Is there a closed form for e^(x^2)? I mean a proof... someone told me that there is no such closed form, but I can't trust him :D he's a teacher :))
128942
Integral of e^(x^2)dx = (Sqrt[Pi]*Erfi[x])/2\nErfi[x] = imaginary error function\nErf[x] is the integral of the Gaussian distribution, given by (2/sqrt(pi)) * Integral from 0 to x of (e^(-t^2))dt
64306
On both equations, you have x^ but don't say what x was raised to. I'm assuming that it is x^2. Is this correct?
340716
Check MathForum.org:\n\nQuestion:\nWhat is the quantity 0^0 (zero to the zeroth power)?  Would it be \n0, since 0 times anything is 0, or would it be 1, since anything to the 0 power is 1?  Or is this undefined, too?\n\nAnswer:\nThis, too, is an indeterminate form.  Its logarithm is 0 * infinity.\n\nMore detail from the website:\nSay you wish to introduce a new element (or maybe two elements) infinity which you wish to append to the real number system.  That is not prohibited.  After all, that is how we got from natural numbers to integers (appending negative integers and zero), and from integers to rationals (appending ratios of integers), and from rationals to reals (appending limits of convergent sequences), and from reals to complexes (appending the square root of -1).  What you end up with is not the real number system, however.  Furthermore, if you wish to define the four operations + - * and / for this new system, you probably want them to be the same on real numbers, and just add on the definitions of things like infinity + r and r/infinity, for any real number r.  Some of these work fine.  It makes sense to define:\n\n        infinity + r = r + infinity = infinity\n  (-infinity) + r = r + (-infinity) = -infinity\n                infinity + infinity = infinity\n          (-infinity) + (-infinity) = -infinity\n                       infinity - r = infinity\n                    (-infinity) - r = -infinity\n                       r - infinity = -infinity\n                    r - (-infinity) = infinity\n             infinity - (-infinity) = infinity\n             (-infinity) - infinity = -infinity\n        infinity * r = r * infinity = infinity  for r > 0\n  (-infinity) * r = r * (-infinity) = -infinity  for r > 0\n        infinity * r = r * infinity = -infinity  for r < 0\n  (-infinity) * r = r * (-infinity) = infinity  for r < 0\n infinity * infinity = (-infinity) * (-infinity) = infinity\n infinity * (-infinity) = (-infinity) * infinity = -infinity\n                       infinity / r = infinity  for r > 0\n                    (-infinity) / r = -infinity  for r > 0\n                       infinity / r = -infinity  for r < 0\n                    (-infinity) / r = infinity  for r < 0\n                       r / infinity = 0\n                    r / (-infinity) = 0\n\nWhere we get into trouble is with defining the following:\n\n        infinity + (-infinity)\n     (-infinity) + infinity\n        infinity - infinity\n     (-infinity) - (-infinity)\n               0 * infinity\n        infinity * 0\n               0 * (-infinity)\n     (-infinity) * 0\n        infinity / infinity\n        infinity / (-infinity)\n     (-infinity) / infinity\n     (-infinity) / (-infinity)\n        infinity / 0 = infinity\n    (-infinity) / 0 = -infinity\n\nThese expressions are called "indeterminate forms."  These can all have a large range of different values, depending on exactly where the "infinity" parts came from.\n\nAs a result, the system you construct is not closed under addition, subtraction, multiplication, or division.\n\nOther indeterminate forms are 0^0 and 1^infinity.\n\nSo the answer, is 0^0 is undefined or indeterminate.
437420
To find a closed form for that infinite sum was known as the Basel Problem. The sum is in fact zeta(2) where zeta is the Riemann Zeta function.\n\nBut how did Euler do it? Well, he was a very clever man. I suggest taking a look at this link for a detailed answer.\n\nhttp://mathworld.wolfram.com/RiemannZetaFunctionZeta2.html
468519
Your limit is not an indeterminate form, so there is no harm in simply plugging in zero for x:\n\n(1+sin(3/0))^0.  The term that is the base of the exponentiation is finite, and any finite value to the 0th power is 1.\n\nSteve\n\nEDIT - I recently thought that perhaps you meant x->infinity.  In that case, take the logarithm of both sides (y the limit), then log y = x*log(1+sin(3/x)) = log(1+sin(3/x)) / (1/x).  using l'Hospital's rule, we have lim x->inf of (-3/x2)cos(3/x)(1/(1+sin(3/x)))/(-1/x2) = 3.  So we have log y = 3, and y = e^3, as u stated.
151880
This integral does exist (it can be evaluated for any endpoints you choose). Since it cannot be represented as a finite combination of elementary functions and operations, it is given its own name, the error function. Many integrals like this have their own names and are quite useful in applied mathematics. See http://mathworld.wolfram.com/Erf.html .\nTo say that it does not exist is unrigorous.
526338
Even 'tho it's obvious, you really should force yourself to say 'derivative with respect to x' because, later on, it won't be quite so obvious what the derivative is being taken with respect to ☺\n\nFirst of all √(4-x²) = (4-x²)^½.  So\n(d/dx)(4-x²)^½ \n½(4-x²)^(-½) (d/dx)(4-x²) (derivative of an expont)\n½(4-x²)^(-½) [(d/dx) 4 - (d/dx)x²] (derivative of difference)\n½(4-x²)^(-½) [0 - 2x] (derivative of constant and exponent)\n½(4-x²)^(-½) (-2x)  (algebra)\n-x(4-x²)^(-½)  (algebra)\n-x/(4-x²)^½ (algebra)\n1x/√(4-x²) (more algebra ☺)  which is the final answer.\n\nSee why you needed to learn algebra? ☺\n\n\nDoug
451577
f(x,y) = x^2*y*(e^(-x^2-y^2))\n\nfx(x,y) = 2x*y*e^(-x^2 - y^2)(1-2x^2) this is wrong:\nf_x(x,y) = x^2y ( e^(-x^2-y^2)) (-2x) + 2xye^(-x^2-y^2))\n= 2xye^(-x^2-y^2) [-x^2 + 1] =0\nif x=0 or y=0 or x^2 =1, \nx=0, y=0, x=1 or x=-1\n\nfy(x,y) = x^2*e^(-x^2 - y^2)(1-2y^2) \nf_y(x,y)=  x^2 y (e^(-x^2-y^2)) (-2y) +  x^2(e^(-x^2-y^2))\n=x^2e^(-x^2-y^2) [-2y^2 + 1] \nx=0, y=1/sqrt(2),y=-1/sqrt(2)\n\nso the critical points are: (0,y)\n(0,0), \n(1,1/sqrt(2)), (1,-1/sqrt(2))\n(-1,1/sqrt(2)),(-1,-1/sqrt(2))\ns
126576
Expressions such as yours is called a power tower. They are differentiable by writing them as exponentials but they are not integrable in closed form. You can find the series form at the source link however.
551630
To extend the above two discussions and bring the matter to a conclusion, more or less.\n\nLet us define a sequence by T0 = 1, Tn = x^Tn-1.  Call the limit of this sequence as n -> infinity  as f(x).\n\nFor any given y = f(x), we see that:\n\n     lim Tn = lim Tn-1 = y\n>>\n    lim Tn = lim(x^Tn-1)=x^limTn-1\n>>\n     y = x^y\n>> \n    x = y ^ (1/y)\n\nFor y >= 1:\nThis can easily be seen to satisfy the  special cases of y = 1, 2 and 4.\n\nx, as a function of y has a maximum at y = e.~ 2.718, which leads to x~1.447.\n\nFor values of  x > e^(1/e),  f(x) does not exist; and the sequence T cannot have a limit > e.\n\nFor y < 1, we can similarly see that limits < 1/e are not possible.  The smallest x for which the limit exists is (1/e)^e ~ 0.066, giving a limit 1/e ~ 0.368
451689
e^x (e^x - 1)= 0\n\nDomain = R\n\nRoot: e^x = 1, x = 0\n\n............-......0.....+\nSign: ----------l------\n...................0             \n\nIts continuous forall x\n\ne^2x-e^x  tends to infinity when x tends to +oo\n\ne^2x - e^x tends to 0- when x tends to -oo\n\nFirst Derivative\n\n2e^2x - e^x = e^x (2e^x - 1)\n\nRoot: 2 e^x = 1\ne^x = 1/2\n\nx = ln 1/2 = -ln 2 = -0,69\n............-......0.....+\nSign: ---------l--------\n...............-ln2\n\n-ln 2 is the x coordinate of a minimum\n\nf(-ln 2) = e^ -2 ln2 - e^-ln 2 = e^- ln 4 - 1/e^ln2\n\nf(-ln 2) = 1/e^ln 4 - 1/2\n\nf(-ln 2) = 1/4 - 1/2 = -1/4\n\n\nSecond Derivative\n\n4 e^2x - e^x = e^x (4 e^x -1)\n\nRoot\n\n4 e^x  = 1\n\ne^x = 1/4\n\nx = - ln 4 = -1.4\n..............-......0....+\nSign: -----------l-------\n..................-ln4\n\n-ln 4 is the x coordinate of an inflexion point.\n\nf(-ln4) = e^ -2 ln4 - e^-ln 4 = e^- ln 16 - 1/e^ln4\n\nf(-ln 4) = 1/e^ln16 - 1/4\n\nf(-ln 4) = 1/16 - 1/4 = 1/16 - 4/16 = -3/16\n\nAna
406268
(1+2/n)^n\n= [(1 + 2/n)^n/2]²\n\n= {[1 + 1/(n/2)]^n/2}²\n= [(1 + 1/p)^p]²\n\nNote: As n→∞, n/2 →∞ ie p→∞\n\nSo lim n→∞(1+2/n)^n \n= lim p→∞[(1 + 1/p)^p]²\n= e²\n\nIn fact you could now argue that lim n→∞(1+p/n)^n = e^p
398375
if ln(x)=y, then by the definition of ln, x=e^y\ne^(ln(x)) for ln(x)=y is e^y\nso e^ln(x)=e^y which we already showed =x which is what we wanted to show. Hope this helps. If not, email me.
439003
Denominator can be written as (x^2-a^2).(x^2+a^2)\n\nThe x^2-a^2 terms cancels when a is not equal to zero leaving the formula as\n\n1/(x^2+a^2)\n\nThe denominator here approaches zero as x approaches a \nso the original formula tends to infinity.  Note that since squares are positive the answer is positive infinity as x approaches a from the left and the right.\n\nOther ways are to use L'Hopitals rule.
19743
UGH!!!  Thx for the effort (I was the one to post that question, as a part of a solution to another problem I found here on Yahoo! Answers.)  Incidentally, I made a mistake when simplifying the other problem and it should be xln(2)-2ln(x)-ln(86400) = 0\n\nI don't know any way to get rid of the ln(x) unless maybe it's using calculus techniques?  I'm just glad this problem is making other ppl trouble about it as much as I did :-).
669874
square root of 2\n\nif 2 = x^(x^(x^...))\n\nthen 2 = x^ 2\n or x = √2\n\n[i hope you were truthful about the first answer gets the points....]\n\n\nI have an interesting follow up question however.  What if the quantity equals 4?  Then 4 = x^4 or x = 4^(1/4) = 2^(1/2) = √2 !!!  \n\nSo.. √2^(√2^(√2^(√2^...))) = 2 and √2^(√2^(√2^(√2^...))) = 4\nso...  2 = 4\nHow can this be??!!!!!\n(obviously its not.  where's the flaw?)
70545
You prove that for any tiny little e>0 you pick, there exists a little d>0 such that \n\nif abs(1/(x^2)-1/x^2))<e, then abs(x-x)<d,\n\nand since x-x=0 (this is a trivial limit unless you've messed up the parentheses in the problem), just let d=e, or e/2, or any value between e and 0.