transgalactic

f(x) and g(x) are differentiable on 0 f(0)=g(0)=0 calculate [math] \lim _{x->0}\frac{cosf(x)-cosg(x)}{x^2} [/math] ?? it looks like cauchys mean value theorem F(x)=cos(f(x)) G(x)=cos(g(x)) what to do next??

function f(x) continues on [a,b] suppose that for every sub part [math][\alpha ,\beta ]\subseteq [a,b][/math] we have[math]\int_{\alpha}^{\beta}f(x)dx>0[/math]. prove that f(x)>=0 for [math] x\in[a,b] [/math] if its wrong give a contradicting example?? i dont have a clue from here to start or how to go. from the given i can conclude that if the sum of all subsections gives us a positive result then the total sum from a to b has to positive too.inte

i got the idea that its the function is bigger then the polynomial by the members with powers higher then "n". but its only a logic how to formulate it into mathematical equations?? the remainder formula is 0<c<x [math] R_n(x)=\frac{f^{n+1}©}{(n+1)!}x^{n+1} [/math] so i guess that for x>0 its: [math] R_n(x)=\frac{f^{n+1}©}{(n+1)!}x^{n+1} [/math] for x<0 [math] R_n(x)=-\frac{f^{n+1}©}{(n+1)!}x^{n+1} [/math] should i prove it by induction?? if so then if n=1 then [math] e^x=1+x+ \frac{f^{1+1}©}{(1+1)!}x^{1+1} [/math] but i dont know whats the value of c?? and still it depends on the values of x ??

so my general function are bigger then their taylor series by the remainder of Rn the functions are greater than the Taylor polynomial up to power n by the members which are in a higher powers then "n" this is the logic how to formulate it into equation??

this is not an induction what is the full proof?

f(x) is differentiable twice at x_0 prove that: [math] f''(x_0 ) = \mathop {\lim }\limits_{h \to \infty } {{f(x_0 + h) - 2f(x_0 ) + f(x_0 - h)} \over {h^2 }} [/math] i tried to solve it but i cant get to the asked expression [math] g'(x_0) = \lim_{h \to 0} \frac{g(x_0 + h) - g(x_0)}{h}\\ [/math] [math] g''(x_0) = \lim_{h \to 0} \frac{g'(x_0 + h) - g'(x_0)}{h}\\ [/math] [math] g''(x_0) = \lim_{h \to 0} \frac{\frac{g(x_0 + 2h) - g(x_0+h)}{h} - \frac{g(x_0 + h) - g(x_0)}{h}}{h}\\ [/math] [math] g''(x_0) = \lim_{h \to 0} \frac{{g(x_0 + 2h) - g(x_0+h)} - {g(x_0 + h) - g(x_0)}}{h^2}\\ [/math] [math] g''(x_0) = \lim_{h \to 0} \frac{{g(x_0 + 2h) - 2g(x_0+h)} { - g(x_0)}}{h^2}\\ [/math] ??

how many times can we differentiate this splitted function on point x=0 ?? [math] f(x) = \{ x^{2n} \sin (\frac{1}{x}) ,x \ne 0 [/math] i know i need to prove it by induction [math]\dfrac{\mathrm{d}^r}{\mathrm{d}x^r}\big(x^{2n}\sin(x^{-1})\big)= \begin{cases} (-1)^rx^{2(n-r)}\sin(x^{-1}+\frac{1}{2}r\pi)+x^{2(n-r)}g_{n,r}(x) & x\neq 0\\ 0& x=0 \end{cases} [/math] for all r≤n, where g_{n,r}(x) is continuous at 0 and vanishes there. what to do next??

i tried to prove for the first derivative but i dont get a final limit here [math] f(x)=e^\frac{-1}{x^2} \\ [/math] [math] f'(0)=\lim_{x->0^+}\frac{e^\frac{-1}{x^2}-0}{x-0} \\ [/math] [math] f'(0)=\lim_{x->0^-}\frac{e^\frac{-1}{x^2}-0}{x-0} [/math] Merged post follows: Consecutive posts mergedthe makloren formula is [math] f(x)=\sum_{n}^{k}\frac{f^{(k)}}{k!}x^k+o(x^n) [/math] how to take the n'th derivative from here?? where to put the formula in and get the n'th derivative?? this function is for approximating so in order for me to get to the n'th members approximation first i need to do manually one by one n times derivative so its not helping me ?? Merged post follows: Consecutive posts mergedi was told "once we can express the function as a power series around zero and it is differentiable at zero, we know it is infinitely differentiable" differentiable around zero part: [math] f'(0)=\lim_{x->0}\frac{e^\frac{-1}{x^2}-0}{x-0} \\ =\lim_{x->0}\frac x{2e^{1/x^2}} [/math] i know that exponentials grow faster then polinomials but the power of the exp is 1/x^2 so the denominator goes to infinity but faster then the numenator so the expression goes to 0. power series around zero part: the power series of e^x [math] g(x)=e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+O(x^3)\\ [/math] i put -1/x^2 instead of x [math] g(x)=e^x=1+\frac{-1}{x^2}+\frac{x^{4}}{2}+\frac{-x^6}{6}+O(-x^6)\\ [/math] what now??

for the second question: [math] (|\sin x|^{n}||\sin x|)^{(n)}=\sum_{k=0}^{n}C^k_n (|\sin x|^{n})^{(n-k)}|sinx|^{(k)}\\ [/math] what to do next??

1=<k=<n find [math] f^{(n)} (x) [/math] of: the first : [math] f(x) = |x|^{n + 1} [/math] the second is: [math] f(x) = |\sin x|^{n + 1} [/math]

if i will prove this for the first derivative and the second derivative and the third derivative but its not prooving for the next endless derivatives its only proving for these cases ??

how many times can we differentiate this splitted function on point x=0 ?? [math] f(x) = \{ x^{2n} \sin (\frac{1}{x}) ,x \ne 0 [/math] on {0,x=0}

prove that this function differentiable endles times on x=0 ?? if i will write the definition of the derivetive i could prove that it differentiates on x=0 how to prove that for endless derivatives of this function ??

sorry forgot the minus [math] \frac{{d^{} }}{{dx^{} }}g_n (x) = \frac{{( - 1)^n {\rm{[}}e^{1/x} ( - \frac{1}{{x^2 }}{\rm{)}}x^{n + 1} - (n + 1)x^n e^{1/x} {\rm{]}}}}{{x^{2n + 2} }}{\rm{ }} [/math] what to do next?? Merged post follows: Consecutive posts mergedwe assume that f_n correct so the f_n+1 is correct too(except the derivative part) but when we take deriavative n+1 we get a different equations the we supposed to get ?? [math] \frac{{d^n }}{{dx^n }}f_{n + 1} (x) = ( - 1)^{n + 1} \frac{{e^{1/x} }}{{x^{n + 2} }} [/math] [math] \frac{{d^{n + 1} }}{{dx^{n + 1} }}f_{n + 1} (x) = ( - 1)^{n + 1} \frac{{ - \frac{1}{{x^2 }}e^{1/x} x^{n + 2} - (n + 2)x^{n + 1} e^{1/x} }}{{x^{2n + 4} }} [/math] Merged post follows: Consecutive posts mergedhow to interpret this prove into d/dx symbols?? Note that the commutator [math] [D,x]=Dx-xD=1 [/math] and so [math] [D^n,x]=D^{n-1}[D,x]+D^{n-2}[D,x]D+\cdots+[D,x]D^{n-1}=nD^{n-1}. [/math] Thus we have Merged post follows: Consecutive posts merged[math] D^{n+1}x^n\exp(x^{-1}) &=[D^{n+1},x]x^{n-1}\exp(x^{-1})+xD^{n+1}x^{n-1}\exp(x^{-1})\\ &=(n+1)D^nx^{n-1}\exp(x^{-1})+xDD^nx^{n-1}\exp(x^{-1})\\ &=(-1)^n(n+1)x^{-n-1}\exp(x^{-1})+xD(-1)^nx^{-n-1}\exp(x^{-1}) [/math] by the inductive hypothesis. Now do the differentiation and you get [math]D^{n + 1} x^n \exp (x^{ - 1} ) = ( - 1)^n (n + 1)x^{ - n - 1} \exp (x^{ - 1} )\quad + ( - 1)^n x(( - n - 1)x^{ - n - 2} \exp (x^{ - 1} ) - x^{ - n - 3} \exp (x^{ - 1} )) = ( - 1)^{n + 1} x^{ - n - 2} \exp (x^{ - 1} )[/math] \\ \end{array} which finishes the induction step.

this is the only equations i can construct [math] \mathop {\mathop {\lim }\limits_{x \to x_0 - } \frac{{f(x_{} ) - f(x_0 )}}{{x - x_0 }} = \mathop {\lim }\limits_{x \to x_0 + } \frac{{f(x_{} ) - f(x_0 )}}{{x - x_0 }}}\limits_{} \\ [/math] [math] \mathop {\mathop {\lim }\limits_{x \to x_0 - } \frac{{g(x_{} ) - g(x_0 )}}{{x - x_0 }} = \mathop {\lim }\limits_{x \to x_0 + } \frac{{g(x_{} ) - g(x_0 )}}{{x - x_0 }}}\limits_{} \\ [/math] "if u(x) = f(x)) then there exists an [math]\epsilon > 0[/math] such that u(y) = f(y) for all [math]y \in [x, x + \epsilon)[/math]. So then [math]u'(x)_+ = f'(x)[/math]." i dont know how to apply this this to my question i showed the differentiation equations i dont know how to continue. Merged post follows: Consecutive posts mergedso u(x) and v(x) around x_0 have to pick different functions because if one is bigger then the other is smaller. so if f(x)=u(x) then g(x)=v(x) and once again because f(x) and g(x) are differentiable then we have one sided derivative. i dont know how to write it thematically but here is a try [math] \forall x_0 \in R [/math] [math] {\rm{ }}\exists \delta {\rm{ > 0 |x - x}}_0 | < \delta {\rm{ }}\mathop {\lim }\limits_{x \to x_0 } u(x) = \mathop {\lim }\limits_{x \to x_0 } v(x) [/math]

you are correct i forgot the denominator part [math] \frac{{d^{} }}{{dx^{} }}g_n (x) = \frac{{( - 1)^n {\rm{[}}e^{1/x} ( - \frac{1}{{x^2 }}{\rm{)}}x^{n + 1} + (n + 1)x^n e^{1/x} {\rm{]}}}}{{x^{2n + 2} }}{\rm{ }} [/math] what to do now?? how to get to the n+1 expression?

ok but its a shot in the dark i dont know what to do after the derivative. from the given derivative of n i find the same derivative of n+1 but it differs the expression we need to get fron the n case ?? [math] \frac{{d^{n + 1} }}{{dx^{n + 1} }}f_{n + 1} (x) = ( - 1)^{n + 1} {\rm{[}}e^{1/x} ( - \frac{1}{{x^2 }}{\rm{)}}x^{n + 2} + (n + 2)x^{n + 1} e^{1/x} {\rm{]}} [/math]

i need to prove it by derivation of n=k case? Merged post follows: Consecutive posts merged[math] {\rm{ }}\frac{{d^n }}{{dx^n }}f_n (x) = g_n (x) \\ [/math] [math] \frac{{d^{n + 1} }}{{dx^{n + 1} }}f_n (x) = \frac{{d^{} }}{{dx^{} }}g_n (x) = ( - 1)^n {\rm{[}}e^{1/x} ( - \frac{1}{{x^2 }}{\rm{)}}x^{n + 1} + (n + 1)x^n e^{1/x} {\rm{] }} \\ [/math] what to next?

so you are saying that because our pattern is true the n=k+1 case is true then n=k is true but our pattern is only a presumption we cant say that its a fact

but how should i go from the n=k case to the n=k+1 case by derivation of n=k expression??

i cant understand how this pattern helps i got my base case [math] \frac{d^n}{dx^n}f_n(x) = g_n(x) [/math] and i need to input n=k+1 in the base case and prove that this k+1 equation is correct using the n=k equation how this pattern helps me in doing that?

if you say that my 4th derivative is correct then i think that the pattern is [math]\frac{{d^{k + 1} }}{{dx^{k + 1} }}f_{n + 1} (x) = x\frac{{d^{k + 1} }}{{dx^{k + 1} }}f_n (x) + (k + 1)\frac{{d^k }}{{dx^k }}f_n (x)[/math] what to do now??

its an obvious result form [math] \begin{aligned} \frac {d}{dx}f_{n+1}(x) &= x\frac {d}{dx}f_{n}(x) + f_n(x) \\ \frac {d^2}{dx^2}f_{n+1}(x) &= x\frac {d^2}{dx^2}f_{n}(x) + 2\frac{d}{dx}f_{n}(x) \\ \frac {d^3}{dx^3}f_{n+1}(x) &= x\frac {d^3}{dx^3}f_{n}(x) + 3\frac {d^2}{dx^2}f_{n}(x) \end{aligned} [/math] the 4th derivative is [math] \frac{{d^4 }}{{dx^4 }}f_{n + 1} (x) = \frac{{d^3 }}{{dx^3 }}f_n (x) + x\frac{{d^4 }}{{dx^4 }}f_n (x) + 3\frac{{d^3 }}{{dx^3 }}f_n (x) = x\frac{{d^4 }}{{dx^4 }}f_n (x) + 4\frac{{d^3 }}{{dx^3 }}f_n (x) \\ [/math] so i cant see why my pattern is wrong??

i think this is the k+1 case correct? [math]\frac{{d^{k + 1} }}{{dx^{k + 1} }}f_{n + 1} (x) = x\frac{{d^{k + 1} }}{{dx^{k + 1} }}f_n (x) + (k + 1)\frac{{d^2 }}{{dx^2 }}f_n (x)[/math]

is that the correct pattern? [math] \frac{{d^3 }}{{dx^3 }}f_{n + 1} (x) = x\frac{{d^3 }}{{dx^3 }}f_n (x) + 3\frac{{d^2 }}{{dx^2 }}f_n (x) \\ [/math] [math] \frac{{d^n }}{{dx^n }}f_{n + 1} (x) = x\frac{{d^n }}{{dx^n }}f_n (x) + 3\frac{{d^{n - 1} }}{{dx^{n - 1} }}f_n (x) \\ [/math]