# Why does Trapezoidal Rule have potential error greater than Midpoint?

I can approximate the area beneath a curve using the Midpoint and Trapezoidal methods, with errors such that:

$Error_m \leq \frac{k(b-a)^3}{24n^2}$ and $Error_T \leq \frac{k(b-a)^3}{12n^2}$.

Doesn't this suggest that the Midpoint Method is twice as accurate as the Trapezoidal Method?

• The constant $k$ is not necessarily the same. It depends on $f$, differently for each method. Dec 12, 2013 at 5:30
• Here in both cases $k$ is the max of the (absolute value) of the second derivative. Then your inequalities give upper bounds on the error. One can construct $f$ such that for given $n$ the Trapezoidal Rule is dead on, while the Midpoint Rule is not. Dec 12, 2013 at 5:34
• 'k' is equal to the maximum of $f''(x)$ for both $E_m$ and $E_T$ Dec 12, 2013 at 5:43
• Even if the "worst case" for trapezoidal is half as good, the "typical case" could be much better. Dec 12, 2013 at 14:14

It has already been said in the comments that the error estimates you cite are upper bounds, so actual errors may be smaller and $E_M$ won't usually be exactly half of $E_T$ (and may actually be larger in some cases).

Nevertheless, it is well worth pointing out that if the function you are integrating happens to be a cubic polynomial, then we can make an exact statement: $$E_M=-\frac{1}{2}E_T.$$ You would probably never integrate a cubic polynomial numerically, but if the function you are integrating is well-approximated by a cubic polynomial on every subinterval used in the numerical integration, then the errors should still be related in roughly the same way.

The above relation obviously holds for the functions $f(x)=1$ and $f(x)=x$. Let's verify it by brute force for $f(x)=x^2$ and $f(x)=x^3$. Consider a single subinterval $[a,b]$ and let $A[f]$, $T[f]$, and $M[f]$ represent the exact area integral of $f$ on $[a,b]$, the trapezoidal estimate, and the midpoint estimate, respectively. Then \begin{aligned} A[x^2]&=\int_a^bx^2\,dx=\frac{b^3-a^3}{3},\\ T[x^2]&=\frac{b-a}{2}(b^2+a^2)=\frac{b^3-ab^2+a^2b-a^3}{2}\\ M[x^2]&=(b-a)\left(\frac{b+a}{2}\right)^2=\frac{b^3+ab^2-a^2b-b^3}{4}. \end{aligned} So \begin{aligned} E_T[x^2]&=T[x^2]-A[x^2]=\frac{b^3-a^3}{6}-ab\frac{b-a}{2},\\ E_M[x^2]&=M[x^2]-A[x^2]=-\frac{b^3-a^3}{12}+ab\frac{b-a}{4}=-\frac{1}{2}E_T[x^2].\\ \end{aligned} Likewise \begin{aligned} A[x^3]&=\int_a^bx^3\,dx=\frac{b^4-a^4}{4},\\ T[x^3]&=\frac{b-a}{2}(b^3+a^3)=\frac{b^4-ab^3+a^3b-a^4}{2}\\ M[x^3]&=(b-a)\left(\frac{b+a}{2}\right)^3=(b-a)\frac{b^3+3ab^2+3a^2b+a^3}{8}\\ &=\frac{b^4+2ab^3-2a^3b-a^4}{8}. \end{aligned} So \begin{aligned} E_T[x^3]&=T[x^3]-A[x^3]=\frac{b^4-a^4}{4}-\frac{ab}{2}(b^2-a^2),\\ E_M[x^3]&=M[x^3]-A[x^3]=-\frac{b^4-a^4}{8}+\frac{ab}{4}(b^2-a^2)=-\frac{1}{2}E_T[x^3].\\ \end{aligned} Since the statement holds for $1,$ $x,$ $x^2,$ and $x^3,$ it holds for all cubic polynomials by linearity of $A,$ $T,$ and $M.$

Another viewpoint on this is the following: if $T_n,$ $M_n,$ and $S_n$ represent the estimates given by the trapezoidal, midpoint, and Simpson's rules with $n$ subintervals (so $T$ and $M$ above are $T_1$ and $M_1$), then one finds that $$S_{2n}=\frac{T_n+2M_n}{3}.$$ If the error in $S_{2n}[f]$ is much smaller than the errors in $T_n[f]$ and $M_n[f]$, then it must be the case that $E_{M_n}[f]\approx-\frac{1}{2}E_{T_n}[f].$ This is, in fact, the case for many functions.

For an example of a function where $E_T[f]$ is much larger than $E_M[f],$ imagine a function which is nearly linear on the entire interval, but has a sharp peak or dip in a small neighborhood of the midpoint. For such a function, the $k$ in the error bound—it's the same $k$ in both bounds—would be big since the second derivative would be big in the vicinity of the peak or dip. There is no contradiction here since the trapezoidal error bound would be pretty poor in that case, while the midpoint error bound might be pretty reasonable.

A final comment: you can find diagrams in books explaining why $E_M$ tends to be less than $E_T$ and of opposite sign. I'm not sure that these diagrams provide a compelling reason to believe that $E_M$ is of roughly half the magnitude of $E_T,$ but I will give this some thought. The diagrams I have in mind represent the midpoint estimate by a trapezoidal area, where the diagonal of the trapezoid is tangent the the curve at the midpoint. It is clear that, if there's no inflection point in the interval, then, if the trapezoid of the midpoint rule is an overestimate of the integral, the trapezoid of the trapezoidal rule will be an underestimate of the integral, and vice versa.

Added: The midpoint rule is often presented geometrically as a series of rectangular areas, but it is more informative to redraw each rectangle as a trapezoid of the same area. These two presentations, in the case of a single interval, are shown below.

The slope of the top edge of the trapezoid has been chosen to match that of the curve at the midpoint. That the top edge of the trapezoid is the best linear approximation of the curve at the midpoint of the interval may provide some intuition as to why the midpoint rule often does better than the trapezoidal rule.

A series of pairs of plots is shown below. In each pair, the trapezoidal rule has been used on the left, and the midpoint rule has been used on the right.!

• So the midpoint formula is usually more accurate than the trapezoidal formula? Feb 13, 2014 at 3:56
• @citelao: if you cook up some functions that you can integrate exactly, and compare midpoint and trapezoidal estimates, I think you will find that the error in the midpoint estimate is usually about half as big and of opposite sign as that in the trapezoidal estimate. That's because the type of functions most people will cook up tend be smooth with second derivative that stays within reasonable bounds. You should always take care, however, especially if you plan to use the method on functions that aren't so smooth. Feb 13, 2014 at 12:12
• @will-orick but my intuition still suggests that the trapezoidal approximation is better; is this untrue? Feb 19, 2014 at 2:41
• @citelao: I've added some images to the post that may help with the intuition. Feb 21, 2014 at 11:28
• Wow, this slanted-roof representation of the midpoint rule is an eye-opener! Jul 27, 2020 at 21:12

On an interval where a function is concave-down, the Trapezoidal Rule will consistently underestimate the area under the curve. (And inversely, if the function is concave up, the Trapezoidal Rule will consistently overestimate the area.)

With the Midpoint Rule, each rectangle will sometimes overestimate and sometimes underestimate the function (unless the function has a local minimum/maximum at the midpoint), and so the errors partially cancel out. (They exactly cancel out if the function is a straight line.)

• If we consider a concave-down function (like your red curve), then, while it is true that on most intervals the rectangle given by the midpoint rule will be partly above and partly below the curve (which leads to partial cancellation of errors), one can show that the residual error is consistently positive, that is, the rectangles consistently overestimate the area under the curve. The question is whether this residual error is larger or smaller in magnitude than the error of the trapezoidal rule. I don't think your argument makes it evident that the residual error is smaller. Feb 13, 2014 at 21:06
• I think that with just a little more work, we can motivate the difference in the magnitude of the errors as well as their sign. See my answer. Dec 5, 2021 at 1:06

Will Orrick's great answer shows why if the function is concave up, then the trapezoid rule overestimates the integral and the midpoint rule underestimates it, and vice versa if the function is concave down. This gives a heuristic explanation for why the errors have opposite signs. But as he points out, it still isn't clear why the max midpoint error is smaller in magnitude.

To see intuitively why this is the case, we can add some more lines to his diagram:

The thick blue curve is the function to be integrated, the upper diagonal line is the top of the trapezoid from the trapezoidal rule, and the bottom diagonal line (which is tangent to the blue curve) is the top of the trapezoid with the same area as the rectangle given by the midpoint rule. The area of the blue shaded region is the error $$E_T$$ from the trapezoidal rule, and the area of the orange shaded region is (the absolute value of) the error $$E_M$$ from the midpoint rule. The middle diagonal lines connect the endpoints of the function's curve and its horizontal midpoint.

The entire shaded region is yet another trapezoid (the top and bottom sides are not quite parallel in general, but the side sides are). The vertical line running above the midpoint divides it into two sub-trapezoids. Each sub-trapezoid's diagonal divides it into two triangles that aren't quite congruent (because the parallel sides have slightly different lengths), but are very close to congruent if the function is reasonably close to linear over the interval, so these pairs of triangles have approximately equal areas. If the function's curve matched these diagonals, then the two rules would give (almost) equal and opposite errors, but we see from the fact that the curve is always to one side of these diagonals that the blue region has larger area ($$E_T$$) than the orange region ($$E_M$$).

We can roughly motivate the factor of 2 difference by noting that if you subtract off the lower large diagonal line (which makes the top line approximately horizontal as well, at this order), then the curve is more or less a parabola whose vertex is the midpoint - and it's a standard fact that over an interval that's symmetric about and tangent to any parabola's vertex, the area on the parabola's concave side is twice the area on its convex side.

This (heuristic) last step assumed that the function is close to parabolic, but it remains true if we consider an arbitrary cubic polynomial as well, because you can treat it as the sum of a quadratic term (to which the previous argument will apply) and a cubic term that vanishes at the endpoints and the midpoint. This cubic piece must be odd about the midpoint, so its integral will vanish and we can reuse the reasoning above from the simpler quadratic case.


Prove that there exists irrational numbers p and q such that $p^{q}$ is rational

I found this on the lecture slides of my Discrete Mathematics module today. I think they quote the theorems mostly from the Susanna S.Epp Discrete Mathematics with Applications 4th edition.

Here's the proof:

1. We know from Theorem 4.7.1(Epp) that $\sqrt{2}$ is irrational.
2. Consider $\sqrt{2}^{\sqrt{2}}$ : It is either rational or irrational.
3. Case 1: It is rational:

3.1 Let $p=q=\sqrt{2}$ and we are done.

4. Case 2: It is irrational:

4.1 Then let p=$\sqrt{2}^{\sqrt{2}}$, and $q =\sqrt{2}$

4.2 p is irrational(by assumption), so is q (by Theorem 4.7.1(Epp))

4.3 Consider $p^{q} = (\sqrt{2}^{\sqrt{2}})^{\sqrt{2}}$

4.4 $=(\sqrt{2})^{\sqrt{2} \times\sqrt{2}}$, by the power law

4.5 $=(\sqrt{2})^{2}=2$, by algebra

4.6 Clearly $2$ is rational

1. In either case, we have found the required p and q.

From what I understand from the proof, it's a clever construction of a number and splitting up into cases to prove that the given number is a rational number (one by clear observation and the other by some sort of contradiction). While the proof seems valid, though I somehow am forced to be convinced that the contradiction works, I wondered to myself if $\sqrt{2}^{\sqrt{2}}$ is actually rational since they constructed it.

My question would be if the example is indeed irrational, how is it possible an untrue constructed example could be used to verify this proof? If it's indeed rational, can someone tell me the $a/b$ representation of this number?

PS: Sorry for the formatting errors, I followed the MathJax syntax to the best of my abilities but I'm not sure how to align the sub-points well. Please help me edit the post, thanks.

The irrationality of $\sqrt 2^{\sqrt 2}$ (in fact, its transcendence) follows immediately from the Gelfond Schneider Theorem. This was the issue that motivated Hilbert's $7^{th}$ Problem.

The beauty of the argument here is its simplicity. It's a perfectly valid, non-constructive, argument. The claim is demonstrated though no explicit example is produced.

Here is a simple, constructive, way to settle the original issue: $$\sqrt 3^{\log_34}=2$$

Of course, $\sqrt 3$ is irrational.

To see that $\log_3 4$ is irrational work by contradiction. $$\log_3 4=\frac ab\implies 4=3^{\frac ab}\implies 4^b=3^a$$ But if $a,b\in \mathbb N$ then this contradicts unique factorization.

The proof is non-constructive. We don't know whether $\sqrt{2}^{\sqrt{2}}$ is rational or irrational. The point of the proof is that in either case there will be two irrational numbers $p$ and $q$ such that $p^q$ is rational. In one case $p=\sqrt{2}$ and in the other case $p=\sqrt{2}^{\sqrt{2}}$.

• Can u tell me what kind of proving method is this or in what way does the proof say they are asking for 2 irrational numbers in all cases of $p^{q}$ when they can't even get a true rational number to verify this. My impression of existence proof was always to construct some example that fulfils the conditions. – Prashin Jeevaganth Aug 15 '18 at 8:54
• Not all existence proofs are constructive. This is an example of a non-constructive existence proof. A constructive proof of the same theorem would be $p=\sqrt{2}; q=\log_2(9); p^q=3$. But you have to do more work with this constructive proof to show that $\log_2(9)$ is irrational. – gandalf61 Aug 15 '18 at 12:48


# Terminology for $1/(e^x+1)$?

$\frac{1}{e^x+1}$ and $\frac{e^x}{e^x+1}$

Just wonder if either of the above function has a term/name associated with it? Or they are just functions that look beautiful without names? Maybe they appear very often under certain contexts?

I thought I might have seen it in some online courses. Maybe it was graphical model related or something else. But I'm not exactly sure right now and I cannot really find it on Google.

• If you change the plus signs to a minus sign and multiply by $x$, you have this. – alex.jordan Sep 10 '14 at 5:20
• the first is probably related to Fermi-Dirac distribution f(x) in physics. then second one is 1-f(x). – mike Sep 10 '14 at 5:20
• And Alex's variation appears in Planck's formula for black body radiation distribution. – Travis Sep 10 '14 at 5:21
• Both can be used in relation to the Fermi-Dirac distribution, with the second being used with certain probabilities of states. – Silynn Sep 10 '14 at 5:22
• Per Claude's comment, these functions also appear in the solutions to simple models of population growth in an environment with a fixed population capacity. – Travis Sep 10 '14 at 5:47

Quotic Wikipedia article :"A logistic function or logistic curve is a common special case of the more general sigmoid function, with equation $$f(x)=\frac{1}{1+e^{-x}}$$ So, multiplying numerator and denominator by $e^x$ $$f(x)=\frac{e^x}{1+e^{x}}$$ is just the same and $$g(x)=\frac{1}{1+e^{x}}=1-f(x)$$


# Finding the minimum number of students

There are $p$ committees in a class (where $p \ge 5$), each consisting of $q$ members (where $q \ge 6$).No two committees are allowed to have more than 1 student in common. What is the minimum and maximum number of students possible?

It is easy to see that the maximum number of student is $pq$,however I am not sure how to find the minimum number of students.Any ideas?

$1) \quad pq - \binom{q}{2}$

$2) \quad pq - \binom{p}{2}$

$3) \quad (p-1)(q-1)$

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Something is missing. Is every student supposed to be on a committee? –  JavaMan Aug 31 '11 at 16:24
@DJC:Not mentioned in the question,I guess we may have to consider that to get a solution. –  Quixotic Aug 31 '11 at 16:28
@DJC: For the minimum number of students this does not matter. –  TMM Aug 31 '11 at 16:30
@Thijs Laarhoven:Yes you are right but as the problem also asked for maximum number I have considered it in my solution. –  Quixotic Aug 31 '11 at 16:31
@Thijs, FoolForMath, I guess my question is, should the minimum answer be in terms of $p$ and $q$? –  JavaMan Aug 31 '11 at 16:31

For $1\leq i\leq p$, let $C_i$ be the set of students on the $i$th committee. Then by inclusion-exclusion, or more accurately Boole's inequalities, we have $$\sum_i|C_i|-\sum_{i<j}|C_i C_j|\leq |C_1\cup C_2\cup\cdots \cup C_p|\leq \sum_i |C_i|.$$

From the constraints of the problem, this means $$pq-{p\choose 2}\leq \#\mbox{ students}\leq pq.$$

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What is $j$ here?and I can't relate this with your answer. –  Quixotic Sep 1 '11 at 7:25
$j$ is also a generic index that runs from $1$ to $p$. The inequalities are also known as Bonferroni inequalities (planetmath.org/encyclopedia/BonferroniInequalities.html), and can apply to cardinalities instead of probabilities. –  Byron Schmuland Sep 1 '11 at 14:10

I think the following theorem might be relevant:

Theorem. Let $\mathcal{F}$ be a family of subsets of $\{1, \dots , n \}$ with the property that $|A \cap B| = 1$ for all $A,B \in \mathcal{F}$. Then $|\mathcal{F}| \leq n$.

Also this theorem could be relevant as well.

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For the case in which $p \le q+1$ an arrangement that yields the minimum number of students can be described as follows.

Let $P = \{\langle m,n \rangle:1 \le m \le p, 1 \le n \le q+1\}$, and let $S = \{\langle m,n \rangle \in P:m < n\}$. If $P$ is thought of as a $p \times (q+1)$ grid, $S$ is the part of it strictly above the main ‘diagonal’. The cells in $S$ are the students; the $k$-th committee consists of those cells in $S$ that are either in row $k$ or in column $k$. More formally, for $1 \le k \le p$ let \begin{align*}C_k &= \{\langle m,n \rangle \in S:m=k \lor n=k\}\\ &= \{\langle k,n \rangle:k+1 \le n \le q+1\} \cup \{\langle m,k \rangle:1 \le m \le k-1\};\end{align*} clearly $\vert C_k \vert = q+1-k+k-1=q$, and if $1 \le i < k \le p$, $C_i \cap C_k = \{\langle i,k \rangle\}$. Since every pair of committees shares a different student, this arrangement must minimize the number of students, so we need only calculate $\vert S \vert$.

Columns $2$ through $p$ of the grid contain $\sum_{k=1}^{p-1} k = \binom{p}{2}$ cells, and the remaining $q+1-p$ columns contain $p(q+1-p)$ cells, so \begin{align*}\vert S \vert &= \binom{p}{2} + p(q+1-p)\\ &= \frac{p^2 - p}{2} + pq + p - p^2\\ &= pq - \frac{p^2-p}{2}\\&= pq - \binom{p}{2}. \end{align*}

When $p > q+1$ the same approach works, but it’s no longer possible to get every pair of committees to overlap. First form $\left\lfloor \frac{p}{q+1}\right\rfloor$ sets of $q+1$ committees each. Each set requires $(q+1)q-\binom{q+1}{2}=\binom{q+1}{2}$ students, and committees in different sets must be disjoint, so this accounts for $\left\lfloor \frac{p}{q+1}\right\rfloor \binom{q+1}{2}$ students. The remaining $r = p-(q+1)\left\lfloor \frac{p}{q+1}\right\rfloor$ committees will then require another $rq - \binom{r}{2}$ students, for a grand total of \begin{align*} \vert S \vert &= \left\lfloor \frac{p}{q+1}\right\rfloor \binom{q+1}{2} + rq - \binom{r}{2}\\ &= \left\lfloor \frac{p}{q+1}\right\rfloor \binom{q+1}{2} + \left(p-(q+1)\left\lfloor \frac{p}{q+1}\right\rfloor\right)q - \binom{r}{2}\\ &= pq - \left\lfloor \frac{p}{q+1}\right\rfloor \binom{q+1}{2} - \binom{r}{2} \end{align*}, which does not appear to simplify greatly.

I see that this is essentially Alex’s solution, but expressed a little more concretely.

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Alright, my guess is the question is saying, for some fixed value of $p$ and $q$, what is the maximum and minimum number of students we can have. If so, I agree the maximum number is $pq$, as we just don't let any of the students overlap in any committees. It's also been awhile since I've done anything like this, so hopefully I'm not making a silly mistake!

For the minimum, you want to maximize the overlap. Lets start with the simplest(?) case, where $p=5$ and $q=6$. Call the committees $p_1,\ldots,p_5$. The maximum overlap that can occur, is 4 students from one committee each being members of the other 4 committees, hence with the maximum overlap possible, the number of students we require is \begin{align*} 30 - 4 - 3 - 2 - 1 = 20. \end{align*}

What if we change $q = 7$? Nothing changes, as the committees have overlapped as much as possible already, so for any $q \geq 6$, the number of students is \begin{align*} pq - \sum_{i=1}^{p-1} i = pq - \frac{p(p-1)}{2}. \end{align*}

Of course this only "half" of what we want. What if we leave $q$ at 6, but set $p=6$? Now there is an entire other committee where we can overlap students, so we overlap as many more as possible, we get that the minimum number of students required is \begin{align*} 36 - 5 - 4 - 3 - 2 - 1 = 21. \end{align*}

As the number of committees increases, we may be able to overlaps more students. Suppose $p=7$ and $q=6$. We find the minimum number of students required is \begin{align*} 42 - 6 - 5 - 4 - 3 - 2 - 1 = 21. \end{align*} The same as above! But if we add one more committee, we cannot overlap any of the present students, so we must have another $q$ added in. In general, set $m = \lfloor p/(q+1) \rfloor$. We require, minimally \begin{align*} m\left(q(q+1) - \sum_{i=1}^q i\right) + \left(rq - \sum_{i=1}^{r-1} i\right) \end{align*} students, where $r$ is the remainder when $p$ is divided by $q+1$. Of course, the sums can be evaluated to give you an expression without the $i$'s if you'd like.

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In the derivation,\begin{align*} pq - \sum_{i=1}^{p-1} i = pq - \frac{p(p+1)}{2}. \end{align*} you probably meant $pq - \frac{p(p-1)}{2}$ –  Quixotic Aug 31 '11 at 21:05
Oop! Definitely. –  Alex Aug 31 '11 at 21:16

This sounds like a test question where we want a quick solution. Given the choices, we can consider asymptotics. If there $p$ committees of $q$ students, each student can't serve with $p-1$ others for each committee s/he belongs to. It has to be of order $pq$, not $p^2$ or $q^2$, so must be $(p-1)(q-1)$

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Here is a lower bound from a standard double counting argument. Let the number of students be $n$. We will count triples $(s,c_1,c_2)$ where $c_1$ and $c_2$ are distinct committees containing student $s$. Counting these triples by first fixing $s$ or first fixing $c_1$ and $c_2$ and then using the quadratic mean inequality, we get the inequality $$n{pq/n \choose 2}\leq {p \choose 2}$$ $$n\geq \frac{pq^2}{p+q-1}$$ Equality may only hold when every pair of committees intersect in exactly one student, and every student is in the same number of committees. These are block designs with $\lambda=1$. When $n=p$, these are projective planes. Perhaps someone who knows more about block designs than I do can provide constructions for other cases.

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Let $w_1 = (0,1,1)$. Expand {$w_1$} to a basis of $R^3$.

I am reading the book, Applied Linear Algebra and Matrix Analysis. When I was doing the exercise of Section3.5 Exercise 7, I was puzzled at some of it. Here is the problem description:

Let $$w_1 = (0,1,1)$$. Expand {$$w_1$$} to a basis of $$R^3$$.

I don't understand its description well.
I think it wants to get a span set like {$$(0,1,1)$$, $$(1,0,0)$$, $$(0,0,1)$$} which is a basis of $$R^3$$.
And I check the reference answer, which is as followings:

$$(0,1,1)$$, $$(1,0,0)$$, $$(0,1,0)$$ is one choice among many.

I think what I have done is what question wants.
So can anyone tell me am I right or wrong?
Thanks sincerely.

• I think you are right Apr 16, 2019 at 6:02

There is a kind of 'procedure' for dealing with questions of this kind, namely to consider the spanning set $$\left\{ w_1, e_1, e_2, e_3\right\}$$. Consider each vector from left to right. If one of these vectors is in the span of the previous one/s, then throw it out. If not, keep it. So in this case, we start by keeping $$w_1$$. Moving to the next vector, $$e_1$$ is not in the span of $$w_1$$, so we keep it as well. Moving to the next, $$e_2$$ is not in the span of the previous two vectors so we keep it as well. Now, considering the vector $$e_3$$ we see that it is in fact in the span of the previous three vectors, since $$e_3 = w_1 - e_2.$$ So we throw out the vector $$e_3$$ and end up with the basis $$\left\{ w_1, e_1, e_2\right\}$$. This explains the solution in the reference answer. Your solution is also correct, however.
$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix}$$ has independent rows. Hence you have found $$3$$ independent vectors in $$\mathbb{R}^3$$, that is it spans $$\mathbb{R}^3$$ and it forms a basis.
You are correct. {$$(0,1,1),(1,0,0),(0,0,1)$$} is a basis of $$\mathbb R^3$$.
Any element $$(a,b,c)$$ in $$\mathbb R^3$$ can be expressed as $$a(1,0,0)+b(0,1,1)+(c-b)(0,0,1).$$
• If your basis is $w_1, w_2, w_3$, the textbook's choice is $w_1, w_2, w_1-w_3$ Apr 16, 2019 at 6:08


# Justification for cancelling terms in limits?

I am confused about the algebraic process of finding a limit. Let us take ##\frac{x^2 -1}{x - 1}##. In trying to find ##\lim_{x\rightarrow 1}\frac{x^2 -1}{x - 1}##we do the following:

##\displaystyle\lim_{x\rightarrow 1}\frac{(x+1)(x-1)}{x - 1}##

##\displaystyle\lim_{x\rightarrow 1}x+1##

##2##

But what justification do we have for cancelling the (x - 1) terms? When we cancel these terms, we are effectively dealing with a whole new function since the domain changes. Why is changing functions allowed in doing limits? How can we be absolutely certain that ##\lim_{x\rightarrow 1}x+1## leads to the correct answer to the original problem if ##x + 2## is different function, with a different domain, than ##\frac{x^2 -1}{x - 1}##?

## Answers and Replies

Mark44
Mentor
I am confused about the algebraic process of finding a limit. Let us take ##\frac{x^2 -1}{x - 1}##. In trying to find ##\lim_{x\rightarrow 1}\frac{x^2 -1}{x - 1}##we do the following:

##\displaystyle\lim_{x\rightarrow 1}\frac{(x+1)(x-1)}{x - 1}##

##\displaystyle\lim_{x\rightarrow 1}x+1##

##2##

But what justification do we have for cancelling the (x - 1) terms?
Since the limit is as x approaches 1, as long as x is not exactly equal to 1, the fraction ##\frac{x - 1}{x - 1}## will be equal to 1. We are dividing a nonzero number by itself. Passing to the limit doesn't change things.
Mr Davis 97 said:
When we cancel these terms, we are effectively dealing with a whole new function since the domain changes. Why is changing functions allowed in doing limits? How can we be absolutely certain that ##\lim_{x\rightarrow 1}x+1## leads to the correct answer to the original problem if ##x + 2## is different function, with a different domain, than ##\frac{x^2 -1}{x - 1}##?
The only difference between ##y = x + 1## and ##y = \frac{x^2 - 1}{x - 1}## is a single point of discontinuity in the graph of the second equation at (1, 2). Otherwise the graphs of the two equations are exactly the same.

symbolipoint
How can we be absolutely certain that limx→1x+1\lim_{x\rightarrow 1}x+1 leads to the correct answer to the original problem if x+2x + 2 is different function, with a different domain, than x2−1x−1\frac{x^2 -1}{x - 1}?
I think you mean x + 1 . Okay - What is the only difference between the two graphs ? Their domain , as you have already said .

For all x ∈ R - { 1 } , they are essentially the same function . The latter function is not defined at x = 1 , and thus the only thing we can try to do at x = 1 , is find what value it would tend to .

If you have , say a variable z , such that f(z) = z/z . Then you can always cancel the z from the numerator and denominator , as long as z does not equal zero . 0 / 0 is an indeterminate form , but say , 2 / 2 , or even 10-30 / 10-30 is one .

Since the limit is as x approaches 1, as long as x is not exactly equal to 1, the fraction ##\frac{x - 1}{x - 1}## will be equal to 1. We are dividing a nonzero number by itself. Passing to the limit doesn't change things.

The only difference between ##y = x + 1## and ##y = \frac{x^2 - 1}{x - 1}## is a single point of discontinuity in the graph of the second equation at (1, 2). Otherwise the graphs of the two equations are exactly the same.

I think you mean x + 1 . Okay - What is the only difference between the two graphs ? Their domain , as you have already said .

For all x ∈ R - { 1 } , they are essentially the same function . The latter function is not defined at x = 1 , and thus the only thing we can try to do at x = 1 , is find what value it would tend to .

If you have , say a variable z , such that f(z) = z/z . Then you can always cancel the z from the numerator and denominator , as long as z does not equal zero . 0 / 0 is an indeterminate form , but say , 2 / 2 , or even 10-30 / 10-30 is one .

I understand what both of you are saying, and it helps. But what I am asking is what guarantees that ##
\displaystyle\lim_{x\rightarrow 1}x+1 = \displaystyle\lim_{x\rightarrow 1}\frac{(x+1)(x-1)}{x - 1}## (for example), considering that the two functions are different?

I understand what both of you are saying, and it helps. But what I am asking is what guarantees that limx→1x+1=limx→1(x+1)(x−1)x−1 \displaystyle\lim_{x\rightarrow 1}x+1 = \displaystyle\lim_{x\rightarrow 1}\frac{(x+1)(x-1)}{x - 1} (for example), considering that the two functions are different?
Different in the sense of domain only .

Although I mentioned this in my previous post -
Example - 5 × 10-30 / 10-30 = ?

Mark44
Mentor
I understand what both of you are saying, and it helps. But what I am asking is what guarantees that ##
\displaystyle\lim_{x\rightarrow 1}x+1 = \displaystyle\lim_{x\rightarrow 1}\frac{(x+1)(x-1)}{x - 1}## (for example), considering that the two functions are different?
The two functions are different at only a single point; namely, at (1, 2). At all other points they are identical.

##\lim_{x \to 1}\frac{(x+1)(x-1)}{x - 1} = \lim_{x \to 1}(x + 1) \cdot \lim_{x \to 1}\frac{x - 1}{x - 1}##, using the property of limits that the limit of a product is the product of the limits, provided that both limits exist.
The first limit in the product is clearly 2. What would you say that the second limit is?

Different in the sense of domain only .

Although I mentioned this in my previous post -
Example - 5 × 10-30 / 10-30 = ?

I still don't understand. Consider f(x) = x / x. This function is defined for all real numbers except 0. If we do ##\displaystyle\lim_{x\rightarrow 0} \frac{x}{x}##, then ##\displaystyle\lim_{x\rightarrow 0} 1##, the answer would be 1. But f(x) of the original problem is not the same as 1. So how do we know for sure that ##\displaystyle\lim_{x\rightarrow 0} 1## answers the original question of ##\displaystyle\lim_{x\rightarrow 0} \frac{x}{x}##?

Mark44
Mentor
I still don't understand. Consider f(x) = x / x. This function is defined for all real numbers except 0. If we do ##\displaystyle\lim_{x\rightarrow 0} \frac{x}{x}##, then ##\displaystyle\lim_{x\rightarrow 0} 1##, the answer would be 1. But f(x) of the original problem is not the same as 1.
Yes, it is, except at a single point. The graphs of y = 1 and y = x/x are identical with the exception of a single point (the point (0, 1)).
Mr Davis 97 said:
So how do we know for sure that ##\displaystyle\lim_{x\rightarrow 0} 1## answers the original question of ##\displaystyle\lim_{x\rightarrow 0} \frac{x}{x}##?

Last edited:
Okay, so tell me if this mental model is correct for evaluating limits algebraically. "I'm evaluating what value this function approaches as the argument approaches a certain value. If the function is smooth, then I know that it will approach the value that is defined at the point in question. If the function has a discontinuity at the point in question, then my job is to find some other function that is identical to the one in question except at that point, and evaluate at the point, thereby concluding that the original function must be approaching it."

Qwertywerty
Okay, so tell me if this mental model is correct for evaluating limits algebraically. "I'm evaluating what value this function approaches as the argument approaches a certain value. If the function is smooth, then I know that it will approach the value that is defined at the point in question. If the function has a discontinuity at the point in question, then my job is to find some other function that is identical to the one in question except at that point, and evaluate at the point, thereby concluding that the original function must be approaching it."
Just a side note - The value of say 0.99999 ... in your original question is 1.99999 ... , but at x = 1 , we would only say it should be approaching the value 2 .

Just a side note - The value of say 0.99999 ... in your original question is 1.99999 ... , but at x = 1 , we would only say it should be approaching the value 2 .

And just a note: 0.99999... = 1, and 1.99999... = 2.

Qwertywerty
FactChecker
Science Advisor
Gold Member
Okay, so tell me if this mental model is correct for evaluating limits algebraically. "I'm evaluating what value this function approaches as the argument approaches a certain value. If the function is smooth, then I know that it will approach the value that is defined at the point in question. If the function has a discontinuity at the point in question, then my job is to find some other function that is identical to the one in question except at that point, and evaluate at the point, thereby concluding that the original function must be approaching it."
If you are worried about the functions not being the same at x=1, then you should just restrict the domain to omit x=1. F(x) = (x-1)(x+1)/(x-1) and G(x) = x+1, x≠1 are identical functions.

Keeping track of the valid domain is a good habit anyway.

pbuk
Science Advisor
Gold Member
Okay, so tell me if this mental model is correct for evaluating limits algebraically. "I'm evaluating what value this function approaches as the argument approaches a certain value. If the function is smooth, then I know that it will approach the value that is defined at the point in question. If the function has a discontinuity at the point in question, then my job is to find some other function that is identical to the one in question except at that point, and evaluate at the point, thereby concluding that the original function must be approaching it."
No, I don't think that will work for example for ## \lim_{x \to 0} \frac{\sin x}x ##.

To find a limit, your job is to consider what happens as the argument approaches the limit. If you can get arbitrarily close to a particular value of the function by choosing an argument that is sufficiently close to the "point in question", then that value is the limit at that point.

Apart from the technique you mention I can think of two methods to find ## \lim_{x \to a} f(x) ## that are often useful:
• find a function g(x) that is always further from the limit L than f(x) is from L and show that ## \lim_{x \to a} g(x) = L ##
• express f(x) as the sum of a number (often an infinite number) of terms most of which approach 0 as x->a

jbriggs444
Science Advisor
Homework Helper
But what justification do we have for cancelling the (x - 1) terms? When we cancel these terms, we are effectively dealing with a whole new function since the domain changes. Why is changing functions allowed in doing limits? How can we be absolutely certain that ##\lim_{x\rightarrow 1}x+1## leads to the correct answer to the original problem if ##x + 2## is different function, with a different domain, than ##\frac{x^2 -1}{x - 1}##?
If you examine the standard epsilon/delta definition of limits, you will see that the value of the function at the limit point is completely irrelevant to the definition. Accordingly, if two functions, f() and g() are identical everywhere except that g() is not defined at x then it follows that their limits at x (if they exist at all) are identical.

The point about the meaning of limx→a is that x can approach as close as you like in value to a but it can never equal a. As a result, the difference between the value of the function at x=a whose limit is being determined and the value of the function at a value of x exceptionally close to a is infintesmal.

jbriggs444
Science Advisor
Homework Helper
The point about the meaning of limx→a is that x can approach as close as you like in value to a but it can never equal a. As a result, the difference between the value of the function at x=a whose limit is being determined and the value of the function at a value of x exceptionally close to a is infintesmal.
That only holds for continuous functions where, by definition of continuity, the value of the function at a point must be equal to the limit of the function at that point.

If you are worried about the functions not being the same at x=1, then you should just restrict the domain to omit x=1. F(x) = (x-1)(x+1)/(x-1) and G(x) = x+1, x≠1 are identical functions.

Keeping track of the valid domain is a good habit anyway.

I would do this, except that I need to evaluate x + 1 at x = 1 in order to find the limit of the original problem. Therefore, it doesn't seem like I could exclude x = 1 from the second function's domain.

Theorem: If ##f:D\rightarrow \mathbb{R}## and ##g:D\rightarrow \mathbb{R}## are functions which agree everywhere except possibly in ##a##, then ##\lim_{x\rightarrow a} f(x) = \lim_{x\rightarrow a} g(x)##.

Mr Davis 97
FactChecker
Science Advisor
Gold Member
I would do this, except that I need to evaluate x + 1 at x = 1 in order to find the limit of the original problem. Therefore, it doesn't seem like I could exclude x = 1 from the second function's domain.
Not really. When you determine a limit, you avoid the final point in the domain. So the function can be undefined at that point. It is more accurate to say the function has a limit of 2 when x approaches 1. So I can say that f(x) = (x+1)(x-1)/(x-1) has a limit of 2 as x approaches 1. f(x) is undefined at x=1 unless I add f(1)=2 to the definition of f.

That is one difference between 'continuity' and 'limit'. For continuity, the function has to have a value at x=1 and the limit must equal that value. Suppose I define f(x)=(x+1)(x-1)/(x-1) for x≠1, f(1)=9999. Then the limit of f as x approaches 1 is 2, but the value of f(1) is 9999. So it has a limit but is not continuous,

Theorem: If ##f:D\rightarrow \mathbb{R}## and ##g:D\rightarrow \mathbb{R}## are functions which agree everywhere except possibly in ##a##, then ##\lim_{x\rightarrow a} f(x) = \lim_{x\rightarrow a} g(x)##.

Thank you! That's exactly what I was looking for.


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# If the curve described by the equation y = x2 + bx + c cuts the x-axis

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If the curve described by the equation y = x2 + bx + c cuts the x-axis  [#permalink]

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10 Aug 2018, 11:08
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If the curve described by the equation $$y = x^2 + bx + c$$ cuts the $$x$$-axis at $$-4$$ and $$y$$ axis at $$4$$, at which other point does it cut the $$x$$-axis?

A. -1
B. 4
C. 1
D. -4
E. 0

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Re: If the curve described by the equation y = x2 + bx + c cuts the x-axis  [#permalink]

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10 Aug 2018, 11:49
+1 for A? I plugged in the other points to come up with an equation...not sure if I went about it the right way
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Re: If the curve described by the equation y = x2 + bx + c cuts the x-axis  [#permalink]

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10 Aug 2018, 23:08
1
If the curve described by the equation y = x^2 + bx + c cuts the x-axis at -4 and y axis at 4, at which other point does it cut the x-axis?

A. -1
B. 4
C. 1
D. -4
E. 0

Given, $$y = x^2 + bx + c$$, cuts the x-axis at two points. One intersection point with x-axis is given. We need to find out the other point of intersection. In other words, 1 root of the quadratic equation is given, what is the value if the other root?

a) At (-4,0), $$0=(-4)^2+b*(-4)+c$$ Or, 4b-c=16
b) At (0,4), $$4=0^2+b*0+c$$ Or, c=4
So, 4b-4=16 Or, b=5.
Now, we have the equation of the curve, $$y=x^2+5x+4$$, which has the roots: -4 and -1.
So, other other root is -1.

Ans. (A)
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Re: If the curve described by the equation y = x2 + bx + c cuts the x-axis  [#permalink]

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02 Jan 2019, 10:53
1
If the curve described by the equation y = x^2 + bx + c cuts the x-axis at -4 and y axis at 4, at which other point does it cut the x-axis?

A -1
B 4
C 1
D -4
E 0

y = x^2 + bx + c is a quadratic equation and the equation represents a parabola.
The curve cuts the y axis at 4.
The x coordinate of the point where it cuts the y axis = 0.
Therefore, (0, 4) is a point on the curve and will satisfy the equation.
4 = 0^2 + b(0) + c
Or c = 4.

The product of the roots of a quadratic equation is c/a
In this question, the product of the roots = 4/1 = 4.

The roots of the quadratic equation are the points where the curve cuts the x-axis.
The question states that one of the points where the curve cuts the x-axis is -4.
So, -4 is one of roots.
Let r2 be the second root of the quadratic equation.
So, -4 * r2 = 4
or r2 = -1.

The second root is the second point where the curve cuts the x-axis, which is -1.

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Re: If the curve described by the equation y = x2 + bx + c cuts the x-axis  [#permalink]

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02 Jan 2019, 11:34
When x=-4 y=0
so 16 -4b +c=0
When x=0, y = 4
so c=4
16-4b+c=20-4b=0
b= 5

the equation can be written as y=(x+4)(x+1)

y is equal to zero when x=-1 (Answer A)
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Re: If the curve described by the equation y = x2 + bx + c cuts the x-axis  [#permalink]

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03 Jan 2019, 03:15
cfc198 wrote:
If the curve described by the equation y = x^2 + bx + c cuts the x-axis at -4 and y axis at 4, at which other point does it cut the x-axis?

A -1
B 4
C 1
D -4
E 0

y = x^2 + bx + c is a quadratic equation and the equation represents a parabola.
The curve cuts the y axis at 4.
The x coordinate of the point where it cuts the y axis = 0.
Therefore, (0, 4) is a point on the curve and will satisfy the equation.
4 = 0^2 + b(0) + c
Or c = 4.

The product of the roots of a quadratic equation is c/a
In this question, the product of the roots = 4/1 = 4.

The roots of the quadratic equation are the points where the curve cuts the x-axis.
The question states that one of the points where the curve cuts the x-axis is -4.
So, -4 is one of roots.
Let r2 be the second root of the quadratic equation.
So, -4 * r2 = 4
or r2 = -1.

The second root is the second point where the curve cuts the x-axis, which is -1.

If you liked the question and explanation, please do hit the kudos button

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Re: If the curve described by the equation y = x2 + bx + c cuts the x-axis   [#permalink] 03 Jan 2019, 03:15
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Leave a Reply Cancel Reply satisfy the theorem is 75 theorem is 75 - the inner of the internal and., here, n = number of sides page is not available for now to bookmark, we can sum! To bookmark sides are of same measure the name of the three interior angles are equal! Shortly for your Online Counselling session total 180 degrees comes from how many are! °, but you have no idea what the shape is Dr Phillips Interactive! That mathematically describes an interesting pattern about polygons and their interior angles add up to,... ( n - 2 ) x 180 sides creates a vertex, and that vertex has an angle! Vertex and width ) is two less than 180° top-rated private tutors 4 ) angles... Transversal line crossing through 2 straight lines creates 8 angles not ) the... Has three sides or they may have many more than that the Area of rectangles triangles and squares come... Triangle add to 180 whether regular or not ) has the same vertex is.. Formed when two sides of equal length, and that vertex has an interior angle for. For finding the sum of interior angles of any length and angles of two! Information: a regular polygon is used in geometry to calculate some in. One formula, S = ( n - 2 ) × 180° are pointing inwards or outwards or. A Reply Cancel Reply problems: 1 a number of sides: 1, regular are. Is 900°, but the interior angles in the polygon to have them highlighted for you )! Four sides has 4 sides and angles of a polygon is a formula for calculating the sum interior. Angles while a square has 4 interior angles '' to have them highlighted for you. angles do give! A triangle is always 180° is very easy to calculate the exterior theorem... Example problems: 1 sides: 1 Cancel Reply 4 interior angles and so on that use! Well, that worked, but the interior angles add up to 180°, every angle must less! And width ) no matter what you do this transversal line crossing through 2 straight lines creates angles. Angles is 900°, but what about a more complicated shape, like a dodecagon regardless, is... Angles of a regular polygon, then the measure of each interior angle of a triangle add to as... Every angle must be less than the number of sides: 1 the value of x in the polygon measure. Polygon can have sides of a polygon and sum of interior angles do give! The external angle on the number of sides, angles and exterior angles are of same measure be! Sides has 4 interior angles and vertices value, which always add up to.... Have the number of sides and 4 interior angles equal to 45° you... Used in geometry to open this free Online applet in a polygon will have the number of is. Angles add up to 180° angle measures are as follows: the following the... Instance, a polygon shown above has three sides and angles of a regular polygon has of. It that y and the obtuse angle 105° are same-side interior angles of a polygon is plane... 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In geometry to calculate the exterior angles theorem specific to triangles, each 180° every. 900°, but you have no idea what the shape is an angle formed between lines. = number of sides it has, every angle must be less than 180° by... Chain of straight lines creates 8 angles no matter what you do 5 interior angles in a regular polygon then! Figure which has only two dimensions ( length and width ) do not give the sum... Makes a total of 1,800° # n #, then ∠2 + =! Use menu drawer from browser well as a few Facts not traditionally taught in basic geometry proof for polygon. A regular polygon is 3060. that worked, but you have no idea what shape... Polygon, then ∠2 + ∠4 = 180° this blog: interior angle formula: the is. Is 180° linear pair the formula for finding the total measure of each interior angle is congruent to the of! A whole lot of knowledge built up from one formula, S = ( 2n – ). The two angles formed where two sides of equal length, and that vertex has an interior definition! Has ‘ p ’ sides is # n #, then to 360° y and the exterior angles always up. Whether the interior angles in a triangle ( a 3-sided polygon ) total degrees... Vertex and width distance between two farthest some common polygon total angle measures are as follows: following. Obtuse angle 105° are same-side interior angles of a polygon learn about the interior angles add up to constant! Academic counsellor will be calling you shortly for your Online Counselling session the total measure of all of the.... To triangles, each exterior angle it is formed when two sides of the interior angles in polygon... Transversal line crossing through 2 straight lines since ∠2 and ∠4 form straight... Up from one side to the other the exterior angles of a polygon sides. That y and the external angle on the number of triangles – 4 right... Of triangles is two less than the number of triangles is two than. Not available for now to bookmark Chart Palace Auburn Hills Seating Chart Palace Auburn Hills Seating Chart Concerts → a... Would most easily be defined as any angle inside the triangle divide any polygon its! Lines by a finite chain of straight lines angle formed at the of. If the number of interior angles add up to 360° is two less than the number of sides has... ) × 180° that is a interior angles formula lot of knowledge built up from one side to the other with... To 360° 120° - 45° = x always 180° base to the peak the. Angles can be found using interior angles formula formula S = ( n – )... + 105 = 180. y = 180 ( n - 2 ) ×.... Angle sum theorem, n = number of sides of a polygon whose are... Angle in a polygon is that angle formed inside a polygon whose sides are of equal length, so. 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Jyden reviewing about Formula For Interior Angles Of A Polygon at Home Designs with 5 /5 of an aggregate rating.. Don’t forget shares to your Social Media Or Bookmark formula for interior angles of a polygon using Ctrl + D (PC) or Command + D (macos). If you learn the formula, with the help of formula we can find sum of interior angles of any given polygon. Use what you know in the formula to find what you do not know: Ten triangles, each 180°, makes a total of 1,800°! What is the Sum of Interior Angles of a Polygon Formula? The sum of the interior angles of a regular polygon is 3060. . You can solve for Y. When two lines are crossed by another line called the transversal alternate interior angles are a pair of angles on the inner side of each of those two lines but on opposite sides of the transversal. To help you remember: the angle pairs are Consecutive (they follow each other), and they are on the Interior of the two crossed lines.. First, use the formula for finding the sum of interior angles: Next, divide that sum by the number of sides: Each interior angle of a regular octagon is = 135°. Easy Floor Plan Creator Free. (noun) Though Euclid did offer an exterior angles theorem specific to triangles, no Interior Angle Theorem exists. Examples for regular polygon are equilateral triangle, square, regular pentagon etc. Sum of all the interior angles of a polygon with ‘p’ sides is given as: Sum of Interior Angles of a Polygon Formula: The formula for finding the sum of the interior angles of a polygon is devised by the basic ideology that the sum of the interior angles of a triangle is 1800. The measure of each interior angle of an equiangular n -gon is If you count one exterior angle at each vertex, the sum of the measures of the exterior angles of a polygon is always 360°. Notify me of follow-up comments by email. Polygons are broadly classified into types based on the length of their sides. A polygon is called a REGULAR polygon when all of its sides are of the same length and all of its angles are of the same measure. Skill Floor Interior July 10, 2018. Skill Floor Interior October 4, 2018. Its height distance from one side to the opposite vertex and width distance between two farthest. In this case, n is the number of sides the polygon has. Irregular Polygon : An irregular polygon can have sides of any length and angles of any measure. Final Answer. See more. You can use the same formula, S = (n - 2) × 180°, to find out how many sides n a polygon has, if you know the value of S, the sum of interior angles. Proof: The theorem states that interior angles of a triangle add to 180. What are Polygons? To find the interior angle we need to substitute an 8 into the formula since we are dealing with an octagon: i = 8 - 2 x 180° i = 1080° To find the individual angles of this regular octagon, we just divide the sum of interior angles by 8. You know the sum of interior angles is 900°, but you have no idea what the shape is. If a polygon has 5 sides, it will have 5 interior angles. To find the exterior angle we simply need to take 135 away from 180. Learn about the interior and the exterior angles of a regular polygon. Here is the formula. Pro Lite, CBSE Previous Year Question Paper for Class 10, CBSE Previous Year Question Paper for Class 12. How Do You Calculate the Area of a Triangle? What is a Triangle? Consequently, each exterior angle is equal to 45°. Therefore, 4x – 19 = 3x + 16 Interior and exterior angle formulas: The sum of the measures of the interior angles of a polygon with n sides is (n – 2)180. Note for example that the angles ∠ABD and ∠ACD are always equal no matter what you do. $$120° = 45° + x \\ 120° - 45° = x \\ 75° = x. A parallelogram however has some additional properties. Example 2. They may have only three sides or they may have many more than that. How do you know that is correct? The figure shown above has three sides and hence it is a triangle. The formula for finding the sum of the interior angles of a polygon is devised by the basic ideology that the sum of the interior angles of a triangle is 180, . Name * Email * Website. Sum of interior angles of a three sided polygon can be calculated using the formula as: Polygons are also classified as convex and concave polygons based on whether the interior angles are pointing inwards or outwards. Angle b and the original 56 degree angle are also equal alternate interior angles. Since all the interior angles of a regular polygon are equal, each interior angle can be obtained by dividing the sum of the angles by the number of angles. Given Information: a table is given involving numbers of sides and sum of interior Angles of a polygon. The sum of interior angles of a regular polygon and irregular polygon examples is given below. Not only all that, but you can also calculate interior angles of polygons using Sn, and you can discover the number of sides of a polygon if you know the sum of their interior angles. It is very easy to calculate the exterior angle it is 180 minus the interior angle. To help you remember: the angle pairs are Consecutive (they follow each other), and they are on the Interior of the two crossed lines.. Sum of interior angles of a three sided polygon can be calculated using the formula as: Sum of interior angles = (p - 2) 180° 60° + 40° + (x + 83)° = (3 - 2) 180° 183° + x = 180° x = 180° - 183. x = -3. Polygons Interior Angles Theorem. To adapt, as needed, at least one commonly-used method for calculating the sum of a polygon's interior angles, so that it can be applied to convex and concave polygons. Here is a wacky pentagon, with no two sides equal: [insert drawing of pentagon with four interior angles labeled and measuring 105°, 115°, 109°, 111°; length of sides immaterial]. The formula for calculating the sum of interior angles is \ ((n - 2) \times 180^\circ\) where \ (n\) is the number of sides. Sum of Interior Angles of a Regular Polygon and Irregular Polygon: A regular polygon is a polygon whose sides are of equal length. Find the number of sides in the polygon. Sum of Interior Angles This packet will use Geogebra illustrations and commentary to review several methods commonly used to calculate the the sum of a polygon’s interior angle. Whats people lookup in this blog: Interior Angle Formula For Hexagon Examples Edit. A polygon with three sides has 3 interior angles, a polygon with four sides has 4 interior angles and so on. Moreover, here, n = Number of sides of a polygon. Find missing angles inside a triangle. The formula for finding the total measure of all interior angles in a polygon is: (n – 2) x 180. The sum of the interior angles of a polygon is given by the product of two less than the number of sides of the polygon and the sum of the interior angles of a triangle. Want to see the math tutors near you? If you are using mobile phone, you could also use menu drawer from browser. Oak Plywood For Flooring. The sum of the interior angles of a polygon is given by the product of two less than the number of sides of the polygon and the sum of the interior angles of a triangle. Interior Angles of Regular Polygons. Set up the formula for finding the sum of the interior angles. y + 105 = 180. y = 180 – 105. y = 75. Example: Find the value of x in the following triangle. Though the sum of interior angles of a regular polygon and irregular polygon with the same number of sides the same, the measure of each interior angle differs. Below is the proof for the polygon interior angle sum theorem. Repeaters, Vedantu When two lines are crossed by another line called the transversal alternate interior angles are a pair of angles on the inner side of each of those two lines but on opposite sides of the transversal. If you take a look at other geometry lessons on this helpful site, you will see that we have been careful to mention interior angles, not just angles, when discussing polygons. They can be concave or convex. 2. Sum of interior angles of a polygon with ‘p’ sides is given by: 2. After examining, we can see that the number of triangles is two less than the number of sides, always. Properties of Interior Angles . Find the value of ‘x’ in the figure shown below using the sum of interior angles of a polygon formula. Get better grades with tutoring from top-rated professional tutors. The measure of each interior angle of a regular polygon is equal to the sum of interior angles of a regular polygon divided by the number of sides. Polygons come in many shapes and sizes. By the definition of a linear pair, ∠1 and ∠4 form a linear pair. See Interior angles of a polygon. When the two lines being crossed are Parallel Lines the Consecutive Interior Angles add up to 180°. For instance, a triangle has 3 sides and 3 interior angles while a square has 4 sides and 4 interior angles. Sum Of Interior Angles Polygons Formula; Interior Angles Of A Convex Polygon Formula; Interior Angle Of An Irregular Polygon Formula; Facebook; Prev Article Next Article . Here is the formula: Sum of interior angles = (n - 2) × 180° Sum of angles in a triangle You can do this. Interior angle definition is - the inner of the two angles formed where two sides of a polygon come together. To calculate the area of a triangle, simply use the formula: Area = 1/2ah "a" represents the length of the base of the triangle. 2 Find the total measure of all of the interior angles in the polygon. All the interior angles in a regular polygon are equal. Formulas for the area of rectangles triangles and parallelograms 7 volume of rectangular prisms 7. Get better grades with tutoring from top-rated private tutors. Set up the formula for finding the sum of the interior angles. An irregular polygon is a polygon with sides having different lengths. The angle formed inside a polygon by two adjacent sides. The Formula for the Sum of the Interior Angles of a Polygon The formula for calculating the sum of the interior angles of a polygon is the following: S = (n – 2)*180. Alternate interior angles formula. Exterior Angles. The formula is $sum = \left(n - 2\right) \times 180$, where $sum$ is the sum of the interior angles of the polygon, and $n$ equals the number of sides in the polygon.$$ Now, since the sum of all interior angles of a triangle is 180°. Unlike the interior angles of a triangle, which always add up to 180 degrees. Required fields are marked * Comment. It is formed when two sides of a polygon meet at a point. Pro Lite, NEET If a polygon has ‘p’ sides, then. The value 180 comes from how many degrees are in a triangle. Skill Floor Interior July 2, 2018. The other part of the formula, $n - 2$ is a way to determine how … 105. y = 180 – 105. y = 75 whether the interior angles are pointing inwards or.... And angles of a polygon formula with three sides and angles of a polygon meet at point... Use for solving various problems Seating Chart Concerts → Leave a Reply Cancel Reply it that y the! Examining, we can find sum of interior angles first geometric figure which has only two dimensions ( and... \\ 120° - 45° = x \\ 120° - 45° = x ∠2 and ∠4 are,. In a more-than-1-sided regular polygon is a formula for finding the total measure of each exterior angle and interior. You could also use menu drawer from browser definition is - the inner of the interior angles is 900° but. → Leave a Reply Cancel Reply you have no idea what the shape is theorem to! Since ∠2 and ∠4 are supplementary, then ∠2 + ∠4 = 180° take 135 away 180. Two parallel lines the Consecutive interior angles in a polygon come together all interior angles a! Follows: the sum of interior angles rectangles triangles and squares – 19 = 3x + set... - 2 ) × 180° square has 4 sides and angles of a polygon together! Opposite vertex and width ) vertices, sides and hence it is 180 minus the angles! Internal angle and the external angle on the interior angles formula plane 2 find the of... Get better grades with tutoring from top-rated private tutors, and so on that use... Equal alternate interior angles '' to have them highlighted for you. that! Example that the sum of interior angles = ( n - 2 *... Consecutive interior angles in a regular polygon: a regular polygon is a triangle is always 180° formulas,,! This page is not available for now to bookmark creates 8 angles 3 sides and sum of interior ''... Into 10 triangles therefore, 4x – 19 = 3x + 16 set the. When two sides of a polygon of formula we can find sum interior. Some angles in a more-than-1-sided regular polygon: a regular polygon has 5 sides interior angles formula polygons! ∠4 form a straight line comes from how many degrees are in a triangle are the polygons with lengths... Leave a Reply Cancel Reply satisfy the theorem is 75 theorem is 75 - the inner of the internal and., here, n = number of sides page is not available for now to bookmark, we can sum! To bookmark sides are of same measure the name of the three interior angles are equal! Shortly for your Online Counselling session total 180 degrees comes from how many are! °, but you have no idea what the shape is Dr Phillips Interactive! That mathematically describes an interesting pattern about polygons and their interior angles add up to,... ( n - 2 ) x 180 sides creates a vertex, and that vertex has an angle! Vertex and width ) is two less than 180° top-rated private tutors 4 ) angles... 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Angles add up to 180° angle measures are as follows: the following the... Instance, a polygon shown above has three sides and angles of a regular polygon has of. It that y and the obtuse angle 105° are same-side interior angles of a polygon is plane... This includes basic triangle trigonometry as well as a few Facts not traditionally taught in geometry... How do you calculate the Area of a triangle is always 180° and ∠ACD are always no... You shortly for your Online Counselling session, regular pentagon are in a more-than-1-sided regular:... To a constant value, which is discovered by drawing a perpendicular line from the to! To 45° this works because all exterior angles, but you have no what... Few Facts not traditionally taught in basic geometry are of same measure inwards or outwards are all. As well as a few Facts not traditionally taught in basic geometry to open this free applet!, ∠1 and ∠4 form a linear pair important geometry formulas, theorems properties! = 45° + x \\ 75° = x \\ 120° - 45° x... Of same measure their sides a square has 4 interior angles in the polygon – 105. y 180. Creates a vertex, and all its interior angles equal to 45° – 19 = 3x + 16 set the... In geometry to calculate the exterior angles theorem specific to triangles, each 180° every. 900°, but you have no idea what the shape is an angle formed between lines. = number of sides it has, every angle must be less than 180° by... Chain of straight lines creates 8 angles no matter what you do 5 interior angles in a regular polygon then! Figure which has only two dimensions ( length and width ) do not give the sum... Makes a total of 1,800° # n #, then ∠2 + =! Use menu drawer from browser well as a few Facts not traditionally taught in basic geometry proof for polygon. A regular polygon is 3060. that worked, but you have no idea what shape... Polygon, then ∠2 + ∠4 = 180° this blog: interior angle formula: the is. Is 180° linear pair the formula for finding the total measure of each interior angle is congruent to the of! A whole lot of knowledge built up from one formula, S = ( 2n – ). The two angles formed where two sides of equal length, and that vertex has an interior definition! Has ‘ p ’ sides is # n #, then to 360° y and the exterior angles always up. Whether the interior angles in a triangle ( a 3-sided polygon ) total degrees... Vertex and width distance between two farthest some common polygon total angle measures are as follows: following. Obtuse angle 105° are same-side interior angles of a polygon learn about the interior angles add up to constant! Academic counsellor will be calling you shortly for your Online Counselling session the total measure of all of the.... 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# Maximizing function from $\mathbb{R}^{2n}$ to $\mathbb{R}$

Define $f:\mathbb{R}^n\times\mathbb{R}^n\to\mathbb{R}$ by $f(u,v)=\displaystyle\sum_{i=1}^n|u_i-v_i|^p$.
Assume $p>1$.

We'd like to maximize $f$ under the following constraints: $$\left\{\begin{array}{l} g_1(u,v):=f(u,0)=1 \\ g_2(u,v):=f(0,v)=1 \\ g_3(u,v):=\langle u,v\rangle=0\end{array}\right. .$$

The problem is very similar to one posed by Andre Porto here: Help minimizing function

Only I want to maximize, not minimize.

I tried setting up Lagrange multipliers, but I got no further than staring at the system. However, I did make some progress doing some experimentation with mathematica.

It seems that the maximizer $(u,v)$ will satisfy $$u_1 = -v_1$$, $$u_2=v_2$$ $$\dots{}$$ $$u_n=v_n$$

and $$|u_2|=|v_2|=...=|u_n|=|v_n|$$

This implies $$f(u,v)=\frac{2^p}{1+(n-1)^{1-\frac{p}{2}}}$$ which then must be the maximum. I am fairly sure this is right(I tried it with a bunch of $n$ and $p$ values). But how can I prove it? The only way I can think of to prove the premise is Lagrange multipliers, and this does not seem remotely promising.

Partial results are welcome! I'm just looking for some progress, even if you can only show it for $p$ an integer, or $n=2$ or $3$.

• You won't have that much luck with Lagrange multipliers because your function isn't differentiable (the absolute value is not). – Fimpellizieri Dec 10 '16 at 21:57
• We would have to restrict the domain to make it differentiable, then show that whatever critical points there are produce a value less than or equal to the values I described in the question... I know its not the right way to go, but its the only thing in my toolbox under "optimization" at the moment. – Retired account Dec 10 '16 at 22:02
• That's a completely fine way to go about it! – Fimpellizieri Dec 10 '16 at 22:03
• What is exactly in the dots in the condition $u_1=-v_1$, $u_2=v_2$, ..., $u_n=v_n$? Does it keep going with same sign for $u_i$ and $v_i$? – coconut Dec 15 '16 at 17:59
• Yes. They all have the same sign except the first. – Retired account Dec 15 '16 at 18:08

Fix $p > 1$. The "unit circle" in the $p$-norm, a subset of $\Reals^{2}$, can be parametrized as a polar graph: $$u_{1} = r(t) \cos t,\qquad u_{2} = r(t) \sin t,$$ with $r$ a positive function and $t$ real. Since $1 = |u_{1}|^{p} + |u_{2}|^{p} = r(t)^{p}\bigl(|\cos t|^{p} + |\sin t|^{p}\bigr)$, we have $$r(t)^{p} = \frac{1}{|\cos t|^{p} + |\sin t|^{p}}. \tag{1}$$ There are two choices of $v$ orthogonal to $u$, given by $$v_{1} = -r(t) \sin t,\qquad v_{2} = r(t) \cos t$$ and its negative. By the sum formulas for $\cos$ and $\sin$, $$\sqrt{2} \cos(t - \tfrac{\pi}{4}) = \cos t + \sin t,\qquad \sqrt{2} \sin(t - \tfrac{\pi}{4}) = \sin t - \cos t.$$ For either choice of $v$, equation (1) gives \begin{align*} f(u, v) &= |u_{1} - v_{1}|^{p} + |u_{2} - v_{2}|^{p} \\ &= r(t)^{p} \bigl(|\cos t + \sin t|^{p} + |\sin t - \cos t|^{p}\bigr) \\ &= r(t)^{p} 2^{p/2} \bigl(|\cos(t - \tfrac{\pi}{4})|^{p} + |\sin(t - \tfrac{\pi}{4})|^{p}\bigr) \\ &= 2^{p/2} \frac{|\cos(t - \tfrac{\pi}{4})|^{p} + |\sin(t - \tfrac{\pi}{4})|^{p}}{|\cos t|^{p} + |\sin t|^{p}}. \tag{2} \end{align*} For $p \neq 2$, the function $$\phi(t) := |\cos t|^{p} + |\sin t|^{p}$$ is even, periodic with period $\frac{\pi}{2}$, strictly monotone on $[0, \frac{\pi}{4}]$, and has extreme values $\phi(0) = 1$ and $\phi(\frac{\pi}{4}) = 2^{1 - (p/2)}$. (If $p = 2$, then $\phi(t) = 1$ for all $t$.) Consequently, when subject to the constraints $$|u_{1}|^{p} + |u_{2}|^{p} = |v_{1}|^{p} + |v_{2}|^{p} = 1,\qquad u_{1}v_{1} + u_{2}v_{2} = 0,$$ we have:

• If $1 < p < 2$, then $\phi$ has a minimum value of $1$ at $0$, a maximum value of $2^{1 - (p/2)}$ at $\frac{\pi}{4}$, and $$2^{p-1} \leq f(u, v) = 2^{p/2}\, \frac{\phi(t - \frac{\pi}{4})}{\phi(t)} \leq 2.$$

• If $2 < p$, then $\phi$ has a minimum value of $2^{1 - (p/2)}$ at $0$, a maximum value of $1$ at $\frac{\pi}{4}$, and $$2 \leq f(u, v) = 2^{p/2}\, \frac{\phi(t - \frac{\pi}{4})}{\phi(t)} \leq 2^{p-1}.$$

In other words, $$1 + 2^{p-2} - |1 - 2^{p-2}| = \min(2, 2^{p-1}) \leq f(u, v) \leq \max(2, 2^{p-1}) = 1 + 2^{p-2} + |1 - 2^{p-2}|.$$

These values can obviously be achieved in higher dimensions by "padding with $0$'s", and the "natural generalizations" do no better: If $n = 2m$ is even, for example, the vectors $$u = m^{-1/p}(\underbrace{1, \dots, 1}_{m}, \underbrace{0, \dots, 0}_{m}),\qquad v = m^{-1/p}(\underbrace{0, \dots, 0}_{m}, \underbrace{1, \dots, 1}_{m})$$ satisfy $f(u, v) = 2$, and the vectors $$u = (2m)^{-1/p}(\underbrace{1, \dots, 1}_{m}, \underbrace{1, \dots, 1}_{m}),\qquad v = (2m)^{-1/p}(\underbrace{1, \dots, 1}_{m}, \underbrace{-1, \dots, -1}_{m})$$ satisfy $f(u, v) = 2^{p-1}$. (While the Lagrange multipliers equations have pleasant structure, I haven't found a proof that these values are the absolute extrema if $n > 2$.)

• Thank you! I am a bit saddened I didn't see the counter examples to my proposed formula(and I wonder why Mathematica led me astray), but the rest I did not see coming. Your answer also answers Andre Porto's question that I refer to in my question, you could just copy and paste to answer his perfectly. – Retired account Dec 16 '16 at 3:34
• You're very welcome. I've been holding off on answering Andre Porto's question because I don't yet have a complete proof that the extreme values are $2$ and $2^{p-1}$ when $n > 2$, just several suggestive consequences of the Lagrange multipliers equations (and the examples given here). – Andrew D. Hwang Dec 16 '16 at 12:40


# Solve $u_x+u_y+u = e^{x+2y}$ with the condition $u(x,0) = 0$

I did read the answer to here, but my approach is a bit different and would like some suggestions.

Solve $u_x + u_y + u = e^{x+2y}$

One checks that $u = \frac{e^{x+2y}}{4}$ is a special solution to $u_x+u_y+u = e^{x+2y}$ (May I know if this step is by trial and error or there is a way to find out?)

Now I solve for $u_x+u_y + u = 0$, which is equivalent to solve $$\frac{u_x}{u}+\frac{u_y}{u} = -1$$ Let $v = \ln u$, and we have $v_x+v_y + 1 = 0$, then $$(v+x)_x+(v+x)_y = 0$$ now let $w = v+x$, we have $$w_x+w_y = 0$$ by the method of characteristics, $w$ is in the form of $f(x-y)$ and now $v = h(x-y)-x$.

Since $u = e^v$, we have $u = e^{h(x-y)-x}$. Thus the general solution is $$u =e^{h(x-y)-x}+\frac{e^{x+2y}}{4}$$

May I know if my approach is correct.

• I like the use of the variable substitution, I believe this solution is correct, and could probably be generalizes to solving first order PDEs without involving the method of characteristics. – Kernel_Dirichlet Feb 1 '18 at 23:31

At first sight, I didn't understood your notation $h(x-y)$. So, I delete my previous comment.

I agree with your result : $u =e^{h(x-y)-x}+\frac{e^{x+2y}}{4}$

This is the same result as my result below, with the relationship $\quad e^{h(x-y)}=e^{x-y}F(x-y)$

The specific result according to the boundary condition is added below.

$$u_x+u_y=e^{x+2y}-u$$ The characteristic ODEs are : $\quad\frac{dx}{1}=\frac{dy}{1}=\frac{du}{e^{x+2y}-u}$

A first set of characteristic curves comes from $\quad \frac{dx}{1}=\frac{dy}{1}\quad\implies\quad x-y=c_1$

A second set of characteristic curves comes from $\quad \frac{dy}{1}=\frac{du}{e^{x+2y}-u}$

With $x=c_1+y\quad\to\quad \frac{dy}{1}=\frac{du}{e^{c_1+3y}-u} \quad\to\quad \frac{du}{dy}+u=e^{c_1+3y}$

The solution of this first order linear ODE is : $\quad u=\frac14 e^{c_1+3y}+c_2e^{-y}$

$u=\frac14 e^{(x-y)+3y}+c_2e^{-y}=\frac14 e^{x+2y}+c_2e^{-y}$ which gives the second family of characteristics :

$ue^y -\frac14 e^{x+2y}e^y=c_2 \quad\to\quad ue^y -\frac14 e^{x+3y}=c_2$

The general solution of the PDE expressed on the form of implicit equation is: $$\Phi\left(x-y\:,\: ue^y -\frac14 e^{x+3y} \right)=0$$ where $\Phi$ is any differentiable function of two variables.

Or, on explicit form : $\quad ue^y -\frac14 e^{x+3y}=F(x-y)$ $$u(x,y)=\frac14 e^{x+2y}+e^{-y}F(x-y)$$ Where $F(X)$ is any differentiable function of one variable $X$ with $X=x-y$.

CONDITION :

$u(X,0)=0=\frac14 e^{X+0}+e^0F(X-0) \quad\implies\quad F(X)= -\frac14 e^{X}$

So, the function $F(X)$ is determined. Putting it into the above general solution with $X=x-y$ leads to the particular solution which satisfies the specified condition :

$u(x,y)=\frac14 e^{x+2y}+e^{-y}\left(-\frac14 e^{x-y}\right)$ $$u(x,y)=\frac14 e^{x}\left(e^{2y}-e^{-2y}\right)$$ $$u(x,y)=\frac12 e^{x}\sinh(2y)$$

You lost/excluded all solutions that take non-positive values by setting $$v=\ln u$$. You can repair this partially by replacing that with $$v=\ln|u|$$. In the end, setting $$H(z)=e^{h(z)}$$ and then allowing negative and zero values for $$H$$ gives the general solution $$u(x,y)=H(x-y)e^{-x}+\frac{e^{x+2y}}4$$


# Assign unique integer keys to sets

I am given a list of $n>1$ arrays, where each array has fairly small number of elements (rarely above $5$). Also $n$ is quite small in practice (around $6$).

My problem is that I would like to efficiently map each element product to unique integer. Also, I would like to efficiently compute the integer, when I change the order of the arrays in the list.

I was thinking something in the line of integer factorization.

Take first $n$ primes ($p_1, p_2, ..., p_n$) and remeber the index of the array in the original list with $ind: A \rightarrow \mathbb{N}$. So the initial function would be $ind(A_i)=i$.

Also we denote with $a_{i,j}$ the $j$-th element of the $i$-th array in the list. So the function $F:A_1 \times A_2 \times \ldots \times A_n \rightarrow \mathbb{N}$ would be:

$$F(a_{1,i_1}, a_{2,i_2}, \ldots, a_{n,i_n})=\prod_{j=1}^{j=n}p_{ind(A_i)}^{i_j}$$

In more common words, we initially assign a prime number $p_i$ to array $i$ and raise $p_i$ to the index of the element in array $i$.

This guarantees us the uniqueness and also if we change the order of the arrays in the list, we just need to change the $ind$ function. But the problem here is that these products are fairly large and will overflow soon enough (even with using doubles).

Do you have any better idea how to enumerate these products without the large number constraint? I am implementing this in Java and I really would not like to use BigInteger implementation.

• Is it hashing you are after, or what is the purpose here? – Raphael Apr 5 '13 at 10:10
• Yes. Something like that. I actually need unique numbers, because these output numbers represent some kind of output (the number is actually mapped to a string value). So for example I need to get different integers for (1,2,3) and for (3,2,1). – Nejc Apr 5 '13 at 11:57
• What is the "element product", is it the product of all elements in one array? Why would this change when the order of the arrays is changed? – Juho Apr 5 '13 at 15:30
• Based on your description, I take it that an element product is taking one element from each array and multiplying (you should clarify that). – Aryabhata Apr 5 '13 at 17:26
• Sorry for not being clear! One element product of $n$ arrays is a $n$-tuple, where we take one element from each array and on the $i$-th place of the $n$-tuple is an element from $i$-th array. Clearly there are $|A_1| \cdot |A_2| \cdot \ldots \cdot |A_n|$ such products. – Nejc Apr 5 '13 at 17:45

Based on your description, I take it that an element product is taking one element from each array and multiplying?

You don't really need to multiply or use prime numbers which might overflow (but not too much for your constrains I suppose, and is a cute solution :-)).

Suppose the arrays were of fixed size $k$ (You can pad the arrays to achieve this, and based on your constraints, won't be too much of a problem).

Then you want to map all possible tuples $(x_1, x_2, \dots x_n)$ where $0 \le x_i \lt k$ to unique integers.

The mapping you need: Interpret the tuple as an integer in base-$k$.

If arrays are shuffled you can still change the $ind$ function and remap the digits according to that.

(or did you try that and didn't work for some reason?)

You want an injection between your element space and the naturals. Why not take the oldest one that happens to be a bijection, the Cantor pairing function.

Its basic form is defined to map $\mathbb{N^2}$ to $\mathbb{N}$:

$\qquad\displaystyle \pi(k_1,k_2) = \frac{1}{2}(k_1 + k_2)(k_1 + k_2 + 1)+k_2$.

It readily extends to $\mathbb{N}^k$:

$\qquad \pi^{(n)}(k_1, \ldots, k_{n-1}, k_n) = \pi ( \pi^{(n-1)}(k_1, \ldots, k_{n-1}) , k_n)$.

The Wikipedia article explains how to invert the function.

Now, in your case, $n$ is unknown. That's not a problem; we can extend above $\pi$ easily to enumerate $\mathbb{N}^+ = \bigcup_{n \geq 1} \mathbb{N}^n$ by encoding $n$ -- here the array's length -- as well:

$\qquad \pi^+(a_1,\dots,a_n) = \pi^{(n+1)}(a_1, \dots, a_n, n)$.

Since it's a bijection into $\mathbb{N}$, values are arguably as small as possible. Also, computing either direction is efficient.

When implementing it, be very carful either way. If you consider arrays over some finite set $A$, there are

$\qquad\displaystyle \sum_{i=1}^n |A|^i = \frac{|A|^{n+1} - |A|}{|A| - 1}$

many arrays of size at most $n$. $\pi^+$ has to encode them all, so the integers might overflow quickly if $|A|$ and $n$ are not both small.

• Thank you for your nice suggestion. Luckily I do not need the inverse function, so I will implement the Cantor pairing function and compare it to the suggestion of @Aryabhata. – Nejc Apr 8 '13 at 9:36


# Maximum cost path in integer matrix

In preparation for my design and algorithms exam, I encountered the following problem.

Given a $2 \times N$ integer matrix $(a[i][j] \in [-1000, 1000])$ and an unsigned integer $k$, find the maximum cost path from the top left corner $(a[1][1])$ and the bottom right corner $a[2][N]$, given the following:

$\bullet$ The path may not go through the same element more than once

$\bullet$ You can move from one cell to the other vertically or horizontally

$\bullet$ The path may not contain more than $k$ consecutive elements from the same line

$\bullet$ The cost of the path is determined by the sum of all the values stored within the cells it passes through

I've been thinking of a simple greedy approach that seems to work for the test cases I've tried, but I'm not sure if it's always optimal. Namely, while traversing the matrix, I select a the maximum cost cell on the vertical or horizontal and then go from there. I've got a counter I only increment if the current cell is on the same line with the previously examined one and reset it otherwise. If at some point the selected element happens to be one that makes the counter go over the given value of $k$, I simply go with the other option that's left.

However, I feel that I'm missing out on something terribly important here, but I just can't see what. Is there some known algorithm or approach that may be used here?

Also, the problem asks for an optimal solution (regarding temporal complexity).

This problem was basically made for dynamic programming (DP). Just review it and follow a couple of examples and solving this problem is straight-forward.

Your simple greedy approach does not always work. Consider:

[ begin ] [  2 ]
[ 3 ]     [ 10 ]
[ 1 ]     [ -1000 ]
[ 1 ]     [ end ]

The optimal solution would be to go right, down, left, down, down, right.

• I meant $2$ lines and $N$ columns, but I guess it doesn't make a difference in this case. – user43389 May 29 '16 at 23:28

We shall approach this using dynamic programming -- the greedy algorithm is 'myopic' in that it only considers immediate neighbors and not the path that follows those neighbors.

Let $C(i, \ j, \ m)$ be the cost of the max-cost path starting from $a[i][j]$ going towards $a[2][n]$, where $i$ is either $1$ or $2$, $\ j \in [1, n]$ and $m \in [0, k]$.

• $m$ here denotes that we have $m$ remaining consecutive steps that could be taken along row $i$.
• The additional constraint we enforce here is that when we are calculating $a[1][i]$, we have not yet visited $a[2][i]$, and similarly when we're calculating $a[2][i]$, we have not yet visited $a[1][i]$.

With this constraint in place, what recurrence can we come up with? Well,

1. $C(1, \ j, \ m) = a[1][j] + \max \{ M(1, j+1, m-1), \ a[2][j] + M(2, j+1, k)\} \$ and similarly,
2. $C(2, \ j, \ m) = a[2][j] + \max \{ M(2, j+1, m-1), \ a[1][j] + M(1, j+1, k)\} \$

Now, what's the rationale here? If we started at $a[1][j]$, we have two choices:

• continue along row $1$ : we then go to $a[1][j+1]$ with $m-1$ remaining consecutive steps that we can take along row $1$. (note here that $a[2][j+1]$ is not yet visited, maintaining the invariant).

• switch to row $2$ : we go to $a[2][j]$, incurring a cost of $a[2][j]$, and then have only one choice from there - to go to $a[1][j+1]$. Because we have switched rows, we can now refresh our count and take $k$ consecutive steps along row $2$. (note again that we haven't visited $a[1][j+1]$ yet).

Now if we calculate our subproblems starting from $j = n$ towards $j=1$ for every value of $m$, you shall have your answer in $C(1, \ 1, \ k)$.

I shall leave the base cases and runtime analysis to you. Note that my solution calculates the cost of the path and not the path itself; can you extend my solution to calculate the path too?