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Question Number 200415 by Mingma last updated on 18/Nov/23

Answered by witcher3 last updated on 18/Nov/23

$$\iff \underset{\mathrm{h}\to 0}{\lim }(-\frac{\mathrm{f}\left(\mathrm{x}\right)-\mathrm{f}(\mathrm{x}-\mathrm{h})}{\mathrm{x}-(\mathrm{x}-\mathrm{h})})=-{\mathrm{f}}^{\prime }\left(\mathrm{x}\right)$$$$\iff -{\mathrm{f}}^{″}\left(\mathrm{x}\right)=\sum _{\mathrm{n}⩾1}\frac{{\mathrm{f}}^{\left(\mathrm{n}\right)}\left(0\right){\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}!}$$$$=\mathrm{f}\left(\mathrm{x}\right)-\mathrm{f}\left(0\right)=-{\mathrm{f}}^{″}\left(\mathrm{x}\right)$$$$\iff {\mathrm{f}}^{″}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(0\right)\Rightarrow {\mathrm{f}}^{″}\left(0\right)=0$$$$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(0\right)+\mathrm{acos}\left(\mathrm{x}\right)+\mathrm{bsin}\left(\mathrm{x}\right)$$$${\mathrm{f}}^{\prime }\left(\mathrm{x}\right)=\mathrm{bcos}\left(\mathrm{x}\right)-\mathrm{asin}\left(\mathrm{x}\right)$$$${\mathrm{f}}^{\prime }\left(0\right)=\mathrm{b}$$$${\mathrm{f}}^{″}\left(0\right)=-\mathrm{a}=0$$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(0\right)+\mathrm{bsin}\left(\mathrm{x}\right)$$$${\mathrm{f}}^2\left(0\right)+{\left(\mathrm{b}\right)}^2=4(\mathrm{b}-1)$$$${\mathrm{b}}^2-4\mathrm{b}+4+{\mathrm{f}}^2\left(0\right)=0$$$${(\mathrm{b}-2)}^2+{\mathrm{f}}^2\left(0\right)=0$$$$\mathrm{b}=2;\mathrm{f}\left(0\right)=0$$$$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=2\sin \left(\mathrm{x}\right)$$$$\mathrm{f}\mathrm{is}\mathrm{uniforlmly}\mathrm{continus}\mathrm{in}\mathbb{R}\mathrm{cause}\mathrm{periodic}$$$$\mathrm{and}\mathrm{continus}\mathrm{in}[0,2\pi ]\Rightarrow \mathrm{uniformly}\mathrm{continus}$$$$$$

Commented by Mingma last updated on 19/Nov/23

Perfect ��

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