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Coins Ethereum (ETH)

# Ethereum Smart Contracts Running Out of Gas Quadruples Following the Istanbul Hard Fork

Ethereum’s successful third upgrade of 2019 Istanbul not without negative consequences.

On Dec. 8, Ethereum successfully completed the Istanbul hard fork at block 9,069,000. This was the third system-wide upgrade in 2019 after St. Petersburg and Constantinople and overall eighth hard fork.

The upgrade included six Ethereum Improvement Proposals viz. 152, 1108, 1344, 1844, 2028, and 2200.

As per ConsenSys blog post, the main issues addressed by these EIPs involves gas costs, interoperability with equipment hash-based proof-of-work (PoW) cryptos like Zcash, and denial-of-service attack resilience.

Just before the Istanbul has to take place, ethereum client Parity released an urgent message asking the client users to upgrade to the latest version of Parity Ethereum, issuing an emergency patch, before the hard fork.

The fix was simple but ethereum core developer Hudson Jameson said if one or two major changes or mining pools would have remained on the old chain, it would have caused confusion and “in a more severe case double spends.”

## Negative Consequences of the Network Upgrade

Now, while the upgrade was technically successful, Antoine Le Calvez data engineer at blockchain analytics company Coin Metrics notes that it wasn’t without negative consequences.

He pointed out that some of the EVM operations were repriced to cost more in fees. This has affected some of the smart contract activity. Interestingly, four of the six EIPs were focused on repricing and reducing the gas cost.

Following the hard fork, on the global level, Le Calvez found that smart contract calls running out of gas have become much more frequent. Failure rates have increased 4 times.

He also found some exchanges are also affected. Cryptocurrency exchange Gemini is one of them that he says hasn’t been able to move user deposits to its hot wallet after the Istanbul upgrade. Each of its attempts is resulting in an out-of-gas error.

However, Gemini says this is because its testnet and mainnet sweepers are a bit different and they are working on replacing them.

“Instead of building a full fidelity test, we calculated the necessary gas limit increase and applied it last week,” explained the exchange. They had to add even more gas after the exchange hit an old Solidity quirk. But now the sweeps are “proceeding normally!”

# Ethereum (ETH) Live Price

## 1 ETH/USD = $540.8404 change ~ 5.00% Coin Market Cap$61.29 Billion

24 Hour Volume

$3.21 Billion 24 Hour VWAP$533

24 Hour Change

\$27.0538
Previous ReviewWho’s The ‘Villain’ Behind MATIC’s 72% Crash?
AnTy has been involved in the crypto space full-time for over two years now. Before her blockchain beginnings, she worked with the NGO, Doctor Without Borders as a fundraiser and since then exploring, reading, and creating for different industry segments.


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# convexity of two convex function combined

If I have such an optimization problem: $$\arg\min_{x,z} \dfrac{||y-x||_2^2}{z}+z(||y-x||_2^2)$$ where $y$ is known and $z>0$.

If $z$ is fixed, then this function w.r.t $x$ is convex. Similarly, if $x$ is given, this function w.r.t $z$ is also convex. However, I wonder:

1. Is the original function w.r.t $x$ and $z$ jointly convex?
2. If so, can I optimize the function alternatively by fixing $x$ solving for $z$ and fixing $z$ solving for $x$ until converging?
3. Are there any relevant papers/materials that I should read for this kind of question?
• I hesitate to give an answer because I'm not sure I'm correct. But WLOG suppose y=0. Then we have $\| x\|_2^2 ( \frac{1}{z} + z)$. This is not convex unless we restrict the domain of z to be positive. In fact, if x is given we have $c(z + \frac{1}{z}$ where c is a constant. In addition to being undefined at 0, it is non-convex, contrary to what you said in your post. – David Kozak Nov 19 '17 at 3:09
• In general, a book on functional analysis ought to answer these types of questions -- Kreyszig's Introductory Functional Analysis with Applications is quite good. – David Kozak Nov 19 '17 at 3:14
• Look at Boyds Convex Analysis – kjetil b halvorsen Nov 19 '17 at 9:42
• Any bilinear function defined on $\mathbb{R}^2$, such as $f(x,y)=xy$, provides simple counterexamples to $(1)$ and $(2)$. – whuber Nov 19 '17 at 16:49
• @kjetil b halvorsen is referring to the freely available "Convex Optimization", by Boyd and Vandenberghe web.stanford.edu/~boyd/cvxbook . Section 3.2 is "Operations that preserve convexity". – Mark L. Stone Nov 19 '17 at 19:11

This is not jointly convex in x and z, even when z > 0.

Consider the example x and z both being 1-D. Let y=0, x=1, z=2. Then the Hessian w.r.t x and z has eigenvalues 5.43 and -0.18, and is therefore indefinite.

I suggest you try using a general purpose nonlinear non-convex optimizer in which you can impose the non-negativity constraint on z. Either use a local optimizer, or you could try using a global optimizer, such as BARON, if the problem dimension is not too high. If using BARON, you will need to place finite lower and upper bounds on all variables - if they are achieved at the optimum, then you need to adjust.

I would suggest trying off the shelf optimizers before trying to gin up your own algorithm. You can try an alternating optimization as you proposed in 2, but I don't think there is any guarantee it will converge, and if it does converge, that it will do so to the correct answer.

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## CNT Bands Challenge Problem

Version 8
by (unknown)
Version 9
by (unknown)

Deletions or items before changed

1 2 3 4 It is important to have a quantitative model describing how an interaction of the CNT with its environment (e.g. supporting substrate, other nanotubes, polymer matrix, etc.) influences CNT ability to conduct current. One possible mathematical formulation of this physical problem can be stated as follows: In a first nearest neighbor pi-orbital tight-binding approximation let us assume that random distortion is described by random shifts $\Delta_\left\{ii\right\}$ of on-site Hamiltonian matrix elements $h_\left\{ii\right\}$. The dispersion $\sigma$ of on-site shifts is defined in its usual way: $\sigma^\left\{2\right\}=<\Delta_\left\{ii\right\}^\left\{2\right\}>-<\Delta_\left\{ii\right\}>^\left\{2\right\}$, where angular brackets denote averaging over all atoms ''i''. - The + The localization length of a randomly distorted CNT is defined as its length, for which the logarithm of the ratio of ideal transmission to the transmission in the CNT subjected to distortion equals 2. - the random short range defects (e.g. ) + Given CNT indexes ($n_1$, $n_2$) and dispersion $\sigma$ find the localization length as function of electron energy. + + + ---- The article contains detailed and self-contained explanation of quantum transmission calculations. As an example providing insight on the properties of Greens functions and contact self-energies, analytical expression for localization length in randomly distorted CNT is derived (Section VIII, Eqs.(41-43)). Download: [[File(Real-Life-Problem_CNT.pdf)]]


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Question

# You want to invest money for your child's education in a certificate of deposit (CD). You want it to be worth \$12,000 in 10 years. How much should you invest if the CD pays interest at a 9\% annual rate compounded a) Annually? b) Continuously?

Decimals
You want to invest money for your child's education in a certificate of deposit (CD). You want it to be worth $$12,000$$ in 10 years. How much should you invest if the CD pays interest at a $$9\%$$ annual rate compounded
a) Annually?
b) Continuously?

2021-05-18
Step 1
First we write the formula $$A= \text{Final amount}\ P= \text{principal amount}\ r= \text{rate of interest}$$ (in decimals)
$$n=$$number of times principal amount compounded in a year and $$t=$$ period of investment
$$A=P\left(1+\frac{r}{n}\right)^{nt}$$
Step 2
Now we put the values $$A= 1200,$$
$$r= 9\%=0.09,\ n =1$$ and $$t=10$$ years
$$12000=P\left(1+\frac{0.09}{1}\right)^{1\times10}$$
$$12000=P(1.09)^{10}$$
$$12000=P\times2.37$$
$$P=\frac{12000}{2.37}=5063.29$$
Step 3
Answer: $$5063.29$$ to be invested to get the final amount $$12000$$ after 10 years compounded annually.


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# The condition for random positive matrice integration

For a $$k \times k$$ positive matrix $$V=(v_{ij}),$$ write $$V=\Gamma'D\Gamma,$$ where $$D=diag(d_1,d_2,\ldots,d_k)$$ with $$d_1>d_2>\cdots>d_k,$$ and $$\Gamma$$ is orthogonal matrix. From the result of Farrell (1985), $$dV=\prod_{i here $$dV=\prod_{i\le j} dv_{ij}$$, $$dD=\prod_{i=1}^k dd_{i},$$ and $$d\Gamma$$ denotes over the space of orthonormal matrices.

Question: What is the sufficient and necessary conditions of a,b,c for the following integration is integrable? $$\int\int\frac{1}{|V|^a|W|^b|I_k+V+W|^c\prod_{i where positive matrix $$W=O'\Lambda O,$$ $$\Lambda=diag(\lambda_1,\lambda_2,\ldots,\lambda_k)$$ with $$\lambda_1>\lambda_2>\cdots>\lambda_k,$$ and $$O$$ is orthogonal matrix.

From (1), it follows that $$\int\int\frac{1}{|D|^a|\Lambda|^b|I_k+\Gamma'D\Gamma+O'\Lambda O|^c} dDd\Lambda d\Gamma dO.$$


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14 Jun 2008, 10:10
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$10,000 is deposited in a certain account that pays r percent annual interest compounded annually, the amount D(t), in dollars, that the deposit will grow to in t years in given by D(t)=10,000(1+(r/100))^t. What amount will the deposit grow to in 3 years? (1) D(1) = 11,000 (2) r=10 Below is a flawed version of the above correct question: 10.$10,000 is deposited in a certain account that pays r percent annual interest compounded annually, the amount D(t), in dollars, that the deposit will grow to in t years in given by D(t)=10,000(1+(r/100))t What amount will the deposit grow to in 3 years?

(1) D(t) = 11,000
(2) r=10

Bunuel
[Reveal] Spoiler: OA

Last edited by Bunuel on 24 Feb 2012, 00:17, edited 5 times in total.
Edited the question and added the OA
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14 Jun 2008, 11:16
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B

statement 1: D(t) = 11000, but t could be any thing, it could be 10 years or 2 years.
insuff

statement 2: suff, because it gives the interest rate and you have the years in the stem, you can find out the total after 3 years.
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14 Jun 2008, 21:25
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quantum wrote:
10. $10,000 is deposited in a certain account that pays r percent annual interest compounded annually, the amount D(t), in dollars, that the deposit will grow to in t years in given by D(t)=10,000(1+(r/100))t What amount will the deposit grow to in 3 years? (1) D(t) = 11,000 (2) r=10 Can anybody explain the logic please? Thank you! This is D indeed. Rephrased, knowing that t=3, the equation is: D(t)=10,000(1+(r/100))3 - If you know either D(t) or r, you can solve this equation for all variables. _________________ Check out the new Career Forum http://gmatclub.com/forum/133 Math Expert Joined: 02 Sep 2009 Posts: 32610 Followers: 5651 Kudos [?]: 68585 [1] , given: 9815 Re:$10,000 is deposited in a certain account that pays r [#permalink]

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23 Feb 2012, 23:59
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Expert's post
quantum wrote:
10. $10,000 is deposited in a certain account that pays r percent annual interest compounded annually, the amount D(t), in dollars, that the deposit will grow to in t years in given by D(t)=10,000(1+(r/100))t What amount will the deposit grow to in 3 years? (1) D(t) = 11,000 (2) r=10 Can anybody explain the logic please? Thank you! Two things: Formula should be $$D(t)=10,000(1+\frac{r}{100})^t$$ and statement (1) should read $$D(1)=11,000$$ (since two statements in DS never contradict and give true information then r=10 must give 11,000 for t=1 as it does). So the question should read:$10,000 is deposited in a certain account that pays r percent annual interest compounded annually, the amount D(t), in dollars, that the deposit will grow to in t years in given by D(t)=10,000(1+(r/100))^t What amount will the deposit grow to in 3 years?

Question: $$D(3)=10,000(1+\frac{r}{100})^3=?$$. Basically the only thing we need is the value of $$r$$.

(1) D(1) = 11,000 --> $$D(1)=10,000(1+\frac{r}{100})^1=11,000$$ --> we can solve for r. Sufficient.

(2) r=10 --> directly gives the value of r. Sufficient.

Hope it's clear.
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23 Feb 2012, 19:53
gmatnub wrote:
B

statement 1: D(t) = 11000, but t could be any thing, it could be 10 years or 2 years.
insuff

statement 2: suff, because it gives the interest rate and you have the years in the stem, you can find out the total after 3 years.

D(t) means a function. Now, constant values (i.e., 11000) can't be equated with functions because as far as I can think, no one can come up with an equation with a variable that always produces the same value for all values of the variable. Hence, in this question, we must assume that D(t) gives 11000 for some value of t. With that thinking, D should be the correct answer.
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23 Feb 2012, 21:35
Chembeti wrote:
gmatnub wrote:
B

statement 1: D(t) = 11000, but t could be any thing, it could be 10 years or 2 years.
insuff

statement 2: suff, because it gives the interest rate and you have the years in the stem, you can find out the total after 3 years.

D(t) means a function. Now, constant values (i.e., 11000) can't be equated with functions because as far as I can think, no one can come up with an equation with a variable that always produces the same value for all values of the variable. Hence, in this question, we must assume that D(t) gives 11000 for some value of t. With that thinking, D should be the correct answer.

Guys $$t$$ is already give, No? $$3$$ years. So $$t$$ should be $$3$$, No? Hence D
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Re: $10,000 is deposited in a certain account that pays r [#permalink] ### Show Tags 24 Feb 2012, 00:10 quantum wrote:$10,000 is deposited in a certain account that pays r percent annual interest compounded annually, the amount D(t), in dollars, that the deposit will grow to in t years in given by D(t)=10,000(1+(r/100))^t. What amount will the deposit grow to in 3 years?

(1) D(1) = 11,000
(2) r=10

Below is a flawed version of the above correct question:
10. $10,000 is deposited in a certain account that pays r percent annual interest compounded annually, the amount D(t), in dollars, that the deposit will grow to in t years in given by D(t)=10,000(1+(r/100))t What amount will the deposit grow to in 3 years? (1) D(t) = 11,000 (2) r=10 Bunuel Hi The answer is indeeded +1 D Statement Break Down : You need to know R. t is already know at 3 years Statement 1 : D(1) : 11,000 Thus you can easily find out 'R' by equating D(1) & the eqn of D(t) Thus sufficient Statement 2 : R know hence sufficient Thus we can find R which is the _________________ Giving +1 kudos is a better way of saying 'Thank You'. Senior Manager Joined: 25 Nov 2011 Posts: 261 Location: India Concentration: Technology, General Management GPA: 3.95 WE: Information Technology (Computer Software) Followers: 3 Kudos [?]: 123 [0], given: 20 Re:$10,000 is deposited in a certain account that pays r [#permalink]

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24 Feb 2012, 00:30
Bunuel wrote:
quantum wrote:
10. $10,000 is deposited in a certain account that pays r percent annual interest compounded annually, the amount D(t), in dollars, that the deposit will grow to in t years in given by D(t)=10,000(1+(r/100))t What amount will the deposit grow to in 3 years? (1) D(t) = 11,000 (2) r=10 Can anybody explain the logic please? Thank you! Two things: Formula should be $$D(t)=10,000(1+\frac{r}{100})^t$$ and statement (1) should read $$D(1)=11,000$$ (since two statements in DS never contradict and give true information then r=10 must give 11,000 for t=1 as it does). So the question should read:$10,000 is deposited in a certain account that pays r percent annual interest compounded annually, the amount D(t), in dollars, that the deposit will grow to in t years in given by D(t)=10,000(1+(r/100))^t What amount will the deposit grow to in 3 years?

Question: $$D(3)=10,000(1+\frac{r}{100})^3=?$$. Basically the only thing we need is the value of $$r$$.

(1) D(1) = 11,000 --> $$D(1)=10,000(1+\frac{r}{100})^1=11,000$$ --> we can solve for r. Sufficient.

(2) r=10 --> directly gives the value of r. Sufficient.

Hope it's clear.

Agreed. But taking it on the face value, Statement 1 can give an impression that, for any value of t, D(t) gives 11000 ,which I think is not correct. Personally I don;t accept this statement, because in DS category, I do following activities:
1. Is the statement is supported for all situations by the premises/facts given in the question stem.
2. Once the answer for the above is Yes, then I will think, does the statement support all situations of the question.

In this exercise, I get a 'No' to the first part from the question. Hence, the answer should be B.

But, as the statement 1, does not hold good on its face value, I just tried mapping it to the question stem for specific scenario and then got the answer D.

Isn't this correct approach?
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Re: $10,000 is deposited in a certain account that pays r [#permalink] ### Show Tags 24 Feb 2012, 01:36 Expert's post Chembeti wrote: Bunuel wrote: quantum wrote: 10.$10,000 is deposited in a certain account that pays r percent annual interest compounded annually, the amount D(t), in dollars, that the deposit will grow to in t years in given by D(t)=10,000(1+(r/100))t What amount will the deposit grow to in 3 years?

(1) D(t) = 11,000
(2) r=10

Can anybody explain the logic please?
Thank you!

Two things:
Formula should be $$D(t)=10,000(1+\frac{r}{100})^t$$ and statement (1) should read $$D(1)=11,000$$ (since two statements in DS never contradict and give true information then r=10 must give 11,000 for t=1 as it does). So the question should read:

$10,000 is deposited in a certain account that pays r percent annual interest compounded annually, the amount D(t), in dollars, that the deposit will grow to in t years in given by D(t)=10,000(1+(r/100))^t What amount will the deposit grow to in 3 years? Question: $$D(3)=10,000(1+\frac{r}{100})^3=?$$. Basically the only thing we need is the value of $$r$$. (1) D(1) = 11,000 --> $$D(1)=10,000(1+\frac{r}{100})^1=11,000$$ --> we can solve for r. Sufficient. (2) r=10 --> directly gives the value of r. Sufficient. Answer: D. Hope it's clear. Agreed. But taking it on the face value, Statement 1 can give an impression that, for any value of t, D(t) gives 11000 ,which I think is not correct. Personally I don;t accept this statement, because in DS category, I do following activities: 1. Is the statement is supported for all situations by the premises/facts given in the question stem. 2. Once the answer for the above is Yes, then I will think, does the statement support all situations of the question. In this exercise, I get a 'No' to the first part from the question. Hence, the answer should be B. But, as the statement 1, does not hold good on its face value, I just tried mapping it to the question stem for specific scenario and then got the answer D. Isn't this correct approach? If we take (1) as it was written: D(t) = 11,000 then it would mean that D(t) has the same value no matter the time period and annual interest, which makes no sense at all. Though technically it still would be sufficient as it would mean that for t=3 the answer is also 11,000. Having said that I recommend not to spend time on flawed version of a question as you won't see such one on the real test. _________________ Senior Manager Joined: 25 Nov 2011 Posts: 261 Location: India Concentration: Technology, General Management GPA: 3.95 WE: Information Technology (Computer Software) Followers: 3 Kudos [?]: 123 [0], given: 20 Re:$10,000 is deposited in a certain account that pays r [#permalink]

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24 Feb 2012, 02:37
Bunuel wrote:

Having said that I recommend not to spend time on flawed version of a question as you won't see such one on the real test.

Sure. I was just thinking whether such questions would appear in a real test, now my doubt is cleared
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Re: $10,000 is deposited in a certain account that pays r [#permalink] 25 Jun 2015, 19:21 Similar topics Replies Last post Similar Topics: 3 Ten years ago, Dorothy deposited a certain amount of money in account 3 29 Jul 2015, 10:02 José deposited a total of$ 10,000 into two interest-bearing accounts 7 29 Dec 2014, 05:28
67 If $1,000 is deposited in a certain bank account and remains 24 01 Nov 2010, 17:32$10,000 is deposited in a certain account that pays r 2 07 Jun 2010, 02:39


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Thread: (Alpha) General relativity is self-inconsistent v2

1. AlphaNumeric: I *still* read that as saying that universal special relativity only exists in the absence of gravity fields. The accuracy of the SR calculations, along with its effective area of symmetry, is dependent upon the curvature in the region. In other words, SR applies everywhere only if spacetime is "flat", otherwise we must consider it an approximation whose error is dependent upon the local curvature fields. Close?

BenTheMan, it appears that my "common misinterpretation" that matter never actually crosses the event horizon is not completely without merit...

USA Today reported on research done at Case Western Reserve which suggests something along the same lines. While it appears that their motivation was to reconcile the apparent loss of information associated with matter falling through the horizon, this interpretation also handles my personal objections and eliminates my confusion.
But a paper accepted recently for publication in the journal, Physical Review D, goes a bit further. Case Western Reserve University researchers Tanmay Vachaspati, Dejan Stojkovic and Lawrence Krauss suggest the event horizon for collapsing stars never quite comes into existence, leaving room for a kind of "pre-Hawking radiation" that might obviate the information paradox. So, the black hole just kind of slowly dissipates before it ever truly forms.
As for an unlucky encyclopedia and its information, "the infalling observer never crosses an event horizon, not because it takes an infinite time, but because there is no event horizon to cross," concludes the study.

2. More references:

http://www.newscientist.com/article/...lly-exist.html

http://www.ccnmag.com/article/astron...es_do_not_form

http://scitation.aip.org/getabs/serv...cvips&gifs=yes

Note that I shared this information without making any personal comments about any other posters. It's not that hard, really.

3. Originally Posted by RJBeery
I was under the assumption that relativistic mass would approach infinity as the horizon was approached. My confusion was in how this "infinite" relativistic mass would be expressed (in terms of infinite energy, or infinite gravitational attraction, etc). I'm wondering now if the problem is somehow resolved because of the curvature of spacetime caused by the black hole, per AlphaNumeric, but I'm still having a problem internalizing what that means.
Hi RJ,
First - a disclaimer. I'm not an expert. I'm a business graduate who likes to splash in the shallows of physics like a day-tripper's kid at the beach. So, what I say could be wrong.

A key is in Alpha's note that SR is a local theory. You can use it for local spacetime, but not necessarily for far-removed spacetime. In practice, "local" for SR means "the nearby spacetime that is close enough to flat".

Compare this with Euclidean geometry on the Earth's surface - On Earth's curved surface, 2D Euclidean geometry is a local theory. It works OK in some local region (eg 40 km East then 30 km North makes 50km at 53° East of North), but not for larger regions (4000km East then 3000km North does not make 5000km at 53° East of North).

What that means in this particular context is that if a massive thing goes past you at high speed, then you can use SR to determine its kinetic energy from its speed relative to you as it passes by (ie when it is local to you). So something passing by with speed approaching c would have arbitrarily large energy.

But, if you measure the speed of that same thing when it is in a different place from you (non-local), SR might not be able to tell you the thing's kinetic energy from that measured speed, because the curvature of space-time might alter the relationship.

In particular:
As some object approaches and crosses the horizon of a black hole, its speed relative to a distant observer will approach and exceed c. But, since there is significant spacetime curvature between the observer and the object, we can't use SR to tell us what the object's energy is.

Beyond that, I can't help you - I very quickly get lost when I try to follow how to use GR quantitatively, so I don't know how to determine what the energy of an object crossing the horizon actually is in a distant observer's reference frame.

4. Originally Posted by RJBeery
SR applies everywhere only if spacetime is "flat", otherwise we must consider it an approximation whose error is dependent upon the local curvature fields. Close?
That's right, but be careful what conclusions you draw from it.

5. Ahh I think I'm close, thanks so much.

6. Originally Posted by RJBeery
BenTheMan, I guess if you or AlphaNumeric define "bullshit" as me asking questions about Physics on a Physics forum to enhance my understanding then I have a lot more bullshit for you. I would also call into question your role as a legitimate moderator if you truly have this attitude.
One should be very careful to draw the distinction between asking questions and making claims.

7. BenTheMan: So it's your position that my claim
Originally Posted by rjbeery
To me, this suggests that we do not fully understand black holes yet so it may be unfair to use our limited knowledge of them to claim contradictions in existing theories.
is "bullshit" even though I have consequently provided evidence to support my perspective? Or is it simply "bullshit" that someone without a Physics degree could possibly have any intuitive insight that would be meaningful in real world Physics? Isn't that the true source of our conflict? You and others are offended that I have the gall to think critically without putting in the life commitment that you have?

8. RJB---

1.) I have started this conversation in an Alpha thread, which is my fault. Such conversations should not be taking place here. Respond, or don't...I don't care. I probably won't read it.

2.)
is "bullshit" even though I have consequently provided evidence to support my perspective?
Quoting things like New Scientist will not help your cause, as they mainly write about crackpot stories. You can find all kinds of news stories to support all kinds of idiotic claims. There ARE physicists who believe black holes don't exist. There are also people who believe that the earth exists on the back of a giant turtle. The fact that some people believe such things doesn't make such beliefs legitimate.

3.)
Originally Posted by RJBeery
My resolution to this is that matter does not actually cross the horizon, ever. Call it a "throw away" comment if you wish but I'm willing to bet I'm not the first person in the world to think of it.
You confidently state that "black holes can't form", even though you have no bones about flaunting your layman status. Problem solved.

All you hard working physicists who have nothing better to do than sit around and solve problems, and think about these problems DAY IN AND DAY OUT can go home now, because I, the armchair physicist, have tidily wrapped things up in an altogether obvious way, that every genius who has ever studied this problem has missed.
Asking questions is encouraged. I've tried to model this forum as a place where people can come and ask questions. Throwing open the doors of the hall and confidently stating that 70 years of science should be overthrown because of your brilliant'' idea is strongly discouraged. Ask Farsight, or Reiku/saxion/gluon, or (...)

We understand many things about black holes, including (more or less) everything that happens at the horizon, where these thought experiments are taking place.

4.)
You and others are offended that I have the gall to think critically without putting in the life commitment that you have?
I can assure you, this is not the case. I and others are offended because you think critically, make a mistake, and then refuse to admit that you're wrong, even when shown the proper logic, or told the correct answer by someone who HAS put in a life commitment to answering these questions.

9. 1. If you hadn't made the "bullshit" comment I wouldn't have responded so rudely.

2. Three of the links I provided were a summary of the paper submitted to and accepted by Physical Review D. The fourth was a link to the actual paper. I was surprised at the prevalence of references to this particular paper which you were apparently unaware of when you made the comment that this concept was a "common misinterpretation".

3. I'm pretty sure I didn't say "confidently that black holes can't form". I said that matter never making across the horizon was a resolution to the problems I had with understanding black holes (while at the same time admitting my lack of knowledge in this area). Then some posters said I don't know enough to even identify problems. Then I pointed out that this very solution not only solves my issues but it also resolves the "loss of information" paradox, and that this POV is seen as a fresh new alternative to the concept of how black holes behave.

If you would've said something like "yes it's possible that matter never makes it across the event horizon but due to your lack of formal education the reason for this is probably much deeper than the reason that you suspect it is," I would have accepted that comment. But the fact is that it was YOU and GUEST432 and ALPHANUMERIC that appeared to be claiming with authority either directly or indirectly that this interpretation was completely FALSE.

4. I invoke the Alpha Rule that requires you to defend your statements. Please provide examples of the following:

Originally Posted by BenTheMan
I and others are offended because you think critically, make a mistake, and then refuse to admit that you're wrong, even when shown the proper logic, or told the correct answer by someone who HAS put in a life commitment to answering these questions.

10. A black hole is a certain type of solution to Einstein's field equations. You claimed (or believed, or felt, or wanted) that there is some sort of paradox (which there isn't), which is resolved if matter cannot cross the event horizon. Please offer us some science. Please outline a detailed review of the simplest case (Schwarzschild) indicating why a test particle can never, ever penetrate the event horizon. This will involve detailed calculation, obviously. The GR I know has no problems in dealing with in falling test particles in a neighbourhood of the event horizon (for example see Kruskal, 1960), so these calculations will be routine. I want to see the armchair physicist do some physics.

If, however, your claims are baseless, then I see a problem with your conformity to these "alpha rules". I have no problem with people making grand claims, as long as they can show something to back them up with. In this case, I'd like to see some physics.

11. Fair warning, I don’t have a lot of time to devote to this old thread. My argument has improved since the OP. (I’ve got it down to the point where the direct objections typically contradict multiple reputable references.) I’ll summarize before responding to any leftover relevant replies.

First, on math: unnecessary. Reams of texts on GR, including sprinkled all over the internet, show its predictions in plain language suitable for laymen. When those predictions clearly contradict one another, that is a problem with those texts at least. I am satisfied that dozens of smart physicists (like Thorne and Taylor and Wheeler) did not all just happen to make the same mistakes. I am satisfied that the true problem lies with GR itself. Your conclusions may differ; that’s fine, we don’t have to agree.

The equivalence principle, in Einstein’s own words, implies in no uncertain terms that the laws of physics are the same in all inertial frames, including those in our real, gravity-endowed universe. This can be true even though a tidal force is always present in our real frames/labs. An inertial frame must meet Thorne’s definition in the OP; i.e. be small enough that the tidal force in the frame is negligible, so that the result of the SR experiment conducted in the frame can agree with the predictions of SR to all the significant digits for the experiment. This is how SR is experimentally confirmed to numerous significant digits in our real labs. As I mentioned to RJBerry above, a speck of dust in a galaxy far, far away from us can throw off any of our SR experiments at some precision. That doesn’t mean that I can’t use the equivalence principle to show a contradiction of GR. The tidal force just needs to be properly accounted for.

In principle, to how many significant digits can an experiment of SR agree with the predictions of SR in a given-sized inertial frame falling through the horizon of a black hole? The answer is that it depends on the mass of the black hole: the larger its mass, the more the significant digits. (The horizon is at r = 2M, so the larger the mass, the greater the circumference of the black hole.) There need not be any spaghettification at the horizon. A given lab falling through the horizon of a black hole, even one larger than a cruise ship, could be a better lab for testing SR (i.e. have a weaker tidal force throughout it) than any lab in which humans have tested SR to date—whether the lab is better depends on the black hole’s mass.

Suppose an experiment of SR is being conducted in the International Space Station, when the ISS (while maintaining its orbit around the Earth) falls through the horizon of a black hole so massive that no instrument in the ISS or yet created by mankind is capable of registering the slightly increased tidal force. Is the experiment invalidated? No, it isn’t, according to equivalence principle. The experiment can continue just fine, and is still valid to a certain number of significant digits. Even as the Earth crosses the black hole’s horizon, the Earth stays in orbit around the Sun, the Moon stays in orbit around the Earth, and nobody need notice anything out of the ordinary, even visually. The spaghettification could be delayed until a billion Earth years later.

If the predictions of GR can be used to show by example that the laws of physics among inertial frames must radically differ, that would be the death of the theory, because GR would then contradict itself (namely, its equivalence principle).

Such an example exists. The tidal force is ruled out as the culprit, because the tidal force in the anomalous frame can be weaker (or stronger, it doesn’t matter) than the tidal force in the “ordinary” frame. In my post to RJBerry above I noted that a law of physics that GR demands in X (an inertial frame falling through the horizon of a black hole) does not apply in an inertial frame that is wholly above the horizon (perhaps orbiting some neutron star such that the tidal force in this frame is stronger than the tidal force in X). In an inertial frame that is wholly above the horizon, even one in which the tidal force is stronger than in X, a free test particle located anywhere in the frame can in principle have the same velocity (limited to some number of significant digits) with respect to the frame as that of any other possible (in principle) free test particle in the frame. That’s a law of physics that does not apply in X, according to GR. In X, a free test particle located below the horizon cannot (even in principle) have the same velocity (to any number of significant digits) with respect to X as that of a free test particle that is above the horizon and escaping to infinity. The particle that is below the horizon must be falling toward the central singularity, according to GR, whereas the escaping particle is rising away from the black hole. GR allows the particle that is above the horizon to escape, with a velocity relative to X that must be less than c as locally measured (or else GR has a different major problem).

I noted above that X need not be arbitrarily small; it could be as large as a cruise ship with my case re GR’s self-inconsistency still made. However, you are free to imagine that X is an arbitrarily small, nonzero-sized frame.

A common objection claims that the escape velocity throughout X must be c, so that no particle in X can be escaping the black hole. I think people glom onto this one when they can’t see any other way to save GR. The horizon (where by definition the escape velocity is c) is a specific altitude that is reached by X as it falls. Then it is logically impossible to preclude X from—when it falls a bit further—covering a region in which a free test particle can either escape (from the upper part) or not escape (from the lower part).

Once people realize that it’s impossible to show that the escape velocity throughout a nonzero-sized X is c, they tend to revert to trying to prove that X must be point-sized. So let me note again that our experiments that have confirmed SR are perfectly valid despite taking place in real (nonzero-sized) labs, because each experiment confirmed SR to only a certain precision (number of significant digits) that was not so fine as to allow the numerical result to be affected by the tidal force in the lab. Which is to say the tidal force in the lab was negligible for the purposes of the experiment. In the self-contradiction of GR shown above, the tidal force is eliminated as a culprit, so the self-contradiction has nothing to do with the size of X per se.

The standby objection is to harp on the lack of math above. So let me emphasize that the law of non-contradiction, a law of logic that trumps anything GR has to say, applies equally well to both worded and mathematical statements. GR texts translate GR’s math into worded predictions. The OP includes the equivalence principle in Einstein’s own words. While you can validly say that I have not shown a self-inconsistency of GR mathematically, claims that I cannot show a self-contradiction without using math, at least a self-contradiction of GR texts, have no merit.

12. Originally Posted by Guest254
A black hole is a certain type of solution to Einstein's field equations. You claimed (or believed, or felt, or wanted) that there is some sort of paradox (which there isn't), which is resolved if matter cannot cross the event horizon. Please offer us some science. Please outline a detailed review of the simplest case (Schwarzschild) indicating why a test particle can never, ever penetrate the event horizon.
Who are you referring to? I made no such claim. Would help if you quoted.

13. Originally Posted by zanket
Who are you referring to? I made no such claim. Would help if you quoted.
I'm referring to RJBeery.

However, in your direction: I believe it's been made clear to you why your conclusions are wrong, but it doesn't seem like you want to listen. Don't take my word for it though: submit your work to a decent theoretical physics journal and let us know how you get on.

14. Originally Posted by Guest254
However, in your direction: I believe it's been made clear to you why your conclusions are wrong, but it doesn't seem like you want to listen.
That's the oldest non-argument in the book.

15. Hey, you're obviously working on another level to me, so I think it best if you send your work to some experts and let them judge its merits. Submit your work to a decent theoretical physics journal, and let us know how you get on.

16. Originally Posted by Guest254
Hey, you're obviously working on another level to me, so I think it best if you send your work to some experts and let them judge its merits. Submit your work to a decent theoretical physics journal, and let us know how you get on.
That's the second-oldest non-argument in the book.

17. Why wont you send your work to a reputable journal?

18. Originally Posted by RJBeery
You and others are offended that I have the gall to think critically without putting in the life commitment that you have?
No, I'm offended you think that without putting in some effort you think you can see further than everyone else on a topic you have read nothing about.

Every single person who has a degree in physics and above worked to get it. Some found the work easy but those people generally spend a lot of time reading and less time going "Huh!?". To do postgrad GR is hard. Not a person who has done it will say otherwise.

But yet you think you don't need to put in any effort to know or do more. If you'd read 'GR' by Wald, cover to cover in a day, put it down and said "It's all pretty straight forward" and then churned out say a 4 line proof to the Positive Energy Theorem, I'd be damn impressed. You've shown that with much less effort than it took me (or anyone else in history) you've grasped the topic. But you haven't. You haven't read any books and it's clear if you did, you'd struggle enormously with them.

Now I know people from both sides of the scale. I know people who effortlessly learnt what still bemuses me, in 1/10 the time I've been thinking about it. Quite a few of them are likely to remain permanent residents of Cambridge, UK. I also know a fair few who fail to grasp why linear equations admit linear combinations of solutions as solutions and it takes 30 minutes of my time to explain it to them. You're the latter and you believe yourself the former. If you were the former people like Ben r Guest or myself would notice, we have seen such people in our friends or our students. They are easy to spot. So its not a case of I'm offended that you can do this stuff with minimal effort compared to us, I'm offended you think you can do this stuff with absolutely zero effort and I'm willing to be you did less than stelllar in high school maths and physics, am I right?

You put in less effort than the brightest of the bright would need to understand GR and despite not being anywhere close to the mediocre of the mediocre, you think you're in a position of knowledge.

I have one or two students like you. They're a little more loud mouth, they put their hand up more to get me to read their 'brilliant method' than to get help and then they sail through to an unspectacular 15/25, or in the case of one prat 2/25 (best mark of the last month for him!).

So tell me, how did you do in high school maths and physics, out of interest?

19. Originally Posted by RJBeery
But the fact is that it was YOU and GUEST432 and ALPHANUMERIC that appeared to be claiming with authority either directly or indirectly that this interpretation was completely FALSE.

4. I invoke the Alpha Rule that requires you to defend your statements. Please provide examples of the following:
Okay, you want an Alpha level response to why GR has SR as a symmetry?

Definition of Lorentz transforms : $\Lambda^{a}_{c}\eta_{ab}\Lambda^{b}_{d} = \eta_{cd}$

Required to prove : $g_{ab}\to g_{ab}$ under $\Lambda$ put at point in space-time.

Introduce new notation : a,b,c,d are curved space indices. $\mu,\nu,\sigma,\tau$ flat space indices.

Define vierbein as the transformation between flat and curved indices : $e^{\mu}_{a}$ thus defining $g_{ab} = e^{\mu}_{a}\eta_{\mu\nu}e^{\nu}_{b}$.

Lorentz transforms $\Lambda$ act only on flat space indices and so if we do a Lorentz transform $e^{\mu}_{a} \to \Lambda_{\sigma}^{\mu}e^{\sigma}_{a}$.

Effect on g : $g_{ab} = (\Lambda_{\sigma}^{\mu}e^{\sigma}_{a})\eta_{\mu\nu }(\Lambda_{\rho}^{\nu}e^{\rho}_{b}) = e^{\sigma}_{a}(\Lambda_{\sigma}^{\mu}\eta_{\mu\nu} \Lambda_{\rho}^{\nu})e^{\rho}_{b} = e^{\sigma}_{a}\eta_{\sigma\rho}e^{\rho}_{b} = g_{ab}$.

Therefore GR has Lorentz symmetry at any point in the space-time where $e^{\mu}_{a}$ is well defined. Shall I continue talking about vierbeins and allowing spinors to couple to space-time or shall I give you 3~6 years to go learn enough GR to understand what I just said?

20. Originally Posted by RJBeery
Consider 2 rockets, A and B, with rocket A an arbitrarily small amount further from the center of the Earth than B. Each rocket is accelerating at 9.0 m/s/s above the Earth's surface at precisely the altitude such that rocket A "escapes to infinity" while B is unable to overcome Earth's gravity, and subsequently falls.
Even though it might be an apt example, noninertial acceleration, like that of a thrusting rocket, opens a big can of worms I don't care to get into.

My gut tells me that you aren't applying the definition of inertial frame correctly because an inertial frame should not have a non-uniform field of gravity applied to it.
My post #51 covers this. Note that a relative gravitational acceleration of two free test particles in different parts of a frame is the definition of the tidal force.

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# matplotlib latex font

Keep in mind that Matplotlib expects a font in True Type format (.ttf). … 예. TeX 수식은 rc 함수를 사용하여 플롯에 삽입 할 수 있습니다.. import matplotlib.pyplot as plt plt.rc(usetex = True) 또는 rcParams 액세스 : . If you need more information just tell me. Changing default font typeface. PLEASE DO NOT ASK FOR HELP ## IF THIS FEATURE DOES NOT DO WHAT YOU EXPECT IT TO. Matplotlib（下称mpl）自带tex引擎，可以解析并显示Latex字符串。虽然中文会覆盖英文字体，但是mpl的数学字体和一般的text字体是独立的，所以我们可以把文字放到Latex字符串里面，并使用\mathrm{text}来使其显示为正体。 > I'm trying to create publication quality plots for inclusion > in LaTeX papers and presentations and have some problems > to get the fonts right. Text handling with matplotlib's LaTeX support is slower than standard text handling, but is more flexible, and produces publication-quality plots. LaTeX 选项通过在rc设置中设置text.usetex:True来激活。 使用 matplotlib 的 LaTeX 支持的文本处理会慢于 matplotlib 的非常强大的 mathtext，但是更灵活，因为可以使用不同的 LaTeX 包（字体包，数学包等）。 结果会十分惊人，特别是当你在图形中使用和主文档相同的字体。 On Tuesday 07 August 2007 07:37:04 am Johan Ekh wrote: > Hi all, > I'm new to Python in general and Matplotlib in particular. You can change math font in your plotting script so that it only takes effect in your script. What can I do? The default font is BitstreamVeraSans Roman, but we want to try out something else.You can pass fontname to .set_xlabel, .set_ylabel, .set_title, or .annotate to specify a particular font. I tried to do everything that I could, but I still unable to run Matplotlib and have a Latex font. update ({'font.size': 12, 'font.family': 'sans', 'text.usetex': False}) Change figure size to make font larger ¶ Instead of changing the font size, we can also make the overall figure smaller (for publication purposes), and then save a pdf file (or some scalable vector format) or use high resolution bitmap formats: Click onto this figure to see the Python code used to generate the PGF file as well as the LaTeX code for the PDF document. This does not change the font for the numbers on the axes. Matplotlib可以无缝的处理LaTex字体，在图中加入数学公式from matplotlib.patches import Polygon import matplotlib.pyplot as plt import numpy as np# 定义一个求积分的函数 def func(x): return 0.3* (x**2) + (0.1*x) + 1# 定义积分区间 a, b = rcParams. Less than ideal, I admit, but sufficient to fool the casual observer: mpl.rcParams['text.latex.preamble'] = [r'\usepackage{helvet}\renewcommand\familydefault{\sfdefault}', r'\usepackage{amsmath}' , r'\usepackage[T1]{fontenc}'] Python + Matplotlib.pyplot で電気力線をポテンシャルによって線色を変えながら描画する 結果図はこちら： 以前プロットした電気力線、 カラーマップの上に黒線で電気力線を描画してもよいけれど、 線の色を電圧に応じて変化させてもいいのではないかと思った。 For example, if we want to add the Helvetica font, we need to check if we have the font in .ttf format installed on our system otherwise we need to download it and install it. 私の場合，過去にpyenvでインストールしたmatplotlibが動かなかった経緯があり，TkAggを指定している． （対応） ~/.marplotlib にmatplotlibrcというファイルを作成し，1行 backend : TkAgg と書き込んで … The axes labels can be found with the search and the fonts match well. I was trying to use xelatex, but it wouldn't run at all. It includes, plt.xticks(fontsize= ) Thank you very much for your help! If you want to use the LaTeX-style font. all, pretty new matplotlib. This option (which is still somewhat experimental) can be activated by setting text.usetex : true in your rc settings. matplotlib Brought to you by: cjgohlke, dsdale, efiring, heeres, and 8 others. In this way, any text in the plot (including TeX formulae) is rendered as text in the final document. can be used. Next, we need to update the font cache from the command line with the following command: The LaTeX option is activated by setting text.usetex: True in your rc settings. However, this caused some of the fonts to look quite different in the plots, and also garbled some of the text where my strings conflicted with TeX syntax. please, note: don't mean change font size font weight. Open an example in Overleaf. rcParams ['text.usetex'] = True. ## LaTeX \usepackage command, please inquire at the ## matplotlib mailing list: #text.latex.preamble : ## IMPROPER USE OF THIS FEATURE WILL LEAD TO LATEX FAILURES ## AND IS THEREFORE UNSUPPORTED. Figure: A LaTeX document which imports a PGF file that was exported from matplotlib. Change the math font temporarily. plt.rcParams['text.usetex'] = True. The document text was easy to change of course, (thanks to LaTeX), but the system complained because the figures I had made using matplotlib contained the wrong font! Note that matplotlib wants the figure # size in inches, so you'll need the … In order to include plots created with matplotlib in TeX documents, they should be saved as pdf or eps files. Matplotlib can use LaTeX to handle the text layout in your figures. The font can also be changed for a specific element in the document. # Plot the median life expectancy by continent ax = df. Operating system: macOS High Sierra version 10.13.5; Matplotlib version: 2.2.2 In this story, I will give you some tricks, 5 powerful tricks in using Matplotlib to create an excellent plot. import matplotlib.pyplot as plt params = {'tex.usetex': True} plt.rcParams.update(params) I had the exact same problem and I fixed it by switching the matplotlib font to Arial. This article shows how to create plots with matplotlib for publications where fonts and font sizes match the LaTeX document and graphics are not blocky, but allow for infinite zooming. ## text.latex.preamble is a single line of LaTeX code that Get latex to tell you its sizes, then use those sizes in matplotlib to create plots that will fit perfectly in your latex file - show_dimensions.tex ... # example, the font size, family, etc. Python下使用matplotlib库时,如何与LaTeX结合起来？ 最近写文章，用Python的matplotlib的库画图，但是纵轴横轴都是些公式，因为会用LaTeX，所以想直接借用LaTeX公式的编辑方法，查文章发… Text handling with matplotlib’s LaTeX support is slower than matplotlib’s very capable mathtext, but is more flexible, since different LaTeX packages (font packages, math packages, etc.) Text handling with matplotlib’s LaTeX support is slower than matplotlib’s very capable mathtext, but is more flexible, since different LaTeX packages (font packages, math packages, etc.) # restore matplotlib. Change the font just for the title or axis labels. In [1]: % matplotlib inline import matplotlib import matplotlib.pyplot as plt import numpy as np plt.setp(ax.get_xticklabels(), Fontsize=) to Set Matplotlib Tick Labels Font Size ax.tick_params(axis='x', Labelsize= ) to Set Matplotlib Tick Labels Font Size In this tutorial article, we will introduce different methods to set tick labels font size in Matplotlib. can be used. The LaTeX option is activated by setting text.usetex: True in your rc settings. The line \usepackage{tgbonum} establishes the font family T e X Gyre Bonum, whose font package name is tgbonum, as the default font for this document. to force matplotlib to produce Type 1 fonts. Filippo. If you face some errors, you need to read the following story. 我正试图在matplotlib中创建一个图形来插入我的LaTeX文档，我需要粗体的'i'和'j'符号，而不是圆点和帽子。 我在我的文档中使用了代码\mathbf{\hat{\textnormal{\bfseries\i}}}，但是由于它使用了amsmath包（或者可能是amssymb包，对不起，我不知道我对LaTeX和matplotlib都很陌生），所以我不能在matplotlib标签中使用它。 I have explained the detailed procedures to deal with LaTeX font in Matplotlib import matplotlib.pyplot as plt plt.rc(usetex = True) or accessing the rcParams: import matplotlib.pyplot as plt params = {'tex.usetex': True} plt.rcParams.update(params) TeX uses the backslash \ for commands and symbols, which can conflict with special characters in Python strings. You need to change the default math font to computer modern. There are two ways to achieve to this. Matplotlib version. matplotlib. En plus de cela, j'utilise ceci: mpl. Using LaTeX font; In default, we can use some nice f o nts that are provided by Matplotlib. matplotlib documentation: 플롯에 TeX 수식 삽입하기. how change font.weight when using [text.usetex]: in following code, using greek symbol (lambda) xaxis weight of symbol seems light , poorly noticeable when pdf file produced latex file.. i don't understand why setting font.weight='bold' doesn't have effect.. any idea how change weight of {lambda} symbol. It is because the default math font has changed since Matplotlib version 2.0. I've spent the last few hours trying to get PGF backend to work. This turns out to be an easy thing to fix, but it took me a while to figure out how to do it, so I thought I would share my step-by-step process here. Adobe Postscript (PSSNFS) font packages # may also be loaded, depending on your font settings Saving and exporting plots that use TeX. But, some symbols are not good enough to be created by Matplotlib. J'ai toujours text.usetex = True dans mon matplotlibrc fichier. import matplotlib as mpl import matplotlib.pyplot as plt import numpy as np # use latex for font rendering mpl.rcParams['text.usetex'] = True x = np.linspace(-50,50,100) y = np.sin(x)**2/x plt.plot(x,y) plt.xlabel(r'$\mathrm{xlabel\;with\;\LaTeX\;font}$') plt.ylabel(r'Not a latex font') plt.show() Esto da el siguiente resultado: To use LaTeX font as your default font in Matplotlib, you can use this code. Latex option is activated by setting text.usetex: True in your script do not for... To run matplotlib and have a LaTeX document which imports a PGF file was! Was exported from matplotlib deal with LaTeX font in True Type format (.ttf ) tried. What you EXPECT it to would n't run at all the numbers on the axes mind! Can use some nice f o nts that are provided by matplotlib slower standard! Xelatex, but is more flexible, and produces publication-quality plots to deal with LaTeX font in... 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# In the following picture, how many pieces can be pasted when the shape

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In the following picture, how many pieces can be pasted when the shape  [#permalink]

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13 Nov 2019, 00:49
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[GMAT math practice question]

In the following picture, how many pieces can be pasted when the shapes shown on the left are pasted as shown on the right?

Attachment:

11.13ps.png [ 20.29 KiB | Viewed 504 times ]

A. $$4$$

B. $$6$$

C. $$8$$

D. $$10$$

E. $$12$$

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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" VP Joined: 19 Oct 2018 Posts: 1158 Location: India Re: In the following picture, how many pieces can be pasted when the shape [#permalink] ### Show Tags 13 Nov 2019, 14:43 Basically, we'll get a regular polygon with interior angle = 360-105-105= 150 (n-2)*180 = n* 150 n=12 MathRevolution wrote: [GMAT math practice question] In the following picture, how many pieces can be pasted when the shapes shown on the left are pasted as shown on the right? Attachment: 11.13ps.png A. $$4$$ B. $$6$$ C. $$8$$ D. $$10$$ E. $$12$$ Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8243 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: In the following picture, how many pieces can be pasted when the shape [#permalink] ### Show Tags 15 Nov 2019, 01:45 1 => When we put two trapezoids together as follows, we have $$∠x = 360° - 2*105° = 75°$$ and $$∠BAC = 30°.$$ Attachment: 11.13ps(a).png [ 14.78 KiB | Viewed 327 times ] Since $$\frac{360°}{30°} = 12$$, we can put in $$12$$ triangles or $$12$$ trapezoids. Therefore, E is the answer. Answer: E _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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Re: In the following picture, how many pieces can be pasted when the shape  [#permalink]

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20 Nov 2019, 02:15
As we can see from the sample figure, it forms a regular polygon with an internal angle of 150 degrees.

(n-2)*180 = 150*n.
On solving we get n = 12.

OA - E

MathRevolution wrote:
[GMAT math practice question]

In the following picture, how many pieces can be pasted when the shapes shown on the left are pasted as shown on the right?

Attachment:
11.13ps.png

A. $$4$$

B. $$6$$

C. $$8$$

D. $$10$$

E. $$12$$
Intern
Joined: 24 Mar 2018
Posts: 10
Re: In the following picture, how many pieces can be pasted when the shape  [#permalink]

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21 Nov 2019, 06:01
75*2 = 150 = 180*(n-2)/n
Hence, n = 12
Re: In the following picture, how many pieces can be pasted when the shape   [#permalink] 21 Nov 2019, 06:01
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text:
# Is calcium chloride an acidic or basic salt?

What I reason: Calcium chloride is the salt of hydrochloric acid and calcium hydroxide. Calcium hydroxide is usually not considered a strong base, and I believe this is because of it's low solubility. $$\ce{HCl}$$ is a strong acid, and so the salt should be slightly acidic. Wikipedia states the $$\mathrm{p}K_\mathrm{a}$$ of $$\ce{CaCl2}$$ is between 8-9, which is in fact slightly acidic, confirming my theory.

What I've experienced: I've prepared a number of aqueous calcium chloride solutions from distilled water, all of which are purple in the presence of universal indicator and tested with a calibrated $$\mathrm{pH}$$ probe to be basic. I've tried multiple sources of calcium chloride and each is basic.

TLDR: Theoretically and literature values state that $$\ce{CaCl2}$$ is acidic, but my empirical evidence shows that it is basic.

Calcium hydroxide has solubility about 1.9 g/L. This is enough to create pH above 11, i.e. a strongly basic solution.

$$\ce{CaCl2}$$ solutions should be very slightly acidic if they were made from pure $$\ce{CaCl2}$$. This might not be the case. In industry calcium chloride is produces by reaction of calcium hydroxide with ammonium chloride, so industrial-grade calcium chloride is likely to be contaminated with calcium hydroxide.

Another potential reason for contamination is hot drying. Calcium chloride is quite hygroscopic, so it needs to be dried before use. When heated to hight enough temperature, moist calcium chloride hydrolyzes https://link.springer.com/article/10.1007/BF02654424

• "Calcium hydroxide has solubility about 1.9 g/L. This is enough to create pH about 10-11" I appreciate to solve and explain it.thanks. – Adnan AL-Amleh May 10 at 17:30
• @AdnanAL-Amleh pH+pOH is roughly 14. For Ca(OH)2 solution [OH] is roughly twice molar concentration of calcium hydroxide, which is roughly 0.025 for saturated solution. $pOH = -log [OH]$; $log_{10} 0.025 = 1.6$; $14-1.6 > 12$. You don't need much base to create a strongly basic solution. – permeakra May 10 at 20:48
• You mean : $\ce{Ca(OH)2_\mathrm{(aq)} <=>Ca(OH)^+_\mathrm{(aq)} + OH^-_\mathrm{(aq)}}$ , so: $$[\ce{OH-}]=[\ce{Ca(OH)2}] = \frac{1.9}{74}=\pu{0.025M}$$ – Adnan AL-Amleh May 10 at 22:27
• @AdnanAL-Amleh Yes. 0.025 M for $[OH-]$ is a low estimate, it is probably higher because $\ce{Ca(OH)+}$ can dissociate too. – permeakra May 10 at 22:55


text: RandomizedSearchCV RandomizedSearchCV solves the drawbacks of GridSearchCV, as it goes through only a fixed number … We want to slow down the learning in b direction, i.e., the vertical direction, and speed up the learning in w direction, i.e., the horizontal direction. Robust logistic regression modelling via the elastic net-type regularization and tuning parameter selection Heewon Park Faculty of Global and Science Studies, Yamaguchi University, 1677-1, Yoshida, Yamaguchi-shi, Yamaguchi Prefecture 753-811, Japan Correspondence heewonn.park@gmail.com Also, elastic net is computationally more expensive than LASSO or ridge as the relative weight of LASSO versus ridge has to be selected using cross validation. Linear regression refers to a model that assumes a linear relationship between input variables and the target variable. When tuning Logstash you may have to adjust the heap size. Train a glmnet model on the overfit data such that y is the response variable and all other variables are explanatory variables. Make sure to use your custom trainControl from the previous exercise (myControl).Also, use a custom tuneGrid to explore alpha = 0:1 and 20 values of lambda between 0.0001 and 1 per value of alpha. See Nested versus non-nested cross-validation for an example of Grid Search within a cross validation loop on the iris dataset. As you can see, for $$\alpha = 1$$, Elastic Net performs Ridge (L2) regularization, while for $$\alpha = 0$$ Lasso (L1) regularization is performed. Tuning the alpha parameter allows you to balance between the two regularizers, possibly based on prior knowledge about your dataset. fitControl <-trainControl (## 10-fold CV method = "repeatedcv", number = 10, ## repeated ten times repeats = 10) The red solid curve is the contour plot of the elastic net penalty with α =0.5. The tuning parameter was selected by C p criterion, where the degrees of freedom were computed via the proposed procedure. In this particular case, Alpha = 0.3 is chosen through the cross-validation. The elastic net regression can be easily computed using the caret workflow, which invokes the glmnet package. viewed as a special case of Elastic Net). multicore (default=1) number of multicore. We apply a similar analogy to reduce the generalized elastic net problem to a gener-alized lasso problem. strength of the naive elastic and eliminates its deﬂciency, hence the elastic net is the desired method to achieve our goal. You can see default parameters in sklearn’s documentation. 5.3 Basic Parameter Tuning. – p. 17/17 Others are available, such as repeated K-fold cross-validation, leave-one-out etc.The function trainControl can be used to specifiy the type of resampling:. The elastic net is the solution β ̂ λ, α β ^ λ, α to the following convex optimization problem: The … 2.2 Tuning ℓ 1 penalization constant It is feasible to reduce the elastic net problem to the lasso regression. multi-tuning parameter elastic net regression (MTP EN) with separate tuning parameters for each omic type. Subtle but important features may be missed by shrinking all features equally. Comparing L1 & L2 with Elastic Net. where and are two regularization parameters. How to select the tuning parameters In this paper, we investigate the performance of a multi-tuning parameter elastic net regression (MTP EN) with separate tuning parameters for each omic type. This is a beginner question on regularization with regression. Once we are brought back to the lasso, the path algorithm (Efron et al., 2004) provides the whole solution path. 2. Fourth, the tuning process of the parameter (usually cross-validation) tends to deliver unstable solutions [9]. If a reasonable grid of alpha values is [0,1] with a step size of 0.1, that would mean elastic net is roughly 11 … I won’t discuss the benefits of using regularization here. Through simulations with a range of scenarios differing in number of predictive features, effect sizes, and correlation structures between omic types, we show that MTP EN can yield models with better prediction performance. Although Elastic Net is proposed with the regression model, it can also be extend to classiﬁcation problems (such as gene selection). References. It is useful when there are multiple correlated features. Output: Tuned Logistic Regression Parameters: {‘C’: 3.7275937203149381} Best score is 0.7708333333333334. Elastic Net: The elastic net model combines the L1 and L2 penalty terms: Here we have a parameter alpha that blends the two penalty terms together. These tuning parameters are estimated by minimizing the expected loss, which is calculated using cross … The Elastic Net with the simulator Jacob Bien 2016-06-27. List of model coefficients, glmnet model object, and the optimal parameter set. For Elastic Net, two parameters should be tuned/selected on training and validation data set. The elastic net regression by default adds the L1 as well as L2 regularization penalty i.e it adds the absolute value of the magnitude of the coefficient and the square of the magnitude of the coefficient to the loss function respectively. Examples For LASSO, these is only one tuning parameter. The lambda parameter serves the same purpose as in Ridge regression but with an added property that some of the theta parameters will be set exactly to zero. Python implementation of "Sparse Local Embeddings for Extreme Multi-label Classification, NIPS, 2015" - xiaohan2012/sleec_python ; Print model to the console. Most information about Elastic Net and Lasso Regression online replicates the information from Wikipedia or the original 2005 paper by Zou and Hastie (Regularization and variable selection via the elastic net). We use caret to automatically select the best tuning parameters alpha and lambda. Suppose we have two parameters w and b as shown below: Look at the contour shown above and the parameters graph. Zou, Hui, and Hao Helen Zhang. ggplot (mdl_elnet) + labs (title = "Elastic Net Regression Parameter Tuning", x = "lambda") ## Warning: The shape palette can deal with a maximum of 6 discrete values because ## more than 6 becomes difficult to discriminate; you have 10. The logistic regression parameter estimates are obtained by maximizing the elastic-net penalized likeli-hood function that contains several tuning parameters. You can use the VisualVM tool to profile the heap. As shown below, 6 variables are used in the model that even performs better than the ridge model with all 12 attributes. So, in elastic-net regularization, hyper-parameter $$\alpha$$ accounts for the relative importance of the L1 (LASSO) and L2 (ridge) regularizations. Through simulations with a range of scenarios differing in. The estimates from the elastic net method are defined by. By default, simple bootstrap resampling is used for line 3 in the algorithm above. All features equally computation issues and show how to select the tuning parameters: (. On training and validation data set, y,... ( default=1 ) tuning parameter selected. Parameters of the elastic net. the elastic-net penalized likeli-hood function that contains several tuning parameters and! Regularization here multiple tuning penalties represent the state-of-art outcome method would represent the state-of-art.. 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text:
# Splitting Fields

This is a problem from Hungerford's book:

Let $E$ be an intermediate field of extension $K\subset F$ and assume that $E=K(u_1, \cdots ,u_r)$ where $u_i$ are (some of the) roots of $f\in K[x]$. Then $F$ is a splitting field of $f$ over $K$ if and only if $F$ is a splitting field of $f$ over E.

This is my attempt.

Let $u_1, \cdots ,v_n$ be the roots of $f$. Then since $F$ is a splitting field of $f$ over $K$, $F=K(v_1, \cdots ,v_n)$ $\implies$ $F=E(v_1, \cdots ,v_n)$. So $F$ is a splitting field of $f$ over $E$ .

Conversely, suppose $v_1, \cdots ,v_n$ are the roots of $f$ in $F$. Then $F=E(v_1, \cdots ,v_n)=K(u_1, \cdots ,u_r,v_1, \cdots, v_n)$. Thus $F$ is a splitting field of $f$ over $K$.

My Question: First I want to know if the proof above is right. The second part of the problem asks to extend the above to splitting fields of arbitrary set of polynomials. This is where I'm lost. I'd like a little help on how to begin.
Thanks.

Update: I think I've got it now. I had some difficulty with the indexing. Thanks.

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There's nothing to do here other than understand and unpack the definition of "splitting field" twice. I recommend you try to do this yourself. If you continue to have difficulty, your next question should be something more elementary about the definition / concept of a splitting field. – Pete L. Clark Mar 2 '11 at 19:32
@Pete: yeah I saw trick after posting the question. Thanks.lol – Nana Mar 2 '11 at 20:02
After two months, perhaps you can start asking questions instead of just copying problems and giving orders? – Arturo Magidin Mar 2 '11 at 20:49
@Arturo: Sorry for the mistake. It wasn't my intention to give orders. It won't happen again. – Nana Mar 2 '11 at 23:05

$\left(\implies\right)$ Let $v_{1},\cdots,v_{n}$ be the roots of $f$ in $F$. Since $F$ is a splitting field of $f$ over $K$, $F=K\left(v_{1,}\cdots,v_{n}\right)$ which implies that $F=E\left(v_{1},\cdots,v_{n}\right)$. So $F$ is a splitting field of $f$ over $E$.
$\left(\Longleftarrow\right)$ Suppose $v_{1},\cdots,v_{n}$ are the roots of $f$ in $F$. Then $F=E\left(v_{1},\cdots,v_{n}\right)$. But each $u_{i}$ is a one of the $v_{j}$, so $F=E\left(v_{1},\cdots,v_{r}\right)=K\left(u_{1},\cdots,u_{r},v_{1},\cdots,v_{n}\right)$.
Thus $F$ is a splitting field of $f$ over $K$.


text:
# 1.4: Failure of Rocks

Let's relate the previously discussed stresses to fracture and faulting. Failure can occur by frictional sliding or fracture. Byerlee's law defines the shear stress needed to cause sliding, essentially it defines the point of failure given an applied normal stress. The law tells us that shear stress ($$\tau$$), increases about linearly with the normal stress. This is modeled with the equation $$\tau=0.85\sigma_{n}$$. Using this calculated shear stress, we can apply this to both frictional sliding and fracture. In the case of failure by frictional sliding, failure is determined by the equation

$\sigma_{s-fric} = C_{o} +\mu_{s}\sigma_{n},$

where $$\sigma_{s-fric}$$ is shear stress, $$C_{o}$$ is the cohesion of fractured rock, $$\mu_{s}$$ is the friction coefficient, and $$\sigma_{n}$$ is the normal stress. Failure occurs when $$\sigma_{s}$$ on the fault plane is bigger than $$\sigma_{s-fail}$$, or $$\sigma_{s}>\sigma_{s-fric}$$. Friction ($$\mu_{s}$$) is typically $$\cong$$ 0.6-0.8 for most rocks, but can be as low as 0.2 for talc and serpentine. The Mohr-Coulomb Fracture Criteria states that failure by fracture is determined by the equation

$\sigma_{s-frac} = C_{1} +\mu_{o}\sigma_{n}.$

where $$\sigma_{s-frac}$$ is the shear stress, $$C_{1}$$ is the cohesion of unfractured rock, and $$\sigma_{n}$$ is the fracture coefficient. Fracture occurs when $$\sigma_{s} > \sigma_{s-frac}$$.

FIGURE failure curves on $$\sigma_{n} vs . \sigma_{s}$$ graph

FIGURE Fault failure point

Calculating Failure Point

Applying what we've learned so far, calculate if the above fault is at or below the failure point. We are given that $$\sigma_{n}$$=200 MPa, $$\sigma_{s}$$=100 MPa, C=10 MPa, and $$\mu_{s}$$=0.8.

\begin{align*} \sigma_{s-fric} &= C_{o} +\mu_{s}\sigma_{n} \\[4pt] &=10 \,MPa + 0.8(200\, MPa) \\[4pt] &=10 + 160 \\[4pt] &=170\, MPa \end{align*}

Since $$\sigma_{s-fric} > \sigma_{s}$$, where $$\sigma_{s-fric}$$ is the stress needed for failure, and $$\sigma_{s}$$ is the applied stress, 170>100, thus the fault is below the failure point.