Mixing
Remainder in taylor's theorem for multivariate functions
$begingroup$
Excerpt is from Courant Intro Analysis book Vol 2 pg 69. I need explanation of the underlined part (entire sentence) including justification. I understand what vanishes to higher order means.
real-analysis calculus
asked Feb 2 at 2:54
helios321helios321
56849
$endgroup$
add a comment |
$begingroup$
Excerpt is from Courant Intro Analysis book Vol 2 pg 69. I need explanation of the underlined part (entire sentence) including justification. I understand what vanishes to higher order means.
real-analysis calculus
asked Feb 2 at 2:54
helios321helios321
56849
$endgroup$
$begingroup$
what is the name of that book?
$endgroup$
– James
Feb 2 at 3:04
$begingroup$
@JimmySabater Hi i edited to include book.
$endgroup$
– helios321
Feb 2 at 3:15
add a comment |
$begingroup$
Excerpt is from Courant Intro Analysis book Vol 2 pg 69. I need explanation of the underlined part (entire sentence) including justification. I understand what vanishes to higher order means.
real-analysis calculus
asked Feb 2 at 2:54
helios321helios321
56849
$endgroup$
Excerpt is from Courant Intro Analysis book Vol 2 pg 69. I need explanation of the underlined part (entire sentence) including justification. I understand what vanishes to higher order means.
real-analysis calculus
real-analysis calculus
asked Feb 2 at 2:54
helios321helios321
56849
asked Feb 2 at 2:54
helios321helios321
56849
edited Feb 2 at 3:15
helios321
asked Feb 2 at 2:54
helios321helios321
56849
asked Feb 2 at 2:54
helios321helios321
56849
asked Feb 2 at 2:54
helios321helios321
56849
56849
$begingroup$
what is the name of that book?
$endgroup$
– James
Feb 2 at 3:04
$begingroup$
@JimmySabater Hi i edited to include book.
$endgroup$
– helios321
Feb 2 at 3:15
add a comment |
$begingroup$
what is the name of that book?
$endgroup$
– James
Feb 2 at 3:04
$begingroup$
@JimmySabater Hi i edited to include book.
$endgroup$
– helios321
Feb 2 at 3:15
$begingroup$
what is the name of that book?
$endgroup$
– James
Feb 2 at 3:04
$begingroup$
what is the name of that book?
$endgroup$
– James
Feb 2 at 3:04
$begingroup$
@JimmySabater Hi i edited to include book.
$endgroup$
– helios321
Feb 2 at 3:15
$begingroup$
@JimmySabater Hi i edited to include book.
$endgroup$
– helios321
Feb 2 at 3:15
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
That is the meaning of the Landau symbol, "little oh". That is, $f(x)= o(g(x))$ means $lim_{xto0}frac{f(x)}{g(x)}=0$. I.e. $f$ goes to zero "at higher order" than $g$.
For instance, $x^n=o(x^m)$ for $ngt m$.
Thus the result you have underlined follows from the fact that the remainder $R_n$ has higher powers of $h$ and $k$ than $d^nf$ does. This information is on the page above.
answered Feb 2 at 3:42
Chris CusterChris Custer
14.3k3827
$endgroup$
$begingroup$
How do you prove that the ratio of the two polynomials in $h$ and $k$ go to $0$? Is there are Hopital rule for multivariable functions? Also I still don't see the relevance of the very last part that $R_n=o(sqrt{(h^2+k^2)^n})$
$endgroup$
– helios321
Feb 2 at 4:05
$begingroup$
For the last part, try squaring both sides. One is a polynomial of degree $2n$, the other $2n+2$. Polynomials of higher degree grow (and decay) faster, a basic fact.
$endgroup$
– Chris Custer
Feb 2 at 4:16
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What's a rough proof sketch of "Polynomials of higher degree grow (and decay) faster"?
$endgroup$
– helios321
Feb 2 at 4:29
$begingroup$
The highest degree term dominates.
$endgroup$
– Chris Custer
Feb 2 at 5:53
$begingroup$
Factor it out; then compare terms of highest degree.
$endgroup$
– Chris Custer
Feb 2 at 8:20
|
show 6 more comments
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1 Answer
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1 Answer
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$begingroup$
That is the meaning of the Landau symbol, "little oh". That is, $f(x)= o(g(x))$ means $lim_{xto0}frac{f(x)}{g(x)}=0$. I.e. $f$ goes to zero "at higher order" than $g$.
For instance, $x^n=o(x^m)$ for $ngt m$.
Thus the result you have underlined follows from the fact that the remainder $R_n$ has higher powers of $h$ and $k$ than $d^nf$ does. This information is on the page above.
answered Feb 2 at 3:42
Chris CusterChris Custer
14.3k3827
$endgroup$
$begingroup$
How do you prove that the ratio of the two polynomials in $h$ and $k$ go to $0$? Is there are Hopital rule for multivariable functions? Also I still don't see the relevance of the very last part that $R_n=o(sqrt{(h^2+k^2)^n})$
$endgroup$
– helios321
Feb 2 at 4:05
$begingroup$
For the last part, try squaring both sides. One is a polynomial of degree $2n$, the other $2n+2$. Polynomials of higher degree grow (and decay) faster, a basic fact.
$endgroup$
– Chris Custer
Feb 2 at 4:16
$begingroup$
What's a rough proof sketch of "Polynomials of higher degree grow (and decay) faster"?
$endgroup$
– helios321
Feb 2 at 4:29
$begingroup$
The highest degree term dominates.
$endgroup$
– Chris Custer
Feb 2 at 5:53
$begingroup$
Factor it out; then compare terms of highest degree.
$endgroup$
– Chris Custer
Feb 2 at 8:20
|
show 6 more comments
$begingroup$
That is the meaning of the Landau symbol, "little oh". That is, $f(x)= o(g(x))$ means $lim_{xto0}frac{f(x)}{g(x)}=0$. I.e. $f$ goes to zero "at higher order" than $g$.
For instance, $x^n=o(x^m)$ for $ngt m$.
Thus the result you have underlined follows from the fact that the remainder $R_n$ has higher powers of $h$ and $k$ than $d^nf$ does. This information is on the page above.
answered Feb 2 at 3:42
Chris CusterChris Custer
14.3k3827
$endgroup$
$begingroup$
How do you prove that the ratio of the two polynomials in $h$ and $k$ go to $0$? Is there are Hopital rule for multivariable functions? Also I still don't see the relevance of the very last part that $R_n=o(sqrt{(h^2+k^2)^n})$
$endgroup$
– helios321
Feb 2 at 4:05
$begingroup$
For the last part, try squaring both sides. One is a polynomial of degree $2n$, the other $2n+2$. Polynomials of higher degree grow (and decay) faster, a basic fact.
$endgroup$
– Chris Custer
Feb 2 at 4:16
$begingroup$
What's a rough proof sketch of "Polynomials of higher degree grow (and decay) faster"?
$endgroup$
– helios321
Feb 2 at 4:29
$begingroup$
The highest degree term dominates.
$endgroup$
– Chris Custer
Feb 2 at 5:53
$begingroup$
Factor it out; then compare terms of highest degree.
$endgroup$
– Chris Custer
Feb 2 at 8:20
|
show 6 more comments
$begingroup$
That is the meaning of the Landau symbol, "little oh". That is, $f(x)= o(g(x))$ means $lim_{xto0}frac{f(x)}{g(x)}=0$. I.e. $f$ goes to zero "at higher order" than $g$.
For instance, $x^n=o(x^m)$ for $ngt m$.
Thus the result you have underlined follows from the fact that the remainder $R_n$ has higher powers of $h$ and $k$ than $d^nf$ does. This information is on the page above.
answered Feb 2 at 3:42
Chris CusterChris Custer
14.3k3827
$endgroup$
That is the meaning of the Landau symbol, "little oh". That is, $f(x)= o(g(x))$ means $lim_{xto0}frac{f(x)}{g(x)}=0$. I.e. $f$ goes to zero "at higher order" than $g$.
For instance, $x^n=o(x^m)$ for $ngt m$.
Thus the result you have underlined follows from the fact that the remainder $R_n$ has higher powers of $h$ and $k$ than $d^nf$ does. This information is on the page above.
answered Feb 2 at 3:42
Chris CusterChris Custer
14.3k3827
answered Feb 2 at 3:42
Chris CusterChris Custer
14.3k3827
answered Feb 2 at 3:42
Chris CusterChris Custer
14.3k3827
answered Feb 2 at 3:42
Chris CusterChris Custer
14.3k3827
14.3k3827
$begingroup$
How do you prove that the ratio of the two polynomials in $h$ and $k$ go to $0$? Is there are Hopital rule for multivariable functions? Also I still don't see the relevance of the very last part that $R_n=o(sqrt{(h^2+k^2)^n})$
$endgroup$
– helios321
Feb 2 at 4:05
$begingroup$
For the last part, try squaring both sides. One is a polynomial of degree $2n$, the other $2n+2$. Polynomials of higher degree grow (and decay) faster, a basic fact.
$endgroup$
– Chris Custer
Feb 2 at 4:16
$begingroup$
What's a rough proof sketch of "Polynomials of higher degree grow (and decay) faster"?
$endgroup$
– helios321
Feb 2 at 4:29
$begingroup$
The highest degree term dominates.
$endgroup$
– Chris Custer
Feb 2 at 5:53
$begingroup$
Factor it out; then compare terms of highest degree.
$endgroup$
– Chris Custer
Feb 2 at 8:20
|
show 6 more comments
$begingroup$
How do you prove that the ratio of the two polynomials in $h$ and $k$ go to $0$? Is there are Hopital rule for multivariable functions? Also I still don't see the relevance of the very last part that $R_n=o(sqrt{(h^2+k^2)^n})$
$endgroup$
– helios321
Feb 2 at 4:05
$begingroup$
For the last part, try squaring both sides. One is a polynomial of degree $2n$, the other $2n+2$. Polynomials of higher degree grow (and decay) faster, a basic fact.
$endgroup$
– Chris Custer
Feb 2 at 4:16
$begingroup$
What's a rough proof sketch of "Polynomials of higher degree grow (and decay) faster"?
$endgroup$
– helios321
Feb 2 at 4:29
$begingroup$
The highest degree term dominates.
$endgroup$
– Chris Custer
Feb 2 at 5:53
$begingroup$
Factor it out; then compare terms of highest degree.
$endgroup$
– Chris Custer
Feb 2 at 8:20
$begingroup$
How do you prove that the ratio of the two polynomials in $h$ and $k$ go to $0$? Is there are Hopital rule for multivariable functions? Also I still don't see the relevance of the very last part that $R_n=o(sqrt{(h^2+k^2)^n})$
$endgroup$
– helios321
Feb 2 at 4:05
$begingroup$
How do you prove that the ratio of the two polynomials in $h$ and $k$ go to $0$? Is there are Hopital rule for multivariable functions? Also I still don't see the relevance of the very last part that $R_n=o(sqrt{(h^2+k^2)^n})$
$endgroup$
– helios321
Feb 2 at 4:05
$begingroup$
For the last part, try squaring both sides. One is a polynomial of degree $2n$, the other $2n+2$. Polynomials of higher degree grow (and decay) faster, a basic fact.
$endgroup$
– Chris Custer
Feb 2 at 4:16
$begingroup$
For the last part, try squaring both sides. One is a polynomial of degree $2n$, the other $2n+2$. Polynomials of higher degree grow (and decay) faster, a basic fact.
$endgroup$
– Chris Custer
Feb 2 at 4:16
$begingroup$
What's a rough proof sketch of "Polynomials of higher degree grow (and decay) faster"?
$endgroup$
– helios321
Feb 2 at 4:29
$begingroup$
What's a rough proof sketch of "Polynomials of higher degree grow (and decay) faster"?
$endgroup$
– helios321
Feb 2 at 4:29
$begingroup$
The highest degree term dominates.
$endgroup$
– Chris Custer
Feb 2 at 5:53
$begingroup$
The highest degree term dominates.
$endgroup$
– Chris Custer
Feb 2 at 5:53
$begingroup$
Factor it out; then compare terms of highest degree.
$endgroup$
– Chris Custer
Feb 2 at 8:20
$begingroup$
Factor it out; then compare terms of highest degree.
$endgroup$
– Chris Custer
Feb 2 at 8:20
|
show 6 more comments
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$begingroup$
what is the name of that book?
$endgroup$
– James
Feb 2 at 3:04
$begingroup$
@JimmySabater Hi i edited to include book.
$endgroup$
– helios321
Feb 2 at 3:15