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\begin{align} & ? + ? + ? = 30 \\ \implies& 3? = 30 \\ \implies& ? = 10 \end{align}

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Hint: is the sum of three odd numbers even or odd?

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Even more simply: You want to find $x$ such that $x = e^{x/10}$. For small $x$, $e^{x/10} \sim 1+x/10$, so, approximately, $x \sim 1+x/10$ or $x \sim 10/9 = 1.111...$. If $f(x) = e^{x/10}$, $f'(x) =e^{x/10}/10 \sim (1.111...)/10 < 1$, so the iteration $x \gets f(x)$ should converge. And the iterations, starting at $10/9$, are $1.1175190687, ... 3 I presume you want the positive solution. Write the equation as$f(x) = 0$where$f(x) = e^{x/10} - x$. A good initial guess is$x_0 = 1$. Newton's method is the iteration $$x_{n+1} = x - \dfrac{f(x)}{f'(x)} = \dfrac{e^{x/10} (10 - x)}{10 - e^{x/10}}$$ You get$x_1 \approx 1.118238267$,$x_2 \approx 1.118325592$, and$x_3is the same (to 10 significant ... 0 Your idea was good but, may be, there was some mistake in the calculation of the derivatives. Using, as Jake Lebovic, $$P = {L \over {1 + C{e^{ - ky}}}}$$ you have by standard differentiation $$\frac {dP}{dy}=\frac{C k L e^{-k y}}{\left(C e^{-k y}+1\right)^2}$$ $$\frac {d^2P}{dy^2}=\frac{2 C^2 k^2 L e^{-2 k y}}{\left(1+C e^{-k y}\right)^3}-\frac{C k^2 L ... 2 This is a Logistic Growth problem. So...$$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 & ... 0 Since2187=3^{7}<7!=5040$then for$n>7$$$3^{n}=3^{7}\cdot3^{n-7}<7!\cdot8\cdot9\cdot\ldots\cdot n=n!$$ since$8,9,\dots n>3$5 We can prove it by induction. For$n\geq 7$, let$S(n)$denote the statement $$S(n) : 3^n < n!.$$ Base case ($n=7$):$S(7)$says that$3^7 = 2187<5040=7!$, and this is true. Induction step: Fix some$k\geq 7$, and assume that$S(k)$is true where $$S(k) : 3^k < k!$$ To be shown is that$S(k+1)$is true where $$S(k+1) : 3^{k+1} < (k+1)!$$ ... -1 $$\int{log\left(\frac{2-a^{-it}}{1-a^{-it}}\right)} dt=$$ $$\frac{1}{log(a)}\left(-iLi_2\left(a^{it}+iLi_2\left(2a^{it}+tlog(a)(-log\left(1-2a^{it}\right)+log(1-a^{it}\right)+log\left(\frac{-1+2a^{it}}{-1+a^{it}}\right)\right)\right)+C$$ With Li_n(x) is the polygarithm function 0 For linearizing any function$f(x)$about a point$x_0$, use: $$f(x)=f(x_0)+(x-x_0)\frac{df}{dx}(x_0)$$ But beware: this approximation works only for points close to$x_0$4 It applies the exponential function$\exp(y) = \sum \frac{1}{n!}y^n$to the differential operator$a\frac{d}{dx}$to give $$\exp\left(a\frac{d}{dx}\right) = \sum \frac{1}{n!}a^n \frac{d^n}{dx^n}.$$ This, when applied to$f(x)$, gives your middle expression. 2$e^{2n+1} - \frac{1}{e} - (e^{2n} - 1)\geq 0e^{2n+1} - \frac{1}{e} \geq e^{2n} - 1\frac{1}{e^{2n} - 1} \geq \frac{1}{e^{2n+1} - \frac{1}{e}}\frac{e^n}{e^{2n} - 1} \geq \frac{e^{n}}{e^{2n+1} - \frac{1}{e}}\frac{e^n}{e^{2n} - 1} \geq \frac{e^{n+1}}{e^{2n+2} - 1}$. 3 here is one way to see$\frac{e^n}{e^{2n} - 1}$is decreasing with$n.$it is enough to show that$\frac{x}{x^2-1}$is decreasing for$x > 1.$we will use the simple facts$\frac1{x}$and its translates are decreasing on$(1, \infty).$we have $$\frac{x}{x^2-1} = \frac12\left(\frac1{x-1}+\frac1{x+1}\right)$$ is the average of two decreasing functions, ... 4 Look at the ratio:$$\frac{a_{n+1}}{a_n} = \frac{e^{n+1}}{e^{2n+2}-1}\frac{e^{2n}-1}{e^n}=\frac{e^{2n+1}-e}{e^{2n+2}-1}.$$Since$e^{2n+1}<e^{2n+2}$and$e>1,$the numerator is less than the denominator. So this ratio is less than$1.2 I would run with \begin{align} a_n-a_{n+1}&=\frac{e^n}{e^{2n}-1}-\frac{e^{n+1}}{e^{2n+2}-1} \\&=\frac{e^n(e^{2n+2}-1)-e^{n+1}(e^{2n}-1)}{(e^{2n}-1)(e^{2n+2}-1)} \\&=\frac{e^2e^{3n}-e^n-ee^{3n}+ee^n}{(e^{2n}-1)(e^{2n+2}-1)} \\&=\frac{e^{3n}(e^2-e)+e^n(e-1)}{(e^{2n}-1)(e^{2n+2}-1)}\geq0. \end{align} So... Alternatively, define ... 2 $$\dfrac{d}{dn}a_n=-\coth(n)\cdot a_n\le 0$$ Soa_{n+r}\le a_n$and this is true, in particular, also for$r=1$. 2 The inequality to be proved is $$\frac{e^{n+1}}{e^{2n+2}-1}\le\frac{e^n}{e^{2n}-1}$$ Write$x=e^n$and note that, for$n>0$,$e^n>1$, so$x>1$and$x^2>1$. Then the inequality to be proved is $$\frac{ex}{e^2x^2-1}\le\frac{x}{x^2-1}$$ that is equivalent to $$ex(x^2-1)\le x(e^2x^2-1)$$ Simplify by$x, getting the equivalent inequality $$... 3 Need to show:$$\frac{e^{n+1}}{e^{2n+2}-1} \leq \frac{e^{n}}{e^{2n}-1}$$that is$$(e^{2n}-1)e^{n+1} \leq (e^{2n+2}-1)e^{n}e^{3n+1}-e^{n+1} \leq e^{3n+2}- e^ne^n(e^{2n+2}-e^{2n+1}+e-1) \geq 0$$Now analysing this shold be easy. Both factors are positive. 2 Assuming n \gt 0 throughout: a_n = \dfrac{e^{n}}{e^{2n}-1} =\dfrac{1}{e^{n}-e^{-n}}. Since e \gt 1, you have e^{n} is an increasing function of n and is greater than 1, so e^{-n} = \dfrac{1}{e^n} is a decreasing function of n and is less than 1, so e^{n}-e^{-n} is a positive and increasing function of n, so a_n = ... 4 Note that 1/a_n=e^n-e^{-n}=2\sinh n. Since the function \sinh is increasing, so is the sequence 1/a_n, and therefore, a_n is decreasing. Another way:$$\frac{a_{n+1}}{a_n}=\frac{e^{n+1}(e^{2n}-1)}{e^n(e^{2n+2}-1)}\leq e\frac{e^{2n}-1}{e^{2n+2}-e^2}=\frac1e<1$$1 If y is the oil in the reservoir and t is the time in years since extraction started, we have$$\frac{dy}{dt} = -0.13y$$Taking R as the original reserve of oil, this is solved by$$y = Re^{-0.13t}$$Then you need to find t such that$$\begin{align}0.21R &= Re^{-0.13t} \\ 0.21 &=e^{-0.13t} \\ \end{align}$$etc. 0 Notice that: t = 0: M(0) = 100 = 100(0.84^0) t = 1: M(1) = 84 = 100(0.84^1) t = 2: M(2) = 70.56 = 100(0.84^2) So that an equation modeling this relationship is M(t) = 100(0.84^t) (similar to what you have written). To find the inverse function, we can 1) replace M(t) with y 2) solve for t 3) replace y with t and replace ... 5 Consider the function$$ g(s) = \exp(-sA)\frac{\partial}{\partial x}\exp\bigl(s(A+xB)\bigr)\bigr|_{x=0} We have, taking derivatives \begin{align*} g'(s) &= -A\exp(-sA)\frac{\partial}{\partial x}\exp\bigl(s(A+xB)\bigr)\bigr|_{x=0}+ \exp(-sA)\frac{\partial}{\partial x}(A+xB)\exp\bigl(s(A+xB)\bigr)\bigr|_{x=0}\\ &= ... 1 Partial answer: consider f(x,\tau)=e^{\tau(A+xB)}. Then\dfrac{d}{dx}e^{A+xB}=\dfrac{\partial f}{\partial x}|_{x=0,\tau=1}=\int_{\tau=0}^{\tau=1}\dfrac{\partial^2 f}{\partial x \partial \tau}|_{x=0}d\tau$$Now \dfrac{\partial^2 f}{\partial x \partial \tau}=\dfrac{\partial}{\partial x}[(A+xB)e^{A+xB}] and then we have to evaluate at x=0. Now we get ... 1 We don't need exact formulas the derivatives of \phi. In (-1,1), every derivative of \phi has the form$$e^{-1/(1-x^2)}p(x)(1-x^2)^{-m}$$for some m\in \{0,1,\dots \}, where p is a polynomial. Functions of this type \to 0 at \pm 1 from within (-1,1). This implies \phi^{(k)}(x) = 0, |x|\ge 1, for all k, proving \phi \in C^\infty(\mathbb ... 5 First, this result is false if \Bbb C is replaced by \Bbb R (even after making reasonable additional hypotheses to avoid the most obvious counterexamples) as Travis remarks, so one should use something specific for the complex numbers. Given a linear operator \phi that we wish to realise as in the image of \exp, one may decompose the vector space ... 1 Nothing new here, but perhaps streamlined a bit: Apply \ln to get$$(1)\,\,\,\,n\ln (1+a_n/n)=\frac{\ln (1+a_n/n)}{a_n/n}\cdot a_n.$$Now as h\to 0, \ln(1+h)/h = (\ln(1+h)-\ln 1)/h \to \ln'(1) = 1\, by definition of the derivative. It follows that the limit in (1) is a. Exponentiating back gives the limit of e^a as desired. 1 This problem can be solved using domain arguments: e^{x} + x = 1 e^{x} = 1 - x ln(e^{x}) = ln(1 - x) x\cdot ln(e) = ln(1 - x) Which leads us to: x = ln(1-x) x cannot be larger than one, because then the expression 1-x will be negative violating the domain of a logarithmic function. x cannot be a positive number between 0 and 1, ... 2 Your argument using L'Hopilat rule is correct you need just to add the condition a>1 For a> 1 it's true that a^x is very larger then x^n to see this you compose with a logarithm:$$\lim_{x\to \infty} \frac{a^x}{x^n}=\lim_{x\to \infty} e^{\displaystyle x\ln(a)-n\ln(x)} =e^{+\infty}=+\infty$$because the linear functions are always larger than ... 1 Since \lim_{n \to \infty} \Bigl(1+\frac{1}{n}\Bigr)^n = e, the upper limit is just e also, and the lower limit, being its reciprocal, is 1/e. Note that for any \varepsilon > 0, you can find an N such that for all k > N, \Bigl(1+\frac{1}{k}\Bigr)^k is within \varepsilon/2 of e. If the first N terms have an average that differs from ... 1 The solution 2 assumes the following interpretation of the problem statement: Suppose you have a radioactive sample of mass m_0 which after 10 years is reduced to mass m_{10} = \frac{m_0}{32}. What is the half-life of the substance? Recall that the half-life of a radioactive substance is the time after which the mass of a sample is halved by ... 2 Just recall that 2^5=32 and then compute 10/5=2 1 Your idea "the function is continuous hence the anti-derivative exits" doesn't work. This is true for real functions, but it is not true anymore in the complex case. For example, complex conjugation z\mapsto \bar z is continuous, but is not the derivative of anything (on any nonempty open set). After a course in complex analysis, the argument that ... 1 You did a taylor expansion for g around x_0, neglected all terms of order 3 or above. Since x_0 is a maximum, you have that g'(x_0)=0 and g''(x_0)\leq 0, so$$g(x)\approx g(x_0)+\frac{1}{2}(x-x_0)^2g''(x_0)=g(x_0)-\frac{1}{2}(x-x_0)^2|g''(x_0)|.$$You used that g''(x_0)\leq0 to write g''(x_0)=-|g''(x_0)|. 5 Hint. You may transform the equation$$ \frac{x-1}{e^x-1} = y \tag1 $$with a little algebra into$$ -(x+y-1)e^{-(x+y-1)}=-ye^{1-y},\tag2 $$set X:=-(x+y-1) obtaining$$ Xe^X=-ye^{1-y} \tag3 $$then use the Lambert function W to get$$ x=1-y-W\left(-ye^{1-y}\right). \tag4 $$4 I would be surprised if they actually used Taylor series. For example, the basic 80x87 "exponential" instruction is F2XM1 which computes 2^x-1 for 0 \le x \le 0.5. I don't think the implementation is documented, but if I were programming this, I might use a minimax polynomial or rational approximation: the following polynomial of degree 9 ... 7 A commonly used set of algorithms is known as CORDIC: COordinate Rotation DIgital Computer. Basically, it uses a bunch of bitshifts, adds, subtracts and look up tables. Its good for cases where you don't have hardware multipliers and what not. You can also truncate series as well. An interesting thing is that you can often do some math and identify which ... 2 The ultimate answer can only be given by the maker of the calculator, but there are several strategies: The value can be calculated through the series expansion. The value can be interpolated from a stored table. The value can be reduced to the value of a smaller/larger argument by basic operations; for example e^{2x} = (e^x)^2 = e^x\times e^x. In ... 0 Note that e^{x}=\sum_{n=0}^{\infty} \frac{x^{n}}{n!} so if you want an approximation of e^{x} you can just truncate the series.$$e^{x}\approx \sum_{n=0}^{N} \frac{x^{n}}{n!}.After some quick searching, it looks like series are probably not (usually) the best way to go. See the accepted answer here: ... 3 \begin{align} a \exp(x)+b x+c&=0 \end{align} Substitute bx+c=y: \begin{align} a \exp((-c+y)/b)+y &= 0 \\ -y &= a\exp(-c/b)\exp(y/b) \\ -y/b\exp(-y/b) &= a/b\exp(-c/b) \end{align} The last equation is in the form u\exp(u)=w, which has a solution in terms of Lambert W function: \begin{align} u&=W(w). \end{align} Hence ... 2 As said, there is an analytical solution in terms of Lambert functionx=-W(d)-\frac{c}{b}$$where$$d=\frac{a }{b}e^{-\frac{c}{b}}$$If you do not want to (or cannot) use Lambert function, then numerical methods are the way to go. Probably the simplest should be Newton method which, starting from a "reasonable" guess x_0, will update it according to ... 1 Good attempt, but you have to consider your support (i.e., the region of integration) for this question. The shaded region is the region in which 0 < x < y. Now consider the following example: The shaded region is$$\mathbb{P}\left(X \leq 2, Y \leq 4\right) = \int_{0}^{2}\int_{x}^{4}f_{X,Y}(x,y)\text{ d}y\text{ d}x\text{.}$$You can generalize ... 1 You have the limits of integration wrong. You should have:$$ F_{X,Y}(x,y)=\int_{v=0}^{y}\int_{u=0}^{x} f(u,v)\; du\; dv $$But your density is zero when u\ge v, so your integral becomes:$$ F_{X,Y}(x,y)=\int_{v=0}^{y}\int_{u=0}^{\mbox{min}(x,v)} e^{-v}\; du\; dv $$So:$$ F_{X,Y}(\infty,\infty)=\int_{v=0}^{\infty}\int_{u=0}^{v} e^{-v}\; du\; ... 0 Substituteu = 1 - y^3$and use the chain rule. 0 $$z=\sqrt{\frac{x^2}{(1-y^3)}}=|x|(1-y^3)^{-0.5}$$ $$\frac{\partial z}{\partial y}=|x|(-.5)(1-y^3)^{-1.5}(-3y^2)=\frac{1.5y^2|x|}{\sqrt{(1-y^3)^3}}=1.5y^2\sqrt{\frac{x^2}{(1-y^3)^3}}$$ 1 Assuming$z\geq 1$, we have: $$\int_{z-1}^{z}\log\left(\frac{1}{z-y}\right) e^{-y^3}\,dy = -\frac{d}{d\alpha}\left.\int_{z-1}^{z}(z-y)^{\alpha}e^{-y^3}\,dy\,\right|_{\alpha=0}$$ hence the original integral depends on the derivatives of incomplete gamma functions. 3 The tangent to$y=e^{bx}$at$x=a$is $$y-e^{ab}=be^{ab}(x-a)\ ,$$ which simplifies to $$y=be^{ab}x+(1-ab)e^{ab}\ .$$ For this to be the required line we need $$be^{ab}=10\ ,\quad (1-ab)e^{ab}=0\ ,$$ so$ab=1$and hence $$b=\frac{10}{e^{ab}}=\frac{10}{e}\ .$$ 0 You defined$d_n$to be the amount of drug in the blood just after the$n$th dose is taken. Where does the expression $$d_n = D\sum_{k = 0}^{n - 1} r^k$$ come from? The$m$th dose contributed$D$milligrams of drug at the moment it was taken. The$n$th dose is taken$n - m$time units later; by then, the contribution of the$m$th dose has decayed to ... 0 Let$d(t)$be the amount of the drug in the blood$t$hours after a steady-state dose for$0\leqslant t<12$. Then$d(t)=d_\infty e^{-\lambda t}$for some$\lambda>0$(this is the definition of exponential decay). Then $$d_{\min} = \lim_{t\uparrow12}d(t)=\lim_{t\uparrow12}d_\infty e^{-\lambda t}=d_\infty e^{-12\lambda}.$$ From the given information ... 0 Hint: If in steady state you have$d_\infty$just after a dose, after$12$hours you have$rd_\infty=d_{min}\$ because it is just before a dose.

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