# Tag Info

8

If such a function exists, then you may add any function whose integral between $0$ and $1$ is zero. Now you should find one particular solution. For this, a constant (with respect to $x$) function suffices.

3

solutions you found are not correct. Correct solutions are $y\le -1$ and $y\ge 2$ but first solution gives no value of $x$ becuase $y$ is always positive. So now $3^x\ge 2$ gives $x\ge log_32$ which is the final answer. Hope this helps !

3

Substitute $t = e^x \implies e^{2x} = (e^x)^2 = t^2$, which translates the expression to $\dfrac{t^2 - 1}{t-1} = \dfrac{(t-1)(t+1)}{(t-1)}$. Then what is the result when replacing $t = e^x$?

2

The first derivative equation is $$k_1(x+1)e^x-k_2e^{k_2x}-k_3=0$$ and the second one, $$k_1(x+2)e^x-k_2^2e^{k_2x}=0.$$ The solution(s) of the latter can be formulated in terms of the Lambert function $W$. https://en.wikipedia.org/wiki/Lambert_W_function#Examples The roots of the second derivative correspond to the extrema of the first derivative, which ...

2

In general, $~\displaystyle\int_0^\infty\exp\Big(-\sqrt[N]x\Big)~dx~=~N!~,~$ so even a relatively simple looking expression like $\displaystyle\int_0^\infty\exp\Big(-x^4\Big)~dx~=~\Big(\tfrac14\Big)!~=~\Gamma\bigg(1+\frac14\bigg)~=~\Gamma\bigg(\frac54\bigg)~$ cannot be expressed in terms of elementary functions, let alone a slightly more complex one, ...

2

Note: $(e^x-1)(e^x+1)=(e^x)^2+e^x-e^x-1=e^{2x}−1$ Also note that the denominator $e^x-1\ne0\implies e^x\ne1\implies x\ne0$ $$\frac{e^{2x}−1}{e^x−1}=\frac{(e^x-1)(e^x+1)}{e^x−1}=e^x+1 \text{ where } x\ne0$$

1

It is worth graphing these functions (and playing with the parameters). For $x>1$ they are both monotone a similar rate of growth to an exponential. The difference is over the range $[0,1]$, where $I_{\frac{1}{4}}$ is more like $x^{\frac{1}{4}}$ and $I_{\frac{-1}{4}}$ is more like $\frac{1}{x^{\frac{1}{4}}}$. But the main point I want to make is that our ...

1

I'm no mathematician, and there may be more concrete definitions, but this is how I think of a function as exponential "growth" and "decay." Theory If an exponential function is "skyrocketing" (for lack of better terminology) and heads towards $\pm\infty$, then it's "growing" (you can think of it as "absolute" growth, and disregard the sign). If an ...

1

You can algebraically rearrange this equation for hours and still get nowhere, because we cannot use algebra to solve it. Frustrating, isn't it? In short, you are better off letting a computer solve this for you via the Lambert W-Function (Wolfram Alpha will do this). Alternatively, you can graph the two sides of the equation and find the approximate ...

1

Answer: $e^{i m \phi}$ is the same thing as $e^{i x}$ when $x = m \phi$. Note that $i m \phi$ stands for the product $i \, m \, \phi$ and not the imaginary part of $\phi$ here. This should be interpreted as a family of periodic functions $e^{i m \phi}$ and $e^{-i m \phi}$ indexed by all non-negative integers $m = 0, 1, 2, \dots$.

1

Note that if $u=3^x$, then $u^2=3^{2x}$. So in terms of $u$ we have $u^2-u\ge2$ or $u^2-u-2\ge0$. You can factor the quadratic polynomial $u^2-u-2$ to get a solution for $u$, then take logarithms to get it in terms of $x$.

1

When you have $-1 > y$ or $y > 2$, replace $y$ with $3^x$, it gives you (remembering that $e > 0$): $$e^{x\ln(3)}>2$$ Which yields to: $$x > \dfrac{\ln(2)}{\ln(3)}$$

1

You know that the basic exponential growth/decay equation is $$A=A_0e^{rt}$$ You are told that when $t=120$ that $$A=\tfrac{1}{2}A_0e^{120r}$$ which you solved correctly for $r$. Now you wish to know the value of $t$ which causes $$0.60A_0=A_0e^{-0.005762265\,t}$$ so ...

1

The Taylor expansion of $\exp\sin x$ around zero is $1+x+x^{2}/2+O(x^{4})$. Therefore, the error is \begin{align*} \left|\exp\sin x-(1+x+x^{2}+x^{3})\right| & =\left|-x^{2}/2+O(x^{3})\right|\\ & \leq|x^{2}|/2+|O(x^{3})|\\ & \approx|x^{2}|/2 & \text{for }|x|\text{ small}. \end{align*}

1

I assume that $y_1$ means $y'$ and $y_2$ means $y''$. So differentiating we get $y'=-2y\frac{1}{\sqrt{1-x^2}}$. Differentiating again we get $y''=-2xy\frac{1}{(1-x^2)^{3/2}}+4y\frac{1}{1-x^2}$. Hence $$(1-x^2)y''-xy'=-2xy\frac{1}{\sqrt{1-x^2}}+4y+2xy\frac{1}{\sqrt{1-x^2}}=4y$$ So the equation holds with $\lambda=4$. Note that this does not agree with the ...

1

On differentiating; $y'=$$y\frac{-2}{√(1-x^2)} which on squaring gives, (1-x^2)y^2=4y^2. Differentiate again and after cancellation of 2y' on both sides, you get (1-x^2)y''-xy'-4y=0 1 The Laurent series at 0 is defined with the help of the Bernoulli numbers. 1 If one knows the following Taylor series expansion, as u \to 0,$$ \log(1+u)=u+O(u^2) $$then one may write, for any fixed real number t, as n \to \infty,$$ \left( 1+\frac{t}{n} \right)^{n}=e^{\large n\log\left(1+\frac{t}{n}\right)}=e^{\large n\left(\frac{t}{n}+O\left(\frac{1}{n^2}\right)\right)}=e^{t+O\left(\frac{1}{n}\right)} $$which gives$$ \... 1$2^x + 4^x= 2\Rightarrow2^x (1 + 2^x ) = 2\Rightarrow 1 + 2^x = 2 ^{1 - x}\Rightarrow1 + 2^x = 2 ^{- x} \times 2$Now Set$ y = 2^x$; then we have$1 + y = y^{-1} \times 2 \Rightarrowy^2 + y -2 = 0$. Which has solutions$y = 1$and$y = -2$.$y = -2 $is unacceptable, because$y = 2^x$is a positive function. So$y = 2^x = 0\$ is ...

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