Geofavor:

I was being precise with the topic/question being argued, not maths in general. Not numbers.

You said earlier that, 5C0 is invalid in this aspect of maths(permutation n combination) because of the comparison of an actual number with zero. You said they shouldn't be compared because 0 is valueless.

Geofavor:

Which was why i reminded you that in this case, it isn't. Why? Because factorial is involved!

possibilita:
yes that's hw 1c is written

agentofchange1:

ok here

possibilita:
thank u very much bros

agentofchange1:


ok here

agentofchange1:


ok here

Antoinne:

There seems to be a mistake in the first two lines in the first picture with y². Would affect the final answer significantly.

jackpot:
Find the inverse laplace transform of 1/(s^5+a^5)

tags: Laplacian, agentofchange1, doubleDx, dejt4u, Alphamaximus, jaryeh

Antoinne:

I took it you'd understand my little explanations.
Let's get the first part sorted. Look at the attached pic.

After resolving the elemental distances, dx & dy and multiplying. Follow the extension down to y = 1-x at z=0, you'll see that the distance dx on the other side is not exactly parallel to the distance you just resolved. I marked it with "A". There's still some kind of triangle there, showing that you elemental distance is not representative of the surface.

If that's too difficult, look at the basic 2D triangle below it. If you took a small portion on the x-axis called dx. Do you think it'll be equal to the length marked "A" on that same triangle merely by tracing it up?

It's the same thing happening on that surface above. You get it?

MaxGraviton:
Please, This is really Urgent. Someone should please help me with the name of that Maths textbook popularly known as BONDE. . . Bonde is the name of the person that wrote the book. But Please I need the actual name. Thanks.

Laplacian:

yes, i think they will b equal. This new diagram luks beta

Antoinne:

Pure Mathematics for Advanced Level
by B.D. Bunday & H. Mulholland

agentofchange1:

considering the denominator
s^5 +a^5
in general
xⁿ + yⁿ
=(x+y) £ y^k* x^(n-1-k) for k=0,1 2, . . . ,(n-1)
where £ = sigma (summation)
does that ring a bell?. I know say you na Prof.

Antoinne:

Come on! No, they will not be equal. Look at the diagram again. "A" will have to be

A.cos(phi) = dx. Not A = dx.

You see it?

Laplacian:

u'r right, they form a right angled triangle...maybe i shoul just take a rest!

jackpot:
Find the inverse laplace transform of 1/(s^5+a^5)

tags: Laplacian, agentofchange1, doubleDx, dejt4u, Alphamaximus, jaryeh

jackpot:
Please help out

Solve the functional equation below for f(x):

1. f(x)-f(-x)= - 1/x

2. f(x)-f(-x)=-1/x, given that f is odd.

tags: agentofchange1, dejt4u, STENON, AmazingAngel, Antoinne, Laplacian, Karmanaut, doubleDx, etc

Laplacian:

let k_n=e^2#ni/5, n=1,...,5 where # is pi
then,
1/(s^5+a^5)=product1/(s-ak_n)....resolve using partial fraction, cover-up rule (all factors are distinct) e.t.c

Laplacian:

if f(x)-f(-x)=-1/x then f(x) can neither be an even function nor have an even part, otherwise we have f(-x)=f(x) or f(x)-f(-x)=0, plug into d given eqn we have 0=-1/x, takin inverse makes x undefined. If f(x) is odd then f(-x)=-f(x), hence 2f(x)=f(x)-f(-x)=-1/x

jackpot:
please, help me out.

[b]1. Consider the parabola: x=(3-2y)(4y+3). Obtain
i. Directrix equation
ii. Focus
iii. Vertex
iv. Length of it's latus-rectum

jackpot:
Well, unlike integers that are either even or odd, a function may be neither even nor odd, eg h(x)=sin x-cos x, x+x².
Also, functions may have non-integral exponent, eg h(x)= x^e=x^{2.718281828. . .} or p(x)= £ x, where £ is the Gamma function or q(x)=x^2/7.

Now, my question is: how are we sure that there are no neither-even-nor-odd functions f satisfying the equation:
f(x)-f(-x)= -1/x ?

tags: Laplacian, agentofchange1, Laplacian , agentofchange1, doubleDx, Antoinne, dejt4u, AlphaMaximus, Laplacian etc

jackpot:
Well, unlike integers that are either even or odd, a function may be neither even nor odd, eg h(x)=sin x-cos x, x+x².
Also, functions may have non-integral exponent, eg h(x)= x^e=x^{2.718281828. . .} or p(x)= £ x, where £ is the Gamma function or q(x)=x^2/7.
Now, my question is: how are we sure that there are no neither-even-nor-odd functions f satisfying the equation:
f(x)-f(-x)= -1/x ?

Antoinne:


Parabola of x = (3-2y)(4y+3)
x = 12y + 9 - 8y² - 6y
x = 9 + 6y - 8y²

Parabola is about y, since y has two values.
We can treat this as a mirror y = 9 + 6x - 8x², and then invert the mirror at the solution, if x is easier to work with.

anyways, for a parabola y = ax² + bx + c with a focus at p distance above the vertex V(h,k).
dy/dx = 2ax + b. At the vertex, you have the least variation, so dy/dx=0
2ax + b = 0
x = -b/2a, this should correspond to h.

y = ax²+bx+c at x = -b/2a, k
k = (4ac - b²)/4a

Assuming parabola at origin and following definition y = 4px²
Move parabola to right of xy plane so that vertex is V(h,k)
4p(y-k) = (x-h)²
Relating coefficients produces p = 1/4a, (focus)

Latus-rectum is length across focus.
if Parabola at origin (0,0)
4py = x²
with y = p on curve
x² = 4p.p = 4p²
x = 2p

Length of latus-rectum
2.x = 4p

Now, from earlier equation y = 9 + 6x - 8x² (mirror)
a = -8, b = 6, c=9

focus, p = 1/4a = 1/(4*-8 )
=-1/32

Vertex, V(h,k)
h = -b/2a = -6/(2*-8 ) = 3/8
k = (4ac - b²)/4a = 81 / 8

Length of latus-rectum = 4p
=abs(4.(-1/32)) = 1 / 8

Directrix equation: y = 325/32, i.e. (1/32 + 81/8 )

We can now go ahead and mirror our solution by just changing all x to y, and y to x
i. Directrix equation: x = 325/32
ii. Focus: -1/32
iii. Vertex: V(81/8, 3/8 )
iv. Length of it's latus-rectum: 1/8

Laplacian:

Assuming parabola at origin and following definition y = 4px²?

jackpot:
Unlike the other solution that is complex upon complex with all the known operations (imaginary, plus, minus, division, multiplication, square root, so many lines , etc) in mathematics coming in a single solution.