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I ‘blogged on this problem here and here. In the session, we used TI-Nspire file QuadExplore.
Next, we explored briefly the same review of trigonometric and polar graphs not as static parent functions under static transformations, but as dynamic curves oscillating between their ceilings and floors. In the session, w... | {
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To demonstrate this final part of the session, I used TI-Nspire file Hidden Conic Behavior.
Here is my PowerPoint file for Powerful Student Proofs. A more detailed sketch of the session and the student proofs is below.
Bending Asymptotes, Bouncing Off Infinity, and Going Beyond
The basic proposal was that adding th... | {
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I show my students’ solutions below. But before you read on, can you give your own solution?
SOLUTION ALERT! Don’t read further if you want to find your own solution.
WHAT I EXPECTED
We’ve had many discussions in class about the power of the Rule of 4–that math ideas can be expressed numerically, graphically, alge... | {
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J used more equations than he needed, but had he restricted himself to just two equations, I’m not sure the lovely pattern would have been so obvious.
This is a very different (and super cool) answer than what I expected my students to produce. Lesson re-learned: Challenge your students, give them room to express cr... | {
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1. On a piece of wax paper, use a pen to draw a line near one edge. (I used a Sharpie on regular copy paper and got enough ink bleed that I’m convinced any standard copy or notebook paper will do. I don’t think the expense of wax paper is required!)
2. All along the line, place additional dots 0.5 to 1 inch apart.
3.... | {
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A parabola is the locus of points equidistant from a given point (focus) and line (directrix).
What makes the parabola interesting, in my opinion, is the interplay between the distance from a line (always perpendicular to some point C on the directrix) and the focus point (theoretically could point in any direction l... | {
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ALGEBRAIC PROOF
While I’m comfortable handling math in almost any representation, my mind most often jumps to algebraic perspectives first. My first inclination was a coordinate proof.
PROOF 1: As all parabolas are similar, it was enough to use a single, upward facing parabola with its vertex at the origin. I plac... | {
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I had a proof, but the algebra seemed more than necessary. Surely there was a cleaner approach.
In the image above, F is the focus, and I is a point on the parabola. If D is the midpoint of $\overline{FC}$, can I conclude $\overline{ID} \perp \overline{FC}$, proving that the perpendicular bisector of $\overline{FC}$... | {
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I created a second version allowing users to set the location of the focus on the positive y-axis and using a slider to determine the distances and constructs the parabola through the definition of the parabola. [In the GeoGebra worksheet (here), you can turn on the hidden circle and lines to see how I constructed it... | {
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1. What is the shortest length of segment CP and where could it be located at that length? What is the longest length of segment CP and where could it be located at that length?
2. Obviously, point C can be anywhere along the directrix. While the focus-to-P segment is theoretically free to rotate in any direction, th... | {
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We had lost the quadratic term, but we still couldn’t sum the series with both a linear and an exponential term. At this point, P asked if we could use the same approach to rewrite the series again. Because the numerators were all odd numbers and each could be written as a sum of 1 and some number of 2s, we got
wher... | {
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# Projectile: $v^*w^*=gk$ for minimum launch velocity
A projectile launched from $O(0,0)$ at velocity $v$ and launch angle $\theta$, passes through $P(k,h)$. The velocity of the projectile at $P$ is $w$. The slope of $OP$ is $\alpha$, i.e. $\tan\alpha=\frac hk$, and the length of $OP$ is $R$.
$\hspace{4cm}$
Let $v^*... | {
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I think the following picture shows a symmetry which could be useful. The minimum possible speed to reach $(k,h)$ is attained when you throw the projectile in the direction bisecting the angle between the $OP$ line and vertical axis. Because this trajectory reaches $P$ with the minimum possible speed, it also solves th... | {
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In the second picture below, when the minimum energy projectile thrown from $A$ reaches $D$, a "projectile" released with zero velocity from $A$ will reach point $G$. In addition, projectiles thrown with parallel velocities (including zero velocity) from the point of origin at the same time remain forming a line parall... | {
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Furthermore, it is also clear that, since there is no acceleration in the $\hat{x}$ direction, $$R\cos\alpha=vt_f\cos\theta~~.\tag{2}$$ From (1) and (2) we deduce $$R=\frac{v^2}{g}\frac{2\cos(\theta)\sin(\theta-\alpha)}{\cos^2\alpha}=\frac{v^2}{g}\frac{(\sin(2\theta-\alpha)-\sin(\alpha))}{\cos^2\alpha}~~,\tag{3}$$ from... | {
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• Interesting! (+1). For the free fall case, terminal velocity is $\sqrt{gR}$. For the inclined case, minimum terminal velocity is $w^*=\sqrt{g(R-h)}$, and the corresponding minimum launch velocity is $v^*=\sqrt{g(R+h)}$, the product of which is $v^*w^*=g\sqrt{R^2-h^2}=gk$. Given the additional insight from your new so... | {
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(NB - velocity here refers to a vector quantity, i.e. both speed and direction.)
Consider the diagram on the right, which is the left diagram scaled by $\frac 2T$.
\begin{align} \triangle O'B'P'= \tfrac 12 \cdot 2v\cdot w\cdot \sin\beta &=\tfrac 12\cdot gT\cdot \tfrac {2R}T\cos\alpha \\ vw\sin\beta&=gR\cos\alpha\\ vw... | {
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$\hspace{5cm}$
Footnote
Using the cosine rule twice,we have
\begin{align} R^2 &=\left(\tfrac {vT}2\right)^2+\left(\tfrac {wT}2\right)^2-2\left(\tfrac {vT}2\right)\left(\tfrac {wT}2\right)\cos\beta\qquad \tag{3}\\ \left(\tfrac 12 gT^2\right)^2 &=\left(\tfrac {vT}2\right)^2+\left(\tfrac {wT}2\right)^2-2\left(\tfrac {v... | {
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** An alternative Second Scenario would be a composite motion of the following:
• $(1)$ vertical launch at $gT$ under gravity for time $T$ (distance travelled: $\frac 12gT^2$), and
• $(2)$ motion at constant velocity $\mathbf w$ for time $T$ (distance travelled: $wT$) and
Alternative Interpretation
An alternative in... | {
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It can be shown that $\tan\theta=\dfrac {R+h}k$, where $R=\sqrt{k^2+h^2}$.
Hence $${v^*}^2=g(R+h)\\ {w^*}^2=g(R-h)$$
LATEST SOLUTION
(Using vectors)
$\hspace{5cm}$
Let $\mathbf v,\mathbf w$ be the initial (launch) and terminal velocity vectors of the projectile, and $\beta$ be the angle between them.
$\hspace{5cm}$... | {
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Distance equations for both equivalent composite motions above are as follows:
\begin{align} \mathbf R&=\mathbf vT+\tfrac 12 \mathbf gT^2\tag{1}\\ \mathbf R&=\mathbf wT-\tfrac 12 \mathbf gT^2\tag{2}\\\\ \tfrac {(1)+(2)}T:\hspace {6.5cm} (\mathbf v+\mathbf w)&=\tfrac 2T \mathbf R\tag{A}\\ \tfrac {(1)-(2)}T:\hspace{6.5c... | {
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# Conjecture:$\int_0^{\pi/2} \left(\frac{\sin(2n+1)x}{\sin x}\right)^\beta dx$ is integer multiple of $\pi/2$,for integer $\beta>2$
I was solving the integral $$I_n=\int_0^{\frac {\pi}{2}} \left(\frac {\sin ((2n+1)x)}{\sin x}\right)^2 dx$$
With $$n\ge 0$$ And $$n\in \mathbb{N}$$
On solving, I got $$I_n =\frac {(2n+1... | {
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$$\def\b{\beta}$$\begin{align*} \newcommand\cmt[1]{{\small\textrm{#1}}} I_n(\b) &= \int_0^{\pi/2} \left(\frac{\sin (2n+1)x}{\sin x}\right)^\b dx \\ &= \frac 1 4 \int_0^{2\pi} \left(\frac{\sin (2n+1)x}{\sin x}\right)^\b dx & \cmt{begin similar to user630708} \\ &= \frac{1}{4i} \oint_\gamma \left(\frac{z^{4n+2}-1}{z^2-1}... | {
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employing $$z=e^{ix}$$ we get
$$4 I_{n,\beta}=\oint_C \left[\frac{z^{4n+2}-1}{z^2-1}\right]^{\beta}\frac{dz}{i z^{2\beta n+1}}$$
where $$C$$ denotes the unit circle in the comlex plane. By the residue theorem (there is one pole inside the contour at $$z=0$$, using f.e. the geometric series you can Show that the Point... | {
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Prove $\frac{1+\cos{(2A)}}{\sin{(2A)}}=\cot{A}$
I am sorry to ask so many of these questions in such as short time span.
But how would I prove this following trigonometric identity. $$\frac{1+\cos(2A)}{\sin(2A)}=\cot A$$ My work thus far is $$\frac{1+\cos^2A-\sin^2A}{2\sin A\cos A}$$ I know $1-\sin^2A=\cos^2A$
So I ... | {
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# Expected number of neighbors
Given a row of 16 houses where 10 are red and 6 are blue, what is the expected number of neigbors of a different color?
-
By "neighbors of a different color"... do you mean given any one house what are the chances that the house to the left and the house to the right of that house (if t... | {
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Similar to my argument above, $$P(X_i = 1) = P(i \text{ is red})P(i+1 \text{ is blue}|i \text{ is red}) + P(i \text{ is blue})P(i+1 \text{ is red}|i \text{ is blue})$$ $$= \frac{m}{m+n} \frac{n}{m+n-1} + \frac{n}{m+n} \frac{m}{m+n-1} = \frac{2mn}{(m+n)(m+n-1)}.$$
Thus $$E[X] = \sum_{i=1}^{m+n-1} \frac{2mn}{(m+n)(m+n-1... | {
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$\displaystyle (n-2r-2)^2 = n+2$
Hence
$\displaystyle r = \frac{n-2 \pm \sqrt{n+2}}{2}$
Which has an integer solution iff $\displaystyle n+2$ is a perfect square.
-
+1 for a different perspective (considering the complement) and for helping me see that the $\frac{120}{15(16)}$ expression in my answer is just $\frac... | {
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# Integration by parts of delta function
I am having the worst time trying to solve this integral,
$$\int g(t)\frac{d}{df}\delta[f(t)]dt,$$
$$= \int g(t)\frac{dt}{df}\frac{d}{df}\delta[f(t)]dt.$$
This should yield,
$$-\bigg[ \frac{dt}{df}\frac{d}{dt} \bigg( \frac{g(t)}{\frac{df}{dt}} \bigg) \bigg]_{f(t)=0}$$
I ha... | {
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Heuristically we have
\begin{align*} \int g \frac{d\delta}{df}(f) \, dt &= \int g \frac{d\delta}{df}(f) \frac{dt}{df} \, df = \int \frac{g}{f'} \frac{d\delta}{df}(f) \, df \\ &= - \int \frac{d}{df}\bigg(\frac{g}{f'}\bigg) \delta(f) \, df \\ &= - \int \frac{dt}{df}\frac{d}{dt}\bigg(\frac{g}{f'}\bigg) \delta(f) \, df \\... | {
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\begin{align*} \int_{\Bbb{R}} g \frac{d\delta}{df}(f) \, dt &= - \sum_{j} \mathrm{sgn} \, f'(t) \cdot \left. \frac{f'(t) g'(t)-g(t) f''(t)}{f'(t)^3} \right|_{t=x_j} \\ &= - \sum_{j} \left. \frac{1}{\left| f'(t) \right|} \bigg( \frac{g(t)}{f'(t)} \bigg)' \right|_{t=x_j} \\ \end{align*}
as desired.
Edit. Following jbc'... | {
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A rigorous and elementary definition was given by Sebastiao e Silva in the 60's and is expounded in the text "An elementary introduction to the theory of ditributions" by Campos Ferreira. Briefly, if a distribution is of finite order, i.e., is an iterated derivative of a continuous function (as $\delta$ and $\delta'$ a... | {
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"lm_q1_score": 0.9838471618625457,
"lm_q1q2_score": 0.8401196384697674,
"lm_q2_score": 0.8539127529517043,
"openwebmath_perplexity": 457.75897812411574,
"openwebmath_score": 0.9996220469474792,
"ta... |
-
Hmm, perhaps you can elaborate on why you think there is an error with the sign. I cannot see any errors, but perhaps I am mistaken. Also, thanks for the reference, in this case this was a physical problem as opposed to a strictly mathematical one, but what is math without rigor! – Shinobii Mar 17 '13 at 19:40
Oh, n... | {
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"openwebmath_score": 0.9996220469474792,
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# Problem on tangents drawn to a circle
I am solving Co-ordinate geometry by S.L. Loney. I am stuck on a problem on circles involving tangents and chords. I am not sure, if my approach is correct to solving this problem. Any inputs, tips that would lead me to correctly solve this problem would help!
Tangents are draw... | {
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Let $$K$$ be an intersection point of lines $$AO$$ and $$5x-3y=10.$$
Thus, $$CK\perp AO$$ and $$AC\perp CO$$, which gives $$CO^2=OK\cdot AO$$ or $$12=\frac{|-10|}{\sqrt{5^2+(-3)^2}}\cdot\sqrt{(5t)^2+(-3t)^2}$$ or $$|t|=\frac{6}{5},$$ which gives $$t=\frac{6}{5}$$ and $$A\left(6,-\frac{18}{5}\right).$$
yes you are thi... | {
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As a supplement to the other answers, I offer the following way to find the intersection point of the tangent lines: it is the pole of the radical axis. So, as per the other answers, subtract the equation of one circle from the other to get the equation $$5x-3y-10=0$$ of the radical axis, on which the two intersection ... | {
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"openwebmath_score": 0.6898491382598877,
"ta... |
# How to simplify absolute value expressions such as: $|x+1|+|x-1|$ algebraically?
After plotting this function on desmos, I understood that this absolute expression is actually a piecewise function.
Hence this: $$f(x)=|x+1|+|x-1|$$is the same as f(x)=\left\{\begin{align}2x\quad&\text{if }x>1\\-2x\quad &\text{if }x<-1... | {
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The blue-orange component is when $$x < -1$$, or $$|x + 1| = -x - 1$$ and $$|x - 1| = -x + 1$$. Thus the sum of these is $$-x - 1 + (-x + 1) = -2x$$.
The red-orange component is when $$-1 \leq x < 1$$, or $$|x + 1| = x + 1$$ and $$|x - 1| = -x + 1$$. Thus the sum of these is $$x + 1 + (-x + 1) = 2$$.
The red-green co... | {
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• Thank you for the answer; I'm trying to understand this. But it'd be helpful if you can elaborate this a bit? – ray_lv Dec 22 '20 at 5:47
• You solved this for 3 cases, I think. But how did you know there should be 3 cases in the first place, and how did you select the domain for each case? – ray_lv Dec 22 '20 at 5:4... | {
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"openwebmath_score": 0.9894774556159973,
"ta... |
Examveda
# One bacterium splits into eight bacteria of the next generation. But due to environment, only 50% of one generation can produced the next generation. If the seventh generation number is 4096 million, what is the number in first generation?
A. 1 million
B. 2 million
C. 4 million
D. 8 million
E. None of ... | {
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# Regression¶
What function we use to estimate a relationship between $$y$$ and the regressor(s) depends on the data that we have. As practice using multiple x-variables, let’s simulate a dataset that is generated by the following equation $$$y = 4 + 0.5 x + 3 x^2 + u$$$
The relationship between $$y$$ and $$x$$ in th... | {
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model = smf.ols('y ~ x + x2', data=df).fit()
print(model.summary())
OLS Regression Results
==============================================================================
Dep. Variable: y R-squared: 0.983
Method: Least Squares F-... | {
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"tag... |
Notes:
[1] Standard Errors assume that the covariance matrix of the errors is correctly specified.
The variable model2 stores a lot of data. Beyond holding summary output for the performance of the regression (which we’ve accessed via the model2.summary() command), we can reference the estimated parameters directly v... | {
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"openwebmath_score": 0.5410914421081543,
"tag... |
model_linear = smf.ols('y ~ x', data=df).fit()
print(model_linear.summary())
OLS Regression Results
==============================================================================
Dep. Variable: y R-squared: 0.056
Method: Least Squ... | {
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"id": null,
"lm_label": "1. YES\n2. YES",
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"lm_q1_score": 0.983847162336162,
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"lm_q2_score": 0.8539127455162773,
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"openwebmath_score": 0.5410914421081543,
"tag... |
Another informative way to jude a modeled relationship is by plotting the residuals. The residuals are the “un-expected” part of the equation. For example, in the linear-fit model, the residual is defined as $$$\hat{u} := y - \hat{y} = y - \hat{c} - \hat{\beta} x$$$$and in the quadratic-fit model the residual,$$\hat{u}... | {
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"openwebmath_score": 0.5410914421081543,
"tag... |
# Looking Beyond OLS¶
OLS works well when the $$y$$ variable in our model is a linear combination of $$x$$ variables. Note that the relationship between $$y$$ and a given regressor may be nonlinear, as in the case of $$y$$ being a function of $$x$$ and $$x^2$$. However, while we may say that $$y$$ is a nonlinear funct... | {
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"openwebmath_score": 0.5410914421081543,
"tag... |
Suppose that we instead took the curve data above and tried to fit it with linear regression.
sns.lmplot(x='x', y='y', data=curve)
<seaborn.axisgrid.FacetGrid at 0x7fe005c01ac0>
Note that in the above plot, there are estimated “probabilities” (the $$\hat{y}$$ values) that are either lower than $$0$$ or higher than ... | {
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"openwebmath_score": 0.5410914421081543,
"tag... |
Now try fitting a model using linear regression (ordinary least squares). When OLS is applied to a $$y$$ variable that only takes value 0 or 1, the model is referred to as a linear probability model (LPM).
lpm = smf.ols('y ~ x', data=df2).fit()
print(lpm.summary())
OLS Regression Results
=... | {
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"openwebmath_score": 0.5410914421081543,
"tag... |
Now fit the model using logistic regression.
logit = smf.logit('y ~ x', data=df2).fit()
print(logit.summary())
Optimization terminated successfully.
Current function value: 0.292801
Iterations 8
Logit Regression Results
==============================================================================
Dep. Variable: ... | {
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"openwebmath_score": 0.5410914421081543,
"tag... |
# Quadratic function of multiple variables
## Definition
Consider variables $x_1,x_2,\dots,x_n$. A quadratic function of the variables $x_1,x_2,\dots,x_n$ is a function of the form:
$\left(\sum_{i=1}^n \sum_{j=1}^n a_{ij} x_ix_j\right) + \left(\sum_{i=1}^n b_ix_i\right) + c$
In vector form, if we denote by $\vec{x}... | {
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"tags": n... |
Item Value Consistency with the case $n = 1$, where $f(x) = ax^2 + bx + c$, $A = (a)$ (a $1 \times 1$ matrix), $\vec{b} = (b)$ (a 1-dimensional vector)
default domain the whole of $\R^n$ the whole of $\R$
range If the matrix $A$ is not positive semidefinite or negative semidefinite, the range is all of $\R$.
If the mat... | {
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If $A$ is negative semidefinite but not negative definite, it depends on whether $\vec{b}$ is in the image of $A$. If yes, replace $A^{-1}\vec{b}$ with the solution $\vec{v}$ to $A\vec{v} = \vec{b}$, so we get a local minimum of $c - \frac{1}{4}\vec{b}^T\vec{v}$ attained at $\frac{-1}{2}\vec{v}$
If $A$ is not negative ... | {
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## Differentiation
### Partial derivatives and gradient vector
#### Case of general matrix
The partial derivative with respect to the variable $x_i$, and therefore also the $i^{th}$ coordinate of the gradient vector, is given by:
$\frac{\partial f}{\partial x_i} = \left(\sum_{j=1}^n (a_{ij} + a_{ji})x_j\right) + b_... | {
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"tags": n... |
$\frac{\partial^2 f}{\partial x_j \partial x_i} = a_{ij} + a_{ji}$
Thus, the Hessian matrix of the quadratic function is given as:
$H(f)(\vec{x}) = A + A^T$
Note that this is independent of the choice of $\vec{x}$. This fact is true only because of the nature of the function: for more general functional forms, the H... | {
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## Points and intervals of interest
For the discussion here, assume that $A$ is symmetric. If it is not, replace $A$ by the matrix $(A + A^T)/2$.
### Critical points
#### Case that the matrix $A$ is invertible
To find the critical points, we need to set the gradient vector equal to zero. This gives the vector equat... | {
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• If $A$ is a symmetric positive-definite matrix, i.e., all its eigenvalues are positive, then the unique critical point $\frac{-1}{2}A^{-1}\vec{b}$ is a point of local minimum and is the unique point of absolute minimum (an alternate derivation of this fact is later in the page).
• If $A$ is negative-definite matrix, ... | {
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• In the case that $A$ is a symmetric positive-semidefinite matrix, i.e., all its nonzero eigenvalues are positive, the function attains a local minimum and also its absolute minimum at all its critical points. Note that, since the function is constant at this minimum value on the affine space, none of these is a point... | {
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"openwebmath_score": 0.9431282877922058,
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## Alternate analysis of extreme values
For the discussion of cases, assume that $A$ is a symmetric matrix. If $A$ is not symmetric, replace it by the symmetric matrix $(A + A^T)/2$.
### Positive definite case
First, we consider the case where $A$ is a symmetric positive definite matrix. In other words, we can write... | {
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# Divisibility property for sequence $a_{n+2}=-2(n-1)(n+3)a_n-(2n+3)a_{n+1}$
Let $$(a_n)$$ be the sequence uniquely defined by $$a_1=0,a_2=1$$ and
$$a_{n+2}=-2(n-1)(n+3)a_n-(2n+3)a_{n+1}$$
Can anybody show (or provide a counterexample) that $$p|a_{p-2}$$ and $$p|a_{p-1}$$ for any prime $$p\geq 5$$ ? I have checked t... | {
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"openwebmath_score": 0.9613379240036011,
"tags"... |
Currently the coefficients contain quadratic polynomials in $n$, but we can make them linear, which makes it a bit easier to work with. With substitution $b_n=\frac{a_n}{(n+2)!}$ and some algebra we get
$$(n+4)b_{n+2}+(2n+3)b_{n+1}+(2n-2)b_n=0$$
Now that the coefficients are linear, let's try to find its generating f... | {
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"lm_q2_score": 0.8459424373085146,
"openwebmath_perplexity": 137.4593926955677,
"openwebmath_score": 0.9613379240036011,
"tags"... |
Now to get information about the coefficients, let's expand it into the series. The power on the right is by Binomial series $$(2x^2+2x+1)^{3/2}=\sum_{k=0}^{\infty}\binom{3/2}{k}2^k(x^2+x)^k$$ Using Binomial theorem for the inner sum and playing with the indicies, we obtain $$[x^n](2x^2+2x+1)^{3/2} = \sum_{n/2 \leq k \... | {
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"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9930961631547577,
"lm_q1q2_score": 0.84010218874087,
"lm_q2_score": 0.8459424373085146,
"openwebmath_perplexity": 137.4593926955677,
"openwebmath_score": 0.9613379240036011,
"tags"... |
• good work indeed ! Apr 14 '18 at 13:36
• Thank you for this thorough effort. May I ask which CAS you used to solve the differential equation ? Apr 15 '18 at 8:00
• @EwanDelanoy I have used Maple dsolve((diff(y(x), x))*(2*x^3+2*x^2+x) = (1/6)*x^2+(2*x^2-x-2)*y(x), y(x)), but it is really just a linear differential equ... | {
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However, there is an alternative approach using the exponential generating function of $$\;a_n,\;$$ namely $$\;A(x):=\sum_{n=0}^\infty a_nx^n/n!.\;$$ Using equation $$(0)$$ for $$\;a_n\;$$ we get the differential equation $$1 = -6A(x) +(3+6x)\frac{d}{dx}A(x) + (1+2x+2x^2)\frac{d^2}{dx^2}A(x).$$ Taking the derivative of... | {
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## Fibonacci multiples
I haven’t written anything here in a while, but hope to write more regularly now that the semester is over—I have a series on combinatorial proofs to finish up, some books to review, and a few other things planned. But to ease back into things, here’s a little puzzle for you. Recall that the Fib... | {
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• Matt Gardner Spencer says:
The key to my proof is using pictures like those here: https://mathlesstraveled.com/2011/05/24/post-without-words-1/ (especially those posted by Xander in the comments) to prove
$F_{mn}=F_n\cdot F_{(m-1)n+1}+F_{n-1}\cdot F_{(m-1)n}$ and then using induction.
• Brent says:
Yes, I can see ... | {
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"... |
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# Introduction to trigonometric substitution
Introduction to trigonometric substitution.
## Want ... | {
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Before differentiating you have: x = 2·sin(θ). After differentiating with respect to θ you have:
dx/dθ = 2·cos(θ) | {
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Finally, you solve for dx:
dx = 2·cos(θ)·dθ
• I've noticed that the question was a lot like the derivative of arcsin(theta) with the exception that instead of 1/sqrt(1-x^2), we had 1/sqrt(4-x^2). The answer, in the end, was just arcsin(x/2). Is there a pattern to it? If it is 1/sqrt(1-x^2), it is arcsin(x/1), and if it... | {
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That means one of the sides, lets say a, is equal to sqrt(c² - b²).
Now we have been given an expression that looks just like that: sqrt(4 - x²), which can be rewritten as sqrt(2² - x²).
That means that for this trig sub triangle, one side (the b side) is equal to x, the other side, the a side is equal to sqrt(2² - x²)... | {
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https://en.wikiversity.org/wiki/Trigonometric_Substitutions
• What if we would've chosen the other side as our x in the beginning?
• Hi, tuf62486!
There are two scenarios, and these are as follows:
- Scenario ONE: https://i.imgur.com/atiQ9yT.png
- Scenario TWO: https://i.imgur.com/1AmZgHZ.png
Explanations:
- Scena... | {
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Must this rng be a ring?
A rng is a ring without the assumption that the ring contains an identity. Consider a finite rng $\mathbf{R}$.
I am investigating conditions that get close forcing an identity but not quite. The closest condition I can think of is the following:
If $a\in \mathbf{R}$ is non-zero then there is... | {
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We say that a commutative rng $R$ has property $\mathcal{P}$ if, for all nonzero $a\in R$, there is some $b\in R$ with $ab\neq 0$. The zero ring has property $\mathcal{P}$, and it has a unit. Let $R$ be a finite nonzero commutative rng such that all smaller commutative rngs with property $\mathcal{P}$ have a unit.
Pic... | {
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It's pretty easy to reduce to when $|R|$ is the power of a prime.
If $|R|=mn$ with $\gcd(m,n)=1$, solve $mx+ny=1$. Show that $$R\to (mR)\times (nR); a\mapsto (mxa,nya)$$ is an isomorphism of rngs.
Now, if $ma\neq 0$, then $mab\neq 0$ for some $b\in R$. But note that $(ma)(mxb)=m(1-ny)(ab)$. So this means that $mR$ ha... | {
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# PCA on correlation or covariance?
What are the main differences between performing principal component analysis (PCA) on the correlation matrix and on the covariance matrix? Do they give the same results?
You tend to use the covariance matrix when the variable scales are similar and the correlation matrix when vari... | {
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Notice also that the outlying individuals (in this data set) are outliers regardless of whether the covariance or correlation matrix is used.
• What is the situation, if I convert the variables to z-scores first? May 18, 2013 at 16:00
• @Jirka-x1 the covariance matrix of standardized variables (i.e. z scores) equals t... | {
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The argument against automatically using correlation matrices is that it is quite a brutal way of standardising your data. The problem with automatically using the covariance matrix, which is very apparent with that heptathalon data, is that the variables with the highest variance will dominate the first principal comp... | {
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TRANSFORMED DATA: If the data have been transformed via normalization, percentiles, or mean-zero standardization (i.e., $$Z$$-scores), so that the range and scale of all the continuous variables is the same, then you could use the Covariance matrix $$\mathbf{C}$$ without any problems. (correlation will mean-zero standa... | {
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For example, if you plug in a $$pct_i$$ value 0.025, you will get $$-1.96=\Phi^{-1}(0.025)$$. Same goes for a plugin value of $$pct_i=0.975$$, you'll get $$1.96=\Phi^{-1}(0.975)$$.
Use of VDW scores is very popular in genetics, where many variables are transformed into VDW scores, and then input into analyses. The adv... | {
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A common answer is to suggest that covariance is used when variables are on the same scale, and correlation when their scales are different. However, this is only true when scale of the variables isn't a factor. Otherwise, why would anyone ever do covariance PCA? It would be safer to always perform correlation PCA.
Im... | {
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To interpret correlation-based and covariance-based PCA, one can then argue that:
1. Covariance-based PCA is equivalent to MLPCA whenever the variance-covariance matrix of the measurement errors is assumed diagonal with equal elements on its diagonal. The measurement error variance parameter can then be estimated by a... | {
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[1] Wentzell, P. D., Andrews, D. T., Hamilton, D. C., Faber, K., & Kowalski, B. R. (1997). Maximum likelihood principal component analysis. Journal of Chemometrics, 11(4), 339-366.
[2] Wentzell, P. D., & Lohnes, M. T. (1999). Maximum likelihood principal component analysis with correlated measurement errors: theoretic... | {
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My suggestion would be to do BOTH a correlation- and covariance-based PCA. If the two give the same (or very similar, whatever this may mean) PCs, then you can be reassured you've got an answer that is meaningul. If they give widely different PCs don't use PCA, because two different answers to one problem is not sensib... | {
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# Health Risk Probability
Question: An actuary is studying the prevalence of three health risk factors, denoted by A, B, and C, within a population of women. For each of the three factors, the probability is 0.1 that a woman in the population only has this risk factor (and no others). For any two of three factors, the... | {
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First, lets write down what we know:
• $P(A \cap B' \cap C') = P(A' \cap B \cap C')= P(A' \cap B' \cap C) = 0.1$
• $P(A \cap B \cap C') = P(A \cap B' \cap C)= P(A' \cap B \cap C) = 0.12$
• $P(A \cap B \cap C|A,B) = \frac{1}{3}.$
As you noted, we are interested in computing the probability $$P(A' \cap B' \cap C'|A') =... | {
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there is a mistake cause 0.6+0.34=0.94 thus P(A’)=0.06. and that changes the whole solution. also in the end 1-0.6-0.3-0.36= -0,26?
• Welcome to stackexchange. That is just a typo in the answer from @megas . He or she uses the correct value in what follows, and reaches the correct conclusion. You should delete this (n... | {
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Trying to understand the basic about recurrence trees
I have little background on recurrence trees, and I am working on the following exercise:
Exercise. Take $$T(n) = 2T(n/2) + 3\log(n)$$. Draw the recurrence trees for $$n=2$$ and $$n=4$$. What can we conclude complexity-wise?
My attempt. For $$n=2$$ we have the fo... | {
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Is my attempt wrong at any point?
• Hint, suppose $n=2^k$ for some natural number $k$. Can you simplify the total sum? Apr 16, 2022 at 0:26
• Taking your hint, we would have that the total sum would be $\sum_{i=0}^{lg 2^k = k} 3*2^i \log(\frac{2^k}{2^i}) = \sum_{i=0}^{k} 3*2^i \log(2^{k-i}) = 3\log(2)\sum_{i=0}^{k} (k... | {
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Suppose $$n=2^k$$ for some integer $$k>0$$. Do not forget $$t_1$$.
$$T(n)=2^kt_1 + \sum_{i=0}^{\log_2(2^k)} 3\cdot2^i \log_2(\frac{2^k}{2^i}) = 2^kt_1 + \sum_{i=0}^{k} 3\cdot2^i (k-i)= 2^kt_1 + 3\sum_{i=0}^{k-1}2^i (k-i)$$
Intuitively, the sum $$\sum_{i=0}^{k-1}2^i (k-i)$$ is not much different from $$2^{k-1}$$, the ... | {
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• I am sorry, I don't understand where $2^k t_1$ comes from in the $T(n)$ deduction. I understand it is related to us getting to the base case $T(1)$, but why does it origin $2^k t_1$ ? (I.e., how do you find the constant that multiplies by $t_1$ in the general formula for $T(n)$) ? Apr 18, 2022 at 9:41
• @roto, please... | {
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# How to find the probability of drawing colored marbles without replacement?
There are $$2$$ red marbles, $$3$$ white marbles and $$5$$ black marbles in a bag. What is the probability of drawing $$1$$ red marble, $$2$$ white marbles and $$3$$ black marbles, if $$6$$ marbles are drawn without replacement?
Answer: $$\... | {
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• if $$k, then subscripts are needed
• if $$k=n$$, then subscripts are not needed
Case $$1$$. $$k
Suppose I want to select $$r=2$$ marbles from a bag with $$1$$ white marble and $$2$$ black marbles, so
• $$k=2$$ and $$n=3$$, so $$k
• So $$\binom{n}{r}=3$$ with combinations: {$$W_1,B_1$$}, {$$W_1,B_2$$}, {$$B_1,B_2$... | {
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1. There are $$10$$ marbles in the bag: marbles $$R_1, R_2$$ are red, marbles $$W_1, W_2, W_3$$ are white, and marbles $$B_1, B_2, B_3, B_4, B_5$$ are black.
2. We draw a uniformly random subset of $$6$$ of the marbles, such as $$\{B_1, B_2, B_3, B_4, B_5, W_1\}$$ or $$\{R_2, W_1, W_3, B_2, B_4, B_5\}$$.
In this case,... | {
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• I understand everything now except the reason why it is necessary to add the subscripts for each colored ball. For example, {$B_1,B_2,B_3,B_4,W_1,W_3$} is distinct from {$B_2,B_1,B_3,B_4,W_1,W_2$}, but the result is the same: $4$ black balls are chosen and $2$ white balls is chosen. So why are the subscripts required... | {
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# Solving Complex Differential Equations using Power Series
I've been looking through old questions for some time and I found an interesting topic (Complex differential equation). I decided to try solving a complex differential equation with a similar premise. I looked at the equation $$z' = \overline{\mathbb{z}} +it$... | {
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This pattern continues for later values of n, giving us the relations
$$C_n = \frac{C_0+i}{n!} , n \in \mathbb{2Z} , n\geq2$$ $$C_n = \frac{C_1-i}{n!} , n \in \mathbb{2Z+1} , n>2$$
With these relations, I was able to then build the final solution of $$z(t)$$ $$z(t) = \displaystyle\sum_{n=0}^{\infty} C_nt^n$$ $$z(t) =... | {
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• I have to commend you for your effort! If this was your first time using TeX and solving any sort of equation like this, it is very impressive! Oct 21, 2019 at 20:36
• Your solution is in fact correct. You can absorb the $i$'s into the arbitrary constants. The constants have to be complex numbers so when you plug in ... | {
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# Moment Of Inertia Of Quarter Circle | {
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and about… • M. There is a plate with moment of inertia Ip = 0. 3, Appendix B 7 5 Nov Shear Stress in Beams 6. The two disks are then linked together without the aid of any external torques, so that $2. The moment of inertia = I = πR 4 /8 In the case of a quarter circle the expression is given as: [Image will be Upload... | {
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50 m from the center, the carousel has an angular velocity of 0. Define following terms 1. Meidensha Electric Manufacturing Company, Ltd. Angular Momentum Any moving body has inertia. 4 A flywheel in the shape of a solid cylinder of radius R = 0. Polar Moment of Inertia The second moment of Area A with respect to the p... | {
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to find a way in AutoCAD to calculate the (momet of inertia). Think of the area of a circle. The handout is available here; ask your classmates for any clarifications and then check in with faculty as needed. The apparatus comes with a set of eight precisely machined masses which can be attached to the copper disc to v... | {
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