Search found 179 matches
- Sat May 12, 2018 4:31 pm
- Forum: Analysis
- Topic: Logrithmic Integral
- Replies: 3
- Views: 5744
Re: Logrithmic Integral
The Fourier series is the way to go here. Recall that $$\cos(2kx) = \dfrac{e^{i2kx} + e^{-i2kx}}{2}$$ Hence, \begin{align*} \sum_{k=1}^{\infty} \dfrac{\cos(2kx)}k &= \sum_{k=1}^{\infty} \dfrac{e^{i2kx} + e^{-i2kx}}{2k} \\&= \dfrac12 \big(-\log (1-e^{i2x} )-\log (1-e^{-i2x} ) \big) \\&= -...
- Sat May 12, 2018 9:22 am
- Forum: Analysis
- Topic: Logarithmic and Trigonometric Integral
- Replies: 2
- Views: 4633
Re: Logarithmic and Trigonometric Integral
A hint is along these lines. Apply the sub $x=\arctan t$ and use the well known fact that
$$\sin \left ( \arctan t \right ) = \frac{t}{\sqrt{t^2+1}}$$
The final answer is $\dfrac{7 \pi^3}{216}$.
$$\sin \left ( \arctan t \right ) = \frac{t}{\sqrt{t^2+1}}$$
The final answer is $\dfrac{7 \pi^3}{216}$.
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: 4167
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: 4356
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: 7189
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: 4405
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: 4529
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: 2966
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: 3330
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 ...