Mixing
Find the solution of $e^{x}+2x=0.96$
$begingroup$
We have the following function $$f(x)=e^{x}+2x.$$
Using the theorem on the derivative of the inverse function and the differentail of the function find the solution of $f(x)=0.96$.
I started by taking $x=f^{-1}(0.96)$.
Now, from teh differential we can write that $f^{-1}(x+Delta x) approx f^{-1}(x)+[f^{-1}(x)]'Delta x$. I do not know if it is correct and what I should do now.
I would be grateful for any hits.
calculus
asked Jan 29 at 22:19
GaharGahar
285
$endgroup$
add a comment |
$begingroup$
We have the following function $$f(x)=e^{x}+2x.$$
Using the theorem on the derivative of the inverse function and the differentail of the function find the solution of $f(x)=0.96$.
I started by taking $x=f^{-1}(0.96)$.
Now, from teh differential we can write that $f^{-1}(x+Delta x) approx f^{-1}(x)+[f^{-1}(x)]'Delta x$. I do not know if it is correct and what I should do now.
I would be grateful for any hits.
calculus
asked Jan 29 at 22:19
GaharGahar
285
$endgroup$
$begingroup$
$f(x)$ is a monic polynomial then there exist one root(or no real root) for $f(x)=const$
$endgroup$
– Samvel Safaryan
Jan 29 at 22:30
$begingroup$
@SamvelSafaryan f is clearly not a polynomial.
$endgroup$
– Shalop
Jan 30 at 1:33
add a comment |
$begingroup$
We have the following function $$f(x)=e^{x}+2x.$$
Using the theorem on the derivative of the inverse function and the differentail of the function find the solution of $f(x)=0.96$.
I started by taking $x=f^{-1}(0.96)$.
Now, from teh differential we can write that $f^{-1}(x+Delta x) approx f^{-1}(x)+[f^{-1}(x)]'Delta x$. I do not know if it is correct and what I should do now.
I would be grateful for any hits.
calculus
asked Jan 29 at 22:19
GaharGahar
285
$endgroup$
We have the following function $$f(x)=e^{x}+2x.$$
Using the theorem on the derivative of the inverse function and the differentail of the function find the solution of $f(x)=0.96$.
I started by taking $x=f^{-1}(0.96)$.
Now, from teh differential we can write that $f^{-1}(x+Delta x) approx f^{-1}(x)+[f^{-1}(x)]'Delta x$. I do not know if it is correct and what I should do now.
I would be grateful for any hits.
calculus
calculus
asked Jan 29 at 22:19
GaharGahar
285
asked Jan 29 at 22:19
GaharGahar
285
asked Jan 29 at 22:19
GaharGahar
285
asked Jan 29 at 22:19
GaharGahar
285
asked Jan 29 at 22:19
GaharGahar
285
285
$begingroup$
$f(x)$ is a monic polynomial then there exist one root(or no real root) for $f(x)=const$
$endgroup$
– Samvel Safaryan
Jan 29 at 22:30
$begingroup$
@SamvelSafaryan f is clearly not a polynomial.
$endgroup$
– Shalop
Jan 30 at 1:33
add a comment |
$begingroup$
$f(x)$ is a monic polynomial then there exist one root(or no real root) for $f(x)=const$
$endgroup$
– Samvel Safaryan
Jan 29 at 22:30
$begingroup$
@SamvelSafaryan f is clearly not a polynomial.
$endgroup$
– Shalop
Jan 30 at 1:33
$begingroup$
$f(x)$ is a monic polynomial then there exist one root(or no real root) for $f(x)=const$
$endgroup$
– Samvel Safaryan
Jan 29 at 22:30
$begingroup$
$f(x)$ is a monic polynomial then there exist one root(or no real root) for $f(x)=const$
$endgroup$
– Samvel Safaryan
Jan 29 at 22:30
$begingroup$
@SamvelSafaryan f is clearly not a polynomial.
$endgroup$
– Shalop
Jan 30 at 1:33
$begingroup$
@SamvelSafaryan f is clearly not a polynomial.
$endgroup$
– Shalop
Jan 30 at 1:33
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If you are approximating something with differentials, start with an $x$ for which you know the value $f(x)$ is near $0.96$.
Hint: Start with $f^{-1}(1) = 0$.
answered Jan 29 at 22:29
Anthony TerAnthony Ter
37116
$endgroup$
1
$begingroup$
Thank you for your hint. So I can write that $f^{-1}(1-0.04) approx f^{-1}(1)+[f^{-1}(1)]'(-0.04)$?
$endgroup$
– Gahar
Jan 29 at 22:35
$begingroup$
Yep, looks good.
$endgroup$
– Anthony Ter
Jan 29 at 22:37
$begingroup$
And now $f^{-1}(0.96) approx f^{-1}(1) + 1/f'(0)(-0.04)$ and since $f'(0)=3$ we have that $f^{-1}(0.96) approx -4/300$. Am I right?
$endgroup$
– Gahar
Jan 29 at 22:40
$begingroup$
Yea, and you can check these types of problems very easily on a calculator by calculating $f(frac{-4}{300})$
$endgroup$
– Anthony Ter
Jan 29 at 22:42
$begingroup$
Ok, thank you very much for your help.
$endgroup$
– Gahar
Jan 29 at 22:43
add a comment |
$begingroup$
This has been written for your curiosity.
In fact, the equation
$$y=e^{a x}+b x$$ shows explicit solution(s) in terms of Lambert function. They write
$$x=frac{y}{b}-frac{1}{a}Wleft(frac{a }{b}e^{frac{a y}{b}}right)$$ So, for your specifdic case $(a=1,b=2,y=frac{24}{25})$
$$x=frac{12}{25}-Wleft(frac{e^{12/25}}{2}right)$$ Since the argument is "small", you could use, as an approximation built around $t=0$,
$$W(t)=t ,frac{1+frac{228 }{85}t+frac{451 }{340}t^2 } {1+frac{313 }{85}t+frac{1193 }{340}t^2+frac{133 }{204}t^3 }$$ This would give, as an approximate solution $xapprox -0.0130442$ while the exact value would be $xapprox -0.0133630 $.
For the specific case you address, plotting the function, you notice that the solution is close to $x=0$. So, make a series expansion
$$e^x+2x=1+3 x+frac{1}{2}x^2+Oleft(x^3right)$$ So, using the first term
$$frac{24}{25}=1+3ximplies x=-frac 1{75} approx -0.0133333 $$ Using the first and second term
$$frac{24}{25}=1+3x+frac{1}{2}x^2implies x=frac{sqrt{223}}{5}-3approx -0.0133631$$
Sooner or later, you will alos learn that, better then with Taylor series, we can approximate functions using Padé approximants. If we use the $[1,n]$ of them which will write as
$$f(x)simeq frac {f(a)+c_{(n)} (x-a)}{1+sum_{k=1}^n d_k (x-a)^k}$$ then, starting with a rational value of $a$, we can get rational approximations of the solution
$$x_{(n)}=a-frac{f(a)}{c_{(n)}}$$
$$left(
begin{array}{ccc}
n & x_{(n)} & x_{(n)} approx \
0 & -frac{1}{75} & -0.01333333333 \
1 & -frac{6}{449} & -0.01336302895 \
2 & -frac{1347}{100801} & -0.01336296267
end{array}
right)$$
answered Jan 30 at 7:37
Claude LeiboviciClaude Leibovici
125k1158136
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you are approximating something with differentials, start with an $x$ for which you know the value $f(x)$ is near $0.96$.
Hint: Start with $f^{-1}(1) = 0$.
answered Jan 29 at 22:29
Anthony TerAnthony Ter
37116
$endgroup$
1
$begingroup$
Thank you for your hint. So I can write that $f^{-1}(1-0.04) approx f^{-1}(1)+[f^{-1}(1)]'(-0.04)$?
$endgroup$
– Gahar
Jan 29 at 22:35
$begingroup$
Yep, looks good.
$endgroup$
– Anthony Ter
Jan 29 at 22:37
$begingroup$
And now $f^{-1}(0.96) approx f^{-1}(1) + 1/f'(0)(-0.04)$ and since $f'(0)=3$ we have that $f^{-1}(0.96) approx -4/300$. Am I right?
$endgroup$
– Gahar
Jan 29 at 22:40
$begingroup$
Yea, and you can check these types of problems very easily on a calculator by calculating $f(frac{-4}{300})$
$endgroup$
– Anthony Ter
Jan 29 at 22:42
$begingroup$
Ok, thank you very much for your help.
$endgroup$
– Gahar
Jan 29 at 22:43
add a comment |
$begingroup$
If you are approximating something with differentials, start with an $x$ for which you know the value $f(x)$ is near $0.96$.
Hint: Start with $f^{-1}(1) = 0$.
answered Jan 29 at 22:29
Anthony TerAnthony Ter
37116
$endgroup$
1
$begingroup$
Thank you for your hint. So I can write that $f^{-1}(1-0.04) approx f^{-1}(1)+[f^{-1}(1)]'(-0.04)$?
$endgroup$
– Gahar
Jan 29 at 22:35
$begingroup$
Yep, looks good.
$endgroup$
– Anthony Ter
Jan 29 at 22:37
$begingroup$
And now $f^{-1}(0.96) approx f^{-1}(1) + 1/f'(0)(-0.04)$ and since $f'(0)=3$ we have that $f^{-1}(0.96) approx -4/300$. Am I right?
$endgroup$
– Gahar
Jan 29 at 22:40
$begingroup$
Yea, and you can check these types of problems very easily on a calculator by calculating $f(frac{-4}{300})$
$endgroup$
– Anthony Ter
Jan 29 at 22:42
$begingroup$
Ok, thank you very much for your help.
$endgroup$
– Gahar
Jan 29 at 22:43
add a comment |
$begingroup$
If you are approximating something with differentials, start with an $x$ for which you know the value $f(x)$ is near $0.96$.
Hint: Start with $f^{-1}(1) = 0$.
answered Jan 29 at 22:29
Anthony TerAnthony Ter
37116
$endgroup$
If you are approximating something with differentials, start with an $x$ for which you know the value $f(x)$ is near $0.96$.
Hint: Start with $f^{-1}(1) = 0$.
answered Jan 29 at 22:29
Anthony TerAnthony Ter
37116
answered Jan 29 at 22:29
Anthony TerAnthony Ter
37116
answered Jan 29 at 22:29
Anthony TerAnthony Ter
37116
answered Jan 29 at 22:29
Anthony TerAnthony Ter
37116
37116
1
$begingroup$
Thank you for your hint. So I can write that $f^{-1}(1-0.04) approx f^{-1}(1)+[f^{-1}(1)]'(-0.04)$?
$endgroup$
– Gahar
Jan 29 at 22:35
$begingroup$
Yep, looks good.
$endgroup$
– Anthony Ter
Jan 29 at 22:37
$begingroup$
And now $f^{-1}(0.96) approx f^{-1}(1) + 1/f'(0)(-0.04)$ and since $f'(0)=3$ we have that $f^{-1}(0.96) approx -4/300$. Am I right?
$endgroup$
– Gahar
Jan 29 at 22:40
$begingroup$
Yea, and you can check these types of problems very easily on a calculator by calculating $f(frac{-4}{300})$
$endgroup$
– Anthony Ter
Jan 29 at 22:42
$begingroup$
Ok, thank you very much for your help.
$endgroup$
– Gahar
Jan 29 at 22:43
add a comment |
1
$begingroup$
Thank you for your hint. So I can write that $f^{-1}(1-0.04) approx f^{-1}(1)+[f^{-1}(1)]'(-0.04)$?
$endgroup$
– Gahar
Jan 29 at 22:35
$begingroup$
Yep, looks good.
$endgroup$
– Anthony Ter
Jan 29 at 22:37
$begingroup$
And now $f^{-1}(0.96) approx f^{-1}(1) + 1/f'(0)(-0.04)$ and since $f'(0)=3$ we have that $f^{-1}(0.96) approx -4/300$. Am I right?
$endgroup$
– Gahar
Jan 29 at 22:40
$begingroup$
Yea, and you can check these types of problems very easily on a calculator by calculating $f(frac{-4}{300})$
$endgroup$
– Anthony Ter
Jan 29 at 22:42
$begingroup$
Ok, thank you very much for your help.
$endgroup$
– Gahar
Jan 29 at 22:43
1
1
$begingroup$
Thank you for your hint. So I can write that $f^{-1}(1-0.04) approx f^{-1}(1)+[f^{-1}(1)]'(-0.04)$?
$endgroup$
– Gahar
Jan 29 at 22:35
$begingroup$
Thank you for your hint. So I can write that $f^{-1}(1-0.04) approx f^{-1}(1)+[f^{-1}(1)]'(-0.04)$?
$endgroup$
– Gahar
Jan 29 at 22:35
$begingroup$
Yep, looks good.
$endgroup$
– Anthony Ter
Jan 29 at 22:37
$begingroup$
Yep, looks good.
$endgroup$
– Anthony Ter
Jan 29 at 22:37
$begingroup$
And now $f^{-1}(0.96) approx f^{-1}(1) + 1/f'(0)(-0.04)$ and since $f'(0)=3$ we have that $f^{-1}(0.96) approx -4/300$. Am I right?
$endgroup$
– Gahar
Jan 29 at 22:40
$begingroup$
And now $f^{-1}(0.96) approx f^{-1}(1) + 1/f'(0)(-0.04)$ and since $f'(0)=3$ we have that $f^{-1}(0.96) approx -4/300$. Am I right?
$endgroup$
– Gahar
Jan 29 at 22:40
$begingroup$
Yea, and you can check these types of problems very easily on a calculator by calculating $f(frac{-4}{300})$
$endgroup$
– Anthony Ter
Jan 29 at 22:42
$begingroup$
Yea, and you can check these types of problems very easily on a calculator by calculating $f(frac{-4}{300})$
$endgroup$
– Anthony Ter
Jan 29 at 22:42
$begingroup$
Ok, thank you very much for your help.
$endgroup$
– Gahar
Jan 29 at 22:43
$begingroup$
Ok, thank you very much for your help.
$endgroup$
– Gahar
Jan 29 at 22:43
add a comment |
$begingroup$
This has been written for your curiosity.
In fact, the equation
$$y=e^{a x}+b x$$ shows explicit solution(s) in terms of Lambert function. They write
$$x=frac{y}{b}-frac{1}{a}Wleft(frac{a }{b}e^{frac{a y}{b}}right)$$ So, for your specifdic case $(a=1,b=2,y=frac{24}{25})$
$$x=frac{12}{25}-Wleft(frac{e^{12/25}}{2}right)$$ Since the argument is "small", you could use, as an approximation built around $t=0$,
$$W(t)=t ,frac{1+frac{228 }{85}t+frac{451 }{340}t^2 } {1+frac{313 }{85}t+frac{1193 }{340}t^2+frac{133 }{204}t^3 }$$ This would give, as an approximate solution $xapprox -0.0130442$ while the exact value would be $xapprox -0.0133630 $.
For the specific case you address, plotting the function, you notice that the solution is close to $x=0$. So, make a series expansion
$$e^x+2x=1+3 x+frac{1}{2}x^2+Oleft(x^3right)$$ So, using the first term
$$frac{24}{25}=1+3ximplies x=-frac 1{75} approx -0.0133333 $$ Using the first and second term
$$frac{24}{25}=1+3x+frac{1}{2}x^2implies x=frac{sqrt{223}}{5}-3approx -0.0133631$$
Sooner or later, you will alos learn that, better then with Taylor series, we can approximate functions using Padé approximants. If we use the $[1,n]$ of them which will write as
$$f(x)simeq frac {f(a)+c_{(n)} (x-a)}{1+sum_{k=1}^n d_k (x-a)^k}$$ then, starting with a rational value of $a$, we can get rational approximations of the solution
$$x_{(n)}=a-frac{f(a)}{c_{(n)}}$$
$$left(
begin{array}{ccc}
n & x_{(n)} & x_{(n)} approx \
0 & -frac{1}{75} & -0.01333333333 \
1 & -frac{6}{449} & -0.01336302895 \
2 & -frac{1347}{100801} & -0.01336296267
end{array}
right)$$
answered Jan 30 at 7:37
Claude LeiboviciClaude Leibovici
125k1158136
$endgroup$
add a comment |
$begingroup$
This has been written for your curiosity.
In fact, the equation
$$y=e^{a x}+b x$$ shows explicit solution(s) in terms of Lambert function. They write
$$x=frac{y}{b}-frac{1}{a}Wleft(frac{a }{b}e^{frac{a y}{b}}right)$$ So, for your specifdic case $(a=1,b=2,y=frac{24}{25})$
$$x=frac{12}{25}-Wleft(frac{e^{12/25}}{2}right)$$ Since the argument is "small", you could use, as an approximation built around $t=0$,
$$W(t)=t ,frac{1+frac{228 }{85}t+frac{451 }{340}t^2 } {1+frac{313 }{85}t+frac{1193 }{340}t^2+frac{133 }{204}t^3 }$$ This would give, as an approximate solution $xapprox -0.0130442$ while the exact value would be $xapprox -0.0133630 $.
For the specific case you address, plotting the function, you notice that the solution is close to $x=0$. So, make a series expansion
$$e^x+2x=1+3 x+frac{1}{2}x^2+Oleft(x^3right)$$ So, using the first term
$$frac{24}{25}=1+3ximplies x=-frac 1{75} approx -0.0133333 $$ Using the first and second term
$$frac{24}{25}=1+3x+frac{1}{2}x^2implies x=frac{sqrt{223}}{5}-3approx -0.0133631$$
Sooner or later, you will alos learn that, better then with Taylor series, we can approximate functions using Padé approximants. If we use the $[1,n]$ of them which will write as
$$f(x)simeq frac {f(a)+c_{(n)} (x-a)}{1+sum_{k=1}^n d_k (x-a)^k}$$ then, starting with a rational value of $a$, we can get rational approximations of the solution
$$x_{(n)}=a-frac{f(a)}{c_{(n)}}$$
$$left(
begin{array}{ccc}
n & x_{(n)} & x_{(n)} approx \
0 & -frac{1}{75} & -0.01333333333 \
1 & -frac{6}{449} & -0.01336302895 \
2 & -frac{1347}{100801} & -0.01336296267
end{array}
right)$$
answered Jan 30 at 7:37
Claude LeiboviciClaude Leibovici
125k1158136
$endgroup$
add a comment |
$begingroup$
This has been written for your curiosity.
In fact, the equation
$$y=e^{a x}+b x$$ shows explicit solution(s) in terms of Lambert function. They write
$$x=frac{y}{b}-frac{1}{a}Wleft(frac{a }{b}e^{frac{a y}{b}}right)$$ So, for your specifdic case $(a=1,b=2,y=frac{24}{25})$
$$x=frac{12}{25}-Wleft(frac{e^{12/25}}{2}right)$$ Since the argument is "small", you could use, as an approximation built around $t=0$,
$$W(t)=t ,frac{1+frac{228 }{85}t+frac{451 }{340}t^2 } {1+frac{313 }{85}t+frac{1193 }{340}t^2+frac{133 }{204}t^3 }$$ This would give, as an approximate solution $xapprox -0.0130442$ while the exact value would be $xapprox -0.0133630 $.
For the specific case you address, plotting the function, you notice that the solution is close to $x=0$. So, make a series expansion
$$e^x+2x=1+3 x+frac{1}{2}x^2+Oleft(x^3right)$$ So, using the first term
$$frac{24}{25}=1+3ximplies x=-frac 1{75} approx -0.0133333 $$ Using the first and second term
$$frac{24}{25}=1+3x+frac{1}{2}x^2implies x=frac{sqrt{223}}{5}-3approx -0.0133631$$
Sooner or later, you will alos learn that, better then with Taylor series, we can approximate functions using Padé approximants. If we use the $[1,n]$ of them which will write as
$$f(x)simeq frac {f(a)+c_{(n)} (x-a)}{1+sum_{k=1}^n d_k (x-a)^k}$$ then, starting with a rational value of $a$, we can get rational approximations of the solution
$$x_{(n)}=a-frac{f(a)}{c_{(n)}}$$
$$left(
begin{array}{ccc}
n & x_{(n)} & x_{(n)} approx \
0 & -frac{1}{75} & -0.01333333333 \
1 & -frac{6}{449} & -0.01336302895 \
2 & -frac{1347}{100801} & -0.01336296267
end{array}
right)$$
answered Jan 30 at 7:37
Claude LeiboviciClaude Leibovici
125k1158136
$endgroup$
This has been written for your curiosity.
In fact, the equation
$$y=e^{a x}+b x$$ shows explicit solution(s) in terms of Lambert function. They write
$$x=frac{y}{b}-frac{1}{a}Wleft(frac{a }{b}e^{frac{a y}{b}}right)$$ So, for your specifdic case $(a=1,b=2,y=frac{24}{25})$
$$x=frac{12}{25}-Wleft(frac{e^{12/25}}{2}right)$$ Since the argument is "small", you could use, as an approximation built around $t=0$,
$$W(t)=t ,frac{1+frac{228 }{85}t+frac{451 }{340}t^2 } {1+frac{313 }{85}t+frac{1193 }{340}t^2+frac{133 }{204}t^3 }$$ This would give, as an approximate solution $xapprox -0.0130442$ while the exact value would be $xapprox -0.0133630 $.
For the specific case you address, plotting the function, you notice that the solution is close to $x=0$. So, make a series expansion
$$e^x+2x=1+3 x+frac{1}{2}x^2+Oleft(x^3right)$$ So, using the first term
$$frac{24}{25}=1+3ximplies x=-frac 1{75} approx -0.0133333 $$ Using the first and second term
$$frac{24}{25}=1+3x+frac{1}{2}x^2implies x=frac{sqrt{223}}{5}-3approx -0.0133631$$
Sooner or later, you will alos learn that, better then with Taylor series, we can approximate functions using Padé approximants. If we use the $[1,n]$ of them which will write as
$$f(x)simeq frac {f(a)+c_{(n)} (x-a)}{1+sum_{k=1}^n d_k (x-a)^k}$$ then, starting with a rational value of $a$, we can get rational approximations of the solution
$$x_{(n)}=a-frac{f(a)}{c_{(n)}}$$
$$left(
begin{array}{ccc}
n & x_{(n)} & x_{(n)} approx \
0 & -frac{1}{75} & -0.01333333333 \
1 & -frac{6}{449} & -0.01336302895 \
2 & -frac{1347}{100801} & -0.01336296267
end{array}
right)$$
answered Jan 30 at 7:37
Claude LeiboviciClaude Leibovici
125k1158136
edited Jan 30 at 8:08
answered Jan 30 at 7:37
Claude LeiboviciClaude Leibovici
125k1158136
answered Jan 30 at 7:37
Claude LeiboviciClaude Leibovici
125k1158136
answered Jan 30 at 7:37
Claude LeiboviciClaude Leibovici
125k1158136
125k1158136
add a comment |
add a comment |
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$begingroup$
$f(x)$ is a monic polynomial then there exist one root(or no real root) for $f(x)=const$
$endgroup$
– Samvel Safaryan
Jan 29 at 22:30
$begingroup$
@SamvelSafaryan f is clearly not a polynomial.
$endgroup$
– Shalop
Jan 30 at 1:33