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yeah, correct. as i said above, i seem to not know how to add
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I agree... as long as you mean the last term is 32(pi)^2 
Ryan "Punctuality is only a virtue for those who aren't smart enough to think of good excuses for being late" John Nichol "Point Of Impact"
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As already pointed out, there's a maths forum but here goes 
y = f(x)
slope M = dy/dx
You have one point on the line - (4PI, f(4PI)), and since it is the tangent to the curve, its slope will be -1/M it has the same slope. Just use the slope-point equation for the straight line (IIRC, it's y - y1 = m(x - x1)).
To find m at the point, simply substitute for x and y in dy/dx.
Please post in the maths forum in future. HTH.
Nostalgia is a very powerful emotion.
-- corrected slope of tangent, which I misread as normal. Thanks to Ryan. modified at 4:57 Wednesday 11th October, 2006
Cheers,
Vikram.
"whoever I am, I'm not other people" - Corinna John.
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Vikram A Punathambekar wrote: Nostalgia is a very powerful emotion
thanks, already got it. as i said above(2 times now lol), i seem to have lost all ability to do simple addition
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Vikram A Punathambekar wrote: since it is the tangent to the curve, its slope will be -1/M.
Not quite. If you know the gradient of a point on the curve, the gradient of the tanget line is equal to the gradient of the curve at the point of the tangent. The gradient of the normal to the line at the given point is -1/m.
The equation of tangent the line is y = f'(x<sub>0</sub>)(x - x<sub>0</sub>) + f(x<sub>0</sub>) where x<sub>0</sub> is the point he wants the tangent at and f'(x) is the first derivative of the curve.
Ryan "Punctuality is only a virtue for those who aren't smart enough to think of good excuses for being late" John Nichol "Point Of Impact"
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Thanks, mate, I corrected my post.
I misread tangent as normal. 
Cheers,
Vikram.
"whoever I am, I'm not other people" - Corinna John.
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Vikram A Punathambekar wrote: Thanks, mate, I corrected my post.
No worries
Ryan "Punctuality is only a virtue for those who aren't smart enough to think of good excuses for being late" John Nichol "Point Of Impact"
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First I'll rephrase the question:
f(x) = g(x).h(x); where
g(x) = x; and
h(x) = 8cos(x) - 2sin(x)
f'(x) = ?
Now some basic calculus results you should know by heart:
sin’(x) = cos(x); and
cos’(x) = -sin(x)
Finally the product rule:
(g(x).h(x))’ = g’(x).h(x) + g(x).h’(x)
So in our case:
g’(x) = 1; and
h’(x) = -8sin(x) – 2cos(x)
Thus:
f’(x) = g’(x).h(x) + g(x).h’(x)
= 1.(8cos(x) - 2sin(x)) + x.(-8sin(x) – 2cos(x))
= 8cos(x) - 2sin(x) – 8.x.sin(x) – 2.x.cos(x)
= cos(x).(8-2x) – sin(x).(2+8x)
Now you can use f’(x) to find the gradient of the tangent and any point.
Once the gradient is known all that needs to be done is to plug it into y=mx+c and choose a c so that it touches the curve at the desired location.
Steve
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Image that we have an array of integers, and we want to know any 2 numbers that their sum is x, using O(n), is it possible ?
------------------------------
"The Soapbox has been so ..."
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sounds like a programming question!
Is this a homework?
(..I don't think i am in The Lounge anymore...")
Anyway, Are there negatives values in that array?
-- modified at 6:10 Wednesday 11th October, 2006
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That's what pop into my mind:
<br />
sumx(A[],x)<br />
{<br />
Array S[x/2];<br />
for(i=0,i<S.lenght;i++) S[i]=0;
for(i=0,i<A.lenght;i++)<br />
{ <br />
a=A[i];<br />
if(a<0 || a>x/2) continue;<br />
if(a<x/2) S[a] |= 1;<br />
if(a>x/2) S[x-a] |= 2;<br />
}<br />
return S;<br />
}<br />
You get an array S.
For a number k<(x/2) you have:
if S[k] is 0 then neither number k, nor (x-k) is in the input array.
if S[k] is 1 then only number k was in the input array.
if S[k] is 2 then only number (x-k) was in the input array.
if S[k] is 3 then k and (x-k) was in the input array. The sum of those two is k+(x-k)=k+x-k=x. bingo! ![Rose | [Rose]](https://www.codeproject.com/script/Forums/Images/rose.gif)
For a number j>(x/2) you have k=(x-j) and continue as above.
The array S can be more comlex, like storing the index of two number in the input array A, if you like. Dealling with negative numbers as well might need a more complex aproach.
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Small bug:
Kastellanos Nikos wrote: if(a<0 || a>x/2) continue;
if(a<0 || a>x) continue;
To put the algorithm in simple words, do a counting sort for numbers <x , so O(n).
lookup is O(1) for a hash table; will have to do x/2 lookups, so O(x).
Overall O(x+n).
The complexity of algorithm is O(x+n); If x is reasonably small, then the algorithm is good enough.
If x>n*log n; a quick sort followed by a binary search is better, but will be managable only in O(n*log n)
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i have written a code to find the determinant of a matrix, i would like to ask is it fishy or can it be devoloped? thanks to you all for replies... my code is written in vb.. and one more thing... i wonder is it the right place to post? thanks anyway...
here is my code...
Sub deneme()
Dim a(1 To 10, 1 To 10) As Double
Dim t As Integer
a1 = 1
t = 0
GoTo git
For i = 1 To 3
For j = 1 To 3
t = t + 1
a(i, j) = t
Next j
Next i
git:
a(1, 1) = -1
a(1, 2) = 0
a(1, 3) = -50
a(2, 1) = 1
a(2, 2) = 4.92
a(2, 3) = -2
a(3, 1) = -1.4
a(3, 2) = 3
a(3, 3) = 1
For i = 1 To 3
For j = 1 To 3
If i = j Then
If a(i, i) = 0 Then GoTo devam
For g = i + 1 To 3
tut = a(g, i)
For b = 1 To 3
a(g, b) = a(g, b) - (a(i, b) * (tut / a(i, i)))
Next b
Next g
End If
Next j
Next i
devam:
For i = 1 To 3
For j = 1 To 3
Cells(i, j) = a(i, j)
Next j
Next i
For i = 1 To 3
a1 = a1 * a(i, i)
Next i
Cells(1, 10) = a1
End Sub
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Well I don't know what your formula does, but why not use the Laplace Expansion?
regards
modified 12-Sep-18 21:01pm.
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first of all thanks for replying... i've examined the laplace expansion. But in high order matrices finding the minor matrices or the cofactors is hard to code for me and i think, i can turn the square matrice to the triangle square matrice. so i just multiply the a[j,j] s then i find the determinant of it.. by the way.. my code is just turn the square matrice to a triangle square matrice. thanks for comments...
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Hi,
I do not know if your code works or not, but if I were you I would use the definition of determinant to compute it.
The hardest part is to code a function to get permutations.
Regards
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thanks for replying my message. my code is working for me for all inputs and it works only square matrices you know... and yeah you are very right... and by the way if you want to try it you can use excel microsoft vb.. thanks for advance...
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But ... determinants are only defined for a square matrix... isn't then the range what you want to compute ?
Regards
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sure i am... i need only square matrices... n unkowns n linear equations. best regards....
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Hey all,
I'm not sure this is the right forum, as I'm not sure what I'm looking for is an algorithm. 
I have an IP address along with a port.
My goal is to determine which running application is using this port.
Googling taught me that this process is called Port-To-Application Mapping (aka PAM),
but I couldn't find any code which does that...
The best thing I found was a program which knows how to do it.
Anyone knows any PAM algorithms?
Thanks in advance,
Shy.
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anyone know step by step mathematical processes for image convolution using laplacian? plz explain...
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Hi,
If I remember well ...
If your image M is considered as a x*y matrix (ie gray scaled) of pixels and a point P is then described as P(i,j) where 0 <= i < n and 0 <= j < n, and the laplacian is considered as a 3*3 matrix L defined as [[a,b,a],[b,c,b],[a,b,a]], the discrete convolution gives you another n*m matrix R where R(x, y) = Sum(i=0, i < 3, Sum(j=0, j < 3, L(i,j)*P(x + i - 1, y + j -1))) (0 outbounds)
Regards
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anyone can help find out any source code for prediction using bayesian belief network algorithm pls...A.S.A.P
respect,relax,respon
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