₦airaland Forum

jackpot: [color=indigo]dear Maths generals, pls help me with this little problem.


[size=16pt]By using the Cartesian-paraboloidal coordinate transformation equations:
(x=uv sin φ, y=uv cos φ, z=¹/₂(u²-v²),
show that the 3-dimensional Laplace's equation
F_xx+F_yy+F_zz=0
reduces to
¹/_(u²+v²) (F_uu+F_vv+ ¹/_u F_u+¹/_v F_v)+¹/_(u²v²)F_φφ = 0.[/size][/color]

Richiez: @benbuks and jackpot let me sit back, relax and learn a bit from you both...cheers!

*blushes*

hi Richiez

are you sure it shouldn't be the other way round?


Lol.

busuyem: Pls house, I'm planning 2 go 4 my MSc. in
Maths. Which option is the best in Maths to
study. I'm contemplating on which to choose
considering the future advantages.
Need ur advice.

benbuks: ..av u heard of sophomore's dream?

benbuks: ....ok try this dear....integrate (x^x^x^x....to infinity.)dx

honey: forgıve my basıc knowledge of mathematıcs............ the three coeffıcıents could all be 0s or 1s havıng (000) (111), and we could have two 1s wıth a sıngle zero, or two 0s wıth a sıngle one...... two 1s wıth a sıngle zero can arranged ın 3 ways, and two 0s wıth a sıngle 1 can be arranged ın 3 ways gıvıng the arrangementr (000) (111) (110) (101) (011) (100) (001) (010), gıvıng a total of 8 ways, out of these arrangements the quadratıc equatıon wıthout real roots ınclude (x²+x+1) and (x[sup]2[/]+1), meanıng the remaınıng 6 arrangements have real roots...... therefore the probabılıty of havıng a quadratıc wıth real roots becomes 6/8 =3/4

Calculusf(x):
...try to show humility sister...i could remember a question benbuks posted,and you asked him not to post simple question again...why can't you solve this,you are just using approximate method...Try to know this that:THERE'S NO MATHS GURU...you can just know up to the level you are and the aspect you are studying......WELL SHA,you tried...

Sir, I already did. It wasn't an approximate method since the summation in my answer is still running from 1 to infinity. The fact is that I left my answer(which is very correct) in series form.

With my method, there is the temptation to conclude that I can integrate just any function!

Boladearo: thats tru, prove it to us queen jackpot

Calculusf(x):
Integrate 7^sinx.dx

Efemena_xy: Eye popping ke?

Depends on the size of your "lunch box" dearie

Even though you got the op done on the cheap at Brazil, they didn't take off "everything", did they?

Idowuogbo: Cemetery

Rip jokes!

Boladearo: yes boss. Queen jackpot, welcum to d ause

honey: take the subtitution e^x=v, then the que. bcums §v.15^v.dx..........dv/dx= e^x, and x= in.v, making dx= dv/v, replacing this in the integral §15^v.dv= 15^v+1/(v+1), replacing v with the original values of x, we have 15^(e^x+1)/(e^x+1)

benbuks: Integrate.(e^x . 3^e^x . 5^e^x)dx..tnx

first of all, your question reduces to

Integrate e^x.15^e^x dx

by making the substitution u=e^x, it further reduces to

Integrate 15^u du = (1/ln15) 15^u + c = (1/ln15) 15^e^x + c.

The next time you post a simple question, I'll whoop your arsse!!!

Calculusf(x):
Integral e^(x^2)

Ortarico: Weird! Use Jacobian of the transformation rule and pls don't ask me cause I dunno it either!

jhydebaba: This bar is freaking dry joor.

Turns right, turns left, checks if no vehicle is coming, sit down on road side and sleep off.

drives closer to him with my Toyota Matrix and jams him inside gutter

bright007: .....OMO THIS ONE NA JAMB KWESSION.

Ellatyra: "WoMEN" has "men"in it too...so?