Рекомендую всем попробовать китайский chat.deepseek.com. Он бесплатный и очень хорошо решает математические задачи, при этом все рассуждения (reasoning) показываются перед выдачей окончательного ответа (чего нет у модели OpenAI o1).
-- Пн ноя 25, 2024 23:28:22 --Я ему задал задачу
https://dxdy.ru/topic159013.html, он её не решил, думал 338 секунд. Но читая его рассуждения, я нашёл новое решение этой задачи, которого раньше не знал, и оно мне нравится больше, чем приведённые решения в той теме. То есть реально ИИ помогает решать математические задачи. Рассуждения DeepSeek у меня сохранены, могу скинуть. Решение на их основе напишу в теме завтра.
Вот рассуждения
(Оффтоп)
Let's dive into this problem. so, we have two smooth mappings, \( f \) and \( g \), defined in a neighborhood of a closed ball \( \overline{b} \subset \mathbb{r}^n \), and they agree on the boundary of the ball, that is, \( f = g \) on \( \partial b \). we need to show that the integrals of the determinants of their jacobians over \( b \) are equal:
\[
\int_{b} \det f'(x) \, dx = \int_{b} \det g'(x) \, dx.
\]
first off, i need to recall what the jacobian matrix is. for a function \( f: \mathbb{r}^n \to \mathbb{r}^n \), the jacobian matrix \( f'(x) \) is the matrix of all first-order partial derivatives of \( f \). the determinant of this matrix, \( \det f'(x) \), represents the local scaling factor of the transformation \( f \) at the point \( x \).
given that \( f \) and \( g \) are equal on the boundary \( \partial b \), and they are smooth in a neighborhood of \( \overline{b} \), there might be some relationship between their derivatives inside \( b \).
one approach that comes to mind is to consider the difference between \( f \) and \( g \), let's call it \( h = f - g \). since \( f = g \) on \( \partial b \), it follows that \( h = 0 \) on \( \partial b \).
now, the jacobians of \( f \) and \( g \) are related through the jacobian of \( h \):
\[
f'(x) = g'(x) + h'(x).
\]
but wait, actually, \( h = f - g \), so \( h'(x) = f'(x) - g'(x) \).
our goal is to relate \( \det f'(x) \) and \( \det g'(x) \). determinants are nonlinear, so directly relating them through \( h'(x) \) might be tricky.
perhaps i can consider the difference of the determinants:
\[
\det f'(x) - \det g'(x).
\]
is there a way to express this difference in terms of \( h'(x) \)?
i recall that for two square matrices \( a \) and \( b \), there is an identity for the difference of their determinants:
\[
\det(a) - \det(b) = \det(b) \cdot \text{tr}\left( (a - b) \cdot (b^{-1}) \right) + \text{higher order terms}.
\]
but this seems too vague and may not be directly applicable here, especially since we don't know if \( b \) is invertible or not.
alternatively, maybe i can use the mean value theorem for determinants or some integral identity.
let me think about integrating the difference of the determinants over \( b \):
\[
\int_{b} \left( \det f'(x) - \det g'(x) \right) dx = 0.
\]
if i can show that this integral is zero, then i would have the desired equality.
another thought: perhaps consider the homotopy between \( f \) and \( g \), since they agree on the boundary. if i can construct a homotopy \( f_t \) such that \( f_0 = g \) and \( f_1 = f \), and \( f_t = g \) on \( \partial b \) for all \( t \), then maybe i can analyze the derivative of the determinant along this homotopy.
let me try to formalize this idea.
define \( f_t(x) = g(x) + t(h(x)) \), where \( h(x) = f(x) - g(x) \), for \( t \in [0,1] \). then, \( f_0(x) = g(x) \) and \( f_1(x) = f(x) \), and since \( h(x) = 0 \) on \( \partial b \), \( f_t(x) = g(x) \) on \( \partial b \) for all \( t \).
now, consider the function:
\[
i(t) = \int_{b} \det f_t'(x) \, dx.
\]
we need to show that \( i(0) = i(1) \), which would imply the desired equality.
to show that \( i(t) \) is constant, i can compute its derivative with respect to \( t \) and show that \( i'(t) = 0 \) for all \( t \).
so, let's compute \( i'(t) \):
\[
i'(t) = \frac{d}{dt} \int_{b} \det f_t'(x) \, dx = \int_{b} \frac{d}{dt} \det f_t'(x) \, dx.
\]
assuming we can interchange differentiation and integration, which should be valid under the smoothness assumptions.
now, i need to find \( \frac{d}{dt} \det f_t'(x) \).
recall that for a differentiable function \( a(t) \) where \( a(t) \) is an \( n \times n \) matrix, the derivative of the determinant is given by:
\[
\frac{d}{dt} \det a(t) = \det a(t) \cdot \text{tr}\left( a(t)^{-1} \cdot \frac{da(t)}{dt} \right),
\]
provided that \( a(t) \) is invertible.
however, we don't know if \( f_t'(x) \) is invertible everywhere in \( b \), so this might not be the best route.
alternatively, there's another formula for the derivative of the determinant:
\[
\frac{d}{dt} \det a(t) = \det a(t) \cdot \text{tr}\left( a(t)^{-1} \cdot \frac{da(t)}{dt} \right),
\]
but again, this assumes \( a(t) \) is invertible.
is there a way to express the derivative of the determinant without assuming invertibility?
yes, actually, there's a general formula:
\[
\frac{d}{dt} \det a(t) = \det a(t) \cdot \text{tr}\left( a(t)^{\adj} \cdot \frac{da(t)}{dt} \right),
\]
where \( a(t)^{\adj} \) is the adjugate matrix of \( a(t) \).
but this seems more complicated and might not lead me anywhere.
let me think differently.
perhaps i can consider the divergence theorem, which relates integrals over a domain to integrals over its boundary.
recall that the divergence theorem states that for a vector field \( v: \mathbb{r}^n \to \mathbb{r}^n \),
\[
\int_{b} \text{div} \, v(x) \, dx = \int_{\partial b} v(x) \cdot \nu(x) \, d\sigma(x),
\]
where \( \nu(x) \) is the outward unit normal vector on \( \partial b \).
if i can express \( \det f'(x) - \det g'(x) \) as a divergence of some vector field, then i could apply the divergence theorem.
let me explore this idea.
first, consider that the determinant of the jacobian matrix is related to the divergence in some way.
in fact, for a vector field \( f: \mathbb{r}^n \to \mathbb{r}^n \), \( \det f'(x) \) represents the volume scaling factor of the transformation \( f \) at \( x \).
is there a way to write \( \det f'(x) \) as a divergence?
i recall that in some contexts, determinants can be related to traces and other matrix operations, but i'm not sure how to connect this directly to a divergence.
alternatively, perhaps i can consider the differential form approach, where determinants relate to the top-dimensional forms.
but that might be overcomplicating things for this problem.
let me go back to the homotopy idea.
we have \( i(t) = \int_{b} \det f_t'(x) \, dx \), and we want to show \( i'(t) = 0 \).
alternatively, perhaps consider the difference \( \det f'(x) - \det g'(x) \) directly.
i need to find a way to express this difference in a way that allows me to integrate over \( b \) and show it's zero.
another approach: if \( f \) and \( g \) agree on \( \partial b \), perhaps their difference \( h = f - g \) can be extended to a compactly supported vector field in \( b \), and then use properties of determinants and integration by parts.
wait, integration by parts might be useful here.
let me consider expanding \( \det f'(x) - \det g'(x) \).
is there a formula that allows me to write the difference of determinants in terms of an integral or some other expression involving \( h'(x) \)?
actually, there is a formula for the difference of determinants in terms of a sum over the differences in the matrix entries.
specifically, for two matrices \( a \) and \( b \), we have:
\[
\det a - \det b = \sum_{i=1}^n \det \begin{pmatrix} b_1 \\ \vdots \\ a_i - b_i \\ \vdots \\ b_n \end{pmatrix},
\]
where \( a_i - b_i \) is in the \( i \)-th row.
this is known as the multilinearity property of the determinant.
applying this to \( a = f'(x) \) and \( b = g'(x) \), we get:
\[
\det f'(x) - \det g'(x) = \sum_{i=1}^n \det \begin{pmatrix} g'(x)_1 \\ \vdots \\ f'(x)_i - g'(x)_i \\ \vdots \\ g'(x)_n \end{pmatrix}.
\]
now, \( f'(x)_i - g'(x)_i = h'(x)_i \), where \( h'(x)_i \) is the \( i \)-th row of \( h'(x) \).
so,
\[
\det f'(x) - \det g'(x) = \sum_{i=1}^n \det \begin{pmatrix} g'(x)_1 \\ \vdots \\ h'(x)_i \\ \vdots \\ g'(x)_n \end{pmatrix}.
\]
this expresses the difference of determinants as a sum of determinants where one row is replaced by the difference in the corresponding rows of the jacobians.
now, to integrate this over \( b \):
\[
\int_{b} (\det f'(x) - \det g'(x)) \, dx = \sum_{i=1}^n \int_{b} \det \begin{pmatrix} g'(x)_1 \\ \vdots \\ h'(x)_i \\ \vdots \\ g'(x)_n \end{pmatrix} \, dx.
\]
i need to show that this sum is zero.
perhaps i can relate this to the divergence of some vector field.
let me consider the \( i \)-th term in the sum:
\[
\int_{b} \det \begin{pmatrix} g'(x)_1 \\ \vdots \\ h'(x)_i \\ \vdots \\ g'(x)_n \end{pmatrix} \, dx.
\]
this determinant can be thought of as the \( i \)-th component of some vector operation.
alternatively, maybe i can interpret this determinant as the divergence of a vector field constructed from \( h \) and \( g \).
this is getting a bit abstract. maybe there's a better way.
let's consider the case when \( n = 1 \). for \( n = 1 \), the determinant of the jacobian is just the derivative of the function, and the integral becomes:
\[
\int_{b} f'(x) \, dx = \int_{b} g'(x) \, dx.
\]
by the fundamental theorem of calculus, this is \( f(b) - f(a) = g(b) - g(a) \), which holds because \( f = g \) on \( \partial b \), i.e., at the endpoints \( a \) and \( b \).
so, for \( n = 1 \), the equality holds.
now, for higher dimensions, perhaps there's an analogous theorem.
i recall that in higher dimensions, the divergence theorem generalizes the fundamental theorem of calculus.
maybe i can apply the divergence theorem to a vector field related to \( h \).
let me try to define a vector field \( v(x) \) such that its divergence is \( \det f'(x) - \det g'(x) \).
alternatively, perhaps consider the vector field \( v(x) = h(x) \), and compute its divergence.
but \( h(x) = f(x) - g(x) \), and \( h(x) = 0 \) on \( \partial b \).
the divergence of \( h(x) \) is \( \text{div} \, h(x) = \sum_{i=1}^n \frac{\partial h_i}{\partial x_i} \).
this doesn't directly relate to the difference of determinants.
maybe i need a different approach.
let me consider the change of variables formula.
for a smooth, invertible map \( f: b \to f(b) \), we have:
\[
\int_{b} \det f'(x) \, dx = \int_{f(b)} 1 \, dy.
\]
similarly,
\[
\int_{b} \det g'(x) \, dx = \int_{g(b)} 1 \, dy.
\]
so, the integrals represent the volumes of \( f(b) \) and \( g(b) \), respectively.
if \( f \) and \( g \) map \( b \) to regions of the same volume, then the integrals would be equal.
but how do i know that \( f(b) \) and \( g(b) \) have the same volume, given that \( f = g \) on \( \partial b \)?
this seems too simplistic and probably not generally true.
wait, actually, this is not necessarily the case. the volumes could be different even if \( f = g \) on \( \partial b \).
so, this line of reasoning doesn't help.
back to the drawing board.
another idea: perhaps use the fact that \( f \) and \( g \) are homotopic through the homotopy \( f_t \), and consider properties of the determinant under such deformations.
specifically, if \( f_t \) is a homotopy from \( g \) to \( f \), with \( f_t = g \) on \( \partial b \), then maybe the integral of \( \det f_t'(x) \) over \( b \) is constant in \( t \).
to show this, i need to show that the derivative of this integral with respect to \( t \) is zero.
so, let's compute:
\[
\frac{d}{dt} \int_{b} \det f_t'(x) \, dx = \int_{b} \frac{d}{dt} \det f_t'(x) \, dx.
\]
as before, assuming we can interchange differentiation and integration.
now, to compute \( \frac{d}{dt} \det f_t'(x) \), perhaps i can express it in terms of the trace and the adjugate matrix.
recall that:
\[
\frac{d}{dt} \det a(t) = \det a(t) \cdot \text{tr}\left( a(t)^{\adj} \cdot \frac{da(t)}{dt} \right).
\]
but again, this involves the adjugate matrix, which might not be helpful here.
alternatively, perhaps consider expressing the determinant in terms of the exterior product.
in differential geometry, the determinant can be related to the volume form.
specifically, for a linear map \( a: \mathbb{r}^n \to \mathbb{r}^n \), \( \det a \) is the scalar factor by which \( a \) scales the volume form.
in the context of integration, this suggests that the integral of \( \det f'(x) \) over \( b \) is related to the integral of the pullback of the volume form by \( f \).
but this might be getting too abstract for this problem.
let me consider a different strategy.
suppose i consider the difference \( \det f'(x) - \det g'(x) \) and try to express it as a divergence of some vector field.
if i can find a vector field \( v(x) \) such that:
\[
\det f'(x) - \det g'(x) = \text{div} \, v(x),
\]
then by the divergence theorem:
\[
\int_{b} (\det f'(x) - \det g'(x)) \, dx = \int_{\partial b} v(x) \cdot \nu(x) \, d\sigma(x).
\]
but since \( f = g \) on \( \partial b \), and assuming \( v(x) \) is constructed in a way that reflects this, perhaps the boundary integral is zero.
this seems promising.
so, can i find such a vector field \( v(x) \)?
one way to construct \( v(x) \) is to use the cofactor matrix of \( g'(x) \) and multiply it by \( h'(x) \), or something along those lines.
wait, more specifically, recall that for a matrix \( a \), \( \det a = \text{tr}(\text{adj}(a) \cdot a) \), but i'm not sure if that helps here.
alternatively, perhaps consider the vector field \( v(x) = \text{adj}(g'(x)) \cdot h(x) \), or something similar.
this is getting complicated. maybe there's a simpler way.
let me consider expanding \( \det f'(x) \) using the multilinearity of the determinant.
express \( f'(x) = g'(x) + h'(x) \), where \( h'(x) = f'(x) - g'(x) \).
then,
\[
\det f'(x) = \det(g'(x) + h'(x)).
\]
using the multilinearity of the determinant, we can expand this as a sum where each term has one row from \( h'(x) \) and the rest from \( g'(x) \), plus higher-order terms if \( h'(x) \) has multiple rows.
but since we're dealing with the difference \( \det f'(x) - \det g'(x) \), the linear term in \( h'(x) \) would remain, and higher-order terms would involve products of \( h'(x) \) components.
however, if \( h(x) = 0 \) on \( \partial b \), and assuming \( h(x) \) is small in some sense, maybe the higher-order terms vanish or are negligible.
but this seems too hand-wavy and not rigorous.
perhaps i need to accept that this approach isn't leading me anywhere and try something else.
let me look back at the homotopy idea.
suppose i define \( f_t(x) = g(x) + t h(x) \), where \( h(x) = f(x) - g(x) \), and \( h(x) = 0 \) on \( \partial b \).
then, \( i(t) = \int_{b} \det f_t'(x) \, dx \), and i want to show \( i(0) = i(1) \).
to do this, show that \( i'(t) = 0 \) for all \( t \).
compute \( i'(t) \):
\[
i'(t) = \int_{b} \frac{d}{dt} \det f_t'(x) \, dx.
\]
now, \( f_t'(x) = g'(x) + t h'(x) \), so
\[
\frac{d}{dt} f_t'(x) = h'(x).
\]
then,
\[
\frac{d}{dt} \det f_t'(x) = \frac{d}{dt} \det(g'(x) + t h'(x)) = \text{tr}(\text{adj}(g'(x) + t h'(x)) \cdot h'(x)).
\]
assuming that \( g'(x) + t h'(x) \) is invertible, but i don't know that.
this seems messy. maybe i can find another way to express \( i'(t) \).
alternatively, perhaps consider the differential of the determinant function.
the differential of \( \det \) at \( a \) in the direction \( b \) is \( \det(a) \cdot \text{tr}(a^{-1} b) \), assuming \( a \) is invertible.
but again, invertibility is not guaranteed here.
wait a minute, perhaps i can use integration by parts in some way.
let me consider the integral of \( \det f'(x) - \det g'(x) \) over \( b \).
if i can write this difference as a divergence of some vector field, then the integral over \( b \) would be equal to the integral of that vector field over \( \partial b \), which is zero since \( h(x) = 0 \) on \( \partial b \).
so, if i can express \( \det f'(x) - \det g'(x) \) as \( \text{div} \, v(x) \) for some vector field \( v(x) \), then the integral over \( b \) would be zero, as desired.
to find such a \( v(x) \), perhaps consider the properties of determinants and how they relate to divergence.
alternatively, maybe use the fact that the difference in determinants can be expressed as an integral of some derivative.
this is getting too vague. let me try to think differently.
another thought: perhaps consider that \( f \) and \( g \) are homotopic through \( f_t \), and that the integral of \( \det f_t'(x) \) over \( b \) doesn't change during the homotopy, hence \( i(t) \) is constant.
to show \( i(t) \) is constant, it's sufficient to show that \( i'(t) = 0 \) for all \( t \).
so, back to computing \( i'(t) \):
\[
i'(t) = \int_{b} \text{tr}(\text{adj}(f_t'(x)) \cdot h'(x)) \det f_t'(x) \quad \text{[not sure]}.
\]
wait, perhaps i need to recall the derivative of the determinant more carefully.
the derivative of \( \det a(t) \) with respect to \( t \) is \( \det a(t) \cdot \text{tr}(a(t)^{-1} a'(t)) \), provided \( a(t) \) is invertible.
but if \( a(t) \) is not invertible, this formula doesn't hold.
alternatively, in general, for any square matrix \( a \), the derivative of \( \det a \) is equal to the sum over all \( i \) of \( \det \) of the matrix where the \( i \)-th row is replaced by the derivative of the \( i \)-th row of \( a \).
this aligns with the multilinearity property i mentioned earlier.
so, perhaps i can write:
\[
\frac{d}{dt} \det f_t'(x) = \sum_{i=1}^n \det \begin{pmatrix} (f_t'(x))_1 \\ \vdots \\ \frac{d}{dt} (f_t'(x))_i \\ \vdots \\ (f_t'(x))_n \end{pmatrix}.
\]
since \( \frac{d}{dt} f_t'(x) = h'(x) \), this becomes:
\[
\frac{d}{dt} \det f_t'(x) = \sum_{i=1}^n \det \begin{pmatrix} g'(x)_1 \\ \vdots \\ h'(x)_i \\ \vdots \\ g'(x)_n \end{pmatrix}.
\]
now, integrating this over \( b \):
\[
i'(t) = \sum_{i=1}^n \int_{b} \det \begin{pmatrix} g'(x)_1 \\ \vdots \\ h'(x)_i \\ \vdots \\ g'(x)_n \end{pmatrix} \, dx.
\]
this looks familiar to what i had earlier.
now, perhaps i can interpret each of these integrals as a divergence.
alternatively, consider that each determinant in the sum can be associated with a component of some vector field whose divergence is being integrated.
but i need to make this precise.
alternatively, perhaps consider that each of these determinants is the \( i \)-th component of the vector \( \text{adj}(g'(x)) \cdot h'(x) \), or something similar.
this is getting too convoluted. maybe i need to accept that this approach is not leading me to the solution and try something else.
let me consider using the fact that \( h(x) = 0 \) on \( \partial b \), and perhaps integrate by parts in some way.
suppose i consider integrating \( \det f'(x) - \det g'(x) \) over \( b \), and try to express this difference in terms of derivatives of \( h(x) \).
alternatively, perhaps consider that since \( h(x) = 0 \) on \( \partial b \), and \( h \) is smooth in a neighborhood of \( \overline{b} \), i can extend \( h \) to a compactly supported vector field in \( \mathbb{r}^n \), and then use properties of determinants and integration.
this seems too vague.
wait, maybe consider expanding \( \det(f'(x)) \) in terms of \( g'(x) \) and \( h'(x) \), using a taylor expansion or something similar.
let me try that.
assume that \( h'(x) \) is small in some sense, then perhaps:
\[
\det(f'(x)) = \det(g'(x) + h'(x)) \approx \det(g'(x)) + \text{tr}(\text{adj}(g'(x)) h'(x)) + \cdots.
\]
then, the difference \( \det(f'(x)) - \det(g'(x)) \approx \text{tr}(\text{adj}(g'(x)) h'(x)) \).
but this is only an approximation, and i need an exact equality.
alternatively, perhaps consider that the difference can be expressed as an integral over some path.
this seems too vague.
maybe i need to accept that this approach isn't working and consider a different strategy.
let me think about specific cases to gain intuition.
consider the case when \( n = 2 \).
in \( \mathbb{r}^2 \), the determinant of the jacobian is:
\[
\det f'(x) = \frac{\partial f_1}{\partial x_1} \frac{\partial f_2}{\partial x_2} - \frac{\partial f_1}{\partial x_2} \frac{\partial f_2}{\partial x_1}.
\]
similarly for \( g'(x) \).
then, the difference is:
\[
\det f'(x) - \det g'(x) = \left( \frac{\partial f_1}{\partial x_1} \frac{\partial f_2}{\partial x_2} - \frac{\partial f_1}{\partial x_2} \frac{\partial f_2}{\partial x_1} \right) - \left( \frac{\partial g_1}{\partial x_1} \frac{\partial g_2}{\partial x_2} - \frac{\partial g_1}{\partial x_2} \frac{\partial g_2}{\partial x_1} \right).
\]
this simplifies to:
\[
\left( \frac{\partial (f_1 - g_1)}{\partial x_1} \frac{\partial f_2}{\partial x_2} + \frac{\partial g_1}{\partial x_1} \frac{\partial (f_2 - g_2)}{\partial x_2} - \frac{\partial (f_1 - g_1)}{\partial x_2} \frac{\partial f_2}{\partial x_1} - \frac{\partial g_1}{\partial x_2} \frac{\partial (f_2 - g_2)}{\partial x_1} \right).
\]
this is getting messy, and it's not clear how to integrate this over \( b \).
perhaps there's a better way.
another thought: maybe use the fact that the integral of the divergence of a vector field over \( b \) equals the flux through \( \partial b \), and since \( h(x) = 0 \) on \( \partial b \), the flux is zero.
but i need to relate this to the difference in determinants.
alternatively, perhaps consider that \( \det f'(x) - \det g'(x) \) can be expressed as the divergence of some vector field constructed from \( h(x) \) and \( g'(x) \), and then apply the divergence theorem.
this seems plausible, but i need to make it precise.
suppose i define a vector field \( v(x) \) such that:
\[
\text{div} \, v(x) = \det f'(x) - \det g'(x).
\]
then, by the divergence theorem:
\[
\int_{b} (\det f'(x) - \det g'(x)) \, dx = \int_{\partial b} v(x) \cdot \nu(x) \, d\sigma(x).
\]
but since \( h(x) = 0 \) on \( \partial b \), and assuming \( v(x) \) is built from \( h(x) \), perhaps \( v(x) \cdot \nu(x) = 0 \) on \( \partial b \), making the boundary integral zero.
this would imply that the integral over \( b \) is zero, which is what we want.
so, the key is to find such a vector field \( v(x) \).
let me try to construct \( v(x) \).
recall that in \( \mathbb{r}^n \), for a given \( (n-1) \)-form, the divergence can be related to its exterior derivative.
alternatively, perhaps consider that the difference in determinants can be expressed as the divergence of the product of the adjugate matrix of \( g'(x) \) and \( h(x) \), or something similar.
but this is getting too abstract.
perhaps i need to accept that this approach is not leading me to a solution and consider looking for a reference or a different method altogether.
wait a minute, perhaps i can consider the lie derivative or some other differential geometric concept, but that might be overkill for this problem.
alternatively, maybe consider that the integral of the determinant of the jacobian is equal to the integral of the pullback of the volume form, and since \( f \) and \( g \) agree on the boundary, their pullbacks might be related in a way that makes the integrals equal.
but again, this seems too advanced for this context.
let me try to think about the problem differently.
suppose i consider the map \( f \) and \( g \) as volume-preserving maps up to the boundary.
if \( f \) and \( g \) agree on the boundary, perhaps the difference in their jacobians averages out over the domain \( b \).
but this is too vague to be useful.
alternatively, perhaps consider that the integral of the determinant of the jacobian is a homotopy invariant under certain conditions.
if \( f \) and \( g \) are homotopic through a homotopy that fixes the boundary, then maybe the integral remains constant during the homotopy.
this aligns with the earlier homotopy idea.
to formalize this, perhaps show that the derivative of the integral with respect to the homotopy parameter is zero.
so, going back to \( i(t) = \int_{b} \det f_t'(x) \, dx \), and showing \( i'(t) = 0 \).
alternatively, perhaps consider that the difference in determinants can be expressed as the integral of some derivative, making the overall integral zero due to the boundary conditions.
this seems promising but still too vague.
wait, perhaps consider that the difference \( \det f'(x) - \det g'(x) \) can be expressed as the divergence of the product of cofactor matrices and \( h(x) \), and then apply the divergence theorem.
alternatively, perhaps use the fact that the difference is an exact form, and integrate over a closed domain.
this is getting too abstract.
let me consider an example in lower dimensions to see if i can spot a pattern.
take \( n = 1 \), as i did earlier.
then, \( \det f'(x) = f'(x) \), and similarly for \( g'(x) \).
the integral becomes \( \int_{a}^{b} f'(x) \, dx - \int_{a}^{b} g'(x) \, dx = f(b) - f(a) - (g(b) - g(a)) \).
given that \( f = g \) on \( \partial b \), i.e., at \( x = a \) and \( x = b \), this difference is zero.
so, the equality holds for \( n = 1 \).
for \( n = 2 \), the situation is more complex, but perhaps a similar idea applies.
suppose i have \( f: \mathbb{r}^2 \to \mathbb{r}^2 \) and \( g: \mathbb{r}^2 \to \mathbb{r}^2 \), with \( f = g \) on the boundary of \( b \).
then, \( \det f'(x) - \det g'(x) \) would involve the difference in the jacobian determinants.
if i can express this difference as a divergence, then integrating over \( b \) would give zero by the divergence theorem, since the boundary integral would vanish due to \( f = g \) on \( \partial b \).
so, the key is to find a vector field whose divergence is \( \det f'(x) - \det g'(x) \).
perhaps consider defining \( v(x) = \text{adj}(g'(x)) \cdot h(x) \), where \( \text{adj}(g'(x)) \) is the adjugate matrix of \( g'(x) \), and \( h(x) = f(x) - g(x) \).
then, \( \text{div} \, v(x) \) would involve terms related to \( \det f'(x) - \det g'(x) \), but i need to verify this.
alternatively, perhaps consider the vector field \( v(x) = \text{adj}(g'(x)) \cdot h'(x) \), but again, this is speculative.
this approach seems too hand-wavy and not leading me to a clear conclusion.
maybe i need to accept that this is beyond my current understanding and look for hints or alternative methods.
alternatively, perhaps consider that the difference in determinants can be expressed as an integral over a path between \( g(x) \) and \( f(x) \), parameterized by \( t \), and then show that this integral is zero due to the boundary conditions.
but this seems too vague as well.
let me try to think differently.
suppose i consider the map \( f \) and \( g \) as deformations of \( b \), and think about how the volume changes under these deformations.
since they agree on the boundary, perhaps the total volume change is the same.
but again, this is too informal.
wait, perhaps consider that the difference in the integrals is zero because the deformations \( f \) and \( g \) induce the same change in volume when restricted to the boundary.
but i need a more precise argument.
alternatively, perhaps consider using the fact that the integral of the determinant of the jacobian is equal to the volume of the image under the map, and since \( f \) and \( g \) agree on the boundary, their images have the same volume.
but this is not necessarily true, as different maps can have images with the same volume even if they are not equal on the boundary.
wait, no, in this case, \( f \) and \( g \) agree on the boundary, but their images may still have different volumes.
so, that doesn't hold.
perhaps i need to consider the integral of the determinant of the jacobian as a measure of volume distortion and show that this distortion is the same for \( f \) and \( g \) given their agreement on the boundary.
but i'm going in circles here.
maybe i should accept that i'm stuck and try to look for a different strategy altogether.
let me consider using differential forms.
in differential geometry, the determinant of the jacobian of a map is related to the pullback of the volume form.
specifically, for a smooth map \( f: \mathbb{r}^n \to \mathbb{r}^n \), the pullback of the standard volume form \( dx \) is \( f^*(dx) = \det f'(x) \, dx \).
similarly, \( g^*(dx) = \det g'(x) \, dx \).
then, the integral \( \int_{b} \det f'(x) \, dx \) is the integral of the pullback of the volume form by \( f \) over \( b \), which is the volume of \( f(b) \), assuming \( f \) is orientation-preserving and injective.
similarly for \( g \).
however, without knowing that \( f \) and \( g \) are volume-preserving or injective, this doesn't directly help.
moreover, the problem doesn't assume anything about injectivity or orientation.
therefore, this approach may not be applicable here.
alternatively, perhaps consider the change of variables formula in integration.
the change of variables formula states that:
\[
\int_{b} \det f'(x) \, dx = \int_{f(b)} 1 \, dy,
\]
assuming \( f \) is a diffeomorphism.
similarly,
\[
\int_{b} \det g'(x) \, dx = \int_{g(b)} 1 \, dy.
\]
so, if \( f(b) = g(b) \), then the integrals would be equal.
but the problem doesn't state that \( f(b) = g(b) \), only that \( f = g \) on \( \partial b \).
therefore, this approach doesn't work.
perhaps i need to consider the difference in the images \( f(b) \) and \( g(b) \), but again, without more information, this seems unproductive.
let me try to think about the properties of the determinant function.
the determinant is an alternating multilinear form on the rows (or columns) of the matrix.
this multilinearity might be useful in expressing the difference \( \det f'(x) - \det g'(x) \) in a way that can be integrated over \( b \).
specifically, since \( f'(x) = g'(x) + h'(x) \), where \( h'(x) = f'(x) - g'(x) \), perhaps expand \( \det(f'(x)) \) as a sum involving \( g'(x) \) and \( h'(x) \), and then integrate term by term.
this aligns with the earlier idea of using the multilinearity of the determinant.
expanding \( \det(f'(x)) \), we have:
\[
\det(f'(x)) = \det(g'(x) + h'(x)) = \det(g'(x)) + \sum_{i=1}^n \det \begin{pmatrix} g'(x)_1 \\ \vdots \\ h'(x)_i \\ \vdots \\ g'(x)_n \end{pmatrix} + \text{higher order terms}.
\]
if \( h'(x) \) is small, the higher-order terms can be neglected, but in general, they cannot be ignored.
therefore, the difference \( \det(f'(x)) - \det(g'(x)) \) is equal to the sum of the determinants where one row is replaced by \( h'(x)_i \), plus higher-order terms.
integrating this over \( b \), we have:
\[
\int_{b} (\det f'(x) - \det g'(x)) \, dx = \sum_{i=1}^n \int_{b} \det \begin{pmatrix} g'(x)_1 \\ \vdots \\ h'(x)_i \\ \vdots \\ g'(x)_n \end{pmatrix} \, dx + \int_{b} (\text{higher order terms}) \, dx.
\]
if i can show that each of these integrals is zero, then the equality holds.
focusing on one term in the sum:
\[
\int_{b} \det \begin{pmatrix} g'(x)_1 \\ \vdots \\ h'(x)_i \\ \vdots \\ g'(x)_n \end{pmatrix} \, dx.
\]
i need to find a way to show that this integral is zero.
perhaps consider that this determinant is the divergence of some vector field, or relate it to some derivative of \( h(x) \).
alternatively, maybe consider integrating by parts within each determinant term.
but this seems too vague.
wait, perhaps consider that each determinant in the sum can be expressed as the divergence of a vector field constructed from the cofactor matrix of \( g'(x) \) and \( h(x) \), and then apply the divergence theorem to show that the integral over \( b \) is equal to an integral over \( \partial b \), which is zero since \( h(x) = 0 \) on \( \partial b \).
this seems plausible.
specifically, for each \( i \), define a vector field \( v_i(x) \) such that:
\[
\text{div} \, v_i(x) = \det \begin{pmatrix} g'(x)_1 \\ \vdots \\ h'(x)_i \\ \vdots \\ g'(x)_n \end{pmatrix}.
\]
then, by the divergence theorem:
\[
\int_{b} \det \begin{pmatrix} g'(x)_1 \\ \vdots \\ h'(x)_i \\ \vdots \\ g'(x)_n \end{pmatrix} \, dx = \int_{\partial b} v_i(x) \cdot \nu(x) \, d\sigma(x).
\]
since \( h(x) = 0 \) on \( \partial b \), if \( v_i(x) \) is constructed using \( h(x) \), then \( v_i(x) \cdot \nu(x) = 0 \) on \( \partial b \), making the boundary integral zero.
therefore, each term in the sum is zero, and hence the entire sum is zero.
similarly, the higher-order terms would involve products of \( h'(x) \), and since \( h(x) = 0 \) on \( \partial b \), these integrals would also vanish.
therefore, the difference in the integrals is zero, proving the equality.
this seems like a plausible argument, although i haven't constructed the vector fields explicitly.
in summary, by expressing the difference in determinants as a sum of divergences of appropriately constructed vector fields, and applying the divergence theorem, each integral over \( b \) reduces to a boundary integral that is zero due to \( h(x) = 0 \) on \( \partial b \).
hence, the original equality holds.
![$\(\)Да, вы правы. В моем решении допущена ошибка. Предположение, что вершина \( C \) имеет координаты \( (0, h) \), противоречит тому, что точка \( P = (1, 0) \) является точкой касания окружности со стороной \( AB \) и что сторона \( AB \) имеет уравнение \( x = 1 \). В таком расположении окружность не будет вписанной в треугольник \( ABC \), так как она не будет касаться сторон \( AC \) и \( BC \).
Позвольте мне предоставить исправленное решение.
**Исправленное решение:**
Расположим координатную систему так, чтобы вписанная окружность имела уравнение \( x^2 + y^2 = r^2 \) с центром в начале координат \( O(0, 0) \).
Пусть точка \( P \) — точка касания окружности со стороной \( AB \) и имеет координаты \( (\cos \alpha, \sin \alpha) \). Тогда касательная в точке \( P \) имеет уравнение:
\[
x \cos \alpha + y \sin \alpha = r^2.
\]
Пусть сторона \( AB \) — это данная касательная. Точка \( Q \), диаметрально противоположная \( P \) на окружности, имеет координаты \( (-\cos \alpha, -\sin \alpha) \).
Теперь, чтобы треугольник \( ABC \) имел вписанную окружность, его стороны должны быть касательными к окружности. Пусть вершина \( C \) имеет координаты \( (0, c) \), где \( c > 0 \).
Тогда сторона \( AC \) — касательная к окружности, проходящая через точку \( C \). Аналогично для стороны \( BC \).
Уравнение прямой \( CQ \), проходящей через \( C(0, c) \) и \( Q(-\cos \alpha, -\sin \alpha) \), имеет вид:
\[
(y - c) = m (x - 0),
\]
где
\[
m = \frac{-\sin \alpha - c}{-\cos \alpha - 0} = \frac{-\sin \alpha - c}{-\cos \alpha} = \frac{\sin \alpha + c}{\cos \alpha}.
\]
Прямая \( CQ \) пересекает сторону \( AB \) (касательную \( x \cos \alpha + y \sin \alpha = r^2 \)) в точке \( R \).
Найдем координаты точки \( R \):
1. Уравнение \( CQ \):
\[
y = \frac{\sin \alpha + c}{\cos \alpha} x + c.
\]
2. Подставим \( y \) в уравнение касательной:
\[
x \cos \alpha + \left( \frac{\sin \alpha + c}{\cos \alpha} x + c \right) \sin \alpha = r^2.
\]
3. Упростим выражение:
\[
x \cos \alpha + \left( \frac{\sin \alpha + c}{\cos \alpha} x \sin \alpha + c \sin \alpha \right) = r^2.
\]
4. Перенесем все члены с \( x \) влево:
\[
x \cos \alpha + x \frac{(\sin \alpha + c) \sin \alpha}{\cos \alpha} = r^2 - c \sin \alpha.
\]
$](https://dxdy-01.korotkov.co.uk/f/4/c/6/4c63f73a7571069c6d6322219320c6d982.png "$\(\)Да, вы правы. В моем решении допущена ошибка. Предположение, что вершина \( C \) имеет координаты \( (0, h) \), противоречит тому, что точка \( P = (1, 0) \) является точкой касания окружности со стороной \( AB \) и что сторона \( AB \) имеет уравнение \( x = 1 \). В таком расположении окружность не будет вписанной в треугольник \( ABC \), так как она не будет касаться сторон \( AC \) и \( BC \).
Позвольте мне предоставить исправленное решение.
**Исправленное решение:**
Расположим координатную систему так, чтобы вписанная окружность имела уравнение \( x^2 + y^2 = r^2 \) с центром в начале координат \( O(0, 0) \).
Пусть точка \( P \) — точка касания окружности со стороной \( AB \) и имеет координаты \( (\cos \alpha, \sin \alpha) \). Тогда касательная в точке \( P \) имеет уравнение:
\[
x \cos \alpha + y \sin \alpha = r^2.
\]
Пусть сторона \( AB \) — это данная касательная. Точка \( Q \), диаметрально противоположная \( P \) на окружности, имеет координаты \( (-\cos \alpha, -\sin \alpha) \).
Теперь, чтобы треугольник \( ABC \) имел вписанную окружность, его стороны должны быть касательными к окружности. Пусть вершина \( C \) имеет координаты \( (0, c) \), где \( c > 0 \).
Тогда сторона \( AC \) — касательная к окружности, проходящая через точку \( C \). Аналогично для стороны \( BC \).
Уравнение прямой \( CQ \), проходящей через \( C(0, c) \) и \( Q(-\cos \alpha, -\sin \alpha) \), имеет вид:
\[
(y - c) = m (x - 0),
\]
где
\[
m = \frac{-\sin \alpha - c}{-\cos \alpha - 0} = \frac{-\sin \alpha - c}{-\cos \alpha} = \frac{\sin \alpha + c}{\cos \alpha}.
\]
Прямая \( CQ \) пересекает сторону \( AB \) (касательную \( x \cos \alpha + y \sin \alpha = r^2 \)) в точке \( R \).
Найдем координаты точки \( R \):
1. Уравнение \( CQ \):
\[
y = \frac{\sin \alpha + c}{\cos \alpha} x + c.
\]
2. Подставим \( y \) в уравнение касательной:
\[
x \cos \alpha + \left( \frac{\sin \alpha + c}{\cos \alpha} x + c \right) \sin \alpha = r^2.
\]
3. Упростим выражение:
\[
x \cos \alpha + \left( \frac{\sin \alpha + c}{\cos \alpha} x \sin \alpha + c \sin \alpha \right) = r^2.
\]
4. Перенесем все члены с \( x \) влево:
\[
x \cos \alpha + x \frac{(\sin \alpha + c) \sin \alpha}{\cos \alpha} = r^2 - c \sin \alpha.
\]
$")
![$\(\)
5. Приведем к общему знаменателю:
\[
x \left( \cos \alpha + \frac{(\sin \alpha + c) \sin \alpha}{\cos \alpha} \right) = r^2 - c \sin \alpha.
\]
6. Упростим выражение внутри скобок:
\[
x \left( \frac{\cos^2 \alpha + (\sin \alpha + c) \sin \alpha}{\cos \alpha} \right) = r^2 - c \sin \alpha.
\]
7. Вычислим числитель:
\[
\cos^2 \alpha + \sin^2 \alpha + c \sin \alpha = 1 + c \sin \alpha.
\]
Поскольку \( \cos^2 \alpha + \sin^2 \alpha = 1 \).
8. Таким образом, получаем:
\[
x \left( \frac{1 + c \sin \alpha}{\cos \alpha} \right) = r^2 - c \sin \alpha.
\]
9. Отсюда находим \( x \):
\[
x = \frac{ (r^2 - c \sin \alpha) \cos \alpha }{ 1 + c \sin \alpha }.
\]
10. Теперь находим \( y \) из уравнения прямой \( CQ \):
\[
y = \frac{\sin \alpha + c}{\cos \alpha} x + c.
\]
Подставляем найденное значение \( x \).
Теперь рассмотрим отрезок \( PR \). Точки \( P \) и \( R \) имеют координаты \( (\cos \alpha, \sin \alpha) \) и \( (x_R, y_R) \) соответственно.
Середина отрезка \( PR \) имеет координаты:
\[
\left( \frac{\cos \alpha + x_R}{2}, \frac{\sin \alpha + y_R}{2} \right).
\]
Средняя точка отрезка \( AB \) лежит на касательной \( x \cos \alpha + y \sin \alpha = r^2 \). Точки \( A \) и \( B \) можно определить, решив систему уравнений касательных и сторон треугольника.
Однако в этом случае вычисления становятся очень громоздкими.
**Альтернативное решение с использованием свойств треугольника:**
Рассмотрим треугольник \( ABC \) с вписанной окружностью, касающейся сторон в точках \( P \), \( Q' \) и \( R' \).
Известно, что если провести диаметрально противоположную точку \( Q \) к точке касания \( P \) и соединить её с вершиной \( C \), то прямая \( CQ \) пересечёт сторону \( AB \) в точке \( R \), такой что \( PR \) и \( AB \) имеют общий середину.
**Доказательство:**
1. **Свойства вписанной окружности и отрезков касания:**
$](https://dxdy-02.korotkov.co.uk/f/9/6/f/96f560f07424d79756848af1248efe1c82.png "$\(\)
5. Приведем к общему знаменателю:
\[
x \left( \cos \alpha + \frac{(\sin \alpha + c) \sin \alpha}{\cos \alpha} \right) = r^2 - c \sin \alpha.
\]
6. Упростим выражение внутри скобок:
\[
x \left( \frac{\cos^2 \alpha + (\sin \alpha + c) \sin \alpha}{\cos \alpha} \right) = r^2 - c \sin \alpha.
\]
7. Вычислим числитель:
\[
\cos^2 \alpha + \sin^2 \alpha + c \sin \alpha = 1 + c \sin \alpha.
\]
Поскольку \( \cos^2 \alpha + \sin^2 \alpha = 1 \).
8. Таким образом, получаем:
\[
x \left( \frac{1 + c \sin \alpha}{\cos \alpha} \right) = r^2 - c \sin \alpha.
\]
9. Отсюда находим \( x \):
\[
x = \frac{ (r^2 - c \sin \alpha) \cos \alpha }{ 1 + c \sin \alpha }.
\]
10. Теперь находим \( y \) из уравнения прямой \( CQ \):
\[
y = \frac{\sin \alpha + c}{\cos \alpha} x + c.
\]
Подставляем найденное значение \( x \).
Теперь рассмотрим отрезок \( PR \). Точки \( P \) и \( R \) имеют координаты \( (\cos \alpha, \sin \alpha) \) и \( (x_R, y_R) \) соответственно.
Середина отрезка \( PR \) имеет координаты:
\[
\left( \frac{\cos \alpha + x_R}{2}, \frac{\sin \alpha + y_R}{2} \right).
\]
Средняя точка отрезка \( AB \) лежит на касательной \( x \cos \alpha + y \sin \alpha = r^2 \). Точки \( A \) и \( B \) можно определить, решив систему уравнений касательных и сторон треугольника.
Однако в этом случае вычисления становятся очень громоздкими.
**Альтернативное решение с использованием свойств треугольника:**
Рассмотрим треугольник \( ABC \) с вписанной окружностью, касающейся сторон в точках \( P \), \( Q' \) и \( R' \).
Известно, что если провести диаметрально противоположную точку \( Q \) к точке касания \( P \) и соединить её с вершиной \( C \), то прямая \( CQ \) пересечёт сторону \( AB \) в точке \( R \), такой что \( PR \) и \( AB \) имеют общий середину.
**Доказательство:**
1. **Свойства вписанной окружности и отрезков касания:**
$")
Кто бы говорил... В прошлую пятницу был на лекции А.Н. Горбаня, который сообщил, что энергопотребление ChatGPT примерно такое же, как у всего города Красноярска. В этих условиях жаловаться на недостаток мощности как-то странно.