text:
Completeness of Propositional Logic: Help understanding a proof.

I'm reading through wikipedia's proof of the completeness of propositional logic and I'm having trouble understanding the last parts of the proof:

At part III, why is "if $G^*$ contains $C$ and $C\implies B$ then it contains $B$" true?

At part IV, it is claimed that as $G^*$ is "truth-like" (I've never seen that term before), then there is a valuation (what does $G^*$-canonical valuation mean?) $v$ such that $v(G^*)=\{1\}$, how can we assure the existence of this valuation?

For your first question: suppose $G^*$ contains $C$ and $C\implies B$ - say, they are both in $G_{17}$. Then does $G_{17}\cup\{B\}$ prove $A$? (Basically, show that - whenever the sentence $B$ happened to show up in your enumeration of sentences - it could not be the case that adding $B$ to $G^*$ would yield a proof of $A$. So, we added $B$ to $G^*$.)

As for your second question: "truthlike" is defined in part III:

This means that it contains $C$ only if it does not contain $\neg C$; If it contains $C$ and contains "If $C$ then $B$" then it also contains $B$; and so forth.

The "and so forth" is a bit vague of course; what's really meant is that $G^*$ is complete and closed under deductions: for every $C$ either $G^*\vdash C$ or $G^*\vdash \neg C$, and if $G^*\vdash C$ then $C\in G^*$.

Now the valuation is just $$\nu(C)=1\iff C\in G^*, \quad\nu(C)=0\iff \neg C\in G^*.$$

(It's also important that $G^*$ is consistent, but this is easy to show. If $G^*$ is inconsistent, then some $G_n$ was inconsistent, but then $G_n\vdash A$ since inconsistent theories prove everything. Oops.)

• Thanks for your answer. I have a few questions, truthlike is a different notion to "complete and closed under deductions", could you explain how it follows from the construction of $G^*$ that it is complete and closed under deductions? Also, I think that if $G^*$ is closed and complete, we have that either $p$ or $\neg p$ are in $G^*$ for all the variables $p$, thus the valuation we need is the extension $v_f:\text{Form}\to\{0,1\}$ of the function $$f:\text{Var}\to \{0,1\}, f(p) = 1 \iff p\in G^*$$ But then, how do we show that $v_f$ satisfies $G^*$?} – YoTengoUnLCD Jan 20 '16 at 20:40
• @YoTengoUnLCD For completeness: suppose $C\not\in G^*$. Then by construction of $G^*$, we must have $G^*\cup\{C\}\vdash A$ (do you see why?). Similarly, if $\neg C\not\in G^*$, then $G^*\cup\{\neg C\}\vdash A$. So $G^*\cup \{C\vee\neg C\}\vdash A$, and so $G^*\vdash A$ - a contradiction. For closure under deductions, it's similar: if $G^*\vdash C$, then $G^*\cup\{C\}\not\vdash A$, so $C$ was put into $G^*$ when it was considered ($C=E_n$ for some $n$; since $G^*\cup\{C\}\not\vdash A$, certainly $G_n\cup\{C\}\not\vdash A$!). Does this help? – Noah Schweber Jan 20 '16 at 21:08
• Yes, it helped, thank you a lot. The only leap I could not complete was that $G^* \cup \{C\vee \neg C\}\vdash A\rightarrow G^*\vdash A$: I understand that $C\vee \neg C$ is a tautology, but I still don't get how to prove that implication. Oh, and also, why does $G^*\cup\{C\}\vdash A$ and $G^*\cup\{\neg C\}\vdash A$ imply that $G^*\cup\{C\vee \neg C\}\vdash A$ ? – YoTengoUnLCD Jan 20 '16 at 23:40
• @YoTengoUnLCD The details will depend on the exact way you're formalizing "$\vdash$", but for instance (using sequent calculus) the first point is cut: if $T\vdash x$ and $T\cup\{x\}\vdash y$, then $T\vdash y$. The second is e.g. $\vee L$ in en.wikipedia.org/wiki/Sequent_calculus#Inference_rules. But I guarantee whatever system you're using allows these. – Noah Schweber Jan 20 '16 at 23:45
• In my course, $A \vdash \alpha$ is defined as "there exists a proof of $\alpha$ from $A$ using 3 axioms and MP (this is an example of a proof (my question) math.stackexchange.com/questions/1620203/…), I'll try to use my system to prove those rules. – YoTengoUnLCD Jan 20 '16 at 23:51


text:
[texhax] RE: [listing.sty] \item[lstinline{foo}] howto?

Morten Høgholm moho01ab at student.cbs.dk
Thu Nov 18 08:50:16 CET 2004

On Thu, 18 Nov 2004 07:31:16 +0100, Goebel, Juergen, OPE26

>> Pavel Minev Penev [mailto:pavpen at berkeley.edu] wrote:
>
>> I don't have time to look at this, but as far as I remember, listings
>> plays with the catcodes of characters at input (what mostly
>> all verbatim
>> environments do), thus, if you are trying to use a listing
>> anywhere, but
>> at the top level of macro expansion, it will not work
>> (because catcodes
>> will already have been defined for the characters to input).
>
> Thanks for the explanation. I'm going to see whether
> the author thinks there is a solution ...

What you can do is something like

\newsavebox\JGbox

\sbox\JGbox{\lstinline!ze_def.h!}
\item[\usebox\JGbox]

--
Morten Høgholm

format: plain


text:
# Binarizing Data in a Network using the sign function

I often see the use of the sign function in machine learning models as a way to binarize data (see eqn 1 here for an example). But the derivative of the sign function is the dirac delta function, so backpropagating through the network will yield either 0 or infinity? I'm confused as to why it makes any sense to still use it?

A more concrete example:

Consider a network where each node in a hidden layer $$n_i$$ measures whether the current training point $$x$$ is within distance $$d_i$$ from some anchor point $$a_i$$. This can be represented as $$sign(||x-a_i||_2 - d_i)$$. This is a rudimentary example of a locality preserving embedding network. Naturally, the loss function will depend on the result of this node in some way. Thereofre when I try to compute the partial derivate of the loss function with respec to $$a_i$$ or $$d_i$$, it will result in a gradient of 0 and learning will not be possible.

Thanks!


text:
Outlook: Eagle Bulk Shipping Inc. Common Stock is assigned short-term Ba1 & long-term Ba1 estimated rating.
Dominant Strategy : Sell
Time series to forecast n: 03 Mar 2023 for (n+16 weeks)
Methodology : Modular Neural Network (Emotional Trigger/Responses Analysis)

## Abstract

Eagle Bulk Shipping Inc. Common Stock prediction model is evaluated with Modular Neural Network (Emotional Trigger/Responses Analysis) and Multiple Regression1,2,3,4 and it is concluded that the EGLE stock is predictable in the short/long term. According to price forecasts for (n+16 weeks) period, the dominant strategy among neural network is: Sell

## Key Points

1. What is the use of Markov decision process?
2. Decision Making
3. Can neural networks predict stock market?

## EGLE Target Price Prediction Modeling Methodology

We consider Eagle Bulk Shipping Inc. Common Stock Decision Process with Modular Neural Network (Emotional Trigger/Responses Analysis) where A is the set of discrete actions of EGLE stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4

F(Multiple Regression)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Modular Neural Network (Emotional Trigger/Responses Analysis)) X S(n):→ (n+16 weeks) $∑ i = 1 n r i$

n:Time series to forecast

p:Price signals of EGLE stock

j:Nash equilibria (Neural Network)

k:Dominated move

a:Best response for target price

For further technical information as per how our model work we invite you to visit the article below:

How do AC Investment Research machine learning (predictive) algorithms actually work?

## EGLE Stock Forecast (Buy or Sell) for (n+16 weeks)

Sample Set: Neural Network
Stock/Index: EGLE Eagle Bulk Shipping Inc. Common Stock
Time series to forecast n: 03 Mar 2023 for (n+16 weeks)

According to price forecasts for (n+16 weeks) period, the dominant strategy among neural network is: Sell

X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)

Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)

Z axis (Grey to Black): *Technical Analysis%

## IFRS Reconciliation Adjustments for Eagle Bulk Shipping Inc. Common Stock

1. If an entity previously accounted at cost (in accordance with IAS 39), for an investment in an equity instrument that does not have a quoted price in an active market for an identical instrument (ie a Level 1 input) (or for a derivative asset that is linked to and must be settled by delivery of such an equity instrument) it shall measure that instrument at fair value at the date of initial application. Any difference between the previous carrying amount and the fair value shall be recognised in the opening retained earnings (or other component of equity, as appropriate) of the reporting period that includes the date of initial application.
2. For a discontinued hedging relationship, when the interest rate benchmark on which the hedged future cash flows had been based is changed as required by interest rate benchmark reform, for the purpose of applying paragraph 6.5.12 in order to determine whether the hedged future cash flows are expected to occur, the amount accumulated in the cash flow hedge reserve for that hedging relationship shall be deemed to be based on the alternative benchmark rate on which the hedged future cash flows will be based.
3. An entity can also designate only changes in the cash flows or fair value of a hedged item above or below a specified price or other variable (a 'one-sided risk'). The intrinsic value of a purchased option hedging instrument (assuming that it has the same principal terms as the designated risk), but not its time value, reflects a one-sided risk in a hedged item. For example, an entity can designate the variability of future cash flow outcomes resulting from a price increase of a forecast commodity purchase. In such a situation, the entity designates only cash flow losses that result from an increase in the price above the specified level. The hedged risk does not include the time value of a purchased option, because the time value is not a component of the forecast transaction that affects profit or loss.
4. Paragraph 5.5.4 requires that lifetime expected credit losses are recognised on all financial instruments for which there has been significant increases in credit risk since initial recognition. In order to meet this objective, if an entity is not able to group financial instruments for which the credit risk is considered to have increased significantly since initial recognition based on shared credit risk characteristics, the entity should recognise lifetime expected credit losses on a portion of the financial assets for which credit risk is deemed to have increased significantly. The aggregation of financial instruments to assess whether there are changes in credit risk on a collective basis may change over time as new information becomes available on groups of, or individual, financial instruments.

*International Financial Reporting Standards (IFRS) adjustment process involves reviewing the company's financial statements and identifying any differences between the company's current accounting practices and the requirements of the IFRS. If there are any such differences, neural network makes adjustments to financial statements to bring them into compliance with the IFRS.

## Conclusions

Eagle Bulk Shipping Inc. Common Stock is assigned short-term Ba1 & long-term Ba1 estimated rating. Eagle Bulk Shipping Inc. Common Stock prediction model is evaluated with Modular Neural Network (Emotional Trigger/Responses Analysis) and Multiple Regression1,2,3,4 and it is concluded that the EGLE stock is predictable in the short/long term. According to price forecasts for (n+16 weeks) period, the dominant strategy among neural network is: Sell

### EGLE Eagle Bulk Shipping Inc. Common Stock Financial Analysis*

Rating Short-Term Long-Term Senior
Outlook*Ba1Ba1
Income StatementBaa2Ba2
Balance SheetCaa2B3
Leverage RatiosBaa2Baa2
Cash FlowBaa2B2
Rates of Return and ProfitabilityCaa2B3

*Financial analysis is the process of evaluating a company's financial performance and position by neural network. It involves reviewing the company's financial statements, including the balance sheet, income statement, and cash flow statement, as well as other financial reports and documents.
How does neural network examine financial reports and understand financial state of the company?

### Prediction Confidence Score

Trust metric by Neural Network: 74 out of 100 with 746 signals.

## References

1. Künzel S, Sekhon J, Bickel P, Yu B. 2017. Meta-learners for estimating heterogeneous treatment effects using machine learning. arXiv:1706.03461 [math.ST]
2. Imbens G, Wooldridge J. 2009. Recent developments in the econometrics of program evaluation. J. Econ. Lit. 47:5–86
3. Candès E, Tao T. 2007. The Dantzig selector: statistical estimation when p is much larger than n. Ann. Stat. 35:2313–51
4. Abadie A, Imbens GW. 2011. Bias-corrected matching estimators for average treatment effects. J. Bus. Econ. Stat. 29:1–11
5. G. Theocharous and A. Hallak. Lifetime value marketing using reinforcement learning. RLDM 2013, page 19, 2013
6. Mnih A, Kavukcuoglu K. 2013. Learning word embeddings efficiently with noise-contrastive estimation. In Advances in Neural Information Processing Systems, Vol. 26, ed. Z Ghahramani, M Welling, C Cortes, ND Lawrence, KQ Weinberger, pp. 2265–73. San Diego, CA: Neural Inf. Process. Syst. Found.
7. Ashley, R. (1983), "On the usefulness of macroeconomic forecasts as inputs to forecasting models," Journal of Forecasting, 2, 211–223.
Frequently Asked QuestionsQ: What is the prediction methodology for EGLE stock?
A: EGLE stock prediction methodology: We evaluate the prediction models Modular Neural Network (Emotional Trigger/Responses Analysis) and Multiple Regression
Q: Is EGLE stock a buy or sell?
A: The dominant strategy among neural network is to Sell EGLE Stock.
Q: Is Eagle Bulk Shipping Inc. Common Stock stock a good investment?
A: The consensus rating for Eagle Bulk Shipping Inc. Common Stock is Sell and is assigned short-term Ba1 & long-term Ba1 estimated rating.
Q: What is the consensus rating of EGLE stock?
A: The consensus rating for EGLE is Sell.
Q: What is the prediction period for EGLE stock?
A: The prediction period for EGLE is (n+16 weeks)


text:
# Tag Archives: Machine Vision

## Configure OpenCV-Python development environment in Windows

In older days, it is relative complex to configure a OpenCV development environment. You have to install an IDE, e.g. Mircosoft Visual Studio, and have to set many configurations of the project, such as the include folder, the header folder and where the IDE can find the third-party OpenCV library. But after Python becomes one of the main streams, things has changed. Python has proven to be a better and easier way to program and test computer vision programs, compared with C++. In this article, I will show you the simple steps that could help you to configure an OpenCV-Python ready environment in Windows.

1. First, you have to download and install Python 2.7 and Numpy. Install them to their default locations, in order to avoid any further issues. The reason to use Python 2.7 instead of Python 3.x is that Python 2.7 is best supported by most third-party libraries, while Python 3 is too new for all the scientific libraries to be fully supported.

2. Download the latest OpenCV release from Sourceforge and extract it into a system folder, e.g., C:\Program Files\opencv.

3. Goto $(opencv folder)\build\python\2.7\x86 or \x64 based on your machine type. Copy cv2.pyd to$(python folder)\lib\site-packages.

That’s all you have to do. Really easy and will only cost you 10 minutes. Now let’s open the Python IDLE and test if the environment is working:

import numpy
import cv2

print cv2.__version__

Run the code and if you got the version printed and no error message, then congratulations: you have succeed. But if you did get error message, then you should check the version of both the Python and the OpenCV is correct. If you have multiple versions of Python, then check if you have run the right version of IDLE.

Before we try more complicated vision algorithms, let’s simply connect a USB camera first and try to get the video stream from the camera and display it on the screen:

import cv2

cap = cv2.VideoCapture(0)

while(True):


text:
# Analysis of a linear-time algorithm for longest palindromic substring

### Background

$\newcommand\ldotd{\mathinner{..}}$Last month, I heard about a new linear-time algorithm to determine the longest palindromic substring called Jeuring's algorithm. It seemed interesting, therefore I had a try to analyze the algorithm. It's not too difficult to show that it's an $\Theta(n)$-time-algorithm, but I want to observe more closely.

I wrote a C++ implementation, and an essay to prove and analyze the algorithm. It's somewhat long, therefore I will only post the critical part of the code, where you can learn the algorithm by heart, see the succeeding section. You can get all my previous work from github. pldrm.nw is the literate programming source file, which produces the pldrm.cc and pldrm.pdf. pldrm.pdf is the essay I've written.

### Problem

Let $A=$ the number of times goto statement is executed in move loop, and $B=$ the number of times move loop is executed where goto statement is NOT executed, $C=$ the number of times a[++j]=min(a[p],l) is executed in move loop. You can see the code in the following section. I found that $A+B+C=2n$ and $B=\sum_{k=2}^{n+1}[b_k=1]$, where $b_k$ is the length of longest tail palindrome of $s[0\ldotd k]$, and $[P]$ is Iverson bracket. For details, you can read my pdf from github. I'm looking for somebody to help me analyze the quantity $A,C$. Any help? Thanks!

### The critical part of the algorithm

Given that $s[1\ldotd n]$ is the string inputted, and $n$ is the length of $s$, where $s[0]=1,s[n+1]=0$. We say $l+r$ is the center of substring $s[l\ldotd r]$. For example, $4$ is the center of $s[2\ldotd2]$ or $s[1\ldotd3]$. $a_k$ is the length of the longest palindrome whose center is $k$, and $a[0\ldotd2n]$ is the array to save $\langle a_k\rangle$. The algorithm is used to determine $a_k$. We call some string $A$ is a tail palindrome of the other string $B$ if and only if $A$ is a palindromic tail substring of $B$. For example, $A=aba$ and $B=aaaababa$, where $A$ is palindromic and $A$ is a tail substring of $B$.

Here's the critical part of the code:

<<main loop>>=
j = 1;
l = 1;
a[0] = 1;
a[1] = 0;
for (int k=2; k<=n+1; k++) {
<<process>>
;
}
@

Process is made up of an infinite loop, which is used to find the longest tail palindrome of $s[0\ldotd k]$. There are two exits of it. One is in extension subroutine, while the other one is in move loop. The way of exit is goto advance;.

<<process>>=
for (;;) {
<<check>>
<<move loop>>
}
@

The check subroutine checks whether a tail palindrome of $s[0\ldotd k-1]$ could be extended to that of $s[0\ldotd k]$. If so, exit from the process loop and advance $k$, otherwise start the move loop.

<<check>>=
if (s[k] == s[k-l-1]) {
l += 2;
}
@

Here's the move loop, which looks short and easy. It's used to find a shorter tail palindrome of $s[0\ldotd k-1]$.

<<move loop>>=
a[++j] = l;
for (p=j-1; --l>=0&&l!=a[p]; p--) {
a[++j] = min(a[p], l);
}
if (l < 0) {
l = 1;
}
@

• Even if looking at your pdf-file, it's hard to understand what is going on. I think your question should be more self-contained and direct: what exactly does the algorithm look like? With some pseudocode then, where's the hard part? – Juho Jul 28 '12 at 9:16
• @Juho I restructed my post so that you can see my problem explicitly. To your disappointment, I have no psuedo code. I thought the C++ code and the comments around code is readable, therefore I pasted the critical part of my code in the main post. Thanks! – Yai0Phah Jul 28 '12 at 10:35
• Pseudocode is generally easier to grasp and therefore to analyse. Also, it would be useful if you defined $A$ and $C$ completely here, for links may break. – Raphael Jul 28 '12 at 11:19
• @Raphael Thanks. the definition of $A,B,C$ are stated here. $B$ is somewhat hard to describe. – Yai0Phah Jul 28 '12 at 14:19


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# Is there a function such that the $n^{th}$ derivative equals the $n^{th}$ power of the first derivative?

I have seen functions whose second derivative is equal to the square of the first derivative. So is there a function whose $$n^{th}$$ derivative is equal to the $$n^{th}$$ power of the first derivative. Or, is there a function such that -

$$\left(\frac{df}{dx}\right)^n=\frac{d^{n}f}{dx^{n}}$$

• this is just an ode.. $y^{(n)} - (y')^n = 0$ – K. Y May 26 at 3:24
• Note that any constant function works. – K. Y May 26 at 3:27
• $f(x)=0$ The closest other functions I can think of are exponential functions. They don't satisfy the equation, but they're pretty close to. – Badr B May 26 at 3:35
• WA can't solve it even for $n=3$... – Micah May 26 at 3:52
• Did you want this to be true for all $n$, or just a given $n$? – Brian Tung May 26 at 5:00

$$f(x) = -((n-1)!)^{1/(n-1)} \ln(x)$$ is a solution.
• @Somos: I think that is an error in the edit by Parcly Taxel. That previously read $-[(n-1)!]^\frac{1}{n-1}\ln x$, and the $(n-1)!$ was replaced by $\Gamma(x)$, rather than $\Gamma(n)$. (Note that $n$ is presumed to be an integer, so the expression did not need to be made continuous.) I'm going to revert the edit. – Brian Tung May 26 at 5:10
• If $y = f'$, the equation says $y^n = y^{(n-1)}$. Constant times a power of $x$ is an obvious thing to try... – Robert Israel May 26 at 14:51


text:
## The mystery of the shifting tropical rain belt

This week’s science+comics interview is brought to you by Yen-Ting Hwang, a UW Atmospheric Sciences student in the final year of her PhD. Ting’s work is focused on large-scale climate dynamics.

The Sahel region in Africa is a semi-arid boundary along the southern edge of the Sahara desert. The people living in this region are balanced on the very edge of habitable terrain, and rely on a short annual rainy season for their crops to grow. The slightest shift in annual climate can result in devastating crop failures and ultimately widespread famine. During the second half of the twentieth century, the region was hit by year after year of drought.

For her final thesis chapter, Ting has tried to understand why that drought might have happened when it did. The life-sustaining annual rainfall in the region occurs as part of a seasonally migrating tropical rainfall pattern. Scientists call this the ITCZ, or the inter-tropical convergence zone, and it’s a rain band that circles the globe like a belt. Solar radiation near the equator is far stronger than at the poles, and when the moist air near the equator is heated, it rises. The rising air eventually reaches an altitude where it’s cooled, and the moisture that was carried aloft condenses to form clouds and rain.

This air is transported poleward at high altitude until it reaches about 30 degrees latitude either north or south, at which point it sinks down toward the earth’s surface. The air then moves back toward the equator, picking up moisture along the way. This cycle is called Hadley Cell circulation.

By piecing together climate observations, scientists know that the drought in the Sahel region between about 1950-1990 was related to a very slight southward shift of the tropical rain belt. This type of shift happens when the temperature difference between the northern and southern hemispheres changes. If the northern hemisphere is cooler, the northern Hadley cell will strengthen as it tries to pull warm air up from the south, and the tropical rainfall band will shift southward.

To answer this question, Ting compared results from twenty different IPCC models (IPCC = Intergovernmental Panel on Climate Change). She looked at a variety of different possible factors that could result in an uneven heating or cooling of the planet between the northern and southern hemispheres – things like clouds and ocean circulation. The factor that showed the strongest correlation was surprising: it was the aerosols!

Aerosols are basically tiny particles that are suspended in the air. Naturally occurring aerosols are always floating around our atmosphere – dust from deserts, smoke from forest fires, and even sea salt. But there are also anthropogenic aerosols – the ones that occur as a result of pollution. There are different types, but Ting found that the strong correlation was with sulfate aerosols, which are white in color. These aerosols have two main effects:

a) They reflect sunlight, which means less sunlight reaches the earth’s surface.
b) Aerosols in the air tend to cause clouds to persist for longer than they would other wise – you might imagine droplets of water staying aloft when they have a handy bit of sulfate to stick to.

Overall, both of these lead to a cooling effect.

Aerosols don’t stick around in the air for long though, and they don’t travel far from their source. It’s a bit counterintuitive to imagine that the smog from an American or European factory could be affecting tropical precipitation. But the slightest shift in the temperature balance between the northern and southern hemispheres is all it takes. Here’s a cartoon showing the basics of how this works:

While the Sahel region was being decimated by drought in the twentieth century, industrialization in developed countries was on the rise.   Since most of the industrialization happening at the time was in the northern hemisphere, that hemisphere was cooler than it otherwise would have been.  And model results indicate that this change was enough to shift the tropical rain band southward.  By the 1980s and 1990s, air pollution was widely recognized as being harmful, and various environmental regulations were put into place.  And, lo and behold: that’s about the same time that the ITCZ shifted back to its previous position.

More recently, Ting has been looking at how we can use climate models to predict how the tropical rainfall band might shift or change given different climate scenarios that are tested by IPCC models.  Of course, the effect of aerosols is only one piece of the puzzle.  The northern hemisphere cooling that Ting was investigating was overlaid on an overall warming trend.  The northern hemisphere was actually warming between 1950-1990, just not as quickly as the southern hemisphere.  In fact, recent studies indicate that the northern hemisphere appears to be warming more quickly, which may have significant consequences for tropical precipitation patterns.

http://www.atmos.washington.edu/~yting/

http://www.atmos.washington.edu/~dargan/

Press releases about some of their other work:

http://www.washington.edu/news/2013/03/11/remote-clouds-responsible-for-climate-models-glitch-in-tropical-rainfall/

http://www.theverge.com/2013/3/25/4129026/clouds-are-hiding-the-the-truth-of-how-much-earths-climate-will-change

http://newscenter.berkeley.edu/2013/04/02/shifting-rainfall-patterns-in-tropics/

Also, this paper is in press, so you can read all the details soon!

Hwang, Y.-T., Frierson, D. M. W., and S. M. Kang. Anthropogenic sulfate aerosol and the southward shift of tropical precipitation in the late 20th century. in press, Geophysical Research Letters

## Reading and writing and super-mussels

##### Don’t tell me how to interpret your results!

The other day, I stumbled on a climate change skeptic website while doing research for my ocean acidification post.  This ‘skeptic’ article included a summary of a peer reviewed article by a prominent scientist on the effect of an acidic ocean environment on mussels near deep sea hydrothermal vents.  The summary included this description of the results of that paper:

[…] there is ample reason to believe that even the worst case atmospheric CO2-induced acidification scenario that can possibly be conceived would not prove a major detriment to most calcifying sea life. Consequently, what will likely happen in the real world should be no problem at all […]

Well, that’s amazing!  If that was the only summary I read, I might be led to believe a few things:

1. There is rock-solid evidence indicating that ocean acidification, though it may be happening, is totally not a big deal.  Dude, look at those mussels!  They are doing great!
2. The mussels adapted to environments even more acidic than the most dire projections by acidification “alarmists”, and
3. the scientist who did that work is well-known and well-respected, and is clearly a climate change skeptic also.

I was curious, so I looked up the original paper to learn more [1]. This is the first sentence of the abstract by Tunnicliffe et al. (2009):

Increasing atmospheric carbon dioxide levels are causing ocean acidification, compromising the ability of some marine organisms to build and maintain support structures as the equilibrium state of inorganic carbon moves away from calcium carbonate.

Seems pretty reasonable to me.  Carbon dioxide is increasing, the ocean is becoming more acidic, and marine organisms are having a hard time because of it.  So I read on:

Few marine organisms tolerate conditions where ocean pH falls significantly below today’s value of about 8.1 and aragonite and calcite saturation values below 1.

So far they don’t seem very “skeptical” of ocean acidification.  Reading further:

We identify four-decade-old mussels, but suggest that the mussels can survive for so long only if their protective shell covering remains intact: crabs that could expose the underlying calcium carbonate to dissolution are absent from this setting.  The mussels’ ability to precipitate shells in such low-pH conditions is remarkable.  Nevertheless, the vulnerability of molluscs to predators is likely to increase in a future ocean with low pH.

Woah.  It actually seems like she’s found this remarkable organism who lives in a unique environment that can survive massively acidic conditions.  But she is NOT extrapolating these results to other types of calcifying organisms.  And in fact, even these tough mussels do in fact suffer from the increased acidity – they have far thinner shells than nearby mussels living in less acidic waters, which leaves them more vulnerable to predators.  If you read on in the paper, you learn that part of the reason these mussels have such incredible survival rates in these conditions is due to a lack of predators:  typical predators in nearby (less-acidic) waters are crabs, but the crabs can’t survive in the very low pH environment.

All this got me thinking about the differences that exist between authors’ intended message and how their work is subsequently portrayed and disseminated to the public.  What was Dr. Tunnicliffe trying to communicate?

I see a couple of different problems:  1)  The language used in peer reviewed literature is technical and often uses a lot of field-specific jargon, and 2) Access to peer-reviewed scientific articles is limited (by cost, mainly).

##### Easy on the jargon!

Was the “climate skeptic” summary accurate?  Was my summary accurate?  Were we both wrong?

I thought Dr. Tunnicliffe’s article was really well-written, clear, and informative, and even so – look how easy it was to subjectively interpret and share her results!

A lot of the time, scientists write peer-reviewed articles for each other.  That’s not to say they’re intentionally being exclusive – it just makes sense to try to aim their writing to other scientists who will use the work as a jumping-off point.  And, as a reader, if you’re not sufficiently familiar with a particular field, it can be difficult to glean the relevant information and to parse the jargon, particularly when you don’t have several hours to critically read it.

So what’s a scientist to do?  I really would like to know. It’s important to have rigorous and peer-reviewed documentation of your research so that others in the same field can build on or question your findings. That’s basically what makes science tick.  But what if authors could also write summaries intended for non-scientists (or scientists in other fields, for that matter) to accompany their more technical papers.  Like Cliff’s Notes of their own work.  So my blog readers could look up the Tunnicliffe paper and easily see for themselves whether my simpleminded take on it was total hooey.  (it might very well be…)

##### If a scientist publishes a peer-reviewed article and no one reads it, is it really science?

Of course, I think I’d be remiss not to mention that I am one of those privileged few who have almost unrestricted access to any peer-reviewed literature I care to get my grubby hands one – while I’m a student at a large American university, I can log into my school’s library website and look at anything I want instantly.  And if I can’t get it instantly in PDF format, I can ask the library to order it in for me, free of charge.  So without even getting to the fact that scientific papers are dense and difficult to synthesize, they’re simply off-limits to almost everyone unless they’re willing to fork over the cash.  For example, the Tunnicliffe paper on the Nature Geoscience website costs \$32.  Whew.

And we wonder why there’s a mistrust of the scientific community:  we write as cryptically as we can (hey – in scientists’ defense – getting all those technical details into a reasonable number of pages for publication is not easy, people), and then we publish in peer-reviewed journals that charge people \$30-40 for the PDF.  Ugh.

But!  The peer-review system is important in maintaining a certain standard and ensuring the credibility of published results.  I don’t know how to change that.  So:  until we figure out how to make peer-reviewed articles more open, maybe some “published-article Cliff’s Notes” or “cheat sheets” wouldn’t hurt.  (I guess now that I’ve said it, it would be shameful of me NOT to do it for my own paper.  At least I’ve only got the one.)

I’d love to have feedback!  It would help me figure out if anyone is thinking the same thing or if I’m totally out to lunch on this one.

More on open access:

Some interesting thoughts on the future of the scientific journal industry (think social media-style):

http://techcrunch.com/2013/02/03/the-future-of-the-scientific-journal-industry/

[1]  Tunnicliffe, Verena, et al. “Survival of mussels in extremely acidic waters on a submarine volcano.Nature Geoscience 2.5 (2009): 344-348.

## What is ocean acidification? (a really brief summary)

Again, as part of my teaching assistant job, I’m going to try to do a little overview of ocean acidification (OA).

### Nuts and bolts

Yes, I’m going to talk about chemistry.  Wait!  Don’t run away!  It won’t be too scary at all.  Let’s start with the sea surface, where CO2 is exchanged between the atmosphere and the ocean.  It’s constantly trying to reach equilibrium which means that if that if CO2 levels in the atmosphere are high, then CO2 gets “pulled” into the ocean.

When CO2 dissolves into the sea water, it can take on one of four forms:  dissolved (aqueous) carbon dioxide (CO2 (aq)), carbonic acid (H2CO3), bicarbonate (HCO3), and carbonate (CO32−).

What determines the relative concentrations of these different forms of carbon, you might ask?  Turns out it is strongly dependent on the temperature and the alkalinity of the water.  Have a look at this diagram, also known as a “Bjerrum plot”:

Looks kinda wacky.  Let’s go through it a step at a time.  The x-axis is indicating the pH, which is a measure of how acidic (or basic) the solution is.  In other words, it’s a measure of the concentration of hydrogen ions ($H^+$) hanging around in the water (that’s what the H in pH stands for). It’s worth noting here that pH is, by definition, a logarithmic scale.  If you move down one number on the pH scale, you’re decreasing by a factor of 10.  If you move up one number, you’re increasing by a factor of 10.  In case you’re curious, here’s the equation:

$pH = -log([H^+])$

So: the pH is the negative log of the hydrogen ion concentration, [H+]. A pH of 7 is neutral (fresh stream water would have a pH of about 7).  If it’s higher, it’s basic, and if it’s lower it’s acidic.

I know I said I wouldn’t use too much chemistry, but for those who might be interested, here’s what’s going on:

CO2 (aq) + H2O H2CO3 HCO3 + H+ CO32− + 2 H+.

As more CO2 is dissolved into the ocean, more hydrogen ions are released, causing that vertical blue line in the Bjerrum plot to move to the left. More hydrogen ions means decreased pH, aka increased acidity. Which is what people are talking about when they say “ocean acidification”.  And when that happens, the equilibrium ratios of the different chemical species of carbon shifts too – carbon dioxide increases, bicarbonate doesn’t change much, and carbonate decreases fairly quickly with even small changes in pH.

### Magnitude of OA

How much has the pH of the oceans changed since the industrial revolution?  It has decreased by approximately 0.1 pH units.  Maybe it sounds small, but remember that pH is a logarithmic scale – so it’s actually a 30% increase in hydrogen ion concentration.

### Impacts of ocean acidification

#### Calcium carbonate shell dissolution

I’ll start off with this:  it’s complicated.  And no one can really say exactly what is going to happen.  But we are able to ascertain certain things.  There have been, and continue to be, many studies focusing on exactly what the impacts of OA will be.  This is often done by taking certain organisms that are believed to be susceptible to acidification, and conducting controlled experiments where they are exposed to increased acidity.

The animals that are likely to be affected are those with shells or plates made of calcium carbonate (CaCO3).  These shells and plates are formed when dissolved ions in seawater precipitate to form CaCO3. In order for those shells to remain intact (ie. not dissolve!) the surrounding seawater needs to be saturated with respect to CO32−.

Detrimental effects have been shown for many organisms, for example: shellfish, foraminifera, coccolithophores, pteropods (e.g. [1],[2],[3]). It’s well worth noting that in certain cases, and for certain species, studies have shown surprising effects in the opposite direction (e.g. [4],[5]).  Like I said:  It’s complicated!

There’s a lot more you can learn about ocean acidification, but it’s more than I want to cover here.  As always, if any of my (7 or so) readers finds errors in this or other posts: let me know so I can fix them!

[1] Orr, James C., et al. “Anthropogenic ocean acidification over the twenty-first century and its impact on calcifying organisms.” Nature 437.7059 (2005): 681-686.
[2] Beaufort, L., et al. “Sensitivity of coccolithophores to carbonate chemistry and ocean acidification.” Nature 476.7358 (2011): 80-83.
[3] Kuffner, Ilsa B., et al. “Decreased abundance of crustose coralline algae due to ocean acidification.” Nature Geoscience 1.2 (2007): 114-117.
[4] Ocean Acidification: A risky shell game (WHOI Oceanus) https://www.whoi.edu/oceanus/viewArticle.do?id=52990
[5] Range, P., et al. “Seawater acidification by CO< sub> 2 in a coastal lagoon environment: Effects on life history traits of juvenile mussels< i> Mytilus galloprovincialis.” Journal of Experimental Marine Biology and Ecology 424 (2012): 89-98.


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# Get started¶

This section focuses on explaining the general functionalities of the package, and how its components interact with each other. Then two tutorials are provided:

For an installation guide, please check here

## Package Description¶

### General package architecture¶

The Causal Discovery Toolbox is a package for causal discovery in the observational setting. Therefore support for data with interventions is not available at the moment, but is considered for later versions.

The package is structured in 5 modules:

1. Causality: cdt.causality implements algorithms for causal discovery, either in the pairwise setting or the graph setting.

2. Independence: cdt.independence includes methods to recover the dependence graph of the data.

3. Data: cdt.data provides the user with tools to generate data, and load benchmark data.

4. Utils: cdt.utils provides tools to the users for model construction, graph utilities and settings.

5. Metrics: cdt.metrics includes scoring metrics for graphs, taking as input networkx.DiGraph

All methods for computation adopt a ‘scikit-learn’ like interface, where .predict() manages to launch the algorithm on the given data to the toolbox, .fit() allows to train learning algorithms Most of the algorithms are classes, and their parameters can be customized in the .__init__() function of the class.

Note

The .predict() function is often implemented in the base class (cdt.causality.graph.GraphModel for causal graph algorithms). .predict() is often a wrapper calling sub-functions depending on the arguments fed to the functions. The sub-functions, such as .orient_directed_graph() for cdt.causality.graph models (which is called when a directed graph is fed as a second argument ), are implemented and documented in the various algorithms.

### Hardware and algorithm settings¶

The toolbox has a SETTINGS class that defines the hardware settings. Those settings are unique and their default parameters are defined in cdt/utils/Settings.

These parameters are accessible and overridable via accessing the class:

>>> import cdt
>>> cdt.SETTINGS

Moreover, the hardware parameters are detected and defined automatically (including number of GPUs, CPUs, available optional packages) at the import of the package using the cdt.utils.Settings.autoset_settings method, ran at startup.

These settings are overriddable in two ways:

1. By changing cdt.SETTINGS attributes, thus changing the SETTINGS for the whole python session.

2. By changing the parameters of the functions/classes used. When their default value in the class definition is None, the cdt.SETTINGS value is taken, by using the cdt.SETTINGS.get_default function. This allows for quick and temporary parameter change.

### The graph class¶

The whole package revolves around using the graph classes of the networkx package. Most of the methods have the option of predicting a directed graph (networkx.DiGraph) or an undirected graph (networkx.Graph).

The networkx library might not be intuitive to use at first, but it comes with many useful tools for graphs. Here is a list of handy function, to understand the output of the toolbox’s outputs:

>>> import networkx as nx
>>> g = nx.DiGraph()  # initialize a directed graph
>>> l = list(g.nodes())  # list of nodes in the graph
>>> a = nx.adj_matrix(g).todense()  # Output the adjacency matrix of the graph
>>> e = list(g.edges())  # list of edges in the graph

Please refer to networkx ‘s documentation for more detailed information: https://https://networkx.github.io/documentation/stable/


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# How do we prove $\cos(\pi/5) - \cos(2\pi/5) = 0.5$ ?.

How do we prove $\cos(\pi/5) - \cos(2\pi/5) = 0.5$ without using a calculator.

Related question: how do we prove that $\cos(\pi/5)\cos(2\pi/5) = 0.25$, also without using a calculator

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A related question: How to prove $\cos \frac{2\pi }{5}=\frac{-1+\sqrt{5}}{4}$?. Perhaps we have some other questions here, where $\cos\frac{\pi}5$ or $\cos\frac{2\pi}5$ is calculated, but I've only found this one. –  Martin Sleziak Apr 12 '12 at 11:44
@MartinSleziak: I have an elementary answer for both questions (both parts of this question, and the related question as well, since my solution uses the polynomial $4x^2-2x-1=t^2-t-1$ for $t=2x$, which has roots $t=\frac{1\pm\sqrt5}{2}$. –  bgins Apr 12 '12 at 16:18
I am not sure if this is what Chandrasekar had in mind, because I don't follow his argument, but I have another post which is either the simplest solution yet, or else is a long-winded restatement of Chandrasekar's more elegant but potentially hard to follow solution (if correct, which I cannot judge). –  bgins Apr 12 '12 at 23:14
Does anyone know if there's a geometric proof of this? Given that this can be expressed in terms of the vertices of a dodecagon it seems like there ought to be one somewhere... –  Steven Stadnicki Apr 12 '12 at 23:32
I gave a proof of a more general statement than your "related question" here: mathoverflow.net/questions/16583/a-trigonometry-problem –  user641 Apr 13 '12 at 10:46

The complex roots of $x^5-1$ are: \begin{align} x_1&=1\\ x_2&=\cos\frac{2\pi}5+i\sin\frac{2\pi}5\\ x_3&=-\cos\frac{\pi}5+i\sin\frac{\pi}5\\ x_4&=-\cos\frac{\pi}5-i\sin\frac{\pi}5\\ x_5&=\cos\frac{2\pi}5-i\sin\frac{2\pi}5 \end{align} using Vieta's formulas you get $$0=x_1+x_2+\dots+x_5=1+2\left(\cos\frac{2\pi}5-\cos\frac\pi5\right)=0,$$ which yields your first equation.

From now on, let $\varphi=\frac{\pi}5$ (for brevity).

We know that $\cos2\varphi-\cos\varphi+\frac12=(2\cos^2\varphi-1)-\cos\varphi+\frac12=2\cos^2\varphi-\cos\varphi-\frac12=0$, i.e. $$2\cos^2\varphi-\cos\varphi=\frac12.$$ Now from $\cos2\varphi=\cos\varphi-\frac12$ you get $$\cos\varphi\cos2\varphi=\cos^2\varphi-\frac{\cos\varphi}2=\frac{2\cos^2\varphi-\cos\varphi}2=\frac14.$$

(Or, as suggested in Chandrasekhar's answer, from $2\cos^2\varphi-\cos\varphi=\frac12$ you can find the value of $\cos\varphi$ by solving the quadratic equation and taking the positive root. Once you know $\cos\varphi$, you can compute $\cos2\varphi$ and many other things. If you try it this way, you can check your result e.g. here.)

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Dear Martin, There seems to be a typo in the real part of $x_4$. Regards, –  Matt E Apr 12 '12 at 10:27
Thanks @MattE, it should be corrected now. –  Martin Sleziak Apr 12 '12 at 10:34

Note that: $\cos{2x} = \cos^{2}{x} - \sin^{2}{x} = 2\:\cos^{2}{x} - 1$. Therefore you have $\cos \frac{2\pi}{5} = 2\:\cos^{2}\frac{\pi}{5} - 1$

Now, \begin{align*} \cos\frac{\pi}{5} - \cos\frac{2\pi}{5} = \cos\frac{\pi}{5} - 2\: \cos^{2}\frac{\pi}{5}+1 \end{align*} This is a quadratic equation of the form $2 x^{2} - x -1 =0$ and solving this will give you the value of $\cos\frac{\pi}{5}$ from which you can find the above value which you need.

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Perhaps it should be mentioned that the quadratic equation has two solutions; but only one of them is positive, so there is no doubt which of the two solutions is $\cos\frac\pi5$. –  Martin Sleziak Apr 12 '12 at 10:38
Chandrasekhar: I think the quadratic equation should be $2x^{2}-x-\frac12=0$ or $4x^2-2x-1=0$. (If I understand correctly, you get the equation from $\cos\frac\pi5-\cos\frac{2\pi}5=\frac12$.) –  Martin Sleziak Apr 12 '12 at 10:51
@Martin: I have just calculated value of $\cos\frac{\pi}{5}$ using that equation. Why should it be $\frac{1}{2}$. That is what i have to prove. –  user9413 Apr 12 '12 at 11:04
It seems that I misunderstood your argumentation (I thought you are using $\cos\frac\pi5-\cos\frac{2\pi}5=\frac12$ to calculate $\cos\frac\pi5$ and then to solve the second part of question.) Nevertheless, the roots of $2x^2-x-1=0$ are $1$ and $-\frac12$, see wolframalpha. –  Martin Sleziak Apr 12 '12 at 11:10
@MartinSleziak: Yeah i know that. $(2x+1) \cdot (x-1)$ –  user9413 Apr 12 '12 at 11:11

If $a=\cos\frac{\pi}5$ and $b=\cos\frac{2\pi}5=2a^2-1$ (by the double-angle identity for cosine), we want to show that $$0=(x-a)(x+b)=x^2-(a-b)x-ab=x^2-\tfrac12x-\tfrac14,$$ i.e., that $a$ and $-b$ are roots of $4x^2-2x-1$. Note also that $4x^2+2x-1=4(x+a)(x-b)$ has roots $-a$ and $b$. What is special about the numbers $\{a,b,-b,-a\}$? They are the $x$-coordinates (real parts) of the nonreal $10$th roots of unity, $$x+iy=e^{\pm\pi i\cdot\frac{k}5} \qquad \text{for} \qquad k\not\equiv0\pmod5.$$ But these satisfy the equation $(x+iy)^5=(-1)^k=\pm1$. Taking the imaginary part, we have \eqalign{ 0 &= \Im\left[(x+iy)^5\right] \\ &= \Im\left[x^5+5x^4(iy)+10x^3(iy)^2+10x^2(iy)^3+5x(iy)^4+(iy)^5\right] \\ &= \Im\left[iy\left(5x^4+10x^2(iy)^2+(iy)^4\right)\right] \\ &= y\left[5x^4-10x^2y^2+y^4\right] \\ \implies 0 &= \left[5x^4-10x^2(1-x^2)+(1-x^2)^2\right] \\ &= 16x^4-12x^2+1 \\ &= 16x^4-8x^2+1 ~-~4x^2 \\ &= \left( 4x^2-1 \right) - \left( 2x \right)^2 \\ &= \left( 4x^2+2x-1\right)\left( 4x^2-2x-1\right) \,. } So far, we have shown this has roots $\{\pm a,\pm b\}$. It only remains to show that $\{-a,b\}$ are the roots of the first factor and that the desired pair $\{a,-b\}$ splits the second quadratic factor into linear terms (as we wrote at the outset).

Perhaps there is a clever way to infer the correct linear order of this set of roots by noticing that $a>\cos\frac\pi4>b$ so that $a^2>\frac12>b^2$ and combining this with our identity $b=2a^2-1$ above.

I propose in stead to use that $a>b$ but then to notice that our quadratic factors are in fact parabolas, with roots that are very easy to order on the $x$-axis. If we let $t=2x$ (which preserves order), then we have $$t^2\pm t-1 = t\,(t\pm1)-1$$ which both pass through $(0,-1)$ and alternately pass through $(\mp1,-1)$, shown respectively in red and blue below; the result follows.

It's only a small extra step to note that the roots of our factor, $t^2-t-1$, are $t=\frac{1\pm\sqrt5}{2}$, or $x=\frac{1\pm\sqrt5}{4}$.

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$$\cos(\pi/5) - \cos(2\pi/5)= {\rm Re}(e^{i\frac{\pi}{5}}-e^{i\frac{2\pi}{5}})=\frac{1}{2}( e^{i\frac{\pi}{5}}-e^{i\frac{2\pi}{5}}+ \overline{e^{i\frac{\pi}{5}}-e^{i\frac{2\pi}{5}}})=\frac{1}{2}( e^{i\frac{\pi}{5}}-e^{i\frac{2\pi}{5}}+ e^{-i\frac{\pi}{5}}-e^{-i\frac{2\pi}{5}})=\frac{e^{\frac{-2i\pi}{5}}}{2}( -e^{i\frac{4\pi}{5}}+e^{i\frac{3\pi}{5}}+ e^{i\frac{\pi}{5}}-1)$$

To simplify the computations, let $\omega=e^{\frac{i \pi}{5}}$. Note that $\omega^5=-1$.

Then

$$\cos(\pi/5) - \cos(2\pi/5)= \frac{-1}{2\omega^2}(\omega^4-\omega^3-\omega+1)= \frac{-1}{2\omega^2}(\omega^4-\omega^3+\omega^2-\omega+1-\omega^2)$$

$$\cos(\pi/5) - \cos(2\pi/5)= \frac{-1}{2\omega^2}(\frac{\omega^5+1}{\omega+1} -\omega^2)=\frac{-1}{2\omega^2}(0 -\omega^2)=\frac{1}{2}$$

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For the first equality.
$\cos(\pi/5) - \cos(2\pi/5)=\cos(3\pi/10-\pi/10) - \cos(3\pi/10+\pi/10)$
$=2\sin(\pi/10)\sin(3\pi/10)=2\sin(\pi/10)\cos(\pi/5)$
$=2\sin(\pi/10)\cos(\pi/10)\cos(\pi/5)/\cos(\pi/10)$
$=\sin(\pi/5)\cos(\pi/5)/\cos(\pi/10)$
$=0.5\sin(2\pi/5)/\cos(\pi/10)=0.5$

Here 's the latter equlity.
$\cos(\pi/5)\cos(2\pi/5) = \sin(\pi/5)\cos(\pi/5)\cos(2\pi/5)/\sin(\pi/5)$
$=0.5\times \sin(2\pi/5)\cos(2\pi/5)/\sin(\pi/5)=0.25\times \sin(4\pi/5)/\sin(\pi/5)$
$=0.25.$
Sorry for the bad composing. I am not familiar with tex.

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the most important one is \frac{a}{b}. I also like to use \cdot or no multiplication symbol at all. –  NeuroFuzzy Apr 13 '12 at 1:37

I was not able (yet) to follow Chandrasekar's solution, but noticed this while trying to understand the argument (how it could possibly lead to the solution, or how exactly he arrives at $2x^2-x-1$ for $x=\cos\frac{\pi}{5}$, which to me seems non-obvious and even a fallacious deduction from his equations and prose -- apologies if I am just being dense)...perhaps it is what Chandrasekar meant all along, but in any case, it does seem to be the most elementary solution available.

Apply the double angle formula $\cos2\theta=\cos^2\theta-\sin^2\theta=2\cos^2\theta-1$ to $\theta=\frac{\pi}{5}$ and $\frac{2\pi}{5}$, with $a=\cos\frac{\pi}{5}$ and $b=\frac{2\pi}{5}$ for convenience, recalling also that $\cos(\pi\pm\theta)=-\cos\theta$: $$b=\cos\frac{2\pi}{5}=2\,\cos^2\frac{\pi}{5}-1=2a^2-1$$ $$-a=\cos\frac{4\pi}{5}=2\,\cos^2\frac{2\pi}{5}-1=2b^2-1$$ Next, subtracting the equations $$\matrix{ 2a^2=1+b\\ 2b^2=1-a}$$ we get

\eqalign{ 2\left(a^2-b^2\right)&=b+a\\ 2\left(a+b\right)\left(a-b\right)&=b+a\\ 2\left(a-b\right)&=1\\ a-b&=\frac12\,. } Furthermore, multiplying, we get $$4(ab)^2=(1+b)(1-a)=1+(b-a)-ab=1+\left(-\tfrac12\right)-ab$$ giving us the quadratic equation $$4(ab)^2+(ab)-\tfrac12=0$$ $$8(ab)^2+2(ab)-1=0$$ $$\left(4ab-1\right)\left(2ab+1\right)=0$$ so that $ab=\frac14$ or $-\frac12$, from which we can choose the former since we know that $0<a<b$.

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This is a very nice solution. –  Martin Sleziak Apr 13 '12 at 5:47

$$\cos(\pi/5)\cos(2\pi/5)=A$$

$$\Longrightarrow\quad A = \frac{\sin(\pi/5)\cos(\pi/5)\cos(2\pi/5)}{\sin(\pi/5)}=\frac{\sin(2\pi/5)\cos(2\pi/5)}{2\sin(\pi/5)}$$

$$A = \frac{\sin(4\pi/5)} {2\cdot 2\cdot\sin(\pi/5)}=\frac{1}{4}$$ since $\sin(4\pi/5)=\sin(\pi/5)$.

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If $\displaystyle x=\frac\pi5,5x=\pi\implies3x=\pi-2x\implies\cos3x=\cos(\pi-2x)\ \ \ \ (1)$

But as $\displaystyle\cos(\pi-2x)=-\cos2x$, (1)$\displaystyle\implies\cos3x+\cos2x=0\ \ \ \ (2)$

If $\displaystyle\cos x=c(2)$ becomes $4c^3+2c^2-3c-1=0\ \ \ \ (3)$

Again from $(1)\implies3x=2n\pi\pm(\pi-2x)$

$\displaystyle'-'\implies x=(2n-1)\pi\implies\cos x=-1$

$\displaystyle'+'\implies x=\frac{(2n+1)\pi}5$ where $n=0,1,2,3,4$

Observe that $\displaystyle\cos\frac{(2\cdot2+1)\pi}5=\cos\pi=-1$(which converges with $'-'$ case )

and $\displaystyle\cos\frac{(2\cdot0+1)\pi}5=\cos\frac{(2\cdot4+1)\pi}5$ as $\displaystyle\cos(2\pi-y)=\cos y$

Similarly, $\displaystyle\cos\frac{(2\cdot1+1)\pi}5=\cos\frac{(2\cdot3+1)\pi}5$

So, the quadratic equation whose roots are $\displaystyle\cos\frac{\pi}5=\cos\frac{9\pi}5,\cos\frac{3\pi}5=\cos\frac{7\pi}5$ will be $\displaystyle\frac{4c^3+2c^2-3c-1}{c+1}=0\iff4c^2-2c-1=0\ \ \ \ (4)$

Now use the fact $\displaystyle c^2=\cos^2x=\frac{1+\cos x}2$

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We can factorize $z^5+1$, using the roots of $-1$, $$z^5+1=(z+1)(z-e^{i\pi/5})(z-e^{i2\pi/5})(z-e^{i3\pi/5})(z-e^{i4\pi/5}).$$ The coefficient of degree $4$ is the sum of the roots, so that $$1-e^{i\pi/5}-e^{i2\pi/5}-e^{i3\pi/5}-e^{i4\pi/5}=0.$$ Taking the real part and noticing that there are complementary angles $$2\ (\cos\frac\pi5-\cos\frac{2\pi}5)=1.$$

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If $\displaystyle x=\frac\pi5,5x=\pi\implies3x=\pi-2x\implies\sin3x=\sin(\pi-2x)\ \ \ \ (1)$

But as $\displaystyle\sin(\pi-y)=\sin y, (1)\displaystyle\implies\sin3x=\sin2x\ \ \ \ (2)$

which $\displaystyle\implies3x=n\pi+(-1)^n2x$

For even $n(=2m)$(say), $\displaystyle x=2m\pi\implies\sin x=0$

For odd $n(=2m+1)$(say), $\displaystyle x=\frac{(2m+1)\pi}5$ where $m=0,1,2,3,4$

Observe that for $\displaystyle m=2, \sin x=\sin\pi=0$

Again from $(2),2\sin x\cos x=3\sin x-4\sin^3x$

If $\displaystyle\sin x\ne0, 2\cos x=3-4\sin^2x=3-4(1-\sin^2x)\iff4\cos^2x-2\cos x-1=0$

whose roots are $\displaystyle\cos\frac{(2\cdot0+1)\pi}5=\cos\frac{(2\cdot4+1)\pi}5$ and $\displaystyle\cos\frac{(2\cdot1+1)\pi}5=\cos\frac{(2\cdot3+1)\pi}5$ (Reason for equality has been explained in my other answer)

Hope you can take it home from here.

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# 22.8 Torque on a current loop: motors and meters

 Page 1 / 4
• Describe how motors and meters work in terms of torque on a current loop.
• Calculate the torque on a current-carrying loop in a magnetic field.

Motors are the most common application of magnetic force on current-carrying wires. Motors have loops of wire in a magnetic field. When current is passed through the loops, the magnetic field exerts torque on the loops, which rotates a shaft. Electrical energy is converted to mechanical work in the process. (See [link] .)

Let us examine the force on each segment of the loop in [link] to find the torques produced about the axis of the vertical shaft. (This will lead to a useful equation for the torque on the loop.) We take the magnetic field to be uniform over the rectangular loop, which has width $w$ and height $l$ . First, we note that the forces on the top and bottom segments are vertical and, therefore, parallel to the shaft, producing no torque. Those vertical forces are equal in magnitude and opposite in direction, so that they also produce no net force on the loop. [link] shows views of the loop from above. Torque is defined as $\tau =\text{rF}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta$ , where $F$ is the force, $r$ is the distance from the pivot that the force is applied, and $\theta$ is the angle between $r$ and $F$ . As seen in [link] (a), right hand rule 1 gives the forces on the sides to be equal in magnitude and opposite in direction, so that the net force is again zero. However, each force produces a clockwise torque. Since $r=w/2$ , the torque on each vertical segment is $\left(w/2\right)F\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta$ , and the two add to give a total torque.

$\tau =\frac{w}{2}F\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta +\frac{w}{2}F\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =\text{wF}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta$

Now, each vertical segment has a length $l$ that is perpendicular to $B$ , so that the force on each is $F=\text{IlB}$ . Entering $F$ into the expression for torque yields

$\tau =\text{wIlB}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta .$

If we have a multiple loop of $N$ turns, we get $N$ times the torque of one loop. Finally, note that the area of the loop is $A=\text{wl}$ ; the expression for the torque becomes

$\tau =\text{NIAB}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta .$

This is the torque on a current-carrying loop in a uniform magnetic field. This equation can be shown to be valid for a loop of any shape. The loop carries a current $I$ , has $N$ turns, each of area $A$ , and the perpendicular to the loop makes an angle $\theta$ with the field $B$ . The net force on the loop is zero.

## Calculating torque on a current-carrying loop in a strong magnetic field

Find the maximum torque on a 100-turn square loop of a wire of 10.0 cm on a side that carries 15.0 A of current in a 2.00-T field.

Strategy

Torque on the loop can be found using $\tau =\text{NIAB}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta$ . Maximum torque occurs when $\theta =\text{90º}$ and $\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =1$ .

Solution

For $\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =1$ , the maximum torque is

${\tau }_{\text{max}}=\text{NIAB}.$

Entering known values yields

$\begin{array}{lll}{\tau }_{\text{max}}& =& \left(\text{100}\right)\left(\text{15.0 A}\right)\left(\text{0.100}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2}\right)\left(2\text{.}\text{00 T}\right)\\ & =& \text{30.0 N}\cdot m.\end{array}$

Discussion

This torque is large enough to be useful in a motor.

Calculate the final velocity attained, when a ball is given a velocity of 2.5m/s, acceleration of 0.67m/s² and reaches its point in 10s. Good luck!!!
2.68m/s
Doc
vf=vi+at vf=2.5+ 0.67*10 vf= 2.5 + 6.7 vf = 9.2
babar
s = vi t +1/2at sq s=58.5 s=v av X t vf= 9.2
babar
how 2.68
babar
v=u+at where v=final velocity u=initial velocity a=acceleration t=time
Eklu
my project is Sol gel process how to prepare this process pls tell me
Bala
the dimension of work and energy is ML2T2 find the unit of work and energy hence drive for work?
KgM2S2
Acquah
Two bodies P and Quarter each of mass 1000g. Moved in the same direction with speed of 10m/s and 20m/s respectively. Calculate the impulse of P and Q obeying newton's 3rd law of motion
kk
Doc
definition for wave?
A disturbance that travel from one medium to another and without causing permanent change to its displacement
Fagbenro
In physics, a wave is a disturbance that transfers energy through matter or space, with little or no associated mass transport (Mass transfer). ... There are two main types ofwaves: mechanical and electromagnetic. Mechanicalwaves propagate through a physical matter, whose substance is being deformed
Devansh
K
Manyo
thanks jare
Doc
Thanks
Note: LINEAR MOMENTUM Linear momentum is defined as the product of a system’s mass multiplied by its velocity: size 12{p=mv} {}
what is physic
zalmia
Study of matter and energy
Fagbenro
physics is the science of matter and energy and their interactions
Acquah
physics is the technology behind air and matter
Doc
Okay
William
hi sir
Bala
how easy to understanding physics sir
Bala
Easy to learn
William
31. Calculate the initial (from rest) acceleration of a proton in a 5.00×106 N/C electric field (such as created by a research Van de Graaff). Explicitly show how you follow the steps in the Problem-Solving Strategy for electrostatics.
A tennis ball is projected at an angle and attains a range of 78. if the velocity is 30metres per second, calculate the angle
Shimolla
what friction
question on friction
Wisdom
the rubbing of one object or surface against another.
author
momentum is the product of mass and it's velocity.
Algayawi
what are bioelements?
Edina
Friction is a force that exist between two objects in contact. e.g. friction between road and car tires.
Eklu
With regards to a shielded cable, is there an induced current on the shield when the center conductor is carrying an AC Current? What is the formula?
what is phenomena
no idea
Awoke
its phenomenon, an observable fact.
author
Mujy achy marks hasil krny k leay kesy tayari krni ho ge?plz help me I'M sooo woried
konsi university m ho and konsa course h
Mohit
what is force
Force is the cause and momentum is its effect.
Salman
A force is a pull or a push on an object, causing an object to move or a moving object to stop.
Eklu
Find the velocity that make one full oscillation in 10 seconds and also makes 1.7meters in the same time
yes
Rafael
OK
Oluwaseun
17m/s
Eklu
17m/s
Gabriel
0.17
Devansh
17m/s
Harrison
how did you solve it
Acquah
t=10s a=1.7m v=? u=0 but v=u+at but u=0 v=at 1.7×10=17 v=17m/s that's how I solved it.
Eklu
kk
Acquah
How is a=1.7m?
El
a is an acceleration
El
eklu..... I respect u....
Doc
its relating to time and oscillation made.
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You mean A as an amplitude?
El
ok
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A voltmeter is a device use in measuring electric potential in a volt.
Find out the equation of a simple harmonic progressive wave that have 0.2m with a frequency of 550hertz.If the wave moves along a positive x axis with the speed of 330m/s


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# Thread: Prove the following statement: If r3 is irrational, then r is irrational. disprove

1. ## Prove the following statement: If r3 is irrational, then r is irrational. disprove

9. a. Prove the following statement: If r3 is irrational, then r is irrational. (by contrapositive)

b. Disprove the converse of the statement in part (a).

a) let r be rational number

$\displaystyle r=\frac{a}{b}$ for $\displaystyle a,b \in \mathbb{Z};b \neq 0$

$\displaystyle r^3=\frac{a^3}{b^3}$

$\displaystyle a^3r^3=b^3 \rightarrow a*a*a*r*r*r=b*b*b$

by definition integers are closed under multiplication, all integers are rational

therefore r^3 is rational

b) converse is if r is irrational then r^3 is irrational

counterexample??

2. ## Re: Prove the following statement: If r3 is irrational, then r is irrational. disprov

If you are told to use the contrapositive, you are NOT trying to reach a contradiction.

These two statements are equivalent: p ->q and ~q -> ~p. So if you can prove the second statement, then you prove the first one.

So since you are trying to show that if r^3 is irrational then r is irrational, one way to do this is to prove its contrapositive, that if r is rational, then r^3 is rational.

There is no disproof involved!

3. ## Re: Prove the following statement: If r3 is irrational, then r is irrational. disprov

To disprove part (b), yes you need a counter-example. So we need to find an irrational number, whose cube is rational.

The obvious candidate is $\sqrt[3]{2}$. Its cube is clearly rational. Can you prove it is irrational?

4. ## Re: Prove the following statement: If r3 is irrational, then r is irrational. disprov

Originally Posted by Prove It
If you are told to use the contrapositive, you are NOT trying to reach a contradiction.

These two statements are equivalent: p ->q and ~q -> ~p. So if you can prove the second statement, then you prove the first one.

So since you are trying to show that if r^3 is irrational then r is irrational, one way to do this is to prove its contrapositive, that if r is rational, then r^3 is rational.

There is no disproof involved!
didn't I prove it by contrapositive? I took p --> q and made it ~q-->~p, I let r be rational then showed r^3 is rational

Originally Posted by Deveno
To disprove part (b), yes you need a counter-example. So we need to find an irrational number, whose cube is rational.

The obvious candidate is $\sqrt[3]{2}$. Its cube is clearly rational. Can you prove it is irrational?
assume the cuberoot of 2 is rational
$\displaystyle \sqrt[3]{2} = \frac{a}{b}; a,b \in \mathbb{Z}, b \neq 0$ and a/b is lowest terms

$\displaystyle 2=\frac{a^3}{b^3}$

$\displaystyle 2a^3=b^3$ the left hand side is even, thus the right side must also be even

$\displaystyle 2a^3=(2m)^3; (2m)=b$

$\displaystyle a^3=\frac{8}{2}m^3$

both have a factor of 2, both are even therefore they were not in there lowest terms, it is not rational, therefore is irrational

5. ## Re: Prove the following statement: If r3 is irrational, then r is irrational. disprov

Originally Posted by Jonroberts74
9. a. Prove the following statement: If r3 is irrational, then r is irrational. (by contrapositive)

b. Disprove the converse of the statement in part (a).

a) let r be rational number

$\displaystyle r=\frac{a}{b}$ for $\displaystyle a,b \in \mathbb{Z};b \neq 0$

$\displaystyle r^3=\frac{a^3}{b^3}$

$\displaystyle a^3r^3=b^3 \rightarrow a*a*a*r*r*r=b*b*b$

by definition integers are closed under multiplication, all integers are rational

therefore r^3 is rational

b) converse is if r is irrational then r^3 is irrational

counterexample??
$\displaystyle r^3=\frac{a^3}{b^3}$

$\displaystyle a^3r^3=b^3 \rightarrow a*a*a*r*r*r=b*b*b$

this was wrong meant

$\displaystyle b^3r^3=a^3 \Rightarrow b*b*b*r*r*r=a*a*a$

,

,

,

,

,

,

,

,

,

### prove that 3 √x is irrational number

Click on a term to search for related topics.


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## Problem of the week

### 2018-23 Game of polynomials

Two players play a game with a polynomial with undetermined coefficients $1 + c_1 x + c_2 x^2 + \dots + c_7 x^7 + x^8.$ Players, in turn, assign a real number to an undetermined coefficient until all coefficients are determined. The first player wins if the polynomial has no real zeros, and the second player wins if the polynomial has at least one real zero. Find who has the winning strategy.


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# Validation of a linear mixed effect Model

I am setting up models of the diversity data (Shannon Index) of bees and hoverflies to find out which in which studied seed mixture (site) the diversity was higher. In addition, I would like to know if this was influenced by the diversity of the plant species or by the flower cover of these. I added the site and the field visits (3 times) as a random effect, because they were chosen randomly and I do not expect a direct effect on the diversity, or this is not of interest.

Since I am not very familiar with model validation and there is no clear explanation online how to interpret the results of a test for overdispersion or zero inflation, I would like to pray for help.

What I have done so far:

my models basically look like this:

lmer(H_bee ~ seedmixture + H_plant + blueh_deck + (1|location) + (1|fieldvisit), data = insect_plant, REML = FALSE).

Some Information: There are two seed mixtures (RH and WD), I have visited 4 locations and those ones in 3 different month (fieldvisit).

Results of the Model

summary(lmer_H_schweb)

Linear mixed model fit by maximum likelihood  ['lmerMod']
Formula: H_bee ~ seedmixture + H_plant + blueh_deck + (1 | location) +      (1 | fieldvisit)
Data: insect_plant

AIC      BIC   logLik deviance df.resid
-87.2    -64.3     50.6   -101.2      189

Scaled residuals:
Min      1Q  Median      3Q     Max
-1.3061 -0.4246 -0.1407  0.0503  5.9872

Random effects:
Groups     Name        Variance Std.Dev.
location   (Intercept) 0.002702 0.05198
fieldvisit (Intercept) 0.002588 0.05087
Residual               0.033112 0.18197
Number of obs: 196, groups:  location, 4; fieldvisit, 3

Fixed effects:
Estimate Std. Error t value
(Intercept)  0.073241   0.060921   1.202
seedmixtureWD   -0.043332   0.031068  -1.395
H_plant      0.034441   0.033140   1.039
blueh_deck  -0.005288   0.005611  -0.943

Correlation of Fixed Effects:
(Intr) StgtWD H_plnt
SaatgutWD  -0.544
H_plant    -0.582  0.341
blueh_deck -0.398  0.433  0.012

I used the DHARMa Package to test for zero-inflation and also overdispersion. The overdispersion Graph looks quite good to me, but the zero-inflation is a bit confusing...

I also used the plot() function with the simulationOutput. Thats the point where I thought that the models are not well. First of all there is the note "Quantile deviations detected" and further more the QQ plot has that big step at the end to another bunch of Datapoints in the plot. Also the red line is not fitting the black points, as I know it from other models.

QQ Plot

Overdispersion test

Zero inflation test

I would be so thankful to get some advice, whether this model is reliable, or how I could get the problem under control. If it were a Poisson distribution, I would switch to a negative binomial distribution. However, I have not found a solution for this in the lmer models.

• Could you please add the results of summary(model) to your question, as text (e.g. in a code block), and delete the image of the fixed-effect table? Images are bad because they're unsearchable and inaccessible to people using screen readers ... Sep 23 at 23:35
• Thank you for the fast feedback. I am new here and did not know how to add code properly in the question, but have added the summary of the model now. Sep 24 at 7:46

You should try to figure out which observations these are and what's going on with them. If you plot(fitted_model, id = 0.05) you'll get a residuals-vs-fitted plot with some outliers (see ?lme4::plot.merMod for details) identified by row number.
For what it's worth, it may not work very well to treat site and field as random effects if they each only have three levels. (It would help to have summary(fitted_model) included in the text of your question [as cut-and-pasted text, not as a screenshot image].)
• Nevertheless, I would worry about the outliers - not so much because I think they will invalidate the model as that you should understand what's going on in your data. You should be able to use the influence() function from the car package to understand what effect these observations are having on your conclusions (or, more crudely, refit the model with those observations excluded and see what happens) Sep 25 at 22:56


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# This algorithm will find a basis for the span of some vectors. How/why does it work?

Say I want to find a basis for $span((1,2,5),(2,4,10),(-3,-5,-13),(2,1,4),(-4,-6,-16))$. Google tells me that to get the answer, I'm supposed to write down the vectors as columns of a matrix:

$$\begin{pmatrix} 1 & 2 & -3 & 2 & -4 \\ 2 & 4 & -5 & 1 & -6 \\ 5 & 10 & -13 & 4 & -16 \\ \end{pmatrix}$$

... then bring said matrix to row echelon form...

$$\begin{pmatrix} 1 & 2 & 0 & -7 & 2 \\ 0 & 0 & 1 & -3 & 2 \\ 0 & 0 & 0 & 0 & 0 \\ \end{pmatrix}$$

... then look at the pivots; in this case the first and third rows contain pivots, therefore the first and third columns of the original matrix are linearly independent while the others are linearly dependent, and therefore those two vectors are a basis of the span.

Okay, that's simple enough to remember and use in the exam but it still feels like magic to me. How does this work? Why do the pivots show which vectors are linearly independent?

• Short answer: Row operations do not change linear relations between columns. Or maybe it's just a shorter way of phrasing your question: "Why do row operations not change linear relations between columns?" Mar 22, 2017 at 18:22
• That's a good way of simplifying the question, yes. I suspect "how are the row spaces of the two matrices related?" is another very closely related question too. Mar 22, 2017 at 20:27
• This method picks out a subset of the vectors that forms a basis, which can be useful, but this is basically a side-effect of computing the null space (kernel) of this matrix. You generally get a more “useful” basis by column reducing the matrix instead. The non-zero columns of the result form a basis. To compare, the row-reduction process gives as a basis $(1,2,5)^T$ and $(-3,-5,-13)^T$, while column-reduction produces $(1,0,1)^T$ and $(0,1,2)^T$ instead. It’s also less sensitive to the order in which you list the vectors.
– amd
Mar 22, 2017 at 23:38

Combine this with the fact that the row rank and column rank (that is, the dimensions of the row and column spaces) of a matrix are equal, and one can see how this method works: In row-reduced echelon form, each pivot column has a $$1$$ in a different row from other pivot columns and zeros elsewhere, so the set of pivot columns are obviously linearly independent. Since elementary row operations don’t affect linear dependency of columns, the corresponding columns of the original matrix are also linearly independent. There are as many pivots as the rank of the matrix, which is equal to the dimension of the column space, so this set of vectors forms a basis for the column space, i.e., a basis for your original set of vectors.


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# zbMATH — the first resource for mathematics

Special tangent valued forms and the Frölicher-Nijenhuis bracket. (English) Zbl 0801.53014
Summary: We define the tangent valued $$\mathcal C$$-forms for a large class of differential geometric categories. We deduce that the Frölicher- Nijenhuis bracket of two tangent valued $$\mathcal C$$ forms is a $$\mathcal C$$-form as well. Then we discuss several concrete cases and we outline the relations to the theory of special connections.
##### MSC:
 53C05 Connections (general theory)
Full Text:


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# Representation in state-space of a system with ideal and real differentiating links in feedback

There are two systems in the state space with an ideal and a real differentiator.

For the first system, the code in Mathematica is as follows.

nsys = NonlinearStateSpaceModel[
x'[t] == D[Power[x[t], 2], t] + 0.01 Sin[10 t], x[t], u[t], x[t], t]
Plot[Evaluate@OutputResponse[{nsys, 1}, 0, {t, 0, 20}], {t, 0, 20}]

What will the code look like for receiving the output signal if instead of the ideal differentiator in feedback the real differentiator?

The equations can be constructed from the block diagram and then directly passed to NDSolve.

eqns = {x'[t] == u[t] + f[t], f'[t] + f[t] == D[Power[x[t], 2], t]} /. u[t] -> 0.01 Sin[10 t];
x[t] /. NDSolve[Join[%, {x[0] == 0.5, f[0] == 0}], {x[t], f[t]}, {t, 0, 5}];
p1 = Plot[%, {t, 0, 5}];

Or we can use SystemsConnectionsModel to build the model and go from there.

add = TransferFunctionModel[{{1, 1}}, s];
int = TransferFunctionModel[1/s, s];
sq = NonlinearStateSpaceModel[{{}, u^2}, {}, u];
diff = TransferFunctionModel[s/(s + 1), s];
{{1, 1} -> {2, 1}, {2, 1} -> {3, 1}, {3, 1} -> {4, 1}, {4, 1} -> {1, 2}},
{{1, 1}}, {{2, 1}}] // SystemsModelMerge
p2 = Plot[Evaluate@OutputResponse[{%, {0.5, 0.5^2}}, 0.01 Sin[10 t], {t, 0, 5}], {t, 0, 5}];

Both approaches should give the same result.

Show[p1, p2]

(The latter approach will not work for the ideal differentiator because NonlinearStateSpaceModel does not support descriptor systems yet.)

• Dear Suba Thomas, the question arose: is it possible to write down the equation of the above systems in the form $x'=Ax+Bu$ (with one state-space variable, not two), or because of the differentiating links is impossible? – dtn Jan 6 at 5:01
• @AndySol You can get an affine representation using AffineStateSpaceModel[eqns, {x[t], f[t]}, u[t], x[t], t]. It will have two states and will be of the form $x'=a(x)+b(x).u$ . The second state comes from the pole in the denominator of the real differentiator. (eqns = {x'[t] == u[t] + f[t], f'[t] + f[t] == D[Power[x[t], 2], t]}) – Suba Thomas Jan 6 at 15:06


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# What test should I use if I want to compare 2 percentages?

I've got this hypothesis "People are better at detecting a lie when they are able to see a persons expression, compared to people that are not". I've tested 2 groups of people: 1) one group of 20 people saw a video of a person stating 3 statements of which one of them is a lie and they had to tell which one of the statements they think is a lie 2) other 20 people did the same but not seeing a person, just listening to him in an audio recording

the results were that 3 out 20 people detected a lie correctly from a video (15%) and 5 people out of other 20 detected a lie correctly from an audio recording (25%), which kind of disproves the hypothesis. Now the question is, what kind of test should i use to compare these results to see if the actual results are significant and not just "accidental"? The closest I could find was the "Two proportions" test, but it is based on having different sizes of samples, whereas i have the same size (20 in each medium)

• I might go here link and look for overlap in the 95% confidence intervals. I would make sure to use the right breed of confidence interval - because they are governed by sample size. – EngrStudent Apr 21 '17 at 11:16
• Equal sample size is not a problem when comparing two proportions. – Michael Chernick Apr 21 '17 at 12:47
• @Velionis Your proposal to the hypothesis testing is not bad. Michael Chermick has pointed out correctly that sample sizes do not cast a shadow when we are using proportions for testing. – Subhash C. Davar Apr 21 '17 at 14:41

 correct incorrect Total video 3 17 20 audio 5 15 20 Total 8 32 40
• Actually, you do have an expected number, and it is the reasons I had added the row and columntotals to the table: $expected value cell_{i,j} = rowtotal_i*columntotal_j/totalN$. You could (1) calculate this expected value for every cell, (2) calculate the difference between observed and expected per cell, (3) square these numbers and then (4) divide by the expected number per cell, (5) sum the values of all four cells, and you'll have calculated the chi-square value. The final step would be to check a chi-square distribution with the appropriate $df$ to obtain a p-value. – IWS Apr 21 '17 at 9:49


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+0

0
67
2

We define a bow-tie quadrilateral as a quadrilateral where two sides cross each other. An example of a bow-tie quadrilateral is shown below.

Seven distinct points are chosen on a circle. We draw all \(\binom{7}{2} = 21\) chords that connect two of these points. Four of these 21 chords are selected at random. What is the probability that these four chosen chords form a bow-tie quadrilateral?

Aug 10, 2020

#1
0


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# derivative of cos

Here is a graph of our situation. We need to go back, right back to first principles, the basic formula for derivatives: dydx = limΔx→0 f(x+Δx)−f(x)Δx. Eg:1. When x = 0.15 (in radians, of course), this expression (which gives us the The differentiation of trigonometric functions is the mathematical process of finding the derivative of a trigonometric function, or its rate of change with respect to a variable. x The area of triangle OAB is: The area of the circular sector OAB is Using these three facts, we can write the following. We differentiate each term from left to right: x(-2\ sin 2y)((dy)/(dx)) +(cos 2y)(1) +sin x(-sin y(dy)/(dx)) +cos y\ cos x, (-2x\ sin 2y-sin x\ sin y)((dy)/(dx)) =-cos 2y-cos y\ cos x, (dy)/(dx)=(-cos 2y-cos y\ cos x)/(-2x\ sin 2y-sin x\ sin y), = (cos 2y+cos x\ cos y)/(2x\ sin 2y+sin x\ sin y), 7. Applications: Derivatives of Logarithmic and Exponential Functions, Differentiation Interactive Applet - trigonometric functions, 1. Derivatives of Csc, Sec and Cot Functions, 3. Derivatives of Inverse Trigonometric Functions, 4.   Here is a different proof using Chain Rule. Derivative of cos(5t). So we can write y = v^3 and v = cos\ So you have the negative two thirds. Below you can find the full step by step solution for you problem. We conclude that for 0 < θ < ½ π, the quantity sin(θ)/θ is always less than 1 and always greater than cos(θ). This can be derived just like sin(x) was derived or more easily from the result of sin(x). θ Derivative of the Logarithmic Function, 6. Using the Pythagorean theorem and the definition of the regular trigonometric functions, we can finally express dy/dx in terms of x. How to find the derivative of cos(2x) using the Chain Rule: F'(x) = f'(g(x)).g'(x) Chain Rule Definition = f'(g(x))(2) g(x) = 2x ⇒ g'(x) = 2 = (-sin(2x)). {\displaystyle x=\cos y\,\!} in from above. The first term is the product of (2x) and (sin x). Then, applying the chain rule to We have 2 products. Equivalently, we can prove the derivative of cos(x) from the derivative of sin(x). y 2 tan 1 u. 1 Use the chain rule… What’s the derivative of SEC 2x? Derivatives of the Sine and Cosine Functions. So, we have the negative two thirds, actually, let's not forget this minus sign I'm gonna write it out here. {\displaystyle \sin y={\sqrt {1-\cos ^{2}y}}\,\!} y is always nonnegative by definition of the principal square root, so the remaining factor must also be nonnegative, which is achieved by using the absolute value of x.). y Solve: cos(x) = sin(x + PI/2) cos(x) = sin(x + PI/2) = sin(u) * (x + PI/2) (Set u = x + PI/2) = cos(u) * 1 = cos(x + PI/2) = -sin(x) Q.E.D. Simple step by step solution, to learn. Let two radii OA and OB make an arc of θ radians. =  2 For the case where θ is a small negative number –½ π < θ < 0, we use the fact that sine is an odd function: The last section enables us to calculate this new limit relatively easily. {\displaystyle x=\cot y} There are rules we can follow to find many derivatives.. For example: The slope of a constant value (like 3) is always 0; The slope of a line like 2x is 2, or 3x is 3 etc; and so on. The second term is the product of (2-x^2) and (cos x). R Derivative is the important tool in calculus to find an infinitesimal rate of change of a function with respect to its one of the independent variable. A Negative sine of X. Find the derivative of y = 3 sin 4x + 5 cos 2x^3. combinations of the exponential functions {e^x} and {e^{ – x < Free derivative calculator - differentiate functions with all the steps. π {\displaystyle \arccos \left({\frac {1}{x}}\right)} You multiply the exponent times the coefficient. ( Using the product rule, the derivative of cos^2x is -sin(2x) Finding the derivative of cos^2x using the chain rule. cos Now, if u = f(x) is a function of x, then by using the chain rule, we have: First, let: u = x^2+ 3 and so y = sin u. , while the area of the triangle OAC is given by. If you're seeing this message, it means we're having trouble loading external resources on our website. Alternatively, the derivative of arcsecant may be derived from the derivative of arccosine using the chain rule. We hope it will be very helpful for you and it will help you to understand the solving process. ( Substituting This website uses cookies to ensure you get the best experience. Here's how to find the derivative of √(sin, Differentiation of Transcendental Functions, 2. Using cos2θ – 1 = –sin2θ, Derivative of cosine; The derivative of the cosine is equal to -sin(x). The process of calculating a derivative is called differentiation. The derivatives of sine and cosine display this cyclic behavior due to their relationship to the complex exponential function. Since each region is contained in the next, one has: Moreover, since sin θ > 0 in the first quadrant, we may divide through by ½ sin θ, giving: In the last step we took the reciprocals of the three positive terms, reversing the inequities. Then, applying the chain rule to {\displaystyle \arcsin \left({\frac {1}{x}}\right)} < x It can be shown from first principles that: Explore animations of these functions with their derivatives here: Differentiation Interactive Applet - trigonometric functions. Simple step by step solution, to learn. Let’s see how this can be done. 2 So, using the Product Rule on both terms gives us: (dy)/(dx)= (2x) (cos x) + (sin x)(2) +  [(2 − x^2) (−sin x) + (cos x)(−2x)], = cos x (2x − 2x) +  (sin x)(2 − 2 + x^2), 6. For any interval over which $$\cos(x)$$ is increasing the derivative is positive and for any interval over which $$\cos(x)$$ is decreasing, the derivative is negative. =cos x(cos x-3\ sin^2x\ cos x) +3(cos^3x\ tan x)sin x-cos^2x, =cos^2x -3\ sin^2x\ cos^2x +3\ sin^2x\ cos^2x -cos^2x, d/(dx)(x\ tan x) =(x)(sec^2x)+(tan x)(1). θ sin You can investigate the slope of the tan curve using an interactive graph. Derivatives of the Sine, Cosine and Tangent Functions. − In this calculation, the sign of θ is unimportant. and  by M. Bourne. The graphs of $$\cos(x)$$ and its derivative are shown below. 1 Below you can find the full step by step solution for you problem. The numerator can be simplified to 1 by the Pythagorean identity, giving us.  x Knowing these derivatives, the derivatives of the inverse trigonometric functions are found using implicit differentiation. And then finally here in the yellow we just apply the power rule. 2 In the diagram, let R1 be the triangle OAB, R2 the circular sector OAB, and R3 the triangle OAC. All derivatives of circular trigonometric functions can be found from those of sin(x) and cos(x) by means of the quotient rule applied to functions such as tan(x) = sin(x)/cos(x). We can differentiate this using the chain rule. Privacy & Cookies | The derivatives of cos(x) have the same behavior, repeating every cycle of 4. , we have: To calculate the derivative of the tangent function tan θ, we use first principles. Therefore, on applying the chain rule: We have established the formula. Our calculator allows you to check your solutions to calculus exercises. Proof of cos(x): from the derivative of sine. Proving the Derivative of Sine. θ on both sides and solving for dy/dx: Substituting Taking the derivative with respect to Derivative Rules. ( IntMath feed |, Use an interactive graph to explore how the slope of sine. This is done by employing a simple trick. slope) equals -2.65. − 0 Alternatively, the derivative of arccosecant may be derived from the derivative of arcsine using the chain rule. sin The current (in amperes) in an amplifier circuit, as a function of the time t (in seconds) is given by, Find the expression for the voltage across a 2.0 mH inductor in the circuit, given that, =0.002(0.10)(120pi) xx(-sin(120pit+pi/6)). Here are useful rules to help you work out the derivatives of many functions (with examples below). Properties of the cosine function; The cosine function is an even function, for every real x, cos(-x)=cos(x). In this tutorial we shall discuss the derivative of the cosine squared function and its related examples. y We hope it will be very helpful for you and it will help you to understand the solving process. For example, the derivative of the sine function is written sin′(a) = cos(a), meaning that the rate of change of sin(x) at a particular angle x = a is given by the cosine of that angle. using the chain rule for derivative of tanx^2. = Below you can find the full step by step solution for you problem. It can be shown from first principles that: (d(sin x))/(dx)=cos x (d(cos x))/dx=-sin x (d(tan x))/(dx)=sec^2x Explore animations of these functions with their derivatives here: Differentiation Interactive Applet - trigonometric functions. Notice that wherever sin(x) has a maximum or minimum (at which point the slope of a tangent line would be zero), the value of the cosine function is zero. Being able to calculate the derivatives of the sine and cosine functions will enable us to find the velocity and acceleration of simple harmonic motion. Use Chain Rule .  1 in from above, we get, Substituting The derivative of sin x is cos x, Derivative Proof of cos(x) Derivative proof of cos(x) To get the derivative of cos, we can do the exact same thing we did with sin, but we will get an extra negative sign. By definition: Using the well-known angle formula tan(α+β) = (tan α + tan β) / (1 - tan α tan β), we have: Using the fact that the limit of a product is the product of the limits: Using the limit for the tangent function, and the fact that tan δ tends to 0 as δ tends to 0: One can also compute the derivative of the tangent function using the quotient rule. ) Derivative of the Exponential Function, 7. g And the derivative of cosine of X so it's minus three times the derivative of cosine of X is negative sine of X. in from above, we get, where e Common trigonometric functions include sin(x), cos(x) and tan(x). 0 sin The Derivative Calculator lets you calculate derivatives of functions online — for free! Derivatives of Sin, Cos and Tan Functions. 2 Find the derivatives of the standard trigonometric functions. (   The brackets make a big difference. The first one, y = cos x2 + 3, or y = (cos x2) + 3, means take the curve y = cos x2 and move it up by 3 units. Answer and Explanation: The derivative of sec2 (x) is 2sec2 (x) tan (x). We hope it will be very helpful for you and it will help you to understand the solving process. π Then. Sitemap | Type in any function derivative to get the solution, steps and graph Calculus can be a bit of a mystery at first. {\displaystyle x=\sin y} Simple, and easy to understand, so dont hesitate to use it as a solution of your homework. Then, $y$ can be written as $y = (cos x)^2$. + =((sin 4x)(2)-(2x+3)(4\ cos 4x))/(sin^2 4x). = Use an interactive graph to investigate it. = A function of any angle is equal to the cofunction of its complement. In single variable calculus, derivatives of all trigonometric functions can be derived from the derivative of cos(x) using the rules of differentiation. , (The absolute value in the expression is necessary as the product of secant and tangent in the interval of y is always nonnegative, while the radical Take the derivative of both sides. y Letting θ  = The chain rule is useful for finding the derivative of a function which could have been differentiated had it been in x, but it is in the form of another expression which could also be differentiated if it stood on its own. cos Since we are considering the limit as θ tends to zero, we may assume θ is a small positive number, say 0 < θ < ½ π in the first quadrant. Write sinx+cosx+tanx as sin(x)+cos(x)+tan(x) 2. r θ = 1 in from above, Substituting The derivative of tan x d dx : tan x = sec 2 x: Now, tan x = sin x cos x. This example has a function of a function of a function. Write secx*tanx as sec(x)*tan(x) 3. Generally, if the function  is any trigonometric function, and  is its derivative, ∫ a cos  n x d x = a n sin  n x + C {\displaystyle \int a\cos nx\,dx={\frac {a}{n}}\sin nx+C} In all formulas the constant a is assumed to be nonzero, and C denotes the constant of integration . = Find the slope of the line tangent to the curve of, (dy)/(dx)=(x(6\ cos 3x)-(2\ sin 3x)(1))/x^2. 2. Simple, and easy to understand, so dont hesitate to use it as a solution of your homework. y sin {\displaystyle {\sqrt {x^{2}-1}}} Differentiate y = 2x sin x + 2 cos x − x2cos x. For example, the derivative of f(x) = sin(x) is represented as f ′(a) = cos(a).  Now (cos x)3 is a power of a function and so we use Differentiating Powers of a Function: Using the Product Rule and Properties of tan x, we have: =[cos^3x\ sec^2x] +tan x[3(cos x)^2(-sin x)], =(cos^3x)/(cos^2x) +(sin x)/(cos x)[3(cos x)^2(-sin x)]. − )  We will use this fact as part of the chain rule to find the derivative of cos(2x) with respect to x. Can we prove them somehow? Author: Murray Bourne | Find the derivative of the implicit function. Explore these graphs to get a better idea of what differentiation means. y The derivative of cos^3(x) is equal to: -3cos^2(x)*sin(x) You can get this result using the Chain Rule which is a formula for computing the derivative of the composition of two or more functions in the form: f(g(x)). Derivatives of Sin, Cos and Tan Functions, » 1. Proof of the Derivatives of sin, cos and tan. The derivative of the sine function is thus the cosine function: $$\frac{d}{dx} sin(x) = cos(x)$$ Take a minute to look at the graph below and see if you can rationalize why cos(x) should be the derivative of sin(x). Derivative of square root of sine x by first principles, derivative of log function by phinah [Solved!]. The tangent to the curve at the point where x=0.15 is shown. , For example, the derivative of the sine function is written sin′(a) = cos(a), meaning that the rate of change of sin(x) at a particular angle x = a is given by the cosine of that angle. → Home | {\displaystyle x} {\displaystyle {\sqrt {x^{2}-1}}} 2 x is always nonnegative by definition of the principal square root, so the remaining factor must also be nonnegative, which is achieved by using the absolute value of x.). − x the fact that the limit of a product is the product of limits, and the limit result from the previous section, we find that: Using the limit for the sine function, the fact that the tangent function is odd, and the fact that the limit of a product is the product of limits, we find: We calculate the derivative of the sine function from the limit definition: Using the angle addition formula sin(α+β) = sin α cos β + sin β cos α, we have: Using the limits for the sine and cosine functions: We again calculate the derivative of the cosine function from the limit definition: Using the angle addition formula cos(α+β) = cos α cos β – sin α sin β, we have: To compute the derivative of the cosine function from the chain rule, first observe the following three facts: The first and the second are trigonometric identities, and the third is proven above. The following derivatives are found by setting a variable y equal to the inverse trigonometric function that we wish to take the derivative of. You can see that the function g(x) is nested inside the f( ) function. 5. {\displaystyle \mathrm {Area} (R_{2})={\tfrac {1}{2}}\theta } Sign up for free to access more calculus resources like . x The differentiation of trigonometric functions is the mathematical process of finding the derivative of a trigonometric function, or its rate of change with respect to a variable. x The derivative of cos x is −sin x (note the negative sign!) Antiderivative of cosine; The antiderivative of the cosine is equal to sin(x). We know that . ) Note that at any maximum or minimum of $$\cos(x)$$ corresponds a zero of the derivative. So the derivative will be equal to. Find the derivative of y = 3 sin3 (2x4 + 1). Many students have trouble with this. cos (5 x) ⋅ 5 = 5 cos (5 x) We just have to find our two functions, find their derivatives and input into the Chain Rule expression. ) . 1 ( Derivative of cos(pi/4). Proving that the derivative of sin(x) is cos(x) and that the derivative of cos(x) is -sin(x). We need to determine if this expression creates a true statement when we substitute it into the LHS of the equation given in the question. y = {\displaystyle 0