wow,,,c as Guys they kill maths father....now i know..smm ppl r really mathematically mad
.

is -1^2=(-1)^2...prove ur answer

"I tell them that if they will occupy
themselves with the study of mathematics
they will find in it the best remedy against
the lust of the flesh.".......@ thomas mann@

Calculusf(x):
...great master but i guess this is newton-raphson iterative method

The prof after the order of Sir Isaac Newton. Yes, you're right. It's otherwise called the nondeterministic polynomial , it's used for solving transcedential equations.

@those using raphson method to solve ques e.g 4^x=8x,,3^x+4^x=5^x.
y is it dat u guys are always sayin dat x1 is 2?
dat means u are also predictin and dat is not an analytical method of solvin it..

i wud like d "5 STAR GENERAL" d citizen to solve it using an analytical method without using raphson as e claims dat it canot only b done graphically.
i rest my case i mean no offense[dont take it personally]

i wud like d "5 STAR GENERAL" d citizen to solve it using an analytical method without using raphson as e claims dat it canot only b done graphically.
i rest my case i mean no offense[dont take it personally]
we are all hear to learn so teach us ur method sir

Mbahchiboy: i wud like d "5 STAR GENERAL" d citizen to solve it using an analytical method without using raphson as e claims dat it canot only b done graphically.
i rest my case i mean no offense[dont take it personally]
we are all hear to learn so teach us ur method sir

Aight Sire, solve it with the best method you know let's learn from you!

doubleDx:

Aight Sire, solve it with the best method you know let's learn from you!

SIR DX i was referin to d 5 star general d citizen,i did mine graphically but u can also paste urs so we could learn from u sir

The five star general,

Here is the clue to the analytical solution to the problem: 3^x + 4^x = 5^x

Divide both side of the equation by 4^x

3^x/4^x + 4^x/4^x = 5^x/4^x

0.75^x + 1 = 1.25^x

(1-0.25)^x + 1 = (1 + 0.25)^x

(1 + (-0.25))^x + 1 = (1 +0.25)^x
U will use binomial theorem to do the expansion and limit the expansion to x^2

(1 +x)^n = 1+ nx + n(n-1)x^2/2!. Use the expansion to get the answer to x by quadratic equation.

Mbahchiboy: SIR DX i was referin to d 5 star general d citizen,i did mine graphically but u can also paste urs so we could learn from u sir

Okay, I get you now! My method is graphical like yours, let's w8 4 other generals to contribute. 1luv

The five star general

Prove that 0! = 1

d citizen:
The five star general

Prove that 0! = 1

...dis ma question u r bringin up..go 2 d privious pages.2 c proof.

d citizen:
The five star general

Prove that 0! = 1

n! = n(n-1)!
:. at n = 1

1! = 1.(1-1)!
Since 1! = 1

then 1 = 1.0!

:. 0! = 1

I said 0 factorial equal to 1

d citizen:
The five star general,

Here is the clue to the analytical solution to the problem: 3^x + 4^x = 5^x

Divide both side of the equation by 4^x

3^x/4^x + 4^x/4^x = 5^x/4^x

0.75^x + 1 = 1.25^x

(1-0.25)^x + 1 = (1 + 0.25)^x

(1 + (-0.25))^x + 1 = (1 +0.25)^x
U will use binomial theorem to do the expansion and limit the expansion to x^2

(1 +x)^n = 1+ nx + n(n-1)x^2/2!. Use the expansion to get the answer to x by quadratic equation.

|sighs|

d citizen:
The five star general

Prove that 0! = 1

CHECK PAGE 6O

to my generals plz help me out
(1)if
coshx=1+x^2/2!+x^4/4!....,
sinhx=x+x^3/3!+x^5/5!.....,
find tanhx?
(2)find x if x^5-32=0
[hint::x must have five root e.g a,b,c,d or e]

Mathematics is never gona to lead you to higher truth..

Mbahchiboy: to my generals plz help me out
(1)if
coshx=1+x^2/2!+x^4/4!....,
sinhx=x+x^3/3!+x^5/5!.....,
find tanhx?
(2)find x if x^5-32=0
[hint::x must have five root e.g a,b,c,d or e]

Hmmmm. . .
1. Your first question demands to be proved using Euler's formula for identities and hyperbolic functions (e^x).
If coshx = 1 + x^2/2! + x^4/4!. . .
Sinhx = x + x^3/3! + x^5/5!. . .

Euler's identity (e^x) for coshx = (e^x + e^-x)/2
sinhx = (e^x - e^-x)/2
:. Tanhx would be = sinhx/coshx
=> (e^x - e^-x)/2 / (e^x + e^-x)/2
=> (e^x - e^-x)/2 * 2/(e^x + e^-x)
=> (e^x - e^-x)/(e^x + e^-x)

So tanhx = (e^x - e^-x)/(e^x + e^-x)

Ortarico:

Hmmmm. . .
1. Your first question demands to be proved using Euler's formula for identities and hyperbolic functions (e^x).
If coshx = 1 + x^2/2! + x^4/4!. . .
Sinhx = x + x^3/3! + x^5/5!. . .

Euler's identity (e^x) for coshx = (e^x + e^-x)/2
sinhx = (e^x - e^-x)/2
:. Tanhx would be = sinhx/coshx
=> (e^x - e^-x)/2 / (e^x + e^-x)/2
=> (e^x - e^-x)/2 * 2/(e^x + e^-x)
=> (e^x - e^-x)/(e^x + e^-x)

So tanhx = (e^x - e^-x)/(e^x + e^-x)

I KNOW DAT BRO

@ortarico
i want it to b express d way i express sinhx and coshx

Mbahchiboy: to my generals plz help me out
(1)if
coshx=1+x^2/2!+x^4/4!....,
sinhx=x+x^3/3!+x^5/5!.....,
find tanhx?
(2)find x if x^5-32=0
[hint::x must have five root e.g a,b,c,d or e]

2. To your second question:
To find the factors of x^5 - 32 = 0, solve x^5 = 32
x^5 = 2^5
x = 2
Therefore, (x - 2) is a factor of (x^5 - 32)
Now simply divide (x^5 - 32) by (x - 2) using long division and you'll get:
(x^5 - 32) = (x - 2) (x^4 + 2x^3 + 4x^2 + 8x + 16)

But (x^5 - 32) should have five values which you denoted to be a, b, c, d and e
:. Further factorization gives their values to be:
(x - 2), (x + 2), (x - 2), (x + 2) and (x - 2)

@ortarico
how did u obtain those values of x after dividin d equ by x-2 to get a new one.
and i dont think x can b -2

Mbahchiboy: @ortarico
i want it to b express d way i express sinhx and coshx

Wow, fine!
Though I only have the basic knowledge of its concept, can it be that way?

Mbahchiboy: @ortarico
how did u obtain those values of x after dividin d equ by x-2 to get a new one.
and i dont think x can b -2

By also factorizing the quotient: (x^4 + 2x^3 + 4x^2 + 8x + 16)
Why do you think it can't?

Ortarico:

By also factorizing the quotient: (x^4 + 2x^3 + 4x^2 + 8x + 16)

WORKINGS PLZ....
CAUSE I DONT THINK IT IS POSSIBLE SIR

Ortarico:

By also factorizing the quotient: (x^4 + 2x^3 + 4x^2 + 8x + 16)
Why do you think it can't?

cause from d ques.
x^5-32=0..
so (-2)^5-32=-64
and dat makes it invalid

Mbahchiboy: cause from d ques.
x^5-32=0..
so (-2)^5-32=-64
and dat makes it invalid

You seem to be right sha. I did it by inspection. It's better if one leaves it as (x - 2) (x^4 + 2x^3 + 4x^2 + 8x + 16) then.

busuyem: Pls, help me with the solution of this question with its explanation:

The difference between the digits of a two-digit number is 1. The number itself is 1 more than 5 times the sum of its digits. If the units digit is greater than the tens digit, find the number.

The two digit no, one will be the unit digit and the other will be the tens digit, according to d question unit digit > tens digit.
Let a = unit digit b = tens digit
.'. a > b
The difference btw d digit of the two digit number is 1 (from the questn)
a-b = 1 (because a is greater that b)......(i)
The number itself is 1 more than 5 times the sum of it digits (from the questn)
b(10)+ a - 1 =5(a + b) ..................(ii)
That ten in bracket is b'cos 'b' is in tens
a= 1 + b .............(iii)
insert the values of 'a' in eqtn ii
10b + (1 + b) - 1 = 5(1 + b + b)
10b + 1 + b - 1 = 5 + 5b + 5b
11b - 10b = 5
b = 5
From eqtn 3
a = b +1
a = 6
So the number is 56.

X^4+2x^3+4x^2+8x+16=0

Ortarico:

You seem to be right sha. I did it by inspection. It's better if one leaves it as (x - 2) (x^4 + 2x^3 + 4x^2 + 8x + 16) then.

...the polynomial is advance,try and read lodovico ferarri's method for quartic equation...don't ask me,if you don't get it ooo coz the formula is very advance...someone said those who study masters in algebra can be taught the formula...BUT YOU CAN USE IT FOR ANY POLYNOMIAL