# Proof by mathematical induction in Z

Is it possible to proof the following by mathematical induction? If yes, how?

$a\in \mathbb{Z} \Rightarrow 3$ | $(a^3-a)$

I should say no, because in my schoolcarrier they always said that mathematical induction is only possible in $\mathbb{N}$. But I never asked some questions why it is only possible in $\mathbb{N}$...

• So, normally it only works on ℕ but with a "trick" you can apply it on ℤ? What do you mean with "assuming for n and proving it for n−1"? – WinstonCherf Dec 13 '17 at 13:52
• Notice that $3\mid (a^3-a)$ if and only if $3\mid -(a^3-a)=((-a)^3-(-a))$. So it suffices to prove the statement for all $a\in \mathbb{N}$. – Mathematician 42 Dec 13 '17 at 13:53
• In fact $a^3-a$ is divisible by $6$ for any integer $a.$ To prove this by induction, first prove it on $\Bbb{N}$ by induction. Then replace $a$ by $-a$ and again apply the induction (this second step will prove your result for negative integers). – Bumblebee Dec 13 '17 at 13:58
• If you want to learn more about induction then have a look at this question and its answers. – drhab Dec 13 '17 at 14:33

In this particular question, you can consider it in two separate cases, first case for $a \ge 0$ and second case for $a < 0$.

Case $a \ge 0$: We will check whether $3 | (a^3-a)$ or not by using induction on $a$. For $a = 0$, we have $3|0$. Now suppose $a \ge 1$ and for all $a$, the argument holds. Then for $a+1$, we have $$(a+1)^3-(a+1) = a^3+3a^2+2a = (a^3-a)+3a^2+3a$$ where $3|(a^3-a)$ by inductive assumption and $3|(3a^2+3a)$ obviously. Therefore, by induction, it holds for all $a \ge 0$.

Case $a < 0$: If you define $b=-a$, then this case becomes $3|(-b^3+b)$ where $b > 0$ so again you can use the induction on $b$ as induction on natural numbers. Proof for this case is similar to the first case.

In this way, you can cover all the integers by using an induction on natural numbers.

• Can you also give the proof please? – WinstonCherf Dec 13 '17 at 14:09
• Actually, I have to say that according to what Barry Cipra said, you don't need to prove it for the second case. But I really suggest you to do it just for practicing induction. – ArsenBerk Dec 13 '17 at 14:23

Technically you need to do two separate inductions. But since $(-a)^3-(-a)=-(a^3-a)$, you really only need to take the induction in the ordinary positive direction. If you do want to do both inductions, you can combine them in a single argument, along the following lines:

The base case is $3\mid0^3-0$, and

$$(a\pm1)^3-(a\pm1)=(a^3\pm3a^2+3a\pm1)-(a\pm1)=(a^3-a)\pm3a^2+3a$$

so $3\mid(a^3-a)$ implies $3\mid((a\pm1)^3-(a\pm1))$.

• Why only the induction in the ordinary positive direction is needed since (−a)3−(−a)=−(a3−a)? Can you please explain that? – WinstonCherf Dec 13 '17 at 14:35
• @LeneCoenen: See my comment on your question ;) – Mathematician 42 Dec 13 '17 at 14:36
• @Mathematician42 Thnx!! – WinstonCherf Dec 13 '17 at 14:38

Induction can be applied on a set if the set involved is equipped with a so-called well-order.

Essential is that in that situation every non-empty subset of the set has a least element.

Note that $\mathbb N$ has a very natural well-order: $0<1<2<\cdots$.

The famiar and well known order $<$ on $\mathbb Z$ is not a well-order. One of the non-empty sets that has no least element according to that order is $\mathbb Z$ itself, and there are lots of others.

This is why on school you were taught that induction was not for $\mathbb Z$.

Overlooked is there that there are well-orders on $\mathbb Z$ also.

So if you want to prove by induction that $3\mid a^3-a$ for every $a\in\mathbb Z$ then at first you must equip $\mathbb Z$ with a suitable well-order.

One (there are more) that can be used for it is:

$$0<'1<'2<'3<'\dots<'-1<'-2<'-3<'\dots$$

If $P(a)$ is true iff $3\mid a^3-a$ then it is enough to prove that:

• $P(0)$
• $P(n)\implies P(n+1)$
• $P(n)\implies P(n-1)$

I should say that it is even more than enough (see the comment of Hagen).

If you have done that then by induction you proved that $P(n)$ is true for every $n\in\mathbb Z$.

• @Avamander Thanks, I repaired. – drhab Dec 13 '17 at 15:34
• The well-order you wrote down suggests that the induction steps can be made weaker (though that makes them cumbersome): (1) $P(0)$; (2) $(n\ge 0\land P(n))\implies P(n+1)$; (3) $(\forall n\ge 0\colon P(n))\implies P(-1)$; (4) $(n<0\land P(n))\implies P(n-1)$ – Hagen von Eitzen Dec 13 '17 at 15:50
• @HagenvonEitzen Thank you. I added a remark on this that refers to your comment. – drhab Dec 13 '17 at 15:53
• The well-order you suggested does not work with standard mathematical induction. Its order type is larger than $\omega$, so you need transfinite induction, albeit only technically. – tomasz Dec 13 '17 at 23:12

The induction principle on $\mathbb{N}$ says: assuming that a property holds for $0$, and that if it holds for $n$ then it holds for $n+1$, then the property is true for all the elements of $\mathbb{N}$. The principle holds because all the elements of $\mathbb{N}$ can be reached by starting from $0$ and applying the operation $n \mapsto n+1$ a finite number of times.

Let's make this a little more abstract. Assuming that a property holds for the initial natural ($0$), and that if it holds for a natural then it also holds for the next natural ($n+1$), then it holds for all naturals.

We can generalize this to other domains than $\mathbb{N}$ by generalizing the notions of “initial” and “next”. Assume that all the elements of a set $D$ can be reached by starting from some initial element and by applying a “derivation” operation a finite number of times. Assuming that a property holds for all the initial elements, and that if it holds for an element then it also holds for a derived element, then the property holds for all the elements.

Application: all the relative integers ($\mathbb{Z}$) can be reached by starting from $0$ (the single initial element) and applying one of the operations $n \mapsto n+1$ or $n \mapsto n-1$ a finite number of times. Therefore, the following induction principle holds on $\mathbb{Z}$: assuming that a property holds for $0$, that if it holds for $n$ then it holds for $n+1$, and that if it holds for $n$ then it holds for $n-1$, then the property holds for all the elements of $\mathbb{Z}$.

Given this principle, proving the property you want is a simple modification from the proof on $\mathbb{N}$.

It's possible to generalize this further by generalizing the notion of “derivation”. An element could be derived from multiple arguments. Assume that there is a family of constructor operations $c_i : D^{a_i} \to D$, where each constructor can take a different number of parameters, such that all elements of $D$ can be reached by applying constructors. The starting point comes from constructors with 0 arguments. Then there is an induction principle on $D$ which states that, assuming that for each constructor $c_i$, if the property holds for $(x_1,\ldots,x_{a_i})$ then it holds for $P(c_i(x_1,\ldots,x_{a_i}))$, then the property holds for all the elements of $D$. The induction principle for $\mathbb{N}$ is a special case with two constructors: $0$ (with 0 arguments) and $n \mapsto n+1$ (with 1 arguments). The induction principle for $\mathbb{Z}$ adds a third constructor $n \mapsto n-1$. You could add a fourth constructor with two arguments $(p,q) \mapsto \begin{cases} p/q & \text{if }q \ne 0 \\ 0 & \text{if } q = 0 \end{cases}$ to get an induction principle for $\mathbb{Q}$.

It's possible to generalize this even further to get induction principles on “larger” spaces (which don't even need to be countable). See drhab's answer.

Technically, induction is a technique applied on the natural numbers.

However, there is nothing stopping you from having two statements applying to natural numbers that you prove seperately, but very similarily:

1. $P(n):3\mid (n^3-n)$
2. $Q(n): 3\mid ((-n)^3 - (-n))$

We can apply induction to prove $P$ and $Q$ for all natural numbers. Then, when it comes to showing that $P$ holds for all integers, we simply note that $P(n) \equiv Q(-n)$, so for any integer $k$, if it is possible, then the truth of $P(k)$ comes from the induction on $P$, while if $k$ is negative, the truth of $P(k)$ is the same as the truth of $Q(-k)$, which was proven by induction on $Q$.

Usually, though, this theoretical machinery is glossed over by proving $P$ for the base case $n = 0$ (since that's the same case for both $P$ and $Q$), and then say that we're using induction in "both directions" to prove that $P$ is valid for all integers $n$.

You can extend the induction principle to work for $\mathbb Z$. The difference is that you instead of implication in the "step" part use equivalence:

If $\phi(0)$ is true and $\forall j\in\mathbb Z: \phi(j)\leftrightarrow\phi(j+1)$ is true then $\forall j\in\mathbb Z:\phi(j)$ is true.

You can also use the normal induction principle twice. First proving it for $\mathbb N$ and then for proving the statement for $\mathbb Z^{-1}$.

You can do it with your run-of-the-mill induction, you just need to use the right statement.

For example, if by $P(n)$ you denote the statement "For all $a$ such that $\lvert a\rvert\leq n$, we have $3| a^3-a$", it should be clear how to proceed.

Since $\mathbb{Z}$ is countable as $\mathbb{N}$ we can extend induction over $\mathbb{Z}$.

BASE CASE:

$$a=1 \implies 3|0$$

INDUCTIVE STEP 1 "UPWARD"

assume: $3|a^3-a$

$$(a+1)^3-(a+1)=a^3+3a^2+3a+1-a-1=a^3-a+3a^2+3a\equiv0\pmod 3$$

thus

$$3|(a+1)^3-(a+1)$$

INDUCTIVE STEP 2 "DOWNWARD"

assume: $3|a^3-a$

$$(a-1)^3-(a-1)=a^3-3a^2+3a-1-a+1=a^3-a-3a^2+3a\equiv0\pmod 3$$

thus

$$3|(a-1)^3-(a-1)$$

Thus: $$\forall a\in \mathbb{Z} \Rightarrow 3|a^3-a$$

• This does not answer the question. You completely missed the point of this question. – Mathematician 42 Dec 13 '17 at 14:02
• Sorry I've extended the result. Now is it ok? – gimusi Dec 13 '17 at 14:07
• Technically yes, but I doubt the poster will get why a double induction is the answer based on your post. Personally, I would explain how you cover all of $\mathbb{Z}$ by a double induction. – Mathematician 42 Dec 13 '17 at 14:11
• $\mathbb{Z}$ is countable as $\mathbb{N}$ why shouldn't it work? Isn't t it sufficient? – gimusi Dec 13 '17 at 14:18
• Yes, I know that, but clearly the poster didn't. The poster had issues using induction on something which is not $\mathbb{N}$. I believe an answer should address that issue first. Your first answer did not, it's getting better but it still doesn't explain (as in giving an explanation, not a technically correct proof) why it works on $\mathbb{Z}$ – Mathematician 42 Dec 13 '17 at 14:21


# Conics Section : Properties

This Note is for Those who love to use Properties and some innovative Techniques while Solving Question of Conics Section ! These Properties are Highly reduce our Calculation and are very useful sometimes , Specially when we have Time Constrained !

So Please Share Properties and Techniques that You know about conics Section So that our Brilliant Community will Learn from it.

So Now Here I Shared some few Techniques ( Properties ) of Conics Section That are Created By Me :)

For Ellipse ( By Me )

$\\ \cfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1\quad \quad \quad (:\quad a>b\quad )$.

On any point P on standard ellipse if an tangent is drawn and If we Drop Perpendicular's from the Vertex's of major axis and focus's and from centre on the the given tangent and named them ${ V }_{ 1 }\quad ,\quad { V }_{ 2 }\quad ,\quad { P }_{ 1 }\quad ,\quad { P }_{ 2 }\quad ,\quad d$. suitably Then :

1)- $\displaystyle{ V }_{ 1 }\quad ,\quad d\quad ,\quad { V }_{ 2 }\quad \longrightarrow \quad AP\\ \\ 2d\quad =\quad { V }_{ 1 }+{ V }_{ 2 }\quad \quad$

2)- $\displaystyle{ P }_{ 1 }\quad ,\quad d\quad ,\quad { P }_{ 2 }\quad \longrightarrow \quad AP\\ \\ 2d\quad ={ \quad P }_{ 1 }+{ P }_{ 2 }$

3)- $\cfrac { { S }_{ 1 }P }{ { S }_{ 2 }P } \quad =\quad \cfrac { \quad P_{ 1 } }{ { \quad P }_{ 2 } }$.

Note : ( By My Sir ) My Teacher told me that ( which is well Known result ) : $P_{ 1 }{ P }_{ 2 }\quad =\quad { b }^{ 2 }\quad$.

My Turn is Over ! Now it's Your Turn ,

So please Post Properties or techniques Related To conics Section That are created by You or may also be you learnt it Somewhere else :)

Reshare This More and More So that it can reaches to everyone , So that we can Learn from Them!

Note by Deepanshu Gupta
5 years, 8 months ago

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There are actually many properties, here are few of them

If the normal at any point P on the ellipse with centre C meet the major & minor axes in G & g respectively & if CF be perpendicular upon this normal then

• PF.PG = $b^{2}$

• PF.Pg = $a^{2}$

• PG.Pg = SP.S'P

• CG.CT = $CS^{2}$

If tangent at the point P of a standard ellipse meets the axes at T & t and CY is perpendicular on it from centre then

• Tt.PY = $a^{2} - b^{2}$

• least value of Tt is $a + b$

This is only for ellipse but there are many for each curve

- 5 years, 8 months ago

Yes I know That That were also Told to me by my Teacher , But Thanks For Sharing It , it will Helpful For others ! Do You Have Your own Properties if You have then Please also Share it with us , Thanks Krishna :)

- 5 years, 8 months ago

Yeah there is one when we have to find the minimum distance between 2 Parabola's it occur at extremities of latus rectum

I don't know how it works but it has worked till now in every problem, I'll try to prove if I got some time meanwhile you can apply and try to prove it :)

Note:- I am not 100% sure with it because I haven't found any case in which this doesn't occur

- 5 years, 7 months ago

It would be much helpful if a diagram is drawn. Thanks.

- 5 years, 8 months ago

Hey, few more, for ellipse:

1. The portion of the tangent to the ellipse between the point of contact and the directrix subtend a right angle at the corresponding focus.

2. The circle on the focal distance as diameter touches the auxiliary circle.

3. Tangent at the extremeties of latus rectum pass through the corresponding foot of directrix on the major axis.

4. Ratio of area of any triangle inscribed in a standard ellipse ( a> b) and that of triangle formed by corresponding points on the auxiliary circle is $\frac {b} {a}$

- 5 years, 8 months ago

It would be much helpful if a diagram is drawn. Thanks.

- 5 years, 8 months ago

Thanks alot :)

- 5 years, 7 months ago

These are good properties of conic sections. Could you add them to the corresponding Conic Section wiki pages? Thanks!

Staff - 5 years, 7 months ago

I have posted my first challenge problem on conic sections. Please post solutions.

Check out my profile page to get other conic challenges.

- 5 years, 6 months ago


# Having sign problem with algebraic fractions

• Jan 26th 2013, 05:48 PM
KevinShaughnessy
Having sign problem with algebraic fractions
Hi,

I'm having a problem with an algebraic fractions equation. It goes:

5/3(v-1) + (3v-1)/(1-v)(1+v) + 1/2(v+1)

The first thing I do is factor out the negative in the first fraction, getting:

-5/3(1-v)

Giving a cd of 6(1-v)(v+1). Now that that is done I multiply the numerators by the necessary factors:

-5*2(v+1)

6(3v-1)

3(1-v)

Which gives me -10v -10 + 18v -6 -3v + 3

Which adds up to 5v - 13

BUT the answer is -5v + 13, and if I factor out the negative in the second equation, all the signs are reversed and the equation works out to the right answer. So I'm confused as to why things didn't work when I factored out the negative in the first fraction.

Thanks,

Kevin
• Jan 26th 2013, 07:53 PM
chiro
Re: Having sign problem with algebraic fractions
Hey KevinShaughnessy.

Can you clarify whether (3v-1)/(1-v)(1+v) is (3v-1)(1+v) / (1-v) or (3v-1) / [(1+v)(1-v)]?
• Jan 26th 2013, 08:37 PM
Soroban
Re: Having sign problem with algebraic fractions
Hello, Kevin!

They approached the problem differently . . . that's all.

Quote:

$\text{Simplify: }\:\frac{5}{3(v-1)} + \frac{3v-1}{(1-v)(1+v)} + \frac{1}{2(v+1)}$

They factored a "minus" out of the second fraction.

. . $\frac{5}{3(v-1)} - \frac{3v-1}{(v-1)(v+1)} + \frac{1}{2(v+1)}$

The LCD is $6(v-1)(v+1)\!:$

. . $\frac{5}{3(v-1)}\cdot {\color{blue}\frac{2(v+1)}{2(v+1)}} - \frac{3v-1}{(v-1)(v+1)}\cdot {\color{blue}\frac{6}{6}} + \frac{1}{2(v+1)}\cdot {\color{blue}\frac{3(v-1)}{3(v-1)}}$

. . $=\;\frac{10(v+1)}{6(v-1)(v+1)} - \frac{6(3v-1)}{6(v-1)(v+1)} + \frac{3(v-1)}{6(v-1)(v+1)}$

. . $=\;\frac{10v + 10 - 18v + 6 + 3v - 3}{6(v-1)(v+1)}$

. . $=\;\frac{13-5v}{6(v-1)(v+1)}$
• Jan 26th 2013, 10:17 PM
KevinShaughnessy
Re: Having sign problem with algebraic fractions
Quote:

Originally Posted by chiro
Hey KevinShaughnessy.

Can you clarify whether (3v-1)/(1-v)(1+v) is (3v-1)(1+v) / (1-v) or (3v-1) / [(1+v)(1-v)]?

It's (3v-1) / [(1+v)(1-v)].
• Jan 26th 2013, 10:19 PM
KevinShaughnessy
Re: Having sign problem with algebraic fractions
Quote:

Originally Posted by Soroban
Hello, Kevin!

They approached the problem differently . . . that's all.

They factored a "minus" out of the second fraction.

. . $\frac{5}{3(v-1)} - \frac{3v-1}{(v-1)(v+1)} + \frac{1}{2(v+1)}$

The LCD is $6(v-1)(v+1)\!:$

. . $\frac{5}{3(v-1)}\cdot {\color{blue}\frac{2(v+1)}{2(v+1)}} - \frac{3v-1}{(v-1)(v+1)}\cdot {\color{blue}\frac{6}{6}} + \frac{1}{2(v+1)}\cdot {\color{blue}\frac{3(v-1)}{3(v-1)}}$

. . $=\;\frac{10(v+1)}{6(v-1)(v+1)} - \frac{6(3v-1)}{6(v-1)(v+1)} + \frac{3(v-1)}{6(v-1)(v+1)}$

. . $=\;\frac{10v + 10 - 18v + 6 + 3v - 3}{6(v-1)(v+1)}$

. . $=\;\frac{13-5v}{6(v-1)(v+1)}$

But aren't the two answers fundamentally different being that one produces a negative number and the other produces the same number but positive?
• Jan 27th 2013, 12:18 AM
earthboy
Re: Having sign problem with algebraic fractions
Quote:

Originally Posted by KevinShaughnessy
But aren't the two answers fundamentally different being that one produces a negative number and the other produces the same number but positive?

your answer was $\frac{5v-13}{6{\color{magenta}(1-v)}(1+v)}$ while most probably the answer in your book is given as $\frac{13-5v}{6{\color{magenta}(v-1)}(1+v)}$, Which you know are the same answers because the answer in your book changed your $(1-v)$ to $(v-1$) by multiplying the numerator and denominator by $-1$.
your answer was $\frac{5v-13}{6{\color{magenta}(1-v)}(1+v)}$ while most probably the answer in your book is given as $\frac{13-5v}{6{\color{magenta}(v-1)}(1+v)}$, Which you know are the same answers because the answer in your book changed your $(1-v)$ to $(v-1$) by multiplying the numerator and denominator by $-1$.


# 'If…then…' and '…if…' and '…only if…' and 'If… only then…' statements?

Suppose you have two statements A and B and "If A then B". I am trying to think of what this implies and alternative ways of writing this.

I think

"If A then B"

= A$\rightarrow$B

= "A is sufficient but not necessary for B. B is neither necessary nor sufficient for A"

= "If not B then not A"

= B'$\rightarrow$ A'

= "B' is sufficient for A' but not necessary"

And it seems to me that 'B if A' is equivalent to 'If A then B' (please correct me if I am wrong!)

When it comes to only if, I think "B only if A" is equivalent to "If A only then B"

I think

"B only if A"

= B$\rightarrow$A

= "B is sufficient for A but not necessary for A. A is necessary for B but not sufficient for B"

="If not A then not B"

=A'$\rightarrow$B'

= "Not A is sufficient for not B and not A is necessary for not B"

I see that I must have made mistakes somewhere because several things are not consistsent. Firstly, I am pretty sure my last statement is wrong but this is how I interpret "If not A then not B". Also, I don't understand how if the '...if...' cases, single headed arrows only implied sufficiency, and here for the '...only if...'single headed arrows seem to be implying something about necessity as well...

Thank you in advance for any clarifications, and also if anyone has a link to a good explanation of these statements I would be very grateful. I am trying to understand them in the context if proofs and proving statements the right way around...

EDIT: Thank you for all the answers and comments so far. Jut thought I would add something that helped me in case someone else comes across my question and also requires help with if/iff/necessary/sufficient etc. I found it easier to visualise the cases. 'B if A' can be represented as a circle A within a circle B. Automatically, if A then B, so A is sufficient but not necessary for B, but B is necessary for A and not B implies not A. However A does not imply B. In a similar way, 'B only if A' is a circle B within a circle A, because B implies A- it is sufficient, but not necessary for A, and A is necessary but not sufficient for B. 'B if and only if A' is the double headed arrow because A and B are the same ring. One is both necessary and sufficient for the other, and one implies the other....

• Yes : $A \to B$ is : "if A then B" and also "B if A" ans "A is a sufficient condition for B". $A \to B$ is also "B is a necessary condition for A" and "A only if B". – Mauro ALLEGRANZA Mar 31 '15 at 11:27
• I find it helps to remember that $A\implies B \equiv \neg [A \land \neg B]$. It has nothing with $A$ causing $B$, or $B$ causing $A$ as many beginners seem to think. There is no causality in mathematics. – Dan Christensen Mar 31 '15 at 15:52

$A\to B$ means "$A$ implies $B$", "$A$ is sufficient for $B$", "if $A$, then $B$", "$A$ only if $B$", and such.

$A\leftarrow B$ means "$A$ is implied by $B$", "$A$ is necessary for $B$", "$A$ whenever $B$", "$A$ if $B$", and such

$A\leftrightarrow B$ means "$A$ is necessary and sufficient for $B$", "$A$ if and only if $B$".

Note: $A\to B$ means "$A$ is sufficient for $B$ and may or may not be necessary for $B$".   It neither affirms nor denies the necessity.

We can also say,

$A\to B$ means "$B$ if $A$", "$B$ whenever $A$", and "$B$ is necessary for $A$",

$A\leftarrow B$ means "$B$ only if $A$", "if $B$, then $A$", and "$B$ is sufficient for $A$".

$A\wedge\neg B$ means "$A$ is not sufficient for $B$", and $\neg A\wedge B$ means "$A$ is not necessary for $B$"

• Brilliant answer! – Konstantin Feb 11 '17 at 17:04

Your thinking is a little off. If $A$ is sufficient for $B$, then indeed $B$ is necessary for $A$. This is because if $B$ is not true, then $A$ is not true.

Long comment

There are no "inconsistency" ...

$A \to B$ is : "if $A$, then $B$" and also "$B$, if $A$" and also "$A$ is a sufficient condition for $B$".

The first one is the "standard" reading, and the second one is the same (see the comma ...).

$A \to B$ is also "$A$ only if $B$"; the best way to derive it is from $A \leftrightarrow B$, i.e. "$A$ if and only if $B$".

This is "($A$, if $B$) and ($A$ only if $B$)", that translates $(B \to A) \land (A \to B)$.

Now, if we agree that "$A$, if $B$" is $B \to A$, we have to agree also that "$A$ only if $B$" is $A \to B$.

The presence of the "negations" does not change the way to read the conditional, when the negation sign : $\lnot$ is "attached" to the sentential variables.

Thus, $\lnot A \to \lnot B$ is : "if not-$A$, then not-$B$", and so on.


Confirm definite integral equals zero $\frac{\sin(x)}{(1-a\cos(x))^{2}}$

Is this statement about the definite integral of a particular function $F$ true? $$\int_0^{2\pi}F(x)\, \mathrm{d}x = \int_0^{2\pi}\frac{\sin(x)}{(1-a\cos(x))^2}\, \mathrm{d}x = 0 \ \text{ for }\ 0<a<1$$

I have evaluated this expression (in WolframAlpha) for various values of a and they all give the value zero. I have read that integrals of the form $$\int G(\cos(x))\sin(x)\, \mathrm{d}x$$ where $G$ is some continuously integrable function are zero over the range $-\pi/2$ to $\pi/2$.

(Edited after comment from Andrey) It seems possible to proceed from here to confirm the postulated statement by symmetry. The function $F$ to be integrated is cyclic with period 2$\pi$ such that $F(x-2\pi) = F(x) =F(x+2\pi)$. Then we just need to prove that the two integrals:- (1) between $-\pi$ and $-\pi/2$, and (2) between $\pi/2$ and $\pi$ are equal in magnitude and opposite in sign.

This would be the case if $F(x) = -F(-x)$. Such is actually the case because the denominator in F() expands to $(1-2a\cos(x)+a^2\cos^2x)$ and has the same values for $(+x)$ and $(-x)$. Whereas the numerator $\sin(x)$ is such that $\sin(x) = -\sin(-x)$.

However I would still like to find an analytical solution.

• It is the case that $F(x)=-F(x)$. The denominator $1-2a\cos x + a^2\cos^2x$ for $x<0$ is the same as $x>0$, since cosine is an even function, i.e. $\cos(-x) = \cos(x)$. – Andrey Kaipov Oct 11 '14 at 22:09
• @Andrey Yes you are right of course. Doh! – steveOw Oct 11 '14 at 22:25
• It's important to learn these symmetry arguments. There is a famous integral from the Putnam that cannot be done any other way: $\int_0^{\pi/2} \frac{1}{1+\tan^{\sqrt{2}} x} = \frac{\pi}{4}$. – Slade Oct 12 '14 at 0:59

Let $u=1-a\cos x$, $du=a\sin x dx$ to get $\displaystyle\frac{1}{a}\int_{b}^{b}\frac{1}{u^2}du=0$ $\;\;\;$ (where $b=1-a$).

• I dont understand the limits b..b. – steveOw Oct 11 '14 at 22:42
• When $x=0, u=1-a(1)=1-a$ and when $x=2\pi, u=1-a(1)=1-a$. – user84413 Oct 11 '14 at 22:51
• @user844413 Wow that is so slick! – steveOw Oct 11 '14 at 23:00

$$\int_a^bF(x)dx = \int_a^bF(a+b-x)dx,$$

we have in this case

$$\int_0^{2\pi}F(x)dx = \int_0^{2\pi}F(2\pi-x)dx,$$

and, from knowledge of the symmetries of the sin and cos functions, we know in this case that $$F(x) = - F(2\pi-x),$$

so, with $I =\int_0^{2\pi}F(x)dx$, we have

$$I=-I,$$

which can only be true if $$I = 0$$

• But how do I know your two integrals are equivalent to start with? I only know: F(x)=-F(-x) and F(x)=-F(2pi-x) and hence F(-x)=F(2pi-x). – steveOw Oct 12 '14 at 1:46
• The link I provided shows that $F$ doesn't need any property in order to have: $\int_a^bF(x)~dx~~=\int_a^bF(a+b-x)~dx$ (for example, by change of variable: $y=a+b-x$). The lucky part for your integral is that we get $-I$ for the second integral. – ir7 Oct 12 '14 at 1:57
• (Aha, I didn't spot the link). The very useful equation in your comment is fundamental to this answer. I suggest it is included in the answer. – steveOw Oct 12 '14 at 15:03
• Ok.It looks better now. Cheers. :) – ir7 Oct 12 '14 at 15:25

This is a special case of a general fact about $u$-substitution. If $G(x)$ is integrable on the interval $[a,b]$, with antiderivative $g(x)$, and $u$ is differentiable, then $$\int_a^b G(u(x))\,u'(x)\, dx = g(u(b))-g(u(a)).$$

If $u(a)=u(b)$, the integral is zero.

The integrand in your example has this form, where $u(x)=\cos(x)$ and $G(u)=\frac{-1}{(1-a\cos(x))^2}$, and $\cos(x)$ has the same value at both limits of integration, so the integral is zero. (You can apply the substitution user84413 suggested, or the simpler $u=\cos x$ to show it.)

You can write down lots of messy-looking integrals that turn out to be zero because they have this form for some $u(x)$ and $G(x)$.

$$\int_1^3 e^{x^2-4x+7}(2-x)\, dx$$

$$\int_0^{2\pi} (\pi-x)\log(2+\sin^2(x-\pi)^2)\, dx$$

$$\int_{\pi/2}^{3\pi/2} (\cos^2 x)^{\sin^2 x}\sin2x\,dx$$

• I like @ir7’s trick for this, which works for these examples. Here’s one where you can’t quickly show that $I=-I$ that way: $\displaystyle\int_1^9 (4\sqrt x-x)^{4\sqrt x-x}(\frac{2}{\sqrt{x}}-1)\,dx$. – Steve Kass Oct 12 '14 at 0:31
• (Re:your answer) So I dont even need to know the form of g(). Nice. – steveOw Oct 12 '14 at 1:01


# Is my proof, by strong induction, of for all $n\in\mathbb{N}$, $G_n=3^n-2^n$ correct?

Let the sequence $G_0, G _1, G_2, ...$ be defined recursively as follows:

$G_0=0, G_1=1,$ and $G_n=5G_{n-1}-6G_{n-2}$ for every $n\in\mathbb{N}, n\ge2$.

Prove that for all $n\in\mathbb{N}$, $G_n=3^n-2^n$.

Proof. By strong induction. Let the induction hypothesis, $P(n)$, be $G_n=3^n-2^n$

Base Case: For $(n=0)$, $P(0)$ is true because $3^0-2^0 =0$

For $(n=1)$, $P(1)$ is true because $3^1-2^1=1$

Inductive Step: Assume that $P(n-1)$ and $P(n-2)$, where $n\ge2$, are true for purposes of induction.

So, we assume that $G_{n-1}=3^{n-1}-2^{n-1}$ and $G_{n-2}=3^{n-2}-2^{n-2}$, and we must show that $G_{ n }=3^{ n }-2^{ n }$.

Since we assumed $P(n-1)$ and $P(n-2)$, we can rewrite $G_n=5G_{n-1}-6G_{n-2}$ as $G_n=5(3^{n-1}-2^{n-1})-{ 6 }(3^{n-2}-2^{n-2})$

So, we get:

$\Rightarrow G_n=5\cdot 3^{ n-1 }-5\cdot 2^{ n-1 }-(\frac { 6 }{ 3 } \cdot 3^{ n-1 }-\frac { 6 }{ 2 } \cdot 2^{ n-1 })$

$\Rightarrow G_n=5\cdot 3^{ n-1 }-5\cdot 2^{ n-1 }-2\cdot 3^{ n-1 }+3\cdot 2^{ n-1 }$

$\Rightarrow G_n=5\cdot 3^{ n-1 }-2\cdot 3^{ n-1 }-5\cdot 2^{ n-1 }+3\cdot 2^{ n-1}$

$\Rightarrow G_n=3\cdot 3^{ n-1 }-2\cdot 2^{ n-1 }$

$\Rightarrow G_n=\frac { 1 }{ 3 } \cdot 3\cdot 3^n-\frac { 1 }{ 2 } \cdot 2\cdot 2^n$

$\Rightarrow G_n=3^n-2^n$

The only real issue I have at this point is that I don't know how to properly conclude this proof with a final statement. A hint/guidance in that regard would be much appreciated.

• It looks great, you have effectly showed that if the property $G_{i}=3^i-2^i$ holds for $i\in\{1,2,\dots n-1\}$ then $G_{n}=3^n-2^n$ also. – Jorge Fernández-Hidalgo Aug 3 '16 at 17:12
• In your description of $P(n)$, the part "for all $n$ $\dots$" is at best superfluous, and at worst confusing or incorrect. Better, for any integer $k\ge 0$ let $P(k)$ be the assertion $G(k)=3^k-2^k$. – André Nicolas Aug 3 '16 at 17:32
• Not necessary. But the deletion of "for all $n$ $\dots$" is. – André Nicolas Aug 3 '16 at 17:47
• @Cherry Thanks for the link. I see that they state strong induction with a base case P(0). But often we don't need base case(s) for strong induction and the induction principle can be stated without any. For example, every integer > 1 is a product of primes. Suppose for induction it is true for all naturals < n. If n is prime we are done, else n is composite so it is a product of smaller naturals n = ab so by induction a,b are products of primes. Appending their products shows that n is a product of primes. No base case! – Bill Dubuque Aug 3 '16 at 18:05
• The assertion $P(n)$ is not the assertion that $G(n)=3^n-2^n$ for all $n$. If we write the latter in symbols, it is $\forall n(G(n)=3^n-2^n)$. Now $n$ is a "dummy variable" which gets quantified out. We want that for any particular $n$, $P(n)$ is the assertion $\dots$. – André Nicolas Aug 3 '16 at 18:12

You have the right idea, but there are some minor points that need correction.

The strong induction principle in your notes is stated as follows:

Principle of Strong Induction $\$ Let $\,P(n)\,$ be a predicate. If

• $\ P(0)$ is true, and

• for all $\,n\in \Bbb N,\ P(0), P(1),\ldots, P(n)\,$ together imply $\,P(n\!+\!1)\,$ then $\,P(n)\,$ is true for all $\,n\in\Bbb N$

Your $\,P(n)\,$ is $\, G_n = 3^n - 2^n.\,$ You have verified that $\,P(0)\,$ is true.

Your induction hypothesis is that $\,P(k)\,$ is true for all $k \le n.\,$ You have essentially shown that $\,P(n\!-\!1),P(n)\,\Rightarrow\,P(n\!+\!1)\,$ but that only works for $\,n\ge 1$ (else $\,P(n-1)\,$ is undefined). Thus you need to separately verify $\,P(1)\,$ (to be pedantic, this is part of the inductive step, not the base case, according to the above formulation of strong induction, though that is a somewhat arbitrary distinction)

It is illuminating to observe that the recurrence in the induction is a special case of

$$a^{n+1}-b^{n+1} =\, (a+b)(a^n-b^n) -ab (a^{n-1} - b^{n-1})$$

which can be verified directly or derived from the fact that $\,a,b\,$ are roots of

$$(x\!-\!a)(x\!-\!b) = x^2\! - (a\!+\!b) x + ab\,\Rightarrow\, x^{n+1}\! = (a\!+\!b)\,x^n - ab\, x^{n-1}$$

The proof will be simpler (and more insightful) if you work with this general case, i.e. prove that $\,f_n = a^n - b^n\,$ satisfies $\,f_{n+1} = (a+b) f_{n} - ab f_{n-1},\ f_0 = 1,\ f_1 = a-b\,$ for all $\,n\ge 0.\,$ Then your problem is just the special case $\,a,b = 3,2,\,$ and the inductive step is much clearer.

• I was under the impression that I showed that $P(n-2),P(n-1)\Rightarrow P(n)$. You stated that I showed $P(n-1),P(n)\Rightarrow P(n+1)$. Why? – Cherry_Developer Aug 3 '16 at 19:57
• @Cherry That's why I said "essentially". Substitute $\,n+1\,$ for $\,n\,$ in your proof to get the upshifted form. I wrote the induction in the above form used in the MIT notes. – Bill Dubuque Aug 3 '16 at 20:01
• Ah ok. I apologize for the slew of questions. I would just much rather struggle with the math, and ask these questions now, than when I actually take discrete math. Thank you for all your help. – Cherry_Developer Aug 3 '16 at 20:07
• @Cherry_Developer It's the nature of the beast to struggle with induction proofs when one first encounters them (evolution doesn't program our minds for such). Many fit into particular patterns that are easier to comprehend in the abstract (such as the above which is essentially exploiting the uniqueness theorem for recurrences).. Another common form of induction is telescopy, e.g. see here for a vivid 2D example. – Bill Dubuque Aug 3 '16 at 20:11

Yes, your proof is perfectly fine. Good job! You can write something like "The assertion follows.". But honestly it isn't necessary since it is in this case pretty simple for readers to see where the proof is complete (after the inductive step).


# Rate of water level for cone shaped water tank

A water tank in the form of an inverted cone is being emptied at the rate of $6$ ft$^3$/min. The altitude of the cone is $24$ ft, and the base radius is $12$ ft. Find how fast the water level is lowering when the water is $10$ ft deep.

I am not how to do this problem, but I've tried this using the volume formula for cone:

$$v={1 \over 3} \pi r^3 h\\ {dv \over dm} = {1 \over 3} \pi (12)^2{dh \over dm}\\ 6 = {1 \over 3} \pi 144 \cdot{dh \over dm}\\ 6 = 48 \pi \cdot {dh \over dm} \\ {1 \over 8 \pi} = {dh \over dm}$$

I am pretty sure that I am wrong.

Could someone help me?

Thanks

The answer is ${6 \over 25 \pi}$ ft /min according to the answer sheet

You can't take $r=12$ in $$v={1 \over 3} \pi r^2 h\\ {dv \over dm} = {2 \over 3} \pi (12){dh \over dm}$$ because the radius of the water is changing as it drains. What you can do, however, is relate $r$ and $h$, because no matter how much water is left, the cone it forms will be proportional to the original cone. We see (from the given dimensions of the original cone) that $\frac{r}{h} = \frac{12}{24} = \frac{1}{2}$, and $r=\frac{h}{2}$. Let's substitute this for $r$ right away: $$v={1 \over 3} \pi r^2 h$$ $$v={1 \over 3} \pi (\frac{h}{2})^2 h$$ $$v={1 \over 3} \pi \frac{h^3}{4}$$ $$\frac{dv}{dm} = \pi (r)\frac{h^2}{4}\frac{dh}{dm}$$

$$6 = \pi (\frac{h}{2})^2\frac{dh}{dm}$$ $$6 = \pi (\frac{h^2}{4})\frac{dh}{dm}$$ and plugging in $h=10$: $$6 = \pi (\frac{100}{4})\frac{dh}{dm}$$ We get $\frac{dh}{dm} = \frac{6}{25\pi}$.

• I added a few more details. Is there anything I should elaborate on? – Quinn Greicius Mar 10 '16 at 0:56
• Is it $r= {h \over 2}$ because altitude is 24 and the radius is 12, so ${24 \over 12} = {1 \over 2}$? – didgocks Mar 10 '16 at 1:00
• Exactly. Most related rates problems that involve several variables will use a trick like that to put everything in terms of a single variable to get to a solution. – Quinn Greicius Mar 10 '16 at 1:01
• Thank you, it was very helpful! – didgocks Mar 10 '16 at 1:02
• I just noticed that $v = {1 \over 3} \pi r^3 h$ is not a correct formula for a cone – didgocks Mar 13 '16 at 14:26

Retain symbols till the last plugin step

$$r = h \cot \alpha \; ; \tan \alpha = \frac12 ;\; V = \pi r^3/3\; \cot \alpha$$

$$V = \pi h^3/3\, \tan ^2\alpha$$

$$dV/dt = \pi h^2 (dh/dt) \tan ^2\alpha$$

Plug in given values to get answer tallying with text.


Doing the math to determine the determinant of the matrix, we get, (8) (3)- … If the generated inverse matrix is correct, the output of the below line will be True. You don’t need to use Jupyter to follow along. Yes! Subtract 1.0 * row 1 of A_M from row 3 of A_M, and     Subtract 1.0 * row 1 of I_M from row 3 of I_M, 5. Here are the steps, S, that we’d follow to do this for any size matrix. Those previous posts were essential for this post and the upcoming posts. In this post, we will be learning about different types of matrix multiplication in the numpy … If you did most of this on your own and compared to what I did, congratulations! \end{bmatrix} Subtract 2.4 * row 2 of A_M from row 3 of A_M    Subtract 2.4 * row 2 of I_M from row 3 of I_M, 7. Now we pick an example matrix from a Schaum's Outline Series book Theory and Problems of Matrices by Frank Aryes, Jr1. Python buffer object pointing to the start of the array’s data. I would even think it’s easier doing the method that we will use when doing it by hand than the ancient teaching of how to do it. The python matrix makes use of arrays, and the same can be implemented. The other sections perform preparations and checks. Plus, if you are a geek, knowing how to code the inversion of a matrix is a great right of passage! NumPy: Determinant of a Matrix. An inverse of a matrix is also known as a reciprocal matrix. The NumPy code is as follows. Python | Numpy matrix.sum() Last Updated: 20-05-2019 With the help of matrix.sum() method, we are able to find the sum of values in a matrix by using the same method. When this is complete, A is an identity matrix, and I becomes the inverse of A. Let’s go thru these steps in detail on a 3 x 3 matrix, with actual numbers. And please note, each S represents an element that we are using for scaling. In Linear Algebra, an identity matrix (or unit matrix) of size $n$ is an $n \times n$ square matrix with $1$'s along the main diagonal and $0$'s elsewhere. data. Subtract 0.472 * row 3 of A_M from row 2 of A_M    Subtract 0.472 * row 3 of I_M from row 2 of I_M. There are also some interesting Jupyter notebooks and .py files in the repo. Returns the (multiplicative) inverse of invertible self. Python statistics and matrices without numpy. Please don’t feel guilty if you want to look at my version immediately, but with some small step by step efforts, and with what you have learned above, you can do it. We start with the A and I matrices shown below. We will see two types of matrices in this chapter. You want to do this one element at a time for each column from left to right. Python Matrix. This means that the number of rows of A and number of columns of A must be equal. Why wouldn’t we just use numpy or scipy? Code faster with the Kite plugin for your code editor, featuring Line-of-Code Completions and cloudless processing. With numpy.linalg.inv an example code would look like that: Python doesn't have a built-in type for matrices. We will be using NumPy (a good tutorial here) and SciPy (a reference guide here). I’ve also saved the cells as MatrixInversion.py in the same repo. It is imported and implemented by LinearAlgebraPractice.py. Also, once an efficient method of matrix inversion is understood, you are ~ 80% of the way to having your own Least Squares Solver and a component to many other personal analysis modules to help you better understand how many of our great machine learning tools are built. The following line of code is used to create the Matrix. See the code below. dtype. Matrix Multiplication in NumPy is a python library used for scientific computing. Below is the output of the above script. Then come back and compare to what we’ve done here. In this post, we create a clustering algorithm class that uses the same principles as scipy, or sklearn, but without using sklearn or numpy or scipy. Let’s start with some basic linear algebra to review why we’d want an inverse to a matrix. If you go about it the way that you would program it, it is MUCH easier in my opinion. A_M and I_M , are initially the same, as A and I, respectively: A_M=\begin{bmatrix}5&3&1\\3&9&4\\1&3&5\end{bmatrix}\hspace{4em} I_M=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}, 1. You can verify the result using the numpy.allclose() function. In other words, for a matrix [[a,b], [c,d]], the determinant is computed as ‘ad-bc’. If a is a matrix object, then the return value is a matrix as well: >>> ainv = inv ( np . One way to “multiply by 1” in linear algebra is to use the identity matrix. 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I want to encourage you one last time to try to code this on your own. which is its inverse. AA^{-1} = A^{-1}A = I_{n} A^{-1}). Let’s first define some helper functions that will help with our work. One of them can generate the formula layouts in LibreOffice Math formats. I love numpy, pandas, sklearn, and all the great tools that the python data science community brings to us, but I have learned that the better I understand the “principles” of a thing, the better I know how to apply it. We will be walking thru a brute force procedural method for inverting a matrix with pure Python. Using the steps and methods that we just described, scale row 1 of both matrices by 1/5.0, 2. \end{bmatrix} I love numpy, pandas, sklearn, and all the great tools that the python data science community brings to us, but I have learned that the better I understand the “principles” of a thing, the better I know how to apply it. Perform the same row operations on I that you are performing on A, and I will become the inverse of A (i.e. 1 & 0 & 0 & 0\\ However, we can treat list of a list as a matrix. In this tutorial, we will make use of NumPy's numpy.linalg.inv() function to find the inverse of a square matrix. So how do we easily find A^{-1} in a way that’s ready for coding? $$If at this point you see enough to muscle through, go for it! The way that I was taught to inverse matrices, in the dark ages that is, was pure torture and hard to remember! 0 & 0 & 1$$ However, we may be using a closely related post on “solving a system of equations” where we bypass finding the inverse of A and use these same basic techniques to go straight to a solution for X. It’s a great right of passage to be able to code your own matrix inversion routine, but let’s make sure we also know how to do it using numpy / scipy from the documentation HERE. 1. The numpy.linalg.det() function calculates the determinant of the input matrix. 0 & 0 & 1 & 0\\ Success! left_hand_side_inverse = left_hand_side.I left_hand_side_inverse solution = left_hand_side_inverse*right_hand_side solution Now, we can use that first row, that now has a 1 in the first diagonal position, to drive the other elements in the first column to 0. Let’s get started with Matrices in Python. The first matrix in the above output is our input A matrix. Learning to work with Sparse matrix, a large matrix or 2d-array with a lot elements being zero, can be extremely handy. The second matrix is of course our inverse of A. \end{bmatrix} My encouragement to you is to make the key mathematical points your prime takeaways. I would not recommend that you use your own such tools UNLESS you are working with smaller problems, OR you are investigating some new approach that requires slight changes to your personal tool suite. Plus, tomorrow… To find out the solution you have to first find the inverse of the left-hand side matrix and multiply with the right side. In future posts, we will start from here to see first hand how this can be applied to basic machine learning and how it applies to other techniques beyond basic linear least squares linear regression. As per this if i need to calculate the entire matrix inverse it will take me 1779 days. A_M has morphed into an Identity matrix, and I_M has become the inverse of A. $$. I don’t recommend using this. Python provides a very easy method to calculate the inverse of a matrix. Inverse of an identity [I] matrix is an identity matrix [I]. Using this library, we can perform complex matrix operations like multiplication, dot product, multiplicative inverse, etc. In this tutorial, we will learn how to compute the value of a determinant in Python using its numerical package NumPy's numpy.linalg.det() function. This blog is about tools that add efficiency AND clarity. Since the resulting inverse matrix is a 3 \times 3 matrix, we use the numpy.eye() function to create an identity matrix. The reason is that I am using Numba to speed up the code, but numpy.linalg.inv is not supported, so I am wondering if I can invert a matrix with 'classic' Python code. So hang on! \begin{bmatrix} In fact, it is so easy that we will start with a 5×5 matrix to make it “clearer” when we get to the coding. PLEASE NOTE: The below gists may take some time to load. To calculate the inverse of a matrix in python, a solution is to use the linear … We will also go over how to use numpy /scipy to invert a matrix at the end of this post. in a single step. How to do gradient descent in python without numpy or scipy. \begin{bmatrix} We’ll do a detailed overview with numbers soon after this. Create a Python Matrix using the nested list data type; Create Python Matrix using Arrays from Python Numpy package; Create Python Matrix using a nested list data type. Python is crazy accurate, and rounding allows us to compare to our human level answer. Be sure to learn about Python lists before proceed this article. \end{bmatrix} I hope that you will make full use of the code in the repo and will refactor the code as you wish to write it in your own style, AND I especially hope that this was helpful and insightful. But it is remarkable that python can do such a task in so few lines of code. There will be many more exercises like this to come. Or, as one of my favorite mentors would commonly say, “It’s simple, it’s just not easy.” We’ll use python, to reduce the tedium, without losing any view to the insights of the method. I know that feeling you’re having, and it’s great! You can verify the result using the numpy.allclose() function. I encourage you to check them out and experiment with them. Base object if memory is from some other object. This is the last function in LinearAlgebraPurePython.py in the repo. Write a NumPy program compute the inverse of a given matrix. \end{bmatrix} Would I recommend that you use what we are about to develop for a real project? To work with Python Matrix, we need to import Python numpy module. Kite is a free autocomplete for Python developers. We will see at the end of this chapter that we can solve systems of linear equations by using the inverse matrix. Here, we are going to reverse an array in Python built with the NumPy module. Why wouldn’t we just use numpy or scipy? I_{2} = Creating a Matrix in NumPy; Matrix operations and examples; Slicing of Matrices; BONUS: Putting It All Together – Python Code to Solve a System of Linear Equations. Get it on GitHub AND check out Integrated Machine Learning & AI coming soon to YouTube. The only really painful thing about this method of inverting a matrix, is that, while it’s very simple, it’s a bit tedious and boring. An object to simplify the interaction of the array with the ctypes module. base. Since the resulting inverse matrix is a 3 \times 3 matrix, we use the numpy.eye() function to create an identity matrix. Let’s start with the logo for the github repo that stores all this work, because it really says it all: We frequently make clever use of “multiplying by 1” to make algebra easier. The main thing to learn to master is that once you understand mathematical principles as a series of small repetitive steps, you can code it from scratch and TRULY understand those mathematical principles deeply. Published by Thom Ives on November 1, 2018November 1, 2018. We then divide everything by, 1/determinant. It should be mentioned that we may obtain the inverse of a matrix using ge, by reducing the matrix $$A$$ to the identity, with the identity matrix as the augmented portion.$$. Let’s first introduce some helper functions to use in our notebook work. 0 & 0 & 0 & 1 $$My approach using numpy / scipy is below. The flip() method in the NumPy module reverses the order of a NumPy array and returns the NumPy array object. 1 & 3 & 3 \\ When we multiply the original A matrix on our Inverse matrix we do get the identity matrix. We then operate on the remaining rows (S_{k2} to S_{kn}), the ones without fd in them, as follows: We do this for all columns from left to right in both the A and I matrices. When you are ready to look at my code, go to the Jupyter notebook called MatrixInversion.ipynb, which can be obtained from the github repo for this project. >>> import numpy as np #load the Library$$. If the generated inverse matrix is correct, the output of the below line will be True. I want to be part of, or at least foster, those that will make the next generation tools. 0 & 1 \\ If you don’t use Jupyter notebooks, there are complementary .py files of each notebook. Data Scientist, PhD multi-physics engineer, and python loving geek living in the United States. It all looks good, but let’s perform a check of A \cdot IM = I. Matrix Operations: Creation of Matrix. An inverse of a square matrix $A$ of order $n$ is the matrix $A^{-1}$ of the same order, such that, their product results in an identity matrix $I_{n}$. In Python, the … Python’s SciPy library has a lot of options for creating, storing, and operating with Sparse matrices. \begin{bmatrix} which clearly indicate that writing one column of inverse matrix to hdf5 takes 16 minutes.


# How do you calculate Arcsin?

Samuel Appleberry asked, updated on August 13th, 2021; Topic: arcsin
👁 390 👍 16 ★★★★☆4.7

Using arcsine to find an angle First, calculate the sine of α by dividng the opposite side by the hypotenuse. This results in sin(α) = a / c = 52 / 60 = 0.8666. Use the inverse function with this outcome to calculate the angle α = arcsin(0.8666) = 60° (1.05 radians).

Add on, is Arctan and tan 1 the same?

The inverse of tangent is denoted as Arctangent or on a calculator it will appear as atan or tan-1. Note: this does NOT mean tangent raised to the negative one power. ... Sine, cosine, secant, tangent, cosecant and cotangent are all functions however, the inverses are only a function when given a restricted domain.

Forbye, how do you find the tangent angle on a calculator? Examples

• Step 1 The two sides we know are Opposite (300) and Adjacent (400).
• Step 2 SOHCAHTOA tells us we must use Tangent.
• Step 3 Calculate Opposite/Adjacent = 300/400 = 0.75.
• Step 4 Find the angle from your calculator using tan-1
• Ever, what is the formula for Arctan?

From this given quantity, 1.732 can be written as a function of tan. 60° = 60 \times \frac{\pi}{180} = 1.047 radians....Solution:

Dimensional Formula Of ResistivityInverse Matrix Formula

How do you do Arctan on TI 84?

Press the calculator's "shift," "2nd" or "function" key, and then press the "tan" key. Type the number whose arctan you want to find. For this example, type in the number "0.577." Press the "=" button.

### What is Arctan 1 in terms of pi?

Since. tan π/4 = tan 45º = 1. The arctangent of 1 is equal to the inverse tangent function of 1, which is equal to π/4 radians or 45 degrees: arctan 1 = tan-1 1 = π/4 rad = 45º

### What is the symbol for Arctan?

Principal valuesNameUsual notationDefinition
arctangenty = arctan(x)x = tan(y)
arccotangenty = arccot(x)x = cot(y)
arcsecanty = arcsec(x)x = sec(y)
arccosecanty = arccsc(x)x = csc(y)

### What is tan 1x?

The Function y = tan -1x = arctan x and its Graph: Since y = tan -1x is the inverse of the function y = tan x, the function y = tan -1x if and only if tan y = x.

### What is Arctan of infinity?

The arctangent is the inverse tangent function. The limit of arctangent of x when x is approaching infinity is equal to pi/2 radians or 90 degrees: The limit of arctangent of x when x is approaching minus infinity is equal to -pi/2 radians or -90 degrees: Arctan

### Where is tan equal to 1?

Basic idea: To find tan-1 1, we ask "what angle has tangent equal to 1?" The answer is 45°. As a result we say that tan-1 1 = 45°. In radians this is tan-1 1 = π/4. More: There are actually many angles that have tangent equal to 1.

### Can Arctan be negative?

The arctangent of a negative number is a negative first quadrant angle, sin-1(-) is in quadrant -I, a clockwise-angle of less than - /2. When you simplify an expression, be sure to use the Arcsine.

### Is Arctan the inverse of tan?

The arctan function is the inverse of the tangent function. It returns the angle whose tangent is a given number. Try this Drag any vertex of the triangle and see how the angle C is calculated using the arctan() function. Means: The angle whose tangent is 0.577 is 30 degrees.

### What is Arcsin on a calculator?

Arcsine definition The arcsine function is the inverse function of y = sin(x). arcsin(y) = sin-1(y) = x + 2kπ

### Does Arcsin cancel out sin?

The arcsine function is the inverse function for the sine function on the interval [ − π / 2 , π / 2 ] . So they "cancel" each other under composition of functions, as follows. The notation for inverse functions, f-1(x) is just that: notation, a shorthand way of writing the inverse of a function f.

### What is Arcsin equal to?

The arcsin function is the inverse of the sine function. It returns the angle whose sine is a given number. ... Means: The angle whose sin is 0.5 is 30 degrees. Use arcsin when you know the sine of an angle and want to know the actual angle.

### How do you use Arctan?

You can use the arctan to determine an angle measure when the side opposite and the side adjacent to the angle are known....When to Use Arctan
• sine = opposite / hypotenuse.
• cosine = adjacent / hypotenuse.
• tangent = opposite / adjacent.
•


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# 𝘶-substitution warmup

AP.CALC:
FUN‑6 (EU)
,
FUN‑6.D (LO)
,
FUN‑6.D.1 (EK)
Before diving into our practice exercise, gain some risk-free experience performing 𝘶-substitution.
Find each indefinite integral.

# Problem 1

integral, cosine, left parenthesis, x, squared, right parenthesis, 2, x, d, x, equals
plus, space, C

# Problem 2

integral, start fraction, 3, x, squared, divided by, left parenthesis, x, cubed, plus, 3, right parenthesis, squared, end fraction, d, x, equals
plus, space, C

# Problem 3

integral, e, start superscript, 4, x, end superscript, d, x, equals
plus, space, C

# Problem 4

integral, x, dot, square root of, start fraction, 1, divided by, 6, end fraction, x, squared, plus, 1, end square root, d, x, equals
plus, space, C

## Want to join the conversation?

• Well, in the problem #2 what happens to the "square power" of "u" when you substitue back the equation x^3 + 3 please? I did not catch what happened to get the given answer. Thank you. :)
• 1/(u^2) == u^(-2). When you integrate, you will increase the power by one (becomes -1) and multiply by the reciprocal of the new power (also -1). Your integral is -1*(u^-1) ==(-1/u).
This problem is tricky because of the properties of exponents, just try rewriting the factors to understand where the exponent went to.
• =∫1/u^2 du shoudn't be = ln(|u^2|)?
• You are reversing the power rule so the answer is -1/u +C. However, integral(1/u) =ln(|u|) + C.
• In problem 2, why the negative?

1/u^2 * du
-1/u
(1 vote)
• Reverse power rule.
∫ u^(-2) du = 1/(-2 + 1) * u^(-2 + 1) + C = -1/u + C.
• I was given the problem:
 ∫ sin³(x)cos(x)dx = ? + C

I entered (sin(x)^4)/4 the first time & was marked wrong. Then I tried entering the exact solution given which is impossible to do as far as I know on my mobile phone: (1/4)sin(x)^4. There is no process or command available to enter it as (1/4)sin^4(x). I'm assuming that is the reason I'm being marked wrong. Is it possible to enter the exponent before a trig function's parentheses? Please advise.
(1 vote)
• sin(x)^4/4 is correct however the exponent is in incorrect spot. The issue with sin(x)^4/4 is it could mistaken with sin(x^4)/4.

You need type sin^4(x)/4 or alternatively (sin(x))^4/4.
• in problem 4 why is xdx= 3?
(1 vote)
• if du = 1/3xdx, you just multiply both sides by 3, and you get 3du = xdx
• how to us u-substitution for the integral of the function 4x/the square root of (1 - x to the 4th)
(1 vote)
• ∫ 4x / sqrt(1 - x^4) dx =
2 ∫ 2x / sqrt(1 - (x^2)^2) dx

Let u = x^2, du = 2x dx, then
2 ∫ 2x dx / sqrt(1 - (x^2)^2) =
2 ∫ du / sqrt(1 - u^2) =
2 arcsin(u) + C =
2 arcsin(x^2) + C.

Hope that I helped.
• In the first question, is it right to take cos(x^2) as u?
• If you choose cos(x^2) as your u, your du ends up being -sin(x^2)*2x*dx. You could rearrange the equation as du/-sin(x^2) = 2x*dx and replace the 2x*dx in the original equation accordingly, but you're still left with the x^2 inside the sine-function. For the u-substitution to work, you need to replace all variables with u and du, so you're not getting far with choosing u = cos(x^2). If you choose, as you should, u = x^2 and your du = 2*x*dx, you'll get int(cos(u)*du) and that's pretty straight-forward to integrate.
• Actually the problem 3 can already be solved by using the integration formula of e.
(1 vote)
• In order for most of these to work, the constant multiple rule must apply to integrals in exactly the same way that it applies to derivatives. Is that assumption correct?
(1 vote)
• Yes the constant multiple rule applies for both derivatives and integrals
(1 vote)
• =∫1/u^2 du = −1/u +C

Anyone could explain why appeared a negative signal on −1/u +C?
(1 vote)
• Think about it this way. Let's say we have the function y=1/x. Now let's think about the graph in the 1st quadrant. The slope is always negative. Therefore, the derivative of the curve at any point on it in the first quadrant should be a negative number. d/dx (1/x) < 0 for x > 0 . This should hopefully provide some intuition for the negative sign.

For the rigorous proof, try finding d/dx (-1/x).
(1 vote)


# Every function $f$ can be factored into $f = i \circ s$, with $i$ injective and $s$ surjective

As in the title, if $$X$$ and $$Y$$ are two arbitrary sets and $$f:X \to Y$$, my proof was by taking $$x_1 \sim x_2 \iff f(x_1) = f(x_2),$$ $$s: X \to X/\sim$$ to be the canonical surjection of $$X$$ into the quotient set of $$X$$ wrt $$\sim$$, i.e. $$s(x) = \{z \in X: f(z) = f(x)\}$$ and $$i: X/\sim \to Y$$ to be the map defined by $$i(Z) = f(z)$$, for any $$z$$ in $$Z$$. Since all $$z$$ in an equivalence class are mapped to the same element of $$Y$$, $$i$$ is well defined.

Is the above correct?

The proposed answer however was another one, namely $$s: X \to f(X)$$ and $$i: f(X) \to Y$$ defined as $$s(x) = f(x)$$ and $$i(w) = w$$.

If my solution was correct, which choice is more "canonical"?

Still, if my solution is correct, to what extent can we say that the decomposition into injective and surjective is "unique"? I would say that $$X / \sim$$ and $$f(X)$$ are "isomorphic" because of something like the first isomorphism theorem in linear algebra..

• I have to say I like more your solution, the proposed answer is just restricting the range of the function, and then using the Inclusion map. Although your answer is basically the same: (I use $s_2$ for the proposed $s$) for each $x∈X$ we have $s_2^{-1}(x)=f^{-1}(x)=[x]_\sim$. So the difference is that while you took the preimages, the proposed solution took the value. The 2 solutions describe the same thing
– ℋolo
Sep 30 '19 at 8:58
• I was a bit puzzled because I did not find my solution anywhere in the web..
– Tom
Sep 30 '19 at 9:02
• But eg wiki en.wikipedia.org/wiki/Bijection,_injection_and_surjection mentions the "proposed" answer and not "mine"..
– Tom
Sep 30 '19 at 9:03
• I am not sure why Wiki chose that way, but, like I said, both ways describe the same thing: going from $x$ to something that describe uniquely $f(x)$, and from there to $f(x)$ itself(note that $f(x)$ is also a way to describe uniquely $f(x)$)(Also note that $|X/\sim|=|f(X)|$ as well as that for every algebra with underline set $f(X)$, there is canonical isomorphic algebra with underline set of $X/\sim$, the canonical bijection between the 2 is the isomorphism)(algebra over a set $A$, is the set $A$ is operators)
– ℋolo
Sep 30 '19 at 12:47

Your answer is correct, assuming by $$i(Z) = f(z)$$, you mean that $$i$$ maps the equivalence class of $$z$$ under $$\sim$$ to $$f(z)$$ (you may want to make this more clear).

One thing to note is that your answer and the answer provided in the solutions are very similar. We can naturally associate the equivalence classes of $$\sim$$ uniquely to elements of the range of $$f$$. If we make this identification, then your answers are really the same.

This lends credence to the idea that this decomposition might be somewhat "unique" in some sense. Let's set up the problem. Suppose $$f : X \to Y$$ satisfies $$f = i_1 \circ s_1 = i_2 \circ s_2$$ where $$s_k : X \to Z_k$$ are surjective and $$i_k : Z_k \to Y$$ are injective, for $$k = 1, 2$$. We can actually show that there is a bijection $$\phi : Z_1 \to Z_2$$ such that $$s_2 = \phi \circ s_1$$ and $$i_1 = i_2 \circ \phi$$.

By this, I mean that there is some $$\phi$$ which provides us a rule for identifying elements of $$Z_1$$ and $$Z_2$$ in such a way that, after identification, the decompositions $$i_1 \circ s_1$$ and $$i_2 \circ s_2$$ become the same decomposition.

So, let's construct this $$\phi$$. As $$i_k$$ is injective, there must exist some left inverses $$j_k : Y \to Z_k$$ (i.e. $$j_k \circ i_k : Z_k \to Z_k$$ is the identity map on $$Z_k$$). Similarly, as $$s_k$$ is surjective, there must exist some right inverses $$t_k : Z_k \to X$$ (i.e. $$s_k \circ t_k : Z_k \to Z_k$$ is the identity map). Define $$\phi = j_2 \circ i_1 : Z_1 \to Z_2.$$ Then $$\phi \circ (s_1 \circ t_2) = j_2 \circ (i_1 \circ s_1) \circ t_2 = j_2 \circ f \circ t_2 = (j_2 \circ i_2) \circ (s_2 \circ t_2),$$ which is the identity on $$Z_2$$. We need to show that $$(s_1 \circ t_2) \circ \phi = s_1 \circ t_2 \circ j_2 \circ i_1$$ is the identity on $$Z_1$$. This is a little less straight forward. \begin{align*} (s_1 \circ t_2) \circ \phi &= (j_1 \circ i_1) \circ (s_1 \circ t_2) \circ \phi \circ (s_1 \circ t_1) \\ &= j_1 \circ (i_1 \circ s_1) \circ t_2 \circ j_2 \circ (i_1 \circ s_1) \circ t_1 \\ &= j_1 \circ f \circ t_2 \circ j_2 \circ f \circ t_1 \\ &= j_1 \circ (i_2 \circ s_2) \circ t_2 \circ j_2 \circ (i_2 \circ s_2) \circ t_1 \\ &= j_1 \circ i_2 \circ s_2 \circ t_1 \\ &= (j_1 \circ i_1) \circ (s_1 \circ t_1), \end{align*} which is the identity on $$Z_1$$, as required. Therefore, $$\phi$$ is invertible with inverse $$s_1 \circ t_2$$.

Obviously, by construction, we have $$i_2 \circ \phi = (i_2 \circ j_2) \circ i_1 = i_1.$$ Our proof also came with an expression for $$\phi^{-1}$$ which we can also use: $$\phi^{-1} \circ s_2 = s_1 \circ (t_2 \circ s_2) = s_1 \implies s_2 = \phi \circ s_1$$ as required.

• Perfectly clear - thank you!
– Tom
Sep 30 '19 at 12:12

Your approach is fine. The relation $$\sim$$ is an equivalence relation on $$X$$ and the quotient set $$X/\sim$$ gives rise to the surjective mapping $$f_1:X\rightarrow X/\sim : x\mapsto \bar x$$ where $$\bar x=\{x'\in X\mid x\sim x'\}$$ is the equivalence class of $$x$$.

Moreover, the mapping $$f_2:X/\equiv \rightarrow f(X): \bar x\mapsto f(x)$$ is injective. This mapping is well-defined, since if $$x\sim x'$$, then $$f(x)=f(x')$$.

Saying ''isomorphic'' is a bit too much, since there is underlying algebraic structure (such as vector spaces).

• in the second line "gives rise to the injective mapping".. you meant surjective?
– Tom
Sep 30 '19 at 8:56
• Try \sim which gives $\sim$ Sep 30 '19 at 8:56


# Prove $\gcd(nn!, n!+1)=1$

For any $n \in \mathbb{N}$, find $\gcd(n!+1,(n+1)!+1)$. First come up with a conjecture, then prove it.

By testing some values, it seems like $\gcd(n!+1,(n+1)!+1) = 1$

I can simplify what's given to me to $\gcd(nn!, n!+1)=1$ but I can't find out how to get it into the form I want it. Can anybody look at what I'm doing and give me any guidance?

$\gcd(n!+1,(n+1)!+1) = 1 \implies \gcd(n!+1,(n+1)n!+1) = 1 \implies \gcd(n!+1,nn!+n!+1) = 1 \implies \gcd(nn!, n!+1) = 1$

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I changed numerous instances of \mathrm{gcd} in this question to \gcd. It's a standard operator name. –  Michael Hardy Apr 11 '12 at 2:42
Thanks for modifying my question to use \gcd and making me realize that command exists. That will help me in the future! Also, thanks to everybody who answered; you all have really helped me! –  Brandon Amos Apr 11 '12 at 11:34
.....and just in case anyone wonders: I just posted 5\gcd(a,b) and 5\mathrm{gcd}(a,b) within a "displayed" $\TeX$ setting in the "answer" box below. Try it and you'll see that they don't both look the same! (One of them has proper spacing between "$5$" and "$\gcd$".) –  Michael Hardy Apr 11 '12 at 21:00

Here is a proof that does not use induction but rather the key property of gcd: $(a,b) = (a-b,b) = (a-kb,b)$ for all $k$.

Take $a=nn!$, $b=n!+1$, $k=n$ and conclude that $(nn!, n!+1)=(nn!-n(n!+1),n!+1)=(-n,n!+1)=1$ since any divisor of $n$ is a divisor of $n!$.

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Or apply the key property again: $(-n, n!+1) = (((n-1)!\times-n) + n! +1 , n! +1) = (1, n!+1) = 1$ –  Aryabhata Apr 11 '12 at 1:43
@Aryabhata This is probably a really simple question, but how did you get from $(-n, n!+1)$ to $(((n-1)!\cdot-n)+n!+1, n!+1)$ from this "key property"? I see how you used it to add $n!+1$ to the first term, but where did the $(n-1)!$ multiplier come from? –  Brandon Amos Apr 11 '12 at 16:37
@user28554: $(a,b) = (ka+b, a)$. The last term should be $-n$, instead of $n! + 1$. We chose $k = (n-1)!$. –  Aryabhata Apr 11 '12 at 16:41
@Aryabhata Ahh, I see now. Thanks! –  Brandon Amos Apr 11 '12 at 16:48
@user28554: You are welcome! –  Aryabhata Apr 11 '12 at 16:49

You don’t need to use induction; you just need to prove the statement in the title. Suppose that $p$ is a prime factor of $nn!$; can $p$ divide $n!+1$?

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Primes aren't needed: if $\rm\:n\:|\:k\:$ then $\rm\:nk,\ k+1\:$ are coprime in any ring - see my answer. –  Bill Dubuque Apr 11 '12 at 3:22
@Bill: I didn’t say that they were. I offered what I consider an easy approach to the problem. In general I write primarily for the questioner, not for possible future readers. –  Brian M. Scott Apr 11 '12 at 3:27

Hint $\$ Put $\rm\:k = n!\$ in: $\rm\ n\:|\:k\:\Rightarrow\:(1+k,nk)= 1\$ by $\rm\: (1-k)\:(1+k) + (k/n)\: nk = 1.\ \$ QED

More generally, note that the above Bezout equation implies $\rm\: 1+k\:$ and $\rm\:nk\:$ are coprime in every ring. Alternatively, with $\rm\:m = k+1,\:$ one can employ Euclid's Lemma (EL) as follows:

$$\rm\:(m,k) =1,\ n\:|\:k\:\Rightarrow (m,n) = 1\:\Rightarrow\:(m,nk)=1\ \ by\ \ EL$$ i.e. $\rm\: mod\ m\!:\ x\:$ is a unit (invertible) iff $\rm\:(x,m) = 1.$ But units are closed under products, divisors, i.e. they form a saturated monoid. So, since $\rm\:k\:$ is a unit so is its divisor $\rm\:n\:$ and so is the product $\rm\:nk.$

-

Note that $$(n-1)!\cdot \underline{n n!} - (n!-1)\underline{(n!+1)} = 1;$$ this Bézout's identity shows that the two underlined quantities must be relatively prime (anything that divides them both must divide the right-hand side). The related identity $$(n-1)! \underline{((n+1)!+1)} - (n!+(n-1)!-1)\underline{(n!+1)} = 1$$ similarly proves that the greatest common divisor of these two underlined terms equals 1.

Of course, discovering these identities in the first place is best done by using the Euclidean algorithm, as in lhf's answer.

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This is precisely what I wrote 5 hours prior, except it explicitly replaces $\rm\:k\:$ by $\rm\: n!\:$ in my Bezout equation. Doing so decreases the generality of the proof, and obscures the key (unit group) structure. –  Bill Dubuque Apr 11 '12 at 20:06
Well then, I won't claim priority. But some might find that my solution is more accessible to the OP than yours. –  Greg Martin Apr 12 '12 at 6:51
My concern is not priority but, rather, pedagogy. I explicitly abstracted $n!$ to any integer $k$ divisible by $n$ in order to make clearer the innate governing multiplicative structure. To succeed in elementary number theory it is essential to learn how to recognize such structure. If students are encouraged to follow shortcuts circumventing such pedagogical routes then they may completely miss the key ideas, and never see the forest for the trees. –  Bill Dubuque Apr 12 '12 at 14:15

I dont know i am right or wrong but i can do this example in following way,\ Let $$\gcd(n\cdot n!,n!+1)=d$$ $$\therefore d\mid n\cdot n!\ ,\ d\mid n!+1$$ $$\Rightarrow d\mid n,\ d\mid n!,\ d\mid n!+1$$ $$\Rightarrow d\mid n!+1-n!$$ Thus $d\mid 1\ \Rightarrow d=1.$

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Why does $d \mid n$ ? Perhaps you mean a prime $d$. –  lhf Apr 11 '12 at 1:47
$8 | 4 \times 4!$, but $8$ does not divide $4$. –  Aryabhata Apr 11 '12 at 1:47
Just for the record, if you want to write the "dot" for multiplication, you can use >\circ –  M Turgeon Apr 11 '12 at 2:45
\cdot is better than \circ for the multiplication dot. –  Greg Martin Apr 11 '12 at 6:33


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Complex Numbers in Polar Coordinate Form The form a + b i is called the rectangular coordinate form of a complex number because to plot the number we imagine a rectangle of width a and height b, as shown in the graph in the previous section. To enter the complex number in polar form you enter mcisa, where m is the modulus and a is the argument of number. Use Integers Or Fractions For Any Numbers In The Expression) Question Viewer The Quotient In Polar Form Is … A complex number in Polar Form must be entered, in Alcula’s scientific calculator, using the cis operator. Find all five values of the following expression, giving your answers in Cartesian form: (-2+5j)^(1/5) [6 marks] Any ideas? Polar to Exponential Form Conversion Calculator. This online calculator will help you to convert rectangular form of complex number to polar and exponential form. You can enter complex numbers in the standard (rectangular) or in the polar form. Complex numbers are written in exponential form .The multiplications, divisions and power of complex numbers in exponential form are explained through examples and reinforced through questions with detailed solutions.. Exponential Form of Complex Numbers A complex number in standard form $$z = a + ib$$ is written in polar form as $z = r (\cos(\theta)+ i \sin(\theta))$ … syms a a=8-7j [theta, r]cart2pol(8, 7) for the polar for but thats it. Polar to Rectangular Online Calculator. Free exponential equation calculator - solve exponential equations step-by-step This website uses cookies to ensure you get the best experience. 9B 10345 ищу Прошивка POLAR 48LTV3101 шасси T. $\begingroup$ Yes, once you calculate the $\tan^{-1}$, you should look at the polar… Given a complex number in rectangular form expressed as $$z=x+yi$$, we use the same conversion formulas as we do to write the number in trigonometric form: At the following model,the arithmetic operations on complex numbers can be easily managed using the Calculators. Example 3.6056cis0.588 . You can use the following trick to allow you to enter angles directly in degrees. Here, both m and n are real numbers, while i is the imaginary number. The succeeding examples illustrate the conversion of the standard complex number z = a + bi to its equivalent polar form (r, θ). The polar form of a complex number expresses a number in terms of an angle $\theta$ and its distance from the origin $r ... Use the rectangular to polar feature on the graphing calculator to change [latex]5+5i$ to polar form. … For complex numbers in rectangular form, the other mode settings don’t much matter. A calculator to calculate the equivalent impedance of a resistor, a capacitor and and inductor in parallel. Polar forms of numbers can be converted into their exponential equivalents relatively easily. The polar form of a complex number expresses a number in terms of an angle $$\theta$$ and its distance from the origin $$r$$. Visualizing complex number multiplication. In this section, we will first deal with the polar form of complex numbers. Dividing complex numbers: polar & exponential form. Key Concepts. Practice: Multiply & divide complex numbers in polar form. I was wondering if anybody knows a way of having matlab convert a complex number in either polar or cartesian form into exponential form and then actually display the answer in the form ' … Trigonometric Form of Complex Numbers Calculator. The calculator gives the impedance as a complex numbers in standard form , its modulus and argument which may be used to write the impedance in exponential and polar forms. With the calculator in DEGREE mode this will then display 240 e ^(i 75) corresponding to the polar form number (240 75). ; The absolute value of a complex number is the same as its magnitude. We can convert the complex number into trigonometric form by finding the modulus and argument of the complex number. Label the x-axis as the real axis and the y-axis as the imaginary axis. Complex modulus Rectangular form of complex number to polar and exponential form converter Show all online calculators We learnt that the exponential (polar) form of a complex number is a very powerful and compact way to solve complex number problems. Type An Exact Answer Using * As Needed. I was having a lot of problems tackling questions based on exponential form calculator but ever since I started using software, math has been really easy for me. Convert the complex number 8-7j into exponential and polar form. Instructions. For example, you can convert complex number from algebraic to trigonometric representation form or from exponential back to algebraic, ect. Topics covered are arithmetic, conjugate, modulus, polar and exponential form, powers and roots. Polar Display Mode “Polar form” means that the complex number is expressed as an absolute value or modulus r and an angle or argument θ. By using this website, you agree to our Cookie Policy. The amplitude of the polar form is the direct amplitude of the ... www.learningaboutelectronics.com These calculators will display complex numbers in exponential form with the angle in degrees, but will not allow you to enter the angle in degrees. asked Dec 25, 2012 in PRECALCULUS by dkinz Apprentice. (This is spoken as “r at angle θ ”.) The calculated values will also be displayed in standard and polar forms among other modular forms. Polar form, where r - absolute value of complex number: is a distance between point 0 and complex point on the complex plane, and φ is an angle between positive real axis and the complex vector (argument). Using this online calculator, you will receive a detailed step-by-step solution to your problem, which will help you understand the algorithm how to convert rectangular form of complex number to polar and exponential form. Below is an interactive calculator that allows you to easily convert complex numbers in polar form to rectangular form, and vice-versa. This is a quick primer on the topic of complex numbers. All Functions Operators + Another way of writing the polar form of the number is using it’s exponential form: me^(ia) . ... 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# Probability chord of bigger circle intersects smaller circle

You are given two concentric circles $$C_1$$ and $$C_2$$ of radius $$r$$ and $$r/2$$ respectively. What is the probability that a randomly chosen chord of $$C_1$$ will intersect $$C_2$$?

Answer: $$1/2, 1/3$$ or $$1/4$$

The first method I used (gives 1/4):

The midpoint of any chord uniquely determines it, as line joining center to midpoint is always perpendicular to chord. So instead of choosing a chord, let's choose points instead that shall be the midpoints of their respective chords. Any point inside inner circle will be a chord that intersects it too, and any point outside will never cut inner circle. Thus probability should be area of inner circle/area of outer circle= $$1/4$$

But then I did it by another method and got another answer (1/3):

Choose a point on bigger circle. Now you can get all chords from $$0$$ to $$π$$ angle. The one which intersect smaller one must lie between tangents to smaller circle from bigger one from that point . We can easily obtain angle between tangents as $$π/3$$ by some trigonometry. We can do same for every other point so answer is $$\frac{π/3}{π}=1/3$$

First I was confused that I was getting two answers. But then I checked the given answer and I saw they were accepting multiple answers.

So I thought about how there could be multiple possible probabilities and only possible reason seemed to be the boundary conditions as I had included diameters in chords in second solution but not in first. However even though there are infinite diameters I still don't think probability should be affected this much as we have infinite points.

Can someone give clarity on this? In particular, what exact conditions are included by which solution, and how will we get the third given answer (1/2)? As far as I can find there are $$3$$ boundary conditions we have to consider carefully-

1. If diameters are included or not
2. Degenerate "chords" that are $$r$$ distance from center, ie they are actually points on the circumference
3. If tangents are included in intersection or not

I believe the discrepancy results from the fact that the problem does not outline how the chords are randomly selected. As you have shown, there are multiple ways to randomly select a chord, and the two ways you have described have different probability distributions.

This can be seen by comparing the PDF of the length of the selected chord in both cases. In your first solution, the probability you select an arbitrary length is uniform. However, in your second solution, shorter chords are more likely to be selected than longer chords. Hence, there is a clear difference in the probability distributions in both selection processes.

As a result, this makes it possible to get different answers. Since the problem failed to elaborate on how chords were chosen, the problem writers were forced to accept multiple answers.

Moreover, the 3 boundary conditions you have outlines will not affect the probability because they have infinitesimal impact on the total probability.

You can get $$\frac{1}{2}$$ by modifying how you select the midpoint of the chord from your first solution. Instead of uniformly selecting a point from the interior of the larger circle, uniformly choose the distance of the midpoint from the center from the range $$[0,r]$$ and then uniformly choose the angle the midpoint makes with some arbitrary point/ray (e.g. define a positive x-axis) from the range $$[0,2\pi)$$. The chord will intersect the smaller circle whenever the chosen distance of the midpoint from the center is $$<\frac{r}{2}$$, which happens exactly $$\boxed{\frac{1}{2}}$$ of the time.

• Thanks for your answer. I also thought the same thing about the boundary conditions but only reason I had my doubts was because we have infinite diameters and infinite tangential chords So they would contribute a lot more than just one chord Jul 27, 2021 at 6:46
• I don't know how to rigorously explain it, but it's kind of like how if you remove a curve of infinite points from a region, the area of the region is still the same even though you removed an infinite number of "things" from the region. Jul 27, 2021 at 6:49
• yeah I think a field of mathematics called "measure theory" is used so quite far beyond high school math Jul 27, 2021 at 6:54


# What is the expected number of coin tosses needed to obtain a head?

Due to my recent misunderstandings regarding the 'expected value' concept I decided to post this question. Although I have easily found the answer on the internet I haven't managed to fully understand it.

I understood that the formula for the expected value is:

$$E(x) = x_1p_1 +x_2*p_2 +...+x_np_n$$

The x's are the possible value that the random variable can take and the p's are the probabilites that this certain value is taken.

So, if I get a head on the first try, then $p_1 = \frac{1}{2} , x_1 = 1$ If I get a head on the second try, then $p_2 = \frac{1}{4} , x_2 = 2$

And then, I woudl have that:

$$E(x) = \frac{1}{2}1+ \frac{1}{4}2 +...$$

So my reasoning led me to an inifnite sum which I don't think I can't evaluate it that easy. In the 'standard' solution of this problem, the expected value is found in a reccurisve manner. So the case in which the head doesn't appear in the first toss is treated reccursively. I haven't understood that step.

My questions are: is my judgement correct? How about that reccursion step? Could somebody explain it to me?

• For fun, I would say 2. =) – Vincent Aug 27 '14 at 15:46
• Yes, I knew that too. :D I just didn't know how we found that answer – Bardo Aug 27 '14 at 15:48

Let $X$ be the number of tosses, and let $e=E(X)$. It is clear that $e$ is finite.

We might get a head on the first toss. This happens with probability $\frac{1}{2}$, and in that case $X=1$.

Or else we might get a tail on the first toss. In that case, we have used up $1$ toss, and we are "starting all over again." So in that case the expected number of additional tosses is $e$. More formally, the conditional expectation of $X$ given that the first toss is a tail is $1+e$.

It follows (Law of Total Expectation) that $$e=(1)\cdot\frac{1}{2}+(1+e)\cdot\frac{1}{2}.$$

This is a linear equation in $e$. Solve.

Remark: The "infinite series" approach gives $$E(X)=1\cdot\frac{1}{2}+2\cdot\frac{1}{2^2}+3\cdot\frac{1}{2^3}+\cdots.$$ This series, and related ones, has been summed repeatedly on MSE.

• Conditioning like I did is a totally standard technique in the calculation of expectation. – André Nicolas Aug 27 '14 at 16:01
• Well, perhaps you can use the series approach for general $p$, and the equation $e=p+(1+e)(1-p)$, and see that they give the same answer. – André Nicolas Aug 27 '14 at 16:09
• Given that the first toss is a tail, $E(X)=1+e$. – André Nicolas Aug 27 '14 at 16:10
• It is not true that with probability $1/2$ you will need $1+e$. What is true is that given the first is tail the total expected number of tosses is $1+e$. – André Nicolas Aug 27 '14 at 16:24
• @Bardo Note that in your previous question in my solution we use essentially the same trick (in a slightly more complex situation). – Aahz Aug 27 '14 at 16:34

Your approach is perfectly fine. The probability of getting the first head in the $n$th trial is $\frac{1}{2^n}$, so we have $$\mathbb{E}(x) = \sum_{ n \geq 1} \frac{n}{2^n}.$$ This infinite sum can be calculated in the following way: first note that $\frac{1}{1-x} = \sum_{n \geq 0} x^n$ for $|x|<1$. Differentiating both sides yields $$\frac{1}{(1-x)^2} = \sum_{n \geq 0} n x^{n-1} = \sum_{n \geq 1} n x^{n-1} = \frac1x \sum_{n \geq 1} n x^n.$$ Pluggin in $x = \frac12$ yields $4 = 2 \sum_{n \geq 1} \frac{n}{2^n} = 2 \mathbb{E}(x)$, or $\mathbb{E}(x) = 2$.

The recursive solution goes, I think, as follows: let $\mathbb{E}(x)$ be the expected number of trials needed. Then the expected number of trials needed after the first trial, given that it was not heads, is also $\mathbb{E}(x)$. In other words, the expected (total) number of trials is $\mathbb{E}(x)+1$ in that case. This gives the equation $$\mathbb{E}(x) = \frac12 + \frac12(\mathbb{E}(x)+1)$$ which gives the same answer $\mathbb{E}(x) = 2$.

• It's good to know that my approach wasn't incorrect, it gives me a little confidence. I am no thinking about that reccursion step. – Bardo Aug 27 '14 at 15:54

To make ends meet... You have been explained by several users that, looking at the toss process itself, one sees that the expectation $E(X)$, that you know is $E(X)=S$, with $$S=\sum_{n\geqslant1}\frac{n}{2^n},$$ solves the relation $$E(X)=1+\frac12E(X).$$ It happens that one can also show directly that $$S=1+\frac12S,$$ this relation following from a shift of indexes. To do so, note that $S=R+T$ with $$R=\sum_{n\geqslant1}\frac{1}{2^n},\qquad T=\sum_{n\geqslant1}\frac{n-1}{2^n},$$ hence, using the change of variable $n=k+1$, $$T=\sum_{k\geqslant0}\frac{k}{2^{k+1}}=\frac12\sum_{k\geqslant0}\frac{k}{2^k}=\frac12\sum_{k\geqslant1}\frac{k}{2^k}=\frac12S,$$ hence the proof would be complete if one knew that $$R=1.$$ To show this, use the same trick once again, that is, note that $$R=\frac12+\sum_{n\geqslant2}\frac{1}{2^n}=\frac12+\sum_{k\geqslant1}\frac{1}{2^{k+1}}=\frac12+\frac12\sum_{k\geqslant1}\frac{1}{2^{k}}==\frac12+\frac12R,$$ hence the proof that $S=2$ is complete.

• Yes, I got it...I knew beforehand the trick of shifting indexes, so I totatly understood your argument..But I am still struggling to undesrtande the simpler argument..Thank you very much! – Bardo Aug 27 '14 at 18:29


# Prove that A Real Root Exists in [-1, 1]

#### anemone

##### MHB POTW Director
Staff member
Given $$\displaystyle f(x)=5tx^4+sx^3+3rx^2+qx+p$$ for $f(x)\in R$. If $r+t=-p$, prove that there is a real root for $f(x)=0$ in $[-1,1]$.

#### Ackbach

##### Indicium Physicus
Staff member
Some ideas:

We rewrite the polynomial as
$$f(x)=5tx^4+sx^3+3rx^2+qx-r-t=0,$$
where we are setting it equal to zero. We evaluate $f(1)$ and $f(-1)$:
\begin{align*}
f(1)&=5t+s+3r+q-r-t=4t+2r+s+q\\
f(-1)&=5t-s+3r-q-r-t=4t+2r-s-q.
\end{align*}
Idea: if $f(1)\cdot f(-1)<0$, then by the Intermediate Value Theorem, we would have shown there is a root in $[-1,1]$. Now
$$f(1)\cdot f(-1)=(4t+2r)^{2}-(s+q)^{2}.$$
Not seeing where to go with this. There's nothing stopping $s=q=0$, with $4t+2r\not=0$, in which case I have not proved what I want to prove.

#### zzephod

##### Well-known member
Given $$\displaystyle f(x)=5tx^4+sx^3+3rx^2+qx+p$$ for $f(x)\in R$. If $r+t=-p$, prove that there is a real root for $f(x)=0$ in $[-1,1]$.
Consider the polynomial $$\displaystyle p(x)=tx^5+(s/4)x^4+rx^3+(q/2)x^2+px-(s/4+q/2)$$.

Then $$\displaystyle p(1)=p(-1)=0$$ and so $$\displaystyle p$$ has an extremum in $$\displaystyle (-1,1)$$, so $$\displaystyle p'(x)$$ has a root in $$\displaystyle (-1,1)$$ ...

.

Last edited:

#### anemone

##### MHB POTW Director
Staff member
Thanks to both, Ackbach and zzephod for participating...

Consider the polynomial $$\displaystyle p(x)=tx^5+(s/4)x^4+rx^3+(q/2)x^2+px-(s/4+q/2)$$.

Then $$\displaystyle p(1)=p(-1)=0$$ and so $$\displaystyle p$$ has an extremum in $$\displaystyle [-1,1]$$, so $$\displaystyle p'(x)$$ has a root in $$\displaystyle [-1,1]$$ ...

.
WoW!!! What an elegant way to approach this problem! Well done, zzephod!

And there is another quite straightforward and beautiful way to tackle it as well...therefore I'll wait for the inputs from other members for now...

#### anemone

##### MHB POTW Director
Staff member
Another method proposed by other to solve this challenge problem is by using the integration method:

Hint:

$$\displaystyle \int_{-1}^1 p(x) dx=0$$