Proof verification of $x_n = sum_{k=1}^n a_kq^k$ is Cauchy given $|a_k| le C, |q| < 1, kinBbb N$
$begingroup$
Given a sequence ${x_n}$:
$$
x_n = sum_{k=1}^n a_kq^k
$$
and:
$$
begin{cases}
|a_k| le C\
|q| < 1\
kinBbb N
end{cases}
$$
Prove ${x_n}$ is a fundamental sequence.
By definition of a fundamental sequence we want to show: $|x_n - x_m| < epsilon$. So:
$$
begin{align}
|x_n - x_m| &= left|sum_{k=1}^n a_kq^k - sum_{k=1}^m a_kq^kright| stackrel{m>n}{=} \
&= left|sum_{k=n+1}^m a_kq^k right| le \
&le sum_{k=n+1}^m left|a_kq^k right| le \ &le sum_{k=n+1}^m left|Cq^k right| stackrel{text{geom. sum}}{=} frac{Cq(q^m - q^n)}{q-1}
end{align}
$$
Let $m > n$:
$$
frac{Cq(q^m - q^n)}{q-1} = frac{Cq(q^n - q^m)}{1-q} le frac{Cq^{n+1}}{1-q}
$$
Let $q$:
$$
q = frac{1}{1+r}, r in Bbb R_{>0}
$$
Then:
$$
frac{Cq^{n+1}}{1-q} = frac{Cq}{(1-q)(1+r)^{n}} stackrel{n > N}{le} frac{Cq}{(1-q)(1+r)^{N}} < epsilon
$$
Finally:
$$
frac{Cq}{(1-q)(1+r)^{N}} < epsilon iff (1+r)^{N} > frac{Cq}{(1-q)epsilon} \
iff N > log_{1+r} frac{Cq}{(1-q)epsilon} implies |x_n - x_m| < epsilon
$$
I'm kindly asking to verify the proof above. Thank you.
calculus sequences-and-series proof-verification cauchy-sequences
asked Dec 19 '18 at 14:17
romanroman
2,36921224
$endgroup$
add a comment |
$begingroup$
Given a sequence ${x_n}$:
$$
x_n = sum_{k=1}^n a_kq^k
$$
and:
$$
begin{cases}
|a_k| le C\
|q| < 1\
kinBbb N
end{cases}
$$
Prove ${x_n}$ is a fundamental sequence.
By definition of a fundamental sequence we want to show: $|x_n - x_m| < epsilon$. So:
$$
begin{align}
|x_n - x_m| &= left|sum_{k=1}^n a_kq^k - sum_{k=1}^m a_kq^kright| stackrel{m>n}{=} \
&= left|sum_{k=n+1}^m a_kq^k right| le \
&le sum_{k=n+1}^m left|a_kq^k right| le \ &le sum_{k=n+1}^m left|Cq^k right| stackrel{text{geom. sum}}{=} frac{Cq(q^m - q^n)}{q-1}
end{align}
$$
Let $m > n$:
$$
frac{Cq(q^m - q^n)}{q-1} = frac{Cq(q^n - q^m)}{1-q} le frac{Cq^{n+1}}{1-q}
$$
Let $q$:
$$
q = frac{1}{1+r}, r in Bbb R_{>0}
$$
Then:
$$
frac{Cq^{n+1}}{1-q} = frac{Cq}{(1-q)(1+r)^{n}} stackrel{n > N}{le} frac{Cq}{(1-q)(1+r)^{N}} < epsilon
$$
Finally:
$$
frac{Cq}{(1-q)(1+r)^{N}} < epsilon iff (1+r)^{N} > frac{Cq}{(1-q)epsilon} \
iff N > log_{1+r} frac{Cq}{(1-q)epsilon} implies |x_n - x_m| < epsilon
$$
I'm kindly asking to verify the proof above. Thank you.
calculus sequences-and-series proof-verification cauchy-sequences
asked Dec 19 '18 at 14:17
romanroman
2,36921224
$endgroup$
add a comment |
$begingroup$
Given a sequence ${x_n}$:
$$
x_n = sum_{k=1}^n a_kq^k
$$
and:
$$
begin{cases}
|a_k| le C\
|q| < 1\
kinBbb N
end{cases}
$$
Prove ${x_n}$ is a fundamental sequence.
By definition of a fundamental sequence we want to show: $|x_n - x_m| < epsilon$. So:
$$
begin{align}
|x_n - x_m| &= left|sum_{k=1}^n a_kq^k - sum_{k=1}^m a_kq^kright| stackrel{m>n}{=} \
&= left|sum_{k=n+1}^m a_kq^k right| le \
&le sum_{k=n+1}^m left|a_kq^k right| le \ &le sum_{k=n+1}^m left|Cq^k right| stackrel{text{geom. sum}}{=} frac{Cq(q^m - q^n)}{q-1}
end{align}
$$
Let $m > n$:
$$
frac{Cq(q^m - q^n)}{q-1} = frac{Cq(q^n - q^m)}{1-q} le frac{Cq^{n+1}}{1-q}
$$
Let $q$:
$$
q = frac{1}{1+r}, r in Bbb R_{>0}
$$
Then:
$$
frac{Cq^{n+1}}{1-q} = frac{Cq}{(1-q)(1+r)^{n}} stackrel{n > N}{le} frac{Cq}{(1-q)(1+r)^{N}} < epsilon
$$
Finally:
$$
frac{Cq}{(1-q)(1+r)^{N}} < epsilon iff (1+r)^{N} > frac{Cq}{(1-q)epsilon} \
iff N > log_{1+r} frac{Cq}{(1-q)epsilon} implies |x_n - x_m| < epsilon
$$
I'm kindly asking to verify the proof above. Thank you.
calculus sequences-and-series proof-verification cauchy-sequences
asked Dec 19 '18 at 14:17
romanroman
2,36921224
$endgroup$
Given a sequence ${x_n}$:
$$
x_n = sum_{k=1}^n a_kq^k
$$
and:
$$
begin{cases}
|a_k| le C\
|q| < 1\
kinBbb N
end{cases}
$$
Prove ${x_n}$ is a fundamental sequence.
By definition of a fundamental sequence we want to show: $|x_n - x_m| < epsilon$. So:
$$
begin{align}
|x_n - x_m| &= left|sum_{k=1}^n a_kq^k - sum_{k=1}^m a_kq^kright| stackrel{m>n}{=} \
&= left|sum_{k=n+1}^m a_kq^k right| le \
&le sum_{k=n+1}^m left|a_kq^k right| le \ &le sum_{k=n+1}^m left|Cq^k right| stackrel{text{geom. sum}}{=} frac{Cq(q^m - q^n)}{q-1}
end{align}
$$
Let $m > n$:
$$
frac{Cq(q^m - q^n)}{q-1} = frac{Cq(q^n - q^m)}{1-q} le frac{Cq^{n+1}}{1-q}
$$
Let $q$:
$$
q = frac{1}{1+r}, r in Bbb R_{>0}
$$
Then:
$$
frac{Cq^{n+1}}{1-q} = frac{Cq}{(1-q)(1+r)^{n}} stackrel{n > N}{le} frac{Cq}{(1-q)(1+r)^{N}} < epsilon
$$
Finally:
$$
frac{Cq}{(1-q)(1+r)^{N}} < epsilon iff (1+r)^{N} > frac{Cq}{(1-q)epsilon} \
iff N > log_{1+r} frac{Cq}{(1-q)epsilon} implies |x_n - x_m| < epsilon
$$
I'm kindly asking to verify the proof above. Thank you.
calculus sequences-and-series proof-verification cauchy-sequences
calculus sequences-and-series proof-verification cauchy-sequences
asked Dec 19 '18 at 14:17
romanroman
2,36921224
asked Dec 19 '18 at 14:17
romanroman
2,36921224
edited Dec 19 '18 at 14:24
roman
asked Dec 19 '18 at 14:17
romanroman
2,36921224
asked Dec 19 '18 at 14:17
romanroman
2,36921224
asked Dec 19 '18 at 14:17
romanroman
2,36921224
2,36921224
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
Well done.
If I want to nitpick, a very minor issue is that it is possible that $q$ can be negative. Hence, making this statement false.$$sum_{k=n+1}^m left|Cq^k right| stackrel{text{geom. sum}}{=} frac{Cq(q^m - q^n)}{q-1}$$
A quick fix is just to change $q$ to $|q|$.
answered Dec 19 '18 at 14:39
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
$begingroup$
indeed, you are right! thanks for the notice
$endgroup$
– roman
Dec 19 '18 at 14:52
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Well done.
If I want to nitpick, a very minor issue is that it is possible that $q$ can be negative. Hence, making this statement false.$$sum_{k=n+1}^m left|Cq^k right| stackrel{text{geom. sum}}{=} frac{Cq(q^m - q^n)}{q-1}$$
A quick fix is just to change $q$ to $|q|$.
answered Dec 19 '18 at 14:39
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
$begingroup$
indeed, you are right! thanks for the notice
$endgroup$
– roman
Dec 19 '18 at 14:52
add a comment |
$begingroup$
Well done.
If I want to nitpick, a very minor issue is that it is possible that $q$ can be negative. Hence, making this statement false.$$sum_{k=n+1}^m left|Cq^k right| stackrel{text{geom. sum}}{=} frac{Cq(q^m - q^n)}{q-1}$$
A quick fix is just to change $q$ to $|q|$.
answered Dec 19 '18 at 14:39
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
$begingroup$
indeed, you are right! thanks for the notice
$endgroup$
– roman
Dec 19 '18 at 14:52
add a comment |
$begingroup$
Well done.
If I want to nitpick, a very minor issue is that it is possible that $q$ can be negative. Hence, making this statement false.$$sum_{k=n+1}^m left|Cq^k right| stackrel{text{geom. sum}}{=} frac{Cq(q^m - q^n)}{q-1}$$
A quick fix is just to change $q$ to $|q|$.
answered Dec 19 '18 at 14:39
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
Well done.
If I want to nitpick, a very minor issue is that it is possible that $q$ can be negative. Hence, making this statement false.$$sum_{k=n+1}^m left|Cq^k right| stackrel{text{geom. sum}}{=} frac{Cq(q^m - q^n)}{q-1}$$
A quick fix is just to change $q$ to $|q|$.
answered Dec 19 '18 at 14:39
Siong Thye GohSiong Thye Goh
103k1468119
answered Dec 19 '18 at 14:39
Siong Thye GohSiong Thye Goh
103k1468119
answered Dec 19 '18 at 14:39
Siong Thye GohSiong Thye Goh
103k1468119
answered Dec 19 '18 at 14:39
Siong Thye GohSiong Thye Goh
103k1468119
103k1468119
$begingroup$
indeed, you are right! thanks for the notice
$endgroup$
– roman
Dec 19 '18 at 14:52
add a comment |
$begingroup$
indeed, you are right! thanks for the notice
$endgroup$
– roman
Dec 19 '18 at 14:52
$begingroup$
indeed, you are right! thanks for the notice
$endgroup$
– roman
Dec 19 '18 at 14:52
$begingroup$
indeed, you are right! thanks for the notice
$endgroup$
– roman
Dec 19 '18 at 14:52
add a comment |
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