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. Hmmm u tried sha buh no be the answer be that.

benbuks: ,...number base conversions involving decimals dont always terminant..so u stop at 4 decimal place..

Thnks so much and God bless u.

hi maths gurus... i need workings on this integral... evaluate the integral § x/(x^3+1)...

Ezechinwa: hi maths gurus... i need workings on this integral... evaluate the integral § x/(x^3+1)...

d gurus in d house will provide d working 4u,...

Ans:
2/(3)^(1/2)arctan(((3)^1/2)/(2x-1))-ln(x+1)+c...i stand 2b corrected..thank u...

wisemania:
d gurus in d house will provide d working 4u,...

Ans:
2/(3)^(1/2)arctan(((3)^1/2)/(2x-1))-ln(x+1)+c...i stand 2b corrected..thank u...

Ur ans is wrong.
Here is some steps.
§x/(x^3+1).
Soln.
Using Partial fractn
§ x+1/3(x^2-x+1) dx - § 1/3(x+1)dx.
1/3§ (x+1)/(x^2-x+1) -1/3§1/(x+1). I'm stopping here. I'm sure u can continue frm here. Cos i'm kind of busy.
Here is the ans. If u move further.
1/6[ln(x^2-x+1)-2ln(x+1)+2rut3 tan^-1(2x-1/rut3)]+ C. Lol. OYO.

jackpot: Hi doubleDx

what's the name and author of that textbook you uploaded while solving that paraboloidal stuff?

It's a combination of Advanced Engineering Mathematics by H.K. Dass and some other stuffs... The scanned pages are from Murray R. Spiegel Mathematical Handbook (Page 126 & 127).

Nice work you've been doing here. Keep it up girl!

jackpot: Hey bro, check this, from where you copied it out. It should be 6¹/₃.

Having said that, put n=1 and n=2 in the T_n equation and equate to 5 and 5 respectively.

You will then get the two equations

a+b=5 . . . . . . . . . . . . . . . . . . . (1)
^a/₃+2b=5. . . . . . . . . . . . . . . . .(2)

solve these simultaneously to get
a=3, b=2

substitute these in the T_k equation(I am calling it T_k instead of T_n because I want to sum from k=1 to n. Remember they are dummy variables after all )

T_k= 3(¹/₃)^k-1 + 2k

sum over all n and simplify (you should use the sum of a GP formula on the first summand and the sum of an AP formula on the second summand here).

The tiny details are left as an easy exercise for the reader

Final answer is

S_n = ⁹/₂ [1-(¹/₃)ⁿ ] + n(n+1)


Thats all!

corrrect ma'am
.
.
Here's lunch
.
.
if T_n+1 = a + bT_n for a sequence T₁, T₂, T₃, . . ., show that T_n+1 - T_n = b^n-1 [a - (1 - b)T₁]

rhydex 247:
Ur ans is wrong.
Here is some steps.
§x/(x^3+1).
Soln.
Using Partial fractn
§ x+1/3(x^2-x+1) dx - § 1/3(x+1)dx.
1/3§ (x+1)/(x^2-x+1) -1/3§1/(x+1). I'm stopping here. I'm sure u can continue frm here. Cos i'm kind of busy.
Here is the ans. If u move further.
1/6[ln(x^2-x+1)-2ln(x+1)+2rut3 tan^-1(2x-1/rut3)]+ C. Lol. OYO.

i tested u!......i kw dis tin da year sha,anywaz i no come dis wrld 2come dey solve maths naw

Swagalord18:
ur very good ooh ...kudos
nd i lyk d way u take ur tym 2 make use of the sub nd sup 4 beta understanding.
.
[size=16pt]Take this as desert[/size]
.
The nth term of the series 5 + 5 + 6¹/₂ + ..... Is given by T_n = a(¹/₃)^n-1 + bn. Find the values of a and b, and the sum of the first n terms of the series

Swagalord18:
corrrect ma'am
.
.
[size=16pt]Here's lunch[/size]
.
.
if T_n+1 = a + bT_n for a sequence T₁, T₂, T₃, . . ., show that T_n+1 - T_n = b^n-1 [a - (1 - b)T₁]

SMH!!!

It's obvious you are not in need of the solutions. I'm equally not here to impress you or anybody.

Take back your meals and feed yourself first.

At least, I know you are very skinny, and in need of sumptuous meals.

I'm out!!!

jackpot: SMH!!!

It's obvious you are not in need of the solutions. I'm equally not here to impress you or anybody.

Take back your meals and feed yourself first.

At least, I know you are very skinny, and in need of sumptuous meals.

I'm out!!!

..this is what i call a psychological phenominal....no tell me say u dey form maths don,..c'mmon baby no fall ma hand na..

jackpot: SMH!!!

It's obvious you are not in need of the solutions. I'm equally not here to impress you or anybody.

Take back your meals and feed yourself first.

At least, I know you are very skinny, and in need of sumptuous meals.

I'm out!!!

i fink ure getting d young man crossed...

Swagalord18: prove that if the first term of an A.P is 1, and the sum of n terms is n², the A.P is 1,3,5.....9

EVEN without solvin much u know it follows a pattern which is::
2n-1.....where n=1,2....

U shouldnt av said this ma'am ....som pple myt feel dat u cant solve it ...

jackpot: SMH!!!

It's obvious you are not in need of the solutions. I'm equally not here to impress you or anybody.

Take back your meals and feed yourself first.

At least, I know you are very skinny, and in need of sumptuous meals.

I'm out!!!

ur here 2 solve maths i thought ..... Datz y i post any question dat gives me dificulty here .....i just use dose words "lunch" "desert" 2 praise u coz i know u'l eat(solve) them ..... And i pose dem 2 u in particular cos i lyk d way u use the sub n sup functions (learnt that from you)... i understand ur solutions more ....well i'm not suprised at d way u reacted ....its normal 4 evri human 2 feel a bit of pride wen they feel they are top in wat they do ...so i wont scold or lambaste u .....afterall ...itz ur right.....
.
After seeing ur post, ..i took it upon myself and put more effort in2 eatin my "sumptuous meal" ..as u called it. ..and whoa!...i solved it without any help ... .....thanks 4 ur help n the previous questionz ...more grease 2 ur elbows...
.

Herez how i did it:
.
T_n-1 - k = b(T_n-k) ....... (1)
.
.
U'l get
.
T_n+1 = k(1-b) + bTn
.
From the relation in d question a = k(1-b) ...input dat and u'l get..; k=a/(1-b)
.
Put this in eq (1) ...u'l get

.
T_n= ^a/_1-b + b^n-1(T₁ - ^a/_1-b)
.
.
The pathway 2 d answer is obvious from here .... .
.
.
Ur still among the pple i look up to ....tho' i never expected this frm u
Thanx once again ma..

Pls pls pls any mathematician in the house?
Help me out, I'm on a hot seat.
(1) DIVIDE 1423 BY 24 IN BASE 5
(2) DIVIDE 2512 BY 4 IN BASE 4

Airty: Pls pls pls any mathematician in the house?
Help me out, I'm on a hot seat.
(1) DIVIDE 1423 BY 24 IN BASE 5
(2) DIVIDE 2512 BY 4 IN BASE 4

....1) lets first convert to base ten.1423(base 5)=
(1x5^3) +(4x5^2) +(2x5^1)+(3x5^0)=125+100+10+3=238
..
24(base 5) =(2x5^1) +(4x5^0)=10+4=14
238/14=17(base ten) =
5|17 =3r2
5|3=0r3
=>1423 /24 =32 .(base 5)

I have 50 naira

I spend. Remaining bal.
20. 30
15. 15
9. 6
6. 0

Tot:50. 51

Pls., where is the extra 1naira coming from?..

Xy'=2x +y
solve d DE

Laplacian: ...there is no analytic solution to:
3^x+4^x=5^x, 4^x=2x, etc...

who tld u dat

Mikebis: who tld u dat

PLZ SHOW US UR METHOD SIR MIKEBIS

benbuks: Xy'=2x +y
solve d DE

Solution.
xy'=2x+y.
divide thru by x.
y'=2+y/x.
y'-y/x=2. Nw recall that dy/dx+P(x)y=Q.
P(x)=-1/x and Q=2.
I.F means integrating factor.
I.F= e^§P(x)dx.
I.F=e^§-1/x dx.
I.F=e^-lnx. I.F=x^-1=1/x.
hence I.F=1/x.
y.I.F= §Q.I.Fdx
y.1/x=§2.1/xdx
y/x=2§1/xdx
y/x=2lnx+C.
y=x(2lnx+C). All is well.

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

This thread is becoming dry this days, guys n girls what's happunin na?.

Hi everyone...I've got good news for you all.
If you want the soft copy of Engineering maths by Alan Jeffrey, send ur request to me via richobiagba@gmail.com please don't post your emails here so that our thread won't look untidy.
Thanks!

Four married couples bought eight seats in a row for a concert. In how many ways can they be seated if
(a) each couple is to sit together
(b) each couple, except a particular newly-wed couple, is not to sit together
(c) a particular couple decided to head home after having a minor argument?

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

benbuks: ..this is what i call a psychological phenominal....no tell me say u dey form maths don,..c'mmon baby no fall ma hand na..

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.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
na joke i dy ooo

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.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
na joke i dy ooo

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 (-_-)

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 (-_-)

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

Alpha Maximus: ...I believe this is a simple combination matter, let's get to it!!!
(A) if each couple is to sit together, then each pair is considered as a single entity in the computation , thus
8C4=8*7*6*5/1*2*3*4=1680/24=70 ways
(B)if each couple except the freshly married novices(excuse my free language) is to sit together, then this would be interpreted as 5 seperate entities in the following computation, thus
8C5=8*7*6*5*4/1*2*3*4*5=6720/120=56 ways
(C)if a particular couple decided to head home over the rented Okokomaiko apartment, then there would be three couples left, translating to 3 seperate entities, thus(and you didn't state the conditions as to which the remaining couples would be seated, so I'm assuming they are together)
8C3=8*7*6/1*2*3 =336/6=56 ways
P.S: if I was wrong, blame it on the mathematics fever, Euler is on the matter!!

que1....if each is considered a single entity then there should be four entities...so dat, 4C1=4
que2...each is not to sit together except 2