₦airaland Forum

wait ooo, did this thread really make all these 150+ pages in just 3 days? Blimey!!!

wow. Just coming back from work and I'm already 11pages behind in less than 24hrs! Blimey!!!


@jaryeh, doubleDx
Sorry, the timing isn't suitable for me to participate walahi. Thanks for the nomination though. My pleasure.


@ Richiez, doubleDx, Alpha Maximus
Great works bros'es!

horlaxymoney: it horlaxymoney....i rep futminna nd estate management buh i b guru/master in mathematics
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Sign me in and c wat i got 4 u...
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Application letter signed? Wil lyk 2 join d thread
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.1/2x + 2/4y = 36
and
3/6y + 4/8x = 72
quadratic equation....
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.giv me a trial

quadratic equation? We give up!!!

By the way, your simultaneous equation no get answer. Simplify the second and compare with the first. This shows that the equations are inconsistent.

doubleDx: This is just round one, other quiz masters will definitely join when they are free! Ain't easy squeezing out time on working days....

on point, Sir!



@Richiez

hope it's not too late to declare interest?

jaryeh: ......and where is Miss Jackpot She should come and join us o. I can't imagine her not being here.

Hi. Reporting on the thread Sir

2 days absence and I've missed quite a lot. Amazing!

Would have loved to participate but I guess the stage is already set.

I see our quiz masters doing the work just fine.

Cheers everybody!!!

icankel: Am sure u'v collected ur AKA-AZU from APGA but remember dat Anambrarian have waken up. I wonder y ppl refused to talk of Ifeanyi Uba even after collecting kerosene, fuel, car, money etc from him dat shows dat Ngige is still d man of d ppl

which Aka azu? I am only commenting and baring my mind on what is directly affecting me.

Ngige is too old and desperate to enter our government house. 11 fucknnn years contesting a single post.

He is already 61. I repeat, he should go back to his village and be eating kolanuts, drinking schnapps and dry-gin, taking snuffs and be settling communal disputes. That's where he is needed for now!

Alpha Maximus: ......lemme try your approach on this question whose answer is obviously '2'
3^x+4^x=5^x
3^x+4^x-5^x=0
X(log3+log4-log5)=0
X(log12/5)=0
X(log2.4)=0
[s]10^0=2.4^x
1=2.4^x
Log1=xlog2.4
X=log1/log2.4[/s]
X=0, which is correct!!!

the cancelled steps makes the "solution" unnecessarily long. From x(log2.4)=0, we have x=0/log2.4 = 0.

I got your sarcasm solution to 4¹/₂ star general benbuks method sha. Glad we are all here to learn. It's obvious the method doesn't work.

At least, we can now detect nonsense anywhere we see incorrect rule of lofarithm being applied.

d citizen:
@jackpot

I need somebody in the house to put jackpot in the eyes of the storm. jackpot identity is questionable. Jackpot, i challenge you to review ur identity with regards to gender. Ur mathematical prowess and sound critical reasoning have put ur gender under helicopter scrutiny.

Msg from the five star general

chei!!! I have dead!!!

so, you don't believe in our ability sha?

**sobs**

**heads off to the nearest Ministry of Women Affairs office to make a report of gender stereotyping against d citizen**

Alpha Maximus: .......
Details
Depth of well: 2500cm=25m
Total distance moved per day =2-1.5=0.5m
Days required for complete ascension:25/0.5=50
Thus , the snail would have 'SA LO'D(yoruba) from the well in 50 days ..I really don't have the right to feel proud of solving this!!!

you are wrong, sweetheart

well, first of all you tried so so much, but I felt you didn't critically examined the snail's movement.

At the end of 46 days, the snail must have moved
46 * 0.5m = 23m
then, on the 47th day, you will agree with me that the snail will first of all move up 2m before sliding down.
So, 23+2=25m, and the snail is already out!

But if you still argue that even though he has reached the top that he must slide down 1.5m, then on the 48th day, the snail must first of all move up 2m, and this gives
47*0.5+2=25.5m

and this time around, the snail don commot from the top tey tey and no need sliding down.

So dear, I'm sorry but you didn't get the answer right, Sir!!!

like seriously, those saying "Onwa ga eti ozo" (the moon will shine again). Please, my people, is moon's light good enough?

Stupid and silly Ngige that blamed Peter Obi and said he should have come out by 3am to receive Fashola's deportees at Upper Iweka.


APGA bu ejesia ogwu na ala Anambra State.

^there you go again!

your line 2 doesn't imply line 3.

ojialo: why don't u go and ask buhari what is he really looking for in Aso rock we can not deny the fact that Chris is a Good man he has his people at heart so let him come and finish what he started obituaries could not finish

i do not support Buhari too.

He has his people(I can vouch that the APC-controlled media constitutes a larger percentage) at heart.

Ngige is already 61. He should go and contest for Igwe ndi Alor. He should be eating kolanuts and snuffs by now. 11years contesting a post you defiled at first through rigging? Gimme a break!

Abiolacicero: see yoúr life? You are On Your Own.

arewatech: Bad market?

Abiolacicero: see yoúr life? You are On Your Own.

APGA is so fly! NGIGE and Co should kiss our arsses!

Kingsley1000: EVEN MY MUM,VICE PRINCIPAL ALSO SAID THE SAME THING,..NGIGE HAS ALREADY WON

maybe you forgot to mention that your mum and her vice principal are from Alor, Ngige's hometown?

Beloved and registered Anambrarian voters whom will determine the outcome of November 16th. Let's shame outsiders(the APC-media) a little bit with our LIKES. They can't dictate how we will govern ourselves.


[size=16pt]
If you think this opinion poll is APC concocted, wack and fake; and that APGA will carry the day, hit LIKE!!![/size]

The problem I have with Ngige is that he is riding on the voter's intelligence. Beware of anybody coming with the "Saviour's Mentality".

Another of his undoing is the wicked lies him and his campaign groups spew on social medias (e.g. Facebook). None of those his Onitsha Pure-Water pelting APGA campaign incidence were true. I'm a witness.

Why would a man who has consistently contested and failed the Governorship's Election for 11years still remain an active contestant. What is he really looking for in the Anambra Government House?

benbuks: ...indomitably propitious.....thanks for that andvanced approach....remember..this thread is a clinic...

yeah.


using product rule to solve the question was d normal thing, but I had other plans

benbuks: ..i kw na.,...prove that tan(3t + pi/4) =dy/dx....

Hi, Sir Benbuks. This is it.


x= e^3tcos 3t, y=e^3tsin 3t

z(t)= x+iy =e^3tcos 3t + i e^3tsin 3t
=e^3t(cos 3t + i sin 3t)
=e^3te^i3t
=e^3(1+i)t
thus
dz/dt= ^d/_dt (x+iy)
=^dx/_dt+i ^dy/_dt
=3(1+i)e^3(1+i)t
=3e^3t[(cos 3t - sin 3t)+i(cos 3t+ sin 3t)]
so,
^dx/_dt=3e^3t(cos 3t - sin 3t)
^dy/_dt=3e^3t(cos 3t + sin 3t)
therefore
^dy/_dx= ^dy/_dt * ^dt/_dx
=^{3e^3t(cos 3t - sin 3t)}/_{3e^3t(cos 3t + sin 3t)}
=^{(cos 3t - sin3t)}/_{(cos 3t + sin 3t)}
and by dividing up and down by cos 3t, we get
dy/dx= ^{(1-tan 3t)}/_{(1+ tan 3t)}
and since 1=tan 45, we have that
^dy/_dx=^{(tan 45 - tan 3t)}/_{[1+(tan 45)(tan 3t)]}
=tan(45+3t)
=tan(3t + pi/4)


That's all!

benbuks: ..no solution to this yet.

dy/dx = dy/dt * dt/dx where dt/dx= 1/(dx/dt).


That kweshun na baby na.

just observed it now. Sun's intensity was reduced to a meagre 5% !

doubleDx: Hi jackpot, could you re-post your approach to that paraboloidal problem? Or FWD it to me here: doubledx9ja(at)gmail.com, I want to cool down and go through it. Was kinda busy, I glanced through it once but did not save it on my PC.

I see you guys devouring Math problems here....Kudos to all the active generals and doctors on the thread! Keep it going people, the thread is getting interesting by the day. I hail ona wella!

done, Sir

Laplacian: @d citizen,
her rationalization approach is error-free, i suggest u use brackets in ur solutions to avoid ambiguity....
...though ur answer is correct but ur 'argument' does not justify u because ur transition (cos@) to d complex domain (e^i@) also changes d condition r<1 to |r|<1...

Hi Sir Laplacian

The answer I gave in limit form may or may not exist. I dey even suspect d thing sef

let @=pi

So, are we saying that
cospi+cos2pi+cos3pi+. . .= - ¹/₂
based on d citizen's result?

Let's examine this series.

Now, hope you know that
cos pi+cos 2pi+cos 3pi+cos 4pi+. . .
=-1+1-1+1-1+. . .

Now some may argue that
-1+1-1+1-1+. . .=(-1+1)+(-1+1)+(-1+1)+. . .
=0+0+0+. . .
=0

Some may equally argue that
-1+1-1+1-1+. . .=-1+(1-1)+(1-1)+(1-1)+. . .
=-1+0+0+0+. . .
=-1

But a series converges if and only if the sequence of partial sums converge.
Combine this with the statement:
"A sequence converges if and only if the subsequential limit set is a singleton."

but the sequence of partal sums is
{-1, 0, -1, 0, -1, 0, -1, . . .}
and the subsequential limit set is {-1, 0}, which is not a singleton.

And as such, the series
cospi+cos2pi+cos3pi+cos4pi+. . .
is not summable(i.e., does not converge.)

Have you heard of Cesaro summability? A series which doesn't converge may still be Cesaro summable. The Cesàro sum of an infinite series is the limit of the arithmetic mean (average) of the first n partial sums of the series, as n goes to infinity.

It turns out that the series -1+1-1+1-1+. . .
is Cesaro-summable to -1/2 (average of the two results 0 and -1)

so, the actual way to put it is that the series
cos@+cos2@+cos3@+. . .
is not summable(i.e., does not converge), but may be Cesaro summable


Now, I hope we all know that a necessary condition for a series to converge is that the n-th term of the series converges to zero.(equivalently, the tail of a convergent series is arbitrary small)

Now, the nth term of the series
cos @+cos2@+cos3@+ . . .
is cos n@, and clearly cos n@ doesn't go to zero as n goes to infinity. In fact, it "dances" between -1 and 1


so my dears and sweeries, I'm officially changing my stand on this issue. Thank God I corrected my self by my self _{pun intended} lol






current stand

The series
cos@+cos2@+cos3@+. . .
is not summable(i.e., it doesn't converge) since its nth term doesn't go to zero.

if @=2k pi for some integers k, then the series equals +ve infinity. In this case, it is not Cesaro summable.

If @ is not an integral multiple of 2pi, then the series equally doesn't converge, but oscillates. In this case, it may be Cesaro summable.








yours trustfully

JACKPOT.

d citizen:
The five star general approach to the problem: cos@+cos2@+ cos3@+

Let C = cos@ +cos2@ +cos3@+

Let S= sin@+sin2@+sin3@+ ( dat is the complex number by of the expression)

C+iS= (cos@ +isin@) (cos2@+isin2@)+ (cos3@ +isin3@)+

Thus, C+iS=e^i@+e^i2@ +ei3@+

C+iS=e^i@/(1 -ei@)(using sum to infinity)

Hi Sir d citizen, first of all, I must start by acknowledging your mathematical prowess and ability to think critically.

But I hope you know that the formula for the sum to infinity of a GP is valid if only absolute value of the common ratio is less than 1; i.e., |r|<1.

But it is easy to see that the GP
e^i@, e^2i@, e^3i@, . . .
has common ratio r= e^i@
and its absolute value is
|r|=|e^i@|=|cos@+i sin@|
=sqrt{cos²@+sin²@}=1
and as such, applying the sum to infinity of a GP there is extremely fallacious.

Consequently, I'm afraid that your solution is wrong.

C+is= e^i@ * (1-e^-i@)/(1-e^i@)(1-e^-i@)

C+is= e^i@-1/2-e^i@-e^-i@

C+iS=cos@+isin@-1/2(1-cos@)

Taking the real part of the expression

Cos@+cos2@+cos3@ = cos@-1/2(1-cos@)

Granted, your solution is already wrong. Now your rationalization(as shown in the bolded) here is also wrong since
^{e^i@}/_(1-e^i@)= -1+¹/_(1-e^i@)
=-1 + ¹/_{{(1-cos@)-i sin@}}
=-1+^{{(1-cos@)+i sin@}}/_{{(1-cos@)²+sin²@}}
=-1+^{{(1-cos@)+i sin@}}/_{2(1-cos@)}
and if you take the real part of this, you should be able to get
-1+ ^(1-cos@)/_2(1-cos@)
=-1+¹/₂
=-¹/₂


My Regards



I remain
your humble servant-ress

JACKPOT

Mbahchiboy: TO MY GENERALS:MAKE UNA DO DIS
1 4 ME::
if S=a(r^n-1)/r-1....
make r d subject of d formular

for n>=6, subject of the formula may not be possible.

@Mbahchiboy, find below what you sought for (i.e., the few digits of 2004²⁰⁰⁴)

jackpot: to get the digits itself, take antilog to base 10 of the mantissa of (2004*Log₁₀2004).

should I throw more light?

Swagalord18: I have some unsolved math problems ....but if i post them now ... U guys will start saying that i'm testing ur ability or this and that...... No problem sha ....i see i'm not welcome here ....4 reasons not clear 2 me ....i'l kindly unfollow ur thread now .....enjoy

Hi Mr Swagalord

pls do post them, but make it open for anybody to solve and eat the sumptuous meal.

You know, I have been eating your sumptuous meals and I'm now obese. Hope you understand?

Mbahchiboy: no qualms 4 d no 1 sir.....
4 d no 2 i got wat u said but i wud like to know how e got those digits,i dont mean d number of digits but d digits itself

to get the digits itself, take antilog to base 10 of the mantissa of (2004*Log₁₀2004).

1 plus the integral part of (2004*Log₁₀2004) gives you the number of digits of 2004^2004.


Elementary reasoning.

Laplacian: ...i think d right 'expression' to use is; the series is not UNIFORMLY CONVERGENT....d relation u gave is really unhelpful & more difficult dan d question itself because;
cosx=1 + x^2/2!+x^4/4! +.....
Cos2x=1 + (2x)^2/2!+(2x)^4/4! +...
Cos3x=1 + (3x)^2/2!+(3x)^4/4! + ...
:
:
so dat we have a matrix of infinit row-column to be summed....

Hi Sir Laplacian,

I think you should change the underlined to minus sign (-)

besides, d given series is convergent for all @ satisfying; cos@-1<sin@*tan(n@)

Explain!

apologies for omiting the solution because it's not easy typing them all out because of too many brackets and fraction over fraction in the solution. I haven't considered uploading it through pix yet.

Hint
To solve this, you may apply DeMoivre's theorem, the sum of a GP, use the necssary complex identities and finally take limit to get

cos@+cos2@+cos3@+...
=Lim_{n --> infty} {^{[sin (n@/2)]}/_[sin(@/2)] cos[(n-1)@/2] - 1 }

Cashio: i wondered really if jackpot is truly a girl/lady cos i have never seen any female with such IQ....thumbs up sis/bro.

***catwalks past him***

no time. Lol

rhydex 247: I assume we are all familiar with Maclaurin series expansion:
f(x)=f(0)+xf'(0)+x^2f"(0)/2!+---
Apply this to cosx, cos2x, then cos3x, etc. and sum them. What do you get?
By the way, I would point out that this series does not converge, but rather with increasing values of 'n' cycles about 0 with an amplitude that depends on the value of @. I'm sure dis helps.

such argument may not hold water since the sister series
sin@+sin2@+sin3@+. . .
for sure, converges for @=0, k(pi) for all integers k.

So, i think convergence depends on the particular value of @.

By the way, lemme attempt solving it.

Swagalord18: smh ...some nairaland guys irritate me ...
See how u draw conclusions ....do u know my story
.
Ddnt u c dat all my questions wer from d same topic ....iznt it obvious that i waz reading that topic ....i gave d questions dat gave me difficulty 2 her cos shez good ..n i understand her solvings...
.
I'm sure if i were d girl here and she ..d guy,, .........ur support will tilt towardz me .....
smh... boys will be boys (-_-)

lol. Una no go kill me with laffta.

Bro, we are good, right?


@Mbachiboy
i see you ooo! Daalu.