Debbyug: I doubt.....when u hv not seen mine....

Debbyug: I doubt.....when u hv not seen mine....

Arithmetic:
§tan^-1{(1-x)/(1+x)}dx.
:::SOLUTION:::
Integrating by part, we have;
u=tan^-1{(1-x)/(1+x)}, v=x, dx=-du(1+x²),
§udv=uv-§vdu,
=tan^-1{(1-x)/(1+x)}.x-§-x/(1+x²)dx.
=xtan^-1{(1-x)/(1+x)}+§x/(1+x²)dx.
For §x/(1+x²)dx, we say, let u=x², dx=du/2x. Substituing, we have;
(1/2)§1/(u+1)du.
1/2ln(u+1)+c, recall, u=x².
=1/2ln(x²+1)+c.
.^.. => xtan^-1{(1-x)/(1+x)} + 1/2ln(x²+1) + c.

benbuks: Integrate arctan([(1-x)/(1+x)]dx

asap.

benbuks: Integrate arctan([(1-x)/(1+x)]dx

asap.

benbuks:

maybe....

workings will be cool bruv.


BTW: Am not a sir ooo

benbuks:

maybe....

workings will be cool bruv.


BTW: Am not a sir ooo

benbuks: manage this


integral of dx/√(12x -9x² )

asap.

Obinoscopy: Dear Maths Guru,

Cauchy Integral Test: Can someone discuss with examples?

benbuks:

ok...uwc.man...whats €-N...?. sir ...
i probably don't know it.

Soneh:
pls I don't get the (a) part, what happened to the powers.
thanks

benbuks:

ok

"""""""""SoLuTiOn"""""""""""""""

take product by the conjugate expression √(n²+ n ) + n at the numerator & denominator

=> Lim _n-->infinity =[[ √(n² +n) - n ] [[√(n² + n) + n ] ] / [√(n² + n ) + n ]

=> Lim _n-->infinity ( n² +n - n² ) / [ √[n ²(1 + 1/n) ] + n ]

=> simplifying & dividing throughout by 'n' we have

n/√( 1 + 1/n ) + 1 since n>0 ,as n--> infinity

we thus have 1/[ √(1+0) +1 ]

=1/2 ......... (proved )

benbuks:

ok

"""""""""SoLuTiOn"""""""""""""""

take product by the conjugate expression √(n²+ n ) + n at the numerator & denominator

=> Lim _n-->infinity =[[ √(n² +n) - n ] [[√(n² + n) + n ] ] / [√(n² + n ) + n ]

=> Lim _n-->infinity ( n² +n - n² ) / [ √[n ²(1 + 1/n) ] + n ]

=> simplifying & dividing throughout by 'n' we have

n/√( 1 + 1/n ) + 1 since n>0 ,as n--> infinity

we thus have 1/[ √(1+0) +1 ]

=1/2 ......... (proved )

benbuks:

as n --> ?...

kalan: Download sample past questions for your delight

http://careerplusonline.net/unilag-pg-entrance-past-questions-and-answers-download

1Book:
...Just upload the final answer where y=? so that it will be clear.

1Book:
...Just upload the final answer where y=? so that it will be clear.

1Book:
...Just upload the final answer where y=? so that it will be clear.

Arithmetic: Question:
Solve:dy/dx+2ytanx=sinx. By 1Book , from Calculus Mental Challenge(CMC).

:::SOLUTION:::.
Integrating both sides,
$(dy/dx+2ytanx)dx=$sinxdx.
$dy/dx.dx+2y$tanxdx=$sinxdx.
$dy+2y(-lncosx)=-cosx+c
NB: y is taken as a constant in $2ytanxdx.
y-2ylncosx=-cosx+c.
y(1-2lncosx)=-cosx+c.
.'.y=(-cosx+c)/(1-2lncosx). Which is the required solution.

Arithmetic: Question:
Solve:dy/dx+2ytanx=sinx. By 1Book , from Calculus Mental Challenge(CMC).

:::SOLUTION:::.
Integrating both sides,
$(dy/dx+2ytanx)dx=$sinxdx.
$dy/dx.dx+2y$tanxdx=$sinxdx.
$dy+2y(-lncosx)=-cosx+c
NB: y is taken as a constant in $2ytanxdx.
y-2ylncosx=-cosx+c.
y(1-2lncosx)=-cosx+c.
.'.y=(-cosx+c)/(1-2lncosx). Which is the required solution.

Arithmetic: Question:
Solve:dy/dx+2ytanx=sinx. By 1Book , from Calculus Mental Challenge(CMC).

:::SOLUTION:::.
Integrating both sides,
$(dy/dx+2ytanx)dx=$sinxdx.
$dy/dx.dx+2y$tanxdx=$sinxdx.
$dy+2y(-lncosx)=-cosx+c
NB: y is taken as a constant in $2ytanxdx.
y-2ylncosx=-cosx+c.
y(1-2lncosx)=-cosx+c.
.'.y=(-cosx+c)/(1-2lncosx). Which is the required solution.

Arithmetic: Question:
Solve:dy/dx+2ytanx=sinx. By 1Book , from Calculus Mental Challenge(CMC).

:::SOLUTION:::.
Integrating both sides,
$(dy/dx+2ytanx)dx=$sinxdx.
$dy/dx.dx+2y$tanxdx=$sinxdx.
$dy+2y(-lncosx)=-cosx+c
NB: y is taken as a constant in $2ytanxdx.
y-2ylncosx=-cosx+c.
y(1-2lncosx)=-cosx+c.
.'.y=(-cosx+c)/(1-2lncosx). Which is the required solution.

efficiencie:

bt if a series y(n) is bounded such that:

if y(i), y(j) ∈y(n) for i,j∈n which are the upper and lower bounds of n=Z, (knowing d@ Z∈(-∞ ∞)) then
y(n)∈(y(i) y(j)) and hence Limy(n)=y(i) as n→i and
Limy(n)=y(j) as n→j

suggesting that a bounded series must have a unique or oscilating limit as in d case of sinusoidal functns.
and a boundless series has no limiting values!

I stand to be correctd tho and i'l get an alternative solutn!!!

ibedun: Selfish woman.

Selfish people - she had to carry the whole thing out of Lagos all the way to Enugu. Probably potentially infecting countless people along the way.

Typical Easterner!!

CFCfan: The Nigerian quartet of Gloria Asumnu, Dominic Duncan, Lawreta Ozoh and Blessing Okagbare clinched the gold with a time of 43.65 seconds!
Congrats to them.

lebesgue:

I do have a pdf copy, but it is very large, so I can't upload or email it. But here is a pdf link (12.9 MB). Don't click this link on your phone! Right-click on it and choose Save link as.

Note: this book is very dense and hard to read; hence I wouldn't recommend it for a beginner.

lebesgue: ^
No sir I am not familiar with it, but I will check it. In the States, we use so many different texts for analysis, and the most popular one is
The Principles of Mathematical Analysis, by Walter Rudin.

lebesgue:
You are most welcome.


What text? If you are referring to the proof, I didn't copy it from a text. I wrote it myself!


No, I am not. I am an undergraduate studying Mathematics, and I intend to be an algebraist.

lebesgue:

Part 3 of 3.

efficiencie:

if x(n) diverges then Lim[x(n)] →(+-)∞ as n→(+-)∞
and if y(n) converges then Lim[y(n)]=a as n→(+-)∞ where a∈R

then

Lim[x(n)+y(n)]=Lim[x(n)] + Lim[y(n)]
=Lim[x(n)] + a
=Lim[x(n)+a]
=Lim[x+a] where x→(+-)∞
=Lim[x]

which is divergent!