# $\color{#D61F06}{\text{Unity}} \ \color{#624F41}{\text{Is}} \ \color{#3D99F6}{\text{Strength}} \ \color{#20A900}{\text{Contest}}$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad$ 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! Note by Nihar Mahajan
5 years, 8 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

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MarkdownAppears as
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MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

## Comments

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Top Newest

Here'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.

- 5 years, 8 months ago

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This is great!

- 5 years, 8 months ago

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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.

- 5 years, 8 months ago

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

- 5 years, 8 months ago

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He is called Isaac Lee. Not very active though.

- 5 years, 8 months ago

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

- 5 years, 8 months ago

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

$\sin { \theta = } \theta$

When $\theta$ is very small.

- 5 years, 8 months ago

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

- 5 years, 8 months ago

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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.

- 5 years, 8 months ago

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Invite Parth too :)

• Kritarth Lohomi : In response to Nihar Mahajan: He is busy..

• Aditya Kumar In response to Kritarth Lohomi: Nice:) Well, this doesn't deserve downvotes.

• Julian Poon In response to Aditya Kumar: Maybe downvotes should not be counted

• Aditya Kumar In response to Julian Poon: Yes but still... it doesn't look good.

• Nihar Mahajan In response to Julian Poon: They are not counted.

• Anik Mandal In response to Kritarth Lohomi: Cool!

• Siddharth Bhatt In response to Kritarth Lohomi: Nice!

- 5 years, 8 months ago

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Here's my entry:

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

- 5 years, 8 months ago

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Nice!

- 5 years, 8 months ago

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

- 5 years, 8 months ago

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

- 5 years, 8 months ago

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

- 5 years, 8 months ago

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$\color{#FFFFFF}{\text{=1 That's a wise one :P}}$

- 5 years, 8 months ago

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

- 5 years, 8 months ago

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

- 5 years, 8 months ago

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Yes. U r right!

- 5 years, 8 months ago

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

- 5 years, 8 months ago

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

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

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

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

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

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

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

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

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

$=\displaystyle\frac11$

$=1$

- 5 years, 8 months ago

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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?

- 5 years, 8 months ago

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Right.

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

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

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

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

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

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

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

$=\displaystyle1$

- 5 years, 8 months ago

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Yeah that was my solution too.Good job.

- 5 years, 8 months ago

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Thanks!

- 5 years, 8 months ago

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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!

- 5 years, 8 months ago

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

- 5 years, 4 months ago

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

- 5 years, 4 months ago

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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! :)

- 5 years, 8 months ago

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

- 5 years, 8 months ago

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

- 5 years, 8 months ago

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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.

- 5 years, 8 months ago

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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.

- 5 years, 8 months ago

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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

- 5 years, 8 months ago

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Lol... I like that

- 5 years, 8 months ago

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Ah.. ok lol

- 5 years, 8 months ago

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Here's my entry

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

- 5 years, 7 months ago

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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".

- 5 years, 7 months ago

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

- 5 years, 8 months ago

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

- 5 years, 8 months ago

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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.

- 5 years, 8 months ago

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

- 5 years, 8 months ago

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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 :)

- 5 years, 8 months ago

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I have sent you some others as message on fb

- 5 years, 8 months ago

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