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Greg Kuperberg
Greg Kuperberg
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Actually Darsh gave an almost full solution. Let me fill in the minor technical details.

  1. We need the following quantitative form of the inverse function theorem. Suppose that $F:\mathbb R^n\to \mathbb R^n$. Assume also that $\|DF(X)^{-1}\|\le C_ 1$$\|DF(X)^{-1}\|\le C_1$, that $\max_ {Y\in B(X, \delta)}\|D^2F(Y)\|\le C_ 2$$\max_{Y\in B(X, \delta)}\|D^2F(Y)\|\le C_2$, and that $C_ 1C_ 2\delta\le\frac 12$$C_1C_2\delta\le\frac 12$. Then $F(B(X,\delta))\supset B(F(X),\frac{\delta}{2C_ 1})$$F(B(X,\delta))\supset B(F(X),\frac{\delta}{2C_1})$.

  2. Take $n=2\ell-1$ and consider the mapping $F:\mathbb R^n\to \mathbb R^n$ given by $F(y_ 1,\dots,y_ n)_k=\sum_{j=1}^n y_ j^k$$F(y_1,\dots,y_n)_k=\sum_{j=1}^n y_j^k$ where $k=1,2,\dots,n$. Take $X=(x_ 1,\dots,x_ n)$$X=(x_1,\dots,x_n)$ where $x_ j=\frac{n+j}{n}$$x_j=\frac{n+j}{n}$ for $j=1,\dots,n$.

  3. Note that in $B(X,1)$, we have $\|D^2F\|\le A^n$ for some absolute $A>1$.

  4. Note also that $DF(X)^*$ is the linear operator that maps the vector $(c_ 1,\dots,c_ n)$$(c_1,\dots,c_n)$ to the vector $p(x_ 1),\dots,p(x_ n)$$p(x_1),\ldots,p(x_n)$ consisting of the values of the polynomial $p(x)=\sum_ {k=1}^n c_k kx^{k-1}$$p(x)=\sum_{k=1}^n c_k kx^{k-1}$. The inverse operator is given by the standard interpolation formula, which allows us to estimate its norm by $B^n$ with some absolute $B>1$.

  5. Thus, taking $\delta=2^{-1}(AB)^{-n}$, we conclude that the image of the ball $B(X,\delta)$ contains a ball of radius $\frac\delta{2A^n}\ge C^{-n}$ with some absolute $C>2$.

  6. In particular, it contains two points with the difference $(0,0,\dots,0,D^{-\ell},D^{-(\ell+1)},\dots,D^{-(2\ell-1)})$ with some absolute $D>1$, which is equivalent to what we need.

Actually Darsh gave an almost full solution. Let me fill in the minor technical details.

  1. We need the following quantitative form of the inverse function theorem. Suppose that $F:\mathbb R^n\to \mathbb R^n$. Assume also that $\|DF(X)^{-1}\|\le C_ 1$, that $\max_ {Y\in B(X, \delta)}\|D^2F(Y)\|\le C_ 2$, and that $C_ 1C_ 2\delta\le\frac 12$. Then $F(B(X,\delta))\supset B(F(X),\frac{\delta}{2C_ 1})$.

  2. Take $n=2\ell-1$ and consider the mapping $F:\mathbb R^n\to \mathbb R^n$ given by $F(y_ 1,\dots,y_ n)_k=\sum_{j=1}^n y_ j^k$ where $k=1,2,\dots,n$. Take $X=(x_ 1,\dots,x_ n)$ where $x_ j=\frac{n+j}{n}$ for $j=1,\dots,n$.

  3. Note that in $B(X,1)$, we have $\|D^2F\|\le A^n$ for some absolute $A>1$.

  4. Note also that $DF(X)^*$ is the linear operator that maps the vector $(c_ 1,\dots,c_ n)$ to the vector $p(x_ 1),\dots,p(x_ n)$ consisting of the values of the polynomial $p(x)=\sum_ {k=1}^n c_k kx^{k-1}$. The inverse operator is given by the standard interpolation formula, which allows us to estimate its norm by $B^n$ with some absolute $B>1$.

  5. Thus, taking $\delta=2^{-1}(AB)^{-n}$, we conclude that the image of the ball $B(X,\delta)$ contains a ball of radius $\frac\delta{2A^n}\ge C^{-n}$ with some absolute $C>2$.

  6. In particular, it contains two points with the difference $(0,0,\dots,0,D^{-\ell},D^{-(\ell+1)},\dots,D^{-(2\ell-1)})$ with some absolute $D>1$, which is equivalent to what we need.

Actually Darsh gave an almost full solution. Let me fill in the minor technical details.

  1. We need the following quantitative form of the inverse function theorem. Suppose that $F:\mathbb R^n\to \mathbb R^n$. Assume also that $\|DF(X)^{-1}\|\le C_1$, that $\max_{Y\in B(X, \delta)}\|D^2F(Y)\|\le C_2$, and that $C_1C_2\delta\le\frac 12$. Then $F(B(X,\delta))\supset B(F(X),\frac{\delta}{2C_1})$.

  2. Take $n=2\ell-1$ and consider the mapping $F:\mathbb R^n\to \mathbb R^n$ given by $F(y_1,\dots,y_n)_k=\sum_{j=1}^n y_j^k$ where $k=1,2,\dots,n$. Take $X=(x_1,\dots,x_n)$ where $x_j=\frac{n+j}{n}$ for $j=1,\dots,n$.

  3. Note that in $B(X,1)$, we have $\|D^2F\|\le A^n$ for some absolute $A>1$.

  4. Note also that $DF(X)^*$ is the linear operator that maps the vector $(c_1,\dots,c_n)$ to the vector $p(x_1),\ldots,p(x_n)$ consisting of the values of the polynomial $p(x)=\sum_{k=1}^n c_k kx^{k-1}$. The inverse operator is given by the standard interpolation formula, which allows us to estimate its norm by $B^n$ with some absolute $B>1$.

  5. Thus, taking $\delta=2^{-1}(AB)^{-n}$, we conclude that the image of the ball $B(X,\delta)$ contains a ball of radius $\frac\delta{2A^n}\ge C^{-n}$ with some absolute $C>2$.

  6. In particular, it contains two points with the difference $(0,0,\dots,0,D^{-\ell},D^{-(\ell+1)},\dots,D^{-(2\ell-1)})$ with some absolute $D>1$, which is equivalent to what we need.

tried to fix LaTeX; added 18 characters in body
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fedja
fedja
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Actually Darsh gave an almost full solution. Let me fill in the minor technical details.

  1. We need the following quantitative form of the inverse function theorem. Suppose that $F:\mathbb R^n\to \mathbb R^n$. Assume also that $\|DF(X)^{-1}\|\le C_1$$\|DF(X)^{-1}\|\le C_ 1$, that $\max_ {Y\in B(X, \delta)}\|D^2F(Y)\|\le C_ 2$, and that $C_1C_2\delta\le\frac 12$$C_ 1C_ 2\delta\le\frac 12$. Then $F(B(X,\delta))\supset B(F(X),\frac{\delta}{2C_1})$$F(B(X,\delta))\supset B(F(X),\frac{\delta}{2C_ 1})$.

  2. Take $n=2\ell-1$ and consider the mapping $F:\mathbb R^n\to \mathbb R^n$ given by $F(y_1,\dots,y_n)_k=\sum_{j=1}^n y_j^k$$F(y_ 1,\dots,y_ n)_k=\sum_{j=1}^n y_ j^k$ where $k=1,2,\dots,n$. Take $X=(x_1,\dots,x_n)$$X=(x_ 1,\dots,x_ n)$ where $x_j=\frac{n+j}{n}$$x_ j=\frac{n+j}{n}$ for $j=1,\dots,n$.

  3. Note that in $B(X,1)$, we have $\|D^2F\|\le A^n$ for some absolute $A>1$.

  4. Note also that $DF(X)^*$ is the linear operator that maps the vector $(c_1,\dots,c_n)$$(c_ 1,\dots,c_ n)$ to the vector $p(x_1),\dots,p(x_n)$$p(x_ 1),\dots,p(x_ n)$ consisting of the values of the polynomial $p(x)=\sum_{k=1}^n c_k kx^{k-1}$$p(x)=\sum_ {k=1}^n c_k kx^{k-1}$. The inverse operator is given by the standard interpolation formula, which allows us to estimate its norm by $B^n$ with some absolute $B>1$.

  5. Thus, taking $\delta=2^{-1}(AB)^{-n}$, we conclude that the image of the ball $B(X,\delta)$ contains a ball of radius $\frac\delta{2A^n}\ge C^{-n}$ with some absolute $C>2$.

  6. In particular, it contains two points with the difference $(0,0,\dots,0,D^{-\ell},D^{-(\ell+1)},\dots,D^{-(2\ell-1)})$ with some absolute $D>1$, which is equivalent to what we need.

Actually Darsh gave an almost full solution. Let me fill in the minor technical details.

  1. We need the following quantitative form of the inverse function theorem. Suppose that $F:\mathbb R^n\to \mathbb R^n$. Assume also that $\|DF(X)^{-1}\|\le C_1$, that $\max_ {Y\in B(X, \delta)}\|D^2F(Y)\|\le C_ 2$, and that $C_1C_2\delta\le\frac 12$. Then $F(B(X,\delta))\supset B(F(X),\frac{\delta}{2C_1})$.

  2. Take $n=2\ell-1$ and consider the mapping $F:\mathbb R^n\to \mathbb R^n$ given by $F(y_1,\dots,y_n)_k=\sum_{j=1}^n y_j^k$ where $k=1,2,\dots,n$. Take $X=(x_1,\dots,x_n)$ where $x_j=\frac{n+j}{n}$ for $j=1,\dots,n$.

  3. Note that in $B(X,1)$, we have $\|D^2F\|\le A^n$ for some absolute $A>1$.

  4. Note also that $DF(X)^*$ is the linear operator that maps the vector $(c_1,\dots,c_n)$ to the vector $p(x_1),\dots,p(x_n)$ consisting of the values of the polynomial $p(x)=\sum_{k=1}^n c_k kx^{k-1}$. The inverse operator is given by the standard interpolation formula, which allows us to estimate its norm by $B^n$ with some absolute $B>1$.

  5. Thus, taking $\delta=2^{-1}(AB)^{-n}$, we conclude that the image of the ball $B(X,\delta)$ contains a ball of radius $\frac\delta{2A^n}\ge C^{-n}$ with some absolute $C>2$.

  6. In particular, it contains two points with the difference $(0,0,\dots,0,D^{-\ell},D^{-(\ell+1)},\dots,D^{-(2\ell-1)})$ with some absolute $D>1$, which is equivalent to what we need.

Actually Darsh gave an almost full solution. Let me fill in the minor technical details.

  1. We need the following quantitative form of the inverse function theorem. Suppose that $F:\mathbb R^n\to \mathbb R^n$. Assume also that $\|DF(X)^{-1}\|\le C_ 1$, that $\max_ {Y\in B(X, \delta)}\|D^2F(Y)\|\le C_ 2$, and that $C_ 1C_ 2\delta\le\frac 12$. Then $F(B(X,\delta))\supset B(F(X),\frac{\delta}{2C_ 1})$.

  2. Take $n=2\ell-1$ and consider the mapping $F:\mathbb R^n\to \mathbb R^n$ given by $F(y_ 1,\dots,y_ n)_k=\sum_{j=1}^n y_ j^k$ where $k=1,2,\dots,n$. Take $X=(x_ 1,\dots,x_ n)$ where $x_ j=\frac{n+j}{n}$ for $j=1,\dots,n$.

  3. Note that in $B(X,1)$, we have $\|D^2F\|\le A^n$ for some absolute $A>1$.

  4. Note also that $DF(X)^*$ is the linear operator that maps the vector $(c_ 1,\dots,c_ n)$ to the vector $p(x_ 1),\dots,p(x_ n)$ consisting of the values of the polynomial $p(x)=\sum_ {k=1}^n c_k kx^{k-1}$. The inverse operator is given by the standard interpolation formula, which allows us to estimate its norm by $B^n$ with some absolute $B>1$.

  5. Thus, taking $\delta=2^{-1}(AB)^{-n}$, we conclude that the image of the ball $B(X,\delta)$ contains a ball of radius $\frac\delta{2A^n}\ge C^{-n}$ with some absolute $C>2$.

  6. In particular, it contains two points with the difference $(0,0,\dots,0,D^{-\ell},D^{-(\ell+1)},\dots,D^{-(2\ell-1)})$ with some absolute $D>1$, which is equivalent to what we need.

replace "implicit" by "inverse" as more appropriate
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fedja
fedja
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Actually Darsh gave an almost full solution. Let me fill in the minor technical details.

  1. We need the following quantitative form of the implicitinverse function theorem. Suppose that $F:\mathbb R^n\to \mathbb R^n$. Assume also that $\|DF(X)^{-1}\|\le C_1$, that $\max_ {Y\in B(X, \delta)}\|D^2F(Y)\|\le C_ 2$, and that $C_1C_2\delta\le\frac 12$. Then $F(B(X,\delta))\supset B(F(X),\frac{\delta}{2C_1})$.

  2. Take $n=2\ell-1$ and consider the mapping $F:\mathbb R^n\to \mathbb R^n$ given by $F(y_1,\dots,y_n)_k=\sum_{j=1}^n y_j^k$ where $k=1,2,\dots,n$. Take $X=(x_1,\dots,x_n)$ where $x_j=\frac{n+j}{n}$ for $j=1,\dots,n$.

  3. Note that in $B(X,1)$, we have $\|D^2F\|\le A^n$ for some absolute $A>1$.

  4. Note also that $DF(X)^*$ is the linear operator that maps the vector $(c_1,\dots,c_n)$ to the vector $p(x_1),\dots,p(x_n)$ consisting of the values of the polynomial $p(x)=\sum_{k=1}^n c_k kx^{k-1}$. The inverse operator is given by the standard interpolation formula, which allows us to estimate its norm by $B^n$ with some absolute $B>1$.

  5. Thus, taking $\delta=2^{-1}(AB)^{-n}$, we conclude that the image of the ball $B(X,\delta)$ contains a ball of radius $\frac\delta{2A^n}\ge C^{-n}$ with some absolute $C>2$.

  6. In particular, it contains two points with the difference $(0,0,\dots,0,D^{-\ell},D^{-(\ell+1)},\dots,D^{-(2\ell-1)})$ with some absolute $D>1$, which is equivalent to what we need.

Actually Darsh gave an almost full solution. Let me fill in the minor technical details.

  1. We need the following quantitative form of the implicit function theorem. Suppose that $F:\mathbb R^n\to \mathbb R^n$. Assume also that $\|DF(X)^{-1}\|\le C_1$, that $\max_ {Y\in B(X, \delta)}\|D^2F(Y)\|\le C_ 2$, and that $C_1C_2\delta\le\frac 12$. Then $F(B(X,\delta))\supset B(F(X),\frac{\delta}{2C_1})$.

  2. Take $n=2\ell-1$ and consider the mapping $F:\mathbb R^n\to \mathbb R^n$ given by $F(y_1,\dots,y_n)_k=\sum_{j=1}^n y_j^k$ where $k=1,2,\dots,n$. Take $X=(x_1,\dots,x_n)$ where $x_j=\frac{n+j}{n}$ for $j=1,\dots,n$.

  3. Note that in $B(X,1)$, we have $\|D^2F\|\le A^n$ for some absolute $A>1$.

  4. Note also that $DF(X)^*$ is the linear operator that maps the vector $(c_1,\dots,c_n)$ to the vector $p(x_1),\dots,p(x_n)$ consisting of the values of the polynomial $p(x)=\sum_{k=1}^n c_k kx^{k-1}$. The inverse operator is given by the standard interpolation formula, which allows us to estimate its norm by $B^n$ with some absolute $B>1$.

  5. Thus, taking $\delta=2^{-1}(AB)^{-n}$, we conclude that the image of the ball $B(X,\delta)$ contains a ball of radius $\frac\delta{2A^n}\ge C^{-n}$ with some absolute $C>2$.

  6. In particular, it contains two points with the difference $(0,0,\dots,0,D^{-\ell},D^{-(\ell+1)},\dots,D^{-(2\ell-1)})$ with some absolute $D>1$, which is equivalent to what we need.

Actually Darsh gave an almost full solution. Let me fill in the minor technical details.

  1. We need the following quantitative form of the inverse function theorem. Suppose that $F:\mathbb R^n\to \mathbb R^n$. Assume also that $\|DF(X)^{-1}\|\le C_1$, that $\max_ {Y\in B(X, \delta)}\|D^2F(Y)\|\le C_ 2$, and that $C_1C_2\delta\le\frac 12$. Then $F(B(X,\delta))\supset B(F(X),\frac{\delta}{2C_1})$.

  2. Take $n=2\ell-1$ and consider the mapping $F:\mathbb R^n\to \mathbb R^n$ given by $F(y_1,\dots,y_n)_k=\sum_{j=1}^n y_j^k$ where $k=1,2,\dots,n$. Take $X=(x_1,\dots,x_n)$ where $x_j=\frac{n+j}{n}$ for $j=1,\dots,n$.

  3. Note that in $B(X,1)$, we have $\|D^2F\|\le A^n$ for some absolute $A>1$.

  4. Note also that $DF(X)^*$ is the linear operator that maps the vector $(c_1,\dots,c_n)$ to the vector $p(x_1),\dots,p(x_n)$ consisting of the values of the polynomial $p(x)=\sum_{k=1}^n c_k kx^{k-1}$. The inverse operator is given by the standard interpolation formula, which allows us to estimate its norm by $B^n$ with some absolute $B>1$.

  5. Thus, taking $\delta=2^{-1}(AB)^{-n}$, we conclude that the image of the ball $B(X,\delta)$ contains a ball of radius $\frac\delta{2A^n}\ge C^{-n}$ with some absolute $C>2$.

  6. In particular, it contains two points with the difference $(0,0,\dots,0,D^{-\ell},D^{-(\ell+1)},\dots,D^{-(2\ell-1)})$ with some absolute $D>1$, which is equivalent to what we need.

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fedja
fedja
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