Mbahchiboy: brb it hasn't come to a fight.........

Mbahchiboy: brb it hasn't come to a fight.........

Mbahchiboy: plz need ur assistance guys::
(1).y"+2y'+y=0.
(2).2004^2004.
plz wen solvin use simple terms 4 my better understanding.
TNX IN ADVANCE

Alpha Maximus: It was the math fever oh!! I'm still in the intensive care unit!!

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 (-_-)

Mbahchiboy: plz need ur assistance guys::
(1).y"+2y'+y=0.
(2).2004^2004.
plz wen solvin use simple terms 4 my better understanding.
TNX IN ADVANCE

Laplacian:
i ve very high esteem 4 ur questns
If x+a^(x^a)=b, from d structure of x we can infer dat d above eqn is a vector eqn, applyin vector tripl produt to d 2nd term on d L.H.S,
x+x(a.a)-a(a.x)=b....eqn1, multiply d given eqn by a. to get,
a.x+a.[a^(x^a)]=a.b, scalar tripl prodct givs
a.x-(x^a).[a^a]=a.b, recall dat
a^a=0, so a.x=a.b, substitt in eqn1, to get; x(1+|a|^2)-a(a.b)=b, so
x=[b+a(a.b)]/(1+|a|^2)

Laplacian:
i ve very high esteem 4 ur questns
If x+a^(x^a)=b, from d structure of x we can infer dat d above eqn is a vector eqn, applyin vector tripl produt to d 2nd term on d L.H.S,
x+x(a.a)-a(a.x)=b....eqn1, multiply d given eqn by a. to get,
a.x+a.[a^(x^a)]=a.b, scalar tripl prodct givs
a.x-(x^a).[a^a]=a.b, recall dat
a^a=0, so a.x=a.b, substitt in eqn1, to get; x(1+|a|^2)-a(a.b)=b, so
x=[b+a(a.b)]/(1+|a|^2)

jackpot: lol. Una no go kill me with laffta.

Bro, we are good, right?


@Mbachiboy
i see you ooo! Daalu.

Alpha Maximus: @benbuks, happy now? The thread has been re-vitalised!!! Very much unlike the Nigerian Federal Varsity Education sector of the nation!!!! Happy solving, I have to take my Mathemacetamol for my Math fever and take some innoculation shots of Calculocyn to prevent re-occurrence of 'strong-head' anti-math bacteria!!

rhydex 247:
@ laplacian.
Here is the soln.
a^(x^a) implies
(a.x)a-(a.a)x. Where x=(x1,x2,x3) let a=(a1,a2,a3).
a.a=(a1,a2,a3)(a1,a2,a3).
a.a=a^21^a^22+a^23.
a.x=(a1,a2,a3)(x1,x2,x3).
a.x=a1x1+a2x2+a3x3.
a^(x^a)=(a1x1+a2x2+a3x3)a-(a^21+a^22+a^23)x. Recall that a=(a1,a2,a3) nd x=(x1,x2,x3).
a^(x^a)=(a1,a2,a3)(a1x1+a2x2+a3x3)-(x1,x2,x3)(a^21+a^22+a^23). When u expand nd open d bracket u get 0.
Hence a^(x^a)=0. Buh 4rm d questn we av x+a^(x^a)=b. Put a^(x^a)=0 in d eqn we av
x+0=b. x=b. Where x=x1,x2,x3. Hence x=(x1,x2,x3)=b. Note ^ means cap but ^ in a^21+a^22+a^23 means raise to power. All is well.

Mbahchiboy: bro it is obvious d guy was just testin her agility.
i support jackpot partially sha........
dis one everybody dy face jackpot abi na high i go call am .....hmmm.
DID U KNOW??................
DAT JACKPOT IS A GUY.
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na joke i dy ooo

benbuks: Sum the series cos@ + cos 2@ + cos3@ +...

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.

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.

Boladearo: A GP first term is a, commom ratio is r, the 6th term is 768. Another gp has first term a, common ratio 6r, the 3rd term is 3456. Find a and r.
Gurus in d ause just use this 4 refreshment, i understand that this is too small, but abeg manage am

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.

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....

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

jackpot: Hi Sir Laplacian,

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

jackpot: lol. Una no go kill me with laffta.

Bro, we are good, right?


@Mbachiboy
i see you ooo! Daalu.

rhydex 247:
Solution.
1). Y''+2Y'+Y=0. Taking an auxiliary equation.
m^2+2m+1=0.
(m+1)(m+1)=0.
m=-1 twice.
The general soln is. Y=Ae^-x+Bxe^-x.

2). 2004^2004. You can't av an analytic solution for this. Buh here is the answer.
2004^2004=1.0069902351614371713287127625540878181640439207...*10^6617. The number length is 6618 decimal digits. All is well.