jackpot: Hahaha. I didnt know your first statement (the part where you replied Maximus) is still on the question. I thought you were boiling
another kettle of fish(answering another
question) with Maximus. I realised it after I'd
posted. I wanted to edit, but you've already replied.
Well, I see.

Don't you think the common difference should be -d (minus d)?
For example, for n=3, we have a=15, d=7.
I think it makes more sense if common
difference is -7.
The same thing goes for other choices of n.

It seems these equations are working. How did you come about them?

i'll post the full solution later (wen am free). I arrived @ it exactly 12:00pm today...it has 2 do wit Pell's equation...i hop u 're familiar wit that?...hey! the error has been duely corrected...thanks...

Laplacian:
@jackpot, 2nioshine&Dx....tanks 4 sharin my burden...believin everyone understnds Dx's solutn, i proceed 2 give my solutn cuz i went farther & wait 4 ur correctns...
Now
d = √[{z² + x² - 2y²}/2]

but d is an integer, so the square root has to go...
Set z=2y+x, then
z²=4y²+4xy+x²
addin
x² - 2y² to both sides and dividin the result by 2 we obtain
[{z² + x² - 2y²}/2]=(x+y)²
hence
d=√{(x+y)²}

d=x+y

but, a=(y²-x²)/d
so that
a=(x-y)*(x+y)/(x+y)=x-y
so for integers x, y & z, the first term a=x-y, and the common differenc d=x+y and lastly z=2y+x

if nw we set x=3, y=4 then z=2*4+3=11, a=4-3=1, d=4+3=7
so the sequence is a=1,
a+d=1+7=8, a+2d=1+14=15

i.e 1, 8, 15.....which is jackpot's result....but i dont knw y it fails for other values of x and y...any suggestion plz?...
Here's the full solution...with a=y-x d=y+x, z=2y+x slotted into either eqn1, eqn2 or eqn3, we arrive arrive @ the followin eqn
2y²-2xy+1=x²
or rearrangin 2 get
3y²+1=x²+2xy+y²
or
3y²+1=(x+y)²
but d=x+y
so
3y²+1=d²
or
d²-3y²=1
remember that since a=y-x
==> a=y+x-2x=d-2y...every other thing follows from the Pell's Equation above...Regards 2 jackpot and 2Dx...

for any confusion, refer 2 page 121

2nioshine: Topic:LAPLACE TRANSFORMATION

Q1
find L{tcosat},where a=constant. from first principle
Q2
find f 'L{e^(4t)cos3t}

Nb:f ' means the derivative of the function

solution to no 1.
L(tcosat) using the shift theorem.
well known that L(cosat)= s/s^2+a^2 where s>0.
now d L(tcosat)= -d/ds(s/s^2+a^2).
Using d quotient rule we av s^2-a^2/(s^2+a^2)^2.

solution to no 2.
L(e^4tcos3t). well known that L(Cos3t)=s/s^2+9.
hence L(e^4tcos3t)= s-4/(s-4)^2+9= s-4/s^2-8s+25.
now d f'L(e^4tcos3t) = f'[ (s-4)/(s^2-8s+25)].
using d quotient rule
we ave s^2-8s+7/(s^2-8s+25)^2.

Alpha Maximus: The product of 19 numbers is the square root of the sum of the inverse of the cube of the result of the sum of the first 13 numbers collective lowest common multiple and the square of the highest common factor of the last 6 numbers... What is the value of the square root of the cube of the subtraction of the other numbers from the largest. Whoever solves this within 3 hrs will get 50,000naira from me, I'm serious. May the odds be in ur favour.

Hunger Games stuff

@akpos4uall

I'm sorry I've been having trouble with my email address. Changing it has become an impossible task.

rhydex 247:

solution to no 1.
L(tcosat) using the shift theorem.
well known that L(cosat)= s/s^2+a^2 where s>0.
now d L(tcosat)= -d/ds(s/s^2+a^2).
Using d quotient rule we av s^2-a^2/(s^2+a^2)^2.

solution to no 2.
L(e^4tcos3t). well known that L(Cos3t)=s/s^2+9.
hence L(e^4tcos3t)= s-4/(s-4)^2+9= s-4/s^2-8s+25.
now d f'L(e^4tcos3t) = f'[ (s-4)/(s^2-8s+25)].
using d quotient rule
we ave s^2-8s+7/(s^2-8s+25)^2.

Thanks for the approach
but note that the first principle required for Q1(not the shifting principle as used)*although a slight flaw in your ans.
And Q2 implies the laplace transform of the derivative i.e Lf'{}.( not so conventional) nd thats d reason i choose the notation f'L{ } ...nice try though...
Sti awaiting replies

2nioshine:
Thanks for the approach
but note that the first principle required for Q1(not the shifting principle as used)*although a slight flaw in your ans.
And Q2 implies the laplace transform of the derivative i.e Lf'{}.( not so conventional) nd thats d reason i choose the notation f'L{ } ...nice try though...
Sti awaiting replies

hmmm. u mean d/dt[L(e^4tcos3t)]. if that is d case solution pending.

Laplacian:
for any confusion, refer 2 page 121

Clear.


Congrats on resolving your 3-yrs old question. See the power of sharing!!!

Any other enigmas for us to brainstorm on?

Laplacian:
for any confusion, refer 2 page 121

Wow, great bruv! Kudos man.... nice work!

@Alpha, I see you.... evil worded problem on rampage

jackpot: Clear.


Congrats on resolving your 3-yrs old question. See the power of sharing!!!

Any other enigmas for us to brainstorm on?

My question. You were in the conspiracy too, weren't you?

^I wouldnt crack my little brain on your question. I would rather leave that priviledge to you.

so dear, write your question as a mathematical equation. Then I will see whether it is solvable

doubleDx:

Wow, great bruv! Kudos man.... nice work!

@Alpha, I see you.... evil worded problem on rampage

Hi DoubleDx. How have you been?

I finally bought the Advanced Engine Maths by H.K.Dass @N1700. It's a nice read so far. His methods of solution are cool.

Thanks for recommending the text.

jackpot: Easy one. . .

But there are many answers to this question sha.

how many do you need? I am serious.


NOTE
I know that the sequence is neither an AP or a GP.

ok @ jack pot quickly give me the solution to the problem within 5mins

find the missing number s; 5,_,1205,_271205,_ ok madam jackpot give me the solutions na

simplify (9-0)(9-1)(9-3)(9-4)...(9-n)

jackpot: Clear.


Congrats on resolving your 3-yrs old question. See the power of sharing!!!

Any other enigmas for us to brainstorm on?

hey, i have more unanswered problems than any user on this thread , few in number theory though (as compared 2 elliptic functions)...here are two interestin ones, i've cracked d first one though, maybe u can generalise my result lik u did 4 alpha maximus, or mayb ur method of solution 'll get 2 b prettier than mine, in any case i'll get 2 learn smthin....subsequently, in all my questions, whenever i refer 2 integers, i mean NON-ZERO integers to avoid the supply of trivial solutions:

1.) find the set of all integer pairs (x, y) such that
x²+y² is divisible by xy+1

2.) if
4ⁿ+2ⁿ+1 is a prime, show that n is some power of 3

Humphrey77: ok @ jack pot quickly give me the solution to
the problem within 5mins

bro, are you really interested in the solution or in speed?


Na exam be this?

jackpot: Hi DoubleDx. How have you been?

I finally bought the Advanced Engine Maths by H.K.Dass @N1700. It's a nice read so far. His methods of solution are cool.

Thanks for recommending the text.

I'm good general jackpot, I hope you are too?...It's been a while! HK Dass Engine is a good math material, nice you bought it!

Hailings!

jackpot: bro, are you really interested in the solution or in speed?


Na exam be this?

no na brain stroming

jackpot: bro, are you really interested in the solution or in speed?


Na exam be this?

give the result of the missing numbers ( happy)

Laplacian:
hey, i have more unanswered problems than any user on this thread , few in number theory though (as compared 2 elliptic functions)...here are two interestin ones, i've cracked d first one though, maybe u can generalise my result lik u did 4 alpha maximus, or mayb ur method of solution 'll get 2 b prettier than mine, in any case i'll get 2 learn smthin....subsequently, in all my questions, whenever i refer 2 integers, i mean NON-ZERO integers to avoid the supply of trivial solutions:

1.) find the set of all integer pairs (x, y) such that
x²+y² is divisible by xy+1

2.) if
4ⁿ+2ⁿ+1 is a prime, show that n is some power of 3

ok. Nice questions

Humphrey77: find the missing number s; 5,_,1205,_271205,_ ok madam jackpot give me the solutions na

rest assured, I'll give you the solution before the day runs out. I am kinda busy now.

state and prove LAGRANGE THEOREM ( ABSTRACT ALGEBRA)

SIMPLIFY (9-0)(9-1)(9-3)...
AND STATE THE BREAKING POINT

:HAPPY DAY

Humphrey77: SIMPLIFY (9-0)(9-1)(9-3)...


:HAPPY

pls simplify it lets c @humphrey

Laplacian:
pls simplify it lets c @humphrey

HERE IT
SOLUTION
(9-0)(9-1)(9-2)(9-3)(9-4)(9-5)(9-6)(9-7)(9-(9-9)(9-10)... AT THE POINT (9-9) WE HAVE ZERO . CLEARLY , (: THE PRODUCT EXPANSION AS IT VAULE EQUAL TO ZERO WITH WHICHTHE BEAKING POINTEXIST AT (9-9) )

THE EXPRESSION (9-0)(9-1)(9-2)(9-3)(9-4)... IS EQAUL TO ZERO

HAPPY

FIND THE VAULE OF K:6"K -31(2"K)+32 IS EQUAL TO ZERO ) : HAPPY ! ** DAY TO MY QUEEN

MY QUEEN GAVE ME THIS QUESTION YESTERDAY AND AS ME TO COMPLETE IT : I said to my self what am i going to do ? told her that am going to post it to my oga on top : complete these : 6,__,204__,__,25779. (hint they are multiple of 3) : HAPPY

Alpha Maximus: The product of 19 numbers is the square root of the sum of the inverse of the cube of the result of the sum of the first 13 numbers collective lowest common multiple and the square of the highest common factor of the last 6 numbers... What is the value of the square root of the cube of the subtraction of the other numbers from the largest. Whoever solves this within 3 hrs will get 50,000naira from me, I'm serious. May the odds be in ur favour.

dude, if u wan post nawti question, at least b real n creative with it. I can see two sums in the question, buh just a conjunction (and)... That aint maths!!