Use Taylor’s Theorem with $n = 2$ to give a second proof of the inequality $e^x$ $ge$ $1+x$ for all real $x$. Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Use Lagrange Remainder Theorem to Prove InequalityFind all the real values of x for which $sum _1^infty x^n/n$ converges.Taylor theorem and a $C^3$ function with the following property…Bounded sequence with subsequences all converging to the same limit means that the sequence itself converges to the same limit.Use the arithmetic-geometric inequality for this list to deduce the arithmetic-geometric inequality for $n$.How do I use the triangle inequality to finish this fact about complex numbers?How to use the theorem of GaußShow that for every $epsilon > 0$ there exists an $M>0$ such that $|f(x)−f(y)|<epsilon$Prove $e^x > 1 + x +frac x^22! + ··· + fracx^nn!$ for $x>0$ with induction.Show that if $x$ is non-negative and $x_n$ is a non-negative sequence converging to $x$ then $sqrtx_ntosqrtx$ as $ntoinfty$
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Use Taylor’s Theorem with $n = 2$ to give a second proof of the inequality $e^x$ $ge$ $1+x$ for all real $x$.
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Use Lagrange Remainder Theorem to Prove InequalityFind all the real values of x for which $sum _1^infty x^n/n$ converges.Taylor theorem and a $C^3$ function with the following property…Bounded sequence with subsequences all converging to the same limit means that the sequence itself converges to the same limit.Use the arithmetic-geometric inequality for this list to deduce the arithmetic-geometric inequality for $n$.How do I use the triangle inequality to finish this fact about complex numbers?How to use the theorem of GaußShow that for every $epsilon > 0$ there exists an $M>0$ such that $|f(x)−f(y)|<epsilon$Prove $e^x > 1 + x +frac x^22! + ··· + fracx^nn!$ for $x>0$ with induction.Show that if $x$ is non-negative and $x_n$ is a non-negative sequence converging to $x$ then $sqrtx_ntosqrtx$ as $ntoinfty$
$begingroup$
I've used Taylors theorem to say $e^x$ = $$e^x=1+x+fracx^22!+fracx^33!+dots$$ but not sure where to go from there. I'm not sure what $n$ refers to in the question so not sure how to use that.
analysis
asked Apr 2 at 14:25
HistoryFan12HistoryFan12
255
$endgroup$
add a comment |
$begingroup$
I've used Taylors theorem to say $e^x$ = $$e^x=1+x+fracx^22!+fracx^33!+dots$$ but not sure where to go from there. I'm not sure what $n$ refers to in the question so not sure how to use that.
analysis
asked Apr 2 at 14:25
HistoryFan12HistoryFan12
255
$endgroup$
add a comment |
$begingroup$
I've used Taylors theorem to say $e^x$ = $$e^x=1+x+fracx^22!+fracx^33!+dots$$ but not sure where to go from there. I'm not sure what $n$ refers to in the question so not sure how to use that.
analysis
asked Apr 2 at 14:25
HistoryFan12HistoryFan12
255
$endgroup$
I've used Taylors theorem to say $e^x$ = $$e^x=1+x+fracx^22!+fracx^33!+dots$$ but not sure where to go from there. I'm not sure what $n$ refers to in the question so not sure how to use that.
analysis
analysis
asked Apr 2 at 14:25
HistoryFan12HistoryFan12
255
asked Apr 2 at 14:25
HistoryFan12HistoryFan12
255
asked Apr 2 at 14:25
HistoryFan12HistoryFan12
255
asked Apr 2 at 14:25
HistoryFan12HistoryFan12
255
asked Apr 2 at 14:25
HistoryFan12HistoryFan12
255
255
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let $f(t) = e^t$.
Pick a point $x in mathbf R$.
According to Taylor's theorem there is a point $c$ in between $x$ and $0$ with the property that $$f(x) = f(0) + f'(0)(x-0) + fracf''(c)2(x-0)^2.$$ That is to say,
$$e^x = 1 + x + frace^c2 x^2.$$
Now observe $dfrace^c2x^2 ge 0$ regardless of the values of $x$ and $c$.
answered Apr 2 at 14:33
Umberto P.Umberto P.
40.6k13370
$endgroup$
add a comment |
$begingroup$
n=2 typically refers to the term that is second order in your variable, in this case, x. So you want to take the Taylor series to the term $x^2/2$. This is also called the Taylor approximation to Second Order. For the sake of limits, it's usually good to include the follow up term in big O notation.
So here we approximate $e^xapprox 1 + x + x^2/2!+O(x^3)$
From here you should be able to set up an inequality that will establish the proof.
answered Apr 2 at 14:31
TurlocTheRedTurlocTheRed
1,034411
$endgroup$
add a comment |
$begingroup$
Taylor theorem with remainder
$$f(x)=f(0)+f'(0)x+...+frac f^(n-1)(0)(n-1)!x^n-1+R_n(x)$$
There are different ways to express this reminder term $R_n(x)$
In your case you have for $f(x)=e^x$ and $n=2$
$$e^x=1+x+R_2(x)$$
You need to prove that $R_2(x)geq 0$.
answered Apr 2 at 14:39
Julian MejiaJulian Mejia
76729
$endgroup$
add a comment |
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3 Answers
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3 Answers
3
active
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votes
$begingroup$
Let $f(t) = e^t$.
Pick a point $x in mathbf R$.
According to Taylor's theorem there is a point $c$ in between $x$ and $0$ with the property that $$f(x) = f(0) + f'(0)(x-0) + fracf''(c)2(x-0)^2.$$ That is to say,
$$e^x = 1 + x + frace^c2 x^2.$$
Now observe $dfrace^c2x^2 ge 0$ regardless of the values of $x$ and $c$.
answered Apr 2 at 14:33
Umberto P.Umberto P.
40.6k13370
$endgroup$
add a comment |
$begingroup$
Let $f(t) = e^t$.
Pick a point $x in mathbf R$.
According to Taylor's theorem there is a point $c$ in between $x$ and $0$ with the property that $$f(x) = f(0) + f'(0)(x-0) + fracf''(c)2(x-0)^2.$$ That is to say,
$$e^x = 1 + x + frace^c2 x^2.$$
Now observe $dfrace^c2x^2 ge 0$ regardless of the values of $x$ and $c$.
answered Apr 2 at 14:33
Umberto P.Umberto P.
40.6k13370
$endgroup$
add a comment |
$begingroup$
Let $f(t) = e^t$.
Pick a point $x in mathbf R$.
According to Taylor's theorem there is a point $c$ in between $x$ and $0$ with the property that $$f(x) = f(0) + f'(0)(x-0) + fracf''(c)2(x-0)^2.$$ That is to say,
$$e^x = 1 + x + frace^c2 x^2.$$
Now observe $dfrace^c2x^2 ge 0$ regardless of the values of $x$ and $c$.
answered Apr 2 at 14:33
Umberto P.Umberto P.
40.6k13370
$endgroup$
Let $f(t) = e^t$.
Pick a point $x in mathbf R$.
According to Taylor's theorem there is a point $c$ in between $x$ and $0$ with the property that $$f(x) = f(0) + f'(0)(x-0) + fracf''(c)2(x-0)^2.$$ That is to say,
$$e^x = 1 + x + frace^c2 x^2.$$
Now observe $dfrace^c2x^2 ge 0$ regardless of the values of $x$ and $c$.
answered Apr 2 at 14:33
Umberto P.Umberto P.
40.6k13370
answered Apr 2 at 14:33
Umberto P.Umberto P.
40.6k13370
answered Apr 2 at 14:33
Umberto P.Umberto P.
40.6k13370
answered Apr 2 at 14:33
Umberto P.Umberto P.
40.6k13370
40.6k13370
add a comment |
add a comment |
$begingroup$
n=2 typically refers to the term that is second order in your variable, in this case, x. So you want to take the Taylor series to the term $x^2/2$. This is also called the Taylor approximation to Second Order. For the sake of limits, it's usually good to include the follow up term in big O notation.
So here we approximate $e^xapprox 1 + x + x^2/2!+O(x^3)$
From here you should be able to set up an inequality that will establish the proof.
answered Apr 2 at 14:31
TurlocTheRedTurlocTheRed
1,034411
$endgroup$
add a comment |
$begingroup$
n=2 typically refers to the term that is second order in your variable, in this case, x. So you want to take the Taylor series to the term $x^2/2$. This is also called the Taylor approximation to Second Order. For the sake of limits, it's usually good to include the follow up term in big O notation.
So here we approximate $e^xapprox 1 + x + x^2/2!+O(x^3)$
From here you should be able to set up an inequality that will establish the proof.
answered Apr 2 at 14:31
TurlocTheRedTurlocTheRed
1,034411
$endgroup$
add a comment |
$begingroup$
n=2 typically refers to the term that is second order in your variable, in this case, x. So you want to take the Taylor series to the term $x^2/2$. This is also called the Taylor approximation to Second Order. For the sake of limits, it's usually good to include the follow up term in big O notation.
So here we approximate $e^xapprox 1 + x + x^2/2!+O(x^3)$
From here you should be able to set up an inequality that will establish the proof.
answered Apr 2 at 14:31
TurlocTheRedTurlocTheRed
1,034411
$endgroup$
n=2 typically refers to the term that is second order in your variable, in this case, x. So you want to take the Taylor series to the term $x^2/2$. This is also called the Taylor approximation to Second Order. For the sake of limits, it's usually good to include the follow up term in big O notation.
So here we approximate $e^xapprox 1 + x + x^2/2!+O(x^3)$
From here you should be able to set up an inequality that will establish the proof.
answered Apr 2 at 14:31
TurlocTheRedTurlocTheRed
1,034411
answered Apr 2 at 14:31
TurlocTheRedTurlocTheRed
1,034411
answered Apr 2 at 14:31
TurlocTheRedTurlocTheRed
1,034411
answered Apr 2 at 14:31
TurlocTheRedTurlocTheRed
1,034411
1,034411
add a comment |
add a comment |
$begingroup$
Taylor theorem with remainder
$$f(x)=f(0)+f'(0)x+...+frac f^(n-1)(0)(n-1)!x^n-1+R_n(x)$$
There are different ways to express this reminder term $R_n(x)$
In your case you have for $f(x)=e^x$ and $n=2$
$$e^x=1+x+R_2(x)$$
You need to prove that $R_2(x)geq 0$.
answered Apr 2 at 14:39
Julian MejiaJulian Mejia
76729
$endgroup$
add a comment |
$begingroup$
Taylor theorem with remainder
$$f(x)=f(0)+f'(0)x+...+frac f^(n-1)(0)(n-1)!x^n-1+R_n(x)$$
There are different ways to express this reminder term $R_n(x)$
In your case you have for $f(x)=e^x$ and $n=2$
$$e^x=1+x+R_2(x)$$
You need to prove that $R_2(x)geq 0$.
answered Apr 2 at 14:39
Julian MejiaJulian Mejia
76729
$endgroup$
add a comment |
$begingroup$
Taylor theorem with remainder
$$f(x)=f(0)+f'(0)x+...+frac f^(n-1)(0)(n-1)!x^n-1+R_n(x)$$
There are different ways to express this reminder term $R_n(x)$
In your case you have for $f(x)=e^x$ and $n=2$
$$e^x=1+x+R_2(x)$$
You need to prove that $R_2(x)geq 0$.
answered Apr 2 at 14:39
Julian MejiaJulian Mejia
76729
$endgroup$
Taylor theorem with remainder
$$f(x)=f(0)+f'(0)x+...+frac f^(n-1)(0)(n-1)!x^n-1+R_n(x)$$
There are different ways to express this reminder term $R_n(x)$
In your case you have for $f(x)=e^x$ and $n=2$
$$e^x=1+x+R_2(x)$$
You need to prove that $R_2(x)geq 0$.
answered Apr 2 at 14:39
Julian MejiaJulian Mejia
76729
answered Apr 2 at 14:39
Julian MejiaJulian Mejia
76729
answered Apr 2 at 14:39
Julian MejiaJulian Mejia
76729
answered Apr 2 at 14:39
Julian MejiaJulian Mejia
76729
76729
add a comment |
add a comment |
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