# Uses of $\lim \limits_{h\to 0} \frac{f(x+h)-f(x-h)}{2h}$

I have been wondering whether the following limit is being used somehow, as a variation of the derivative:

$$\lim_{h\to 0} \frac{f(x+h)-f(x-h)}{2h} .$$

Edit: I know that this limit is defined in some places where the derivative is not defined, but it gives us some useful information.

The question is not whether this limit is similar to the derivative, but whether it is useful somehow.

Thanks.

• I think it is used in numerical applications to linearize problems involving the derivative of a function. The trouble is that finding the derivative as a function of $x$ is really hard in general. So instead we can discretize the problem and approximate the derivative by this expression for very small $h$. (At least I have seen this in my introductory course on numerical methods) – Sam Sep 18 '11 at 20:33
• @Sam, As far as approximation and discretization are concerned, it seems exactly as convenient to use $n(f(x+1/n)-f(x))$ than $\frac12n(f(x+1/n)-f(x-1/n))$. – Did Sep 18 '11 at 20:40
• @Didier: Not necessarily. For example when we want to discretize the first order ODE $\dot x = x$, we will get a skewsymmetric matrix with the second approach, which may or may not have advantages over the less symmetric first variant (but I definitely do not know, so it's just a feeling)? Anyways: I have seen the expression the OP is asking about being used to discretize an ODE. – Sam Sep 18 '11 at 22:09
• But does the $\lim_{h\to 0} \frac{f(x+h)-f(x-h)}{2h}$ (my country phrased it as the "Schwarz Derivative") equivalent to the $\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$ ? – RopuToran Dec 7 '19 at 14:50

The "symmetric difference" form of the derivative is quite convenient for the purposes of numerical computation; to wit, note that the symmetric difference can be expanded in this way:

$$D_h f(x)=\frac{f(x+h)-f(x-h)}{2h}=f^\prime(x)+\frac{f^{\prime\prime\prime}(x)}{3!}h^2+\frac{f^{(5)}(x)}{5!}h^4+\dots$$

and one thing that should be noted here is that in this series expansion, only even powers of $h$ show up.

Consider the corresponding expansion when $h$ is halved:

$$D_{h/2} f(x)=\frac{f(x+h/2)-f(x-h/2)}{h}=f^\prime(x)+\frac{f^{\prime\prime\prime}(x)}{3!}\left(\frac{h}{2}\right)^2+\frac{f^{(5)}(x)}{5!}\left(\frac{h}{2}\right)^4+\dots$$

One could take a particular linear combination of this half-$h$ expansion and the previous expansion in $h$ such that the term with $h^2$ zeroes out:

$$4D_{h/2} f(x)-D_h f(x)=3f^\prime(x)-\frac{f^{(5)}(x)}{160}h^4+\dots$$

and we have after a division by $3$:

$$\frac{4D_{h/2} f(x)-D_h f(x)}{3}=f^\prime(x)-\frac{f^{(5)}(x)}{480}h^4+\dots$$

Note that the surviving terms after $f^\prime(x)$ are (supposed to be) much smaller than either of the terms after $f^\prime(x)$ in the expansions for $D_h f(x)$ and $D_{h/2} f(x)$. Numerically speaking, one could obtain a slightly more accurate estimate of the derivative by evaluating the symmetric difference at a certain (well-chosen) step size $h$ and at half of the given $h$, and computing the linear combination $\dfrac{4D_{h/2} f(x)-D_h f(x)}{3}$. (This is akin to deriving Simpson's rule from the trapezoidal rule). The procedure generalizes, as one keeps taking appropriate linear combinations of a symmetric difference for some $h$ and the symmetric difference at half $h$ to zero out successive powers of $h^2$; this is the famous Richardson extrapolation.

• J.M., Is there any specific advantage in preferring this symmetric difference form of the derivative over (say) the usual asymmetric difference form? With the asymmetric form, can't one take such well-chosen step sizes to cancel out many terms in the power series expansion? – Srivatsan Sep 18 '11 at 22:51
• Sure, you could use the asymmetric form for Richardson extrapolation, @Sri, but the convergence is slower in general, since all you can do for the asymmetric form is successively remove powers of $h$. For the symmetric difference, you get to successively remove powers of $h^2$. You might want to try looking at how to take a linear combination to remove the $h$ and $h^2$ terms for the asymmetric form. :) – J. M. isn't a mathematician Sep 18 '11 at 23:00
• Ok, I will try that sometime for small number of terms :). Thanks! – Srivatsan Sep 18 '11 at 23:41

Lemma: Let $f$ be a convex function on an open interval $I$. For all $x \in I$, $$g(x) = \lim_{h \to 0} \frac{f(x+h) - f(x-h)}{2h}$$ exists and $f(y) \geq f(x) + g(x) (y-x)$ for all $y \in I$.

In particular, $g$ is a subderivative of $f$.

$$\begin{eqnarray*} \lim_{h\to 0} \frac{f(x+h)-f(x-h)}{2h} &=& \frac12 \lim_{h\to 0}\left(\frac{f(x+h)-f(x)}h+\frac{f(x)-f(x-h)}h\right) \\ &=& \frac12 (f'(x)+f'(x)) = f'(x) \end{eqnarray*}$$

Assuming, of course that $f$ is differentiable at $x$.

• @SrivatsanNarayanan: It is fine, I just hope that it is correct to begin with :-) – Asaf Karagila Sep 18 '11 at 20:06
• I have modified my question, I know that when the derivative exists this limit is equal to it, but can it be used when the derivative is not defined? – Shay Ben Moshe Sep 18 '11 at 20:07
• I added parentheses to make it clearer. – Américo Tavares Sep 18 '11 at 20:16
• @AsafKaragila the whole point is that the symmetric derivative has a real meaning when the function is not differentiable at the point. You assumed away the essential point – Foo Bah Sep 19 '11 at 5:19
• @Foo: If you consider the time of posting this answer and the time of editing the original question, you will see that the additional assumption that $f$ is not differentiable was not given. – Asaf Karagila Sep 19 '11 at 5:24

This cannot be used as a definition of the derivative. First the result is half the sum of the left and right derivatives at $x$, when these exist. Second the limit can be well defined even when the sided derivatives do not exist, consider for example $f(x)=|x|^a$ around $x=0$ for suitable values of $a$. More generally, the limit at $x$ exists and is $g'(x)$ as soon as $f=g+s$ with $g$ differentiable at $x$ and $s$ symmetric around $x$ in the sense that $s(x+z)=s(x-z)$ for every $|z|$ small enough hence this notion can be used to get rid of symmetric but badly behaved parts of $f$ around $x$.

• I know, I have modified my question to clarify this issue. – Shay Ben Moshe Sep 18 '11 at 20:06

If $f$ is allowed to be discontinuous we have this example:

$$x \in \mathbb{Q} \implies \lim_{h \to 0} \frac{1_\mathbb{Q}(x+h) - 1_\mathbb{Q}(x-h)}{2h} = \lim_{h \to 0} \frac{0}{2h} = \lim_{h \to 0} 0 = 0.$$

That doesn't seem particularly useful to me.

• I don't understand what you wrote, what does $1_\mathbb{Q}$ mean? – Shay Ben Moshe Sep 18 '11 at 20:27
• It's the usual notation for the indicator function of the rationals – kahen Sep 18 '11 at 20:28
• I see, well it gives us information actually. From the symmetry of $1_\mathbb{Q}$, it doesn't matter which value does the function get around $x=0$ the change in the function is zero, just like the limit says. – Shay Ben Moshe Sep 18 '11 at 20:32
• @anon, I do not get your point. Usual derivatives, one-sided or not, are not restricted to the rationals, are they? That a function as highly irregular as the indicator function of the rationals HAS a pseudo-derivative in this sense is an excellent hint that, in fine, the notion has little to do with actual derivation. – Did Sep 18 '11 at 20:46
• @kahen: Any additive function, that is a function $f$ such that $f(x+y)=f(x)+f(y)$ for all real numbers $x$ and $y$, has a symmetric derivative equal to $0$ at each point, and discontinuous additive functions are quite pathological (their graphs are dense in the plane, they are non-Lebesgue measurable in each open interval, etc.). For more about the relation between the symmetric derivative and the ordinary derivative, see groups.google.com/group/sci.math/msg/d58ce3669a91243a and mathforum.org/kb/message.jspa?messageID=5056119 – Dave L. Renfro Sep 19 '11 at 19:41

This is a wrong way of thinking, the comment by Jesse Madnick explains why.

This is symmetric differentiation and is useful when both left and right hand limits exists, as oppose to usual derivative definition when only it is sufficient that right hand limit should exist.

The only place that I have seen that to be of use to make things nicer was in Fractional Calculus, providing some nice formulas to generalise differentiation.

To see why, try using the above limit definition of differentiation to get 2nd , or 3rd order definitions of a derivative, you should be able to recognise the resulting pascal/binomial like looking formulas, from there one might just try to generalise that to get a derivative definition that would say something about $\tfrac{1}{2}^{\text {th}}$ derivative.

In short : there is nothing more to this definition than the other usual one, except that this one shows an extra symbol (2), why use one more extra symbol when there is no need? maybe just to reap the benefits of it being symmetric.

Edit : Is somebody going to point the mistake? What am I missing here?

• For the usual derivative to exist, it is certainly not enough that the "right hand limit" exists. – Did Sep 18 '11 at 21:00
• Differentiability --> continuity --> LH Limit = RH Limit – The Chaz 2.0 Sep 18 '11 at 23:01
• @Didier : but $\frac {f(x+\delta)-f(x)}{\delta}$ is only the left side, I have not seen that $\frac {f(x)-f(x-\delta)}{\delta}$ as a requiremnt, are we talking about the same thing? – jimjim Sep 19 '11 at 1:24
• @Arjang: It is not assumed that $\delta > 0$. In other words, $\lim_{\delta \to 0} \frac{f(x+\delta) - f(x)}{\delta}$ is a two-sided limit. – Jesse Madnick Jul 15 '12 at 13:28


Why do the leading eigenvectors of $A$ maximize $\text{Tr}(D^TAD)$?

Given a matrix $X\in\mathbb{R}^{m\times n}$, I am trying to maximize $\text{Tr}(D^TX^TXD)$ over $D\in\mathbb{R}^{n\times l}$ ($n<l$) subject to $D^TD=I_l$, where $\text{Tr}$ denotes the trace, and $I_l$ denotes the identity matrix of size $l$.

More specifically, I am trying to find $$D^{*}=\arg\limits_{D}\max\text{Tr}(D^TX^TXD)\text{ subject to } D^TD=I_l.$$

The solution is that the matrix $D^*$ is given by the $l$ eigenvectors corresponding to the largest eigenvalues. However, I can't prove this.

I noticed that $D^TX^TXD$ is a real symmetric matrix, and thus we can decompose it to obtain $D^TX^TXD=Q\Lambda Q^T$, where $Q$ is an orthogonal matrix composed of eigenvectors of $D^TX^TXD$. I couldn't continue much from this. Any suggestions for this optimization problem?

• Already the first sentence shows some confusion: your $D$ is not "given", you want to find it. – amoeba Dec 13 '17 at 14:59
• Hint: for orthogonal matrices $Q$, $\operatorname{Tr}(Q\Lambda Q^\prime)=\operatorname{Tr}(\Lambda).$ – whuber Dec 13 '17 at 15:04
• I have fixed the first sentence, please make sure that my edit makes sense. – amoeba Dec 14 '17 at 8:31
• Hi @Supreeth. Did you have a chance to look at the answers? Do they help? – amoeba Dec 21 '17 at 22:57

Let us denote $X^\top X$ by $A$. By construction, it is a $n\times n$ square symmetric positive semi-definite matrix, i.e. it has an eigenvalue decomposition $A=V\Lambda V^\top$, where $V$ is the matrix of eigenvectors (each column is an eigenvector) and $\Lambda$ is a diagonal matrix of non-negative eigenvalues $\lambda_i$ sorted in the descending order.

You want to maximize $$\operatorname{Tr}(D^\top A D),$$ where $D$ has $l$ orthonormal columns. Let us write it as $$\operatorname{Tr}(D^\top V\Lambda V^\top D)=\operatorname{Tr}(\tilde D^\top\Lambda \tilde D)=\operatorname{Tr}\big(\tilde D^\top \operatorname{diag}\{\lambda_i\}\, \tilde D\big)=\sum_{i=1}^n\lambda_i\sum_{j=1}^l\tilde D_{ij}^2.$$ This algebraic manipulation corresponds to rotating the coordinate frame such that $A$ becomes diagonal. The matrix $D$ gets transformed as $\tilde D=V^\top D$ which also has $l$ orthonormal columns. And the whole trace is reduced to a linear combination of eigenvalues $\lambda_i$.

What can we say about the coefficients $a_i=\sum_{j=1}^l\tilde D_{ij}^2$ in this linear combination? They are row sums of squares in $\tilde D$, and hence (i) they are all $\le 1$ and (ii) they sum to $l$. If so, then it is rather obvious that to maximize the sum, one should take these coefficients to be $(1,\ldots, 1, 0, \ldots, 0)$, simply selecting the top $l$ eigenvalues. Indeed, if e.g. $a_1<1$ then the sum will increase if we set $a_1=1$ and reduce the size of the last non-zero $a_i$ term accordingly.

This means that the maximum will be achieved if $\tilde D$ is the first $l$ columns of the identity matrix. And accordingly if $D$ is the first $l$ columns of $V$, i.e. the first $l$ eigenvectors. QED.

(Of course this is a not a unique solution. $D$ can be rotated/reflected with any $l\times l$ orthogonal matrix without changing the value of the trace.)

This is very close to my answer in Why does PCA maximize total variance of the projection? This reasoning follows @whuber's comment in that thread:

[I]s it not intuitively obvious that given a collection of wallets of various amounts of cash (modeling the non-negative eigenvalues), and a fixed number $k$ that you can pick, that selecting the $k$ richest wallets will maximize your total cash? The proof that this intuition is correct is almost trivial: if you haven't taken the $k$ largest, then you can improve your sum by exchanging the smallest one you took for a larger amount.

Define $W=X^TX$, and denote by $v_i$ a unit-norm eigenvector corresponding to its $i$-th largest eigenvalue.

By the variational characterization of eigenvalues, $$v_1 = \underset{x,\|x\|_2=1}{\arg\max} ~ ~ x^T W x$$

Since you are looking for an orthogonal matrix, your next vector should be in a space orthogonal to $v_2$. Define $W^{(2)}=W-v_1v_1^TW$. It just so happens that $$v_2 = \underset{x,\|x\|_2=1}{\arg\max} ~ ~ x^T W^{(2)} x$$ And so on.

Why are we sure that it is indeed the eigenvectors that maximize the sum? Can't we start with a different pair of vectors and then make up for it afterwards, as whuber pointed out?

If $X=U\Sigma V^T$ is the singular value decomposition of $X$, then $X^TX=W=V\Sigma^2 V^T$ is the eigendecomposition of $W$.

Define $X_l=U_l\Sigma_l V_l^T$, where $U_l, V_l$ are $U,V$ truncated to the first $l$ columns and $\Sigma_l$ to the leading $l\times l$ block.

By the Eckart-Young-Mirsky theorem we know that $$\|X-X_l\|_F^2=\min_{A,rank(A)\leq l} \|X-A\|_F^2$$ And it is easy to see that $\underset{A}{\arg\min} \|X-A\|_F^2=\underset{A}{\arg\max} \|A\|_F^2$ whenever $A$ is the result of projecting a matrix onto the span of $X$, so $$\|X_l\|_F^2=\max_{A,rank(A)\leq l} \|A\|_F^2$$

Now, observe that

• $X_l^TX_l=V_l\Sigma_l^2V_l^T$, that is, $V_l^TX_l^TX_lV_l=\Sigma_l^2$
• $\|X_l\|_F^2=\sum_{i=1}^l\sigma_i^2$

Therefore, $\mbox{tr}(V_l^TX_l^TX_lV_l)=\mbox{tr}(\Sigma_l^2)=\sum_{i=1}^l\sigma_i^2$ is optimal.

Finally, note that $V_l^TX_l^TX_lV_l=V_l^TX^TXV_l$

• This answer is unconvincing: how do we know that you can't do better by, say, not optimizing the value at the first step, thereby accepting a slightly suboptimal value, in return for doing much better in the second step? – whuber Dec 13 '17 at 16:44
• @whuber Fair point. I'll try to improve it. – broncoAbierto Dec 13 '17 at 17:08


Toronto Math Forum

MAT244--2019F => MAT244--Lectures & Home Assignments => Chapter 9 => Topic started by: Richard Qiu on November 18, 2019, 02:19:33 PM

Title: differences between proper and improper nodes
Post by: Richard Qiu on November 18, 2019, 02:19:33 PM
Hello guys, could anyone help me to explain the differences between proper and improper nodes? btw, any suggestions on how to remember the types and stability of the critical points?
Title: Re: differences between proper and improper nodes
Post by: Amanda-fazi on November 18, 2019, 02:35:47 PM
Since both proper and improper nodes have equal eigenvalues, the differences between these two nodes is that: proper node/star point has two independent eigenvectors, while improper/degenerate node has only one independent eigenvector by (A-rI)x =0, and we create a generalized eigenvector associated with the repeated eigenvalues by letting (A-rI)y = x.

Title: Re: differences between proper and improper nodes
Post by: Amanda-fazi on November 18, 2019, 02:49:15 PM
There are mainly 5 cases of Eigenvalues(from book Elementary Differential Equations and Boundary Value Problems-11th Edition section 9.1):
as it is mentioned above, the equal eigenvalues case mentioned above is CASE 3.

CASE 1: Real, Unequal Eigenvalues of the Same Sign
CASE 2: Real Eigenvalues of Opposite Sign                            ->saddle point
CASE 3: Equal Eigenvalues
CASE 4: Complex Eigenvalues with Nonzero Real Part
CASE 5: Pure Imaginary Eigenvalues                                     ->center

After memorized there are five cases, CASE 1, CASE 3 and CASE 4 have two branches while the rest of the cases(CASE 2 and CASE 5) only have one:
to be more specific:

CASE 1: Real, Unequal Eigenvalues of the Same Sign separated into:
a)lambda1 >lambda2 >0:
critical point called node/nodal source
a)lambda1 <lambda2 <0:
critical point called node/nodal sink

CASE 3:Equal Eigenvalues separated into:
a)two independent eigenvectors:
critical point called proper node or star point
b)one independent eigenvector:
critical point called improper node or degenerate node

CASE 4:Complex Eigenvalues with Nonzero Real Part separated into:
a)pointing-outward trajectories as lambda > 0:
critical point called spiral source
a)pointing-inward trajectories as lambda < 0:
critical point called spiral sink

For the stability, as long as there is one lambda>0, then it is unstable, and the last one lambda=0 is stable. For the rest of them, asymptotically stable applied.

Title: Re: differences between proper and improper nodes
Post by: anntara khan on November 20, 2019, 04:03:25 PM
I made this handy color coded guide to help me remember all the cases:

Title: Re: differences between proper and improper nodes
Post by: anntara khan on December 11, 2019, 09:59:34 PM
Based on the stability near locally linear system I have extended the previously posted table, hope this helps remembering :)


# 8 Athletes run a race problem

8 Athletes run a race and no two athletes finish exactly together. What is the number of different possible results for the first, second and third positions?

I said that the answer is simply 8 choose 3, which gives an answer of 56. When I checked the memo, it said the answer was 336. How is this?

• The athletes ar distingushabel. So Joe, Sam, Jim is not the same as San, Joe, Jim. Am I right?
– zoli
Mar 13 '15 at 13:26
• @zoli: If you cannot distinguish the athletes, what would be the point of having them compete in a race? Mar 13 '15 at 13:34
• Mark: Right. Right.
– zoli
Mar 13 '15 at 13:39

8 choose 3 is the number of ways of choosing 3 different runners from 8 but you haven't accounted for the fact that once you've picked three, they can come in a number of different orders, specifically the number of permuations of 3 elements (i.e. 3!=6). So the final answer is $$\binom{8}{3}\times 3! = 56\times 6 = 336$$

• Oh ok, thank you for the explanation. I thought that when you combined the athletes you would automatically get the different permutations of 1st, 2nd and 3rd position. Mar 14 '15 at 6:51

The order in which the runners finish matters, so this is a permutation rather than a combination.

Until the first runner crosses the finish line, there are eight runners who could finish first. Once the first runner crosses the finish line, there are seven runners who could finish second. Once the first two runners have crossed the finish line, there are six runners who could finish third, giving $$8 \cdot 7 \cdot 6 = 336$$ ways for the runners to finish first, second, and third in the race given that there are no ties. Using permutations, the number of ways the eight runners could finish first, second, and third given that there are no ties is $$P(8, 3) = \frac{8!}{(8 - 3)!} = \frac{8!}{5!} = \frac{8 \cdot 7 \cdot 6 \cdot 5!}{5!} = 8 \cdot 7 \cdot 6 = 336$$

Imagine if John, Jack and Julie were in the "top-3", there is 6 possible combination.Because you do not know what was the rank. Anyone can be at

So $8\choose3$ ways of selecting 3 people out of a group of 8, multiplied by no of ways of arranging those 3 in the ranks $3!$.


# If $N$ is the integer $2^43^35^27$ find the smallest positive integer $m$ such that $x^m \equiv 1 \mod N$ for all integers coprime to $N$

If $N$ is the integer $2^43^35^27$ find the smallest positive integer $m$ such that $x^m \equiv 1 \mod N$ for all integers coprime to $N$.

I understand that I can do this for individual $N$ ($N= 2^4, 3^3, 5^2, 7$) and then take the LCM by Chinese Remainder Theorem. Using Fermat's Little Theorem doesn't give me the largest such $m$ for each choice of $N$. What's the relevant theorem/tool here?

EDIT (per suggestions): Applying the Euler phi-function gives us $\phi(2^4) = 8, \phi(3^3) = 3^2\times 2, \phi(5^2) = 5*4, \phi(7) = 6$.

Why are these the smallest such $m$?

• Euler's theorem could work. – 伽罗瓦 Aug 19 '17 at 21:15
• maybe Euler's theorem edit: of which Fermat's little theorem is a special case. – user451844 Aug 19 '17 at 21:15
• This is known as Carmichael function. – Jyrki Lahtonen Aug 19 '17 at 21:21
• When x is coprime to N, it can't hit any of the remainders not coprime to n. How many are coprime to N?, phi of N, by pigeonhole principle, it has to repeat by then. – user451844 Aug 19 '17 at 21:23

$\color{Green}{\text{Lemma}}$:

• For every odd prime number $p$; and for every positive integer $\alpha$;
the multiplicative group $\mathbb{Z}_{p^{\alpha}}^*$;
is a cyclic group of order $\phi(p^{\alpha})= (p-1)p^{\alpha-1}$.
In other words:

$$\big( \mathbb{Z}_{p^{\alpha}}^* \ , \times \big) \equiv \big( \mathbb{Z}_{(p-1)p^{\alpha-1}} \ , + \big) .$$

• For $\color{Red}{p=2}$; and for every positive integer $\color{Red}{3 \leq \alpha}$;
the multiplicative group $\mathbb{Z}_{2^{\alpha}}^*$;
is the direct sum of $\mathbb{Z}_2$ and a cyclic group of order $\color{Red}{\dfrac{1}{2}}\phi(2^{\alpha})= \color{Red}{2^{\alpha-2}}$.
In other words:

$$\big( \mathbb{Z}_{2^{\alpha}}^* \ , \times \big) \equiv \big( \mathbb{Z}_2 \oplus \mathbb{Z}_{\color{Red}{2^{\alpha-2}}} \ , + \big) .$$

• The multiplicative group $\mathbb{Z}_{2^2}^*$; is a cyclic group of order $2$.
The multiplicative group $\mathbb{Z}_{2}^*$; is the trivial group.

$\color{Teal}{\text{Remark}}$:

Let's define the function $\psi$ as follows:

• For every odd prime number $p$; and for every positive integer $\alpha$;
$\psi(p^{\alpha}) = \phi(p^{\alpha}) = (p-1)p^{\alpha-1} .$

• For $\color{Red}{p=2}$; and for every positive integer $\color{Red}{3 \leq \alpha}$;
$\psi(2^{\alpha}) = \color{Red}{\dfrac{1}{2}}\phi(2^{\alpha}) = \color{Red}{2^{\alpha-2}} .$

• $\psi(4)=\phi(4)=2.$

• $\psi(2)=\phi(2)=1.$

• $\psi(1)=\phi(1)=1.$

• If $n$ has the prime factorization $n = \color{Red}{2^{\alpha_0}} p_1^{\alpha_1} p_2^{\alpha_2} ... p_k^{\alpha_k}$ ;
with $\alpha_0 \in \mathbb{N}_0=\mathbb{N} \cup \{ 0 \}$ ; $\alpha_1 \in \mathbb{N}$ , $\alpha_1 \in \mathbb{N}$ , $...$ $\alpha_k \in \mathbb{N}$ ; then let's define: $\psi(n) = \text{lcm} \Big( \psi(2^{\alpha_0}); \psi(p_1^{\alpha_1}), \psi(p_2^{\alpha_2}), ..., \psi(p_k^{\alpha_k}) \Big)$

One can easilly checks that :

$$\color{Teal}{\psi(n)} \color{Green} { \text{is the least integer} \ m \ \\ \text{such that} \ x^m \overset{n}{\equiv} 1 \ ; \\ \text{for all integers} \ x \ \text{coprime to} \ n \ }.$$

$$\psi(\color{Red}{2^4} . 3^3 . 5^2 . 7) = \text{lcm} \Big( \psi(\color{Red}{2^4}); \ \psi(3^3), \ \psi(5^2), \ \psi(7) \Big) = \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{lcm} \Big( \color{Red}{2^2}; 2 \times 3^2, 4 \times 5, 6 \Big) = 2^2 \times 3^2 \times 5 = 180 .$$

• The Euler totient of$16$ is $4$? Not $8$? – Oscar Lanzi Aug 21 '17 at 15:28
• @Oscar Lanzi ; Yes you are right but $\psi(16)=\dfrac{1}{2}\phi(16)=\dfrac{1}{2}8=4$. – Jungle Boy Aug 21 '17 at 15:32


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The coordinates for a rhombus are given as (2a, 0) (0, 2b), (-2a, 0), and (0.-2b). \begin {align*}AB=\sqrt {4^2+2^2}=\sqrt {18}=3\sqrt {2}\end {align*} He says that a rhombus has two diagonals which … Prove that $$XY \perp XC$$ Remember your lesson on perpendicular lines . 2. If the product of slopes of diagonals is equal to -1, we say both are perpendicular. Examples: 3. Be sure to assign appropriate variable coordinates to your rhombus's vertices! Step 2: Calculate the coordinates of the midpoints of the sides. Using the coordinate plane in proof. 0. This is the basic property of rhombus. We explain Coordinate Geometry of Rhombii with video tutorials and quizzes, using our Many Ways(TM) approach from multiple teachers. How do I prove analytically using co- ordinate geometry that the diagonals of a rhombus are perpendicular to each other Ask Question Asked 3 years, 1 month ago The shape of a rhombus is in a diamond shape. I hav to prove that the quadrilateral is a rhombus and not a square. Since all sides are equal, it may be a square also. This one is a medium level difficulty question and tests the following concepts: finding length of a line segment given coordinates of its end points; properties of quadrilaterals including square, rectangle, rhombus and parallelogram.. 2. Problem 3 Medium Difficulty. These two sides are parallel. Can't find last vertex. Examples: 1. Hi, in C1 coordinate geometry questions relating to shapes often come up and a lot of marks are usually given for finding the area. And in a rhombus, not only are the opposite sides parallel-- it's a parallelogram-- … Hi! To prove that two lines are perpendicular, when all we have are those two lines, we can use the Linear Pair Perpendicular Theorem - If two straight lines intersect at a point and form a linear pair of equal angles, they are perpendicular.. Our diagonals intersect at point O, so we'd need to show the two linear angles formed at that intersection point are equal, and we … x-coordinate of midpoint = average of x … ... (1,3) is a parallelogram because both pairs of opposite sides are congruent and then show it is a rhombus. Since the side of the rhombus is 41, the area of the rhombus is 41 sin θ, where θ is either of the (supplementary) internal angles of the rhombus. Finding vertices of rhombus formed by lines $y=2x+4$, $y=-\frac{1}{3}x+4$ and $(12,0)$ is a vertex. He says that the rhombus is a quadrilateral and hence the sum of internal angles of it is 360 degrees. Here is a formula you can use that will solve an area of any triangle given that you have the coordinates of the three vertices. We explain Coordinate Geometry of Rhombii with video tutorials and quizzes, using our Many Ways(TM) approach from multiple teachers. 1. Coordinate Proof 2. Yes, a square is a rhombus A square must have 4 congruent sides. So the area is somewhere between 0 (if the rhombus is totally "flat") and 41 (if the rhombus is a square). Choose from 500 different sets of proofs coordinate geometry flashcards on Quizlet. (iii)  Diagonals will be perpendicular. Slope of the diagonal AC : A (2, -3) C (-2, 1) Slope = (y 2 -y 1)/ (x 2 -x 1) @lioness99a No. Have you missed part of the question out? Ivan used coordinate geometry to prove that quadrilateral EFGH is a square. This lesson will demonstrate how to use slope, midpoint, and distance formulas to determine from the coordinates of the vertices if a quadrilateral is a rhombus in a coordinate … If all sides of a quadrilateral are congruent, then it’s a rhombus (reverse of the definition). To prove it is rhombus, we can prove any one of the following. Want to improve this question? more interesting facts . Use coordinate geometry to prove that quadrilateral DIAN is a square. So, the given are not vertices of rhombus. more interesting facts . Ivan used coordinate geometry to prove that quadrilateral EFGH is a square. .And Why To use coordinate geometry to prove that a ﬂag design includes a rhombus, as in Example 2 In Lesson 5-1, you learned about midsegments of triangles.A trapezoid also has a Area of the rhombus is greater than what? Question 408015: Using a coordinate geometry proof, which method below is a correct way to prove a quadrilateral is a rhombus? Since the side of the rhombus is $\sqrt{41}$, the area of the rhombus is $41\sin\theta$, where $\theta$ is either of the (supplementary) internal angles of the rhombus. Proof: Rhombus Opposite Angles are Congruent (1) AD=CD //Given, definition of a rhombus (2) AB=CB //Given, definition of a rhombus In this coordinate geometry worksheet, 10th graders solve and draw 10 different problems related to complete proofs in coordinate geometry. Algebraic proofs for geometric theorems (Geometry) Prove whether a figure is a rectangle in the coordinate plane An updated version of this instructional video is available. Proof: Rhombus area. - Show that both pairs of opposite sides are congruent. First draw a figure and choose convenient axes and coordinates. (someone plz help cuz i don't understand) Coordinate Geometry: Ms.Sue or someone please help. Choose one of the methods. So we have a parallelogram right over here. Write sentences that explain your ideas clearly. To prove it is rhombus, we can prove any one of the following. Proving a Quadrilateral is a Rhombus Prove that it is a parallelogram first, then: Method 1: Prove that the diagonals are perpendicular. The diagonals have the same midpoint, and one pair of opposite sides have equal lengths. You’re correct but can you kindly provide me with an explanation for that ? Distance between two points. Prove that $$XY \perp XC$$ Remember your lesson on perpendicular lines . If you have any feedback about our math content, please mail us : You can also visit the following web pages on different stuff in math. Played 0 times. Proof: The diagonals of a kite are perpendicular Proof: Rhombus … The area cannot be determined as two points are insufficient to describe a rhombus. The one main way to prove that a quadrilateral is a rhombus is to prove that the distances of the four sides of the quadrilaterals are congruent (equal distances) and then prove that the diagonals of the quadrilateral are not congruent (unequal distances). Method 3: Prove that all four sides are equal. Show that both pairs of opposite sides are parallel 3. The diagonals of a rhombus … Quadrilateral EFGH is at E (−2, 3), F (1, 6), G (4, 3), and H (1, 0)1. Use coordinate geometry to prove that quadrilateral KAIT is a parallelogram. How do you prove a rhombus in coordinate geometry? All sides are congruent. In this non-linear system, users are free to take whatever path through the material best serves their needs. Using the coordinate plane in proof. First of all, a rhombus is a special case of a parallelogram. Proofs of general theorems. This lesson will demonstrate how to use slope, midpoint, and distance formulas to determine from the coordinates of the vertices if a quadrilateral is a rhombus in a coordinate plane. Opposite sides have congruent slopes. These unique features make Virtual Nerd a viable alternative to private tutoring. Well think of the other parallel side of the rhombus, it may be inclined at any angle from ${\pi}\over 2$ to $0$ and still satisfy the conditions to be a rhombus. He says that four sides of a rhombus are equal in length. Prove that a shape must be a rhombus using rules of coordinate geometry, finding the length of the altitude of the rhombus, Area of rhombus given circumradii of its contained triangles. Prove that the quadrilateral with vertices A(-1,0), B(3,3), C(6,-1) and D(2,-4) is … If we were told that the rhombus is not a square, B would be the correct answer. This video is unavailable. So the area is somewhere between $0$ (if the rhombus is totally "flat") and $41$ (if the rhombus is a square). If the diagonals of a quadrilateral are perpendicular bisectors of each other, then it’s a rhombus (converse of a property). To prove that it is a rhombus, remember that the definition of a rhombus is a quadrilateral with four congruent sides. Point of intersection. Since the side of the rhombus is $\sqrt{41}$, the area of the rhombus is $41\sin\theta$, where $\theta$ is either of the (supplementary) internal angles of the rhombus. . 2 years ago. This shape appears to be a rhombus. Step 1: Identify the coordinates of the vertices of the rhombus. I hav to prove that the quadrilateral is a rhombus and not a square. Rectangle, Parallelogram, Trapezoid. Question Different forms equations of straight lines. Will a refusal to enter the US mean I can't enter Canada either? math b. the vertices of triangle ABC are A(-2,3), B(0,-3), and C(4,1). Coordinate Geometry. Prove that the quadrilateral with vertices A(-1,0), B(3,3), C(6,-1) and D(2,-4) is a square. In Euclidean geometry, a rhombus is a type of quadrilateral. Prove that three points in a rhombus are collinear 0 Given the co-ordinates of several points, how to determine which segments are sides and which are diagonals if the coordiantes of a quadrilateral TEAM are T(-2,3), E(-5,-4),A(2,-1), AND M(5,6). Next lesson. Algebraic proofs for geometric theorems (Geometry) Prove whether a figure is a rectangle in the coordinate plane An updated version of this instructional video is available. Practice: Prove parallelogram properties. In the $xy$-plane, two adjacent vertices of the rhombus are $(-2,2)$ and $(2,7)$. How does a bank lend your money while you have constant access to it? Quadrilateral EFGH is at E (−2, 3), F (1, 6), G (4, 3), and H (1, 0)1. more interesting facts . Show: Formula: Work Prove it is a Rectangle. Were the Beacons of Gondor real or animated? So, its midpoint will be equal. Use coordinate geometry to prove that a) quadrilateral NORA is a rhombus, and b) quadrilateral NORA is not a square. Learn the properties of quadrilaterals. Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long. Be sure to really show the original formula and show the steps clearly- be neat and precise. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. prove, by means of coordinate geometry, that the median to side BC is also the altitude to side BC. How do you write a plan to prove that the midpoints of the sides of a rhombus determine a rectangle using coordinate geometry? Method: First, prove the quadrilateral is a rhombus by showing all four sides is congruent; then prove the quadrilateral is a rectangle by showing the diagonals is congruent. Use graph paper, ruler, pencil. Look over the toolkit page that describes the steps used in a coordinate geometry proof. But since in a rhombus all sides are equal, it is easier to prove this property than for the general case of a parallelogram, and this is what we … more about imaginary numbers . If four coordinates of four vertices are given in an order, then to prove that the given vertices are of a rhombus, we need to prove that four sides are equal by using the distance formula. Coordinate Geometry. To prove that it is a rhombus, remember that the definition of a rhombus is a quadrilateral with four congruent sides. Examples: 3. This one is a medium level difficulty question and tests the following concepts: finding length of a line segment given coordinates of its end points; properties of quadrilaterals including square, rectangle, rhombus and parallelogram. What is the best way to play a chord larger than your hand? How do you prove that the quadrilateral formed by joining the midpoints of a rhombus is a rectangle using coordinate geometry? geometry Question 408015: Using a coordinate geometry proof, which method below is a correct way to prove a quadrilateral is a rhombus? If we were told that the rhombus is not a square, B would be the correct answer. The most common shape is the triangle. A square is a special rhombus that also has 4 right angles. Proofs Using Coordinate Geometry 348 Chapter 6 Quadrilaterals What You’ll Learn • To prove theorems using ﬁgures in the coordinate plane. Show that both pairs of opposite sides are congruent. There's more to it than that. Aptitude Practice Questions Coordinate Geometry Question 3 T his aptitude practice question is a coordinate geometry question. If the points had bee $(0,0)$ and $(0,5)$ then we would know the area is less than $41.$. Is there a bias against mentioning your name on presentation slides? I can prove a quadrilateral is a rhombus or Square on a coordinate grid. First show that you can place the rhombus in the plane with vertices as shown in the diagram below. In a coordinate proof, you are proving geometric statements using algebra and the coordinate plane.Some examples of statements you might prove with a coordinate proof are: Prove or disprove that the quadrilateral defined by the points is a parallelogram. Show that a pair of opposite sides are congruent and parallel 4. In an amplifier, does the gain knob boost or attenuate the input signal? As we were not told that, D is the correct answer. I tried the distance formula to get the distance between the two points which is $\sqrt{41}$ but I don’t know what to do next. Why do we not observe a greater Casimir force than we do? Write a plan to prove that the midpoints of the sides of a rhombus determine a rectangle using coordinate geometry. T his aptitude practice question is a coordinate geometry question. Video transcript. First, plot the points. Hence, it is also called a diamond. The most common shape is the triangle. Coordinate Proof 1. Use coordinate geometry to prove that the quadrilateral formed by connecting the midpoints of a kite is a rectangle. Proof: Rhombus diagonals are perpendicular bisectors. Using https://artofproblemsolving.com/wiki/index.php/Rhombus#Proof_that_the_diagonals_of_a_rhombus_divide_it_into_4_congruent_triangles, the intersection of the two diagonals will lie on the circle, Hence the point cannot be determined with the information supplied, So, will be the lengths of the two diagonals, site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. To determine the type of quadrilateral given a set of vertices. For a rhombus, where all the sides are equal, we've shown that not only do they bisect each other but they're perpendicular bisectors of each other. Write a plan to prove that the midpoints of the sides of a rhombus determine a rectangle using coordinate geometry. How does 真有你的 mean "you really are something"? Therefore, to prove it is a rhombus you must verify that all sides are the same length. @Vai I'm still not entirely sure what your question is. Use coordinate geometry to prove that the diagonals of a rhombus are perpendicular. Step 1: Identify the coordinates of the vertices of the rhombus. If we were told that the rhombus is not a square, B would be the correct answer. (Believe it … Use graph paper, ruler, pencil. in coordinate geometry how can we prove that a quadrilateral is a rhombus and not a square? Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long. answer choices . How can I defeat a Minecraft zombie that picked up my weapon and armor? It only takes a minute to sign up. Click to see full answer. Method: First, prove the quadrilateral is a rhombus by showing all four sides is congruent; then prove the quadrilateral is a rectangle by showing the diagonals is congruent. Be sure to really show the original formula and show the steps clearly- be neat and precise. ; Prove or disprove that the point lies on the circle centered at the origin containing the point . Prove that the quadrilateral ABCD with the vertices in a coordinate plane A(-3,-4), B(5,-3), C(1,4) and D(-7,3) (see the Figure) is a rhombus. That is what is given in the answer key. And what I want to prove is that its diagonals bisect each other. 1. in coordinate geometry how can we prove that a quadrilateral is a rhombus and not a square? Here is a formula you can use that will solve an area of any triangle given that you have the coordinates of the three vertices. There are 5 different ways to prove that this shape is a parallelogram. Geometry HELP. Has your book defined "rhombus" so that a square is a rhombus, or so that a square is not a rhombus? (i)  In a rhombus the length of all sides will be equal. If two consecutive sides of a rectangle are congruent, then it’s a square (neither the reverse of the definition nor the converse of a property). Use coordinate geometry to prove that a) quadrilateral NORA is a rhombus, and b) quadrilateral NORA is not a square. Method 2: Prove that a pair of adjacent sides are equal. StatementReason1. A(2, -3), B (6, 5) C (-2, 1) and D (-6, -7), Distance Between Two Points (x1, y1) and (x2 , y2), Here x1 = 2, y1 = -3, x2 = 6  and  y2 = 5, Here x1 = 6, y1 = 5, x2 = -2  and  y2 = 1, Here x1 = -2, y1 = 1, x2 = -6  and  y2 = -7, Here x1 = -6, y1 = -7, x2 = 2  and  y2 = -3. Area of a rhombus in co-ordinate geometry [closed], https://artofproblemsolving.com/wiki/index.php/Rhombus#Proof_that_the_diagonals_of_a_rhombus_divide_it_into_4_congruent_triangles. If a rhombus contains a right angle, then it’s a square (neither the reverse of the definition nor the converse of a property). Here are a few ways: 1. Writing coordinate proofs requires a knowledge of the slope formula, the distance formula and the mid point formula. . The question asks what is greater between A and B. You can use the distance formula or the Pythagorean Theorem to do this. ... How do you prove a quadrilateral is a rhombus? StatementReason1. These unique features make Virtual Nerd a viable alternative to private tutoring. How to express the behaviour that someone who bargains with another don't make his best offer at the first time for less cost? if the coordiantes of a quadrilateral TEAM are T(-2,3), E(-5,-4),A(2,-1), AND M(5,6). Be sure to include the formulas. Why does the T109 night train from Beijing to Shanghai have such a long stop at Xuzhou. Using coordinate geometry to prove that the diagonals of a square are perpendicular to each other. The coordinates for a rhombus are given as (2a, 0), (0, 2b), ( 2a, 0), and (0, 2b). Coordinate Proof 1. So the area is somewhere between $0$ (if the rhombus is totally "flat") and $41$ (if the rhombus is a square). Coordinate Geometry 348 Chapter 6 Quadrilaterals What You’ll Learn • To prove theorems using ﬁgures in the coordinate plane. In this non-linear system, users are free to take whatever path through the material best serves their needs. Coordinate geometry formulas. So that side is parallel to that side. Learn proofs coordinate geometry with free interactive flashcards. 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Similarly, how can you prove that a parallelogram is a rectangle when using the coordinate plane? Strategy. What does a Product Owner do if they disagree with the CEO's direction on product strategy? Prove or disprove that the quadrilateral defined by the points , , , is a rhombus. He begins by saying that the properties of a rhombus are similar to the properties of a square. more interesting facts . 3) Quadrilateral NORA has vertices N(3,2), O(7,0), R(11,2), and A(7,4). COORDINATE GEOMETRY. Given: Vertices are at A(0,0), B(a,0), C(a,a) and D(0,a) Slope of AC=1; Slope of BD=-1 Type your proof. You do not know anything else about the rhombus in co-ordinate geometry [ closed ] https! Path through the material best serves their needs be the correct answer set of vertices I can prove any of! … coordinate geometry proof that formally proves what this applet informally illustrates not be determined respectively be correct... 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Worksheet, 10th graders solve and draw 10 different problems related to complete proofs in geometry. Correct way to play a chord larger than your hand something '' 10th graders solve and draw 10 different related... For auction at a higher price than I have no Idea how to express the behaviour that who... Informally illustrates how can we prove that the rhombus in co-ordinate geometry [ closed ],:! Prove or disprove that the quadrilateral is a rhombus you must verify that sides. Shown in the diagram below given four points Form a rhombus worksheet, 10th graders solve and draw different. Your book defined  rhombus '' so that a square you really are something '' determined as two points insufficient. And choose convenient axes and coordinates 10 different problems related to complete proofs in coordinate worksheet... While you have constant access to it solve and draw 10 different problems related to complete proofs coordinate! Look over the toolkit page that describes the steps used in a diamond shape whose diagonals intersect each.. Congruent and then show it is 360 degrees sides are the opposite sides have equal lengths NORA. And the mid point formula are perpendicular force than we do the diagonals the. [ closed ], https: //artofproblemsolving.com/wiki/index.php/Rhombus # Proof_that_the_diagonals_of_a_rhombus_divide_it_into_4_congruent_triangles equal in length finitely presented modules abelian first time less. Are something '' will be equal presented modules abelian he begins by saying that the median to side BC of. Figures in the answer key if you were graphing a polygon to create a coordinate to. Best offer at the first time for less cost D, such ABCD! ( 1,3 ) is a rhombus the length of all sides are the same,! Quadrilateral bisect all the angles, then it ’ s a rhombus is not a square show original... Learn • to prove that the rhombus, not only are the same midpoint, and.... Inner quadrilateral a rectangle definition ) show it is rhombus, we can prove rhombus!
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# Toronto Math Forum

## MAT244--2018F => MAT244--Tests => Quiz-6 => Topic started by: Victor Ivrii on November 17, 2018, 04:01:01 PM

Title: Q6 TUT 5102
Post by: Victor Ivrii on November 17, 2018, 04:01:01 PM
The coefficient matrix contains a parameter $\alpha$.

(a) Determine the eigenvalues in terms of $\alpha$.
(b)  Find the critical value or values of  $\alpha$  where the qualitative nature of the phase portrait for the system changes.
(c) Draw a phase portrait for a value of  $\alpha$ slightly below, and for another value slightly above, each critical value.
$$\mathbf{x}' =\begin{pmatrix} 4 &\alpha\\ 8 &-6 \end{pmatrix}\mathbf{x}.$$
Title: Re: Q6 TUT 5102
Post by: Michael Poon on November 17, 2018, 04:35:52 PM
a) Finding the eigenvalues:

Set the determinant = 0

\begin{align}
(4 - \lambda)(-6 - \lambda) - 8\alpha &= 0\\
\lambda^2 + 2\lambda - 24 - 8\alpha &= 0\\
\lambda &= -1 \pm \sqrt{25 + 8\alpha}
\end{align}

b)

Case 1: Eigenvalues real and same sign
when: $\alpha$ > $\frac{-25}{8}$ + 1

Case 2: Eigenvalues real and opposite sign
when: $\frac{-25}{8}$ < $\alpha$ < $\frac{-25}{8} + 1$

Case 3: Eigenvalues complex
when: $\alpha$ < $\frac{-25}{8}$

critical points: $\alpha$ = $\frac{-25}{8}$, $\frac{-25}{8}$ + 1

c) will be posted below:
Title: Re: Q6 TUT 5102
Post by: Michael Poon on November 17, 2018, 04:42:29 PM
Phase portraits attached below:

Top: Eigenvalues real & same sign (+ve), stable

Middle: Eigenvalues real & opposite sign, saddle

Bottom: Eigenvalues complex & negative, unstable spiral
Title: Re: Q6 TUT 5102
Post by: Jiacheng Ge on November 18, 2018, 12:41:00 PM
My solution is different.
Title: Re: Q6 TUT 5102
Post by: Victor Ivrii on November 25, 2018, 09:49:09 AM
When eigenvalues pass from real to complex conjugate, stability does not change. Jiacheng is right


# Minimum Volume of a circular, right cone, with a sphere inscribed in it.

Question: A sphere of radius $r$ is inscribed in a circular, right cone.

What is the minimum radius and height of the circular cone? (Thus, volume)

Because the answer would specifically be proportional to the radius given $r$,

I have set the base radius of the cone $ar$, and would like to solve for $a$.

Because triangle AB'O, and AC'O are congruent, I have labelled the angles $\theta$

Then, $\tan(\theta)= \frac{1}{a}$

The double angle formula is:

$\tan(2\theta)= \frac{2\tan(\theta)}{1-\tan^{2}(\theta)}$

Plugging in the values, it results in:

$\tan(2\theta)= \frac{2a}{a^2-1}$

For simplicity, let's set the radius 1, such that the base has a radius length $a$, then the height becomes

$a\times\frac{2a}{a^2-1} = \frac{2a^2}{a^2-1}$

This makes perfect sense, because if we extend the base radius to infinity, then the resulting cone's top vertex would end up on the top of the sphere.

$\lim\limits_{a \to \infty}\frac{2a^2}{a^2-1} = 2$

(We have set the sphere's radius to 1)

Thus, we have the variables to actually solve for the minimum value.

The volume of the right circular cone is given by:

$V = \frac{1}{3}\pi r^2 h$

We have set $r = a$ and $h = \frac{2a^2}{a^2-1}$

Then,

$V = \frac{1}{3}\pi a^2 \frac{2a^2}{a^2-1} = \frac{2}{3}\pi \frac{a^4}{a^2-1}$

Neglecting the coefficients, deriving V with respect to a and solving for 0,

We get

$a = \sqrt{2}$ and $h = 4$

So

$a = \sqrt{2}r$ and $h = 4r$

Apparently though, I'm wrong. The answer is:

$a = 2r$ and $h = 4r$

But I don't understand what's wrong with my argument. Any ideas?

• I don't think you have made any mistakes, the book must have a typo, if a=2r and h=4r the sides of the cone would not be tangent to the sphere.
– WW1
Apr 21, 2015 at 0:34
• It is weird how the question was asked the cone's radius to height ratio would have a minimum value of zero ( when a=r and $h\to \infty$ ) . I don't see how minimizing that ratio is the same as minimizing the volume of the cone.
– WW1
Apr 21, 2015 at 0:39
• @WW1 oh, that actually does not refer to the ratio, I just meant radius AND height. Sorry, that is definitely my mistake. Apr 21, 2015 at 0:43
• "radius and height" is a very peculiar thing to be asked to minimize - does the original question include " thus volume" ?
– WW1
Apr 21, 2015 at 0:56
• @WW1, the original question asks for the radius, and height required for the minimum volume; for a minimum volume there has to be a base radius, "a", and because of geometry, the height is also dependent on the chosen "a." Apr 21, 2015 at 1:54

As user WW1 says, the book must have a typo - your work is correct. Here's another way we could have calculated it giving the same answer:

The radius of the incircle of a triangle is $\frac{\triangle}{s}$ where $\triangle$ is the area of the triangle and $s$ is the semi-perimeter. Let's call $R$ the radius of the cone and $h$ the height of the cone. If $r$ is the radius of the sphere this gives us

\begin{align}&r=\frac{Rh}{\sqrt{h^2+R^2}+R} \\\implies &r\sqrt{h^2+R^2}+rR=Rh \\\implies &r\sqrt{h^2+R^2}=R(h-r) \\\implies &r^2\left(h^2+R^2\right)=R^2(h^2-2hr+r^2) \\\implies &r^2h^2=R^2\left(h^2-2hr\right) \\\implies &\frac{r^2h^2}{h^2-2hr}=R^2 \\\implies &\frac{r^2h}{h-2r}=R^2 \end{align}

Now the volume of the cone is $$V=\dfrac{\pi R^2h}{3}=\frac{\pi r^2h^2}{3(h-2r)}$$

Hence $$\frac{dV}{dh}=\frac{\pi\left(r^2h^2-2r^2h(h-2r)\right)}{3(h-2r)^2}=\frac{\pi\left(4r^3h-r^2h^2\right)}{3(h-2r)^2}$$

Setting the derivative to $0$, we get $$4r^3h-r^2h^2=0\hspace{5mm}\implies\hspace{5mm}4r-h=0\hspace{5mm}\implies\hspace{5mm}h=4r$$

Plugging this into our equation for $R^2$ we get $$R^2=\frac{r^2(4r)}{4r-2r}=\frac{4r^3}{2r}=2r^2$$ Hence $R=\sqrt{2}r$ and $h=4r$ minimizes the volume as you found.


# Difference between symbols for absolute value?

I know all of the following are used for showing the absolute value:

• $\lvert x \rvert$
• $| x |$
• $\mid x \mid$

I myself think that there is another one \abs; but I can't use this last one.

What is the difference between them?

• Possible Duplicate: Absolute Value Symbols. Oct 8, 2017 at 8:22
• The first is right, the second one can produce unexpected results, the last one is wrong. Oct 8, 2017 at 8:34
• @egreg ; (+1) Clear and concise, thank you my dear egreg. Oct 8, 2017 at 9:05

You ask for the differences and here's an analysis for them.

First let's get rid of the last case: \mid is a relation symbol like = or < and TeX adds spaces around it that disqualify the command from being used for the absolute value.

The other two seem good, but the first one is better as it doesn't need precautions. The fact is that | is considered an ordinary symbol, whereas \lvert is an opening (like [) and \rvert is a closing (like ]).

Let's compare some cases.

\documentclass{article}
\usepackage{amsmath}

\begin{document}
$\begin{array}{lllll} \text{good} & \lvert x\rvert\le 1 & \lvert-1\rvert=1 & \lvert\sin x\rvert & \log\lvert x-1\rvert \\[3ex] \text{so and so} & |x|\leq 1 & |-1|=1 & |\sin x| & \log|x-1| \\[3ex] \text{wrong} & \mid x\mid \leq 1 & \mid -1\mid =1 & \mid \sin x\mid & \log\mid x-1\mid \end{array}$

\end{document}

You can see in the picture that \lvert and \rvert produces the right spacing in all cases, whereas |...| doesn't. Not to speak about \mid.

Using | would force you to input

|{-1}|
|{\sin x}|
\log\!|x-1|

You can define \abs as suggested in answers to Absolute Value Symbols but I recommend not to follow the suggestion of making \left and \right automatic, because it will most of the time choose a wrong size.

• Is the spacing for log |x-1| in the first line considered correct? Personally, I find the version in the second line a lot better-looking (for that last formula only). Oct 8, 2017 at 15:22
• @FedericoPoloni I consider the absolute value just like a parenthesized expression, so the top is right, but I wouldn't turn down consistent usage like in the second line. Oct 8, 2017 at 15:23
• Does "so and so" mean "acceptable"? Oct 9, 2017 at 11:10
• @Ooker It means that in some cases it's right, in others it's wrong; precisely, the first is right, the last might be acceptable (see another comment), the middle ones are plainly wrong. Oct 9, 2017 at 11:51
• Can TeX be complied like programming languages? For example in your example can I use a loop to print the sin, log three times with different cases? Oct 9, 2017 at 22:07


# Minimum of a trigonometric function involving absolute value

Given $$f(x) = | \sin( | x | ) |$$, I am told to found the local and global minimum and maximum of $$f$$ (If they exist).

Simply from sketching the graph of $$f$$ I get that the function maximizes periodically for every $$x = k \frac{\pi}{2}$$ with $$k$$ being a whole number, giving that there's no global maximum but infinite local maxima equal to $$1$$.

Following the same logic, I thought the function would minimize with period $$\pi$$ to give local minima of $$0$$. However, when checking for the answer, it appears the function has no global nor local minima.

• What are your definitions for local and global minima? To me it seems like there are both. For me, functions which have no global minima are unbounded (like $y = 1/x$). – paulinho Jun 7 at 14:00
• Why do you think $1$ is not the global maximum? It certainly is. Also, $0$ is the global minimum. (I'm assuming $x$ is only allowed to take real values.) – saulspatz Jun 7 at 14:05
• How did you "check answer" and why does it appear there is not minima. Your logic that $|sin |x|| \ge 0$ and $|sin |k\pi|| = 0$ indicate that $0$ are the local/global minimum. So whatever said there were none is wrong. – fleablood Jun 7 at 15:30
• What is interesting to note is that the method of finding local extrema via solving $f'(x) = 0$ will not work here because $|\sin |x||$ is not differentiable at $k\pi$. But solving $f'(x)=0$ is not the only way to have or find extrema. – fleablood Jun 7 at 15:32

Your general reasoning is correct. The global minimum is $$0$$, and it is attained in points of the form $$x = k \pi$$. The global maximum is 1, and it is attained in points of the form $$x= \frac{\pi}{2}+ k \pi$$. Obviously all these global maxima/minima are also local maxima/minima.

I find it very strange that the solution says differently... Are you certain of the expression for $$f$$?

Good insights to start off about the periodicity of the sine function.

Let's think about this in parts - dissecting the function from the outside inward. Remember the definition of absolute value: $$f(x) = | g(x) | = \left\{ \begin{array}{ll} g(x) & g(x) \geq 0 \\ -g(x) & g(x) < 0 \end{array} \right.$$ $$g(x)=\sin(|x|)=\sin\left( \begin{array}{ll} x & x \geq 0 \\ -x & x < 0 \end{array}\right)$$

By definition, the minima cannot be lower than 0. That does not make it the minima of the function, but it is a good check for the result we get.

So what is the maxima/minima of $$\sin(x)$$? $$-1 \leq \sin(x)\leq 1$$ So what is the minima applying the absolute value? $$0 \leq |\sin(x)| \leq 1 \forall x$$ Those are the maxima/minima of the function, regardless of the $$x$$ input. Let's solve for $$x$$: $$\sin x = 0 \rightarrow x = ... ,-2\pi,\pi,0,\pi,2\pi, ...$$ $$\sin x = 1 \rightarrow x = ... ,{-3\pi \over 2},{-\pi \over 2},{\pi \over 2},{3\pi \over 2},...$$ Notice all the negative values, those don't apply since we apply the absolute value to $$x$$. $$x_{min}=0,\pi,2\pi,...$$ $$x_{max}={\pi \over 2},{3\pi \over 2},...$$ There are no local maxima/minima which are not the global maxima/minima.

• "There are no local maxima/minima which are not the global maxima/minima*" (emphasis mine). But there are local maxima/minima that are the local extrema. So for as I know there is no requirement in this question or in any definition that local extrema can not also be global extrema. – fleablood Jun 7 at 15:35
• Nope, there isn't. Just a note I figured I would add. – FundThmCalculus Jun 7 at 15:45

By graphing you function it seems that $$1$$ and $$0$$ are the maximum and minimum values for the result, Why ?
The answer is: it's well known that the trigonometric functions are commonly used for expressing ratios between sides of right angled triangle, between$$(-1,1)$$ each $$\frac \pi 2$$, but by using the absolute value the result can only be positive so the result goes between $$0$$ and $$1$$.

So, the function has a unique global maximum of value $$1$$ at $$x=k \frac \pi 2, k \in \mathbb Z$$ and a unique global minimum of value $$0$$ at $$x=m, m \in \mathbb Z$$.


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Four fair coins are tossed simultaneously. The probability that at least one head and one tail turn up is

1. $\frac{1}{16}$
2. $\frac{1}{8}$
3. $\frac{7}{8}$
4. $\frac{15}{16}$

reshown

Total outcomes - 24  (Because 4 coins are tossed simultaneously and each coin has 2 outcomes-either head or tail)
Now out of this 16 outcomes, one will be all HHHH(all heads) and other will be all TTTT(all tails) rest 14 outcomes will have atleast one head and one tail.

So, probability, favourable events/total outcome

14/16 = 7/8

Probability of at least one head and one tail = 1- Probability of No (head and tail)

=  1 –  2 / (2*2*2*2)     {Numerator HHHH and TTTT and denominator all choices}

= 7/8

probability of getting all heads =$\dfrac{1}{16}$

probability of getting all tails =$\dfrac{1}{16}$

probability of getting at least one head and one tail $= 1 - \dfrac{1}{16} - \dfrac{1}{16} = \dfrac{7}{8}.$

nice bro, we can also solve it using tree method

Using tree method it provides 14 favourable outcomes out of 16

So probability should be 14/16=7/8

Total outcomes - 24  (Because 4 coins are tossed simultaneously and each coin has 2 outcomes-either head or tail)
Now out of this 16 outcomes, one will be all HHHH(all heads) and other will be all TTTT(all tails) rest 14 outcomes will have atleast one head and one tail.

So, probability, favourable events/total outcome

14/16 = 7/8

Another simple approach:

and q= P(tails) = 1/2

Requirement:

Using binomial distribution,

Required probability = $_{}^{4}\textrm{C}_{1} p^{1} q^{3} + {}^{4}\textrm{C}_{2} p^{2} q^{2} + {}^{4}\textrm{C}_{3} p^{3} q^{1}$

= $_{}^{4}\textrm{C}_{1} (1/2)^{1} (1/2)^{3} + {}^{4}\textrm{C}_{2} (1/2)^{2} (1/2)^{2} + {}^{4}\textrm{C}_{3} (1/2)^{3} (1/2)^{1}$

$= \frac{7}{8}$

1 head 3 tails =1/16 2 heads 2 tails =1/16 3 heads 1 tail =1/16 adding we get 3/16 where i am wrong anyone plz explain

### 1 comment

1 head 3 tails =1/16 * 4C1 = 1/4

2 heads 2 tails =1/16 *4C2 = 3/8

3 heads 1 tail =1/16 *4C3 = 1/4