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Newton's Generalization of the Binomial Theorem and big o notation

0

$begingroup$

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

share|cite|improve this question

asked Mar 19 at 12:12

Almot1960Almot1960

2,325825

$endgroup$

add a comment | 

0

$begingroup$

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

share|cite|improve this question

asked Mar 19 at 12:12

Almot1960Almot1960

2,325825

$endgroup$

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

share|cite|improve this question

asked Mar 19 at 12:12

Almot1960Almot1960

2,325825

$endgroup$

share|cite|improve this question

asked Mar 19 at 12:12

Almot1960Almot1960

2,325825

share|cite|improve this question

asked Mar 19 at 12:12

Almot1960Almot1960

2,325825

asked Mar 19 at 12:12

Almot1960Almot1960

2,325825

asked Mar 19 at 12:12

Almot1960Almot1960

2,325825