# Tag Info

## Hot answers tagged calculus

12

We have that $$\left(1+\frac{1}{n}\right)^n=\mathrm{e}^{n\log(1+\frac{1}{n})}=\mathrm{e}^{1-\frac{1}{2n}+{\mathcal O}(n^{-2})}=\mathrm{e}\left(1-\frac{1}{2n}+{\mathcal O}\Big(\frac{1}{n^2}\Big)\right),$$ since $$\log \Big(1+\frac{1}{n}\Big)=\frac{1}{n}-\frac{1}{2n^2}+{\mathcal O}\Big(\frac{1}{n^3}\Big),$$ and hence $$... 11 This integral can be evaluated in a closed form for arbitrary real exponents, and does not seem to be related to Herglotz-like integrals. Assume a,b\in\mathbb{R}. Note that$$\int_0^\infty\frac{\ln\left(\frac{1+x^a}{1+x^b}\right)}{\ln x}\frac{dx}{1+x^2}=\int_0^\infty\frac{\ln\left(\frac{1+x^a}2\right)}{\ln ...

11

$$x^3-y^3=(x-y)(x^2+xy+y^2)\implies x-y=\frac{x^3-y^3}{x^2+xy+y2}$$ Now we put $$x=\sqrt[3]{8+h}\;,\;\;y=2\implies \sqrt[3]{8+h}-2=\frac{8+h-8}{(8+y)^{2/3}+2\sqrt[3]{8+h}+4}\implies$$ $$\frac{\sqrt[3]{8+h}-2}h=\frac1{(8+h)^{2/3}+2\sqrt[3]{8+h}+4}\xrightarrow[h\to 0]{}\frac1{8^{2/3}+2\sqrt[3]8+4}=\ldots$$

5

This solution also appears on MathOverflow. We can think of $I_{n}$ as being a classical partition function for $n$ beads on a circle which cannot pass through each other, with logarithmic interaction potential between each bead and its next-to-nearest neighbors on either side. For $I_{2n}$ the beads fall into two colors" which do not have logarithmic ...

4

One neat way to solve some limits is to use Taylor series. You see that $$\sqrt[3]{8 + h} -2 = 2\sqrt[3]{1 + h/8} -2 \approx 2\left(1 + \frac{h}{24}\right) -2 = \frac{h}{12}$$ so the limit becomes $$\lim_{h \to 0} \frac{h/12}{h} = \frac{1}{12}.$$

3

Write $L=\lim_{n\to\infty}a_n$. This means that for every $\epsilon>0$ there exists a whole number $N$ such that for all integers $n\ge N$, $$|L-a_n|<\varepsilon$$ Now, to prove $\lim_{n\to\infty}a_{n+1}=L$ as well, consider any $n\ge N$ and note that $n+1\ge N$. So by our initial statement, $|L-a_{n+1}|<\varepsilon$.

3

In a double integral, you are actually integrating a differential two-form: $$\int_R \mathrm{f}(x,y) \ \mathrm{d}x \wedge \mathrm{d}y$$ Here, $\mathrm{d}x$ and $\mathrm{d}y$ are the basis differential one-forms and $\mathrm{d}x \wedge \mathrm{d}y$ is their exterior product.

2

First of all in order for $\log_a x$ to make sense we need that $a>0$ and $a\ne 1$. Case 1. $a>1$, then $\log a >0$, and hence $\frac{1}{\log a}\ge\frac{1}{\log x}$, and thus $$\log_a x+\log_x x=\frac{\log x}{\log a}+1\ge \frac{\log x}{\log x}+1=2,$$ since $0<a\le x$ implies that $\log a\le \log x$. Then minimum is attained for $x=a$. The ...

2

The standard Wirtinger's inequality requires that $\int_a^b f\,dx=0$, which implies that $c_0=0$, and hence the difficulty you are encounter does not exist. If instead you assume that $f(a)=f(b)=0$, then you should exploit this by using sine series, i.e., for $a=0$ and $b=\pi$, $$f(x)=\sum_{n=1}^\infty a_n\sin nx,$$ and hence $$\int_0^\pi f^2= ... 2 The matrix exponential is given by:$$\tag 1 e^{At} = \sum_{k=0}^{n-1} \alpha_k A^k$$where the \alpha_i's are determined from the set of equations given by the eigenvalues of A, as:$$\tag 2 e^{\lambda_i t} = \sum_{k=0}^{n-1} \alpha_k \lambda_i^k$$We are given:$$e^{At} = 1/2\begin{bmatrix}e^{2t}+e^{-t} & e^{2t} - e^{-t} \\ e^{2t}-e^{-t} & ...

2


1

Here is a proof not involving Fourier Series at all. Since $f(a) = 0$ we can write, using the fundamental theorem of calculus: $$f(t) = \int_a^t \frac{\mathrm{d}f(\tau)}{\mathrm{d}\tau}\mathrm{d}\tau,\ \ \ t\in[a,b],$$ hence $$\left|f(t)\right| \le \int_a^t \left|\frac{\mathrm{d}f(\tau)}{\mathrm{d}\tau}\right|\mathrm{d}{\tau}.$$ Cauchy-Schwarz ...

1

No need to compute two different integrals. Note that $\cos 5x = \operatorname{Re} e^{5ix}$. Put $$f(z) = \frac{e^{5iz}}{z^4+1}$$ and integrate over the boundary of a half-disc in the upper half-plane. On the semi-circle $C_R^+$: $$\left| \int_{C_R^+} \frac{e^{5iz}}{z^4+1} \right| \le \frac{1}{R^4-1} \cdot \pi R \to 0$$ as $R \to \infty$ since \$|e^{5iz}| ...

Only top voted, non community-wiki answers of a minimum length are eligible