Let $f:R^n\to R^n$ be a smooth function and $g:R^n\to R$ $$g(x_1,\dots,g_n)=x_1^5+\dots+x_n^5$$ Assume that $g\circ f\equiv 0$. Prove that $\det Df\equiv 0$.  The only idea I have is to apply the chain rule $d(g\circ f)_p=dg_{f(p)}\cdot df_p$, where each term is the corresp. matrix of partial derivatives. Then since $g\circ f=0$, $d(g\circ f)_p=0$ for any $p$. But this doesn't imply that either of the two matrices on the RHS is zero. Is it still true that the determinant of one of them is zero? Why do we need the explicit form of $g$?
121498
Let $f:\mathbb{R}^n \rightarrow \mathbb{R}^n$ be a smooth function and let $g:\mathbb{R}^n \rightarrow \mathbb{R}$ be defined by $g(x_1,...,x_n)=x_1^5+...+x_n^5$. Suppose $g\circ f\equiv 0$. Show that $\det Df\equiv 0$.  I was going to start the solution from $n=2$ and then use induction on $n$.  So, for $n=2$, $f:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ is a smooth function and $g:\mathbb{R}^2 \rightarrow \mathbb{R}$ where $g(x_1,x_2)=x_1^5+x_2^5$. Then $Df=\begin{pmatrix}   f_{x_1x_1} &amp; f_{x_1x_2}  \\   f_{x_2x_1} &amp; f_{x_2x_2}   \end{pmatrix}$  But now I am stuck in proving the result. Am I in the correct track? Can anyone plese give me a hint to continue?  Ok so since $g\circ f$ is $0$, the image of $f$ is the line $x_1+x_2=0$ which has no min, max or saddle.   If det$Df(x_0)&gt;0$ and $f_{x_1x_1}(x_0)&gt;0$ OR det$Df(x_0)&gt;0$ and $f_{x_1x_1}(x_0)&lt;0$, we have contradiction.   But if det$Df(x_0)&gt;0$ and $f_{x_1x_1}(x_0)=0$, then det$Df(x_0)=0$ or, det$Df(x_0)&lt;0$(Since $f_{x_1x_2}=f_{x_2x_1}$). So this is impossible.  If det$Df(x_0)&lt;0$ we have saddle and it is impossible.   So finally we conclude that det$Df(x)=0$ for all $x$. Am I correct?
31554
Let $n&gt;1$ , consider $\mathbb R^n$ with standard volume form $\omega _0 =dx_1 \wedge ...\wedge dx_n$ ; let $r,R$ be two positive real numbers with $R&gt;r$ . Then is it true that for every $a,b \in B(0;r)$ , $\exists$ a diffeomorphism $f:\mathbb R^n \to \mathbb R^n $ such that $f(a)=b ; f(x)=x , \forall x \in \mathbb R^n $ with $||x||&gt;R$ and $f^*\omega _0=\omega _0$ ?   As $f^*(dx_1 \wedge ...\wedge dx_n)=(det f' )(dx_1 \wedge ...\wedge dx_n)$ , so basically given $R&gt;r&gt;0$ and $||a||,||b||&lt;r$ , we have to find a diffeomorphism $f : \mathbb R^n \to \mathbb R^n  $ with $f(a)=b$ ; $f|_{\mathbb R^n \setminus B[0;R]}=Id$ and $det f'(p)=1 , \forall p \in \mathbb R^n $ .  Definitely this cannot work out in $\mathbb R$ , because the only differentiable function satisfying $det f'(p)=1 , \forall p$ and $f(a)=b$ is $f(x)=x+b-a$ and this function never gives identity unless $a =b$
82871
The question is:  Let $\phi ： \mathbb{R}^n \to \mathbb{R}^n$ be a $C^1$ map and let $y = \phi(x)$ be the change of variables. Show that $$dy_1\wedge\dots\wedge dy_n = (\operatorname{det}D\phi(x))\cdot dx_1\wedge\dots\wedge dx_n.$$  I tried $n = 2, 3$ but still didn't get any rule. Do we need to do this under integral?
57458
Let $f,g:\mathbb R^n \rightarrow \mathbb R^n$, $(x_1,\dots, x_n)\mapsto f,g(x_1,\dots, x_n)$ be smooth functions with compact support.  Use the following notation: for an index $\alpha = (\alpha_1,\dots, \alpha_n)$ where $\alpha_i$ is a natural number, we denote by $D^{\alpha}$ the derivative with respect to $x_1$ $\alpha_1$ times, with respect to $x_2$, $\alpha_2$ times, and so on until $x_n$, $\alpha_n$ times.  Can we then say that $$\int_{\mathbb R^n} D^{\alpha} f(x_1,\dots, x_n) g(x_1,\dots, x_n)dx_1 \cdots dx_n = (-1)^{|\alpha|} \int_{\mathbb R^n} f(x_1,\dots, x_n) D^{\alpha} g(x_1,\dots, x_n)dx_1 \cdots dx_n$$ where $|\alpha| = \sum_{i=1}^n \alpha_i$.  Thanks you!
85287
Suppose $g\in C^1 [a,b]$. Prove that for all $\epsilon &gt; 0$, there is $\delta &gt; 0$ such that $|{g'(c)} - {{g(d)-g(c)} \over {d-c}} |{&lt; \epsilon }$ for all points $c,d \in [a,b]$ with $0 &lt;|d-c|&lt; \delta$   First, I don't know what $C^1 [a,b]$ means.  Some ideas:  By Mean value theorem, $g'(c) =  {{g(b)-g(a)} \over {b-a}} $ since $c\in[a,b]$.   To show $|{g'(c)} - {{g(d)-g(c)} \over {d-c}} |{&lt; \epsilon }$ whenever $0 &lt;|d-c|&lt; \delta$ for $ c,d \in [a,b]$. I guess I have to use the definition of limit of continuous function. But, I don't know how to connect all of these ideas.
96608
$f(r,\theta)=(r\cos\theta,r\sin\theta)$ Then  $1.$ $Df(r,\theta)$ is always nonzero for any $(r,\theta)\in\mathbb{R}^2,r\ne 0$  $2.$ $f$ is injective on $\{(r,\theta):r\ne 0\}$  $3.$ On any neighborhood of $(r,\theta):r\ne 0$, $f$ is one one.  $4$ $Df(r,\theta)=r^2I\forall (r,\theta):r\ne 0$  I have calculated that $Df(r,\theta)$ is the matrix $$\begin{bmatrix} \cos\theta &amp;-r \sin\theta \\ \sin\theta &amp; r\cos\theta \end{bmatrix}$$ so its determinant is r which is nonzero.so by inverse mapping theorem $3$ is true, $1$ is true also, I don't know about $2$ and $4$, could any one tell me how to prove or disprove , for $4$ I have taken $(r,\theta)=(r,\pi)$ and saw the contradiction.
78184
I have tried an attempt on the following:  Let $F:U\in \Bbb{R}^n\to \Bbb{R}$ and $f:\Bbb{R}\to \Bbb{R}$ where $f$ is an even function. Now $F(\mathbf{x})=f(|\mathbf{x}|)$, where $|\ . |$ is the Euclidean norm. Given $f$ is $C^r$, I have to show that $F$ is $C^r$.  My attempt was to argument by induction. I showed that $F$ is $C^1$, basically using univariate chain rule on $\frac{\partial f\left(\sqrt{x_1^2+\dots+x_n^2}\right)}{\partial x_1}$ and arguing that it would be continuous using the fact that $f$ is $C^r$. Since the argument works for all $x_i$, it establishes first-differntiability for $F$ and continuity of the operator $DF(\mathbf{x})$. Then I moved on to show $F$ is $C^r$ assuming $F$ is $C^{r-1}$ and $f$ is $C^r$. Now what I did was partially differentiate:  $$ {\partial \over \partial x_2} \left[\frac{\partial ^{k-1}f}{\partial x_1^{k-1}}\right] = \underbrace{{\partial \over \partial |\mathbf{x}|} \left[\frac{\partial ^{k-1}f}{\partial x_1^{k-1}}\right]}_{(1)} . \underbrace{{\partial |\mathbf{x}| \over \partial x_2}}_{(2)} $$  My argument runs as follows: this partial derivative is continuous since $f$ is $C^r$ and hence $(1)$ is continuous. Continuity of $(2)$ is establishd in a similar manner, since $\sqrt{\mathbf{x}}$ is $C^r$ if $\mathbf{x} \neq \mathbf{0}$. Then the product would be continuous. I am uneasy on multivariate arguments and dont know if my approach is completely off track. I didnt use the even function property, so I am sure I might me missing something. Any hints and references would be welcome. Thanks!
46151
I have to prove that determinant of skew- symmetric matrix of odd order is zero and also that its adjoint doesnt exist. I am sorry if the question is duplicate or already exists.I am not getting any start.I study in Class 11 so please give the proof accordingly. Thanks!
99258
Calculate the differential of the function $f: \Bbb R^n \to \Bbb R$ given by $$f(x) = x^T A x$$ with $A$ symmetric. Also, differentiate this function with respect to $x^T$.     How exactly does this work in the case of vectors and matrices? Could anyone please help me out?
71400
Suppose I had a function that satisfied the property $f(x)=f(x-1)+g(x)$.  For any $x\in\mathbb N$, it is easy enough to see that this boils down to the statement  $$f(x)=f(0)+\sum_{k=1}^xg(k)$$  If we return to our functional equation and differentiate it $n$ times, we get  $$f^{(n)}(x)=f^{(n)}(x-1)+g^{(n)}(x)$$  Once again, for any $x\in\mathbb N$, we have  $$f^{(n)}(x)=f^{(n)}(0)+\sum_{k=1}^xg^{(n)}(k)$$  One could then integrate both sides.  For example, this asserts that  $$\sum_{k=1}^xg(k)=xf'(0)+\int_0^x\sum_{k=1}^tg'(k)dt$$  If the sum inside the intagral is generalized to any upper bound.  How would I justify this for discrete functions?    An example of this is letting $f_p(x)=\sum_{k=1}^xk^p$, as it seems to follow the above result.  We can see from the above that  $$f_p(x)=a_px+p\int_0^xf_{p-1}(t)dt$$  $$a_p=1-p\int_0^1f_{p-1}(t)dt$$  which appears to be a true recursive formula.    Under what conditions can I extend discrete functions $f(x)$ and $g(x)$ into continuous differentiable functions and apply this method?
44911
Given a function $g$ which has derivative $g'$ for all x $\in {R}$ and satisfying $g'(0)=2$ and $g(x+y)=e^yg(x)+e^xg(y)$ for all $x,y\in {R}$ Show that $g'(x)+g(x)-2e^x=0$  $\dfrac{g(x+y)}{e^{x+y}}=\dfrac{g(x)}{e^{x}}+\dfrac{g(y)}{e^{y}}$ Putting $x=0$, $g'(y)=2e^y+g(x)$ Also putting $y=0$,we get, $0=e^xg(0)\implies g(0)=0$ I had a strong hunch that $g(x)=e^x-e^{-x}$ but it does not satisfy $g(x+y)=e^yg(x)+e^xg(y)$ Please help.
16531
Consider the function $g:\mathbb{R^2} \rightarrow \mathbb{R}$ defined by   $g(x,y) = \left\{\begin{matrix}  \dfrac{x^3y}{x^2+y^2}&amp; \textrm{if } (x,y) \neq (0,0) \\   0 &amp; \textrm{if }(x,y) = (0,0) \end{matrix}\right.$  How do I prove this is continuous?  I know that I have to:  Consider $x_0 \in \mathbb{R^2}$, and let $\epsilon &gt; 0$ be given. I am confused on how to pick a $\delta$ that will work such that $d(x,x_o) &lt; \delta$.
36032
I do not understand a remark in Adams' Calculus (page 628 $7^{th}$ edition). This remark is about the derivative of a determinant whose entries are functions as quoted below.      $``$Since every term in the expansion of a determinant of any order is a product involving one element from each row, the general product rule implies that the derivative of an $n\times n$ determinant whose elements are functions will be the sum of $n$ such $n\times n$ determinants, each with the elements of one of the rows differentiated. For the $3\times 3$ case we have      $$\frac{d}{dt}\begin{vmatrix}   a_{11}(t) &amp; a_{12}(t) &amp; a_{13}(t) \\   a_{21}(t) &amp; a_{22}(t) &amp; a_{23}(t) \\   a_{31}(t) &amp; a_{32}(t) &amp; a_{33}(t) \end{vmatrix}=\begin{vmatrix}   a'_{11}(t) &amp; a'_{12}(t) &amp; a'_{13}(t) \\   a_{21}(t) &amp; a_{22}(t) &amp; a_{23}(t) \\   a_{31}(t) &amp; a_{32}(t) &amp; a_{33}(t) \end{vmatrix}+\begin{vmatrix}   a_{11}(t) &amp; a_{12}(t) &amp; a_{13}(t) \\   a'_{21}(t) &amp; a'_{22}(t) &amp; a'_{23}(t) \\   a_{31}(t) &amp; a_{32}(t) &amp; a_{33}(t) \end{vmatrix}+\begin{vmatrix}   a_{11}(t) &amp; a_{12}(t) &amp; a_{13}(t) \\   a_{21}(t) &amp; a_{22}(t) &amp; a_{23}(t) \\   a'_{31}(t) &amp; a'_{32}(t) &amp; a'_{33}(t) \end{vmatrix}."$$   It is not difficult to check this equality by simply expanding both sides. However, the remark sounds like using some clever trick to get this result. Can anyone explain it to me, please? Thank you!
98874
Let $f\in L^1_\mathbb C(\mathbb R^n)$. I once read, in one of my old exam, that if $\hat{f}(\mathbb R^n)\subset\mathbb R_+$, then $\hat{f}\in L^1(\mathbb R^n)$. As far as I remember, the professor gave an identity using mollifiers but I'm not 100% sure. Any hints on how to prove that ?  Thanks!
10225
Suppose $f$ and $g$ are continuously differentiable functions such that $f(x) = g'(x)$ and $g(x) = f'(x)$ and that any product of $f, f', g$ and $g'$ is commutative for all $x ∈ R$. Show that $ f^2 − g^2 = C $ for some real constant C  I have actually no clue how to solve this, and would be really greatfull for all the help i can get
40843
Let  $$f_n := \begin{vmatrix} a_1 &amp; b_1 \\ c_1 &amp; a_2 &amp; b_2 \\ &amp; c_2 &amp; \ddots &amp; \ddots \\ &amp; &amp; \ddots &amp; \ddots &amp; b_{n-1} \\ &amp; &amp; &amp; c_{n-1} &amp; a_n \end{vmatrix}$$  Apparently, the determinant of the tridiagional matrix above is given by the recurrence relation   $$f_n = a_n f_{n-1} - c_{n-1} b_{n-1}f_{n-2}$$  with initial values $f_0 = 1$ and $f_{-1} = 0$ (according to Wikipedia). Can anyone please explain to me how they came to this recurrence relation?  I don't really understand how to derive it.
79378
Let $A=(a_{ij})$ be a $n \times n$ matrix with $a_{ij}=\gcd(i,j) , \forall i,j=1,2, \cdots, n$ , then how do we prove $\det A=\prod_{i=1}^n \phi(i)$ ? , where $\phi$ is the Euler's phi function