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

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!

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Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## 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.

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This is great!

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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.

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Can you tell your friend's name? Is he/she on brilliant?

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He is called Isaac Lee. Not very active though.

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

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There's an even easier method, and that is using the approximation

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

When \(\theta\) is very small.

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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.

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Invite Parth too :)

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\[\int_{-1}^{1}\frac{d}{dx}\left(\frac{1.5}{2^{\frac{1}{x}} + 1}\right)dx\]

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

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

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

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Here's my entry:

\[1=\lim_{x\to0}\sum_{n=1}^{\infty}\sum_{r=0}^{\infty}\frac{(-1)^r}{n+r}\dbinom{n+r}{r}x^{n+r-1}\]

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Nice!

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My entry: \[\left( \frac { \int _{ 0 }^{ \infty }{ \frac { a^{ 3 } }{ { e }^{ a }-1 } } da }{ \int _{ 0 }^{ \infty }{ { b }^{ 3 }{ e }^{ -b } } db } \right) .\frac { 90 }{ { \pi }^{ 4 } } =1\]

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Lol!! Numerator is Riemann zeta function and denominator is Gamma function.

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Yes. U r right!

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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!

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Here's my entry, \[\displaystyle{ \lim_{x \to 0}} \log_{tan^{2} x} \left(tan^{2} 2x \right) = 1\]

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

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

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

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

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

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

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

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

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

\(=\displaystyle\frac11\)

\(=1\)

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

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

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

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

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

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

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

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

\(=\displaystyle1\)

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my late entry @Nihar Mahajan \[\prod_{k=2}^\infty \left(\zeta(k)\sum_{n=1}^\infty \dfrac{\mu(n)}{n^k}\right) =1\]

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

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@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! :)

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Here's my entry

\[\large{\frac { 7 }{ 26 } \left[ \frac { d }{ dx } { \left[ \tan ^{ -1 }{ \left( \frac { 2{ x }^{ 4 } }{ 3{ x }^{ 2 }-7x+1 } \right) } +\sin ^{ -1 }{ \left( \frac { 26x+4 }{ 7 } \right) } \right] }_{ x=0 } \right] =1}\]

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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".

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I just think that this would be fun if it implies the rule (but I don't think it is): \[\displaystyle \ce{^16_8A} \ce{^20_10B} = 1\]

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Is the equation in your entry somehow related to radioactivity? Because I see a nice correlation between A and B (according to your notation and how it's written). B reduces to A after occurrence of one \(\alpha\)-decay (followed by a \(\gamma\)-decay, as usual to stabilize energy levels).

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.

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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.

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

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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 = 1

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Lol... I like that

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Ah.. ok lol

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Here is my entry \[-(\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{\ldots}}}})(\large \lim_{n\to\infty}(\frac {n^2(n sin(\frac {1}{n})-2)}{1+3n^2}))=1\]

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

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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 :)

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I have sent you some others as message on fb

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This is actually a bit forced and not really original. The nested radical is actually a problem by Ramanujan which is well known to equal \(3\). So, you just multiply it with \(\frac {-1}3\) to get \(-1\), the \(\frac {-1}3\) coming very trivially from the limit. The negative sign at the very left results in getting \((-1)\times (-1)=1\).

Seeing that your profile picture is a picture of Ramanujan, this post looks quite apt, haha.

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Lol I have not copied exactly the same I have made edits in it :-p

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