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Prove $P_n(x) = sum_{k=0}^{n}b_k(x-b)^k $ satisfies $limlimits_{xto b}dfrac{f(x)-P_n(x)}{(x-b)^n} = 0 $ iff...
Prove $P_n(x) = sum_{k=0}^{n}b_k(x-b)^k$ satisfies $$limlimits_{xto b}dfrac{f(x)-P_n(x)}{(x-b)^n} = 0 $$ iff $P_n = T_n$ Where $T_n$ is the nth Taylor polynomial of $f$ about $b$.
I know from Taylor's Theorem that $$limlimits_{xto b}dfrac{f(x)-T_n(x,b)}{(x-b)^n} = 0$$ I can see the forward proof intuitively from Taylor's theorem, but intuition isn't reliable.
real-analysis limits polynomials
asked Nov 20 '18 at 3:52
t.perez
599
add a comment |
Prove $P_n(x) = sum_{k=0}^{n}b_k(x-b)^k$ satisfies $$limlimits_{xto b}dfrac{f(x)-P_n(x)}{(x-b)^n} = 0 $$ iff $P_n = T_n$ Where $T_n$ is the nth Taylor polynomial of $f$ about $b$.
I know from Taylor's Theorem that $$limlimits_{xto b}dfrac{f(x)-T_n(x,b)}{(x-b)^n} = 0$$ I can see the forward proof intuitively from Taylor's theorem, but intuition isn't reliable.
real-analysis limits polynomials
asked Nov 20 '18 at 3:52
t.perez
599
add a comment |
Prove $P_n(x) = sum_{k=0}^{n}b_k(x-b)^k$ satisfies $$limlimits_{xto b}dfrac{f(x)-P_n(x)}{(x-b)^n} = 0 $$ iff $P_n = T_n$ Where $T_n$ is the nth Taylor polynomial of $f$ about $b$.
I know from Taylor's Theorem that $$limlimits_{xto b}dfrac{f(x)-T_n(x,b)}{(x-b)^n} = 0$$ I can see the forward proof intuitively from Taylor's theorem, but intuition isn't reliable.
real-analysis limits polynomials
asked Nov 20 '18 at 3:52
t.perez
599
Prove $P_n(x) = sum_{k=0}^{n}b_k(x-b)^k$ satisfies $$limlimits_{xto b}dfrac{f(x)-P_n(x)}{(x-b)^n} = 0 $$ iff $P_n = T_n$ Where $T_n$ is the nth Taylor polynomial of $f$ about $b$.
I know from Taylor's Theorem that $$limlimits_{xto b}dfrac{f(x)-T_n(x,b)}{(x-b)^n} = 0$$ I can see the forward proof intuitively from Taylor's theorem, but intuition isn't reliable.
real-analysis limits polynomials
real-analysis limits polynomials
asked Nov 20 '18 at 3:52
t.perez
599
asked Nov 20 '18 at 3:52
t.perez
599
asked Nov 20 '18 at 3:52
t.perez
599
asked Nov 20 '18 at 3:52
t.perez
599
asked Nov 20 '18 at 3:52
t.perez
599
599
add a comment |
add a comment |
1 Answer
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Hint: Assuming $f$ is $n+1$-th continuously differentiable, then
begin{align}
f(x)=& f(b)+f'(b)(x-b)+ldots+ frac{f^{(n)}(b)}{n!}(x-b)^n+frac{f^{(n+1)}(xi(x))}{(n+1)!}(x-b)^{n+1}\
=& T_n(x)+frac{f^{(n+1)}(xi(x))}{(n+1)!}(x-b)^{n+1}
end{align}
where $b<xi(x)<x$. By the hypothesis, we have
begin{align}
frac{f(x)-P_n(x)}{(x-b)^n} = frac{T_n(x)-P_n(x)}{(x-b)^n}+frac{f^{(n+1)}(xi(x))}{(n+1)!}(x-b).
end{align}
Since the left hand side tends to zero as $xrightarrow b$ then so does the right hand side. In particular, since
begin{align}
left|frac{f^{(n+1)}(xi(x))}{(n+1)!}(x-b)^{n+1}right|leq M|x-b|rightarrow 0
end{align}
as $xrightarrow b$, then
begin{align}
frac{T_n(x)-P_n(x)}{(x-b)^n} rightarrow 0
end{align}
as $xrightarrow b$.
answered Nov 20 '18 at 4:33
Jacky Chong
17.7k21128
Thank you so much, this is very helpful! How exactly does this prove $T_n(x) = P_n(x)$? I see the denominator in the last line tends to $0$, but couldnt that mean the numerator tends to anything? Could I use l'hopital to prove $T_n(x) = P_n(x)$ in the last step?
– t.perez
Nov 20 '18 at 4:47
@t.perez suppose they are not equal, then what can you conclude?
– Jacky Chong
Nov 20 '18 at 5:06
you could conclude that the limit DNE if they're not equal since the denominator would equal 0, so they must be equal. Does that sound correct?
– t.perez
Nov 20 '18 at 5:23
Seems like you got it.
– Jacky Chong
Nov 20 '18 at 5:23
thank you so much! you are very helpful
– t.perez
Nov 20 '18 at 5:31
add a comment |
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Hint: Assuming $f$ is $n+1$-th continuously differentiable, then
begin{align}
f(x)=& f(b)+f'(b)(x-b)+ldots+ frac{f^{(n)}(b)}{n!}(x-b)^n+frac{f^{(n+1)}(xi(x))}{(n+1)!}(x-b)^{n+1}\
=& T_n(x)+frac{f^{(n+1)}(xi(x))}{(n+1)!}(x-b)^{n+1}
end{align}
where $b<xi(x)<x$. By the hypothesis, we have
begin{align}
frac{f(x)-P_n(x)}{(x-b)^n} = frac{T_n(x)-P_n(x)}{(x-b)^n}+frac{f^{(n+1)}(xi(x))}{(n+1)!}(x-b).
end{align}
Since the left hand side tends to zero as $xrightarrow b$ then so does the right hand side. In particular, since
begin{align}
left|frac{f^{(n+1)}(xi(x))}{(n+1)!}(x-b)^{n+1}right|leq M|x-b|rightarrow 0
end{align}
as $xrightarrow b$, then
begin{align}
frac{T_n(x)-P_n(x)}{(x-b)^n} rightarrow 0
end{align}
as $xrightarrow b$.
answered Nov 20 '18 at 4:33
Jacky Chong
17.7k21128
Thank you so much, this is very helpful! How exactly does this prove $T_n(x) = P_n(x)$? I see the denominator in the last line tends to $0$, but couldnt that mean the numerator tends to anything? Could I use l'hopital to prove $T_n(x) = P_n(x)$ in the last step?
– t.perez
Nov 20 '18 at 4:47
@t.perez suppose they are not equal, then what can you conclude?
– Jacky Chong
Nov 20 '18 at 5:06
you could conclude that the limit DNE if they're not equal since the denominator would equal 0, so they must be equal. Does that sound correct?
– t.perez
Nov 20 '18 at 5:23
Seems like you got it.
– Jacky Chong
Nov 20 '18 at 5:23
thank you so much! you are very helpful
– t.perez
Nov 20 '18 at 5:31
add a comment |
Hint: Assuming $f$ is $n+1$-th continuously differentiable, then
begin{align}
f(x)=& f(b)+f'(b)(x-b)+ldots+ frac{f^{(n)}(b)}{n!}(x-b)^n+frac{f^{(n+1)}(xi(x))}{(n+1)!}(x-b)^{n+1}\
=& T_n(x)+frac{f^{(n+1)}(xi(x))}{(n+1)!}(x-b)^{n+1}
end{align}
where $b<xi(x)<x$. By the hypothesis, we have
begin{align}
frac{f(x)-P_n(x)}{(x-b)^n} = frac{T_n(x)-P_n(x)}{(x-b)^n}+frac{f^{(n+1)}(xi(x))}{(n+1)!}(x-b).
end{align}
Since the left hand side tends to zero as $xrightarrow b$ then so does the right hand side. In particular, since
begin{align}
left|frac{f^{(n+1)}(xi(x))}{(n+1)!}(x-b)^{n+1}right|leq M|x-b|rightarrow 0
end{align}
as $xrightarrow b$, then
begin{align}
frac{T_n(x)-P_n(x)}{(x-b)^n} rightarrow 0
end{align}
as $xrightarrow b$.
answered Nov 20 '18 at 4:33
Jacky Chong
17.7k21128
Thank you so much, this is very helpful! How exactly does this prove $T_n(x) = P_n(x)$? I see the denominator in the last line tends to $0$, but couldnt that mean the numerator tends to anything? Could I use l'hopital to prove $T_n(x) = P_n(x)$ in the last step?
– t.perez
Nov 20 '18 at 4:47
@t.perez suppose they are not equal, then what can you conclude?
– Jacky Chong
Nov 20 '18 at 5:06
you could conclude that the limit DNE if they're not equal since the denominator would equal 0, so they must be equal. Does that sound correct?
– t.perez
Nov 20 '18 at 5:23
Seems like you got it.
– Jacky Chong
Nov 20 '18 at 5:23
thank you so much! you are very helpful
– t.perez
Nov 20 '18 at 5:31
add a comment |
Hint: Assuming $f$ is $n+1$-th continuously differentiable, then
begin{align}
f(x)=& f(b)+f'(b)(x-b)+ldots+ frac{f^{(n)}(b)}{n!}(x-b)^n+frac{f^{(n+1)}(xi(x))}{(n+1)!}(x-b)^{n+1}\
=& T_n(x)+frac{f^{(n+1)}(xi(x))}{(n+1)!}(x-b)^{n+1}
end{align}
where $b<xi(x)<x$. By the hypothesis, we have
begin{align}
frac{f(x)-P_n(x)}{(x-b)^n} = frac{T_n(x)-P_n(x)}{(x-b)^n}+frac{f^{(n+1)}(xi(x))}{(n+1)!}(x-b).
end{align}
Since the left hand side tends to zero as $xrightarrow b$ then so does the right hand side. In particular, since
begin{align}
left|frac{f^{(n+1)}(xi(x))}{(n+1)!}(x-b)^{n+1}right|leq M|x-b|rightarrow 0
end{align}
as $xrightarrow b$, then
begin{align}
frac{T_n(x)-P_n(x)}{(x-b)^n} rightarrow 0
end{align}
as $xrightarrow b$.
answered Nov 20 '18 at 4:33
Jacky Chong
17.7k21128
Hint: Assuming $f$ is $n+1$-th continuously differentiable, then
begin{align}
f(x)=& f(b)+f'(b)(x-b)+ldots+ frac{f^{(n)}(b)}{n!}(x-b)^n+frac{f^{(n+1)}(xi(x))}{(n+1)!}(x-b)^{n+1}\
=& T_n(x)+frac{f^{(n+1)}(xi(x))}{(n+1)!}(x-b)^{n+1}
end{align}
where $b<xi(x)<x$. By the hypothesis, we have
begin{align}
frac{f(x)-P_n(x)}{(x-b)^n} = frac{T_n(x)-P_n(x)}{(x-b)^n}+frac{f^{(n+1)}(xi(x))}{(n+1)!}(x-b).
end{align}
Since the left hand side tends to zero as $xrightarrow b$ then so does the right hand side. In particular, since
begin{align}
left|frac{f^{(n+1)}(xi(x))}{(n+1)!}(x-b)^{n+1}right|leq M|x-b|rightarrow 0
end{align}
as $xrightarrow b$, then
begin{align}
frac{T_n(x)-P_n(x)}{(x-b)^n} rightarrow 0
end{align}
as $xrightarrow b$.
answered Nov 20 '18 at 4:33
Jacky Chong
17.7k21128
answered Nov 20 '18 at 4:33
Jacky Chong
17.7k21128
answered Nov 20 '18 at 4:33
Jacky Chong
17.7k21128
answered Nov 20 '18 at 4:33
Jacky Chong
17.7k21128
17.7k21128
Thank you so much, this is very helpful! How exactly does this prove $T_n(x) = P_n(x)$? I see the denominator in the last line tends to $0$, but couldnt that mean the numerator tends to anything? Could I use l'hopital to prove $T_n(x) = P_n(x)$ in the last step?
– t.perez
Nov 20 '18 at 4:47
@t.perez suppose they are not equal, then what can you conclude?
– Jacky Chong
Nov 20 '18 at 5:06
you could conclude that the limit DNE if they're not equal since the denominator would equal 0, so they must be equal. Does that sound correct?
– t.perez
Nov 20 '18 at 5:23
Seems like you got it.
– Jacky Chong
Nov 20 '18 at 5:23
thank you so much! you are very helpful
– t.perez
Nov 20 '18 at 5:31
add a comment |
Thank you so much, this is very helpful! How exactly does this prove $T_n(x) = P_n(x)$? I see the denominator in the last line tends to $0$, but couldnt that mean the numerator tends to anything? Could I use l'hopital to prove $T_n(x) = P_n(x)$ in the last step?
– t.perez
Nov 20 '18 at 4:47
@t.perez suppose they are not equal, then what can you conclude?
– Jacky Chong
Nov 20 '18 at 5:06
you could conclude that the limit DNE if they're not equal since the denominator would equal 0, so they must be equal. Does that sound correct?
– t.perez
Nov 20 '18 at 5:23
Seems like you got it.
– Jacky Chong
Nov 20 '18 at 5:23
thank you so much! you are very helpful
– t.perez
Nov 20 '18 at 5:31
Thank you so much, this is very helpful! How exactly does this prove $T_n(x) = P_n(x)$? I see the denominator in the last line tends to $0$, but couldnt that mean the numerator tends to anything? Could I use l'hopital to prove $T_n(x) = P_n(x)$ in the last step?
– t.perez
Nov 20 '18 at 4:47
Thank you so much, this is very helpful! How exactly does this prove $T_n(x) = P_n(x)$? I see the denominator in the last line tends to $0$, but couldnt that mean the numerator tends to anything? Could I use l'hopital to prove $T_n(x) = P_n(x)$ in the last step?
– t.perez
Nov 20 '18 at 4:47
@t.perez suppose they are not equal, then what can you conclude?
– Jacky Chong
Nov 20 '18 at 5:06
@t.perez suppose they are not equal, then what can you conclude?
– Jacky Chong
Nov 20 '18 at 5:06
you could conclude that the limit DNE if they're not equal since the denominator would equal 0, so they must be equal. Does that sound correct?
– t.perez
Nov 20 '18 at 5:23
you could conclude that the limit DNE if they're not equal since the denominator would equal 0, so they must be equal. Does that sound correct?
– t.perez
Nov 20 '18 at 5:23
Seems like you got it.
– Jacky Chong
Nov 20 '18 at 5:23
Seems like you got it.
– Jacky Chong
Nov 20 '18 at 5:23
thank you so much! you are very helpful
– t.perez
Nov 20 '18 at 5:31
thank you so much! you are very helpful
– t.perez
Nov 20 '18 at 5:31
add a comment |
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