\(\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\)

Hello every \(-e^{i\pi}\) ,

I am holding a contest named "Unity Is Strength".The aim of this contest is to motivate the community to create **beautiful mathematical expressions** that equal 1 i.e unity.

**What you have to do?**

You have to generate an expression that equals \(1\) and post it on this note in the following format:

"Here's my entry: <expression> = 1"

**Rules And Regulations:**

The expression must be

**original**and it**must equal one**.Every member is allowed only

**1**entry. You can update your entry by editing your comment, or deleting it and adding a new one.Submission of famous identities

**is prohibited**. For example : Proposing Euler's identity in this form: \(-e^{i\pi}=1\) is not allowed.Just modifying the known identities/equations to make them equal 1

**is prohibited**. For example: We all know in \(\Delta ABC , \sum \tan A = \prod \tan A\) , so submitting it in this form: \(\dfrac{\sum \tan A}{\prod \tan A}=1\) is not allowed.Trivial proposals like \(2-1=1\) or \(\dfrac{a\times 1}{a}=1\) or putting some numbers to force the expression to be equal to \(1\) are

**not expected**.Irrelevant/abusive comments would be deleted.

Please follow the rules and regulations.

**Details about the contest:**

This contest starts from the time this note is posted and ends on 18 October 2015 at 3:30 pm GMT. The entries after this time won't be accepted.

The entry with the maximum value of \(upvotes\) will be declared the winning entry. The member who posted that entry will be the Winner of the Unity Is Strength Contest. So please

**Do Upvote**all the entries you like, and even if you downvote any entries ,it**won't affect**the score of that/those individuals.The results will be announced at any day in the next week of the last date of submission.

The contest starts now. Please like,reshare this note so that it will reach most of the brilliantians!

Best Of Luck!

## Comments

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TopNewestHere's my entry :

\[\Re\left(\frac{2}{1-e^{i\theta}}\right)=1~~\forall~\theta\in\Bbb R\setminus\{x\mid x=2\pi n~,~n\in\Bbb Z\}\]

where \(\Re(z)\) denotes the real part of \(z\in \Bbb C\).

This is something that I noticed a few months ago while solving a few trigonometry problems on Brilliant. I'm not sure whether this complies with Rule #5 or not but I'm posting it anyway. – Prasun Biswas · 1 year, 11 months ago

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– Julian Poon · 1 year, 11 months ago

This is great!Log in to reply

Here's my submission.

\[\lim _{ n\rightarrow \infty }{ \frac { \sin \left( 2\times 10^{ -n }° \right) -\sin \left( 2\times 10^{ -(n+1) }° \right) }{ \pi \times 10^{ -(n+2) } } } =1\]

This was not found by me, it was found by my friend about a year ago. – Julian Poon · 1 year, 11 months ago

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– Nihar Mahajan · 1 year, 11 months ago

Can you tell your friend's name? Is he/she on brilliant?Log in to reply

Isaac Lee. Not very active though. – Julian Poon · 1 year, 11 months ago

He is calledLog in to reply

– Surya Prakash · 1 year, 11 months ago

It's very easy though it seems to be tough. It can be achieved by using L'Hospital Rule.Log in to reply

\[\sin { \theta = } \theta \]

When \(\theta\) is very small. – Julian Poon · 1 year, 11 months ago

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– Surya Prakash · 1 year, 11 months ago

yes!! it's very easy. You are right. But we should convert degrees to radians.Log in to reply

Here is my entry

\[\dfrac{\left(\displaystyle\int_0^1\ln\left(\frac{1}{x}\right)^{49} \dfrac{x^3}{1-2x^4} \text{dx} \right)\times{2^{101}}}{49! \text{Li}_{50}}=1\]

\(\text{Li}_{x} \) is the polylogarithm function. – Kritarth Lohomi · 1 year, 11 months ago

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\[\int_{-1}^{1}\frac{d}{dx}\left(\frac{1.5}{2^{\frac{1}{x}} + 1}\right)dx\] – Andrew Ellinor · 1 year, 11 months ago

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– Aditya Agarwal · 1 year, 11 months ago

Isn't it an improper integral? It is discontinuous at \(x=0\) .Log in to reply

– Andrew Ellinor · 1 year, 11 months ago

Yes, it is an improper integral. But that doesn't mean it can't be evaluated.Log in to reply

– Nihar Mahajan · 1 year, 11 months ago

\(\color{white}{\text{=1 That's a wise one :P}}\)Log in to reply

Here's my entry:

\[1=\lim_{x\to0}\sum_{n=1}^{\infty}\sum_{r=0}^{\infty}\frac{(-1)^r}{n+r}\dbinom{n+r}{r}x^{n+r-1}\] – Kenny Lau · 1 year, 11 months ago

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– Julian Poon · 1 year, 11 months ago

Nice!Log in to reply

My entry: \[\left( \frac { \int _{ 0 }^{ \infty }{ \frac { a^{ 3 } }{ { e }^{ a }-1 } } da }{ \int _{ 0 }^{ \infty }{ { b }^{ 3 }{ e }^{ -b } } db } \right) .\frac { 90 }{ { \pi }^{ 4 } } =1\] – Aditya Kumar · 1 year, 11 months ago

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– Surya Prakash · 1 year, 11 months ago

Lol!! Numerator is Riemann zeta function and denominator is Gamma function.Log in to reply

– Aditya Kumar · 1 year, 11 months ago

Yes. U r right!Log in to reply

My submission: \[\Large\int^1_0\frac{\ln(\sin^{-1}[\int^3_e\Gamma(x)dx]+e)}{e^{\delta(\int^\pi_0[\cot x]dx)}}dx\]

where \([.]\) denotes the floor function/ greatest integer function, \(\delta(.)\) denotes the Delta Function and \(\Gamma(.)\) denotes the Gamma Function. All the best! – Aditya Agarwal · 1 year, 11 months ago

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Here's my entry, \[\displaystyle{ \lim_{x \to 0}} \log_{tan^{2} x} \left(tan^{2} 2x \right) = 1\] – Athiyaman Nallathambi · 1 year, 11 months ago

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\(=\displaystyle\lim_{x\to0}\frac{\log(\tan^22x)}{\log{\tan^2x}}\)

\(=\displaystyle\lim_{t\to0}\frac{\log\left(\frac{2t}{1-t^2}\right)^2}{\log{t^2}}\)

\(=\displaystyle\lim_{t\to0}\frac{\log\left(\frac{2t}{1-t^2}\right)}{\log{t}}\)

\(=\displaystyle\lim_{t\to0}\frac{\frac{2(1-t^2)-2t(-2t)}{(1-t^2)^2}/\left(\frac{2t}{1-t^2}\right)}{1/t}\)

\(=\displaystyle\lim_{t\to0}t\frac{2+2t^2}{(1-t^2)^2}/\left(\frac{2t}{1-t^2}\right)\)

\(=\displaystyle\lim_{t\to0}t\frac{2+2t^2}{1-t^2}/(2t)\)

\(=\displaystyle\lim_{t\to0}\frac{2+2t^2}{2(1-t^2)}\)

\(=\displaystyle\lim_{t\to0}\frac{1+t^2}{1-t^2}\)

\(=\displaystyle\frac11\)

\(=1\) – Kenny Lau · 1 year, 11 months ago

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– Athiyaman Nallathambi · 1 year, 11 months ago

Well yeah you could do that. But wouldnt it be easier to apply L'Hopital rule right after applying change of base theorem?Log in to reply

\(\quad\displaystyle\lim_{x\to0}\log_{\tan^2x}(\tan^22x)\)

\(=\displaystyle\lim_{x\to0}\frac{\log(\tan^22x)}{\log{\tan^2x}}\)

\(=\displaystyle\lim_{x\to0}\frac{\log(\tan2x)}{\log{\tan x}}\)

\(=\displaystyle\lim_{x\to0}\frac{2\sec^2x/(\tan2x)}{\sec^2x/(\tan x)}\)

\(=\displaystyle\lim_{x\to0}\frac{2/(\tan2x)}{1/(\tan x)}\)

\(=\displaystyle\lim_{x\to0}\frac{2\tan x}{\tan2x}\)

\(=\displaystyle\lim_{x\to0}\frac{2\sec^2x}{2\sec^2x}\)

\(=\displaystyle1\) – Kenny Lau · 1 year, 11 months ago

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– Athiyaman Nallathambi · 1 year, 11 months ago

Yeah that was my solution too.Good job.Log in to reply

– Kenny Lau · 1 year, 11 months ago

Thanks!Log in to reply

my late entry @Nihar Mahajan \[\prod_{k=2}^\infty \left(\zeta(k)\sum_{n=1}^\infty \dfrac{\mu(n)}{n^k}\right) =1\] – Aareyan Manzoor · 1 year, 8 months ago

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– Nihar Mahajan · 1 year, 8 months ago

No problem. Though I am unable to actually evaluate the LHS , the equation looks interesting!Log in to reply

@Sravanth Chebrolu Sorry to say , but I have confirmed that your proposal is not original with some people. Your proposal is not accepted. I am deleting it. Don't worry , you can post another , you have whole week in hand! :) – Nihar Mahajan · 1 year, 11 months ago

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Here's my entry

\[\large{\frac { 7 }{ 26 } \left[ \frac { d }{ dx } { \left[ \tan ^{ -1 }{ \left( \frac { 2{ x }^{ 4 } }{ 3{ x }^{ 2 }-7x+1 } \right) } +\sin ^{ -1 }{ \left( \frac { 26x+4 }{ 7 } \right) } \right] }_{ x=0 } \right] =1}\] – Lakshya Sinha · 1 year, 11 months ago

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– Nihar Mahajan · 1 year, 11 months ago

The contest is already over and I am in process of posting the results. Don't delete this , I would mention this as a "late entry".Log in to reply

I just think that this would be fun if it implies the rule (but I don't think it is): \[\displaystyle \ce{^16_8A} \ce{^20_10B} = 1\] – Jimmy PrevailLone · 1 year, 11 months ago

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Then again, looking at the numbers, this might be something else because B doesn't satisfy the "n/p ratio > 1.5" criteria for radioactivity to occur. – Prasun Biswas · 1 year, 11 months ago

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– Nihar Mahajan · 1 year, 11 months ago

Even I don't understand your notation.Wait , are you using the notation for atomic mass , atomic number of an element? If so , then it is not allowed.Log in to reply

– Julian Poon · 1 year, 11 months ago

I don't really understand your notation. Mind if you elaborate on the notation used? Thanks a lot!Log in to reply

– Jimmy PrevailLone · 1 year, 11 months ago

This was just a silly equation which won't make any sense here. So A is Oxygen, denoted by O and B is Neon, denoted by Ne. When we connect both, we get ONe = 1Log in to reply

– Nehemiah Osei · 1 year, 11 months ago

Lol... I like thatLog in to reply

– Julian Poon · 1 year, 11 months ago

Ah.. ok lolLog in to reply

Here is my entry \[-(\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{\ldots}}}})(\large \lim_{n\to\infty}(\frac {n^2(n sin(\frac {1}{n})-2)}{1+3n^2}))=1\] – Atul Shivam · 1 year, 11 months ago

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– Kenny Lau · 1 year, 11 months ago

You mean \(n\to\infty\) right? And use "\left(...\right)" and also "\displaystyle" to make things neater :)Log in to reply

– Nihar Mahajan · 1 year, 11 months ago

Sorry , but your submission includes the famous Ramanujan's nested radicals which is not allowed as stated in rule number 3. Don't delete this , let it stay :)Log in to reply

– Atul Shivam · 1 year, 11 months ago

I have sent you some others as message on fbLog in to reply

Seeing that your profile picture is a picture of Ramanujan, this post looks quite apt, haha. – Prasun Biswas · 1 year, 11 months ago

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– Atul Shivam · 1 year, 11 months ago

Lol I have not copied exactly the same I have made edits in it :-pLog in to reply