proof 1/F in F-distribution?
463856
Gosh this is a bit vague - but I think you're referring to why we use 1/F when getting a lower confidence interval or doing a 2-sided test.\n\nThe definition of an F m,n-dsitribution is:\n\n[Chi-square(m)/ m] ÷ [Chi-square (n) / n]\n\nTherefore if we did 1 / Fm,n this would be:\n\n[Chi-square(n)/ n] ÷ [Chi-square (m) / m]\n\nSince "dividing by a fraction is the same as turning it over and mulitplying by it".\n\nSo suppose we want a lower 5% limit for F4,5, ie:\n\nP(F4,5 < ? ) = 0.05\n\nWe can invert both sides:\n\nP(1/F4,5 > 1/?) = 0.05\n\n(Note the sign changes, eg 2 < 3 but 1/2 > 1/3)\n\nBut using the result from earlier 1/F4,5 = F5,4:\n\nSO we now have:\n\nP(F5,4 > 1/?) = 0.05\n\nSo what we do is look at the upper 5% point in the F5,4 -tables which is 6.256.\n\nSo 1/? = 6.256\n\nHence ? = 1/6.256\n\nSo the shortcut rule is to get the value we do "one over the F distribution with the number swapped round"\n\nHope this helps.
538748
Yes, but you need to account for two cases. The full solution:\n\n(x-1) / (x-10) < 0\n\nx-1 < 0 [if x > 10] OR\nx-1 > 0 [if x < 10] (by "flip the sign" rule; x cannot be 10)\n\nFirst inequality can never be satisfied.\nSecond inequality means 10 > x > 1.\n\nYou do not use foil for that.\n\nClassical (and Kemmy) probably interpreted it as a number/phase line, which can be faster in a simple case like this. For more difficult inequalities it's best to write out steps.
384711
f(x) = (x+5)/(x+1)\n=> d/dx (f(x)) = -4/(x+1)^2\n\nBy Taylor's series expansion,\nf(x+deltax) = f(x) + d/dx(f(x))*deltax (ignoring higher terms)\n\nSubstituting, \nf(1.1) = f(1) + f'(1)*0.1\n= 3 - 1*0.1\n= 2.9
394920
1) [1/x] <= 1/x, for all x>0\n2) [1/x] = 1/x, iff 1/x is an integer, so x=1/n for some integer n.\n3) [1/x] < 1/x, for all x != 1/n for any n (x>0).\n4) So, f(x) = [x*[1/x]] = [x*1/x] = [1] = 1 for all x=1/n for some integer n.\n5) Also, x*[1/x] < x*1/x = 1 for all x != 1/n for any n (x>0),\nthen f(x) = [x*[1/x]] = [<1] = 0 for all x != 1/n for any n (x>0).\n\nlim f(x)(x->a) = b, iff for ALL sequences {a_n->a} f(a_n)->b.\n\nLet's take 2 sequences:\n{a_n = 1/n} f(a_n)=1 for all n (see 4), so f(a_n)->1.\n{b_n = 2/(2*n+1)} f(b_n)=0 for all n (see 5), so f(b_n)->0.\nEasy to see that {a_n}->0, {b_n}->0.\n\nThus, there is NO limit of f(x) when x->0.\n\nPS: "iff" means "if and only if"
837082
Hint: Let C=X&#92;O and let f(x)=dist(x,C). The other way is trivial.
245944
The integral of a function over a definite interval.
840688
Not sure what you were doing. I would rewrite f(x) in this form:\n\nf(x) = (2x-1)^(-1/5)\n\nThen you're dealing with a u^n situation, so\n\nf'(x) = (-1/5)(2x-1)^(-6/5)*(2)\n\nf'(x) = (-2/5)(2x-1)^(-6/5) \n\nThat's equivalent to what you had, but a little simpler.\n\nLet the underboss know that this is a "site" and I did answer your question.
646401
it's the inverse x^-1=1/x
311134
f(x) = 1/(x+5)(x+2)\n\nso -5 and -2 are restricted from the domain since then we would have the denominator = 0 and the function undefined.
733701
let us denote the marginal density of X_i by f_i, that is the density of X_1 is f_1.\nlet us denote the density of the triplet (X_1, X_2 X_3) by f.\n\nas X_1, X_2, X_3 are stochasticaly independent,\n\nf(x_1, x_2, x_3) = f_1(x_1)  * f_2(x_2) *  f3(x_3)\n\nlet g be a bounded borelian function on R ^3\n\nset:\nphi(x_1, x_2, x_3) = (x_1 + x_2 + x_3) ^ 2 / 6\n\nE(g o phi (X_1, X_2, X_3)) = &#92;int g o phi(x_1, x_2, x_3) f(x_1, x_2, x_3) dx_1 dx_2 dx_3\n\n = &#92;int g( ( x_1 + x_2 + x_3) ^ 2 / 6) f(x_1, x_2, x_3) dx_1dx_2dx_3\n\nthen you may conclude using Fubini theorem
936881
There is a local maximum at x = c on f(x) if\nf'(c) = 0 AND f"(c) < 0\n\nOccasionally also when f"(c) = zero .. this occurs when there is a multiple root at that value of x.\nThen further testing is needed. In this case:\n1. if If f(c-) < f(c) AND f(c+)< f(c) then it is a local maximum;OR\n2. if f'(c-) >0 and f'(c+) <0 then it is a local maximum; OR\n3. if f"(c-) <0 and f"(c+) <0 then it is a local maximum
122198
http://en.wikipedia.org/wiki/Cumulative_distribution_function
832569
df/dx=(df/dg)*(dg/dx)=x’=1, dg/dx=(dg/df)*(df/dx)=x’=1; both are const\nIt means g’(b)=1
954661
Suppose f : D → R has a limit at x0. Prove that | f | : D → R has a limit at x0 and that limx→x0 | f(x) | = | limx→x0 f(x) |.\n\n\nLet limx→x0 f(x) = c\n\nNow | f(x) | = f(x) for f(x) ≥ 0 = c\n= -f(x) for f(x) < 0 = -c\n\nSo limx→x0 | f(x) |  \n\n= limx→x0 f(x) for f(x) ≥ 0 = c\nlimx→x0 -f(x) for f(x) <0 = - limx→x0 f(x) x < 0 = -c\n\n= |c|\n\n= | limx→x0 f(x) | .................... QED
880673
hint: mean value theorem
700967
Restating your question (for syntax purposes):\n\nF(x) = sqrt(x-1) / e^x \n\nThe critical values are defined to be when F'(x) = 0 or when F'(x) is undefined.  Let's solve for F'(x).\n\nTo solve for F'(x) we need to use the quotient rule.\n\nF'(x) = [1/(2sqrt(x-1))*e^x - sqrt(x-1)*e^x]/(e^x)^2\nF'(x) = (e^x - 2*(x-1)*e^x)/[(e^2x)*(2*sqrt(x-1))]\nF'(x) = 1/2 * [1 - 2(x-1)]/(e^x * sqrt(x-1))\nF'(x) = 1/2 * [2 - 2x]/(e^x * sqrt(x-1))\n\nWe then calculate when F'(x) = 0 or when F'(x) is undefined.\n0 = 1/2 * [2 - 2x]/(e^x * sqrt(x-1))\n0 = [2 - 2x]/(e^x * sqrt(x-1))\n\nSo \n2 - 2x = 0\ne^x * sqrt(x-1) = 0\n\n2x = 2\nsqrt(x-1) = 0\n\nx = 1\nx-1 = 0\n\nWhich yields x = 1 as a critical value.
728458
Here is how you do it (I'll use ^ for the power sign).\n- Divide by P:    (1+r/n)^n = F/P\n- Take the 1/n the power:     (1+r/n) = (F/P) ^ (1/n)\n- Solve for r:   r= n* ((F/P) ^ (1/n) -1 )\n\nIn this case F=20000, P = 12600, n=12 therefore r = 0.47 or 47%\n\nSeeing this answer I wonder if I didn't misinterpret your notation, but with the method given, you can solve the question for yourself.