Need derviation of some formulas! Help!

1. Euler's Formula - cosx+isinx=eixcos x + isin x = {e}^{ix}

2. Relation between Mean, Mode, Median - Mode=3Median2MeanMode = 3*Median - 2*Mean

3. Poisson's Distribution - P(X=r)=eλ.λrr!P(X=r) = {e}^{-\lambda}.\frac{{\lambda}^{r}}{r!} where λ(mean)=np(n=No.oftrials,p=probabilityofsuccess)\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- sinα+sin(α+β)+......sin(α+(n1)β)=sinnβ2sinβ2.sin(α+(n1)β2)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 sinnβ2sinβ2.cos(α+(n1)β2)\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. :)

Note by Kartik Sharma
6 years, 6 months ago

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To prove (i)(i) we have z=cos(x)+isin(x)z=cos(x)+isin(x) where zz is a function of xx

Differentiating zz with respect to xx we get :


Since i2=1{i}^{2}=-1 hence we have :


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


Seperating variables and integrating both sides we have :

dzz=idx \int { \frac { dz }{ z } } =\int { idx }


When x=0,z=1x=0,z=1 hence we get C=0C=0

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

z=eix\Rightarrow z={e}^{ix}

Finally :


Ronak Agarwal - 6 years, 6 months ago

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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?

Kartik Sharma - 6 years, 6 months ago

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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.

Ronak Agarwal - 6 years, 5 months ago

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

sneha sharma - 6 years, 5 months ago

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

Deepanshu Gupta - 6 years, 5 months ago

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To prove 11 Euler's Formula, simple use the expansion of exe^x , i.e. ex=1+x+x22!+x33!+x44!+x55!+.....e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+.....

    eix=1+x.i+x2.i22!+x3.i33!+x4.i44!+x5.i55!+.....\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!}+.....

=1+x.ix22!i.x33!+x44!+i.x55!+........ \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!}+........

=(1x22!+x44!x66!+.....)+i.(xx33!+x55!x77!+......) \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!}+......)

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

Because we know that cosx=1x22!+x44!x66!+.....cosx=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+..... sinx=xx33!+x55!x77!+......sinx=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+......

Hence proved that cosx+isinx=eix\boxed{cosx+isinx=e^{ix}}

Sandeep Bhardwaj - 6 years, 6 months ago

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

Kartik Sharma - 6 years, 6 months ago

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

[i=1naibi]2[i=1nai2]2[i=1nbi2]2 [\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 a1b1=a2b2=.....=anbn \frac{a_{1}}{b_{1}} = \frac{a_{2}}{b_{2}} = .....=\frac{a_{n}}{b_{n}}

LetA=[i=1nai2]2,B=[i=1nbi2]2,C=[i=1naibi]2 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 C2AB C^{2} \leq AB

If B = 0 then bi=0b_{i} = 0 for i = 1 , 2, . , n .Hence C = 0 and is true .Therefore it is sufficient to consider the case B0 B\neq 0 This implies B > 0 .

0i=1n(BaiCbi)2=i=1n(B2ai22BCaibi+C2bi2 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(ABC2) = B(AB - C^{2})

Sinc eB> 0 we get ABC2>0 AB - C^{2} > 0 which is the inequality C2ABC^{2} \leq AB. Moreover , equality holds if i=1n(BaiCbi)2=0\sum_{i = 1}^{n} (Ba_{i} - Cb_{i})^{2} = 0

This is equivalent to

aibi=CB \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 c2+d2=(a2+b2)3c^{2} + d^{2} = (a^{2} + b^{2})^{3},

prove that a3c+b3d \frac{a^{3}}{c} + \frac{b^{3}}{d} if ad = bc .

Using above inequality we get ,

(a2+b2)2=(a3cac+b3dbd)2(a^{2} + b^{2})^{2} = (\sqrt{\frac{a^{3}}{c}}\sqrt{ac} + \sqrt{\frac{b^{3}}{d}}\sqrt{bd} )^{2}

(a3c+b3d)(ac+bd) \leq ( \frac{a^{3}}{c} + \frac{b^{3}}{d})(ac + bd) , where the equality holds

Thus (a3c+b3d)(ac+bd)(a2+b2)3 \leq ( \frac{a^{3}}{c} + \frac{b^{3}}{d})(ac + bd) \geq ( a^{2} + b^{2})^{3}

=(a2+b2)1/2(a2+b2)3/2 = ( a^{2} + b^{2})^{1/2} ( a^{2} + b^{2})^{3/2}

=(a2+b2)1/2(c2+d2)1/2ac+bd = ( 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 ac=bd \frac{a}{c} = \frac{b}{d}

Combining both we get a3c+b3d1\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 , α=a,β=b \alpha =a , \beta = b

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

=cos(ab/2)cos(a+b/2)+cos(a+b/2)cos(a+3b/2)....+cos[(a+(n3/2)b)]cos[a+(n1/2)b] = 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(ab/2)cos(a+(n1/2)b] = cos(a - b/2) - cos(a + (n - 1/2)b]

=2sin[a+((n1)/2)b]sin(nb/2) = 2sin[a +((n-1)/2)b]sin(nb/2)

X=sin[a+n12b]sinnb2sinb2 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

U Z - 6 years, 6 months ago

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

Kartik Sharma - 6 years, 6 months ago

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I am giving proof of 6th6th formula : S(let)=sinα+sin(α+β)+......sin(α+(n1)β)=sinnβ2sinβ2.sin(α+(n1)β2)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 2.sinA.sinB=cos(AB)cos(A+B)\boxed{2.sinA.sinB=cos(A-B)-cos(A+B)}     2.sinα.sinβ2=cos(αβ2)cos(α+β2)\implies 2.sin\alpha.sin\frac{\beta}{2}=cos(\alpha-\frac{\beta}{2})-cos(\alpha+\frac{\beta}{2})

    2.sin(α+β).sinβ2=cos(α+β2)cos(α+3β2)\implies 2.sin(\alpha+\beta).sin\frac{\beta}{2}=cos(\alpha+\frac{\beta}{2})-cos(\alpha+\frac{3\beta}{2})

    2.sin(α+2β).sinβ2=cos(α+2ββ2)cos(α+2.β+β2)\implies 2.sin(\alpha+2\beta).sin\frac{\beta}{2}=cos(\alpha+2\beta-\frac{\beta}{2})-cos(\alpha+2.\beta+\frac{\beta}{2})






    2.sin(α+(n1)β).sinβ2=cos(α+(n1)ββ2)cos(α+(n1).β+β2)\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.sinβ2×S=2sin(α+(n12).β).sinn.β2\implies 2.sin\frac{\beta}{2} \times S=2sin(\alpha+(\frac{n-1}{2}).\beta).sin\frac{n.\beta}{2}     S=sinα+sin(α+β)+......sin(α+(n1)β)=sinnβ2sinβ2.sin(α+(n1)β2)\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 cosα+cos(α+β)+......+cos(α+(n1)β)=sinnβ2sinβ2.cos(α+(n1)β2)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 2.cosA.cosB=sin(A+B)sin(AB)\boxed{2.cosA.cosB=sin(A+B)-sin(A-B)}

enjoy !

Sandeep Bhardwaj - 6 years, 6 months ago

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Thanks for sharing! You are quite good at Latex

Kartik Sharma - 6 years, 6 months ago

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

Kartik Sharma - 6 years, 6 months ago

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

Kartik Sharma - 6 years, 6 months ago

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

Kartik Sharma - 6 years, 6 months ago

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

Saurabh Chauhan - 6 years, 5 months ago

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ART OF IGNORING! (I also know it)

Kartik Sharma - 6 years, 5 months ago

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HUH! I bet you're gonna read this.

Saurabh Chauhan - 6 years, 5 months ago

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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.

Sharky Kesa - 6 years, 6 months ago

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

Kartik Sharma - 6 years, 6 months ago

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Maybe, there should be wiki pages for these.

Sharky Kesa - 6 years, 6 months ago

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@Sharky Kesa YEAH! Agree!

Kartik Sharma - 6 years, 6 months ago

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