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15

Observe that if $ f(x)=f'(x) $ then $$ \left(\frac{f(x)}{e^x}\right)'=\frac{f'(x)-f(x)}{e^x}=0 $$ Hence $\dfrac{f(x)}{e^x}$ is constant...


11

Hint: Mutiply by $e^x$ and rearrange to get a quadratic equation in $e^x$.


10

$$\lim_{n\to\infty} \frac{3^n}{2^n+3^n}=\lim_{n\to \infty}\frac{1}{(\frac{2}{3})^n+1} =1.$$Since $\frac{2}{3}<1$ , so $(\frac{2}{3})^n\to 0$ as $n\to \infty$.


9

Hint. One may recall that for a differentiable function around $x_o$ we have $$ \frac{f(x_0+h)-f(x_0) }{h} \to f'(x_0),\quad h\to 0. $$ Apply it to $f(x)=e^x$.


7

Hint: $$\frac{2^{2x}}{5^{x-1}}=5\frac{2^{2x}}{5^x}=5\frac{4^{x}}{5^x}=5\left(\frac{4}{5}\right)^x $$


7

No. Suppose that $f \not\equiv 0$. We solve the ODE: $$f(x) = f'(x) \implies \frac{f'(x)}{f(x)} = 1 \implies \int \frac{f'(x)}{f(x)}\,{\rm d}x = \int \,{\rm d}x \implies \ln |f(x)| = x+c, \quad c \in \Bbb R$$With this, $|f(x)| = e^{x+c} = e^ce^x$. Call $e^c = A > 0$. Eliminating the absolute value, we have $f(x) = Ae^x$, with $A \in \Bbb R \setminus ...


5

You got a very good start. Now rewrite the numerator $n$ as $(n-1)+1$. Our sum is equal to $$\sum_1^\infty \frac{n-1}{(n-1)!}+\sum_1^\infty \frac{1}{(n-1)!}.$$ By the method you already used, the first sum is $\sum_2^\infty \frac{1}{(n-2)!}$. This sum is $e$. More simply, the second sum is $e$.


5

At the beginning of the year, in your AP Calculus AB class, you found derivatives by using one of two definitions: 1) The derivative of a function at a specific x-value: $\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}=f'(a)$ 2) The derivative of a function "in general", i.e. the "derivative function": $\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}=f'(x)$ The ...


5

Let $r=s/t$ where $s\in\mathbb{Z}$ and $r\in\mathbb{N}$. Assume $e^r=p/q$ where $p,q\in\mathbb{N}$. Then $$ pqt^nJ_n(r)= p^2 t^n A_n\left(\frac{s}{t}\right)+ q^2 t^n B_n\left(\frac{s}{t}\right)\in\mathbb{Z} \tag{1} $$ Note that $$ 0<J_n(x)\leq\frac{x^{2n}}{n!}\int_{-x}^xe^tdt=\frac{2 x^{2n}\sinh x}{n!} $$ So $$ 0<pqt^nJ_n(r)\leq \frac{2 pq t^n ...


5

Hint: The series representation for the exponential will work. Use the fact that if $x$ is positive then $$e^x\gt \frac{x^3}{3!}.$$


4

$$e^x-e^{-x}=\frac{3}{2}$$ Multiply by $e^x$. $${\left(e^{x}\right)^2}-1=\frac{3e^x}{2}$$ Let $u=e^x$ $$u^2-\frac{3u}{2}-1=0$$ $$2u^2-3u-2=0$$ Now solve for $u$, and back substitute into $u=e^x$. Consider only positive solutions for $ u$.


4

Hint: If $f(x)=e^x$, then the given limit is $f'(2)$.


4

Consider $g(x) = f(x)\exp(-x)$. Then we have $g'(x) = f'(x)\exp(-x)-f(x)\exp(-x) = 0$. Thus, $g\equiv c$ for some constant $c$. Hence $f(x) = c\exp(x)$.


4

This might not be rigorous, but note that one has $$(x+a)^x=x^{x+a}\Rightarrow x\ln(x+a)=(x+a)\ln x\Rightarrow \frac{\ln(x+a)}{x+a}=\frac{\ln x}{x}.$$ Then, let $f(x)=\frac{\ln x}{x}$. One has $f'(x)=0\iff x=e$, and considering the graph of $y=f(x)$ should give you the answer.


4

Recall the definition of exponential function of one variable in terms of power series: $$ e^x := \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots $$ Similarly, we can define the matrix exponent of $n\times n$ matrix $A$: $$ e^A : = I + A + \frac{1}{2!}A^2 + \frac{1}{3!}A^3 + \dots = \sum_{n=0}^{\infty}\frac{A^n}{n!}, ...


3

I think other answers given here assume the existence of a nice function $e^{x}$ and this makes the proof considerably simpler. However I believe that it is better to approach the problem of solving $f'(x) = f(x)$ without knowing anything about $e^{x}$. When we go down this path our final result is the following: There exists a unique function ...


3

You have $\dfrac{dy}{dx}=y$. Often one writes $\dfrac{dy} y = dx$ and then evaluates both sides of $\displaystyle\int\frac{dy} y = \int dx$, etc. However, for a question like this perhaps one should be more careful. If $f(x)\ne 0$ for all $x$, then one has $\dfrac{f'(x)}{f(x)}=1$ for all $x$. This implies $$ \frac{d}{dx} \log |f(x)| = 1 $$ for all $x$. ...


3

First limit is easy. Let $$f(n) = \left(1 + \frac{1}{n!}\right)^{n}$$ We will use the fact that the sequence $a_{n} = (1 + (1/n))^{n}$ is strictly increasing for $n \geq 3$ and tends to a positive limit (when $n \to \infty$) denoted by $e$. Also $2 < e < 3$. These facts can be proven (and given in many textbooks) without any standard theory of ...


3

HINT: $$ \lim_{n \to \infty} \left(1 + \frac{1}{n!}\right)^n=\lim_{n \to \infty} \left(\left(1 + \frac{1}{n!}\right)^{n!}\right)^{n/n!} $$ and the inner limit is $e$, hence the final one is 1. The second case is similar.


3

By using the concept of AM > or equal to GM Since both functions $2^x$ and $4^x$ are positive and their product is finite value ($=4$) we can apply AM>GM . (2^x + 4/2^x)/2 > or equal to (2^x*4/2^x)^1/2 => 2^x + 4/2^x > or equal to $4 . $ Hence the range is $[4,\infty).$


3

Write $$f(x)=x^{\sqrt x}$$ Then $$g(x)=\ln f(x)=\sqrt x\ln x=\frac{\ln x}{x^{-1/2}}$$ Now use l'Hopital to compute $$\lim_{x\to0^+}g(x)$$ Since $x\mapsto e^x$ is continuous, $$\lim_{x\to 0^+}f(x)=e^{\lim_{x\to0^+}g(x)}$$


3

Suppose $n\cdot 2^n=160$. Since $160=2^5\cdot 5$, you know that $5$ divides $n$, so $n=5m$. Then $$ 5m\cdot2^{5m}=160 $$ becomes $m\cdot 2^{5m}=32$ and so $m=1$. However, the general solution of $x\cdot 2^x=a$, for arbitrary $a$, cannot be determined “explicitly”, without using “higher level” functions such as Lambert's $W$.


3

You can modify your equation as such $$ \ln(2)n e^{\ln(2)n} = 160\ln(2) $$ So that $$ n = \frac{1}{\ln(2)} W(160\ln(2))=5$$ Where $W$ is the Lambert W function. I do not know any identity on W that would lead the result 5 without numerically computing W (which is usually done using root-finding).


3

The math is right, but if you are using implicit differentiation the point is not to solve for $y$. Instead you would differentiate each term with respect to $x$, assuming that $y$ is some function of $x$ whose derivative is $dy/dx$. For instance, the term $y^2e^x$ yields $$2y\frac{dy}{dx}e^x+y^2e^x$$ Hopefully this helps as a starting point. (You will ...


3

We have $\sin(z) = \sin(x+iy) = \frac{e^{ix-y}-e^{-ix+y}}{2i}$ $\frac{e^{ix}e^{-y}-e^{-ix}e^{y}}{2i}$ So $|\sin(z)| = \left|\frac{e^{ix}e^{-y}-e^{-ix}e^{y}}{2i}\right|$ $=\frac{\left|e^{ix}e^{-y}-e^{-ix}e^{y}\right|}{2}$ $\geq \left|\frac{\left|e^{ix}e^{-y}\right|-\left|e^{-ix}e^{y}\right|}{2}\right|$ Because $|e^{ix}| = 1$, we can simplify, $= ...


3

Quite simply, the reason why the official solution is $6$ is because the instructions specifically state to round the result to the nearest (whole) year. There is no implied directive to do the rounding in the context of the minimum time needed for the investment to actually achieve a doubling of its value. You are simply asked to calculate the time, and ...


3

This is the same as solving $$2\sinh x=\frac 32$$ so you could use the logarithmic formula for $\operatorname{arsinh} x$.


3

$$\large e^{5i\pi/6}=e^{5i\pi/6-i\pi+i\pi}=e^{-i\pi/6+i\pi}=e^{-i\pi/6}e^{i\pi}=-e^{-i\pi/6}$$


3

Both are true. Note that $$\lim_{\circ\to\infty}\left(1+\frac{1}{\circ}\right)^\circ=e$$ and that $t^2\to\infty$ as $t\to\pm\infty$.


3

$$\frac{\ln(1+x)}x=1-\frac x2+\frac{x^2}3-\frac{x^3}4+o(x^3)$$ Substitute $-\frac x2+\frac{x^2}3-\frac{x^3}4$ to $u$ in the development of $\mathrm e^u$, first computing the succesive powers of $u$: \begin{align*} u^2&=\frac{x^2}4-\frac{x^3}3+o(x^3),\\ u^3&-\frac{x^3}8+o(x^3), \end{align*} so that $$(1+x)^{\tfrac 1x}=\mathrm ...



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