Evaluate $$\lim \limits_{n\to \infty }\sin^2 \left(\pi \sqrt{(n!)^2-(n!)}\right)$$ I tried it by Stirling’s Approximation $$n! \approx \sqrt{2\pi n}.n^n e^{-n}$$ but it leads us to nowhere. Any hint will be of great help.

Evaluate: $$\sum_{n=0}^{\infty} \frac {(-1)^n}{4^{4n+1}(4n+1)} $$ I rewrote the sum as $$\sum_{n=0}^{\infty} \frac {1}{4^{8n-7}(8n-7)} – \sum_{n=0}^{\infty} \frac {1}{4^{8n-3}(8n-3)}$$ Now, I tried to express this as a Geometric Series and Partial Fraction but was unable to do so. I also tried to use Riemann Sum, but I don’t know how to apply it here. Any help will […]

Solve $\cos x+8\sin x-7=0$ My attempt: \begin{align} &8\sin x=7-\cos x\\ &\implies 8\cdot \left(2\sin \frac{x}{2}\cos \frac{x}{2}\right)=7-\cos x\\ &\implies 16\sin \frac{x}{2}\cos \frac{x}{2}=7-1+2\sin ^2\frac{x}{2}\\ &\implies 16\sin \frac{x}{2}\cos \frac{x}{2}=6+2\sin^2 \frac{x}{2}\\ &\implies 8\sin \frac{x}{2}\cos \frac{x}{2}=3+\sin^2 \frac{x}{2}\\ &\implies 0=\sin^2 \frac{x}{2}-8\sin \frac{x}{2}\cos \frac{x}{2}+3\\ &\implies 0=\sin \frac{x}{2}\left(\sin \frac{x}{2}-8\cos \frac{x}{2}\right)+3 \end{align} I’m not sure how to proceed from here (if this process is even […]

Let $k\in\mathbb{N}$ be odd and $N\in\mathbb{N}$. You may assume that $N>k^2/4$ although I don’t think that is relevant. Let $\zeta:=\exp(2\pi i/k)$ and $\alpha_v:=\zeta^v+\zeta^{-v}+\zeta^{-1}$. As it comes from the trace of a positive matrix I know that the following is real: $$\sum_{v=1}^{\frac{k-1}{2}}\sec^2\left(\frac{2\pi v}{k}\right)\frac{(\overline{\alpha_v}^N+\alpha_v^N\zeta^{2N})(\zeta^{2v}-1)^2}{\zeta^N\zeta^{2v}}.$$ I am guessing, and numerical evidence suggests, that in fact $$\frac{(\overline{\alpha_v}^N+\alpha_v^N\zeta^{2N})(\zeta^{2v}-1)^2}{\zeta^N\zeta^{2v}}$$ is real […]

How can I prove that, for $a,b \in \mathbb{Z}$ we have $$ 0 \leq \left \lfloor{\frac{2a}{b}}\right \rfloor – 2 \left \lfloor{\frac{a}{b}}\right \rfloor \leq 1 \, ? $$ Here, $\left \lfloor\,\right \rfloor$ is the floor function. I tried the following: say that $\frac{2a}{b} = x$, and $ \left \lfloor{\frac{2a}{b}}\right \rfloor = m$, with $0 \leq x […]

Lately, I’ve been very confused about the weird properties of limits. For example, I was very surprised to find out that $\lim_{n \to \infty} (3^n+4^n)^{\large \frac 1n}=4$ , because if you treat this as an equation, you can raise both sides to the $n$ power, subtract, and reach the wrong conclusion that $\lim_{n \to \infty} […]

first of all I am sorry if the level of this question is nowhere near the usual level of questions on this site because my math knowledge is still very basic. I hope you won’t mind. I found this problem on some site: “A basket of oranges costs 20 dollars, a basket of pears costs […]

From the Generating function for Legendre Polynomials: $$\Phi(x,h)=(1-2xh+h^2)^{-1/2}\quad\text{for}\quad \mid{h}\,\mid\,\lt 1$$ My text states that: For $x=1$ $$\Phi(1,h)=\color{red}{(1-2h+h^2)^{-1/2}=\frac{1}{1-h}}=1+h+h^2+\cdots$$ My question is about the justification of the equality marked $\color{red}{\mathrm{red}}$. Since although $$\Phi(1,h)=(1-2h+h^2)^{-1/2}=\Big((1-h)(1-h)\Big)^{-1/2}$$$$=\Big((1-h)^2\Big)^{-1/2}=\color{#180}{\frac{1}{1-h}}=1+h+h^2+\cdots\tag{1}$$ as required. I could also write $$\Phi(1,h)=(1-2h+h^2)^{-1/2}=\Big((h-1)(h-1)\Big)^{-1/2}$$$$=\Big((h-1)^2\Big)^{-1/2}=\color{blue}{\frac{1}{h-1}}=\frac{1}{h}\left(1-\frac{1}{h}\right)^{-1}\ne 1+h+h^2+\cdots\tag{2}$$ Why is it that $\Phi(1,h)$ is equal to $(1)$ but not equal to $(2)$? I think […]

Let $f(x) = \sqrt{(x-a)^2 + b^2} +\sqrt{(x-c)^2 + d^2}$, where all coefficients are real. It can be shown using a distance formula that the minimal value of $f(x)$ is $D = \sqrt{(a-c)^2+(|b|+|d|)^2}$. Show that result without derivatives and without a distance formula. At what value of $x$ does the minimum of $f(x)$ occur? Hint: this […]

Given the following $f(x) – f'(x) = x^3 + 3x^2 + 3x +1$ Calculate $f(9) = ?$ I have tried to play with different number of derivatives. Also tried to solve it by equations. Maybe there is some geometric meaning that could shade the light ? I feel it is no complex problem at all. […]

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