Endpoints behavior of binomial series
$begingroup$
The binomial series is known a ∑(1+x)^k
The last step for me to understand this series is when x equals to either 1 or -1.
Basically, I could compute that k must be greater than -1 to converge when x = -1 or 1.
I just don't understand what the internet and book propose--the series converges at 1 if -1 < k < 0 and at both endpoints if k > 0.
I could only come up with -1 < k. I can't even find the bit of k < 0.
calculus analysis power-series taylor-expansion
asked Jan 15 at 7:12
Andes LamAndes Lam
63
$endgroup$
add a comment |
$begingroup$
The binomial series is known a ∑(1+x)^k
The last step for me to understand this series is when x equals to either 1 or -1.
Basically, I could compute that k must be greater than -1 to converge when x = -1 or 1.
I just don't understand what the internet and book propose--the series converges at 1 if -1 < k < 0 and at both endpoints if k > 0.
I could only come up with -1 < k. I can't even find the bit of k < 0.
calculus analysis power-series taylor-expansion
asked Jan 15 at 7:12
Andes LamAndes Lam
63
$endgroup$
$begingroup$
Yes, sorry for the typo, shd be a series rather than a sequence
$endgroup$
– Andes Lam
Jan 15 at 7:30
add a comment |
$begingroup$
The binomial series is known a ∑(1+x)^k
The last step for me to understand this series is when x equals to either 1 or -1.
Basically, I could compute that k must be greater than -1 to converge when x = -1 or 1.
I just don't understand what the internet and book propose--the series converges at 1 if -1 < k < 0 and at both endpoints if k > 0.
I could only come up with -1 < k. I can't even find the bit of k < 0.
calculus analysis power-series taylor-expansion
asked Jan 15 at 7:12
Andes LamAndes Lam
63
$endgroup$
The binomial series is known a ∑(1+x)^k
The last step for me to understand this series is when x equals to either 1 or -1.
Basically, I could compute that k must be greater than -1 to converge when x = -1 or 1.
I just don't understand what the internet and book propose--the series converges at 1 if -1 < k < 0 and at both endpoints if k > 0.
I could only come up with -1 < k. I can't even find the bit of k < 0.
calculus analysis power-series taylor-expansion
calculus analysis power-series taylor-expansion
asked Jan 15 at 7:12
Andes LamAndes Lam
63
asked Jan 15 at 7:12
Andes LamAndes Lam
63
edited Jan 15 at 7:31
Andes Lam
asked Jan 15 at 7:12
Andes LamAndes Lam
63
asked Jan 15 at 7:12
Andes LamAndes Lam
63
asked Jan 15 at 7:12
Andes LamAndes Lam
63
63
$begingroup$
Yes, sorry for the typo, shd be a series rather than a sequence
$endgroup$
– Andes Lam
Jan 15 at 7:30
add a comment |
$begingroup$
Yes, sorry for the typo, shd be a series rather than a sequence
$endgroup$
– Andes Lam
Jan 15 at 7:30
$begingroup$
Yes, sorry for the typo, shd be a series rather than a sequence
$endgroup$
– Andes Lam
Jan 15 at 7:30
$begingroup$
Yes, sorry for the typo, shd be a series rather than a sequence
$endgroup$
– Andes Lam
Jan 15 at 7:30
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
1 + a.sum(k=0,n) a$^k$ = sum(k=0,n) a$^k$ + a$^{n+1}$
sum(k=0,n) a$^k$ = (1 - a$^{n+1}$)/(1 - a)
Sum converges when |a| < 1.
Your series converges when |1 + x| < 1
which is equivalent to -1 < 1 + x < 1.
In the last part are you confusing k with x?
I have understood you're talking about a geometric series.
If that is wrong, then write explicitly what the binomial series is.
answered Jan 15 at 9:52
William ElliotWilliam Elliot
8,0102720
$endgroup$
$begingroup$
Sorry but i am referring to the converging point when x = 1 and x = -1
$endgroup$
– Andes Lam
Jan 18 at 0:44
$begingroup$
It still remains a mystery what binomial series you are considering. Give an explicitc fo
$endgroup$
– William Elliot
Jan 18 at 4:15
$begingroup$
It remains a mystery what the series you are considering. Give a clear, full and explicit description of the series. @AndesLam
$endgroup$
– William Elliot
Jan 18 at 4:18
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
1 + a.sum(k=0,n) a$^k$ = sum(k=0,n) a$^k$ + a$^{n+1}$
sum(k=0,n) a$^k$ = (1 - a$^{n+1}$)/(1 - a)
Sum converges when |a| < 1.
Your series converges when |1 + x| < 1
which is equivalent to -1 < 1 + x < 1.
In the last part are you confusing k with x?
I have understood you're talking about a geometric series.
If that is wrong, then write explicitly what the binomial series is.
answered Jan 15 at 9:52
William ElliotWilliam Elliot
8,0102720
$endgroup$
$begingroup$
Sorry but i am referring to the converging point when x = 1 and x = -1
$endgroup$
– Andes Lam
Jan 18 at 0:44
$begingroup$
It still remains a mystery what binomial series you are considering. Give an explicitc fo
$endgroup$
– William Elliot
Jan 18 at 4:15
$begingroup$
It remains a mystery what the series you are considering. Give a clear, full and explicit description of the series. @AndesLam
$endgroup$
– William Elliot
Jan 18 at 4:18
add a comment |
$begingroup$
1 + a.sum(k=0,n) a$^k$ = sum(k=0,n) a$^k$ + a$^{n+1}$
sum(k=0,n) a$^k$ = (1 - a$^{n+1}$)/(1 - a)
Sum converges when |a| < 1.
Your series converges when |1 + x| < 1
which is equivalent to -1 < 1 + x < 1.
In the last part are you confusing k with x?
I have understood you're talking about a geometric series.
If that is wrong, then write explicitly what the binomial series is.
answered Jan 15 at 9:52
William ElliotWilliam Elliot
8,0102720
$endgroup$
$begingroup$
Sorry but i am referring to the converging point when x = 1 and x = -1
$endgroup$
– Andes Lam
Jan 18 at 0:44
$begingroup$
It still remains a mystery what binomial series you are considering. Give an explicitc fo
$endgroup$
– William Elliot
Jan 18 at 4:15
$begingroup$
It remains a mystery what the series you are considering. Give a clear, full and explicit description of the series. @AndesLam
$endgroup$
– William Elliot
Jan 18 at 4:18
add a comment |
$begingroup$
1 + a.sum(k=0,n) a$^k$ = sum(k=0,n) a$^k$ + a$^{n+1}$
sum(k=0,n) a$^k$ = (1 - a$^{n+1}$)/(1 - a)
Sum converges when |a| < 1.
Your series converges when |1 + x| < 1
which is equivalent to -1 < 1 + x < 1.
In the last part are you confusing k with x?
I have understood you're talking about a geometric series.
If that is wrong, then write explicitly what the binomial series is.
answered Jan 15 at 9:52
William ElliotWilliam Elliot
8,0102720
$endgroup$
1 + a.sum(k=0,n) a$^k$ = sum(k=0,n) a$^k$ + a$^{n+1}$
sum(k=0,n) a$^k$ = (1 - a$^{n+1}$)/(1 - a)
Sum converges when |a| < 1.
Your series converges when |1 + x| < 1
which is equivalent to -1 < 1 + x < 1.
In the last part are you confusing k with x?
I have understood you're talking about a geometric series.
If that is wrong, then write explicitly what the binomial series is.
answered Jan 15 at 9:52
William ElliotWilliam Elliot
8,0102720
answered Jan 15 at 9:52
William ElliotWilliam Elliot
8,0102720
answered Jan 15 at 9:52
William ElliotWilliam Elliot
8,0102720
answered Jan 15 at 9:52
William ElliotWilliam Elliot
8,0102720
8,0102720
$begingroup$
Sorry but i am referring to the converging point when x = 1 and x = -1
$endgroup$
– Andes Lam
Jan 18 at 0:44
$begingroup$
It still remains a mystery what binomial series you are considering. Give an explicitc fo
$endgroup$
– William Elliot
Jan 18 at 4:15
$begingroup$
It remains a mystery what the series you are considering. Give a clear, full and explicit description of the series. @AndesLam
$endgroup$
– William Elliot
Jan 18 at 4:18
add a comment |
$begingroup$
Sorry but i am referring to the converging point when x = 1 and x = -1
$endgroup$
– Andes Lam
Jan 18 at 0:44
$begingroup$
It still remains a mystery what binomial series you are considering. Give an explicitc fo
$endgroup$
– William Elliot
Jan 18 at 4:15
$begingroup$
It remains a mystery what the series you are considering. Give a clear, full and explicit description of the series. @AndesLam
$endgroup$
– William Elliot
Jan 18 at 4:18
$begingroup$
Sorry but i am referring to the converging point when x = 1 and x = -1
$endgroup$
– Andes Lam
Jan 18 at 0:44
$begingroup$
Sorry but i am referring to the converging point when x = 1 and x = -1
$endgroup$
– Andes Lam
Jan 18 at 0:44
$begingroup$
It still remains a mystery what binomial series you are considering. Give an explicitc fo
$endgroup$
– William Elliot
Jan 18 at 4:15
$begingroup$
It still remains a mystery what binomial series you are considering. Give an explicitc fo
$endgroup$
– William Elliot
Jan 18 at 4:15
$begingroup$
It remains a mystery what the series you are considering. Give a clear, full and explicit description of the series. @AndesLam
$endgroup$
– William Elliot
Jan 18 at 4:18
$begingroup$
It remains a mystery what the series you are considering. Give a clear, full and explicit description of the series. @AndesLam
$endgroup$
– William Elliot
Jan 18 at 4:18
add a comment |
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$begingroup$
Yes, sorry for the typo, shd be a series rather than a sequence
$endgroup$
– Andes Lam
Jan 15 at 7:30