Let $\mathcal{F}_n$ denote the $n$ -th Fibonacci number and $\mathcal{L}_n$ the $n$ – th Lucas. Prove that

$$\prod_{n=1}^{\infty} \left ( 1 + \frac{1}{\mathcal{F}_{2^n +1} \mathcal{L}_{2^n+1}} \right ) = \frac{3}{\varphi^2}$$

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- Sun Mar 12, 2023 3:03 pm
- Forum: Calculus
- Topic: An infinite product
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- Sun Mar 12, 2023 2:52 pm
- Forum: Meta
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### Re: Forum upgrade to latest version

Greetings everyone, we are pleased to announce that the forum software has been upgraded to the latest version hardening the security of our website. You will notice that many cosmetic things have been restored to normal. This new version is compatible with php 8.2 that our server is currently runni...

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- Forum: Meta
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### Forum upgrade to latest version

Greetings, we have updated the forum to its latest version phpbb 3.3.x. You will find that many bugs have been fixed in this latest version. We would like to also inform you that the ability to add tags has been restored and now it's working flawlessly. You can select among many different tags to ca...

- Sun Apr 10, 2022 6:24 am
- Forum: Complex Analysis
- Topic: Contour integral
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### Re: Contour integral

It follows from Taylor's theorem that $f(z)=\sum \limits_{n=0}^{\infty} c_n z^n$ and that the convergence is uniform. Thus, \begin{align*} \frac{1}{2\pi i }\oint \limits_{|z|=1} \frac{\overline{f(z)}}{z-\alpha} \,\mathrm{d}z &=\frac{1}{2\pi i }\oint \limits_{|z|=1} \sum_{n=0}^{\infty} \frac{\ove...

- Fri Nov 06, 2020 11:59 am
- Forum: Linear Algebra
- Topic: Rank of product of matrices
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**4366**

### Re: Rank of product of matrices

It holds that $${\rm nul} (T_1 T_2) \leq {\rm nul} (T_1) + {\rm nul} (T_2)$$ where $T_1, \; T_2$ are the corresponding linear transformations. Proof: The proof of the lemma is based on the rank - nullity theorem. Based upon the above lemma we have that \begin{align*} {\rm rank} \left ( T_1 T_2 \rig...

- Fri Nov 06, 2020 11:57 am
- Forum: Linear Algebra
- Topic: On permutation
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**3167**

### Re: On permutation

The sum of $D(\sigma)$ over the even permutations minus the one over the odd permutations is the determinant of the matrix $A$ with entries $a_{i,j}=\vert i-j\vert$ and this determinant is known to be

$$\det A = (-1)^{n-1} (n-1) 2^{n-2}$$

$$\det A = (-1)^{n-1} (n-1) 2^{n-2}$$

- Fri Nov 06, 2020 6:50 am
- Forum: Competitions
- Topic: An equality with matrices
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**3301**

### Re: An equality with matrices

Let $A, B$ be elements of an arbitrary associative algebra with unit. Then: \begin{align*} \left ( A^{-1} +\left ( B^{-1} - A \right )^{-1} \right )^{-1} &= \left ( A^{-1} \left ( B^{-1} - A \right )\left ( B^{-1} - A \right )^{-1} + A^{-1} A \left ( B^{-1} - A \right )^{-1} \right )^{-1} \\ &am...

- Fri Nov 06, 2020 6:36 am
- Forum: Algebraic Structures
- Topic: Sum equals to zero
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**2242**

### Re: Sum equals to zero

Let us suppose that $|\mathcal{G}| = \kappa$ and $x= \frac{1}{\kappa} \sum \limits_{g \in \mathcal{G}} g $. We note that for every $h \in \mathcal{G}$ the depiction $\varphi: \mathcal{G} \rightarrow \mathcal{G}$ such that $\varphi(g)=h g $ is $1-1$ and onto. Thus: \begin{align*} x^2 &=\left ( \f...

- Fri Nov 06, 2020 5:28 am
- Forum: Algebraic Structures
- Topic: Isomorphic groups
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**2577**

### Re: Isomorphic groups

Using $x^{-1}yx = y^{-1}$ or equivalently $yx = xy^{-1}$ we can write each element of $\mathcal{Q}_{2^n}$ in the form $x^ry^s$ where $r,s \in \mathbb{N} \cup \{0\}$. Using $x^2 = y^{2^{n-2}}$ we may assume that $r\in \{0,1\}$. Using $y^{2^{n-1}} = 1$ we may also assume that $s\in \{0,1,\ldots,2^{n...

- Tue Jun 09, 2020 11:33 am
- Forum: Functional Analysis
- Topic: Inner product space
- Replies:
**1** - Views:
**3050**

### Re: Inner product space

**Hint:**Equality holds when vectors are parallel i.e, $u=kv$, $k \in \mathbb{R}^+$ because $u \cdot v= \|u \| \cdot \|v\| \cos \theta$ when $\cos \theta=1$, the equality of the Cauchy-Schwarz inequality holds.