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Let $fin C^2(mathbb R)$. I have to prove that there is $cin [a,b]$ s.t. $f(b)=f(a)+f'(a)(b-a)+fracf''(c)2(b-a)^2.$
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Let $fin C^2(mathbb R)$. I have to prove that there is $cin [a,b]$ s.t. $$f(b)=f(a)+f'(a)(b-a)+fracf''(c)2(b-a)^2.$$
I know that we can apply Rolle's theorem twice with $$g(x):=f(x)-f(a)-f'(a)(x-a)+fracf(a)+f'(a)(b-a)-f(b)(b-a)^2(x-a)^2,$$
but set such a function looks so unnatural for me (I would never think to set such a function), I was wondering if there where a more intuitive way to do it.
The aim of the exercise is to prove Taylor theorem, so I can't use Taylor polynomial.
real-analysis
asked Jul 23 at 21:05
user386627
714214
add a comment |Â
up vote
3
down vote
favorite
Let $fin C^2(mathbb R)$. I have to prove that there is $cin [a,b]$ s.t. $$f(b)=f(a)+f'(a)(b-a)+fracf''(c)2(b-a)^2.$$
I know that we can apply Rolle's theorem twice with $$g(x):=f(x)-f(a)-f'(a)(x-a)+fracf(a)+f'(a)(b-a)-f(b)(b-a)^2(x-a)^2,$$
but set such a function looks so unnatural for me (I would never think to set such a function), I was wondering if there where a more intuitive way to do it.
The aim of the exercise is to prove Taylor theorem, so I can't use Taylor polynomial.
real-analysis
asked Jul 23 at 21:05
user386627
714214
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $fin C^2(mathbb R)$. I have to prove that there is $cin [a,b]$ s.t. $$f(b)=f(a)+f'(a)(b-a)+fracf''(c)2(b-a)^2.$$
I know that we can apply Rolle's theorem twice with $$g(x):=f(x)-f(a)-f'(a)(x-a)+fracf(a)+f'(a)(b-a)-f(b)(b-a)^2(x-a)^2,$$
but set such a function looks so unnatural for me (I would never think to set such a function), I was wondering if there where a more intuitive way to do it.
The aim of the exercise is to prove Taylor theorem, so I can't use Taylor polynomial.
real-analysis
asked Jul 23 at 21:05
user386627
714214
Let $fin C^2(mathbb R)$. I have to prove that there is $cin [a,b]$ s.t. $$f(b)=f(a)+f'(a)(b-a)+fracf''(c)2(b-a)^2.$$
I know that we can apply Rolle's theorem twice with $$g(x):=f(x)-f(a)-f'(a)(x-a)+fracf(a)+f'(a)(b-a)-f(b)(b-a)^2(x-a)^2,$$
but set such a function looks so unnatural for me (I would never think to set such a function), I was wondering if there where a more intuitive way to do it.
The aim of the exercise is to prove Taylor theorem, so I can't use Taylor polynomial.
real-analysis
asked Jul 23 at 21:05
user386627
714214
edited Jul 23 at 21:13
asked Jul 23 at 21:05
user386627
714214
asked Jul 23 at 21:05
user386627
714214
asked Jul 23 at 21:05
user386627
714214
714214
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
7
down vote
accepted
First, we have:
beginalign*
f(b) &= f(a) + int_a^b f'(t),dt \
&= f(a) + int_a^b left( f'(a) + int_a^t f''(u),du right) ,dt \
&= f(a) + (b-a) f'(a) + int_a^b int_a^t f''(u),du,dt \
&= f(a) + (b-a)f'(a) + int_a^b int_u^b f''(u),dt,du \
&= f(a) + (b-a)f'(a) + int_a^b (b-u) f''(u),du.
endalign*
Now, since $f''$ is continuous on $[a,b]$, it has a minimum value $m$ and a maximum value $M$ on this interval. Then
$$frac12(b-a)^2 m le int_a^b (b-u) f''(u),du le frac12(b-a)^2 M.$$
Therefore, $frac2(b-a)^2 int_a^b (b-u) f''(u)$ lies between $m$ and $M$, so by the intermediate value theorem, there is some $cin [a,b]$ such that $f''(c) = frac2(b-a)^2 int_a^b (b-u) f''(u)$. It then follows that for this $c$,
$$f(b) = f(a) + (b-a) f'(a) + frac12 (b-a)^2 f''(c)$$
as desired.
answered Jul 23 at 21:30
Daniel Schepler
6,7331513
Waouuu ! wonderful :) Moreover, I understand where come from the rest integral in Taylor. Your answer is perfect and helped me A LOT !!!
– user386627
Jul 23 at 21:32
Nice, after the first application of FTC I always used integration by parts to get to the last equality: never thought about iterating FTC and Fubini: it seems also much more natural.
– Bob
Jul 23 at 21:55
add a comment |Â
up vote
1
down vote
Perhaps I can explain what's going on in that auxiliary function.
The way one usually goes about proving the MVT, is that you "slant" your original function by a linear function so that the difference is zero at the end points. This sets you up to use Rolle's Theorem.
It's also useful to notice here that we use a linear polynomial because a linear polynomial is the bare minimum of what we would need to get two desiderata: we want the difference to be $0$ at $a$ and $0$ at $b$.
What's going on in that function is that it's like a "second-order slant". The first tier of this slant is similar to the first application: we want the difference to be zero at the endpoints. This will give us a point in the middle where the first derivative will be zero. But now we want another point where the derivative will be zero so that we can invoke Rolle's Theorem again. Why don't we be easy on ourselves and just stipulate where that other zero will be? Let's construct the slanting function so that the difference will be $0$ at $a$, $0$ at $b$, and so that the derivative will be $0$ at $a$. This is three criteria we want. We can use a quadratic to get these things we want. And the quadratic you have is the one that will do that.
In general, if you want $n+1$ criteria on a polynomial and its subsequent derivatives to satisfy, you can find an $n$ degree polynomial that will do the trick.
This could extend to prove Taylor's Theorem in general. The slanting function would still make the difference $0$ at the endpoints. But you would demand a much higher degree of vanishing for the derivatives at $a$.
If you really want to use induction here, you might try proving this lemma first:
Lemma: Suppose you have $n+1$ real numbers $a_0, a_1, ldots, a_n-1$ and $b_0$. Then there exists a polynomial $p$ of degree $n$ on $[a,b]$ such that
$$p(a)=a_0,, p'(a)=a_1,,ldots,,, p^(n-1)(a)=a_n-1, text and,, p(b)=b_0,.$$
From there you would consider $f-p$ and apply Rolle's theorem to your heart's desire.
answered Jul 23 at 21:55
Robert Wolfe
5,30722261
Thank you for the explanation. I't very useful :)
– user386627
Jul 23 at 22:05
add a comment |Â
up vote
0
down vote
Consider the taylor expansion of $f(x)$ at $x_0=a$:
$f(x)=f(a)+frac f'(a) 1!(x-a)^1 +R_2(x)$
,
Applying Lagrange`s form for the remainder we will get :
$R_1=frac f''(c_x)(x-a)^2 2!$, looking at $x=b$ we get the form needed. ($c_x in(a,b)$)
Edit: Since you can`t use the taylor expansion, You can define a maybe more intiutive function :
$g(x)=f(x)-(f(a)+f'(a)(x-a))$, This $g(x)$ is actually the remainder function of the 2nd degree expansion of the Taylor expansion.
define $gamma(t)=(x-a)^2$ , and apply Cauchy`s mean value theorem, to get :
$exists rin(x,a): frac g(x)-g(a) gamma(x)-gamma(a)=frac g'(r) gamma'(r)$, it is easy to see $g(a)=gamma(a)=0$, so rearranging the terms we get :
$g(x)=frac gamma(x) g'(r) gamma'(x) = frac (x-a)^2(f'(r)-f'(a)) 2(r-a)$, and applying the mean value again to get $exists cin(x,r) : g(x)=fracf''(c)(x-a)^2 2$.
$implies fracf''(c)(x-a)^2 2=f(x)-(f(a)+f'(a)(x-a))$, set $x=b$ and rearrange the terms to get the form needed.
I still had to use mean-value theorem twice but I hope this answer is any help.
answered Jul 23 at 21:12
Sar
40410
The aim of the exercise is to prove Taylor theorem, so I can't use it.
– user386627
Jul 23 at 21:13
I`ve added a proof with a (maybe) more intuitive function.
– Sar
Jul 23 at 21:47
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
First, we have:
beginalign*
f(b) &= f(a) + int_a^b f'(t),dt \
&= f(a) + int_a^b left( f'(a) + int_a^t f''(u),du right) ,dt \
&= f(a) + (b-a) f'(a) + int_a^b int_a^t f''(u),du,dt \
&= f(a) + (b-a)f'(a) + int_a^b int_u^b f''(u),dt,du \
&= f(a) + (b-a)f'(a) + int_a^b (b-u) f''(u),du.
endalign*
Now, since $f''$ is continuous on $[a,b]$, it has a minimum value $m$ and a maximum value $M$ on this interval. Then
$$frac12(b-a)^2 m le int_a^b (b-u) f''(u),du le frac12(b-a)^2 M.$$
Therefore, $frac2(b-a)^2 int_a^b (b-u) f''(u)$ lies between $m$ and $M$, so by the intermediate value theorem, there is some $cin [a,b]$ such that $f''(c) = frac2(b-a)^2 int_a^b (b-u) f''(u)$. It then follows that for this $c$,
$$f(b) = f(a) + (b-a) f'(a) + frac12 (b-a)^2 f''(c)$$
as desired.
answered Jul 23 at 21:30
Daniel Schepler
6,7331513
Waouuu ! wonderful :) Moreover, I understand where come from the rest integral in Taylor. Your answer is perfect and helped me A LOT !!!
– user386627
Jul 23 at 21:32
Nice, after the first application of FTC I always used integration by parts to get to the last equality: never thought about iterating FTC and Fubini: it seems also much more natural.
– Bob
Jul 23 at 21:55
add a comment |Â
up vote
7
down vote
accepted
First, we have:
beginalign*
f(b) &= f(a) + int_a^b f'(t),dt \
&= f(a) + int_a^b left( f'(a) + int_a^t f''(u),du right) ,dt \
&= f(a) + (b-a) f'(a) + int_a^b int_a^t f''(u),du,dt \
&= f(a) + (b-a)f'(a) + int_a^b int_u^b f''(u),dt,du \
&= f(a) + (b-a)f'(a) + int_a^b (b-u) f''(u),du.
endalign*
Now, since $f''$ is continuous on $[a,b]$, it has a minimum value $m$ and a maximum value $M$ on this interval. Then
$$frac12(b-a)^2 m le int_a^b (b-u) f''(u),du le frac12(b-a)^2 M.$$
Therefore, $frac2(b-a)^2 int_a^b (b-u) f''(u)$ lies between $m$ and $M$, so by the intermediate value theorem, there is some $cin [a,b]$ such that $f''(c) = frac2(b-a)^2 int_a^b (b-u) f''(u)$. It then follows that for this $c$,
$$f(b) = f(a) + (b-a) f'(a) + frac12 (b-a)^2 f''(c)$$
as desired.
answered Jul 23 at 21:30
Daniel Schepler
6,7331513
Waouuu ! wonderful :) Moreover, I understand where come from the rest integral in Taylor. Your answer is perfect and helped me A LOT !!!
– user386627
Jul 23 at 21:32
Nice, after the first application of FTC I always used integration by parts to get to the last equality: never thought about iterating FTC and Fubini: it seems also much more natural.
– Bob
Jul 23 at 21:55
add a comment |Â
up vote
7
down vote
accepted
up vote
7
down vote
accepted
First, we have:
beginalign*
f(b) &= f(a) + int_a^b f'(t),dt \
&= f(a) + int_a^b left( f'(a) + int_a^t f''(u),du right) ,dt \
&= f(a) + (b-a) f'(a) + int_a^b int_a^t f''(u),du,dt \
&= f(a) + (b-a)f'(a) + int_a^b int_u^b f''(u),dt,du \
&= f(a) + (b-a)f'(a) + int_a^b (b-u) f''(u),du.
endalign*
Now, since $f''$ is continuous on $[a,b]$, it has a minimum value $m$ and a maximum value $M$ on this interval. Then
$$frac12(b-a)^2 m le int_a^b (b-u) f''(u),du le frac12(b-a)^2 M.$$
Therefore, $frac2(b-a)^2 int_a^b (b-u) f''(u)$ lies between $m$ and $M$, so by the intermediate value theorem, there is some $cin [a,b]$ such that $f''(c) = frac2(b-a)^2 int_a^b (b-u) f''(u)$. It then follows that for this $c$,
$$f(b) = f(a) + (b-a) f'(a) + frac12 (b-a)^2 f''(c)$$
as desired.
answered Jul 23 at 21:30
Daniel Schepler
6,7331513
First, we have:
beginalign*
f(b) &= f(a) + int_a^b f'(t),dt \
&= f(a) + int_a^b left( f'(a) + int_a^t f''(u),du right) ,dt \
&= f(a) + (b-a) f'(a) + int_a^b int_a^t f''(u),du,dt \
&= f(a) + (b-a)f'(a) + int_a^b int_u^b f''(u),dt,du \
&= f(a) + (b-a)f'(a) + int_a^b (b-u) f''(u),du.
endalign*
Now, since $f''$ is continuous on $[a,b]$, it has a minimum value $m$ and a maximum value $M$ on this interval. Then
$$frac12(b-a)^2 m le int_a^b (b-u) f''(u),du le frac12(b-a)^2 M.$$
Therefore, $frac2(b-a)^2 int_a^b (b-u) f''(u)$ lies between $m$ and $M$, so by the intermediate value theorem, there is some $cin [a,b]$ such that $f''(c) = frac2(b-a)^2 int_a^b (b-u) f''(u)$. It then follows that for this $c$,
$$f(b) = f(a) + (b-a) f'(a) + frac12 (b-a)^2 f''(c)$$
as desired.
answered Jul 23 at 21:30
Daniel Schepler
6,7331513
answered Jul 23 at 21:30
Daniel Schepler
6,7331513
answered Jul 23 at 21:30
Daniel Schepler
6,7331513
answered Jul 23 at 21:30
Daniel Schepler
6,7331513
6,7331513
Waouuu ! wonderful :) Moreover, I understand where come from the rest integral in Taylor. Your answer is perfect and helped me A LOT !!!
– user386627
Jul 23 at 21:32
Nice, after the first application of FTC I always used integration by parts to get to the last equality: never thought about iterating FTC and Fubini: it seems also much more natural.
– Bob
Jul 23 at 21:55
add a comment |Â
Waouuu ! wonderful :) Moreover, I understand where come from the rest integral in Taylor. Your answer is perfect and helped me A LOT !!!
– user386627
Jul 23 at 21:32
Nice, after the first application of FTC I always used integration by parts to get to the last equality: never thought about iterating FTC and Fubini: it seems also much more natural.
– Bob
Jul 23 at 21:55
Waouuu ! wonderful :) Moreover, I understand where come from the rest integral in Taylor. Your answer is perfect and helped me A LOT !!!
– user386627
Jul 23 at 21:32
Waouuu ! wonderful :) Moreover, I understand where come from the rest integral in Taylor. Your answer is perfect and helped me A LOT !!!
– user386627
Jul 23 at 21:32
Nice, after the first application of FTC I always used integration by parts to get to the last equality: never thought about iterating FTC and Fubini: it seems also much more natural.
– Bob
Jul 23 at 21:55
Nice, after the first application of FTC I always used integration by parts to get to the last equality: never thought about iterating FTC and Fubini: it seems also much more natural.
– Bob
Jul 23 at 21:55
add a comment |Â
up vote
1
down vote
Perhaps I can explain what's going on in that auxiliary function.
The way one usually goes about proving the MVT, is that you "slant" your original function by a linear function so that the difference is zero at the end points. This sets you up to use Rolle's Theorem.
It's also useful to notice here that we use a linear polynomial because a linear polynomial is the bare minimum of what we would need to get two desiderata: we want the difference to be $0$ at $a$ and $0$ at $b$.
What's going on in that function is that it's like a "second-order slant". The first tier of this slant is similar to the first application: we want the difference to be zero at the endpoints. This will give us a point in the middle where the first derivative will be zero. But now we want another point where the derivative will be zero so that we can invoke Rolle's Theorem again. Why don't we be easy on ourselves and just stipulate where that other zero will be? Let's construct the slanting function so that the difference will be $0$ at $a$, $0$ at $b$, and so that the derivative will be $0$ at $a$. This is three criteria we want. We can use a quadratic to get these things we want. And the quadratic you have is the one that will do that.
In general, if you want $n+1$ criteria on a polynomial and its subsequent derivatives to satisfy, you can find an $n$ degree polynomial that will do the trick.
This could extend to prove Taylor's Theorem in general. The slanting function would still make the difference $0$ at the endpoints. But you would demand a much higher degree of vanishing for the derivatives at $a$.
If you really want to use induction here, you might try proving this lemma first:
Lemma: Suppose you have $n+1$ real numbers $a_0, a_1, ldots, a_n-1$ and $b_0$. Then there exists a polynomial $p$ of degree $n$ on $[a,b]$ such that
$$p(a)=a_0,, p'(a)=a_1,,ldots,,, p^(n-1)(a)=a_n-1, text and,, p(b)=b_0,.$$
From there you would consider $f-p$ and apply Rolle's theorem to your heart's desire.
answered Jul 23 at 21:55
Robert Wolfe
5,30722261
Thank you for the explanation. I't very useful :)
– user386627
Jul 23 at 22:05
add a comment |Â
up vote
1
down vote
Perhaps I can explain what's going on in that auxiliary function.
The way one usually goes about proving the MVT, is that you "slant" your original function by a linear function so that the difference is zero at the end points. This sets you up to use Rolle's Theorem.
It's also useful to notice here that we use a linear polynomial because a linear polynomial is the bare minimum of what we would need to get two desiderata: we want the difference to be $0$ at $a$ and $0$ at $b$.
What's going on in that function is that it's like a "second-order slant". The first tier of this slant is similar to the first application: we want the difference to be zero at the endpoints. This will give us a point in the middle where the first derivative will be zero. But now we want another point where the derivative will be zero so that we can invoke Rolle's Theorem again. Why don't we be easy on ourselves and just stipulate where that other zero will be? Let's construct the slanting function so that the difference will be $0$ at $a$, $0$ at $b$, and so that the derivative will be $0$ at $a$. This is three criteria we want. We can use a quadratic to get these things we want. And the quadratic you have is the one that will do that.
In general, if you want $n+1$ criteria on a polynomial and its subsequent derivatives to satisfy, you can find an $n$ degree polynomial that will do the trick.
This could extend to prove Taylor's Theorem in general. The slanting function would still make the difference $0$ at the endpoints. But you would demand a much higher degree of vanishing for the derivatives at $a$.
If you really want to use induction here, you might try proving this lemma first:
Lemma: Suppose you have $n+1$ real numbers $a_0, a_1, ldots, a_n-1$ and $b_0$. Then there exists a polynomial $p$ of degree $n$ on $[a,b]$ such that
$$p(a)=a_0,, p'(a)=a_1,,ldots,,, p^(n-1)(a)=a_n-1, text and,, p(b)=b_0,.$$
From there you would consider $f-p$ and apply Rolle's theorem to your heart's desire.
answered Jul 23 at 21:55
Robert Wolfe
5,30722261
Thank you for the explanation. I't very useful :)
– user386627
Jul 23 at 22:05
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Perhaps I can explain what's going on in that auxiliary function.
The way one usually goes about proving the MVT, is that you "slant" your original function by a linear function so that the difference is zero at the end points. This sets you up to use Rolle's Theorem.
It's also useful to notice here that we use a linear polynomial because a linear polynomial is the bare minimum of what we would need to get two desiderata: we want the difference to be $0$ at $a$ and $0$ at $b$.
What's going on in that function is that it's like a "second-order slant". The first tier of this slant is similar to the first application: we want the difference to be zero at the endpoints. This will give us a point in the middle where the first derivative will be zero. But now we want another point where the derivative will be zero so that we can invoke Rolle's Theorem again. Why don't we be easy on ourselves and just stipulate where that other zero will be? Let's construct the slanting function so that the difference will be $0$ at $a$, $0$ at $b$, and so that the derivative will be $0$ at $a$. This is three criteria we want. We can use a quadratic to get these things we want. And the quadratic you have is the one that will do that.
In general, if you want $n+1$ criteria on a polynomial and its subsequent derivatives to satisfy, you can find an $n$ degree polynomial that will do the trick.
This could extend to prove Taylor's Theorem in general. The slanting function would still make the difference $0$ at the endpoints. But you would demand a much higher degree of vanishing for the derivatives at $a$.
If you really want to use induction here, you might try proving this lemma first:
Lemma: Suppose you have $n+1$ real numbers $a_0, a_1, ldots, a_n-1$ and $b_0$. Then there exists a polynomial $p$ of degree $n$ on $[a,b]$ such that
$$p(a)=a_0,, p'(a)=a_1,,ldots,,, p^(n-1)(a)=a_n-1, text and,, p(b)=b_0,.$$
From there you would consider $f-p$ and apply Rolle's theorem to your heart's desire.
answered Jul 23 at 21:55
Robert Wolfe
5,30722261
Perhaps I can explain what's going on in that auxiliary function.
The way one usually goes about proving the MVT, is that you "slant" your original function by a linear function so that the difference is zero at the end points. This sets you up to use Rolle's Theorem.
It's also useful to notice here that we use a linear polynomial because a linear polynomial is the bare minimum of what we would need to get two desiderata: we want the difference to be $0$ at $a$ and $0$ at $b$.
What's going on in that function is that it's like a "second-order slant". The first tier of this slant is similar to the first application: we want the difference to be zero at the endpoints. This will give us a point in the middle where the first derivative will be zero. But now we want another point where the derivative will be zero so that we can invoke Rolle's Theorem again. Why don't we be easy on ourselves and just stipulate where that other zero will be? Let's construct the slanting function so that the difference will be $0$ at $a$, $0$ at $b$, and so that the derivative will be $0$ at $a$. This is three criteria we want. We can use a quadratic to get these things we want. And the quadratic you have is the one that will do that.
In general, if you want $n+1$ criteria on a polynomial and its subsequent derivatives to satisfy, you can find an $n$ degree polynomial that will do the trick.
This could extend to prove Taylor's Theorem in general. The slanting function would still make the difference $0$ at the endpoints. But you would demand a much higher degree of vanishing for the derivatives at $a$.
If you really want to use induction here, you might try proving this lemma first:
Lemma: Suppose you have $n+1$ real numbers $a_0, a_1, ldots, a_n-1$ and $b_0$. Then there exists a polynomial $p$ of degree $n$ on $[a,b]$ such that
$$p(a)=a_0,, p'(a)=a_1,,ldots,,, p^(n-1)(a)=a_n-1, text and,, p(b)=b_0,.$$
From there you would consider $f-p$ and apply Rolle's theorem to your heart's desire.
answered Jul 23 at 21:55
Robert Wolfe
5,30722261
edited Jul 23 at 22:15
answered Jul 23 at 21:55
Robert Wolfe
5,30722261
answered Jul 23 at 21:55
Robert Wolfe
5,30722261
answered Jul 23 at 21:55
Robert Wolfe
5,30722261
5,30722261
Thank you for the explanation. I't very useful :)
– user386627
Jul 23 at 22:05
add a comment |Â
Thank you for the explanation. I't very useful :)
– user386627
Jul 23 at 22:05
Thank you for the explanation. I't very useful :)
– user386627
Jul 23 at 22:05
Thank you for the explanation. I't very useful :)
– user386627
Jul 23 at 22:05
add a comment |Â
up vote
0
down vote
Consider the taylor expansion of $f(x)$ at $x_0=a$:
$f(x)=f(a)+frac f'(a) 1!(x-a)^1 +R_2(x)$
,
Applying Lagrange`s form for the remainder we will get :
$R_1=frac f''(c_x)(x-a)^2 2!$, looking at $x=b$ we get the form needed. ($c_x in(a,b)$)
Edit: Since you can`t use the taylor expansion, You can define a maybe more intiutive function :
$g(x)=f(x)-(f(a)+f'(a)(x-a))$, This $g(x)$ is actually the remainder function of the 2nd degree expansion of the Taylor expansion.
define $gamma(t)=(x-a)^2$ , and apply Cauchy`s mean value theorem, to get :
$exists rin(x,a): frac g(x)-g(a) gamma(x)-gamma(a)=frac g'(r) gamma'(r)$, it is easy to see $g(a)=gamma(a)=0$, so rearranging the terms we get :
$g(x)=frac gamma(x) g'(r) gamma'(x) = frac (x-a)^2(f'(r)-f'(a)) 2(r-a)$, and applying the mean value again to get $exists cin(x,r) : g(x)=fracf''(c)(x-a)^2 2$.
$implies fracf''(c)(x-a)^2 2=f(x)-(f(a)+f'(a)(x-a))$, set $x=b$ and rearrange the terms to get the form needed.
I still had to use mean-value theorem twice but I hope this answer is any help.
answered Jul 23 at 21:12
Sar
40410
The aim of the exercise is to prove Taylor theorem, so I can't use it.
– user386627
Jul 23 at 21:13
I`ve added a proof with a (maybe) more intuitive function.
– Sar
Jul 23 at 21:47
add a comment |Â
up vote
0
down vote
Consider the taylor expansion of $f(x)$ at $x_0=a$:
$f(x)=f(a)+frac f'(a) 1!(x-a)^1 +R_2(x)$
,
Applying Lagrange`s form for the remainder we will get :
$R_1=frac f''(c_x)(x-a)^2 2!$, looking at $x=b$ we get the form needed. ($c_x in(a,b)$)
Edit: Since you can`t use the taylor expansion, You can define a maybe more intiutive function :
$g(x)=f(x)-(f(a)+f'(a)(x-a))$, This $g(x)$ is actually the remainder function of the 2nd degree expansion of the Taylor expansion.
define $gamma(t)=(x-a)^2$ , and apply Cauchy`s mean value theorem, to get :
$exists rin(x,a): frac g(x)-g(a) gamma(x)-gamma(a)=frac g'(r) gamma'(r)$, it is easy to see $g(a)=gamma(a)=0$, so rearranging the terms we get :
$g(x)=frac gamma(x) g'(r) gamma'(x) = frac (x-a)^2(f'(r)-f'(a)) 2(r-a)$, and applying the mean value again to get $exists cin(x,r) : g(x)=fracf''(c)(x-a)^2 2$.
$implies fracf''(c)(x-a)^2 2=f(x)-(f(a)+f'(a)(x-a))$, set $x=b$ and rearrange the terms to get the form needed.
I still had to use mean-value theorem twice but I hope this answer is any help.
answered Jul 23 at 21:12
Sar
40410
The aim of the exercise is to prove Taylor theorem, so I can't use it.
– user386627
Jul 23 at 21:13
I`ve added a proof with a (maybe) more intuitive function.
– Sar
Jul 23 at 21:47
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Consider the taylor expansion of $f(x)$ at $x_0=a$:
$f(x)=f(a)+frac f'(a) 1!(x-a)^1 +R_2(x)$
,
Applying Lagrange`s form for the remainder we will get :
$R_1=frac f''(c_x)(x-a)^2 2!$, looking at $x=b$ we get the form needed. ($c_x in(a,b)$)
Edit: Since you can`t use the taylor expansion, You can define a maybe more intiutive function :
$g(x)=f(x)-(f(a)+f'(a)(x-a))$, This $g(x)$ is actually the remainder function of the 2nd degree expansion of the Taylor expansion.
define $gamma(t)=(x-a)^2$ , and apply Cauchy`s mean value theorem, to get :
$exists rin(x,a): frac g(x)-g(a) gamma(x)-gamma(a)=frac g'(r) gamma'(r)$, it is easy to see $g(a)=gamma(a)=0$, so rearranging the terms we get :
$g(x)=frac gamma(x) g'(r) gamma'(x) = frac (x-a)^2(f'(r)-f'(a)) 2(r-a)$, and applying the mean value again to get $exists cin(x,r) : g(x)=fracf''(c)(x-a)^2 2$.
$implies fracf''(c)(x-a)^2 2=f(x)-(f(a)+f'(a)(x-a))$, set $x=b$ and rearrange the terms to get the form needed.
I still had to use mean-value theorem twice but I hope this answer is any help.
answered Jul 23 at 21:12
Sar
40410
Consider the taylor expansion of $f(x)$ at $x_0=a$:
$f(x)=f(a)+frac f'(a) 1!(x-a)^1 +R_2(x)$
,
Applying Lagrange`s form for the remainder we will get :
$R_1=frac f''(c_x)(x-a)^2 2!$, looking at $x=b$ we get the form needed. ($c_x in(a,b)$)
Edit: Since you can`t use the taylor expansion, You can define a maybe more intiutive function :
$g(x)=f(x)-(f(a)+f'(a)(x-a))$, This $g(x)$ is actually the remainder function of the 2nd degree expansion of the Taylor expansion.
define $gamma(t)=(x-a)^2$ , and apply Cauchy`s mean value theorem, to get :
$exists rin(x,a): frac g(x)-g(a) gamma(x)-gamma(a)=frac g'(r) gamma'(r)$, it is easy to see $g(a)=gamma(a)=0$, so rearranging the terms we get :
$g(x)=frac gamma(x) g'(r) gamma'(x) = frac (x-a)^2(f'(r)-f'(a)) 2(r-a)$, and applying the mean value again to get $exists cin(x,r) : g(x)=fracf''(c)(x-a)^2 2$.
$implies fracf''(c)(x-a)^2 2=f(x)-(f(a)+f'(a)(x-a))$, set $x=b$ and rearrange the terms to get the form needed.
I still had to use mean-value theorem twice but I hope this answer is any help.
answered Jul 23 at 21:12
Sar
40410
edited Jul 23 at 21:43
answered Jul 23 at 21:12
Sar
40410
answered Jul 23 at 21:12
Sar
40410
answered Jul 23 at 21:12
Sar
40410
40410
The aim of the exercise is to prove Taylor theorem, so I can't use it.
– user386627
Jul 23 at 21:13
I`ve added a proof with a (maybe) more intuitive function.
– Sar
Jul 23 at 21:47
add a comment |Â
The aim of the exercise is to prove Taylor theorem, so I can't use it.
– user386627
Jul 23 at 21:13
I`ve added a proof with a (maybe) more intuitive function.
– Sar
Jul 23 at 21:47
The aim of the exercise is to prove Taylor theorem, so I can't use it.
– user386627
Jul 23 at 21:13
The aim of the exercise is to prove Taylor theorem, so I can't use it.
– user386627
Jul 23 at 21:13
I`ve added a proof with a (maybe) more intuitive function.
– Sar
Jul 23 at 21:47
I`ve added a proof with a (maybe) more intuitive function.
– Sar
Jul 23 at 21:47
add a comment |Â
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