תרגיל 1
הגדירו lim ( x , y ) → ( a , b ) f ( x , y ) = L \displaystyle \lim_{ (x,y) \to (a,b) }f(x,y)=L ( x , y ) → ( a , b ) lim f ( x , y ) = L
תהי f : R 2 → R f:\mathbb{R}^2\to \mathbb{R} f : R 2 → R ( a , b ) (a,b) ( a , b ) lim ( x , y ) → ( a , b ) f ( x , y ) = L \displaystyle \lim_{ (x,y) \to (a,b) }f(x,y)=L ( x , y ) → ( a , b ) lim f ( x , y ) = L ε > 0 \epsilon>0 ε > 0 δ > 0 \delta>0 δ > 0 ∣ x − a ∣ < δ ∧ ∣ y − b ∣ < δ |x-a|<\delta \land |y-b|<\delta ∣ x − a ∣ < δ ∧ ∣ y − b ∣ < δ
∣ f ( x , y ) − L ∣ < ε |f(x,y)-L|<\epsilon ∣ f ( x , y ) − L ∣ < ε
תהי f ( x , y ) f(x,y) f ( x , y ) R 2 \mathbb{R}^2 R 2 f f f ( 0 , 0 ) (0,0) ( 0 , 0 ) רמז: משפט לגרנז'
הוכחה: יהי ε > 0 \epsilon>0 ε > 0 K > 0 K>0 K > 0 x , y ∈ R x,y \in \mathbb{R} x , y ∈ R ∣ ∂ f ∂ x ( x , y ) ∣ ≤ K , ∣ ∂ f ∂ y ( x , y ) ∣ ≤ K \left\vert\parder{f}{x}(x,y)\right\vert\leq K,\; \left\vert\parder{f}{y}(x,y)\right\vert\leq K ∂ x ∂ f ( x , y ) ≤ K , ∂ y ∂ f ( x , y ) ≤ K δ = ε 2 K > 0 \delta=\frac{\epsilon}{2K}>0 δ = 2 K ε > 0 x , y ∈ R x,y \in \mathbb{R} x , y ∈ R ∣ x ∣ < δ |x|<\delta ∣ x ∣ < δ ∣ y ∣ < δ |y|<\delta ∣ y ∣ < δ g ( k ) = f ( x , k ) − f ( x , 0 ) g(k)=f(x,k)-f(x,0) g ( k ) = f ( x , k ) − f ( x , 0 ) h ( ℓ ) = f ( ℓ , 0 ) − f ( 0 , 0 ) h(\ell)=f(\ell,0)-f(0,0) h ( ℓ ) = f ( ℓ , 0 ) − f ( 0 , 0 ) x x x y y y f f f h , g h,g h , g [ 0 , x ] [0,x] [ 0 , x ] [ x , 0 ] [x,0] [ x , 0 ] c 1 , c 2 ∈ ( 0 , x ) c_{1},c_{2} \in(0,x) c 1 , c 2 ∈ ( 0 , x )
∂ f ∂ x ( c 1 , 0 ) = h ′ ( c 1 ) = h ( x ) − h ( 0 ) x = f ( x , 0 ) − f ( 0 , 0 ) x \parder{f}{x}(c_{1},0)=h'(c_{1})=\frac{h(x)-h(0)}{x}=\frac{f(x,0)-f(0,0)}{x} ∂ x ∂ f ( c 1 , 0 ) = h ′ ( c 1 ) = x h ( x ) − h ( 0 ) = x f ( x , 0 ) − f ( 0 , 0 ) f ( x , 0 ) − f ( 0 , 0 ) x ≤ K ⟹ f ( x , 0 ) − f ( 0 , 0 ) ≤ K x < K δ \frac{f(x,0)-f(0,0)}{x}\leq K \implies f(x,0)-f(0,0)\leq Kx<K \delta x f ( x , 0 ) − f ( 0 , 0 ) ≤ K ⟹ f ( x , 0 ) − f ( 0 , 0 ) ≤ K x < Kδ באותו האופן עם g g g
f ( x , y ) − f ( x , 0 ) < K δ f(x,y)-f(x,0)< K\delta f ( x , y ) − f ( x , 0 ) < Kδ לכן מתקיים:
∣ f ( x , y ) − f ( 0 , 0 ) ∣ ≤ ∣ f ( x , y ) − f ( x , 0 ) ∣ + ∣ f ( x , 0 ) − f ( 0 , 0 ) ∣ < 2 K δ = ε |f(x,y)-f(0,0)|\leq |f(x,y)-f(x,0)|+|f(x,0)-f(0,0)|<2K\delta=\epsilon ∣ f ( x , y ) − f ( 0 , 0 ) ∣ ≤ ∣ f ( x , y ) − f ( x , 0 ) ∣ + ∣ f ( x , 0 ) − f ( 0 , 0 ) ∣ < 2 Kδ = ε לכן, לפי הגדרת הגבול מתקיים כי f ( x , y ) → ( x , y ) → ( 0 , 0 ) f ( 0 , 0 ) f(x,y)\rarrowlim{(x,y)\to(0,0)}f(0,0) f ( x , y ) ( x , y ) → ( 0 , 0 ) f ( 0 , 0 ) f f f ∎
תרגיל 2
יהי α > 0 \alpha>0 α > 0
f ( x , y ) = { ( x 2 ) α y ( x 2 + y 2 ) 2 α ( x , y ) ≠ ( 0 , 0 ) 0 ( x , y ) = ( 0 , 0 ) f(x,y)=\begin{cases}
\frac{(x^2)^\alpha y}{(x^2+y^2)^{2\alpha}} & (x,y)\neq(0,0) \\
0 & (x,y)=(0,0)
\end{cases} f ( x , y ) = { ( x 2 + y 2 ) 2 α ( x 2 ) α y 0 ( x , y ) = ( 0 , 0 ) ( x , y ) = ( 0 , 0 ) קבעו מהם הערכים של α > 0 \alpha>0 α > 0
f ( r cos θ , r sin θ ) = ( r 2 cos 2 θ ) α r sin θ ( r 2 ( cos 2 θ + sin 2 θ ) ) 2 α = r 2 α + 1 cos α θ sin θ r 4 α = r 2 α + 1 4 α cos α θ sin θ f(r\cos\theta,r\sin\theta)=\frac{(r^2\cos^2\theta)^{\alpha}r\sin\theta}{(r^2(\cos^2\theta+\sin^2\theta))^{2\alpha}}=\frac{r^{2\alpha+1}\cos^\alpha\theta \sin\theta}{r^{4\alpha}}=r^{\frac{2\alpha+1}{4\alpha}}\cos^\alpha\theta \sin\theta f ( r cos θ , r sin θ ) = ( r 2 ( cos 2 θ + sin 2 θ ) ) 2 α ( r 2 cos 2 θ ) α r sin θ = r 4 α r 2 α + 1 cos α θ sin θ = r 4 α 2 α + 1 cos α θ sin θ נגדיר F ( r ) = r 2 α + 1 4 α , G ( θ ) = cos α θ sin θ F(r)=r^{\frac{2\alpha+1}{4\alpha}},G(\theta)=\cos^\alpha\theta \sin\theta F ( r ) = r 4 α 2 α + 1 , G ( θ ) = cos α θ sin θ G ( θ ) G(\theta) G ( θ ) R \mathbb{R} R cos β , sin β \cos\beta,\sin\beta cos β , sin β R \mathbb{R} R
F ( r ) = r 2 α + 1 4 α → r → 0 ∀ α : 2 α + 1 4 α ≥ 1 0 F(r)=r^{\frac{2\alpha+1}{4\alpha}} \rarrowlim{\begin{matrix}
r\to 0 \\
\forall\alpha:\; \frac{2\alpha+1}{4\alpha}\geq 1
\end{matrix}} 0 F ( r ) = r 4 α 2 α + 1 r → 0 ∀ α : 4 α 2 α + 1 ≥ 1 0 נשים לב כי f ( r cos θ , r sin θ ) = F ( r ) G ( θ ) f(r\cos\theta,r\sin\theta)=F(r)G(\theta) f ( r cos θ , r sin θ ) = F ( r ) G ( θ ) 0 < α < 1 2 0<\alpha< \frac{1}{2} 0 < α < 2 1 f → ( x , y ) → ( 0 , 0 ) 0 f\rarrowlim{(x,y)\to(0,0)}0 f ( x , y ) → ( 0 , 0 ) 0 0 < α < 1 2 0<\alpha< \frac{1}{2} 0 < α < 2 1 f f f
יהי α > 0 \alpha>0 α > 0
f ( x , y ) = { ( x 2 ) α y ( x 2 + y 2 ) 2 α ( x , y ) ≠ ( 0 , 0 ) 0 ( x , y ) = ( 0 , 0 ) f(x,y)=\begin{cases}
\frac{(x^2)^\alpha y}{(x^2+y^2)^{2\alpha}} & (x,y)\neq(0,0) \\
0 & (x,y)=(0,0)
\end{cases} f ( x , y ) = { ( x 2 + y 2 ) 2 α ( x 2 ) α y 0 ( x , y ) = ( 0 , 0 ) ( x , y ) = ( 0 , 0 ) קבעו מהם הערכים של α > 0 \alpha>0 α > 0
לפי משפט, גזירות ⟸ רציפות ולכן, f f f α > 1 2 \alpha> \frac{1}{2} α > 2 1 0 < α < 1 2 0<\alpha< \frac{1}{2} 0 < α < 2 1
∂ f ∂ x ( 0 , 0 ) = lim h → 0 f ( h , 0 ) h = lim h → 0 ( h 2 ) α ⋅ 0 h 4 α + 1 = lim h → 0 0 = 0 \parder{f}{x}(0,0)=\lim_{ h \to 0 } \frac{f(h,0)}{h}=\lim_{ h \to 0 } \frac{(h^2)^{\alpha}\cdot 0}{h^{4\alpha+1}}=\lim_{ h \to 0 } 0=0 ∂ x ∂ f ( 0 , 0 ) = h → 0 lim h f ( h , 0 ) = h → 0 lim h 4 α + 1 ( h 2 ) α ⋅ 0 = h → 0 lim 0 = 0 ∂ f ∂ y ( 0 , 0 ) = lim h → 0 f ( 0 , h ) h = lim n → 0 0 2 α h h 4 α + 1 = lim h → 0 0 = 0 \parder{f}{y}(0,0)=\lim_{ h \to 0 } \frac{f(0,h)}{h}=\lim_{ n \to 0 } \frac{0^{2\alpha}h}{h^{4\alpha+1}}=\lim_{ h \to 0 } 0=0 ∂ y ∂ f ( 0 , 0 ) = h → 0 lim h f ( 0 , h ) = n → 0 lim h 4 α + 1 0 2 α h = h → 0 lim 0 = 0 כעת, נשתמש בהגדרת גזירות:
Δ f = f ( h , k ) = ∂ f ∂ x ( 0 , 0 ) ⋅ h + ∂ f ∂ y ( 0 , 0 ) ⋅ k + α ( h , k ) h 2 + k 2 = α ( h , k ) h 2 + k 2 \Delta f=f(h,k)=\parder{f}{x}(0,0)\cdot h+\parder{f}{y}(0,0)\cdot k+\alpha(h,k)\sqrt{ h^2+k^2 }=\alpha(h,k)\sqrt{ h^2+k^2 } Δ f = f ( h , k ) = ∂ x ∂ f ( 0 , 0 ) ⋅ h + ∂ y ∂ f ( 0 , 0 ) ⋅ k + α ( h , k ) h 2 + k 2 = α ( h , k ) h 2 + k 2 α ( h , k ) = f ( h , k ) h 2 + k 2 = h 2 α y ( h 2 + k 2 ) 2 α + 1 2 \alpha(h,k)= \frac{f(h,k)}{\sqrt{ h^2+k^2 }}=\frac{h^{2\alpha}y}{(h^2+k^2)^{2\alpha+\frac{1}{2}}} α ( h , k ) = h 2 + k 2 f ( h , k ) = ( h 2 + k 2 ) 2 α + 2 1 h 2 α y נבדוק מתי זה מתכנס ל-0 0 0
α ( r cos θ , r sin θ ) = r 2 α + 1 cos 2 α θ sin θ r 4 α + 1 = r 2 α + 1 4 α + 1 cos 2 α θ sin θ = F ( r ) G ( θ ) \alpha(r\cos\theta,r\sin\theta)=\frac{r^{2\alpha+1}\cos^{2\alpha}\theta \sin\theta}{r^{4\alpha+1}}=r^{\frac{2\alpha+1}{4\alpha+1}} \cos^{2\alpha}\theta \sin\theta =F(r)G(\theta) α ( r cos θ , r sin θ ) = r 4 α + 1 r 2 α + 1 cos 2 α θ sin θ = r 4 α + 1 2 α + 1 cos 2 α θ sin θ = F ( r ) G ( θ ) נשים לב כי לאור העובדה ש-cos θ , sin θ \cos\theta,\sin\theta cos θ , sin θ G ( θ ) G(\theta) G ( θ )
F ( r ) → r → 0 α : 2 α + 1 4 α + 1 ≥ 1 0 F(r)\rarrowlim{\begin{matrix}
r\to 0 \\
\alpha:\;\;\frac{2\alpha+1}{4\alpha+1}\geq 1
\end{matrix}}0 F ( r ) r → 0 α : 4 α + 1 2 α + 1 ≥ 1 0 עבור α ≤ 0 \alpha\leq 0 α ≤ 0 f f f 0 0 0 ( 0 , 0 ) (0,0) ( 0 , 0 ) α > 0 \alpha>0 α > 0 f f f ( 0 , 0 ) (0,0) ( 0 , 0 )
תרגיל 3
ג'ורג'י חפר בגינה ומצא את הפונקציה f ( x , y ) = { ( x − 1 ) 2 y ( x − 1 ) 2 + y 2 ( x , y ) ≠ ( 1 , 0 ) 0 ( x , y ) = ( 1 , 0 ) \displaystyle f(x,y)=\begin{cases} \frac{(x-1)^2y}{(x-1)^2+y^2} & (x,y)\neq(1,0)\\0 & (x,y)=(1,0) \end{cases} f ( x , y ) = { ( x − 1 ) 2 + y 2 ( x − 1 ) 2 y 0 ( x , y ) = ( 1 , 0 ) ( x , y ) = ( 1 , 0 ) ( 1 , 0 ) (1,0) ( 1 , 0 ) f f f
נסתכל על הנגזרות החלקיות החלקיות בנקודה:
∂ f ∂ x ( 1 , 0 ) = lim h → 0 ( 1 + h − 1 ) 2 ⋅ 0 ( 1 + h − 1 ) 2 + 0 2 = lim h → 0 0 = 0 \parder{f}{x}(1,0)=\lim_{ h \to 0 } \frac{(1+h-1)^2\cdot 0}{(1+h-1)^2+0^2}=\lim_{ h \to 0 } 0=0 ∂ x ∂ f ( 1 , 0 ) = h → 0 lim ( 1 + h − 1 ) 2 + 0 2 ( 1 + h − 1 ) 2 ⋅ 0 = h → 0 lim 0 = 0 ∂ f ∂ y ( 1 , 0 ) = lim h → 0 ( 1 − 1 ) 2 h ( 1 − 1 ) 2 + h 2 = lim h → 0 0 = 0 \parder{f}{y}(1,0)=\lim_{ h \to 0 } \frac{(1-1)^2h}{(1-1)^2+h^2}=\lim_{ h \to 0 } 0=0 ∂ y ∂ f ( 1 , 0 ) = h → 0 lim ( 1 − 1 ) 2 + h 2 ( 1 − 1 ) 2 h = h → 0 lim 0 = 0 כעת נמצא את α ( h , k ) \alpha(h,k) α ( h , k )
Δ f = f ( 1 + h , k ) = ∂ f ∂ x ( 1 , 0 ) ⋅ h + ∂ f ∂ y ( 1 , 0 ) ⋅ k + α ( h , k ) h 2 + k 2 = α ( h , k ) h 2 + k 2 \Delta f=f(1+h,k)=\parder{f}{x}(1,0)\cdot h+\parder{f}{y}(1,0)\cdot k+\alpha(h,k)\sqrt{ h^2+k^2 }=\alpha(h,k)\sqrt{ h^2+k^2 } Δ f = f ( 1 + h , k ) = ∂ x ∂ f ( 1 , 0 ) ⋅ h + ∂ y ∂ f ( 1 , 0 ) ⋅ k + α ( h , k ) h 2 + k 2 = α ( h , k ) h 2 + k 2 α ( h , k ) = f ( 1 + h , k ) h 2 + k 2 = h 2 k h 2 + k 2 h 2 + k 2 = h 2 k ( h 2 + k 2 ) 1.5 \alpha(h,k)=\frac{f(1+h,k)}{\sqrt{ h^2+k^2 }}=\frac{\frac{h^2k}{h^2+k^2}}{\sqrt{ h^2+k^2 }}=\frac{h^2k}{(h^2+k^2)^{1.5}} α ( h , k ) = h 2 + k 2 f ( 1 + h , k ) = h 2 + k 2 h 2 + k 2 h 2 k = ( h 2 + k 2 ) 1.5 h 2 k α ( r cos θ , r sin θ ) = r 3 cos 2 θ sin θ ( r 2 ( cos 2 θ + sin 2 θ ) ) 1.5 = r 3 r 3 cos 2 θ sin θ = F ( r ) G ( θ ) \alpha(r\cos\theta,r\sin\theta)=\frac{r^3\cos^2\theta \sin\theta}{(r^2(\cos^2\theta+\sin^2\theta))^{1.5}}=\frac{r^3}{r^3} \cos^2\theta \sin\theta=F(r)G(\theta) α ( r cos θ , r sin θ ) = ( r 2 ( cos 2 θ + sin 2 θ ) ) 1.5 r 3 cos 2 θ sin θ = r 3 r 3 cos 2 θ sin θ = F ( r ) G ( θ ) נשים לב כי עבור העקום - ציר ה-x x x ( 0 , 0 ) (0,0) ( 0 , 0 ) θ = 0 \theta=0 θ = 0 α ( r cos θ , r sin θ ) = 0 → r cos θ → 0 0 \alpha(r\cos\theta,r\sin\theta)=0 \rarrowlim{r\cos\theta\to0}0 α ( r cos θ , r sin θ ) = 0 r c o s θ → 0 0 θ = π 4 \theta=\frac{\pi}{4} θ = 4 π cos 2 ( π 4 ) sin ( π 4 ) = ( 2 2 ) 2 2 2 = 1 2 2 2 = 2 4 ≠ 0 \cos^2\left( \frac{\pi}{4} \right)\sin\left( \frac{\pi}{4} \right)=\left( \frac{\sqrt{ 2 }}{2} \right)^2 \frac{\sqrt{ 2 }}{2}=\frac{1}{2} \frac{\sqrt{ 2 }}{2}=\frac{\sqrt{ 2 }}{4}\neq 0 cos 2 ( 4 π ) sin ( 4 π ) = ( 2 2 ) 2 2 2 = 2 1 2 2 = 4 2 = 0 α ( r cos θ , r sin θ ) = 0 → r cos θ → 0 0 \alpha(r\cos\theta,r\sin\theta)=0 \cancel{ \rarrowlim{r\cos\theta\to0} }0 α ( r cos θ , r sin θ ) = 0 r c o s θ → 0 0 α \alpha α ( 0 , 0 ) (0,0) ( 0 , 0 ) f f f ( 1 , 0 ) (1,0) ( 1 , 0 )
יהי u ^ = ( a , b ) \hat{u}=(a,b) u ^ = ( a , b ) ∂ f ∂ u ( 1 , 0 ) \parder{f}{u}(1,0) ∂ u ∂ f ( 1 , 0 ) u ^ \hat{u} u ^ ∂ f ∂ u ( 1 , 0 ) = ∇ ⃗ f ( 1 , 0 ) ⋅ u ^ \parder{f}{u}(1,0)=\vec{\nabla}f(1,0)\cdot \hat{u} ∂ u ∂ f ( 1 , 0 ) = ∇ f ( 1 , 0 ) ⋅ u ^
מכיוון שהפונקציה f f f ( 1 , 0 ) (1,0) ( 1 , 0 )
∂ f ∂ u ( 1 , 0 ) = lim h → 0 f ( 1 + a h , b h ) − f ( 1 , 0 ) h = lim h → 0 f ( 1 + a h , b h ) h = lim h → 0 1 h ⋅ a 2 b 2 h 4 h 2 ( a 2 + b 2 ) = \parder{f}{u}(1,0)=\lim_{ h \to 0 } \frac{f(1+ah,bh)-f(1,0)}{h}=\lim_{ h \to 0 } \frac{f(1+ah,bh)}{h}=\lim_{ h \to 0 } \frac{1}{h}\cdot \frac{a^2b^2h^4}{h^2(a^2+b^2)}= ∂ u ∂ f ( 1 , 0 ) = h → 0 lim h f ( 1 + ah , bh ) − f ( 1 , 0 ) = h → 0 lim h f ( 1 + ah , bh ) = h → 0 lim h 1 ⋅ h 2 ( a 2 + b 2 ) a 2 b 2 h 4 = = lim h → 0 a 2 b 2 h a 2 + b 2 = ↓ חשבון גבולות a 2 b 2 a 2 + b 2 lim h → 0 h = 0 =\lim_{ h \to 0 } \frac{a^2b^2h}{a^2+b^2} \underset{\begin{matrix}
\downarrow \\
\R{\text{חשבון גבולות}}
\end{matrix}}{=} \frac{a^2b^2}{a^2+b^2}\lim_{ h \to 0 } h=0 = h → 0 lim a 2 + b 2 a 2 b 2 h ↓ חשבון גבולות = a 2 + b 2 a 2 b 2 h → 0 lim h = 0 לכן מתקיים כי:
∂ f ∂ u ( 1 , 0 ) = 0 \parder{f}{u}(1,0)=0 ∂ u ∂ f ( 1 , 0 ) = 0 נבדוק כעת מהו ∇ ⃗ f ( 1 , 0 ) \vec{\nabla}f(1,0) ∇ f ( 1 , 0 )
∇ ⃗ f ( 1 , 0 ) = ( 0 0 ) \vec{\nabla}f(1,0)=\begin{pmatrix}
0 \\
0
\end{pmatrix} ∇ f ( 1 , 0 ) = ( 0 0 ) ∇ ⃗ f ( 1 , 0 ) ⋅ u ^ = 0 ⋅ a + 0 ⋅ b = 0 \vec{\nabla}f(1,0)\cdot \hat{u}=0\cdot a+0\cdot b =0 ∇ f ( 1 , 0 ) ⋅ u ^ = 0 ⋅ a + 0 ⋅ b = 0 ראינו בחישוב הנ"ל כי ללא תלות ב-a a a b b b ∂ f ∂ u ( 1 , 0 ) = 0 \parder{f}{u}(1,0)=0 ∂ u ∂ f ( 1 , 0 ) = 0
∀ u ^ ∈ R 2 : ∂ f ∂ u ( 1 , 0 ) = 0 = ∇ ⃗ f ( 1 , 0 ) ⋅ u ^ \forall \hat{u}\in \mathbb{R}^2:\;\ \parder{f}{u}(1,0)=0=\vec{\nabla}f(1,0)\cdot \hat{u} ∀ u ^ ∈ R 2 : ∂ u ∂ f ( 1 , 0 ) = 0 = ∇ f ( 1 , 0 ) ⋅ u ^
האם f f f f x y ( 1 , 0 ) = f y x ( 1 , 0 ) f_{xy}(1,0)=f_{yx}(1,0) f x y ( 1 , 0 ) = f y x ( 1 , 0 ) f f f
∂ 2 f ∂ x ∂ y = ∂ ∂ x ( 0 ) = 0 \parder{^2f}{x \partial y}=\parder{}{x}(0)=0 ∂ x ∂ y ∂ 2 f = ∂ x ∂ ( 0 ) = 0 ∂ 2 f ∂ y ∂ x = ∂ ∂ y ( 0 ) = 0 \parder{^2f}{y \partial x}=\parder{}{y}(0)=0 ∂ y ∂ x ∂ 2 f = ∂ y ∂ ( 0 ) = 0 נשים לב כי הנגזרות המעורבות אכן שוות.הערה: פה גם גילינו כי הנ"ח מסדר שני באמת רציפות ולכן משפט שוורץ תקף.
תרגיל 4
תהי f ( x , y ) f(x,y) f ( x , y ) R 2 \mathbb{R}^2 R 2 f ( x , y ) = f ( y , x ) f(x,y)=f(y,x) f ( x , y ) = f ( y , x ) R 2 \mathbb{R}^2 R 2 f f f ( 0 , 0 ) (0,0) ( 0 , 0 ) ( − 1 , 1 ) (-1,1) ( − 1 , 1 )
נסמן u ⃗ = ( − 1 , 1 ) \vec{u}=(-1,1) u = ( − 1 , 1 ) u ^ = u ⃗ ∥ u ^ ∥ = u ⃗ 1 2 + ( − 1 ) 2 = 1 2 u ⃗ \hat{u}=\frac{\vec{u}}{\left\lVert \hat{u} \right\rVert}=\frac{\vec{u}}{\sqrt{ 1^2+(-1)^2 }}=\frac{1}{\sqrt{ 2 }}\vec{u} u ^ = ∥ u ^ ∥ u = 1 2 + ( − 1 ) 2 u = 2 1 u u ^ = ( − 1 2 , 1 2 ) \hat{u}=\left( -\frac{1}{\sqrt{ 2 }}, \frac{1}{\sqrt{ 2 }} \right) u ^ = ( − 2 1 , 2 1 ) f f f ∂ f ∂ u ( 0 , 0 ) = ∇ ⃗ f ( 0 , 0 ) ⋅ u ^ \parder{f}{u}(0,0)=\vec{\nabla}f(0,0)\cdot \hat{u} ∂ u ∂ f ( 0 , 0 ) = ∇ f ( 0 , 0 ) ⋅ u ^
∂ f ∂ x ( 0 , 0 ) = lim h → 0 f ( h , 0 ) − f ( 0 , 0 ) h = ↓ f ( h , k ) = f ( k , h ) lim h → 0 f ( 0 , h ) − f ( 0 , 0 ) h = ∂ f ∂ y ( 0 , 0 ) \parder{f}{x}(0,0)=\lim_{ h \to 0 } \frac{f(h,0)-f(0,0)}{h} \underset{\begin{matrix}
\downarrow \\
f(h,k)=f(k,h)
\end{matrix}}{=}\lim_{ h \to 0 } \frac{f(0,h)-f(0,0)}{h}=\parder{f}{y}(0,0) ∂ x ∂ f ( 0 , 0 ) = h → 0 lim h f ( h , 0 ) − f ( 0 , 0 ) ↓ f ( h , k ) = f ( k , h ) = h → 0 lim h f ( 0 , h ) − f ( 0 , 0 ) = ∂ y ∂ f ( 0 , 0 ) לכן מתקיים:
∂ f ∂ u ( 0 , 0 ) = − 1 2 ∂ f ∂ x ( 0 , 0 ) + 1 2 ∂ f ∂ y ( 0 , 0 ) = 1 2 ( ∂ f ∂ y ( 0 , 0 ) − ∂ f ∂ x ( 0 , 0 ) ) = 1 2 ⋅ 0 = 0 \parder{f}{u}(0,0)=-\frac{1}{\sqrt{ 2 }}\parder{f}{x}(0,0)+\frac{1}{\sqrt{ 2 }}\parder{f}{y}(0,0)=\frac{1}{\sqrt{ 2 }}\left( \parder{f}{y}(0,0) -\parder{f}{x}(0,0)\right)=\frac{1}{\sqrt{ 2 }}\cdot 0=0 ∂ u ∂ f ( 0 , 0 ) = − 2 1 ∂ x ∂ f ( 0 , 0 ) + 2 1 ∂ y ∂ f ( 0 , 0 ) = 2 1 ( ∂ y ∂ f ( 0 , 0 ) − ∂ x ∂ f ( 0 , 0 ) ) = 2 1 ⋅ 0 = 0