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7

On $\mathbb R$, we have $f(x) = f(0) e^x$. So now apply the identity theorem to $F(z)=f(z)-f(0)e^z$, which has all the real numbers as roots. Since $\mathbb R$ definitely has a limit point, we will have $F(z)=0 \quad \forall z \in \mathbb C$, which implies $f(z)=f(0)e^z.$ And the derivative condition would give us $f(0)=1.$


7

$G(x)=\int_0^{x}(2t+1)^3dt$ so $G'(x)=(2x+1)^3$ and $F(x)=G(x^7)-G(x^4)$ taking derivatives $$F'(x)=G'(x^7)7x^6-G'(x^4)4x^3=7(2x^7+1)^3x^6-4(2x^4+1)^3x^3$$


5

After applying the $u$-substitution you suggestion, we have $$\require{cancel} \int_0^{\pi / 4} (\cos 2x)^{11 / 2} \cos x \,dx = \int_0^{1 / \sqrt{2}} (1 - 2 t^2)^{11 / 2} dt.$$ For a small difference in aesthetics we may as well make the linear substitution $t = \frac{1}{\sqrt{2}} s$, giving $$\require{cancel} \int_0^{\pi / 4} (\cos 2x)^{11 / 2} \cos x \,dx ...


4

Use successive integrations by parts. Here is the first: $$\int (1-2t^2)^{11/2} dt=t(1-2t^2)^{11/2}-\int \frac{11}{2}t (-4t)(1-2t^2)^{9/2}dt$$ $$=t(1-2t^2)^{11/2}-11\int (1-2t^2-1)(1-2t^2)^{9/2}dt$$ $$=t(1-2t^2)^{11/2}-11\int (1-2t^2)^{11/2}dt+11\int (1-2t^2)^{9/2}dt$$ Hence $$12\int (1-2t^2)^{11/2} dt=t(1-2t^2)^{11/2}+11\int (1-2t^2)^{9/2}dt$$ When you ...


4

The goal is to represent $$\dfrac{dt}{dx} = \dfrac19 \cdot \dfrac{(3x^2-9x+1)^2}{(x^3-x+1)^2} = \dfrac{d(u/v)}{dx} = \dfrac{vdu/dx - u dv/dx}{v^2}$$ Hence, it is tempting to choose $v=(x^3-x+1)$. This means we need $u$ such that $$(x^3-x+1)u'(x) - (3x^2-1)u = (x^2-3x+1/3)^2$$ Hence, $u(x)$ must be a quadratic as $u(x) = ax^2+bx+c$. This gives us ...


3

$$\left|\int_0^1 x^nf(x)\right| \le \int_0^1 |f(x)|x^n \le M\int_0^1 x^n = M \frac 1{n+1}$$ Where $M$ is the maximum of $|f(x)|$ in $[0,1]$ which exists thanks to the Weierstrass theorem. Now take the limit for $n \to \infty$ and use the squeeze theorem to conclude.


2

The following method feels more systematic to me: $$\begin{align} \int\frac{\mathrm{d}x}{x^2\left(1+x^4\right)^{3/4}} &=\frac14\int\frac{4x^3\,\mathrm{d}x}{x^5\left(1+x^4\right)^{3/4}}\\ &=\frac14\int\frac{\mathrm{d}t}{t^{5/4}\left(1+t\right)^{3/4}};~~~\small{x^4=t}\\ ...


2

Consider $g(u) = \frac12 f(u^{-1/2}) u^{-3/2}$. Then for each $n$: $$ 0 = \int_a^1 t^{-2n} f(t) dt = \int_1^{a^{-2}} u^n g(u) du $$ so $g = 0$ (a classical application of the Stone Weiertrass theorem) and $f = 0$.


2

There’s no reason to think that $f$ is either always positive or always negative. HINT: Prove and use the following facts: If $f(a)\ne 0$, then there is an $\delta>0$ such that $f(x)$ and $f(a)$ have the same sign whenever $|x-a|<\delta$. If $f(a)=0$, $|f(x)-f(a)|=\big||f(x)|-f(a)\big|$ for all $x$.


2

This is a Logistic Growth problem. So... $$P'(y) = kP(L - P)$$ and $$P(y) = {L \over {1 + C{e^{ - ky}}}}$$ If we add in the specifics of your problem... $$\eqalign{ & P = {{250} \over {1 + 4{e^{ - 0.01y}}}} \cr & P'(y) = kP(L - P) = 0.01P(250 - P) = 2.5P(y) - .01P{(y)^2} \cr & P'' = 2.5 - .02P = 0 \cr & P = 125 \cr & ...


2

If $f(t)=(2t-1)^3,a(x)=x^4,b(x)=x^7, F'(x)=f(b(x))b'(x)-f(a(x))a'(x).$


2

The second part of the assertion is false: $\mathrm e^{-t^2} >0$ and it's a continuous, hence its integral is $>0$ as soon as $x\neq 0$.


2

I did not completely finished it but I think it is doable. Consider differentiating $(x^3-x+1)^{-1}$, the derivative is $-3x^2(x^3-x^2+1)^{-2}$. The denominator is right, so quotient rule will be natural to use. Next we try to guess what the antiderivative of the whole expression is. Consider quotient rule, $(\frac{u}{v})'=\frac{u'v-v'u}{v^2}$. Therefore ...


2

from $ \frac {a_n} {a_{n-1}} \le \frac {b_n} {b_{n-1}} $ you get $$ a_N = \frac {a_0} {b_0} \times b_0 \times \prod_{n = 1}^{N} \frac {a_{n}} {a_{n-1}} \le \frac {a_0} {b_0} \times b_0 \times \prod_{n = 1}^{N} \frac {b_{n}} {b_{n-1}} = \frac {a_0} {b_0} \times b_N $$ so $$\sum b_N<\infty \implies \sum a_N<\infty $$


2

Denote the differential operator by $D$. The equation $Df=\lambda f$ is equivalent to $f'=\lambda f$. Multiplying by $e^{-\lambda x}$ we obtain: $$f'(x)e^{-\lambda x}-e^{-\lambda x}\lambda f(x)=0$$ which is the same as $$(f(x)e^{-\lambda x})'=0$$ Which means that there exists a constant $c$ such that $$f(x)e^{-\lambda x}=c \ \text{for all $x$}$$ Hence ...


2

Define an auxiliary function $$g(x)=\int_{-1}^x (\cos(t^2)+t) \, dt$$ By the fundamental theorem of calculus, $g'(x)=\cos(x^2)+x$. However, note that $$h(x)=g(\sin(x))$$ Hence by the chain rule: $$h'(x)=g'(\sin(x))\cos(x)=(\cos(\sin ^2(x))+\sin(x))\cos(x)$$


2

Suppose $a_n + b_n \to L \in \mathbb {R}.$ Let $M$ be any subsequential limit of $b_n.$ Then there is a subsequence $b_{n_k} \to M.$ Thus $a_{n_k} = -b_{n_k} + (b_{n_k}+a_{n_k}) \to -M + L.$ There are $2750$ possibilities for $M,$ hence for $-M + L.$ This shows $a_n$ has at least $2750$ subsequential limits, contradiction.


1

Some tricks to help you: a. count squares b. calculate the area of known figures, such as triangles and rectangles. For example, the integral in (b) can be calculate as the same of the area of a trapezoid($(1+3)\cdot 2/2$), a rectangle ($1\cdot 3$), and a triangle ($3\cdot 2/2$) to give a grand total of $10$. That should cover it, good luck.


1

So in the end I solved this using a pretty basic method: the triangle inequality (proven in problem 4-9 b) in the book). So the whole situation can be described with a picture: We see that the points $(x,0)$, $(y,0)$ and $(z, f(z))$ represent a triangle. And according to the triangle inequality we have, for every $z$ $$ \sqrt{(f(z))^2+(y-z)^2} \leq ...


1

You need to use the fact that if a sequence converges, any subsequence converges to the same limit. Since $\{y_{3n}\}$ converges to say $y$, then $\{y_{6n}\}$ must also converge to $y$. But $\{y_{6n}\}$ is a subsequence of $\{y_{2n}\}$ and since that sequence converges, it must converges to $y$. A similar argument will show that $\{y_{2n+1}\}$ will ...


1

For the integrand \begin{equation*} \frac{1}{1-3\sin(x)} \end{equation*} substitute $u=\tan(\frac{x}{2})$ & $du=\frac{1}{2}dx\sec^2(\frac{x}{2}).$ Then transform the integrand using \begin{equation*} \sin(x)=\frac{2u}{u^2+1},~\cos(x)=\frac{1-u^2}{u^2+1} \& dx=\frac{2}{u^2+1}du \end{equation*} to get \begin{equation*} \int ...


1

Hint: $$ \frac 1n \cdot \left(f\left(\frac{1}{n}\right)+f\left(\frac{2}{n}\right)+f\left(\frac{3}{n}\right)+ \ldots +f\left(\frac{n}{n}\right)\right) $$ is a Riemann sum for $\int_0^1 f(x) \, dx$ and the limit of $$ n \cdot \left(1-f\left(\frac{1}{\sqrt{n}}\right)\right) $$ can be calculated easily, for example with L'Hospital or using the Taylor series of ...


1

the half period of the $sin$ is the distance between consecutive local max and min. therefore half period $T/2 = 45^\circ = \pi/4 \to T = \pi/2.$ the sinusoidal is oscillating about the average value $\frac 12\left(7-(-1)\right) = 4$ and the amplitude is $7-4 = 3.$ putting all these together, we get $$y = 4 - 3\sin 4(t - \pi/24)=4-3\sin\left(4t - ...


1

Assume that $f'$ is bounded by some $M$, you get $$\left|\int _{\frac{k-1}n}^{\frac kn} f(t) dt - \frac 1n f\left( \frac kn \right)\right| =\left| \int _{\frac{k-1}n}^{\frac kn} \left[f(t) - f\left( \frac kn \right)\right]dt\right| \le \int _{\frac{k-1}n}^{\frac kn} \left| f(t) - f\left( \frac kn \right) \right|dt \\ \le M\int _{\frac{k-1}n}^{\frac kn} ...


1

If $f\colon[0,1]\to\mathbb{R}$ is Riemann integrable, then $$ \lim_{n\to\infty}\frac1n\sum_{k=1}^nf\Bigl(\frac kn\Bigr)=\int_0^1f(x)\,dx. $$ Depending on the definition of Riemann integral you are using, this is either an immediate consequence of the definition or a theorem. If $f$ is bounded, say $|f(x)|\le M$, then $$ \Bigl|\frac1n\sum_{k=1}^nf\Bigl(\frac ...


1

If $f$ is increasing then $$a_n=\sum_{k=1}^n\left( \underbrace{f(k)-\int_{k-1}^k f(x)dx}_{\ge0}\right)$$ so $(a_n)$ is increasing and $$f(k)-\int_{k-1}^k f(x)dx=\int_{k-1}^k (f(k)-f(x))dx\le f(k)-f(k-1)$$ so by telescoping we get $$a_n\le f(n)-f(0)$$ so if $f$ has a finite limit at $+\infty$ then $a_n$ is bounded above and then it's convergent.


1

I have written an answer to explain what is going on, but a little mistake made me loose the LaTeX original file. Fortunately, I have it on PDF ; I include it as a picture : Take care that your $F$ is transformed in $f$ from the $7$-th line! I add a picture which summarizes what I wrote : Hope it will be helpful!


1

$\newcommand{\Reals}{\mathbf{R}}$In case it's a helpful mnemonic, the implicit function theorem (in this setting) is essentially a non-linear version of Gaussian elimination from linear algebra. Specifically, let $m$ and $k$ be positive integers, and put $n = m + k$. Suppose $A$ is an $m \times n$ matrix, and let $F:\Reals^{n} \to \Reals^{m}$ be ...


1

You have $f'(x) = (x^2-4)/(2+\cos^2(x))$ The denominator, $2+\cos^2 x$, is strictly positive since it's always $\ge 2$. Therefore $f$ increases where the numerator is positive and decreases where the numerator is negative. The numerator is $x^2-4=(x-2)(x+2)$ and that is positive when $x>2$ and when $x<-2$, and is negative when $-2<x<2$. So ...


1

\begin{align} F'(x) &= \exp - \frac {x^2}2 = \sum_{n = 0}^\infty \frac{(-x^2/2)^n}{n!} \\ &= \sum_{n = 0}^\infty \frac{(-1)^n}{2^nn!} x^{2n} \\ \implies F(x) &= \sum_{n = 0}^\infty \frac{(-1)^n}{2^nn!} \frac{x^{2n+1}}{2n+1} \end{align}



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