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8

By integral test, $$\int_{n+1} ^\infty \frac{1}{x^2} dx \le \sum_{k = n+1}^\infty \frac{1}{k^2} \le \int_n ^\infty \frac{1}{x^2} dx$$ Thus $$\frac{1}{n+1} \le \sum_{k = n+1}^\infty \frac{1}{k^2} \le \frac{1}{n}.$$

5

First, the derivative is a linear operator. This means that $$(f(x) + g(x))' = f'(x) + g'(x)$$ and $$(cf)'(x) = cf'(x)$$ So, in the case that you give, $h(x) = 4x^2 - 2$, you can look at this as $$h(x) = f(x) + g(x) \quad \text{where} \quad f(x) = 4x^2 \quad \text{and} \quad g(x) = -2$$ Then, you have \begin{align*} h'(x) &= (f(x) + g(x))' \\ ...

5

Hint: try to minimize the function $$f(x) = \exp(-x) - 1 + x$$ using calculus. You will find that its unique minimum occurs at $x = 0$.

4

The integral diverges: $$\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} e^{e^{x+y}+x+y} dx dy\ge \int_{1}^{+\infty} \int_{1}^{+\infty} e^{e^{x+y}+x+y} dx dy\ge \int_{1}^{+\infty} \int_{1}^{+\infty} e^{e^{1+1}+1+1} dx dy=+\infty.$$

3

Outline: It can be done. We use "bump" functions, please see Wikipedia. Let $f_0(x)=a_0$ on the interval $(-1/2,1/2)$ and make it decay to identically $0$ beyond $(-1,1)$ by multiplying $a_0$ by a suitable bump function which is $1$ on $(-1/2,1/2)$ and decays to $0$ beyond $(-1,1)$. Let $f_1(x)=a_1x$ on $(-1/4,1/4)$, and let it decay to $0$ by multiplying ...

3

Your approach is absolutely fine and the result that you have obtained is correct. If you handed me a homework like this I would be satisfied, since it is clear that you have understood the procedure of implicit differentiation. Still, the problem is somewhat pedantic and asks you to do a final superfluous simplification: note that $a$ is absent from all of ...

3

Your proof looks correct (but you slightly abused notation at the end - you should say $\to 1 \cdot 1^2 \cdots$, not $= 1 \cdot 1^2 \cdots$). You can shorten it if you use $\lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2}$. $$\frac{\sin x}{x} \frac{1- \cos x }{x^2 \cos x} \frac{1}{\sqrt{1+ \tan x} + \sqrt{1+ \sin x}}$$ $$=\frac{\sin x}{x}\frac{1-\cos ... 3 Taking the limit along the axes gives zero. Taking |x|^c=|y|^d=t\to 0 gives$$ \frac{t^\frac{a}{c}\cdot t^\frac{b}{d}}{t+t}=\frac{t}{2t}=\frac12. $$3 \exp(-x) is convex, so it's above its tangents, especially its tangent at 0. 3 Another approach is to recall that$$\left(1-\frac xn\right)^n$$is a inceasing function of n and is bounded above by its limit, e^{-x}. Therefore$$e^{-x}\ge \left(1-\frac xn\right)^n\ge 1-x$$and we're done. 3 Hint By definition,$$\sum_{k=1}^{n} \frac{1}{k^2} = H_n^{(2)}$$where appears the harmonic number. For large values of n,$$H_n^{(2)}=\frac{\pi ^2}{6}-\frac{1}{n}+\frac{1}{2 n^2}+O\left(\left(\frac{1}{n}\right)^3\right)$$Then, the result. 2 Hint: If f(x) is twice-differentiable, then f(x) is concave (down) if and only if f ′′(x) is non-positive. (see) And we have, for your function:$$ f''(x)=60x^2(x^2+3x+2) $$Can you verify this and find the intervals wher it is not positive? 2 (4x^2-2)'=(4x^2)'-(2)'=4(x^2)'-0=4\cdot 2x=8x. Remember that the derivative of a constant is always 0. 2$$ \begin{split} \frac{df(x)}{dx} = \frac{d}{dx} \left[ 4x^2 - 2\right] = \frac{d\left[ 4x^2 \right]}{dx} - \frac{d [2]}{dx} = 4 \frac{d\left[ x^2 \right]}{dx} - 0 = 4 \cdot 2x = 8x \end{split} $$2 Try this:$$f(x) = \begin{cases}x & x\in \mathbb Q\\ 0& \text{otherwise}\end{cases}$$2 For \forall x\in [1,2): [x]=1. Therefore \forall \epsilon>0 always exists \delta>0, for example \delta=0.3 such that 0=|[x]-1|<\epsilon,\forall x: |x-\frac{3}{2}|<\delta=0.3 which means for all x\in(\frac{3}{2}-\delta,\frac{3}{2}+\delta)=(1.2,1.8) 2 Like you said,$$\begin{align*}\int \arcsin x\,dx&=x\arcsin x-\int\frac{x}{\sqrt{1-x^2}}\,dx\\[1ex] &=x\arcsin x+\frac{1}{2}\int \frac{dw}{\sqrt w}&\text{where }w=1-x^2\end{align*}$$Proceed with the remaining integral, then back-substitute. 1 The formula of integration by parts is$$ \int f'(x)g(x)\,dx=f(x)g(x)-\int f(x)g'(x)\,dx $$and in your case f'(x)=1 (so you can take f(x)=x) and g(x)=\arcsin(x). Thus the correct steps are$$ \int 1\cdot\arcsin(x)\,dx=x\arcsin(x)-\int x\frac{1}{\sqrt{1-x^2}}\,dx $$Now the integral is elementary, because$$ ...

1

Let $\sqrt{x}=t$: the limit becomes \begin{align*} \lim_{t\to1}\frac{t^4-t}{1-t} &=-\lim_{t\to1}\frac{t(t^3-1)}{t-1}\\ &=-\lim_{t\to1}\frac{t(t-1)(t^2+t+1)}{t-1}\\ &=-\lim_{t\to1}\ t(t^2+t+1)\\ &=-1\cdot(1+1+1)\\ &=\boxed{-3} \end{align*}

1

On your second example, it is far less scary than it looks. The first thing you generally want to do is try to get rid of the fractions by multiplying through by the denominators, and in this case, that gives you: $$(2x^2-4x+1)e^x = 0(2x^2+1)^2 = 0$$ Now, remember that the only way a product can be zero is if at least one of its factors is 0: $$(2x^2 - 4x ... 1 By now, I've found a closed-form by doing some integral evaluation, a lot of hypergeometric, polylogarithm and polygamma manipulation.$$ S = \sqrt{\pi}\left(\frac{\pi}{12}\zeta(3)+\frac{1}{192\sqrt3}\psi^{(3)}\left(\tfrac13\right)-\frac{\pi^4}{72\sqrt3}-1\right). $$1 You are correct. The gradient at a point will give you the direction of maximum increase in the value of the function. Its direction will be \frac{\nabla f}{|\nabla f|} In your case:$$\nabla f = (12x^2yz^2+2z^3+yz,4x^3z^2+xz,8x^3yz+6xz^2+xy)$$Therefore,$$\nabla f(1,-1,-1) = (-13,3,13)$$As you have already calculated. It's direction would be ... 1 This is a linear differential equation. Therefore it is defined on the largest intervals for which the functions used in the differential equation are all defined. Hence solutions are defined on intervals for which \cos x doesn't vanish. If you want a solution defined at 0, the interval is (-\frac{\pi}{2},\frac{\pi}{2}) 1 Hints: For the first one, try approaching (1,0) first along the x axis, and then along the curve y = \sqrt {1-x^2}/2. The second one looks false to me: Here we can look at a curve of the form y = \sqrt {1-x^2} -f(x), where f(x) is positive but \to 0 very fast as x\to 1^-. I'm thinking f(x) = (1-x^2)^{2m} should be small enough to make ... 1 You start with f(x) = x(4-x). The derivative is$$f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x},$$assuming the limit exists. To calculate the f(x + \Delta x) part, you substitute x + \Delta x everywhere you see x in x(4-x). Some perhaps clearer examples:$$f(7a) = 7a(4-7a)f(2 \pi e^3) = 2 \pi e^3 (4 - 2 \pi ...

1

Suppose $a_{n+1}=xa_n+ya_{n-1}$ Then $4=2x+y$ and $9=3x+y$ $\implies$ $x=5,y=-6$ Hence $a_{n+1}=5a_n-6a_{n-1}$, whose general term is $a_n=p2^n+q3^n$ When $a_1=a_2=1$, $2p+3q=1$ and $4p+9q=1$, Hence $p=1 ,q=-{1\over 3}$

1

By the fundamental theorem of calculus we have $$f(4) - f(2) = \int_{t=2}^{4}f'(t) < 0,$$ so your guess is right.

1

Look at the plot of $f'$ between $2$ and $4$. It's negative between those points. This means that the slope of $f(x)$ is negative as we move rightward across the interval $(2,4)$. This means that $f(x)$ will $\mathbf{decrease}$ as $x$ moves rightward from $2$ to $4$.

1

Hint: Let $g(x)=2x$. Then $f(x)=4x^2=\left(g(x)\right)^2$ Therefore $f'(x)=2\cdot g(x)\cdot g'(x)$

1

There's a separate rule (the sum rule) for taking the derivative of any sum of terms: (f(x) + g(x))' = f'(x) + g'(x) Use this to split the sum into its terms and take the derivative of each term.

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