**1. Euler's Formula** - \(cos x + isin x = {e}^{ix}\)

**2. Relation between Mean, Mode, Median** - \(Mode = 3*Median - 2*Mean\)

**3. Poisson's Distribution** - \(P(X=r) = {e}^{-\lambda}.\frac{{\lambda}^{r}}{r!}\) where \(\lambda(mean) = np(n = No. of trials, p = probability of success)\)

**4. RMS - Arithmetico-Geometric and Harmonic Mean inequality**

**5. Cauchy-Shwartz's inequality**

**6. Sum of sines with angles in AP**- \(sin\alpha + sin(\alpha + \beta) + ...... sin(\alpha + (n-1)\beta) = \frac{sin\frac{n\beta}{2}}{sin\frac{\beta}{2}}. sin(\alpha + (n-1)\frac{\beta}{2})\)

and similarly

**Sum of cosines with angles in AP** - which gives \(\frac{sin\frac{n\beta}{2}}{sin\frac{\beta}{2}}. cos(\alpha + (n-1)\frac{\beta}{2})\)

THANKS a lot in anticipation! And I believe that you have the answers and you must also be in need of some of the formulas. You are free to share them here to clear all your doubts. :)

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TopNewestTo prove \((i)\) we have \(z=cos(x)+isin(x)\) where \(z\) is a function of \(x\)

Differentiating \(z\) with respect to \(x\) we get :

\(\frac{dz}{dx}=-sin(x)+icos(x)\)

Since \({i}^{2}=-1\) hence we have :

\(\frac{dz}{dx}={i}^{2}sin(x)+icos(x)=i(cos(x)+isin(x))\)

Since \(cos(x)+isin(x)=z\) hence we have :

\(\frac{dz}{dx}=iz\)

Seperating variables and integrating both sides we have :

\( \int { \frac { dz }{ z } } =\int { idx } \)

\(ln(z)=ix+C\)

When \(x=0,z=1\) hence we get \(C=0\)

\(\Rightarrow ln(z)=ix\)

\(\Rightarrow z={e}^{ix}\)

Finally :

\(cos(x)+isin(x)=e^{ix}\)

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I Love Your way To prove This equation ! Hats Off!! I 'am Amazed How Can any person down-votes it ??

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Seriously is my proof that much bad, I wonder about the reason for 4 downvotes.

Sorry if you didn't like the proof.

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I think when x=0 , then at the same time z cannot be equal to one.

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Wow! Amazing! This made me become your fan! Thank You! Try other proofs too! And yes can you give me an explanation on why Taylor series makes sense?

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To prove \(1\) Euler's Formula, simple use the expansion of \(e^x\) , i.e. \[e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+.....\]

\(\implies e^{ix}=1+x.i+\frac{x^2.i^2}{2!}+\frac{x^3.i^3}{3!}+\frac{x^4.i^4}{4!}+\frac{x^5.i^5}{5!}+.....\)

\( \quad \quad \quad \quad =1+x.i-\frac{x^2}{2!}-i.\frac{x^3}{3!}+\frac{x^4}{4!}+i.\frac{x^5}{5!}+........\)

\( \quad \quad \quad \quad =(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+.....)+i.(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+......)\)

\(\quad \quad \quad \quad =cosx +i.sinx\)

Because we know that\[cosx=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+.....\] \[sinx=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+......\]Hence proved that \[\boxed{cosx+isinx=e^{ix}}\]

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Can you tell me why does Taylor series make sense? How to find them? *if you don't mind

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5) and 6) are simple, i don't know about rest

\( [\sum_{i = 1}^{n} a_{i}b_{i}]^{2} \leq [\sum_{i = 1}^{n} a_{i}^{2}]^{2}[\sum_{i = 1}^{n} b_{i}^{2}]^{2}\)

and the equality holds if \( \frac{a_{1}}{b_{1}} = \frac{a_{2}}{b_{2}} = .....=\frac{a_{n}}{b_{n}}\)

Let\( A = [\sum_{i = 1}^{n} a_{i}^{2}]^{2} , B =[\sum_{i = 1}^{n} b_{i}^{2}]^{2} , C = [\sum_{i = 1}^{n} a_{i}b_{i}]^{2}\)

then \( C^{2} \leq AB\)

If B = 0 then \(b_{i} = 0\) for i = 1 , 2, . , n .Hence C = 0 and is true .Therefore it is sufficient to consider the case \( B\neq 0 \) This implies B > 0 .

\( 0 \leq \sum_{i = 1}^{n} (Ba_{i} - Cb_{i})^{2} = \sum_{i = 1}^{n} ( B^{2}a_{i}^{2} - 2BCa_{i}b_{i} + C^{2}b_{i}^{2}\)

distributing the sum,

\( = B(AB - C^{2})\)

Sinc eB> 0 we get \( AB - C^{2} > 0\) which is the inequality \(C^{2} \leq AB\). Moreover , equality holds if \(\sum_{i = 1}^{n} (Ba_{i} - Cb_{i})^{2} = 0\)

This is equivalent to

\( \frac{a_{i}}{b_{i}} = \frac{C}{B}\) for i = 1 , 2 , .... , n

Simple application and beautiful application

If a, b , c and d are positive real numbers such that \(c^{2} + d^{2} = (a^{2} + b^{2})^{3}\),

prove that \( \frac{a^{3}}{c} + \frac{b^{3}}{d}\) if ad = bc .

Using above inequality we get ,

\((a^{2} + b^{2})^{2} = (\sqrt{\frac{a^{3}}{c}}\sqrt{ac} + \sqrt{\frac{b^{3}}{d}}\sqrt{bd} )^{2}\)

\( \leq ( \frac{a^{3}}{c} + \frac{b^{3}}{d})(ac + bd)\) , where the equality holds

Thus \( \leq ( \frac{a^{3}}{c} + \frac{b^{3}}{d})(ac + bd) \geq ( a^{2} + b^{2})^{3}\)

\( = ( a^{2} + b^{2})^{1/2} ( a^{2} + b^{2})^{3/2}\)

\( = ( a^{2} + b^{2})^{1/2}( c^{2} + d^{2})^{1/2} \geq ac + bd\)

again using the inequality , the equality holds in the last step ( above) if \( \frac{a}{c} = \frac{b}{d}\)

Combining both we get \(\frac{a^{3}}{c} + \frac{b^{3}}{d} \geq 1 \) and the equality holds if ad = bc.

6) One of my favorite

let s be equal to the given series , \( \alpha =a , \beta = b\)

\( 2Xsin(b/2) = 2sinasin(b/2) + 2sin(a + b)sin(b/2) ............\)

\( = cos( a - b/2) - cos( a + b/2) + cos(a + b/2) - cos( a + 3b/2) .... + cos[(a + (n - 3/2)b)] - cos[a + (n - 1/2)b]\)

\( = cos(a - b/2) - cos(a + (n - 1/2)b]\)

\( = 2sin[a +((n-1)/2)b]sin(nb/2)\)

\( X = \huge{\boxed{\frac{sin[a + \frac{n - 1}{2}b]sin\frac{nb}{2}}{sin\frac{b}{2}}}}\)

Similarly for the cosine series it can be done , try it ,its interesting and enjoyable.

Lovely Application of this can be found in THIS QUESTION

Enjoy@Kartik Sharma

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Nice Thanks! I solved that problem! Maybe I would give my solution too!

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Comment deleted Nov 12, 2014

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Thanks! Nice! Try other ones too!

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I am giving proof of \(6th\) formula : \[S(let)=sin\alpha + sin(\alpha + \beta) + ...... sin(\alpha + (n-1)\beta) = \frac{sin\frac{n\beta}{2}}{sin\frac{\beta}{2}}. sin(\alpha + (n-1)\frac{\beta}{2})\]

We know that \[\boxed{2.sinA.sinB=cos(A-B)-cos(A+B)}\] \(\implies 2.sin\alpha.sin\frac{\beta}{2}=cos(\alpha-\frac{\beta}{2})-cos(\alpha+\frac{\beta}{2})\)

\(\implies 2.sin(\alpha+\beta).sin\frac{\beta}{2}=cos(\alpha+\frac{\beta}{2})-cos(\alpha+\frac{3\beta}{2})\)

\(\implies 2.sin(\alpha+2\beta).sin\frac{\beta}{2}=cos(\alpha+2\beta-\frac{\beta}{2})-cos(\alpha+2.\beta+\frac{\beta}{2})\)

\(\bullet\)

\(\bullet\)

\(\bullet\)

\(\bullet\)

\(\bullet\)

\(\implies 2.sin(\alpha+(n-1)\beta).sin\frac{\beta}{2}=cos(\alpha+(n-1)\beta-\frac{\beta}{2})-cos(\alpha+(n-1).\beta+\frac{\beta}{2})\)

By adding, we get\(2.sin\frac{\beta}{2}[sin\alpha+sin(\alpha+\beta)+.....+sin(\alpha+(n-1)\beta)]=cos(\frac{\alpha-\beta}{2})-cos(\alpha+(\frac{n-1}{2}).\beta)\)

\(\implies 2.sin\frac{\beta}{2} \times S=2sin(\alpha+(\frac{n-1}{2}).\beta).sin\frac{n.\beta}{2}\) \[\implies S=sin\alpha + sin(\alpha + \beta) + ...... sin(\alpha + (n-1)\beta) = \frac{sin\frac{n\beta}{2}}{sin\frac{\beta}{2}}. sin(\alpha + (n-1)\frac{\beta}{2})\]

In the similar way,we can prove\[cos\alpha + cos(\alpha + \beta) + ......+cos(\alpha + (n-1)\beta) = \frac{sin\frac{n\beta}{2}}{sin\frac{\beta}{2}}. cos(\alpha + (n-1)\frac{\beta}{2})\]To prove this, we will use the formula \(\boxed{2.cosA.cosB=sin(A+B)-sin(A-B)}\)

enjoy !

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Thanks for sharing! You are quite good at Latex

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7. Proof for the integrating factor- \(I = {e}^{\int{p(x)} dx}\) where \(\frac{dy}{dx} + p(x)y = Q\) is the differential equation.Log in to reply

If you have some more of which you need derivations of, you can share them here!

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@Calvin Lin @Michael Mendrin @Sandeep Bhardwaj @Sanjeet Raria @Sharky Kesa @Krishna Ar @Satvik Golechha @Finn Hulse @Daniel Liu @megh choksi

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and more.. (I can't write names of all of 'em).

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You sir, can sometimes be so overt that it's covert.

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ART OF IGNORING! (I also know it)

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HUH! I bet you're gonna read this.

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AoPS (Art of Problem Solving) has the answer to your question. That's where I learnt a lot of my formulae.

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Yeah! But it would be better if I get them here.

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Maybe, there should be wiki pages for these.

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