Let f:C-C be continuous function.If  f^2,f^3  be  analytic show  f is analytic?
401175
note that f^3/f^2 = f except for the points where f(z) = 0 (thos will be singularities).  But you can show that those singularities are removable (that will be your job - just remember that the order of the zero of f^3 will be larger than the order of the zero of f^2...)  Since they are removable, you can extend this to an entire function.  But, this function must be f, since they agree on a set with a limit point (namely, the set of f(z) s.t f(z) is not zero)... We should note that none of this will work if f is constant, but the constant case should be easy, right?
547668
--- An informal definition of a continuous function is a graph that can be drawn without picking up your pencil.\n--- This graph, however, is NOT differentiable at the point x = 0 because it is a cusp, i.e., the slope of the graph from the left is not equal to the slope of the graph from the right.\n--- If you take the derivative, you get -(2/3)x^(-1/3). If you substitute in x = 0, you get an undefined derivative. i.e., not differentiable at x = 0.
730983
First, define a new function g(x)=f(x)-x. f(x) has a fixed point iff ∃x: g(x)=0. By the fundamental theorem of calculus:\ng(x) = g(0) + [0, x]∫g'(a) da\n\nLet c be sup(|f'(x)|). Then f'(x)≤c<1, and so g'(a)≤c-1<0. So for x>0:\ng(x) < g(0) + [0, x]∫c-1 da = g(0) + x(c-1).\nAnd for x<0:\ng(x) > g(0) + [0, x]∫c-1 da = g(0) + x(c-1).\n\nHowever, [x→∞]lim g(0) + x(c-1) = -∞ (since c-1 is negative), and since g(x) is bounded above by this function for x>0, [x→∞]lim g(x) must also be -∞. Conversely, [x→-∞]lim g(0) + x(c-1) = ∞, and since g(x) is bounded below by this functon for x<0, [x→-∞]lim g(x) is ∞. But this means that g(x) is positive for some negative value of x and negative for some positive value of x, thus by the intermediate value theorem, ∃x: g(x)=0 and therefore f(x) has a fixed point. To see that it's the only one, consider that if g(x) had two zeros then by Rolle's theorem there would exist a point where g'(x) is zero, which contradicts the fact that g'(x) is strictly negative everywhere.\n\nNote that this proof remains valid if c is merely the supremum of f'(x), instead of |f'(x)|.\n\nFor the counterexample with |f'(x)|<1 (but not sup(|f'(x)|)<1), consider f(x)={1/x+x if x≥1, 2 if x<1}. This function is everywhere differentiable ([x→1-]lim f(x) = [x→1+]lim f(x) = 2, and [x→1-]lim f'(x) = [x→1+]lim f'(x) = 0), and ∀x, |f'(x)|<1, but f(x) has no fixed points.
1045860
Function y = f(x) is continuous at point x=a if the following three conditions are satisfied :\n\ni.) f(a) is defined ,\nii.)lim f(x) exists (i.e., is finite) ,\nand\niii.) lim (x ->a)  f(x)  = f(a) \n\nApplying them to your case:\n(1) f(0) is defined, being a^2 and a  in two cases.\n(2)limit also exists here\n(3) f(0)=f(0-)=f(0+)\ntherefore,\na^2=a\nor a(a-1)=0\nor a= 0 or 1.
472136
False. For example: F(x)=x^2 increases for x>0.\n\nF(x)=x^3 always increases.
837082
Hint: Let C=X&#92;O and let f(x)=dist(x,C). The other way is trivial.
888635
Let f(x) = 3x^4 - 25 + 5/4. Consider an arbitrary real valued x say x = c. We need to determine if the left-hand behavior and the right-hand behavior of the function f at x = c are the same or not. \n\nBecause f(x) is a polynomial function (of degree 4) on x, it follows that the function f is continuous on R (the set of real numbers).  So, if we take the limit on the left (and right) of x, we will get the same value for both sides. \n\nThat is, the behavior on the left (Left-hand Limit) will be: as x approaches  c from the left, we get f(x) approaches f(c) = 3c^4 - 25 + 5/4. Similarly, the behavior on the right (Right-hand Limit) will be: as x approaches c from the right, we also get f(x) approaches f(c) = 3c^4 - 25 + 5/4. \n\nTherefore, both the left and right-hand limit yields to f(c), i.e. as x approaches c, we always obtain f(c).\n\nHence, both left and right-hand behavior of f are the same for all real valued x.
354577
The function f(x)=Sigma of((f^n)(c)/n!)*((x-c)^n) is what you start with.  You can replace the Sigma(sum) with an integral.  Derivatives are the basically the opposite of Integrals.  If you derive the integral, you end up back at the start: ((f^n)(c)/n!)*((x-c)^n).  It might be easier to use a graph to see what happens. \n\nBut basically, intervals are merely used to look at a part of whatever is on the graph.  It doesn't matter what you do to that part, you are still looking at that part.  That's why the intervals don't change.\n\nDoes that help you or confuse you?
460866
Not true. It is true for continuous homomorphisms. In that case a homomorphism is of the form f(x)=ax for some constant a. THen a^n=1, so a=+-1.
589780
The definition of continuity that the preimage of any open set is open, is elegant but not directly amenable to intuition in my opinion. Maybe an easier way of putting it (logically equivalent) is the following. It's much closer to the the usually concept of continuity but applies to any topological spaces:\n\nA function f:X-->Y is continuous if and only if it preserves limit point relationships, that is: If A is any subset of X and p a limit point of A, then f(p) is a limit point, or point, of f(A).\n\nIf every open set S containing f(p) has an open preimage f^(-1)(S), then p being the limit point of some set A means that points of A are contained in f^(-1)(S) --> points of f(A) are contained in S. The converse is also easily proved so this is an equivalent definition of continuity.\n\nThe concept of a limit point p of a set A is relatively easy to grasp. It just means there's no open neighborhood of p which doesn't contain points of A - there's no finite margin separating p from points of A.\n\nIf f is continuous, for any point p we can typically construct a sequence A having p as its unique limit point, then the definition of continuity tells us that the sequence of image points under f is f(p).
203181
I'll interpret this as f(x,y)=x^3 y/(x^2 + y^2). I would rewrite it as f(x,y)=xy [x^2/(x^2 +y^2)]. As (x,y)->(0,0), the first term goes to 0 (clearly), while the second term is bounded (by 1), so the limit is 0.
782051
Given 2 functions f and g with the domains Df and Dg respectively.\n\nAny composite function of f and g will as domain have Dfg=Df ∩ Dg &#92;{ those interval or point where the composite functions has poles coming from the way the functions is composited}\n\nEX 1)\nf=x^2 ; Df={x in R l 0<x<20} by own choice\ng=x ..; Dg={x in Rl x belongs to [-5,5[} by own choice\n\na composite function fg=f/g+1/f-g\nfinding the domain Dfg:\nDfg={ x in R l x belongs to Df ∩ Dg &#92; {those interval or point where the composite functions has poles coming from the way the functions are composited} }\n\nDf∩Dg=\n{x in R l 0<x<20}∩{x in Rl x belongs to [-5,5[}=\n{x in R l 0<x<5}\n--------------------\n\n\nfinding: \nPole={those interval or point where the composite functions has poles coming from the way the functions are composited}\n\nlook at the composite function: fg=f/g+1/f-g.\nWe easily see the composite function has pole for g=0=x => x=0 and for f-g=0=x^2-x=x(x-1)=> x=0 or x=1=>\nPole={ x in R l x=0 or x=1}\n----------------------------------\nnow we can find the domain for the composit function fg\n\nDfg=Df ∩ Dg &#92; Pole ={x in R l 0<x<5} &#92; { x in R l x=0 or x=1} }= {x in R l 0<x<5 & x≠1}\n=======================
824521
this just means Y is a function of x\n\ny = x^2(2-3)\ny = -x^2
818927
if you want f(3)\nyou will have f (2 +1) which leads us to know that 2 is x...\nsubstitute all x's with 2's and then your answer will be revealed
864853
a)  f(0) = f(0-0) = f(0) - f(0) = 0.\n\nb)  Fix k in the reals, and define f(x) = kx.\n\nc)  Suppose you have f(x)=f(y) for some reals x and y.  Then f(x)-f(y) = 0, and since f preserves subtraction, f(x-y) = 0.  Since 0 is the only element which gets sent to zero, this implies that x-y = 0, and thus x = y.\n\nI have a challenge for you: find a nontrivial function f which preserves subtraction but is not injective.  There are such functions, but they are not necessarily easy to find.  Cheers!
914146
For x < 1 \nf(x) = |x-1| + 2\nWhen x → 1-, f(x) →2\nWhen x < 1, |x-1| + 2 = 1 - x + 2 = 3 - x. So slope is -1.\n\nFor x ≥ 1\nFor continuity f(1) = 2.\nFor differentiability f'(1) = -1\n\nSo a + b = 2\n2a + b = -1\n\nThus a = -3, b = 5\n\nf(x) = |x-1| + 2, for x < 1 \n..... = 5x - 3x², for x ≥ 1\nis continuous and differentiable for all x
432891
If g(x) is differentiable, then it certainly is continuous, and hence integrable.  You can define H(x) as the following definite integral:\n\nH(x) = 1 + Integral from 0 to x of { g(t) dt }