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Question Number 227872    Answers: 0   Comments: 0

a=(dv/dt)=((v−u)/t) a=((v−u)/t) v−u=at v=u+at.....eq(i) 6.Distance(S).it is the how far a body t

$$\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{v}−\mathrm{u}}{\mathrm{t}} \\ $$$$\mathrm{a}=\frac{\mathrm{v}−\mathrm{u}}{\mathrm{t}} \\ $$$$\mathrm{v}−\mathrm{u}=\mathrm{at} \\ $$$$\mathrm{v}=\mathrm{u}+\mathrm{at}.....\mathrm{eq}\left(\mathrm{i}\right) \\ $$$$\mathrm{6}.\mathrm{Distance}\left(\mathrm{S}\right).\mathrm{it}\:\mathrm{is}\:\mathrm{the}\:\mathrm{how}\:\mathrm{far}\:\mathrm{a}\:\mathrm{body}\:\mathrm{t}\: \\ $$

Question Number 227871    Answers: 0   Comments: 0

Question Number 227870    Answers: 0   Comments: 0

Question Number 227868    Answers: 0   Comments: 1

Question Number 227865    Answers: 0   Comments: 0

Question. Suppose that we consider the expression (1+(1/n))^n . as n→∞ the sequence converges to a unique irrational constant denoted by ′′e′′ def. lim_(n→∞) (1+(1/n))^n =e Let Sequence A_h =(1+(1/h))^h , h>1 Let (1, 1+(1/n) , 1+(1/n) , .... , +1+(1/n)_( _(n times) ) ) ((1+n(1+(1/n)))/(n+1))>(1+(1/n))^(n/(n+1)) ⇒1+(1/(n+1))>(1+(1/n))^(n/(n+1)) (AM>GM ) ∴ A_n <A_(n+1) A_n is monotonic increase by the Bernoulli Inequality 1+nx<(1+x)^n 1+n∙(1/n)<(1+(1/n))^n and (1+(1/n))^n =1+n((1/n))+((n(n−1))/(2!))((1/n))^2 +...<1+1+(1/2)+((1/2))^2 +((1/2))^3 ....=3 ∴ (1+(1/n))^n bounded above 3 2<(1+(1/n))^n <3 I can accept the logic so far but how can we be absolutely certain that the limit ′′lim_(n→∞) (1+(1/n))^n converges to that specific value 2.718281828... Note that deriving ′′e′′ through the infinite series Σ_(h=0) ^∞ (1/(h!)) is strictly forbidden here. Using another defintion of ′′e′′ to prove this limit would be a premature conclusion and a logical contradiction..... After all the interval between 2 and 3 is densely packed with infinitely many rational and irrational numbers. why must it be that particular one.....???

$$\mathrm{Question}. \\ $$$$\: \\ $$$$\mathrm{Suppose}\:\mathrm{that}\:\mathrm{we}\:\mathrm{consider}\:\mathrm{the}\:\mathrm{expression}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}} .\: \\ $$$$\mathrm{as}\:{n}\rightarrow\infty\:\mathrm{the}\:\mathrm{sequence}\:\mathrm{converges}\:\mathrm{to}\:\mathrm{a}\:\mathrm{unique} \\ $$$$\mathrm{irrational}\:\mathrm{constant}\:\mathrm{denoted}\:\mathrm{by}\:''\boldsymbol{\mathrm{e}}'' \\ $$$$\mathrm{def}.\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}} ={e} \\ $$$$\mathrm{Let}\:\mathrm{Sequence}\:{A}_{{h}} =\left(\mathrm{1}+\frac{\mathrm{1}}{{h}}\right)^{{h}} \:,\:{h}>\mathrm{1}\: \\ $$$$\: \\ $$$$\mathrm{Let}\:\left(\mathrm{1},\underset{\underset{{n}\:\mathrm{times}} {\:}} {\underbrace{\:\mathrm{1}+\frac{\mathrm{1}}{{n}}\:,\:\mathrm{1}+\frac{\mathrm{1}}{{n}}\:,\:....\:,\:+\mathrm{1}+\frac{\mathrm{1}}{{n}}}}\:\right) \\ $$$$\frac{\mathrm{1}+{n}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)}{{n}+\mathrm{1}}>\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{\frac{{n}}{{n}+\mathrm{1}}} \Rightarrow\mathrm{1}+\frac{\mathrm{1}}{{n}+\mathrm{1}}>\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}/\left({n}+\mathrm{1}\right)} \:\:\left(\mathrm{AM}>\mathrm{GM}\:\right) \\ $$$$\therefore\:{A}_{{n}} <{A}_{{n}+\mathrm{1}} \:\: \\ $$$${A}_{{n}} \:\mathrm{is}\:\mathrm{monotonic}\:\mathrm{increase} \\ $$$$\mathrm{by}\:\mathrm{the}\:\mathrm{Bernoulli}\:\mathrm{Inequality}\:\mathrm{1}+{nx}<\left(\mathrm{1}+{x}\right)^{{n}} \: \\ $$$$\mathrm{1}+{n}\centerdot\frac{\mathrm{1}}{{n}}<\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}} \:\mathrm{and}\: \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}} =\mathrm{1}+{n}\left(\frac{\mathrm{1}}{{n}}\right)+\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}!}\left(\frac{\mathrm{1}}{{n}}\right)^{\mathrm{2}} +...<\mathrm{1}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} ....=\mathrm{3} \\ $$$$\therefore\:\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}} \:\mathrm{bounded}\:\mathrm{above}\:\:\mathrm{3}\: \\ $$$$\mathrm{2}<\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}} <\mathrm{3}\:\: \\ $$$$\: \\ $$$$\mathrm{I}\:\mathrm{can}\:\mathrm{accept}\:\mathrm{the}\:\mathrm{logic}\:\mathrm{so}\:\mathrm{far} \\ $$$$\mathrm{but}\:\mathrm{how}\:\mathrm{can}\:\mathrm{we}\:\mathrm{be}\:\mathrm{absolutely}\:\mathrm{certain} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{limit}\:''\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}} \:\mathrm{converges}\:\mathrm{to}\:\mathrm{that}\:\mathrm{specific}\:\mathrm{value} \\ $$$$\mathrm{2}.\mathrm{718281828}...\:\mathrm{Note}\:\mathrm{that}\:\mathrm{deriving}\:''\boldsymbol{\mathrm{e}}''\:\mathrm{through}\:\mathrm{the}\:\mathrm{infinite}\:\mathrm{series} \\ $$$$\underset{{h}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{h}!}\:\:\mathrm{is}\:\mathrm{strictly}\:\mathrm{forbidden}\:\mathrm{here}.\:\mathrm{Using}\:\mathrm{another} \\ $$$$\mathrm{defintion}\:\mathrm{of}\:\:''\boldsymbol{\mathrm{e}}''\:\mathrm{to}\:\mathrm{prove}\:\mathrm{this}\:\mathrm{limit}\:\mathrm{would}\:\mathrm{be}\:\mathrm{a}\:\mathrm{premature}\:\mathrm{conclusion} \\ $$$$\mathrm{and}\:\mathrm{a}\:\mathrm{logical}\:\mathrm{contradiction}..... \\ $$$$\mathrm{After}\:\mathrm{all}\:\mathrm{the}\:\mathrm{interval}\:\mathrm{between}\:\mathrm{2}\:\mathrm{and}\:\mathrm{3}\:\mathrm{is}\:\mathrm{densely}\:\mathrm{packed}\:\mathrm{with} \\ $$$$\mathrm{infinitely}\:\mathrm{many}\:\mathrm{rational}\:\mathrm{and}\:\mathrm{irrational}\:\mathrm{numbers}. \\ $$$$\mathrm{why}\:\mathrm{must}\:\mathrm{it}\:\mathrm{be}\:\mathrm{that}\:\mathrm{particular}\:\mathrm{one}.....??? \\ $$

Question Number 227855    Answers: 0   Comments: 1

Question Number 227853    Answers: 1   Comments: 0

prove:Σ_(i=1) ^n (ln((i+1)/i))^2 <(n/(n+1))

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{prove}:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{ln}\frac{{i}+\mathrm{1}}{{i}}\right)^{\mathrm{2}} <\frac{{n}}{{n}+\mathrm{1}} \\ $$

Question Number 227862    Answers: 2   Comments: 0

if a_(n+1) =(3/(4−a_n )) for n≥1 and a_1 =0, find a_n in terms of n.

$${if}\:{a}_{{n}+\mathrm{1}} =\frac{\mathrm{3}}{\mathrm{4}−{a}_{{n}} }\:{for}\:{n}\geqslant\mathrm{1}\:{and}\:{a}_{\mathrm{1}} =\mathrm{0}, \\ $$$${find}\:{a}_{{n}} \:{in}\:{terms}\:{of}\:{n}. \\ $$

Question Number 227847    Answers: 1   Comments: 4

Question Number 227838    Answers: 0   Comments: 0

Question Number 227837    Answers: 0   Comments: 2

Can someone spot a mistake in my result or have a different approach? Question below

$${Can}\:{someone}\:{spot}\:{a}\:{mistake}\:{in}\:{my}\:{result} \\ $$$${or}\:{have}\:{a}\:{different}\:{approach}?\: \\ $$$${Question}\:{below} \\ $$

Question Number 227831    Answers: 0   Comments: 4

If the roof beam load is given by 10kN/m and the roof unit area load is 1.2kN/m^2 : (i) Determine the imposed loads on beams 1 and 2 (BM−1 and BM−2) (ii) Load to be carried by column 1 and 2 (CT−1 and CT−2)

$${If}\:{the}\:{roof}\:{beam}\:{load}\:{is}\:{given}\:{by} \\ $$$$\mathrm{10}{kN}/{m}\:{and}\:{the}\:{roof}\:{unit}\:{area}\:{load} \\ $$$${is}\:\mathrm{1}.\mathrm{2}{kN}/{m}^{\mathrm{2}} : \\ $$$$\left({i}\right)\:{Determine}\:{the}\:{imposed}\:{loads}\:{on} \\ $$$${beams}\:\mathrm{1}\:{and}\:\mathrm{2}\:\left({BM}−\mathrm{1}\:{and}\:{BM}−\mathrm{2}\right) \\ $$$$\left({ii}\right)\:{Load}\:{to}\:{be}\:{carried}\:{by}\:{column}\:\mathrm{1}\:{and} \\ $$$$\mathrm{2}\:\left({CT}−\mathrm{1}\:{and}\:{CT}−\mathrm{2}\right) \\ $$

Question Number 227824    Answers: 2   Comments: 0

A team that is 100 meters long is moving forward in a straight line at a constant speed. A messenger runs at a constant speed from the rear of the team to the front to deliver a message. Then without changing the speed he runs back to the rear of the team. By the time he returns to the rear the team has advanced 240 meters. How far has the messenger traveled?

$$\mathrm{A}\:\mathrm{team}\:\mathrm{that}\:\mathrm{is}\:\mathrm{100}\:\mathrm{meters}\:\mathrm{long}\:\mathrm{is} \\ $$$$\mathrm{moving}\:\mathrm{forward}\:\mathrm{in}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line}\: \\ $$$$\mathrm{at}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{speed}.\:\mathrm{A}\:\mathrm{messenger}\: \\ $$$$\mathrm{runs}\:\mathrm{at}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{speed}\:\mathrm{from}\:\mathrm{the}\: \\ $$$$\mathrm{rear}\:\mathrm{of}\:\mathrm{the}\:\mathrm{team}\:\mathrm{to}\:\mathrm{the}\:\mathrm{front}\:\mathrm{to}\: \\ $$$$\mathrm{deliver}\:\mathrm{a}\:\mathrm{message}.\:\mathrm{Then}\:\mathrm{without}\: \\ $$$$\mathrm{changing}\:\mathrm{the}\:\mathrm{speed}\:\mathrm{he}\:\mathrm{runs}\:\mathrm{back}\:\mathrm{to}\: \\ $$$$\mathrm{the}\:\mathrm{rear}\:\mathrm{of}\:\mathrm{the}\:\mathrm{team}.\:\mathrm{By}\:\mathrm{the}\:\mathrm{time}\:\mathrm{he}\: \\ $$$$\mathrm{returns}\:\mathrm{to}\:\mathrm{the}\:\mathrm{rear}\:\mathrm{the}\:\mathrm{team}\:\mathrm{has} \\ $$$$\mathrm{advanced}\:\mathrm{240}\:\mathrm{meters}.\:\mathrm{How}\:\mathrm{far}\:\mathrm{has}\: \\ $$$$\mathrm{the}\:\mathrm{messenger}\:\mathrm{traveled}? \\ $$

Question Number 227823    Answers: 2   Comments: 1

Question Number 227811    Answers: 1   Comments: 1

Question Number 227807    Answers: 1   Comments: 0

The issue related to posting of borderless tables is fixed in latest version.

$$\mathrm{The}\:\mathrm{issue}\:\mathrm{related}\:\mathrm{to}\:\mathrm{posting}\:\mathrm{of} \\ $$$$\mathrm{borderless}\:\mathrm{tables}\:\mathrm{is}\:\mathrm{fixed}\:\mathrm{in}\:\mathrm{latest}\:\mathrm{version}. \\ $$

Question Number 227806    Answers: 1   Comments: 0

Question Number 227805    Answers: 1   Comments: 0

∫_0 ^(+∞) sin(x)sin(x^2 )dx

$$\int_{\mathrm{0}} ^{+\infty} {sin}\left({x}\right){sin}\left({x}^{\mathrm{2}} \right){dx} \\ $$

Question Number 227796    Answers: 2   Comments: 0

Question Number 227795    Answers: 2   Comments: 0

Question Number 227792    Answers: 1   Comments: 0

Prove that, ((a/b)+(b/a))^2 +((b/c)+(c/b))^2 +((c/a)+(a/c))^2 −4 =((a/b)+(b/a))((b/c)+(c/b))((c/a)+(a/c))

$$\:\:{Prove}\:{that}, \\ $$$$\:\left(\frac{{a}}{{b}}+\frac{{b}}{{a}}\right)^{\mathrm{2}} +\left(\frac{{b}}{{c}}+\frac{{c}}{{b}}\right)^{\mathrm{2}} +\left(\frac{{c}}{{a}}+\frac{{a}}{{c}}\right)^{\mathrm{2}} −\mathrm{4} \\ $$$$\:=\left(\frac{{a}}{{b}}+\frac{{b}}{{a}}\right)\left(\frac{{b}}{{c}}+\frac{{c}}{{b}}\right)\left(\frac{{c}}{{a}}+\frac{{a}}{{c}}\right) \\ $$

Question Number 227787    Answers: 0   Comments: 3

Question Number 227789    Answers: 2   Comments: 1

Question Number 227782    Answers: 2   Comments: 1

Question Number 227781    Answers: 0   Comments: 0

Prove that there exists a bijection between the closed interval [0,1] and the open interval (0,1) f(x)= { (((1/2)x , x=((1/2))^n )),((x otherwise )) :} f(x) maps [0,1] one to one and onto [0,1) let g(x)=1−x g(x) maps [0,1) one to onen onto (0,1] f(x) maps (0,1] one to one and onto (0,1) ∴f(g(f(x))) maps [0,1] one to one and onto (0,1) ■

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{there}\:\mathrm{exists}\:\mathrm{a}\:\mathrm{bijection}\:\mathrm{between}\:\mathrm{the} \\ $$$$\mathrm{closed}\:\mathrm{interval}\:\left[\mathrm{0},\mathrm{1}\right]\:\mathrm{and}\:\mathrm{the}\:\mathrm{open}\:\mathrm{interval}\:\left(\mathrm{0},\mathrm{1}\right) \\ $$$${f}\left({x}\right)=\begin{cases}{\frac{\mathrm{1}}{\mathrm{2}}{x}\:,\:{x}=\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} }\\{{x}\:\:\:\mathrm{otherwise}\:}\end{cases}\:\:{f}\left({x}\right)\:\mathrm{maps}\:\left[\mathrm{0},\mathrm{1}\right]\:\mathrm{one}\:\mathrm{to}\:\mathrm{one}\:\mathrm{and}\:\mathrm{onto}\:\left[\mathrm{0},\mathrm{1}\right)\:\:\:\:\: \\ $$$$\mathrm{let}\:{g}\left({x}\right)=\mathrm{1}−{x}\:\:\mathrm{g}\left({x}\right)\:\mathrm{maps}\:\left[\mathrm{0},\mathrm{1}\right)\:\mathrm{one}\:\mathrm{to}\:\mathrm{onen}\:\mathrm{onto}\:\left(\mathrm{0},\mathrm{1}\right] \\ $$$${f}\left({x}\right)\:\mathrm{maps}\:\left(\mathrm{0},\mathrm{1}\right]\:\mathrm{one}\:\mathrm{to}\:\mathrm{one}\:\mathrm{and}\:\mathrm{onto}\:\left(\mathrm{0},\mathrm{1}\right) \\ $$$$\therefore{f}\left(\mathrm{g}\left({f}\left({x}\right)\right)\right)\:\mathrm{maps}\:\left[\mathrm{0},\mathrm{1}\right]\:\mathrm{one}\:\mathrm{to}\:\mathrm{one}\:\mathrm{and}\:\mathrm{onto}\:\left(\mathrm{0},\mathrm{1}\right) \\ $$$$\: \\ $$$$\blacksquare \\ $$

Question Number 227737    Answers: 0   Comments: 0

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