Search found 177 matches
Re: Analysis
solve the question Let $f:[0, 2\pi] \rightarrow [0, 2\pi]$ be continuous such that $f(0)=f(2\pi)$. Show that there exists $x \in [0, 2\pi]$ such that $$f(x)= f(x+\pi)$$ Hello my friend, consider the function $g(t)=f(t) - f(t+\pi) \; , \; t \in [0, 2\pi]$. Clearly, $g$ is continuous as a sum of cont...
- Sun Jan 14, 2018 8:30 pm
- Forum: Real Analysis
- Topic: Zero function from an inequality
- Replies: 1
- Views: 3707
Re: Zero function from an inequality
We note that \begin{align*} \left ( e^{-2x} f^2(x)\right ) ' &= 2e^{-2x}\left ( f(x)f'(x) - f^2(x) \right ) \\ &\leq 2e^{-2x}\left ( |f(x)f'(x)| - f^2(x) \right ) \\ &= 2e^{-2x}|f(x)|\left ( |f'(x)| - |f(x)| \right ) \\ &\leq 0 \end{align*} hence the function $e^{-2x} f^2(x)$ is decr...
- Sun Jan 14, 2018 8:21 pm
- Forum: Real Analysis
- Topic: Existence of $c$
- Replies: 1
- Views: 3884
Re: Existence of $c$
Let $I(x) = \bigintsss_{0}^{x} f(t) \, {\rm d}t$ and $G(x) = \bigintsss_{0}^{x} I(t) \, {\rm d}t$. Integrating by parts $(1)$ reveals that $$ \int_{0}^{1} I(t) \, {\rm d}t =0 \implies G(1) =0$$ Now let us consider the function $\displaystyle K(x) = \frac{G(x)}{x^2}$ . It holds that $K(1)=0$. As we c...
- Sat Jan 13, 2018 12:40 pm
- Forum: Calculus
- Topic: \(\int_{-1}^1 \log(1-x)\,\log(1+x)\, dx\)
- Replies: 5
- Views: 6717
Re: \(\int_{-1}^1 \log(1-x)\,\log(1+x)\, dx\)
Evaluate \[\displaystyle\int_{-1}^1 \log(1-x)\,\log(1+x)\, dx\,.\] Yet another solution. Making use of the symmetry, we get that: \begin{align*} \int_{-1}^{1} \log(1-x) \log(1+x) \, {\rm d}x &= 2 \int_{0}^{1} \log(1-x)\log(1+x)\, {\rm d}x \\ &=2\int_{0}^{1} \log(1-x) \sum_{n=1}^{\infty} \fr...
- Wed Jan 03, 2018 8:34 pm
- Forum: Linear Algebra
- Topic: On linear operators
- Replies: 0
- Views: 3861
On linear operators
Let $\alpha \in \mathbb{R} \setminus \{0\} $ and suppose that $F ,G$ are linear operators from $\mathbb{R}^n$ into $\mathbb{R}^n$ satisfying \begin{equation*}F\circ G - G \circ F =\alpha F \end{equation*} Show that for all k \in \mathbb{N} one has \[F^k \circ G - G \circ F ^k= \alpha k F^k\] Show th...
- Thu Dec 07, 2017 8:14 am
- Forum: Linear Algebra
- Topic: Dimension of intersection of subspaces
- Replies: 1
- Views: 3994
Dimension of intersection of subspaces
If
\begin{align*}
W_1 &=\{(x, y, z)| x + y- z =0\}\\
W_2 &= \{(x, y, z)| 3x +y- 2z =0\}\\
W_3 &=\{(x, y, z)| x -7y+3z =0\}
\end{align*}
then find $\dim (W_1 \cap W_2 \cap W_3 )$ and $\dim (W_1 \cap W_2 ).$
\begin{align*}
W_1 &=\{(x, y, z)| x + y- z =0\}\\
W_2 &= \{(x, y, z)| 3x +y- 2z =0\}\\
W_3 &=\{(x, y, z)| x -7y+3z =0\}
\end{align*}
then find $\dim (W_1 \cap W_2 \cap W_3 )$ and $\dim (W_1 \cap W_2 ).$
- Mon Dec 04, 2017 7:42 am
- Forum: Real Analysis
- Topic: Uniform convergence
- Replies: 0
- Views: 2653
Uniform convergence
Prove that the series of functions $\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^3 x^2 + n}$ converges uniformly for all $x \geq 0$.
- Thu Nov 30, 2017 10:00 pm
- Forum: Real Analysis
- Topic: Multiple integral
- Replies: 1
- Views: 3091
Re: Multiple integral
Since $\mathcal{S}$ is a positive symmetric matrix , so is $\mathcal{S}^{-1}$. For a positive symmetric matrix $\mathcal{A}$ there exists an $\mathcal{R}$ positive symmetric matrix such that $\mathcal{A} = \mathcal{R}^2$. Applying this to $\mathcal{S}^{-1}$ our integral becomes \[\mathcal{M} = \int ...
- Wed Nov 29, 2017 9:20 am
- Forum: Algebraic Structures
- Topic: True or false statements
- Replies: 1
- Views: 3297
True or false statements
Let $n \in \mathbb{Z}$ such that $n \geq 2$. Let $\mathcal{S}_n$ be the permutation group on $n$ letters and $\mathcal{A}_n$ be the alternating group. We also denote $\mathbb{C}^*$ the group of non zero complex numbers under multiplication. Which of the following are correct statements? For every in...
- Thu Nov 02, 2017 9:35 pm
- Forum: Real Analysis
- Topic: On the evaluation of the Fresnel integral
- Replies: 0
- Views: 2512
On the evaluation of the Fresnel integral
We are aware of the Fresnel integral \begin{equation} \int_0^\infty \sin x^2 \, {\rm d}x = \frac{1}{2} \sqrt{\frac{\pi}{2}} \end{equation} The most common proof goes with complex analysis. Try to provide a proof with Real Analysis. There are at least $2$ proofs. The one is more elegant than the othe...