text:
Next: 7. Still More Area Up: 6. Limits of Sequences Previous: 6.6 Geometric Series   Index

# 6.7 Calculation of

6.81   Example. We will calculate , where is a positive number. Let . Then (see the figure)

and hence

Thus

i.e.
 (6.82)

Since

it follows from the squeezing rule that
 (6.83)

Notice that in this example the squeezing rule has allowed us to prove the existence of a limit whose existence was not obvious.

6.84   Example. We will show that for all
 (6.85)

Let , and let . Let

By (6.82), we have

so

It follows from (6.83) that
 (6.86)

From the picture, we see that

i.e.

Hence

By (6.86), we have

so by the squeezing rule, , i.e.

This completes the proof of (6.85).

The reason we assumed to be positive in the previous example was to guarantee that has a logarithm. We could extend this proof to work for arbitrary , but we suggest an alternate proof for negative in exercise 6.97

6.87   Example (Numerical calculation of ) It follows from the last example that

I wrote a Maple procedure to calculate by using this fact. The procedure limcalc(n) below calculates

and I have printed out the results for n = 1,2,...,6.

> limcalc := n -> (1+ .01^n)^(100^n);

> limcalc(1);

> limcalc(2);

> limcalc(3);

> limcalc(4);

> limcalc(5);

> limcalc(6);

6.88   Exercise. From my computer calculations it appears that

Explain what has gone wrong. What can I conclude about the value of from my program?

6.89   Example. Actually, Maple is smart enough to find the limit, and does so with the commands below. The command evalf returns the decimal approximation of its argument.

> limit( (1+1/n)^n,n=infinity);

> evalf(%);

6.90   Entertainment ( .) Find the limit of the sequence , or else show that the sequence diverges.

6.91   Example (Compound interest.) The previous exercise has the following interpretation.

Suppose that dollars is invested at % annual interest, compounded times a year. The value of the investment at any time is calculated as follows:

Let , and let be the value of the investment at time Then

 (6.92)

and in general
 (6.93)

The value of the investment does not change during the time interval
For example, if denotes the value of one dollar invested for one year at % annual rate of interest with the interest compounded times a year, then

Thus it follows from our calculation that if one dollar is invested for one year at % annual rate of interest, with the interest compounded infinitely often'' or continuously'', then the value of the investment at the end of the year will be

If the rate of interest is 100%, then the value of the investment is dollars, and the investor should expect to get \$2.71 from the bank.

This example was considered by Jacob Bernoulli in 1685. Bernoulli was able to show that .[8, pp94-97]

6.94   Exercise. Calculate the following limits.
A
a)
.
b)
.

6.95   Exercise.
a)
Use the formula for a finite geometric series,

to show that
 (6.96)

b)
Let Use inequality (6.96) to show that

for all such that .
c)
Prove that for all .

6.97   Exercise. A Let . Use exercise 6.95 to show that

(Hence we have .)

Hint: Note that for all real numbers .

Next: 7. Still More Area Up: 6. Limits of Sequences Previous: 6.6 Geometric Series   Index
Ray Mayer 2007-09-07


text:
## Rumor source detection for rumor spreading on random increasing trees.(English)Zbl 1306.05223

Summary: In recent papers, D. Shah and T. Zaman [“Detecting sources of computer viruses in networks: theory and experiment”, in: Proceedings of the ACM SIGMETRICS international conference on measurement and modeling of computer systems, SIGMETRICS 10. New York, NY: Association for Computing Machinery (ACM). 203–2014 (2010; doi:10.1145/1811099.1811063); “Rumors in a network: who’s the culprit?”, IEEE Trans. Inform. Theory 57, No. 8, 5163–5181 (2011; doi:10.1109/TIT.2011.2158885)] proposed the rumor center as an effective rumor source estimator for rumor spreading on random graphs. They proved for a very general random tree model that the detection probability remains positive as the number of nodes to which the rumor has spread tends to infinity. Moreover, they derived explicit asymptotic formulas for the detection probability of random $$d$$-regular trees and random geometric trees. In this paper, we derive asymptotic formulas for the detection probability of grown simple families of random increasing trees. These families of random trees contain important random tree models as special cases, e.g., binary search trees, recursive trees and plane-oriented recursive trees. Our results show that the detection probability varies from 0 to 1 across these families. Moreover, a brief discussion of the rumor center for unordered trees is given as well.

### MSC:

 05C80 Random graphs (graph-theoretic aspects) 05C05 Trees 60C05 Combinatorial probability
Full Text:

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## one size fits all vs goal posts [NCA / SHAM]

Hi all,

also remember that when we do BE we treat T and R in the same fashion. So regardless of how terrible we bias some estimate we may do that equally for the two treatments we are comparing. This will work as long as the phenomena that we discuss (funky data points, imputation, deletion, addition, intellingent fixes) occur with equal probability across groups.
We may bias some estimate (like lambda) rather terribly either in positive or negative direction, but that does not automatically imply that the expected point estimate changes and therefore it may be entirely safe in term of patient's risk. "It depends" as they say.

For example, there was a post recently (Mittyri? Hötzi? Someone else?) who wrote that when we do log down in BE we assume a first order elimination. That is much a personal interpretation, I think. When I do log or linear down in BE, I am personally only saying I am willing to make the same error for T and R regardless of how the drug is eliminated, full stop and that's how I am not necessarily strongly favouring one method over the other. Even if the SPC or FOI info indicates first order elimination I am perfectly fine with lin. down especially if this is what the CRO usually does well. Wouldn't want them to get out of their comfort zone.
From the top of my head, I believe I have never been in a situation where I needed to defend an AUC value per se, I was only in situations where regulators questioned the proof of BE.

For dose-finding or superiority it may be a somewhat different situation.

» Another point is that I used PHX NCA like a standard candle (maybe regulators do it as well?). But it turns out that in some rare extreme cases it is worth to think about alternative approaches. Would it be accepted? How to be sure that the method stated in SAP would work better than PHX algo? Is it possible to explain in the report any disagreement with PHX output?

Regulators are using PHX now????

This week's list of things I absolutely detest: Corona virus, the which function in R, WIA-WIA interfaces for scanning under Windows 10, the Bee Gees, the smell of my fridge.

Best regards,
ElMaestro


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# A vessel contains milk and water in the ratio 3:2

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Intern
Joined: 22 Sep 2011
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A vessel contains milk and water in the ratio 3:2 [#permalink]

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26 Sep 2011, 10:49
1
KUDOS
A vessel contains milk and water in the ratio 3:2. The volume of the contents is increased by 50% by adding water to this. From this solution 30 litre of the mixture is withdrawn and then replaced with water. The resultant ratio of milk to water in the final solution is 3:7. Find the original volume of the solution?

Also, please update how much time it took to resolve the problem. I want to check my level..whether it is really difficult and time consuming or not.
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Re: Ratio Proportion - tough problem 750+ [#permalink]

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27 Sep 2011, 04:22
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Expert's post
jrymbei wrote:
Guys, Here's another problem I found while practicing today. Initially I was not able to figure out the problem itself and it took a long time to resolve this one..

A vessel contains milk and water in the ratio 3:2. The volume of the contents is increased by 50% by adding water to this. From this solution 30 litre of the mixture is withdrawn and then replaced with water. The resultant ratio of milk to water in the final solution is 3:7. Find the original volume of the solution?

Also, please update how much time it took to resolve the problem. I want to check my level..whether it is really difficult and time consuming or not.

The problem isn't very tough if your understand your mixtures well.
Work with each statement one by one.
Check out these post to understand the equations used:
http://www.veritasprep.com/blog/2011/03 ... -averages/
http://www.veritasprep.com/blog/2011/04 ... -mixtures/

Step 1:
"A vessel contains milk and water in the ratio 3:2. The volume of the contents is increased by 50% by adding water to this."

You have milk:water = 3:2
volume of this solution:volume of water added = 2:1 (Since volume is increased by 50%)
Let's focus on milk. Solution 1 with 3/5 milk is mixed with solution 2, which has no milk, in the ratio 2:1
Fraction of milk in final mixture of step 1 = [(3/5)*2 + 0*1]/(2+1) = 2/5
Say the volume of this final mixture is V litres.

Step 2:
"From this solution 30 litres of the mixture is withdrawn and then replaced with water. "

This just means that (V - 30) lt of our final mixture above is mixed with 30 lt of water.

Step 3:
"The resultant ratio of milk to water in the final solution is 3:7. "

This tells us that when we mix V-30 lt of mixture in which milk is 2/5 with 30 lt water, we get fraction of milk = 3/10

(V - 30)/30 = (0 - 3/10)/(3/10 - 2/5)
We get V = 120 lt

Since V was the volume of the final mixture in step 1, the volume of solution 1 in step 1 must have been 80 lt (because when you increase it by 50% i.e. 40 lt, you get 120 lt)

The original volume of the solution = 80 lt
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 05 Sep 2011 Posts: 2 Followers: 0 Kudos [?]: 3 [2] , given: 0 Re: Ratio Proportion - tough problem 750+ [#permalink] ### Show Tags 27 Sep 2011, 10:30 2 This post received KUDOS Hey - here's another approach, hope it helps. Ratio of milk to water = 3:2...so we can take the volume of milk to be 3V and the volume of water to be 2V, and the total volume to be 3V + 2V = 5V. Volume is increased by 50% by adding water. Therefore volume goes from 5V to 7.5V (an increase of 2.5V). Since only water has been added the NEW volume of water in solution is now 2V + 2.5V = 4.5V. And the new ratio of milk to water is 3:4.5 Now 30 liters are withdrawn. Since we know the new ratio we can calculate that the volume of milk withdrawn is (30/7.5)*3 = 12 liters, which means 18 liters of water have been withdrawn (30 liters in total, minus 12 liters of milk). Resultant volume of milk = 3V -12 Resultant volume of water = 2V + 2.5V -18 + 30 (Remember 30 liters of solution have been withdrawn, 18L milk and 12L water, and replaced with 30 liters of just water). The ratio of these two will equal 3:7. Solving: (3V-12)/(4.5V +12) = 3/7 V = 16 & the original volume is 5V = 5*16 = 80. Please let me know what you think of this approach. Didn't take long at all. Intern Joined: 22 Sep 2011 Posts: 28 Followers: 0 Kudos [?]: 21 [0], given: 20 Re: Ratio Proportion - tough problem 750+ [#permalink] ### Show Tags 27 Sep 2011, 14:51 I understood both the method by Ned and Karishma but the approach mentioned by Ned seems to more easy to interpret. Kudos to both of you for your help. Just want to know is there any good guide where I can have better concept on mixtures where volume and concentrations are involved. I am having tough time to resolve this kind of problem. Besides, do you think this kind of problems are for higher scoring, I mean is this problem is of good standard or we can expect more tougher then this one. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6971 Location: Pune, India Followers: 2031 Kudos [?]: 12770 [0], given: 221 Re: Ratio Proportion - tough problem 750+ [#permalink] ### Show Tags 27 Sep 2011, 22:37 jrymbei wrote: I understood both the method by Ned and Karishma but the approach mentioned by Ned seems to more easy to interpret. Kudos to both of you for your help. Just want to know is there any good guide where I can have better concept on mixtures where volume and concentrations are involved. I am having tough time to resolve this kind of problem. Besides, do you think this kind of problems are for higher scoring, I mean is this problem is of good standard or we can expect more tougher then this one. This problem is definitely not very easy but it has been made tough by adding steps to it. Also, you need to re-trace the steps to get to 80 lts. That makes it a little time consuming and tedious. GMAT generally does not give you long tedious questions though they are tricky. The moment the trick clicks, you need 30 secs to solve the question. There is a fun element in the actual GMAT questions - they make you think hard, not calculate a lot. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: Ratio Proportion - tough problem 750+ [#permalink]

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28 Sep 2011, 04:44
I agree with you Karishma. Is there any guide for quick tricks. I had a long way to go to bring myself to a competent level.To me it's like a never ending story, I need real good score since my acad was not worth mentioning Lots of studies needs to be done.
Re: Ratio Proportion - tough problem 750+   [#permalink] 28 Sep 2011, 04:44
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# A vessel contains milk and water in the ratio 3:2

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text:
# Function Rules

Learners solve and complete 20 different problems that include creating function rules for each situation. First, they find the number given for the function represented by each rule. Pupils write a rule for the function represented by each table.


text:
# Which quadrant does the given angle 127 lie?

Dec 11, 2016

IInd Quadrant

#### Explanation:

${127}^{o}$ is more than ${90}^{o}$ and less than ${180}^{o}$


text: Take for example the script http://math.stanford.edu/~stange/scripts/tate_via_nets.gp What I do is copying the script to a txt file and name it tatepairing.gp. Then I go to my sage notebook and choose above the option 'gp'. In the first block I write "\r filename" where filename is the whole path of the file tatepairing.gp (so C:....\tatepairing.gp) and evaluate it. If you go to the link you see a function tate_pairing_alg with input an elliptic curve, two points on it and an integer. Therefore I define (over $\mathbb{F}_5$) e = ellinit([0,0,0,4,0]*Mod(1,5)) P1= [2,4] P2=[1,0]


text:
Journal article Open Access

# Intégration massive d'électronique de puissance : synchronisation et stabilité des grands systèmes électriques

Debry, Marie-Sophie; Denis, Guillaume; Panciatici, Patrick; Prevost, Thibault; Xavier, Florent; Guillaud, Xavier; Kestelyn, Xavier; Menze, Andreas

### Citation Style Language JSON Export

{
"publisher": "Zenodo",
"DOI": "10.5281/zenodo.824917",
"container_title": "Revue de l'Electricit\u00e9 et de l'Electronique",
"title": "Int\u00e9gration massive d'\u00e9lectronique de puissance : synchronisation et stabilit\u00e9 des grands syst\u00e8mes \u00e9lectriques",
"issued": {
"date-parts": [
[
2016,
12,
22
]
]
},
"abstract": "
Renewable generation is mainly connected through converters. It can provide more and more services to the grid such as voltage support or frequency control. However, these services may not be sufficient for extremely high penetrations. As the share of such generating units is growing rapidly, some synchronous areas could in the future be occasionally operated without synchronous machines. In such conditions, system stability will have to be ensured with the same level of reliability as today. Today, operation of power systems is based on the presence of synchronous machines. Frequency is linked to the balance between consumption and generation of electricity via the rotating masses equation. This will not be inherently valid for grids without synchronous machines. The issue of operating a network with 100\u00a0% power electronics is quite well solved for small isolated systems. The same doesn\u2019t apply for large transmission systems where grid topology and power injections are highly variable and are not known at any time by all system components or even by a centralized entity. This paper describes the research that needs to be achieved to remove barriers to high penetrations of converters.
",
"author": [
{
"family": "Debry, Marie-Sophie"
},
{
"family": "Denis, Guillaume"
},
{
"family": "Panciatici, Patrick"
},
{
"family": "Prevost, Thibault"
},
{
"family": "Xavier, Florent"
},
{
"family": "Guillaud, Xavier"
},
{
"family": "Kestelyn, Xavier"
},
{
"family": "Menze, Andreas"
}
],
"page": "54-59",
"volume": "2016-5",
"type": "article-journal",
"id": "824917"
}
101
96
views


text:
# RENEWABLE ENERGY'S MARKETS AND INDUSTRIES TRENDS

## GLOBAL OVERVIEW

We need a low-carbon world, and it means we need to encourage a switch away from fossil fuels and towards clean power. Renewable energy comes from natural sources such as wind, sunlight, tides, rain and geothermal heat.

SUPPLYgoGREEN technologies for a green world works to spread existing environmental technologies and sustainable solutions, bringing all together.

Source: Renewables 2017 - REN21 (Renewable Energy Police Network for the 21st. Century)


text:
# stephanieconcepcionramirez

index
writings

*

*

stephaniecramirez.com←

I remember wanting so much more from clouds

than the droplets that take up so much space

Aimlessly and freely making way through the inexhaustible sky

I remember thinking I would one day get to the bottom of their plump dense curves

A formulaic response and quick turn to reality was enough to snap me out

of my wondering

The daydreams of being submerged in all they could be composed of

Engulfed in possibly knowing the magic of what keeps them afloat

despite their ability to hold the thunder storms over the horizon

I remember being filled with hope and holding my breath as I flew towards them

getting closer and closer to my disappointment for falling for the illusion of a cloud

Austin, TX. 2018