A difficulty in understanding a step in the proof of Thm. 11.5.6 in Petrovic. The Next CEO of Stack OverflowA difficulty in understanding the proof of completeness of $l_2$.A difficulty in understanding theorem 4.2 in Israel Gohberg.showing that a continuous function attains each of its values(or each real number in general) exactly 3 times.Estimating the error of approximation for $sqrtx$ for $|x-1| leq 0.5$. No.4.5.3 PetrovicA difficulty in understanding a proof for L'Hospital's rule (in Petrovic)A difficulty in understanding the proof of 7.2.3 Petrovic.A difficulty in understanding a step in a solution.A difficulty in understanding a statement in example 10.6.6 Petrovic.A difficulty in understanding theorem 10.6.7 in Petrovic.(n-dimensional intermediate value theorem)A difficulty in understanding the n-dimensional second order derivative.
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A difficulty in understanding a step in the proof of Thm. 11.5.6 in Petrovic.
The Next CEO of Stack OverflowA difficulty in understanding the proof of completeness of $l_2$.A difficulty in understanding theorem 4.2 in Israel Gohberg.showing that a continuous function attains each of its values(or each real number in general) exactly 3 times.Estimating the error of approximation for $sqrtx$ for $|x-1| leq 0.5$. No.4.5.3 PetrovicA difficulty in understanding a proof for L'Hospital's rule (in Petrovic)A difficulty in understanding the proof of 7.2.3 Petrovic.A difficulty in understanding a step in a solution.A difficulty in understanding a statement in example 10.6.6 Petrovic.A difficulty in understanding theorem 10.6.7 in Petrovic.(n-dimensional intermediate value theorem)A difficulty in understanding the n-dimensional second order derivative.
$begingroup$
The theorem and its proof is given below:
But I do not understand how the last equality come from the previous one, could anyone explain this for me please?
real-analysis calculus analysis multivariable-calculus
asked Mar 20 at 11:06
hopefullyhopefully
315215
$endgroup$
add a comment |
$begingroup$
The theorem and its proof is given below:
But I do not understand how the last equality come from the previous one, could anyone explain this for me please?
real-analysis calculus analysis multivariable-calculus
asked Mar 20 at 11:06
hopefullyhopefully
315215
$endgroup$
add a comment |
$begingroup$
The theorem and its proof is given below:
But I do not understand how the last equality come from the previous one, could anyone explain this for me please?
real-analysis calculus analysis multivariable-calculus
asked Mar 20 at 11:06
hopefullyhopefully
315215
$endgroup$
The theorem and its proof is given below:
But I do not understand how the last equality come from the previous one, could anyone explain this for me please?
real-analysis calculus analysis multivariable-calculus
real-analysis calculus analysis multivariable-calculus
asked Mar 20 at 11:06
hopefullyhopefully
315215
asked Mar 20 at 11:06
hopefullyhopefully
315215
asked Mar 20 at 11:06
hopefullyhopefully
315215
asked Mar 20 at 11:06
hopefullyhopefully
315215
asked Mar 20 at 11:06
hopefullyhopefully
315215
315215
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You have
$$f(x) cdot (f(x) - f(a)) - f(a)cdot (f(x) - f(a)) = (f(x) - f(a)) cdot (f(x) - f(a)) = ||f(x) - f(a)||^2$$
So, the author brings the term $f(a)cdot (f(x) - f(a))$ from the RHS to the LHS and then just writes the scalar product of $(f(x) - f(a))$ with itself as its squared Euclidian norm.
answered Mar 20 at 11:11
trancelocationtrancelocation
13.4k1827
$endgroup$
add a comment |
$begingroup$
I'll revrite @trancelocation's answer in a more pedagogical way:
beginalign*f(x) cdot (f(x) - f(a)) &= colorredf(a)cdot (f(x) - f(a)) + (f(x) - f(a)) cdot Df(b)(x-a)\
f(x) cdot underline(f(x) - f(a))-colorredf(a)cdot underline(f(x) - f(a))&=(f(x) - f(a)) cdot Df(b)(x-a)\
(f(x)-f(a)) cdot underline(f(x) - f(a)) & =(f(x) - f(a)) cdot Df(b)(x-a)\
|f(x)-f(a)|^2&=(f(x) - f(a)) cdot Df(b)(x-a)
endalign*
and lastly Cauchy Schwartz implies
$$|f(x)-f(a)|^2leq |(f(x) - f(a))| cdot |Df(b)(x-a)|$$
answered Mar 20 at 11:17
b00n heTb00n heT
10.5k12335
$endgroup$
add a comment |
$begingroup$
You just have to pass one term to the LHS.
$$
f(x)cdot (f(x)-f(a))= f(a)cdot(f(x)-f(a))+(f(x)-f(a))Df(b)(x-a) Leftrightarrow
$$
$$
f(x)cdot (f(x)-f(a))-f(a)cdot(f(x)-f(a))=(f(x)-f(a))Df(b)(x-a)Leftrightarrow
$$
$$
(f(x)-f(a))cdot (f(x)-f(a)) = (f(x)-f(a))Df(b)(x-a)Leftrightarrow
$$
$$
|f(x)-f(a)|^2 = (f(x)-f(a))Df(b)(x-a) Rightarrow
$$
$$
|f(x)-f(a)|^2 = |(f(x)-f(a))Df(b)(x-a)| Rightarrow
$$
$$
|f(x)-f(a)|^2 leq |f(x)-f(a)| |Df(b)(x-a)|
$$
answered Mar 20 at 11:18
PierreCarrePierreCarre
1,665212
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You have
$$f(x) cdot (f(x) - f(a)) - f(a)cdot (f(x) - f(a)) = (f(x) - f(a)) cdot (f(x) - f(a)) = ||f(x) - f(a)||^2$$
So, the author brings the term $f(a)cdot (f(x) - f(a))$ from the RHS to the LHS and then just writes the scalar product of $(f(x) - f(a))$ with itself as its squared Euclidian norm.
answered Mar 20 at 11:11
trancelocationtrancelocation
13.4k1827
$endgroup$
add a comment |
$begingroup$
You have
$$f(x) cdot (f(x) - f(a)) - f(a)cdot (f(x) - f(a)) = (f(x) - f(a)) cdot (f(x) - f(a)) = ||f(x) - f(a)||^2$$
So, the author brings the term $f(a)cdot (f(x) - f(a))$ from the RHS to the LHS and then just writes the scalar product of $(f(x) - f(a))$ with itself as its squared Euclidian norm.
answered Mar 20 at 11:11
trancelocationtrancelocation
13.4k1827
$endgroup$
add a comment |
$begingroup$
You have
$$f(x) cdot (f(x) - f(a)) - f(a)cdot (f(x) - f(a)) = (f(x) - f(a)) cdot (f(x) - f(a)) = ||f(x) - f(a)||^2$$
So, the author brings the term $f(a)cdot (f(x) - f(a))$ from the RHS to the LHS and then just writes the scalar product of $(f(x) - f(a))$ with itself as its squared Euclidian norm.
answered Mar 20 at 11:11
trancelocationtrancelocation
13.4k1827
$endgroup$
You have
$$f(x) cdot (f(x) - f(a)) - f(a)cdot (f(x) - f(a)) = (f(x) - f(a)) cdot (f(x) - f(a)) = ||f(x) - f(a)||^2$$
So, the author brings the term $f(a)cdot (f(x) - f(a))$ from the RHS to the LHS and then just writes the scalar product of $(f(x) - f(a))$ with itself as its squared Euclidian norm.
answered Mar 20 at 11:11
trancelocationtrancelocation
13.4k1827
answered Mar 20 at 11:11
trancelocationtrancelocation
13.4k1827
answered Mar 20 at 11:11
trancelocationtrancelocation
13.4k1827
answered Mar 20 at 11:11
trancelocationtrancelocation
13.4k1827
13.4k1827
add a comment |
add a comment |
$begingroup$
I'll revrite @trancelocation's answer in a more pedagogical way:
beginalign*f(x) cdot (f(x) - f(a)) &= colorredf(a)cdot (f(x) - f(a)) + (f(x) - f(a)) cdot Df(b)(x-a)\
f(x) cdot underline(f(x) - f(a))-colorredf(a)cdot underline(f(x) - f(a))&=(f(x) - f(a)) cdot Df(b)(x-a)\
(f(x)-f(a)) cdot underline(f(x) - f(a)) & =(f(x) - f(a)) cdot Df(b)(x-a)\
|f(x)-f(a)|^2&=(f(x) - f(a)) cdot Df(b)(x-a)
endalign*
and lastly Cauchy Schwartz implies
$$|f(x)-f(a)|^2leq |(f(x) - f(a))| cdot |Df(b)(x-a)|$$
answered Mar 20 at 11:17
b00n heTb00n heT
10.5k12335
$endgroup$
add a comment |
$begingroup$
I'll revrite @trancelocation's answer in a more pedagogical way:
beginalign*f(x) cdot (f(x) - f(a)) &= colorredf(a)cdot (f(x) - f(a)) + (f(x) - f(a)) cdot Df(b)(x-a)\
f(x) cdot underline(f(x) - f(a))-colorredf(a)cdot underline(f(x) - f(a))&=(f(x) - f(a)) cdot Df(b)(x-a)\
(f(x)-f(a)) cdot underline(f(x) - f(a)) & =(f(x) - f(a)) cdot Df(b)(x-a)\
|f(x)-f(a)|^2&=(f(x) - f(a)) cdot Df(b)(x-a)
endalign*
and lastly Cauchy Schwartz implies
$$|f(x)-f(a)|^2leq |(f(x) - f(a))| cdot |Df(b)(x-a)|$$
answered Mar 20 at 11:17
b00n heTb00n heT
10.5k12335
$endgroup$
add a comment |
$begingroup$
I'll revrite @trancelocation's answer in a more pedagogical way:
beginalign*f(x) cdot (f(x) - f(a)) &= colorredf(a)cdot (f(x) - f(a)) + (f(x) - f(a)) cdot Df(b)(x-a)\
f(x) cdot underline(f(x) - f(a))-colorredf(a)cdot underline(f(x) - f(a))&=(f(x) - f(a)) cdot Df(b)(x-a)\
(f(x)-f(a)) cdot underline(f(x) - f(a)) & =(f(x) - f(a)) cdot Df(b)(x-a)\
|f(x)-f(a)|^2&=(f(x) - f(a)) cdot Df(b)(x-a)
endalign*
and lastly Cauchy Schwartz implies
$$|f(x)-f(a)|^2leq |(f(x) - f(a))| cdot |Df(b)(x-a)|$$
answered Mar 20 at 11:17
b00n heTb00n heT
10.5k12335
$endgroup$
I'll revrite @trancelocation's answer in a more pedagogical way:
beginalign*f(x) cdot (f(x) - f(a)) &= colorredf(a)cdot (f(x) - f(a)) + (f(x) - f(a)) cdot Df(b)(x-a)\
f(x) cdot underline(f(x) - f(a))-colorredf(a)cdot underline(f(x) - f(a))&=(f(x) - f(a)) cdot Df(b)(x-a)\
(f(x)-f(a)) cdot underline(f(x) - f(a)) & =(f(x) - f(a)) cdot Df(b)(x-a)\
|f(x)-f(a)|^2&=(f(x) - f(a)) cdot Df(b)(x-a)
endalign*
and lastly Cauchy Schwartz implies
$$|f(x)-f(a)|^2leq |(f(x) - f(a))| cdot |Df(b)(x-a)|$$
answered Mar 20 at 11:17
b00n heTb00n heT
10.5k12335
answered Mar 20 at 11:17
b00n heTb00n heT
10.5k12335
answered Mar 20 at 11:17
b00n heTb00n heT
10.5k12335
answered Mar 20 at 11:17
b00n heTb00n heT
10.5k12335
10.5k12335
add a comment |
add a comment |
$begingroup$
You just have to pass one term to the LHS.
$$
f(x)cdot (f(x)-f(a))= f(a)cdot(f(x)-f(a))+(f(x)-f(a))Df(b)(x-a) Leftrightarrow
$$
$$
f(x)cdot (f(x)-f(a))-f(a)cdot(f(x)-f(a))=(f(x)-f(a))Df(b)(x-a)Leftrightarrow
$$
$$
(f(x)-f(a))cdot (f(x)-f(a)) = (f(x)-f(a))Df(b)(x-a)Leftrightarrow
$$
$$
|f(x)-f(a)|^2 = (f(x)-f(a))Df(b)(x-a) Rightarrow
$$
$$
|f(x)-f(a)|^2 = |(f(x)-f(a))Df(b)(x-a)| Rightarrow
$$
$$
|f(x)-f(a)|^2 leq |f(x)-f(a)| |Df(b)(x-a)|
$$
answered Mar 20 at 11:18
PierreCarrePierreCarre
1,665212
$endgroup$
add a comment |
$begingroup$
You just have to pass one term to the LHS.
$$
f(x)cdot (f(x)-f(a))= f(a)cdot(f(x)-f(a))+(f(x)-f(a))Df(b)(x-a) Leftrightarrow
$$
$$
f(x)cdot (f(x)-f(a))-f(a)cdot(f(x)-f(a))=(f(x)-f(a))Df(b)(x-a)Leftrightarrow
$$
$$
(f(x)-f(a))cdot (f(x)-f(a)) = (f(x)-f(a))Df(b)(x-a)Leftrightarrow
$$
$$
|f(x)-f(a)|^2 = (f(x)-f(a))Df(b)(x-a) Rightarrow
$$
$$
|f(x)-f(a)|^2 = |(f(x)-f(a))Df(b)(x-a)| Rightarrow
$$
$$
|f(x)-f(a)|^2 leq |f(x)-f(a)| |Df(b)(x-a)|
$$
answered Mar 20 at 11:18
PierreCarrePierreCarre
1,665212
$endgroup$
add a comment |
$begingroup$
You just have to pass one term to the LHS.
$$
f(x)cdot (f(x)-f(a))= f(a)cdot(f(x)-f(a))+(f(x)-f(a))Df(b)(x-a) Leftrightarrow
$$
$$
f(x)cdot (f(x)-f(a))-f(a)cdot(f(x)-f(a))=(f(x)-f(a))Df(b)(x-a)Leftrightarrow
$$
$$
(f(x)-f(a))cdot (f(x)-f(a)) = (f(x)-f(a))Df(b)(x-a)Leftrightarrow
$$
$$
|f(x)-f(a)|^2 = (f(x)-f(a))Df(b)(x-a) Rightarrow
$$
$$
|f(x)-f(a)|^2 = |(f(x)-f(a))Df(b)(x-a)| Rightarrow
$$
$$
|f(x)-f(a)|^2 leq |f(x)-f(a)| |Df(b)(x-a)|
$$
answered Mar 20 at 11:18
PierreCarrePierreCarre
1,665212
$endgroup$
You just have to pass one term to the LHS.
$$
f(x)cdot (f(x)-f(a))= f(a)cdot(f(x)-f(a))+(f(x)-f(a))Df(b)(x-a) Leftrightarrow
$$
$$
f(x)cdot (f(x)-f(a))-f(a)cdot(f(x)-f(a))=(f(x)-f(a))Df(b)(x-a)Leftrightarrow
$$
$$
(f(x)-f(a))cdot (f(x)-f(a)) = (f(x)-f(a))Df(b)(x-a)Leftrightarrow
$$
$$
|f(x)-f(a)|^2 = (f(x)-f(a))Df(b)(x-a) Rightarrow
$$
$$
|f(x)-f(a)|^2 = |(f(x)-f(a))Df(b)(x-a)| Rightarrow
$$
$$
|f(x)-f(a)|^2 leq |f(x)-f(a)| |Df(b)(x-a)|
$$
answered Mar 20 at 11:18
PierreCarrePierreCarre
1,665212
answered Mar 20 at 11:18
PierreCarrePierreCarre
1,665212
answered Mar 20 at 11:18
PierreCarrePierreCarre
1,665212
answered Mar 20 at 11:18
PierreCarrePierreCarre
1,665212
1,665212
add a comment |
add a comment |
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