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Newton's Generalization of the Binomial Theorem and big o notation
The Next CEO of Stack Overflowmaclaurin series and induction over binomial theoremGeneralization Of The Binomial TheoremTaylor expansion of $(1+x)^α$ to binomial series – why does the remainder term converge?Using Newton's Generalized Binomial TheoremUse induction and Newton's binomial formula to show that $binomn0+binomn1+cdot+binomnn=2^n, forall nin mathbb N$Proof of Newton's Binomial ExpansionOn a connection between Newton's binomial theorem and general Leibniz rule using a new method.Newton's Generalization of the Binomial TheoremVerifying Newton's Binomial Series for $(1+x)^−2$Proof of Newton's Generalized Binomial Theorem (without Calculus)
$begingroup$
we know that :
(Newton's Generalization of the Binomial Theorem)
Let $x,y∈mathbbR$ where $0≤∣x∣<∣y∣$ and let $α∈mathbbR$. Then the expansion of the binomial $(x+y)^α$ is given by the infinite series
$$(x+y)^α=binomalpha0x^0y^α+binomalpha1x^1y^alpha -1+binomalpha2x^αy^alpha-2+...=∑_k=0^inftybinomalphakx^ky^α−k$$
Now let $y=1$
$$beginalign(x+1)^α&=underbracebinomalpha0x^0+cdots+binomalphanx^n+binomalphan+1x^n+1+cdots\
&=∑_k=0^inftybinomalphakx^k
endalign$$
So we have :
$$(x+1)^α-Big(underbracebinomalpha0x^0+cdots+binomalphanx^nBig)=x^n∑_k=n+1^inftybinomalphakx^k-n$$
And we know that :
$$x^n∑_k=n^inftybinomalphakx^k-n$$
where again the series converges for $0≤∣x∣<1$ .
Now how prove that :
$$x^n∑_k=n^inftybinomalphakx^k-n=O(x^n)$$
calculus algebra-precalculus
$endgroup$
add a comment |
$begingroup$
we know that :
(Newton's Generalization of the Binomial Theorem)
Let $x,y∈mathbbR$ where $0≤∣x∣<∣y∣$ and let $α∈mathbbR$. Then the expansion of the binomial $(x+y)^α$ is given by the infinite series
$$(x+y)^α=binomalpha0x^0y^α+binomalpha1x^1y^alpha -1+binomalpha2x^αy^alpha-2+...=∑_k=0^inftybinomalphakx^ky^α−k$$
Now let $y=1$
$$beginalign(x+1)^α&=underbracebinomalpha0x^0+cdots+binomalphanx^n+binomalphan+1x^n+1+cdots\
&=∑_k=0^inftybinomalphakx^k
endalign$$
So we have :
$$(x+1)^α-Big(underbracebinomalpha0x^0+cdots+binomalphanx^nBig)=x^n∑_k=n+1^inftybinomalphakx^k-n$$
And we know that :
$$x^n∑_k=n^inftybinomalphakx^k-n$$
where again the series converges for $0≤∣x∣<1$ .
Now how prove that :
$$x^n∑_k=n^inftybinomalphakx^k-n=O(x^n)$$
calculus algebra-precalculus
$endgroup$
add a comment |
$begingroup$
we know that :
(Newton's Generalization of the Binomial Theorem)
Let $x,y∈mathbbR$ where $0≤∣x∣<∣y∣$ and let $α∈mathbbR$. Then the expansion of the binomial $(x+y)^α$ is given by the infinite series
$$(x+y)^α=binomalpha0x^0y^α+binomalpha1x^1y^alpha -1+binomalpha2x^αy^alpha-2+...=∑_k=0^inftybinomalphakx^ky^α−k$$
Now let $y=1$
$$beginalign(x+1)^α&=underbracebinomalpha0x^0+cdots+binomalphanx^n+binomalphan+1x^n+1+cdots\
&=∑_k=0^inftybinomalphakx^k
endalign$$
So we have :
$$(x+1)^α-Big(underbracebinomalpha0x^0+cdots+binomalphanx^nBig)=x^n∑_k=n+1^inftybinomalphakx^k-n$$
And we know that :
$$x^n∑_k=n^inftybinomalphakx^k-n$$
where again the series converges for $0≤∣x∣<1$ .
Now how prove that :
$$x^n∑_k=n^inftybinomalphakx^k-n=O(x^n)$$
calculus algebra-precalculus
$endgroup$
we know that :
(Newton's Generalization of the Binomial Theorem)
Let $x,y∈mathbbR$ where $0≤∣x∣<∣y∣$ and let $α∈mathbbR$. Then the expansion of the binomial $(x+y)^α$ is given by the infinite series
$$(x+y)^α=binomalpha0x^0y^α+binomalpha1x^1y^alpha -1+binomalpha2x^αy^alpha-2+...=∑_k=0^inftybinomalphakx^ky^α−k$$
Now let $y=1$
$$beginalign(x+1)^α&=underbracebinomalpha0x^0+cdots+binomalphanx^n+binomalphan+1x^n+1+cdots\
&=∑_k=0^inftybinomalphakx^k
endalign$$
So we have :
$$(x+1)^α-Big(underbracebinomalpha0x^0+cdots+binomalphanx^nBig)=x^n∑_k=n+1^inftybinomalphakx^k-n$$
And we know that :
$$x^n∑_k=n^inftybinomalphakx^k-n$$
where again the series converges for $0≤∣x∣<1$ .
Now how prove that :
$$x^n∑_k=n^inftybinomalphakx^k-n=O(x^n)$$
calculus algebra-precalculus
calculus algebra-precalculus
asked Mar 19 at 12:12
Almot1960Almot1960
2,325825
2,325825
add a comment |
add a comment |
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active
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