$ f $ dibedakan dalam $ (0,0). $
Definisi: Let $V\subseteq{\mathbb{R}^{m}}$ satu set terbuka, $a\in V$ y $f\colon V\to\mathbb{R}^{n}$sebuah fungsi. Kami akan mengatakan itu$f$ dibedakan dalam $a,$ jika ada transformasi linier $f'(a)\colon\mathbb{R}^{m}\to\mathbb{R}^{n}$sedemikian rupa sehingga \ begin {persamaan} f (a + h) = f (a) + f '(a) (h) + r (h), \ qquad \ lim_ {h \ rightarrow 0} {\ dfrac {r (h )} {\ lVert h \ rVert}} = 0. \ end {persamaan}
Membiarkan $ a \in \mathbb {R}$menjadi. Tentukan fungsinya$ f \colon \mathbb {R}^ {2} \to \mathbb {R} $ diberikan oleh
\ begin {persamaan} f (x, y) = \ kiri \ {\ begin {matriks} \ dfrac {x \ sin ^ {2} (x) + axy ^ {2}} {x ^ {2} + 2y ^ {2} + 3y ^ {4}} & (x, y) \ neq (0,0) \\ 0 & (x, y) = (0,0) \ end {matrix} \ kanan. \ end {persamaan}
Temukan nilai $ a $ maka $ f $ dibedakan oleh $ (0,0). $
Upaya saya:
Kami mengamati itu
\ begin {persamaan} \ dfrac {\ partial f} {\ partial x} (0,0) = 0 = \ dfrac {\ partial f} {\ partial y} (0,0). \ end {persamaan}
Jika $(x,y)\in\mathbb{R}^{2}\setminus\{(0,0)\},$ kemudian
\ begin {persamaan} \ dfrac {\ partial f} {\ partial x} (x, y) = \ dfrac {\ sin ^ {2} (x) (2y ^ {2} + 3y ^ {4} -x ^ {2}) + x \ sin (2x) (x ^ {2} + 2y ^ {2} + 3y ^ {4}) + ay ^ {2} (2y ^ {2} + 3y ^ {4} -x ^ {2})} {(x ^ {2} + 2y ^ {2} + 3y ^ {4}) ^ {2}} \ end {persamaan}
\ begin {persamaan} \ dfrac {\ partial f} {\ partial y} (x, y) = \ dfrac {2axy (x ^ {2} -3y ^ {4}) - 4xy \ sin ^ {2} (x ) (1 + 3y ^ {2})} {(x ^ {2} + 2y ^ {2} + 3y ^ {4}) ^ {2}} \ end {persamaan}
Jika $\dfrac{\partial f}{\partial y}(x,y)=0,$ kemudian
\begin{align} 2axy(x^{2}-3y^{4})-4xy\sin^{2}(x)(1+3y^{2})=0&\quad\Longleftrightarrow\quad a(x^{2}-3y^{4})=2\sin^{2}(x)(1+3y^{2})\\ &\quad\Longleftrightarrow\quad a=\dfrac{2\sin^{2}(x)(1+3y^{2})}{x^{2}-3y^{4}} \end{align}
\ begin {persamaan} f (x, y) = \ kiri \ {\ begin {matriks} x \ sin ^ {2} (x) & (x, y) \ neq (0,0) \\ 0 & (x , y) = (0,0) \ end {matrix} \ kanan. \ end {persamaan}
\ begin {persamaan} \ dfrac {\ partial f} {\ partial x} (0,0) = 0 = \ dfrac {\ partial f} {\ partial y} (0,0) \ end {persamaan}
Dari sini berikut itu $\dfrac{\partial f}{\partial x}(x,y)$ dan $\dfrac{\partial f}{\partial y}(x,y)$ disambung oleh $(0,0)$ y $f$ dibedakan oleh $(0,0).$
Apakah argumen saya benar? Setiap saran diterima.
Jawaban
Kami punya itu
$$\dfrac{\partial f}{\partial x}(0,0)=\lim_{h\to 0}\frac{\dfrac{h\sin^{2}(h)}{h^{2}}}{h} =\lim_{h\to 0}\dfrac{h\sin^{2}(h)}{h^3}=1$$
$$\dfrac{\partial f}{\partial y}(0,0)=\lim_{k\to 0}\frac{\dfrac{0}{2k^{2}+3k^4}}{k} =0$$
maka menurut definisi kita perlu memeriksanya
$$\lim_{(h,k)\to (0,0)}\frac{\dfrac{h\sin^{2}(h)+ahk^{2}}{h^{2}+2k^{2}+3k^{4}}-h}{\sqrt{h^2+k^2}} =\lim_{(h,k)\to (0,0)} \dfrac{h\sin^{2}(h)+ahk^{2}-h^3-2hk^2-3hk^4}{(h^{2}+2k^{2}+3k^{4})\sqrt{h^2+k^2}}=0$$
yang memang benar oleh $a=2$
$$\dfrac{h\sin^{2}(h)+ahk^{2}-h^3-2hk^2-3hk^4}{(h^{2}+2k^{2}+3k^{4})\sqrt{h^2+k^2}}=\dfrac{h(h^2+O(h^4))+2hk^{2}-h^3-2hk^2-3hk^4}{(h^{2}+2k^{2}+3k^{4})\sqrt{h^2+k^2}}=$$
$$=\dfrac{-3hk^4+O(h^5)}{(h^{2}+2k^{2}+3k^{4})\sqrt{h^2+k^2}}$$
kemudian gunakan koordinat kutub.
Pendekatan yang agak berbeda:
Agar dapat terdiferensiasi, suatu fungsi harus kontinu dan memiliki turunan kontinu (atau memiliki turunan dengan singularitas esensial). Kontinuitas mensyaratkan bahwa batas saat Anda mendekati titik tersebut harus sama, terlepas dari arah pendekatan Anda.
Misalkan kita mendekati sepanjang garis $x=y=\epsilon$. Kemudian kami memiliki (menggunakan fakta bahwa$\frac{d}{da}\sin^2(a)=\sin(2a)$: $$g(\epsilon)=f(\epsilon,\epsilon) = \frac{\epsilon\sin^2(\epsilon)+a\epsilon^3}{\epsilon^2+2\epsilon^2+3\epsilon^4} = \frac{\sin^2(\epsilon)+a\epsilon^2}{3\epsilon+3\epsilon^3}=\frac{1}{3}\frac{\sin^2(\epsilon)+a\epsilon^2}{\epsilon+\epsilon^3}$$ $$g'(\epsilon)=\frac{1}{3}\frac{(\epsilon+\epsilon^3)(\sin(2\epsilon)+2a\epsilon)-(\sin^2(\epsilon)+a\epsilon^2)(1+3\epsilon^2)}{\epsilon^2+2\epsilon^4+\epsilon^6} = \frac{1}{3}\frac{\epsilon\sin(2\epsilon)+2a\epsilon^2+\epsilon^3\sin(2\epsilon)+2a\epsilon^4-\sin^2(\epsilon)-a\epsilon^2-3\epsilon^2\sin^2(\epsilon)-3a\epsilon^5}{\epsilon^2+2\epsilon^4+\epsilon^6}$$ $$\lim_{\epsilon\rightarrow0}g'(\epsilon)=\frac{1}{3}\lim_{\epsilon\rightarrow0}\frac{\epsilon\sin(2\epsilon)+2a\epsilon^2+\epsilon^3\sin(2\epsilon)+2a\epsilon^4-\sin^2(\epsilon)-a\epsilon^2-3\epsilon^2\sin^2(\epsilon)-3a\epsilon^5}{\epsilon^2+2\epsilon^4+\epsilon^6} = \frac{1}{3} \lim_{\epsilon\rightarrow0} \frac{\sin(2\epsilon)+2\epsilon\cos(2\epsilon)+4a\epsilon+3\epsilon^2\sin(2\epsilon)+2\epsilon^3\cos(2\epsilon)+8a\epsilon^3-\sin(2\epsilon)-2a\epsilon-6\epsilon\sin^2(\epsilon)-3\epsilon^2\sin(2\epsilon)-15a\epsilon^4}{2\epsilon+8\epsilon^3+6\epsilon^5} = \frac{1}{3} \lim_{\epsilon\rightarrow0} \frac{2\epsilon\cos(2\epsilon)+2a\epsilon+2\epsilon^3\cos(2\epsilon)+8a\epsilon^3-6\epsilon\sin^2(\epsilon)-15a\epsilon^4}{2\epsilon+8\epsilon^3+6\epsilon^5} = \frac{1}{3} \lim_{\epsilon\rightarrow0} \frac{2\cos(2\epsilon)+2a+2\epsilon^2\cos(2\epsilon)+8a\epsilon^2-6\sin^2(\epsilon)-15a\epsilon^3}{2+8\epsilon^2+6\epsilon^4} = \frac{1}{3} \frac{2+2a}{2} = \frac{1+a}{3}$$
Misalkan kita mendekati sepanjang garis $-x=y=\epsilon$. Kemudian kami memiliki:$$h(\epsilon)=f(-\epsilon,\epsilon) = \frac{-\epsilon\sin^2(-\epsilon)-a\epsilon^3}{\epsilon^2+2\epsilon^2+3\epsilon^4} = \frac{-\epsilon\sin^2(\epsilon)-a\epsilon^3}{\epsilon^2+2\epsilon^2+3\epsilon^4}= -g(\epsilon)$$ $$h'(\epsilon)=-g'(\epsilon)$$ $$\lim_{\epsilon\rightarrow0}h'(\epsilon)=-\lim_{\epsilon\rightarrow0}g'(\epsilon)=-\frac{1+a}{3}$$
Di kedua arah, batas turunannya ada, dan oleh karena itu karena arah pendekatannya tidak masalah, kami mensyaratkan bahwa batasnya sama. $\frac{1+a}{3}=-\frac{1+a}{3}$, yang artinya $a=-1$.