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5

Let's assume that $k \ll N$. By using the Taylor series expansion $\ln(1-x) = -x+O(x^2)$, we have: $$\displaystyle\sum_{i = 1}^{k-1}\ln\left(1-\dfrac{i}{2N}\right) \approx \displaystyle\sum_{i = 1}^{k-1}-\dfrac{i}{2N} = -\dfrac{1}{2N}\cdot\dfrac{(k-1)k}{2} = -\dfrac{\binom{k}{2}}{2N}.$$ Then, by exponentiating both sides, and using the Taylor series ...


4

Stone-Weierstrass theorem. Functions of the given form are an algebra ...


3

$$\sum_{k=1}^{2j}(-1)^{k-1}k^2 = \sum_{h=1}^{j}\left[(2h-1)^2-(2h)^2\right]=-\sum_{h=1}^{j}(4h-1)=-j(2j+1) \tag{1}$$ hence: $$2^i\prod_{j=1}^{i}\sum_{k=1}^{2j}(-1)^{k-1}k^2=(-1)^i\prod_{j=1}^{i}(2j)(2j+1)=(-1)^i (2i+1)!\tag{2} $$ and the other product can be managed in a similar way.


2

I assume you want to find the greatest common divisor. The following pseudocode algorithm works, and you can decide whether it meets your requirements: Initialize $g$ to be the empty product For $i$ from $0$ to $N_a$ do For $j$ from $0$ to $N_b$ do Set $temp = \gcd(a_i,b_j)$ Set $g = g \times temp$ (append a term onto the product representation for ...


2

Just write down everything carefully... There's no funny business going on. The map $\phi_B$ is surjective: this means that for every element $(\beta_i) \in \prod_i [B,A_i]$, i.e. for every collection of morphisms $\beta_i : B \to A_i$, there exists some $\beta : B \to A$ such that $\phi_B(\beta) = (\beta_i)$. In other words, there exists a map $\beta : B ...


1

If $A$ and $B$ are Hermitian matrices with $AB = 0$, then you know that any eigenvector $x$ of $A$ associated with $\lambda \neq 0$ satisfies $Bx = 0$. That's about all you can say about eigenvalues. As for eigenvectors: commuting Hermitian matrices share an orthonormal basis of eigenvectors. If there are no repeated eigenvectors, then any eigenvector of ...


1

As Mattos said, the problem uses $x^{2} + 1$ and you used $x^{2} - 1$. Otherwise you were ok to the point where you said you were confident. At that point you have two fractions to add. You need to get a common denominator and add the numerators, not multiply the numerators.


1

So if $f(x) = x(x^2 +1)^{\frac{1}{2}}$ Then : $f'(x) = [x(x^2 +1)^{\frac{1}{2}}]' = x'(x^2 +1)^{\frac{1}{2}} + x[(x^2 +1)^{\frac{1}{2}}]' = (x^2 +1)^{\frac{1}{2}} + x[\sqrt{x^2+1}]' = (x^2 +1)^{\frac{1}{2}} + x \frac{1}{2\sqrt{x^2+1}}[x^2+1]' = (x^2 +1)^{\frac{1}{2}} + x \frac{1}{2\sqrt{x^2+1}}2x = (x^2 +1)^{\frac{1}{2}} + 2x^2 \frac{1}{2\sqrt{x^2+1}} $ If ...


1

The simple induction would work with no problem if we were given that $\sum a_k\le 1/2$. Which is enough for the presumed application to infinite products, but doesn't solve the current problem. Hint for the problem as given: $$\log(1+t)\le t\quad(t\ge0)$$ and $$\log(1+2t)\ge t\quad(0\le t\le 1).$$ You can prove both these by writing the logarithm as the ...


1

Hint notice the powers of $2$ the power in $a_2$ is $a_2=a_1.2-1$ in $a_3=a_2.2+1$ then in $a_4=a_3.2-1$ and then this goes on like $a_n=a_{n-1}.2\pm 1$ where $n$ decides whether it's plus or minus depending on odd or even number which I have explained above. So now this sum of powers should be divisible by $19$ to get an integer so some of first powers are ...



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