Finding roots with Newton's method












0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your method doesn´t look like the Newton–Raphson method.
    $endgroup$
    – callculus
    Jan 5 at 16:24










  • $begingroup$
    Right, removed the tag. (There's no tag for Newton's method)
    $endgroup$
    – user3132457
    Jan 5 at 16:25










  • $begingroup$
    But what method is it?
    $endgroup$
    – callculus
    Jan 5 at 16:26










  • $begingroup$
    It is Newton's method. Yeah it doesn't look like the one you showed but that's how we learned to do it.
    $endgroup$
    – user3132457
    Jan 5 at 16:27












  • $begingroup$
    Do you have a reference?
    $endgroup$
    – callculus
    Jan 5 at 16:30
















0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your method doesn´t look like the Newton–Raphson method.
    $endgroup$
    – callculus
    Jan 5 at 16:24










  • $begingroup$
    Right, removed the tag. (There's no tag for Newton's method)
    $endgroup$
    – user3132457
    Jan 5 at 16:25










  • $begingroup$
    But what method is it?
    $endgroup$
    – callculus
    Jan 5 at 16:26










  • $begingroup$
    It is Newton's method. Yeah it doesn't look like the one you showed but that's how we learned to do it.
    $endgroup$
    – user3132457
    Jan 5 at 16:27












  • $begingroup$
    Do you have a reference?
    $endgroup$
    – callculus
    Jan 5 at 16:30














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 16:25







user3132457

















asked Jan 5 at 15:58









user3132457user3132457

1598




1598












  • $begingroup$
    Your method doesn´t look like the Newton–Raphson method.
    $endgroup$
    – callculus
    Jan 5 at 16:24










  • $begingroup$
    Right, removed the tag. (There's no tag for Newton's method)
    $endgroup$
    – user3132457
    Jan 5 at 16:25










  • $begingroup$
    But what method is it?
    $endgroup$
    – callculus
    Jan 5 at 16:26










  • $begingroup$
    It is Newton's method. Yeah it doesn't look like the one you showed but that's how we learned to do it.
    $endgroup$
    – user3132457
    Jan 5 at 16:27












  • $begingroup$
    Do you have a reference?
    $endgroup$
    – callculus
    Jan 5 at 16:30


















  • $begingroup$
    Your method doesn´t look like the Newton–Raphson method.
    $endgroup$
    – callculus
    Jan 5 at 16:24










  • $begingroup$
    Right, removed the tag. (There's no tag for Newton's method)
    $endgroup$
    – user3132457
    Jan 5 at 16:25










  • $begingroup$
    But what method is it?
    $endgroup$
    – callculus
    Jan 5 at 16:26










  • $begingroup$
    It is Newton's method. Yeah it doesn't look like the one you showed but that's how we learned to do it.
    $endgroup$
    – user3132457
    Jan 5 at 16:27












  • $begingroup$
    Do you have a reference?
    $endgroup$
    – callculus
    Jan 5 at 16:30
















$begingroup$
Your method doesn´t look like the Newton–Raphson method.
$endgroup$
– callculus
Jan 5 at 16:24




$begingroup$
Your method doesn´t look like the Newton–Raphson method.
$endgroup$
– callculus
Jan 5 at 16:24












$begingroup$
Right, removed the tag. (There's no tag for Newton's method)
$endgroup$
– user3132457
Jan 5 at 16:25




$begingroup$
Right, removed the tag. (There's no tag for Newton's method)
$endgroup$
– user3132457
Jan 5 at 16:25












$begingroup$
But what method is it?
$endgroup$
– callculus
Jan 5 at 16:26




$begingroup$
But what method is it?
$endgroup$
– callculus
Jan 5 at 16:26












$begingroup$
It is Newton's method. Yeah it doesn't look like the one you showed but that's how we learned to do it.
$endgroup$
– user3132457
Jan 5 at 16:27






$begingroup$
It is Newton's method. Yeah it doesn't look like the one you showed but that's how we learned to do it.
$endgroup$
– user3132457
Jan 5 at 16:27














$begingroup$
Do you have a reference?
$endgroup$
– callculus
Jan 5 at 16:30




$begingroup$
Do you have a reference?
$endgroup$
– callculus
Jan 5 at 16:30










1 Answer
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$begingroup$

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 $a$
$$b=min(a, {frac{1}{c}}) = min(0.015, {frac{1}{2.38}}) = 0.015$$






share|cite|improve this answer









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    $begingroup$

    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 $a$
    $$b=min(a, {frac{1}{c}}) = min(0.015, {frac{1}{2.38}}) = 0.015$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      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 $a$
      $$b=min(a, {frac{1}{c}}) = min(0.015, {frac{1}{2.38}}) = 0.015$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        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 $a$
        $$b=min(a, {frac{1}{c}}) = min(0.015, {frac{1}{2.38}}) = 0.015$$






        share|cite|improve this answer









        $endgroup$



        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 $a$
        $$b=min(a, {frac{1}{c}}) = min(0.015, {frac{1}{2.38}}) = 0.015$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 5 at 16:23









        user3132457user3132457

        1598




        1598






























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