Proving difference between two functions is small when $x$ is small…Upperbound approximation to the sum of Euler's totient functionAsymptotic expansion for the solution of linear KDV eq.monotonic decrising p normProof of inequality using Taylor series and geometric seriesProduct of two monotonic real functionsMonotonicity of the difference of monotonic composite functionsProve that a sequence is monotonically decreasing/increasingFinding lower bound of summation in inequalityExamine the following sequences of numbers $(a_n)$ with $n in mathbbN$ for the properties of monotonicity and boundedness.Approximating $pi$ with arctangent
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Proving difference between two functions is small when $x$ is small…
Upperbound approximation to the sum of Euler's totient functionAsymptotic expansion for the solution of linear KDV eq.monotonic decrising p normProof of inequality using Taylor series and geometric seriesProduct of two monotonic real functionsMonotonicity of the difference of monotonic composite functionsProve that a sequence is monotonically decreasing/increasingFinding lower bound of summation in inequalityExamine the following sequences of numbers $(a_n)$ with $n in mathbbN$ for the properties of monotonicity and boundedness.Approximating $pi$ with arctangent
$begingroup$
I'm really struggling to prove the following claim and I was wondering whether anyone could help me.
Claim:$$ 0<|x| leq 10^-4 implies bigg fracxe^x-1 - frac1sum_k=1^3 fracx^k-1k! bigg leq 10^-12$$
My attempt at a proof: I have already proven the following inequality,
$$bigg frace^x-1x - sum_k=1^n fracx^k-1k! bigg leq frac(n+1)!$$
So let $n=3$, then
beginalign*
bigg fracxe^x-1 - frac1sum_k=1^3 fracx^k-1k! bigg &= bigg fracx(e^x-1)(sum_k=1^3 fracx^k-1k!) cdot left( frace^x-1x - sum_k=1^3 fracx^k-1k! right) bigg \
&= bigg fracx(e^x-1)(sum_k=1^3 fracx^k-1k!) bigg cdot bigg frace^x-1x - sum_k=1^3 fracx^k-1k! bigg \
&leq bigg fracx(e^x-1)(sum_k=1^3 fracx^k-1k!) bigg cdot fracx(3+1)! \
&leq bigg fracx(e^x-1)(sum_k=1^3 fracx^k-1k!) bigg cdot frac(10^-4)^3 e^10^-44! \
&= 10^-12 cdot bigg fracx(e^x-1)(sum_k=1^3 fracx^k-1k!) bigg cdot frac e^10^-44!
endalign* Hence, it suffices to show that for $0 < |x| leq 10^-4$,
beginalign*
f(x) := bigg fracx(e^x-1)(sum_k=1^3 fracx^k-1k!) bigg < frac4!e^10^-4
endalign*
Since $f(x)$ is monotonically decreasing on the intervals $[-10^-4, 0)$ and $(0, 10^-4]$, we know that for $0 <|x| leq 10^-4$,
beginalign*
f(x) leq f left( -10^-4 right) approx 1.0001 < frac4!e^10^-4
endalign* Therefore, for $0 < |x| leq 10^-4$,
beginalign*
bigg fracxe^x-1 - frac1sum_k=1^3 fracx^k-1k! bigg leq 10^-12
endalign* $square$
My big problem here is that I don't actually know how to prove my claim that $f(x)$ is monotonically decreasing. Does anyone know how to prove this?
Or even better, does anyone know of a more direct proof that avoids this problem altogether?
Thanks!
taylor-expansion approximation monotone-functions
asked Mar 29 at 22:19
jasonjason
1029
$endgroup$
add a comment |
$begingroup$
I'm really struggling to prove the following claim and I was wondering whether anyone could help me.
Claim:$$ 0<|x| leq 10^-4 implies bigg fracxe^x-1 - frac1sum_k=1^3 fracx^k-1k! bigg leq 10^-12$$
My attempt at a proof: I have already proven the following inequality,
$$bigg frace^x-1x - sum_k=1^n fracx^k-1k! bigg leq frac(n+1)!$$
So let $n=3$, then
beginalign*
bigg fracxe^x-1 - frac1sum_k=1^3 fracx^k-1k! bigg &= bigg fracx(e^x-1)(sum_k=1^3 fracx^k-1k!) cdot left( frace^x-1x - sum_k=1^3 fracx^k-1k! right) bigg \
&= bigg fracx(e^x-1)(sum_k=1^3 fracx^k-1k!) bigg cdot bigg frace^x-1x - sum_k=1^3 fracx^k-1k! bigg \
&leq bigg fracx(e^x-1)(sum_k=1^3 fracx^k-1k!) bigg cdot fracx(3+1)! \
&leq bigg fracx(e^x-1)(sum_k=1^3 fracx^k-1k!) bigg cdot frac(10^-4)^3 e^10^-44! \
&= 10^-12 cdot bigg fracx(e^x-1)(sum_k=1^3 fracx^k-1k!) bigg cdot frac e^10^-44!
endalign* Hence, it suffices to show that for $0 < |x| leq 10^-4$,
beginalign*
f(x) := bigg fracx(e^x-1)(sum_k=1^3 fracx^k-1k!) bigg < frac4!e^10^-4
endalign*
Since $f(x)$ is monotonically decreasing on the intervals $[-10^-4, 0)$ and $(0, 10^-4]$, we know that for $0 <|x| leq 10^-4$,
beginalign*
f(x) leq f left( -10^-4 right) approx 1.0001 < frac4!e^10^-4
endalign* Therefore, for $0 < |x| leq 10^-4$,
beginalign*
bigg fracxe^x-1 - frac1sum_k=1^3 fracx^k-1k! bigg leq 10^-12
endalign* $square$
My big problem here is that I don't actually know how to prove my claim that $f(x)$ is monotonically decreasing. Does anyone know how to prove this?
Or even better, does anyone know of a more direct proof that avoids this problem altogether?
Thanks!
taylor-expansion approximation monotone-functions
asked Mar 29 at 22:19
jasonjason
1029
$endgroup$
add a comment |
$begingroup$
I'm really struggling to prove the following claim and I was wondering whether anyone could help me.
Claim:$$ 0<|x| leq 10^-4 implies bigg fracxe^x-1 - frac1sum_k=1^3 fracx^k-1k! bigg leq 10^-12$$
My attempt at a proof: I have already proven the following inequality,
$$bigg frace^x-1x - sum_k=1^n fracx^k-1k! bigg leq frac(n+1)!$$
So let $n=3$, then
beginalign*
bigg fracxe^x-1 - frac1sum_k=1^3 fracx^k-1k! bigg &= bigg fracx(e^x-1)(sum_k=1^3 fracx^k-1k!) cdot left( frace^x-1x - sum_k=1^3 fracx^k-1k! right) bigg \
&= bigg fracx(e^x-1)(sum_k=1^3 fracx^k-1k!) bigg cdot bigg frace^x-1x - sum_k=1^3 fracx^k-1k! bigg \
&leq bigg fracx(e^x-1)(sum_k=1^3 fracx^k-1k!) bigg cdot fracx(3+1)! \
&leq bigg fracx(e^x-1)(sum_k=1^3 fracx^k-1k!) bigg cdot frac(10^-4)^3 e^10^-44! \
&= 10^-12 cdot bigg fracx(e^x-1)(sum_k=1^3 fracx^k-1k!) bigg cdot frac e^10^-44!
endalign* Hence, it suffices to show that for $0 < |x| leq 10^-4$,
beginalign*
f(x) := bigg fracx(e^x-1)(sum_k=1^3 fracx^k-1k!) bigg < frac4!e^10^-4
endalign*
Since $f(x)$ is monotonically decreasing on the intervals $[-10^-4, 0)$ and $(0, 10^-4]$, we know that for $0 <|x| leq 10^-4$,
beginalign*
f(x) leq f left( -10^-4 right) approx 1.0001 < frac4!e^10^-4
endalign* Therefore, for $0 < |x| leq 10^-4$,
beginalign*
bigg fracxe^x-1 - frac1sum_k=1^3 fracx^k-1k! bigg leq 10^-12
endalign* $square$
My big problem here is that I don't actually know how to prove my claim that $f(x)$ is monotonically decreasing. Does anyone know how to prove this?
Or even better, does anyone know of a more direct proof that avoids this problem altogether?
Thanks!
taylor-expansion approximation monotone-functions
asked Mar 29 at 22:19
jasonjason
1029
$endgroup$
I'm really struggling to prove the following claim and I was wondering whether anyone could help me.
Claim:$$ 0<|x| leq 10^-4 implies bigg fracxe^x-1 - frac1sum_k=1^3 fracx^k-1k! bigg leq 10^-12$$
My attempt at a proof: I have already proven the following inequality,
$$bigg frace^x-1x - sum_k=1^n fracx^k-1k! bigg leq frac(n+1)!$$
So let $n=3$, then
beginalign*
bigg fracxe^x-1 - frac1sum_k=1^3 fracx^k-1k! bigg &= bigg fracx(e^x-1)(sum_k=1^3 fracx^k-1k!) cdot left( frace^x-1x - sum_k=1^3 fracx^k-1k! right) bigg \
&= bigg fracx(e^x-1)(sum_k=1^3 fracx^k-1k!) bigg cdot bigg frace^x-1x - sum_k=1^3 fracx^k-1k! bigg \
&leq bigg fracx(e^x-1)(sum_k=1^3 fracx^k-1k!) bigg cdot fracx(3+1)! \
&leq bigg fracx(e^x-1)(sum_k=1^3 fracx^k-1k!) bigg cdot frac(10^-4)^3 e^10^-44! \
&= 10^-12 cdot bigg fracx(e^x-1)(sum_k=1^3 fracx^k-1k!) bigg cdot frac e^10^-44!
endalign* Hence, it suffices to show that for $0 < |x| leq 10^-4$,
beginalign*
f(x) := bigg fracx(e^x-1)(sum_k=1^3 fracx^k-1k!) bigg < frac4!e^10^-4
endalign*
Since $f(x)$ is monotonically decreasing on the intervals $[-10^-4, 0)$ and $(0, 10^-4]$, we know that for $0 <|x| leq 10^-4$,
beginalign*
f(x) leq f left( -10^-4 right) approx 1.0001 < frac4!e^10^-4
endalign* Therefore, for $0 < |x| leq 10^-4$,
beginalign*
bigg fracxe^x-1 - frac1sum_k=1^3 fracx^k-1k! bigg leq 10^-12
endalign* $square$
My big problem here is that I don't actually know how to prove my claim that $f(x)$ is monotonically decreasing. Does anyone know how to prove this?
Or even better, does anyone know of a more direct proof that avoids this problem altogether?
Thanks!
taylor-expansion approximation monotone-functions
taylor-expansion approximation monotone-functions
asked Mar 29 at 22:19
jasonjason
1029
asked Mar 29 at 22:19
jasonjason
1029
asked Mar 29 at 22:19
jasonjason
1029
asked Mar 29 at 22:19
jasonjason
1029
asked Mar 29 at 22:19
jasonjason
1029
1029
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Write $|fracxe^x-1 - fracxsum_k=1^3 fracx^kk!|
= |x||fracsum_k=1^3 fracx^kk! - (e^x-1)(e^x-1)sum_k=1^3 fracx^kk!| $
You can use the $|x|<10^-4$ bound given, and also instead an easier inequality for the numerator using the series sum for $e^x$. Then it is clear that the denominator is obviously increasing in $x$.
So I guess this comes under "direct proof". I think your function $f(x)$ is probably decreasing though.
answered Mar 29 at 23:53
George DewhirstGeorge Dewhirst
7514
$endgroup$
$begingroup$
Thanks for the response! I'm not quite sure how the denominator is obviously increasing in $x$ nor how this fact if true would help establish an upper bound? Wouldn't we want the denominator to be decreasing in $x$? And I'm not sure exactly what the easier inequality for the numerator you're talking about would be?
$endgroup$
– jason
Mar 30 at 9:23
$begingroup$
The numerator is equal to the sum: $sum_k=4^infty fracx^kk!$ which you could bound. This is using the power series expansion for $e^x$. Denominator is increasing, that's to say that the expression as a whole is decreasing.
$endgroup$
– George Dewhirst
Mar 30 at 13:50
add a comment |
$begingroup$
For $|x|<epsilon=10^-4$, by the mean value theorem $$left|frace^x-1xright|=e^yge e^-epsilonge1-epsilon,$$ and $$1+fracx2+fracx^26ge1-fracepsilon2,$$ so it follows that $$left|fracxe^x-1frac11+fracx2+fracx^26right|lefrac1(1-epsilon)^2le2lefrac4!e^epsilon$$
answered Mar 30 at 8:38
ChrystomathChrystomath
2,013513
$endgroup$
$begingroup$
Hi thanks for getting back to me! This argument is fine for $x geq 0$, but I don't think any similar argument works for $x < 0$...
$endgroup$
– jason
Mar 30 at 9:30
$begingroup$
I thought $0<x$. I'll modify for $x<0$.
$endgroup$
– Chrystomath
Mar 31 at 6:04
add a comment |
Your Answer
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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active
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active
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votes
$begingroup$
Write $|fracxe^x-1 - fracxsum_k=1^3 fracx^kk!|
= |x||fracsum_k=1^3 fracx^kk! - (e^x-1)(e^x-1)sum_k=1^3 fracx^kk!| $
You can use the $|x|<10^-4$ bound given, and also instead an easier inequality for the numerator using the series sum for $e^x$. Then it is clear that the denominator is obviously increasing in $x$.
So I guess this comes under "direct proof". I think your function $f(x)$ is probably decreasing though.
answered Mar 29 at 23:53
George DewhirstGeorge Dewhirst
7514
$endgroup$
$begingroup$
Thanks for the response! I'm not quite sure how the denominator is obviously increasing in $x$ nor how this fact if true would help establish an upper bound? Wouldn't we want the denominator to be decreasing in $x$? And I'm not sure exactly what the easier inequality for the numerator you're talking about would be?
$endgroup$
– jason
Mar 30 at 9:23
$begingroup$
The numerator is equal to the sum: $sum_k=4^infty fracx^kk!$ which you could bound. This is using the power series expansion for $e^x$. Denominator is increasing, that's to say that the expression as a whole is decreasing.
$endgroup$
– George Dewhirst
Mar 30 at 13:50
add a comment |
$begingroup$
Write $|fracxe^x-1 - fracxsum_k=1^3 fracx^kk!|
= |x||fracsum_k=1^3 fracx^kk! - (e^x-1)(e^x-1)sum_k=1^3 fracx^kk!| $
You can use the $|x|<10^-4$ bound given, and also instead an easier inequality for the numerator using the series sum for $e^x$. Then it is clear that the denominator is obviously increasing in $x$.
So I guess this comes under "direct proof". I think your function $f(x)$ is probably decreasing though.
answered Mar 29 at 23:53
George DewhirstGeorge Dewhirst
7514
$endgroup$
$begingroup$
Thanks for the response! I'm not quite sure how the denominator is obviously increasing in $x$ nor how this fact if true would help establish an upper bound? Wouldn't we want the denominator to be decreasing in $x$? And I'm not sure exactly what the easier inequality for the numerator you're talking about would be?
$endgroup$
– jason
Mar 30 at 9:23
$begingroup$
The numerator is equal to the sum: $sum_k=4^infty fracx^kk!$ which you could bound. This is using the power series expansion for $e^x$. Denominator is increasing, that's to say that the expression as a whole is decreasing.
$endgroup$
– George Dewhirst
Mar 30 at 13:50
add a comment |
$begingroup$
Write $|fracxe^x-1 - fracxsum_k=1^3 fracx^kk!|
= |x||fracsum_k=1^3 fracx^kk! - (e^x-1)(e^x-1)sum_k=1^3 fracx^kk!| $
You can use the $|x|<10^-4$ bound given, and also instead an easier inequality for the numerator using the series sum for $e^x$. Then it is clear that the denominator is obviously increasing in $x$.
So I guess this comes under "direct proof". I think your function $f(x)$ is probably decreasing though.
answered Mar 29 at 23:53
George DewhirstGeorge Dewhirst
7514
$endgroup$
Write $|fracxe^x-1 - fracxsum_k=1^3 fracx^kk!|
= |x||fracsum_k=1^3 fracx^kk! - (e^x-1)(e^x-1)sum_k=1^3 fracx^kk!| $
You can use the $|x|<10^-4$ bound given, and also instead an easier inequality for the numerator using the series sum for $e^x$. Then it is clear that the denominator is obviously increasing in $x$.
So I guess this comes under "direct proof". I think your function $f(x)$ is probably decreasing though.
answered Mar 29 at 23:53
George DewhirstGeorge Dewhirst
7514
answered Mar 29 at 23:53
George DewhirstGeorge Dewhirst
7514
answered Mar 29 at 23:53
George DewhirstGeorge Dewhirst
7514
answered Mar 29 at 23:53
George DewhirstGeorge Dewhirst
7514
7514
$begingroup$
Thanks for the response! I'm not quite sure how the denominator is obviously increasing in $x$ nor how this fact if true would help establish an upper bound? Wouldn't we want the denominator to be decreasing in $x$? And I'm not sure exactly what the easier inequality for the numerator you're talking about would be?
$endgroup$
– jason
Mar 30 at 9:23
$begingroup$
The numerator is equal to the sum: $sum_k=4^infty fracx^kk!$ which you could bound. This is using the power series expansion for $e^x$. Denominator is increasing, that's to say that the expression as a whole is decreasing.
$endgroup$
– George Dewhirst
Mar 30 at 13:50
add a comment |
$begingroup$
Thanks for the response! I'm not quite sure how the denominator is obviously increasing in $x$ nor how this fact if true would help establish an upper bound? Wouldn't we want the denominator to be decreasing in $x$? And I'm not sure exactly what the easier inequality for the numerator you're talking about would be?
$endgroup$
– jason
Mar 30 at 9:23
$begingroup$
The numerator is equal to the sum: $sum_k=4^infty fracx^kk!$ which you could bound. This is using the power series expansion for $e^x$. Denominator is increasing, that's to say that the expression as a whole is decreasing.
$endgroup$
– George Dewhirst
Mar 30 at 13:50
$begingroup$
Thanks for the response! I'm not quite sure how the denominator is obviously increasing in $x$ nor how this fact if true would help establish an upper bound? Wouldn't we want the denominator to be decreasing in $x$? And I'm not sure exactly what the easier inequality for the numerator you're talking about would be?
$endgroup$
– jason
Mar 30 at 9:23
$begingroup$
Thanks for the response! I'm not quite sure how the denominator is obviously increasing in $x$ nor how this fact if true would help establish an upper bound? Wouldn't we want the denominator to be decreasing in $x$? And I'm not sure exactly what the easier inequality for the numerator you're talking about would be?
$endgroup$
– jason
Mar 30 at 9:23
$begingroup$
The numerator is equal to the sum: $sum_k=4^infty fracx^kk!$ which you could bound. This is using the power series expansion for $e^x$. Denominator is increasing, that's to say that the expression as a whole is decreasing.
$endgroup$
– George Dewhirst
Mar 30 at 13:50
$begingroup$
The numerator is equal to the sum: $sum_k=4^infty fracx^kk!$ which you could bound. This is using the power series expansion for $e^x$. Denominator is increasing, that's to say that the expression as a whole is decreasing.
$endgroup$
– George Dewhirst
Mar 30 at 13:50
add a comment |
$begingroup$
For $|x|<epsilon=10^-4$, by the mean value theorem $$left|frace^x-1xright|=e^yge e^-epsilonge1-epsilon,$$ and $$1+fracx2+fracx^26ge1-fracepsilon2,$$ so it follows that $$left|fracxe^x-1frac11+fracx2+fracx^26right|lefrac1(1-epsilon)^2le2lefrac4!e^epsilon$$
answered Mar 30 at 8:38
ChrystomathChrystomath
2,013513
$endgroup$
$begingroup$
Hi thanks for getting back to me! This argument is fine for $x geq 0$, but I don't think any similar argument works for $x < 0$...
$endgroup$
– jason
Mar 30 at 9:30
$begingroup$
I thought $0<x$. I'll modify for $x<0$.
$endgroup$
– Chrystomath
Mar 31 at 6:04
add a comment |
$begingroup$
For $|x|<epsilon=10^-4$, by the mean value theorem $$left|frace^x-1xright|=e^yge e^-epsilonge1-epsilon,$$ and $$1+fracx2+fracx^26ge1-fracepsilon2,$$ so it follows that $$left|fracxe^x-1frac11+fracx2+fracx^26right|lefrac1(1-epsilon)^2le2lefrac4!e^epsilon$$
answered Mar 30 at 8:38
ChrystomathChrystomath
2,013513
$endgroup$
$begingroup$
Hi thanks for getting back to me! This argument is fine for $x geq 0$, but I don't think any similar argument works for $x < 0$...
$endgroup$
– jason
Mar 30 at 9:30
$begingroup$
I thought $0<x$. I'll modify for $x<0$.
$endgroup$
– Chrystomath
Mar 31 at 6:04
add a comment |
$begingroup$
For $|x|<epsilon=10^-4$, by the mean value theorem $$left|frace^x-1xright|=e^yge e^-epsilonge1-epsilon,$$ and $$1+fracx2+fracx^26ge1-fracepsilon2,$$ so it follows that $$left|fracxe^x-1frac11+fracx2+fracx^26right|lefrac1(1-epsilon)^2le2lefrac4!e^epsilon$$
answered Mar 30 at 8:38
ChrystomathChrystomath
2,013513
$endgroup$
For $|x|<epsilon=10^-4$, by the mean value theorem $$left|frace^x-1xright|=e^yge e^-epsilonge1-epsilon,$$ and $$1+fracx2+fracx^26ge1-fracepsilon2,$$ so it follows that $$left|fracxe^x-1frac11+fracx2+fracx^26right|lefrac1(1-epsilon)^2le2lefrac4!e^epsilon$$
answered Mar 30 at 8:38
ChrystomathChrystomath
2,013513
edited Mar 31 at 6:26
answered Mar 30 at 8:38
ChrystomathChrystomath
2,013513
answered Mar 30 at 8:38
ChrystomathChrystomath
2,013513
answered Mar 30 at 8:38
ChrystomathChrystomath
2,013513
2,013513
$begingroup$
Hi thanks for getting back to me! This argument is fine for $x geq 0$, but I don't think any similar argument works for $x < 0$...
$endgroup$
– jason
Mar 30 at 9:30
$begingroup$
I thought $0<x$. I'll modify for $x<0$.
$endgroup$
– Chrystomath
Mar 31 at 6:04
add a comment |
$begingroup$
Hi thanks for getting back to me! This argument is fine for $x geq 0$, but I don't think any similar argument works for $x < 0$...
$endgroup$
– jason
Mar 30 at 9:30
$begingroup$
I thought $0<x$. I'll modify for $x<0$.
$endgroup$
– Chrystomath
Mar 31 at 6:04
$begingroup$
Hi thanks for getting back to me! This argument is fine for $x geq 0$, but I don't think any similar argument works for $x < 0$...
$endgroup$
– jason
Mar 30 at 9:30
$begingroup$
Hi thanks for getting back to me! This argument is fine for $x geq 0$, but I don't think any similar argument works for $x < 0$...
$endgroup$
– jason
Mar 30 at 9:30
$begingroup$
I thought $0<x$. I'll modify for $x<0$.
$endgroup$
– Chrystomath
Mar 31 at 6:04
$begingroup$
I thought $0<x$. I'll modify for $x<0$.
$endgroup$
– Chrystomath
Mar 31 at 6:04
add a comment |
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