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I found a form of the lagrange theorem that I don't know,

I didn't find something similar on the internet.

Suppose that $f:[a,b]\times[c,d] \subset \mathbb{R}^2 \longrightarrow \mathbb{R}$ is a continuous function. Consider the function

\begin{align} I:[c,d] &\longrightarrow \, \mathbb{R}\\ y\,\,\, &\longmapsto\, I(y) = \int_{a}^{b} f(x,y)\, dx\\ \end{align}

Then they say:

Take a y $\in $ [c,d]. From the Lagrange theorem it follows that $\exists \,\theta \in (0,1) $ so that: \begin{align} \frac{I(y+k)-I(y)}{k} &= \int_a^b \frac{f(x,y+k)-f(x,y)}{k} \, dx \\ &= \int_a^b \frac{\partial f}{\partial y}(x,y+\theta k) \, dx \end{align}

The only form of the lagrange theorem that I know is this one: $$ f'(c) = \frac{f(b)-f(a)}{b-a}. $$

Does someone know how they get an angle theta?

Chappers
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Ayoub Rossi
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  • I don't think that's an angle but rather "a small" number. Observe that the second integral in the first line in that part, when $;k\to 0;$ , is the definition of the the partial derivative of $;f;$ wrt $;y;$ ... – DonAntonio Apr 22 '17 at 14:18
  • @DonAntonio I see, we always use theta as an angle at that's why I was confused, so instead of writing the k $\rightarrow$ 0 they use that theta? – Ayoub Rossi Apr 22 '17 at 14:19

2 Answers2

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Note that the following is equivalent to MVT:

Let $a\in \Bbb R$ and $k>0$. If $f: [a,a+k] \to \Bbb R$ is continuous on $[a,a+k]$ and differentiable on $(a,a+k)$, then there exists $\theta \in (0,1)$ such that

$$f'(a + \theta k) = \frac{f(a+k)-f(a)}{k}$$

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It's actually the same thing in different notation: if we write $f(x,y) = F_x(y)$, then the quotient inside the integral is $$ \frac{F_x(y+k)-F_x(y)}{k}. $$ By Lagrange (AKA the mean value theorem), there is a $c$ with $y<c<y+k$ so $$ \frac{F_x(y+k)-F_x(y)}{k} = F_x'(c). $$ Any such $c$ can be written as $(1-\theta)y+\theta (y+k) = y + \theta k$ for some $0<\theta<1$, and this is what $\theta $ is: a real number between $0$ and $1$, which depends on $y,k$ and $F_x$.

Chappers
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