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calculus
|
The Limit Comparison Test V1
|
https://math.stackexchange.com/questions/2089685/the-limit-comparison-test-v1
|
<p>I'm currently learning Calculus 2, more specifically I'm learning about sequences and series. I'm not enjoying this section as much as I thought I would, this is because I'm having to learn all these different tests to determine the convergence and being shown no justification as to why it works. I've been shown the proofs, but the proofs are not justifying to my mind why they even work. </p>
<p><strong>Limit Comparison Test:</strong> </p>
<p>Suppose that we have two series $\displaystyle\sum a_n $ and $\displaystyle\sum b_n$ with $a_n\geq0$,$b_n>0 $ $\forall n$. Define, $$ c = \displaystyle\lim_{n\to \infty} \frac{a_n}{b_n} $$ if $c$ is positive (i.e. $c>0$) and is finite (i.e. $c<\infty$) then either both series converge or both series diverge. </p>
<p><em>The first question I'd like to ask is, why does this even work?</em> </p>
<p>The second question I'd like to ask is, under what conditions does this work? Why do I ask this? Consider the two following series: $ \displaystyle\sum_{n=1}^{\infty} \frac{1}{n^3}$ and $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2}$.Both are p-series and a p-series converges when $p>1$ and diverges when $p\leq1$. Therefore, the two series above converges. Trying to verify this with the limit comparison test would go something like this $$ \displaystyle\lim_{n\to \infty} \frac{n^2}{n^3} = \displaystyle\lim_{n\to \infty} \frac{1}{n} =0$$ $c\not>0$ which is implying that both series don't converge. So, what is going on?</p>
<p>One small final question, I'd really like to improve in this part of the course and be in a position where I don't have to remember all these annoying tests and just be able to derive certain things from logic. Would this be too much to hope for considering I'm only doing a Calculus course and not something like Real-Analysis. </p>
|
<p>The limit comparison test is very powerful. One of my favorite applications is this: Determine the convergence/divergence of
$$\sum_{n=1}^\infty \frac 1{n^{1+1/n}}.$$</p>
<p>The heuristic is very simple. For large $n$, we're saying that basically $a_n = cb_n$ (for some positive number $c$). Ignoring small values of $n$, then $\sum a_n = c\sum b_n$, so obviously both series converge together or diverge together. The rigorous proof is only slightly more intricate, sandwiching $a_n$ between $c'b_n$ and $c''b_n$ for $0<c'<c<c''$.</p>
<p>Your logic in your second question is flawed. We're not saying things are "if and only if." We're saying that <em>provided</em> $\lim a_n/b_n = c$ <em>and</em> $0<c<\infty$, we can make an inference. We (ostensibly) know nothing if $c=0$ or $c=\infty$.</p>
<p>You should try to develop intuition based on this sort of comparison. Have a short list of series you know are convergent and divergent. Then try to say to yourself, "When $n$ is large, what do these terms look like?" (I.e., what are a convenient $b_n$ and what is the $c$?) Try this out with the one I gave at the outset.</p>
| 200
|
calculus
|
Can I find the surface area for cone by the surface area formula?
|
https://math.stackexchange.com/questions/4165567/can-i-find-the-surface-area-for-cone-by-the-surface-area-formula
|
<p>From section 8.2 in Stewart's calculus, I think I understand the derivation of the surface area formula <span class="math-container">$\int_{a}^{b} 2 \pi y \sqrt{1 +y'} dx$</span>. It's developed from the surface area of frustums, which is developed from the surface area of a cone, which was found by a sector's area.</p>
<p>So, my question is, can I use that surface area formula to find the surface area of a cone by calculus (rather than how I previously found it by a sector's area) -- or is it logically wrong since I'm already using what I want to find?</p>
<p>I did try it and it did work (using <span class="math-container">$y=rx/h$</span>) and so on, but I can't help but feel it's circular reasoning. I don't need help in the computational aspects, rather, I need help to the question "is if okay to apply the surface area formula to the cone (when said surface area formula depends on a cone's surface area)" Or, is it okay in this instance since the formulas for cone's surface area were derived differently?</p>
<p>--</p>
<p>Using specifics, the surface area formula seems based upon frustums with surface areas <span class="math-container">$\pi (r_1 + r_2) \ell$</span> or <span class="math-container">$2\pi r \ell$</span> where <span class="math-container">$r$</span> is the average radius <span class="math-container">$r=\frac{1}{2}(r_1+r_2)$</span>.</p>
<p>Now is it okay to use the line <span class="math-container">$y=\frac{r}{h}x$</span> in the surface area formula <span class="math-container">$$\int_{0}^{h} 2\pi \frac{r}{h}x\sqrt{1+\left(\frac{r}{h}\right)^2} dx $$</span> to prove a cone's surface area is <span class="math-container">$\pi r \ell$</span>? Since I find a formula almost identical to what created the surface area formula (like if <span class="math-container">$r_1=0$</span>), I'm wondering if this method of finding a cone's surface area is logically valid or am I being circular in my reasoning?</p>
| 201
|
|
calculus
|
Problem about Conservative Vector
|
https://math.stackexchange.com/questions/4093830/problem-about-conservative-vector
|
<p>Let <span class="math-container">$F:\mathbb R^n\to \mathbb R^n$</span> where <span class="math-container">$F(x_1,x_2,...,x_n)=(x_1,x_2,...,x_n)$</span>. show that <span class="math-container">$F$</span> is a conservative vector fields, that is, there is a potential function <span class="math-container">$f$</span> where <span class="math-container">$\nabla f = F$</span> .</p>
|
<p><span class="math-container">$f(x_1,x_2,\ldots,x_n)=\frac{1}{2}(x_1^2+x_2^2+\ldots+x_n^2)$</span></p>
| 202
|
calculus
|
Proving the Monotonicity of a function?
|
https://math.stackexchange.com/questions/850467/proving-the-monotonicity-of-a-function
|
<p>Given a function of numerous variables, say $f(x,y,z)$, what are the usual approaches one can take to prove that $f(x,y,z)$ is monotonically increasing, or decreasing in $x$?</p>
<p>I am aware that one can calculate the functions derivative and attempt to prove that it is positive or negative for any $y$ and any $z$. However sometimes the derivatives are quite complex, and challenging to analyze.</p>
<p>Is there any other way to prove the monotonicity of a function?</p>
|
<p>It depends on how your function is defined. Many times proving directly is the easiest. </p>
<p>i.e. Arbitrarily fix $x_1 < x_2$ and $y_{*}, z_{*}$ in your spaces $X$, $Y$, $Z$ that compose the domain of your function $X\times Y\times Z$. Show that $f(x_{1},y_{*},z_{*})<f(x_{2},y_{*}, z_{*})$. (same as what Jika said)</p>
<p>However, sometimes functions are defined in weird ways or they have many terms and it is easier to do something like this:</p>
<p>"Show that for any $x_{o}\in\mathbb{R}$, the function $$g_{x_{o}}(x)=f(x,y_{*},z_{*})-f(x_{o},y^{*},z^{*})$$ is only positive on $(x_{o}, \infty)$."</p>
<p>This method is particularly nice if it turns out that when you subtract the functions, you get something readily recognizable, like the power series expansion for a trig function—in which case, you might know something about whether $g_{x_o}$ is positive on $(x_{o}, \infty)$.</p>
| 203
|
calculus
|
$\lim_{h \to 0} \int_{x}^{x+h} \ln(t) dt$
|
https://math.stackexchange.com/questions/3463830/lim-h-to-0-int-xxh-lnt-dt
|
<p><span class="math-container">$\lim_{h \to 0} \int_{x}^{x+h} \ln(t) dt$</span></p>
<p>Unless I'm missing something, isn't this just <span class="math-container">$0$</span> due to how the integral is just <span class="math-container">$\int_{x}^{x}=0$</span> </p>
<p>I'm sure I could integrate the inside and then evaluate that as <span class="math-container">$h \to 0$</span>, but this answer seems to deceptively easy if correct</p>
|
<p>Yes, you're correct. You can argue it that way, or even if you go as far as integrating first you'll find the same result:</p>
<p><span class="math-container">\begin{eqnarray*}
\lim_{h\to 0} \int_x^{x+h}\ln(t)dt & = & \lim_{h\to 0} \left . t\ln(t) - t \right |_x^{x+h} \\
& = & \lim_{h\to 0}(x+h)\ln(x+h) - (x+h) - x\ln(x) + x \\
& = & \lim_{h\to 0} x \ln\left (\frac{x+h}{x}\right ) + h\ln(x+h) - h \\
& = & x \ln(1) + 0 - 0 \;\; =\;\; 0.
\end{eqnarray*}</span></p>
| 204
|
calculus
|
Functional Gaussian Integral
|
https://math.stackexchange.com/questions/4169820/functional-gaussian-integral
|
<p>I am trying to reproduce the result below</p>
<p><span class="math-container">$$\int\mathcal{D}V~e^{-\int dx~[a V^{2}(x)+iV(x)\int_{-\infty}^{\infty}dt~ \bar{\psi}^{a}(x,t)\gamma_{ab}\psi^{b}(x,t)]}= e^{-\frac{1}{4a}\int dx\int\int_{-\infty}^{\infty}dt dt'(\bar{\psi}^{a}(x,t)\psi^{a}(x,t))(\bar{\psi}^{b}(x,t')\psi^{b}(x,t'))}$$</span>
Where <span class="math-container">$\gamma_{ab}$</span> is a <span class="math-container">$2\times 2$</span> identity matrix and <span class="math-container">$\psi\bar{\psi}=-\bar{\psi}\psi$</span> are Grassman variables and <span class="math-container">$a$</span> is a constant.</p>
<p>I tried to reproduce it but could not succeed. Any comments or suggestions are welcome. I am new in this site, I would request not to close or downvote this question.</p>
<p>Here is my attempt:
Let <span class="math-container">$f(x)=\int_{-\infty}^{\infty}dt~\bar{\psi}^{a}(x,t)\gamma_{ab}\psi^{b}(x,t)$</span>. Then we can write</p>
<p><span class="math-container">$$\int\mathcal{D}V~e^{-\int dx~[a V^{2}(x)+iV(x)\int_{-\infty}^{\infty}dt~ \bar{\psi}^{a}(x,t)\gamma_{ab}\psi^{b}(x,t)]} = \int\mathcal{D}V~e^{-\int dx~[a V^{2}(x)+iV(x)f(x)]}$$</span>
Completing the square gives, <span class="math-container">$[aV^{2}(x)+iV(x)f(x)]=a(V+\frac{if}{2a})^{2}+\frac{f^{2}}{4a}$</span>. Then
<span class="math-container">$$\int\mathcal{D}V~e^{-\int dx~[a V^{2}(x)+iV(x)f(x)}=e^{-\frac{f^{2}}{4a}}\int\mathcal{D}V~e^{-\int~dr V^{2}(r)dr}
$$</span>
I got the factor <span class="math-container">$e^{-\frac{f^{2}}{4a}}$</span> and the integral <span class="math-container">$\int\mathcal{D}V~e^{-\int~dr V^{2}(r)dr}$</span> gives a constant contribution.</p>
| 205
|
|
calculus
|
Apply function fractional times
|
https://math.stackexchange.com/questions/1250978/apply-function-fractional-times
|
<p>For example, one can apply $\cos x$ to number $a$ one time to get $\cos a$, two times $\cos \cos a$, three times $\cos \cos \cos a$, and so on. Is there a way to define fractional application for $\cos$? Or for any other function? Maybe exists general theory for that?</p>
|
<p>Obviously $f(x,n)$ defined as taking $g$ $n$ times of $x$ is a function $(\mathbb R,\mathbb N)\to\mathbb R$. Any extension of $f$ to $(\mathbb R, \mathbb Q)$, could be considered a way to apply $g$ a rational number of times, if you extend it to $\mathbb R^2$, you could consider it a definition of applying $g$ any (real) number of times.</p>
<p>Extending functions to a larger domain is often done, a very well-known example being the factorial function being extended from $\mathbb N$ to $\mathbb C$ by the gamma function. But considering it applying a function a rational/real/complex/... number of times isn't so common, there's rarely any thing to be gained from viewing it that way.</p>
<p>In the specific case of $\cos$, I don't recall ever seeing such an extension, but we could define one, by saying that for any non-natural number of applications the result is $0$. It's not interesting but now we have a way to apply $\cos$ any number of times we want.</p>
| 206
|
calculus
|
Fourier Series of $\cos^{n}x$
|
https://math.stackexchange.com/questions/627718/fourier-series-of-cosnx
|
<p>I need help evaluating the integrals in Fourier Series.</p>
<p>For example, for the function <span class="math-container">$\cos^{2}x$</span>, I can evaluate <span class="math-container">$a_0$</span>, <span class="math-container">$a_n$</span>, and <span class="math-container">$b_n$</span>, where <span class="math-container">$a_n$</span> is the coefficients of the cosine terms and <span class="math-container">$b_n$</span> the coefficients of the sine terms. In this case, because it is an even function, only cosine terms will exist, and the integral for calculating it will become:</p>
<p><span class="math-container">$$a_n=(1/2 {\pi})\int_{-\pi}^{\pi} cos^2xcosnx dx$$</span></p>
<p>How can this be solved?</p>
<p>Similarly, could someone please show the step-by-step calculation for the Fourier Series of <span class="math-container">$\cos^nx$</span>?</p>
|
<p>Hint:</p>
<p>$\cos^{2}(x)=\frac{1+\cos(2x)}{2}$ </p>
<p>and </p>
<p>$\cos(x)\cos(y)=\frac{1}{2}\big(\cos(x-y)-\cos(x+y)\big)$</p>
| 207
|
calculus
|
The Fundamental Theorem of Calculus Questions?
|
https://math.stackexchange.com/questions/4115061/the-fundamental-theorem-of-calculus-questions
|
<p><strong>Background Information:</strong>
One of the most important ideas that Green discussed in his Essay is the connection between
what happens within a body and the properties of that body’s surface. He realized that, because the
boundary of an object is one dimension lower than the interior, the connection can be represented by
transforming k−dimensional integrals to simpler integrals involving a (k−1)−dimensional object. In
fact, the origin of Green’s Theorem goes to the very heart of Calculus: the Fundamental Theorem.</p>
<p><strong>The Question:</strong>
Write out the Fundamental Theorem of Calculus. How might we see this as reducing a
k−dimensional integral to a (k − 1) dimensional integral?</p>
<p><strong>The work so Far:</strong>
The Fundamental Theorem of Calculus states that when taking the anti derivative of an integrand you can evaluate the integral by taking the endpoints of the integral and subtracting using the anti derivative. The formula for the Fundamental Theorem of Calculus is as follows: <span class="math-container">$$\int^{x=b}_{x=a} \, f(x) \, dx = F(b) - F(a).$$</span></p>
|
<p>Well, we know that the boundary of <span class="math-container">$[a,b]$</span> is <span class="math-container">$\{a,b\}$</span>. Then, in some sense,
<span class="math-container">$$\int_a^b f = \int_{\partial[a,b]} F = \int_{\{a,b\}} F.$$</span>
Put another way,
<span class="math-container">$$\int_{[a,b]} \partial f = \int_{\partial[a,b]} f.$$</span>
In particular, this is true if we define <span class="math-container">$\int_{\{a,b\}} f := f(b) - f(a)$</span>. Such a definition is reasonable if you think about what an integral represents: the area (2d concept) under a curve (1d) vs the distance (like length, the 1d analogue of area) between points (0d). In this sense, the FTOC is just a specific instance of Stokes' theorem:
<span class="math-container">$$\int_{\partial\Omega} \omega = \int_\Omega d\omega.$$</span></p>
| 208
|
calculus
|
What is the $\lim_{h\to0}$ of the average value of $f(x)$ on the interval $[x, x+h]$
|
https://math.stackexchange.com/questions/2121801/what-is-the-lim-h-to0-of-the-average-value-of-fx-on-the-interval-x-x
|
<p>If $f(x)$ is a continuous function on the interval $[x, x+h]$, find $$\lim_{h\to 0} f(x)_{avg}$$ </p>
<p>I suspect I'm using the limit definition of the derivative, and to obtain the average value I've integrated over $[{x, x+ h}]$: $$\frac{\int_x^{x+h}f(x + h) - \int_x^{x+h}f(x )}{(x+h) - x} $$</p>
<p>What is the next step, or what have I done incorrectly? I'm not sure what the goal is.</p>
|
<p>I will note you did the first step wrong. Note that you should have</p>
<p>$$g(x)=\int_a^xf(t)\ dt$$</p>
<p>Differentiating:</p>
<p>$$g'(x)=\lim_{h\to0}\frac{\int_a^{x+h}f(t)\ dt-\int_a^xf(t)\ dt}h=\lim_{h\to0}\frac1h\int_x^{x+h}f(t)\ dt$$</p>
<p>If we suppose the following statement, perhaps as axiom, that $\int_x^{x+h}f(t)\ dt=\int_x^{x+h}f_{avg}(x,x+h)\ dt$ where $f_{avg}(a,b)$ is the average value of $f$ for $a\le x\le b$, then,</p>
<p>$$g'(x)=\lim_{h\to0}\frac1h\int_x^{x+h}f_{avg}(x,x+h)\ dt=\lim_{h\to0}\frac1hf_{avg}(x,x+h)\int_x^{x+h}1\ dt\\=\lim_{h\to0}\frac1hf_{avg}(x,x+h)\cdot h$$</p>
<p>This integral is a trivial one that may be geometrically solved for, and from here, we see the $h$'s cancel, leaving us with</p>
<p>$$g'(x)=\lim_{h\to0}f_{avg}(x,x+h)\stackrel?=f_{avg}(x,x)=f(x)$$</p>
<p>which I suppose is what you want. Somewhere along the line though, we need continuity, which I suppose could come through with a limit property:</p>
<p>if $f(x)$ is continuous, then $\lim f(g(x))=f(\lim g(x))$.</p>
| 209
|
calculus
|
Meaning of $\int_{a}^{b}dx$
|
https://math.stackexchange.com/questions/4164659/meaning-of-int-abdx
|
<p>I know the meaning of <span class="math-container">$\int_{a}^{b}f(x)dx$</span>, which is <span class="math-container">$F(b)-F(a)$</span>. Geometrically, it gives us the area under the graph from <span class="math-container">$x=a$</span> to <span class="math-container">$x=b$</span></p>
<p>But what does <span class="math-container">$\int_{a}^{b}dx$</span> mean mathematically with respect to <span class="math-container">$f(b)-f(a)$</span> and what is its geometrical meaning?</p>
|
<p><span class="math-container">$\int_a^bdx$</span> would simply be the same as <span class="math-container">$\int_a^b 1\ dx$</span>, so it would be the area under the curve <span class="math-container">$f(x)=1$</span>, which is indeed equal to the length of the segment from <span class="math-container">$a$</span> to <span class="math-container">$b$</span>, i.e., <span class="math-container">$b-a$</span>.</p>
| 210
|
calculus
|
Riemann sum problem.
|
https://math.stackexchange.com/questions/2985141/riemann-sum-problem
|
<p>I had a practice midterm that had the following question: </p>
<p><span class="math-container">$A = \lim_{x \to\infty} R_n = \lim_{x \to\infty} (\sum_{i=1}^{n} f(x_i)\triangle x)$</span></p>
<p>Use this definition to find an expression for the area under the graph of <span class="math-container">$f(x) = \frac{log x}{x^2}$</span></p>
<p>So what I did: </p>
<p><span class="math-container">$\triangle x = \frac{2}{n}$</span> </p>
<p><span class="math-container">$R_n = \lim_{x \to\infty} \sum_{i=1}^{n} (\frac{log(1+\frac{2i}{n})}{(1+\frac{2i}{n})^2})*\frac{2}{n}$</span></p>
<p>According to any of the work I have done with practice questions this was the same approach I applied to all of them and got the right answer. </p>
<p>Now, in the answers version of the practice exam this was the answer: </p>
<p><span class="math-container">$R_n = \lim_{x \to\infty} \sum_{i=1}^{n} (2in*\frac{log(n+2i)-log(n)}{(n+2i)^2})$</span></p>
<p>How is he obtaining this final answer? </p>
|
<p>I get <span class="math-container">$\lim_{n\to\infty}\sum_{i=1}^n 2n\cdot \frac{\log(n+2i)-\log n}{(n+2i)^2}$</span>.</p>
<p>This is straight forward using <span class="math-container">$\log\frac ab=\log a-\log b$</span>, plus a little algebra. </p>
| 211
|
calculus
|
Derivative using Fundamental Theorem of Calculus when integrand has product of two functions?
|
https://math.stackexchange.com/questions/4174763/derivative-using-fundamental-theorem-of-calculus-when-integrand-has-product-of-t
|
<p>I want to find the derivative of the following:</p>
<p><span class="math-container">$$exp \left( -\int_{t-\tau(t)}^t \frac{\mu(x)U(x)}{S} \,dx \right)$$</span></p>
<p>I tried to use the Fundamental theorem of calculus of the form:</p>
<p><span class="math-container">$$\frac{d}{dx}\int_0^x t^3 \,dx = f(x)\frac{dx}{dx} - f(0)\frac{d0}{dx} = x^3$$</span></p>
<p>(from Wikipedia) and I got <em>something</em>, but I'm not confident I followed all the rules correctly. I know to start with the chain rule to deal with the <span class="math-container">$exp()$</span>, but then finding the derivative of the integral is mixing me up. My question for this part is, is it correct to start like this:</p>
<p><span class="math-container">$$-\left(\frac{\mu(t)U(t)}{S}\frac{dt}{dt} - \frac{\mu(t-\tau(t))U(t-\tau(t))}{S}\frac{d(t-\tau(t))}{dt}\right)$$</span></p>
<p>or do I need to deal with the product of functions first before applying the FTC (e.g., via substitution or integrating by parts).</p>
<p>Sorry if this is a silly question. It's been a long time since I took calculus and I couldn't find a similar example online.</p>
<p>Thanks!</p>
|
<p>As you say, the <span class="math-container">$\exp$</span> part is straightforward, so let's look at the derivative of
<span class="math-container">$$
Q = \int_{t-\tau(t)}^t \frac{\mu(x)U(x)}{S} \,dx.
$$</span></p>
<p>A standard thing to do here is to write this as a sum of two integrals, splitting at some arbitrary (but fixed) point <span class="math-container">$b$</span>:
<span class="math-container">\begin{align}
Q
&= \int_{t-\tau(t)}^t \frac{\mu(x)U(x)}{S} \,dx\\
&= \int_{t-\tau(t)}^b \frac{\mu(x)U(x)}{S} \,dx
+ \int_b^t \frac{\mu(x)U(x)}{S} \,dx\\
&= -\int_b^{t-\tau(t)} \frac{\mu(x)U(x)}{S} \,dx
+ \int_b^t \frac{\mu(x)U(x)}{S} \,dx\\
\end{align}</span>
Now you've got two terms, each of which is amenable to taking derivatives. For the first, you get the derivative
<span class="math-container">$$
-\frac{\mu(t-v(t))U(t-v(t))}{S}\left(1 - v'(t) \right).
$$</span>
where that final factor comes from applying the chain rule. For the second, you get just
<span class="math-container">$$
\frac{\mu(t)U(t)}{S}.
$$</span></p>
<p>And that's the end of the story.</p>
<p>To answer the question you asked, though, suppose we define
<span class="math-container">$$
H(x) = \frac{\mu(x)U(x)}{S}
$$</span>
Then you have
<span class="math-container">$$
Q = \int_{t-v(t)}^t H(x) dx
$$</span>
to which you can apply the FTC, and the derivative involves a difference of expressions involving <span class="math-container">$H$</span>, which you can then expand back out --- no need to integrate a product at all.</p>
| 212
|
calculus
|
Solving for $i$, given $S=\sum_{n=1}^m \frac{A_n}{(1+i)^{t_n}}$
|
https://math.stackexchange.com/questions/4174327/solving-for-i-given-s-sum-n-1m-fraca-n1it-n
|
<p>I'm faced with a problem that is unfortunately beyond my current mathematical skills.</p>
<p>I have an equation that goes like this:</p>
<p><span class="math-container">$$
S=\sum_{n=1}^m \frac{A_n}{(1+i)^{t_n}}
$$</span></p>
<p>My goal is to transform it so that I arrive at formula to calculate <code>i</code>. I wish I could provide my "research", but I have none as it was limited to trying to use online software in order to transform this. Otherwise it is way beyond what I'm capable of so the furthest I got is to write it down on paper.</p>
<p>EDIT</p>
<p>Since there were a couple of absolutely valid questions I thought a bit of context would be helpful. It's a formula for annual percentage rate as defined by FCA (UK's financial regulatory body). And so:</p>
<ul>
<li>S is the total "payable" amount</li>
<li>n is the payment sequence</li>
<li>A is the payment amount (<span class="math-container">$A_n$</span> is the payment amount for the specific payment)</li>
<li>t is the "time period" measured in years (<span class="math-container">$t_n$</span> is the time between the specific payment date and the beginning of the commitment, e.g. for 6th payment when on a monthly basis it's going to equal 0.5, for 18th it's going to be 1.5)</li>
</ul>
|
<p>As said in comments, solving for <span class="math-container">$i$</span> the equation <span class="math-container">$$S=\sum_{n=1}^m \frac{A_n}{(1+i)^{t_n}}$$</span> will require numerical methods.</p>
<p>However, since <span class="math-container">$i \ll 1$</span>, we can try to obtain <em>approximations</em>.</p>
<p>We have, by Taylor expansion or the binomial theorem
<span class="math-container">$$\frac{1}{(1+i)^{t_n}}=\sum_{k=0}^\infty \binom{-t_n}{k}\,i^k$$</span> So, truncating after the fourth power for example and summing over <span class="math-container">$n$</span>
<span class="math-container">$$S= a_0+a_1\,i+a_2\,i^2+a_3\,i^3+a_4\,i^4+O\left(i^5\right) $$</span></p>
<p>Using for easier notations <span class="math-container">$$\Sigma_k=\sum_{n=1}^m A_n\, t_n^k$$</span></p>
<p><span class="math-container">$$a_0=\Sigma_0\qquad a_1=-\Sigma_1\qquad a_2=\frac 1{2!}\left(\Sigma_1+\Sigma_2 \right)\qquad a_3=-\frac 1{3!}\left(2\Sigma_1+3\Sigma_2+\Sigma_3 \right)$$</span>
<span class="math-container">$$a_4=\frac 1{4!}\left(\Sigma_1+11 \Sigma_2+6 \Sigma_3+\Sigma_4\right)$$</span>
Now, using series reversion, the <em>approximation</em>
<span class="math-container">$$i_{(4)}=x-\frac{a_2 }{a_1}x^2+\frac{2 a_2^2-a_1 a_3}{a_1^2}x^3+\frac{-5 a_2^3+5 a_1 a_2 a_3-a_1^2 a_4}{a_1^3}x^4+
O\left(x^5\right)\qquad\text{with}\qquad x=\frac{S-a_0}{a_1 }$$</span></p>
<p>For a first illustration, using
<span class="math-container">$$m=6 \qquad A_n=1234+56n \qquad t_n=p_n \qquad S=7000$$</span>this would give <span class="math-container">$i_{(4)}=0.0301004$</span> while the solution is <span class="math-container">$i=0.0301207$</span>.</p>
<p>Try it with some data of yours; if it is more or less acceptable, we could improve it using mor terms for a better and better accuracy.</p>
<p><strong>Edit</strong></p>
<p>As said in comments, if you want a very accurate solution, use Newton method with <span class="math-container">$i_0=\frac{S-a_0}{a_1 }$</span>. For the worked example (and a ridiculous number of figures), the iterates will be
<span class="math-container">$$\left(
\begin{array}{cc}
k & I_k \\
0 & 0.025945022825051725837 \\
1 & 0.030036719253941629330 \\
2 & 0.030120651266811630596 \\
3 & 0.030120685453289100601 \\
4 & 0.030120685453294768411
\end{array}
\right)$$</span></p>
| 213
|
calculus
|
Why do we use only the positive root when differentiating an inverse trig function whose inside is linear?
|
https://math.stackexchange.com/questions/4171429/why-do-we-use-only-the-positive-root-when-differentiating-an-inverse-trig-functi
|
<p>I'm taking calc 1, and I'm struggling with these types of problems. Example: differentiate <span class="math-container">$y=\sin^{-1}(-4x-1)$</span></p>
<p>I think I understand how to solve these problems, but my answers typically have <span class="math-container">$\pm$</span> roots, like in this example: <span class="math-container">${y}'=\frac{-4}{\pm \sqrt{-16x^{2}-8x}}$</span></p>
<p>However, I'm taking this course through Outlier and their system always marks this wrong. They want only the <em>positive</em> root in the denominator. i.e. <span class="math-container">${y}'=\frac{-4}{\sqrt{-16x^{2}-8x}}$</span></p>
<p>Why is this - is this just an oversight in their system, or am I missing some logic here that lets me eliminate the negative root as a possible answer?</p>
<p>Note: domain is all real numbers</p>
<p><strong>Here is my workflow:</strong></p>
<p>First, invert the function...</p>
<p><span class="math-container">${y}=\sin^{-1}(-4x-1)\Rightarrow-4x-1=\sin y$</span></p>
<p>Then, differentiate the equality...</p>
<p><span class="math-container">${\frac{\mathrm{d} }{\mathrm{d} x}}[-4x-1]=-4$</span></p>
<p><span class="math-container">${\frac{\mathrm{d} }{\mathrm{d} x}}[\sin y]=\cos y(y')$</span></p>
<p>Then, use trig identities to get terms we can substitute to get terms of <span class="math-container">$x$</span></p>
<p><span class="math-container">$\sin^{2} y+\cos^{2} y=1$</span></p>
<p><span class="math-container">$\cos^{2} y=1-\sin^{2} y$</span></p>
<p><span class="math-container">$\cos y=\pm \sqrt{1-\sin^{2} y}$</span></p>
<p>Finally, plug it all back in to the expression and simplify terms to get the answer...</p>
<p><span class="math-container">${\frac{\mathrm{d} }{\mathrm{d} x}}=\frac{-4}{\pm \sqrt{-16x^{2}-8x}}$</span></p>
| 214
|
|
calculus
|
decomposing a fraction into partial fractions
|
https://math.stackexchange.com/questions/686382/decomposing-a-fraction-into-partial-fractions
|
<p>could someone please help me to decompose the following fraction into partial fractions?</p>
<p>$$\frac{1}{(a-x)(b-x)^{1/2}}$$</p>
<p>where a and b are just constants.</p>
<p>Thanks</p>
|
<p>Usually, a partial fraction decomposition is only possible for rational functions. The square root inside the denominator would prevent this kind of decomposition. In</p>
<p>$$\frac{a-b}{(x-a)\sqrt{x-b}}=\sqrt{x-b}\frac{(x-b)-(x-a)}{(x-a)(x-b)}$$</p>
<p>for $x>b$ one can decompose the second factor, but the square root remains.</p>
| 215
|
calculus
|
Multiplying top and bottom by $ \cos (x) $ to solve integral?
|
https://math.stackexchange.com/questions/4140640/multiplying-top-and-bottom-by-cos-x-to-solve-integral
|
<p>Please take a look at this integral. Why is this method not a valid way of solving this integral?</p>
<p><span class="math-container">$\displaystyle \int \frac{1}{\sin (x) \cos(x)} \ dx = \int \frac{\cos (x)}{\sin (x) \cos^2(x)} \ dx = \int \frac{\cos(x)}{\sin (x) (1-\sin^2 (x))} \ dx = \int \frac{1}{u(1-u^2)} \ du$</span></p>
<p>And by utilizing partial fractions we arrive at</p>
<p><span class="math-container">$ \displaystyle \int \frac{1}{\sin (x) \cos(x)} \ dx = \ln(|\sin (x)|)+\frac{1}{2}\ln(|1-\sin(x)|)-\frac{1}{2}\ln(|1+\sin (x)|)$</span></p>
<p>I am guessing the incorrect step was multiplying top and bottom of the fraction <span class="math-container">$\frac{1}{\sin (x) \cos(x)}$</span> by <span class="math-container">$\cos x$</span>. But <strong>why</strong> is this step incorrect? I have not changed the value of the function since it was just a multiplication by 1 and in addition to that I do not believe that any domain issues arise or division by zero issues arise. As we have already started with a function that has both <span class="math-container">$\sin x$</span> and <span class="math-container">$\cos x$</span> in the dominator, we then assume this function would only be defined for values when <span class="math-container">$\sin(x) \not= 0 $</span> and <span class="math-container">$ \cos(x) \not = 0 $</span> therefore multiplying by <span class="math-container">$\frac{\cos (x)}{\cos(x)}$</span> shouldn't cause any further issues?</p>
|
<p>You have done the partial fractions incorrectly. It should be:</p>
<p><span class="math-container">$$\frac{1}{u(1-u^2)} = \frac{1}{u} - \frac{1}{2(u+1)} \color{red}{-} \frac{1}{2(u-1)}$$</span></p>
| 216
|
calculus
|
Trig substitution reversion issue $\pm$
|
https://math.stackexchange.com/questions/4140799/trig-substitution-reversion-issue-pm
|
<p>I am working through the 100 integrals video on YouTube and I came across this question. I solved it correctly, but I want some clarification on a step that I made.</p>
<p><span class="math-container">$$\displaystyle\int \frac{e^x\sqrt{e^x-1}}{e^x+3} \ dx$$</span></p>
<p><span class="math-container">$$ u = e^x +3 \implies \int\frac{\sqrt{u-4}}{u} \ du = 2\int \frac{\sqrt{\frac{u}{4}-1}}{u} \ du $$</span></p>
<p><span class="math-container">$$ u = 4\sec^2 (\theta) \implies 2\int \frac{\tan (\theta)\times8\sec^2 (\theta)\tan (\theta)}{4\sec^2 (\theta)} \ d\theta = 4 \int \tan^2 (\theta) \ d\theta = 4(\tan (\theta) - \theta)$$</span></p>
<p>I was now faced with the task of reverse substituting to get my answer into a function of <span class="math-container">$x$</span>. So in order to achieve that I did</p>
<p><span class="math-container">$$ u = 4\sec^2 (\theta) \implies u= 4(1+tan^2 (\theta)) \implies \tan (\theta) = \pm \sqrt{\frac{u-4}{4}}$$</span></p>
<p>This is where the issues in my mind started to arise. How do we decide on which value of <span class="math-container">$\tan (\theta)$</span> to take? When I solved this question, I decided to take the positive value of <span class="math-container">$tan(\theta)$</span> and arrived at the correct answer, but I couldn't come up with any sound mathematical reasoning as to why taking the positive value was the correct thing to do here.</p>
|
<p>Briefly, your original integrand is non-negative for all real <span class="math-container">$x \geq 0$</span>, so the antiderivative sought is increasing. Only the <span class="math-container">$+$</span> branch of square root gives an increasing function of <span class="math-container">$x$</span>.</p>
<p>In more detail, your work shows (including the prospective sign ambiguity and omitting the constants of integration)
<span class="math-container">\begin{align*}
\frac{e^{x} \sqrt{e^{x} - 1}}{e^{x} + 3}\, dx
&= \frac{\sqrt{u - 4}}{u}\, du\qquad u = e^{x} + 3 \\
&= 4\int \tan^{2}\theta\, d\theta\qquad u = 4\sec^{2}\theta \\
&= 4(\tan\theta - \theta) \\
&= 4\biggl[\pm\sqrt{\frac{u - 4}{4}} \mp \arctan\sqrt{\frac{u - 4}{4}}\biggr] \\
&= \pm4\biggl[\sqrt{\frac{e^{x} - 1}{4}} - \arctan\sqrt{\frac{e^{x} - 1}{4}}\biggr].
\end{align*}</span>
Since <span class="math-container">$v - \arctan v$</span> is an increasing function of <span class="math-container">$v$</span>, the expression in square brackets is increasing in <span class="math-container">$x$</span>. This imposes the plus sign.</p>
| 217
|
calculus
|
Hyperbolic Trig Proofs/Definitions
|
https://math.stackexchange.com/questions/1834088/hyperbolic-trig-proofs-definitions
|
<p>My first post! Hello World!</p>
<p>I was looking back at my notes from Calculus I & II (my how the time has passed!)
I came back across Hyperbolic Trig Functions, sinh, cosh, etc.</p>
<p>I remember being presented the identities, how to use them, derivatives, integrals, etc. I was wondering if anyone could provide me the proof or how these identities are derived, where exactly does e^x come in to play? I've done some searching and haven't come up with anything, so if an answer already exists can you point me in the right direction? No need to solve a problem that has already been solved!</p>
<p>Thank you and I look forward to asking/answering many questions through my Math school journey!</p>
<p>Update (6-22-16): I meant how are the identities obtained? Where does e^x come into play?</p>
|
<p>$\cosh(x)=\frac{e^x+e^{-x}}{2}$ </p>
<p>$\sinh(x)=\frac{e^x-e^{-x}}{2}$</p>
<p>by definition. Most other identities follow from basic calculus. </p>
| 218
|
calculus
|
How does $e^{-\ln x} = e^{\ln(1/x)}$
|
https://math.stackexchange.com/questions/4159565/how-does-e-ln-x-e-ln1-x
|
<p>I understand the inverse of e^{x} is the natural logarithm. However I don't understand how the following expression is true:</p>
<p><span class="math-container">$e^{-\ln x} = e^{\ln(1/x)}$</span></p>
<p>Any assistance is appreciated.</p>
|
<p>One of the properties of logarithms is the following:</p>
<p><span class="math-container">$$\log({x^k}) = k\log{x}$$</span></p>
<p>Therefore when you have <span class="math-container">$-\ln x$</span>, you essentially go backwards:</p>
<p><span class="math-container">$$-\ln x = -1 \times \ln x = \ln(x^{-1}) = \ln \left( \frac{1}{x} \right) $$</span></p>
| 219
|
calculus
|
What is meant by $f(x)$ is function of $x$. Or $f(x)$ as a function of $y$?
|
https://math.stackexchange.com/questions/4180376/what-is-meant-by-fx-is-function-of-x-or-fx-as-a-function-of-y
|
<p>I am so confused about the terminology and vocabulary here. I tried googling it but couldn't find anything satisfactory. I have a test tomorrow. I would be glad if someone could explain what this conceptually means.</p>
|
<p>I'm guessing you're currently in high school so without beating around the bush and/or being pedantic and asking you for definitions (which is futile as it is obvious you have this question <em>because</em> you don't know your definitions in the first place) <em>and</em> since you have an exam coming up very soon, I'll give you a straight answer.</p>
<p><span class="math-container">$f(x)$</span> simply means the output value <strong>using the function</strong> <span class="math-container">$f$</span> when <span class="math-container">$x$</span> is the input value. For example: <span class="math-container">$f(x)=2x+3$</span> then for input value <span class="math-container">$x=2$</span>, the output value <strong>using the function</strong> <span class="math-container">$f$</span> is <span class="math-container">$f(2)=2(2)+3=7$</span>. You'd usually (conventionally) want to plot these output values of function <span class="math-container">$f$</span> on <span class="math-container">$Y$</span> axis and input values on <span class="math-container">$X$</span> axis on a graph paper. So we set <span class="math-container">$y = f(x)$</span>. And hence, now <span class="math-container">$y$</span> is simply another name for <span class="math-container">$f(x)$</span>.</p>
| 220
|
calculus
|
Why is $\lim_{\delta x\to0} \frac{\delta x}{\delta x} = 1$
|
https://math.stackexchange.com/questions/3118997/why-is-lim-delta-x-to0-frac-delta-x-delta-x-1
|
<p>Why is <span class="math-container">$\lim_{\delta x\to0} \frac{\delta x}{\delta x} = 1$</span>, considering that both are infinitesimally small but may be different from each other?</p>
<p>Also, if so, why can I not replace <span class="math-container">$\frac{\delta f}{\delta x} = \frac{\frac{1}{x + \delta x} - \frac{1}{x}}{\delta x}
= 1$</span> directly instead of having to reduce it first, since it already amounts to <span class="math-container">$\lim_{\delta x\to0} \frac{\delta x}{\delta x}$</span> ?</p>
|
<blockquote>
<p>Why is <span class="math-container">$\lim_{\delta x\to0} \frac{\delta x}{\delta x} = 1$</span>, considering that both are infinitesimally small but may be different from each other?</p>
</blockquote>
<p>No, they are never different from each other. <span class="math-container">$\delta x = \delta x$</span>. It is the same variable. Furthermore for <span class="math-container">$\delta x \neq 0$</span> we have <span class="math-container">$\frac{\delta x}{\delta x} = 1$</span>, hence you get <span class="math-container">$\lim_{\delta x\to 0} 1 = 1$</span>.</p>
<blockquote>
<p>Also, if so, why can I not replace <span class="math-container">$\frac{\delta f}{\delta x} = \frac{\frac{1}{x + \delta x} - \frac{1}{x}}{\delta x}
= 1$</span> directly instead of having to reduce it first, since it already amounts to <span class="math-container">$\lim_{\delta x\to0} \frac{\delta x}{\delta x}$</span> ?</p>
</blockquote>
<p>It looks like you are considering the derivative of <span class="math-container">$f(x)=\frac 1 x$</span>. Fixing <span class="math-container">$x$</span> you set <span class="math-container">$\delta f = f(x+\delta) - f(x)$</span> for every <span class="math-container">$\delta x\neq 0$</span>. Note that <span class="math-container">$\delta f$</span> and <span class="math-container">$\delta x$</span> are different things and hence the quotient is not just <span class="math-container">$1$</span>. The derivative is then
<span class="math-container">$$
f'(x) = \lim_{\delta x\to 0} \frac{\delta f}{\delta x} = \lim_{\delta x\to 0} \frac{\frac{1}{x+\delta x}-\frac{1}{x}}{\delta x},
$$</span>
which is not related to <span class="math-container">$\frac{\delta x}{\delta x}$</span> at all.</p>
| 221
|
calculus
|
Bound to $\sum_i^n \sqrt{a_i}$
|
https://math.stackexchange.com/questions/4180683/bound-to-sum-in-sqrta-i
|
<p>I am trying to find a bound to this: <span class="math-container">$\sum_i^n \sqrt{a_i}$</span> when <span class="math-container">$a_i$</span> are positive integers.
I think that the following is true, but can't prove it.
<span class="math-container">$$\sum_i \sqrt{a_i} \le (\sum_i a_i)^{3/4}$$</span>
I need a tighter bound, can anyone help ?</p>
| 222
|
|
calculus
|
Trap Rule for sin(x)
|
https://math.stackexchange.com/questions/393619/trap-rule-for-sinx
|
<blockquote>
<p>Use the trapezoidal rule with $N=6$ to approximate the arc length of the curve $f(x) = \sin(x)$ from $x=0$ to $x=\pi$.</p>
</blockquote>
<p>So I found that $\Delta x = \frac{\pi}{6}$ which means that my interval points are $0,\frac{\pi}{3}, \frac{\pi}{2}, \frac{2\pi}{3}, \frac{5\pi}{6}$ and $\pi$.</p>
<p>Is that right?</p>
<p>And then after that I just use the trap rule forumula computing the value of $\sin$ of all my interval points.</p>
<p>Is my thinking correct?</p>
|
<p>The formula for step size is given by: </p>
<p>$$\displaystyle h = \frac{b-a}{N} = \frac{\pi - 0}{6} = \frac{\pi}{6}$$</p>
<p>We are also given that $x_0 = 0$.</p>
| 223
|
calculus
|
force is generally a function of $\mathbf{r}(t)$, $\mathbf{v}(t)$ and $t$
|
https://math.stackexchange.com/questions/3001165/force-is-generally-a-function-of-mathbfrt-mathbfvt-and-t
|
<blockquote>
<p>Force is generally a function of <span class="math-container">$\mathbf{r}(t)$</span>, <span class="math-container">$\mathbf{v}(t)$</span> and <span class="math-container">$t$</span>. <span class="math-container">$$1-)\begin{cases}
\mathbf F: \mathbb R^3\times\mathbb R^3 \times \mathbb R \ \rightarrow \mathbb R^3\\
(\mathbf r, \mathbf v, t) \longmapsto \mathbf F(\mathbf r,\mathbf v,t)
\end{cases}$$</span></p>
</blockquote>
<p>But I do't understand it . Because function from <span class="math-container">$\mathbb{R^7}$</span> to <span class="math-container">$\mathbb{R^3}$</span> I don't know how is that .</p>
<hr>
<p>Can we conclude that <span class="math-container">$\mathbf F(\mathbf r(t),\mathbf v(t),t)=\mathbf F(t)$</span> ??
Notice that :
<span class="math-container">$$\begin{cases}
\mathbf r: \mathbb R \rightarrow \mathbb R^3\\
t \longmapsto \mathbf r(t)
\end{cases}$$</span></p>
<p><span class="math-container">$$\begin{cases}
\mathbf v: \mathbb R \rightarrow \mathbb R^3\\
t \longmapsto \mathbf v(t)
\end{cases}$$</span></p>
|
<p>Imagine a world where the sun is moving rapidly in our chosen coordinate system and has a huge current loop around the equator. The earth has a huge negative charge and is moving in the gravitational and magnetic fields of the sun. At any time <span class="math-container">$t$</span> we need <span class="math-container">$\mathbf r, \mathbf v, t$</span> to compute the force on the earth. That is seven real numbers, three for each vector and one for the time, so the domain of the force function is <span class="math-container">$\Bbb R^7$</span>. It is not a geometric space as the coordinates have different meaning and we can't rotate among them freely, but it still has the structure of <span class="math-container">$\Bbb R^7$</span>. We can take any point in the space and use the function to get the force. The force is a vector, so it has three components. This is the real physical input to the problem-the law that gives us the force on the earth. </p>
<p>Now we can use Newton's laws to integrate the differential equation and find <span class="math-container">$\mathbf v(t), \mathbf r(t)$</span>. Having done that, we can plug into <span class="math-container">$\mathbf F(\mathbf r,\mathbf v,t)$</span> and compute the force at any time along the trajectory. That gives us a function <span class="math-container">$\mathbf F(t)$</span>. In one sense it is an abuse to use the same symbol <span class="math-container">$\mathbf F$</span> because one is a function of seven variables and one is a function of one variable. In another sense it is not an abuse because we are computing the same force in both cases. The seven dimensional version is much more general. It can compute the force for any position, velocity, and time. The one dimensional version can only compute the force for positions along the trajectory at the time when the earth was there.</p>
| 224
|
calculus
|
Can L'Hopital's rule be applied only for a part of a function?
|
https://math.stackexchange.com/questions/2600840/can-lhopitals-rule-be-applied-only-for-a-part-of-a-function
|
<p>For example, in
$\lim_{x\to 0_+} (x^2 \ln x+bx+c) $
can it be applied only for $x^2\ln x$? (of course not in this form)</p>
|
<p>If you have $\lim(f(x)+g(x)+\cdots)$, then you can always compute the limit term-wise, i.e.</p>
<p>$$\lim f(x)+\lim g(x)+\cdots$$</p>
<p>as long as all the single limits exist. So in order to apply l'Hospital to only one of the terms, first use this rule to get</p>
<p>$$\lim (x^2\ln(x)+bx+x)=\lim(x^2\ln(x))+\lim(bx+x).$$</p>
<p>Then apply l'Hospital to whatever part you want.</p>
| 225
|
calculus
|
Can't see why one of these functions is conservative, and the other isn't.
|
https://math.stackexchange.com/questions/4182392/cant-see-why-one-of-these-functions-is-conservative-and-the-other-isnt
|
<p>I am really confused here: Why one of these functions is conservative, while the other not?</p>
<p><span class="math-container">$F_{1} = \frac{-y \hat i + x \hat j}{x^2+y^2}$</span></p>
<p><span class="math-container">$F_{2} = \frac{x \hat i + y \hat j}{x^2+y^2}$</span></p>
<p>Suppose both these vector functions are applied on a region <span class="math-container">$D=R^{2}-(0,0)$</span>.</p>
<blockquote>
<p><span class="math-container">$\nabla \times F_{1} =\nabla \times F_{2} = 0 $</span></p>
<p>Both apply at the same region.</p>
</blockquote>
<p>Even so, every exercise i see about this region says something like:: "F1 isn't a gradient, the region isn't simple connected/ F2 is a gradient". So:</p>
<p>Why (1) in one case the fact that D is "a ill region" import, and in the other not?</p>
<p>(2)As i pointed, both vectors have the same properties, but one is gradient, and the other not. So what is preciselly being the differential here that says when one is a gradient and when one isn't?</p>
|
<p>You are checking whether a vector field is a gradient by seeing if it is curl free. Over a simply-connected domain, being curl free and being a gradient are equivalent. The purpose of this exercise is that these two concepts, equivalent over nice domains, are no longer equivalent on the punctured plane. Just because something is curl free, does <em>not</em> mean it is a gradient. To prove that <span class="math-container">$F_2$</span> is a gradient, instead of checking the curl, you should try and explicitly find what it is the gradient of. To see that <span class="math-container">$F_1$</span> isn't a gradient, just integrate it around a circle of radius 1 and you'll get a nonzero value (if <span class="math-container">$F_1$</span> was a vector field, then of course the line integral around any closed curve would be zero by the Fundamental Theorem of Calculus).</p>
| 226
|
calculus
|
Moment of Inertia around z axis
|
https://math.stackexchange.com/questions/1617665/moment-of-inertia-around-z-axis
|
<p>Hello I am having difficulty with the following;</p>
<p>I am wanting to find I, the moment of inertia about the z axis of the region that is bounded by the paraboloid $z=x^{2}+y^{2}$ and the $z=1$ plane, where the density is proportional to the distance from the z axis.</p>
<p>Here is what I have tried:</p>
<p>I thought maybe I could use the formula</p>
<p>$I= \iiint_{D}(x^2+y^2)\rho(x,y,z)dV$</p>
<p>and use that since $\rho(x,y,z)$ is proportional to distance from the $z$ axis, then for some constant $K$ we have $\rho(x,y,z)=K$(distance from z axis)</p>
<p>But I am not sure which distance from the z axis to use, or which formula to use.</p>
<p>Anyway, assuming the above is correct, then I would have</p>
<p>$I= \iiint_{D}K(x^2+y^2)(distance from z -axis)dV$</p>
<p>and the region D would be determined by knowing we are bound by the paraboloid and z=1 plane.</p>
<p>But I am stuck on it so far, can anyone please help me? Is it the correct approach? and how should I find distance to z axis? Maybe I don't need triple integrals? Anyways, I have tried my best to work on it, but I am not making any progress. I think if someone helped to explain then I could understand it for the future</p>
<p>Thank you</p>
<p>Update:</p>
<p>I am still working on this problem days after. I really wish I could just get some help so I can study it and move on!</p>
<p>Here is what else I have tried;</p>
<p>If we try cylindrical coordinates,</p>
<p>then $z=r^{2}$ and z goes to the plane ie $z =1$</p>
<p>\theta, would go from $0$ to $2pi$, and $r$ from $0$ to $1$</p>
<p>(however I still don't know how I could represent $K\rho $</p>
<p>but I would then have $\iiint_{D'} (r^3)(K(\rho))dzdrd\theta$
my only guess would be that K(\pho)=K(r)</p>
<p>giving $$K\iiint_{D'}r^{4}dzdrd\theta$$
But I don't know if so far it is the right approach? Please anyone?</p>
|
<p>Work in cylindrical coordinates $(r,\theta,z)$. The element of volume is $r\,dr\,d\theta\,dz$. The distance of a point to the $z$ axis is just $r$, and the density $\lambda r$. The paraboloid is $r^2=z$ or $r=\sqrt z$.</p>
<p>So the mass is</p>
<p>$$M=\int_{z=0}^1\int_{\theta=0}^{2\pi}\int_{r=0}^{\sqrt z}\lambda r\,r\,dr\,d\theta\,dz=\lambda2\pi\int_{z=0}^1\left.\frac{r^3}3\right|_0^{\sqrt z}dz=\lambda2\pi\left.\frac2{3\cdot5}z^{5/2}\right|_0^1=\lambda\frac{4\pi}{15}.$$</p>
<p>And the moment of inertia around $z$</p>
<p>$$I_{zz}=\int_{z=0}^1\int_{\theta=0}^{2\pi}\int_{r=0}^{\sqrt z}r^2\lambda r\,r\,dr\,d\theta\,dz=\lambda2\pi\int_{z=0}^1\left.\frac{r^5}5\right|_0^{\sqrt z}dz=\lambda2\pi\left.\frac2{5\cdot7}z^{7/2}\right|_0^1=\lambda\frac{4\pi}{35}.$$</p>
<p>Then</p>
<p>$$I_{zz}=\frac37M.$$</p>
| 227
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calculus
|
First fundamental theorem of calculus where the bounds are not 0 to x.
|
https://math.stackexchange.com/questions/4099535/first-fundamental-theorem-of-calculus-where-the-bounds-are-not-0-to-x
|
<p>Suppose <span class="math-container">$F(x) = \int_{3x+8}^{x^{2}+5x+1}\csc^{2}\left(t\right)dt$</span>. How would one find <span class="math-container">$F'(x)$</span> using the first fundamental theorem of calculus? I am aware of how to do this when the bounds are 0 to f(x) through use of chain rule, but I don't know how to do this when the first bound is not 0.</p>
|
<p>Hint: <span class="math-container">$\int_a^bf(x) dx=\int_a^0f(x) dx+\int_0^bf(x) dx=\int_0^bf(x) dx-\int_0^af(x) dx$</span></p>
| 228
|
calculus
|
The intuitive meaning of integrals
|
https://math.stackexchange.com/questions/2992196/the-intuitive-meaning-of-integrals
|
<p>I am an engineering student and i always encounter problems that needs integrals I know that integral is area under the curve , etc.... but till now i could not develop and intuitive meaning for integration. does integration rely only on the idea of area under the curve. do the physics laws that are based on integration like for example the that the work is the integral of the F in infinitesimal distance was proved by the idea of the area under the curve or is there an intuitive way that i could understand the integration through it? </p>
| 229
|
|
calculus
|
Area under Curve Limits
|
https://math.stackexchange.com/questions/2484422/area-under-curve-limits
|
<p>If S be the area of the region enclosed by $y=e^{-x^{2}}$, y=0, x=0 and x=1. </p>
<p>Then
(A) $S \ge \frac {1}{e}$ (B) $S \ge 1-\frac {1}{e}$<br>
(C) $S \le \frac {1}{4}(1+\frac{1}{√e})$ (D) $S \le \frac {1}{√2}+\frac{1}{√e}(1-\frac{1}{√2})$ </p>
<p>The correct answer is A,B and D it is multiple choice
I can find B using the concept $e^{-x^{2}}>e^{-x}$ for x $\in$ (0,1) and integrating from 0 to $\frac{1}{e}$ but not able to deduce A and D. </p>
|
<p>(A) follows from (B) as $S\ge1-\frac1e>\frac1e$ (this follows from $e>2$)</p>
<p>Don't know about (D) though... </p>
| 230
|
calculus
|
how to find a function f(n) (continuous on R) such that $(-1)^{f(n)}$ is positive when $n=1, 2, 5, 6, 9, 10....$, and <0 for other natural number?
|
https://math.stackexchange.com/questions/4087649/how-to-find-a-function-fn-continuous-on-r-such-that-1fn-is-positiv
|
<p>Further more, can we have a general way to find <span class="math-container">$f(n)$</span> which is negative whenever we design?
(note: we just take <span class="math-container">$n$</span> as natural number)</p>
<p>I think some function with <span class="math-container">$\sin$</span>, <span class="math-container">$\cos$</span> will satisfy this.</p>
|
<p>To put it another way: you want continuous <span class="math-container">$f:\Bbb R\to\Bbb R$</span> such that <span class="math-container">$f(n)$</span> is an even integer for <span class="math-container">$n=1,2,5,6,9,10,\ldots$</span> and an odd integer for <span class="math-container">$n=3,4,7,8,11,12,\ldots$</span></p>
<p>Looking at the graph of this, it is clear than a <span class="math-container">$\cos$</span> function will do the trick, with maxima at <span class="math-container">$x=1.5,5.5,9.5,\ldots$</span> Can you see how to do this?</p>
| 231
|
calculus
|
Find max vertical distance
|
https://math.stackexchange.com/questions/164982/find-max-vertical-distance
|
<p>What is the maximum vertical distance between the line
$y = x + 20$
and the parabola
$y = x^2$ for $−4 ≤ x ≤ 5?$</p>
<p>What steps do I take to solve this? Do I have to use the distance formula and what do I do with the points it gave me?</p>
<p>If anyone could just bounce me in the right direction that would be neat. I can probably work an answer from there!</p>
<p>Also what's the distance formula to use here?</p>
|
<p>The vertical distance at $x=a$ is the difference in $y$-coordinates at $x=a$, so it’s $|(x+20)-x^2|$. Now $x^2-x-20=(x+4)(x-5)$, so it’s negative between $x=-4$ and $x=5$. Thus, on the interval $[-4,5]$ we have $|(x+20)-x^2|=x+20-x^2$, not $x^2-x-20$.</p>
<p>Now let $f(x)=x+20-x^2$ and find the maximum of $f(x)$ on the interval $[-4,5]$.</p>
| 232
|
calculus
|
Limit $\lim _{ \theta \to 0 }{ \frac { cos2\theta -cos\theta }{ \theta } } $
|
https://math.stackexchange.com/questions/1111672/limit-lim-theta-to-0-frac-cos2-theta-cos-theta-theta
|
<p>$$\lim _{ \theta \to 0 }{ \frac { cos2\theta -cos\theta }{ \theta } } $$</p>
<p>Steps I took:</p>
<p>$$\lim _{ \theta \rightarrow 0 }{ \frac { 1-2sin^{ 2 }\theta -cos\theta }{ \theta } } =$$</p>
<p>$$\lim _{ \theta \rightarrow 0 }{ \frac { -2sin^{ 2 }\theta }{ \theta } } +\lim _{ \theta \rightarrow 0 }{ \frac { 1-cos\theta }{ \theta } } $$</p>
<p>$$\lim _{ \theta \rightarrow 0 }{ \frac { 1-cos\theta }{ \theta } =0 } $$</p>
<p>$$\lim _{ \theta \rightarrow 0 }{ \frac { -2(sin\theta )(sin\theta ) }{ \theta } = } $$</p>
<p>$$\lim _{ \theta \rightarrow 0 }{ { \quad -2(sin\theta ) }\cdot 1=0 } $$</p>
<p>$$0+0=0$$</p>
<p>Something seems off about the way I went about this but I can't figure it out.</p>
|
<p>your proof is correct. but if you are going to use $\lim_{\theta \to 0}\frac{1-\cos \theta}{\theta} = 0,$ you could have split $\cos(2\theta) - \cos \theta$ as $(1-\cos \theta) -(1 - \cos 2 \theta)$ at the beginning itself.</p>
| 233
|
calculus
|
Integral Issues.
|
https://math.stackexchange.com/questions/1156239/integral-issues
|
<p>$\displaystyle \int \cosh ^2t\,\sinh ^5t \; \textrm{d}t \,$</p>
<p>Can't for the life of me figure this one out. I have tried various substitutions. The pythagorean hyperbolic identity, the double variable identity. Nothing. Could someone give me a push please. </p>
|
<p>With some manipulation using $\cosh^2x-\sinh^2x=1\implies \sinh^4t=(\cosh^2t-1)^2$:
$$\cosh ^2t\,\sinh ^5t =\sinh t \cosh^6 t-2 \sinh t \cosh^4 t+\sinh t \cosh^2 t$$
Now try $x=\cosh t,{\rm d}x/{\rm d}t=\sinh t$</p>
| 234
|
calculus
|
How to prove this in smart way
|
https://math.stackexchange.com/questions/1161800/how-to-prove-this-in-smart-way
|
<p>How to prove this in a a smart way?</p>
<blockquote>
<p>If $y= \sin (m \sin^{-1} (x))$, then $(1-x^2)y^{(n+2)}-(2n+1)x{y^{(n+1)}}+(m^2-n^2)y^{(n)}=0$ derivative.</p>
</blockquote>
<p>I have been able to prove it by differentiating it twice and using Leibniz theorem, but thats very long, is there a nice way to prove this?</p>
|
<p>$$y'=m(1-x^2)^{-1/2}\cos(m\sin^{-1}(x))$$
$$y''=mx(1-x^2)^{-3/2}\cos(m\sin^{-1}(x))-m^2(1-x^2)^{-1}\sin(m\sin^{-1}(x)),$$
so that $$(1-x^2)y''-xy'+m^2y=0.$$
this establishes the base case of the recurrence.</p>
<p>Now derive</p>
<p>$$(1-x^2)y^{(n+2)}-(2n+1)xy^{(n+1)}+(m^2-n^2)xy^{(n)}=0$$ and get</p>
<p>$$-2xy^{(n+2)}+(1-x^2)y^{(n+3)}-(2n+1)y^{(n+1)}-(2n+1)xy^{(n+2)}+(m^2-n^2)y^{(n+1)}=0,$$
or after regrouping (denoting $n^+:= n+1$)
$$(1-x^2)y^{(n^++2)}-(2n^++1)xy^{(n^++1)}+\left(m^2-(n^+)^2\right)y^{(n^+)}=0.$$</p>
| 235
|
calculus
|
Equation to a level surface
|
https://math.stackexchange.com/questions/1188100/equation-to-a-level-surface
|
<p>Could someone please help me with the following question:</p>
<blockquote>
<p>Consider the function $g(x,y,z)=\ln(x^2-y+z^2)$. Find an equation of the level surface of the function through the point $(-1,2,1)$ which does not have $\ln$. Hint: first find $g(-1,2,1).$</p>
</blockquote>
<p>When I sub in the points I get $$g(-1,2,1)=\ln(1-2+1)=\ln(0)=\text{undefined}$$. Where am I going wrong?</p>
<p>Thanks,
bbelson01</p>
|
<p>$$\ln(x^2-y+z^2)=\ln(x_0^2-y_0+z_0^2)$$
can be rewritten
$$y=x^2+z^2+(y_0-x_0^2-z_0^2).$$</p>
<p>It remains the same paraboloid of revolution, with the apex moving along the axis $y$.</p>
| 236
|
calculus
|
Finding the parametrization for a sphere?
|
https://math.stackexchange.com/questions/1423532/finding-the-parametrization-for-a-sphere
|
<p>Find a parametrization for the circle centered around the origin, of radius 3 and contained in the xz-plane.</p>
<p>So from what I gathered you use the formula of sphere $x^2+y^2+z^2= r^2$ to solve this problem. So you know what the radius is 3 yet how does one find xyz just from having the radius?</p>
|
<p>A three dimensional surface in a two dimensional plane!? For a constant <span class="math-container">$y$</span> value we can define a small circle and the plane it is contained in.</p>
<p>We can use spherical coordinates. Choose a particular latitude, translate the circle with arbitrary displacements <span class="math-container">$(k_1,k_2,k_3)=0 .$</span> In the picture R=1 but you can set it <span class="math-container">$R=3$</span>.</p>
<pre><code> k1 = 1; k2 = 2; k3 = 3; ph = .785; R = 1;
X = {(R Cos[th]) Cos[ph] + k1,
R Sin[th] + k2, (R Cos[th]) Sin[ph] + k3}
ParametricPlot3D[X, {th, 0, 2 Pi}, PlotStyle -> {Red, Thick}]
</code></pre>
<p><a href="https://i.sstatic.net/0EKEh.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/0EKEh.png" alt="enter image description here" /></a></p>
| 237
|
calculus
|
Computing $\bigtriangledown^2(1/r)$
|
https://math.stackexchange.com/questions/1445477/computing-bigtriangledown21-r
|
<p>Given that:</p>
<p>$$\vec{r} = x\hat{i}+y\hat{j}+z\hat{k}$$</p>
<p>and $r$ is the magnitude of $\vec{r}$</p>
<p>Then what is:</p>
<p>$$\bigtriangledown^2(1/r)$$</p>
<p><strong>EDIT:</strong>
I know that $\bigtriangledown^2F(x)$ is the divergence of the gradient of $F(x)$ thus my attempt to solve the question was to compute the gradient of $1/r$ and then compute the divergence of that, yet my attempt is not yielding the correct answer.</p>
<p>The answer should be 0</p>
|
<p>Hint: In Cartesian coordinates</p>
<p>$$\bigtriangledown^2f(x,y,z)=\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}+\frac{\partial^2f}{\partial z^2}$$and here$$r=|\vec{r}|=\sqrt{x^2+y^2+z^2}$$
$$f=\frac{1}{r}=(x^2+y^2+z^2)^{\frac{-1}{2}}$$
$$\frac{\partial f}{\partial x}=-\frac{x}{(x^2+y^2+z^2)^{\frac{3}{2}}}$$
$$\frac{\partial^2 f}{\partial x^2}=\frac{2x^2-y^2-z^2}{(x^2+y^2+z^2)^{\frac{5}{2}}},\frac{\partial^2 f}{\partial y^2}=\frac{2y^2-x^2-z^2}{(x^2+y^2+z^2)^{\frac{5}{2}}},\frac{\partial^2 f}{\partial z^2}=\frac{2z^2-y^2-x^2}{(x^2+y^2+z^2)^{\frac{5}{2}}}$$
$$\bigtriangledown^2f(x,y,z)=\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}+\frac{\partial^2f}{\partial z^2}=0$$</p>
<p>See the following links which verify the results:</p>
<p><a href="http://www.wolframalpha.com/input/?i=del%20%28x%5E2%2By%5E2%2Bz%5E2%29%5E%28-1%2F2%29" rel="nofollow">Gradient</a> of $f$</p>
<p><a href="http://www.wolframalpha.com/input/?i=div%20%7B-x%2F%28x%5E2%2By%5E2%2Bz%5E2%29%5E%283%2F2%29%2C%20-y%2F%28x%5E2%2By%5E2%2Bz%5E2%29%5E%283%2F2%29%2C%20-z%2F%28x%5E2%2By%5E2%2Bz%5E2%29%5E%283%2F2%29%7D" rel="nofollow">Divergence</a> of $\bigtriangledown f$</p>
| 238
|
calculus
|
Finding discontinuities points
|
https://math.stackexchange.com/questions/1536722/finding-discontinuities-points
|
<blockquote>
<p>find discontinuities points of the function $f(x)=x-\lfloor{x}\rfloor$</p>
</blockquote>
<p>I know that there is no limit $f(x)=\lfloor{x}\rfloor$ when $x\in \mathbb{N}$ Is it sufficient to say that therefore there are discontinuities points when $x\in \mathbb{N}$?</p>
|
<p>In every open interval $(n,n+1)$, we have $\lfloor x\rfloor=n$, hence $f(x)=x-\lfloor x\rfloor=x-n$, which is well-known to be a continuous function.</p>
<p>Hence the only discontinuities are at $n$, as $\lim_{x\to n^{-}}f(x)=\lim_{x\to n^{-}}(x-n+1)=1$ while $\lim_{x\to n^{+}}f(x)=\lim_{x\to n^{+}}(x-n)=0$.</p>
| 239
|
calculus
|
How to plot $f(x)=x^{2/3}$
|
https://math.stackexchange.com/questions/1691903/how-to-plot-fx-x2-3
|
<p>I'm using Leithold's book to teach calculus. In a exercise Leithold asks how to draw $f(x)=x^{2/3}$. I don't know how to plot this function since I can't use the derivative methods he develop afterwards. Until this page of the book Leithold only covers limits, continuity, tangents and basic derivatives. He didn't talk about concavity, inflection points, absolute values, etc. yet.</p>
<p>So How do I plot this function using only the definitions he made until this exercise?</p>
|
<p>Using continuity, you can find that f is continuous at the origin:</p>
<p>$$
\lim_{x \to 0} f(x) = f(0) = 0 \\
$$</p>
<p>Using limits, you can find what happens at the ends:</p>
<p>$$
\lim_{x \to -\infty} f(x) = +\infty \\
\lim_{x \to +\infty} f(x) = + \infty \\
$$</p>
<p>Using limits, you can find the inclination of the tagent line at the origin:</p>
<p>$$
\begin{eqnarray}
f'(0) &=& \lim_{\Delta x \to 0} \frac{f(0 + \Delta x) - f(0)}{\Delta x} \\
&=& \lim_{\Delta x \to 0} \frac{(\Delta x)^{2/3} - 0}{\Delta x} \\
&=& \lim_{\Delta x \to 0} \frac{1}{(\Delta x)^{1/3}} \\
&=& \infty
\end{eqnarray}
$$</p>
<p>Using more limits, you can find the inclination of the tangent line at any point:</p>
<p>$$
\begin{eqnarray}
f'(x) &=& \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} \\
&=& \lim_{\Delta x \to 0} \frac{(x + \Delta x)^{2/3} - x^{2/3}}{\Delta x} \\
&=& \lim_{\Delta x \to 0} \frac{[(x + \Delta x)^{2/3} - x^{2/3}][(x + \Delta x)^{4/3} + (x + \Delta x)^{2/3}x^{2/3} + x^{4/3}]}{\Delta x[(x + \Delta x)^{4/3} + (x + \Delta x)^{2/3}x^{2/3} + x^{4/3}]} \\
... \\
&=& \lim_{\Delta x \to 0} \frac{2x + \Delta x}{(x + \Delta x)^{4/3} + (x + \Delta x)^{2/3}x^{2/3} + x^{4/3}} \\
&=& \frac{2x}{x^{4/3} + x^{2/3}x^{2/3} + x^{4/3}} \\
&=& \frac{2}{3x^{1/3}}
\end{eqnarray}
$$</p>
<p>You now know the graph tends to $+\infty$ at the ends, that it is continuous at the origin but the tangent line has an inclination of $\pi/2$, and that the inclination is negative when $x < 0$ and positive when $x > 0$. That should be plenty to draw a nice graph.</p>
| 240
|
calculus
|
Calculus problem - Unknown variable in a quadratic
|
https://math.stackexchange.com/questions/1980858/calculus-problem-unknown-variable-in-a-quadratic
|
<p>Is there an $a$ such that $\lim_{x \rightarrow -3} \frac{10x^2+ax+a+8}{x^2+x-6}$ exists?</p>
<p>I can't seem to find how to actually solve it other than guessing, and I'm not sure there actually is a solution.</p>
|
<p>Hint:</p>
<p>Consider $2$ cases, when the numerator is evaluated to $0$ and when it is not at $x=-3$.</p>
<p>For the case when the numerator is $0$,</p>
<p>You can use L'hopital's rule and evaluate $$\lim_{x \rightarrow -3} \frac{20x+a}{2x+1}$$</p>
<p>Alternatively, $$10x^2+49x+57=(x+3)(10x+19)$$ Since we know that the numerator evaluates to $0$ at $x=-3$, $x+3$ is a factor.</p>
| 241
|
calculus
|
$dy/dx$ problems, please help
|
https://math.stackexchange.com/questions/2056607/dy-dx-problems-please-help
|
<p>Find $dy/dx$ given $y\cos(xy)=3$.
Also find $dy/dx$ given $y=(2+\sin x)^{\cos x}$</p>
<p>I'm having a hard time solving for $dy/dx$ given $y\cos(xy)= 3$. Because of the $3$, wouldn't the right side of the equation equal $0$? And dividing $0$ by the derivative of the left side to get $dy/dx$ alone also equal $0$? </p>
<p>Also, I solved for $dy/dx$ as best as I could but when I checked using the Mathway app, it couldn't compute the solution. So I have no way of checking to see if I'm on the right track. For $dy/dx$ I got $(\cos^{2}x)/(2+\sin x) -\sin(x)\cdot\ln(2+\sin x)$ all multiplied by $y$.</p>
|
<p>For $y\cos(xy)=3$ one must find $\dfrac{dy}{dx}$ by implicit differentiation. For students who find implicit differentiation difficult I recommend first considering both $x$ and $y$ as functions of some third variable such as $t$ and </p>
<p>$(1)$differentiate both sides of the equation with respect to $t$, </p>
<p>$(2)$ multiply both sides by $dt$ then </p>
<p>$(3)$ divide both sides by $dx$</p>
<p>$(4)$ Solve the resulting equation for $\dfrac{dy}{dx}$</p>
<p>\begin{eqnarray}
\cos(xy)\frac{dy}{dt}-y\sin(xy)\left(y\frac{dx}{dt}+x\frac{dy}{dt}\right)&=&0\tag{1}\\
\cos(xy)dy-y\sin(xy)\left(y\,dx+x\,dy\right)&=&0\tag{2}\\
\cos(xy)\frac{dy}{dx}-y\sin(xy)\left(y+x\frac{dy}{dx}\right)&=&0\tag{3}\\
\left[\cos(xy)-xy\sin(xy)\right]\frac{dy}{dx}&=&y^2\sin(xy)\tag{4}\\
\frac{dy}{dx}&=&\dfrac{y^2\sin(xy)}{\cos(xy)-xy\sin(xy)}
\end{eqnarray}</p>
<p>You appear to have done the second problem correctly.</p>
| 242
|
calculus
|
How to show $\lim_{n \rightarrow \infty} \frac{[a^{n+1}]}{[a^n]}=a$?
|
https://math.stackexchange.com/questions/2083127/how-to-show-lim-n-rightarrow-infty-fracan1an-a
|
<p>How to show that $\lim_{n \rightarrow \infty} \frac{[a^{n+1}]}{[a^n]}=a$, where
$[a]$ = integer part of a?<br>
Here $a>1$. But I suspect it is true for all $a \ne 0$. </p>
|
<p>For $|a|>1$,</p>
<p>$$\frac{[a^{n+1}]}{[a^n]}=\frac{a^{n+1}-\{a^{n+1}\}}{a^n-\{a^n\}}=a\frac{1-\dfrac{\{a^{n+1}\}}{a^{n+1}}}{1-\dfrac{\{a^{n}\}}{a^n}}\to a.$$</p>
<p>As the fractional parts are bounded, the numerator and denominator both tend to $1$.</p>
<hr>
<p>This can be extended to $|a|\ge1$ as with $|a|=1$, the fractional parts are all $0$.</p>
<hr>
<p>For $0\le a<1$, the limit is undefined (none of the ratios are defined).</p>
<hr>
<p>For $-1<a<0$, the limit is $-1$ (all ratios are $-1$).</p>
| 243
|
calculus
|
How to solve this complicated integral
|
https://math.stackexchange.com/questions/2246369/how-to-solve-this-complicated-integral
|
<p>I am trying to compute the following integral:
$$
I = \int^\infty_1\frac{\operatorname{frac}(x)\cos(a\ln x)}{x^b}\,dx
$$
where $\operatorname{frac}(x) = x - \operatorname{int}(x)$ is the fractional part of $x$, $a > 0$ and $b > 1$.</p>
<p>This is what I got so far.</p>
<p>Let $\operatorname{int}(x) = n$ so that $\operatorname{frac}(x) = x - n$ and $n \le x < n+1$.</p>
<p>The integral $I$ can then be formulated as:
$$
I = \sum^\infty_1 \int^{n+1}_n\frac{(x-n)\cos(a\ln x)}{x^b}\,dx
$$</p>
<p>And I got stuck at computing the rather complicated integral
$$
J = \int^{n+1}_n\frac{(x-n)\cos(a\ln x)}{x^b}\,dx
$$</p>
|
<p><strong>Hint</strong>:</p>
<p>$$\int x^c\cos(a\ln x)\,dx=\int e^{(c+1)t}\cos(at)\,dt$$ can be integrated analytically, for example by means of complex numbers.</p>
<p>Indeed,</p>
<p>$$\int e^{bt}\cos(at)=\int\Re(e^{(b+ia)t})=\Re\left(\int e^{(b+ia)t}\right)=\Re\left(\frac{e^{(b+ia)t}}{b+ia}\right),$$ which can be expanded back in terms of real functions.</p>
<p>Your $I$ integral can now be expressed as a complicated summation, for which I wouldn't hope a closed formula.</p>
| 244
|
calculus
|
A clue to solve this equation
|
https://math.stackexchange.com/questions/2271578/a-clue-to-solve-this-equation
|
<p>how to prove that if $f(x,y)=0$ and $g(x,z)=0$ and if $f$ and $g$ are differentiable,then:</p>
<p>$$\dfrac{\partial f}{\partial y}.\dfrac {\partial g}{\partial x}dy=\dfrac{\partial f}{\partial x}.\dfrac {\partial g}{\partial z}dz$$</p>
<p>I think $y$ and $z$ should be dependent, however there is no mentioning for that in the question.</p>
|
<p>We have</p>
<p>$$\dfrac{\partial f}{\partial x}dx+\dfrac{\partial f}{\partial y}dy=0,\\
\dfrac{\partial g}{\partial x}dx+\dfrac{\partial g}{\partial z}dz=0.$$</p>
<p>Then eliminating $dx$,</p>
<p>$$\frac{\dfrac{\partial f}{\partial y}}{\dfrac{\partial f}{\partial x}}dy=\frac{\dfrac{\partial g}{\partial z}}{\dfrac{\partial g}{\partial x}}dz.$$</p>
| 245
|
calculus
|
Prove that the following expression is always less than x for all values of x and k.
|
https://math.stackexchange.com/questions/2273552/prove-that-the-following-expression-is-always-less-than-x-for-all-values-of-x-an
|
<p>Prove that </p>
<p>$$\frac{x^2+kx}{2x+k}$$</p>
<p>is less than x for all values of x and k where x>0, k>0 and k is a constant.</p>
<p>How would I prove this? I have differentiated it with respect to x and noticed that the derivative is always less than 1 for all values of x and k, this means that if the value of x increases the value out always increases less than the increase in x. </p>
|
<p>Multiplying by $2x+k$ positive, $$x^2+kx<x(2x+k)$$</p>
<p>then</p>
<p>$$x^2>0.$$</p>
<hr>
<p>Alternatively,</p>
<p>$$\frac{x^2+kx}{2x+k}-x=-\frac{x^2}{2x+k}<0.$$</p>
| 246
|
calculus
|
Simple Algebra ,Radicals ,Prime Numbers
|
https://math.stackexchange.com/questions/2275235/simple-algebra-radicals-prime-numbers
|
<p>a,b are prime numbers
c∈ℕ</p>
<p>2√a + 7√b = c√3</p>
<p>a²+b²+c²=?</p>
<p>I don't really know how to solve it</p>
|
<p>$$2\sqrt{\frac a3}+7\sqrt{\frac b3}\in\mathbb N$$ is only possible if the radicals have rational values (no linear combination of irrationals gives an integer).</p>
<p>Then, only one prime gives the square of a rational when divided by $3$: obviously $3$. From this, $c=9$.</p>
| 247
|
calculus
|
Quick Questions for Evaluating an Integral
|
https://math.stackexchange.com/questions/2439131/quick-questions-for-evaluating-an-integral
|
<p>The calculus shown below is confusing to me. I understand the first step, moving m outside the integral and rewriting in terms of dt, but how does the rest of the evaluation work?</p>
<p>$$\int m \frac{d^2x}{dt^2}dx = m\int\frac{d^2x}{dt^2}\frac{dx}{dt}dt = \frac{m}{2}\int\frac{d}{dt}\left(\frac{dx}{dt}\right)^2dt = \frac{m}{2}\left(\frac{dx}{dt}\right)^2+\mathrm{constant}$$</p>
<p>Thank you in advance. For reference, this is from Quantum Chemistry by McQuarrie.</p>
|
<p>With $v=\dfrac{dx}{dt}$,</p>
<p>$$\int\frac{d^2x}{dt^2}dx=\int\frac{dv}{dt}dx=\int (dv)\frac{dx}{dt}=\int v\,dv=\frac12v^2.$$</p>
| 248
|
calculus
|
Nature of roots of a hectic polynomial
|
https://math.stackexchange.com/questions/2667240/nature-of-roots-of-a-hectic-polynomial
|
<blockquote>
<p>Let $p(x)$ be a $100$-degree polynomial with $100$ real and distinct roots, say $\alpha_1,\alpha_2,\cdots,\alpha_{100}$, and so $$p(x)=A(x-\alpha_1)(x-\alpha_2)\cdots(x-\alpha_{100}),$$
where $A\in\mathbb{R}\setminus\{0\}$ and $α_{i}\neq 0$ for all $i\in[1,100]$.
Find nature of roots of the equation </p>
</blockquote>
<p>$$x^2p''(x)+3xp'(x)+p(x)=0$$</p>
<p>and also find nature of roots of the equation </p>
<p>$$10p(x)p''(x)=99(p'(x))^2.$$</p>
<p>Try:</p>
<p>$$(x^2p''(x)+2xp'(x)+xp'(x)+p(x)=0,\\
\frac{d}{dx}\bigg(x^2p'(x)\bigg)+\frac{d}{dx}\bigg(xp(x)\bigg)=0.$$</p>
<p>Could some help me to solve it, thanks in advance.</p>
|
<p>You have
$$((px)'x)'=(px)''x+(px)'=(p''x+2p')x+(p'x+p)=p''x^2+3p'x+p.$$</p>
<p>The roots of $(px)'$ are the extrema of $px$, which are real and comprised in the $100$ intervals $(\alpha_k,\alpha_{k+1})$, where we define $\alpha_0:=0$.</p>
<p>Then again, the roots of $((px)'x)'$ are real and comprised in the $100$ intervals $(\beta_k,\beta_{k+1}$, where the $\beta_k$ are the above roots plus $\beta_0:=0$.</p>
<p>As an illustration, a simpler case with $\alpha_1=-1,\alpha_2=1,\alpha_3=2$ and the polynomials $\color{blue}p,\color{lightgreen}{(px)'},\color{magenta}{((px')x)'}$:</p>
<p><a href="https://i.sstatic.net/wCs6i.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/wCs6i.png" alt="enter image description here"></a></p>
<hr>
<p>The second case can be solved from $(p^\alpha(x))''$.</p>
| 249
|
calculus
|
Solving exponential equation with two variables
|
https://math.stackexchange.com/questions/2713725/solving-exponential-equation-with-two-variables
|
<p>Given are two equations:</p>
<p>$$v_1 = v_0 (1 - e^{-\frac{t_1}{\tau}})$$</p>
<p>$$v_2 = v_0 (1 - e^{-\frac{t_2}{\tau}})$$</p>
<p>We know that</p>
<p>$$t_2 > t_1$$
$$v_2 > v_1$$
$$\tau > 0$$
$$v_0 > 0$$
$$\tau, v_0 \in ℝ$$</p>
<p>Given $t_1, v_1, t_2, v_2$, how can we solve for $\tau, v_0$?</p>
|
<p>Let $p:=e^{-t_1/\tau}$ so that $e^{-t_2/\tau}=p^\alpha$, where $\alpha$ is known. The equation can be written</p>
<p>$$v_1(1-p^\alpha)=v_2(1-p).$$</p>
<p>$\alpha$ can be an integer $>4$ so that there are certainly cases such that there is no closed-form solution. (In fact, there are only closed-form solutions for a few rational values of $\alpha$.)</p>
| 250
|
calculus
|
Error in evaluation of $\displaystyle\lim_{x\to 0} \frac{x\cos x - \ln (1+x)}{x^2}$
|
https://math.stackexchange.com/questions/2723256/error-in-evaluation-of-displaystyle-lim-x-to-0-fracx-cos-x-ln-1xx
|
<p>Evaluate $$\displaystyle\lim_{x\to 0} \frac{x\cos x - \ln (1+x)}{x^2}$$</p>
<p>Here's my method but that results into an error. </p>
<p>\begin{align}
\lim_{x\to 0} \frac{x\cos x - \ln (1+x)}{x^2}
&=\lim_{x\to 0}\frac{\cos x}{x} - \lim_{x\to 0}\left(\frac{1}{x}\right)\lim_{x\to 0}\left(\frac{\ln(1+x)}{x}\right)\\
&= \frac{\cos x}{x} - \frac{1}{x} \\
&= \frac{\cos x -1}{x}\\
&= 0 \quad \text{(either by L'hôpital or some manipulations)}
\end{align}</p>
<p>I have used the fact that </p>
<p>$$\lim_{x\to 0} \frac{\ln(1+x)}{x}=1$$</p>
<p>And the answer seems to be $\frac{1}{2}$.
I can do it in different ways (L'hôpital precisely) but please point out my error. </p>
|
<p>The error lies in these steps:</p>
<p>$$\lim_{x\to 0}\frac{\cos x}{x} - \lim_{x\to 0}\left(\frac{1}{x}\right)\lim_{x\to 0}\left(\frac{\ln(1+x)}{x}\right)\color{red}{=}\frac{\cos x}{x} - \frac{1}{x} \\$$
This is <strong>not correct</strong> because once you split the limit, you need to put the values. Evidently it fails here because you get $\infty - \infty$ as the answer which is not well defined. </p>
<p>Here after applying your limits after splitting, we cannot manipulate further, because we give the expression a value.</p>
<p>To solve the limit, consider using <em>L'hopital</em> or <em>taylor expansions</em> to get limit as $\tfrac{1}{2}$.</p>
| 251
|
calculus
|
Easy question: Why is $+ C$ outside the brackets?
|
https://math.stackexchange.com/questions/2741914/easy-question-why-is-c-outside-the-brackets
|
<p>$$100(-10te^-0.1t + 10 \int e^{-0.1t}dt) = 100(-10te^-0.1t -100e^{-0.1t})+C$$</p>
<p>Why is the $+C$ outside of the brackets if the integration was done inside? I'm looking at my math book and I'm baffled.</p>
<p>Thanks for the help.</p>
|
<p>As far as I can see, you are multiplying the integral by a constant. The $+C$ is just another constant, it can take any value. So it does not matter if you have $+1000C$ or $-0.00001C$, because they are just constants, and you can denote them as $+C$.</p>
| 252
|
calculus
|
Show that there exists a $x \in \mathbb{R}$ such that
|
https://math.stackexchange.com/questions/2840854/show-that-there-exists-a-x-in-mathbbr-such-that
|
<p>No idea where to start on this question. Any help is appreciated:</p>
<blockquote>
<p>$$\text{Show that there exists a $x \in \mathbb{R}$ such that } x^{21}+\frac{200}{1+x^4+\cos^2x}=120$$</p>
</blockquote>
<p>Thank you</p>
|
<p><strong>Hint:</strong></p>
<p>Define $f : \mathbb{R} \to \mathbb{R}$ as $f(x) = x^{21}+\frac{200}{1+x^4+\cos^2x}$.</p>
<p>Clearly $f$ is continuous and $\lim_{x\to\pm\infty} f(x) = \pm\infty$.</p>
<p>Hence $f$ is surjective.</p>
| 253
|
calculus
|
Curve Length Of A Unit Sphere Which Intersect With A Plane
|
https://math.stackexchange.com/questions/2862222/curve-length-of-a-unit-sphere-which-intersect-with-a-plane
|
<p>Find the curve length of the intersection between the unit sphere $x^2+y^2+z^2=1$ and the plane $x+y=1$</p>
<p>I have read <a href="https://math.stackexchange.com/questions/2004224/parametrization-of-the-intersection-between-a-sphere-and-a-plane">this</a> and <a href="https://math.stackexchange.com/questions/2339243/when-find-the-equation-of-intersection-of-plane-and-sphere?noredirect=1&lq=1">this</a> but I still do not manage, I will go <a href="http://tutorial.math.lamar.edu/Classes/CalcIII/CalcIII.aspx" rel="nofollow noreferrer">over this</a> I hope it will address the subject.</p>
<p>So we a unit sphere $x^2+y^2+z^2=1$, I know that the parametrization of a sphere is $(r \sin\rho \cos \theta, r\sin \rho \sin \theta, r\cos \rho)$</p>
<p>Now the plane is $x+y=1$ which can be written as $x=1-y$ and to satisfy that the sphere intersect with the plane with need $(1-y)^2+y^2+z^2=1$ which is $1-2y+y^2+y^2+z^2=1$ or $2y(y-1)+z^2=0$</p>
<p>I need to move to the parametrization to find the curve length, how do I do it?</p>
|
<p>Actually you don't need to do any integration.
First of all, you know the curve is a circle, so to find the perimeter, you only need to know the radius.</p>
<p>The idea is that you can find the distance between the plane and the centre of the ball (I will leave it as an exercise). Let me call it $D$. You know that radius of the ball is 1, so the radius of the circle is given by
$$\sqrt{1-D^2}$$
So the curve length is $2\pi \sqrt{1-D^2}$.</p>
| 254
|
calculus
|
How to prove the following statements about tangent lines to $y=ax^2+bx+c$?
|
https://math.stackexchange.com/questions/2958859/how-to-prove-the-following-statements-about-tangent-lines-to-y-ax2bxc
|
<p>Consider the graph of the equation <span class="math-container">$y=ax^2+bx+c$</span>, <span class="math-container">$a≠0$</span>. Prove the following:</p>
<p>a. If <span class="math-container">$a$</span> and <span class="math-container">$c$</span> have the same sign, that is <span class="math-container">$ac > 0$</span>, then there are exactly two tangent lines to the graph that pass through the origin.</p>
<p>b. If <span class="math-container">$a$</span> and <span class="math-container">$c$</span> have opposite signs, that is <span class="math-container">$ac < 0$</span>, then no tangent line to the graph passes through the origin.</p>
<p>c. What happens if <span class="math-container">$c=0$</span>?</p>
<p>I understand the concepts above if I draw it out, but I just don't know how to go about proving it with math. Thanks in advanced!</p>
|
<p>A line through the origin has the equation <span class="math-container">$y=mx$</span>. It tangents the parabola if it makes a "double" intersection with it.</p>
<p>In other words,
<span class="math-container">$$mx=ax^2+bx+c$$</span></p>
<p>must have a double root. This occurs when the discriminant</p>
<p><span class="math-container">$$(b-m)^2-4ac$$</span> is zero and is only possible when <span class="math-container">$ac\ge0$</span>, giving the solutions</p>
<p><span class="math-container">$$m=b\pm2\sqrt{ac}.$$</span></p>
<p>You can conclude for c.</p>
<hr>
<p>Alternatively, the tangent at <span class="math-container">$x=t$</span> has the equation</p>
<p><span class="math-container">$$y=(2at+b)(x-t)+at^2+bt+c.$$</span></p>
<p>It passes through the origin if</p>
<p><span class="math-container">$$0=-(2at+b)t+at^2+bt+c$$</span> or</p>
<p><span class="math-container">$$-at^2+c=0.$$</span></p>
<p>The discussion of the number of solutions in <span class="math-container">$t$</span> is easy.</p>
| 255
|
calculus
|
How fast is the area of rectangle increasing?
|
https://math.stackexchange.com/questions/3077488/how-fast-is-the-area-of-rectangle-increasing
|
<p>The length of a rectangle is increasing at a rate of 8 cm/s and
its width is increasing at a rate of <span class="math-container">$3$</span> cm/s . When the length is
20 cm and the width is 10 cm, how fast is the area of the rectangle
increasing?</p>
<p>So on internet I found a solution but I didn't do that way and I am still thinking that I am not wrong but the answer is not the same. I am gonna write both the solutions which I found on int and by myself and I will be waiting your help.</p>
<p>Which I found on the int:
<span class="math-container">$A=lw$</span> then take derivative <span class="math-container">$\frac{dA}{dt}= \frac{dl}{dt}.w + l.\frac{dw}{dt}$</span></p>
<p>using given number <span class="math-container">$\frac{dA}{dt}= (8)(10) + (20)(3)$</span></p>
<p>My answer: given numbers--> <span class="math-container">$\frac{dl}{dt}= 8$</span>, <span class="math-container">$\frac{dw}{dt}=3$</span>, <span class="math-container">$l=20$</span>, <span class="math-container">$w =10$</span></p>
<p>so <span class="math-container">$A=wl$</span> when I wanna write <span class="math-container">$w$</span> in terms of <span class="math-container">$l$</span> ----> <span class="math-container">$l=2w$</span></p>
<p>so <span class="math-container">$A=2w*w$</span> when I take derivative of it ---> <span class="math-container">$\frac{dA}{dw}= 4w $</span></p>
<p>according to chain rule <span class="math-container">$\frac{dA}{dt}= \frac{dA}{dw}\frac{dw}{dt}$</span> </p>
<p>when I put the numbers ----> <span class="math-container">$4w*3$</span> and we know that <span class="math-container">$w=10$</span> </p>
<p>It should be 120. I think I found my mistake but still couldn't understand why. I write <span class="math-container">$w$</span> in terms of l but if I do the other way then the result is 160. What am I doing wrong?</p>
|
<p>One millisecond later, the sides are <span class="math-container">$20.008$</span> and <span class="math-container">$10.003$</span> and the relation <span class="math-container">$l=2w$</span> is no more true.</p>
<p>The rate of increase of the area must be close to</p>
<p><span class="math-container">$$\frac{20.008\cdot10.003-20\cdot10}{10^{-3}}=140.024.$$</span></p>
<p>With one microsecond, we get</p>
<p><span class="math-container">$$\frac{20.000008\cdot10.000003-20\cdot10}{10^{-6}}=140.000024.$$</span></p>
<p>This confirms the answer <span class="math-container">$140$</span>.</p>
<p>The reason why your method doesn't work is because</p>
<p><span class="math-container">$$\frac{20}{10}\ne\frac{8}{3}.$$</span></p>
| 256
|
calculus
|
Differential Notation Misunderstanding
|
https://math.stackexchange.com/questions/3104366/differential-notation-misunderstanding
|
<p>Consider I have a function <span class="math-container">$v=e^u$</span> where u is from the set of all Real numbers. Now, if I take the derivative here, I can get <span class="math-container">$dv/du = e^u$</span>. If I multiply both sides by the <span class="math-container">$du$</span>, I will get <span class="math-container">$dv=e^u*du$</span>. However, if I interpret this <span class="math-container">$dv$</span> and <span class="math-container">$du$</span> as <span class="math-container">$\Delta v$</span> and as <span class="math-container">$\Delta u$</span>, I can write this statement: <span class="math-container">$\Delta v = e^{\Delta u}$</span>. If we take the limit as <span class="math-container">$u$</span> approaches <span class="math-container">$0$</span>, then we will get <span class="math-container">$dv=e^{du}$</span> However, this is not the same as <span class="math-container">$dv=e^u*du$</span>. What went wrong?</p>
<p>Thanks :)</p>
|
<p><span class="math-container">$$\Delta v=e^{u+\Delta u}-e^u=e^u(e^{\Delta u}-1),$$</span> not <span class="math-container">$$e^{\Delta u}.$$</span></p>
<p>By the way, </p>
<p><span class="math-container">$$\lim_{\Delta u\to0}\frac{e^{\Delta u}-1}{\Delta u}=1,$$</span> and this justifies</p>
<p><span class="math-container">$$\frac{dv}{du}=e^u=v.$$</span></p>
| 257
|
calculus
|
Finding the interval of when this function decreases
|
https://math.stackexchange.com/questions/3175234/finding-the-interval-of-when-this-function-decreases
|
<p>From an old math exam I found the question to find the interval for when a function is decreasing(so it can be used for the Integration test). But I can't seem to figure it out.</p>
<p>The function in question is:</p>
<p><span class="math-container">$f(x) =\dfrac{\sqrt{x}}{(x^\frac{3}{2} +2)^2}$</span></p>
<p>There is apparently an effective way to this because it was a small question with just a few points.</p>
<p>So can anybody show me what i am missing?</p>
|
<p>You are looking for the interval where the derivative is negative.</p>
<p>I will use a little trick, for comfort: as <span class="math-container">$x\ge0$</span>, I will replace <span class="math-container">$x$</span> by <span class="math-container">$z^2$</span> to get rid of the half-exponents. As the relation <span class="math-container">$x=z^2$</span> is monotonous, this will not cause trouble.</p>
<p>Now,</p>
<p><span class="math-container">$$\left(\frac{z}{(z^3+2)^2}\right)'=\frac{(z^3+2)^2-6z^3(z^3+2)}{(z^3+2)^4}$$</span> and after simplification the numerator is</p>
<p><span class="math-container">$$2-5z^3.$$</span></p>
<p>Hence</p>
<p><span class="math-container">$$z\ge\sqrt[3]{\frac 25}$$</span> or
<span class="math-container">$$x\ge\left(\frac 25\right)^{2/3}.$$</span></p>
| 258
|
calculus
|
convergent series for $\sum_{n=1}^{\infty}\frac{n!}{n^n}$
|
https://math.stackexchange.com/questions/3179505/convergent-series-for-sum-n-1-infty-fracnnn
|
<p>Help me please , I am not able to solve this problem.I have tried in many ways to figure out such as Ration test , Integral test , Comparison test , Limit Comparison Test , Root Test but i can't find the way out . This is my first question and i'm not good at English. If there is something wrong or you are not comfortable with my language usage I'm so sorry.</p>
|
<p>The usual approach to factorial-based problems is to use <a href="https://en.wikipedia.org/wiki/Stirling%27s_approximation" rel="nofollow noreferrer">Stirling's approximation</a> <span class="math-container">$n!\approx\sqrt{2\pi n}n^ne^{-n}$</span>, which shows this series converges provided <span class="math-container">$\sum_{n\ge 1}\sqrt{2\pi n}e^{-n}$</span> does, which is the case. If you didn't know this result, another option is to note that <span class="math-container">$$\frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}}=\left(1+\frac{1}{n}\right)^{-n}\approx\frac{1}{e},$$</span>completing the problem by the ratio test.</p>
| 259
|
calculus
|
What do the following parametric curves represent?
|
https://math.stackexchange.com/questions/3205317/what-do-the-following-parametric-curves-represent
|
<p>(a) <span class="math-container">$x(v)= 3, y(v)= 4, z(v)= v$</span> for <span class="math-container">$−\infty < v < \infty$</span>,</p>
<p>(b) <span class="math-container">$x(t)= 3\cos(t), y(t)= 2\sin(t), z(t)= 3t−1$</span> for <span class="math-container">$0 \leq t < 2\pi$</span>.</p>
<p>I have no idea where to start. Our lecturer in class only went through parametric curves with 2 parameters. Any help is much appreciated. Thank you</p>
|
<p>The graph of <span class="math-container">$$x(v)= 3, y(v)= 4, z(v)= v$$</span> is a vertical line passing through the point <span class="math-container">$(3,4,0)$</span> since <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are fixed and <span class="math-container">$z$</span> runs freely from <span class="math-container">$-\infty $</span> to <span class="math-container">$\infty$</span></p>
<p>The graph of </p>
<p><span class="math-container">$$x(t)= 3\cos(t), y(t)= 2\sin(t), z(t)= 3t−1 ,0 \leq t < 2\pi $$</span> is a spiral curve whose projection on the <span class="math-container">$x-y$</span> plane is an ellipse and its <span class="math-container">$z$</span> coordinate runs from <span class="math-container">$-1$</span> to <span class="math-container">$6\pi -1$</span></p>
<p>A three D grapher will help you to visualize the curve in part <span class="math-container">$b$</span> better. </p>
| 260
|
calculus
|
Can I use the partial implicit differentiation with $x = e^xy$?
|
https://math.stackexchange.com/questions/3575733/can-i-use-the-partial-implicit-differentiation-with-x-exy
|
<p>I want to know if I can use the partial implicit differentiation with this problem.</p>
<p>What is the derivative of <span class="math-container">$x = e^{xy}$</span>?</p>
|
<p>Considering <span class="math-container">$y=f(x)$</span>, you get:
<span class="math-container">$$(x)'_x=(e^{xy})'_x \Rightarrow \\
1=e^{xy}\cdot (y+xy')\Rightarrow \\
y'=\frac{1-ye^{xy}}{xe^{xy}}$$</span>
Wolfram <a href="https://www.wolframalpha.com/input/?i=derivative%20x%3De%5Exy" rel="nofollow noreferrer">answer</a>.</p>
<p>Alternatively:
<span class="math-container">$$ F(x,y)=x-e^{xy}=0\\
\frac{dy}{dx}=-\frac{F_x}{F_y}=-\frac{1-ye^{xy}}{-xe^{xy}}=\frac{1-ye^{xy}}{xe^{xy}}$$</span></p>
| 261
|
calculus
|
How to solve $\frac{y}{y'}=ln(y)$ for $y$?
|
https://math.stackexchange.com/questions/3613859/how-to-solve-fracyy-lny-for-y
|
<p><span class="math-container">$\frac{y}{y'}=\ln(y)$</span></p>
<p><span class="math-container">$ydx=\ln(y)dy$</span></p>
<p><span class="math-container">$dx=\frac{\ln y}{y} dy$</span></p>
<p>]<span class="math-container">$\ln(y) =z$</span> => <span class="math-container">$dz=dy/y$</span></p>
<p>then <span class="math-container">$x+C=z^2$</span></p>
<p>Am I doing this right?</p>
|
<p>Almost: <span class="math-container">$x+C=\int zdz=\frac12z^2$</span>.</p>
| 262
|
calculus
|
Intuition behind integrating and differentiating determinants?
|
https://math.stackexchange.com/questions/3689270/intuition-behind-integrating-and-differentiating-determinants
|
<p><a href="https://byjus.com/jee/differentiation-integration-of-determinants/" rel="nofollow noreferrer">https://byjus.com/jee/differentiation-integration-of-determinants/</a></p>
<p>I saw this and I can't understand how this formula was derived, like why can we integrate row wise and add up determinants? Is there an intuition/ proof behind this?</p>
|
<p><strong>Hint</strong>: expanding
<span class="math-container">$$
\det \left(
\begin{matrix}
f_1(x) & g_1(x) \\
f_2(x) & g_2(x)
\end{matrix}
\right) = f_1(x)g_2(x)-f_2(x)g_1(x)
$$</span>
and differentiating,
<span class="math-container">$$f_1'(x)g_2(x)+f_1(x)g_2'(x)-f_2'(x)g_1(x)-f_2(x)g_1'(x),$$</span>
you get the derivative of <span class="math-container">$\det \left(
\begin{matrix}
f_1(x) & g_1(x) \\
f_2(x) & g_2(x)
\end{matrix}
\right)$</span>. You can rewrite the result as
<span class="math-container">$$
\det
\left(
\begin{matrix}
f_1'(x) & g_1'(x) \\
f_2(x) & g_2(x)
\end{matrix}
\right)+
\det \left(
\begin{matrix}
f_1(x) & g_1(x) \\
f_2'(x) & g_2'(x)
\end{matrix}
\right).
$$</span>
Use the same idea with integration.</p>
| 263
|
calculus
|
Questions about parametric equations
|
https://math.stackexchange.com/questions/3720326/questions-about-parametric-equations
|
<p>Consider the parametric equations: <span class="math-container">$$x=t^3-3t, \; \; y=t^2+t+1.$$</span></p>
<ol>
<li>What is the lowest point on this parametric curve?</li>
<li>For what values of <span class="math-container">$t$</span> does the curve move left, move right, move up and move down?</li>
<li>When is the curve concave up?</li>
<li>Find the area contained inside the loop of this curve if the curve intersects itself at the point <span class="math-container">$(-2,3)$</span>.</li>
</ol>
<p>Well for 1. I get <span class="math-container">$(2,1)$</span> after calculating <span class="math-container">$\frac{dy}{dx}=0$</span> for <span class="math-container">$t$</span>.<br />
I’m not sure on 2.<br />
For 3. we just find <span class="math-container">$t$</span> such that <span class="math-container">$\frac{d^2y}{dx^2}>0$</span>.<br />
For 4. I’m not sure on this one too.<br />
Thanks for the help.</p>
|
<p><span class="math-container">$$\dot x=3t^2-3$$</span> is negative for <span class="math-container">$-1<t<1$</span>, meaning that the curve is traversed from right to left in this range, and conversely.</p>
<p><span class="math-container">$$\dot y=2t+1$$</span> is negative when <span class="math-container">$t\le-\dfrac12$</span> and the curve is traversed from top to bottom in this range and conversely. The lowest point is reached at <span class="math-container">$t=-\dfrac12$</span>.</p>
<p>For the final question, use a curvilinear integral,</p>
<p><span class="math-container">$$A=\oint x\,dy=\int x\dot y\,dt.$$</span></p>
| 264
|
calculus
|
Odd Intuitive proof for L'Hospital's rule
|
https://math.stackexchange.com/questions/3751075/odd-intuitive-proof-for-lhospitals-rule
|
<p>A professor of mine intuitively showed why L'Hospital's rule works for the <span class="math-container">$0/0$</span> case (by some simplifying assumptions). I understood that. He then contended that this is enough to prove that the rule works for the <span class="math-container">$\infty / \infty$</span> case. This ,he claimed, is because if the function for which we want the limit (<span class="math-container">$ =f(x)/g(x) $</span>) is of the <span class="math-container">$\infty / \infty$</span> form , we can write it as <span class="math-container">$ (1/g(x)) \div (1/f(x)) $</span> and now the limit has both numerator and denominator approaching <span class="math-container">$0$</span>. (this has also been argued <a href="https://www.youtube.com/watch?v=Hu0z-sFfF8Y" rel="nofollow noreferrer">elsewhere</a> 11:35 onwards) But I did not understand this last bit. I concede that both numerator and denominator behave so but the L'hopital rule we intuitively "proved" demanded that we should differentiate both the numerator and denominator <em>if</em> both approach zero. But differentiating both in this new form does not result in <span class="math-container">$f'(x) / g'(x)$</span> .</p>
<p>I understand this was not intended to be a rigorous proof, but this does not seem to work intuitively either. Have I missed something or is the proof odd ?</p>
|
<p>Yes, even the intuitive approach in this case will require some more work. We'll need the fact that <span class="math-container">$\left(\frac{1}{f}\right)'=-\frac{f'}{f^2}$</span>. Now apply the <span class="math-container">$0/0$</span> rule to <span class="math-container">$\frac{1/g}{1/f}$</span> to get that the limit of <span class="math-container">$\frac{f}{g}$</span> is the same as that of <span class="math-container">$\frac{g'/g^2}{f'/f^2}=\frac{g'}{f'}\frac{f^2}{g^2}$</span> (if this exists):</p>
<p><span class="math-container">$$\lim_{x\to x_0}\frac{f}{g}=\lim_{x\to x_0}\frac{g'}{f'}\frac{f^2}{g^2}.$$</span></p>
<p>Now multiply both sides by <span class="math-container">$\frac{f'g}{g'f}$</span> to get the desired result.</p>
<p>But do keep in mind that we didn't take care of the subtleties, like why we can just multiply by stuff in the limit. But you wanted an intuitive proof, not a rigorous one, so there it is.</p>
| 265
|
calculus
|
Differentiating $V_c=V_s(1-e^{-t/T})$
|
https://math.stackexchange.com/questions/3870146/differentiating-v-c-v-s1-e-t-t
|
<p>I have a formula for an electronic circuit as follows</p>
<p><span class="math-container">$$V_c=V_s(1-e^{-t/T})$$</span>
Apparently this differentiates to <span class="math-container">$$(V_s/T) e^{-t/T}$$</span></p>
<p>I say apparently because I looked up the answer which is a bit naughty but I can't figure it out. Even with the answer I'm unable to determine how to get from a to b.</p>
<p>Can someone explain the process here and dumb it down a bit please. I'm a middle aged man and well out of practice</p>
<p>Thanks for help</p>
|
<p><span class="math-container">\begin{align}
\frac{d}{dt} V_s(1-e^{-t/T})
&= V_s \frac{d}{dt} (1-e^{-t/T})
&\text{$V_s$ is a constant factor, can be pulled out of the derivative}
\\
&= V_s \frac{d}{dt}(- e^{-t/T}) & \text{$1$ is additive constant, has derivative zero}
\\
&= -V_s \frac{d}{dt} e^{-t/T} & \text{$-1$ is constant factor, can be pulled out}
\end{align}</span></p>
<p>To differentiate <span class="math-container">$e^{-t/T}$</span>, use the chain rule. With <span class="math-container">$f(x) = e^x$</span> and <span class="math-container">$g(t) = -t/T$</span>, we have <span class="math-container">$f(g(t)) = e^{-t/T}$</span>. Using the fact that <span class="math-container">$f'(x) = e^x$</span> and <span class="math-container">$g'(t) = -1/T$</span>, we have
<span class="math-container">$$\frac{d}{dt} e^{-t/T} = f'(g(t)) \cdot g'(t) = e^{-t/T} \cdot (-1/T).$$</span>
Plugging this back in above gives you the answer.</p>
| 266
|
calculus
|
For what values of $c$ does the curve $ y = cx^{3} + e^{x} $ have inflection points?
|
https://math.stackexchange.com/questions/3924969/for-what-values-of-c-does-the-curve-y-cx3-ex-have-inflection-poi
|
<p>For what values of <span class="math-container">$c$</span> does the curve <span class="math-container">$ y = cx^{3} + e^{x} $</span> have inflection points?</p>
<p>at first I found first derivative <span class="math-container">$ f^{'}(x) = 3cx^2 + e^{x} $</span></p>
<p>then second derivative <span class="math-container">$ f^{''} (x) = 6cx + e^{x} $</span></p>
<p>now the second derivative should go from positive to negative or from negative to positive yes?
and I wrote this system
<span class="math-container">$$ -e^{x} < 6cx < 0 $$</span></p>
<p>so give me a hint i don't know that to do</p>
|
<p>Write the equation <span class="math-container">$y''=6cx+e^x=0$</span> as a system
<span class="math-container">$$\begin{cases}
y=e^x\\
y=-6cx\\
\end{cases}
$$</span></p>
<p>we see that for <span class="math-container">$c>0$</span> there is one and only one solution: one inflection point.</p>
<p>The tangent passing through the origin <span class="math-container">$(0,0)$</span> has equation <span class="math-container">$y=ex$</span>, thus</p>
<p>for negative <span class="math-container">$c$</span> we have no solution when <span class="math-container">$-6c<e$</span> that is <span class="math-container">$-\frac{e}{6}<c<0$</span></p>
<p>for <span class="math-container">$c<-\frac{e}{6}$</span> we have two intersection that is two inflection points.</p>
<p>For <span class="math-container">$c=-\frac{e}{6}$</span> second derivative is <span class="math-container">$y''=e^x-ex$</span> which is zero at <span class="math-container">$x=1$</span>. This is not an inflection point since third derivative <span class="math-container">$y^{(3)}=e^x-e$</span> is zero too at <span class="math-container">$x=1$</span>.</p>
<p>It happens like in <span class="math-container">$y=x^4$</span> where second derivative <span class="math-container">$y''=12x^2$</span> is zero at <span class="math-container">$x=0$</span> but there is no inflection point since <span class="math-container">$y^{(3)}(0)=0$</span>.</p>
<p><a href="https://i.sstatic.net/darJS.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/darJS.png" alt="enter image description here" /></a></p>
| 267
|
calculus
|
Find the derivatives to $f(x)=4/x^2$ and $g(t)=(t-5)/(1+\sqrt{t}\,)$
|
https://math.stackexchange.com/questions/979347/find-the-derivatives-to-fx-4-x2-and-gt-t-5-1-sqrtt
|
<p>I have these two assignments: </p>
<blockquote>
<p>Find the derivatives to (a) $f(x)=4/x^2$ and b) $g(t)=(t-5)/(1+\sqrt{t}\,)$ by using the definition $$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}=f'(x)$$</p>
</blockquote>
<p>a) This is my attempt at (a); am I correct?
$$\lim_{h \to 0} \frac{\left(\displaystyle\frac{4}{(x+h)^2}-\frac{4}{x^2}\right)}{h}=\frac{1}{h}\left(\displaystyle\frac{4}{(x+h)^2}-\frac{4}{x^2}\right)$$
Then I found the common denominator
$$\begin{align*}
\frac{4}{(x+h)^2}-\frac{4}{x^2}
&=\frac{4x^2}{(x+h)^2x^2}-\frac{4(x+h)^2}{(x+h)^2x^2} \\[6pt]
&=\frac{4x^2-4x^2-4h^2-8xh}{(x+h)^2x^2} \\[6pt]
&=\frac{-4h^2-8xh}{(x+h)^2x^2}
\end{align*}$$
Then I expand the denominator to
$$\frac{-4h^2-8xh}{x^4+2x^3h+h^2x^2}
=\frac{\left(-4h^2-8h\right)}{x^3+2x^3h+h^2x^2}\frac{1}{h}
=\frac{\left(-4h^2-8h\right)}{hx^3+2x^3h+h^2x^2}$$
And $h$ goes out with:
$$\frac{-4h^2-8}{x^3+2x^3h+h^2x^2}
= \lim_{h \to 0}\frac{-4\cdot0^2-8}{x^3+2x^3\cdot 0+0^2\cdot x^2}
= -\frac{8}{x^3}$$
I dont know how to do part (b)</p>
|
<ol>
<li>First function</li>
</ol>
<p>$$f'(x)=\lim_{h \to 0} \frac{\left(\frac{4}{(x+h)^2}-\frac{4}{x^2}\right)}{h}=\frac{1}{h}\left(\frac{4}{(x+h)^2}-\frac{4}{x^2}\right)=\lim_{h \to 0}\frac{1}{h}\frac{-4h^2-8xh}{(x+h)^2x^2}=\lim_{h \to 0}\frac{1}{h}\frac{-4h(h+2x)}{(x+h)^2x^2}= \lim_{h \to 0}\frac{-4(h+2x)}{(x+h)^2x^2}= \frac{-8x}{x^4}= -\frac{8}{x^3}, $$</p>
<p>for all $x\neq 0$.</p>
<ol start="2">
<li>Second function (in what follows we consider $t\geq 0$)</li>
</ol>
<p>$$f'(t)=\lim_{h \to 0} \frac{1}{h}\left(\frac{t+h-5}{1+\sqrt{t+h}}-\frac{t-5}{1+\sqrt{t}}
\right) =
\lim_{h \to 0} \frac{1}{h}\left(\frac{(t+h-5)(1+\sqrt{t})-(t-5)(1+\sqrt{t+h})}{(1+\sqrt{t+h})(1+\sqrt{t})}\right);$$</p>
<p>Now we simplify the numerator arriving at</p>
<p>$$f'(t)=
\lim_{h \to 0}
\frac{1}{h}\left(\frac{-(t-5)(\sqrt{t+h}-\sqrt{t})+h(1+\sqrt{t})}{(1+\sqrt{t+h})(1+\sqrt{t})}\right)= \lim_{h \to 0} \frac{1}{h}\frac{-(t-5)(\sqrt{t+h}-\sqrt{t})}{(1+\sqrt{t+h})(1+\sqrt{t})}+\frac{1}{h}\frac{h(1+\sqrt{t})}{(1+\sqrt{t+h})(1+\sqrt{t})}=
\lim_{h \to 0} \frac{1}{h}\frac{-(t-5)(\sqrt{t+h}-\sqrt{t})}{(1+\sqrt{t+h})(1+\sqrt{t})}+\frac{1}{(1+\sqrt{t+h})};
$$</p>
<p>All we need is to use the "trick":</p>
<p>$$\frac{1}{h}\frac{-(t-5)(\sqrt{t+h}-\sqrt{t})}{(1+\sqrt{t+h})(1+\sqrt{t})}=
\frac{1}{h}\frac{-(t-5)(\sqrt{t+h}-\sqrt{t})}{(1+\sqrt{t+h})(1+\sqrt{t})}\frac{(\sqrt{t+h}+\sqrt{t})}{(\sqrt{t+h}+\sqrt{t})}=
\frac{1}{h}\frac{-(t-5)h}{(1+\sqrt{t+h})(1+\sqrt{t})}\frac{1}{(\sqrt{t}+\sqrt{t+h})}=
-\frac{(t-5)}{(1+\sqrt{t+h})(1+\sqrt{t})(\sqrt{t}+\sqrt{t+h})}.
$$</p>
<p>In summary:</p>
<p>$$f'(t)= \lim_{h \to 0} -\frac{(t-5)}{(1+\sqrt{t+h})(1+\sqrt{t})(\sqrt{t}+\sqrt{t+h})}+\frac{1}{(1+\sqrt{t+h})}= \\
-\frac{(t-5)}{2(1+\sqrt{t})^2\sqrt{t}}+\frac{1}{(1+\sqrt{t})}. $$ </p>
| 268
|
calculus
|
How to find the limit
|
https://math.stackexchange.com/questions/3813828/how-to-find-the-limit
|
<p>How can we find the limit
<span class="math-container">$$\lim_{x\to 0} \frac{(e^x-1-x)^2}{x(\sin x -x)}$$</span>?</p>
|
<p><span class="math-container">$$\lim_{x\to 0} \frac{(e^x-1-x)^2}{x(\sin x -x)} = (\lim_{x\to 0} \frac{e^x-1-x}{x^2})^2\cdot\lim_{x\to 0} \frac{x^3}{\sin x -x} = (\frac{1}{2})^2\cdot(-6).$$</span></p>
<p>The last two limits are calculated with L'Hospital's rule.</p>
| 269
|
calculus
|
Find the only f=vt that has f(2t)=4f(t)
|
https://math.stackexchange.com/questions/4086587/find-the-only-f-vt-that-has-f2t-4ft
|
<p>I'm just starting calculus 1 and I don't know how to solve this. Can someone please help?</p>
<p>The problem below involves linear functions <span class="math-container">$f(t) = vt + C$</span>. Find the constants v and C.</p>
<p>Find the only <span class="math-container">$f=vt$</span> that has <span class="math-container">$f(2t)=4f(t)$</span>. show that every <span class="math-container">$f=\frac{1}{2}at^2$</span> has this property.</p>
|
<p>Assuming <span class="math-container">$v$</span> is constant: <span class="math-container">$f(2t) = 4f(t)$</span> implies <span class="math-container">$v(2t) = 4(vt)$</span> and therefore <span class="math-container">$2vt = 0$</span>.</p>
<p>This yields either <span class="math-container">$t$</span> or <span class="math-container">$v$</span> must be <span class="math-container">$0$</span>, but since the equation has to hold for all <span class="math-container">$t$</span> this means <span class="math-container">$v$</span> must be <span class="math-container">$0$</span>.</p>
<p>Now let <span class="math-container">$f(t) = \frac{1}{2}at^2$</span>. <span class="math-container">$f(2t) = \frac{1}{2}a(2t)^2 = \frac{1}{2}a(4t^2) = 4(\frac{1}{2}at^2) = 4f(t)$</span>, regardless of <span class="math-container">$a$</span>.</p>
<p>Hope this helps!</p>
| 270
|
calculus
|
$2^x+2^{-x} = 5$, solve $4^x+4^{-x}$ using the rules of exponents
|
https://math.stackexchange.com/questions/2017827/2x2-x-5-solve-4x4-x-using-the-rules-of-exponents
|
<blockquote>
<p>$2^x+2^{-x} = 5$</p>
<p>Solve:</p>
<p>$4^x+4^{-x}$</p>
</blockquote>
<p>I know I can solve this by solving the equation $2^x+2^{-x} = 5$ and then replacing $x$ on the second one with the result, but I found that to be too lengthy and overcomplicated.</p>
<p>Is there a faster and simpler way of solving this using some rules?</p>
<p>The solution is $4^x+4^{-x} = 23$</p>
|
<p>$4^x+4^{-x}=2^{2x}+2^{-2x}=(2^x+2^{-x})^2-2\cdot 2^x \cdot 2^{-x}=25-2=23$</p>
| 271
|
calculus
|
How to prove that sin(1/x) is continuous at x≠0
|
https://math.stackexchange.com/questions/3672511/how-to-prove-that-sin1-x-is-continuous-at-x%e2%89%a00
|
<p>Can someone help with the proof that sin(1/x) is continuous for all x≠0.(By the help of epsilon delta defination)</p>
<p>I am sharing what I have tried so far not much though.
I have figured out that modulus value of</p>
<p>sin(1/x)-sin(1/a) is less than modulus value of</p>
<p>(1/x)-(1/a) for all a≠0.From here I am thinking to figure out the required "delta" for the epsilon. But I am unable to do so. Please help me with the solution.</p>
|
<p><span class="math-container">$|\frac 1 x -\frac 1 a|=\frac {|x-a|} {|a||x|} \leq \frac {|x-a|} {|a|(|a|-|x-a|)} <\frac {|x-a|} {|a|(|a|/2)}$</span> if <span class="math-container">$|x-a| <|a|/2$</span>. Hence <span class="math-container">$|\frac 1 x -\frac 1 a|<\epsilon$</span> if <span class="math-container">$|x-a| <|a|/2$</span> and <span class="math-container">$|x-a| <\epsilon (|a|^{2} /2)$</span>. Take <span class="math-container">$\delta \in (0, \min \{|a|/2, \epsilon (|a|^{2} /2)\})$</span>.</p>
| 272
|
calculus
|
Function increase or decrease
|
https://math.stackexchange.com/questions/1732642/function-increase-or-decrease
|
<p>The question is</p>
<blockquote>
<p><span class="math-container">$$\text{Let } f(r) = r^{1/3} + \frac 1r \text{ for } r>0$$</span>
a) Determine where the function <span class="math-container">$f$</span> is increasing or decreasing.</p>
<p>b) Determine where the function <span class="math-container">$f$</span> is concave upward or downward.</p>
</blockquote>
<p>I've worked out</p>
<p><span class="math-container">$$f'(r)= \frac 13r^{-2/3} -\frac 1{r^2}$$</span></p>
<p>but don't know how to work out the values of <span class="math-container">$r$</span>.</p>
<p>What do I do next?</p>
|
<p>I'll help you with part a). We can see that $f'(r)$ is defined for all $r>0$, so we just need to find where $f'(r)=0$. Let's start by rewriting that as</p>
<p>$$\frac 13r^{-2/3} -r^{-2}=0$$</p>
<p>Move the second term to the other side to get</p>
<p>$$\frac 13r^{-2/3}=r^{-2}$$</p>
<p>Taking the reciprocal of each side, remembering that $\dfrac 1{a^{-b}}=a^b$,</p>
<p>$$3r^{2/3}=r^2$$</p>
<p>Take both sides to the third power to get</p>
<p>$$3^3r^2=r^6$$</p>
<p>Divide by $r^2$ to get</p>
<p>$$3^3=r^4$$</p>
<p>and finally</p>
<p>$$r=3^{3/4}=\sqrt[4]{27}$$</p>
<p>This is where the derivative is zero. Now look at the intervals $(0,3^{3/4})$ and $(3^{3/4},\infty)$ to find where the derivative is positive and where it is negative, i.e. where the function increases and where it decreases. Check all this with a graph:</p>
<p><a href="https://i.sstatic.net/lnhXU.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/lnhXU.jpg" alt="enter image description here"></a></p>
<p>Let me know if you need more help.</p>
| 273
|
calculus
|
Is there anywhere where there are in-depth walkthroughs of problems on Stewart's Calculus?
|
https://math.stackexchange.com/questions/4190525/is-there-anywhere-where-there-are-in-depth-walkthroughs-of-problems-on-stewarts
|
<p>I'm currently taking calc 2 and using Stewart's calculus. My major qualms with the book is the lack of examples. On a scale of 1-10, the practice problems in the chapters are like 1-3, then the example problems are all very difficult without walkthroughs at the end of the book. I'm struggling to figure out where I went wrong when I get a wrong answer. Is there any site which aggregates in-depth solutions of difficult problems?</p>
<p>I'm currently on arc-lengths and surface area, and I keep getting crazy integrals which I have no idea how to solve. For example, I I've recently gotten <span class="math-container">$∫\sqrt{(1+cos^2)}$</span>, <span class="math-container">$∫\sqrt{(1+(1-1/x)^2}$</span>. I'm not sure if I'm just setting up the problems wrong, or what...</p>
<p>I got <span class="math-container">$∫\sqrt{(1+cos^2)}$</span> from y = <span class="math-container">$\sin x$</span> and <span class="math-container">$∫\sqrt{(1+(1-1/x)^2}$</span> from y = x - <span class="math-container">$\ln{x}$</span>.</p>
| 274
|
|
calculus
|
Taking the Derivative of Both Sides of an Equation
|
https://math.stackexchange.com/questions/3240322/taking-the-derivative-of-both-sides-of-an-equation
|
<p>If we have an equation like </p>
<p>y = x^2</p>
<p>This implies that </p>
<p>y’ = 2x</p>
<p>If we have an equation like </p>
<p>x = 4x^2</p>
<p>and we take the derivative of both sides we get</p>
<p>1 = 8x</p>
<p>With the solution x = 1/8, which is not the solution to the original equation. This is instead the value where the slopes of both sides of the equation are equal.</p>
<p>So when (and why) is an equation resulting from the derivation of both sides of some original equation implied by the original equation.</p>
|
<p>Your problem comes from thinking that you take derivatives of equations. But you don't. You take derivatives of <em>functions</em>.</p>
<p>For example, you can think of the equation
<span class="math-container">$$
x^2 -1 = (x-1)(x+1)
$$</span>
as telling you two different ways to write the same function of <span class="math-container">$x$</span>. When you differentiate the two expressions using the rules for derivatives you will get the same answer, in two different forms.</p>
<p>The equation in your question,
<span class="math-container">$$
x = 4x^2 ,
$$</span>
does not say that two functions are equal. It asks for the value of <span class="math-container">$x$</span> at which those two functions happen to have the same value. That's just a number. It makes no sense to take the derivative.</p>
| 275
|
calculus
|
Determining the uniform convergence
|
https://math.stackexchange.com/questions/3028636/determining-the-uniform-convergence
|
<p>Show that the series ,whose partial sum of n terms is <span class="math-container">$S_n=\frac{x}{(1+nx^2)}$</span>, converges uniformly for all real x.</p>
<p>I found that the series is pointwise convergent to 0 for all x.
For showing uniform convergence, I found out that the function S attains maximum value at <span class="math-container">$x=\frac{1}{\sqrt{n}}$</span>..i.e. <span class="math-container">$M= \frac{1}{\sqrt{n}}$</span>.which tends to 0 as n tends to infinity.So it is proved that it is uniformly convergent.</p>
<p>However i am doubtful if this works for all real x or for particular closed interval.</p>
|
<p>Note the sum of the first <span class="math-container">$n$</span> terms as <span class="math-container">$S_n(x)$</span>. We have </p>
<p><span class="math-container">$$ S_n(x) = \frac{x}{1+nx^2}$$</span>
and <span class="math-container">$ \lim_{n \to \infty} S_n(x) = 0 \forall x \in \mathbb{R}.$</span> Therefore the function <span class="math-container">$S_n$</span>(x) converges point wise to <span class="math-container">$0$</span> on <span class="math-container">$\mathbb{R}$</span>.
To show uniform convergence to <span class="math-container">$0$</span> on <span class="math-container">$D \subset \mathbb{R}$</span> we must show that
<span class="math-container">$$ \forall \epsilon, \exists N \in \mathbb{N} : \sup_{x \in D}\vert S_n(x) - 0 \vert < \epsilon.$$</span></p>
<p>You have shown that <span class="math-container">$$ \sup_{x \in \mathbb{R}} \vert S_n (x)\vert = S_n(\frac{1}{\sqrt{n}}) = \frac{1}{n + 1} \rightarrow 0 \text{ as } n \to \infty.$$</span> </p>
<p>Remember by definition that <span class="math-container">$ \lim_{n \to \infty } a_n = L \in \mathbb{R} \iff \forall \epsilon , \exists N \in \mathbb{N},n > N : \vert a_n - L \vert < \epsilon$</span>. Therefore take <span class="math-container">$a_n = \sup_{x \in \mathbb{R}} \vert S_n(x)\vert$</span>.</p>
<p>By definition <span class="math-container">$S_n$</span> converges uniformly to <span class="math-container">$0$</span> on <span class="math-container">$\mathbb{R}$</span>.</p>
| 276
|
calculus
|
Confused about variables
|
https://math.stackexchange.com/questions/4192699/confused-about-variables
|
<p>This is probably a very dumb question but after trying to review some calculus after years not using it, I am confused by variables in the equation for a tangent line.
So I watched the very first lecture on calculus by MIT ( <a href="https://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006" rel="nofollow noreferrer">https://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006</a>) and at around <span class="math-container">$0:31$</span>, I am very confused about the <span class="math-container">$X$</span> and <span class="math-container">$X_0$</span> he uses.</p>
<p>Basically, there is a function <span class="math-container">$y = \frac{1}{x}$</span> and we are trying to find the area under all triangles formed by the tangent line of this function. Differentiation was used in order to find the function <span class="math-container">$y=\frac{1}{x^2}$</span> which shows the slope of the original function for any <span class="math-container">$X$</span>. I perfectly understand the logic of these steps etc. but I am extremely buffled by separating the <span class="math-container">$X$</span> and <span class="math-container">$X_0$</span> -> the lecturer even pointed out that it often confuses people but didn´t elaborate more on it.</p>
<p>Basically, the problem for me starts right when trying to find the equation for the tangent line, given the derivative we found. The equation for a line is <span class="math-container">$Y - y_0 = m(X-x_0)$</span>
For some reason, I am tempted to just write: <span class="math-container">$Y - y_0 = (\frac{1}{x^2})*(X-X)$</span> which obviously is a nonsense since we´d obtain a zero in the parenthasis. But I just cannot justify in my mind the fact that we use x0 instead of <span class="math-container">$X$</span> to model the slope <span class="math-container">$m$</span>. I´d say that the slope will be a function of <span class="math-container">$X$</span> and therefore, I´d never ever think of using something like the <span class="math-container">$x_0$</span>.
Can please someone help to clarify this for me? thank you!</p>
|
<p>Well, the underlying concept is simple. We take a certain point <span class="math-container">$(x_0, y_0)$</span> lying on the given curve. At this point, slope of tangent is obviously given by:
<span class="math-container">$$m=\left(\frac {dy}{dx}\right)_{x=x_0}$$</span>
Hence, for your given curve, at <span class="math-container">$x=x_0$</span>, <span class="math-container">$$m=-\frac {1}{{x_0}^2}$$</span>
Thus, keeping in mind that <span class="math-container">$(x_0, y_0)$</span> are <strong>not</strong> variables, rather parameters, we can write equation of tangent:
<span class="math-container">$$\frac {y-y_0}{x-x_0}=m=-\frac {1}{{x_0}^2}$$</span>
Here <span class="math-container">$x,y$</span> represent the variables in which we represemt the line.The slope, <span class="math-container">$m$</span>, and one point on the line, that is, the point of tangency <span class="math-container">$(x_0,y_0)$</span>, parameters for us.</p>
| 277
|
calculus
|
Integral over circle area limited by two straight lines
|
https://math.stackexchange.com/questions/4193374/integral-over-circle-area-limited-by-two-straight-lines
|
<p>I need to integrate a function over the area limited by the circle and two straight lines, i.e.<span class="math-container">$x^2+y^2<R^2$</span> and <span class="math-container">$x<-b, y>a$</span>. For this I integrate over <span class="math-container">$y$</span> from <span class="math-container">$a$</span> to <span class="math-container">$\sqrt{R^2-x^2}$</span> and <span class="math-container">$x$</span> from <span class="math-container">$-\sqrt{R^2-a^2}$</span> to <span class="math-container">$-b$</span>. The problem is the result I get is not equal to <span class="math-container">$0$</span> if I substitute <span class="math-container">$a=R$</span> or <span class="math-container">$b=R$</span>, which is what I would have expected it to be correct, since then the area limited by these curves dissapear. Are these limits wrong and should I add some Heaviside function to meet my expectations or the integration result is correct and I should expect this kind of behaviour?</p>
|
<p>Consider the lines <span class="math-container">$x = -b$</span> and <span class="math-container">$y = a.$</span>
These two lines are perpendicular and divide the plane into three quadrants
around the point <span class="math-container">$(-b,a).$</span></p>
<p>The two conditions <span class="math-container">$x < -b$</span> and <span class="math-container">$y > a$</span> together say that we can integrate only in the quadrant that is above and to the left of the point <span class="math-container">$(-b,a).$</span></p>
<p>If <span class="math-container">$a$</span> and <span class="math-container">$b$</span> are both positive, it should be clear that this quadrant intersects the disk <span class="math-container">$x^2 + y^2 < R^2$</span> only if <span class="math-container">$(-b,a)$</span> is inside the disk.
That is, if <span class="math-container">$(-b)^2 + a^2 = a^2 + b^2 \geq R^2$</span> then there is no area to integrate over, and the integral is zero.</p>
<p>If <span class="math-container">$a = R$</span> or <span class="math-container">$b = R$</span> then you will always have <span class="math-container">$a^2 + b^2 \geq R^2$</span>,
so your original thinking that the integral should be zero was correct.
But if you blindly apply the formulas for the lower and upper bounds you may not get a zero integral. You should also see then that the lower bound is greater than the upper bound in at least one of the integrals, which is a symptom of the error.</p>
<p>The general procedure for integrating over a region is,
first determine that you have a region of integration at all,
then determine what the region is,
then find formulas for the bounds of the integrals.</p>
<p>I suppose you could multiply the entire integral by a Heaviside function whose value is <span class="math-container">$1$</span> when <span class="math-container">$a^2 + b^2 < R^2$</span> and <span class="math-container">$0$</span> when <span class="math-container">$a^2 + b^2 > R^2,$</span>
and this would give you the correct zero result when <span class="math-container">$(-b,a)$</span> is outside the disk, but I think it is simpler to just put a condition on the answer.</p>
<p>If you allow <span class="math-container">$a < 0$</span> or <span class="math-container">$b < 0$</span> then the situation becomes more complicated.
You have a region of integration if <span class="math-container">$(-b,a)$</span> is in the first quadrant below the line <span class="math-container">$y=R$</span>, in the third quadrant to the right of the line <span class="math-container">$x=-R,$</span> or in the fourth quadrant. In some cases you may have to change the order of integration or even split the integral into two parts that are added together at the end in order to get correct results.</p>
| 278
|
calculus
|
$\sqrt{a} +\sqrt{b} = 20$. What is the maximum value of $a-5b$?
|
https://math.stackexchange.com/questions/1749111/sqrta-sqrtb-20-what-is-the-maximum-value-of-a-5b
|
<p>It is given that <span class="math-container">$$\sqrt{a} +\sqrt{b} = 20$$</span>
Where a and b are real numbers.</p>
<p>What is the maximum value of <span class="math-container">$a-5b$</span>?</p>
|
<p>$$\sqrt{b}=20-\sqrt{a}$$
$$b=(20-\sqrt{a})^2=400-40\sqrt{a}+a$$
$$a-5b=a-5(400-40\sqrt{a}+a)=-4a+200\sqrt{a}-2000$$</p>
<p>$$\frac{d}{da}(a-5b)=-4+\frac{100}{\sqrt{a}}$$</p>
<p>Since $\sqrt{a}<20$, </p>
<p>$$\frac{d}{da}(a-5b)>0$$</p>
<p>Hence, $a-5b$ is maximum when $a$ is maximum, viz., when $a=400$ (and $b=0$)</p>
<p>Hence, the maximum value of $a-5b$ is $400$.</p>
| 279
|
calculus
|
How to obtain the integral representation of Modified Bessel function $I_0(2)$?
|
https://math.stackexchange.com/questions/2903380/how-to-obtain-the-integral-representation-of-modified-bessel-function-i-02
|
<p>It is known that</p>
<p>$\displaystyle I_0(2)=\sum_{k=0}^{\infty}\frac{1}{(k!)^2} = \frac{1}{\pi}\int_{0}^{\pi}e^{2\cos\theta}d\theta$
(<a href="http://mathworld.wolfram.com/ModifiedBesselFunctionoftheFirstKind.html" rel="nofollow noreferrer">http://mathworld.wolfram.com/ModifiedBesselFunctionoftheFirstKind.html</a>)
But how do people derive the equation?</p>
|
<p>Exchange the order of integration:
$$\frac{1}{\pi}\int_{0}^{\pi}e^{2\cos\theta}d\theta$$
$$=\frac{1}{\pi}\int_{0}^{\pi}\sum_{n=0}^{\infty}\frac{\cos^n \theta}{n!}2^nd\theta$$
$$=\frac{1}{\pi}\sum_{n=0}^{\infty}\frac{2^n}{n!}\int_{0}^{\pi}\cos^n \theta d\theta.$$
For the integral $\int_{0}^{\pi}\cos^n \theta d\theta$, it is zero when $n$ is odd and
$$\int_{0}^{\pi}\cos^{2n} \theta d\theta=2\int_{0}^{\pi/2}\cos^{2n} \theta d\theta=
\pi\frac{(2n-1)!!}{(2n)!!}.$$
So $$\frac{1}{\pi}\int_{0}^{\pi}e^{2\cos\theta}d\theta
=\sum_{n=0}^{\infty}\frac{2^{2n}}{(2n)!}\frac{(2n-1)!!}{(2n)!!}=\sum_{n=0}^{\infty}\frac{1}{(n!)^2}.$$</p>
| 280
|
calculus
|
Proof that $f(x)=4x^4-2x+1$ has no real roots.
|
https://math.stackexchange.com/questions/2909560/proof-that-fx-4x4-2x1-has-no-real-roots
|
<p>My thought was to:<br>
1) hypothesis there are 2 real roots for this equation,<br>
2) apply Rolle's theorem and come to a reductio ad absurdum<br>
and then if there aren't 2 real roots, it has to be 1. If there is 1 real root, this means that it has to have 3 non-real roots. But non real roots come in pairs, so either is 2 real- 2 non real, either 0 real- 4 non real. Therefore, it has no real roots.</p>
<p>In case my thought is correct, the problem is that $f'(x)=0 => x^3=1/8$ doesn't lead me in reductio ad absurdum, because it has 1 real and 2 non- real roots.<br>
I'm stuck and I'm about to punch the desk. Please, release me from this martyrdom.</p>
|
<ul>
<li>$f'(x)=2(8x^3-1)$, so there's a single critical point: $\; x=\frac12$.</li>
<li>$f''(x)=48x^2\ge 0$, so by the second derivative test, this critical point is a <em>minimum</em>, and this minimum is an absolute minimum.</li>
<li>$f(\frac12)=\frac14>0$.</li>
</ul>
| 281
|
calculus
|
Summation of $n^2x^n$ terms
|
https://math.stackexchange.com/questions/2923002/summation-of-n2xn-terms
|
<p>How does one evaluate the following summation of $n^2$ terms by $x^n$ terms.
I have tried to do it, but couldn't figure it out as it is not the same as summing up $nx^n$ terms.</p>
<p>$$\sum_{n=0}^\infty n^2 x^n$$</p>
|
<p>$\displaystyle \frac{1}{1-x} = \sum_{n=0}^{\infty}x^n$</p>
<p>Differentiating (and multiplying with $x$)we have,</p>
<p>$\displaystyle \frac{x}{(1-x)^2}=\sum_{n=0}^{\infty}nx^n$</p>
<p>Differentiating(and multiplying with $x$) we have, </p>
<p>$\displaystyle \frac{[(1-x)^2(1)-(x)2(1-x)(-1)]x}{(1-x)^4}= \frac{x^2+x}{(1-x)^3}=
\sum_{n=0}^{\infty}n^2x^n$</p>
| 282
|
calculus
|
Finding range of $a$
|
https://math.stackexchange.com/questions/2928763/finding-range-of-a
|
<blockquote>
<p>If <span class="math-container">$$f(x) =
\begin{cases}
|x-2|+a^2-9a-9, &\text{if }x<2\\
2x-3, &\text{if } x\geqslant2
\end{cases}$$</span> has local minima at <span class="math-container">$x=2$</span>, then range of <span class="math-container">$a$</span> is… ?</p>
</blockquote>
<p>My failed Attempt: </p>
<p>Wrote
<span class="math-container">$$f(x) =
\begin{cases}
-x+a^2-9a-7, &\text{if }x<2\\
2x-3, &\text{if } x\geqslant2
\end{cases}$$</span></p>
<p>The problem here is that I don't know whether the function is continious is required or not for finding maxima and minima. Not that I could check it if it was </p>
<p>Then My next step would be differentiating </p>
<p><span class="math-container">$$f(x) =
\begin{cases}
-1, &\text{if }x<2\\
2, &\text{if } x\geqslant2
\end{cases}$$</span></p>
<p>That means the function is decreasing before <span class="math-container">$x=2$</span> and then increasing. This is the best I could think for this question but it's not providing me any hint on how to proceed </p>
|
<p>A non-continuous function can also have the minimum at <span class="math-container">$x=2$</span></p>
<p>Here you want a local minimum at <span class="math-container">$x=2$</span> thus
<span class="math-container">$$
|2-2|+a^2-9a-9\ge 2\cdot2-3
$$</span></p>
<p><span class="math-container">$$
a^2-9a-9\ge 1
$$</span>
<span class="math-container">$$(a+1)(a-10)\ge 0$$</span></p>
<p>You get <span class="math-container">$$a\in(-\infty,-1]\cup[10,\infty)$$</span></p>
<p>You can look <a href="https://www.desmos.com/calculator/ihzzriphej" rel="noreferrer">here on desmos</a> for the simulation</p>
| 283
|
calculus
|
How to integral $\int\limits_{0}^{\pi \over 6} {x \over \sqrt{1-2\sin{x}}}dx$ ..?
|
https://math.stackexchange.com/questions/2931296/how-to-integral-int-limits-0-pi-over-6-x-over-sqrt1-2-sinxdx
|
<p><span class="math-container">$$\int\limits_{0}^{\pi \over 6} {x \over \sqrt{1-2\sin{x}}}dx$$</span></p>
<p>I attempted lots of permutations but I can't solve it..
moreover, I don't know its convergence or divergence... please help!</p>
|
<p><em>This is not a serious answer. Just done for the fun of it.</em></p>
<p>As Ekesh answered, there is no solution even using special functions.</p>
<p>For the fun of it, I tried to see what would give the magnificent approximation
<span class="math-container">$$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$</span> proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician that is to say more than <span class="math-container">$1400$</span> years ago.</p>
<p>This leads to a quite complex integral
<span class="math-container">$$I=\int\limits_{0}^{\pi \over 6}\frac{x}{\sqrt{1-\frac{32 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}}}$$</span> The antiderivative can be computed in terms of various (ugly) elliptic integrals and the integral is such that
<span class="math-container">$$\frac{216}{\pi^2}I=16(4+3 i)\, \Pi \left(\frac{1-3i}{5};\sin
^{-1}\left(\sqrt{\frac{i-1}{6}}\right)|\frac{18-24i}{25}\right)-3 \left((3-4 i)+20(1+ i) F\left(\sin
^{-1}\left(\sqrt{\frac{i-1}{6}}\right)|\frac{18-24i}{25}\right)+4(1+3 i) E\left(\sin
^{-1}\left(\sqrt{\frac{i-1}{6}}\right)|\frac{18-24i}{25}\right)\right)$$</span>
This is not nice at all but it evaluates to <span class="math-container">$I\approx 0.380381$</span> (relative error <span class="math-container">$=0.36$</span>%).</p>
| 284
|
calculus
|
If $\int_0^x f^2(t)dt \le f(x)$ for all $x \in [0,1]$, then $\min_{[0,1]} f(x) \le 1$?
|
https://math.stackexchange.com/questions/2944797/if-int-0x-f2tdt-le-fx-for-all-x-in-0-1-then-min-0-1-fx
|
<p>Suppose that <span class="math-container">$f$</span> is a continuous function on <span class="math-container">$[0,1]$</span> and
<span class="math-container">$$\int_0^x [f(t)]^2dt \le f(x) \quad \text{for all} \quad x \in[0,1].$$</span>
Prove or disprove
<span class="math-container">$$\min_{0\le x\le 1} f(x) \le 1.$$</span>
In case the desired inequality does not hold, what is the best upper bound?</p>
<p>Thanks.</p>
|
<p>I assume that <span class="math-container">$f^2(t)$</span> means <span class="math-container">$\big(f(t)\big)^2$</span>. I have a very weak bound
<span class="math-container">$$\min_{x\in[0,1]}\,f(x)<2\sqrt{2}\,,$$</span>
and do not know how to improve it. Maybe somebody can use my proof to get a better bound.</p>
<p>Suppose on the contrary that there exists a function <span class="math-container">$f:[0,1]\to\mathbb{R}$</span> satisfying
<span class="math-container">$$\int_0^x\,\big(f(t)\big)^2\,\text{d}t\leq f(x)\text{ for all }x\in[0,1]\tag{*}$$</span>
such that
<span class="math-container">$$k:=\min_{x\in[0,1]}\,f(x)\geq 2\sqrt{2}\,.$$</span>
From (*), we get that
<span class="math-container">$$f(x)\geq \max\big\{k^2x,k\}\text{ for all }x\in[0,1]\,.$$</span>
We use (*) once again and find that
<span class="math-container">$$f(x)\geq \int_0^{\frac1k}\,k^2\,\text{d}x+\int_{\frac{1}{k}}^x\,(k^2t)^2\,\text{d}t=\frac{2k}{3}+\frac{k^4x^3}{3}$$</span>
for <span class="math-container">$x\in\left[\dfrac1k,1\right]$</span>. </p>
<p>Define polynomials <span class="math-container">$P_0$</span>, <span class="math-container">$P_1$</span>, <span class="math-container">$P_2$</span>, <span class="math-container">$\ldots$</span> as follows:
<span class="math-container">$$P_0(z):=1\,,\,\,P_1(z):=z\,,\,\,P_2(x):=\frac{2}{3}+\frac{z^3}{3}\,,$$</span>
and
<span class="math-container">$$P_r(z)=1+\int_{1}^z\,\big(P_{r-1}(\zeta)\big)^2\,\text{d}\zeta$$</span>
for <span class="math-container">$r=3,4,5,\ldots$</span>. It can be proven by induction that, for each <span class="math-container">$r=0,1,2,\ldots$</span>, <span class="math-container">$$f(x)\geq k\,P_r(kx)$$</span> for all <span class="math-container">$x\in\left[\dfrac1k,1\right]$</span>.</p>
<p>We can prove by induction that <span class="math-container">$0\leq P_r(z)\leq 1$</span> for every <span class="math-container">$z\in[0,1]$</span>. That is, the constant term of <span class="math-container">$P_r$</span> is nonnegative for every <span class="math-container">$r=0,1,2,\ldots$</span>. It is obvious that the coefficients of higher-order terms in <span class="math-container">$P_r$</span> are nonnegative. Furthermore, the degree of <span class="math-container">$P_r$</span> is <span class="math-container">$2^r-1$</span>, and the coefficient <span class="math-container">$\lambda_r$</span> of the <span class="math-container">$(2^r-1)$</span>-st degree term of <span class="math-container">$P_r$</span> is given by the recurrence relation
<span class="math-container">$$\lambda_r=\frac{1}{2^r-1}\lambda_{r-1}^2\,.$$</span>
In other words,
<span class="math-container">$$\lambda_r=\frac{1}{\prod\limits_{j=1}^r\,(2^j-1)^{2^{r-j}}}\geq \frac{1}{\prod\limits_{j=2}^r\,2^{j\cdot2^{r-j}}}=\frac{1}{2^{3\cdot 2^{r-1}-r-2}}\,.$$</span>
That is,
<span class="math-container">$$f(1)\geq k\,P_r(k)\geq \frac{k^{2^r}}{2^{3\cdot2^{r-1}-r-2}}$$</span>
for every <span class="math-container">$r=0,1,2,\ldots$</span>. However, as <span class="math-container">$k\geq 2\sqrt{2}$</span>, <span class="math-container">$k^{2^r}$</span> grows faster than <span class="math-container">$2^{3\cdot2^{r-1}-r-2}$</span>, namely,
<span class="math-container">$$\lim_{r\to\infty}\,\frac{k^{2^r}}{2^{3\cdot2^{r-1}-r-2}}=\infty\,.$$</span>
This yields a contradiction. </p>
<p><strong>Remark.</strong> I believe that the polynomials <span class="math-container">$P_r$</span> converge pointwise, as <span class="math-container">$r\to\infty$</span>, to <span class="math-container">$P$</span> on <span class="math-container">$(-2,+2)$</span>, where <span class="math-container">$$P(z):=\frac{1}{2-z}\text{ for }z\in(-2,+2)\,.$$</span>
I conjecture also that, for <span class="math-container">$z\geq 2$</span>, <span class="math-container">$\lim\limits_{r\to\infty}\,P_r(z)=\infty$</span>. If this is true, then it follows immediately that <span class="math-container">$\min\limits_{x\in[0,1]}\,f(x)<2$</span>.</p>
<p><strong>Counterexample.</strong> Interestingly, let <span class="math-container">$k\in(1,2)$</span> and define
<span class="math-container">$$f(x):=\max\left\{k,\frac{k}{2-kx}\right\}\text{ for all }x\in[0,1]\,.$$</span>
Then, we see that <span class="math-container">$\min\limits_{x\in[0,1]}\,f(x)=k$</span>. Furthermore,
<span class="math-container">$$f(x)=k\geq k^2x=\int_0^x\,k^2\,\text{d}t=\int_0^x\,\big(f(t)\big)^2\,\text{d}t$$</span>
for <span class="math-container">$x\in\left[0,\dfrac1k\right]$</span>. For <span class="math-container">$x\in\left[\dfrac1k,1\right]$</span>, we have
<span class="math-container">$$f(x)=\frac{k}{2-kx}=k+\int_{\frac1k}^x\,\left(\frac{k}{2-kt}\right)^2\,\text{d}t=\int_0^x\,\big(f(t)\big)^2\,\text{d}t\,.$$</span>
This shows that, if <span class="math-container">$\min\limits_{x\in[0,1]}\,f(x)<c$</span> for every such function <span class="math-container">$f$</span>, then <span class="math-container">$c\geq 2$</span>.</p>
| 285
|
calculus
|
Evaluate limit using L'Hospital
|
https://math.stackexchange.com/questions/2946383/evaluate-limit-using-lhospital
|
<p>Evaluate <span class="math-container">$\lim_{x\to0} \frac{\sin(x^{30})}{\sin^{30}(5x)} $</span></p>
<p>I have tried applying L'Hospital's rule, but it took me a lot of time to factor the derivative. Is there any way can resolve this problem. Thanks.</p>
<p>The answer is <span class="math-container">$\frac{1}{5^{30}}$</span> but my result is 1.</p>
|
<p>Use Taylor series so you don't have to differentiate <span class="math-container">$30$</span> times. We have <span class="math-container">$\sin(t)=t-\frac16t^3+\cdots$</span>. Inserting <span class="math-container">$t=x^{30}$</span> gives us the numerator, and inserting <span class="math-container">$t=5x$</span> and raising the result to the power of <span class="math-container">$30$</span> gives the denominator:
<span class="math-container">$$
\frac{\sin(x^{30})}{\sin^{30}(5x)}=\frac{x^{30}-\frac13x^{90}+\cdots}{(5x)^{30}-30\cdot\frac16(5x)^{32}+\cdots}
$$</span>
Now divide both the numerator and the denominator by <span class="math-container">$x^{30}$</span>, and the limit will be much easier to compute.</p>
| 286
|
calculus
|
Maximum value on a circle
|
https://math.stackexchange.com/questions/2949154/maximum-value-on-a-circle
|
<p>I need to find the maximum value of a function on a circle: Let <span class="math-container">$C$</span> denote the circle of radius <span class="math-container">$6$</span> centered at the origin in the <span class="math-container">$xy$</span>-plane. Find the maximum value of <span class="math-container">$x^2y$</span> on <span class="math-container">$C$</span>. Where do I even start with this?</p>
|
<p>Hint: For <span class="math-container">$(x,y)$</span> on the circle of radius <span class="math-container">$6$</span>, we have
<span class="math-container">$$
x^2=36-y^2
$$</span>
So you can find a single variable function to maximize.</p>
| 287
|
calculus
|
Evaluation of Integration using limit as a sum
|
https://math.stackexchange.com/questions/2952267/evaluation-of-integration-using-limit-as-a-sum
|
<blockquote>
<p>Evaluation of <span class="math-container">$\displaystyle \int^{2}_{1}\frac{1}{x}dx$</span> using limit as a sum</p>
</blockquote>
<p>Try: Using The formula <span class="math-container">$$\int^{b}_{a}f(x)dx = \lim_{h\rightarrow 0}h\times \sum^{n-1}_{r=1}f(a+rh)$$</span></p>
<p>where <span class="math-container">$nh=b-a$</span></p>
<p>above <span class="math-container">$a=1,b=2$</span> and <span class="math-container">$\displaystyle f(x)=\frac{1}{x}$</span> and <span class="math-container">$nh=1$</span></p>
<p><span class="math-container">$$\int^{2}_{1}\frac{1}{x}dx = \lim_{h\rightarrow 0} \sum^{n-1}_{r=1}f(a+rh)=\lim_{h\rightarrow 0}h\cdot \sum^{n-1}_{r=0}f(1+rh) $$</span></p>
<p><span class="math-container">$$\lim_{h\rightarrow 0}h\cdot \sum^{n-1}_{r=0}\frac{1}{1+rh}=\lim_{h\rightarrow 0}\bigg[\frac{h}{1+h}+\frac{h}{1+2h}+\cdots \cdots +\frac{h}{1+(r-1)h}\bigg]$$</span>
i did not know how i proceed, struck here</p>
<p>could some help me. thanks</p>
|
<p>Consider the points <span class="math-container">$x=c^k$</span> with <span class="math-container">$c^n=2$</span>.</p>
<p><span class="math-container">$$\int_1^2\frac{dx}{x}\approx\sum_{k=0}^{n+1} \frac{\Delta c^k}{c^k}=\sum_{k=1}^n \frac{ c^{k+1}-c^k}{c^k}=n(c-1)=n\left(\sqrt[n]2-1\right).$$</span></p>
<p>Then,</p>
<p><span class="math-container">$$\lim_{n\to\infty}n\left(\sqrt[n]2-1\right)=\lim_{h\to0}\frac{2^h-1}h=\left.(2^h)'\right|_{h=0}=\log2.$$</span></p>
<hr>
<p>Note that this is in fact a discrete version of an exponential change of variable, <span class="math-container">$x=e^t$</span>, giving</p>
<p><span class="math-container">$$\int_1^2\frac{dx}x=\int_{\log1}^{\log2}\frac{e^t\,dt}{e^t}=\int_0^{\log2}dt.$$</span></p>
<p>The latter integral can be trivially computed as a sum.</p>
| 288
|
calculus
|
Show that $c$ is in interval $[e,3]$ for $c\cdot \ln{c} + c − 6 = 0$
|
https://math.stackexchange.com/questions/2959847/show-that-c-is-in-interval-e-3-for-c-cdot-lnc-c-%e2%88%92-6-0
|
<p>Show that there is a unique number <span class="math-container">$c \in \mathbb{R}$</span> that fulfills the equation and that this number is in the interval <span class="math-container">$[e, 3]$</span>.</p>
<p><span class="math-container">$$ c\cdot \ln{c} + c − 6 = 0$$</span></p>
<p>At first I was thinking about using linear approximation, but then I would need an <span class="math-container">$a$</span> and <span class="math-container">$x$</span>. So I'm quite confused as how to start solving this problem.</p>
|
<p>Note <span class="math-container">$c>0$</span>.</p>
<p>Set <span class="math-container">$e^y:=c$</span>, then</p>
<p>1)<span class="math-container">$ye^y +e^y -6=0$</span>,</p>
<p><span class="math-container">$e^y(1+y)-6=0$</span>.</p>
<p><span class="math-container">$f(y):=(1+y)e^y-6.$</span></p>
<p><span class="math-container">$f(0)=-5;$</span> </p>
<p><span class="math-container">$f'(y)=e^y+(1+y)e^y \gt 0$</span> for <span class="math-container">$y \ge 0$</span>.</p>
<p>Strictly increasing for <span class="math-container">$y \ge 0.$</span></p>
<p><span class="math-container">$y=1$</span>: <span class="math-container">$f(1)= 2e-6 <0$</span>.</p>
<p><span class="math-container">$y=\log 3$</span>:</p>
<p><span class="math-container">$ f(\log 3)=(1+\log 3)3-6>0$</span>, </p>
<p>since <span class="math-container">$\log 3 >1$</span>.</p>
<p>Recalling <span class="math-container">$c=e^y:$</span></p>
<p><span class="math-container">$c \in [e,3]$</span>.</p>
<p>2) Let <span class="math-container">$y \le 0$</span>.</p>
<p>A) <span class="math-container">$-1 \le y \le 0.$</span></p>
<p><span class="math-container">$f(y)=(1+y)e^y -6= (1-|y|)e^{-|y|}-6 \le -5.$</span></p>
<p><span class="math-container">$f(0)=-5$</span>; <span class="math-container">$f(-1)= -6.$</span></p>
<p>B) <span class="math-container">$y \lt -1.$</span></p>
<p><span class="math-container">$f(y)= (1+y)e^y -6 \lt -6.$</span></p>
<p>Combining:</p>
<p><span class="math-container">$f(y) \lt 0$</span> for <span class="math-container">$y \le 0.$</span></p>
<p><span class="math-container">$f(y)$</span> is strictly increasing for <span class="math-container">$y \ge 0$</span>.</p>
<p>Hence <span class="math-container">$f(y)$</span> has exactly one zero, <span class="math-container">$y \in \mathbb{R}$</span>.</p>
| 289
|
calculus
|
Absolute conditional minimum of function in n-dimensional space
|
https://math.stackexchange.com/questions/2971893/absolute-conditional-minimum-of-function-in-n-dimensional-space
|
<p>Function</p>
<p><span class="math-container">$$F(x_1,x_2,...,x_n) = \sum_{i=1}^n x_i$$</span></p>
<p>on the constraint</p>
<p><span class="math-container">$$G(x_1,x_2,...,x_n)=\prod_{i=1}^n x_i-1$$</span></p>
|
<p>The inequality of arithmetic and geometric means says:</p>
<p><span class="math-container">$(x_1x_2....x_n)^{1/n} \le \frac{F(x_1,...,x_n)}{n}$</span>.</p>
<p>If <span class="math-container">$x_1x_2....x_n=1$</span>, then we get</p>
<p><span class="math-container">$n=F(1,...,1) \le F(x_1,...,x_n)$</span>.</p>
| 290
|
calculus
|
calculate $\lim_{x\to\infty} x + \sqrt[3]{1-x^3}$
|
https://math.stackexchange.com/questions/2979793/calculate-lim-x-to-infty-x-sqrt31-x3
|
<p>So I multiplied by the conjugate and got <span class="math-container">$$\lim_{x\to\infty} \frac{x^2-(1-x^3)^\frac{2}{3} + x(1-x^3)^\frac{1}{3}-(1-x^3)}{x-(1-x^3)^\frac{2}{3}}$$</span></p>
<p>and this is where I got stuck.</p>
|
<blockquote>
<p>So I multiplied by the conjugate and got</p>
</blockquote>
<p>What conjugate expression was that exactly...?</p>
<p>You want to get rid of the cube root by using:
<span class="math-container">$$a+b=\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2-ab+b^2}=\frac{a^3+b^3}{a^2-ab+b^2}$$</span>
with, in your case, <span class="math-container">$a=x$</span> and <span class="math-container">$b=\sqrt[3]{1-x^3}$</span>.</p>
| 291
|
calculus
|
$|\int \limits_a^b f(x) dx|\leq\int \limits_a^b |f(x)|dx$ for f continuous
|
https://math.stackexchange.com/questions/2981463/int-limits-ab-fx-dx-leq-int-limits-ab-fxdx-for-f-continuous
|
<p>How to prove <span class="math-container">$|\int \limits_a^b f(x) dx|\leq\int \limits_a^b |f(x)|dx$</span> for f continuous? This is a step in the solution of a problem from Mendelson's introduction to topology. This book assumes the reader has only a background in first-year calculus, not measure theory or advanced calculus.</p>
|
<p>You can write: <span class="math-container">$$f(x)=f_+(x)-f_-(x)$$</span> where <span class="math-container">$f_+(x)=\max(f(x),0)$</span> and <span class="math-container">$f_-(x)=-\min(f(x),0)$</span></p>
<p>Note that: <span class="math-container">$$|f(x)|=f_+(x)+f_-(x)$$</span></p>
<p>The functions <span class="math-container">$f_+$</span> and <span class="math-container">$f_-$</span> are both continuous and nonnegative so that: <span class="math-container">$$\left|\int_a^bf(x)dx\right|=\left|\int_a^bf_+(x)dx-\int_a^bf_-(x)dx\right|\leq \int_a^bf_+(x)dx+\int_a^bf_-(x)dx=\int_a^b|f(x)|dx$$</span></p>
| 292
|
calculus
|
Number of solutions.
|
https://math.stackexchange.com/questions/2982498/number-of-solutions
|
<blockquote>
<p>For each positive real number <span class="math-container">$\lambda$</span>, let <span class="math-container">$A_\lambda$</span> be the set of all natural numbers <span class="math-container">$n$</span> such that <span class="math-container">$|\sin\sqrt{n+1}-\sin\sqrt n|<\lambda$</span>. Let <span class="math-container">$A_\lambda^c$</span> be the complement of <span class="math-container">$A_\lambda$</span> in the set of all natural numbers. Then<br>
(A) <span class="math-container">$A_{1/2},A_{1/3},A_{2/5}$</span> are all finite sets<br>
(B) <span class="math-container">$A_{1/3}$</span> is a finite set but <span class="math-container">$A_{1/2},A_{2/5}$</span> are infinite sets<br>
(C) <span class="math-container">$A_{1/2}^c,A_{1/3}^c,A_{2/5}^c$</span> are all finite sets<br>
(D) <span class="math-container">$A_{1/3},A_{2/5}$</span> are finite sets and <span class="math-container">$A_{1/2}$</span> is an infinite set</p>
</blockquote>
<p>How to actually start the problem? I tried using finding monotonocity of the function by replacing <span class="math-container">$n$</span> by <span class="math-container">$x$</span>. But am not able to comment.</p>
<p>At infinity the limit is 0. How to use that.</p>
|
<p><span class="math-container">$$\sin \sqrt{n+1} - \sin \sqrt{n}=2 \sin \frac12 (\sqrt{n+1}-\sqrt{n}) \cos \frac12 (\sqrt{n+1}+\sqrt{n})= \\ = 2 \sin \frac{1}{2(\sqrt{n+1}+\sqrt{n})} \cos \frac12 (\sqrt{n+1}+\sqrt{n})$$</span></p>
<p>From this we can see that the limit for <span class="math-container">$n \to \infty$</span> is really <span class="math-container">$0$</span>, as the OP claims.</p>
<p>Now we need to show that there's an infinite set of <span class="math-container">$n$</span> for any <span class="math-container">$\lambda \in \mathbb{R}^+$</span> such that:</p>
<p><span class="math-container">$$2 \left|\sin \frac{1}{2(\sqrt{n+1}+\sqrt{n})} \cos \frac12 (\sqrt{n+1}+\sqrt{n}) \right| < \lambda$$</span></p>
<p>This actually directly follows from the definition of the limit at <span class="math-container">$\infty$</span> and our claim that it's <span class="math-container">$0$</span>. But let's consider it more carefully:</p>
<p>We have:</p>
<p><span class="math-container">$$2 \left|\sin \frac{1}{2(\sqrt{n+1}+\sqrt{n})} \cos \frac12 (\sqrt{n+1}+\sqrt{n}) \right| < \\ < 2 \left|\sin \frac{1}{2(\sqrt{n+1}+\sqrt{n})}\right| < \frac{1}{\sqrt{n+1}+\sqrt{n}} < \frac{1}{2\sqrt{n}}$$</span></p>
<p>Now pick:</p>
<p><span class="math-container">$$N = \begin{cases} 1, & \lambda > \frac12 \\ \frac{1}{4 \lambda^2}, & \lambda \leq \frac12 \end{cases}$$</span></p>
<p>For any <span class="math-container">$n>N$</span> we have:</p>
<p><span class="math-container">$$\frac{1}{2\sqrt{n}} < \lambda$$</span></p>
<p><span class="math-container">$$|\sin \sqrt{n+1} - \sin \sqrt{n}| < \lambda$$</span></p>
<p>Since there's an infinite set of <span class="math-container">$n$</span> larger than some finite <span class="math-container">$N$</span>, we have proven that for every <span class="math-container">$\lambda \in \mathbb{R}^+$</span> the set <span class="math-container">$A_\lambda$</span> is infinite.</p>
<hr>
<p>Conversely, we also need to prove that the complement of this set is finite.</p>
<p>Consider:</p>
<p><span class="math-container">$$|\sin \sqrt{n+1} - \sin \sqrt{n}| \geq \lambda$$</span></p>
<p>We have already proved that for any <span class="math-container">$n>N$</span> defined as above, the opposite inequality holds. Which means, we are only left with the set of <span class="math-container">$n \leq N$</span> to pick from, which is finite by definition.</p>
<p>So we have proved that every complementary set is finite, which leaves the option (C).</p>
| 293
|
calculus
|
How to differentiate this double integral from Christopher Bishop's Pattern Recognition book.
|
https://math.stackexchange.com/questions/2982727/how-to-differentiate-this-double-integral-from-christopher-bishops-pattern-reco
|
<p>In Bishop's book Pattern Recognition and Machine Learning, the following can be found on page 46:</p>
<p>(1):
<span class="math-container">$$
J[f] = \iint\{f(\mathbf{x}) - t\}^2p(\mathbf{x},t)\mathrm{d}t \mathrm{d}\mathbf{x}
$$</span></p>
<p>He then differentiates this expression with respect to <span class="math-container">$f(\mathbf{x})$</span>:</p>
<p>(2):
<span class="math-container">$$
\frac{\delta J[f]}{\delta f(\mathbf{x})} = 2 \int \{f(\mathbf{x}) - t\}p(\mathbf{x}, t)\mathrm{d}t
$$</span></p>
<p>My question is how to do this differentiation. One of the things I'm not getting is how the integral for <span class="math-container">$\mathbf{x}$</span> disappears even though we're not differentiating w.r.t. <span class="math-container">$\mathbf{x}$</span>.</p>
<p>If someone could provide a step by step derivation from (1) to (2) that would be very appreciated.</p>
<p>Thanks :)</p>
<p>EDIT:</p>
<p>So I've tried to use the Euler-Lagrange equation like this:</p>
<p><span class="math-container">$$
L = \int{\{f(\mathbf{x}) - t\}^2p(\mathbf{x},t)\mathrm{d}t}
$$</span>
<span class="math-container">$$
\frac{\delta J}{\delta f(\mathbf{x})} = \frac{\partial L}{\partial f} - \frac{d}{\mathrm{d}\mathbf{x}}\frac{\partial L}{\partial f^\prime}
$$</span>
(3):
<span class="math-container">$$
\frac{\partial L}{\partial f} = \frac{\partial}{\partial f} \int{\{f(\mathbf{x}) - t\}^2p(\mathbf{x},t)\mathrm{d}t} = 2\int{\{f(\mathbf{x}) - t\} p(\mathbf{x}, t)}\mathrm{d}t
$$</span></p>
<p>This implies that <span class="math-container">$\frac{d}{\mathrm{d}\mathbf{x}}\frac{\partial L}{\partial f^\prime} = \frac{d}{\mathrm{d}\mathbf{x}}\frac{\partial}{\partial f^\prime}\int{\{f(\mathbf{x}) - t\}^2p(\mathbf{x},t)\mathrm{d}t} = 0$</span>. But I have no idea of how to prove this. Any ideas?</p>
| 294
|
|
calculus
|
Approximate $(0.99)^{300}$ without calculator
|
https://math.stackexchange.com/questions/2987471/approximate-0-99300-without-calculator
|
<blockquote>
<p>Approximate <span class="math-container">$(0.99)^{300}$</span> without calculator.</p>
</blockquote>
<p>This question is in my textbook but i don't know how to approximate without calculator. How can i evaluate without calculator? Thanks in advance.</p>
|
<p><span class="math-container">$$300 \ln (1-1/100) \approx 300 (-1/100-1/20000) \approx -3$$</span></p>
<p><span class="math-container">$$e^{-3} = (3-(3-e))^{-3} \approx \frac{1}{27} \left(1+(3-e)\right)=\frac{4-e}{27}=0.0475...$$</span></p>
<p><span class="math-container">$$0.99^{300}=0.0490...$$</span></p>
<p>As for "without calculator", using <span class="math-container">$e=2.718...$</span> is enough. If you remember how to divide by hand.</p>
| 295
|
calculus
|
Prove that, $\int^{\infty}_0\frac{x\sin(rx)}{a^2+x^2}dx=\frac{\pi}{2}e^{-ar}$
|
https://math.stackexchange.com/questions/2992228/prove-that-int-infty-0-fracx-sinrxa2x2dx-frac-pi2e-ar
|
<p>The question is: prove that</p>
<p><span class="math-container">$$\int^{\infty}_0\frac{x\sin(rx)}{a^2+x^2}dx=\frac{\pi}{2}e^{-ar}$$</span></p>
<p>This is what I've got so far:</p>
<p>Let <span class="math-container">$I(r)=\int^{\infty}_0\frac{x\sin(rx)}{a^2+x^2}dx$</span></p>
<p><span class="math-container">$I'(r)=\int^{\infty}_0\frac{x^2\cos(rx)}{a^2+x^2}dx$</span> = <span class="math-container">$\int^{\infty}_0\cos(rx)dx-\int^{\infty}_0\frac{a^2\cos(rx)}{a^2+x^2}dx$</span></p>
<p><span class="math-container">$I''(r)=\int^{\infty}_0-x\sin(rx)dx+\int^{\infty}_0\frac{xa^2\sin(rx)}{a^2+x^2}dx$</span></p>
<p><span class="math-container">$I''(r)=\int^{\infty}_0-x\sin(rx)dx+a^2I(r)$</span></p>
<p>I think I'm supposed to form a differential equation and solve it by letting <span class="math-container">$I(r)=c_1e^r+c_2e^{-r}$</span>, but the problematic part is this: <span class="math-container">$\int^{\infty}_0-x\sin(rx)dx$</span>. I don't believe it simplifies to any constant and the differential equation doesn't solve nicely. </p>
<p>Hints/suggestions appreciated!</p>
| 296
|
|
calculus
|
Show $f(x) >0$ for $x>x_0$ if its $f' >f$ and $f(x_0)=0$
|
https://math.stackexchange.com/questions/3011769/show-fx-0-for-xx-0-if-its-f-f-and-fx-0-0
|
<blockquote>
<p>Let <span class="math-container">$f: \mathbb{R} \rightarrow \mathbb{R}$</span> be a differentiable function. Suppose that <span class="math-container">$f'(x)>f(x)$</span> for all <span class="math-container">$x \in \mathbb{R}$</span>, and <span class="math-container">$f(x_0)=0$</span> for some <span class="math-container">$x_0 \in R$</span>. Prove that <span class="math-container">$f(x)>0$</span> for all <span class="math-container">$x>x_0$</span>. As an application of this, show that <span class="math-container">$ae^x=a+x+x^2/2$</span>.</p>
</blockquote>
<p>Here is my attempt:
We know that <span class="math-container">$\lim_{h \to 0}{\frac{f(x+h)-f(x)}{h}}-f(x)\geq 0$</span>. So let <span class="math-container">$x=x_o+h$</span>, <span class="math-container">$\lim_{h \to 0}{\frac{f(x_0+h)-f(x_0)}{h}}-f(x_0)\geq 0$</span>. So <span class="math-container">$\lim_{h \to 0}{\frac{f(x_0+h)}{h}}\geq 0$</span>. And I'm stuck.</p>
|
<p>Proceed by contradiction. First since <span class="math-container">$f'(x_0) > f(x_0) = 0$</span>, there is some <span class="math-container">$t > 0$</span> s.t. <span class="math-container">$f (x)>0$</span> on <span class="math-container">$(x_0, x_0+t)$</span>. Assume <span class="math-container">$f(x) \leqslant 0$</span> for some <span class="math-container">$z > x_0$</span>, and <span class="math-container">$z$</span> is the minimal one, i.e. <span class="math-container">$f(y) >0$</span> when <span class="math-container">$x_0 < y < z$</span>, hence <span class="math-container">$f'(y) >0$</span> on <span class="math-container">$(x_0, z)$</span> as well, and by the MVT, <span class="math-container">$f(z) > f(x_0) = 0$</span>, contradiction. </p>
| 297
|
calculus
|
Getting the rate of drain from a tank
|
https://math.stackexchange.com/questions/3012499/getting-the-rate-of-drain-from-a-tank
|
<p>A tank with a top radius of <strong>1m</strong>, a bottom radius of <strong>0.5m</strong> and a height of <strong>2m</strong> is initially filled with water. Water drains through a square hole of side <strong>3cm</strong> in the bottom.</p>
<p>How do I get the rate of drain,
<span class="math-container">\begin{equation}
\frac{\partial V}{\partial t}
\end{equation}</span>
of the tank? </p>
|
<p>Conservation of mass and constant density tells us change in volume in the tank equals change in volume out of tank.</p>
<p>Thinking area times velocity gives you rate of volume change out of tank, by dimensional analysis.</p>
<p>Bernouli s principle says</p>
<p><span class="math-container">$\frac{1}{2}v^2=gh$</span></p>
<p><span class="math-container">$v=\sqrt{2gh}$</span></p>
<p>Area=side length of square hole squared.</p>
<p>Keep in mind h varies with time.</p>
<p>Volume in tank:
<span class="math-container">$V=\frac{\pi}{3}(h+2)(r_1+\frac{h(r_2-r_1)}{2})^2-\frac{\pi}{3}2r_1^2$</span></p>
<p>So :</p>
<p><span class="math-container">$A\sqrt{2gh}=\frac{dV}{dt}$</span></p>
<p>Will give you a differential equation in h as function of t.</p>
<p>Insert back into expression for V to get it as function of t. Differentiate to find alternative expression for derivative.</p>
| 298
|
calculus
|
Can the lower limit of $\frac{d}{dx} \int^x_a f(t)dt = f(x)$ be $-\infty$?
|
https://math.stackexchange.com/questions/3030569/can-the-lower-limit-of-fracddx-intx-a-ftdt-fx-be-infty
|
<p>I'm self studying math, based on the fundamental theorem of Calculus, <span class="math-container">$$\frac{d}{dx} \int^x_a f(t)dt = f(x)$$</span> can the lower limit be <span class="math-container">$-\infty$</span>?</p>
|
<p>Yes, lower limit can be <span class="math-container">$-\infty$</span> but only <em>provided</em> the resulting improper integral converges. This only means if you first integrate from <span class="math-container">$a$</span> to <span class="math-container">$x,$</span> get that answer, and then let <span class="math-container">$a \to -\infty,$</span> that limit exists.</p>
<p>I'm assuming <span class="math-container">$f(t)$</span> is defined and continuous (or sufficiently nice) on <span class="math-container">$(-\infty,x).$</span></p>
| 299
|
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