Finding roots with Newton's method
I have
$$
f(x)=x^7-12
$$
and need to find the roots with $epsilon=0.0001$ accuracy. I have done the first steps but can't continue further. Here's what I've done:
Ive narrowed down the range where I there is a root and found those $x$ values:
$$f(1.4)=-1.45, f(1.43)=0.227$$
So I have $$Omega a=[1.4;1.43]$$
$$f prime(x) = 7x^6, f prime prime(x) = 42x^5$$
$${frac{1}{|f prime(x)|}} = max{frac{1}{|7x^6|}} = (x=1.4) = {frac{1}{52.7}} = a_1,$$
$${frac{|f prime prime(x)|}{1}} = max|21x^5| = (x=1.43)= 125.5 = a_2$$
$$c = a_1a_2 = {frac{125.5}{52.7}} = 2.38$$
Choose $$a = {frac{1.4+1.43}{2}}=1.41$$
$$b=min(a, {frac{1}{c}}) = min(1.41, {frac{1}{2.38}})$$
The minimum is the value of $c$, and our teacher said that we need to apply such steps after which the minimum is the value of $a$. What am I doing wrong?
roots
asked 2 days ago
user3132457user3132457
1336
|
show 1 more comment
I have
$$
f(x)=x^7-12
$$
and need to find the roots with $epsilon=0.0001$ accuracy. I have done the first steps but can't continue further. Here's what I've done:
Ive narrowed down the range where I there is a root and found those $x$ values:
$$f(1.4)=-1.45, f(1.43)=0.227$$
So I have $$Omega a=[1.4;1.43]$$
$$f prime(x) = 7x^6, f prime prime(x) = 42x^5$$
$${frac{1}{|f prime(x)|}} = max{frac{1}{|7x^6|}} = (x=1.4) = {frac{1}{52.7}} = a_1,$$
$${frac{|f prime prime(x)|}{1}} = max|21x^5| = (x=1.43)= 125.5 = a_2$$
$$c = a_1a_2 = {frac{125.5}{52.7}} = 2.38$$
Choose $$a = {frac{1.4+1.43}{2}}=1.41$$
$$b=min(a, {frac{1}{c}}) = min(1.41, {frac{1}{2.38}})$$
The minimum is the value of $c$, and our teacher said that we need to apply such steps after which the minimum is the value of $a$. What am I doing wrong?
roots
asked 2 days ago
user3132457user3132457
1336
Your method doesn´t look like the Newton–Raphson method.
– callculus
2 days ago
Right, removed the tag. (There's no tag for Newton's method)
– user3132457
2 days ago
But what method is it?
– callculus
2 days ago
It is Newton's method. Yeah it doesn't look like the one you showed but that's how we learned to do it.
– user3132457
2 days ago
Do you have a reference?
– callculus
2 days ago
|
show 1 more comment
I have
$$
f(x)=x^7-12
$$
and need to find the roots with $epsilon=0.0001$ accuracy. I have done the first steps but can't continue further. Here's what I've done:
Ive narrowed down the range where I there is a root and found those $x$ values:
$$f(1.4)=-1.45, f(1.43)=0.227$$
So I have $$Omega a=[1.4;1.43]$$
$$f prime(x) = 7x^6, f prime prime(x) = 42x^5$$
$${frac{1}{|f prime(x)|}} = max{frac{1}{|7x^6|}} = (x=1.4) = {frac{1}{52.7}} = a_1,$$
$${frac{|f prime prime(x)|}{1}} = max|21x^5| = (x=1.43)= 125.5 = a_2$$
$$c = a_1a_2 = {frac{125.5}{52.7}} = 2.38$$
Choose $$a = {frac{1.4+1.43}{2}}=1.41$$
$$b=min(a, {frac{1}{c}}) = min(1.41, {frac{1}{2.38}})$$
The minimum is the value of $c$, and our teacher said that we need to apply such steps after which the minimum is the value of $a$. What am I doing wrong?
roots
asked 2 days ago
user3132457user3132457
1336
I have
$$
f(x)=x^7-12
$$
and need to find the roots with $epsilon=0.0001$ accuracy. I have done the first steps but can't continue further. Here's what I've done:
Ive narrowed down the range where I there is a root and found those $x$ values:
$$f(1.4)=-1.45, f(1.43)=0.227$$
So I have $$Omega a=[1.4;1.43]$$
$$f prime(x) = 7x^6, f prime prime(x) = 42x^5$$
$${frac{1}{|f prime(x)|}} = max{frac{1}{|7x^6|}} = (x=1.4) = {frac{1}{52.7}} = a_1,$$
$${frac{|f prime prime(x)|}{1}} = max|21x^5| = (x=1.43)= 125.5 = a_2$$
$$c = a_1a_2 = {frac{125.5}{52.7}} = 2.38$$
Choose $$a = {frac{1.4+1.43}{2}}=1.41$$
$$b=min(a, {frac{1}{c}}) = min(1.41, {frac{1}{2.38}})$$
The minimum is the value of $c$, and our teacher said that we need to apply such steps after which the minimum is the value of $a$. What am I doing wrong?
roots
roots
asked 2 days ago
user3132457user3132457
1336
asked 2 days ago
user3132457user3132457
1336
edited 2 days ago
user3132457
asked 2 days ago
user3132457user3132457
1336
asked 2 days ago
user3132457user3132457
1336
asked 2 days ago
user3132457user3132457
1336
1336
Your method doesn´t look like the Newton–Raphson method.
– callculus
2 days ago
Right, removed the tag. (There's no tag for Newton's method)
– user3132457
2 days ago
But what method is it?
– callculus
2 days ago
It is Newton's method. Yeah it doesn't look like the one you showed but that's how we learned to do it.
– user3132457
2 days ago
Do you have a reference?
– callculus
2 days ago
|
show 1 more comment
Your method doesn´t look like the Newton–Raphson method.
– callculus
2 days ago
Right, removed the tag. (There's no tag for Newton's method)
– user3132457
2 days ago
But what method is it?
– callculus
2 days ago
It is Newton's method. Yeah it doesn't look like the one you showed but that's how we learned to do it.
– user3132457
2 days ago
Do you have a reference?
– callculus
2 days ago
Your method doesn´t look like the Newton–Raphson method.
– callculus
2 days ago
Your method doesn´t look like the Newton–Raphson method.
– callculus
2 days ago
Right, removed the tag. (There's no tag for Newton's method)
– user3132457
2 days ago
Right, removed the tag. (There's no tag for Newton's method)
– user3132457
2 days ago
But what method is it?
– callculus
2 days ago
But what method is it?
– callculus
2 days ago
It is Newton's method. Yeah it doesn't look like the one you showed but that's how we learned to do it.
– user3132457
2 days ago
It is Newton's method. Yeah it doesn't look like the one you showed but that's how we learned to do it.
– user3132457
2 days ago
Do you have a reference?
– callculus
2 days ago
Do you have a reference?
– callculus
2 days ago
|
show 1 more comment
1 Answer
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Okay, I got what's wrong. I chose $a$ incorrectly. Should be
$$ a={frac{1.43-1.4}{2}} = {frac{0.03}{2}} = 0.015 $$
In that case we can have $b$ to be the value of $a$
$$b=min(a, {frac{1}{c}}) = min(0.015, {frac{1}{2.38}}) = 0.015$$
answered 2 days ago
user3132457user3132457
1336
add a comment |
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Okay, I got what's wrong. I chose $a$ incorrectly. Should be
$$ a={frac{1.43-1.4}{2}} = {frac{0.03}{2}} = 0.015 $$
In that case we can have $b$ to be the value of $a$
$$b=min(a, {frac{1}{c}}) = min(0.015, {frac{1}{2.38}}) = 0.015$$
answered 2 days ago
user3132457user3132457
1336
add a comment |
Okay, I got what's wrong. I chose $a$ incorrectly. Should be
$$ a={frac{1.43-1.4}{2}} = {frac{0.03}{2}} = 0.015 $$
In that case we can have $b$ to be the value of $a$
$$b=min(a, {frac{1}{c}}) = min(0.015, {frac{1}{2.38}}) = 0.015$$
answered 2 days ago
user3132457user3132457
1336
add a comment |
Okay, I got what's wrong. I chose $a$ incorrectly. Should be
$$ a={frac{1.43-1.4}{2}} = {frac{0.03}{2}} = 0.015 $$
In that case we can have $b$ to be the value of $a$
$$b=min(a, {frac{1}{c}}) = min(0.015, {frac{1}{2.38}}) = 0.015$$
answered 2 days ago
user3132457user3132457
1336
Okay, I got what's wrong. I chose $a$ incorrectly. Should be
$$ a={frac{1.43-1.4}{2}} = {frac{0.03}{2}} = 0.015 $$
In that case we can have $b$ to be the value of $a$
$$b=min(a, {frac{1}{c}}) = min(0.015, {frac{1}{2.38}}) = 0.015$$
answered 2 days ago
user3132457user3132457
1336
answered 2 days ago
user3132457user3132457
1336
answered 2 days ago
user3132457user3132457
1336
answered 2 days ago
user3132457user3132457
1336
1336
add a comment |
add a comment |
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Your method doesn´t look like the Newton–Raphson method.
– callculus
2 days ago
Right, removed the tag. (There's no tag for Newton's method)
– user3132457
2 days ago
But what method is it?
– callculus
2 days ago
It is Newton's method. Yeah it doesn't look like the one you showed but that's how we learned to do it.
– user3132457
2 days ago
Do you have a reference?
– callculus
2 days ago