text stringlengths 1 2.12k | source dict |
|---|---|
\begin{equation*} P = f (t) = 12 + 2t^3. \end{equation*}
1. Graph the function on the domain $[0, 5]\text{.}$
2. Find a formula for the inverse function, $t = f^{-1}(P)\text{.}$ What is the meaning of the inverse function in this context?
3. Sketch a graph of the inverse function.
Solution
1. The graph of $f$ is shown in Figure194, with $t$ on the horizontal axis and $P$ on the vertical axis.
2. We solve $P = 12 + 2t^3$ for $t$ in terms of $P\text{.}$ \begin{align*} 2t^3 \amp = P - 12\amp\amp\text{Substract 12 from both sides.}\\ t^3 \amp = \frac{P - 12}{2}\amp\amp\text{Divide both sides by 2.}\\ t \amp = \sqrt[3]{\frac{P - 12}{2}}\amp\amp\text{Take cube roots.} \end{align*} The inverse function is $t = f^{-1}(P) =\sqrt[3]{\dfrac{P - 12}{2}}\text{.}$ It tells us the number of years it takes for the pheasant population to grow to size $P\text{.}$
3. The graph of $f^{-1}$ is shown in Figure195, with $P$ on the horizontal axis and $t$ on the vertical axis.
The formula $T = f(L) = 2\pi \sqrt{\dfrac{L}{32}}$ gives the period in seconds, $T\text{,}$ of a pendulum as a function of its length in feet, $L\text{.}$
1. Graph the function on the domain $[0, 5]\text{.}$
2. Find a formula for the inverse function, $L = f^{-1}(T )\text{.}$ What is the meaning of the inverse function in this context?
3. Sketch a graph of the inverse function.
### SubsectionWhen Is the Inverse a Function?
We can always find the inverse of a function $y=f(x)$ simply by solving for $x$ thus interchanging the role of the input and output variables. In the preceding examples, this process created a new function. However, this process does not always lead to be a function. | {
"domain": "unl.edu",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109520836027,
"lm_q1q2_score": 0.8520821559508629,
"lm_q2_score": 0.8652240860523328,
"openwebmath_perplexity": 217.23917974865927,
"openwebmath_score": 0.844607412815094,
"tags": null,
"url": "https://mathbooks.unl.edu/PreCalculus/Inverse-Functions.html"
} |
For example, to find the inverse of $y = f (x) = x^2\text{,}$ we solve for $x$ to get $x = \pm\sqrt{y}\text{.}$ When we regard $y$ as the input and $x$ as the output, the relationship does not describe a function. This can be seen as plugging in $y=4 \text{,}$ for example, gives two outputs $x=2$ and $x=-2\text{.}$ The graphs of $f$ and its inverse are shown in Figure197. (Note that for the graph of the inverse, we plot $y$ on the horizontal axis and $x$ on the vertical axis.) Because the graph of the inverse does not pass the vertical line test, it is not a function.
For many applications, it is important to know whether or not the inverse of $f$ is a function. This can be determined from the graph of $f\text{.}$ When we interchange the roles of the input and output variables, horizontal lines of the form $y = k$ become vertical lines.
Thus, if the graph of the inverse is going to pass the vertical line test, the graph of the original function must pass the horizontal line test, namely, that no horizontal line should intersect the graph in more than one point. Notice that the graph of $f(x) = x^2$ does not pass the horizontal line test, so we would not expect its inverse to be a function.
###### Horizontal Line Test
If no horizontal line intersects the graph of a function more than once, then its inverse is also a function.
We have been talking about how to tell if the inverse of a function is also a function, but in practice this is not the language typicaly used. Usually we ask this same question in the form "Is the function invertible?" The following definition explains this relationship:
###### Invertible Function
If $y=f(x)$ is a function such that its inverse, $x=f^{-1}(y)\text{,}$ is also a function then we say that $f(x)$ is an invertible function.
###### Example200
Which of the functions in Figure201 are invertible?
Solution | {
"domain": "unl.edu",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109520836027,
"lm_q1q2_score": 0.8520821559508629,
"lm_q2_score": 0.8652240860523328,
"openwebmath_perplexity": 217.23917974865927,
"openwebmath_score": 0.844607412815094,
"tags": null,
"url": "https://mathbooks.unl.edu/PreCalculus/Inverse-Functions.html"
} |
###### Example200
Which of the functions in Figure201 are invertible?
Solution
In each case, apply the horizontal line test to determine whether the function is invertible. Because no horizontal line intersects their graphs more than once, the functions pictured in Figures201(a) and (c) are invertible. The functions in Figures201(b) and (d) are not invertibe.
### SubsectionMathematical Properties of the Inverse Function
The inverse function $f^{-1}$ undoes the effect of the function $f\text{.}$ In Example152, the function $f(t) = 6 + 2t$ multiplies the input by $2$ and then adds $6$ to the result. The inverse function $f^{-1}(H) = \dfrac{H -6}{2}$ undoes those operations in reverse order: It subtracts $6$ from the input and then divides the result by $2\text{.}$
If we apply the function $f$ to a given input value and then apply the function $f^{-1}$ to the output from $f\text{,}$ the end result will be the original input value. For example, if we choose $t = 5$ as an input value, we find that \begin{align*} f(\alert{5})\amp= 6 + 2(\alert{5}) = \blert{16}\amp\amp\text{ Multiply by 2, then add 6.}\\ \text{and } f^{-1}(\blert{16}) \amp = \frac{\blert{16} - 6}{2} = \alert{5}.\amp\amp\text{Subtract 6, then divide by 2.} \end{align*}
We return to the original input value, $5\text{,}$ as illustrated in Figure202.
Example203 illustrates the fact that if $f^{-1}$ is the inverse function for $f\text{,}$ then $f$ is also the inverse function for $f^{-1}\text{.}$
###### Example203
Consider the function $f(x) = x^3 + 2$ and its inverse, $f^{-1}(y) = \sqrt[3]{y - 2}\text{.}$ | {
"domain": "unl.edu",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109520836027,
"lm_q1q2_score": 0.8520821559508629,
"lm_q2_score": 0.8652240860523328,
"openwebmath_perplexity": 217.23917974865927,
"openwebmath_score": 0.844607412815094,
"tags": null,
"url": "https://mathbooks.unl.edu/PreCalculus/Inverse-Functions.html"
} |
Consider the function $f(x) = x^3 + 2$ and its inverse, $f^{-1}(y) = \sqrt[3]{y - 2}\text{.}$
1. Show that the inverse function undoes the effect of $f$ on $x = 2\text{.}$
2. Show that $f$ undoes the effect of the inverse function on $y = -25\text{.}$
Solution
1. First evaluate the function $f$ for $x = 2\text{:}$
Then evaluate the inverse function $f^{-1}$ at $y = 10\text{:}$
\begin{equation*} f^{-1}(\blert{10}) = \sqrt[3]{\blert{10} - 2} = \sqrt[3]{8}= \alert{2}. \end{equation*}
We started and ended with $2\text{.}$
2. First evaluate the function $f^{-1}$ for $y = -25\text{:}$
\begin{equation*} f^{-1}(\alert{-25}) = \sqrt[3]{-25 - 2} = \blert{-3}. \end{equation*}
Then evaluate the function $f$ for $x = -3\text{:}$
\begin{equation*} f (\blert{-3}) = (\blert{-3})^3 + 2 = \alert{-25}. \end{equation*}
We started and ended with $-25\text{.}$
###### Example204
1. Find a formula for the inverse of the function $f(x)=\dfrac{2}{x - 1}\text{.}$
2. Show that $f^{-1}$ undoes the effect of $f$ on $x = 3\text{.}$
3. Show that $f$ undoes the effect of $f^{-1}$ on $y = -2\text{.}$
Solution
1. To find the inverse, we solve for $x\text{:}$
\begin{equation*} \begin{aligned} y \amp= \frac{2}{x-1} \\ y(x-1) \amp= 2 \\ x-1 \amp= \frac{2}{y} \\ x \amp =\frac{2}{y}+1 \end{aligned} \end{equation*}
Therefore $f^{-1}(y)=\frac{2}{y}+1\text{.}$
2. First evaluate the function $f$ for $x=3\text{:}$
Then evaluate the inverse function $f^{-1}$ at $y=1\text{:}$
We started and ended with $3\text{.}$
3. First evaluate the inverse function $f^{-1}$ for $y=-2\text{:}$
Then evaluate the original function $f$ at $x=0\text{:}$
We started and ended with $-2\text{.}$
### SubsectionSymmetry | {
"domain": "unl.edu",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109520836027,
"lm_q1q2_score": 0.8520821559508629,
"lm_q2_score": 0.8652240860523328,
"openwebmath_perplexity": 217.23917974865927,
"openwebmath_score": 0.844607412815094,
"tags": null,
"url": "https://mathbooks.unl.edu/PreCalculus/Inverse-Functions.html"
} |
We started and ended with $-2\text{.}$
### SubsectionSymmetry
So far we have been careful to keep track of the input and output variables when we work with inverse functions. This is important when we are dealing with applications; the names of the variables are usually chosen because they have a meaning in the context of the application, and it would be confusing to change them.
However, we can also study inverse functions purely as mathematical objects. There is a relationship between the graph of a function and the graph of its inverse that is easier to see if we plot them both on the same set of axes.
A graph does not change if we change the names of the variables, so we can let $x$ represent the input for both functions, and let $y$ represent the output. Consider the function $C = h(F)$ from Example161, and its inverse function, $F = h^{-1}(C)\text{.}$ The formulas for these functions are \begin{align*} C \amp = h(F) = \frac{5}{9}(F - 32)\\ F \amp = h^{-1}(C) = 32 + \frac{9}{5}C. \end{align*} But their graphs are the same if we write them as \begin{align*} y \amp = h(x) =\frac{5}{9}(x - 32)\\ y \amp= h^{-1}(x) = 32 + \frac{9}{5}x. \end{align*} The graphs are shown in Figure205.
Now, for every point $(a, b)$ on the graph of $f\text{,}$ the point $(b, a)$ is on the graph of the inverse function. Observe in Figure205 that the points $(a, b)$ and $(b, a)$ are always located symmetrically across the line $y=x\text{.}$ The graphs are symmetric about the line $y=x$, which means that if we were to place a mirror along the line $y=x\text{,}$ each graph would be the reflection of the other.
###### Example206
Graph the function $f (x) = 2\sqrt{x + 4}$ on the domain $[-4, 12]\text{.}$ Graph its inverse function $f^{-1}$ on the same grid.
Solution | {
"domain": "unl.edu",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109520836027,
"lm_q1q2_score": 0.8520821559508629,
"lm_q2_score": 0.8652240860523328,
"openwebmath_perplexity": 217.23917974865927,
"openwebmath_score": 0.844607412815094,
"tags": null,
"url": "https://mathbooks.unl.edu/PreCalculus/Inverse-Functions.html"
} |
Solution
The graph of $f$ has the same shape as the graph of $y = \sqrt{x}\text{,}$ shifted $4$ units to the left and stretched vertically by a factor of $2\text{.}$ Figure207a shows the graph of $f\text{,}$ along with a table of values. By interchanging the rows of the table, we obtain points on the graph of the inverse function, shown in Figure207b.
If we use $x$ as the input variable for both functions, and $y$ as the output, we can graph $f$ and $f^{-1}$ on the same grid, as shown in Figure208. The two graphs are symmetric about the line $y = x\text{.}$
Graph the function $f (x) = x^3 + 2$ and its inverse $f^{-1}(x) = \sqrt[3]{x - 2}$ on the same set of axes, along with the line $y = x\text{.}$ | {
"domain": "unl.edu",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109520836027,
"lm_q1q2_score": 0.8520821559508629,
"lm_q2_score": 0.8652240860523328,
"openwebmath_perplexity": 217.23917974865927,
"openwebmath_score": 0.844607412815094,
"tags": null,
"url": "https://mathbooks.unl.edu/PreCalculus/Inverse-Functions.html"
} |
# Evaluate $\int_{1}^{\infty} \frac{\sqrt{x - 1}}{(x + 1)^{2}} ~ \mathrm{d}{x}$.
I need to solve the following integral: $$I = \int_{1}^{\infty} \frac{\sqrt{x - 1}}{(x + 1)^{2}} ~ \mathrm{d}{x}.$$ Wolfram Alpha gives the answer as $\dfrac{\pi}{2 \sqrt{2}}$.
I think it’s achievable by complex analysis, but I really have no idea how. Also, is there a special name for this integral, i.e., does it have some known physical significance?
• The integrand has a closed-form antiderivative. Nov 13 '15 at 2:55
• You don't need to use complex analysis. Nov 13 '15 at 3:03
• if you want to use contour integration, take a keyhole around with a slit at the line $(1,\infty)$ Nov 13 '15 at 10:30
Let $u=\sqrt{x-1}$, $du=\frac{1}{2\sqrt{x-1}}$, then
$$I=2\int_0^\infty\frac{u^2}{(u^2+2)^2}du$$
We can apply partial fractions here.
$$=2\int_0^\infty\left(\frac{1}{u^2+2}-\frac{2}{(u^2+2)^2}\right)du$$
$$=2\int_0^\infty\frac{1}{u^2+2}du-4\int_0^\infty\frac{1}{(u^2+2)^2}du$$
The first integrand is almost $\tan^{-1}$, so we can factor the 2 and apply the substitution $s=\frac{u}{\sqrt{2}}$.
$$I=\sqrt 2\int_0^\infty\frac{1}{s^2+1}ds-4\int_0^\infty\frac{1}{(u^2+2)^2}du$$
$$=\frac{\pi}{\sqrt 2}-4\int_0^\infty\frac{1}{(u^2+2)^2}du$$
Now to tackle the second integral we can use a trig sub.
Let $u=\sqrt{2}\tan (t)$, $du=\sqrt 2 \sec^2(t)dt$.
$$I=\frac{\pi}{\sqrt 2}-4\sqrt 2 \int_0^{\pi/2}\frac{\sec ^2 t}{4\sec^4t}dt$$
$$=\frac{\pi}{\sqrt 2}-\sqrt 2\int _0^{\pi/2}\cos^2(t) dt$$
$$=\frac{\pi}{\sqrt 2}-\sqrt 2 \int_0^{\pi/2}\left(\frac{1}{2}\cos(2t)+\frac{1}{2} \right)dt$$
If we substitute $v=2t$ and split the integral up we get
$$I=\frac{\pi}{\sqrt 2}-\frac{1}{2\sqrt 2} \int_0^\pi \cos(v)dv-\frac{1}{\sqrt 2}\int_0^{\pi/2}1dt$$
The first integral clearly goes to 0 and the second integral becomes $\frac{\pi}{2\sqrt 2}$.
Therefore
$$I=\frac{\pi}{2\sqrt 2}$$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109520836026,
"lm_q1q2_score": 0.8520821508171078,
"lm_q2_score": 0.8652240808393984,
"openwebmath_perplexity": 289.02061330389927,
"openwebmath_score": 1.000000238418579,
"tags": null,
"url": "https://math.stackexchange.com/questions/1526717/evaluate-int-1-infty-frac-sqrtx-1x-12-mathrmdx/1526750"
} |
Therefore
$$I=\frac{\pi}{2\sqrt 2}$$
• There's a much easier way to find the $u$ integral. Using IBP gives $$\int u\cdot\frac{2u}{(u^2+2)^2}du = -u\cdot\frac{1}{u^2+2} + \int \frac{1}{u^2+2}\,du = -\frac{u}{u^2+2} + \arctan u$$ Nov 18 '15 at 5:51
Using a keyhole countour with the origin of the keyhole at $z=1$ and the small circle enclosing the value $z=1$ and using $$f(z) = \frac{\exp(1/2\log(z-1))}{(1+z)^2}$$ with the branch cut of the logarithm on the positive real axis and the argument from $0$ to $2\pi$ we get for the integral $$I=\int_1^\infty \frac{\sqrt{x-1}}{(1+x)^2} dx$$ that $$I(1-\exp(\pi i)) = 2\pi i\mathrm{Res}_{z=-1} f(z)$$ or $$I = \pi i\mathrm{Res}_{z=-1} f(z).$$
Now the logarithmic term is certainly analytic in a neighborhood of $z=-1$ and we have $$\mathrm{Res}_{z=-1} f(z) = \left.\left(\exp(1/2\log(z-1))\right)'\right|_{z=-1} = \left. \left(\exp(1/2\log(z-1))\right) \frac{1}{2}\frac{1}{z-1}\right|_{z=-1} \\= \exp(1/2(\log 2 + \pi i)) \times -\frac{1}{2}\frac{1}{2} = -\frac{1}{4} \sqrt{2} i.$$
This yields $$I = -\frac{1}{4} \sqrt{2} i \times\pi i=\frac{\sqrt{2}\pi}{4}.$$
Remark. The estimates for the circular components are done using ML same as at this MSE link. We get for the large circle parameterized by $z=R e^{it}$ $$\lim_{R\rightarrow \infty} 2\pi R \frac{\sqrt{R+1}}{(R-1)^2} = 0.$$
The small circle is a parameterized with $z=1+\epsilon e^{it}$ and we get $$\lim_{\epsilon\rightarrow 0} 2\pi\epsilon \frac{\sqrt{\epsilon}}{4} = 0.$$
$$I=\int_{1}^{\infty}\frac{\sqrt{x-1}}{(x+1)^2}dx$$ Use substitution $\frac{1}{x+1}=t$ which implies $\frac{dx}{(x+1)^2}=-dt$ So
$$I=-\int_{0.5}^{0}\sqrt{\frac{1}{t}-2}\:dt=\int_{0}^{0.5}\frac{\sqrt{1-2t}}{\sqrt{t}}dt$$
Again use substitution $\sqrt{t}=y$ which implies $\frac{dt}{\sqrt{t}}=2dy$
$$I=2\int_{0}^{\sqrt{0.5}}\sqrt{1-2y^2}dy=2\sqrt{2}\int_{0}^{\sqrt{0.5}}\sqrt{(\sqrt{0.5})^2-t^2}$$
Use standard integral $$\int\sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}sin^{-1}(\frac{x}{a})$$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109520836026,
"lm_q1q2_score": 0.8520821508171078,
"lm_q2_score": 0.8652240808393984,
"openwebmath_perplexity": 289.02061330389927,
"openwebmath_score": 1.000000238418579,
"tags": null,
"url": "https://math.stackexchange.com/questions/1526717/evaluate-int-1-infty-frac-sqrtx-1x-12-mathrmdx/1526750"
} |
we get $$I=\frac{\pi}{2\sqrt{2}}$$
Hint Substitution $x=t^2+1$...
This is the general antiderivative. Just take the limits and you're good. $$\int{\frac{\sqrt{x-1}}{(x+1)^2}dx}$$ Substitute $u = \sqrt{x-1}$ $$= \int{\frac{u}{(u^2+2)^2}du}$$ $$= 2\int\left(\frac{u}{u^2+2}-\frac{2}{(u^2+2)^2}\right)du$$ $$= \int\frac{u}{\frac{u^2}{2}+1}du-4\int\frac{2}{(u^2+2)^2}du$$ Substitute $s = \frac{u}{\sqrt{2}}$ and $ds = \frac{1}{\sqrt{2}}$ $$= \sqrt{2}\int\frac{1}{s^2+1}ds-4\int\frac{2}{(u^2+2)^2}du$$ $$= \sqrt{2}\arctan(s)-4\int\frac{2}{(u^2+2)^2}du$$ Substitute $u = \sqrt{2}\tan (p)\quad$ and $\quad du = \sqrt{2}\sec^2(p) dp\quad$ and $\quad(u^2 + 2)^2 = (2\tan^2(p)+2)^2 = 4\sec^4(p)\quad$ and $\quad p=\arctan\frac{u}{\sqrt{2}}$ $$= \sqrt{2}\arctan(s)-\sqrt{2}\int\cos^2(p)du$$ $$= \sqrt{2}\arctan(s)-\sqrt{2}\int\left(\frac{1}{2} \cos(2p) + \frac{1}{2}\right)du$$ $$= \sqrt{2}\arctan(s)-\frac{1}{\sqrt{2}}\int\cos(2p) - \frac{1}{\sqrt{2}}\int du$$ $$= \sqrt{2}\arctan(s)-\frac{p}{\sqrt{2}} - \frac{\sin(p)\cos(p)}{\sqrt{2}}$$ $$= \sqrt{2}\arctan(s)-\frac{\arctan\frac{u}{\sqrt{2}}}{\sqrt{2}} - \frac{\sin(\arctan\frac{u}{\sqrt{2}})\cos(\arctan\frac{u}{\sqrt{2}})}{\sqrt{2}}$$ Note that $\cos(\arctan(z)) = \frac{1}{\sqrt{z^2 + 1}}$ and $\sin(\arctan(z)) = \frac{z}{\sqrt{z^2+1}}$ $$= \frac{2\sqrt{2}(u^2+2)\arctan(s)+\sqrt{2}(u^2+2)\arctan\frac{u}{\sqrt{2}}+2u}{2(u^2+2)}$$ $$=\frac{\sqrt{2}(u^2+2)\arctan \frac{u}{\sqrt{2}}-2u}{2(u^2+2)}$$ $$=\frac{\sqrt{2}(x+1)\arctan \frac{\sqrt{x-1}}{\sqrt{2}}-2\sqrt{x-1}}{2(x+1)}$$ $$=\frac{\arctan \frac{\sqrt{x-1}}{\sqrt{2}}}{\sqrt{2}}-\frac{\sqrt{x-1}}{x+1}$$
I think I have a different answer.
Let $\frac{x+1}{2} = u$, so that $dx = 2 du$ and $x-1 = 2(u-1)$. Then | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109520836026,
"lm_q1q2_score": 0.8520821508171078,
"lm_q2_score": 0.8652240808393984,
"openwebmath_perplexity": 289.02061330389927,
"openwebmath_score": 1.000000238418579,
"tags": null,
"url": "https://math.stackexchange.com/questions/1526717/evaluate-int-1-infty-frac-sqrtx-1x-12-mathrmdx/1526750"
} |
Let $\frac{x+1}{2} = u$, so that $dx = 2 du$ and $x-1 = 2(u-1)$. Then
\begin{align} \int_1^\infty \frac{\sqrt{x-1}}{(x+1)^2} dx &= \int_1^\infty \frac{\sqrt{2} (u-1)^{\frac{1}{2}}}{4u^2} 2 du \\ &= \frac{1}{\sqrt{2}} \int_1^\infty u^{-2} (u-1)^\frac{1}{2} du \end{align} Now let $u = \frac{1}{t}$, so that $du = \frac{-1}{t^2} dt$. Continuing. \begin{align} \phantom{\int_1^\infty \frac{\sqrt{x-1}}{(x+1)^2} dx} &= \frac{1}{\sqrt{2}} \int_0^1 t^2 (1-t)^\frac{1}{2} \frac{1}{\sqrt{t}} \frac{1}{t^2} dt \\ &= \frac{1}{\sqrt{2}} \int_0^1 t^{\frac{-1}{2}} (1-t)^\frac{1}{2} dt \\ &= \frac{1}{\sqrt{2}} \int_0^1 t^{\frac{1}{2}-1} (1-t)^{\frac{3}{2} - 1} dt \\ &= \frac{1}{\sqrt{2}} B ( \frac{1}{2} . \frac{3}{2} ) \\ &= \frac{\Gamma(\frac{1}{2}) \Gamma(\frac{3}{2})} {\sqrt{2}\Gamma(2)} \\ &= \frac{\pi}{2\sqrt{2}}. \end{align}
Original image of work by hand.
• Honestly, with 1500 points I expect you know the customs here, and I see many answers where you used LaTeX. You could make an effort and write your answer correctly. Nov 14 '15 at 0:48 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109520836026,
"lm_q1q2_score": 0.8520821508171078,
"lm_q2_score": 0.8652240808393984,
"openwebmath_perplexity": 289.02061330389927,
"openwebmath_score": 1.000000238418579,
"tags": null,
"url": "https://math.stackexchange.com/questions/1526717/evaluate-int-1-infty-frac-sqrtx-1x-12-mathrmdx/1526750"
} |
# [SOLVED]a linear equation in 3 variables
#### karush
##### Well-known member
write a linear equation in 3 variables that is satisfied by all 3 of the given ordered triples
$(1,1,1), (0,2,0), (1,0,0)$
the examples in book are all on the $ax+by+cz=d$ equation but with just ordered triples there is no $d$ or can it be found from them... otherwise I would solve this by simultaneously.
the answer to this is $2x+y-z=2$
#### chisigma
##### Well-known member
write a linear equation in 3 variables that is satisfied by all 3 of the given ordered triples
$(1,1,1), (0,2,0), (1,0,0)$
the examples in book are all on the $ax+by+cz=d$ equation but with just ordered triples there is no $d$ or can it be found from them... otherwise I would solve this by simultaneously.
the answer to this is $2x+y-z=2$
You have the linear system...
$\displaystyle a + b + c = d$
$\displaystyle 2 b = d$
$\displaystyle a = d\ (1)$
... the solution of which is $a=d,\ b= \frac{d}{2},\ c=- \frac{d}{2}$, so that You can write...
$\displaystyle a x + \frac{d}{2} y - \frac{d}{2} z = d \implies x + \frac{y}{2} - \frac{z}{2} = 1\ (2)$
Kind regards
$\chi$ $\sigma$
#### HallsofIvy
##### Well-known member
MHB Math Helper
write a linear equation in 3 variables that is satisfied by all 3 of the given ordered triples
$(1,1,1), (0,2,0), (1,0,0)$
the examples in book are all on the $ax+by+cz=d$ equation but with just ordered triples there is no $d$ or can it be found from them... otherwise I would solve this by simultaneously.
the answer to this is $2x+y-z=2$
I would have done this just a little differently from chisigma. First, note that there is no "unique" solution. for any ax+ by+ cz= d describing the line, you could multiply through by any number and get a new equation for the same line. You could. for example, divide both sides by d to get (a/d)x+ (b/d)y+ (c/d)z= 1, then let a'= a/d, b'= b/d, and c'= c/d so that the equation is a'x+ b'y+ c'z= 1, with only three unknown values. | {
"domain": "mathhelpboards.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109503004293,
"lm_q1q2_score": 0.8520821390067531,
"lm_q2_score": 0.8652240704135291,
"openwebmath_perplexity": 468.6514466117882,
"openwebmath_score": 0.9999923706054688,
"tags": null,
"url": "https://mathhelpboards.com/threads/a-linear-equation-in-3-variables.5803/"
} |
I suspect that the answer you are asked for is the one that has all integer coefficients with no common factor.
Now, from ax+ by+ cz= d, (1, 1, 1) on the line means we must have a+ b+ c= d. (0, 2, 0) on the line means 0a+ 2b+ 0c= 2b= d. (1, 0, 0) on the line means 1a+ 0b+ 0c= a= d. Since both 2b and a are equal to d, a= 2b and we can replace both a and d with 2b in the first equation: 2b+ b+ c= 2b so that b+ c= 0 or b= -c.
Now we can write everything in terms of b: a= 2b, c= -b, d= 2b, and so the equation is
2bx+ by- bz= 2b. Here, b can be any (non-zero) number and still give an equation describing the line but dividing both sides of the equation by b gives
2x+ y- z= 2, the simplest form for the equation as all numbers are integers and they do not all have a common factor so cannot be reduced.
#### Prove It
##### Well-known member
MHB Math Helper
write a linear equation in 3 variables that is satisfied by all 3 of the given ordered triples
$(1,1,1), (0,2,0), (1,0,0)$
the examples in book are all on the $ax+by+cz=d$ equation but with just ordered triples there is no $d$ or can it be found from them... otherwise I would solve this by simultaneously.
the answer to this is $2x+y-z=2$
There will be a unique plane which satisfies these three points simultaneously. The plane will have the same coefficients as its normal vector, and the normal vector will also be normal to any vectors in the plane. So if we can get two vectors that lie in the plane, taking their cross product will give the normal vector. Call two of these vectors \displaystyle \begin{align*} \mathbf{a} \end{align*} and \displaystyle \begin{align*} \mathbf{b} \end{align*}. We could have
\displaystyle \begin{align*} \mathbf{a} &= ( 1 - 0, 1 - 2, 1 - 0) \\ &= ( 1 , -1, 1 ) \\ \\ \mathbf{b} &= (1 - 0, 0 - 2, 0 - 0 ) \\ &= ( 1 , -2, 0) \end{align*} | {
"domain": "mathhelpboards.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109503004293,
"lm_q1q2_score": 0.8520821390067531,
"lm_q2_score": 0.8652240704135291,
"openwebmath_perplexity": 468.6514466117882,
"openwebmath_score": 0.9999923706054688,
"tags": null,
"url": "https://mathhelpboards.com/threads/a-linear-equation-in-3-variables.5803/"
} |
\displaystyle \begin{align*} \mathbf{n} &= \mathbf{a} \times \mathbf{b} \\ &= \left| \begin{matrix} \mathbf{i} & \phantom{-}\mathbf{j} & \mathbf{k} \\ 1 & -1 & 1 \\ 1 & -2 & 0 \end{matrix} \right| \\ &= \mathbf{i} \, \left| \begin{matrix} -1 & 1 \\ -2 & 0 \end{matrix} \right| - \mathbf{j} \, \left| \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} \right| + \mathbf{k} \, \left| \begin{matrix} 1 & -1 \\ 1 & -2 \end{matrix} \right| \\ &= \mathbf{i} \, \left[ -1 \cdot 0 - 1 \cdot (-2) \right] - \mathbf{j} \, \left( 1 \cdot 0 - 1 \cdot 1 \right) + \mathbf{k} \, \left[ 1 \cdot (-2) - (-1) \cdot 1 \right] \\ &= 2\,\mathbf{i} + \mathbf{j} - \mathbf{k} \end{align*}
The plane will have the same coefficients as the normal vector, so that gives \displaystyle \begin{align*} 2x + y - z = d \end{align*}. Since we have three points that lie on the plane, any of them can be substituted to find \displaystyle \begin{align*} d \end{align*}. If we substitute \displaystyle \begin{align*} (1, 0, 0) \end{align*} that gives
\displaystyle \begin{align*} 2 \cdot 1 + 0 - 0 &= d \\ 2 &= d \end{align*}
and thus the plane is \displaystyle \begin{align*} 2x + y - z = 2 \end{align*}. | {
"domain": "mathhelpboards.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109503004293,
"lm_q1q2_score": 0.8520821390067531,
"lm_q2_score": 0.8652240704135291,
"openwebmath_perplexity": 468.6514466117882,
"openwebmath_score": 0.9999923706054688,
"tags": null,
"url": "https://mathhelpboards.com/threads/a-linear-equation-in-3-variables.5803/"
} |
I'm having an issue plotting torus links in Mathematica. What I'm trying to generate is a (2,8) torus link like the one in the picture in Wikipedia:
using equations such as those outlined here.
However, I appear to only get half of the respective torus link.
a = 1; d = 4; p = 2; q = 8;
ParametricPlot3D[{(a*Sin[q*t] + d)*Sin[p*t], (a*Sin[q*t] + d)*
Cos[p*t], a*Cos[q*t]}, {t, 0, 2*Pi}, PlotStyle -> Orange,
PlotRange -> All] /. Line[pts_, rest___] :> Tube[pts, 0.2, rest]
(a and d are parameters that control the width of the tube and the distance of the curve from the origin respectively, outlined here.)
Can anyone explain why I am only seeing half of the link?
• You're only plotting one set of parametric equations and you can see in the image that you don't expect there to be any intersections so the smooth parametric equations clearly can't generate the structure you want. On the other hand, it ought to be clear that the others may be generated via a simple rotation in the x-y plane. – b3m2a1 Aug 2 '18 at 8:09
The second torus is created by a rotation of $$\pi/4$$ (around {0, 0, 1}):
R = RotationMatrix[Pi/4, {0, 0, 1}];
torus = {(a*Sin[q*t] + d)*Sin[p*t], (a*Sin[q*t] + d)*Cos[p*t], a*Cos[q*t]};
ParametricPlot3D[{torus, R.torus} // Evaluate, {t, 0, 2*Pi}, PlotStyle -> {Orange, Blue}, PlotRange -> All]
Here's I think a generalization of what Ulrich gave to an arbitrary $(p, q)$ torus (if I read Wikipedia right):
plotPQTorus[{p_, q_}, a : _?NumericQ : 1, d : _?NumericQ : 4,
ops : OptionsPattern[]] :=
Block[{t},
ParametricPlot3D[
Evaluate@
Table[
RotationMatrix[
i*2 \[Pi]/q, {0, 0, 1}].{(a*Sin[q*t] + d)*
Sin[p*t], (a*Sin[q*t] + d)*Cos[p*t], a*Cos[q*t]},
{i, 0, If[Divisible[q, p], p - 1, 0]}
],
{t, 0, 2*Pi},
PlotRange -> All,
ops
] /. Line[pts_, rest___] :> Tube[pts, 0.2, rest]
]
Here are a few plots:
Table[plotPQTorus[{p, Fibonacci[q]}, Boxed -> False,
Axes -> None], {p, 1, 4}, {q, 2, 6, 2}] // Grid | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9621075690244281,
"lm_q1q2_score": 0.8520727559437188,
"lm_q2_score": 0.8856314858927011,
"openwebmath_perplexity": 2329.600533817695,
"openwebmath_score": 0.2998276650905609,
"tags": null,
"url": "https://mathematica.stackexchange.com/questions/179382/issue-plotting-torus-link"
} |
Table[plotPQTorus[{p, Fibonacci[q]}, Boxed -> False,
Axes -> None], {p, 1, 4}, {q, 2, 6, 2}] // Grid
Note that relatively prime things are single connected loops (as Wikipedia suggests they should be)
The (1,4) torus knot is known to KnotData[], so we can use that to build the (2,8) knot:
knot14 = First[KnotData[{"TorusKnot", {1, 4}}, "ImageData"]]; | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9621075690244281,
"lm_q1q2_score": 0.8520727559437188,
"lm_q2_score": 0.8856314858927011,
"openwebmath_perplexity": 2329.600533817695,
"openwebmath_score": 0.2998276650905609,
"tags": null,
"url": "https://mathematica.stackexchange.com/questions/179382/issue-plotting-torus-link"
} |
# Logic problem
##### Active member
Consider the following sequence of statements:
$$S_1: \text{at least 1 of the statements }S_1-S_n \text{ is false}\\ S_2: \text{at least 2 of the statements }S_1-S_n \text{ are false}\\ \vdots \\ S_n: \text{at least } n \text{ of the statements }S_1-S_n \text{ are false}$$
Where $n$ is some integer.
Question: for which $n$ are these statements self-consistent? In those cases: what is the truth value of each statement?
I got this off of a blog I tend to frequent. I will wait before posting the solution this time.
EDIT:
Changed the question; I had written the statements wrong
Last edited:
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Suppose $k$ out of $n$ statements are true.
Then $S_1$ up to $S_k$ have to be true and the rest has to be false.
This appears to be consistent for any $n$ and any $0\le k \le n$.
##### Active member
Suppose $k$ out of $n$ statements are true.
Then $S_1$ up to $S_k$ have to be true and the rest has to be false.
This appears to be consistent for any $n$ and any $0\le k \le n$.
Sorry about that, you were absolutely right about the question as phrased.
However, this new version should prove to be a bit more interesting. This is what I had meant; I had accidentally written "true" instead of "false".
#### Klaas van Aarsen
##### MHB Seeker
Staff member
If $S_n$ is true, then $n$ statements are false, including $S_n$.
Therefore $S_n$ is false.
We now know that at least $1$ statement is false.
Therefore $S_1$ is true.
For $n=1$ this is a contradiction, and for $n=2$ this is a consistent solution.
For $n \ge 3$ we can say, that if $S_{n-1}$ were true, then $n-1$ statements are false.
Since $S_1$ is true, this implies that $S_{n-1}$ is false.
Therefore $S_{n-1}$ is false.
So at least $2$ statements are false.
Therefore $S_2$ is true.
For $n=3$ this is a contradiction, and for $n=4$ this is a consistent solution.
Etcetera. | {
"domain": "mathhelpboards.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9621075744568837,
"lm_q1q2_score": 0.8520727491378914,
"lm_q2_score": 0.8856314738181875,
"openwebmath_perplexity": 447.92497156639786,
"openwebmath_score": 0.5636366605758667,
"tags": null,
"url": "https://mathhelpboards.com/threads/logic-problem.5096/"
} |
Etcetera.
In other words, we get a consistent consistent solution if and only if $n$ is even.
In that case $S_1$ up to $S_{n/2}$ are true and $S_{n/2+1}$ up to $S_{n}$ are false. $\qquad \blacksquare$
##### Active member
If $S_n$ is true, then $n$ statements are false, including $S_n$.
Therefore $S_n$ is false.
We now know that at least $1$ statement is false.
Therefore $S_1$ is true.
For $n=1$ this is a contradiction, and for $n=2$ this is a consistent solution.
For $n \ge 3$ we can say, that if $S_{n-1}$ were true, then $n-1$ statements are false.
Since $S_1$ is true, this implies that $S_{n-1}$ is false.
Therefore $S_{n-1}$ is false.
So at least $2$ statements are false.
Therefore $S_2$ is true.
For $n=3$ this is a contradiction, and for $n=4$ this is a consistent solution.
Etcetera.
In other words, we get a consistent consistent solution if and only if $n$ is even.
In that case $S_1$ up to $S_{n/2}$ are true and $S_{n/2+1}$ up to $S_{n}$ are false. $\qquad \blacksquare$
Couldn't have phrased it better myself.
The source, for anybody interested:
The Parity Paradox – Futility Closet
I highly recommend the website as a time-wasting tool. | {
"domain": "mathhelpboards.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9621075744568837,
"lm_q1q2_score": 0.8520727491378914,
"lm_q2_score": 0.8856314738181875,
"openwebmath_perplexity": 447.92497156639786,
"openwebmath_score": 0.5636366605758667,
"tags": null,
"url": "https://mathhelpboards.com/threads/logic-problem.5096/"
} |
# Thread: Is it a field? set of real numbers in the form...
1. ## Is it a field? set of real numbers in the form...
Hi,
problem:
Let $Q(\sqrt{2})$ be the set of all real numbers of the form
$\alpha+\beta\sqrt{2}$, where $\alpha\;and\;\beta$ are rational.
(a) Is $Q(\sqrt{2})$ a field?
(b) What if $\alpha\;and\;\beta$ are required to be integers?
attempt:
(a)
I look at the axioms for a field and check whether any of them fail. I am unable to find one that does.
Is it ok to say that I can express any real number in the form $\alpha+\beta\sqrt{2}$ as long as $\alpha\;and\;\beta$ are rational?
(b)
I am unable to find an axiom that "fails", but say I want to write the real number $\frac{\sqrt{2}}{2}$. I don't think I can do this if $\alpha\;and\;\beta$ are integers..
Thanks
2. Originally Posted by Mollier
Hi,
problem:
Let $Q(\sqrt{2})$ be the set of all real numbers of the form
$\alpha+\beta\sqrt{2}$, where $\alpha\;and\;\beta$ are rational.
(a) Is $Q(\sqrt{2})$ a field?
(b) What if $\alpha\;and\;\beta$ are required to be integers?
attempt:
(a)
I look at the axioms for a field and check whether any of them fail. I am unable to find one that does.
Is it ok to say that I can express any real number in the form $\alpha+\beta\sqrt{2}$ as long as $\alpha\;and\;\beta$ are rational?
(b)
I am unable to find an axiom that "fails", but say I want to write the real number $\frac{\sqrt{2}}{2}$. I don't think I can do this if $\alpha\;and\;\beta$ are integers..
Thanks
just go through each property for a field and make sure that they all hold. you have to verify these explicitly.
and no, you cannot say any real number can be expressed in that form. how would you express $\pi$ that way, for instance?
3. Originally Posted by Mollier
Hi,
problem:
Let $Q(\sqrt{2})$ be the set of all real numbers of the form
$\alpha+\beta\sqrt{2}$, where $\alpha\;and\;\beta$ are rational.
(a) Is $Q(\sqrt{2})$ a field?
(b) What if $\alpha\;and\;\beta$ are required to be integers?
attempt: | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9621075701109193,
"lm_q1q2_score": 0.8520727481932139,
"lm_q2_score": 0.885631476836816,
"openwebmath_perplexity": 174.76001380594448,
"openwebmath_score": 0.8847749829292297,
"tags": null,
"url": "http://mathhelpforum.com/advanced-algebra/122477-field-set-real-numbers-form.html"
} |
attempt:
(a)
I look at the axioms for a field and check whether any of them fail. I am unable to find one that does.
Is it ok to say that I can express any real number in the form $\alpha+\beta\sqrt{2}$ as long as $\alpha\;and\;\beta$ are rational?
(b)
I am unable to find an axiom that "fails", but say I want to write the real number $\frac{\sqrt{2}}{2}$. I don't think I can do this if $\alpha\;and\;\beta$ are integers..
Thanks
Firstly, if every real number was of the form $\alpha+\beta \sqrt{2}$ with $\alpha, \beta \in \mathbb{Q}$ then we can easily define a bijection $\mathbb{R} \rightarrow \mathbb{Q} \times \mathbb{Q}$, $\alpha + \beta \sqrt{2} \mapsto (\alpha, \beta)$. As there exists a bijection between $\mathbb{Q}$ and $\mathbb{Q} \times \mathbb{Q}$ we have that $|\mathbb{R}| = |\mathbb{Q}|$, a contradiction.
Secondly, in both of your problems you know that "most" of the axioms for fields holds (both are clearly rings, and multiplication in both cases is commutative). The one to concentrate on will be multiplicative inverses.
What is the inverse of $\alpha+\beta \sqrt{2}$ with $\alpha$ and $\beta$ rationals?
What if both were integers, does that imply that the inverse has integer coefficients? (Alternatively, think about what the problem is saying...take $\beta = 0$. Then if your ring was a field every element in this subset has an inverse in the ring. This subset is just the integers. So, every integer has an inverse of the form $\alpha + \beta \sqrt{2}$. That is, every rational number of the form $\frac{1}{\gamma}, \gamma \in \mathbb{Z} \setminus \{0\}$ can be written as $\alpha + \beta \sqrt{2}, \alpha, \beta \in \mathbb{Z}$. As we are, allegedly, in a field we have that $\frac{n}{\gamma}$ is also of this form, with $n \in \mathbb{N}$. Thus, every rational number is of the form $\alpha + \beta \sqrt{2}$. This is silly!) | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9621075701109193,
"lm_q1q2_score": 0.8520727481932139,
"lm_q2_score": 0.885631476836816,
"openwebmath_perplexity": 174.76001380594448,
"openwebmath_score": 0.8847749829292297,
"tags": null,
"url": "http://mathhelpforum.com/advanced-algebra/122477-field-set-real-numbers-form.html"
} |
4. Originally Posted by Jhevon
and no, you cannot say any real number can be expressed in that form. how would you express $\pi$ that way, for instance?
Thanks
Originally Posted by Swlabr
The one to concentrate on will be multiplicative inverses.
What is the inverse of $\alpha+\beta \sqrt{2}$ with $\alpha$ and $\beta$ rationals?
What if both were integers, does that imply that the inverse has integer coefficients? (Alternatively, think about what the problem is saying...take $\beta = 0$. Then if your ring was a field every element in this subset has an inverse in the ring. This subset is just the integers. So, every integer has an inverse of the form $\alpha + \beta \sqrt{2}$. That is, every rational number of the form $\frac{1}{\gamma}, \gamma \in \mathbb{Z} \setminus \{0\}$ can be written as $\alpha + \beta \sqrt{2}, \alpha, \beta \in \mathbb{Z}$. As we are, allegedly, in a field we have that $\frac{n}{\gamma}$ is also of this form, with $n \in \mathbb{N}$. Thus, every rational number is of the form $\alpha + \beta \sqrt{2}$. This is silly!)
It would be $\frac{1}{\alpha+\beta\sqrt{2}}$ .
Now, the inverse would have to be in $Q(\sqrt{2})$, so it would have to be in the form $\alpha+\beta\sqrt{2}$.
Since $\frac{1}{\alpha+\beta\sqrt{2}}$ is not in that form, $Q(\sqrt{2})$ is not a field.
Is this totally off? I think I need to get (a) right, before I move on to (b).
Thanks guys!
5. Originally Posted by Mollier
Thanks
It would be $\frac{1}{\alpha+\beta\sqrt{2}}$ .
Now, the inverse would have to be in $Q(\sqrt{2})$, so it would have to be in the form $\alpha+\beta\sqrt{2}$.
Since $\frac{1}{\alpha+\beta\sqrt{2}}$ is not in that form, $Q(\sqrt{2})$ is not a field.
Is this totally off? I think I need to get (a) right, before I move on to (b). | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9621075701109193,
"lm_q1q2_score": 0.8520727481932139,
"lm_q2_score": 0.885631476836816,
"openwebmath_perplexity": 174.76001380594448,
"openwebmath_score": 0.8847749829292297,
"tags": null,
"url": "http://mathhelpforum.com/advanced-algebra/122477-field-set-real-numbers-form.html"
} |
Thanks guys!
You want to try and write $\frac{1}{\alpha + \beta \sqrt{2}}$ as $\alpha^{\prime} + \beta^{\prime}\sqrt{2}$. I mean, $\sqrt{2} = \frac{2}{\sqrt{2}}$...numbers are strange things...just because something isn't obviously in a certain form does not mean you cannot write it in that form.
I shall give you a tantalising hint: It is a field.
6. $
\frac{1}{\alpha+\beta\sqrt{2}}=\frac{\alpha-\beta\sqrt{2}}{\alpha^2-2\beta^2}=\left(\frac{\alpha}{\alpha^2-2\beta^2}\right) - \left(\frac{\beta}{\alpha^2-2\beta^2}\right)\sqrt{2}
$
$\alpha'=\left(\frac{\alpha}{\alpha^2-2\beta^2}\right)$
$\beta'=\left(\frac{\beta}{\alpha^2-2\beta^2}\right)$
Since both $\alpha'\;and\;\beta'$ are rationals, $Q(\sqrt{2})$ is a field?
Thanks mate!
7. Originally Posted by Mollier
$
\frac{1}{\alpha+\beta\sqrt{2}}=\frac{\alpha-\beta\sqrt{2}}{\alpha^2-2\beta^2}=\left(\frac{\alpha}{\alpha^2-2\beta^2}\right) - \left(\frac{\beta}{\alpha^2-2\beta^2}\right)\sqrt{2}
$
$\alpha'=\left(\frac{\alpha}{\alpha^2-2\beta^2}\right)$
$\beta'=\left(\frac{\beta}{\alpha^2-2\beta^2}\right)$
Since both $\alpha'\;and\;\beta'$ are rationals, $Q(\sqrt{2})$ is a field?
Thanks mate!
Yeah, that's it! Although to be thorough you would need to make sure there is no division by 0 going on.
8. So, $\alpha^2-2\beta^2\neq 0$,
then $\beta \neq \frac{1}{\sqrt{2}}\alpha$, but $\beta$ is then not a rational number anyways..
As for (b), can I use a similar argument by saying that since $\alpha'$ and $\beta'$ are rational numbers, $Q(\sqrt{2})$ is not a field when $\alpha,\beta\in Z$?
Thank you very much.
9. Originally Posted by Mollier
So, $\alpha^2-2\beta^2\neq 0$,
then $\beta \neq \frac{1}{\sqrt{2}}\alpha$, but $\beta$ is then not a rational number anyways..
As for (b), can I use a similar argument by saying that since $\alpha'$ and $\beta'$ are rational numbers, $Q(\sqrt{2})$ is not a field when $\alpha,\beta\in Z$? | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9621075701109193,
"lm_q1q2_score": 0.8520727481932139,
"lm_q2_score": 0.885631476836816,
"openwebmath_perplexity": 174.76001380594448,
"openwebmath_score": 0.8847749829292297,
"tags": null,
"url": "http://mathhelpforum.com/advanced-algebra/122477-field-set-real-numbers-form.html"
} |
Thank you very much.
For part (b) you need to find a counter-example - $\alpha^{\prime}$ and $\beta^{\prime}$ may look like non-integers, but this may not always be the case.
So, to complete the problem simply take an integer value of $\alpha$ and an integer value of $\beta$ such that $\alpha^{\prime} \notin \mathbb{Z}$ or $\beta^{\prime} \notin \mathbb{Z}$. For simplicity, let one of them be zero.
10. Originally Posted by Mollier
So, $\alpha^2-2\beta^2\neq 0$,
then $\beta \neq \frac{1}{\sqrt{2}}\alpha$, but $\beta$ is then not a rational number anyways..
No, this is not a proof. You want to assume that $\alpha^2 -2\beta^2=0$ and look for a contradiction.
11. $\alpha^2-2\beta^2\neq 0$ because of the irrationality of $\sqrt{2}$, it can be easily proved in number theory.
12. Originally Posted by Swlabr
No, this is not a proof. You want to assume that $\alpha^2 -2\beta^2=0$ and look for a contradiction.
$\alpha^2-2\beta^2=0 \Rightarrow \alpha=\sqrt{2}\beta$
Contradiction since $\alpha\in Q$
13. Originally Posted by Mollier
$\alpha^2-2\beta^2=0 \Rightarrow \alpha=\sqrt{2}\beta$
Contradiction since $\alpha\in Q$
Precisely!
Sorry to be pedantic about it, but in maths the proof is the king...
14. Please don't apologize. I have a BS in mechanical engineering from a horrible school, so I have now decided to actually learn some math. You being "pedantic" is exactly what I need! Thanks! | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9621075701109193,
"lm_q1q2_score": 0.8520727481932139,
"lm_q2_score": 0.885631476836816,
"openwebmath_perplexity": 174.76001380594448,
"openwebmath_score": 0.8847749829292297,
"tags": null,
"url": "http://mathhelpforum.com/advanced-algebra/122477-field-set-real-numbers-form.html"
} |
# Math Help - rational numbers
1. ## rational numbers
Find two rational numbers with denominators 11 and 13, respectively, and a sum of 7/143.
I got x/11 + y/13=7/143.
Two unknowns and one equation. Kind of stuck.
2. Might help to multiply through by 143 and clear the fractions. Note that 11 x 13 = 143.
Then you're into a conventional linear Diophantine equation for which there are techniques (which I'd need to look up). What's your level?
3. Don't forget $x$ and $y$ are integers. The equation they satisfy is $13x+11y=7$.
You should first find one particular solution $(x_0,y_0)$ (you may start by finding $x,y\in\mathbb{Z}$ such that $13x+11y=1$ (why is it possible ?)), and then look for the others: they satisfy $13x+11y=13x_0+11y_0$, hence $13(x-x_0)=11(y_0-y)$ and you should then be able to conclude that $x=x_0+11k$ and $y=y_0-13k$ for some integer $k$.
Laurent.
4. Originally Posted by Matt Westwood
Might help to multiply through by 143 and clear the fractions. Note that 11 x 13 = 143.
Then you're into a conventional linear Diophantine equation for which there are techniques (which I'd need to look up). What's your level?
What do u mean level? I am taking Elem. Number Theory and this is some of the recm. hw problems to do for the section of euclidean algorithm
5. Originally Posted by rmpatel5
What do u mean level? I am taking Elem. Number Theory and this is some of the recm. hw problems to do for the section of euclidean algorithm
6. Originally Posted by Laurent
Don't forget $x$ and $y$ are integers. The equation they satisfy is $13x+11y=7$.
You should first find one particular solution $(x_0,y_0)$ (you may start by finding $x,y\in\mathbb{Z}$ such that $13x+11y=1$ (why is it possible ?)), and then look for the others: they satisfy $13x+11y=13x_0+11y_0$, hence $13(x-x_0)=11(y_0-y)$ and you should then be able to conclude that $x=x_0+11k$ and $y=y_0-13k$ for some integer $k$. | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9890130589886011,
"lm_q1q2_score": 0.8520725381869193,
"lm_q2_score": 0.8615382076534743,
"openwebmath_perplexity": 409.548473341809,
"openwebmath_score": 0.8476439714431763,
"tags": null,
"url": "http://mathhelpforum.com/number-theory/47919-rational-numbers.html"
} |
Laurent.
I am really lost with what you do. I know why i can say 13x + 11y=1 because the gcd(11,13)=1 and there is a thm that says ax+by=1. Should i just slove the two system of equations ?
7. Originally Posted by rmpatel5
gcd(11,13)=1 and there is a thm that says ax+by=1.
There is a theorem that says that there exists x and y such that 11x+13y=1. Now, you should find an example of such a couple (x,y). Either you guess it, or you make use of the Euclidean algorithm as you may have done in class.
8. Originally Posted by rmpatel5
I am really lost with what you do. I know why i can say 13x + 11y=1 because the gcd(11,13)=1 and there is a thm that says ax+by=1. Should i just slove the two system of equations ?
Ok, so solve for x and y in 13x+11y=1, using the Euclidean algorithm (http://en.wikipedia.org/wiki/Euclidean_algorithm). Actually, do the Euclidian algorithm over 13 and 11, and observe...
Then, multiply by 7 :
13*(7x)+11*(7y)=7
Uh.
9. Use the Euclidean Algorithm to get the gcd of 11 and 13 (yes, you know this is 1, but the calculations you did in the algorithm help you find out what values of a and b give you 11a + 13b = 1.
You now want two numbers x and y such that 11x + 13y = 7.
Well, you just got 11a + 13b = 1, so multiply everything by 7:
$11 \times 7a + 13 \times 7b = 7$
So 7a and 7b are the numbers you want for x and y. Job done.
10. You do not need to use Euclidean algorithm.
It is quite obvious that $13(6) + 11(-7) = 1$.
Thus, $13(42) + 11(-49) = 7$.
This means the solutions to $13x+11y=7$ are: $x=42 - 11t \mbox{ and }y=-49 + 42t$.
11. 6 and -5 work as well, so are there many solutions to this problem. I got them by doing the EA | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9890130589886011,
"lm_q1q2_score": 0.8520725381869193,
"lm_q2_score": 0.8615382076534743,
"openwebmath_perplexity": 409.548473341809,
"openwebmath_score": 0.8476439714431763,
"tags": null,
"url": "http://mathhelpforum.com/number-theory/47919-rational-numbers.html"
} |
Topic: Discrete-space Fourier transform computation
## Question
Compute the discrete-space Fourier transform of the following signal:
$f[m,n]= 2^{-(n+m)} u[n] u[m]$
First off notice that this equation can easily be separated into two functions $g[m]=2^{-m}u[m]$ and $h[n]=2^{-n}u[n]$ where $f[m,n]=g[m]h[n]$. Then since both equations are the same except for a change of variable where m=n or vice verse we can show the DTFT of either g[m] or h[n] and it should suffice for the other. If they were not the same we would have to evaluate both of them.
\begin{align} G[u]&=\sum_{m=-\infty}^{\infty}2^{-m}u[m]e^{-j u m} \\ &=\sum_{m=0}^{\infty}( \frac{1}{2e^{ju}} )^{m} \\ &=\frac{1}{1-\frac{1}{2e^{ju}}} \\ &=\frac{-2e^{ju}}{1-2e^{ju}} \\ \end{align}
Thus
$F[u,v]= \frac{4e^{j(u+v)}}{(1-2e^{ju})(1-2e^{jv})}$
\begin{align} F [u,v] &= \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} f[m,n]e^{-j(mu + nv)}\\ &= \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} 2^{-(n+m)} u[n] u[m] e^{-j(mu + nv)}\\ &= \sum_{m=-\infty}^{\infty} 2^{-m} u[m] e^{-j(mu)} \sum_{n=-\infty}^{\infty} 2^{-n} u[n] e^{-j(nv)}\\ &= \sum_{m= 0}^{\infty} 2^{-m} e^{-j(mu)} \sum_{n=0}^{\infty} 2^{-n} e^{-j(nv)}\\ &= \sum_{m= 0}^{\infty} (2e^{ju})^{-m} \sum_{n=0}^{\infty} (2e^{jv})^{-n}\\ &= \frac{1}{1-\frac{1}{2e^{ju}}}\cdot\frac{1}{1-\frac{1}{2e^{jv}}}\\ &= \frac{1}{1-\frac{1}{2}e^{-ju}}\cdot\frac{1}{1-\frac{1}{2}e^{-jv}}\\ &= \frac{1}{(1-\frac{1}{2}e^{-ju})(1-\frac{1}{2}e^{-jv})} \end{align}
--Xiao1 23:03, 19 November 2011 (UTC)
Instructor's comments: Solutions 1 and solutions 2 are the two main ways to answer the question. The second one is a bit longer to write, in my opinion, but it does not require knowing the separation property of the continuous-space Fourier transform. Note however that, in the exam, if you do not have the separation property in the table, you will need to prove it in order to get full credit. -pm | {
"domain": "projectrhea.org",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9890130554263734,
"lm_q1q2_score": 0.8520725333598925,
"lm_q2_score": 0.8615382058759129,
"openwebmath_perplexity": 2124.7650561476125,
"openwebmath_score": 1.0000097751617432,
"tags": null,
"url": "https://www.projectrhea.org/rhea/index.php/Compute_DSFT_product_two_step_functions_ECE438F11"
} |
# Absolute Maximum And Minimum Calculator On Interval | {
"domain": "rciretedimpresa.it",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9890130576932457,
"lm_q1q2_score": 0.8520725317968266,
"lm_q2_score": 0.86153820232079,
"openwebmath_perplexity": 373.11161112116935,
"openwebmath_score": 0.6303600668907166,
"tags": null,
"url": "http://hefc.rciretedimpresa.it/absolute-maximum-and-minimum-calculator-on-interval.html"
} |
20 at x = –0. > @ 2 f x x x 6 2 on 1,27 11. Occurence of absolute maxima: If f(x) is continuous in a closed interval I, then the absolute maximum of f(x) in I is the maximum value of f(x) on all local maxima and endpoints on I. These values correspond to the probability of observing such an extreme value by chance. If we find all possible local extrema, then the global maximum, if it exists, must be the value of f(x) on the interval 1 ≤ x ≤ 4, and the maximum is f(4) = 9. Find more Mathematics widgets in Wolfram|Alpha. F INDING a maximum or a minimum has its application in pure mathematics, where for example we could find the largest rectangle that has a given perimeter. Sample size calculation for trials for superiority, non-inferiority, and equivalence. It is a measure of your capacity for aerobic work and can be a predictor of your potential as an endurance athlete. 80 at x = –1. If the absolute maximum occurs at an interior point, then we have found an absolute maximum in the open interval. The absolute maximum value of f x x x( ) 3 12 M 32 on the closed interval > @M2, 4 occurs at x = A) 4 B) 2 C) 1 D) 0 E) M2 2. C) Absolute maximum only. For example, you could say,“The pulse rates are between 56 and 92 bpm. The closed interval method is a way to solve a problem within a specific interval of a function. Inequalities and ‘Maximum-Minimum’ Problems Henry Liu, 26 February 2007 There are many olympiad level problems in mathematics which belong to areas that are not covered well at all at schools. It is important to understand the difference between the two types of minimum/maximum (collectively called extrema) values for many of the applications in this chapter and so we use a variety of examples to help with this. Absolute zero is defined as the point where no more heat can be removed from a system, according to the absolute or thermodynamic temperature scale. Four Function and. Left bounds go on left side of min/max and right bounds go on right side. (3) Because | {
"domain": "rciretedimpresa.it",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9890130576932457,
"lm_q1q2_score": 0.8520725317968266,
"lm_q2_score": 0.86153820232079,
"openwebmath_perplexity": 373.11161112116935,
"openwebmath_score": 0.6303600668907166,
"tags": null,
"url": "http://hefc.rciretedimpresa.it/absolute-maximum-and-minimum-calculator-on-interval.html"
} |
Function and. Left bounds go on left side of min/max and right bounds go on right side. (3) Because the system has been carefully planned using modern top-down programming techniques, it is relatively easy to modify and extend. The largest of the values from steps 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value. We take the derivative using the quotient rule: f0(x) =. The MAX function is a built-in function in Excel that is categorized as a Statistical Function. Relative Minimum - The lowest point on an interval of a curve. Once you've found the max-imum and minimum on this line (as well as on the other two lines that make up the boundary), compare all the values you've checked to find out that the absolute maximum and the absolute minimum are f(1 4,2) = 673 64 (←absolute maximum) f(1 2, 4. An absolute maximum occurs at c if for all x in the domain of f. The absolute is measured in liters of oxygen per minute. f x x() 3 d. Meaning answers obtained by looking at the graph in the calculator will not earn V = - 4x +3; [1, 3] BIUA. Since the function is not defined for some open interval around either c or d, a local maximum or local minimum cannot occur at this point. An absolute maximum or minimum can occur, however, because the definition requires that the point simply be in the domain of the function. Find the absolute maximum and the absolute minimum values of the function shown below, on the given interval. • Find the values of f at the endpoints of the interval. Absolute Maximum/Minimum Values of Multivariable Functions - Part 1 of 2 To find absolute max/min values of a continuous function g on a closed bounded set D: 1. The graph of y = cos x. Absolute Maximum and Absolute Minimum. Free Maximum Calculator - find the Maximum of a data set step-by-step This website uses cookies to ensure you get the best experience. Find the absolute maximum and absolute minimum values of f(x) = x2 −4 x2 +4 on the interval [−4,4]. | {
"domain": "rciretedimpresa.it",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9890130576932457,
"lm_q1q2_score": 0.8520725317968266,
"lm_q2_score": 0.86153820232079,
"openwebmath_perplexity": 373.11161112116935,
"openwebmath_score": 0.6303600668907166,
"tags": null,
"url": "http://hefc.rciretedimpresa.it/absolute-maximum-and-minimum-calculator-on-interval.html"
} |
Find the absolute maximum and absolute minimum values of f(x) = x2 −4 x2 +4 on the interval [−4,4]. Subtract 6 6 from 1 1. It is important to understand the difference between the two types of minimum/maximum (collectively called extrema) values for many of the applications in this chapter and so we use a variety of examples to help with this. Find the absolute maximum and absolute minimum values of f on the given interval. MIN([DISTINCT] expr) Minimum value returned by expr MOD(x,y) Remainder of x divided by y MONTHS_BETWEEN(end_date, start_date) Number of months between the 2 dates (integer) NEW_TIME(date, zone1, zone2) Convert between GMT and US time zones (but not CET) NEXT_DAY(date,day_of_week) '12-OCT-01','Monday' will return the next Mon after 12 Oct NLS. 4 is the lower limit. (d) Find the absolute minimum value of f x over the closed interval 5 ddx 5. Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. For, sin (x + ) = cos x. denbal87 New member. Therefore, let's consider the function over the closed interval If the maximum value occurs at an interior point, then we have found the value in the open interval that maximizes the area of the garden. f(x)=x+ 9 x on [0. On the graph above of the function f on the closed interval [a, e], the point (a, f (a)) represents the absolute minimum, and the point (d, f (d)) represents the absolute maximum. Question 203087: Find the absolute maximum and absolute minimum values of the function below. An absolute minimum is the lowest y value or output value a. Thread starter denbal87; Start date Nov 4, 2014; D. The largest value found in steps 2 and 3 above will be the absolute maximum and the smallest value will be the absolute minimum. This lesson will focus on the maximum and minimum points. These points are sometimes referred to as max, min, extreme values, or extrema. (d) Find the absolute minimum value of f x over the closed interval 5 ddx 5. Let's find, for | {
"domain": "rciretedimpresa.it",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9890130576932457,
"lm_q1q2_score": 0.8520725317968266,
"lm_q2_score": 0.86153820232079,
"openwebmath_perplexity": 373.11161112116935,
"openwebmath_score": 0.6303600668907166,
"tags": null,
"url": "http://hefc.rciretedimpresa.it/absolute-maximum-and-minimum-calculator-on-interval.html"
} |
(d) Find the absolute minimum value of f x over the closed interval 5 ddx 5. Let's find, for example, the absolute extrema of h (x)=2x^3+3x^2-12x h(x) = 2x3 +3x2 −12x. Sure, there are other, more precise, definitions, but that will work for what we want to do. 67, and a relative maximum of 4. Use the Stefan Boltzmann relationship (I = σT 4 where σ = 5. Q: Determine the absolute maximum and minimum values of the function on the given interval. summary_interval # How often to write checkpoints (rounds up to the nearest statistics # interval). I have an assignment where I have to write a program that accepts a series of integers from the keyboard from 1 to 100 using a single sentinel controlled loop, meaning I will need to also assign a sentinel number that when entered, will stop the loop and display the results. This is defined everywhere and is zero at $\ds x=\pm \sqrt{3}/3$. By using this website, you agree to our Cookie Policy. For each x value: Determine the value of f '(x) for values a little smaller and a little larger than the x value. The absolute max occurs at S = The absolute min occurs at S =. Decide whether you have a minimum or a maximum. )Given the function 𝑓(𝑥= 𝑥2+ 𝑥+ , chose values for a, b, and c in that could work for the graph shown. Which method do you prefer? f (x) = 1 + 3x^2 - 2x^3. the absolute (global) maximum 3. Meaning answers obtained by looking at the graph in the calculator will not earn V = - 4x +3; [1, 3] BIUA. VO2 max in men is approximately 40-60% higher in males than females. On the interval, fnmin then finds all local extrema of the function as left and right limits at a jump and as zeros of the function's first derivative. If you're behind a web filter, please make sure that the domains *. Question 203087: Find the absolute maximum and absolute minimum values of the function below. We take the derivative using the quotient rule: f0(x) =. The debt to equity ratio is a financial, liquidity ratio that compares a company’s total debt | {
"domain": "rciretedimpresa.it",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9890130576932457,
"lm_q1q2_score": 0.8520725317968266,
"lm_q2_score": 0.86153820232079,
"openwebmath_perplexity": 373.11161112116935,
"openwebmath_score": 0.6303600668907166,
"tags": null,
"url": "http://hefc.rciretedimpresa.it/absolute-maximum-and-minimum-calculator-on-interval.html"
} |
=. The debt to equity ratio is a financial, liquidity ratio that compares a company’s total debt to total equity. To find the extreme values of a function (the highest or lowest points on the interval where the function is defined), first calculate the derivative of the function and make a study of sign. The absolute maximum on the interval is 138 at x=-2. For example, the following are all equivalent confidence intervals: 20. Find the absolute maximum and absolute minimum values of f on the given interval. Approximating Relative Extrema. Also the lowest value of either the X of the Y is placed first in the set. the absolute (global) maximum 3. Theorem (Extreme Value Theorem) If f is continuous on a closed interval [a,b], then f attains an absolute maximum value f(c) and an absolute minimum value f(d) at some. Example 1: Consider the three curves shown below. In worst case, if all intervals are from ‘min’ to ‘max’, then time complexity becomes O((max-min+1)*n) where n is number of intervals. Find the extreme values of f on the boundary of D. Occurence of absolute minima: If f(x) is continuous in a closed interval I, then the absolute minimum of f(x) in I is the minimum value of f(x) on all local. Three major examples are geometry, number theory, and functional equations. Go to window and set your X minimum to -1 and your X maximum to 5. Extreme Values of Functions Definitions An extreme value of a function is the largest or smallest value of the function in some interval. We usually distinguish between local and global (or absolute) extreme values. An extreme value, or extremum (plural extrema), is the smallest (minimum) or largest (maximum) value of a function, either in an arbitrarily small neighborhood of a point in the function's domain — in which case it is called a relative or local extremum — or on a given set contained in the domain (perhaps all of it) — in which case it is called an absolute or global extremum (the latter. Calculus I: Candidates Test for | {
"domain": "rciretedimpresa.it",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9890130576932457,
"lm_q1q2_score": 0.8520725317968266,
"lm_q2_score": 0.86153820232079,
"openwebmath_perplexity": 373.11161112116935,
"openwebmath_score": 0.6303600668907166,
"tags": null,
"url": "http://hefc.rciretedimpresa.it/absolute-maximum-and-minimum-calculator-on-interval.html"
} |
which case it is called an absolute or global extremum (the latter. Calculus I: Candidates Test for Global Extrema 1) If a continuous function f is defined on a finite, closed interval, such as −1≤x≤4 or [−1,4], or, more generally, a≤x≤b or [a,b], then f always has a global minimum value and a global maximum value on that interval. At t =0 the position of the object is 5. Examples: Input: v = {{1, 2}, {2, 4}, {3, 6}} first_page Given an absolute sorted array and a number K, find the pair whose sum is K. Use a graphing calculator to approximate the intervals where each function is increasing and Increase and Decrease Absolute minimum:. Answer: First, find the critical points by finding where the derivative equals zero: f0(x) = (x2 +4)(2x)−(x2 −4. If f has a local maximum or minimum at c and f'(c) exists, then f'(c) = 0. For a strictly unimodal function with an extremum inside the interval, it will find that extremum, while for an interval containing multiple extrema (possibly including the interval boundaries), it will converge to one of them. f(x) = x4 – A: Plot the graph for f(x) in the interval for [-2, 0]. Find the absolute maximum and absolute minimum values of f on the given interval. pow(x, 2) + math. Visual Magnitude Calculator: Computes the visual magnitude of a star from its absolute magnitude and distance. If f has a local maximum or minimum at c and f'(c. An absolute maximum or minimum can occur, however, because the definition requires that the point simply be in the domain of the function. Extreme Values A Global Maximum A function f has a global (absolute) maximum at x =c if f (x)≤ f (c) for all x∈Df. Absolute Maximum - The highest point on a curve. Note: From our definition of absolute maxima and minima, if $(a, f(a))$ is an absolute max/min, then it is also a local max/min too. Let's Practice:. between -30 to 20 function is decreasing because there are no local minima and maxima in between them. f(x)= 490x x2 +49 on [0,10] 2 Fall 2016, Maya. Stack | {
"domain": "rciretedimpresa.it",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9890130576932457,
"lm_q1q2_score": 0.8520725317968266,
"lm_q2_score": 0.86153820232079,
"openwebmath_perplexity": 373.11161112116935,
"openwebmath_score": 0.6303600668907166,
"tags": null,
"url": "http://hefc.rciretedimpresa.it/absolute-maximum-and-minimum-calculator-on-interval.html"
} |
are no local minima and maxima in between them. f(x)= 490x x2 +49 on [0,10] 2 Fall 2016, Maya. Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. Here again we are giving definitions that appeal to your geometric intuition. f(x)=xln(2x)on[0. Local Extreme Values of a Function Let c be an interior point of the domain of the. The absolute is measured in liters of oxygen per minute. ) Find the absolute maximum and minimum values of the function on the given interval. Q: Determine the absolute maximum and minimum values of the function on the given interval. Extreme value theorem tells us that a continuous function must obtain absolute minimum and maximum values on a closed interval. VO2max stands for maximal oxygen uptake and refers to the amount of oxygen your body is capable of utilizing in one minute. so minimum value of f(x) in interval [0. For example, consider the functions shown in Figure(d), (e), and (f). Find more Mathematics widgets in Wolfram|Alpha. It's easy to find one with neither absolute extrema. Identify the location and value of the absolute maximum and absolute minimum of a function over the domain of the function graphically or by using a graphing utility. From the graph you can see that is has a maximum at (3, 27) and a minimum at (1. Find the absolute maximum and absolute minimum values of f on the given interval. The maximum value of a function that has a derivative at all points in an For a function f(x) that has a derivative at every point in an interval [a, b], the maximum or minimum values can be found by using the following procedure: 1. By using this website, you agree to our Cookie Policy. Similar topics can also be found in the Calculus section of the site. So the absolute max value is 19 and the absolute min value is 1. The minimum and maximum describe the spread of the data. By using this website, you agree to our Cookie Policy. When calculating Intervals the X values are placed | {
"domain": "rciretedimpresa.it",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9890130576932457,
"lm_q1q2_score": 0.8520725317968266,
"lm_q2_score": 0.86153820232079,
"openwebmath_perplexity": 373.11161112116935,
"openwebmath_score": 0.6303600668907166,
"tags": null,
"url": "http://hefc.rciretedimpresa.it/absolute-maximum-and-minimum-calculator-on-interval.html"
} |
this website, you agree to our Cookie Policy. When calculating Intervals the X values are placed "on top of" the Y values. Enter DNE if the absolute maximum or minimum does not exist. Pick the largest and smallest. Find the absolute maximum and minimum values on theinterval: f(x) = x - 2cosx [-pi, pi] f ' (x) = 1+2sinx f (-pi)= -pi - 2cos(-pi) = -pi - 2. Thus if one has a sample {, …,}, and one picks another observation +, then this has / (+) probability of being the largest value seen so far. At t =0 the position of the object is 5. MIN([DISTINCT] expr) Minimum value returned by expr MOD(x,y) Remainder of x divided by y MONTHS_BETWEEN(end_date, start_date) Number of months between the 2 dates (integer) NEW_TIME(date, zone1, zone2) Convert between GMT and US time zones (but not CET) NEXT_DAY(date,day_of_week) '12-OCT-01','Monday' will return the next Mon after 12 Oct NLS. A higher debt to equity ratio indicates that more creditor financing (bank loans) is used than investor financing (shareholders). The range spread then uses the range to find a percentage that the maximum is greater than the minimum, using the minimum as a base. The absolute minimum on the interval is -237 at x=3. f x x( ) cos 2 b. Explain your reasoning. maximum" functions. Similarly, the global minimum is located at the lowest point. A relative minimum is a point that is lower than all the other points around it. The absolute max occurs at S = The absolute min occurs at S =. question_answer. Find the maximum and minimum values of the function f(x) = ln x/x on the interval [1, 3]. Hit graph and then hit 2nd Trace I think to bring up a menu that has minimum and maximum in there. Typical values for are 0. Absolute Maximum And Minimum Calculator. So, f(b) is a relative maximum of f. The maximum will occur at the highest value and the minimum will occur at the lowest value. If you're seeing this message, it means we're having trouble loading external resources on our website. To find the maximum | {
"domain": "rciretedimpresa.it",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9890130576932457,
"lm_q1q2_score": 0.8520725317968266,
"lm_q2_score": 0.86153820232079,
"openwebmath_perplexity": 373.11161112116935,
"openwebmath_score": 0.6303600668907166,
"tags": null,
"url": "http://hefc.rciretedimpresa.it/absolute-maximum-and-minimum-calculator-on-interval.html"
} |
it means we're having trouble loading external resources on our website. To find the maximum and/or minimum on an interval, check the values at the critical points and at the ends of the interval. Find the the critical points of f on D. Fermat's Theorem. From this list of values we see that the absolute maximum is 8 and will occur at $$t = 2$$ and the absolute minimum is -3 which occurs at $$t = 1$$. You can take notes in the margins or on the flip-side of each sheet. Finding Extrema on a closed interval: 1. Subtract 6 6 from 1 1. In the single-variable case, it is known, by the Extreme Value Theorem, that if f is continuous on a closed interval [a;b], then it has has an absolute maximum and an absolute minimum on [a;b]. Find more Mathematics widgets in Wolfram|Alpha. Since the function is concave down at x=1 and has a critical point at x=1 (zero slope) then the function has a local maximum at x=1. find the absolute maximum and absolute minimum values of the fuction f(x)=2x-13ln(3x) on interval [1,8] 2. ^2 = 825 Thus function has absolute minimum value at x = 2 and absolute maximum value at x = 5 in the interval [-1, 5]. Wolfram alpha paved a completely new way to get knowledge and information. Enter the equation in the Y= section for Y1. minimum" or "4. f(x)=x^3-3x+1; [0,3]. Local Extreme Values of a Function Let c be an interior point of the domain of the. Let f be a function defined and. Endpoint Discontinuities: only one of the one-sided limits exists. f(x) = x + (4/x) on the interval [1,5]. A relative (or local) maximum occurs at c if for all x in an open interval containing c. 11) A) Absolute minimum only. But there is one very important condition that guarantees both an absolute minimum and an absolute maximum. If f has a local maximum or minimum at c and f'(c) exists, then f'(c) = 0. Absolute Maximum and Absolute Minimum. So the absolute max value is 19 and the absolute min value is 1. org are unblocked. Absolute Maximum/Minimum Values of Multivariable | {
"domain": "rciretedimpresa.it",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9890130576932457,
"lm_q1q2_score": 0.8520725317968266,
"lm_q2_score": 0.86153820232079,
"openwebmath_perplexity": 373.11161112116935,
"openwebmath_score": 0.6303600668907166,
"tags": null,
"url": "http://hefc.rciretedimpresa.it/absolute-maximum-and-minimum-calculator-on-interval.html"
} |
and the absolute min value is 1. org are unblocked. Absolute Maximum/Minimum Values of Multivariable Functions - Part 1 of 2 To find absolute max/min values of a continuous function g on a closed bounded set D: 1. Get the free "Function Extrema - Math 101" widget for your website, blog, Wordpress, Blogger, or iGoogle. so minimum value of f(x) in interval [0. Sometimes it's important to consider points which are only largest or smallest in small parts of a graph. (c) Find all intervals on which the graph of f is concave up and also has positive slope. Thus, to find the absolute maximum (absolute minimum) value of the function, we choose the largest and smallest amongst the numbers f(a), f(c 1 ), f(c 2. have both an absolute maximum and an absolute minimum. From the graph you can see that is has a maximum at (3, 27) and a minimum at (1. Finding Extrema on a closed interval: 1. f(5) = (5)^4 + 8*(5)^3 -32*(5)^2 = 825 Thus function has absolute minimum value at x = 2 and absolute maximum value at x = 5 in the interval [-1, 5]. Local minima and maxima (First Derivative Test) by Paul Garrett is licensed under a Creative Commons Attribution-Noncommercial-ShareAlike 4. In a blank cell, enter this formula =Max(ABS(A1:D10)), see screenshot: 2. The student familiar with the sum formula can easily prove that. Then press Ctrl+Shift+Enter keys, and the largest absolute values will be displayed in the. False There is a local maximum at x = 0. The largest of the values from steps 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value. Fold Unfold. Examples: Input: v = {{1, 2}, {2, 4}, {3, 6}} first_page Given an absolute sorted array and a number K, find the pair whose sum is K. This lesson will focus on the maximum and minimum points. Find more Mathematics widgets in Wolfram|Alpha. Relative Minimum - The lowest point on an interval of a curve. # The text file for eval. f(x) = @ 10. absolute minimum value at x = 2 is -48. The first | {
"domain": "rciretedimpresa.it",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9890130576932457,
"lm_q1q2_score": 0.8520725317968266,
"lm_q2_score": 0.86153820232079,
"openwebmath_perplexity": 373.11161112116935,
"openwebmath_score": 0.6303600668907166,
"tags": null,
"url": "http://hefc.rciretedimpresa.it/absolute-maximum-and-minimum-calculator-on-interval.html"
} |
of a curve. # The text file for eval. f(x) = @ 10. absolute minimum value at x = 2 is -48. The first derivative test: Let f (x) be a function and x = c a critical point of f. f(t) = 4t + 4 cot(t/2), [π/4, 7π/4] I'm stuck after I take the first derivative. Occurence of absolute minima: If f(x) is continuous in a closed interval I, then the absolute minimum of f(x) in I is the minimum value of f(x) on all local. Since the function is not defined for some open interval around either c or d, a local maximum or local minimum cannot occur at this point. Occurence of absolute maxima: If f(x) is continuous in a closed interval I, then the absolute maximum of f(x) in I is the maximum value of f(x) on all local maxima and endpoints on I. Mean number of days ≥ 30, 35 or 40 °C The average number of days in the period when the daily maximum air temperature was equal to, or exceeded 30, 35 or 40 °C. Look at the graph of f (x) = x 3 + 4x 2 - 12x over the interval [0, 3], Figure 1a. Free functions extreme points calculator - find functions extreme and saddle points step-by-step This website uses cookies to ensure you get the best experience. We take the derivative using the quotient rule: f0(x) =. A higher debt to equity ratio indicates that more creditor financing (bank loans) is used than investor financing (shareholders). f x x3 2 on 3,1 > @ 15. Question 203087: Find the absolute maximum and absolute minimum values of the function below. Since the function is not defined for some open interval around either c or d, a local maximum or local minimum cannot occur at this point. The minimum value for this range is the mean subtracted by the confidence interval and the maximum value is calculated by the mean added by the confidence interval. f(x) = x4 – A: Plot the graph for f(x) in the interval for [-2, 0]. f(x) = (x^2 - 1)^3, [-1, 5] 14. 2 Maximum and Minimum on an Interval. Use a graphing calculator to approximate the intervals where each function is increasing and Increase and | {
"domain": "rciretedimpresa.it",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9890130576932457,
"lm_q1q2_score": 0.8520725317968266,
"lm_q2_score": 0.86153820232079,
"openwebmath_perplexity": 373.11161112116935,
"openwebmath_score": 0.6303600668907166,
"tags": null,
"url": "http://hefc.rciretedimpresa.it/absolute-maximum-and-minimum-calculator-on-interval.html"
} |
graphing calculator to approximate the intervals where each function is increasing and Increase and Decrease Absolute minimum:. Find the absolute maximum and absolute minimum values of f on the given interval. In that case, the point right on the border might be the maximum or minimum of the curve. From the graph you can see that is has a maximum at (3, 27) and a minimum at (1. f(c) is another relative minimum of f. It could very well continue to increase or decrease once we leave the interval. Example: Find the absolute maximum and minimum of:. Note: From our definition of absolute maxima and minima, if $(a, f(a))$ is an absolute max/min, then it is also a local max/min too. Question 203087: Find the absolute maximum and absolute minimum values of the function below. Occurence of absolute maxima: If f(x) is continuous in a closed interval I, then the absolute maximum of f(x) in I is the maximum value of f(x) on all local maxima and endpoints on I. 1 absolute maximum or minimum. The maximum will occur at the highest f (x) f (x) value and the minimum will occur at the lowest f (x) f (x) value. Find the the critical points of f on D. We say that the function f(x) has a global maximum at x=x 0 on the interval I, if for all. Absolute Maximum and Absolute Minimum. In this case, “absolute extrema” is just a fancy way of saying the single highest point and single lowest point in the interval. D) Absolute minimum and. The largest of the values from steps 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value. The restrictions stated or implied for such functions will determine the domain from which you must work. 587 Get more help from Chegg Get 1:1 help nowfrom expert CalculustutorsSolve itwith our calculusproblem solver and calculator. For example, consider the functions shown in Figure(d), (e), and (f). Compare the f (x) f ( x) values found for each value of x x in order to determine the absolute maximum and minimum over the | {
"domain": "rciretedimpresa.it",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9890130576932457,
"lm_q1q2_score": 0.8520725317968266,
"lm_q2_score": 0.86153820232079,
"openwebmath_perplexity": 373.11161112116935,
"openwebmath_score": 0.6303600668907166,
"tags": null,
"url": "http://hefc.rciretedimpresa.it/absolute-maximum-and-minimum-calculator-on-interval.html"
} |
values found for each value of x x in order to determine the absolute maximum and minimum over the given interval. Evaluate the function to find the y -values at all critical numbers and at each endpoint. (To make the distinction clear, sometimes the ‘plain’ maximum and minimum are called absolute maximum and minimum. Absolute minimum definition is - the smallest value that a mathematical function can have over its entire curve. f ( 1) = − 5 f ( 1) = - 5. Sal finds the absolute maximum value of f(x)=8ln(x)-x² over the interval [1,4]. earn credit. Python List max() Method - Python list method max returns the elements from the list with maximum value. True You are given that the function f ( x ) = 2 ( x + 3 ) x 2 + x − 2 has an absolute maximum on the interval − 2 < x < 1. Find the absolute maximum and absolute minimum values of f on the given interval. We usually distinguish between local and global (or absolute) extreme values. The smallest y -value is the absolute minimum and the largest y -value is the. On the interval, fnmin then finds all local extrema of the function as left and right limits at a jump and as zeros of the function's first derivative. The maximum will occur at the highest value and the minimum will occur at the lowest value. Roll your mouse over the Extreme Value Theorem to check your answers. A closed interval like [2, 5] includes the endpoints 2 and 5. For this data set, the minimum (lowest) value is 56 and the maximum (highest) value is 92. It could very well continue to increase or decrease once we leave the interval. Endpoint Discontinuities: only one of the one-sided limits exists. A point at which a function attains its minimum value among all points where it is defined is a global (or absolute) minimum. maximum: ( (62 -√3763)/9, (-461347 +7526√3763)/243)) ≈ (. If an absolute maximum or minimum does not exist, enter NONE. F INDING a maximum or a minimum has its application in pure mathematics, where for example we could find the largest | {
"domain": "rciretedimpresa.it",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9890130576932457,
"lm_q1q2_score": 0.8520725317968266,
"lm_q2_score": 0.86153820232079,
"openwebmath_perplexity": 373.11161112116935,
"openwebmath_score": 0.6303600668907166,
"tags": null,
"url": "http://hefc.rciretedimpresa.it/absolute-maximum-and-minimum-calculator-on-interval.html"
} |
or a minimum has its application in pure mathematics, where for example we could find the largest rectangle that has a given perimeter. Every function that’s continuous on a closed interval has an absolute maximum value and an absolute minimum value (the absolute extrema) in that interval — in other words, a highest and lowest point — though there can be a tie for the highest or lowest value. In that case, the point right on the border might be the maximum or minimum of the curve. Absolute maximum is highest of and. f(x) = 4x^3 - 6x^2 - 144x + 9, [-4, 5] 13. Find the extreme values of f on the boundary of D. Before Using this Calculator. The following small array formulas can help you to find out the largest absolute value and the smallest absolute value. Explain the meaning of the result. Get the free "Max/Min Finder" widget for your website, blog, Wordpress, Blogger, or iGoogle. f(x) = x + (4/x) on the interval [1,5]. Learn more about population standard deviation, or explore other statistical calculators, as well as hundreds of other calculators addressing math, finance, health, fitness, and more. Calculates the root of the equation f(x)=0 from the given function f(x) and its derivative f'(x) using Newton method. There are two kinds of extrema (a word meaning maximum or minimum): global and local, sometimes referred to as "absolute" and "relative", respectively. Example: Find the absolute maximum and minimum of:. f(c) is another relative minimum of f. The student familiar with the sum formula can easily prove that. Return to Contents. It is a greatest value in a set of points but not highest when compared to all values in a set. ) Find the absolute max/min values of f(x) = x2 4 x2+4 on the interval [ 4;4]. We say that the function f(x) has a global maximum at x=x 0 on the interval I, if for all. Wolfram alpha paved a completely new way to get knowledge and information. exp(exp) * math. Bolzano's proof consisted of showing that a continuous function on a closed | {
"domain": "rciretedimpresa.it",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9890130576932457,
"lm_q1q2_score": 0.8520725317968266,
"lm_q2_score": 0.86153820232079,
"openwebmath_perplexity": 373.11161112116935,
"openwebmath_score": 0.6303600668907166,
"tags": null,
"url": "http://hefc.rciretedimpresa.it/absolute-maximum-and-minimum-calculator-on-interval.html"
} |
exp(exp) * math. Bolzano's proof consisted of showing that a continuous function on a closed interval was bounded, and then showing that the function attained a maximum and a minimum value. For, sin (x + ) = cos x. You can take notes in the margins or on the flip-side of each sheet. Maxima and minima are points where a function reaches a highest or lowest value, respectively. Find the local maximum and minimum values of using both the First and Second Derivative Tests. Generating random numbers Problem. Let's find, for example, the absolute extrema of h (x)=2x^3+3x^2-12x h(x) = 2x3 +3x2 −12x. Four Function and. (b) Use calculus to find the exact maximum and minimum values. This important theorem can guide our investigations when we search for absolute extreme values of a function. 4 is the lower limit. Mean number of days ≥ 30, 35 or 40 °C The average number of days in the period when the daily maximum air temperature was equal to, or exceeded 30, 35 or 40 °C. in some open interval containing c. The Organic Chemistry Tutor 200,049 views 1:10:05. Over the long term about one day in ten can be expected to have a (maximum or minimum) temperature exceed the decile 9 value. Sample size calculation for trials for superiority, non-inferiority, and equivalence. To define these terms more formally: a function f has an absolute maximum at x = b if f ( b )≥ f ( x ) for all x in the domain of f. Therefore, let’s consider the function over the closed interval If the maximum value occurs at an interior point, then we have found the value in the open interval that maximizes the area of the garden. You have to use a graphing calculator to find that out. Calculate Field honors the Transfer Field Domain Descriptions environment. Therefore, we are trying to determine the maximum value of A(x) for x over the open interval $$(0,50)$$. The calculator will find the critical points, local and absolute (global) maxima and minima of the single variable function. Enter the equation in the Y= | {
"domain": "rciretedimpresa.it",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9890130576932457,
"lm_q1q2_score": 0.8520725317968266,
"lm_q2_score": 0.86153820232079,
"openwebmath_perplexity": 373.11161112116935,
"openwebmath_score": 0.6303600668907166,
"tags": null,
"url": "http://hefc.rciretedimpresa.it/absolute-maximum-and-minimum-calculator-on-interval.html"
} |
absolute (global) maxima and minima of the single variable function. Enter the equation in the Y= section for Y1. (c) Find all intervals on which the graph of f is concave up and also has positive slope. Step 6: As mentioned earlier, A (x) is a continuous function over the closed, bounded. If f has a local maximum or minimum at c and f'(c. (c) For any. Some problems may have two or more constraint equations. On a closed interval these points are referred to as absolute or global minimum/maximum points. In worst case, if all intervals are from 'min' to 'max', then time complexity becomes O((max-min+1)*n) where n is number of intervals. The “V” = volume per time. Local Extreme Values of a Function Let c be an interior point of the domain of the. Also time complexity of above solution depends on lengths of intervals. Find the absolute maximum and the absolute minimum values of the function shown below, on the given interval. 2 Maximum and Minimum on an Interval. The golden-section search is a technique for finding an extremum (minimum or maximum) of a function inside a specified interval. Evaluate the function to find the y -values at all critical numbers and at each endpoint. The smallest y -value is the absolute minimum and the largest y -value is the. Hill Sphere Calculator: Computes the Hill Sphere of an object: Sail Calculator: Computes the maximum velocity possible from acceleration caused by light. The calculators will allow you to convert any heart rate between 63% and 102% of your maximum heart rate to a percentage of your VO2max , or any percentage of VO2max. Look at the graph of f (x) = x 3 + 4x 2 - 12x over the interval [0, 3], Figure 1a. save_path = FLAGS. Example 2: Locate the value(s) where the function attains an absolute maximum and the value(s) where the function attains an absolute minimum, if they exist. f(x)=x+ 9 x on [0. f(t) = 4t + 4 cot(t/2), [π/4, 7π/4] I'm stuck after I take the first derivative. Explain your reasoning. Sal finds the absolute | {
"domain": "rciretedimpresa.it",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9890130576932457,
"lm_q1q2_score": 0.8520725317968266,
"lm_q2_score": 0.86153820232079,
"openwebmath_perplexity": 373.11161112116935,
"openwebmath_score": 0.6303600668907166,
"tags": null,
"url": "http://hefc.rciretedimpresa.it/absolute-maximum-and-minimum-calculator-on-interval.html"
} |
7π/4] I'm stuck after I take the first derivative. Explain your reasoning. Sal finds the absolute maximum value of f(x)=8ln(x)-x² over the interval [1,4]. question_answer. Calculus I: Candidates Test for Global Extrema 1) If a continuous function f is defined on a finite, closed interval, such as −1≤x≤4 or [−1,4], or, more generally, a≤x≤b or [a,b], then f always has a global minimum value and a global maximum value on that interval. The debt to equity ratio is a financial, liquidity ratio that compares a company’s total debt to total equity. 47, an absolute minimum of –5. 2 Maximum and Minimum on an Interval. You should substitute those and pick the greatest for the maximum value. No Local Extrema Compare the f (x) f (x) values found for each value of x x in order to determine the absolute maximum and minimum over the given interval. Find the absolute maximum and absolute minimum values of f on the given interval. It also has its application to commercial problems, such as finding the least dimensions of a carton that is to contain a given volume. (1 point) Find the absolute maximum and absolute minimum values of the function f (x) = x3 + 6x2 — 63x + 8 over each of the indicated intervals. maximum" functions. f (c) is called the global (absolute) maximum value. Then press Ctrl+Shift+Enter keys, and the largest absolute values will be displayed in the. This is achieved with a minimum of manual intervention. A relative minimum is a point that is lower than all the other points around it. The graph of y = cos x is the graph of y = sin x shifted, or translated, units to the left. Learn more about population standard deviation, or explore other statistical calculators, as well as hundreds of other calculators addressing math, finance, health, fitness, and more. f(x)= 490x x2 +49 on [0,10] 2 Fall 2016, Maya. An open interval like (2, 5) excludes the endpoints. X Values: [a,b] Y Values: / [c,d] Result: [e,f] [1,3] / [2,4] [. In this module you will be asked to calculate | {
"domain": "rciretedimpresa.it",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9890130576932457,
"lm_q1q2_score": 0.8520725317968266,
"lm_q2_score": 0.86153820232079,
"openwebmath_perplexity": 373.11161112116935,
"openwebmath_score": 0.6303600668907166,
"tags": null,
"url": "http://hefc.rciretedimpresa.it/absolute-maximum-and-minimum-calculator-on-interval.html"
} |
[a,b] Y Values: / [c,d] Result: [e,f] [1,3] / [2,4] [. In this module you will be asked to calculate the sample size for 6 situations. Therefore, let's consider the function over the closed interval If the maximum value occurs at an interior point, then we have found the value in the open interval that maximizes the area of the garden. Local Extreme Values of a Function Let c be an interior point of the domain of the. (a) Use a graph to estimate the absolute maximum and minimum values of the function to two decimal places. 20 at x = –0. By using this website, you agree to our Cookie Policy. Find the values of f f f at the critical numbers of f f f in (a, b). To find the maximum and/or minimum on an interval, check the values at the critical points and at the ends of the interval. Explain your reasoning. Significance Levels The significance level for a given hypothesis test is a value for which a P-value less than or equal to is considered statistically significant. The triangular distribution, along with the PERT distribution, is also widely used in project management (as an input into PERT and hence critical path method (CPM)) to model events which take place within an interval defined by a minimum and maximum value. The concern is that a dose given too soon after the previous dose may reduce the response. Pick the largest and smallest. Find the local or absolute minimum or maximum of an equation using a graphing calculator; Determine the intervals on which a function is increasing, decreasing, or constant using a graphing calculator (for precalculus) Determine an appropriate viewing rectangle for the graph of an equation; Match an equation to its graph; Graph an equation on. Finding the absolute max and min is a snap. (a) When is the object at rest? (b) Evaluate 6 1 ∫ vt dt(). An absolute minimum occurs at c if for all x in the domain of f. This is achieved with a minimum of manual intervention. Sample size calculation for trials for superiority, non-inferiority, | {
"domain": "rciretedimpresa.it",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9890130576932457,
"lm_q1q2_score": 0.8520725317968266,
"lm_q2_score": 0.86153820232079,
"openwebmath_perplexity": 373.11161112116935,
"openwebmath_score": 0.6303600668907166,
"tags": null,
"url": "http://hefc.rciretedimpresa.it/absolute-maximum-and-minimum-calculator-on-interval.html"
} |
minimum of manual intervention. Sample size calculation for trials for superiority, non-inferiority, and equivalence. Please answer the following questions about the function Instructions: If you are asked to. The range spread then uses the range to find a percentage that the maximum is greater than the minimum, using the minimum as a base. Final the absolute maximum and minimum values on the given interval. We usually distinguish between local and global (or absolute) extreme values. Find the extreme values of f on the boundary of D. 23] is -80 at x = 20 and maximum value of f(x) in interval [0,23] is at zero which is 0. Sometimes it's important to consider points which are only largest or smallest in small parts of a graph. Fold Unfold. Find the maximum / minimum absolute values with Formulas. f (1) = −5 f ( 1) = - 5. Here is the code to do that. Find the absolute maximum and absolute minimum values of f on the given interval. org are unblocked. Enter DNE if the absolute maximum or minimum does not exist. 388360 # Get 3 integers from 0 to 100 # Use max=101 because it will. Finding Extrema on a closed interval: 1. To find the maximum and/or minimum on an interval, check the values at the critical points and at the ends of the interval. You have 3 solutions: x=0, x=1, and this one. )Given the function 𝑓(𝑥= 𝑥2+ 𝑥+ , chose values for a, b, and c in that could work for the graph shown. If we break down the formula we can see why it gets its strange name. Before Using this Calculator. These points are sometimes referred to as max, min, extreme values, or extrema. In a blank cell, enter this formula =Max(ABS(A1:D10)), see screenshot: 2. Therefore, we are trying to determine the maximum value of A(x) for x over the open interval $$(0,50)$$. Answer: First, find the critical points by finding where the derivative equals zero: f0(x) = (x2 +4)(2x)−(x2 −4. ) Find the absolute max/min values of f(x) = x2 4 x2+4 on the interval [ 4;4]. The generic word for minimum or maximum is | {
"domain": "rciretedimpresa.it",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9890130576932457,
"lm_q1q2_score": 0.8520725317968266,
"lm_q2_score": 0.86153820232079,
"openwebmath_perplexity": 373.11161112116935,
"openwebmath_score": 0.6303600668907166,
"tags": null,
"url": "http://hefc.rciretedimpresa.it/absolute-maximum-and-minimum-calculator-on-interval.html"
} |
values of f(x) = x2 4 x2+4 on the interval [ 4;4]. The generic word for minimum or maximum is extremum. An absolute minimum is the lowest y value or output value a. An absolute maximum or minimum can occur, however, because the definition requires that the point simply be in the domain of the function. F INDING a maximum or a minimum has its application in pure mathematics, where for example we could find the largest rectangle that has a given perimeter. 80 at x = –1. ) Find the absolute max/min values of f(x) = x2 4 x2+4 on the interval [ 4;4]. Find the maximum / minimum absolute values with Formulas. a local (relative) maximum 6. Which method do you prefer? f (x) = 1 + 3x^2 - 2x^3. f(x) = x^2 + 250/x on the open interval (0,infinity ) I know that the absolute max is the answer NONE but I can not figure out the absolute min can someone help please thanks. The “V” = volume per time. # The text file for eval. Where does it flatten out? Where the slope is zero. There is 95% confidence that the constructed interval includes the population mean. In worst case, if all intervals are from 'min' to 'max', then time complexity becomes O((max-min+1)*n) where n is number of intervals. It then evaluates the function at these extrema and at the endpoints of the interval, and determines the minimum over all these values. So, f(b) is a relative maximum of f. Recommended and Minimum Ages and Intervals Between Doses of Routinely Recommended Vaccines1,2,3,4 Vaccine and dose number minimum interval between doses is equal to the greatest interval of any of the individual components. For a strictly unimodal function with an extremum inside the interval, it will find that extremum, while for an interval containing multiple extrema (possibly including the interval boundaries), it will converge to one of them. So the absolute max value is 19 and the absolute min value is 1. The range spread then uses the range to find a percentage that the maximum is greater than the minimum, using the | {
"domain": "rciretedimpresa.it",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9890130576932457,
"lm_q1q2_score": 0.8520725317968266,
"lm_q2_score": 0.86153820232079,
"openwebmath_perplexity": 373.11161112116935,
"openwebmath_score": 0.6303600668907166,
"tags": null,
"url": "http://hefc.rciretedimpresa.it/absolute-maximum-and-minimum-calculator-on-interval.html"
} |
then uses the range to find a percentage that the maximum is greater than the minimum, using the minimum as a base. Absolute Maximum and Absolute Minimum This page is intended to be a part of the Real Analysis section of Math Online. We do not know that a function necessarily has a maximum value over an open interval. In this module you will be asked to calculate the sample size for 6 situations. These values correspond to the probability of observing such an extreme value by chance. Solution for Find the absolute maximum and minimum of the function f (x) = x³ - x+2 on the interval [0, 3] %D. Find the absolute maximum and absolute minimum values of f on the given interval. If f has a local maximum or minimum at c and f'(c) exists, then f'(c) = 0. For instance, in the example at. f (c) is called the global (absolute) maximum value. Thread starter denbal87; Start date Nov 4, 2014; D. If you finish a job in less than 25% of the time allotted, you will be paid a Time Bonus, so try to finish as quickly as possible! The maximum Time Bonus is a 25% boost to your Base Reward. Calculate Field honors the Transfer Field Domain Descriptions environment. Explain your reasoning. Enter the equation in the Y= section for Y1. In the single-variable case, it is known, by the Extreme Value Theorem, that if f is continuous on a closed interval [a;b], then it has has an absolute maximum and an absolute minimum on [a;b]. ) On the other hand, it is possible to see directly that. How to calculate a confidence interval? First, you need to calculate the mean of your sample set. An absolute minimum occurs at c if for all x in the domain of f. An absolute maximum is the highest y value or output value a graph has over a specific interval. 140 of 155. The golden-section search is a technique for finding an extremum (minimum or maximum) of a function inside a specified interval. Let this index be 'max_index', return max_index + min. Infinite Discontinuities: both one-sided limits are infinite. | {
"domain": "rciretedimpresa.it",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9890130576932457,
"lm_q1q2_score": 0.8520725317968266,
"lm_q2_score": 0.86153820232079,
"openwebmath_perplexity": 373.11161112116935,
"openwebmath_score": 0.6303600668907166,
"tags": null,
"url": "http://hefc.rciretedimpresa.it/absolute-maximum-and-minimum-calculator-on-interval.html"
} |
'max_index', return max_index + min. Infinite Discontinuities: both one-sided limits are infinite. This lesson will focus on the maximum and minimum points. Bolzano's proof consisted of showing that a continuous function on a closed interval was bounded, and then showing that the function attained a maximum and a minimum value. (1 point) Let g(s) = i on the interval [0, 1. These extreme values are obtained, either on a relative extremum point within the interval, or on the endpoints of the interval. (𝑓𝑥)=5 Two solutions. Pick the largest and smallest. This corresponds to zero Kelvin, or minus 273. We know that the absolute max/min values of f(x) will occur either at an endpoint or a critical number. summary_interval # How often to write checkpoints (rounds up to the nearest statistics # interval). Calculate the range for your confidence statistics. Audio dithering. This is defined everywhere and is zero at $\ds x=\pm \sqrt{3}/3$. Looking first at $\ds x=\sqrt{3}/3$, we see that $\ds f(\sqrt{3}/3)=-2\sqrt{3}/9$. The maximum will occur at the highest f (x) f ( x) value and the minimum will occur at the lowest f (x) f ( x) value. The sample maximum and minimum provide a non-parametric prediction interval: in a sample from a population, or more generally an exchangeable sequence of random variables, each observation is equally likely to be the maximum or minimum. Sometimes it's important to consider points which are only largest or smallest in small parts of a graph. f x x3 2 on 3,1 > @ 15. find the absolute maximum and absolute minimum values of the fuction f(x)=2x-13ln(3x) on interval [1,8] 2. Time needed: 10 minutes. The calculators will allow you to convert any heart rate between 63% and 102% of your maximum heart rate to a percentage of your VO2max , or any percentage of VO2max. X Values: [a,b] Y Values: / [c,d] Result: [e,f] [1,3] / [2,4] [. On a closed interval these points are referred to as absolute or global minimum/maximum points. f(5) = (5)^4 + 8*(5)^3 | {
"domain": "rciretedimpresa.it",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9890130576932457,
"lm_q1q2_score": 0.8520725317968266,
"lm_q2_score": 0.86153820232079,
"openwebmath_perplexity": 373.11161112116935,
"openwebmath_score": 0.6303600668907166,
"tags": null,
"url": "http://hefc.rciretedimpresa.it/absolute-maximum-and-minimum-calculator-on-interval.html"
} |
these points are referred to as absolute or global minimum/maximum points. f(5) = (5)^4 + 8*(5)^3 -32*(5)^2 = 825 Thus function has absolute minimum value at x = 2 and absolute maximum value at x = 5 in the interval [-1, 5]. By practicing these kinds of problems you can understand this topic clearly. Thus if one has a sample {, …,}, and one picks another observation +, then this has / (+) probability of being the largest value seen so far. Calculus Refresher by Paul Garrett. Relative Minimum - The lowest point on an interval of a curve. It is a measure of your capacity for aerobic work and can be a predictor of your potential as an endurance athlete. The maximum acceleration attained on the interval 03ddt by the particle whose velocity is given by v t t t t( ) 3 12 4 M 32 is A)9 B)12 C)14 D)21 E)40 3. f(x) = x^2 + 250/x on the open interval (0,infinity ) I know that the absolute max is the answer NONE but I can not figure out the absolute min can someone help please thanks. Local Extreme Values of a Function Let c be an interior point of the domain of the. An extreme value, or extremum (plural extrema), is the smallest (minimum) or largest (maximum) value of a function, either in an arbitrarily small neighborhood of a point in the function's domain — in which case it is called a relative or local extremum — or on a given set contained in the domain (perhaps all of it) — in which case it is called an absolute or global extremum (the latter. To find the extreme values of a function (the highest or lowest points on the interval where the function is defined), first calculate the derivative of the function and make a study of sign. It can be used as a worksheet function (WS) in Excel. A continuous function f(x) on a closed and bounded interval [a,b] has both an absolute min-imum and an absolute maximum on the interval. Locate the maximum or minimum points by using the TI-83 calculator under and the “3. (c) Find all intervals on which the graph of f is concave up and | {
"domain": "rciretedimpresa.it",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9890130576932457,
"lm_q1q2_score": 0.8520725317968266,
"lm_q2_score": 0.86153820232079,
"openwebmath_perplexity": 373.11161112116935,
"openwebmath_score": 0.6303600668907166,
"tags": null,
"url": "http://hefc.rciretedimpresa.it/absolute-maximum-and-minimum-calculator-on-interval.html"
} |
TI-83 calculator under and the “3. (c) Find all intervals on which the graph of f is concave up and also has positive slope. We know that the absolute max/min values of f(x) will occur either at an endpoint or a critical number. Q: Determine the absolute maximum and minimum values of the function on the given interval. Find the absolute maximum and minimum values on theinterval: f(x) = x - 2cosx [-pi, pi] f ' (x) = 1+2sinx f (-pi)= -pi - 2cos(-pi) = -pi - 2. 5] The calculator performs interval arithmetic operations and computers interval version of mathematical functions. Finding Minimums and Maximums. These extreme values are obtained, either on a relative extremum point within the interval, or on the endpoints of the interval. By using this website, you agree to our Cookie Policy. Then press Ctrl+Shift+Enter keys, and the largest absolute values will be displayed in the. a local (relative) minimum 5. To find the maximum and/or minimum on an interval, check the values at the critical points and at the ends of the interval. absolute minimum value at x = 2 is -48. We know that the absolute max/min values of f(x) will occur either at an endpoint or a critical number. The largest of the values from steps 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value. Example 2: Locate the value(s) where the function attains an absolute maximum and the value(s) where the function attains an absolute minimum, if they exist. We usually distinguish between local and global (or absolute) extreme values. An extreme value, or extremum (plural extrema), is the smallest (minimum) or largest (maximum) value of a function, either in an arbitrarily small neighborhood of a point in the function's domain — in which case it is called a relative or local extremum — or on a given set contained in the domain (perhaps all of it) — in which case it is called an absolute or global extremum (the latter. f(x) = 4x^3 – 6x^2 – 144x + 9, [-4, 5] 13. The | {
"domain": "rciretedimpresa.it",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9890130576932457,
"lm_q1q2_score": 0.8520725317968266,
"lm_q2_score": 0.86153820232079,
"openwebmath_perplexity": 373.11161112116935,
"openwebmath_score": 0.6303600668907166,
"tags": null,
"url": "http://hefc.rciretedimpresa.it/absolute-maximum-and-minimum-calculator-on-interval.html"
} |
is called an absolute or global extremum (the latter. f(x) = 4x^3 – 6x^2 – 144x + 9, [-4, 5] 13. The student familiar with the sum formula can easily prove that. (c) For any. If the absolute maximum occurs at an interior point, then we have found an absolute maximum in the open interval. These are the examples in the topic increasing and decreasing intervals. Maximum intervals — they don't exist. Let f be a function defined and. It is a greatest value in a set of points but not highest when compared to all values in a set. Sample size calculation for trials for superiority, non-inferiority, and equivalence. Local Extreme Values of a Function Let c be an interior point of the domain of the. Thus, to find the absolute maximum (absolute minimum) value of the function, we choose the largest and smallest amongst the numbers f(a), f(c 1 ), f(c 2. Absolute minimum/maximum _____ d. Before going to class, some students have found it helpful to print out Purplemath's math lesson for that day's topic. These extreme values are obtained, either on a relative extremum point within the interval, or on the endpoints of the interval. For each x value: Determine the value of f '(x) for values a little smaller and a little larger than the x value. f(x) = x4 – A: Plot the graph for f(x) in the interval for [-2, 0]. If you finish a job in less than 25% of the time allotted, you will be paid a Time Bonus, so try to finish as quickly as possible! The maximum Time Bonus is a 25% boost to your Base Reward. Generating random numbers Problem. Use the Stefan Boltzmann relationship (I = σT 4 where σ = 5. minimum" or "4. Find the maximum / minimum absolute values with Formulas. in some open interval containing c. The local maximum and minimum are the lowest values of a function given a certain range. A closed interval like [2, 5] includes the endpoints 2 and 5. B) No absolute extrema. (To make the distinction clear, sometimes the ‘plain’ maximum and minimum are called absolute maximum and | {
"domain": "rciretedimpresa.it",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9890130576932457,
"lm_q1q2_score": 0.8520725317968266,
"lm_q2_score": 0.86153820232079,
"openwebmath_perplexity": 373.11161112116935,
"openwebmath_score": 0.6303600668907166,
"tags": null,
"url": "http://hefc.rciretedimpresa.it/absolute-maximum-and-minimum-calculator-on-interval.html"
} |
the distinction clear, sometimes the ‘plain’ maximum and minimum are called absolute maximum and minimum. Step 6: As mentioned earlier, A (x) is a continuous function over the closed, bounded. Finding the absolute max and min is a snap. Let this index be ‘max_index’, return max_index + min. checkpoint_interval # Where to write out summaries. It is important to understand the difference between the two types of minimum/maximum (collectively called extrema) values for many of the applications in this chapter and so we use a variety of examples to help with this. (c) Find all intervals on which the graph of f is concave up and also has positive slope. Maxima and minima are points where a function reaches a highest or lowest value, respectively. Let Purplemath help you always be prepared! Go to the lessons!. sin(x * y) I have an interval for x [-1, 1] and y [-1, 1]. 487] Calculating confidence intervals: Calculating a confidence interval involves determining the sample mean, X̄, and the population standard deviation, σ, if possible. summary_interval = FLAGS. The maximum will occur at the highest value and the minimum will occur at the lowest value. A point at which a function attains its maximum value among all points where it is defined is called a global (or absolute) maximum. Since the function is not defined for some open interval around either c or d, a local maximum or local minimum cannot occur at this point. f x x x 32 3 on 3,1> @ 12. Similar topics can also be found in the Calculus section of the site. between -30 to 20 function is decreasing because there are no local minima and maxima in between them. Let this index be ‘max_index’, return max_index + min. Therefore, f achieves its absolute minimum of −14 at x = −1 and its absolute maximum of 6 at both x = 1 and x = 4. 80 at x = –1. The absolute maximum value of f x x x( ) 3 12 M 32 on the closed interval > @M2, 4 occurs at x = A) 4 B) 2 C) 1 D) 0 E) M2 2. Mean number of days ≥ 30, 35 or 40 °C The average | {
"domain": "rciretedimpresa.it",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9890130576932457,
"lm_q1q2_score": 0.8520725317968266,
"lm_q2_score": 0.86153820232079,
"openwebmath_perplexity": 373.11161112116935,
"openwebmath_score": 0.6303600668907166,
"tags": null,
"url": "http://hefc.rciretedimpresa.it/absolute-maximum-and-minimum-calculator-on-interval.html"
} |
@M2, 4 occurs at x = A) 4 B) 2 C) 1 D) 0 E) M2 2. Mean number of days ≥ 30, 35 or 40 °C The average number of days in the period when the daily maximum air temperature was equal to, or exceeded 30, 35 or 40 °C. A point at which a function attains its minimum value among all points where it is defined is a global (or absolute) minimum. 14 If the first dose of recombinant zoster vaccine (Shingrix) is administered to someone 18-49 years of age, the dose does not need to be repeated. From the graph you can see that is has a maximum at (3, 27) and a minimum at (1. Free Maximum Calculator - find the Maximum of a data set step-by-step This website uses cookies to ensure you get the best experience. An absolute maximum or minimum can occur, however, because the definition requires that the point simply be in the domain of the function. 1 absolute maximum or minimum. f (1) = −5 f ( 1) = - 5. Question 203087: Find the absolute maximum and absolute minimum values of the function below. In this module you will be asked to calculate the sample size for 6 situations. summary_interval # How often to write checkpoints (rounds up to the nearest statistics # interval). 2 Maximum and Minimum on an Interval. Theorem (Extreme Value Theorem) If f is continuous on a closed interval [a,b], then f attains an absolute maximum value f(c) and an absolute minimum value f(d) at some. The absolute minimum of the function f(x) = x2-9 on the interval - 4 Sxs 3 has a value of (Simplify your answer. In the single-variable case, it is known, by the Extreme Value Theorem, that if f is continuous on a closed interval [a;b], then it has has an absolute maximum and an absolute minimum on [a;b]. Any global maximum or minimum must of course be a local maximum or minimum. (1 point) Let g(s) = i on the interval [0, 1. Absolute & Local Minimum and Maximum Values - Relative Extrema, Critical Numbers / Points Calculus - Duration: 1:10:05. The student familiar with the sum formula can easily prove that. so | {
"domain": "rciretedimpresa.it",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9890130576932457,
"lm_q1q2_score": 0.8520725317968266,
"lm_q2_score": 0.86153820232079,
"openwebmath_perplexity": 373.11161112116935,
"openwebmath_score": 0.6303600668907166,
"tags": null,
"url": "http://hefc.rciretedimpresa.it/absolute-maximum-and-minimum-calculator-on-interval.html"
} |
Calculus - Duration: 1:10:05. The student familiar with the sum formula can easily prove that. so minimum value of f(x) in interval [0. pow(y, 2)) * -1 return math. Extreme values ©2010 Iulia & Teodoru Gugoiu - Page 1 of 2 3. For instance, in the example at. How to use absolute minimum in a sentence. the absolute minimum age of 50 years when evaluating records retrospectively. Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. (a) Interval : [—8, 0]. The smallest y -value is the absolute minimum and the largest y -value is the. 1 absolute maximum or minimum. Fermat's Theorem. Generating random numbers Problem. Help with finding absolute max/min values for a function. in some open interval containing c. Local minima and maxima (First Derivative Test) by Paul Garrett is licensed under a Creative Commons Attribution-Noncommercial-ShareAlike 4. Let f be a function defined and. $f(x) = x^5 - x^3 + 2$, $-1 \leqslant x \leqslant 1$. 1 absolute maximum or minimum. The absolute max occurs at S = The absolute min occurs at S =. (a) fxc 0 at x 3, 1, 4 f c changes from positive to negative at 3 and 4. For each x value: Determine the value of f '(x) for values a little smaller and a little larger than the x value. between -30 to 20 function is decreasing because there are no local minima and maxima in between them. have both an absolute maximum and an absolute minimum. Find the extreme values of f on the boundary of D. An absolute extremum is an absolute maximum or an absoute minimum, and absolute extrema are absolute maximum and absolute minimum. Zoom in on the interval [-2,2] using the x-axis. In worst case, if all intervals are from 'min' to 'max', then time complexity becomes O((max-min+1)*n) where n is number of intervals. The MAX function is a built-in function in Excel that is categorized as a Statistical Function. Then press Ctrl+Shift+Enter keys, and the largest absolute values will be displayed in the. minimum” | {
"domain": "rciretedimpresa.it",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9890130576932457,
"lm_q1q2_score": 0.8520725317968266,
"lm_q2_score": 0.86153820232079,
"openwebmath_perplexity": 373.11161112116935,
"openwebmath_score": 0.6303600668907166,
"tags": null,
"url": "http://hefc.rciretedimpresa.it/absolute-maximum-and-minimum-calculator-on-interval.html"
} |
Then press Ctrl+Shift+Enter keys, and the largest absolute values will be displayed in the. minimum” or “4. Find the absolute maximum and absolute minimum values of f on the given interval. Find the absolute maximum and minimum values on theinterval: f(x) = x - 2cosx [-pi, pi] f ' (x) = 1+2sinx f (-pi)= -pi - 2cos(-pi) = -pi - 2. The calculator will find the critical points, local and absolute (global) maxima and minima of the single variable function. Find the absolute maximum and. By using this website, you agree to our Cookie Policy. denbal87 New member. | {
"domain": "rciretedimpresa.it",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9890130576932457,
"lm_q1q2_score": 0.8520725317968266,
"lm_q2_score": 0.86153820232079,
"openwebmath_perplexity": 373.11161112116935,
"openwebmath_score": 0.6303600668907166,
"tags": null,
"url": "http://hefc.rciretedimpresa.it/absolute-maximum-and-minimum-calculator-on-interval.html"
} |
# How to round 0.4999… ? Is it 0 or 1?
If you want to round the repeating decimal 0.4999... to a whole number what is the right answer? Is it 0 or 1?
On the one hand you only look at the first digit when you round numbers which in this case is 4 so the answer should be 0.
On the other hand 0.4999... is only another representation for 0.5 which makes the result 1.
My question
Which of these rules wins? (My gut feeling is that the most consistent result should be 1)
Edit
Just for clarification: What is meant by rounding here is replacing a fractional decimal number by one with fewer digits with the "Round half up"-Tie-breaking method.
-
It appears that the rule is inconsistent. – Isaac Solomon Oct 21 '12 at 9:04
Doesn't that depend on what type of rounding you are using? If you're using Bankers Rounding, you would round towards 0. – Fake Name Oct 21 '12 at 12:22
@FakeName: Fair enough. What is obviously meant here is replacing a fractional decimal number by one with fewer digits with the "Round half up"-Tie-breaking method: en.wikipedia.org/wiki/Rounding – vonjd Oct 21 '12 at 13:50
This question should NOT be closed; it has a definitive answer - with references "out there" to support it. At the point when this was closed, the issue on the table (discussion!) was if 0.499... was 0.5 or not: reference is en.wikipedia.org/wiki/0.999... – Richard Sitze Oct 22 '12 at 0:08
I agree with Richard. Given that we have a method for rounding midpoint values, and the author of the question has supplied a method, it follows that this question has an answer. – Doug Spoonwood Oct 22 '12 at 2:52
It's $1$, because $0.49\ldots$ is the same as $0.5$. If rounding is to be well-defined, it can't map one real number to two integers, so whatever it maps $0.49\ldots$ to, it better maps it to the same integer as $0.5$. You could round both to $0$, of course, but that wouldn't then be the way we usually round. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.974821163419856,
"lm_q1q2_score": 0.8520690000927736,
"lm_q2_score": 0.8740772482857833,
"openwebmath_perplexity": 432.2086184790631,
"openwebmath_score": 0.8157265186309814,
"tags": null,
"url": "http://math.stackexchange.com/questions/217893/how-to-round-0-4999-is-it-0-or-1"
} |
What this shows you is that rounding doesn't commute with limits, i.e. there's a difference between find the limit of a sequence and then rounding, and rounding first and then finding the limit. As you correctly observed, all values $a_n = 0.4\underbrace{9\ldots 9}_{n\text{ times}}$ are rounded down to zero. Thus, $$\lim_{n\to\infty} \text{round }(a_n) = 0$$ On the other hand, $\lim_{n\to\infty} a_n = 0.5$, and thus $$\text{round }\left(\lim_{n\to\infty} a_n\right) = 1$$
There's another word for functions which don't commute with limit - they're called non-continuous. So what you have discovered is simply that rounding is not a continuous function.
-
Perhaps I'm nitpicking, but is the result of rounding the limit perhaps $1$ instead? – pimvdb Oct 21 '12 at 10:05
@pimvdb: I went ahead and edited the answer, since it's obviously what was intended. (And so many people seem to agree.) – ShreevatsaR Oct 21 '12 at 14:30
+1 especially for pointing out the connection with continuity. – Neal Oct 21 '12 at 16:00
@pimvdb Indeed the limit should be $1$, and I wouldn't call pointing out blatant errors nitpicking ;-) Thanks for noticing, and thanks to ShreevatsaR for fixing this! – fgp Oct 21 '12 at 19:42
@jmoreno - but it is, in fact the same as 0.5: en.wikipedia.org/wiki/0.999... – Richard Sitze Oct 22 '12 at 0:01 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.974821163419856,
"lm_q1q2_score": 0.8520690000927736,
"lm_q2_score": 0.8740772482857833,
"openwebmath_perplexity": 432.2086184790631,
"openwebmath_score": 0.8157265186309814,
"tags": null,
"url": "http://math.stackexchange.com/questions/217893/how-to-round-0-4999-is-it-0-or-1"
} |
For $0\le x\le 1$, we round $x$ to $1$ if $x\ge \frac12$ and to $0$ if $x<\frac12$ (though there are many conventions, see e.g. Wikipedia on rounding; the section "Table-maker's dilemma" a bit further down may also be interesting). Since $0.4\bar9=\frac12$, we should round to $1$. Another way of looking at this is that we always consider only the standard decimal expansion (i.e. we prefer $\bar0$ over $\bar 9$), and we are allowed to treat the first decimal as $4$ only if we know that it cannot turn out as $5$ "later". Thus if an inexact measurement gives us that $0.495\le x\le0.5$, we cannot say definitely, what $\operatorname{round}(x)$ should be (we could if the measurement resulted in $0.495<x<0.5$). This is not different from the fact that we cannot say definitely what $\operatorname{round}(x)$ should be if our measuremen merely says that $0.4997<x<0.5003$.
-
You're right. Since $.4\bar{9}=.5,$ if you want your rounding function to be well-defined you'll have to require an exception: round based on the first digit after the one you're rounding to, unless it's a $4$ followed by infinitely many $9$s.
-
It is not obvious that 0.5 should be rounded to 1. Obviously 0.49999... should be treated the same because it is the same number. Wikipedia rounding article
-
$1 - 0.5 =0.5 \quad$ and $\quad 1 - 0.4999\ldots = 0.500\ldots = 0.5 \quad$ so $\quad 0.5 = 0.4\overline9$
Well, the above line says the proof / intuition. From our information... $0.5$ well rounds to $1$.
-
Are you sure that $1-0.4999\ldots\ne 0.500\ldots1$? – MJD Nov 18 '12 at 16:46
@MJD: Not sure if serious... – Clive Newstead Nov 18 '12 at 16:47
@MJD: In 0.500...1, the zeroes never end, so the $1$ never comes ;) – Parth Kohli Nov 18 '12 at 16:52
I find very convenient to define rounding as follows: $$round(x) = \left \lfloor x + \frac{1}{2}\right \rfloor$$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.974821163419856,
"lm_q1q2_score": 0.8520690000927736,
"lm_q2_score": 0.8740772482857833,
"openwebmath_perplexity": 432.2086184790631,
"openwebmath_score": 0.8157265186309814,
"tags": null,
"url": "http://math.stackexchange.com/questions/217893/how-to-round-0-4999-is-it-0-or-1"
} |
Considering this rule, what follows is understanding the context. If you are working with a computer, you have only a fixed number of decimal positions to work with, in which case, if x=0.4999999999999999 (16 decimal positions), then
round(x)=floor(0.4999999999999999 + 0.5)=
=floor(0.9999999999999999)=
=0
Of course, if you are talking about a real number, then 0.4999... equals 0.5, in which case $round(x)=1$.
So, the real question here is: are you working with a fixed number of decimal positions or not? If the answer is 'yes', then definitely round(x)=0. If the answer is 'no', then you are talking about a number which is equal to 0.5, and $round(x)=1$
-
@fgp answered this well but there is another aspect that I don't think has been touched on in any of the answers. What is the type of thing that we should consider as an input to the process of rounding? If it is a real number, we are unable to apply the rule without a specific digit sequence representing that number. If instead we interpret rounding as an operation on a possibly infinite digit sequence, we can now apply the rule, but its result has no meaning for general real numbers. One way out which has been discussed is to make a bijection between digit sequences and reals by disallowing digit sequences ending in $999...$, and this view effectively unasks the question by taking the position that $0.4999...$ is not a valid representation of a real number, but I think this is an overly burdensome technicality. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.974821163419856,
"lm_q1q2_score": 0.8520690000927736,
"lm_q2_score": 0.8740772482857833,
"openwebmath_perplexity": 432.2086184790631,
"openwebmath_score": 0.8157265186309814,
"tags": null,
"url": "http://math.stackexchange.com/questions/217893/how-to-round-0-4999-is-it-0-or-1"
} |
I think an easier way to look at it is just to say that rounding is an operation that applies to finite digit sequences and not to real numbers in general nor to infinite digit sequences. When we say something like "$\pi$ rounds down to $3$" it is a lazy way of saying that all sufficiently precise approximations do. But when we say "$0.5$ rounds up to $1$", exactly that is meant, because we don't mean to suggest that $0.4 + \sum_{k=2}^{n}{9 \cdot 10^{-k}}$ will round up to $1$ for sufficiently large $n$. And without context, if it is said about a real number $x$ that "$x$ rounds up to $1$" then that can be read as "$x$ has a finite decimal representation which rounds up to $1$, or $x$ has no finite decimal representation and all sufficiently precise finite decimal approximations to $x$ round up to $1$". In this way we can meaningfully discuss rounding real numbers in terms of rounding finite digit sequences, and without explicitly addressing the problem of sequences ending in $999...$.
-
If you consistently use the round-up half method you don't always look at the first digit of the numeral when rounding numbers. You go by the method. The method allows that if we have anything else than a half-way value, then you can round by the rule "if (working in base 10) the first digit of the numeral after the decimal point belongs to {0, 1, 2, 3, 4}, then round down, if the first digit of the decimal point belongs to {5, 6, 7, 8, 9}, then round up." But, that rule does not say anything about half-way values. Going by q=FLOOR(y + .5) we can reason as follows: q=FLOOR(.499... + 5)=FLOOR(.999...)=FLOOR(1)=1, since .999...=1. So, that method yields that .499... rounds to 1.
-
It is the same as 0.5 by definition of decimal numbering. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.974821163419856,
"lm_q1q2_score": 0.8520690000927736,
"lm_q2_score": 0.8740772482857833,
"openwebmath_perplexity": 432.2086184790631,
"openwebmath_score": 0.8157265186309814,
"tags": null,
"url": "http://math.stackexchange.com/questions/217893/how-to-round-0-4999-is-it-0-or-1"
} |
-
It is the same as 0.5 by definition of decimal numbering.
-
And rounding takes by definition the first digit which is 4 in this case. So this is exactly where the two definitions collide! – vonjd Oct 21 '12 at 18:36
@vonjd I disagree that rounding by definition takes the first digit and then rounds. I ask you to try and round 1/3 without converting it into a decimal. Do you not round 1/3 down to 0, since it comes as clear that 1/3 lies closer to 0 than to 1? Or I suggest you round 12/11 to the nearest whole number. Or round 155/66 to the nearest whole number. Finding the multiples of a denominator d will help here, and what half of d equals. If the denominator equals 7, then if the numerator of "the remainder" equals 1, 2, or 3 we round down, if the numerator equals 4, 5, or 6 we round up. – Doug Spoonwood Oct 21 '12 at 19:17
E. G. Say we have 13/7. 13/7=7/7+6/7, correct? So, 13/7 has "remainder" of 6/7, or in other words it has a fractional part (search that term on Wikipedia) of 6/7. Thus, 13/7 rounds up to 14/7 (6/7 lies closer to 7/7 than to 0/7). You could also try and rounding numbers in binary, trinary, or any n-ary base system. E. G. round 1.34523 in base 7. In base 7, 1.4142 rounds up to 2. So, I simply do not see how the digit makes the difference here. More goes on than that! – Doug Spoonwood Oct 21 '12 at 19:23
@vonjd this is an exception. rounding takes first digit EXCEPT for values like this, where fraction is written in limitless form. – Suzan Cioc Oct 21 '12 at 19:25
This is just an obersvation that I was missing in the many answers already given. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.974821163419856,
"lm_q1q2_score": 0.8520690000927736,
"lm_q2_score": 0.8740772482857833,
"openwebmath_perplexity": 432.2086184790631,
"openwebmath_score": 0.8157265186309814,
"tags": null,
"url": "http://math.stackexchange.com/questions/217893/how-to-round-0-4999-is-it-0-or-1"
} |
This is just an obersvation that I was missing in the many answers already given.
While there is absolutely no doubt that the mathematical function $\Bbb R\to\Bbb Z$ defined by $x\mapsto\lfloor x+\frac12\rfloor$ when applied to the real (in fact rational) number represented by the repetitive infinite decimal sequence $0.4\overline9$ takes as value the integer$~1$, saying that this is what happens when rounding the repeating decimal $0.4999\ldots$ is misleading. That is because "rounding" suggests an operation done during actual computation. This ignores the often overlooked fact that it is impossible to do arithmetic computations with infinite decimals. This does not mean of course that there isn't a class of infinite decimals with which computations are possible; for instance if all decimals are repeating, then one can replace them by rational numbers and compute with those, which is definitely possible. However the misconception is that, because we can do arithmetic on numbers with any fixed number of decimal places and the procedures for doing so are always basically the same, therefore one can apply these same procedures to infinite decimals. However just looking at these procedures will reveal that this is plainly false: all of these procedures need at some point (often right at the beginning) to locate the final decimal position, and for an infinite decimal such a position simply does not exist. This means that in practice whenever one needs rounding decimal numbers as a computational operation, then those decimals are necessarily finite, and the number $0.4999\ldots$ simply cannot occur. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.974821163419856,
"lm_q1q2_score": 0.8520690000927736,
"lm_q2_score": 0.8740772482857833,
"openwebmath_perplexity": 432.2086184790631,
"openwebmath_score": 0.8157265186309814,
"tags": null,
"url": "http://math.stackexchange.com/questions/217893/how-to-round-0-4999-is-it-0-or-1"
} |
In fact there is no effective system for computing with (general) true real numbers at all, for the simple reason that the majority of the real numbers cannot be represented in any way (a concrete representation having to be, by the nature of things in the real world, finite). One can however restrict to the (countable) subset of real numbers that do allow some kind of representation (which is less brutal that restricting to some arbitrarily chosen finite subset of numbers declared representable, as is done in numeric computation), and then arithmetic and other kinds of computation do become possible, and indeed can be done without having to introduce approximations. This has been studied in constructive mathematics, but it turns out that for this purpose representing real numbers by infinite decimals, which might be produced on demand by a finite algorithm, is not a viable option. Even just finding the first decimal of the sum of two numbers so given by infinite decimals cannot be assured by an algorithm; this is in fact closely related to issues invoked in this question and the related $0.9999\ldots$ problematic. This in contrast to computations with formal power series with integer (or rational) coefficients: these suffer from the same "most elements cannot be represented" problem as the real numbers, but if one sticks to series whose coefficients can be computed by an algorithm, then all arithmetic (and many other) operations on them can be performed algorithmically without problem. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.974821163419856,
"lm_q1q2_score": 0.8520690000927736,
"lm_q2_score": 0.8740772482857833,
"openwebmath_perplexity": 432.2086184790631,
"openwebmath_score": 0.8157265186309814,
"tags": null,
"url": "http://math.stackexchange.com/questions/217893/how-to-round-0-4999-is-it-0-or-1"
} |
A more workable way to represent real numbers is by a sequence of definite approximations (rational numbers with a given maximal error), with some kind of guarantee that the error tends to $0$. But even then there are sacrifices one has to make: it is impossible to computations with discontinuous functions (like the floor or round-to-integer operations): in the constructive theory of real numbers there only exist continuous functions. So all of this is rather far removed from what one does in numeric computation, where real numbers are necessarily approximated, the number of decimals finite, and the question of how to round $0.4\overline9$ moot; the rule for rounding of looking at the first digit to be discarded in the rounding is perfectly valid in practice.
-
If we round c to d, then c approximately equals d. If d also consists of an integer, then c lies closer to d than any other integer i. In other words for all integers i such that i does not equal d, ABS(ci-)>ABS(cd-), where ABS(ci-) indicates the number by which c and i differ by (the absolute value of the difference of c an i). This implies that we can use the distance metric between real numbers to see how rounding works as follows.
Consider the set S=={[l, lu+2/), (lu+2/, u]} where ul-=1. In other words, l and u differ by 1, and we consider the set of all numbers between x and y except for the midpoint m==lu+2/ between them. We can use the metric function D(x,y)=|xy-|=ABS(xy-), where x belongs to S and y belongs to {l, u}, to determine how to round all members of S by following rule:
If D(x, l)>D(x, u), then round to u. If D(x, u)>D(x, l), then round to l.
Since .5=.49999..., it follows that D(.49999..., 1)=D(.49999...., 0)=.4999...=.5. Thus, the distance function D(x, y)=ABS(xy-) does not give us any guide as to how to round here.
Consequently, we have several different choices that we can make which will not go against our intuition of rounding as follows: | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.974821163419856,
"lm_q1q2_score": 0.8520690000927736,
"lm_q2_score": 0.8740772482857833,
"openwebmath_perplexity": 432.2086184790631,
"openwebmath_score": 0.8157265186309814,
"tags": null,
"url": "http://math.stackexchange.com/questions/217893/how-to-round-0-4999-is-it-0-or-1"
} |
1. Choose to round .49999... to 1 (maybe you value larger numbers in general),
2. Choose to round .49999... to 0 (maybe you value smaller numbers in general),
3. Choose to allow .49999... to get rounded to both as one pleases or sees fit at particular points in time, or
4. Choose to round .49999... to neither 0 nor 1 (maybe you don't want to make a choice here).
None of those possibilities win or lose until we have a standard to measure a winner or loser for the rules. There also doesn't exist a "right answer" here unless we have a means to determine what properties a "right answer" has in this context. I suspect that many people would resist 3. with the possibility that one could round .4999... to 1 at one time, and then round .4999... to 0 at another as one pleases, thus indicating arbitrariness. But, it does not follow that such comes as an improper procedure unless one first rates such "arbitrariness" as bad, and something so utterly awful and terrible that we or others must never, ever engage in such "arbitrariness". Perhaps many people would see 4. as "evasive", but unless there exists a logical basis to picking either 0 or 1, anyone accused of "evasiveness" by picking 4. could very well respond to such a charge, that anyone who sees choosing 4. as indicating "evasiveness", insists dogmatically that all such questions absolutely must have answers, and/or that the people claiming "evasiveness" have copped out of making choices on the basis of logic as much as possible.
So, what sort of choice do you want to make?
-
As a programmer, 0.4999... is closer to 0 than it is to 1, so I would always round down to 0.
Keep it simple! Rounding is always to the closest whole number. You only need to apply special cases when there are two closest whole numbers.
But there are situations where this would be the wrong choice, so you need to do some research every time you apply rounding function to a value. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.974821163419856,
"lm_q1q2_score": 0.8520690000927736,
"lm_q2_score": 0.8740772482857833,
"openwebmath_perplexity": 432.2086184790631,
"openwebmath_score": 0.8157265186309814,
"tags": null,
"url": "http://math.stackexchange.com/questions/217893/how-to-round-0-4999-is-it-0-or-1"
} |
For example, the tax department in the country I live in does not always acknowledge cents. Under legislation, there are times when the only valid rounding function is floor.
-
I think we all agree that $0.4999$ should round down to $0$. The question is whether $.499999999......$ should (with infinitely many $9$s). – Jason DeVito Oct 21 '12 at 18:16
How close $0.4999...$ is to $0$ or $1$ is independent on whether or not you are a programmer. In fact, it is the same distance from $0$ and $1$. If the sequence of digits were to terminate after finitely many (e.g. in computing) then fine, but here we have infinite precision and the number is exactly equal to $0.5$. – Clive Newstead Oct 21 '12 at 18:20
@AbhiBeckert: The case in point is that it is indeed equal to 0.5, as is 0.999... equal to 1: en.wikipedia.org/wiki/0.999... – vonjd Oct 21 '12 at 18:32
Dear Abhi, I have downvoted your answer, since it it seems to be premised on a confusion with regard to the fact that $0.4999\cdots$ (infinitely many $9$'s) is equal to $0.5$. Regards, – Matt E Oct 21 '12 at 19:56
@AbhiBeckert: If you accept that $0.4999...$ is a real number, and you also claim it's less than $0.5$, then (1) you should be able to say what is the distance between them (for any two real numbers $x<y$, their distance is another real number $y-x$: and no, "infinitesimal" is not a number), and (2) you should be able to exhibit a number that lies between them (for any $x<y$, there exist infinitely many real numbers between, e.g. $x<(x+y)/2<y$). The fact that there is no number between $0.4999...$ and $0.5$ should convince you that they are two representations of exactly the same number. – ShreevatsaR Oct 23 '12 at 5:16 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.974821163419856,
"lm_q1q2_score": 0.8520690000927736,
"lm_q2_score": 0.8740772482857833,
"openwebmath_perplexity": 432.2086184790631,
"openwebmath_score": 0.8157265186309814,
"tags": null,
"url": "http://math.stackexchange.com/questions/217893/how-to-round-0-4999-is-it-0-or-1"
} |
# How to translate mathematical intuition into a rigorous proof?
I've seen a lot of questions about how to develop mathematical intuition, but often I have the opposite problem.
Several times I have run into a situation where I want to solve a math problem, and I play around with it until my intuition reaches the point where I have a good idea of how the complete proof might be structured. But then when it comes to actually write out the proof, I have trouble translating this intuition into a rigorous proof. I want to give the following example, from this PDF of Putnam training problems.
1.13. Prove that for every $n\ge 2$, the expansion of $(1+x+x^2)^n$ contains at least one even coefficient.
When first thinking about this problem, I wanted to look at the polynomials $\pmod{2}$ so that "even coefficient" is simplified to "zero coefficient."
Then from doing a few computations, I notice a pattern.
\begin{align*} (1+x+x^2)^2&=(1+x+x^2)+x(1+x+x^2)+x^2(1+x+x^2)\\ &=1+x+x^2\\ &+x+x^2+x^3\\ &+x^2+x^3+x^4\\ &=1+x^2+x^4 \end{align*}
I notice that multiplying a polynomial by $(1+x+x^2)$ is like adding that polynomial with itself three times, each shifted over. The $x$ shifts it over by one place and the $x^2$ shifts it over by two places.
Continuing this pattern, we can get the coefficients of $(1+x+x^2)^3$ as follows:
\begin{array}{ccccccc} 1&0&1&0&1&&\\ &1&0&1&0&1&\\ &&1&0&1&0&1\\ \hline 1&1&0&1&0&1&1 \end{array}
So $(1+x+x^2)^3\equiv 1+x+x^3+x^5+x^6 \pmod{2}$
Let's make a triangle of a few more results:
\begin{array}{cccccccccccc} 0:&&&&&&1&&&&&\\ 1:&&&&&1&1&1&&&&\\ 2:&&&&1&0&1&0&1&&&\\ 3:&&&1&1&0&1&0&1&1&&\\ 4:&&1&0&0&0&1&0&0&0&1&\\ 5:&1&1&1&0&1&1&1&0&1&1&1\\ \end{array}
We see that the structure of our problem is essentially the same as an elementary cellular automaton (specifically, rule 150). I will come back to this later. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211582993981,
"lm_q1q2_score": 0.8520689940182796,
"lm_q2_score": 0.8740772466456689,
"openwebmath_perplexity": 227.65682371870014,
"openwebmath_score": 0.8129568099975586,
"tags": null,
"url": "https://math.stackexchange.com/questions/2830788/how-to-translate-mathematical-intuition-into-a-rigorous-proof/2831036"
} |
I notice another peculiar pattern with the $1$st, $2$nd, and $4$th rows: There are only $1$s on the ends and in the center. Perhaps this pattern continues for all powers of $2$. And it does, which is not hard to prove with induction.
Claim: $\forall n\ge 0,\ (1+x+x^2)^{2^n}\equiv 1+x^{2^n}+x^{2^{n+1}}\pmod{2}$
Base case: $(1+x+x^2)^{2^0}\equiv 1+x+x^2\equiv 1+x^{2^0}+x^{2^1}\pmod{2}$
Inductive Step:
\begin{align*} (1+x+x^2)^{2^n}&\equiv ((1+x+x^2)^{2^{n-1}})^2\\ &\equiv (1+x^{2^{n-1}}+x^{2^n})^2\\ &\equiv 1+x^{2^n}+x^{2^{n+1}}\pmod{2} \end{align*}
This is where the intuition comes in that is hard for me to express rigorously. On one of the $2^n$th rows, look at the $0$ equidistant from the leftmost $1$ and the $1$ in the center. If the rightmost $1$ wasn't there, then that $0$ would stay a $0$ forever because of symmetry: any effect from the left is cancelled by an effect from the right.
However, the rightmost $1$ does exist, so it will eventually change that $0$. But since effects are local in this automaton, it will require over $2^n$ more rows until the rightmost $1$ affects the $0$ we are looking at (there are over $2^n$ spaces between them). In other words, we have shown that the $0$ will stay a $0$ for all rows up to the $2^{n+1}$th row. Applying this logic on each power of two row, we show that there is always a zero coefficient from row $2$ onwards.
Is there a more rigorous way I can express those last two paragraphs? And in general, what advice do you have for translating intuition to a rigorous proof? | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211582993981,
"lm_q1q2_score": 0.8520689940182796,
"lm_q2_score": 0.8740772466456689,
"openwebmath_perplexity": 227.65682371870014,
"openwebmath_score": 0.8129568099975586,
"tags": null,
"url": "https://math.stackexchange.com/questions/2830788/how-to-translate-mathematical-intuition-into-a-rigorous-proof/2831036"
} |
• In your example you changed the question. Originally it asked for a $0$ in every row. Your version only talks about rows of the form $2^n$, but you show a very special form for those rows. Your induction is fine and proves what you wanted to prove, but it is not the original question. It might be a step along the way I can imagine (but I am speculating) that you can show a partial row starting with $1$, a bunch of $0$s and a $1$ will always have a $0$ somewhere. If you can do that you are done in conjunction with what you have done. – Ross Millikan Jun 25 '18 at 4:15
• @RossMillikan I did not change the question in any way. The observation I made about the $2^n$ rows was a stepping stone in my proof. Look at the last three paragraphs. – Riley Jun 25 '18 at 4:18
• You said you would come back to rule 150 but I don't think you did – Mark Jun 25 '18 at 4:49
• @Mark I meant that I will come back to framing the problem in terms of cellular automata, which I did when discussing the "effects" of the $1$s, and the locality of the automaton. – Riley Jun 25 '18 at 4:51
• [Just a comment as the question is about the automaton-intuition being turned into a proof.] Another way of thinking about this expression is, the coefficient of $x^m$ in the expansion shows the number of ways to achieve a total of $m$ by rolling a die with sides $0, 1, 2$ a total of $n$ times. To see why, try making a three by three times table model to see the result of squaring the expression; then, erase the base of $x$. – Benjamin Dickman Jun 25 '18 at 5:17
You have proved that these binary strings are symmetric so it suffices to prove the result for only the "half-strings." Imagine cutting your triangle down the middle. Let $x_n$ be the $n^{th}$ bit string, which is length $n+1$.
We have $x_0 = 1$, $x_1 = 11$, $x_2 = 101$, etc.
Now you can address the $j^{th}$ bit inside $x_n$ as $x_{n,j}$.
This addressing system will make it easier to talk more concretely about the action of the automaton. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211582993981,
"lm_q1q2_score": 0.8520689940182796,
"lm_q2_score": 0.8740772466456689,
"openwebmath_perplexity": 227.65682371870014,
"openwebmath_score": 0.8129568099975586,
"tags": null,
"url": "https://math.stackexchange.com/questions/2830788/how-to-translate-mathematical-intuition-into-a-rigorous-proof/2831036"
} |
You can pin down a definition of "locality of action" by saying that there is some function $f(a, b, c)$ such that $\forall n, j: x_{n,j} = f(x_{n-1, j-1}, x_{n-1,j}, x_{n-1,j+1})$ (with some additional nuisance conditions about the boundary). This means that the $j^{th}$ bit in the current string can be computed by neary-by bits in the previous string. Here $f$ represents the action of this rule 150 you mention.
Now use $f$ again to show $x_{n,j} = f(f(x_{n-2, j-2}, x_{n-2,j-1}, x_{n-2, j}), f(x_{n-2, j-1}, x_{n-2,j}, x_{n-2, j+1}), f(x_{n-2, j}, x_{n-2,j+1}, x_{n-2, j+2}))$
Now you can see when we repeat this procedure $k$ times we will end up with a trinary abstract syntax tree depth $k$. To avoid having to talk about the exact structure of the expression at the $k^{th}$ level, we can reason about this substitution process purely formally as a process operating on the (countably infinite) set of symbols $\{ \underline{f}, \underline{(}, \underline{)} \} \cup \{ \underline{x_{i,j}} \}_{(i,j) \in \mathbb{N}^2}$ (here the commas are not meant to be part of the symbol set, they are just there to separate symbols from each other). You can prove by induction, that for any $k$, that $x_{n,j}$ is computed by an expression using the symbols $\{ \underline{f}, \underline{(}, \underline{)} \} \cup \{\underline{ x_{n-k, j-k}},\underline{ x_{n-k, j-k+1}}, …,\underline{ x_{n-k, j+k-1}},\underline{ x_{n-k, j+k} }\}$ (the $2k+1$ bits in the $(n-k)^{th}$ string centered on the $j^{th}$ bit) intermixed with function applications of $f$.
Fix $n$ and let $k_n$ be the distance to the previous power of $2$. (So if $n=13$, $k_n=5$ since the previous power of $2$ is $8$). And define $j_n \equiv (n-k_n)/2$. This is the index of the central bit of the $(n-k_n)^{th}$ bit string. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211582993981,
"lm_q1q2_score": 0.8520689940182796,
"lm_q2_score": 0.8740772466456689,
"openwebmath_perplexity": 227.65682371870014,
"openwebmath_score": 0.8129568099975586,
"tags": null,
"url": "https://math.stackexchange.com/questions/2830788/how-to-translate-mathematical-intuition-into-a-rigorous-proof/2831036"
} |
You have demonstrated that those bits at indices $\{ (n-k_n, m) \}_{1 \le m \lt n }$ are all zero since $n-k_n$ is a power of $2$. So any well-formed expression using only the symbols $\{ \underline {f}, \underline{(}, \underline{)}\} \cup \{ \underline{x_{n-k_n, j_n-k_n}}, \underline{x_{n-k_n, j_n-k_n+1}}, …,\underline{x_{n-k_n, j_n+k_n-1}}, \underline{x_{n-k_n, j_n+k_n}}\}$ will evaluate to zero by $f(0,0,0) = 0$, and in particular the expression representing $x_{n,j_n}$ (created from the $k_n$ iterations of the formal symbol substitution procedure) evaluates to $0$.
So as $n$ increments, $x_{n,j_n}$ traces out the path of the central bits you were referring to, all of which are zero.
General thoughts about formalization: There are certain technical devices which are work horses of the formalization process. Two of them I have used in this proof:
• indexing time and space, with a numeric coordinate system
• encoding a time evolving system as function application
A third device which I used is not as common, and comes from computer science and logic:
Something that I have been realizing is that these devices are not obvious (at least to me). Humanity was around for a long time before the Cartesian coordinate system was discovered and used to formalize geometrical intuition. It took even longer to develop the idea that almost every mathematical object could be encoded as an intricate tower of sets (ZFC).
I guess for me it's a matter of building up a library of these devices and binding them to the correct intuitions by repeated use.
In certain areas like computer programming, there are very few tools for converting intuition into proof. There were a couple of very good devices created like
• loop invariants
• Hoare logic
but multithreaded and heap-allocating programs are still an active area of research. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211582993981,
"lm_q1q2_score": 0.8520689940182796,
"lm_q2_score": 0.8740772466456689,
"openwebmath_perplexity": 227.65682371870014,
"openwebmath_score": 0.8129568099975586,
"tags": null,
"url": "https://math.stackexchange.com/questions/2830788/how-to-translate-mathematical-intuition-into-a-rigorous-proof/2831036"
} |
but multithreaded and heap-allocating programs are still an active area of research.
• By the way, I have not verified the claim about the sequence actually evolving as the cellular automata $f$ but I'm trusting it's true. – Mark Jun 25 '18 at 7:01
• I think your definition of $j$ doesn't quite match the way you use it. You're using it as if $j$ represents the $x$ coordinate in the triangle rather than the $j$th bit. Also I think you need to consider whole strings instead of half strings for the automation to work. And I think you need to define $x_{n,j}=0$ when $j$ is out of bounds. Otherwise a great answer. – Riley Jun 25 '18 at 12:50
• Yeah, for it to make sense, we need to start counting things at $0$ as we do in computer programs. $x_{n,j} \cong x[n][j]$. As for the whole strings vs half strings, maybe it will be convenient to use negative indices to talk about the left half of the triangle. – Mark Jun 25 '18 at 14:48
Here is one answer to the question. If $$\, n\ge 2\,$$ then the first $$4$$ entries in each row, namely, $$\, 1 0 1 0, 1 1 0 1, 1 0 0 0, 1 1 1 0, \dots, \,$$ repeat in a period of $$4$$. There is a $$0$$ in each of these first $$4$$ entries. This was not the proof you wanted because it is too simple.
Your proof is essentially the following. You noticed something about a $$0$$ in row $$\, 2^n \,$$ and that this $$0$$ continues up to the row $$\, 2^{n+1}. \,$$ We formalize that observation by working with polynomials over the field with two elements. Define the Laurent polynomials $$\, y(n) := x^n + x^{-n} \,$$ and since the characteristic of the field is $$2$$, $$\, y(0) = 1 + 1 = 0. \,$$ Using simple algebra of exponents we get that $$\, y(n)y(m) = y(n+m) + y(n-m), \,$$ and the special cases $$\, y(n)^2 = y(2n) \,$$ and $$\, y(2^n) = y(1)^{2^n}.$$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211582993981,
"lm_q1q2_score": 0.8520689940182796,
"lm_q2_score": 0.8740772466456689,
"openwebmath_perplexity": 227.65682371870014,
"openwebmath_score": 0.8129568099975586,
"tags": null,
"url": "https://math.stackexchange.com/questions/2830788/how-to-translate-mathematical-intuition-into-a-rigorous-proof/2831036"
} |
Define the row polynomials $$\, r(m) := (1 + y(1))^m. \,$$ The general row is $$\, r(m) = r(2^n+k) \,$$ where $$\, 0\le k < 2^n. \,$$ Notice that $$\, r(2^n) = 1 + y(1)^{2^n} = 1 + y(2^n). \,$$ Also $$\, r(m) = 1 + \sum_{i=1}^m c_i y(i), \,$$ where $$\, c_i \,$$ is the coefficient of $$\, y(i). \,$$ You noticed that $$\, c_{2^{n-1}} = 0 \,$$ but how to prove it? We calculate that $$\, r(m) = (1 \!+\! y(2^n)) r(k) = (1 \!+\! y(2^n))(1 \!+\! cy(2^{n-1}) \!+\! t) = (1 \!+\! y(2^n))(1 \!+\! t) \!+\! c y(3\,2^{n-1}) \,$$ where $$\, t \,$$ is all the terms with $$\, y(i), \,$$ and $$\, i\ne 2^{n-1}, \,i < 2^n. \,$$ But $$\, y(2^n)y(i) = y(2^n-i) \!+\! y(2^n+i) \,$$ and since $$\, i \ne 2^{n-1} \,$$ we get no $$\, y(2^{n-1}) \,$$ terms in the product $$\, (1 \!+\! y(2^n))(1 \!+\! t). \,$$ That proves it.
This answers your particular question, but what about intuition versus rigor? I think there are no good answer to that question. You have to always look for any patterns or regularities and simple or edge cases. You have to be lucky and it helps to have a lot of background experience to draw upon.
• I recognize that there is a simpler answer to the question. It is given in the PDF I liked to. I am not asking for the solution, I am asking about how to translate my ideas into a rigorous proof. – Riley Jun 25 '18 at 4:06
• I thought I was translating your ideas combined with ideas of my own into a proof. But intuitions are tricky to express clearly. Maybe you wanted an "automaton theoretic" proof? – Somos Dec 3 '18 at 19:55 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211582993981,
"lm_q1q2_score": 0.8520689940182796,
"lm_q2_score": 0.8740772466456689,
"openwebmath_perplexity": 227.65682371870014,
"openwebmath_score": 0.8129568099975586,
"tags": null,
"url": "https://math.stackexchange.com/questions/2830788/how-to-translate-mathematical-intuition-into-a-rigorous-proof/2831036"
} |
# Equivalence of inner products?
So, let $\langle \cdot, \cdot \rangle_{1}$ and $\langle \cdot, \cdot \rangle_{2}$ be inner products on a real vector space $V$. Assume that \begin{align} \langle v, v \rangle_{1} = \langle v, v \rangle_{2} \end{align} for any $v \in V$.
Is it true that $\langle v, w \rangle_{1} = \langle v, w \rangle_{2}$ for all $v,w \in V$?
I can only think of two inner products on a real vector space $V$ are the standard dot product and the scaled dot product. This holds for these two since the scaled dot product would require all scalars to be 1. Can anyone think of another inner product on $V$ to test this with, or should I start trying to prove it?
• Take a basis $v_1,\dots ,v_n$ for $V$. Then $\langle v_i,v_i\rangle_1=\langle v_i,v_i\rangle_2$ for all $1\leq i\leq n$ implies that $\langle v,w\rangle_1=\langle v,w\rangle_2$.
– DKal
Mar 17 '14 at 21:09
For every $v,w \in V$ and $c$ real $$\langle v+cw,v+cw\rangle_1=\langle v+cw,v+cw\rangle_2$$ or $$\langle v,v\rangle_1+2c\langle v,w\rangle_1+c^2\langle w,w\rangle_1= \langle v,v\rangle_2+2c\langle v,w\rangle_2+c^2\langle w,w\rangle_2$$ and hence, as $\langle v,v\rangle_1=\langle v,v\rangle_2$ and $\langle w,w\rangle_1=\langle w,w\rangle_2$, then $$\langle v,w\rangle_1=\langle v,w\rangle_2.$$
Even though Yiorgos S. Smyrlis' answer is simpler, I think it is useful having a little different proof, just to add another point of view of the same argument, which is always good :)
Consider the norm $\| \cdot \|_1$ induced by $\langle \cdot, \cdot \rangle_1$ and the norm $\| \cdot \|_2$ induced by $\langle \cdot , \cdot \rangle_2$. By hypothesis and definition of induced norm $$\| v \|_1^2 = \langle v, v \rangle_1= \langle v, v \rangle_2= \| v \|_2^2$$ $\forall v \in V$. So the two norms are the same (the power doesn't make any problem thanks to the positivity of the norm.) | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211597623863,
"lm_q1q2_score": 0.8520689889017713,
"lm_q2_score": 0.8740772400852111,
"openwebmath_perplexity": 145.5758585312135,
"openwebmath_score": 0.9495441317558289,
"tags": null,
"url": "https://math.stackexchange.com/questions/716005/equivalence-of-inner-products"
} |
Then we want to do the reverse reasoning, if two product-induced-norms coincide on every element of the space, what can we say about inner products?
It's immediate (and left as little exercise for the reader - just write down the definitions and use bilinearity-) to prove that given a real inner product and its norm the following equivalence is true $$\langle x,y \rangle = \dfrac{1}{4} \left( \|x+y \|^2 - \|x-y\|^2 \right)$$ This formula is called the polarization identity (for more see here)
So by the above identity and the hypothesis we have, $$\langle x,y \rangle_1 = \dfrac{1}{4} \left( \|x+y \|_1^2 - \|x-y\|_1^2 \right) = \dfrac{1}{4} \left( \|x+y \|_2^2 - \|x-y\|_2^2 \right) = \langle x,y \rangle_2$$ $\forall x,y \in V$ and so we are done.
I wanted to stress the depth correlation between an inner product and its induced norm, and show some useful tools. Hope it helps :) | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211597623863,
"lm_q1q2_score": 0.8520689889017713,
"lm_q2_score": 0.8740772400852111,
"openwebmath_perplexity": 145.5758585312135,
"openwebmath_score": 0.9495441317558289,
"tags": null,
"url": "https://math.stackexchange.com/questions/716005/equivalence-of-inner-products"
} |
# How many random walk steps until the path self-intersects?
Take a random walk in the plane from the origin, each step of unit length in a uniformly random direction.
Q. How many steps on average until the path self-intersects?
My simulations suggest ~$$8.95$$ steps.
Red: origin. Top: the $$8$$-th step self-intersects. Bottom: the $$11$$-th step self-intersects. (Not to same scale.)
I suspect this is known in the SAW literature (SAW=Self-Avoiding Walk), but I am not finding this explicit number.
Related: self-avoidance time of random walk.
Added. Here is a histogram of the number of steps to self-intersection.
$$10000$$ random trials.
• This is asking for a uniformly random direction in the plane. By comparison, in 1-d the corresponding number is $(1/2)2 + (1/4)3 + (1/8)4 + \cdots = 3$; in 3-d the corresponding number is $\infty$. – Matt F. Mar 10 '19 at 4:40
• I added a second example. I have thousands of these. :-) – Joseph O'Rourke Mar 10 '19 at 18:56
• One can also ask for the asymptotic as $n$ goes to infinity of the probability that a length $n$ random walk does not self-intersect. The form of this asymptotic might be close to the case of random walks on grids, which if I recall right is conjectured but not known. – Will Sawin Mar 12 '19 at 17:42
• FWIW I ran this on $10^7$ random configurations and got a mean of ~$8.886$ steps. – Chip Hurst Mar 13 '19 at 14:22
• I ran $10^9$ simulations over night and the mean stayed at ~$8.886$. Here's a link of the simulation tallies: wolfr.am/C3YsNZmE. The largest run consisted of 98 lines. – Chip Hurst Mar 14 '19 at 12:16
Here are some computational insights.
I ran $$N = 10^k$$ simulations for $$3 \leq k \leq 12$$. For $$N = 10^9$$ I recorded some explicit instances and recorded which pairs intersected, as opposed to just the lengths. For all other runs, I simply summed the results.
First and foremost here's a summary of the runs: | {
"domain": "mathoverflow.net",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211612253743,
"lm_q1q2_score": 0.8520689789888074,
"lm_q2_score": 0.8740772286044095,
"openwebmath_perplexity": 486.9452189925769,
"openwebmath_score": 0.8837517499923706,
"tags": null,
"url": "https://mathoverflow.net/questions/325052/how-many-random-walk-steps-until-the-path-self-intersects"
} |
First and foremost here's a summary of the runs:
$$\begin{array}{ |l|l|l|l| } \hline N & \text{mean} & \text{std/sqrt(}N\text{)} & \text{total number of segments} \\ \hline 10^3 & 8.732 & 0.16106 & 8732 \\ 10^4 & 8.781 & 0.0528827 & 87810 \\ 10^5 & 8.86218 & 0.0166654 & 886218 \\ 10^6 & 8.89447 & 0.00524551 & 8894468 \\ 10^7 & 8.88463 & 0.00165934 & 88846265 \\ 10^8 & 8.88595 & 0.000524796 & 888595095 \\ 10^9 & 8.88615 & 0.000165965 & 8886153545 \\ 10^{10} & 8.88631 & - & 88863113226 \\ 10^{11} & 8.88614 & - & 888613831649 \\ 10^{12} & 8.88614 & - & 8886139852418 \\ \hline \end{array}$$
We can apply the central limit theorem to this (sub)sequence of means. Even though $$\text{std/sqrt(}N\text{)}$$ was not calculated for $$N = 10^{12}$$, based on prior terms ~$$0.00000524$$ seems like a sufficient estimate.
From here I think we can be fairly confident that the first few digits of the mean are $$\bf{8.8861}$$. Here are the first few standard deviations ($$68-95-99.7$$ rule):
$$m \pm 1\sigma \rightarrow [8.886135, 8.886145] \\ m \pm 2\sigma \rightarrow [8.886129, 8.886150] \\ m \pm 3\sigma \rightarrow [8.886124, 8.886156]$$
In fact it takes $$8$$ standard deviations to lose the digit $$1$$:
$$m \pm 8\sigma \rightarrow [8.886098, 8.886181]$$
Finally, based on an inverse symbolic search, we can find that $${\bf\frac{1795}{202}} = 8.886138613...$$ falls within $$0.25\sigma$$ of $$m = 8.886139852418$$ and so maybe this has potential to be the closed form.
For $$N = 10^9$$, here is the histogram of the number of steps to self-intersection, whose shape is quite similar to the one posted by OP:
If we plot look at the log plot, we can see the tail follows a decaying exponential quite closely (in red): | {
"domain": "mathoverflow.net",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211612253743,
"lm_q1q2_score": 0.8520689789888074,
"lm_q2_score": 0.8740772286044095,
"openwebmath_perplexity": 486.9452189925769,
"openwebmath_score": 0.8837517499923706,
"tags": null,
"url": "https://mathoverflow.net/questions/325052/how-many-random-walk-steps-until-the-path-self-intersects"
} |
From here we might think we could use a gamma distribution or something similar to estimate the data, but I was only able to find a tight approximation with one distribution: the inverse Gaussian distribution $$\operatorname{IG}(8.886, 26)$$. Here's its PDF overlaid on the histogram:
I also recorded which segments intersected each other for $$N = 10^9$$. The data shows segment $$n$$ most often intersects with segment $$n-2$$, followed by $$n-3$$, etc which I think makes sense. Borrowing terminology from MattF's answer, the quick intersections are the most probable.
In the following diagram an $$(x, y)$$ pair indicates how often segment $$x$$ intersected with segment $$y$$. Each diagonal approximately decays exponentially, all with the same rate.
For each of the 4 cores I ran the simulations on, I recorded a single sequence of each length. The most interesting ones of course are the long ones.
Here's an instance of length 96. It's riddled with near misses and is ultimately squashed by a quick intersection.
Here the suspicious looking part is indeed intersection free -- it's about $$0.01 \,\text{rad}$$:
Lastly I looked at the histogram of angles in my saved simulations:
I think this bias (introduced by stopping once an intersection occurs) says that the best way to stay intersection free is to not curl back up towards yourself.
I have compiled all of my data into a Mathematica notebook which can be downloaded here. | {
"domain": "mathoverflow.net",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211612253743,
"lm_q1q2_score": 0.8520689789888074,
"lm_q2_score": 0.8740772286044095,
"openwebmath_perplexity": 486.9452189925769,
"openwebmath_score": 0.8837517499923706,
"tags": null,
"url": "https://mathoverflow.net/questions/325052/how-many-random-walk-steps-until-the-path-self-intersects"
} |
I have compiled all of my data into a Mathematica notebook which can be downloaded here.
• Great data! I especially like the somewhat surprising angles histogram (but I accept your explanation). – Joseph O'Rourke Mar 25 '19 at 19:27
• Thanks. Yes I was surprised by that histogram too. Originally I thought there might have been a bug in my code. I reran my code without the intersection stopping condition and the histogram came out uniform. An example of that is in the linked notebook. – Chip Hurst Mar 25 '19 at 19:31
• And by the way, I made a couple edits since posting -- I don't know if you saw them -- a conjectured closed form and an approximate distribution fit. – Chip Hurst Mar 25 '19 at 19:32
• $2+11\sqrt{29/74}$ is even closer to the empirical average. – Matt F. Mar 25 '19 at 20:16
• The tail behaviour means that the intersection probability tends towards a constant $p$ which you could get from the slope. The behaviour is then modelled reasonably well by $$\Pr(\text{length} = k \mid k > k_0) = \Pr(\text{length} > k_0) \, (1-p)^{k-k_0-1} p$$ once the walk is past $k_0$ steps. – student Mar 26 '19 at 14:34
We can get an upper bound of $$15.52$$ by considering just the cases of interesction of next-to-consecutive steps, which we call quick intersections.
Let $$\alpha$$ be the angle between step $$AB$$ and step $$BC$$, and let $$u=\alpha/\pi$$. So $$u$$ is (initially) uniformly distributed between $$0$$ and $$1$$. Then, as Gerhard Paseman and Bill Bradley determined, the probability that steps $$AB$$ and $$CD$$ intersect is $$P(\text{intersection|u})= \begin{cases} \frac14-\frac u4\ \ \text{ if } u\le \frac13\\ \frac12-u\ \ \text{ if } \frac13\le u \le \frac12\\ \phantom{1-2}0 \ \ \text{ if } \frac12 \le u\\ \end{cases}$$ So the (initial) overall probability of an intersection is $$\int P(\text{intersection}|u)du=1/12$$. | {
"domain": "mathoverflow.net",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211612253743,
"lm_q1q2_score": 0.8520689789888074,
"lm_q2_score": 0.8740772286044095,
"openwebmath_perplexity": 486.9452189925769,
"openwebmath_score": 0.8837517499923706,
"tags": null,
"url": "https://mathoverflow.net/questions/325052/how-many-random-walk-steps-until-the-path-self-intersects"
} |
This means that for the next step, we have $$P(u|\text{no intersection last time})=\frac{12}{11} \begin{cases} \frac34+\frac u4\ \ \text{ if } u\le \frac13\\ \frac12 + u\, \ \ \text{ if } \frac13\le u \le \frac12\\ \phantom{1-2}1 \ \ \text{ if } \frac12 \le u\\ \end{cases}$$ \begin{align} P(\text{intersection|no intersection last time}) &=\frac{12}{11}\int_{u=0}^1 \begin{cases} (\frac34+\frac u4)(\frac14-\frac u4)\ \ \text{ if } u\le \frac13\\ (\frac12+u)(\frac12-u)\ \ \, \text{ if } \frac13\le u \le \frac12\\ \ \ \phantom{1-2}1(0)\ \ \ \ \ \ \ \ \ \ \ \text{ if } \frac12 \le u \end{cases} du\\ &=\frac{29}{396} \end{align} So the probability of no quick intersection after $$n$$ steps is $$\frac{11}{12}\left(\frac{367}{396}\right)^{n-3}$$ and the expected number of steps until a quick intersection is $$3\left(\frac{1}{12}\right)+\sum_{n=4}^\infty n\left( \frac{11}{12}\left(\frac{367}{396}\right)^{n-4}\! - \frac{11}{12}\left(\frac{367}{396}\right)^{n-3} \right)=\frac{450}{29} \simeq 15.52$$
• Nice analysis, MattF! – Joseph O'Rourke Mar 23 '19 at 0:11
Inspired by Bill Bradley's post, I've decided to lower bound the probability of an intersection which (using the method suggested in Bill's post) gives a slightly less weak upper bound for the expected length of a random walk.
Using Bill's picture and rescaling so that angles are measured as nonnegative fractions of 2pi, we see one domain of self intersection is when $$\alpha \lt \beta$$ and $$\alpha + 2\beta \lt 1/2$$. This region is a triangle of base 1/4, and height 1/6, with an area of 1/48. Missing the measure 0 set where $$\alpha=\beta$$, we reflect across that line, and then once more across the line $$\alpha+\beta=1$$ to get the other three regions having an intersection, giving the probability of self intersection of three-step paths to be 4/48=1/12. | {
"domain": "mathoverflow.net",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211612253743,
"lm_q1q2_score": 0.8520689789888074,
"lm_q2_score": 0.8740772286044095,
"openwebmath_perplexity": 486.9452189925769,
"openwebmath_score": 0.8837517499923706,
"tags": null,
"url": "https://mathoverflow.net/questions/325052/how-many-random-walk-steps-until-the-path-self-intersects"
} |
Now let's extend this to paths with four steps. The number of cases to determine the probability exactly is more than I care to tackle, so I will cheat and underestimate it. I do this by considering just two cases: that an intersection occurs in the first three steps, or that it does not but occurs in the last three steps.
If we considered these cases as disjoint, we would get a probability of 1/12 + 1/12 = 1/6. They aren't, so we will consider the case that $$\beta \leq 1/4$$ and $$\alpha \gt 1/4$$ or more visibly use 3/4 of the chance that the last three steps self-intersect. This gives a lower bound of (1/12) times (1+3/4), or 7/48, which gives 1/7 as a weak lower bound.
Using this and Bill's estimating method gives 1 + 7*(4-1)=22 as a weak upper bound on the number of steps needed to self-intersect. If we could improve the lower bound from 1/7 to 1/6 for self intersection of a four step path, this would lead to an upper bound of 19 instead of 22.
But if we can cheat once, we can cheat again. The first cheat went from two neighboring angles to three, so we will go from three to five and lower bound the crossing probability of six segments as (7/4) times the (lower bound of the) probability of four segments. We share the middle angle so that we can nicely separate the domains of the middle angle turning sharp left or sharp right. We do a similar cut and trim and end up with a lower bound of 49/192 =1/4+1/192 for the probability of 6 segments crossing. This gives a slightly better upper bound of 1 +4*5 or 21 for the step length of a walk that does not cross itself.
To justify the cheat, one has to do a lot more work to make sure that the probability grows by 7/4 at this step. However, I won't do that. I use the second cheat instead to motivate the idea that 22 is an upper bound that can be improved with a simple argument, and hope to inspire someone to come up with a simple and more honest and precise estimate. | {
"domain": "mathoverflow.net",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211612253743,
"lm_q1q2_score": 0.8520689789888074,
"lm_q2_score": 0.8740772286044095,
"openwebmath_perplexity": 486.9452189925769,
"openwebmath_score": 0.8837517499923706,
"tags": null,
"url": "https://mathoverflow.net/questions/325052/how-many-random-walk-steps-until-the-path-self-intersects"
} |
Gerhard "Long As It's Not Graded" Paseman, 2019.03.20.
• I would like to point out that many of my signatures encode some double meaning. In this particular case, I want to make it clear that this analysis fails on other manifolds. In particular, Joseph's problem makes sense on a torus or pseudo sphere, even on non-oriented manifolds or graphs with a rigid geometric structure. I encourage people to consider the version for a torus. Gerhard "Promises To Not Grade Work" Paseman, 2019.03.20. – Gerhard Paseman Mar 20 '19 at 17:21
• The version of Joseph's question on a compact space is likely in the graph theory literature. Consider a cover of the space by uniform mostly disjoint regions, and make a graph with a region a vertex and a line joining two regions if a step takes one from one region to another. One can now approximate a random walk by a transition matrix, or use the theory of random walks in graphs to compute a desired expectation. Thus the actual expectation can be considered as a limit of expectations using various partitions of the space. Gerhard "Still Interested In Torus Version" Paseman, 2019.03.20. – Gerhard Paseman Mar 20 '19 at 18:20
(Thanks for noticing my earlier error, Gerhard!)
Here's a loose but very simple upper bound. Consider three consecutive unit steps, which might look like this:
We take $$\alpha\in[0, \pi]$$ and $$\beta\in[0,2\pi)$$. (By symmetry we can reflect $$\alpha\in (\pi, 2\pi)$$ back to $$\pi-\alpha$$.) Then to overlap, we need $$\alpha < \pi/2$$, $$\beta<\pi/2$$ and $$\alpha + \beta < \pi$$ The first two events are independent, with probabilities 1/2 and 1/4.
For the third event, consider a case where the three segments just barely cross, e.g.:
If we decrease the angles $$\alpha, \beta$$, the steps will still intersect, so it's sufficient to consider this case. | {
"domain": "mathoverflow.net",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211612253743,
"lm_q1q2_score": 0.8520689789888074,
"lm_q2_score": 0.8740772286044095,
"openwebmath_perplexity": 486.9452189925769,
"openwebmath_score": 0.8837517499923706,
"tags": null,
"url": "https://mathoverflow.net/questions/325052/how-many-random-walk-steps-until-the-path-self-intersects"
} |
Because the bottom and left edges are both length one, the upper and rightmost angles are both $$\beta$$. Since we have a triangle, $$\alpha+2\beta=\pi$$, so the probability of this event is 2/3. (There's a symmetric case where the roles of $$\alpha$$ and $$\beta$$ are reversed.) It's possibly clearer to see this by examining a phase space diagram ("blue" = "line segments intersect"):
The blue region takes up 2/3rds of the (uniformly sampled) space. Conditioning on everything, the probability of overlap is 1/2 * 1/4 * 2/3 = 1/12
We can apply this argument to multiple non-overlapping pairs of angles. Therefore, the expected number of steps until self-intersection is upper-bounded by $$\leq 1 + 2\times 12 = 25$$
• Something does not look right. Do you want to find it, or shall I expand on my comments above? Gerhard "Pretty Sure About One Twelfth" Paseman, 2019.03.20. – Gerhard Paseman Mar 20 '19 at 15:49
• Looks good now. I like the picture and the estimate. Gerhard "Blue Is A Nice Color" Paseman, 2019.03.21. – Gerhard Paseman Mar 21 '19 at 23:55 | {
"domain": "mathoverflow.net",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211612253743,
"lm_q1q2_score": 0.8520689789888074,
"lm_q2_score": 0.8740772286044095,
"openwebmath_perplexity": 486.9452189925769,
"openwebmath_score": 0.8837517499923706,
"tags": null,
"url": "https://mathoverflow.net/questions/325052/how-many-random-walk-steps-until-the-path-self-intersects"
} |
# How do I take a fraction to a negative power?
I ran into this issue during my homework. Using the rules of logarithms, I need to prove that $$-2\ln\bigg(\frac{2}{\sqrt{6}}\bigg)=\ln3-\ln2$$ So here were my steps:
1. First step: $$-2\ln\bigg(\frac{2}{\sqrt{6}}\bigg)=\ln\left(\bigg(\frac{2}{\sqrt{6}}\bigg)^{-2}\right)$$ And that's as far as I got, because now I want to use the form $$\ln(a/b) = \ln(a) - \ln(b)$$, but first I need to reduce the fraction because it is raised to the $$-2$$.
How do I evaluate $$\bigg(\frac{2}{\sqrt{6}}\bigg)^{-2}$$ ?
Thanks
• Note that $x^{-a} = \frac{1}{x^a}$ – desiigner Jul 11 '19 at 19:53
By definition $$a^{-k} = \frac 1{a^k}$$
So $$\left(\frac{2}{\sqrt{6}}\right)^{-2} =\frac 1{\left(\frac{2}{\sqrt{6}}\right)^{2}}=$$
$$\frac 1{\left(\frac {2^2}{\sqrt 6^2}\right)}=\frac {\sqrt 6^2}{2^2}=\frac 64=\frac 32$$
It will help to realize that $$(\frac ab)^{-1} = 1/(a/b) = \frac ba$$ and that $$(\frac ab)^k = \frac {a^k}{b^k}$$ to realize that that means $$\left(\frac ab\right)^{-k} = \frac 1{\left(\frac ab\right)^k}= \frac 1{\left(\frac {a^k}{b^k}\right)} = \frac {b^k}{a^k}.$$
(Also $$(\frac ab)^{-k} = [(\frac ab)^{-1}]^k = (\frac ba)^k=\frac {b^k}{a^k}$$ or that $$(\frac ab)^{-k} = \frac {a^{-k}}{b^{-k}} = (1/a^k)/(1/b^k) = \frac {b^k}{a^k}$$.)
In any event
$$\left(\frac {2}{\sqrt 6}\right)^{-2} = \left(\frac {\sqrt 6}{ 2}\right)^2 = \frac {\sqrt 6^2}{2^2} = \frac 64 = \frac 32.$$
• Perfect, this is the explanation I was looking for. So now after a few steps, I'll get ln(3/2) = ln(3) - ln(2). – Miguel Aragon Jul 11 '19 at 21:36
\begin{align} -2\ln \left( \frac{2}{\sqrt 6} \right) &= -2\big( \ln(2)-\ln(\sqrt{6}) \big) \\ &= -2\ln(2)+2\ln(6^{1/2}) \\ &= -2\ln(2)+\ln(2\cdot 3) \\ &=-2\ln(2)+ \big( \ln(2)+\ln(3)\big) \\ &=\ln(3)-\ln(2) \end{align} And for your specific question, remember that $$\left( \frac{a}{b} \right)^{-n}=\left( \frac{b}{a} \right)^n$$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9787126531738088,
"lm_q1q2_score": 0.8520645706247889,
"lm_q2_score": 0.8705972768020108,
"openwebmath_perplexity": 289.97974398157544,
"openwebmath_score": 0.9951451420783997,
"tags": null,
"url": "https://math.stackexchange.com/questions/3290293/how-do-i-take-a-fraction-to-a-negative-power"
} |
Well you may start by distributing the index since $$2$$ and $$\sqrt 6$$ are positive. Thus $$\left(\frac{2}{\sqrt{6}}\right)^{-2}=\frac{2^{-2}}{{\sqrt 6}^{-2}}.$$
Then recall that for any nonzero number $$a$$ and any negative integer $$-n,$$ we have $$a^{-n}=\frac {1}{a^n}.$$ Applying this to your expression, we have $$\frac{2^{-2}}{{\sqrt 6}^{-2}}=\frac{\frac {1}{2^2}}{\frac{1}{{\sqrt 6}^2}}=\frac{\frac {1}{4}}{\frac{1}{6}}=\frac{6}{4}=\frac 32.$$
$$-2 \ln(\frac{2}{\sqrt{6}}) = \ln(\frac{2}{\sqrt{6}})^{-2} = \ln \frac{1}{(\frac{2}{\sqrt{6}})^{2}} = \ln \frac{6}{4} = \ln\frac{3}{2} = \ln 3 - \ln 2$$
$$-2 \ln \left( \frac{2}{\sqrt{6}} \right) = \ln 3 - \ln 2$$
if and only if
$$\ln \left( \frac{2}{\sqrt{6}} \right)^{-2} = \ln \left( \frac{3}{2}\right)$$
if and only if
$$\ln \left[ \frac{1}{\left(\frac{2}{\sqrt{6}}\right)^{2}} \right] = \ln \left( \frac{3}{2}\right)$$
And so, we have
$$\ln \left( \frac{6}{4} \right) = \ln \left( \frac{3}{2}\right)$$
which is true. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9787126531738088,
"lm_q1q2_score": 0.8520645706247889,
"lm_q2_score": 0.8705972768020108,
"openwebmath_perplexity": 289.97974398157544,
"openwebmath_score": 0.9951451420783997,
"tags": null,
"url": "https://math.stackexchange.com/questions/3290293/how-do-i-take-a-fraction-to-a-negative-power"
} |
And so, we have
$$\ln \left( \frac{6}{4} \right) = \ln \left( \frac{3}{2}\right)$$
which is true.
• you started by assuming the statement to be proven, so in reality you haven't shown anything. You need to run the argument in reverse – peek-a-boo Jul 11 '19 at 20:27
• One way to prove something is to take the LHS = RHS and proceed with a series of iff statements until something obvious is obtained. I neglected to include them initially. – mlchristians Jul 11 '19 at 20:29
• $4 = 10 \implies 7-3 =7+3 \implies-3= 3 \implies (-3)^2 = 3^2 \implies 9=9$ which is true. Therefore $4 = 10$. Or if penguins were elephants then they would both eat food. Which is true. So penguins are elephants. – fleablood Jul 11 '19 at 20:37
• "One way to prove something is to take the LHS = RHS and proceed with a series of iff statements until something obvious is obtained." But if you do that you MUST specify the steps are actually if and only if steps. If you do not SPECIFICALLY state that it is reasonable to assume they are only forward implications and the proof is not valid. – fleablood Jul 11 '19 at 20:38
• Your first comment is not if and only if; and neither is you second. Forward implications is one-way; necessary and sufficient (i.e., if and only if are both ways). In the case of this problem, start from the given equation and work it to where I finished. Any problem with that. Now, if you want to do the work, start w/ the last statement and by a series of implications, show it implies the initial equation. I chose iff and only if instead. – mlchristians Jul 11 '19 at 20:45
For a non-zero real number, we have $$1=x^0=x^{2-2}=x^2\cdot x^{-2}\tag1$$ from How to understand why $$x^0=1$$ where $$x$$ is any real number? | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9787126531738088,
"lm_q1q2_score": 0.8520645706247889,
"lm_q2_score": 0.8705972768020108,
"openwebmath_perplexity": 289.97974398157544,
"openwebmath_score": 0.9951451420783997,
"tags": null,
"url": "https://math.stackexchange.com/questions/3290293/how-do-i-take-a-fraction-to-a-negative-power"
} |
Putting $$x=2/\sqrt6$$ into $$(1)$$, we get $$1=\left(\frac2{\sqrt6}\right)^2\cdot\left(\frac2{\sqrt6}\right)^{-2}\tag2$$ and since we know that for a positive integer $$n$$ and positive real numbers $$a,b$$, $$\left(\frac ab\right)^n=\underbrace{\frac ab\cdot \frac ab\cdot\ldots\cdot\frac ab}_{n\,\text{times}}=\underbrace{\frac{a\cdot a\cdot\ldots\cdot a}{b\cdot b\cdot\ldots\cdot b}}_{n\,\text{times}}=\frac{a^n}{b^n},\tag3$$ we have that $$\left(\frac2{\sqrt6}\right)^2=\frac{2^2}{\left(\sqrt6\right)^2}=\frac46=\frac23\tag4.$$ Finally, putting $$(4)$$ back into $$(2)$$ yields $$1=\frac23\cdot\left(\frac2{\sqrt6}\right)^{-2}\implies\left(\frac2{\sqrt6}\right)^{-2}=\frac1{2/3}=\frac32\tag5$$ as the answer. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9787126531738088,
"lm_q1q2_score": 0.8520645706247889,
"lm_q2_score": 0.8705972768020108,
"openwebmath_perplexity": 289.97974398157544,
"openwebmath_score": 0.9951451420783997,
"tags": null,
"url": "https://math.stackexchange.com/questions/3290293/how-do-i-take-a-fraction-to-a-negative-power"
} |
# Any “shortcuts” to proving that $\frac{\sin(x)}2+\sin^2(\frac x2)\tan(\frac x2)\to\tan(\frac x2)$
I was working on simplifying some trig functions, and after a while of playing with them I simplified $$\frac{\sin(x)}{2}+\sin^2\left(\frac{x}{2}\right)\tan\left(\frac{x}{2}\right) \rightarrow \tan\left(\frac{x}{2}\right)$$
The way I got that result, however, was with what I think a very "roundabout" way. I first used the half-angle formulaes, then used $$x=\pi/2-\beta$$, and that simplified to $$\frac{\cos(\beta)}{1+\sin(\beta)}$$ where I again used the coordinate change to get $$\frac{\sin(x)}{1+\cos(x)}\rightarrow\tan\left(\frac{x}{2}\right)$$
I tried using the online trig simplifiers but none succeeded. Of course, after you know the above identity, it's easy to prove by proving that $$\frac{\sin(x)}{2}=\tan\left(\frac{x}{2}\right)-\sin^2\left(\frac{x}{2}\right)\tan\left(\frac{x}{2}\right)$$
Is there a more direct way to get the identity? I guess what I'm asking is, am I missing any "tricks" or software that I could have on my toolbelt so that next time I don't spend hours trying to simplify trig identities?
With $$s:=\sin\frac x2,c:=\cos\frac x2$$, $$\sin x=2sc$$ and
$$\frac122sc+s^2\frac sc=\frac{sc^2+s^3}c=\frac sc.$$
• I accepted this answer since it makes all steps explicit – Esteban Sep 30 '20 at 15:03
• @Esteban: thank you for the explanation. – Yves Daoust Sep 30 '20 at 15:20
Use $$\sin(x) = 2 \sin(x/2) \cos(x/2)$$ then we have $$\begin{eqnarray*} \frac{\sin(x)}{2}+\sin^2\left(\frac{x}{2}\right)\tan\left(\frac{x}{2}\right) &=& \frac{\sin(x/2)}{\cos(x/2)} \underbrace{\left( \cos^2(x/2) + \sin^2(x/2) \right)}_{=1} \\ &=& \tan (x/2). \end{eqnarray*}$$
• I actually tried this route but got stuck, unfortunately. So I think it's worth commenting that the "trick" here to get to your right hand expression is to multiply the 2sin(x/2)cos(x/2)/2 term by cos(x/2)/cos(x/2). – Esteban Sep 30 '20 at 14:51 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9787126500692716,
"lm_q1q2_score": 0.8520645646359986,
"lm_q2_score": 0.8705972734445508,
"openwebmath_perplexity": 379.3691216335252,
"openwebmath_score": 0.9426677823066711,
"tags": null,
"url": "https://math.stackexchange.com/questions/3843961/any-shortcuts-to-proving-that-frac-sinx2-sin2-frac-x2-tan-frac-x2/3843975"
} |
Fairly obvious: $$\sin^2\left(\frac{x}{2}\right)\tan\left(\frac{x}{2}\right)=(1-\cos^2\left(\frac{x}{2}\right))\tan\left(\frac{x}{2}\right)=\tan\left(\frac{x}{2}\right)-\cos\left(\frac{x}{2}\right)\sin\left(\frac{x}{2}\right)=\tan\left(\frac{x}{2}\right)-\frac{\sin x}{2}$$
• All the tried trig simplifiers beg to differ it was fairly obvious, unfortunately. I like the route you took. – Esteban Sep 30 '20 at 14:54
$$\frac{\sin x}{2} = \sin\frac{x}{2}\cos\frac{x}{2} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} \cos^2\frac{x}{2} = \tan\frac{x}{2} \left(1 - \sin^2\frac{x}{2}\right)$$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9787126500692716,
"lm_q1q2_score": 0.8520645646359986,
"lm_q2_score": 0.8705972734445508,
"openwebmath_perplexity": 379.3691216335252,
"openwebmath_score": 0.9426677823066711,
"tags": null,
"url": "https://math.stackexchange.com/questions/3843961/any-shortcuts-to-proving-that-frac-sinx2-sin2-frac-x2-tan-frac-x2/3843975"
} |
#### [SOLVED] Determining precisely where $\sum_{n=1}^\infty\frac{z^n}{n}$ converges?
Inspired by the exponential series, I'm curious about where exactly the series $\displaystyle\sum_{n=1}^\infty\frac{z^n}{n}$ for $z\in\mathbb{C}$ converges.
I calculated $$\limsup_{n\to\infty}\sqrt[n]{\frac{1}{n}}=\limsup_{n\to\infty}\frac{1}{n^{1/n}}$$ and $$\lim_{n\to\infty} n^{1/n}=e^{\lim_{n\to\infty}\log(n)/n}=e^{\lim_{n\to\infty}1/n}=e^0=1.$$ So the radius of convergence is $1$, so the series converges on all $z$ inside $S^1$. But is there a way to tell for which $z$ on the unit circle the series converges? I know it converges for $z=-1$, but diverges for $z=1$, but I don't know about the rest of the circle. For what other $z$ does this series converge? Thanks.
#### @Pookaros 2017-05-10 23:18:42
Even though the 1st answer is the more easy and appropriate here is a more analytic one. $\sum_{n=1}^{k} (z^n)/n =(z/1-z) $$\sum_{n=1}^{k-1} [(1-z^n)/n(n+1) + (1-z^k)/k] (we can prove that with induction till k). We can also see that 0 \leqslant |(1-z^k)/k|\leqslant 2/k\rightarrow0 for k\rightarrow$$\infty$ also
0 $\leqslant$ |(1-z^n)/n(n+1)|$\leqslant$ 2/n(n+1)
$\sum_{n=1}^{\infty}2/n(n+1)$ $(converges)$ so $\sum_{n=1}^{\infty} (z^n)/n$ $(converges)$ as well
#### @Martin Argerami 2012-03-11 00:56:15
Fix $z$ in the unit circle, i.e. $|z|=1$. We want to apply Dirichlet's test: if $\{a_n\}$ are real numbers and $\{b_n\}$ complex numbers such that:
1. $a_1\geq a_2\geq\cdots$
2. $\lim_{n\to\infty}a_n=0$
3. There exists $M>0$ such that $\left|\sum_{n=1}^Nb_n\right|\leq M$ for every $N\in\mathbb{N}$;
then $\sum_{n=1}^\infty a_nb_n$ converges. Here $a_n=1/n$, $b_n=z^n$. The first two conditions are clearly satisfied, and for the third one: $$\left|\sum_{n=1}^Nz^{ n }\right|=\left|\frac{z-z^{N+1}}{1-z}\right|\leq\frac{2}{|1-z|}$$ for all $N\in\mathbb{N}$. This shows that the third condition is satisfied for every $z\ne1$ in the circle. | {
"domain": "tutel.me",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9787126475856414,
"lm_q1q2_score": 0.8520645608307629,
"lm_q2_score": 0.870597271765821,
"openwebmath_perplexity": 522.6435572542091,
"openwebmath_score": 0.9750022292137146,
"tags": null,
"url": "https://tutel.me/c/mathematics/questions/118686/determining+precisely+where+sum_n1inftyfracznn+converges"
} |
In conclusion, the series converges for every $z$ with $|z|\leq1$ other than $z=1$, and it diverges for $|z|>1$.
#### @philmcole 2017-12-06 12:32:40
Can you explain this step: $|\sum_{n=1}^Nz^{ n }| = |\frac{z-z^{N+1}}{1-z}|$ ?
Here.
#### @philmcole 2017-12-06 13:48:17
Yeah, but where do you get that additional $z$ in the nominator from? By the formula you linked it should be $|\frac{1-z^{N+1}}{1-z}|$, no?
#### @Martin Argerami 2017-12-06 13:49:24
If you start the sum from $0$, yes.
#### @Pacciu 2012-03-11 23:23:01
The followin theorem on power series is due to E. Picard:
Let $(a_n)$ be a sequence of real numbers.
If the sequence $(a_n)$ is nonnegative, decreases and tends to zero when $n\to \infty$, then the complex power series $\sum a_n\ z^n$ converges in the closed unit disc $\overline{D}(0;1)$ with the only possible exception of the point $1$.
The proof of Picard's theorem relies on Abel's summation by parts formula, as far as I remember.
Now, the coefficients of your series, i.e. $a_n=1/n$, satisfy the assumptions of Picard's theorem, hence your series converges at least in $\overline{D}(0;1)\setminus \{1\}$; on the other hand, the series diverges when $z=1$ (for it becomes the harmonic series).
Therefore the convergence set of $\sum 1/n\ z^n$ is $\overline{D}(0;1)\setminus \{1\}$.
### Determining precisely where $\sum_{n=0}^\infty \left(\frac{z^n}{n}\right)$ converges uniformly
• 2017-09-13 22:36:35
• Pookaros
• 77 View
• 0 Score
• Tags: complex-analysis
### [SOLVED] To prove the statements related to power series
• 2017-01-27 02:53:47
• Kavita
• 653 View
• 1 Score
• Tags: complex-analysis | {
"domain": "tutel.me",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9787126475856414,
"lm_q1q2_score": 0.8520645608307629,
"lm_q2_score": 0.870597271765821,
"openwebmath_perplexity": 522.6435572542091,
"openwebmath_score": 0.9750022292137146,
"tags": null,
"url": "https://tutel.me/c/mathematics/questions/118686/determining+precisely+where+sum_n1inftyfracznn+converges"
} |
# Probability: Taking Cups From A Box
I just saw the question in an exam paper. My friend and I have different answer and I am not sure which solution is correct, so I hope someone can help me.
There are 16 cups in a box, 11 are blue and 5 are white. Six cups are taken out from it at the same time. What is the probability of taking at least 4 white cups?
My solution:
$$P=\frac{\binom54\binom{11}2+\binom55\binom{11}1}{\binom{16}6}$$
My friend's solution:
$$P=\frac{5!\times11\times6+5!\times11\times10\times_6\text{P}_2}{_{16}\text{P}_6}$$
My friend use a different approach and his answer is different from mine. I wonder if I am correct and why. Please show me the reason, thank you!
-
Your method is correct. – André Nicolas Apr 15 '13 at 8:02
@AndréNicolas Can you tell me why I am correct? After my friend told me his solution, I am so confused with the question ... – redcap Apr 15 '13 at 8:22 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9911526438717138,
"lm_q1q2_score": 0.8520580062857801,
"lm_q2_score": 0.8596637577007393,
"openwebmath_perplexity": 100.20600061453757,
"openwebmath_score": 0.8496515154838562,
"tags": null,
"url": "http://math.stackexchange.com/questions/362140/probability-taking-cups-from-a-box"
} |
Your method is correct. There are $\binom{16}{6}$ equally likely ways to pick a collection of $6$ cups. (We can imagine the cups have secret ID numbers written in invisible ink.)
Your numerator counts the favourables. For example, the are $\binom{5}{4}$ ways to pick $4$ whites. For each of these ways there are $\binom{11}{2}$ ways to pick the accompanying blues, for a total of $\binom{5}{4}\binom{11}{2}$.
Your friend's proposed solution can be made to work, though it is perhaps a little trickier to do it correctly. The friend observed that counting order of picking, there are ${}_{16}P_6$ equally likely ways of picking $6$ cups. Now we count the number of favourables. Your friend's first term in the count of favourables is correct, the second is not.
There are $\binom{6}{2}$ ways of picking the locations of the $4$ blue cups. For each of these ways, the first chosen location can be filled in $11$ ways, and the second in $10$ ways. Then the remaining locations can be filled with white cups in ${}_5P_4$ ways. Multiply to get the total with $4$ white and $2$ blue. There are two mistakes in the proposed expression there. First, instead of $\binom{6}{2}$ there is ${}_6P_2$. Secondly, there is the $4!$ which may be a typo. For the placement of white cups in the $4$ available slots, either use the permutation symbol ${}_5P_4$, or decide the $4$ whites can be chosen in $\binom{5}{4}$ ways, and then permuted in $4!$ ways. That part turns out to be $(5)4!$, which is the same as $5!$.
You are welcome. I had figured the $4!$ was an unimportant error. But that still leaves an overcount in that part by a factor of $2$. That error is somewhat more structural. – André Nicolas Apr 15 '13 at 9:30 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9911526438717138,
"lm_q1q2_score": 0.8520580062857801,
"lm_q2_score": 0.8596637577007393,
"openwebmath_perplexity": 100.20600061453757,
"openwebmath_score": 0.8496515154838562,
"tags": null,
"url": "http://math.stackexchange.com/questions/362140/probability-taking-cups-from-a-box"
} |
# Parameterize unitary without transpose
For all unitary matrices, i.e. $A \overline{A}^T = I$, there is a skew-Hermitian matrix $X$ so that $A = exp(X)$. So the unitary group has $n^2$ dimensions.
Is there any similar parameterisation of all matrices with $A \overline{A} = I$? What can be said about the number of dimensions? (Btw. is there any name for this type of matrices in literature?)
• Note that If $M$ is a real square matrix of the relevant size, then $A = \exp(iM)$ has the property that $\overline{A} = \exp(-iM)$, so that $A\overline{A} = I$. – Geoff Robinson Oct 19 '15 at 22:56
• @QiaochuYuan: Right, this collection is not closed under multiplication. Does this already exclude any notion of dimensionality and parameterisation? – Sebastian Schlecht Oct 20 '15 at 8:38
• @GeoffRobinson: Right, but are these all matrices with this property? – Sebastian Schlecht Oct 20 '15 at 9:05
• @GeoffRobinson: The answer is yes if we add $A$ invertible, as $\exp: M(N,\mathbb{C}) \rightarrow GL(N,\mathbb{C})$ is surjective and $\exp(M) \overline{\exp(M)}=I$ follows $M = -\overline{M}$. – Sebastian Schlecht Oct 20 '15 at 9:13
• Well,$A$ certainly needs to be invertible in the question. – Geoff Robinson Oct 20 '15 at 10:00
The set $V= \{ A\in M_n(\mathbb{C})\ |\ A\bar A = I\}$ is a smooth submanifold of $M_n(\mathbb{C})$ with real dimension $n^2$.
Proof: Consider the involution $\iota:\mathrm{GL}(n,\mathbb{C})\to \mathrm{GL}(n,\mathbb{C})$ defined by $$\iota(A) = (\bar A)^{-1}.$$ This is an anti-holomorphic involution of the complex manifold $\mathrm{GL}(n,\mathbb{C})$, and its fixed locus is precisely $V$. Thus, $V$ is a totally real submanifold of $\mathrm{GL}(n,\mathbb{C})$ with (real) dimension $n^2$. | {
"domain": "mathoverflow.net",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.991152643087697,
"lm_q1q2_score": 0.8520580056117895,
"lm_q2_score": 0.8596637577007394,
"openwebmath_perplexity": 125.47789443929702,
"openwebmath_score": 0.9622570872306824,
"tags": null,
"url": "https://mathoverflow.net/questions/221340/parameterize-unitary-without-transpose"
} |
Also: While it's true that the map $f:M_n(\mathbb{R})\to \mathrm{GL}(n,\mathbb{C})$ defined by $$f(a) = \exp(ia)$$ has its image in $V$, it is not a 'parametrization' everywhere, i.e., $f$ is not a local diffeomorphism everywhere. While this is true on a neighborhood of $0\in M_n(\mathbb{R})$, at other places, the map $f$ definitely is not a local diffeomorphism. For example, if $a$ has $n$ eigenvalues of the form $2k_i\pi$ (where $k_1,\ldots, k_n$ are integers, not all zero), then $f(a) = I$, but the space of such real matrices $a$ has positive dimension for any $n$-tuple $(k_1,\ldots, k_n)$ for which not all of the $k_i$ are equal.
It turns out that $f$ is surjective. The proof uses the fact that, for $A\in V$, we have $A\bar A = \bar A A = I$, so, in particular, $A$ and $\bar A$ commute and hence can be put simultaneously in Jordan normal form by a real conjugation. Then, breaking $\mathbb{C}^n$ into a sum of complexifications of real subspaces according to the eigenvalues of $A$ and using the semi-simple/nilpotent decomposition appropriately, one can reduce to dealing with the upper triangular case, and the result can then be proved using simple facts about power series in commuting nilpotent variables. Details upon request (see below).
Of course, all of this is probably a special case of known facts about affine symmetric spaces. I have now realized that $V$ is simply the Cartan embedding for the affine symmetric space $\mathrm{GL}(n,\mathbb{C})/\mathrm{GL}(n,\mathbb{R})$. This Cartan embedding $$\sigma:\mathrm{GL}(n,\mathbb{C})/\mathrm{GL}(n,\mathbb{R})\longrightarrow V\subset \mathrm{GL}(n,\mathbb{C})$$ is given by $\sigma(B\cdot\mathrm{GL}(n,\mathbb{R})) = B\,(\,\overline B\,)^{-1}$. The exponential map we have been discussing is just the geodesic mapping of this affine symmetric space, so, probably all of this follows from general theory. | {
"domain": "mathoverflow.net",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.991152643087697,
"lm_q1q2_score": 0.8520580056117895,
"lm_q2_score": 0.8596637577007394,
"openwebmath_perplexity": 125.47789443929702,
"openwebmath_score": 0.9622570872306824,
"tags": null,
"url": "https://mathoverflow.net/questions/221340/parameterize-unitary-without-transpose"
} |
Details: Let $A\in\mathrm{GL}(n,\mathbb{C})$ satisfy $A\overline{A} = I$. Let $\lambda\in\mathbb{C}$ be an eigenvalue of $A$ of multiplicity $m\le n$ and let $$V_\lambda = \{ v\in\mathbb{C}^n \ |\ (A{-}\lambda)^mv = 0\ \}.$$ be the associated generalized eigenspace. Of course, $\mathbb{C}^n$ is the direct sum of these generalized eigenspaces. Since $\overline{A}$ commutes with $A$, it preserves each of these subspaces. Moreover, one clearly has $$\overline{V_\lambda} = V_{1/\bar\lambda},$$ so each of the spaces $V_\lambda+V_{1/\bar\lambda}$ is invariant under conjugation and hence is the complexification of a real subspace $W_\lambda\subset \mathbb{R}^n$. It follows that, by conjugating by a real matrix, we can assume that $A$ (and hence $\overline A$) is in block diagonal form, so it suffices to treat the case where either $A$ has a single eigenvalue $\lambda$ satisfying $\lambda\bar\lambda=1$, so that $\mathbb{C}^n=V_\lambda$, or else $A$ has an eigenvalue $\lambda$ satisfying $\lambda\bar\lambda > 1$ and $\mathbb{C}^n=V_\lambda\oplus V_{1/\bar\lambda}$.
In either case, we can assume that $\lambda = r\ge 1$ is real, since, if $\lambda = r e^{i\theta}$, we can replace $A$ by $e^{-i\theta}A$ and show that $e^{-i\theta}A$ is in the image of $f$ (since $I$ commutes with everything and $e^{i\theta} I = \exp(i\theta I)$.) | {
"domain": "mathoverflow.net",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.991152643087697,
"lm_q1q2_score": 0.8520580056117895,
"lm_q2_score": 0.8596637577007394,
"openwebmath_perplexity": 125.47789443929702,
"openwebmath_score": 0.9622570872306824,
"tags": null,
"url": "https://mathoverflow.net/questions/221340/parameterize-unitary-without-transpose"
} |
First, consider the case $\lambda = 1$. Then $A = C + i S$ where $C$ and $S$ are real matrices and $((C-I) + iS)^n = (A-I)^n = 0$. Moreover, $N=C-I$ and $S$ commute since $I = A\bar A = C^2+S^2 +i(SC-CS) = I + 2N + N^2 + S^2$. Note that $N$ and $S$ are nilpotent commuting matrices. Since they satisfy $$2N + N^2 = -S^2,$$ it follows that $N = p(-S^2)$ where $p(t) = -\tfrac12 t^2 + \cdots$ is the (unique) power series (with real coefficients) that satisfies $2p(t) + p(t)^2 = -t^2$. Since $S$ is nilpotent, it follows that $N = p(-S^2)$ expresses $N$ canonically as a polynomial in $S$. Now let $q(t)$ be the (unique) power series with real coefficients that satisfies $\sin q(t) = t$. Note that we must have $p(-q(t)^2) = \cos(t)-1$. Now, because $S$ is nilpotent, we have $S = \sin q(S)$, where $q(S)$ is a real polynomial in $S$. Putting all of this together, we have $$A = I + p(-S^2) + iS = \cos(q(S)) + i\sin(q(S)) = \exp(iq(S)).$$ Finally, before leaving this special case, let us note that, because $A$ satisfies its characteristic polynomial $(A-I)^n = 0$, it follows that $\overline A = A^{-1}$ is expressed as a universal polynomial in $A$ with real coefficients, so $iS = \tfrac12(A-\overline{A})$ is also expressed as a universal polynomial with real coefficients. Since $q(t)$ is an odd power series, it satisfies $iq(it) = f(t)$ where $f$ has real coefficients, so there is actually a formula of the form $A = \exp( i g_n(A))$ for all $A \in \mathrm{GL}(n,\mathbb{C})$ satisfying $A\bar A = I$ and $(A-I)^n=0$ for some universal polynomial $g_n(t)$ with real coefficients that also has the property that $g_n(A)$ is a real $n$-by-$n$ matrix for all such $A$. (This remark will be used below.) | {
"domain": "mathoverflow.net",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.991152643087697,
"lm_q1q2_score": 0.8520580056117895,
"lm_q2_score": 0.8596637577007394,
"openwebmath_perplexity": 125.47789443929702,
"openwebmath_score": 0.9622570872306824,
"tags": null,
"url": "https://mathoverflow.net/questions/221340/parameterize-unitary-without-transpose"
} |
Now, finally, let us assume that $A$ has eigenvalues $\lambda = e^t$ and $1/\lambda = e^{-t}$ for some $t>0$. Then $V_\lambda$ and $\overline{V_\lambda} = V_{1/\lambda}$ are disjoint, complementary complex subspaces of $\mathbb{C}^n$ and so we must have that there exists a matrix $Q\in M_n(\mathbb{R})$ such that $$V_\lambda = \{ v + iQv \ | \ v\in \mathbb{R}^n \}.$$ Because $V_\lambda$ is closed under multiplication by $i$, it follows that $Q^2 = -I$. Setting $$S = \cosh t + i\sinh t\,Q = \exp(i t Q),$$ we see that $Sw= e^t w$ for all $w \in V_\lambda$ and $Sw = e^{-t}w$ for all $w \in V_{1/\lambda}$, so $S$ is the semi-simple part of $A$. Hence $S$ can be written as $S = s_\lambda(A)$, where $s_\lambda(t)$ a polynomial in $t$ with real coefficients (that depend on $\lambda$). Thus, $\bar S = s_\lambda(\bar A)$ can be written as a universal polynomial in $A$, implying that $Q$ itself can be written as a universal polynomial in $A$ and hence, in particular, it commutes with $A$ and $\bar A$. Now, writing $$A = \exp(i t Q) B,$$ we see that $B\bar B = I$ and that $B$ can be written as a polynomial in $A$. Moreover, the eigenvalues of $B$ are now all equal to $1$, so $(B-I)^n=0$, and so $B = \exp(i g_n(B))$ (as per above), where $g_n(B)$ is a polynomial in $A$, which, therefore, commutes with $Q$ (which is a polynomial in $A$). Finally, we have $$A = \exp(i t Q)\exp(i g_n(B)) = \exp(i (t Q+g_n(B))),$$ as desired. | {
"domain": "mathoverflow.net",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.991152643087697,
"lm_q1q2_score": 0.8520580056117895,
"lm_q2_score": 0.8596637577007394,
"openwebmath_perplexity": 125.47789443929702,
"openwebmath_score": 0.9622570872306824,
"tags": null,
"url": "https://mathoverflow.net/questions/221340/parameterize-unitary-without-transpose"
} |
• Thank you very much for your insightful answer. The only main missing part for me is the surjectivity of $f$. – Sebastian Schlecht Oct 20 '15 at 16:43
• Robert, I learn so much new material from your answers! Just great! – Suvrit Oct 21 '15 at 13:24
• Robert, great answer. So a bit about the details, for simplicity assuming $A$ is diagonalisable. Commutativity gives as $A = U^{-1} \Lambda U$ for real $U$. $A$ is invertible, hence there is $\exp(i L) = \Lambda$. So, $A = \exp( i U^{-1} L U)$. Why is $U^{-1} L U$ real? – Sebastian Schlecht Oct 22 '15 at 20:05
• If $A$ is diagonal, then $A\bar A = I$ implies that all of the eigenvalues of $A$ are complex numbers of unit modulus, so each of them is of the form $e^{ir}$ where $r$ is real. – Robert Bryant Oct 22 '15 at 20:40
• I'm confused, I only wanted $A$ diagonalisable and not $A$ diagonal. – Sebastian Schlecht Oct 22 '15 at 21:24
Qiaochu Yuan is a little hard about our variety $V$; Kummer showed that an element of $V$ is in the form $U\bar{U}^{-1}$ (a generalization of this result is the Hilbert's theorem 90). Close to the subject, cf. "Ikramov, On the matrix equation $X\bar{X}=A$" http://link.springer.com/article/10.1134%2FS1064562409010153#page-1
Robert wrote a nice answer; we can give more elementary proofs as follows. $\mathbb{C}$ is seen as $\mathbb{R}^2$.
Proposition 1. $dim(V)=n^2$ when $V$ is considered as a real algebraic set.
Proof. According to Kummer, if $R\bar{R}=I$, then there is $T\in GL_n(\mathbb{C})$ s.t. $R=T\overline{T}^{-1}$. Let $f:T\in GL_n\rightarrow T\overline{T}^{-1}$; note that $f$ is a pseudo-parametrization of our set; in fact $f$ is a submersion. It remains to calculate the rank of $Df_T:H\in M_n\rightarrow (H-T\overline{T}^{-1}\overline{H})\overline{T}^{-1}$. Note that $rank(Df_T)=rank(H\rightarrow T^{-1}H-\overline{T}^{-1}\overline{H})$. | {
"domain": "mathoverflow.net",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.991152643087697,
"lm_q1q2_score": 0.8520580056117895,
"lm_q2_score": 0.8596637577007394,
"openwebmath_perplexity": 125.47789443929702,
"openwebmath_score": 0.9622570872306824,
"tags": null,
"url": "https://mathoverflow.net/questions/221340/parameterize-unitary-without-transpose"
} |
Let $T^{-1}=U+iV,H=X+iY$, where $U,V,X,Y$ are real matrices. Then $rank(Df_A)=rank(g:(X,Y)\in \mathbb{R}^{2n^2}\rightarrow UY+VX\in \mathbb{R}^{n^2}$; we show that $g$ is onto. Let $A\in M_n$; since $U+iV$ is invertible, there is $\lambda\in \mathbb{R}$ s.t. $U+\lambda V$ is invertible. Finally $g(\lambda(U+\lambda V)^{-1}A,(U+\lambda V)^{-1}A)=A$ and we are done.
Proposition 2. Any $R\in V$ can be written as $exp(iA)$ where $A$ is a real matrix.
Proof. Let $R=U+iV$, where $U,V$ are real. $R\bar{R}=I$ is equivalent to $U^2+V^2=I,VU=UV$; that is equivalent to: there is a real matrix $A$ s.t. $U=\cos(A),V,=\sin(A)$ and $R=\exp(iA)$.
EDIT. (cf. Sebastian's comment). $e^Me^{\bar{M}}=I$ does not imply that $e^{M+\bar{M}}=I$.
Proof. Take $M=\begin{pmatrix}i\pi&1\\0&-i\pi\end{pmatrix}$. | {
"domain": "mathoverflow.net",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.991152643087697,
"lm_q1q2_score": 0.8520580056117895,
"lm_q2_score": 0.8596637577007394,
"openwebmath_perplexity": 125.47789443929702,
"openwebmath_score": 0.9622570872306824,
"tags": null,
"url": "https://mathoverflow.net/questions/221340/parameterize-unitary-without-transpose"
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.