Newton sums in easier notation

an+bn+cn=(a+b+c)(an1+bn1+cn1)(ab+bc+ac)(an2+bn2+cn2)+abc(an3+bn3+cn3)\LARGE a^n+b^n+c^n=(a+b+c)(a^{n-1}+b^{n-1}+c^{n-1})-(ab+bc+ac)(a^{n-2}+b^{n-2}+c^{n-2})+abc(a^{n-3}+b^{n-3}+c^{n-3})

an+bn=(a+b)(an1+bn1)ab(an2+bn2)\huge a^n+b^n=(a+b)(a^{n-1}+b^{n-1})-ab(a^{n-2}+b^{n-2})

an+1an=(a+1a)(an1+1an1)(an2+1an2)\huge a^n+\frac{1}{a^n}=(a+\frac{1}{a})(a^{n-1}+\frac{1}{a^{n-1}})-(a^{n-2}+\frac{1}{a^{n-2}})

For exercises: Try this problem

Try this very nice set by Aditya.

And this

Note by Sanjeet Raria
5 years, 1 month ago

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Can we use newton sums for more then three unknowns also . . I mean can we do this:- (a4+b4+c4+d4)=(a+b+c+d)(a3+b3+c3+d3)(ab+bc+bd+ac+cd+ad)(a2+b2+c2+c2)+abcd(a+b+c+d)(a^4+b^4+c^4+d^4)=(a+b+c+d)(a^3+b^3+c^3+d^3)-(ab+bc+bd+ac+cd+ad)(a^2+b^2+c^2+c^2)+abcd(a+b+c+d)

Aman Sharma - 5 years ago

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(an+bn+cn+dn)=(a+b+c+d)(an1+bn1+cn1+dn1)(ab+ac+ad+bc+bd+cd)(an2+bn2+cn2+dn2)+(abc+acd+bcd+abd)(an3+bn3+cn3+dn3)abcd(an4+bn4+cn4+dn4)(a^n+b^n+c^n+d^n)=(a+b+c+d)(a^{n-1}+b^{n-1}+c^{n-1}+d^{n-1})-(ab+ac+ad+bc+bd+cd) (a^{n-2}+b^{n-2}+c^{n-2}+d^{n-2})+(abc+acd+bcd+abd) (a^{n-3}+b^{n-3}+c^{n-3}+d^{n-3})-abcd (a^{n-4}+b^{n-4}+c^{n-4}+d^{n-4})

Sanjeet Raria - 5 years ago

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Ok thank you so much.........a last question consider the following polynomial:- x3+3x2+9x+1x^3+3x^2+9x+1 Suppose we are asked to find the sum of cubes of roots of this polynomial Let (\a,b,c)\ be the roots...now as you know we can easily find values of (\a+b+c,ab+bc+ac,abc)\ by using vieta's formula.........is it correct to use newton sums here..... Also i searched about newton sums on internet and found that newton sums are relations between cofficiants of a polynomial and roots of polynomial is it correct .......also in all these identities i observed that sign of terms on RHS are ulternating.. can we use it as a generalisation

(I am realy sorry to disturb you..... :-( i am very new with maths and want learn it in more depth so please help me)

Aman Sharma - 5 years ago

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@Aman Sharma Yeah you can use them anywhere & this would be the best approach.

Feel free to ask or share anything regarding math. It never bores me.

Sanjeet Raria - 5 years ago

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@Sanjeet Raria In this website i found a different explaination about newton sums please check it out

Aman Sharma - 5 years ago

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@Aman Sharma Yes.. That's why I've named it "in easy notation"

Sanjeet Raria - 5 years ago

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@Sanjeet Raria :-) ok thank you so much for helping me going to try out these identities in the questions you attatched with your note ......it is a very helpfull note realy

Aman Sharma - 5 years ago

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@Aman Sharma Yes you can generalize this pattern.

Sanjeet Raria - 5 years ago

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@Sanjeet Raria can you tell how did you derive this pattern

avn bha - 4 years, 11 months ago

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Remember it is cyclic. U missed abc+Bca.....

Dinesh Chavan - 5 years ago

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You've to count every case, including the third symmetric sum.

Sanjeet Raria - 5 years ago

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@Sanjeet Raria please help

Aman Sharma - 5 years ago

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Can you please give an example of how to use these identities....i am struggling with algebra...... and sorry to disturb you

Aman Sharma - 5 years, 1 month ago

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I attached many questions which uses these identities. Just see my note again. Sorry i could not provide the questions regarding the second identity because i didn't come across any. But one can find questions regarding this in Quadratic equations chapter. Questions are asked to find the symmetric expression of roots etc. @Aman Sharma @Pranjali Bhargava

Sanjeet Raria - 5 years, 1 month ago

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But how can we solve the questions related to this using exponential function's graph. Like if we have 3^x + 4^x + 5^x = 6^x (this is a ques. of A. Dasgupta), then divide on both sides by 6^x...without using the identity

Pranjali Bhargava - 5 years, 1 month ago

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Yeah.. There's only graphical approach to such questions that you mentioned, applying algebra would be very lengthy & less reliable.

Sanjeet Raria - 5 years, 1 month ago

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