text:
Engineering Experience 4: Design a Small Solar Vehicle/Nl/2012: Team AM6

Voor het vak engineering experience 4 moeten we dit semester een zonnewagen/Small Solar Vehicle (SSV) bouwen. Elk team (8 personen) krijgt een zonnepaneel en een DC-motor, hieruit wordt de zonnewagen opgebouwd. Na 10 weken racen alle teams tegen elkaar, tijdens deze race is het niet enkel de bedoeling om de snelste zonnewagen te bouwen maar ook om de origineelste en meest innovatieve wagen te laten zien.

Procesverslag
a)	Planning

De voorstelling, algemene planning en takenverdeling zijn uiteengezet in de volgende documenten. -	Grantt chart -	PVA -	Samenwerkingscontract -	WBS

b) Realisatie van de SSV

Op de wikipage staat een algemene, wekelijkse beschrijving van opdrachten, vorderingen, problemen en oplossingen. *De wikipage: http://en.wikiversity.org/wiki/Engineering_Experience_4:_Design_a_Small_Solar_Vehicle/Nl/2012:_Team_AM6 De vergaderverslagen zijn uitgebreider en bevatten een aanwezigheidslijst, de feedback van de coach, afspraken en een korte termijn takenverdeling. *zie document: vergaderverslagen. In de laatste weken van het project werden de taken groot zodat we de taken makkelijk konden verdelen. Er was geen nood aan aparte vergaderverslagen voor duidelijkheid.

c) Bouwproces SSV Het in elkaar steken van de wagen gebeurde in één keer zonder tussenstappen. De onderdelen: -	De basis uit plexiglas o	Getekend in solid edge, bewerkt in fablab -	De wielen uit plexiglas o	Gemaakt in fablab -	De assen -	De houder van het zonnepaneel -	Het zonnepaneel -	Staafjes in het midden voor lijnsbegeleiding -	Staafje en blokje op 10cm voor tijdsmeting tijdens race -	DC-motor -	tandwielen

d) Beschrijving SSV en argumentatie

-	Basis Het basismateriaal van de zonnewagen is plexiglas. Dit materiaal is relatief licht, goedkoop en vooral makkelijk bewerkbaar in FabLab. De basis van de wagen bestaat uit een raketvormig stuk plexiglas met een lengte van ongeveer 20 centimeter en een dikte van vier millimeter. Aangezien het gaat om een kleine wagen met een kleine massa, is de plaat dik genoeg om als basis te dienen. In deze raketvormige basis zijn er vijf uitsnijdingen. Vier waar de wielen dookomen en een waar een tandwiel doorkomt.

-	Wielen Oorspronkelijk was het idee om kleine CD’s als wielen te gebruiken. Deze hebben een diameter van acht centimeter. Al gauw bleken deze moeilijk monteerbaar op de assen en doordat ze zeer dun zijn, zijn ze instabiel. Uiteindelijk zijn de wielen uitgesneden in plexiglas. De diameter van acht centimeter bleef behouden. Doordat de wielen op maat gemaakt zijn, passen deze perfect op de assen. Samen met de vier millimeter dikte en de goede afsluiting rond de assen, is dit beter materiaal om te gebruiken als wielen. Om de wielen en assen te kunnen monteren, zijn er vier kleine plaatjes aan de onderkant van de zonnewagen gelijmd. Twee vooraan en twee achteraan. In elk plaatje zit een gat, met daarin een rollager. Deze dient om de wrijving tussen de assen en de plaatjes zo klein mogelijk te houden.

-	Tandwielen en assen. Zowel de tandwielen en de assen zijn aangekocht via internet. Een groot aanbod met duidelijke beschrijvingen maakte het mogelijk om tandwielen en assen te bestellen. De tandwieloverbrenging is berekend op 7, maar toegepast met 5,8. Dit doordat het grootste passende tandwiel slechts 58 tanden had, en het kleinste er al 10 had. Indien er gebruik wordt gemaakt van een overbrengen van 5,8 is er minder verlies dan dat er meerdere tandwielen gecombineerd worden om een overbrenging van 7 te verwezenlijken. Het gaat hier om plastieken tandwielen. Het tandwiel van 10 tanden is met behulp van een klein tussenstuk rechtstreeks geplaatst op de motor. Het tandwiel met 58 tanden is nauw aansluitend met een stalen as van vier millimeter dikte. Deze as zit door de achterste rollagers, bevestigd aan het frame. Op deze as zitten ook twee wielen gelijmd en geklemd. De andere twee wielen zijn op dezelfde manier vooraan bevestigd op een kortere as. Doordat er achteraan meer plaats is, is er gekozen voor achterwielaandrijving.

-	Lijnbegeleiding Om ervoor te zorgen dat de wagen op een rechte lijn blijft, is er op het traject een begeleidingsrail voorzien. Om deze ten volste te benutten is er zowel vooraan als achteraan de wagen begeleiding voorzien. De begeleiders bestaan uit twee stalen staafjes vooraan en twee achteraan. Aan deze staafjes zijn kleine plexiglazen wieltjes bevestigd. Doordat aan weerszijde van de rail een wieltje zit zal de wagen op de baan blijven.

-	Zonnepaneelbevestiging Aan het zonnepaneel is met ducktape een GPS-houder geplakt. Dit om het zonnepaneel niet te beschadigen. De andere kant van de GPS-houder bevat een zuignap die aan het frame kan bevestigd worden. Door gebruik te maken van de GPS-houder is er steeds een mogelijkheid om het zonnepaneel te verwijderen en om de hoek t.o.v. de zon te veranderen.

e) aanvulling

1.	Inleiding Enerzijds maakten we berekeningen en simulaties voor het project. Dit verslag bevat alle overige informatie over het project, namelijk de beschrijving, de planning, de evolutie en het uiteindelijke resultaat.

2.	Planning De wijzigingen in de Gantt Chart komen voort uit: o	Afwezigheden en ziekte van teamleden o	Onvoorziene taken o	??? De problemen qua organisatie en samenwerking: o	Communicatie ging moeilijk gezien de groep vrij groot was. Ook bijeenkomen was moeilijker dan bij kleine groepen. De takenverdeling en georganiseerd werken was zeer belangrijk om dit op te lossen. o	Opvolging en controle: Soms werd de taak te laat of niet uitgevoerd. In toekomstige projecten kunnen we eventueel de groepsleider minder werk geven en meer een controlerende rol laten uitoefenen.

3.	Oude vaardigheden o	Werken in groep: vergaderen, taakverdeling, communicatie, etc o	Plannen o	Technisch tekenen Nieuwe vaardigheden o	Werken volgens functie met bepaalde verantwoordelijkheid o	Werken in een grotere groep o	Opvolging van teamleden / feedback aan elkaar geven

4.	Conclusie o	Het eindoordeel over het proces van het team: Bij het project hadden we soms problemen maar door deze tijdig aan te kaarten op bijeenkomsten, vonden we in overleg steeds een oplossing waar ieder teamlid het mee eens was. o	Het eindoordeel over het product: Het wagentje rijdt vlot en blijft op de baan. Volgens het plan van aanpak streven we naar de snelste wagen en we slaagden er dus in een snelle SSV te bouwen. o	Verbeteringen in de toekomst: 	Bij een grote groep is er nood aan een persoon die de rest opvolgt en bijstuurt. 	Vergaderverslagen zijn zeer handig gezien er in een grote groep met een groot project vaak kleine dingen vergeten worden. Ook zijn de afwezigen snel terug op de hoogte en kunnen zij terug meehelpen.

5.	Literatuur / Referenties

Week 1
In week 1 zijn de verschillende functies verdeeld:

Karlien zal samen met Glenn het team proberen in goede banen te leiden door de gewenste doelstellingen en deadlines te behalen. Pieter en Christophe zullen afwisselend de vergaderverslagen opstellen terwijl Milan de vergaderingen zal organiseren. De projectmap zal beheerd worden door Anton. Als laatste zullen Brecht en Cesar de website beheren.

Ook werden de nodige documenten opgesteld:
 * Plan van aanpak
 * Gantt chart
 * WBS
 * Samenwerkingscontract

En werd er een logo ontworpen dat gebaseerd is op de gekozen teamnaam: The Mean Green Solar Machine.

Week 2
In week 2 hebben we ons zonnepanneel en de DC motor gekregen. Na een seminarie hebben we op het zonnepaneel enkele testen uitgevoerd, hieruit bleek dat de 'm' van ons zonnepaneel gelijk is aan 1,16.

(I = Isc - Is*(e^(U/(mNUr)-1))

Deze m heeft steeds een waarde tussen 1 en 5, hoe hoger de m-waarde, hoe beter het zonnepaneel werkt. Een m-waarde van 1,16 wilt dus zeggen dat ons zonnepaneel niet meer optimaal werkt dit komt omdat dit zonnepaneel al enkele keren is gebruikt tijdens vorige jaren voor het EE4 project.

Deze week hebben we ook feedback gekregen van onze mentor over het plan van aanpak, de WBS en de gantt chart, op enkele kleine puntjes na waren deze documenten zeer goed opgesteld. Ook hebben we er een nieuw teamlid bij: Marie Ingabire, zij zal de realisatie van de technische tekening in goede banen leiden.

Week 3
Deze week werd een seminarie gegeven over het Design voor de SSV daarbij ook een inleiding over simulaties voor ons autootje. Op het design werd er onmiddellijk gebrainstormd en al enkele prototypes werden uitgetekend. Deze zullen we dan willen uittesten in de simulaties. Het ruw model bestaat uit 4 wielen die zullen gebouwd worden uit 4 kleine CD's. Het onderstel heeft een driehoekige vorm waarvan de achterwielen in de hoekpunten liggen. De voorste wielen vormen samen het voorste hoekpunt. Ze delen dus het voorste punt via hun as. In het midden bouwen we een dragende staaf waarop we het zonnepaneel zullen plaatsen. Van hierop is het vrij voor beweging om een zo goed mogelijk vermogen uit het paneeltje te krijgen. Het frame zal bestaan uit plexiglas, hout of plastiek. Hiervoor zal er eerst nog wat opzoekwerk verricht moeten worden op vlak van gewicht, prijs, sterkte etc. Creatieve ideeën: Het zonnepaneel zou automatisch kunnen met de zon meebewegen.

Verder werd nog extra info gehaald over de DC motor uit de datasheets. Deze informatie kan dan later verwerkt worden binnen het simulatie programma.

Er zal nu zo snel mogelijk een simulatie gemaakt worden met de geschatte waarden.

Week 4
Deze week werd een seminarie gegeven over matlab en simulink. Deze programmas kunnen we dan verder gebruiken voor de simulatie van de auto. Verdere verdeling van taken is dus wel vereist aangezien de taken van vorige week nog niet afgeraakt zijn. Simulink blijft nog altijd een beetje tegen steken. Er vormen zich problemen dat we niet verstaan. Daarbij was niet iedereen op tijd tijdens het seminarie. Al bij al niet zo'n goede week. Volgende week zal er betering in moeten zitten. Taken zijn herverdeeld en er wordt gericht om alle taken af te krijgen tegen volgende week. Dit is nodig want de simulatie moet af zijn volgende week!

Even een tandje bijsteken!

Week 5
In week 5 werd er een seminarie gegeven door het umicore solar team over hun avontuur met de solar car. Ook werd er een seminarie gegeven over fablab, een labo waar groept studenten geavanceerde machines mogen gebruiken. In het FabLab kunnen we gebruik maken van een lasercutter/graveerder, een freesmachine, een 3D-scanner en een 3D-printer. Vooral de lasercutter is interessant voor ons project. Deze kunnen we gebruiken om ons frame uit te snijden. Het idee is nu dat we dit uit plexiglas gaan maken. Het is een redelijk lichte materie en is gemakkelijk te gebruiken in deze FabLab machines.

Week 6
Deze week moesten we ons voorlopig verslag aan onze begeleider voorstellen. We hebben ons blijkbaar gemist in de hellingshoek van het wagentje en zullen dit moeten aanpassen. Verder nog wat uitwerking van de uitleg hier en daar en vervolledigen van ons Sankey Diagram en dan staan we er op papier al heel goed voor. Het is ook de peer assessment week. hier worden de teamleden voor het eerst beoordeeld door mede teamleden.

Over het design van het wagentje: Materiaal voor het frame wordt Plexiglas. Dit is een gemakkelijk uit te snijden materiaal, inclusief voor tandwielen. Het frame zal een driehoekige structuur hebben en daarbij door 4 wielen ondersteund zijn. De as aan de voorkant zal een statische as zijn waar de wieltjes (lees 'kleine cdtjes') rond zullen draaien door middel van een kogellager. De achterste as, de aandrijfas, zal een vrij roterende zijn in kogellagers waaraan de achterwielen aan vastzitten. Op deze as zal dus ook de aandrijving van motor op wielen komen te zitten. In het midden installeren we een draagcillinder waarop het zonnepaneel vrij kan roteren om een zo goed mogelijke positie in te nemen. Het frezen van het frame en onderdelen zal best voor de paasvakantie moeten gebeuren! Alsook het bestellen van onze onderdelen. Extratje dat we zouden willen toevoegen aan onze auto zijn ledlampjes die zullen gevoed worden door een aparte batterij. Dit kunnen we implementeren als we nog onder het voorziene gewicht zitten.

Week 7
Deze week feedback gekregen op ons eindverslag: - We gebruiken nog wat teveel een spreektaal, minder informeel zijn dus - in handberekeingen waren nogwat fouten te verbeteren alsook in de berekende maximum snelheid in simulink - We komen een optimaal rendement uit van 84% wat goed is, maar voor de berekeningen gebruiken we beter een wat lagere waarde - Daarbij in Simulink waren we nog vergeten de uphill functie erbij te berekenen, silly us.

Dus over het algemeen een goed beeld en zo was ook de quotering die we op dit verslag kregen. So keep up the good work.

In de namiddag (16u) zijn we dan nog eens samengekomen wanneer iedereen vrij was. Hierin hebben we de zoektocht opgezet naar het nodige materiaal voor onze zonnewagen te kunnen realiseren. We zochten naar kogellagers, assen en tandwielen. Dit vooral eerst in winkels in de buurt (denk aan metal leuven, de brico...) maar dat was niet zo'n succes. Tandwielen vinden die ons de juist tandwieloverbrenging zouden geven waren ook niet gemakkelijk te vinden. De onderdelen werden later deze week online besteld. Tegelijkertijd werd er ook gewerkt aan het tekenen van het wagentje in SolidEdge zodat we deze later zouden kunnen brengen naar het FabLab. Daar kunnen we onze stukken uitsnijden uit plexiglas en kan de opbouw van de auto beginnen!

Week 8
Met nog wat feedback op hoever we staan van onze coach, zetten we ons verder in voor het uittekenen in 3D hoe ons autotje er zal uitzien. Tegelijkertijd herbekijken we elk nog eens onze onderdelen van het eindverslag dat nog wat verbetering vereiste. De wagen zal bestaan uiteen raket-achtige vorm als uitsnijding van het plexiglas. Hierin wordt plaats gemaakt voor wielen die door de plaat zullen komen, inclusief het aandrijftandwiel. Deze zit vast op een as van 4mm dikte en heeft ongeveer een lengte van 20cm. De wielen worden vastgemaakt op deze as die zal vrij roteren in ashouders door middel van kogellagers. Dezelfde opstelling wordt voor de voorste as gebruikt. Verder wordt er nog niet aan een bevestiging gedacht voor de motor, aangezien deze heel precies moet zijn om de tandwielen goed te laten passen.

Week 9
Maandag werden de tekeningen naar het fablab genomen dat gelegen is aan de faculteit van werktuigkunde. Daar beschikte het lab over 3 lasercutters, een 3D-printer en andere handige hulpmiddelen zoals boren, zagen, lijm... Het was wel al redelijk druk in het lab aangezien de studenten van werktuigkunde, neem ik aan, ook bezig zijn met een project waar ze een squash bal zo ver mogelijk moeten kunnen projecteren. Hiervoor gebruiken zij ook de lasercutter aangezien ze praktisch zelf alles op maat moeten uitsnijden in hout, gaande van tandwielen tot de tandwielkasten zelf. Wanneer er eindelijk een pc vrijkomt, wordt de kans onmiddellijk gegrepen om de lasercutter aan zijn werk te zetten op een 600x300x4 plaat van plexi. Na wat uitleg over hoe je van je 3D tekening naar een effectieve print komt begint de lasercutter aan een razende snelheid aan de gravering en heel kort daarna al aan het uitsnijden van alle benodigde delen. In slechts zo'n 5 min was heel het proces al klaar en kon de Mean Green Solar Machine in elkaar geknutseld worden. Met behulp van de lijmen, de bestelde rollagers en asklemmen werd de auto geassembleerd en getest. Dit gebeurde op het circuit dat toen op de -2 parking opgesteld stond. Zonder aandrijving kwam het autootje vanzelf al eigenlijk redelijk ver! Met behulp van een batterij werd er ook een de motor getest. De batterij zorgde alleszins voor overbodige dosis aan energie aangezien de wagen er vandoor schoot toen het werd losgelaten. Het reed zelfs zo brutaal dat er 1 van de assen terug was losgekomen doordat de wagen niet meer op het traject was gebleven. Dit wees erop dat onze wagen nog niet volmaakt was en dus werd er teruggekeerd naar de tekentafel om hiervoor een oplossing te vinden.

Week 10
Na het incidentje van vorige week werd het plan nog eens herbekeken om het mogelijk te maken dat het wagentje zo nauw mogelijk het traject zal volgen. Er werd geopteerd voor een vrij simpele oplossing. Er zullen 4 staafjes verticaal naar beneden gericht worden die van het frame naar beneden zullen komen. Deze worden verstevigd in de horizontale manier door er schijfjes rond te monteren zodat het alligneringsstaafje op zijn plaats zou blijven zitten. Dit betekende wel dat er terug een kort bezoek aan het fablab nodig was. Opnieuw na een eindje wachten werden de nieuwe onderdelen uitgesneden uit de resterende plexiglas plaat die nog wat plaats overhad. De staafjes werden gemonteerd en het alligneersysteem werd opnieuw getest en leek correct te werken. Alleen wordt er nog wat gevreesd dat het wagentje een beetje teveel kan omhoog springen bij een van de hoeken van het traject enzo uit het L-profiel springen. De wagen zal dan nog altijd moeten getest worden op enkel zonne-energie op dat het genoeg energie zal leveren op het stuk omhoog van een halve meter over vier meter te overwinnen. Afwachten op goed weer dus en hopelijk hebben we dan de kans om de ultieme test voor het wagentje uit te voeren!


text:
User:Egm4313.s12.team8.colocar/R3.7

Problem
Expand the series on both sides of (1)-(2) p.7-12b to verify equality.

$$ \displaystyle \sum_{j=2}^5 c_j j (j-1) x^{j-2} = \sum_{k=0}^{3} c_{k+2} (k+2) (k+1) x^{k} = \sum_{j=0}^{3} c_{j+2} (j+2) (j+1) {x^j}$$

$$ \displaystyle \sum_{j=1}^5 c_j \cdot j \cdot x^{j-1} = \sum_{k=0}^{4} c_{k+1} (k+1) x^{k} = \sum_{j=0}^{4} c_{j+1} (j+1) x^{j} $$

Solution
Negate middle portion of (1) and (2) as they display the transitional meaning to the function.

1  $$ \displaystyle \sum_{j=2}^5 c_j j (j-1) x^{j-2} = \sum_{j=0}^{3} c_{j+2} (j+2) (j+1) {x^j}$$

 $$ \displaystyle (c_2 2 (2-1) x^{2-2})+(c_3 3 (3-1) x^{3-2})+(c_4 4 (4-1) x^{4-2})+(c_5 5 (5-1) x^{5-2}) = (c_{0+2}(0+2)(0+1){x^0})+(c_{1+2}(1+2)(1+1){x^1})+(c_{2+2}(2+2)(2+1){x^2})+(c_{3+2}(3+2)(3+1){x^3}) $$

 $$ \displaystyle (c_2 2 (1) x^{0})+(c_3 3 (2) x^{1})+(c_4 4 (3) x^{2})+(c_5 5 (4) x^{3}) = (c_{2}(2)(1){x^0})+(c_{3}(3)(2){x^1})+(c_{4}(4)(3){x^2})+(c_{5}(5)(4){x^3}) $$

 $$ \displaystyle (2 c_2)+(6 c_3 x)+(12 c_4 x^{2})+(20 c_5 x^{3}) = 2c_2+ 6c_3 x + 12c_4 x^{2}+ 20c_5 x^{3} $$

2  $$ \displaystyle \sum_{j=1}^5 c_j \cdot j \cdot x^{j-1} = \sum_{j=0}^{4} c_{j+1} (j+1) x^{j} $$

 $$ \displaystyle (c_1 (1) x^{1-1})+(c_2 (2) x^{2-1})+(c_3 (3) x^{3-1})+(c_4 (4) x^{4-1})+(c_5 (5) x^{5-1}) = (c_{0+1} (0+1) x^{0})+(c_{1+1} (1+1) x^{1})+(c_{2+1} (2+1) x^{2})+(c_{3+1} (3+1) x^{3})+(c_{4+1} (4+1) x^{4}) $$

 $$ \displaystyle (c_1 x^{0})+(c_2 (2) x^{1})+(c_3 (3) x^{2})+(c_4 (4) x^{3})+(c_5 (5) x^{4}) = (c_{1} (1))+(c_{2} (2) x)+(c_{3} (3) x^{2})+(c_{4} (4) x^{3})+(c_{5} (5) x^{4}) $$

 $$ \displaystyle c_1+ 2c_2 x + 3c_3 x^{2} + 4c_4 x^{3}+ 5c_5 x^{4} = c_1+ 2c_2 x + 3c_3 x^{2} + 4c_4 x^{3}+ 5c_5 x^{4}  $$

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: User:Egm4313.s12.team8.colocar/R3.7


text:
University of Florida/Egm4313/s12.team11.imponenti/R3.8

Problem 3.8
solved by Luca Imponenti

Problem Statement
Find a (real) general solution. State which rule you are using. Show each step of your work.
 * $$y''+4y'+4y=e^{-x}cos(x)\!$$

Homogeneous Solution
To find the homogeneous solution, $$y_h\!$$, we must find the roots of the equation

$$\lambda^2+4\lambda+4=0\!$$

$$(\lambda+2)(\lambda+2)=0\!$$

$$\lambda=-2\!$$

We know the homogeneous solution for the case of a double root to be

$$y_h=c_1e^{\lambda x}+c_2xe^{\lambda x}\!$$

 * $$y_h=c_1e^{-2x}+c_2xe^{-2x}\!$$

Particular Solution
We have the following excitation
 * $$r(x)=e^{-x}cos(x)\!$$

From table 2.1, K 2011, pg. 82, we have
 * $$y_p(x)=e^{-x}[Kcos(x)+Msin(x)]\!$$

Since this does not correspond to our homogeneous solution we can use the '''Basic Rule (a), K 2011, pg. 81''' to solve for the particular solution

$$y_p''+4y_p'+4y_p=r(x)\!$$

where

$$y_p'=e^{-x}[-Ksin(x)+Mcos(x)]-e^{-x}[Kcos(x)+Msin(x)]\!$$

$$y_p'=e^{-x}[-Ksin(x)+Mcos(x)-Kcos(x)-Msin(x)]\!$$

and

$$y_p''=e^{-x}[-Kcos(x)-Msin(x)+Ksin(x)-Mcos(x)]-e^{-x}[Ksin(x)+Mcos(x)-Kcos(x)-Msin(x)]\!$$

$$y_p''=e^{-x}[-Kcos(x)-Msin(x)+Ksin(x)-Mcos(x)+Ksin(x)-Mcos(x)+Kcos(x)+Msin(x)]\!$$

$$y_p''=e^{-x}[2Ksin(x)-2Mcos(x)]\!$$

Plugging these equations back into the differential equation

$$y_p''+4y_p'+4y_p=r(x)\!$$

$$e^{-x}[2Ksin(x)-2Mcos(x)]+4e^{-x}[-Ksin(x)+Mcos(x)-Kcos(x)-Msin(x)]+4e^{-x}[Kcos(x)+Msin(x)]=e^{-x}cos(x)\!$$

$$e^{-x}[2Ksin(x)-2Mcos(x)-4Ksin(x)+4Mcos(x)-4Kcos(x)-4Msin(x)+4Kcos(x)+4Msin(x)]=e^{-x}cos(x)\!$$

$$2Mcos(x)-2Ksin(x)=cos(x)\!$$

from the above equation it is obvious that $$K=0\!$$ and $$M=\frac{1}{2}\!$$

therefore the particular solution to the differential equation is

 * $$y_p(x)=\frac{1}{2}e^{-x}sin(x)\!$$

General Solution
The general solution will be the summation of the homogeneous and particular solutions

$$y(x)=y_h(x)+y_p(x)\!$$

$$y(x)=c_1e^{-2x}+c_2xe^{-2x}+\frac{1}{2}e^{-x}sin(x)\!$$

$$y(x)=e^{-2x}(c_1+c_2x)+\frac{1}{2}e^{-x}sin(x)\!$$

The coefficients $$c_1\!$$ and $$c_2\!$$ can be readily solved for given either initial conditions or boundary value conditions.

Problem Statement
Find a (real) general solution. State which rule you are using. Show each step of your work.
 * $$y''+y'+(\pi ^2+\frac{1}{4})y=e^{-\frac{x}{2}}sin(\pi x)\!$$

Homogeneous Solution
To find the homogeneous solution, $$y_h\!$$, we must find the roots of the equation

$$\lambda^2+\lambda+(\pi ^2+\frac{1}{4})=0\!$$

$$\lambda = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\!$$ with $$a=1, b=1,  c=(\pi ^2+\frac{1}{4})\!$$

$$\lambda = \frac{-1 \pm \sqrt{1^2-4*1*(\pi ^2+\frac{1}{4})}}{2*1}\!$$

$$\lambda =-\alpha \pm i\omega=-\frac{1}{2} \pm \pi i\!$$

We know the homogeneous solution for the case of a double root to be

$$y_h=e^{-\alpha x}[Acos(\omega x)+Bsin(\omega x)]\!$$

 * $$y_h=e^{-\frac{x}{2}}[Acos(\pi x)+Bsin(\pi x)]\!$$

Particular Solution
We have the following excitation
 * $$r(x)=e^{-\frac{x}{2}}sin(\pi x)\!$$

From table 2.1, K 2011, pg. 82, we have
 * $$y_p(x)=e^{-\frac{x}{2}}[Kcos(\pi x)+Msin(\pi x)]\!$$

Since this corresponds to our homogeneous solution we must use the '''Modification Rule (b), K 2011, pg. 81''' to solve for the particular solution

so $$y_p(x)=xe^{-\frac{x}{2}}[Kcos(\pi x)+Msin(\pi x)]\!$$

differentiating

$$y_p'=\pi xe^{-\frac{x}{2}}[-Ksin(\pi x)+Mcos(\pi x)]+(e^{\frac{x}{2}}-\frac{1}{2}xe^{-\frac{x}{2}})[Kcos(\pi x)+Msin(\pi x)]\!$$

$$y_p'=e^{-\frac{x}{2}}(\pi x[-Ksin(\pi x)+Mcos(\pi x)]+(1-\frac{1}{2}x)[Kcos(\pi x)+Msin(\pi x)])\!$$

$$y_p'=e^{-\frac{x}{2}}[-\pi xKsin(\pi x)+\pi xMcos(\pi x)+Kcos(\pi x)+Msin(\pi x)-\frac{1}{2}xKcos(\pi x)-\frac{1}{2}xMsin(\pi x)]\!$$

and

$$y_p''=e^{-\frac{x}{2}}[-\pi ^2xKcos(\pi x)-\pi Ksin(\pi x)-\pi ^2xMsin(\pi x)+\pi Mcos(\pi x)-\pi Ksin(\pi x)+\pi Mcos(\pi x)+\frac{1}{2}\pi xKsin(\pi x)-\frac{1}{2}Kcos(\pi x)-\frac{1}{2}\pi xMcos(\pi x)-\frac{1}{2}Msin(\pi x)]-\frac{1}{2}e^{-\frac{x}{2}}[-\pi xKsin(\pi x)+\pi xMcos(\pi x)+Kcos(\pi x)+Msin(\pi x)-\frac{1}{2}xKcos(\pi x)-\frac{1}{2}xMsin(\pi x)]\!$$

$$y_p''=e^{-\frac{x}{2}}[-\pi ^2xKcos(\pi x)-\pi Ksin(\pi x)-\pi ^2xMsin(\pi x)+\pi Mcos(\pi x)-\pi Ksin(\pi x)+\pi Mcos(\pi x)+\frac{1}{2}\pi xKsin(\pi x)-\frac{1}{2}Kcos(\pi x)-\frac{1}{2}\pi xMcos(\pi x)-\frac{1}{2}Msin(\pi x)+\frac{1}{2}\pi xKsin(\pi x)-\frac{1}{2}\pi xMcos(\pi x)-\frac{1}{2}Kcos(\pi x)-\frac{1}{2}Msin(\pi x)+\frac{1}{4}xKcos(\pi x)+\frac{1}{4}xMsin(\pi x)]\!$$

grouping cosine and sine terms we get

$$y_p''=e^{-\frac{x}{2}}[(2\pi M- \pi xM+(\frac{1}{4}-\pi ^2)xK-K)cos(\pi x)+((\frac{1}{4}-\pi ^2)xM-M+\pi xK-2\pi K)sin(\pi x)]\!$$

and

$$y_p'=e^{-\frac{x}{2}}[(\pi xM+K-\frac{1}{2}xK)cos(\pi x)+(-\pi xK +M-\frac{1}{2}xM)sin(\pi x)]\!$$

next we substitute the above equations into the ODE

$$y_p''+y_p'+(\pi ^2+\frac{1}{4})y_p=r(x)\!$$

$$e^{-\frac{x}{2}}[(2\pi M- \pi xM+(\frac{1}{4}-\pi ^2)xK-K)cos(\pi x)+((\frac{1}{4}-\pi ^2)xM-M+\pi xK-2\pi K)sin(\pi x)]+e^{-\frac{x}{2}}[(\pi xM+K-\frac{1}{2}xK)cos(\pi x)+(-\pi xK +M-\frac{1}{2}xM)sin(\pi x)]+(\pi ^2+\frac{1}{4})xe^{-\frac{x}{2}}[Kcos(\pi x)+Msin(\pi x)]=e^{-\frac{x}{2}}sin(\pi x)\!$$

$$e^{-\frac{x}{2}}[(2\pi M- \pi xM+(\frac{1}{4}-\pi ^2)xK-K)cos(\pi x)+((\frac{1}{4}-\pi ^2)xM-M+\pi xK-2\pi K)sin(\pi x)+(\pi xM+K-\frac{1}{2}xK)cos(\pi x)+(-\pi xK +M-\frac{1}{2}xM)sin(\pi x)+(\pi ^2+\frac{1}{4})x(Kcos(\pi x)+Msin(\pi x))]=e^{-\frac{x}{2}}sin(\pi x)\!$$

$$e^{-\frac{x}{2}}[(2\pi M- \pi xM+(\frac{1}{4}-\pi ^2)xK-K+\pi xM+K-\frac{1}{2}xK+(\pi ^2+\frac{1}{4})xK)cos(\pi x)+((\frac{1}{4}-\pi ^2)xM-M+\pi xK-2\pi K-\pi xK +M-\frac{1}{2}xM+(\pi ^2+\frac{1}{4})xM)sin(\pi x)]=e^{-\frac{x}{2}}sin(\pi x)\!$$

$$2\pi Mcos(\pi x)-2\pi Ksin(\pi x)=sin(\pi x)\!$$

after cancelling terms; we can equate cosine and sine coefficients to get two equations

$$2\pi M=0\!$$

$$-2\pi K=1\!$$

so $$M=0\!$$ and $$K=-\frac{1}{2\pi}\!$$

and the particular solution to the ODE is

 * $$y_p(x)=-\frac{1}{2\pi}xe^{-\frac{x}{2}}cos(\pi x)\!$$

General Solution
The general solution will be the summation of the homogeneous and particular solutions

$$y=y_h+y_p\!$$

$$y=e^{-\frac{x}{2}}[Acos(\pi x)+Bsin(\pi x)]-\frac{1}{2\pi}xe^{-\frac{x}{2}}cos(\pi x)\!$$

$$y=e^{-\frac{x}{2}}[Acos(\pi x)+Bsin(\pi x)-\frac{1}{2\pi}xcos(\pi x)]\!$$

Egm4313.s12.team11.imponenti 04:28, 21 February 2012 (UTC)


text:
User:Knssilm

I'm a student, a professional photographer and an amateur mycologist (not in that order). I was previously an IT professional but am currently seeking a degree in medicine and emergency care. More info on myself can be found on my Wikipedia User Page

Where I Am
I am lurking the School of Medicine and currently looking to contribute in the Division of Anatomy.

title: User:Knssilm


text:
Engineering Experience 4: Design a Small Solar Vehicle/2012: Team PM12

Team Member

 * Lihua Chen
 * Chaoyi Chen
 * Shoujun Zhang
 * Tianhua Xu
 * Yibo Sun
 * Weihong Zhao
 * Xue Feng
 * Limin Liang

Team Leader
*Draw up the plan for the project and the agenda of the following meeting. *Lead the cooperation of members to achieve the schedule of the whole of the project. *Join the key part of the project with other members. *Contact with the coach and the teacher of team.
 * Lihua Chen

Supervisor
*Point out the strength and weakness in every meeting. *Criticize team leader and other members. *Check whether the agenda has been completed.
 * Chaoyi Chen

Secretary
*Record the meeting reports *Update the wiki page at least once a week *Make budgets
 * Limin Liang

Mechanical Designer
*Consult previous colleagues *Design mechanical structure *Assemble mechanical structure
 * Shoujun Zhang(principal), Yibo Sun, Xue Feng

Analyzer & Simulator
*Calculate the ratio of gears or center of gravity by hand *Simulate with Matlab
 * Tianhua Xu(principal), Chaoyi Chen, Weihong Zhao

Team Rule

 * 1)The meeting for the EE4 discussion will be held every Wednesday at 3 PM at the canteen of Group T.
 * 2)If someone cannot attend the following meeting, he/she has to ask for a leave in advance before midnight of Tuesday.
 * 3)For the one who is late for the meeting, he/she has to treat other team members to something to drink. If someone is late for   more than 3 times, the leader has the right to tell the coach.
 * 4)For every meeting, the leader has to distribute specific assignment to every member and every member has to finish his/her work *in time, otherwise the member will be criticized in next meeting.
 * 5)For every meeting, the secretary ought to make notes very carefully and sort out into a detailed meeting record in time.
 * 6)In order to make the meeting more high-efficient, the leader should make full preparation in advanced.
 * 7)When the meeting begins, the team members are expected to change their phone into vibration mode, so as not to interfere the    meeting conducting smoothly.
 * 8)Every member should pay attention to the meeting, chatting is forbidden when the meeting is on.
 * 9)Show esteem to each other. Don’t try to refute others when they are sharing their opinions.
 * 10)For a key decision, it can only be decided by voting after consulting the coach.

Plan of Approach
Our group's name is Rainbow. During EE4, we will build a small solar vehicle. At last we will hand in a final report. In this plan of approach, we will show you our plan, our goals, limitations and supposition.

Introduction
This plan of approach is the framework of small solar vehicle. We need to build small solar vehicle following this plan, so it is very important for us to make our goals. And we need to preliminary design our solar vehicle in this plan. We will calculate the probably cost as well. But it does not mean that we will make solar vehicle follow this plan absolutely without any changes. Our group will be divided into several parts and we will think about what problems will occur and try to solve them.

Project Description
At first, we will make a Work Breakdown Structure(WBS). It’s a working procedure of the project. It clarifies what we should do in the next few weeks. Then we will make a Gantt Chart according to the WBS, this is the guideline for our project which will be followed closely by each team member. Each line of the Gantt Chart should be executed by a team member. This implies that every one of the team knows their tasks and should stick to that. To build the SSV we need materials, to decide which material we are going to use we need some tests first. After that we will make a list of material and ask everyone’s approval and decide it finally. In one word, we will focus on the Engineer and Enterprising.

Plan
Firstly, we should get the solar battery from our coach. Secondly, design the chassis according to the weight and size of the battery. Then decide the number of the wheels and their size after calculating the driving force and different kinds of resistance. Then, we can design the general structure of the car. After that, we ought to apply it into tests and find problems it may encounter, collect them and improve them.

Goals

 * To build a SSV and make it run
 * To make it as light as possible so that it can run faster, taking stabilization into account.
 * To adjust the center of gravity to make it fly farther.
 * To post our logo in the car.
 * Finally, our solar car Rainbow can complete the race excellently.

Problems and Restrictions
It is our first time to design such a SSV, lacking of experience. And it is inevitable to encounter some difficulties, both in mechanical designing and simulation, for instance, the position of center of gravity and ratio of mutual gears. And the ability of searching for ideal material is also very significant for our whole project. Finally, teamwork is a key factor effecting our success.

Suppositions

 * 1. The car may not run along a straight line.
 * 2. The motor cannot drive the car, because the car is too heavy.
 * 3. One motor can make two or more wheels work.
 * 4. The car is solid enough.

Engineering

 * character for DC-motors: power, losses.
 * data of solar cell: short circuit current, open circuit voltage, efficiency, influence of light intensity on the parameters.
 * dynamic calculations: strength functioning at a mobile vehicle, speed, acceleration, torque, role resistance.
 * calculations concerning maths: panel orientation in function of the position of the sun.
 * simple calculations concerning aerodynamics: resistance in function of the speed.

Enterprising

 * strategic marketing plan: product, price, place, promotion.
 * budget management: decrease cost as much as possible, raise cost performance as much as possible.


text: University of Florida/Egm4313/s12.team14.report3


text: User:OpenScientist/Open grant writing/Wissenswert 2011/Documentation/Categorizer


text:
User:Engsihn

Ghararm Sihn:

I am a student at Howard Community College, taking Intro Engineering course. I am going to decide Engineering as major.

Introduction to Engineering
 * Project 1 - 3d Scan
 * /my work/

 * Project 2 - Kinect Robotic Car
 * /myWork2/

 * Project 3 - Spaghetti Bridge
 * /myWork3/

title: User:Engsihn


text:
Gene transcriptions/Elements/Angiotensinogen core promoters

Angiotensinogen core promoter element 1 (AGCE1) is an example of a core promoter element that may occur in a DNA sequence for only one gene (the human angiotensinogen gene angiotensin).

Proof of concept
For the original research project focused on the transcription of A1BG, proof of concept may be simply that AGCE1 occurs in the core promoter.

Core promoters
The core promoter is the minimal portion of the promoter required to properly initiate gene transcription. It contains a binding site for RNA polymerase (RNA polymerase I, RNA polymerase II, or RNA polymerase III). Any specific gene’s core promoter is ultimately targeted by a vast network of regulatory factors that contribute to the initiation of transcription by RNA polymerase.

AGCE1 can occur upstream from -25 to -1 base pairs (bp)s of the transcription start site.

Consensus sequences
The consensus sequence is 3'-A/C-T-C/T-5'. By convention, this is 5'-MTY-3' on the mRNA, with T replaced by uracil (U).

The core nucleotides for AGCE1 include 3'-A/C-T-C/T-G-T-G-5', "located between the TATA box and transcription initiation site (positions −25 to −1) is an authentic regulator of human AG transcription."

Human genes
"[T]he core promoter element 1 (AGCE1) of the human angiotensinogen gene (AGT) ... acts as a critical regulator of AGT transcription". GeneID: 183 angiotensinogen (serpin peptidase inhibitor, clade A, member 8) [ Homo sapiens ] has the official symbol AGT.

Gene transcriptions
Within the variation allowed by the consensus sequence, it is likely that the transcriptional activities of an E-box-like motif of two juxtaposed AGCE1 elements have the following ratios: "CTC/CTC and ATC/ATC homozygotes or the CTC/ATC heterozygotes are 2.5 times higher than that of the ATT/ATT homozygote and that the CTC/ATT and ATC/ATT heterozygotes are 1.75 times higher than that of the ATT/ATT homozygote."

Angiotensinogen
Angiotensinogen is an α-2-globulin produced constitutively and released into the circulation mainly by the liver. It is a member of the serpin family, although it is not known to inhibit other enzymes, unlike most serpins. Plasma angiotensinogen levels are increased by plasma corticosteroid, estrogen, thyroid hormone, and angiotensin II levels.

Angiotensin I ([CAS registry number] CAS# 11128-99-7) is formed by the action of renin on angiotensinogen. Renin cleaves the peptide bond between the leucine (Leu) and valine (Val) residues on angiotensinogen, creating the ten-amino acid peptide (des-Asp) angiotensin I. Angiotensin I is converted to angiotensin II (AII) through removal of two C-terminal residues by the enzyme angiotensin-converting enzyme (ACE, or kinase).

A1BG gene transcriptions
An AGCE1 occurs at 3'-ATC-5', in the nucleotides before the transcription start site for the transcription of A1BG, which ends 5 bp upstream from the TSS and is the only AGCE1 within -25 and -1 bp of the TSS. Usually, the AGCE1 spans -20 to -18 bp, when a TATA box is present.

However, for the transcription of A1BG, there is no TATA box and the core nucleotides 3'-GTG-5' do not occur on either side of 3'-ATC-5'.

Hypotheses

 * 1) AGCE1 does not participate in the transcription of A1BG.


text:
User:Egm4313.s12.team2/Report3

Effort Distribution
--Nicolas Ellis 02:45, 22 February 2012 (UTC)

title: User:Egm4313.s12.team2/Report3


text:
User:Bema Self

Greetings all! I'm new and still exploring Wikiversity. I'll come back and fill out my about me section later on. Stay tuned =) --Bema Self 01:28, 21 February 2012 (UTC)

title: User:Bema Self


text:
User talk:Bema Self

What Do You Want to Learn?
I have a multitude of experiences and a vast wealth of knowledge. Let me know what you want to learn, and I'll do my best to make it happen! --Bema Self 01:29, 21 February 2012 (UTC)

title: User talk:Bema Self


title: File:VHDL.Signal.1.A.20120221.pdf


text:
Wikiversity:Colloquium/archives/February 2012

If you think MediaWiki does not offer what you need to boost your learning activities...
...have a look at the current discussion in the mailing list about the Sandbox Server Some ideas from the past what the current software (installation) misses can be read here:
 * Vision/2009
 * Technical needs
 * Topic:Sandbox_Server_0.5, Erkan Yilmaz uses the Chat + Identi.ca 14:23, 2 February 2012 (UTC)

Elementary algebra
Being new here, can I know what is needed to be filled here (in very brief), and I am ready, Elementary algebra. Recently, I improved the Pythagorean theorem to a good level, if you can help me here, that would be very thankful --Extra999 13:57, 9 February 2012 (UTC)
 * Elementary algebra: you could ask e.g. User:Joshua.thomas.bird who created it (what he would like to see there? but since it's a wiki you can also "be bold"). See also here if this shares more light into the darkness and what a change of that page would mean for the other learning resources. Erkan Yilmaz uses the Chat + Identi.ca 18:24, 9 February 2012 (UTC)

MediaWiki 1.19
(Apologies if this message isn't in your language.) The Wikimedia Foundation is planning to upgrade MediaWiki (the software powering this wiki) to its latest version this month. You can help to test it before it is enabled, to avoid disruption and breakage. More information is available in the full announcement. Thank you for your understanding.

Guillaume Paumier, via the Global message delivery system (wrong page? You can fix it.). 14:57, 12 February 2012 (UTC)

wikis which use a (more) formal (peer) review process
Does anyone know some? I found so far scholarpedia, nupedia (is no wiki). There was a request in the German WV and I've collected what I found so far here - but I think there are certainly more. Thanks, Erkan Yilmaz uses the Chat + Identi.ca 01:19, 13 February 2012 (UTC)

Vision for Wikiversity in 2012?
Share your dreams: Vision/2012, Erkan Yilmaz uses the Chat + Identi.ca 01:25, 14 February 2012 (UTC)
 * As a relative newcomer, I would really love to see a push to explore ways that we can facilitate more conversation and discussion as well as simply getting more people through the door. Its a bit disheartening to see a complete lack of conversation going on and there still doesn't seem to be a critical mass of people to make significant strides forward. I'm looking forward to working on some new learning projects, hopefully some of which will be more collaborative in nature. Anyways, I'm glad to be here and I'm interested to see what others have to contribute! MyNameWasTaken 02:31, 14 February 2012 (UTC)
 * What is something useful we could collaborate on? Cheat sheets for example? --HappyCamper 08:12, 14 February 2012 (UTC)
 * Start wiki editing again as I have now completed my N.Diploma in Electrical Engineering. I'll have time finally (",) --Thuvack | talk 13:14, 15 February 2012 (UTC)
 * Thuvack, you are back!!!!! :):):) You are much needed here! --HappyCamper 17:21, 15 February 2012 (UTC)
 * Great, will be awaiting m ore of your blog posts :-) Erkan Yilmaz uses the Chat + Identi.ca 21:29, 15 February 2012 (UTC)
 * I could be interested in working on just about anything really. It's my understanding now that the goal in developing resources here is not simply to present material but to facilitate collaborative construction and development including personal learning. I would love to pick a basic topic that I know only a little about and work with one or more interested people to develop a learning project while exploring the topic for personal education. This stands in sharp contrast to my first project, which is essentially a port of a pre-existing course structure with external readings. MyNameWasTaken 18:56, 15 February 2012 (UTC)
 * Well, for collaboration most often it's just needed to make contact (be it with feedback on the talk page: of a learning resource or a user's), moste often newcomers won't contact first because of shyness. What helps here in my view is a more prominent advertizing of the chat (since people tend to ask their questions more easy in a non-permanent medium, e.g. on-wiki it stays basically forever), Erkan Yilmaz uses the Chat + Identi.ca 20:07, 15 February 2012 (UTC)
 * And another note: asking persons individually will work better than a general mention to all (people tend to not feel adressed somehow or act), Erkan Yilmaz uses the Chat + Identi.ca 21:29, 15 February 2012 (UTC)
 * Cheat sheets are a nice thing, since they save time for people, so there's an immediate win
 * Though they need more specific domain knowledge and not everybody could help create for each topic perhaps (without learning basics)? Erkan Yilmaz uses the Chat + Identi.ca 21:29, 15 February 2012 (UTC)
 * But anyone can just add something to it. Same with Formal letter writing. --HappyCamper 05:58, 16 February 2012 (UTC)

separate vision: immediately implementable + rest
There are now 2 sections: Vision/2012, would be great if you could add e.g. one or more things you miss, Erkan Yilmaz uses the Chat + Identi.ca 20:24, 15 February 2012 (UTC)
 * Perhaps we should also find an easier way of summarizing what is happening on Wikiversity. See UTPA STEM for example. --HappyCamper 06:03, 16 February 2012 (UTC)
 * Sure, will help. A use case could be: someone viewing activity + creating a page + then:
 * 1. telling the "class instructor" (in the hope there will be added more info from the insider himself. Could later also be used as a step in the WV:Feedback. On the other hand: could also backfire sometimes, because people want to stay at begin under the radar, so an email may be better?)
 * 2. making a note in the colloquium (to attract more awareness, see "under the radar" above), Erkan Yilmaz uses the Chat + Identi.ca 18:34, 16 February 2012 (UTC)

but who...
So, another question: who would do these things then? in the end: we can talk big about this and that, but who realizes it and why? Erkan Yilmaz uses the Chat + Identi.ca 18:32, 16 February 2012 (UTC)

Created new category, Freedom of speech
Created new category, for Freedom of speech. This is in conjunction with crosswiki sister project coordination at Commons:Category:Freedom of speech. Please feel free to help populate it, that'd be most appreciated. ;) Cheers, -- Cirt (talk) 06:04, 17 February 2012 (UTC)
 * Category:Freedom of speech

Mobile Version??
I see Wikipedia projects have got mobile versions. When will wikiversity have one?? --Thuvack | talk 18:53, 17 February 2012 (UTC)
 * Well, the needed extension MobileFrontend is already installed at WV, perhaps more info here or here, Erkan Yilmaz uses the Chat + Identi.ca 20:18, 17 February 2012 (UTC)
 * Better: [//en.m.wikiversity.org/wiki/Wikiversity:Colloquium Mobile Colloquium]. -- dark lama  21:05, 17 February 2012 (UTC)

Personal learning environment (PLE) retry question after 4 months
Ca. 4 months ago I asked this. I was wondering if in the new year there would be more feedback... (feedback can be by anyone), Erkan Yilmaz uses the Chat + Identi.ca 05:54, 13 February 2012 (UTC)
 * Sorry for not responding to your first inquiry but I get bogged down with all the other junk I do. Anyways, I don't use my PLE very much anymore unfortunately. I do like how fluid the wiki environment is and how easy it is to structure thoughts and content. I've noticed though that my Google Docs has become my new PLE. It's not quite the same but it roughly serves the same function. With regards to WV in general I haven't been as active as I would like to be. I've been focusing on school lately so that doesn't leave me with much time to do all the things I used to do. :( Devourer09  ( t · c ) 17:39, 21 February 2012 (UTC)


title: File:Table2.1.JPG


text:
University of Florida/Egm4313/s12 Report 3, Problem 3.5

Problem 3.5
Solved by: Andrea Vargas

Problem Statement
Given $$y''-3y'+2y=4x^2-6x^5\!$$

1. Obtain the coefficients of $$ x,x^2,x^3,x^5 \!$$ 2.Verify all equations by long-hand expansion. Use the series before adjusting the indexes.

3. Put the system of equations in an upper triangular matrix.

4. Solve for $$c_0,c_1,c_2,c_3,c_4,c_5\!$$ by using back substitution.

5. Using the initial conditions $$ y(0)=1, y'(0)=0 \!$$ find $$y(x)\!$$ and plot it

Solution
1. To obtain the equations for the coefficients we use the following equation from (1) p7-11:

$$\sum_{j=0}^{3}[c_{j+2}(j+2)(j+1)-3c_{j+1}(j+1)+2c_j]x^j-3c_5(5)x^4+2[c_4x^4+c^5x^5]=4x^2-6x^5 \!$$

Finding the coefficients of $$x\!$$ where $$j=1\!$$:

$$[c_3(3)(2)-3(c_2)(2)+2c_1]x^1=[6c_3-6c_2+2c_1]x\!$$

Finding the coefficients of $$x^2\!$$ where $$j=2\!$$:

$$[c_4(4)(3)-3(c_3)(3)+2c_2]x^2=[12c_4-9c_3+2c_2]x^2\!$$

Finding the coefficients of $$x^3\!$$ where $$j=3\!$$:

$$[c_5(5)(4)-3(c_4)(4)+2c_3]x^3=[20c_4-12c_3+2c_2]x^3\!$$

Finding the coefficients of $$x^5\!$$ where $$j=5\!$$:

$$c_5x^5\!$$

2. To verify all the above equations by long hand expansion we use the following equation from (4) p7-12: $$\sum_{j=2}^{5}c_j\times j\times(j-1)\times x^{j-2}-3\sum_{j=1}^5c_j\times j\times x^{j-1}+2\sum_{j=0}^5c_j\times x^j=4x^2-6x^5\!$$

Finding the coefficients when $$j=0\!$$:

$$2c_0\!$$

Finding the coefficients when $$j=1\!$$:

$$-3c_1+2c_1x_1\!$$

Finding the coefficients when $$j=2\!$$:

$$2c_2-6c_2x+2c_2x^2\!$$

Finding the coefficients when $$j=3\!$$:

$$6c_3x-9c_3x^2+2c_3x^3\!$$

Finding the coefficients when $$j=5\!$$:

$$20c_5x^3-15c_5x^4+2c_5x^5\!$$

By collecting these terms we can compare them to the equations of part 1.

Coefficients of $$x\!$$:

$$[c_3(3)(2)-3(c_2)(2)+2c_1]x^1=[6c_3-6c_2+2c_1]x\!$$

Coefficients of $$x^2\!$$:

$$[c_4(4)(3)-3(c_3)(3)+2c_2]x^2=[12c_4-9c_3+2c_2]x^2\!$$

Coefficients of $$x^3\!$$:

$$[c_5(5)(4)-3(c_4)(4)+2c_3]x^3=[20c_4-12c_3+2c_2]x^3\!$$

Coefficients of $$x^5\!$$:

$$c_5x^5\!$$

We can see that we obtain the same system of equations to solve for the coefficients with both methods.

3.Constructing the coefficients matrix: $$\begin{bmatrix} 2&-3&2&0&0&0 \\ 0&2&-6&6&0&0 \\  0&0&2&-9&12&0 \\  0&0&0&2&-12&20 \\  0&0&0&0&2&-15 \\  0&0&0&0&0&2 \end{bmatrix}\!$$

Then, the system becomes:

$$\begin{bmatrix} 2&-3&2&0&0&0 \\ 0&2&-6&6&0&0 \\  0&0&2&-9&12&0 \\  0&0&0&2&-12&20 \\  0&0&0&0&2&-15 \\  0&0&0&0&0&2 \end{bmatrix} \begin{bmatrix} c_0 \\ c_1 \\ c_2 \\ c_3 \\ c_4 \\ c_5 \\ \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\ 4 \\ 0 \\ 0 \\ -6 \\ \end{bmatrix}\!$$ 4. Solving for the coefficients: $$2c_5=-6 \rightarrow c_5=-3\!$$ $$ 2c_4-(15)(-3)=0\rightarrow c_4=\frac{-45}{2}\!$$ $$ 2c_3-12(-\frac{45}{2})+20(-3)=0 \rightarrow c_3=-105 \!$$ $$ 2c_2-9(-105)+12(-\frac{45}{2})=4 \rightarrow c_2=-\frac{671}{2}\!$$ $$ 2c_1-6(-\frac{671}{2})+6(-105)=0 \rightarrow c_1=-\frac{1383}{2} \!$$ $$ 2c_0-3(-\frac{1383}{2})+2(-\frac{671}{2})=0 \rightarrow c_0=-\frac{2807}{4} \!$$

Particular solution
This yields the particular solution: $$y_p(x)=-3x^5-\frac{45}{2}x^4-105x^3-\frac{671}{2}x^2-\frac{1383}{2}x-\frac{2807}{4}\!$$

Homogeneous Solution
$$y{}''_h-y{}'_h+2y_h=0\!$$ $$\lambda^2-3\lambda+2=0\!$$ Then, we can find the characteristic equation: $$(\lambda-2)(\lambda-1)\!$$ $$\lambda=2,1\!$$ Then the solution for the homogeneous equation becomes: $$y_h(x)=C_1e^{2x}+C_2e^{x}\!$$

General Solution
Using the given initial conditions $$ y(0)=1, y'(0)=0 \!$$ we find the overall solution:

$$ y(x)=y_h+y_p \!$$ $$ y(x)=C_1e^{2x}+C_2e^x-3x^5-\frac{45}{2}x^4-105x^3-\frac{671}{2}x^2-\frac{1383}{2}x-\frac{2807}{4} \!$$ $$ y'(x)=2C_1e^{2x}+C_2e^x-15x^4-90x^3-315x^2-671x-\frac{1383}{2}\!$$ Using the initial conditions to solve for $$ C_1 \!$$ and $$ C_2 \!$$ $$ 1=C_1+C_2-\frac{2807}{4} \!$$ $$ 0=C_1+C_2-\frac{1383}{2} \!$$ $$ C_1=-\frac{45}{4} \; \; \; C_2=714 \!$$ The general solution becomes

$$ y(x)=-\frac{45}{4}e^{2x}+714e^x-3x^5-\frac{45}{2}x^4-105x^3-\frac{671}{2}x^2-\frac{1383}{2}x-\frac{2807}{4} \!$$

Plot
Below is a plot of the solution:

--Egm4313.s12.team11.vargas.aa 05:56, 21 February 2012 (UTC)


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Survey research and design in psychology/Tutorials/Correlation/Instructor notes

General advice to instructors

 * 1) Point out the practice correlation quiz
 * 2) Explain the general steps in conducting a correlational analysis (5 mins)
 * 3) Go over each cell of this table which shows (10 mins): types of correlation in relation to the level of measurement
 * 4) Demonstrate and assist students to conduct the types of correlation exercises (65 mins)
 * 5) Nominal by nominal
 * 6) Gender and God - do this step by step
 * 7) Smoking and Story - run via copied syntax and interpret output
 * 8) Demonstrate Correlation explore and Correlation guess. Challenge students to get 50% (very difficult). (10 mins)
 * 9) Note that one of the plots doesn't display on Correlation guess?
 * 10) Demonstrate and allow class to explore effects of outliers. (10 mins)
 * 11) An alternative is to conduct this as a whiteboard exercise (works well esp. if Java isn't working or time is tight)
 * 12) Demonstrate correlations and non-linear relations (5 mins). If time allows (generally not), set the task as a problem - to draw scatterplots and work out the correlations for each nominated pair and then to explain what is going on (10 mins).


text:
File:Fig3.2.1.png

Licensing
for school project

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Customer service

Customer service is an essential aspect of marketing, which encompasses several principal elements of business, that focus on providing resources to customers who purchase from or do business with their enterprise.

Pages

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title: Customer service


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User:Proof2012.1

''An inspiring from Sondow's formula we have work out the following ... '' ---

Sondow's formula

$$\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{2n(2n+1)}=\gamma$$

Definitions

Let

$$X_A=1-2\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right)+ 3\left(\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+\frac{1}{5\cdot6}-\frac{1}{6\cdot7}\right)-\cdots$$

$$z=\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{n(n+1)(n+2)}$$

$$E=\sum_{n=1}^{\infty}\frac{1}{2^n-1}=\frac{1}{2^1-1}+\frac{1}{2^2-1}+\frac{1}{2^3-1}+\cdots=1.60669...$$

and

$$E_s=\sum_{n=1}^{\infty}\frac{1}{2^n+1}=\frac{1}{2^1+1}+\frac{1}{2^2+1}+\frac{1}{2^3+1}+\cdots=0.76449...$$

--

Overall summary of this paper

Pair 1

---

$$E=\frac{1\cdot2^1}{(2^1-1)(2^2-1)}+\frac{2\cdot2^2}{(2^2-1)(2^3-1)}+ \frac{3\cdot2^3}{(2^3-1)(2^4-1)}+\cdots$$

$$E_s=\frac{1\cdot2^1}{(2^1+1)(2^2+1)}+\frac{2\cdot2^2}{(2^2+1)(2^3+1)}+ \frac{3\cdot2^3}{(2^3+1)(2^4+1)}+\cdots$$

and

Pair 2

---

$$1=\frac{2^1}{(2^1-1)(2^2-1)}+\frac{2^2}{(2^2-1)(2^3-1)}+ \frac{2^3}{(2^3-1)(2^4-1)}+\cdots$$

$$\frac{1}{3}=\frac{2^1}{(2^1+1)(2^2+1)}+\frac{2^2}{(2^2+1)(2^3+1)}+ \frac{2^3}{(2^3+1)(2^4+1)}+\cdots$$

Pair 3

---

$$\frac{E-1-2\ln2-X_A}{2}+\frac{43}{18}=\frac{2\cdot2}{(2^1-1)(2^2-1)}+\frac{4\cdot2^3}{(2^3-1)(2^4-1)} +\frac{6\cdot2^5}{(2^5-1)(2^6-1)}+\cdots$$

$$\frac{X_A+2\ln2-3+E}{2}+\frac{11}{18}=\frac{3\cdot2^2}{(2^2-1)(2^3-1)}+\frac{5\cdot2^4}{(2^4-1)(2^5-1)} +\frac{7\cdot2^6}{(2^6-1)(2^7-1)}+\cdots$$

Pair 4

$$E=\sum_{n=1}^{k\to\infty}\frac{2^n}{2^n-1}-k$$

$$E_s=k-\sum_{n=1}^{k\to\infty}\frac{2^n}{2^n+1}$$

Pair 5

-

$$E-1=\frac{1}{2(2^1-1)}+\frac{1}{2^2(2^2-1)}+\frac{1}{2^3(2^3-1)}+\cdots$$

$$1-E_s=\frac{1}{2(2^1+1)}+\frac{1}{2^2(2^2+1)}+\frac{1}{2^3(2^3+1)}+\cdots$$

-

Random 6

$$\frac{E+E_s}{2}-1=\frac{1}{2(2^2-1)}+\frac{1}{2^2(2^4-1)}+\frac{1}{2^3(2^6-1)}+\cdots$$

$$\frac{3}{2}-2z=E_s\cdot$$

-

---

An inspiring formula from Sondow

$$\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{2n(2n+1)}=\gamma$$

It is from this formula that give me an idea to write ...

Page 1

--

Definition

Let

$$X_2=\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{n(n+1)}$$

$$X_2=\frac{1}{1\cdot2}+2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}\right)+ 3\left(\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\right)+\cdots$$

I have noticed that the sum of the following fractions give the reciprocal power of 2.

e.g,

$$\frac{1}{2\cdot3}+\frac{1}{3\cdot4}=\frac{1}{4}=\frac{1}{2^2}$$

$$\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}=\frac{1}{8}=\frac{1}{2^3}$$

So the above series $$X_2$$ can be rewritten as

$$X_2=\frac{1}{2^1}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+\cdots+\frac{n}{2^n}$$

This series can be evaluated by using Arithmetic-Geometri formula

$$X_2=\sum_{n=1}^{\infty}\frac{n}{2^n}=2$$

or

$$\frac{X_2}{2}=\sum_{n=1}^{\infty}\frac{n}{2^{n+1}}=1$$

Page 2.

-

Definition

Let

$$Y_0=\frac{1}{1\cdot2}-2\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)- 3\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)-\cdots$$

$$Y_1$$ be the Vacca series,

$$Y_1=\gamma=\frac{1}{2}-\frac{1}{3}+2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}\right)+3\left(\frac{1}{8}-\cdots-\frac{1}{15}\right)+\cdots$$

$$Y_2=2\ln2-1=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+\cdots$$

Evaluation

--

Evaluate $$Y_2-Y_0\cdot$$

$$Y_2=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+\frac{1}{5\cdot6}-\frac{1}{6\cdot7}+\frac{1}{7\cdot8}-\frac{1}{8\cdot9}+\cdots$$

Expanded $$Y_0\cdot$$

$$Y_0=\frac{1}{1\cdot2}-\frac{2}{2\cdot3}+\frac{2}{3\cdot4}-\frac{3}{4\cdot5}+\frac{3}{5\cdot6}-\frac{3}{6\cdot7}+\frac{3}{7\cdot8}-\frac{4}{8\cdot9}+\cdots$$

$$Y_2-Y_0=0+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{2}{4\cdot5}-\frac{2}{5\cdot6}+\frac{2}{6\cdot7}-\frac{2}{7\cdot8}+\frac{3}{8\cdot9}-\frac{3}{9\cdot10}+\cdots$$

Factorise [$$Y_2-Y_0\cdot$$] and we have

$$Y_2-Y_0=1\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)+ 2\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)+ 3\left(\frac{1}{8\cdot9}-\cdots-\frac{1}{15\cdot16}\right)+\cdots$$

Evaluate $$Y_1-[Y_2-Y_0]\cdot$$

$$Y_1=\gamma=1\left(\frac{1}{2}-\frac{1}{3}\right)+2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}\right)+3\left(\frac{1}{8}-\cdots-\frac{1}{15}\right)+\cdots$$

$$Y_2-Y_0=1\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)+ 2\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)+ 3\left(\frac{1}{8\cdot9}-\cdots-\frac{1}{15\cdot16}\right)+\cdots$$

e.g,

$$\frac{1}{2}-\frac{1}{2\cdot3}=\frac{1}{3}$$

$$-\frac{1}{3}+\frac{1}{3\cdot4}=-\frac{1}{4}$$

$$Y_1-(Y_2-Y_0)=1\left(\frac{1}{3}-\frac{1}{4}\right)+2\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}\right)+3\left(\frac{1}{9}-\cdots-\frac{1}{16}\right)+\cdots$$

Therefore we have

$$Y_1-Y_2+Y_0=1\left(\frac{1}{3}-\frac{1}{4}\right)+2\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}\right)+3\left(\frac{1}{9}-\cdots-\frac{1}{16}\right)+\cdots$$

I have noticed that if I added this series [$$Y_1-Y_2+Y_0$$] to $$Y_1$$ I will end up with series $$\frac{X_2}{2}$$ which I have evaluated in page 1.

Evaluate of [$$Y_1-Y_2+Y_0\cdot$$] + [$$Y_1\cdot$$]

$$Y_1-Y_2+Y_0=1\left(\frac{1}{3}-\frac{1}{4}\right)+2\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}\right)+3\left(\frac{1}{9}-\cdots-\frac{1}{16}\right)+\cdots$$

$$Y_1=1\left(\frac{1}{2}-\frac{1}{3}\right)+2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}\right)+3\left(\frac{1}{8}-\cdots-\frac{1}{15}\right)+\cdots$$

e.g,

$$1\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{2}-\frac{1}{3}\right)=\frac{1}{2}-\frac{1}{4}$$

$$2\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}\right)= \frac{1}{4}-\frac{1}{8}$$

and so on ...

$$Y_1-Y_2+Y_0+Y_1=1\left(\frac{1}{2}-\frac{1}{4}\right)+2\left(\frac{1}{4}-\frac{1}{8}\right)+3\left(\frac{1}{8}-\frac{1}{16}\right)+\cdots$$

Simplify, we have

$$2Y_1-Y_2+Y_0=1\left(\frac{1}{4}\right)+2\left(\frac{1}{8}\right)+3\left(\frac{1}{16}\right)+\cdots$$

Where 4, 8, 16, ... are power of 2. Rewrite the series and we have

$$2Y_1-Y_2+Y_0=1\left(\frac{1}{2^2}\right)+2\left(\frac{1}{2^3}\right)+3\left(\frac{1}{2^4}\right)+\cdots$$

Recall from page 1, where $$X_2\cdot$$

$$2=\frac{1}{2^1}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+\cdots$$

Divide 2 by both sides and we have

$$1=\frac{1}{2^2}+\frac{2}{2^3}+\frac{3}{2^4}+\frac{4}{2^5}+\cdots$$

Therefore we have

$$2Y_1-Y_2+Y_0=1\cdot$$

From here we solve for $$Y_0$$

$$Y_0=1-2Y_1+Y_2\cdot$$

Recall from definition, where

$$Y_1=\gamma\cdot$$  and

$$Y_2=2\ln2-1\cdot$$

Substitute these values and we solve for $$Y_0$$

$$Y_0=1-2\gamma+2\ln2-1\cdot$$

Therefore we have,

$$Y_0=2\ln2-2\gamma\cdot$$

So we have worked out the close-form for $$Y_0$$

Recall

$$Y_0=\frac{1}{1\cdot2}-2\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)- 3\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)-\cdots$$

and therefore we have

$$2\ln2-2\gamma=\frac{1}{1\cdot2}-2\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)- 3\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)-\cdots$$

'''Let us rewrite this series in a neater form ''' $$2\ln2-2\gamma=\frac{1}{1\cdot2}-2\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)- 3\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)-\cdots$$

Add -1 to both sides

$$2\ln2-2\gamma-1=-\frac{1}{1\cdot2}-2\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)- 3\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)-\cdots$$

Then multiply by -1 to both sides and we have

$$2\gamma+1-2\ln2=\frac{1}{1\cdot2}+2\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)+ 3\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)+ \cdots$$

From this series, we can derive some other interesting close-form for other series.

Recall from definition,

$$Y_2=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+\frac{1}{5\cdot6}-\frac{1}{6\cdot7}+\frac{1}{7\cdot8}-\frac{1}{8\cdot9}+\cdots$$

Let us add these two series [$$2\gamma+1-2\ln2\cdot$$] + [$$Y_2\cdot$$]

Evaluation of [$$2\gamma+1-2\ln2\cdot$$] + [$$Y_2\cdot$$]

-

Expand [$$2\gamma+1-2\ln2\cdot$$] and we have

$$2\gamma+1-2\ln2=\frac{1}{1\cdot2}+\frac{2}{2\cdot3}-\frac{2}{3\cdot4}+\frac{3}{4\cdot5}-\frac{3}{5\cdot6}+\frac{3}{6\cdot7}-\frac{3}{7\cdot8}+\frac{4}{8\cdot9}-\cdots$$

[$$2\gamma+1-2\ln2\cdot$$] + [$$Y_2\cdot$$] and we have

$$2\gamma+1-2\ln2+Y_2=1+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{2}{4\cdot5}-\frac{2}{5\cdot6}+\frac{2}{6\cdot7}-\frac{2}{7\cdot8}+\cdots$$

Factorise the terms

$$2\gamma+1-2\ln2+Y_2=1+1\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)+2\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)+\cdots$$

Recall $$Y_2=2\ln2-1\cdot$$,

substitute $$Y_2\cdot$$ into [$$2\gamma+1-2\ln2+Y_2\cdot$$] and we have,

$$2\gamma-1=1\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)+2\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)+\cdots$$

Recall from definition

$$Y_1=\gamma=1\left(\frac{1}{2}-\frac{1}{3}\right)+2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}\right)+3\left(\frac{1}{8}-\cdots-\frac{1}{15}\right)+\cdots$$

Let us evaluate [$$Y_1\cdot$$] - [$$2\gamma-1\cdot$$]

e.g,

$$\frac{1}{2}-\frac{1}{2\cdot3}=\frac{1}{3}$$

$$-\frac{1}{3}+\frac{1}{3\cdot4}=-\frac{1}{4}$$, and so on ...

and we have

$$1-\gamma=1\left(\frac{1}{3}-\frac{1}{4}\right)+2\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}\right)+3\left(\frac{1}{9}-\cdots-\frac{1}{16}\right)+\cdots$$

Question to be asked

We know the close-form of the following series.

$$\gamma=1\left(\frac{1}{2}-\frac{1}{3}\right)+2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}\right)+3\left(\frac{1}{8}-\cdots-\frac{1}{15}\right)+\cdots$$

and

$$1-\gamma=1\left(\frac{1}{3}-\frac{1}{4}\right)+2\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}\right)+3\left(\frac{1}{9}-\cdots-\frac{1}{16}\right)+\cdots$$

except

$$U_X=1\left(1-\frac{1}{2}\right)+2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\right)+3\left(\frac{1}{7}-\cdots-\frac{1}{14}\right)+\cdots$$

Before we move on to the next page we can derive another close-form from series [$$2\gamma+1-2\ln2$$]

Let introduce another new series namely $$Y_a\cdot$$, where

$$Y_a=\frac{1}{1\cdot2}-2\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)+ 3\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)-\cdots$$

Recall [$$2\gamma+1-2\ln2\cdot$$]

$$2\gamma+1-2\ln2=\frac{1}{1\cdot2}+2\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)+ 3\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)+ \cdots$$

Evaluate [$$2\gamma+1-2\ln2\cdot$$] + [$$Y_a\cdot$$]

--

As for this series $$Y_a$$, I don't remember how I derived it.

Evaluation of [$$2\gamma+1-2\ln2\cdot$$] + [$$Y_a\cdot$$], and we have

$$2\gamma+1-2\ln2+Y_a=1+6\left(\frac{1}{4\cdot5}-\cdot-\frac{1}{7\cdot8}\right)+10\left(\frac{1}{16\cdot17}-\cdots-\frac{1}{31\cdot32}\right)+\left(\frac{1}{64\cdot65}-\cdots-\frac{1}{127\cdot128}\right)+\cdots$$

Substitue in the value for $$Y_a\cdot$$

$$2\gamma+1-2\ln2+2\ln2-1=1+6\left(\frac{1}{4\cdot5}-\cdot-\frac{1}{7\cdot8}\right)+10\left(\frac{1}{16\cdot17}-\cdots-\frac{1}{31\cdot32}\right)+\left(\frac{1}{64\cdot65}-\cdots-\frac{1}{127\cdot128}\right)+\cdots$$

Simplify

$$2\gamma=1+6\left(\frac{1}{4\cdot5}-\cdots-\frac{1}{7\cdot8}\right)+10\left(\frac{1}{16\cdot17}-\cdots-\frac{1}{31\cdot32}\right)+14\left(\frac{1}{64\cdot65}-\cdots-\frac{1}{127\cdot128}\right)+\cdots$$

Divide 2 by both sides and we have

$$\gamma=\frac{1}{2}+3\left(\frac{1}{4\cdot5}-\cdots-\frac{1}{7\cdot8}\right)+5\left(\frac{1}{16\cdot17}-\cdots-\frac{1}{31\cdot32}\right)+7\left(\frac{1}{64\cdot65}-\cdots-\frac{1}{127\cdot128}\right)+\cdots$$

Page 3.

--

Definition

Let

$$X_E=1-2\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right)- 3\left(\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+\frac{1}{5\cdot6}-\frac{1}{6\cdot7}\right)-\cdots$$

Recall from page 2,

$$Y_2=2\ln2-1=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+\cdots$$

$$Y_1=\gamma=\frac{1}{2}-\frac{1}{3}+2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}\right)+3\left(\frac{1}{8}-\cdots-\frac{1}{15}\right)+\cdots$$

Evaluation

--

As for the series $$Y_2$$ I am going to rewrite it as,

$$2-2\ln2=1-\frac{1}{1\cdot2}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{4\cdot5}-\cdots$$

We will called it $$Y_3\cdot$$, so we have

$$Y_3=1-\frac{1}{1\cdot2}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{4\cdot5}-\cdots$$

Evaluate $$Y_3-X_E\cdot$$

-

Let us expand $$X_E\cdot$$

$$X_E=1-\frac{2}{1\cdot2}+\frac{2}{2\cdot3}-\frac{3}{3\cdot4}+\frac{3}{4\cdot5}- \frac{3}{5\cdot6}+\frac{3}{6\cdot7}-\frac{4}{7\cdot8}+\cdots$$

Recall $$Y_3\cdot$$

$$Y_3=1-\frac{1}{1\cdot2}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{4\cdot5}-\cdots$$

Let us evaluate $$Y_3-X_E\cdot$$,

$$Y_3-X_E=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{2}{3\cdot4}-\frac{2}{4\cdot5}+ \frac{2}{5\cdot6}-\frac{2}{6\cdot7}+\frac{3}{7\cdot8}-\cdots$$

Factorise and we have

$$Y_3-X_E=\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right) +2\left(\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+\frac{1}{5\cdot6}-\frac{1}{6\cdot7}\right)+3\left(\frac{1}{7\cdot8}-\cdots-\frac{1}{14\cdot15}\right)+\cdots$$

We will called it $$Y_4=Y_3-X_E\cdot$$

Recall

$$Y_1=1\left(\frac{1}{2}-\frac{1}{3}\right)+2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}\right)+3\left(\frac{1}{8}-\cdots-\frac{1}{15}\right)+\cdots$$

Let us evaluate [$$Y_4\cdot$$] + [$$Y_1\cdot$$] and we have

E.g,

$$\frac{1}{2}+\frac{1}{2}=1$$

$$-\frac{1}{3}-\frac{1}{2\cdot3}=-\frac{1}{2}$$

and so on ...

and again we add [$$Y_4+Y_1\cdot$$] to $$Y_1\cdot$$ and we have

$$Y_4+Y_1=1\left(1-\frac{1}{2}\right)+2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\right)+3\left(\frac{1}{7}-\cdots-\frac{1}{14}\right)+\cdots$$

$$Y_4+2Y_1=1\left(1-\frac{1}{3}\right)+2\left(\frac{1}{3}-\frac{1}{7})+(\frac{1}{7}-\frac{1}{15}\right)+\cdots$$

Let us expand on the RHS, and we have

$$Y_4+2Y_1=1-\frac{1}{3}+\frac{2}{3}-\frac{2}{7}+\frac{3}{7}-\frac{3}{15}+\cdots$$

Simplify and we have

$$Y_4+2Y_1=1+\frac{1}{3}+\frac{1}{7}+\frac{1}{15}+\cdots$$

Change the denominator to power of 2

$$Y_4+2Y_1=\frac{1}{2^1-1}+\frac{1}{2^2-1}+\frac{1}{2^3-1}+\frac{1}{2^4-1}+\cdots$$

On the RHS series, we have Erdos-Borwein constant (E), and we have

$$Y_4+2Y_1=E\cdot$$

Now we substiute the values for $$Y_4\cdot$$ and $$Y_1\cdot$$

Where $$Y_4=2-2\ln2-X_E\cdot$$ and $$Y_1=\gamma\cdot$$ and we have

$$2-2\ln2-X_E+2\gamma=E\cdot$$

Solve for $$X_E\cdot$$

$$2-2\ln2+2\gamma-E=X_E\cdot$$

So now we have worked out the close-form $$X_E\cdot$$ (sort of!)

Recall from definition

$$X_E=1-2\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right)- 3\left(\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+\frac{1}{5\cdot6}-\frac{1}{6\cdot7}\right)-\cdots$$

$$2-2\ln2+2\gamma-E=1-2\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right)- 3\left(\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+\frac{1}{5\cdot6}-\frac{1}{6\cdot7}\right)-\cdots$$

Tidy this up,

Add -2 and multiply by -1 to both sides we have,

$$E-2\gamma+2\ln2=1+2\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right)+ 3\left(\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+\frac{1}{5\cdot6}-\frac{1}{6\cdot7}\right)-\cdots$$

But this is not the series that we are interested in ...

Recall

$$Y_4+Y_1=1\left(1-\frac{1}{2}\right)+2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\right)+3\left(\frac{1}{7}-\cdots-\frac{1}{14}\right)+\cdots$$

This is the one we are interested in.

$$Y_4+Y_1=1\left(1-\frac{1}{2}\right)+2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\right)+3\left(\frac{1}{7}-\cdots-\frac{1}{14}\right)+\cdots$$

$$-2\ln2+2-X_E+\gamma=1\left(1-\frac{1}{2}\right)+2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\right)+3\left(\frac{1}{7}-\cdots-\frac{1}{14}\right)+\cdots$$

and where $$X_E=2-2\ln2+2\gamma-E\cdot$$, substitute this in the above and we have

$$-2\ln2+2-2+2\ln2-2\gamma+E+\gamma=1\left(1-\frac{1}{2}\right)+2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\right)+3\left(\frac{1}{7}-\cdots-\frac{1}{14}\right)+\cdots$$

Simplify and we have

$$E-\gamma=1\left(1-\frac{1}{2}\right)+2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\right)+3\left(\frac{1}{7}-\cdots-\frac{1}{14}\right)+\cdots$$

Page 4.

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Recall from the definition on Page 1

$$X_2=\frac{1}{1\cdot2}+2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}\right)+ 3\left(\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\right)+\cdots$$

$$X_2=2\cdot$$

and on page 2.

$$2\gamma+1-2\ln2=\frac{1}{1\cdot2}+2\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)+3\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)+\cdots$$

Let us called the above series $$Y_3\cdot$$, so

$$Y_3=2\gamma+1-2\ln2\cdot$$

Let us evaluate [$$X_2\cdot$$] + [$$X_3\cdot$$], and we have

$$X_2+X_3=1+2\left(\frac{2}{2\cdot3}\right)+3\left(\frac{2}{4\cdot5}+\frac{2}{6\cdot7}\right)+\cdots$$

Divide both sides by 2 and we have

$$\frac{X_2+X_3}{2}=\frac{1}{2}+2\left(\frac{1}{2\cdot3}\right)+3\left(\frac{1}{4\cdot5}+\frac{1}{6\cdot7}\right)+\cdots$$

Substitute in $$X_2\cdot$$ and $$X_3\cdot$$

$$\frac{2+2\gamma+1-2\ln2}{2}=\frac{1}{2}+2\left(\frac{1}{2\cdot3}\right)+3\left(\frac{1}{4\cdot5}+\frac{1}{6\cdot7}\right)+\cdots$$ Simplify and we have

$$\frac{3}{2}+\gamma-\ln2=\frac{1}{2}+2\left(\frac{1}{2\cdot3}\right)+3\left(\frac{1}{4\cdot5}+\frac{1}{6\cdot7}\right)+\cdots$$

We can re-write series $$\left[\frac{3}{2}+\gamma-\ln2\right]$$ as

$$2+\gamma-\ln2=1+2\left(\frac{1}{2}-\frac{1}{3}\right)+3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}\right)+\cdots$$

Let us evaluate [$$X_2-X_3\cdot$$] and we have

$$X_2-X_3=0+2\left(\frac{2}{3\cdot4}\right)+3\left(\frac{2}{5\cdot6}+\frac{2}{7\cdot8}\right)+\cdots$$

Divide both sides by 2 and we have

$$\frac{X_2-X_3}{2}=2\left(\frac{1}{3\cdot4}\right)+3\left(\frac{1}{5\cdot6}+\frac{1}{7\cdot8}\right)+\cdots$$

Substitute in $$X_2\cdot$$ and $$X_3\cdot$$

$$\frac{2-2\gamma-1+2\ln2}{2}=2\left(\frac{1}{3\cdot4}\right)+3\left(\frac{1}{5\cdot6}+\frac{1}{7\cdot8}\right)+\cdots$$

Simplify and we have,

$$\frac{1}{2}-\gamma+\ln2=2\left(\frac{1}{3\cdot4}\right)+3\left(\frac{1}{5\cdot6}+\frac{1}{7\cdot8}\right)+\cdots$$

We can re-write series $$\left[\frac{1}{2}-\gamma+\ln2\right]$$ as

$$1-\gamma+\ln2=1\left(1-\frac{1}{2}\right)+2\left(\frac{1}{3}-\frac{1}{4}\right)+3\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}\right)+\cdots$$

Evaluate $$\left[2+\gamma-\ln2\right]-\left[1-\gamma+\ln2\right]$$ and we have

$$2-\gamma+\ln2=2\left(\frac{1}{3}\right)+3\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{7}\right)+ 4\left(\frac{1}{9}-\frac{1}{10}+\cdots+\frac{1}{15}\right)+5\left(\frac{1}{11}-\frac{1}{12}+\cdots+\frac{1}{31}\right)+\cdots$$

Page 5.

Recall definition from page 3

$$X_E=1-2\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right)- 3\left(\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+\frac{1}{5\cdot6}-\frac{1}{6\cdot7}\right)-\cdots$$

page 2

$$Y_0=\frac{1}{1\cdot2}-2\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)- 3\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)-\cdots$$

and

page 1

$$X_2=\frac{1}{2^1}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+\cdots$$

Where,

$$X_E=2\gamma-2\ln2+2-E\cdot$$ ,

$$Y_0=2\ln2-2\gamma\cdot$$

and

$$X_2=2\cdot$$

We begin be evaluating $$[X_E] + [Y_0]\cdot$$

-

$$X_E+Y_0=\frac{3}{2}-2\left(\frac{1}{1\cdot2}-\frac{1}{3\cdot4}\right)- \left(\frac{1}{3\cdot4}-\frac{1}{7\cdot8}\right)-\left(\frac{1}{7\cdot8}-\frac{1}{15\cdot16}\right)-\cdots$$

Substitute in $$X_E\cdot$$ and $$Y_0\cdot$$

$$2\gamma-2\ln2+2-E+2\ln2-2\gamma=\frac{3}{2}-2\left(\frac{1}{1\cdot2}-\frac{1}{3\cdot4}\right)- \left(\frac{1}{3\cdot4}-\frac{1}{7\cdot8}\right)-\left(\frac{1}{7\cdot8}-\frac{1}{15\cdot16}\right)-\cdots$$

simplify and we have

$$2-E=\frac{3}{2}-2\left(\frac{1}{1\cdot2}-\frac{1}{3\cdot4}\right)- \left(\frac{1}{3\cdot4}-\frac{1}{7\cdot8}\right)-\left(\frac{1}{7\cdot8}-\frac{1}{15\cdot16}\right)-\cdots$$

Where $$X_2\cdot$$ we multiply it by -1 to have

$$-X_2=-\frac{1}{2^1}-\frac{2}{2^2}-\frac{3}{2^3}-\frac{4}{2^4}-\cdots$$

Let use evaluate [$$-X_2\cdot$$]+[$$2-E\cdot$$]

e.g,

$$\frac{3}{2}-\frac{1}{2}=1$$

$$\frac{1}{1\cdot2}-\frac{1}{3\cdot4}+\frac{1}{2^2}=\frac{2}{3}$$

$$\frac{1}{3\cdot4}-\frac{1}{7\cdot8}+\frac{1}{2^3}=\frac{4}{21}$$

and so on ...

$$-X_2+2-E=1-2\left(\frac{2}{3}\right)-3\left(\frac{4}{21}\right)-4\left(\frac{8}{105}\right)-\cdots$$

I have noticed that

$$\frac{2}{3}=\frac{2}{(2^1-1)(2^2-1)}$$

$$\frac{4}{21}=\frac{4}{(2^2-1)(2^3-1)}$$

and so on ...

Substitute in $$X_2\cdot$$

$$-2+2-E=1-\frac{2\cdot2}{(2^1-1)(2^2-1)}-\frac{3\cdot2^2}{(2^2-1)(2^3-1)}-\frac{4\cdot2^3}{(2^3-1)(2^4-1)}-\cdots$$

Simplify and we have

$$-E=1-\frac{2\cdot2}{(2^1-1)(2^2-1)}-\frac{3\cdot2^2}{(2^2-1)(2^3-1)}-\frac{4\cdot2^3}{(2^3-1)(2^4-1)}-\cdots$$

Tidy it up by add -1 and then multiply by -1 to both sides and we have

$$E+1=\frac{2\cdot2}{(2^1-1)(2^2-1)}+\frac{3\cdot2^2}{(2^2-1)(2^3-1)}+\frac{4\cdot2^3}{(2^3-1)(2^4-1)}+\cdots$$

Page 6.

-

Recall from page 2

$$Y_1=\gamma=1\left(\frac{1}{2}-\frac{1}{3}\right)+2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}\right)+3\left(\frac{1}{8}-\cdots-\frac{1}{15}\right)+\cdots$$

The result from page 3

We called this series $$Y_E\cdot$$

$$Y_E=E-\gamma=1\left(1-\frac{1}{2}\right)+2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\right)+3\left(\frac{1}{7}-\cdots-\frac{1}{14}\right)+\cdots$$

Result from page 2

We call this series $$Y_X\cdot$$

$$Y_X=2\gamma-1=1\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)+2\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)+\cdots$$

and from page 1

$$X_2=\frac{1}{2^1}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+\cdots$$

We rewrite this series as,

$$\frac{X_2}{2}=\frac{1}{2^2}+\frac{2}{2^3}+\frac{3}{2^4}+\frac{4}{2^5}+\cdots$$

We begin by evaluating [$$Y_E-Y_1\cdot$$]

- $$Y_E-Y_1=1\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right)+2\left(\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+\frac{1}{5\cdot6}-\frac{1}{6\cdot7}\right)+\cdots$$

Substittue in $$Y_E\cdot$$ and $$Y_1\cdot$$ and we have

$$E-2\gamma=1\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right)+2\left(\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+\frac{1}{5\cdot6}-\frac{1}{6\cdot7}\right)+\cdots$$

and we called this series $$Y_Y\cdot$$

$$Y_Y=E-2\gamma\cdot$$

Now we evaluate $$Y_X\cdot$$ + $$Y_Y\cdot$$

and we have

$$Y_X+Y_Y=1\left(\frac{1}{1\cdot2}-\frac{1}{3\cdot4}\right) +2\left(\frac{1}{3\cdot4}-\frac{1}{7\cdot8}\right)+\left(\frac{1}{7\cdot8}-\frac{1}{15\cdot16}\right)+\cdots$$

Substiute in $$Y_X\cdot$$ and $$Y_Y\cdot$$

$$E-2\gamma+2\gamma-1=1\left(\frac{1}{1\cdot2}-\frac{1}{3\cdot4}\right) +2\left(\frac{1}{3\cdot4}-\frac{1}{7\cdot8}\right)+\left(\frac{1}{7\cdot8}-\frac{1}{15\cdot16}\right)+\cdots$$

Simplify and we have

$$E-1=1\left(\frac{1}{1\cdot2}-\frac{1}{3\cdot4}\right) +2\left(\frac{1}{3\cdot4}-\frac{1}{7\cdot8}\right)+\left(\frac{1}{7\cdot8}-\frac{1}{15\cdot16}\right)+\cdots$$

We called this series $$Y_Z\cdot$$

$$Y_Z=E-1\cdot$$

Let us now evaluate $$\left[Y_Z+\frac{X_2}{2}\right]$$

Recall

$$X_2=\frac{1}{2^1}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+\cdots$$

$$Y_Z+\frac{X_2}{2}=\frac{2}{3}+2\left(\frac{4}{21}\right)+3\left(\frac{8}{105}\right)+\cdots$$

Substitute in for $$Y_Z\cdot$$and $$\frac{X_2}{2}\cdot$$ and we have

$$E-1+1=\frac{2}{3}+2\left(\frac{4}{21}\right)+3\left(\frac{8}{105}\right)+\cdots$$

Simplify and we have

$$E=\frac{1\cdot2}{(2^1-1)(2^2-1)}+\frac{2\cdot2^2}{(2^2-1)(2^3-1)}+\frac{3\cdot2^3}{(2^3-1)(2^4-1)}+\cdots$$

Page 7

-

Recall from Page 5, the two results

$$E+1=\frac{2\cdot2}{(2^1-1)(2^2-1)}+\frac{3\cdot2^2}{(2^2-1)(2^3-1)}+\frac{4\cdot2^3}{(2^3-1)(2^4-1)}+\cdots$$

Let A = E + 1

and Page 6

$$E=\frac{1\cdot2}{(2^1-1)(2^2-1)}+\frac{2\cdot2^2}{(2^2-1)(2^3-1)}+\frac{3\cdot2^3}{(2^3-1)(2^4-1)}+\cdots$$

Let B = E

We begin by evaluating [A] - [B]

$$A-B=\frac{2}{(2^1-1)(2^2-1)}+\frac{2^2}{(2^2-1)(2^3-1)}+\frac{2^3}{(2^3-1)(2^4-1)}+\cdots$$

Substitute in for [A] and [B]

$$E+1-E=\frac{2}{(2^1-1)(2^2-1)}+\frac{2^2}{(2^2-1)(2^3-1)}+\frac{2^3}{(2^3-1)(2^4-1)}+\cdots$$

Simplify and we have

$$1=\frac{2}{(2^1-1)(2^2-1)}+\frac{2^2}{(2^2-1)(2^3-1)}+\frac{2^3}{(2^3-1)(2^4-1)}+\cdots$$

Recall from page 6

$$Y_Z=E-1\cdot$$

$$E-1=1\left(\frac{1}{1\cdot2}-\frac{1}{3\cdot4}\right) +2\left(\frac{1}{3\cdot4}-\frac{1}{7\cdot8}\right)+\left(\frac{1}{7\cdot8}-\frac{1}{15\cdot16}\right)+\cdots$$

Expand and collect the same denominators,

$$E-1=\frac{1}{1\cdot2}-\frac{1}{3\cdot4} +\frac{2}{3\cdot4}-\frac{2}{7\cdot8}+\frac{3}{7\cdot8}-\frac{3}{15\cdot16}+\cdots$$

Simplify and we have

$$E-1=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{7\cdot8}+\frac{1}{15\cdot16}+\cdots$$

Change the denominators to power of 2, and we have

$$E-1=\frac{1}{2^1(2^1-1)}+\frac{1}{2^2(2^2-1)}+\frac{1}{2^3(2^3-1)}+\frac{1}{2^4(2^4-1)}+\cdots$$

Page 8

--

Definition

Let

$$X_A=1-2\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right)+ 3\left(\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+\frac{1}{5\cdot6}-\frac{1}{6\cdot7}\right)-\cdots$$

and recall $$X_E\cdot$$ series from page 3

$$X_E=1-2\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right)- 3\left(\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+\frac{1}{5\cdot6}-\frac{1}{6\cdot7}\right)-\cdots$$

$$X_E=2\gamma-2\ln2+2-E\cdot$$

and recall $$Y_0$$ from Page 2

$$Y_0=\frac{1}{1\cdot2}-2\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)- 3\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)-\cdots$$

$$Y_0=2\ln2-2\gamma\cdot$$

and $$Y_a$$ from page 2

$$Y_a=\frac{1}{1\cdot2}-2\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)+ 3\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)-\cdots$$

$$Y_a=2\ln2-1\cdot$$

$$P_X=\frac{8}{9}=\frac{2}{2^2}+\frac{4}{2^4}+\frac{6}{2^6}+\cdots$$

$$P_Y=\frac{11}{18}=\frac{3}{2^3}+\frac{5}{2^5}+\frac{7}{2^7}+\cdots$$

We begin by evaluating $$X_A\cdot$$+$$X_E\cdot$$

-

$$X_A+X_E=2-4\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right)- 8\left(\frac{1}{7\cdot8}-\cdots-\frac{1}{14\cdot15}\right)-12\left(\frac{1}{31\cdot32}-\cdots-\frac{1}{62\cdot63}\right)-\cdots$$

Divide both sides by 2, we have

$$\frac{X_A+X_E}{2}=1-2\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right)- 4\left(\frac{1}{7\cdot8}-\cdots-\frac{1}{14\cdot15}\right)-6\left(\frac{1}{31\cdot32}-\cdots-\frac{1}{62\cdot63}\right)-\cdots$$

and evaluating of $$Y_0\cdot$$+$$Y_a\cdot$$

-

$$Y_0+Y_a=1-4\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)- 8\left(\frac{1}{8\cdot9}-\cdots-\frac{1}{15\cdot16}\right)-12\left(\frac{1}{32\cdot33}-\cdots-\frac{1}{63\cdot64}\right)-\cdots$$

Divide by 2

$$\frac{Y_0+Y_a}{2}=\frac{1}{2}-2\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)- 4\left(\frac{1}{8\cdot9}-\cdots-\frac{1}{15\cdot16}\right)-6\left(\frac{1}{32\cdot33}-\cdots-\frac{1}{63\cdot64}\right)-\cdots$$

Let $$M=\frac{X_A+X_E}{2}\cdot$$ and $$N=\frac{Y_0+Y_a}{2}\cdot$$

Evaluate M + N

--

$$M+N=\frac{3}{2}-2\left(\frac{1}{1\cdot2}-\frac{1}{3\cdot4}\right)- 4\left(\frac{1}{7\cdot8}-\frac{1}{15\cdot16}\right)-6\left(\frac{1}{31\cdot32}-\frac{1}{63\cdot64}\right)-\cdots$$

Recall

$$P_X=\frac{8}{9}=\frac{2}{2^2}+\frac{4}{2^4}+\frac{6}{2^6}+\cdots$$

Now we evaluate [M + N] - [$$P_X\cdot$$]

$$M+N-P_X=\frac{3}{2}-2\left(\frac{2}{3}\right)-4\left(\frac{8}{105}\right) -6\left(\frac{32}{1953}\right)-\cdots$$

Tidy it up, add $$-\frac{3}{2}$$ and then multiply by -1 to both sides and we have

$$-M-N+P_X+\frac{3}{2}=\frac{2\cdot2}{(2^1-1)(2^2-1)}+\frac{4\cdot2^3}{(2^3-1)(2^4-1)} +\frac{6\cdot2^5}{(2^5-1)(2^6-1)}+\cdots$$

Now substitute the values of M, N, $$P_X$$ on to the LHS and we have

$$\frac{-X_A-X_E-Y_a-Y_0}{2}+\frac{8}{9}+\frac{3}{2}=\frac{2\cdot2}{(2^1-1)(2^2-1)}+\frac{4\cdot2^3}{(2^3-1)(2^4-1)} +\frac{6\cdot2^5}{(2^5-1)(2^6-1)}+\cdots$$

$$\frac{-X_A-2\gamma+2\ln2-2+E-2\ln2+1-2\ln2+2\gamma}{2}+\frac{8}{9}+\frac{3}{2}=\frac{2\cdot2}{(2^1-1)(2^2-1)}+\frac{4\cdot2^3}{(2^3-1)(2^4-1)} +\frac{6\cdot2^5}{(2^5-1)(2^6-1)}+\cdots$$

$$\frac{E-1-2\ln2-X_A}{2}+\frac{43}{18}=\frac{2\cdot2}{(2^1-1)(2^2-1)}+\frac{4\cdot2^3}{(2^3-1)(2^4-1)} +\frac{6\cdot2^5}{(2^5-1)(2^6-1)}+\cdots$$

Evaluating $$X_E-X_A\cdot$$

--

$$X_E-X_A=-6\left(\frac{1}{3\cdot4}-\frac{1}{6\cdot7}\right)- 10\left(\frac{1}{15\cdot16}-\cdots-\frac{1}{30\cdot31}\right)-14\left(\frac{1}{63\cdot64}-\cdots-\frac{1}{126\cdot127}\right)-\cdots$$

Divide both sides by 2, we have

$$\frac{X_E-X_A}{2}=-3\left(\frac{1}{3\cdot4}-\frac{1}{6\cdot7}\right)- 5\left(\frac{1}{15\cdot16}-\cdots-\frac{1}{30\cdot31}\right)-7\left(\frac{1}{63\cdot64}-\cdots-\frac{1}{126\cdot127}\right)-\cdots$$

and evaluating of $$Y_0-Y_a\cdot$$

---

$$Y_0+Y_a=-6\left(\frac{1}{4\cdot5}-\frac{1}{7\cdot8}\right)- 10\left(\frac{1}{16\cdot17}-\cdots-\frac{1}{31\cdot32}\right)-14\left(\frac{1}{64\cdot65}-\cdots-\frac{1}{127\cdot128}\right)-\cdots$$

Divide by 2

$$\frac{Y_0-Y_a}{2}=-3\left(\frac{1}{4\cdot5}-\frac{1}{7\cdot8}\right)- 5\left(\frac{1}{16\cdot17}-\cdots-\frac{1}{31\cdot32}\right)-7\left(\frac{1}{64\cdot65}-\cdots-\frac{1}{127\cdot128}\right)-\cdots$$

Let $$P=\frac{X_E-X_A}{2}\cdot$$ and $$Q=\frac{Y_0-Y_a}{2}\cdot$$

Evaluate P + Q

$$P+Q=-3\left(\frac{1}{4\cdot5}-\frac{1}{7\cdot8}\right)- 5\left(\frac{1}{16\cdot17}-\frac{1}{31\cdot32}\right)-7\left(\frac{1}{63\cdot64}-\frac{1}{127\cdot128}\right)-\cdots$$

Recall

$$P_Y=\frac{11}{18}=\frac{3}{2^3}+\frac{5}{2^5}+\frac{7}{2^7}+\cdots$$

Now we evaluate [P + Q] - [$$P_Y$$]

$$P+Q-P_Y=-3\left(\frac{4}{21}\right)-5\left(\frac{16}{465}\right) -7\left(\frac{64}{8001}\right)-\cdots$$

Tidy it up, multiply by -1 to both sides and we have

$$-P-Q+P_Y=3\left(\frac{4}{21}\right)+5\left(\frac{16}{465}\right) +7\left(\frac{64}{8001}\right)+\cdots$$

Now substitute for the values of P, Q, $$P_Y\cdot$$ on to the LHS and we have

$$\frac{-2\gamma+2\ln2-2+E+X_A-2\ln2+2\gamma+2\ln2-1}{2}+\frac{11}{18}=3\left(\frac{4}{21}\right)+5\left(\frac{16}{465}\right) +7\left(\frac{64}{8001}\right)+\cdots$$

Simplify and we have

$$\frac{X_A+2\ln2-3+E}{2}+\frac{11}{18}=\frac{3\cdot2^2}{(2^2-1)(2^3-1)}+\frac{5\cdot2^4}{(2^4-1)(2^5-1)} +\frac{7\cdot2^6}{(2^6-1)(2^7-1)}+\cdots$$

Page 9

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Recall from page 8 the two results

Let this series be R

$$\frac{E-1-2\ln2-X_A}{2}+\frac{43}{18}=\frac{2\cdot2}{(2^1-1)(2^2-1)}+\frac{4\cdot2^3}{(2^3-1)(2^4-1)} +\frac{6\cdot2^5}{(2^5-1)(2^6-1)}+\cdots$$

and

and let this series be T

$$\frac{X_A+2\ln2-3+E}{2}+\frac{11}{18}=\frac{3\cdot2^2}{(2^2-1)(2^3-1)}+\frac{5\cdot2^4}{(2^4-1)(2^5-1)} +\frac{7\cdot2^6}{(2^6-1)(2^7-1)}+\cdots$$

Evaluate R + T

$$R+T=\frac{2\cdot2^1}{(2^1-1)(2^2-1)}+\frac{3\cdot2^2}{(2^2-1)(2^3-1)} +\frac{4\cdot2^3}{(2^3-1)(2^4-1)}+\cdots$$

$$\frac{2E-4}{2}+\frac{54}{18}=\frac{2\cdot2^1}{(2^1-1)(2^2-1)}+\frac{3\cdot2^2}{(2^2-1)(2^3-1)} +\frac{4\cdot2^3}{(2^3-1)(2^4-1)}+\cdots$$

Simplify and we have

$$E+1=\frac{2\cdot2^1}{(2^1-1)(2^2-1)}+\frac{3\cdot2^2}{(2^2-1)(2^3-1)} +\frac{4\cdot2^3}{(2^3-1)(2^4-1)}+\cdots$$

This is the result from page 5

Page 10.

-

Show that,

$$\frac{1}{9}=\frac{2}{(2^1-1)(2^2-1)(2^3-1)}+\frac{2^2}{(2^2-1)(2^3-1)(2^4-1)}+\frac{2^3}{(2^3-1)(2^4-1)(2^5-1)}+\cdots$$

and

$$\frac{1}{45}=\frac{2}{(2^1+1)(2^2+1)(2^3+1)}+\frac{2^2}{(2^2+1)(2^3+1)(2^4+1)}+\frac{2^3}{(2^3+1)(2^4+1)(2^5+1)}+\cdots$$

The above two series are found by trial and error method

Page 11.

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Definition

Let

$$\frac{X_2}{2}=\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{2n(n+1)}=1$$

and

Sondow's formula

$$S_2=\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{2n(2n+1)}=\gamma$$

Evaluate $$\left(\frac{X_2}{2}-S_2\right)$$

$$\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{2n(n+1)}- \sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{2n(2n+1)}$$

$$\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{2n(n+1)}-\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{2n(2n+1)}=1-\gamma$$

simplify and we have

$$\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{(2n+1)(2n+2)}=1-\gamma$$

Evaluate $$\left(\frac{S_2}{2}-\frac{X_2}{4}\right)$$

$$\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{2n(2n+1)}-\frac{1}{2}\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{2n(n+1)}=\gamma-\frac{1}{2}$$

simplify and we have

$$\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{2n(2n+1)(2n+2)}=\gamma-\frac{1}{2}$$

---

Page 12

-

An inspiring formula from Sondow

$$\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{2n(2n+1)}=\gamma$$

Let

$$z=\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{n(n+1)(n+2)}$$

Expand the series

$$z=\frac{1}{1\cdot2\cdot3}+2\left(\frac{1}{2\cdot3\cdot4}+ \frac{1}{3\cdot4\cdot5}\right)+3\left(\frac{1}{4\cdot5\cdot6}+\cdots+ \frac{1}{7\cdot8\cdot9}\right))+\cdots$$

Re-write it as

$$z-\frac{1}{1\cdot2\cdot3}=2\left(\frac{1}{2\cdot3\cdot4}+ \frac{1}{3\cdot4\cdot5}\right)+3\left(\frac{1}{4\cdot5\cdot6}+\cdots+ \frac{1}{7\cdot8\cdot9}\right))+\cdots$$

Recall from Page 1

$$X_2=\frac{1}{2^1}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+\cdots=2$$

Re-write it as

$$\frac{X_2}{2}-\frac{1}{4}=\frac{2}{2^3}+\frac{3}{2^4}+\frac{4}{2^5}+\cdots=\frac{3}{4}$$

Let us evaluate $$\left(\frac{X_2}{2}-\frac{1}{4}\right)-\left(z-\frac{1}{6}\right)$$

Substitute in $$X_2=2$$

$$\frac{3}{4}-z+\frac{1}{6}=2\left(\frac{1}{2^3}-\frac{1}{2\cdot3\cdot4}- \frac{1}{3\cdot4\cdot5}\right)+3\left(\frac{1}{2^4}-\frac{1}{4\cdot5\cdot6}-\cdots-\frac{1}{7\cdot8\cdot9}\right))-\cdots$$

Simplify and we have

$$\frac{11}{12}-z=2\left(\frac{1}{15}\right)+3\left(\frac{2}{45}\right)+ 4\left(\frac{4}{153}\right)+\cdots$$

I have noticed that

$$\frac{2^0}{(2^1+1)(2^2+1)}=\frac{1}{15}$$

$$\frac{2^1}{(2^2+1)(2^3+1)}=\frac{2}{45}$$

$$\frac{2^2}{(2^3+1)(2^4+1)}=\frac{4}{153}$$

and son on ...

Re-write the above series as

$$\frac{11}{12}-z=\frac{2\cdot2^0}{(2^1+1)(2^2+1)}+\frac{3\cdot2^1}{(2^2+1)(2^3+1)}+ \frac{4\cdot2^2}{(2^3+1)(2^4+1)}+\cdots=z_1$$

Let $$z_1 = \frac{11}{12}-z$$

Let us introduce $$z_2$$

$$z_2=\frac{2^0}{(2^1+1)(2^2+1)}+\frac{2^1}{(2^2+1)(2^3+1)}+ \frac{2^2}{(2^3+1)(2^4+1)}+\cdots=\frac{1}{6}$$

[The series $$z_2$$ I found this through trial and error method, but I have verify it.]

Evaluate $$z_1-z_2$$

--

Recall

$$z_1=\frac{2\cdot2^0}{(2^1+1)(2^2+1)}+\frac{3\cdot2^1}{(2^2+1)(2^3+1)}+ \frac{4\cdot2^2}{(2^3+1)(2^4+1)}+\cdots$$

$$z_1-z_2=\frac{1\cdot2^0}{(2^1+1)(2^2+1)}+\frac{2\cdot2^1}{(2^2+1)(2^3+1)}+ \frac{2^2}{(2^3+1)(2^4+1)}+\cdots$$

Substitute in for the values of $$z_1$$ and $$z_2$$

$$\frac{3}{4}-z=\frac{1\cdot2^0}{(2^1+1)(2^2+1)}+\frac{2\cdot2^1}{(2^2+1)(2^3+1)}+ \frac{3\cdot2^2}{(2^3+1)(2^4+1)}+\cdots$$

Multiply 2 by both sides and we have

$$\frac{3}{2}-2z=\frac{1\cdot2^1}{(2^1+1)(2^2+1)}+\frac{2\cdot2^2}{(2^2+1)(2^3+1)}+ \frac{3\cdot2^3}{(2^3+1)(2^4+1)}+\cdots$$

Nice pair

---

Let

$$E_s=\sum_{n=1}^{\infty}\frac{1}{2^n+1}=\frac{1}{2^1+1}+\frac{1}{2^2+1}+\frac{1}{2^3+1}+\cdots=0.7644...$$

Let $$E_s=\frac{3}{2}-2z$$

$$E_s=\frac{1\cdot2^1}{(2^1+1)(2^2+1)}+\frac{2\cdot2^2}{(2^2+1)(2^3+1)}+ \frac{3\cdot2^3}{(2^3+1)(2^4+1)}+\cdots$$

Recall from Page 6

$$E=\frac{1\cdot2^1}{(2^1-1)(2^2-1)}+\frac{2\cdot2^2}{(2^2-1)(2^3-1)}+ \frac{3\cdot2^3}{(2^3-1)(2^4-1)}+\cdots$$

[Therefore we have a nice pair!]

A quick summary of Page 12

-

Definition

Where E is Erdos-Borwein constant and $$E_s$$ is ? constant.

$$E=\sum_{n=1}^{\infty}\frac{1}{2^n-1}=\frac{1}{2^1-1}+\frac{1}{2^2-1}+\frac{1}{2^3-1}+\cdots=1.60669...$$

$$E_s=\sum_{n=1}^{\infty}\frac{1}{2^n+1}=\frac{1}{2^1+1}+\frac{1}{2^2+1}+\frac{1}{2^3+1}+\cdots=0.76449...$$

We have

$$E=\frac{1\cdot2^1}{(2^1-1)(2^2-1)}+\frac{2\cdot2^2}{(2^2-1)(2^3-1)}+ \frac{3\cdot2^3}{(2^3-1)(2^4-1)}+\cdots$$

$$E_s=\frac{1\cdot2^1}{(2^1+1)(2^2+1)}+\frac{2\cdot2^2}{(2^2+1)(2^3+1)}+ \frac{3\cdot2^3}{(2^3+1)(2^4+1)}+\cdots$$

and

$$1=\frac{2^1}{(2^1-1)(2^2-1)}+\frac{2^2}{(2^2-1)(2^3-1)}+ \frac{2^3}{(2^3-1)(2^4-1)}+\cdots$$

$$\frac{1}{3}=\frac{2^1}{(2^1+1)(2^2+1)}+\frac{2^2}{(2^2+1)(2^3+1)}+ \frac{2^3}{(2^3+1)(2^4+1)}+\cdots$$

Continuing from Page 12

-

Let

Recall from Page 2

$$1-\gamma=1\left(\frac{1}{3}-\frac{1}{4}\right)+2\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}\right)+3\left(\frac{1}{9}-\cdots-\frac{1}{16}\right)+\cdots$$

New definition ($$U_s$$)

$$U_s=1\left(\frac{1}{4}-\frac{1}{5}\right)+2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{8}-\frac{1}{9}\right)+3\left(\frac{1}{10}-\cdots-\frac{1}{17}\right)+\cdots$$

Evaluate $$[1-\gamma]+U_s$$

-

$$1-\gamma+U_s=1\left(\frac{1}{3}-\frac{1}{5}\right)+2\left(\frac{1}{5}-\frac{1}{9}\right)+ 3\left(\frac{1}{9}-\frac{1}{17}\right)+\cdots$$

$$1-\gamma+U_s=\frac{1}{3}-\frac{1}{5}+\frac{2}{5}-\frac{2}{9}+\frac{3}{9}-\frac{3}{17}+\cdots$$

$$1-\gamma+U_s=\frac{1}{3}+\frac{1}{5}+\frac{1}{9}+\frac{1}{17}+\cdots$$

Where

$$\frac{1}{3}=\frac{1}{2^1+1}$$

$$\frac{1}{5}=\frac{1}{2^2+1}$$

$$\frac{1}{9}=\frac{1}{2^3+1}$$

and so on ...

Re-write it as

$$1-\gamma+U_s=\frac{1}{2^1+1}+\frac{1}{2^2+1}+\frac{1}{2^3+1}+\frac{1}{2^4+1}\cdots$$

Recall from Definition from Page 12 (summary)

$$E_s=\frac{1}{2^1+1}+\frac{1}{2^2+1}+\frac{1}{2^3+1}+\cdots$$

Therefore we have

$$1-\gamma+U_s=E_s\cdot$$

Make $$U_s\cdot$$ the subject

$$U_s=E_s+\gamma-1\cdot$$

Evaluation

Re-write $$U_s=E_s+\gamma-1$$ as $$U_s=E_s+\gamma-\frac{3}{3}$$

Recall from page 12 (summary)

$$\frac{1}{3}=\frac{2^1}{(2^1+1)(2^2+1)}+\frac{2^2}{(2^2+1)(2^3+1)}+ \frac{2^3}{(2^3+1)(2^4+1)}+\cdots$$

Multiply by 3 and we have

$$\frac{3}{3}=\frac{3\cdot2^1}{(2^1+1)(2^2+1)}+\frac{3\cdot2^2}{(2^2+1)(2^3+1)}+ \frac{3\cdot2^3}{(2^3+1)(2^4+1)}+\cdots$$ Recall from page 12 (summary)

$$E_s=\frac{1\cdot2^1}{(2^1+1)(2^2+1)}+\frac{2\cdot2^2}{(2^2+1)(2^3+1)}+ \frac{3\cdot2^3}{(2^3+1)(2^4+1)}+\cdots$$

Let us evaluate $$E_s-\frac{3}{3}$$ and we have

$$E_s-\frac{3}{3}=\frac{(1-3)\cdot2^1}{(2^1+1)(2^2+1)}+\frac{(2-3)\cdot2^2}{(2^2+1)(2^3+1)}+ \frac{(3-3)\cdot2^3}{(2^3+1)(2^4+1)}+\frac{(4-3)\cdot2^4}{(2^4+1)(2^5+1)}\cdots$$

Let $$z_3=E_s-\frac{3}{3}$$

$$z_3=\frac{(1-3)\cdot2^1}{(2^1+1)(2^2+1)}+\frac{(2-3)\cdot2^2}{(2^2+1)(2^3+1)}+ \frac{(3-3)\cdot2^3}{(2^3+1)(2^4+1)}+\frac{(4-3)\cdot2^4}{(2^4+1)(2^5+1)}\cdots$$

$$U_s=\gamma+z_3$$

Recall $$U_s$$

$$U_s=1\left(\frac{1}{4}-\frac{1}{5}\right)+2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{8}-\frac{1}{9}\right)+3\left(\frac{1}{10}-\cdots-\frac{1}{17}\right)+\cdots$$

Finally we have

$$\gamma+z_3=1\left(\frac{1}{4}-\frac{1}{5}\right)+2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{8}-\frac{1}{9}\right)+3\left(\frac{1}{10}-\cdots-\frac{1}{17}\right)+\cdots$$

Continuing from Page 12

Recall from definitions

Let

$$T_1=E$$ and  $$T_2=E_s$$

$$T_1=\sum_{n=1}^{\infty}\frac{1}{2^n-1}$$

$$T_2=\sum_{n=1}^{\infty}\frac{1}{2^n+1}$$

We have calculated the following series by trial and error method

Let $$T_3=E$$

$$T_3=\sum_{n=1}^{k\to\infty}\frac{2^n}{2^n-1}-k$$

and $$T_4=E_s$$

$$T_4=k-\sum_{n=1}^{k\to\infty}\frac{2^n}{2^n+1}$$

Here we will show that our $$T_3$$ and $$T_4$$ are correct.

Let us show that $$T_1+T_2=T_3+T_4$$

Evaluate of $$T_1+T_2$$

---

$$T_1+T_2=\sum_{n=1}^{\infty}\frac{1}{2^n-1}+\sum_{n=1}^{\infty}\frac{1}{2^n+1}$$

simplify and we have

$$T_1+T_2=\sum_{n=1}^{\infty}\frac{2^{n+1}}{2^{2n}-1}$$

Evaluate of $$T_3+T_4$$

--

$$T_3+T_4=\sum_{n=1}^{k\to\infty}\frac{2^n}{2^n-1}-k+k-\sum_{n=1}^{k\to\infty}\frac{2^n}{2^n+1}$$

simplify

$$T_3+T_4=\sum_{n=1}^{k\to\infty}\frac{2^n}{2^n-1}-\sum_{n=1}^{k\to\infty}\frac{2^n}{2^n+1}$$

$$T_3+T_4=\sum_{n=1}^{k\to\infty}\frac{2^n(2^n+1-2^n+1)}{(2^n-1)(2^n+1)}$$

$$T_3+T_4=\sum_{n=1}^{k\to\infty}\frac{2^n(2)}{2^{2n}-1}$$

Therefore we have

$$T_3+T_4=\sum_{n=1}^{k\to\infty}\frac{2^{n+1}}{2^{2n}-1}$$

Finally we have show that

$$T_1+T_2=T_3+T_4$$

Page 13

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Recall from Page 7

$$E-1=\frac{1}{2(2^1-1)}+\frac{1}{2^2(2^2-1)}+\frac{1}{2^3(2^3-1)}+\cdots$$

Using trial and error method we found close-form for series $$1-E_s$$

$$1-E_s=\frac{1}{2(2^1+1)}+\frac{1}{2^2(2^2+1)}+\frac{1}{2^3(2^3+1)}+\cdots$$

Evaluate $$\left[E-1\right]-\left[1-E_s\right]$$

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Simplify and we have

$$\frac{E+E_s}{2}-1=\frac{1}{2(2^2-1)}+\frac{1}{2^2(2^4-1)}+\frac{1}{2^3(2^6-1)}+\cdots$$

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