what are identity functions?
437573
Functions that maps x to x\nWhere x is an element of the function
984829
basically, think of a function as a machine:\n\ntake this an an example: f(x) = 2x +3\nsee the x in brackets?\n\nif x = 1, then f(1) = 2(1) + 3 = 5\nif x = 2, then f(2) = 2(2) + 3 = 7\nif x = 3, then f(3) = 2(3) + 3 = 9\n\nsee whats happening? the answer DEPENDS on number inputted in the brackets.\n\nso therefore, the definition you quoted "a consistent, predictable, relationship between the independant and dependant variable" is true.
733318
I think your identity should be :\ncsc^2(x) / [1 + tan^2(x)] = cot^2(x)\nas you forgot to square the tan(x).\n\nWorking on the 1 + tan^2(x) gives:\n1 + sin^2(x) / cos^2(x)\n= [sin^2(x) + cos^2(x)] / cos^2(x)\n= 1 / cos^2(x)\n\nNow we have : csc^2(x) / [1 / cos^2(x)]\n= [1 / sin^2(x)] / [1 / cos^2(x)]\n= cos^2(x) / sin^2(x)\n= cot^2(x), which was to be shown.
856009
false,\nuse the identity and -identity,\nboth are invertible, but the sum is the zero matrix, that is NOT invertible.....
296652
Where's the function?
1033116
A X = D\nX = A^(-1) D. (A inverse * D)\n Do it ur self.
200427
tangent = opposite/adjacent (toa)\nsine = opposite/hypotenuse (soh)\ncosine = adjacent/hypotenuse (cah)\n\nTom's old aunt sat on her coffin and howled.\n\nI learned that 10 years ago and I still remember it for some reason. Then cotangent is just the flip of tangent, secant is the flip of cosine, and cosecant is the flip of sine.\n\nAll, sine, tangent, cosine (all students take calculus) tells you which of the trig functions are positive in the various quadrants I-IV, in that order.\n\nThese are the only things in trig that you need to sit down and memorize (more or less). Almost else can be justified using this information, specifically the trig identities. The best way to remember these guys is to derive them yourself. If you try to memorize them all, you're bound to forget them since they don't have convenient mnemonics, but if you derive them yourself and understand why they hold, you tend to remember them.\n\nSpecifically you can prove easily that tan = sin/cos, and that sin^2 + cos^2 = 1. Once you've proven these, you can prove a ton of other identities.
80570
f(x) = x^2 - 1\nIf you replace x with x + h, then\nf(x + h) = (x + h)^2 - 1\nYou will NOT do any operations when changing the function. Instead, just SUBSTITUTE. (A technique is putting parentheses in the variable to change, except if it already has its own parentheses. Then you may exchange the variables inside the parentheses with what you want to substitute)\nf(x) = x^2 - 1\nPutting the parentheses, we get\nf(x) = (x)^2 - 1\nThen replace x with x + h\nf(x + h) = (x + h)^2 - 1\nHope this helps ;-)
948694
The anti-derivative of a function f(x) is another function F(x) so that F´(x) = f(x). \n\nIf f(x) = x^n (n<-1 or n>-1) then the anti-derivative is F(x) = (x^(n+1))/(n+1)  + C ... we need to add an arbitrary constant.\n\nIn your case we have:  f(x) = x^1  and  F(x) = (x^2)/2 + C
204094
ok, i think i have an idea for this one. \nSketch of a proof:\nSuppose that a is an element of order k. Also suppose that f is an isomorphism. Then we know that f(a*a*a...k times)=f(a)*f(a)*f(a)*...k times (homomorphism definition). We know that f(a)^k is the identity because isomorphisms take the identity to the identity. Now see that k must be the first number with the quality f(a)^k=e. This comes as a result of the isomorphism having a kernel with one element. If we suppose for the sake of contradiction that f(a)^m=e with m<k, then we would have f(a*a*a*a...m times)=e, so that means that f(a^m)=e at the same time as f(a^k)=e with a^m not equal a^k, this is a contradiction.\n\nQED...ish
114851
It is a function, that express the relation between the amount of the input data and the amount of primitive operations that will be needed to complete a task.\n\nFor example if you have to find the sum of N numbers, one trivial solution would be to scan all of them and collect the sum incrementaly. Then you will be traversed one time each number, so the computational complecity will be f(N) - you see it is linear function.\n\nAnother example is the sorting N numbers. The famous quick sort algorithm have f(log(2,N)) - it is logarithmic.\n\nOfthen this function is showed as theta from the greek alphabet :)
1039086
<identity impersonate="true" />
465001
Zero is known as the additive identity because when you add it to anything (say "x"), it doesn't change the value of "x".
760867
f(x) = (x+3)(x^2 + 16)
207605
bool function(int a, int b, int c)\n{\nreturn a || b || c;\n}
946513
Subject: Re: Proving trig identiies...small error..thisis the right one\nFrom: \nDate: Monday June 14, 2004 at 08:35:54\n\nRHS= 2csc^2x=2(cot^2x+1) \n\nnote :-(csc^2x=cot^2x+1) \n\n=2cot^2x+2 \n\n=2cos^2x/sin^2x +2 \n\nnote :-((cot^2x=cos^2x/sin^2x) \n\n=2cos^2x/(1-cos^2x) +2 \n\nnote :-((sin^2x=1-cos^2x) \n\n=2cos^2x +2(1-cos^2x) /(1-cos^2x) \n\n=2cos^2x +2 -2cos^2x) /(1-cos^2x) \n\n=2/(1-cos^2x) \n\n=2/(1-cosx)(1+cosx) \n\nnote :-((1-cos^2x=(1-cosx)(1+cosx)) \n\n\nTHEREFORE lHS =2/(1-cosx)(1+cosx) (1-cos^2x=(1-cosx)(1+cosx)) \nnow for the RHS we get \n\nRHS= 1/1+COSX + 1/1-COSX \n\n=(1-cosx) + ( 1 + cosx)/(1-cosx)(1-cosx) \n\nnote :-(( through expansion) \n\n=2/(1-cosx)(1+cosx) \n\nnote :-(cosx on the numerator cancel out) \n\ntherefore RHS= LHS \n\nTHEREFORE 2csc^2x= 1/1+cosx + 1/1-cosx \n\nas required\n\n\n\nCurrent Thread\n   Proving trig identiies   a girl - Monday June 14, 2004 at 08:06:52 \n   Re: Proving trig identiies   - Monday June 14, 2004 at 08:31:28 \n--> Re: Proving trig identiies...small error..thisis the right one   - Monday June 14, 2004 at 08:35:54 <--
809444
ohh ya,,i know this,,\n\n1)f(x)=((x)^2)+2\n\n2)f(x)=( ( (x-3)^(-2) ) +2)\n\nbring on more of domain and range questions,,i love those problems,,