# Tag Info

31

The limit is not defined because in order for the limit to exist, the value of the function for every possible path to $(0,0)$ must tend to the same finite value. When $y = x^2$, you have not necessarily shown that the limit is in fact $0$. When you transformed to polar coordinates and then took the limit as $r \to 0$, you are assuming that $\theta$ is a ...

17

There are many cases where this will happen. By simple algebra, the relation $(uv)'=u'v'$ can be written as $$\Bigl(\frac{u'}{u}-1\Bigr)\Bigl(\frac{v'}{v}-1\Bigr)=1$$ and this leads to $$\ln u(t)=\int\frac{v'}{v'-v}\,dt\ .$$ You will have to make sure that the integral exists and gives a suitable result, but this should give you a solution for many ...

11

Let $z=e^{i x}$; then $dx = -i dz/z$ and the integral is equal to $$-i 8 \oint_{|z|=1} dz \frac{z^3}{z^8 + 6 z^4+1}$$ By the residue theorem, the integral is then equal to $i 2 \pi$ times the sum of the residues at each pole inside the unit circle. The residue at each pole $z_k$ is equal to $$-i 8 \frac{z_k^3}{8 z_k^7 + 24 z_k^3} = -i \frac1{z_k^4+3}$$ ...

11

Let $\alpha >0$, and consider the path $\gamma_\alpha(t) = (t,\alpha t^2)$. Then we have $f \circ \gamma_\alpha (t) = {\alpha t^4 \over t^4+ \alpha^2 t^4 }$, and the limit as $t \to 0$ is ${1 \over \alpha}$. The limit exists along all of these paths, but is different. If the limit exists, its value must be independent of how $(x,y) \to 0$.

10

You haven't taken into account what happens if $\theta$ is variable as a function of $r$ when $r$ goes to $0$. Choose $\theta$ so that $\sin \theta = r$, i.e. $\theta$ is approximately $r$ and you will get $\cos \theta$ is about 1 for small $r$, and then the limit will not be zero, so the limit doesn't exist. If you want to use polar coordinates to show ...

7

We have $\displaystyle\cos^4x+\sin^4x=(\cos^2x+\sin^2x)^2-2\cos^2x\sin^2x=1-2\cos^2x\sin^2x$ $\displaystyle=\frac{2-\sin^22x}2=\frac{2(1+\tan^22x)-\tan^22x}{2\sec^22x}=\frac{\tan^22x+2}{2\sec^22x}$ $$\int\frac{dx}{\cos^4x+\sin^4x}=\int\frac{2\sec^22x}{\tan^22x+2}dx$$ Setting $\tan2x=u,$ ...

5

No, there is no such $C$. Assume that there is and take $y = 0$. Then $|e^x - 1| \le C|x|$ for all $x$ which would for example imply that $e^x/x^2 \to 0$ as $x \to \infty$ and that is false. On the other hand, if $x$ and $y$ are restricted to a compact set, then the mean value theorem implies the existence of such a $C$. In other words, $f$ is locally ...

4

Setting $\displaystyle e^{\dfrac{i\pi t}2}=u\implies u^2=e^{i\pi t};u^3=e^{\dfrac{3i\pi t}2};u^4=e^{2i\pi t}$ $$\frac{e^{\dfrac{it\pi}2}-e^{\dfrac{3it\pi}2}}{1 - e^{2\pi it}} =\frac{u-u^3}{1-u^4}=\frac{u(1-u^2)}{(1-u^2)(1+u^2)}=\frac u{1+u^2}$$ if $\displaystyle1-u^2\ne0\iff u^2\ne1\iff e^{i\pi t}\ne1=e^{2m\pi i}\iff t\ne2m$ where $m$ is any integer ...

4

The answer is No: Let $\sum_na_n$ an absolutely convergent series and denote $S$ its sum and let $\sum_n c_n$ a conditionally convergent series and denote $S'$ its sum then we define the series $\sum_n b_n$ by $$b_1=c_1-S'+S\quad;\quad b_n=c_n\;\forall n\ge 2$$ hence $$\sum_{n=1}^\infty a_n=\sum_{n=1}^\infty b_n$$ but $\sum_{n=1}^\infty|b_n|=\infty$.

4

I know at least two, maybe even three contexts where "derivatives" (in a loose sense) exist: If $F : A \to B$ is a (covariant) functor between abelian categories which is left exact, and the categories in question are well behaved, then we can talk about the right derived functors $R^iF$ of $F$. These are functors such that whenever $0 \to A \to B \to C ... 4 (1) The conceptual difference is huge. A series $$F(s,t):=\sum_{m,\>n}a_{mn} s^m t^n\tag{1}$$ has${1\over2}(r+1)(r+2)$"free" coefficients$a_{mn}$up to total degree$r$, whereas a series $$f(z):=\sum_{k=1}^\infty c_k z^k=\sum_{k=1}^\infty c_k (s+it)^k\tag{2}$$ has only$r+1$"free" coefficients$a_k$up to total degree$r$. The essential difference ... 3 Let$u$and$v$be functions of$t$. then$(uv)'=u'v'\iff u'v+uv'=u'v'\iff u'(v-v')+v'u=0\iff u'+\frac {v'}{v-v'}u=0$Let u be the unkown function, then multiply both sides by$e^{\int \frac{v'}{v-v'}}dt$:$u'(e^{\int \frac{v'}{v-v'}dt})+(e^{\int \frac{v'}{v-v'}dt}\frac {v'}{v-v'})u=0 (*)$Notice that$(e^{\int \frac{v'}{v-v'}dt})'=e^{\int ...

3

Notice that the given sequence is a partial sum of a geometric sequence hence $$\mu_n(x):=f(x)-f_n(x)=\sum_{k=n+1}^\infty(x/2)^k=(x/2)^{n+1}\frac{1}{1-x/2}$$ so we have that $$\sup_{x\in(-2,2)}|f(x)-f_n(x)|\ge\lim_{x\to2}\mu_n(x)=+\infty$$ and thus the sequence cannot converge uniformly.

3

You are given that $f'''$ is continuous. Since $f'''(x_0) \not= 0$ there exists a neighborhood $I$ of $x_0$ on which $f'''$ is either always positive or always negative. In (say) the former case, $f''' > 0$ so that $f''$ is increasing on $I$, and thus $f'' < 0$ on an interval $(x_0-\epsilon,x_0)$ and $f'' > 0$ on an interval $(x_0,x_0+\epsilon)$. ...

3

There are enumerations with and without this property. The first claim is easy (just let $q_n=\frac1{n^2}$ for odd $n$ and enumerate the rest among the even indices somehow). To enumerate $\mathbb Q\setminus\{0\}$ such that $|q_n|>\frac 1n$ for all indices $n$, start with any enumeration $p_n$ of $\mathbb Q\setminus\{0\}$ and recursively define $q_n$ as ...

3

Suppose there is an epsilon ball around $w$ which is disjoint from the image of $f$. Consider $g(z)=\frac{1}{f(z)-w}$ which is analytic except for at $z_n$ and $0$. Since $g$ is bounded above by $\frac{1}{\epsilon}$ these singularities are removable. Hence $g$ extends to an analytic function which is zero at $z_n$ (and at $0$). Hence $g$ is zero on a set ...

3

Let us start from your conclusion $s_{n}<r_n$ for $n\geq M$. This can be written in the following equivalent form $$\forall\, n\geq M,\quad\frac{1}{n+1}<\frac{a_n}{n}-\frac{a_{n+1}}{n+1}$$ This implies, by adding these inequalities $$\forall\, m\geq M,\quad\sum_{k=M+1}^{m-1}\frac{1}{k+1}<\frac{a_{M+1}}{M+1}-\frac{a_{m}}{m} <\frac{a_{M+1}}{M+1} ... 3$$ \int_{-\infty}^\infty \int_{-\infty}^\infty \frac{f(x) \overline{g(y)}}{|x-y|^{1/4}} \, dy dx = \int_{-\infty}^\infty \hat f(\xi) \overline{\hat g(\xi)} k(\xi) \, d\xi ,$$where k(\xi) is the Fourier transform of x\mapsto |x|^{-1/4}, that is, k(\xi) = C |\xi|^{-3/4} for some constant C > 0. So the answer is "yes." 3 Maybe, it would be better for you to choose a slightly more advanced topic (like introduction to topological manifolds using Lee's book, or Boothy's book) instead of multi-variable analysis for next semester. You should be able to cover multi-variable mathematical analysis yourself by using a suitable textbook (like Zorich's Mathematical Analysis II, which ... 3 This only holds for a certain subset of \mathbb R^2. To find it, solve the equation xy=x+y, since this will be the border between S_1:=\{xy < x+y\} and S_2:=\{xy > x+y\}. Your equation is valid only on \overline{S_2} Let us now find S_2:$$xy = x + y \Rightarrow_{x\ne 0} y = 1 + \frac yx \Rightarrow y(1-\frac1x) = 1 \Rightarrow y = ...

3

I'm not sure why you were asked to "compute" those forms in all their lengthy glory. The key thing to realize is this: As you suggested, $\partial S$ is the union of those two $3$-spheres (appropriately oriented—I'll get to that in a moment). Since $dt=0$ on $\partial S$ (as each component is contained in a slice with $t=\text{constant}$), any term in ...

2

Let $x\not=0$. Use the binomial theorem: \begin{align} \frac{\frac{1}{(x+h)^n}-\frac{1}{x^n}}{h} &=-\frac{(x+h)^n - x^n}{(x+h)^n x^n h}\\ &=-\frac{1}{h (x+h)^n x^n} \left(\sum_{k=0}^n {n\choose k} x^k h^{n-k} - x^n\right)\\ &=-\frac{1}{h (x+h)^n x^n} \sum_{k=0}^{n-1} {n\choose k} x^k h^{n-k}\\ &=-\frac{1}{(x+h)^n x^n} \sum_{k=0}^{n-1} ...

2

WLOG: $x<y$. As you probably know, for a continuously differentiable function, you have $$f(x) - f(y) = f'(\zeta)(x-y)$$ for some value $\zeta\in[x,y]$. This means, of course, that $|e^x - e^y| = |e^\zeta||x-y|$ for some $\zeta\in[x,y]$. Since this means that $e^\zeta \geq e^x$, this also means that $$|e^x - e^y| \geq e^x |x-y|,$$ and this means that ...

2

I'm not sure if that is what you are after, but: If $\omega$ is a primitive $2k$-th root of unity, then the solutions of $z^k+1 = 0$ are $\omega^{2n+1},\, 0 \leqslant n < k$, so $$z^k + 1 = \prod_{n=0}^{k-1} (z-\omega^{2n+1}),$$ and hence \begin{align} \prod_{n=1}^{k-1} (z-\omega^{2n+1}) &= \frac{z^k+1}{z-\omega}\\ &= \frac{z^k - ... 2 I think this is a valid argument: Assume not; that is, assume that there are a,b with b>a such that u(a)=u(b)=0 but we do not have v(d)=0 for any d\in (a,b). Then the functionf(x)=\frac{u(x)}{v(x)}$$is continuous on the interval [a,b] and differentiable on (a,b), and further we have f(a)=f(b)=0. Therefore by Rolle's Theorem, there must ... 2 For z \neq z_0 and z_0 \neq 0:$$\begin{align} \sum_{k=0}^{n-1} z^k z_0^{n-1-k} & = z_0^{n-1} \sum_{k=0}^{n-1} \left( \frac{z}{z_0} \right)^k \\ & = z_0^{n-1} \frac{\left(\frac{z}{z_0}\right)^n - 1}{\frac{z}{z_0} - 1} \\ & = \frac{\frac{z^n}{z_0} - z_0^{n-1}}{\frac{z}{z_0} - 1} \\ & = \frac{z^n - z_0^n}{z - z_0} \end{align}$$Therefore ... 2 Not really, sorry. You need to show, for given \varepsilon>0, that there is N\in \mathbb{N}, such that$$\frac{x}{x+n}< \varepsilon$$for all n\ge N, independently of x. Which, I'm afraid, is not even true, because if you have any N you can choose x  very large (larger than N) such that the fraction becomes >\frac{1}{2}, say. What ... 2 Well, if \sum a_n x^n converges absolutely for all x then \sum a_n x^{n-1} converges absolutely for all x. This implies that \sum a_n (|x|+|y|)^n converges absolutely for all |x|+|y|. Now$$\begin{align}\sum |a_n|(|x|^{n-1}+|x|^{n-1}|y|+...+|y|^{n-1})&\leq \sum |a_n|\left(\sum_{k=0}^{n-1}\binom{n-1}{k}|x|^{n-1-k}|y|^k\right)\\&\leq ...

2

For all $x,y\in \mathbb R$ the following holds: \begin{align} \sin(x+iy)&=\dfrac{e^{i(x+iy)}-e^{-i(x+iy)}}{2i}\\ &=\dfrac{e^{-y+ix}-e^{y-ix}}{2i}\\ &=\dfrac{e^{-y}(\cos(x)+i\sin(x))-e^{y}(\cos(x)-i\sin(x))}{2i}\\ &=\dfrac{(e^{-y}-e^y)\cos(x)+i(e^{{-y}}+e^y)\sin(x)}{2i}\\ ...

2

You have the right approach. Suppose first $f$ is surjective. If $g\in W^*$ and $f^*(g)=0$, then $g(f(v))=0$ for all $v\in V$ and since $f$ is surjective this means $g(w)=0$ for all $w\in W$, hence $g=0$. Thus $f^*$ is injective. Now suppose $f$ is not surjective, so its image is a proper linear subspace. Let $w\notin f(V)$ and extend $w$ and a basis for ...

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