}}$.
I tried to solve it using power series, but I can not finish with any convincing argument to said that the sum is an integer. (Also the use of a computer isn't allowed.)
| This is the answer to the first version of the question:
We want to compute:
$$ \sum_{k=1}^{2014}\frac{\frac{3k}{2015}}{\frac{3k}{2015}+\frac{7}{2}}=\sum_{k=1}^{2014}\frac{6k}{6k+5\cdot 7\cdot 13\cdot 31} $$
but that number cannot be an integer because $6\cdot 2001+5\cdot 7\cdot 13\cdot 31$ is a prime.
Anyway, the value of the LHS is about $2015\int_{0}^{1}\frac{3x}{3x+\frac{7}{2}}\,dx = 2015\left(1-\frac{7}{6}\log\frac{13}{7}\right)$.
We may also estimate the difference between our sum and $2015\left(1-\frac{7}{6}\log\frac{13}{7}\right)$ through the Hermite-Hadamard inequality, since $\frac{3x}{3x+\frac{7}{2}}$ is a concave function on $[0,1]$. That gives another way for proving that our sum is not an integer, since it gives that our sum is between $559.4$ and $559.8$.
This is the answer to the second version of the question. If
$$ g(t) = \frac{3^t}{3^t+3^{1/2}} $$
we have:
$$ g(t)+g(1-t) = \frac{3^t}{3^t+3^{1/2}}+\frac{3^{1-t}}{3^{1-t}+3^{1/2}}=\frac{1}{1+3^{1/2-t}}+\frac{1}{1+3^{t-1/2}}=\color{red}{1}$$
hence the claim is trivial.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1474849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 5,
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Using generating functions to find the coefficient I was looking at an example in my textbook where it was asked to find in how many ways can a police captain distribute 24 rifle shells to four police officers so that each officer gets at least three shells, but not more than eight. Using generating functions: ${f(x)=(x^3+x^4+x^5+...+x^8)^4}$, and we will be looking to find the coefficient of ${x^{24}}$ in $f(x)$
We know that $f(x)= x^{12}((1-x^6)/(1-x))^4$
Therefore the answer will be the coefficient of $x^{12}$ in the expansion of $(1-x^6)^4*(1-x)^{-4}$
Which in turn is equal to $A$:
$[1-\dbinom{4}{1}x^6+\dbinom{4}{2}x^{12}-\dbinom{4}{3}x^{18}+x^{24}]*[\dbinom{-4}{0}+\dbinom{-4}{1}(-x)+\dbinom{-4}{2}(-x)^2+...]$
Up to this point I perfectly understand the problem and the proposed solution; but the textbook goes on saying that the above expression is equivalent to $B$:
$[\dbinom{-4}{12}(-1)^{12}-\dbinom{4}{1}\dbinom{-4}{6}(-1)^6+\dbinom{4}{2}\dbinom{-4}{0}]$
I do not understand how they went from $A$ to $B$
I know that $C$: $[\dbinom{-4}{12}(-1)^{12}]$ is the coefficient of $x^{24}$ in: $x^{12}[\dbinom{-4}{0}+\dbinom{-4}{1}(-x)+\dbinom{-4}{2}(-x)^2+...]$
But I do not seem to make the link between $C$ and the following subtraction and addition, namely:
$[C-\dbinom{4}{1}\dbinom{-4}{6}(-1)^6+\dbinom{4}{2}\dbinom{-4}{0}]$
| If $(1-x^6)^4$ has $x^k (12\ge k\ge 0)$ term then we select coefficient of $x^{12-k}$ in $(1-x)^{-4}$
Multiplying both coefficients results in the coefficient of $x^{k} + x^{12-k} = x^{12}$ which is the required coefficient we need.
Other terms will have contribution $0$ in finding the coefficient of $x^{12}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1476081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
No limit for $\frac{x^2 y}{x^2-y^2}$ when $(x,y) \rightarrow (0,0)$ I'm trying to calculate
$$\lim_{(x,y)\to(0,0)}\dfrac{x^2 y}{x^2-y^2}$$
I've tried a lot of paths: $y=x$, $y=x+1$, $y=x^2$... Always I end up with $0$.
The answer of my homework and the Wolfram Alpha say me it doesn't exist.
What I'm doing wrong?
I appreciate any help.
| This solution is more like what you were trying.
Start with $x=2y$, you will get
$$\lim_{(x,y)\to(0,0)}\dfrac{x^2 y}{x^2-y^2}
=\lim_{(x,y)\to(0,0)}\dfrac{4y^2 y}{4y^2-y^2} = \lim_{(x,y)\to(0,0)}\dfrac{4y^3}{3y^2} = \lim_{(x,y)\to(0,0)}\dfrac{4y}{3}=0$$
But for $x=y^2-y$ you will get
$$\lim_{(x,y)\to(0,0)}\dfrac{x^2 y}{x^2-y^2} = \lim_{(x,y)\to(0,0)}\dfrac{(y^2-y)^2y}{(y^2-y)^2-y^2} = \lim_{(x,y)\to(0,0)}\dfrac{y^5-2y^4+y^3}{y^4-2y^3} = \lim_{(x,y)\to(0,0)}\dfrac{y^2-2y+1}{y-2} = -\dfrac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1483887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluation of $ \int\frac{1}{2x\sqrt{1-x}\cdot \sqrt{2-x+\sqrt{1-x}}}dx$
Evaluation of $\displaystyle \int\frac{1}{2x\sqrt{1-x}\cdot \sqrt{2-x+\sqrt{1-x}}}dx$
$\bf{My\; Try:}$ Let $x=\sin^2 \theta\;,$ Then $dx = 2\sin \theta \cdot \cos \theta d\theta$
So Integral $$\displaystyle I = \int\frac{2\sin \theta \cos \theta }{2\sin^2 \theta \cos \theta\sqrt{2-\sin^2 \theta+\cos \theta}}d\theta = \int\frac{1}{\sin \theta\sqrt{\cos^2 \theta+\cos \theta+1}}d\theta$$
So $$\displaystyle I = \int\frac{\sin \theta}{(1-\cos^2 \theta)\sqrt{\cos^2 \theta+\cos \theta+1}}d\theta\;,$$ Now put $\cos \theta = t\;,$ Then $\sin \theta d\theta = -dt$. So $$\displaystyle I = \int\frac{1}{(1-t^2)\sqrt{t^2+t+1}}dt$$
How can we continue?
| I give you a step or two, and know you can finish from them:
Make a partial fraction decomposition of the rational part, and write your integral as
$$
\frac{1}{2}\int\frac{1}{(1-t)\sqrt{1+t+t^2}}\,dt+\frac{1}{2}\int\frac{1}{(1+t)\sqrt{1+t+t^2}}\,dt
$$
In the first integral, let
$$
u=\frac{1-t}{\sqrt{1+t+t^2}},
$$
and in the second, let
$$
u=\frac{1+t}{\sqrt{1+t+t^2}}.
$$
You will end up with integrals of the form
$$
\int\frac{1}{c-u^2}\,du
$$
which I'm certain you can handle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1485581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Let $k$ be a natural number . Then $3k+1$ , $4k+1$ and $6k+1$ cannot all be square numbers.
Let $k$ be a natural number. Then $3k+1$ , $4k+1$ and $6k+1$ cannot all be square numbers.
I tried to prove this by supposing one of them is a square number and by substituting the corresponding $k$ value. But I failed to prove it.
If we ignore one term, we can make the remaining terms squares. For example, $3k+1$ and $4k+1$ are both squares if $k=56$ (they are $13^2$ and $15^2$); $3k+1$ and $6k+1$ are both squares if $k=8$ (they are $5^2$ and $7^2$); $4k+1$ and $6k+1$ are both squares if $k=20$ (they are $9^2$ and $11^2$).
| I think this will work, but I don't quite have time today to fill in all details.
If $3k+1=x^2,4k+1=y^2,6k+1=z^2$, then $2x^2+z^2=3y^2$. Letting $a=x/y,b=z/y$, this gives $2a^2+b^2=3$. We can parametrize the whole space of solutions like one usually does for Pythagoream triples: $(1,1)$ is an obvious point on the curve. Now let $t$ be any real and let $Y=t(X-1)+1$ be a line through it. It intersects the curve $2a^2+b^2=3$ at point $(1,1)$ and other point $(a,b)$ satisfying $2a^2+b^2=3$ and $b=t(a-1)+1$. Substituting, we get $a^2(2+t^2)+a(2t-2t^2)+(t^2-2t-2)=0$. If we interpret this as a quadratic in $a$, $1$ is one solution, so by Viete's formulas, the other is $a=(t^2-2t-2)/(t^2+2)$. Then $b=(-t^2-4t+2)/(t^2+2)$. Now we look at $x/z=a/b=(t^2-2t-2)/(-t^2-4t+2)$. Letting $t=n/m$ be rational, this gives $x/z=(n^2-2nm-2m^2)/(-n^2-4nm+2m^2)$.
Clearly $x,z$ are relatively prime, since $2x^2-z^2=1$. I claim $n^2-2nm-2m^2,-n^2-4nm+2m^2$ have $\gcd$ dividing $6$. This is because their sum is $6nm$, and if $p\mid n$ and $p\mid n^2-2nm-2m^2$, then $p\mid m$ and vice versa.
(to be continued)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "62",
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Linear Algebra - change of basis matrix - quick method In the class, the instructor supplied the following exercise:
Given the following vectors:
$$
\begin{align}
\psi_1(x) &= \frac{1}{\sqrt 2} \\
\psi_2(x) &= \sqrt{\frac{3}{2}} \cdot x \\
\varphi_1(x) &= \frac{\sqrt 3}{2} \cdot x + \frac{1}{2} \\
\varphi_2(x) &= \frac{\sqrt 3}{2} \cdot x - \frac{1}{2}
\end{align}
$$
Then $(\psi_1, \psi_2)$ and $(\varphi_1, \varphi_2)$ form a basis.
Find a matrix which converts a vector $\begin{pmatrix} c_1 \\ c_2 \end{pmatrix}_{\psi}$ in base $\psi$ to a vector $\begin{pmatrix} d_1 \\ d_2 \end{pmatrix}_{\varphi}$ in base $\varphi$.
The solution according to the instructor:
The instructor guessed the following:
$$
\begin{align}
\varphi_1(x) &= \frac{1}{\sqrt 2} \cdot \big(\; \psi_1(x) + \psi_2(x) \;\big) \\
\varphi_1(x) &= \frac{1}{\sqrt 2} \cdot \big( -\psi_1(x) + \psi_2(x) \;\big)
\end{align}
$$
So the required matrix $P$ is:
$$
P = \begin{pmatrix}
1 & 1\\
-1 & 1\\
\end{pmatrix} \cdot \frac{1}{\sqrt 2}
$$
and the conversion is done as follows: $\vec d = P \cdot \vec c$.
Or, when elaborating:
$$
\begin{pmatrix} d_1 \\ d_2 \end{pmatrix} =
\begin{pmatrix}
\frac{1}{\sqrt 2} & \frac{1}{\sqrt 2}\\
-\frac{1}{\sqrt 2} & \frac{1}{\sqrt 2}\\
\end{pmatrix}
\cdot
\begin{pmatrix} c_1 \\ c_2 \end{pmatrix}
$$
Now, this guess seems easy. But I'm having difficulties with:
*
*How did he know that he needs to find $\varphi_i$ as a linear combination of $\psi_i$ in order to find the matrix in question?
*What should I do if guessing isn't as easy as in the example?
Solve it using the standard method of $P = M_\varphi^E M_E^\psi $ ? ($E$ is standard basis ; $E = (1, x)$)
It takes precious time. Isn't there a quicker method?
Edit:
I solved it as follows:
$$
P = M_\varphi^E \cdot M_E^\psi \equiv P_\varphi^\psi
$$
when:
*
*The basis in the subscript is the range, and the basis in the superscript is the domain
*Matrix M is a also a change of basis matrix (identity map $Id$)
*E is standard basis
Therefore, according to the definition of change-of-basis matrix:
$$
P_\varphi^\psi = \begin{pmatrix}
| & | \\
[\psi_1]_\varphi & [\psi_2]_\varphi \\
| & | \\
\end{pmatrix} = \begin{pmatrix}
\lambda_1 & \lambda_3 \\
\lambda_2 & \lambda_4 \\
\end{pmatrix}
$$
Now, according to the definition of coordinate vector:
$$
\begin{align*}
& [\psi_1]_\varphi =
\begin{pmatrix} \lambda_1 \\ \lambda_2 \end{pmatrix} =
\lambda_1 \cdot \varphi_1 + \lambda_2 \cdot \varphi_2 \\
&
[\psi_2]_\varphi =
\begin{pmatrix} \lambda_3 \\ \lambda_4 \end{pmatrix} =
\lambda_3 \cdot \varphi_1 + \lambda_4 \cdot \varphi_2
\\
\end{align*}
$$
According to all the actions that I did until now - we know that we need to find $\psi$ as a function of $\phi$ - which is the opposite of what the instructor did.
Let's solve the above equations:
$$
\begin{align*}
\left[ \frac{1}{\sqrt 2} \right]_\varphi &= \frac{1}{\sqrt 2} \cdot (\varphi_1 - \varphi_2) \\
\left[ \sqrt{\frac{3}{2}} \cdot x \right]_\varphi &= \frac{1}{\sqrt 2} \cdot (\varphi_1 + \varphi_2)
\end{align*}
$$
$$
\Longrightarrow P_\varphi^\psi = \begin{pmatrix}
1 & 1 \\
-1 & 1 \\
\end{pmatrix} \cdot \frac{1}{\sqrt 2}
$$
Which is exactly the same matrix that the instructor received. He told me that there's symmetry because that the matrix is a Unitary operator. When I showed this (my solution) to the instructor, he said that I mixed the positions of the bases vectors, so I got a wrong result. But again I don't understand what's wrong - everything is done that same as I did in Linear Algebra course.
| Using the “standard method:” $$
M_E^\psi=\pmatrix{\frac1{\sqrt2} & 0 \\ 0 & \sqrt\frac32} \\
M_\varphi^E=\pmatrix{\frac12 & -\frac12 \\ \frac{\sqrt3}2 & \frac{\sqrt3}2}^{-1}=\pmatrix{1 & \frac1{\sqrt3} \\ -1 & \frac1{\sqrt3}} \\
M_\varphi^E M_E^\psi = \pmatrix{1 & \frac1{\sqrt3} \\ -1 & \frac1{\sqrt3}}\pmatrix{\frac1{\sqrt2} & 0 \\ 0 & \sqrt\frac32}=\pmatrix{\frac1{\sqrt2} & \frac1{\sqrt2} \\ -\frac1{\sqrt2} & \frac1{\sqrt2}}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1493915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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General form of $\sqrt{a - b} - \sqrt{a + b}$?
General form of $\sqrt{a - b} - \sqrt{a + b}$?
What I would do is: let
$x = \sqrt{a - b} - \sqrt{a + b}$
$x^2 = 2a - 2\sqrt{a^2 - b^2}$
Then since $a + b > a - b$
$x = -\sqrt{2a - 2\sqrt{a^2 - b^2}}$
This is particularly useful in solving problem with nested square roots (like, let $a = 11, b = 2\sqrt{10}$
I am wondering, is there any other derivation and MORE WANTED: any simpler formula?
| You want to calculate $u-v$ for $u$ and $v$ such that $u^2=a-b$ and $v^2=a+b$. Let us assume that $a>b$, i.e., $u,v>0$.
Then you have $u^2v^2=(a-b)(a+b)=a^2-b^2$ which leaves you with
$$uv=\sqrt{a^2-b^2}.$$
Then you have
$$(u-v)^2=u^2+v^2-2uv=2a-2\sqrt{a^2+b^2}$$
and, knowing that $u<v$ and $u-v<0$, you get
$$u-v = \sqrt{2a-2\sqrt{a^2+b^2}}.$$
Similarly, you can use
$$(u+v)^2=u^2+v^2+2uv = 2a+2\sqrt{a^2+b^2}$$
to calculate
$$u+v=\sqrt{2a+2\sqrt{a^2+b^2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1494304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Taylor polynomial of degree 2 of $e^{x^2+x}$ I want to find the Taylor polynomial of degree 2 of $e^{x^2+x}$ and this is what the answer should be:
$$e^{x^2+x} = e^{x^2}e^{x} = (1 + x^2 + O(x^4)) (1 + x + \cfrac{x^2}{2} + O(x^3)) = 1 + x + \cfrac{3x^2}{2} + O(x^3)$$
The thing is that I don't understand what happens after the last equals sign. If anyone would care to explain that would be much appreciated.
| You multiply the two factors with the usual rules for polynomials (distributivity/associativity), with the additional rules that $O(x^k)\cdot \alpha x^\ell = O(x^{k+\ell})$ and $O(x^k)\cdot O(x^\ell) = O(x^{k+\ell})$:
$$
\begin{align}
(1 + x^2 + O(x^4)) (1 + x + \frac{x^2}{2} + O(x^3))
= 1 + x + \frac{x^2}{2} + O(x^3) + x^2 \cdots
\end{align}
$$
except that you stop whenever you get a power bigger than what you get in any of the $O(\cdot)$ so far. This is why here I stopped the expansion and didn't include any $x^3$, $x^4$, or $x^5$: there is no point, as I have a $O(x^3)$ already which "swallows" all these terms. You can go further in the expansion, but in the end any $x^k$ term for $k\geq 3$ will disappear, since they are all hidden in the $O(x^3)$.
Then, you gather the terms:
$$
\begin{align}
1 + x + \frac{x^2}{2} + O(x^3) + x^2= 1+x +\frac{3}{2}x^2 + O(x^3)
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1494402",
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"source": "stackexchange",
"question_score": "1",
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Find three consecutive entries of a row of Pascal triangle that are in the ratio of 1 : 2 : 3
Find three consecutive entries of a row of Pascal triangle that are in the ratio of 1 : 2 : 3
This means that:
$$\begin{align}
2\binom{n}{k} =\binom{n}{k+1}\\
3\binom{n}{k} =\binom{n}{k+2}\\
\end{align}$$
I simplefied these equations:
$$\begin{align}
2\binom{n}{k} =\binom{n}{k+1}\\
\end{align}$$
Is the same as:
$$\begin{align}
2k+2 = n-k\\
3k+2 = n
\end{align}$$
And:
$$\begin{align}
3\binom{n}{k} =\binom{n}{k+2}\\
\end{align}$$
Is the same as:
$$\begin{align}
3(k+1)(k+2) = (n-k-1)(n-k)
\end{align}$$
I do not know how I could go further so that I would end up with the value of n and k. This is because I can not simplify 3(k+1)(k+2) = (n-k-1)(n-k) enough.
How can I proceed so that I can get the value of n and k for:
$$\begin{align}
3(k+1)(k+2) = (n-k-1)(n-k) \\
3k+2 = n
\end{align}$$
| Since $n=3k+2$, you can have
$$3(k+1)(k+2)=(\color{red}{3k+2}-k-1)(\color{red}{3k+2}-k)$$
Now solve this for $k$. (don't forget to check if each $k$ is sufficient.)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Questions concerning $ \det ({}^A_{C\,}{}^B_D) = \det ({}^D_{B\,}{}^C_A)$ Let $A, B, C, D $ be $n \times n $ matrices. Using Schur complements I have found that
$$ \begin{pmatrix} A & B \\ C & D \end{pmatrix} =
\begin{pmatrix} A & 0 \\ 0 & I \end{pmatrix}
\begin{pmatrix} I & 0 \\ C & I \end{pmatrix}
\begin{pmatrix} I & A^{-1}B \\ 0 & D-CA^{-1}B \end{pmatrix} $$
and
$$ \begin{pmatrix} D & C \\ B & A \end{pmatrix} =
\begin{pmatrix} I & CA^{-1} \\ 0 & I \end{pmatrix}
\begin{pmatrix} D-CA^{-1}B & 0 \\ 0 & A \end{pmatrix}
\begin{pmatrix} I & 0 \\ A^{-1}B & I \end{pmatrix}, $$
from which the determinant equality follows, as long as $ A^{-1} $ exists. However, how do I tackle this when $A$ is singular? Here, I could just switch decompositions, but then I will get the same problem when $D$ is singular. Maybe one could derive two more decompositions using Schur complements, involving $B^{-1}$ and $C^{-1}$, respectively, and then one could say that the equality holds if at least one of the submatrices is nonsingular? Then, if all four submatrices are singular, the determinant must be zero -- from which equality follows trivially.
Also, does the equality hold when $A, B, C, D$ are not necessarily square but of matching sizes? Here, it doesn't seem like the decompositions will be valid, as $A$ or $D$ aren't necessarily square matrices (although then $B$ and $C$ must be).
| One approach is to take advantage of the continuity of the determinant. In particular, if $A$ is singular, we have
$$
\det \pmatrix{A &B\\C&D} = \lim_{t \to 0} \det\pmatrix{A + tI & B\\C&D}
= \lim_{t \to 0} \det \pmatrix{D&C\\B&A + tI} =
\det \pmatrix{D&C\\B&A}
$$
Alternatively, you could have noted that
$$
\pmatrix{D&C\\B&A} = \pmatrix{0&I\\I&0} \pmatrix{A&B\\C&D} \pmatrix{0&I\\I&0}
$$
from there, we would only need to show that
$$
\det \pmatrix{0&I_n\\I_n&0} = (-1)^n
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1500366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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} |
Express a quadratic form in three variable in the format $x^tAx$ using a substitution $x=Py$ I was asked to determine if a quadratic form is positive definite. To do so I must convert in the format $x^tAx$ using a substitution x=Py. So that "it can be written in diagonal form".
$$Q(x,y,z) = 3x^2 + 8xz+2y^2+z^2$$
My idea is:
$$(x,y,z) \begin{pmatrix} 3x+4z \\ 2y \\ 4x+z^2 \end{pmatrix} $$
$$\begin{pmatrix} 3&0&4 \\0&2&0 \\ 4&0&1 \end{pmatrix} $$
Is this correct? And then it would be positive definite if all eigen values of this are equal to or larger than 0?
| The matrix should be:
$$\mathbf{A}=\left(\begin{matrix}3&0&4\\0&2&0\\4&0&1\end{matrix}\right)$$
Multiplying $\mathbf{x^TAx}$ out where $\mathbf{x}=\left(\begin{matrix}x&y&z\end{matrix}\right)^T$ gives:
$$\left(\begin{matrix}x&y&z\end{matrix}\right)\left(\begin{matrix}3&0&4\\0&2&0\\4&0&1\end{matrix}\right)\left(\begin{matrix}x\\y\\z\end{matrix}\right)=\left(\begin{matrix}x&y&z\end{matrix}\right)\left(\begin{matrix}3x+4z\\2y\\4x+z\end{matrix}\right)\\=3x^2+4xz+2y^2+4xz+z^2=3x^2+8xz+2y^2+z^2$$
We can orthogonally diagonalise $\mathbf{A}=\mathbf{PDP^T}$ as $\mathbf{A}$ is symmetric. If we let $\mathbf{y}=\mathbf{P^Tx}$, then $$\mathbf{x^TPDP^Tx}=\mathbf{y^TDy}$$
which is in the form $\lambda_1y_1^2+\lambda_2y_2^2+\lambda_3y_3^2$, where $\lambda_i$ are the eigenvalues and $\mathbf{y}=\left(\begin{matrix}y_1&y_2&y_3\end{matrix}\right)^T$.
Hence, a quadratic form being positive definite is equivalent to nonnegative eigenvalues of $\mathbf{A}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1500922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove by induction: $2!\cdot 4!\cdot 6!\cdot\cdot\cdot (2n)!\ge ((n+1)!)^n$ Prove by induction: $2!\cdot 4!\cdot 6!\cdot\cdot\cdot (2n)!\ge ((n+1)!)^n$
For $n=1$ inequality holds.
$(*)$For $n=k$
$2!\cdot\cdot\cdot (2k)!\ge ((k+1)!)^k$
Multiplying LHS and RHS with $(2k+2)!$ gives
$$2!\cdot\cdot\cdot (2k)!(2k+2)!\ge ((k+1)!)^k(2k+2)!$$
Assume (by contradiction)$$2!\cdot\cdot\cdot (2k)!(2k+2)!< ((k+1)!)^k(2k+2)!$$
$$2!\cdot\cdot\cdot (2k)!(2k+2)!-((k+1)!)^k(2k+2)!<0$$
$$(2k+2)!(2!\cdot\cdot\cdot (2k)!-((k+1)!)^k)<0$$
$(2!\cdot\cdot\cdot (2k)!-((k+1)!)^k)\ge 0$ by $(*)$, thus inequality holds $\forall n\in\mathbb{N}$
Is this proof correct?
|
Assume (by contradiction)
$$2!⋅⋅⋅(2k)!(2k+2)!<((k+1)!)k(2k+2)!$$
Why would you assume that, you just showed the contrary one line above ?
Once you're here
$$2!\cdot\cdot\cdot (2k)!(2k+2)!\ge ((k+1)!)^k(2k+2)!$$
you just have to show that $((k+1)!)^k(2k+2)! \ge (k+2)!^{k+1}$.
$(k+2)!^{k+1} = (k+1)!^k \times (k+2)^k \times (k+2)!$ and $(k+2)^k \times (k+2)! \le (2k+2)!$ (because $k+2 < k+2+i$ for all $i$). Thus, you have your result.
| {
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"source": "stackexchange",
"question_score": "1",
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} |
Find $s^4-18s^2-8s$
Let $a,b,c$ be the roots of $x^3-9x^2+11x-1=0$, and let $s=\sqrt{a}+\sqrt{b}+\sqrt{c}$. Find $s^4-18s^2-8s$.
$s^4 - 18s^2 - 8s = (s)(s + 4)(s - 2 + \sqrt{6})(s - 2 - \sqrt{6})$
$P(x) = (x - a)(x - b)(x - c)$
From Vieta's formulas:
$a + b+ c = 9$
$abc = 1$
$ab + ac + bc = 11$
Actually, if $y = \sqrt{x}$ then if $y$ is a root, the function must map $y \to x$, which means, let $x \to x^2$ to get:
$P_2(x) = x^6 - 9x^4 + 11x^2 - 1 = 0$.
Then $s = \sum \text{roots } P_2(x) = 0$.
But then $s^4 - 18s^2 - 8s = 0$? What am I doing wrong?
| Let us instead see how the cubic whose roots are squares of the roots of $x^3-sx^2+ tx- 1$ (why?) would look like. Using Vieta, this is easily written as:
$$x^3-(s^2-2t)x^2+(t^2-2s)x-1$$
Comparing with our given cubic, we get $s^2-2t = 9$ and $t^2-2s = 11$ which on eliminating $t$ gives us $s^4 -18s^2-8s = -37$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1502141",
"timestamp": "2023-03-29T00:00:00",
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Prove: ${n\choose 0}-\frac{1}{3}{n\choose 1}+\frac{1}{5}{n\choose 2}-...(-1)^n\frac{1}{2n+1}{n\choose n}=\frac{n!2^n}{(2n+1)!!}$
Prove: $${n\choose 0}-\frac{1}{3}{n\choose 1}+\frac{1}{5}{n\choose 2}-...+(-1)^n\frac{1}{2n+1}{n\choose n}=\frac{n!2^n}{(2n+1)!!}.$$
Here, $(2n+1)!!$ is an "odd factorial": $(2n+1)!! = 1 \cdot 3 \cdot 5 \cdot \cdots \cdot (2n+1$).
How to prove this equation?
Is it possible to use induction?
$${n\choose 0}-\frac{1}{3}{n\choose 1}+\frac{1}{5}{n\choose 2}-...(-1)^n\frac{1}{2n+1}{n\choose n}=\sum\limits_{k=0}^{n}{n\choose k}(-1)^k\frac{1}{2k+1};$$
$$(2n+1)!!=\frac{(2n)!(2n+1)}{2^nn!}\Rightarrow \frac{n!2^n}{(2n+1)!!}=\frac{2^{2n}(n!)^2}{(2n)!(2n+1)};$$
$$\sum\limits_{k=0}^{n}{n\choose k}(-1)^k\frac{1}{2k+1}=\frac{2^{2n}(n!)^2}{(2n)!(2n+1)}$$
What now?
| One may first observe that your initial sum may be rewritten as
$$
\begin{align}
\sum_{k=0}^n \frac{(-1)^k}{2k+1}\binom{n}{k}
&=\sum_{k=0}^n (-1)^k\left(\int_0^1 x^{2k}\:dx\right)\binom{n}{k}\\\\
&=\int_0^1\sum_{k=0}^n (-1)^k\binom{n}{k}x^{2k}\:dx\\\\
&=\int_0^1(1-x^2)^ndx.
\end{align}
$$ Then, integrating by parts, we have
$$
\begin{align}
\int_0^1(1-x^2)^ndx
&=\left. x(1-x^2)^n\right|_0^1+2n\int_0^1x^2(1-x^2)^{n-1}dx\\\\
&=0-2n\int_0^1(1-x^2-1)(1-x^2)^{n-1}dx\\\\
&=-2n\int_0^1(1-x^2)^ndx+2n\int_0^1(1-x^2)^{n-1}dx.
\end{align}
$$ which is equivalent to
$$
\int_0^1(1-x^2)^ndx=\frac{2n}{2n+1}\int_0^1(1-x^2)^{n-1}dx
$$ giving
$$
\int_0^1(1-x^2)^ndx=\frac{2}{3}\frac{4}{5}\cdots\frac{2n}{2n+1}=\frac{2^{2n}(n!)^2}{(2n+1)(2n)!}
$$ Finally
$$
\sum_{k=0}^n \frac{(-1)^k}{2k+1}\binom{n}{k}=\frac{2^{2n}(n!)^2}{(2n+1)(2n)!}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1503530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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} |
How to find $abc$ if one is given the values of $a + b + c,$ $a^2 +b^2+c^2, \ a^3+b^3+c^3$ and $ac+bc+ab,$ If I am given the values of $a + b + c,$ $a^2 +b^2+c^2, \ a^3+b^3+c^3$ and $ac+bc+ab,$ how do I find value of $abc \ ?$
I expanded $(a+b+c)^3$ to get $a^3+b^3+c^3 + 3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b) +6abc $ but how do I manipulate $3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b)$ to obtain a fixed value from it?
Appreciate all advice, thank you.
| Hint:
$$
a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1506183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Number of squarefree positive integers less than $100$
An integer is called squarefree if it is not divisible by the square
of a positive integer greater than $1$. Find the number of squarefree
positive integers less than $100$.
My attempt:
I apply the inclusion-exclusion principle directly.
*
*Total number of integers = 99
*Number of integers divisible by $2^2$ = $24$
*Number of integers divisible by $3^2$ = $11$
*Number of integers divisible by $4^2$ = $6$
*Number of integers divisible by $5^2$ = $3$
*Number of integers divisible by $6^2$ = $2$
*Number of integers divisible by $7^2$ = $2$
*Number of integers divisible by $8^2$ = $1$
*Number of integers divisible by $9^2$ = $1$
*Number of integers divisible by $2^2$ and $3^2$ = $2$
*Number of integers divisible by $2^2$ and $4^2$ = $1$
Then the required solution would be $99-(24+11+6+3+2+2+1+1)+2+1=52$. But the solution is $61$. Where is my mistake?
| An integer is square-free if and only if none of its prime factors appear with an exponent $\ge 2$ in its prime factorization.
The only primes that can appear with an exponent of 2 or more in a number less than $100$ are those that are less than $\sqrt{100}=10$, that is, $2$, $3$, $5$, and $7$.
So the numbers you need to exclude are just those that are multiples of $4$, $9$, $25$ or $49$.
Neither of the multiples of $25$ and $49$ are multiples of any of the other squares, so they can be subtracted separately. All we need to care about is the double-counting of multiples of $36$, so:
*
*Start with $99$ possible numbers.
*Subtract $24$ multiples of $4$.
*Subtract $11$ multiples of $9$.
*Add back $2$ because $36$ and $72$ were each subtracted twice.
*Subtract $3$ multiples of $25$.
*Subtract $2$ multiples of $49$.
$$ 99 - 24 - 11 + 2 - 3 - 2 = 61 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1507658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to have idea to prove trigonometric identities Hi to explain this better I'll take an example.
I have this identity that's giving me a hard time.
$$\frac{\cos^2(a)-\sin^2(b)}{\sin^2(a)\sin^2(b)} = \cot^2(a)\cot^2(b)-1$$
This is what i would do
$$\cos^2(a)/(\sin^2(a)·\sin^2(b))-\sin^2(b)/(\sin^2(a)·\sin^2(b)) \\
\cot^2(a)·1/\sin^2(b)-1/\sin^2(a)$$
then, we know that
$$1=\cos^2(b)+\sin^2(b) \\
\vdots \\
\cot^2(a)·\cot^2(b)+\cot^2(a)-1/\sin^2(a)$$
Which of course is wrong but just wanted to show you guys how my mind thinks.
Is there any right way of solving this or do I just have to keep trying.
THANKS.
| First use $\cot x=\frac{\cos x}{\sin x}$ and work with the RHS .
$$\cot^2 a \cdot \cot^2 b-1=\frac{\cos^2 a \cdot \cos^2 b}{\sin^2 a \cdot \sin^2 b}-1=\frac{\cos^2a \cdot \cos^2 b-\sin^2 a \cdot \sin^2 b}{\sin^2 a \cdot \sin^2 b}$$
All that remains to prove is that
$$\cos^2a \cdot \cos^2 b-\sin^2 a \cdot \sin^2 b=\cos^2 a-\sin^2 b$$ but this follows using the classical $\cos^2x+\sin^2x=1$ as follows :
$$\cos^2a \cdot \cos^2 b-\sin^2 a \cdot \sin^2 b=\cos^2 a \cdot (1-\sin^2 b)-(1-\cos^2a) \cdot \sin^2 b=\cos^2 a-\sin^2 b$$ as wanted .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1509681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A Limit of Complex Trig Please help with a messy trig. I only want the process of evaluating the limits! Appreciate!
$$\lim_{x \to 0} \frac{(x-\sin x)\sin x}{(1-\cos x)^2}$$
or the alternative form
$$\lim_{x \to 0} \frac{(x-\sin x)(1+\cos x)}{(1-\cos x)(\sin x)}$$
Wolframalpha gave $2/3$
The original question(with image) is asking the area ratio. $\frac{ABD}{ADBC}$
colored segment ABD over the triangle ABC minus segment ABD
| If you have seen what they are, and know how to use them, one easy and systematic way is to use Taylor expansions. Recalling that when $x\to0$, one has
$$
\sin x = x - \frac{x^3}{6} + o(x^3)
$$
and
$$
\cos x = 1 - \frac{x^2}{2} + o(x^2)
$$
you obtain
$$
\frac{(x-\sin x)\sin x}{(1-\cos x)^2} = \frac{\left(\frac{x^3}{6}+o(x^3)\right)(x+o(x))}{\left(\frac{x^2}{2}+o(x^2)\right)^2}
\operatorname*{\sim}_{x\to 0} \frac{\frac{x^3}{6}\cdot x }{\frac{x^4}{4}} = \frac{2}{3}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1510265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating the indefinite trigonometric integral $\int \frac{dx}{(3+2\sin x)^2}$
$$\int \frac{dx}{(3+2 \sin x)^2}$$
ATTEMPT:-
Re Writing the integral as:
$I=\int \frac{2\cos x \sec x \,dx}{2(3+2 \sin x)^2}$ and using by parts:-
$\int u\,dv= uv-\int v\,du$
Here $u=\sec x \implies du=\sec x \tan x \,dx$
$\quad$ $dv=\frac{2\cos x \,dx}{2(3+2sinx)^2} \implies v=\frac{-1}{2(3+2\sin x)}$
$I=\frac{-\sec x\,dx}{2(3+2\sin x)} +\int \frac{\sec x \tan x \,dx}{2(3+2\sin x)}$
Let $I'=\int \frac{\sec x \tan x \,dx}{2(3+2\sin x)}$
$\implies I'=\int \frac{\sin x \,dx}{2\cos^2x(3+2\sin x)}$
$\implies I'=\int \frac{\sin \,dx}{2(1-\sin^2x)(3+2\sin x)}$
$\implies I'=\int \frac{-dx}{4(1+\sin x)} +\frac{3dx}{5(2\sin x+3)} + \frac{-dx}{20(\sin x-1)}$ which can be easily done by weierstrass's substitution.
But I am not able to modify my answer as given in the text.
Text Ans:-$\frac{2\cos x\,dx}{5(3+2\sin x)^2} + \frac{2}{5\sqrt{5}} \arctan\frac{(3\tan(\frac{x}{2})+2)}{\sqrt{5}} +c.$
My Ans:-$\frac{-\sec x\,dx}{2(3+2\sin x)} + \frac{6}{5\sqrt{5}} \arctan\frac{(3\tan(\frac{x}{2})+2)}{\sqrt{5}}-10\tan x+15\sec x-15.$
| Let $\displaystyle I = \int\frac{1}{(3+2\sin x)^2}dx\;,$ Now Put $\displaystyle \frac{2+3\sin x}{(3+2\sin x)} = t\;,$
Then $$\displaystyle \frac{(3+2\sin x)\cdot 3\cos x-(2+3\sin x)\cdot 2\cos x}{(3+2\sin x)^2}dx = dt$$
so we get $$\frac{5\cos x}{(3+2\sin x)^2}dx = dt\Rightarrow \frac{1}{(3+2\sin x)^2}dx = \frac{1}{5\cos x}dt.$$
So We get $$I = \frac{1}{5}\int \frac{1}{\cos x}dt$$
Now Above we have $$\frac{2+3\sin x}{3+2\sin x}=t\Rightarrow \sin x = \frac{2-3t}{2t-3}.$$
So we get $$\cos x= \sqrt{1-\sin^2 x} = \frac{\sqrt{5}\cdot \sqrt{1-t^2}}{2t-3}.$$
So we get $$I = \frac{1}{5\sqrt{5}}\int\frac{(2t-3)}{\sqrt{1-t^2}}dt = \frac{2}{5\sqrt{5}}\int\frac{t}{\sqrt{1-t^2}}dt-\frac{3}{5\sqrt{5}}\int\frac{1}{\sqrt{1-t^2}}dt$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Why is this sum what it is? I am trying to prove that
$$\frac{1}{1} + \frac{1}{1+2} + \frac{1}{1+2+3} + \cdots + \frac{1}{1+2+3+\cdots+n} = \frac{2n}{n+1}$$
I know the denominators have a closed form $\sum_{i=1}^{k} i = \frac{k(k+1)}{2}$ implying
$$\frac{1}{1} + \frac{1}{1+2} + \frac{1}{1+2+3} + \cdots + \frac{1}{1+2+3+\cdots+n} = 2\sum_{k=1}^{n}\frac{1}{k(k+1)}$$
Can't figure out where to go from here.
| You can also just do a straight proof using induction and your formula $\sum_{i=1}^{k} i = \frac{k(k+1)}{2}$
Just add $\frac{2}{(n+1)(n+2)} + \frac{2n}{n+1}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Is the argument used by the Numberphile video to show that $1 + 2 + 3 + 4 + \dots = -1/12$ valid? At the end of the Wikipedia article on:
$$1+2+3+4 +\dots$$
an argument is present that the sum adds up to $-\frac{1}{12}$.
Here is the numberphile video.
Here's my attempt to fill in the details from the argument:
(1) Let $S_1 = 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + \dots$
(2) Let $S_2 = 1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + \dots$
(3) $2S_2 = 1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + \dots$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\, 1 - 2 + 3 - 4 + 5 - 6 + 7 + \dots$
(4) $\,\,\,\,\,\,\,\,\,\, = 1 -1 + 1 -1 + 1 - 1 + 1 - 1 + \dots$
(5) $S_1$ is the Grandi's series whose Cesaro's sum is $1/2$
(6) So, $S_2 = 1/4$
(7) Let $S = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + \dots$
(8) $S - S_2 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + \dots$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-1 + 2 - 3 + 4 - 5 + 6 -7 + 8 + \dots$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,= 0 + 4 + 0 + 8 + 0 + 12 + 0 + 16 + \dots$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,= 4 + 8 + 12 + 16 + 20 + 24+ 28 + \dots$
(9) $4S = 4 + 8 + 12 + 16 + 20 + 24 + 28 + \dots$
(10) $S - S_2 = S - 1/4 = 4S$
(11) Solving for $S - 1/4 = 4S$ gives $S = -\frac{1}{12}$
Is this argument valid? Is there a flaw in the logic that leads to the right answer in the wrong way?
| We need to be very careful with blind manipulations like the ones in the video - in particular, you can get situations like:
$$S = 1 + 2 + 3 + 4 + \dots$$
$$-S = S - 2S = (1 + 2 + 3 + 4 + \dots) - (2 + 4 + 6 + 8 + \dots)$$
$$= 1 + (2 - 2) + (3 - 4) + (4 - 6) + (5 - 8) + \dots = 1 + 0 - 1 - 2 - 3 - 4 +\dots = 1 - S$$
So $-S = 1- S$ and $0 = 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $\int_{0}^{\frac{\pi}{4}}\tan^6(x) \sec(x) dx = I$ then express $\int_{0}^{\frac{\pi}{4}} \tan^8(x) \sec(x) dx$ in terms of $I$ how can I proceed with this exercise?
If
$$\int_{0}^{\frac{\pi}{4}} \tan^6(x) \sec(x) dx = I$$
then express
$$\int_{0}^{\frac{\pi}{4}} \tan^8(x) \sec(x) dx$$
in terms of $I$.
What I've got so far:
$$\int_{0}^{\frac{\pi}{4}} \tan^8(x) \sec(x) dx = \int_{0}^{\frac{\pi}{4}} \tan^2(x) \tan^6(x) \sec(x) dx = \int_{0}^{\frac{\pi}{4}} \left( \sec^2(x) - 1 \right) \tan^6(x) \sec(x) dx = \\ = - \int_{0}^{\frac{\pi}{4}} \tan^6(x) \sec(x) dx + \int_{0}^{\frac{\pi}{4}} \tan^6(x) \sec^3(x) dx = -I + \cdots$$
Any help is highly appreciated.
$\\$
(Exercise 50 from Stewart's Calculus book section chapter 7.2 7th edition)
| Let's be a little nit more general and look at the Integrals
$$
I_{m}=\int \tan^m(x)\sec(x)dx\\
I_{m-2}=\int \tan^{m-2}(x)\sec(x)dx
$$
now substitute $x=\arctan(y)$ . We get
$$
I_m=\int\frac{y^m}{\sqrt{1+y^2}}dy=\int y^{m-1}\frac{y}{\sqrt{1+y^2}}dy
$$
Integrating by parts:
$$
I_m=y^{m-1}\sqrt{1+y^2}-(m-1)\int y^{m-2}\sqrt{1+y^2}dy=\\
y^{m-1}\sqrt{1+y^2}-(m-1)\int y^{m-2}\frac{1+y^2}{\sqrt{1+y^2}}dy=\\
y^{m-1}\sqrt{1+y^2}-(m-1)(I_m+I_{m-2})
$$
Therefore
$$
I_m=\frac{1}{m}y^{m-1}\sqrt{1+y^2}-\frac{m-1}{m}I_{m-2}
$$
putting $m=8$, and plugging in the appropriate endpoints of integration $y=0$ and $y=1$ u can find your question as a special case of the above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1515905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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} |
Residues of poles Find $Res_{f}\left ( z_{0} \right )$, where, $f\left ( z \right )=\frac{1}{z^{4}+4}$, for $z_{0}=1+i$
The definition for
$$Res_{f}\left ( 1+i \right ) =\lim_{z \to z_{0}} \left\{\left ( z-\left ( 1+i \right ) \right ) \cdot \frac{1}{z^{4}+4} \right\}$$
and the roots for
$$z^{4}+4$$ are $\sqrt{+2i}$, $\sqrt{-2i}$, $- \sqrt{+2i}$, $- \sqrt{-2i}$
I'm a bit stuck here. Could someone give me a push?
| Using $1 + i = \sqrt{2} \, e^{\tan^{-1}(1)} = \sqrt{2} \, e^{\pi i/4}$ and $\sqrt{i} = e^{\pi \, i/4}$ then
\begin{align}
\lim_{z = 1+i} f &= \lim_{1+i} \left\{ \frac{z - \sqrt{2} \, e^{\pi i/4}}{z^{4}+4} \right\} \\
&= \lim \left\{ \frac{z - \sqrt{2} \, e^{\pi i/4}}{(z^{2} + 2i)(z^{2}- 2 i) } \right\} \\
&= \lim \left\{ \frac{z - \sqrt{2} \, e^{\pi i/4}}{(z^{2} + 2 i) (z - \sqrt{2} e^{\pi i/4})( z + \sqrt{2} e^{\pi i/4})} \right\} \\
&= \frac{1}{(2 \, e^{\pi i/2} + 2 i)(2 \, \sqrt{2} \, e^{\pi i/4}) } \\
&= \frac{1}{8 \sqrt{2}} \, \frac{1}{ e^{3 \pi i/4}} \\
&= - \frac{1 + i}{16}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1517029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Using L'Hopital to solve $\lim_{x\to +\infty}\frac{\frac{-1}{x^2}}{\sin^2\left(\frac{2}{x}\right)}$
Use L'Hopital to calculate
$$\lim_{x\to
+\infty}\frac{\frac{-1}{x^2}}{\sin^2\left(\frac{2}{x}\right)}$$
Right now this yields $\frac{0}{0}$ so let's go ahead and use L'Hopital:
$$\lim_{x\to
+\infty}\frac{\frac{2}{x^3}}{2\cdot \sin\left(\frac{2}{x}\right)\cdot\cos \left(\frac{2}{x}\right)\cdot\left(\frac{-2}{x^2}\right)}$$
This just won't do. Perhaps we should flip the numerator with the denominator instead:
$$-\frac{\csc^2\left(\frac{2}{x}\right)}{x^{2}}$$
This yields -$\frac{\infty}{\infty}$, so we can go ahead and apply L'Hopital:
$$-\frac{2\cdot-\csc\left(\frac{2}{x}\right)\cdot\cot\left(\frac{2}{x}\right)\cdot\frac{-2}{x^2}}{2x}$$
If I evaluate this I will get $\frac{0}{\infty}$
I have the feeling I'm not supposed to keep going this path, and there's a simpler solution (using L'Hopital). What can I do to solve this?
| Use the substitution $u=\frac{2}{x}$
Then the limit becomes
\begin{align}
\lim_{u\to 0} \frac{-\frac{1}{4}u^2}{\sin^2{u}}&=\lim_{u\to 0}\frac{-\frac{1}{2}u}{2\sin u\cos u}\\ \\
&=-\frac{1}{4}\lim_{u\to 0}\frac{u}{\sin u\cos u}\\ \\
&=-\frac{1}{4}\lim_{u\to 0}\frac{1}{\cos^2 u-\sin^2 u}\\ \\
&=-\frac{1}{4}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1521841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Find the sum of the series $\frac{1}{3}+\frac{1}{4}.\frac{1}{2!}+\frac{1}{5}.\frac{1}{3!}+\cdots$
Find the sum of the series $$\frac{1}{3}+\frac{1}{4}.\frac{1}{2!}+\frac{1}{5}.\frac{1}{3!}+\cdots$$
I found that the n-th term of the series is $\frac{1}{n+2}.\frac{1}{n!}$ , but I am unable to find the sum.
| HINT:
$$\frac{1}{(n+2)n!}=\frac{1}{(n+1)!}-\frac{1}{(n+2)!}$$
Now telescope.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Use induction to prove that $\sum_{k=1}^{2^{n}}\frac{1}{k} \geq 1 +\frac{n}{2}$? For the base case I have $n = 1$, so
${1\over1} + {1\over2} = {3\over2}, 1 + {1\over2} = {3\over2}$. Hence this is true. For the induction hypothesis we can assume n=k is true also. For the induction step I plug in $n=k+1$ but i'm not sure what to do with it.
| If $n=1$, then
$$
\sum_{k=1}^{2^{n}}\frac{1}{k} = 1 + \frac{1}{2} = 1 + \frac{n}{2};
$$
if $n \geq 1$ is an integer such that
$$
\sum_{k=1}^{2^{n}}\frac{1}{k} \geq 1 + \frac{n}{2},
$$
then
\begin{align*}
\sum_{k=1}^{2^{n+1}}\frac{1}{k}
&= \sum_{k=1}^{2^{n}}\frac{1}{k} + \sum_{k=2^{n}+1}^{2^{n+1}}\frac{1}{k}\\
&\geq 1 + \frac{n}{2} + \sum_{k=2^{n}+1}^{2^{n+1}}\frac{1}{k}\\
&= 1 + \frac{n}{2} + \frac{1}{2^{n}+1} + \cdots + \frac{1}{2^{n+1}}\\
&\geq
1 + \frac{n}{2} + \frac{(2^{n+1} - 2^{n})}{2^{n}+1}\\
&= 1 + \frac{n}{2} + \frac{2^{n}}{2^{n}+1}\\
&\geq 1 + \frac{n}{2} + \frac{2^{n}}{2\cdot 2^{n}}\\
&= 1 + \frac{n}{2} + \frac{1}{2}\\
&= 1 + \frac{n+1}{2}.
\end{align*}
| {
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"url": "https://math.stackexchange.com/questions/1522059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $ax+by=1 $ implies $\frac{1}{x^2+y^2} \leq a^2+b^2$
Show that for real positive x,y,a and b: $ax+by=1 \implies \frac{1}{x^2+y^2} \leq a^2+b^2$
I used to use contradiction but no results ;
We show that
$$a^2+b^2 < \frac{1}{x^2+y^2} \implies ax+by \ne1$$
$$a^2+b^2 < \frac{1}{x^2+y^2} \iff (a^2+b^2)(x^2+y^2)<0 \iff (ax-by)^2-(ay-bx)^2 <0$$
I'm stuck here, I thought about proving it by putting something in a denominator which is not zero so it's not $1$.
| Hint:
Use Cauchy-Schwarz inequality.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Primitive Pythagorean triple divisible by 3 Prove that for any primitive Pythagorean triple (a, b, c), exactly one of a and b must be a multiple
of 3, and c cannot be a multiple of 3.
My attempt:
Let a and b be relatively prime positive integers.
If $a\equiv \pm1 \pmod{3}$ and $b\equiv \pm1 \pmod{3}$,
$c^2=a^2+b^2\equiv 1+1\equiv 2 \pmod{3}$
This is impossible as the only quadratic residues modulo 3 are 0 and 1.
So far, so good.
If one of a, b is $\equiv 0 \pmod{3}$ and the other is $\equiv \pm1 \pmod{3}$,
$c^2=a^2+b^2\equiv 0+1\equiv 1 \pmod{3}$
This is the part I don't understand. Just because $c^2\equiv 1\pmod{3}$ doesn't mean that $c^2$ must be a perfect square. For example, $a=12$ and $b=13$ satisfy the above conditions but $c^2=a^2+b^2=313$, which isn't a perfect square.
| Your first step was a good start: It proves that at least one of $a$ and $b$ must be a multiple of $3$. We will have to eliminate the case of both $a$ and $b$ being multiples of $3$, but that is easy: If $a=3p$ and $b=3q$ then
$$c^2 = a^2+b^2 = (3p)^2 + (3q)^2 = 9(p^2+q^2) = 9m$$ so $c$ is also a multiple of $3$, in which case the triplet is not primitive, as $3$ divides all three of them.
So exactly one of $a$ and $b$ must be a multiple of $3$.
Your second step has shown that since exactly one of $a$ and $b$ is a multiple of $3$, $c \equiv 1 \pmod{3}$. You don't care that not every number of the form $3k+1$ is a square; all you care about is that the number that happens to be $c^2$ must be expressible as $3k+1$. And this completes the proof that $c$ is not a multiple of $3$.
| {
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"url": "https://math.stackexchange.com/questions/1523230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A Limit Question of 0/0 Uncertainty This questions of mine, I couldn't solve whatever I did. Please do not use L'hospital as it has not been taught to us and I don't think that kind of answer will be accepted in the exam.
$$\lim\limits_{x\to 0} \frac{3x + \sin^2x}{\sin2x - x^3}$$
| L'Hôpitals rule is not necessary here. Just divide the numerator and the denominator by $2x$ and exploit the fact that $$\lim\limits_{x\to 0}\frac{\sin x}{x}=1$$
Here are the steps
$$\lim\limits_{x\to 0} \frac{3x + \sin^2x}{\sin2x - x^3}$$
$$=\lim\limits_{x\to 0} \frac{\frac{3}{2} + \frac{\sin^2x}{2x}}{\frac{\sin2x}{2x} - \frac{x^2}{2}}$$
$$=\frac{\frac{3}{2} + \frac12\lim\limits_{x\to 0} \frac{\sin^2x}{x}}{\lim\limits_{x\to 0} \frac{\sin2x}{2x} -\lim\limits_{x\to 0} \frac{x^2}{2}}$$
$$=\frac{\frac{3}{2} + \frac12\left(\lim\limits_{x\to 0} \frac{\sin x}{x}\right)\left(\lim\limits_{x\to 0} \sin x\right)}{\lim\limits_{x\to 0} \frac{\sin2x}{2x} -\lim\limits_{x\to 0} \frac{x^2}{2}}$$
$$=\frac{\frac{3}{2} + \frac12\cdot 1\cdot 0}{1-0}$$
$$=\frac32$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1525647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Proof for convergence of this series I want to determine if $\displaystyle\sum_{n=1}^\infty\frac{(n^2+n+1)}{(n^4+2n+2)}$ converges.
The ratio test in this case is inconclusive so i tried to do something like this.
$\displaystyle\sum_{n=1}^\infty\frac{(n^2+n+1)}{(n^4+2n+2)}\leq\sum_{n=1}^\infty\frac{(n^2+n+1)}{(n^4)}$
$\displaystyle\int_{1}^{\infty}\sum_{n=1}^\infty\frac{(n^2+n+1)}{(n^4)}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}$.
Since the second series converges so does the first one.
Is this valid, or how can I else prove if the first series converges?
| Yes, you have
$$0\leq \frac{(n^2+n+1)}{(n^4+2n+2)}\leq \frac{(n^2+n+1)}{(n^4)} = \frac{1}{n^2} + \frac{1}{n^3} + \frac{1}{n^4}.$$
And
$$
\sum\frac{1}{n^2} + \frac{1}{n^3} + \frac{1}{n^4}
$$
is a sum of convergent $p$-series, so it is convergent. Now the original series is convergent by the comparison test.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do I find $M>0$ for $\lim _{ x\rightarrow \infty }{ \frac { x^{ 2 }+2x-3 }{ x^2-1 }=1 } $ Given the following limit, find such an $M>0$ that for every $x>M$, the expression is $\frac { 1 }{ 3 } $ close to the limit. In other words find $M>0$ such that for every $x>M:\left| f(x)-L \right| <\frac { 1 }{ 3 } $ for the following function:
$$\lim _{ x\rightarrow \infty }{ \frac { x^{ 2 }+2x-3 }{ x^2-1 }=1 } $$
So, it seems to me that this question is asking for me to prove that:
If $|f(x)-L|<\varepsilon$, then $x>M$
Steps I took:
$$\left| \frac { x^{ 2 }+2x-3 }{ x^{ 2 }-1 } -1 \right| <\frac { 1 }{ 3 } $$
$$\Longrightarrow \left| \frac { (x+3)(x-1) }{ (x-1)(x+1) } -1 \right| <\frac { 1 }{ 3 } $$
$$\Longrightarrow \left| \frac { (x+3) }{ (x+1) } -\frac { (x+1) }{ (x+1) } \right| <\frac { 1 }{ 3 } $$
$$\Longrightarrow \left| \frac { 2 }{ x+1 } \right| <\frac { 1 }{ 3 } $$
$$\Longrightarrow \frac { 2 }{ \left| x+1 \right| } <\frac { 1 }{ 3 } $$
Without loss of generality, assume that $x>-1$
Then, we get: $$\frac { 2 }{ x+1 } <\frac { 1 }{ 3 } $$
$$\Longrightarrow 2<\frac { 1 }{ 3 } (x+1)$$
$$\Longrightarrow 2<\frac { 1 }{ 3 } x+\frac { 1 }{ 3 } $$
$$\Longrightarrow \frac { 1 }{ 3 } x>2-\frac { 1 }{ 3 } $$
$$\Longrightarrow x>5$$
I don't really know what to do once I hit the "Without loss of generality, assume that..." line. I would like to know if I had the right idea from the beginning of what to prove and where I went wrong in my proof (that is, if it is wrong).
| In response to @Cherry_Developer's comment, take $M=5$. Then, for $x>M$, we have:
\begin{align}
\left|f(x)-L\right|=\left| \frac { x^{ 2 }+2x-3 }{ x^{ 2 }-1 } -1 \right| = \left| \frac { 2 }{ x+1 } \right| < \frac{2}{5+1} = \frac{1}{3}.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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An inequality using convex functions: If $\frac{a^2}{1+a^2}+\frac{b^2}{1+b^2}+\frac{c^2}{1+c^2} = 1$ then $abc \le \frac1{2\sqrt 2}$ I see this in a Chinese convex analysis book.
Suppose that $a$,$b$,$c$ are positive real numbers satisfying
\begin{equation}
\frac{a^2}{1+a^2}+\frac{b^2}{1+b^2}+\frac{c^2}{1+c^2} = 1.
\end{equation}
Show that $abc \le \dfrac1{2\sqrt 2}$.
Since it's from a convex analysis book, I tried proving this using Jensen's inequality. However, I can't think of a suitable convex function. Therefore, I tried AM–GM, but I can't get a product $abc$.
$$(abc)^2\le\frac18\iff8(abc)^2\le1$$
$$\iff\left(\frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}\right)^3 (abc)^2 \le 1$$
Finally, I used Lagrange multiplier to solve the problem, but I think there is some more elementary solution.
$$f(a,b,c)=abc$$
$$g(a,b,c)=\frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}-2=0$$
$$\nabla f(a,b,c)=(bc,ca,ab)$$
$$\nabla g(a,b,c)=\left(-\frac{2a}{(1+a^2)^2},-\frac{2b}{(1+b^2)^2},-\frac{2c}{(1+c^2)^2}\right)$$
$$\because \nabla f = \lambda \nabla g$$
$$\therefore bc = -\frac{2a\lambda}{(1+a^2)^2} \iff abc = -\frac{2a^2\lambda}{(1+a^2)^2}$$
$$abc = -\frac{2a^2\lambda}{(1+a^2)^2} = -\frac{2b^2\lambda}{(1+b^2)^2} = -\frac{2c^2\lambda}{(1+c^2)^2}$$
$$\frac{a}{1+a^2}=\frac{b}{1+b^2}=\frac{c}{1+c^2}=\mu$$
$$a+\frac1a=b+\frac1b=c+\frac1c$$
$$\because \frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}=2$$
$$\therefore \frac{\mu}{a}+\frac{\mu}{b}+\frac{\mu}{c}=2$$
$$\because \frac{a^2}{1+a^2}+\frac{b^2}{1+b^2}+\frac{c^2}{1+c^2} = 1$$
$$\therefore a\mu + b\mu + c\mu = 1$$
$$\frac1a+\frac1b+\frac1c=2(a+b+c)$$
$$3(a+b+c)=a+\frac1a+b+\frac1b+c+\frac1c=3\left(a+\frac1a\right)$$
$$b+c=\frac1a$$
Similarly, $c+a=\dfrac1b$ and $a+b=\dfrac1c$. Substitute $a=\dfrac1{b+c}$ into the other two equations.
$$c+\frac1{b+c}=\frac1b$$
$$\frac1{b+c}+b=\frac1c$$
$$b(b+c)c+b=b+c$$
$$c+b(b+c)c=b+c$$
Subtracting one equation from another, we get $b=c$. Similarly, we have $a=b=c$. It remains to substitute it back to the original constraint and calculate the product.
Any alternative solution is appreciated.
| By C-S $$1=\sum_{cyc}\frac{a^2}{1+a^2}\geq\frac{(a+b+c)^2}{3+a^2+b^2+c^2},$$
which by AM-GM gives
$$3\geq2(ab+ac+bc)\geq6\sqrt[3]{a^2b^2c^2}$$
and we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Evaluating $\lim_{x\to0}\left(\frac{1+x2^x}{1+x3^x}\right)^\frac1{x^2}$ Find the following limit
$$
\lim_{x\to0}\left(\frac{1+x2^x}{1+x3^x}\right)^\frac1{x^2}
$$
I have used natural logarithm to get
$$
\exp\lim_{x\to0}\frac1{x^2}\ln\left(\frac{1+x2^x}{1+x3^x}\right)
$$
After this, I have tried l'opital's rule but I was unable to get it to a simplified form.
How should I proceed from here? Any here is much appreciated!
| Note that $\lim\limits_{x\to0} xn^x=0$.
Next, use the L'Hospital's rule twice to get
$$\lim_{x\to0}\frac{\ln \left({1+xn^x}\right)}{x^2}=
\lim_{x\to0}\frac{\frac{x n^x \ln ^2n+2 n^x \ln n}{x n^x+1}-\frac{\left(n^x+x n^x \ln n\right)^2}{\left(x n^x+1\right)^2}}{2}
=\frac{2 \ln n-1}{2}
$$
Now
$$\lim_{x\to0}\frac{\ln\left(\frac{1+x2^x}{1+x3^x}\right)}{x^2}=
\lim_{x\to0}\frac{\ln\left({1+x2^x}\right)-\ln\left({1+x3^x}\right)}{x^2}
=\frac{2 \ln 2-1}{2} - \frac{2 \ln 3-1}{2}
= \ln \frac{2}{3}$$
Finally, take exponent to get
$$\lim_{x\to0}\left(\frac{1+x2^x}{1+x3^x}\right)^\frac1{x^2}=\frac{2}{3}$$
Note
$$\frac{d}{dx} n^x = n^x \ln n$$
$$\frac{d}{dx}\ln \left({1+xn^x}\right)=\frac{n^x+x n^x\ln n }{1+xn^x}$$
$$\frac{d^2}{dx}\ln \left({1+xn^x}\right)=\frac{d^2}{dx}\frac{n^x+x n^x\ln n }{1+xn^x}=
\frac{(2n^x\ln n +x n^x(\ln n)^2)(1+xn^x)-(n^x+x n^x\ln n)^2 }{(1+xn^x)^2}=
\frac{x n^x \ln ^2n+2 n^x \ln n}{x n^x+1}-\frac{\left(n^x+x n^x \ln n\right)^2}{\left(x n^x+1\right)^2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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integrate sin(x). OK. I have a doubt with this: I know $-\cos(x) + k =\int \sin x\,dx$ but doing
$\sin(x)=2\sin\frac{x}{2}\cos\frac{x}{2}$ I get $\int \sin x\,dx = \int 2\sin\frac{x}{2}\cos\frac{x}{2}\,dx $ if $ u = \sin \frac{x}{2}$ then $du = \cos \frac{x}{2} \frac{dx}{2}$ then $$\int \sin x\,dx = \int 2\sin \frac{x}{2}\cos \frac{x}{2}\, dx =4\int \sin \frac{x}{2} \cos\frac{x}{2}\frac{dx}{2}=4\int u\,du= 4\frac{u^2}{2}= 2u^2 = 2\sin^2\frac{x}{2}$$ then if $\theta = \frac{x}{2} \rightarrow \cos 2\theta =-2\sin^2(\theta)+k \rightarrow \frac{\cos(2\theta)}{2}=\cos^2 \theta -1+\frac{k}{2} $ if $k=0$, then $$\frac{\cos 2\theta}{2}=\cos^2 \theta-1.$$ Now, why isn't it a trigonometric identity? or Is it? Because I think that I found one.
| You made a mistake. If $-2\sin^2(x) + k = \cos(2x)$, then plug in $0$ to get $0 + k = 1$.
Also, with $k=1$, this identity is already known. You didn't discover something new.
See this Wikipedia article. Here you see that
$$
\cos(2x) = 2\cos^2(x) - 1.
$$
| {
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Find the joint PMF of X and Y, are they independent? A fair die is rolled, and then a coin with probability $p$ of Heads is flipped as many times as the die roll says, e.g., if the result of the die roll is a 3, then the coin is flipped 3 times. Let $X$ be the result of the die roll and $Y$ be the number of times the coin lands Heads.
Find the joint PMF of $X$ and $Y$. Are they independent?
My issue is with finding the joint PMF. I started by finding the supports of both variables. The support of X is {1,2,3,4,5,6} and the support of Y is {0,1,2,3,4,5,6}. Since the value of the roll dictates how many heads can be flipped, the joint support is of size 27. I am trying to construct a table of all the different probabilities but I can't seem to figure out what the specific probabilities are. Once I find the joint PMF and marginals I can figure out if the variables are independent or not.
Here is my work so far...
$$\begin{array}{c|c|c|c|c|c|c|}
& \text{1} & \text{2} & \text{3} & \text{4} & \text{5} & \text{6} & f_Y(y) \\ \hline
\text{0} & p & p & p & p & p & p & 6p \\ \hline
\text{1} & p & p & p & p & p & p & 6p \\ \hline
\text{2} & 0 & p & p & p & p & p & 5p \\ \hline
\text{3} & 0 & 0 & p & p & p & p & 4p \\ \hline
\text{4} & 0 & 0 & 0 & p & p & p & 3p \\ \hline
\text{5} & 0 & 0 & 0 & 0 & p & p & 2p \\ \hline
\text{6} & 0 & 0 & 0 & 0 & 0 & p & 1p \\ \hline
f_X(x) & 0 & 0 & 0 & 0 & 0 & p & 1p \\ \hline
\end{array}$$
| We use Baym's rule $$P(X=x, Y=y) = P(Y=y| X=x) \cdot P(X=x)$$
Since $P(X=x) = \frac{1}{6}$ and $Y\sim Bin(x, p)$ we obtain
$$P(X=x, Y=y) = \frac{1}{6}\; \binom{x}{y} p^y (1-p)^{x-y} $$
The two random variables $X$ and $Y$ are not (!) independent, because if $y=6$ we know that $x=6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1534426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.