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27 May 2016, 03:09
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David used part of $100,000 to purchase a house. Of the remaining portion, he invested 1/3 of it at 4 percent simple annual interest and 2/3 of it at 6 percent simple annual interest. If after a year the income from the two investments totaled$320, what was the purchase price of the house?

(A) $96,000 (B)$94,000
(C) $88,000 (D)$75,000

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27 May 2016, 03:31
4
4
Let the Price of house was x.

Then

$$\frac{1}{3}$$(10000-x) * $$\frac{4}{100}$$ --> 1st investment

$$\frac{2}{3}$$(10000-x) * $$\frac{6}{100}$$ --> 2nd investment

$$\frac{4}{300}$$(10000-x) *(1+3) = 320

(10000-x) = 20 * 300 = 6000

x = 94000

Option B is correct
General Discussion
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Joined: 05 May 2016
Posts: 26
Location: United States
WE: Web Development (Computer Software)
Re: David used part of $100,000 to purchase a house. Of the remaining port [#permalink] Show Tags 14 Jun 2016, 11:57 3 nalinnair wrote: David used part of$100,000 to purchase a house. Of the remaining portion, he invested 1/3 of it at 4 percent simple annual interest and 2/3 of it at 6 percent simple annual interest. If after a year the income from the two investments totaled $320, what was the purchase price of the house? (A)$96,000
(B) $94,000 (C)$88,000
(D) $75,000 (E)$40,000

An alternative solution:

"Of the remaining portion" is split two ways - 1/3 & 2/3. The interest earned is an integer amount. So, we can pick some smart numbers for the remaining amount and calculate the interest rate. Use answer choices to pick a number that is divisible by 3.

Of all the answer choices, (A) and (D) do not offer a number divisible by 3. (E) can be eliminated easily because when you invest 1/3 of $60,000 ($20,000) @ 4% interest rate, you earn $800 interest, which is definitely more than$320. But, if you calculate 2/3 of $60,000 ($40,000) @ 6% interest rate, you earn $2400 interest. Total interest when$60,000 is the remaining amount: $800 +$2400 = $3200. We need$320. It means the remaining amount should be $6000. Therefore, (B) is the answer. It may seem like a little bit lengthy approach, but with enough practice you can save yourself some trouble solving algebraic equations and save some time Happy solving! Intern Joined: 26 Sep 2017 Posts: 14 Re: David used part of$100,000 to purchase a house. Of the remaining port  [#permalink]

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24 Nov 2017, 03:13
Let the Price of house was x.

Then

$$\frac{1}{3}$$(10000-x) * $$\frac{4}{100}$$ --> 1st investment

$$\frac{2}{3}$$(10000-x) * $$\frac{6}{100}$$ --> 2nd investment

$$\frac{4}{300}$$(10000-x) *(1+3) = 320

(10000-x) = 20 * 300 = 6000

x = 94000

Option B is correct

Could you please clarify why there is no (1+%) in your base formula. As I remember simple interest formula should be interest*(1+years*%/100).
What am I doing wrong?
Intern
Joined: 16 Oct 2017
Posts: 30
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Concentration: Healthcare, Finance
David used part of $100,000 to purchase a house. Of the remaining port [#permalink] Show Tags Updated on: 25 Nov 2017, 23:29 nalinnair wrote: David used part of$100,000 to purchase a house. Of the remaining portion, he invested 1/3 of it at 4 percent simple annual interest and 2/3 of it at 6 percent simple annual interest. If after a year the income from the two investments totaled $320, what was the purchase price of the house? (A)$96,000
(B) $94,000 (C)$88,000
(D) $75,000 (E)$40,000

Let $$x$$ be the amount of money left after purchase of the house. So $$\frac{1}{3} * x$$ was invested under $$\frac{104}{100}$$ percent and $$\frac{2}{3} * x$$ was invested under $$\frac{106}{100}$$. and $$x + 320$$ we will get our investment back with the interests

$$\frac{x}{3} * \frac{104}{100} + \frac{2}{3}x * \frac{106}{100}$$ =$$x + 320$$
$$\frac{104x + 2 * 106x}{300} = x + 320$$
$$316x = 300x + 320 * 300$$
$$16x = 32 * 10 * 300$$
$$x = 6000$$

To get the price of the house we should substract the money left after after purchasing from 100,000 :

$$100000 - 6000 = 94000$$

Originally posted by Vorovski on 24 Nov 2017, 04:17.
Last edited by Vorovski on 25 Nov 2017, 23:29, edited 1 time in total.
SC Moderator
Joined: 22 May 2016
Posts: 1830
David used part of $100,000 to purchase a house. Of the remaining port [#permalink] Show Tags 25 Nov 2017, 13:46 1 lor12345 wrote: adiagr wrote: Let the Price of house was x. Then $$\frac{1}{3}$$(10000-x) * $$\frac{4}{100}$$ --> 1st investment $$\frac{2}{3}$$(10000-x) * $$\frac{6}{100}$$ --> 2nd investment $$\frac{4}{300}$$(10000-x) *(1+3) = 320 (10000-x) = 20 * 300 = 6000 x = 94000 Option B is correct Hi adiagr, Could you please clarify why there is no (1+%) in your base formula. As I remember simple interest formula should be interest*(1+years*%/100). What am I doing wrong? Thanks you in advance lor12345 , I think adiagr might not be around anymore. It looks as if you just got the formulas slightly mixed up. There is no (1 + %) in his formula because$320 is interest only. See his RHS: it's $320. That's interest only. (We weren't given$6,320, which in this case is principal + interest.)

Simple interest with principal PLUS accrued interest ("A") formula is indeed

A = P(1 + rt)

But interest only for simple interest formula is

I = (P)(r)(t)

adiagr used this formula, and defined P as ($100,000 - x)*(1/3) in one case, and ($100,000 - x)*(2/3) in the other.

True, both are multiplied by the respective interest rates, but without the 1 because his x is derived from subtraction from $100,000. If you use a multiplier with a 1, you include interest earned on$100,000. That is not how it worked. David earned interest on some small part of $100,000. If you want to include 1 in the multiplier, it seems easiest if you separate the principal (which earned interest) from the$100,000 (most of which did not). It's plenty hard either way, IMO.

See Vorovski , above, who used a (1+%) multiplier by considering ONLY x as principal (and not $100,000 - x, until the end). (Whew. I backsolved (used the answer choices) in about 90 seconds. See how I used the answer choices, here.) Hope that helps. _________________ In the depths of winter, I finally learned that within me there lay an invincible summer. -- Albert Camus, "Return to Tipasa" Intern Joined: 26 Sep 2017 Posts: 14 Re: David used part of$100,000 to purchase a house. Of the remaining port  [#permalink]

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26 Nov 2017, 04:19
genxer123 wrote:
lor12345 wrote:
Let the Price of house was x.

Then

$$\frac{1}{3}$$(10000-x) * $$\frac{4}{100}$$ --> 1st investment

$$\frac{2}{3}$$(10000-x) * $$\frac{6}{100}$$ --> 2nd investment

$$\frac{4}{300}$$(10000-x) *(1+3) = 320

(10000-x) = 20 * 300 = 6000

x = 94000

Option B is correct

Could you please clarify why there is no (1+%) in your base formula. As I remember simple interest formula should be interest*(1+years*%/100).
What am I doing wrong?

lor12345 , I think adiagr might not be around anymore. It looks as if you just got the formulas slightly mixed up.

There is no (1 + %) in his formula because $320 is interest only. See his RHS: it's$320. That's interest only. (We weren't given $6,320, which in this case is principal + interest.) Simple interest with principal PLUS accrued interest ("A") formula is indeed A = P(1 + rt) But interest only for simple interest formula is I = (P)(r)(t) adiagr used this formula, and defined P as ($100,000 - x)*(1/3) in one case, and ($100,000 - x)*(2/3) in the other. True, both are multiplied by the respective interest rates, but without the 1 because his x is derived from subtraction from$100,000. If you use a multiplier with a 1, you include interest earned on $100,000. That is not how it worked. David earned interest on some small part of$100,000.

If you want to include 1 in the multiplier, it seems easiest if you separate the principal (which earned interest) from the $100,000 (most of which did not). It's plenty hard either way, IMO. See Vorovski , above, who used a (1+%) multiplier by considering ONLY x as principal (and not$100,000 - x, until the end).

(Whew. I backsolved in about 90 seconds. Much faster for me.)

Hope that helps.

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Joined: 02 Jan 2017
Posts: 81
Location: Pakistan
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GMAT 1: 650 Q47 V34
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David used part of $100,000 to purchase a house. Of the remaining port [#permalink] Show Tags 16 Dec 2017, 21:53 nalinnair wrote: David used part of$100,000 to purchase a house. Of the remaining portion, he invested 1/3 of it at 4 percent simple annual interest and 2/3 of it at 6 percent simple annual interest. If after a year the income from the two investments totaled $320, what was the purchase price of the house? (A)$96,000
(B) $94,000 (C)$88,000
(D) $75,000 (E)$40,000

100,000 = 100 K
h = house

[100k - h]/3 * (4/100) + [100k - h]/3* (12/100) = 320

Taking [100k - h]/3 * (4/100) as a factor

[100k - h]/3 * (4/100) *[ 1+3] = 320

[100k - h]/3 * (4/100)*4 = 320

[100k - h]*16/100 =320 *3

100k - h = (320*3*100)/16

h= 96000
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Joined: 17 Jun 2017
Posts: 12

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05 Mar 2018, 03:09
nalinnair wrote:
David used part of $100,000 to purchase a house. Of the remaining portion, he invested 1/3 of it at 4 percent simple annual interest and 2/3 of it at 6 percent simple annual interest. If after a year the income from the two investments totaled$320, what was the purchase price of the house?

(A) $96,000 (B)$94,000
(C) $88,000 (D)$75,000
(E) $40,000 Does the term income indicates that we need to use (100000-x)/3 * 4/100 + (100000-x)* 2/3 * 6/100 = 320? INSTEAD OF (100000-x)/3 * 104/100 + (100000-x)* 2/3 * 106/100 =320 ? please help Director Joined: 26 Oct 2016 Posts: 664 Location: United States Concentration: Marketing, International Business Schools: HBS '19 GMAT 1: 770 Q51 V44 GPA: 4 WE: Education (Education) Re: David used part of$100,000 to purchase a house. Of the remaining port  [#permalink]

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18 Mar 2018, 19:15
1
Of every 3 dollars invested, $1 earns 4% interest and$2 earns 6% interest.
Average interest earned by every $3 = (1*4 + 2*6)/3 = 16/3% = 16/300. From here we can plug in the answers, which represent the purchase price of the house. Answer choice C: 88,000 Amount invested = 100,000 - 88,000 = 12000. Interest earned = (16/300)(12000) = 640. Since the interest earned is too great, the purchase price of the house must INCREASE, so that the amount invested DECREASES. Answer choice B: 94,000 Amount invested = 100,000 - 94,000 = 6000. Interest earned = (16/300)(6000) = 320. Success! The correct answer is B. _________________ Thanks & Regards, Anaira Mitch Director Joined: 09 Mar 2016 Posts: 655 David used part of$100,000 to purchase a house. Of the remaining port  [#permalink]

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20 Mar 2018, 13:50
1
generis wrote:
lor12345 wrote:
Let the Price of house was x.

Then

$$\frac{1}{3}$$(10000-x) * $$\frac{4}{100}$$ --> 1st investment

$$\frac{2}{3}$$(10000-x) * $$\frac{6}{100}$$ --> 2nd investment

$$\frac{4}{300}$$(10000-x) *(1+3) = 320

(10000-x) = 20 * 300 = 6000

x = 94000

Option B is correct

Could you please clarify why there is no (1+%) in your base formula. As I remember simple interest formula should be interest*(1+years*%/100).
What am I doing wrong?

lor12345 , I think adiagr might not be around anymore. It looks as if you just got the formulas slightly mixed up.

There is no (1 + %) in his formula because $320 is interest only. See his RHS: it's$320. That's interest only. (We weren't given $6,320, which in this case is principal + interest.) Simple interest with principal PLUS accrued interest ("A") formula is indeed A = P(1 + rt) But interest only for simple interest formula is I = (P)(r)(t) adiagr used this formula, and defined P as ($100,000 - x)*(1/3) in one case, and ($100,000 - x)*(2/3) in the other. True, both are multiplied by the respective interest rates, but without the 1 because his x is derived from subtraction from$100,000. If you use a multiplier with a 1, you include interest earned on $100,000. That is not how it worked. David earned interest on some small part of$100,000.

If you want to include 1 in the multiplier, it seems easiest if you separate the principal (which earned interest) from the $100,000 (most of which did not). It's plenty hard either way, IMO. See Vorovski , above, who used a (1+%) multiplier by considering ONLY x as principal (and not$100,000 - x, until the end).

(Whew. I backsolved in about 90 seconds. Much faster for me.)

Hope that helps.

Hi generis how is life ?:)

so in general 1+% means a principal account +acrrued percent right ?

but we are interested only in finding accrued percent so that's why we use only 0.04 and 0.06 without "+1" , correct ?
Director
Joined: 09 Mar 2016
Posts: 655

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20 Mar 2018, 18:41
rocko911 wrote:
nalinnair wrote:
David used part of $100,000 to purchase a house. Of the remaining portion, he invested 1/3 of it at 4 percent simple annual interest and 2/3 of it at 6 percent simple annual interest. If after a year the income from the two investments totaled$320, what was the purchase price of the house?

(A) $96,000 (B)$94,000
(C) $88,000 (D)$75,000
(E) $40,000 Does the term income indicates that we need to use (100000-x)/3 * 4/100 + (100000-x)* 2/3 * 6/100 = 320? INSTEAD OF (100000-x)/3 * 104/100 + (100000-x)* 2/3 * 106/100 =320 ? please help rocko911 , yes, exactly. The word "income" tells us we are solving for interest only. Interest is defined in one way as the income earned on a principal amount of money. The original principal amount of money is NOT defined as "income." In addition, here, we are NOT told something such as: "after a year, the total amount of money from David's investment is$_____"

So the word "income" lets us know that we need to use the "interest only" formula.

The income earned from the money invested
= interest = $320 Interest only = (Principal Amt) * $$\frac{rate}{100}$$ * time OR, Interest only = (Principal Amt * rate * time), if the interest rate is in decimal form Your first equation is correct. Hope that helps. _________________ In the depths of winter, I finally learned that within me there lay an invincible summer. -- Albert Camus, "Return to Tipasa" SC Moderator Joined: 22 May 2016 Posts: 1830 David used part of$100,000 to purchase a house. Of the remaining port  [#permalink]

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20 Mar 2018, 19:05
1
dave13 wrote:
generis wrote:

There is no (1 + %) in his formula because $320 is interest only. See his RHS: it's$320. That's interest only.
(We weren't given $6,320, which in this case is principal + interest.) Simple interest with principal PLUS accrued interest ("A") formula is indeed A = P(1 + rt) But interest only for simple interest formula is I = (P)(r)(t) (Whew. I backsolved in about 90 seconds. Much faster for me.) Hope that helps. Hi generis how is life ?:) well explained:) i actually had the same question but after reading your post got it, so in general 1+% means a principal account +acrrued percent right ? but we are interested only in finding accrued percent so that's why we use only 0.04 and 0.06 without "+1" , correct ? Hi dave13 - life is good. You? Yes, you are 100% correct. If you see a 1.xx for an interest rate, it almost always means the principal has been included... If you see a 0.xx for an interest rate, it means the principal has not been included. Nice work. Below is an explanation for WHY the "1" usually means that we're talking about principal, too. The 1 means 100% of Principal= $$\frac{100}{100} P= 1P$$ For SIMPLE INTEREST (not compound) interest only formula is $$i = P * r * t$$ $$i$$ = interest $$P$$ = principal $$r$$ = rate in decimal form $$t$$ = time Total Amount(principal plus interest) has a different formula. But it borrows from the interest only formula. The "1" stands for 100% of original Principal, or $$\frac{100}{100}P$$ or $$1P$$ Total amount (principal plus interest) earned formula is: $$A = P(1 + rt)$$ Breaking it down, Total Amount, A A = Principal + Interest A = Principal + (P*r*t) (from above for interest only) A = P + (P*r*t) Now factor out P A = P (1 + rt) If this makes no sense to you, ignore it. You have the basic concept! _________________ In the depths of winter, I finally learned that within me there lay an invincible summer. -- Albert Camus, "Return to Tipasa" SC Moderator Joined: 22 May 2016 Posts: 1830 David used part of$100,000 to purchase a house. Of the remaining port  [#permalink]

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21 Mar 2018, 01:07
2
dave13 wrote:
Let the Price of house was x.
Then
$$\frac{1}{3}$$(10000-x) * $$\frac{4}{100}$$ --> 1st investment

$$\frac{2}{3}$$(10000-x) * $$\frac{6}{100}$$ --> 2nd investment

$$\frac{4}{300}$$(10000-x) *(1+3) = 320

(10000-x) = 20 * 300 = 6000

x = 94000

Option B is correct

generis, how de got this $$\frac{4}{300}$$(100,000-x) * (1+3) = 320[/m]

dave13 - important steps were not spelled out.

This one is a bear to explain.
I tacked (*1) to the first expression in the equation
because (*1) changes nothing but makes the factoring out easier to see.
I have broken the steps down below.

That said, I would not and did not use much algebra to solve.

$$x$$ = price of house
$$(100,000 - x)$$= amount invested
End game: Find a common factor

Original equation (not in solution above)
($$\frac{1}{3}*(100,000-x)*\frac{4}{100}) + (\frac{2}{3}*(100,000-x)*\frac{6}{100})= 320$$

$$(\frac{100,000-x}{3} * \frac{4}{100} * 1) + ((\frac{1}{3}*\frac{2}{1})*(100,000-x)*\frac{6}{100})= 320$$

$$(\frac{100,000-x}{3} * \frac{4}{100} * 1) = (\frac{1}{3}*(100,000-x)*\frac{2}{1}*\frac{6}{100}) = 320$$

$$(\frac{100,000-x}{3} * \frac{4}{100} * 1) + (\frac{100,000-x}{3}*\frac{12}{100}) = 320$$

$$(\frac{100,000-x}{3} * \frac{4}{100} * 1) + (\frac{100,000-x}{3}*\frac{4*3}{100}) = 320$$

$$(\frac{100,000-x}{3} * \frac{4}{100} * 1) + (\frac{100,000-x}{3}*\frac{4}{100} * 3) = 320$$

COMMON FACTOR in both terms: $$\frac{100,000-x}{3}*\frac{4}{100}$$

COMMON FACTOR, factored OUT of blue equation above
$$(\frac{100,000-x}{3} * \frac{4}{100} * 1) + (\frac{100,000-x}{3}*\frac{4}{100} * 3) = 320$$
$$(\frac{100,000-x}{3} *\frac{4}{100})$$ * $$(1 + 3) = 320$$

When you factor it out of the first expression, you "leave" 1; out of the second expression, you "leave" 3.
(Try multiplying it back the other way, e.g. A(b+c) = Ab + Ac)

The rest:
$$(\frac{100,000-x}{3} *\frac{4}{100})* (1 + 3) = 320$$

$$(\frac{100,000-x}{3} *\frac{4}{100})* (4) = 320$$

$$\frac{100,000-x}{3} * (\frac{4}{100} * 4) = 320$$

$$(\frac{100,000-x}{3} * \frac{16}{100})= 320$$

Clear the 3 in the denominator. Multiply both sides by 3

$$100,000-x * \frac{16}{100}= 960$$

$$100,000 - x = 960 * \frac{100}{16}$$

$$100,000 - x = \frac{960}{16}*100$$

$$100,000 - x = 60 * 100$$

$$100,000 - x = 6,000$$

$$100,000 - 6,000 = x$$

$$94,000 = x$$

If you cannot follow that algebra, do not worry.
It is not necessary to solve the problem with that particular version of algebra.
You can use another kind of algebra. Or very little.

I backsolved.
Quote:
David used part of $100,000 to purchase a house. Of the remaining portion, he invested 1/3 of it at 4 percent simple annual interest and 2/3 of it at 6 percent simple annual interest. If after a year the income from the two investments totaled$320, what was the purchase price of the house?

(A) $96,000 (B)$94,000
(C) $88,000 (D)$75,000
(E) $40,000 Start with C)$88,000

If the house cost $88,000, then (100,000 - 88,000) =$12,000 is left over

1/3 of that $12,000 is invested at 4 percent$4,000 earns .04
2/3 of that $12,000 is invested at 6 percent$8,000 earns .06

Combined, do they earn $320 in a year? $$.04(4,000) + .06(8,000) =$$ $$160 + 480 = 600$$ NO. If the house costs$88,000, the leftover money is $12,000. It earns$600 in one year. Too much.
The base principal (the leftover money ) is too great.
The house needs to be more expensive.

Try B) $94,000 If the house cost$94,000, then (100,000 - 94,000) = $6,000 is left over 1/3 of that$6,000 is invested at 4 percent
$2,000 earns .04 2/3 of that$6,000 is invested at 6 percent
$4,000 earns .06 Combined, do they earn$320 in a year?

$$.04(2,000) + .06(4,000) =$$
$$80 + 240 = 320$$

BINGO

Hope that helps, dave13
_________________

In the depths of winter, I finally learned
that within me there lay an invincible summer.

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text:
# Synopsis: Chiral contortions

#### Magnetic Control of Crystal Chirality and the Existence of a Large Magneto-Optical Dichroism Effect in CuB2O4

M. Saito, K. Ishikawa, K. Taniguchi, and T. Arima

Published September 12, 2008

Molecules that have opposite chirality are mirror images of one another, like two identical, but counter-winding staircases. Despite the subtle difference in symmetry, the two chiral forms of a molecule can have very distinct optical and catalytic properties.

The ordering of spins in a magnet can also have a definite chirality. In principle, one could imagine a host of applications if a magnetic field could switch the spins in a material between two chiralities, but such substances are rare and the effect is typically weak in those that do. Writing in Physical Review Letters, Mitsuru Saito and colleagues of Tohoku University in Sendai, Japan, report they can induce a huge change in the optical response of , —a low-temperature antiferromagnet—by rotating the spins into a chiral structure with a magnetic field.

In zero magnetic field, the spin arrangement in ${\text{CuB}}_{2}{\text{O}}_{4}$ is a mixture of opposite chiralities. However, if a relatively small (500 G) magnetic field is applied along a certain crystallographic axis, the spins rotate into a definite chiral configuration. If the field is rotated by ninety degrees, the chirality changes.

Saito et al. confirm the magnetically switched chirality by looking for the difference in the absorption of left and right circularly polarized light for the two different field configurations. The large dichroism that they observe is four orders of magnitude larger than what has previously been reported and may put real magneto-optical applications for this material within reach. – Jessica Thomas

format: plain


text:
# Markdown footnotes in Drafts

Footnotes in Markdown are pretty easy to just type in directly, and that’s what I did for quite a while when writing posts in Drafts on my iPad. A couple of weeks ago, I decided I wanted to add a bit of automation to avoid retyping the footnote marker. Seemed like an easy task, but it took me three tries to get what I wanted.

First, let’s go over what I mean by “footnotes in Markdown.” As defined by John Gruber, Markdown doesn’t support footnotes, but many popular implementations, like MultiMarkdown and PHP Markdown Extra (which I use to build ANIAT pages) do. I’ve even heard tell that Gruber himself uses an off-brand extension to plain ol’ Markdown to handle footnotes on Daring Fireball.

All of these implementations use a slight variation on Markdown’s reference link syntax. Here’s an example:

The low angle feels better when I'm working while sitting
cross-legged on a bed or floor.[^chicago]

[^chicago]: Twenty-five or -six to four.

The idea is to use an reference string that starts with a caret inside square brackets where you want the footnote marker to appear. Then elsewhere in the text (and it could be anywhere, but I like putting it right after the paragraph with the marker) comes that same marker string followed by a colon and the footnote text. The goal of my automation was to type the marker only once and to format the parts on separate lines.

My first thought was to use a TextExpander snippet with a fill-in. Drafts supports TextExpander and so do many other text editors, so this would be a portable solution. Here’s the snippet:

The idea here is to work easily in the most common situation, which is when I’m typing along and think “here’s a good spot for a footnote.” After typing the snippet’s abbreviation, I’m swept away to the TextExpander app, where I fill in the part of the marker string that comes between the caret and the closing bracket.

After typing in the marker text and tapping the Done button, I’m taken back to Drafts, where the snippet has been put in place. And here’s where this solution is less than ideal. Moving from Drafts to TextExpander and back takes Drafts out of edit mode—when I return from the round-trip to TextExpander, there’s no cursor blinking in the draft, so I can’t just start typing in the footnote text. I have to tap after the colon to reestablish editing mode and set the cursor where it needs to be to type the footnote text. The need to shift back into editing mode (which will happen with any TextExpander snippet that uses fill-ins) is annoying and made me search for another solution.

My second solution was a two-step Drafts action that used the Prompt and Insert Text step types. The Prompt step asks for the marker,

and the Insert Text step uses that marker to insert text in a manner very similar to the TextExpander snippet

In practice, there were two problems with this solution, both having to do with Drafts’s mode shifting. As with the TextExpander snippet, after the text is inserted, I need to tap after the colon to put Drafts back into editing mode and set the cursor. But even before that, when the Prompt window appears, the text field in it doesn’t have focus, so I have to reach up from the keyboard to tap in it before typing. So this was actually worse than the TextExpander solution.

To get what I really wanted, automation that didn’t take me away from the keyboard to tap the screen, I needed to do some scripting. What I came up with was a single-step Drafts action that works like this:

When I want to insert a footnote, I type the whole marker string—brackets, caret, and all—where the footnote marker should go.

The low angle feels better when I'm working while sitting
cross-legged on a bed or floor.[^chicago]|

(The vertical bar shows the cursor position.)

Then I invoke the Footnote action, which is in my “Blogging” action set and has both a keyboard button with an asterisk glyph and an external keyboard shortcut of ⇧⌘8 (⇧8 is the asterisk). This adds a blank line, a copy of the marker string, a colon, and a space. And it sets the cursor after the space, so I’m ready to type in the footnote text.

The low angle feels better when I'm working while sitting
cross-legged on a bed or floor.[^chicago]

[^chicago]: |

This solution requires a little bit more typing—I have to type the brackets and caret myself—but requires no tapping on the screen. The action is defined this way,

and its script looks like this:

javascript:
1:  // Cursor location
2:  var insertionPoint = editor.getSelectedRange()[0];
3:
4:  // Get footnote marker from text before cursor
5:  var nearby = draft.content.substring(insertionPoint - 25, insertionPoint);
6:  var regex = /($\^[^$]+\])\$/;
7:  var fnMarker = nearby.match(regex)[1];
8:
9:  // Insert a blank line and the marker,

Line 5 grabs the 25 characters before the cursor and puts that string into the variable nearby.1 Lines 6 and 7 then search that string for the bracket-caret-string-bracket text that defines a Markdown footnote marker and put it into the variable fnMarker.


text:
Study the Speed-time Graph of a Car Given Alongside and Answer the Following Questions: - Science

Graph

Study the speed-time graph of a car given alongside and answer the following questions:

(i) What type of motion is represented by OA ?
(ii) What type of motion is represented by AB ?
(iii) What type of motion is represented by BC ?
(iv) What is the acceleration of car from O to A ?
(v)What is the acceleration of car from A to B ?
(vi) What is the retardation of car from B to C ?

Solution

(i) OA represents uniform acceleration
(ii) AB represents constant speed.
(iii) BC represents uniform retardation.
(iv) Acceleration of car from O to A = slope of line OA

a = ((40-0))/((10-0))
= 4  "m/s"^2
(v) Acceleration of car from A to B is zero as it has uniform speed .
(vi) Retardation of car from B to C = slope of line BC

a = ((40-0))/((50-30)) "m/s"^2
= 2  "m/s"^2

Concept: Velocity - Time Graphs
Is there an error in this question or solution?

APPEARS IN

Lakhmir Singh Science Class 9 Physics
Chapter 1 Motion
Hots Questions | Q 62 | Page 44


text:
# If f(x) = x2 then find $$\frac{f(2+h) – f(2)}{h}$$

Solution

Given

f(x) = x2

We have to find $$\frac{f(2+h) – f(2)}{h}$$ so,

f(2+h) = (2+h)2 = 4 + 4h + h2

f(2) = 4

Now,

$$= \frac{f(2+h) – f(2)}{h}$$

$$= \frac{4 + 4h + h^2 – 4}{h}$$

$$= \frac{h(4 + h)}{h}$$

= 4 + h


text:
# Difference between revisions of "Ferrofluid monolayers: monodisperse particles"

## Non Branched systems

We investigate the peculiarities brought by the Quasi-2D geometry into the aggregation ferrofluid microstructure in particular in the range of parameters for which the chain-like structures have been observed implicitly or explicitly in 3D experiments, simulations and theoretical models. In difference to previous papers studying aggregates in q2D monolayers, we focus our attention on the detailed analysis and extensive comparison of the results coming from MD simulations with a newly developed theoretical approach, which give us cluster sizes and distributions as functions of the magnetic interaction and area fraction of ferroparticles in monolayers.

A thorough comparison of the predictions of the theory with the values obtained through equilibrium Molecular Dynamic simulations have shown that: in the range of the space of parameters ${\displaystyle \lambda }$ (dipolar coupling), and area fraction ${\displaystyle \phi }$ in which the theory could be expected to be valid (regions with low number of branched structures), the theory is able to closely predict the simulation results, and allow us to calculate the partition functions for these clusters in constrained geometries and describe the influence of the latter constraints on the entropy of a ferrofluid.

The comparison shows that excluded area interactions play a crucial role in quasi 2D, where the constrains imposed by the Q2D monolayer geometry change the entropy of clusters. The refinement of the present theory by dropping out several of the limiting constraints is currently being worked out.

## Branched and Network systems

At high values of the dipolar coupling and area fraction parameters, the systems exhibit kind of network and branched structures which are currently characterized via simulation and theory.


text:
1. ABSTRACTS.
The abstracts for Sections 1-6 will be written in English, by using the LATEX template below.
CAIM Abstracts Template (for sections 1-6)

\documentclass[11pt, a4paper]{article}
\begin{document}
\title{Title of the abstract}
\author{Author$^1$, Author$^2$ \\ $^1$Affiliation, $^2$Affiliation }
\date{}
\maketitle

Text of abstract.
\end{document}

The abstracts for Section 7 will be written by using Microsoft Word (in English or Romanian).
For the security of data transmission, the authors are kindly asked to send the abstracts also by e-mail at the address caim.2013@yahoo.com (for Sections 1-6 in tex and pdf format).
The authors are also asked to name the files containing the abstracts with the name of the Corresponding Author (for example IONESCU-Aurelian.tex, IONESCU-Aurelian.pdf, or, for Section 7, IONESCU-Aurelian.doc).
The Scientific Committee will select the works that will be communicated and will announce the authors in at most two weeks from the moment of receiving the abstracts about the acceptance (or non-acceptance) of the paper.

2. PAPERS.
After the Conference, the Scientific Committee will select from the presented papers, the papers to be published in ROMAI Journal (http://rj.romai.ro) for Sections 1-6, and in ROMAI Educational Journal (http://rej.romai.ro) for Section 7.
The Scientific Committee has the right to reject any work which exhibits an inappropriate scientific or linguistic level.
The authors should obey the standard copyright regulations and are responsible for the scientific and linguistic accuracy of the text.

Instructions for Authors
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The papers will be sent to caim2013@yahoo.com


text:
PROBLEMS FOR DECEMBER
 location:

## PROBLEMS FOR DECEMBER

Professor E.J. Barbeau
Department of Mathematics
University of Toronto
Toronto, ON M5S 3G3
no later than January 31, 2002.

Note. The incentre of a triangle is the centre of the inscribed circle that touches all three sides. A set is connected if, given two points in the set, it is possible to trace a continuous path from one to the other without leaving the set.

121.
Let $n$ be an integer exceeding 1. Let ${a}_{1},{a}_{2},\dots ,{a}_{n}$ be posive real numbers and ${b}_{1},{b}_{2},\dots ,{b}_{n}$ be arbitrary real numbers for which

$\sum _{i\ne j}{a}_{i}{b}_{j}=0 .$

Prove that $\sum _{i\ne j}{b}_{i}{b}_{j}<0 .$

122.
Determine all functions $f$ from the real numbers to the real numbers that satisfy

$f\left(f\left(x\right)+y\right)=f\left({x}^{2}-y\right)+4f\left(x\right)y$

for any real numbers $x$, $y$.

123.
Let $a$ and $b$ be the lengths of two opposite edges of a tetrahedron which are mutually perpendicular and distant $d$ apart. Determine the volume of the tetrahedron.

124.
Prove that

$\frac{\left({1}^{4}+\frac{1}{4}\right)\left({3}^{4}+\frac{1}{4}\right)\left({5}^{4}+\frac{1}{4}\right)\dots \left({11}^{4}+\frac{1}{4}\right)}{\left({2}^{4}+\frac{1}{4}\right)\left({4}^{4}+\frac{1}{4}\right)\left({6}^{4}+\frac{1}{4}\right)\dots \left({12}^{4}+\frac{1}{4}\right)}=\frac{1}{313} .$

125.
Determine the set of complex numbers $z$ which satisfy

$\mathrm{Im} \left({z}^{4}\right)=\left(\mathrm{Re} \left({z}^{2}\right)\right){}^{2} ,$

and sketch this set in the complex plane. (Note: Im and Re refer respectively to the imaginary and real parts.)

126.
Let $n$ be a positive integer exceeding $1$, and let $n$ circles (i.e., circumferences) of radius 1 be given in the plane such that no two of them are tangent and the subset of the plane formed by the union of them is connected. Prove that the number of points that belong to at least two of these circles is at least $n$.

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text:
# How long can a moon remain non-spherical, and what might it look like when full?

I'm trying to "build" a world in which the moon is always visible at night, and only at night, to the majority of areas on that world. I've been told that an L2 Legrange point could be a viable configuration for this to be true, and that essentially means that the moon orbits the world at the same rate that the world orbits its sun, and remains constantly on the opposite side of the world from the sun. I am unsure as to the moon's direction of orbit and direction and rate of spin.

It seems that this configuration would result in the moon always appearing full, so I was thinking of ways for it to appear as a crescent while actually being "full" and still fitting the other conditions. I'm wondering if a moon that is not a true sphere, but rather a spherical crescent might work, as I expect it would have close enough orbital mechanics to work out mostly the same. I imagine that if it spins in the right direction at the same rate that it orbits the world, it could always appear as a crescent.

I have many questions, but the most important is: Is this viable, and if so, for how long? Will the orbit likely decay or will the moon become a sphere again (in my mind, it was carved by some sort of collision), and how long would both take if so?

I'm guessing the answer will change depending on the size, distance, etc. of all objects involved (moon, world, and sun). Perhaps the moon can be small enough that its own gravity won't cause it to reform as a sphere at all, and if so, can it still be big enough to be seen as a disc in the sky on a world that is habitable by humans and other life? Can a world of an appropriate size be sufficient for its own plant and animal life to have evolved by the point humans find it? And would anything else have to be specific, like the world's axial tilt or its ratio of days to a year?

I'm new to this, and any input would be greatly appreciated!

• The L2 Lagrangian point is unstable, and the slightest perturbation will cause your moon to drift away. Only the L4 and L5 points are stable, and they won’t do what you need. To quote Wikipedia: “Any object orbiting at L1, L2, or L3 will tend to fall out of orbit; it is therefore rare to find natural objects there, and spacecraft inhabiting these areas must employ station keeping in order to maintain their position.” Jun 9, 2018 at 7:01
• Unless I'm mistaken, you won't have a perpetual full moon - assuming the relative sizes of the planet and moon are like on Earth, you should have a perpetual eclipse instead. After all, the planet is (by definition) directly between the star and L2. Jun 9, 2018 at 10:02
• It is my impression that a moon's orbit can only be stable if the year of the planet is at least 9 times as long a month that it takes the moon to orbit the planet. So the moon will have to orbit the planet several times during a year and been in the day sky. But I did make some suggestions for the moon to be invisible when in the day sky answering another question. Jun 9, 2018 at 17:58
• Here is the question with my answer how to make the planet's moon invisible whenever it is in the day sky, so it is only seen at night. worldbuilding.stackexchange.com/questions/109024/… Jun 9, 2018 at 18:04
• And also this: worldbuilding.stackexchange.com/questions/95251/… Jun 9, 2018 at 18:12

...the moon is always visible at night, and only at night,

A stable orbit that does this is impossible (there are however some compromises that are possible outlined below). Someone, somewhere on the planet will see the moon during the day. The 2nd lagrangian point (L2) is indeed the orbit that would place a planet in constant conjunction (i.e. located between) with another planet: Image taken from wikipedia. The orbit that follows the path at L2 would indeed keep the interior planet between it and the star. However, the orbit is unstable: any perturbation away from L2 will continue to grow in time, causing eventual misalignment. If you are willing to ignore this fact, then it's mission accomplished. But if you need something with a bit more verisimilitude, I have a few potential candidates:

### 1. The geostationary Moon

From Wikipedia:

Earth's moon currently orbits at a distance of $3.8\times 10^8$m and $27.3$ days. We could get the orbital period down to a single Earth day if it orbited closer at $4.2\times 10^7$ m, or about $1/9$ the distance. The advantage of doing this would be the moon would always be in the same place in the sky, day or night. It would travel through its full phase (new moon to full moon) in a single day. And it would only be visible within some approximately $180^\circ$ arc of longitude. There would be a side to the planet that never saw the moon. A closer Moon would produce stronger tidal effects, unless you also reduced it's radius by 1/9th. Alternatively, you could lengthen the Earth day to match the lunar month and keep everything else constant. On Earth, the only geostationary satellites that are stable are the ones at longitudes 75-West and 105-West, and a Moon would eventually settle down there over geologic time. These longitudes are particular to Earth's gravitational field, and would be different for another planet.

### 2. The Sun Synchronous Moon

(source unknown):

A sun synchronous orbit maintains the same angle to the sun throughout the year. In other words, an orbiting body will appear in the sky in the same location at the same solar time each day. Everybody on the planet sees the orbiting body, but at a different time depending on their longitude and latitude. On Earth, sun synchronous satellites have short periods around 1-3 hours. For example, if a person on the East coast of the United States sees a sun synchronous satellite appear on the horizon at dawn, then every day it will appear at dawn (the same time relative to the Sun). Furthermore, since there is about a 3 hour time difference between the East coast and the West coast of the United States, a person on the West coast will also see the satellite at dawn when it comes around it's second orbit.

It is possible to extend that 3-hour orbital period a little bit if you fiddle with the mass and oblateness of the Earth, but it can't be extended to a single Earth day (I tried) without some nonsensical values.

Sun synchronous orbits need to be high inclination (near polar) orbits. However just like the geosynchronous orbit, half the locations on Earth will only see the moon during the day, and the other half only see the moon during the night. There is however a pleasant compromise: an orbit that is aligned perpendicular to the sun will follow the day-night terminator and will be visible to everyone on the planet only at dawn and dusk.

In a sun synchronous orbit, the moon would always appear in the same phase to the occupants of the planet, since it always appears at the same relative angle to the sun.

### Summary

The L2 location gets you what you want, but the orbit is not stable to perturbations. If your planet is in a system with other outer planets, like Earth, then this will occur, and my intuition tells me it would take less than 100 planet revolutions if the solar system is like Earths.
There are some stable orbits that offer compromises, but someone on the planet will see the moon during the day, at least some of the time.

• I may be satisfied with the L2 arrangement and just say that the system is perfect enough and without chance disturbances, that it never decays. Alternatively, for purposes of storytelling, the moon showing always and only night doesn't have to have always been true for that planet, as long as it is true for the century or so that humans find and inhabit it. What does really matter is that the planet is habitable and Earth-like, with its own evolved life, tides, seasons, and weather patterns. And that once humans are there, everyone around the world East-West sees the moon every night. Jun 9, 2018 at 21:53


text:
## How is difficulty calculated?

63

21

Can anyone explain me in the plain English how difficulty is calculated. I have a very approximate understanding that it is calculated based on the amount of hash power in all the bitcoin community over a specific period of time. But this is very vague.

Also I understand it can change very rapidly. Can it only increase? Is there any formula how to calculate it or predict it?

Thanks for a detailed answer, Meni Rosenfeld. Just to make sure I got everything right. I am summing up all the time, it took to generate the last 2016 blocks. And then apply the formula.

2I think follow up questions are better as comments to the answer. Basically yes, but no summing is actually needed - you can just take the timestamps of the last block and of the one 2016 blocks before, and subtract. – Meni Rosenfeld – 2012-12-19T14:18:30.203

79

The Bitcoin difficulty started at 1 (and can never go below that). Then for every 2016 blocks that are found, the timestamps of the blocks are compared to find out how much time it took to find 2016 blocks, call it T. We want 2016 blocks to take 2 weeks, so if T is different, we multiply the difficulty by (2 weeks / T) - this way, if the hashrate continues the way it was, it will now take 2 weeks to find 2016 blocks.

For example, if it took only 10 days it means difficulty is too low and thus will be increased by 40%.

The difficulty can increase or decrease depending on whether it took less or more than 2 weeks to find 2016 blocks. Generally, the difficulty will decrease after the network hashrate drops.

If the correction factor is greater than 4 (or less than 1/4), then 4 or 1/4 are used instead, to prevent the change to be too abrupt.

There is a bug in the implementation, due to which the calculation is based on the time to find the last 2015 blocks rather than 2016. Fixing it would require a hard fork and is thus deferred for now.

It is possible to give a rough estimate for the next difficulty change, based on the time to find the recent blocks. Nobody can make longer-term predictions for the future difficulty reliably, but anyone is free to speculate based on exchange rate trends, Moore's law and other hardware advances.

Is that bug still there? Shouldn't there be a GitHub issue for that so that people can keep track of the status... With such bugs in the protocol, it will be hard for the Bitcoin ecosystem to move away from the situation in which the Satoshi client devs set all the rules. – Steven Roose – 2013-06-11T22:23:06.347

1@StevenRoose: AFAIK it is, but I will leave it to people who are more involved with the core code to comment... This is adequate for a separate SE question. – Meni Rosenfeld – 2013-06-12T15:19:32.937

3Good answer, but one small, but capital point is eluded: how do the nodes in the network agree on what is the difficulty ? – deadalnix – 2014-03-14T18:35:56.613

6@deadalnix: The difficulty of a block is a deterministic calculation based on the data of the previous blocks. All nodes independently do the same calculation and get the same result. – Meni Rosenfeld – 2014-03-15T21:36:15.107

@MeniRosenfeld thank you for that answer. But to me, it just move the problem around. As I understand you answer, the computation is made from the timestamp available in the blocks. But then, how does a p2p network agree on the timestamp ? – deadalnix – 2014-03-16T01:18:52.630

3@deadalnix: The timestamp is a part of the block, which means that whoever found the block decides what to put in it. The timestamp must be no sooner than the median of the past 11 blocks. Also, if a node receives a block with a timestamp more than 2 hours in the future it will reject it and not propagate it. – Meni Rosenfeld – 2014-03-16T16:10:23.307

1References to the Bitcoin specifications or codebase would be much appreciated. Revising your answer is strongly preferred over shoving stuff into the comments. – Indolering – 2014-07-18T18:25:45.933

@Indolering: The stuff in the comments is not part of the core answer. They are replies to specific followup questions. – Meni Rosenfeld – 2014-07-20T04:38:26.167

@Meni, so if I want to calculate the round completion should I do CURRENT_BLOCK_COUNT_IN_THIS_ROUND/2015 instead of CURRENT_BLOCK_COUNT_IN_THIS_ROUND/2016 ??? – mixdev – 2014-08-05T03:08:13.183

1@mixdev: The "round" (difficulty retarget) is done every 2016 blocks. But the recalculation currently is not prev_targetTIME_FOR_LAST_2016_Blocks/1209600, it's prev_targetTIME_FOR_LAST_2015_Blocks/1209600. – Meni Rosenfeld – 2014-08-05T09:35:28.060

So this basically means that errors will add up as every time only the last epoch time is considered? This would mean that stating in 2140 the last bitcoin will be mined is wrong as it could be years or even decades before that. – tobi – 2018-06-05T08:51:21.977

@tobi: The off-by-one bug is responsible for blocks being found 0.05% slower on average than the "official" 2016 per two weeks. There is a separate issue that causes another 0.05% slowdown. Together they amount to only 1 month slowdown over a century. However, there is a separate effect that causes blocks to be found faster due to the increase in hashrate. This is more significant and the last coin will be mined probably 1-2 years too early because of this. – Meni Rosenfeld – 2018-06-05T09:50:02.670

@tobi: The total amount of speedup due to hashrate increase can be calculated as roughly log (difficulty) * 2 weeks. This is the main reason block halving parties are at a different date each time - November in 2012, July in 2016 etc. – Meni Rosenfeld – 2018-06-05T09:53:02.120

yes I was referring to the speedup due to increasing hashrate and not the 2015 blocks bug – tobi – 2018-06-05T12:24:46.420

2@tobi: Oh ok. Previous comments were about the bug, and "errors" hinted that it's the bug, so I assumed we're talking about that. So, yes. If we assume that around 2140 the hashrate will be around *1B what it is now, the schedule will be ahead by 96 weeks, or almost two years. But there is yet another effect - a delay caused by the fact that in the beginning, the difficulty was 1 even though the hashrate wasn't enough to justify it. – Meni Rosenfeld – 2018-06-05T13:15:37.803

16

Meni's answer is good. I just want to give some practical detail method about difficulty calculation, perhaps helpful for future views of this question's answer.

Let's take a look at Satoshi's genesis block header (part of related info):

$bitcoin-cli getblockhash 0 000000000019d6689c085ae165831e934ff763ae46a2a6c172b3f1b60a8ce26f$ bitcoin-cli getblockheader 000000000019d6689c085ae165831e934ff763ae46a2a6c172b3f1b60a8ce26f
{
...
"height": 0,
...
"bits": "1d00ffff",
"difficulty": 1,
...
}

As we can see above, the genesis block has a '1' difficulty and '1d00ffff' bits. The bitcoin bits means the 'target' hash value, the new generated block must meet a condition: block header's double SHA-256 hash value must less than this 'target' value.

The '1d00ffff' bits value in genesis block means the 'target' value:

[0x00000000,0xffff,{0x00..0x00}]
{0x00..0x00} at above has 26 bytes 0x00.

Then, to find a new block, you must search that 32 bits nNonce value (and nTimes and the hashMerkleRoot also) until the block hash value has 4 bytes zero leading. By the way, the nNonce is one of the fields in block header structure:

 struct header_structure{            // BYTES   NAME
uint32_t nVersion;              // 4       version
uint8_t hashPrevBlock[32];      // 32      previous block header hash
uint8_t hashMerkleRoot[32];     // 32      merkle root hash
uint32_t nTime;                 // 4       time
uint32_t nBits;                 // 4       target
uint32_t nNonce;                // 4       nonce
};

Because SHA-256 algorithm (as well as any cryptographically secure hash algorithm) produces output that will appear like an uniformly random sequence, the practical 'trial and error' method is the only way to find a new block to meet the condition. The probability to find a block with the 4 bytes zero leading hash value is 1/(2^32), that means the average 'trial and error" numbers are exactly 2^32 (i.e. 4G).

For human easy understanding about this 'target' hash value, We define the term 'difficulty', which means the average 'trial and error" numbers to find a block to meet the 'target' condition. And we define the 'difficulty' unit: 1 'difficulty' = 4G hashes

Then, till today, the bitcoin blockchain height reach 501509, let's take a look at its header:

\$ bitcoin-cli getblockheader 0000000000000000006c5532f4fd9ee03e07f94df165c556b89c495e97680147
{
...
"height": 501509,
...
"bits": "18009645",
"difficulty": 1873105475221.611,
...
}

The block 501509's bits = 0x18009645, it's the compact format of 256 bits integer, its 256 bits format is:

[0x00000000,0x00000000,0x009645,{0x00..0x00}]
{0x00..0x00} at above has 21 bytes 0x00.
that is  0x009645 * (256 ^ 21）
The genesis block's target is  ( 0x00ffff * 256 ^ 26 ）which is the difficulty unit '1.0'.
So, the difficulty
= (0x00ffff * 256 ^ 26）/ (0x009645 * 256 ^ 21)
= 65535/38469 * (256^5)
= 1.703579505575918 * 2^40
= 1873105475221.611

So far, you have all the detail about how to calculate the 'difficulty'. In some cases, we also use the simple format 1.7T to say the difficulty, in above example：

 (1.703579505575918 * 2^40) = 1.703579505575918T
1T = 2^40 = 1024^4

1d is 29 in Dec(not 26). SHS is SHA – Boris Ivanov – 2018-08-06T18:31:37.997

thanks @BorisIvanov, the typo error SHS has been fixed. But 1d indeed means 26 bytes zero tail instead of 29, please read the example detail showed above. – gary – 2018-08-07T08:13:25.227

ah yeah. Significand – Boris Ivanov – 2018-08-07T15:29:10.150

3

I would like to give my 2 cents here, by expliciting the relationship between the probability of mining a block given the current target t and the corresponding difficulty d as it is calculated in bitcoin core.

So cryptographic hash functions are idealized by the random oracle abstraction [https://en.wikipedia.org/wiki/Random_oracle]. We can therefore model the output of the doubleSHA256 hash function used in PoW as a uniform variable in the space {0,1}^256, i.e. arrays of 256 bits. Thus the probability of a single hash h being a valid hash is:

p = P(h < t) = t /( 2^{256} - 1 )

On the other hand d is calculated as follows, just as @gary explained before only transformed into decimals:

d = ( (2^{16} - 1) * 2^{8*26} ) / t = ( (2^{16} -1) * 2^{208} ) / t

The implementation is in [https://github.com/bitcoin/bitcoin/blob/master/src/rpc/blockchain.cpp], line 60, function GetDifficulty. Actually if someone can explain how exactly the code maps to the formula above, that would be helpful. Combining those two formulas we obtain:

d = ( (2^{16} -1) * 2^{208} ) / ( p * (2^{256} - 1) ) ~ 2^{-32} / p

Analyzing this last expression the difficulty is the ratio between the probability of obtaining a hash lower than 2^{224} (which is the lowest decimal number that has a binary representation using 256 bits starting with 32 zero bits) and the probability of obtaining a valid hash based on the current target t. This is a direct implication of defining, in the genesis block, as difficulty 1 the one associated to the hexadecimal target 0x1d00ffff, expressed in what I think is called the 32-bit compact form for 256-bit numbers.

A nice question I believe is why this specific compact form was chosen for representing the target.

Upvoted! The compact form provides 3 most significant bytes for the target, in the min difficulty the 3 most significant bytes are 00ffff. – James C. – 2019-01-27T14:48:54.383


text:
# How do you integrate int 1/(2x^3)dx?

Nov 2, 2016

$\int \frac{1}{2 {x}^{3}} \mathrm{dx} = \frac{- 1}{4 {x}^{2}} + C$

#### Explanation:

Using the power rule we have:

$\int \frac{1}{2 {x}^{3}} \mathrm{dx} = \int \frac{1}{2} {x}^{-} 3 \mathrm{dx}$

$= \frac{1}{2} \left({x}^{-} \frac{2}{-} 2\right) + C$
$= \frac{- 1}{4 {x}^{2}} + C$


text:
# Comparative statics

In this graph, comparative statics shows an increase in demand causing a rise in price and quantity. Comparing two equilibrium states, comparative statics doesn't describe how the increases actually occur.

In economics, comparative statics is the comparison of two different economic outcomes, before and after a change in some underlying exogenous parameter.[1]

As a type of static analysis it compares two different equilibrium states, after the process of adjustment (if any). It does not study the motion towards equilibrium, nor the process of the change itself.

Comparative statics is commonly used to study changes in supply and demand when analyzing a single market, and to study changes in monetary or fiscal policy when analyzing the whole economy. Comparative statics is a tool of analysis in microeconomics (including general equilibrium analysis) and macroeconomics. Comparative statics was formalized by John R. Hicks (1939) and Paul A. Samuelson (1947) (Kehoe, 1987, p. 517) but was presented graphically from at least the 1870s.[2]

For models of stable equilibrium rates of change, such as the neoclassical growth model, comparative dynamics is the counterpart of comparative statics (Eatwell, 1987).

## Linear approximation

Comparative statics results are usually derived by using the implicit function theorem to calculate a linear approximation to the system of equations that defines the equilibrium, under the assumption that the equilibrium is stable. That is, if we consider a sufficiently small change in some exogenous parameter, we can calculate how each endogenous variable changes using only the first derivatives of the terms that appear in the equilibrium equations.

For example, suppose the equilibrium value of some endogenous variable ${\displaystyle x}$  is determined by the following equation:

${\displaystyle f(x,a)=0\,}$

where ${\displaystyle a}$  is an exogenous parameter. Then, to a first-order approximation, the change in ${\displaystyle x}$  caused by a small change in ${\displaystyle a}$  must satisfy:

${\displaystyle B{\text{d}}x+C{\text{d}}a=0.}$

Here ${\displaystyle {\text{d}}x}$  and ${\displaystyle {\text{d}}a}$  represent the changes in ${\displaystyle x}$  and ${\displaystyle a}$ , respectively, while ${\displaystyle B}$  and ${\displaystyle C}$  are the partial derivatives of ${\displaystyle f}$  with respect to ${\displaystyle x}$  and ${\displaystyle a}$  (evaluated at the initial values of ${\displaystyle x}$  and ${\displaystyle a}$ ), respectively. Equivalently, we can write the change in ${\displaystyle x}$  as:

${\displaystyle {\text{d}}x=-B^{-1}C{\text{d}}a.}$

Dividing through the last equation by da gives the comparative static derivative of x with respect to a, also called the multiplier of a on x:

${\displaystyle {\frac {{\text{d}}x}{{\text{d}}a}}=-B^{-1}C.}$

### Many equations and unknowns

All the equations above remain true in the case of a system of ${\displaystyle n}$  equations in ${\displaystyle n}$  unknowns. In other words, suppose ${\displaystyle f(x,a)=0}$  represents a system of ${\displaystyle n}$  equations involving the vector of ${\displaystyle n}$  unknowns ${\displaystyle x}$ , and the vector of ${\displaystyle m}$  given parameters ${\displaystyle a}$ . If we make a sufficiently small change ${\displaystyle {\text{d}}a}$  in the parameters, then the resulting changes in the endogenous variables can be approximated arbitrarily well by ${\displaystyle {\text{d}}x=-B^{-1}C{\text{d}}a}$ . In this case, ${\displaystyle B}$  represents the ${\displaystyle n}$ ×${\displaystyle n}$  matrix of partial derivatives of the functions ${\displaystyle f}$  with respect to the variables ${\displaystyle x}$ , and ${\displaystyle C}$  represents the ${\displaystyle n}$ ×${\displaystyle m}$  matrix of partial derivatives of the functions ${\displaystyle f}$  with respect to the parameters ${\displaystyle a}$ . (The derivatives in ${\displaystyle B}$  and ${\displaystyle C}$  are evaluated at the initial values of ${\displaystyle x}$  and ${\displaystyle a}$ .) Note that if one wants just the comparative static effect of one exogenous variable on one endogenous variable, Cramer's Rule can be used on the totally differentiated system of equations ${\displaystyle B{\text{d}}x+C{\text{d}}a\,=0}$ .

### Stability

The assumption that the equilibrium is stable matters for two reasons. First, if the equilibrium were unstable, a small parameter change might cause a large jump in the value of ${\displaystyle x}$ , invalidating the use of a linear approximation. Moreover, Paul A. Samuelson's correspondence principle[3][4][5]:pp.122–123. states that stability of equilibrium has qualitative implications about the comparative static effects. In other words, knowing that the equilibrium is stable may help us predict whether each of the coefficients in the vector ${\displaystyle B^{-1}C}$  is positive or negative. Specifically, one of the n necessary and jointly sufficient conditions for stability is that the determinant of the n×n matrix B have a particular sign; since this determinant appears as the denominator in the expression for ${\displaystyle B^{-1}}$ , the sign of the determinant influences the signs of all the elements of the vector ${\displaystyle B^{-1}C{\text{d}}a}$  of comparative static effects.

#### An example of the role of the stability assumption

Suppose that the quantities demanded and supplied of a product are determined by the following equations:

${\displaystyle Q^{d}(P)=a+bP}$
${\displaystyle Q^{s}(P)=c+gP}$

where ${\displaystyle Q^{d}}$  is the quantity demanded, ${\displaystyle Q^{s}}$  is the quantity supplied, P is the price, a and c are intercept parameters determined by exogenous influences on demand and supply respectively, b < 0 is the reciprocal of the slope of the demand curve, and g is the reciprocal of the slope of the supply curve; g > 0 if the supply curve is upward sloped, g = 0 if the supply curve is vertical, and g < 0 if the supply curve is backward-bending. If we equate quantity supplied with quantity demanded to find the equilibrium price ${\displaystyle P^{eqb}}$ , we find that

${\displaystyle P^{eqb}={\frac {a-c}{g-b}}.}$

This means that the equilibrium price depends positively on the demand intercept if gb > 0, but depends negatively on it if gb < 0. Which of these possibilities is relevant? In fact, starting from an initial static equilibrium and then changing a, the new equilibrium is relevant only if the market actually goes to that new equilibrium. Suppose that price adjustments in the market occur according to

${\displaystyle {\frac {dP}{dt}}=\lambda (Q^{d}(P)-Q^{s}(P))}$

where ${\displaystyle \lambda }$  > 0 is the speed of adjustment parameter and ${\displaystyle {\frac {dP}{dt}}}$  is the time derivative of the price — that is, it denotes how fast and in what direction the price changes. By stability theory, P will converge to its equilibrium value if and only if the derivative ${\displaystyle {\frac {d(dP/dt)}{dP}}}$  is negative. This derivative is given by

${\displaystyle {\frac {d(dP/dt)}{dP}}=-\lambda (-b+g).}$

This is negative if and only if gb > 0, in which case the demand intercept parameter a positively influences the price. So we can say that while the direction of effect of the demand intercept on the equilibrium price is ambiguous when all we know is that the reciprocal of the supply curve's slope, g, is negative, in the only relevant case (in which the price actually goes to its new equilibrium value) an increase in the demand intercept increases the price. Note that this case, with gb > 0, is the case in which the supply curve, if negatively sloped, is steeper than the demand curve.

## Without constraints

Suppose ${\displaystyle p(x;q)}$  is a smooth and strictly concave objective function where x is a vector of n endogenous variables and q is a vector of m exogenous parameters. Consider the unconstrained optimization problem ${\displaystyle x^{*}(q)=\arg \max p(x;q)}$ . Let ${\displaystyle f(x;q)=D_{x}p(x;q)}$ , the n by n matrix of first partial derivatives of ${\displaystyle p(x;q)}$  with respect to its first n arguments x1,...,xn. The maximizer ${\displaystyle x^{*}(q)}$  is defined by the n×1 first order condition ${\displaystyle f(x^{*}(q);q)=0}$ .

Comparative statics asks how this maximizer changes in response to changes in the m parameters. The aim is to find ${\displaystyle \partial x_{i}^{*}/\partial q_{j},i=1,...,n,j=1,...,m}$ .

The strict concavity of the objective function implies that the Jacobian of f, which is exactly the matrix of second partial derivatives of p with respect to the endogenous variables, is nonsingular (has an inverse). By the implicit function theorem, then, ${\displaystyle x^{*}(q)}$  may be viewed locally as a continuously differentiable function, and the local response of ${\displaystyle x^{*}(q)}$  to small changes in q is given by

${\displaystyle D_{q}x^{*}(q)=-[D_{x}f(x^{*}(q);q)]^{-1}D_{q}f(x^{*}(q);q).}$

Applying the chain rule and first order condition,

${\displaystyle D_{q}p(x^{*}(q),q)=D_{q}p(x;q)|_{x=x^{*}(q)}.}$

(See Envelope theorem).

### Application for profit maximization

Suppose a firm produces n goods in quantities ${\displaystyle x_{1},...,x_{n}}$ . The firm's profit is a function p of ${\displaystyle x_{1},...,x_{n}}$  and of m exogenous parameters ${\displaystyle q_{1},...,q_{m}}$  which may represent, for instance, various tax rates. Provided the profit function satisfies the smoothness and concavity requirements, the comparative statics method above describes the changes in the firm's profit due to small changes in the tax rates.

## With constraints

A generalization of the above method allows the optimization problem to include a set of constraints. This leads to the general envelope theorem. Applications include determining changes in Marshallian demand in response to changes in price or wage.

## Limitations and extensions

One limitation of comparative statics using the implicit function theorem is that results are valid only in a (potentially very small) neighborhood of the optimum—that is, only for very small changes in the exogenous variables. Another limitation is the potentially overly restrictive nature of the assumptions conventionally used to justify comparative statics procedures.

Paul Milgrom and Chris Shannon[6] pointed out in 1994 that the assumptions conventionally used to justify the use of comparative statics on optimization problems are not actually necessary—specifically, the assumptions of convexity of preferred sets or constraint sets, smoothness of their boundaries, first and second derivative conditions, and linearity of budget sets or objective functions. In fact, sometimes a problem meeting these conditions can be monotonically transformed to give a problem with identical comparative statics but violating some or all of these conditions; hence these conditions are not necessary to justify the comparative statics. Stemming from the article by Milgrom and Shannon as well as the results obtained by Veinott[7] and Topkis[8] an important strand of operational research was developed called monotone comparative statics. In particular, this theory concentrates on the comparative statics analysis using only conditions that are independent of order-preserving transformations. The method uses lattice theory and introduces the notions of quasi-supermodularity and the single-crossing condition. The wide application of monotone comparative statics to economics includes production theory, consumer theory, game theory with complete and incomplete information, auction theory, and others.[9]

## Notes

1. ^ (Mas-Colell, Whinston, and Green, 1995, p. 24; Silberberg and Suen, 2000)
2. ^ Fleeming Jenkin (1870), "The Graphical Representation of the Laws of Supply and Demand, and their Application to Labour," in Alexander Grant, Recess Studies and (1872), "On the principles which regulate the incidence of taxes," Proceedings of the Royal Society of Edinburgh 1871-2, pp. 618-30., also in Papers, Literary, Scientific, &c, v. 2 (1887), ed. S.C. Colvin and J.A. Ewing via scroll to chapter links.
3. ^ Samuelson, Paul, "The stability of equilibrium: Comparative statics and dynamics", Econometrica 9, April 1941, 97-120: introduces the concept of the correspondence principle.
4. ^ Samuelson, Paul, "The stability of equilibrium: Linear and non-linear systems", Econometrica 10(1), January 1942, 1-25: coins the term "correspondence principle".
5. ^ Baumol, William J., Economic Dynamics, Macmillan Co., 3rd edition, 1970.
6. ^ Milgrom, Paul, and Shannon, Chris. "Monotone Comparative Statics" (1994). Econometrica, Vol. 62 Issue 1, pp. 157-180.
7. ^ Veinott (1992): Lattice programming: qualitative optimization and equilibria. MS Stanford.
8. ^ See: Topkis, D. M. (1979): “Equilibrium Points in Nonzero-Sum n-Person Submodular Games,” SIAM Journal of Control and Optimization, 17, 773–787; as well as Topkis, D. M. (1998): Supermodularity and Complementarity, Frontiers of economic research, Princeton University Press, ISBN 9780691032443.
9. ^ See: Topkis, D. M. (1998): Supermodularity and Complementarity, Frontiers of economic research, Princeton University Press, ISBN 9780691032443; and Vives, X. (2001): Oligopoly Pricing: Old Ideas and New Tools. MIT Press, ISBN 9780262720403.


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# Q & A: do stringy dimensions violate Special Relativity?

Q:
I had a discussion with my highschool physics teacher and he was trying to explain to me the difference between 'extrinsic' curvature and 'intrinsic' curvature in spacetime. One example that really helped me was a "closed"/cylinder universe. We picture it as 'curved' because to picture it we intend to embed it in a higher dimensional spacetime. Yet in-reality, the closed spacetime can be everywhere flat. Because it is everywhere flat, we can use special relativity. My teacher and I worked it out (and he found some papers as well), that even though there is no local preferred frame since special relativity works in flat spacetimes, the topology of the spacetime (I don't know what that means besides closed+finite in a direction vs open+infinite in a direction) gives it a preferred _global_ frame. That is, two clocks can move inertially and meet up at two different events, and not agree on the time spent between meeting. This preferred "global" frame is (as we were able to work out, and as the paper stated), the one in which a surface of constant time goes around the closed direction of the universe and meets back up (in other frames, it will instead "helix" around in spacetime). I'm sorry, for all the background, but I'm really enjoying this, and wanted to make it clear the background behind this question which would probably sound very odd otherwise. The question is: Since spacetime in string theory has many curled up dimensions, while there is no preferred local frame (locally SR remains exact -- local poincare invariance is exact), there will be a global preferred frame in the same sense as above. Since strings can wind all the way around these dimensions, wouldn't their dynamics be affected by this global preferred frame. Furthermore, since we zoom out to see 4-dimensional spacetime in ordinary life, wouldn't these curled dimensions (since we kind of 'average' over all of them, their global stucture matters when we zoom out) induce a preferred frame in the remaining 4-dimensions? So does string theory predict what appears to be 'effectively' modification to SR when we try to describe the world as only 4-dimensional? My teacher said we'd have to go to experts for this one, so we are both very interested in your response. Thank you!
- Samantha Hersh (age 18)
Harperville, Mississippi, US
A:
Samantha- Your background introduction is beautifully clear and spot-on. Please consider coming to study physics at UIUC. Your teacher sounds amazing as well.

As for your question, it's too hard for me, so I'll forward it to a string theorist. I'm sure the answer is that SR emerges as correct on spacetime scales large compared to the Planck scale in regions which are intrinsically flat, but I'm not sure what the limits are on that conclusion. Certainly if there are some other extended dimensions you can get effects which would never show up in plain SR. There may be other symptoms as well, which I hope we'll both soon learn about.

Mike W.

I guess what we're talking about here is an example where spacetime is S^1xR. There is a frame in which space is S^1 and time is R (the reals). The best way to think of this is as a 'compactification' of flat spacetime, RxR.  We regard the circle S^1 as R/Z (reals mod integers), which means if we take a coordinate x along R, we regard a point x and a point x+2\pi m R as equivalent (for m an integer and R the radius of the circle).

This compactification breaks the Lorentz invariance of the 'covering space' RxR. To see this, start in the frame that I was discussing above, but impose the identification (x,t) ~ (x+2\pi R, t). If we boost to another frame, this identification becomes

(x',t') ~ (x',t') + 2\pi R (\gamma,v\gamma)

where v is the velocity of the frame (in units c=1). Note that (\gamma, v\gamma) is a unit spacelike Lorentz vector (which in the origin frame was just (1,0)). Thus, to specify the compactification, we have to select a fixed unit spacelike Lorentz vector, and this is where the notion of a 'preferred frame' comes from.

In string theory, what often happens is that we soup up this example to include extra large flat dimensions (that are not compact). So spacetime might be of the form S^1xR^3xR (compact direction, space, time). The 5-dimensional covering space has 4+1 Lorentz symmetry, but the compactified space has only 3+1 Lorentz symmetry (one can still boost along the large flat directions) -- one spatial direction is distinguishable from the other 3).

R

(published on 01/25/2010)