can anyone help me to find the square root of non perfect squares? [closed]

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Can anyone give me good and more easier method to find the square roots of non perfect squares .It finds very hard for me to calculate without calculator in my exams

square-numbers

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edited Jul 29 at 19:13

Will Jagy

96.8k594195

asked Jul 29 at 14:06

Akash. B

134

closed as off-topic by Hurkyl, Did, user21820, amWhy, Claude Leibovici Aug 3 at 12:17

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Hurkyl, Did, user21820, amWhy, Claude Leibovici

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Do you actually have to calculate something like $sqrt61$ by hand ?
  – Peter
  Jul 29 at 14:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Peter Indians have to.
  – MalayTheDynamo
  Jul 29 at 14:26
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Who are these "Guys" in the title?
  – Did
  Jul 29 at 14:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  How accurate must the result be ?
  – Peter
  Jul 29 at 15:03
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Instead of getting help to find an easier method, you'd need to get help concerning how to speak to someone properly. What did you actually mean by "Guys"? On such an occasion you won't have been warned for that.
  – Cargobob
  Jul 29 at 15:14
  
  

  
  
  
  

 | 
show 2 more comments

up vote
0
down vote

favorite

1

Can anyone give me good and more easier method to find the square roots of non perfect squares .It finds very hard for me to calculate without calculator in my exams

square-numbers

share|cite|improve this question

edited Jul 29 at 19:13

Will Jagy

96.8k594195

asked Jul 29 at 14:06

Akash. B

134

closed as off-topic by Hurkyl, Did, user21820, amWhy, Claude Leibovici Aug 3 at 12:17

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Hurkyl, Did, user21820, amWhy, Claude Leibovici

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Do you actually have to calculate something like $sqrt61$ by hand ?
  – Peter
  Jul 29 at 14:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Peter Indians have to.
  – MalayTheDynamo
  Jul 29 at 14:26
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Who are these "Guys" in the title?
  – Did
  Jul 29 at 14:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  How accurate must the result be ?
  – Peter
  Jul 29 at 15:03
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Instead of getting help to find an easier method, you'd need to get help concerning how to speak to someone properly. What did you actually mean by "Guys"? On such an occasion you won't have been warned for that.
  – Cargobob
  Jul 29 at 15:14
  
  

  
  
  
  

 | 
show 2 more comments

up vote
0
down vote

favorite

1

up vote
0
down vote

favorite

1

1

Can anyone give me good and more easier method to find the square roots of non perfect squares .It finds very hard for me to calculate without calculator in my exams

square-numbers

share|cite|improve this question

edited Jul 29 at 19:13

Will Jagy

96.8k594195

asked Jul 29 at 14:06

Akash. B

134

Can anyone give me good and more easier method to find the square roots of non perfect squares .It finds very hard for me to calculate without calculator in my exams

square-numbers

share|cite|improve this question

edited Jul 29 at 19:13

Will Jagy

96.8k594195

asked Jul 29 at 14:06

Akash. B

134

share|cite|improve this question

share|cite|improve this question

edited Jul 29 at 19:13

Will Jagy

96.8k594195

edited Jul 29 at 19:13

Will Jagy

96.8k594195

edited Jul 29 at 19:13

Will Jagy

96.8k594195

96.8k594195

asked Jul 29 at 14:06

Akash. B

134

asked Jul 29 at 14:06

Akash. B

134

asked Jul 29 at 14:06

Akash. B

134

134

closed as off-topic by Hurkyl, Did, user21820, amWhy, Claude Leibovici Aug 3 at 12:17

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Hurkyl, Did, user21820, amWhy, Claude Leibovici

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Hurkyl, Did, user21820, amWhy, Claude Leibovici Aug 3 at 12:17

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Hurkyl, Did, user21820, amWhy, Claude Leibovici

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Do you actually have to calculate something like $sqrt61$ by hand ?
  – Peter
  Jul 29 at 14:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Peter Indians have to.
  – MalayTheDynamo
  Jul 29 at 14:26
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Who are these "Guys" in the title?
  – Did
  Jul 29 at 14:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  How accurate must the result be ?
  – Peter
  Jul 29 at 15:03
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Instead of getting help to find an easier method, you'd need to get help concerning how to speak to someone properly. What did you actually mean by "Guys"? On such an occasion you won't have been warned for that.
  – Cargobob
  Jul 29 at 15:14
  
  

  
  
  
  

 | 
show 2 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Do you actually have to calculate something like $sqrt61$ by hand ?
  – Peter
  Jul 29 at 14:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Peter Indians have to.
  – MalayTheDynamo
  Jul 29 at 14:26
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Who are these "Guys" in the title?
  – Did
  Jul 29 at 14:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  How accurate must the result be ?
  – Peter
  Jul 29 at 15:03
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Instead of getting help to find an easier method, you'd need to get help concerning how to speak to someone properly. What did you actually mean by "Guys"? On such an occasion you won't have been warned for that.
  – Cargobob
  Jul 29 at 15:14
  
  

  
  
  
  

Do you actually have to calculate something like $sqrt61$ by hand ?
– Peter
Jul 29 at 14:16

Do you actually have to calculate something like $sqrt61$ by hand ?
– Peter
Jul 29 at 14:16

@Peter Indians have to.
– MalayTheDynamo
Jul 29 at 14:26

@Peter Indians have to.
– MalayTheDynamo
Jul 29 at 14:26

2

2

Who are these "Guys" in the title?
– Did
Jul 29 at 14:42

Who are these "Guys" in the title?
– Did
Jul 29 at 14:42

How accurate must the result be ?
– Peter
Jul 29 at 15:03

How accurate must the result be ?
– Peter
Jul 29 at 15:03

2

2

Instead of getting help to find an easier method, you'd need to get help concerning how to speak to someone properly. What did you actually mean by "Guys"? On such an occasion you won't have been warned for that.
– Cargobob
Jul 29 at 15:14

Instead of getting help to find an easier method, you'd need to get help concerning how to speak to someone properly. What did you actually mean by "Guys"? On such an occasion you won't have been warned for that.
– Cargobob
Jul 29 at 15:14

 | 
show 2 more comments

5 Answers
5

active

oldest

votes

up vote
3
down vote



accepted

It reminds me of a small trick when I first learned calculus:

$$fracsqrtx+h-sqrtxhapprox frac12sqrt x$$
for small $h$, due to the first principle of calculus.

By rearranging,
$$sqrtx+happroxsqrt x+frach2sqrt x$$

For example,
$$sqrt4.1=sqrt4+0.1approxsqrt4+frac0.12sqrt4=2.025$$

which has a percentage error of $0.0076$%.

---

Indeed, there is a step by step procedure to take square root by hand.

Given a number $x$, estimate the number of digits of $sqrt x$ (this is not so hard for small $x$).

Then, let $sqrt x=overlineABCcdots.abccdots$ (that’s a decimal dot).

Due to my poor english, I would instead show a worked example.

Compute $sqrt123$ rounded down to the nearest tenth.

Let $sqrt123=overlineAB.a+cdots$

1. Find the largest integer $A$ such that
   $(10A)^2le 123$

We get $A=1$.

2. Find the largest integer $B$ such that $(10A+B)^2le 123$

We get $B=1$.

3. Find the largest integer $a$ such that $left(10A+B+fraca10right)^2le123$

We get $a=0$.

So $sqrt123$ rounds down to $11.0$.

share|cite|improve this answer

edited Jul 29 at 14:50

answered Jul 29 at 14:26

Szeto

3,8781421

add a comment | 

up vote
2
down vote

one can do Newton's method by hand. Some judgment is required in the number of digits in the division step.

For $sqrt 61$ the estimate $8$ is probably closer than $7$

$$ frac618 = 7.625 $$
$$ frac8 + 7.6252 = 7.8125 $$
$$ $$
$$ $$
$$ frac617.8125 = 7.808 $$
$$ frac7.8125 + 7.8082 = 7.81025 $$
$$ $$
$$ $$
$$ frac617.81025 approx 7.810249352 $$
$$ frac7.81025 + 7.8102493522 approx 7.810249675 $$
$$ $$
$$ $$

The attractive aspect is that the number of correct digits roughly doubles ate each step
$$8$$
$$7.8125$$
$$7.81025$$
$$7.810249675906661$$
$$7.810249675906654$$

share|cite|improve this answer

edited Jul 29 at 19:10

answered Jul 29 at 15:11

Will Jagy

96.8k594195

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I included a "tweak" to Newton's Method in my A............+1
  – DanielWainfleet
  Jul 30 at 2:56
  

  
  
  
  

add a comment | 

up vote
1
down vote

If you're happy with fractions, I like the "Babylonian method" (you can Google it).

To estimate $sqrtk$, start with a guess $x_0$ then use the iteration $x_n+1=frac12(x_n+frackx_n)$.

For example, estimating $sqrt5$ :
$$beginalign
x_0&=2
\ x_1 &=frac12(2+frac52) = frac94
\ x_2 &=frac12(frac94+frac5frac94) =frac12(frac94+frac209) = frac16172
\ & mathrmetc.
endalign$$

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answered Jul 29 at 15:40

Malkin

1,099421

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up vote
1
down vote

If you want $sqrtN=sqrta^2+bquad$ with $a$ being the greatest integer possible such that $bge 0$.

Then by factoring $a$ you get by Taylor expansion: $$x_0=aleft(1+frac ba^2right)^frac 12approx aleft(1+frac b2a^2right)approx a+frac b2a$$

Starting from this seed you can apply the Newton method to $f(x)=x^2-N$.

This gives $$x_n+1=x_n-fracf(x_n)f'(x_n)$$

In particular $requirecancelf(x_0)=(a+frac b2a)^2-(a^2+b)=cancela^2+cancel2afrac b2a+(frac b2a)^2-cancela^2-cancelb$

Thus a good approximation after just $1$ iteration is given by

$$x_1=left(a+frac b2aright)-fracleft(frac b2aright)^22left(a+frac b2aright)$$

Once simplified it can be written: $$dfrac8a^4+8ba^2+b^24a(2a^2+b)$$

This generally is enough to have a few significant digits right:

$sqrt5=sqrt2^2+1approxdfrac16172approx 2.236111quad[2.236067]$

$sqrt91=sqrt9^2+10approxdfrac590686192approx 9.539405quad[9.539392]$

I think the advantage of this method with $a,b$ is that the formula in yellow is not too difficult to remember for your examination and it gives you a fair result already.

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answered Jul 29 at 17:12

zwim

11k627

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up vote
0
down vote

Method 1: The Generalized Binomial Theorem (Newton): If $|x|<1$ then $(1+x)^r=1+A_1x+A_2x^2+A_3x^3+...$ where $A_1=r$ and $A_n+1 =A_ncdot frac (r-n)n.$ For example $sqrt 61=8(1+r)^1/2$ with $r=-3/64.$

Method 2: About 19 centuries ago Heron (Hero) of Alexandria wrote of letting $x_n+1=x_n-frac x_n^2-A2x_n,$ when $A>0$ and $x_1>0,$ to make a sequence $(x_1,x_2,x_3,...)$ converging to $sqrt A.$ This is a special instance of Newton's Method, as already covered in another answer. A "tweak" to this is to, instead, let $x_n+1=x_n-frac 2x(x_n^2-A)3x_n^2+A$, in which the number of decimal places of accuracy approximately triples at each iteration.

share|cite|improve this answer

answered Jul 30 at 2:53

DanielWainfleet

31.4k31542

add a comment | 

5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

It reminds me of a small trick when I first learned calculus:

$$fracsqrtx+h-sqrtxhapprox frac12sqrt x$$
for small $h$, due to the first principle of calculus.

By rearranging,
$$sqrtx+happroxsqrt x+frach2sqrt x$$

For example,
$$sqrt4.1=sqrt4+0.1approxsqrt4+frac0.12sqrt4=2.025$$

which has a percentage error of $0.0076$%.

---

Indeed, there is a step by step procedure to take square root by hand.

Given a number $x$, estimate the number of digits of $sqrt x$ (this is not so hard for small $x$).

Then, let $sqrt x=overlineABCcdots.abccdots$ (that’s a decimal dot).

Due to my poor english, I would instead show a worked example.

Compute $sqrt123$ rounded down to the nearest tenth.

Let $sqrt123=overlineAB.a+cdots$

1. Find the largest integer $A$ such that
   $(10A)^2le 123$

We get $A=1$.

2. Find the largest integer $B$ such that $(10A+B)^2le 123$

We get $B=1$.

3. Find the largest integer $a$ such that $left(10A+B+fraca10right)^2le123$

We get $a=0$.

So $sqrt123$ rounds down to $11.0$.

share|cite|improve this answer

edited Jul 29 at 14:50

answered Jul 29 at 14:26

Szeto

3,8781421

add a comment | 

up vote
3
down vote



accepted

It reminds me of a small trick when I first learned calculus:

$$fracsqrtx+h-sqrtxhapprox frac12sqrt x$$
for small $h$, due to the first principle of calculus.

By rearranging,
$$sqrtx+happroxsqrt x+frach2sqrt x$$

For example,
$$sqrt4.1=sqrt4+0.1approxsqrt4+frac0.12sqrt4=2.025$$

which has a percentage error of $0.0076$%.

---

Indeed, there is a step by step procedure to take square root by hand.

Given a number $x$, estimate the number of digits of $sqrt x$ (this is not so hard for small $x$).

Then, let $sqrt x=overlineABCcdots.abccdots$ (that’s a decimal dot).

Due to my poor english, I would instead show a worked example.

Compute $sqrt123$ rounded down to the nearest tenth.

Let $sqrt123=overlineAB.a+cdots$

1. Find the largest integer $A$ such that
   $(10A)^2le 123$

We get $A=1$.

2. Find the largest integer $B$ such that $(10A+B)^2le 123$

We get $B=1$.

3. Find the largest integer $a$ such that $left(10A+B+fraca10right)^2le123$

We get $a=0$.

So $sqrt123$ rounds down to $11.0$.

share|cite|improve this answer

edited Jul 29 at 14:50

answered Jul 29 at 14:26

Szeto

3,8781421

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

It reminds me of a small trick when I first learned calculus:

$$fracsqrtx+h-sqrtxhapprox frac12sqrt x$$
for small $h$, due to the first principle of calculus.

By rearranging,
$$sqrtx+happroxsqrt x+frach2sqrt x$$

For example,
$$sqrt4.1=sqrt4+0.1approxsqrt4+frac0.12sqrt4=2.025$$

which has a percentage error of $0.0076$%.

---

Indeed, there is a step by step procedure to take square root by hand.

Given a number $x$, estimate the number of digits of $sqrt x$ (this is not so hard for small $x$).

Then, let $sqrt x=overlineABCcdots.abccdots$ (that’s a decimal dot).

Due to my poor english, I would instead show a worked example.

Compute $sqrt123$ rounded down to the nearest tenth.

Let $sqrt123=overlineAB.a+cdots$

1. Find the largest integer $A$ such that
   $(10A)^2le 123$

We get $A=1$.

2. Find the largest integer $B$ such that $(10A+B)^2le 123$

We get $B=1$.

3. Find the largest integer $a$ such that $left(10A+B+fraca10right)^2le123$

We get $a=0$.

So $sqrt123$ rounds down to $11.0$.

share|cite|improve this answer

edited Jul 29 at 14:50

answered Jul 29 at 14:26

Szeto

3,8781421

It reminds me of a small trick when I first learned calculus:

$$fracsqrtx+h-sqrtxhapprox frac12sqrt x$$
for small $h$, due to the first principle of calculus.

By rearranging,
$$sqrtx+happroxsqrt x+frach2sqrt x$$

For example,
$$sqrt4.1=sqrt4+0.1approxsqrt4+frac0.12sqrt4=2.025$$

which has a percentage error of $0.0076$%.

---

Indeed, there is a step by step procedure to take square root by hand.

Given a number $x$, estimate the number of digits of $sqrt x$ (this is not so hard for small $x$).

Then, let $sqrt x=overlineABCcdots.abccdots$ (that’s a decimal dot).

Due to my poor english, I would instead show a worked example.

Compute $sqrt123$ rounded down to the nearest tenth.

Let $sqrt123=overlineAB.a+cdots$

1. Find the largest integer $A$ such that
   $(10A)^2le 123$

We get $A=1$.

2. Find the largest integer $B$ such that $(10A+B)^2le 123$

We get $B=1$.

3. Find the largest integer $a$ such that $left(10A+B+fraca10right)^2le123$

We get $a=0$.

So $sqrt123$ rounds down to $11.0$.

share|cite|improve this answer

edited Jul 29 at 14:50

answered Jul 29 at 14:26

Szeto

3,8781421

share|cite|improve this answer

share|cite|improve this answer

edited Jul 29 at 14:50

edited Jul 29 at 14:50

edited Jul 29 at 14:50

answered Jul 29 at 14:26

Szeto

3,8781421

answered Jul 29 at 14:26

Szeto

3,8781421

answered Jul 29 at 14:26

Szeto

3,8781421

3,8781421

add a comment | 

add a comment | 

up vote
2
down vote

one can do Newton's method by hand. Some judgment is required in the number of digits in the division step.

For $sqrt 61$ the estimate $8$ is probably closer than $7$

$$ frac618 = 7.625 $$
$$ frac8 + 7.6252 = 7.8125 $$
$$ $$
$$ $$
$$ frac617.8125 = 7.808 $$
$$ frac7.8125 + 7.8082 = 7.81025 $$
$$ $$
$$ $$
$$ frac617.81025 approx 7.810249352 $$
$$ frac7.81025 + 7.8102493522 approx 7.810249675 $$
$$ $$
$$ $$

The attractive aspect is that the number of correct digits roughly doubles ate each step
$$8$$
$$7.8125$$
$$7.81025$$
$$7.810249675906661$$
$$7.810249675906654$$

share|cite|improve this answer

edited Jul 29 at 19:10

answered Jul 29 at 15:11

Will Jagy

96.8k594195

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I included a "tweak" to Newton's Method in my A............+1
  – DanielWainfleet
  Jul 30 at 2:56
  

  
  
  
  

add a comment | 

up vote
2
down vote

one can do Newton's method by hand. Some judgment is required in the number of digits in the division step.

For $sqrt 61$ the estimate $8$ is probably closer than $7$

$$ frac618 = 7.625 $$
$$ frac8 + 7.6252 = 7.8125 $$
$$ $$
$$ $$
$$ frac617.8125 = 7.808 $$
$$ frac7.8125 + 7.8082 = 7.81025 $$
$$ $$
$$ $$
$$ frac617.81025 approx 7.810249352 $$
$$ frac7.81025 + 7.8102493522 approx 7.810249675 $$
$$ $$
$$ $$

The attractive aspect is that the number of correct digits roughly doubles ate each step
$$8$$
$$7.8125$$
$$7.81025$$
$$7.810249675906661$$
$$7.810249675906654$$

share|cite|improve this answer

edited Jul 29 at 19:10

answered Jul 29 at 15:11

Will Jagy

96.8k594195

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I included a "tweak" to Newton's Method in my A............+1
  – DanielWainfleet
  Jul 30 at 2:56
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

one can do Newton's method by hand. Some judgment is required in the number of digits in the division step.

For $sqrt 61$ the estimate $8$ is probably closer than $7$

$$ frac618 = 7.625 $$
$$ frac8 + 7.6252 = 7.8125 $$
$$ $$
$$ $$
$$ frac617.8125 = 7.808 $$
$$ frac7.8125 + 7.8082 = 7.81025 $$
$$ $$
$$ $$
$$ frac617.81025 approx 7.810249352 $$
$$ frac7.81025 + 7.8102493522 approx 7.810249675 $$
$$ $$
$$ $$

The attractive aspect is that the number of correct digits roughly doubles ate each step
$$8$$
$$7.8125$$
$$7.81025$$
$$7.810249675906661$$
$$7.810249675906654$$

share|cite|improve this answer

edited Jul 29 at 19:10

answered Jul 29 at 15:11

Will Jagy

96.8k594195

one can do Newton's method by hand. Some judgment is required in the number of digits in the division step.

For $sqrt 61$ the estimate $8$ is probably closer than $7$

$$ frac618 = 7.625 $$
$$ frac8 + 7.6252 = 7.8125 $$
$$ $$
$$ $$
$$ frac617.8125 = 7.808 $$
$$ frac7.8125 + 7.8082 = 7.81025 $$
$$ $$
$$ $$
$$ frac617.81025 approx 7.810249352 $$
$$ frac7.81025 + 7.8102493522 approx 7.810249675 $$
$$ $$
$$ $$

The attractive aspect is that the number of correct digits roughly doubles ate each step
$$8$$
$$7.8125$$
$$7.81025$$
$$7.810249675906661$$
$$7.810249675906654$$

share|cite|improve this answer

edited Jul 29 at 19:10

answered Jul 29 at 15:11

Will Jagy

96.8k594195

share|cite|improve this answer

share|cite|improve this answer

edited Jul 29 at 19:10

edited Jul 29 at 19:10

edited Jul 29 at 19:10

answered Jul 29 at 15:11

Will Jagy

96.8k594195

answered Jul 29 at 15:11

Will Jagy

96.8k594195

answered Jul 29 at 15:11

Will Jagy

96.8k594195

96.8k594195

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  I included a "tweak" to Newton's Method in my A............+1
  – DanielWainfleet
  Jul 30 at 2:56
  

  
  
  
  

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  I included a "tweak" to Newton's Method in my A............+1
  – DanielWainfleet
  Jul 30 at 2:56
  

  
  
  
  

I included a "tweak" to Newton's Method in my A............+1
– DanielWainfleet
Jul 30 at 2:56

I included a "tweak" to Newton's Method in my A............+1
– DanielWainfleet
Jul 30 at 2:56

add a comment | 

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1
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If you're happy with fractions, I like the "Babylonian method" (you can Google it).

To estimate $sqrtk$, start with a guess $x_0$ then use the iteration $x_n+1=frac12(x_n+frackx_n)$.

For example, estimating $sqrt5$ :
$$beginalign
x_0&=2
\ x_1 &=frac12(2+frac52) = frac94
\ x_2 &=frac12(frac94+frac5frac94) =frac12(frac94+frac209) = frac16172
\ & mathrmetc.
endalign$$

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answered Jul 29 at 15:40

Malkin

1,099421

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up vote
1
down vote

If you're happy with fractions, I like the "Babylonian method" (you can Google it).

To estimate $sqrtk$, start with a guess $x_0$ then use the iteration $x_n+1=frac12(x_n+frackx_n)$.

For example, estimating $sqrt5$ :
$$beginalign
x_0&=2
\ x_1 &=frac12(2+frac52) = frac94
\ x_2 &=frac12(frac94+frac5frac94) =frac12(frac94+frac209) = frac16172
\ & mathrmetc.
endalign$$

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answered Jul 29 at 15:40

Malkin

1,099421

add a comment | 

up vote
1
down vote

up vote
1
down vote

If you're happy with fractions, I like the "Babylonian method" (you can Google it).

To estimate $sqrtk$, start with a guess $x_0$ then use the iteration $x_n+1=frac12(x_n+frackx_n)$.

For example, estimating $sqrt5$ :
$$beginalign
x_0&=2
\ x_1 &=frac12(2+frac52) = frac94
\ x_2 &=frac12(frac94+frac5frac94) =frac12(frac94+frac209) = frac16172
\ & mathrmetc.
endalign$$

share|cite|improve this answer

answered Jul 29 at 15:40

Malkin

1,099421

If you're happy with fractions, I like the "Babylonian method" (you can Google it).

To estimate $sqrtk$, start with a guess $x_0$ then use the iteration $x_n+1=frac12(x_n+frackx_n)$.

For example, estimating $sqrt5$ :
$$beginalign
x_0&=2
\ x_1 &=frac12(2+frac52) = frac94
\ x_2 &=frac12(frac94+frac5frac94) =frac12(frac94+frac209) = frac16172
\ & mathrmetc.
endalign$$

share|cite|improve this answer

answered Jul 29 at 15:40

Malkin

1,099421

share|cite|improve this answer

share|cite|improve this answer

answered Jul 29 at 15:40

Malkin

1,099421

answered Jul 29 at 15:40

Malkin

1,099421

answered Jul 29 at 15:40

Malkin

1,099421

1,099421

add a comment | 

add a comment | 

up vote
1
down vote

If you want $sqrtN=sqrta^2+bquad$ with $a$ being the greatest integer possible such that $bge 0$.

Then by factoring $a$ you get by Taylor expansion: $$x_0=aleft(1+frac ba^2right)^frac 12approx aleft(1+frac b2a^2right)approx a+frac b2a$$

Starting from this seed you can apply the Newton method to $f(x)=x^2-N$.

This gives $$x_n+1=x_n-fracf(x_n)f'(x_n)$$

In particular $requirecancelf(x_0)=(a+frac b2a)^2-(a^2+b)=cancela^2+cancel2afrac b2a+(frac b2a)^2-cancela^2-cancelb$

Thus a good approximation after just $1$ iteration is given by

$$x_1=left(a+frac b2aright)-fracleft(frac b2aright)^22left(a+frac b2aright)$$

Once simplified it can be written: $$dfrac8a^4+8ba^2+b^24a(2a^2+b)$$

This generally is enough to have a few significant digits right:

$sqrt5=sqrt2^2+1approxdfrac16172approx 2.236111quad[2.236067]$

$sqrt91=sqrt9^2+10approxdfrac590686192approx 9.539405quad[9.539392]$

I think the advantage of this method with $a,b$ is that the formula in yellow is not too difficult to remember for your examination and it gives you a fair result already.

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answered Jul 29 at 17:12

zwim

11k627

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up vote
1
down vote

If you want $sqrtN=sqrta^2+bquad$ with $a$ being the greatest integer possible such that $bge 0$.

Then by factoring $a$ you get by Taylor expansion: $$x_0=aleft(1+frac ba^2right)^frac 12approx aleft(1+frac b2a^2right)approx a+frac b2a$$

Starting from this seed you can apply the Newton method to $f(x)=x^2-N$.

This gives $$x_n+1=x_n-fracf(x_n)f'(x_n)$$

In particular $requirecancelf(x_0)=(a+frac b2a)^2-(a^2+b)=cancela^2+cancel2afrac b2a+(frac b2a)^2-cancela^2-cancelb$

Thus a good approximation after just $1$ iteration is given by

$$x_1=left(a+frac b2aright)-fracleft(frac b2aright)^22left(a+frac b2aright)$$

Once simplified it can be written: $$dfrac8a^4+8ba^2+b^24a(2a^2+b)$$

This generally is enough to have a few significant digits right:

$sqrt5=sqrt2^2+1approxdfrac16172approx 2.236111quad[2.236067]$

$sqrt91=sqrt9^2+10approxdfrac590686192approx 9.539405quad[9.539392]$

I think the advantage of this method with $a,b$ is that the formula in yellow is not too difficult to remember for your examination and it gives you a fair result already.

share|cite|improve this answer

answered Jul 29 at 17:12

zwim

11k627

add a comment | 

up vote
1
down vote

up vote
1
down vote

If you want $sqrtN=sqrta^2+bquad$ with $a$ being the greatest integer possible such that $bge 0$.

Then by factoring $a$ you get by Taylor expansion: $$x_0=aleft(1+frac ba^2right)^frac 12approx aleft(1+frac b2a^2right)approx a+frac b2a$$

Starting from this seed you can apply the Newton method to $f(x)=x^2-N$.

This gives $$x_n+1=x_n-fracf(x_n)f'(x_n)$$

In particular $requirecancelf(x_0)=(a+frac b2a)^2-(a^2+b)=cancela^2+cancel2afrac b2a+(frac b2a)^2-cancela^2-cancelb$

Thus a good approximation after just $1$ iteration is given by

$$x_1=left(a+frac b2aright)-fracleft(frac b2aright)^22left(a+frac b2aright)$$

Once simplified it can be written: $$dfrac8a^4+8ba^2+b^24a(2a^2+b)$$

This generally is enough to have a few significant digits right:

$sqrt5=sqrt2^2+1approxdfrac16172approx 2.236111quad[2.236067]$

$sqrt91=sqrt9^2+10approxdfrac590686192approx 9.539405quad[9.539392]$

I think the advantage of this method with $a,b$ is that the formula in yellow is not too difficult to remember for your examination and it gives you a fair result already.

share|cite|improve this answer

answered Jul 29 at 17:12

zwim

11k627

If you want $sqrtN=sqrta^2+bquad$ with $a$ being the greatest integer possible such that $bge 0$.

Then by factoring $a$ you get by Taylor expansion: $$x_0=aleft(1+frac ba^2right)^frac 12approx aleft(1+frac b2a^2right)approx a+frac b2a$$

Starting from this seed you can apply the Newton method to $f(x)=x^2-N$.

This gives $$x_n+1=x_n-fracf(x_n)f'(x_n)$$

In particular $requirecancelf(x_0)=(a+frac b2a)^2-(a^2+b)=cancela^2+cancel2afrac b2a+(frac b2a)^2-cancela^2-cancelb$

Thus a good approximation after just $1$ iteration is given by

$$x_1=left(a+frac b2aright)-fracleft(frac b2aright)^22left(a+frac b2aright)$$

Once simplified it can be written: $$dfrac8a^4+8ba^2+b^24a(2a^2+b)$$

This generally is enough to have a few significant digits right:

$sqrt5=sqrt2^2+1approxdfrac16172approx 2.236111quad[2.236067]$

$sqrt91=sqrt9^2+10approxdfrac590686192approx 9.539405quad[9.539392]$

I think the advantage of this method with $a,b$ is that the formula in yellow is not too difficult to remember for your examination and it gives you a fair result already.

share|cite|improve this answer

answered Jul 29 at 17:12

zwim

11k627

share|cite|improve this answer

share|cite|improve this answer

answered Jul 29 at 17:12

zwim

11k627

answered Jul 29 at 17:12

zwim

11k627

answered Jul 29 at 17:12

zwim

11k627

11k627

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Method 1: The Generalized Binomial Theorem (Newton): If $|x|<1$ then $(1+x)^r=1+A_1x+A_2x^2+A_3x^3+...$ where $A_1=r$ and $A_n+1 =A_ncdot frac (r-n)n.$ For example $sqrt 61=8(1+r)^1/2$ with $r=-3/64.$

Method 2: About 19 centuries ago Heron (Hero) of Alexandria wrote of letting $x_n+1=x_n-frac x_n^2-A2x_n,$ when $A>0$ and $x_1>0,$ to make a sequence $(x_1,x_2,x_3,...)$ converging to $sqrt A.$ This is a special instance of Newton's Method, as already covered in another answer. A "tweak" to this is to, instead, let $x_n+1=x_n-frac 2x(x_n^2-A)3x_n^2+A$, in which the number of decimal places of accuracy approximately triples at each iteration.

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answered Jul 30 at 2:53

DanielWainfleet

31.4k31542

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up vote
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Method 1: The Generalized Binomial Theorem (Newton): If $|x|<1$ then $(1+x)^r=1+A_1x+A_2x^2+A_3x^3+...$ where $A_1=r$ and $A_n+1 =A_ncdot frac (r-n)n.$ For example $sqrt 61=8(1+r)^1/2$ with $r=-3/64.$

Method 2: About 19 centuries ago Heron (Hero) of Alexandria wrote of letting $x_n+1=x_n-frac x_n^2-A2x_n,$ when $A>0$ and $x_1>0,$ to make a sequence $(x_1,x_2,x_3,...)$ converging to $sqrt A.$ This is a special instance of Newton's Method, as already covered in another answer. A "tweak" to this is to, instead, let $x_n+1=x_n-frac 2x(x_n^2-A)3x_n^2+A$, in which the number of decimal places of accuracy approximately triples at each iteration.

share|cite|improve this answer

answered Jul 30 at 2:53

DanielWainfleet

31.4k31542

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up vote
0
down vote

up vote
0
down vote

Method 1: The Generalized Binomial Theorem (Newton): If $|x|<1$ then $(1+x)^r=1+A_1x+A_2x^2+A_3x^3+...$ where $A_1=r$ and $A_n+1 =A_ncdot frac (r-n)n.$ For example $sqrt 61=8(1+r)^1/2$ with $r=-3/64.$

Method 2: About 19 centuries ago Heron (Hero) of Alexandria wrote of letting $x_n+1=x_n-frac x_n^2-A2x_n,$ when $A>0$ and $x_1>0,$ to make a sequence $(x_1,x_2,x_3,...)$ converging to $sqrt A.$ This is a special instance of Newton's Method, as already covered in another answer. A "tweak" to this is to, instead, let $x_n+1=x_n-frac 2x(x_n^2-A)3x_n^2+A$, in which the number of decimal places of accuracy approximately triples at each iteration.

share|cite|improve this answer

answered Jul 30 at 2:53

DanielWainfleet

31.4k31542

Method 1: The Generalized Binomial Theorem (Newton): If $|x|<1$ then $(1+x)^r=1+A_1x+A_2x^2+A_3x^3+...$ where $A_1=r$ and $A_n+1 =A_ncdot frac (r-n)n.$ For example $sqrt 61=8(1+r)^1/2$ with $r=-3/64.$

Method 2: About 19 centuries ago Heron (Hero) of Alexandria wrote of letting $x_n+1=x_n-frac x_n^2-A2x_n,$ when $A>0$ and $x_1>0,$ to make a sequence $(x_1,x_2,x_3,...)$ converging to $sqrt A.$ This is a special instance of Newton's Method, as already covered in another answer. A "tweak" to this is to, instead, let $x_n+1=x_n-frac 2x(x_n^2-A)3x_n^2+A$, in which the number of decimal places of accuracy approximately triples at each iteration.

share|cite|improve this answer

answered Jul 30 at 2:53

DanielWainfleet

31.4k31542

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 2:53

DanielWainfleet

31.4k31542

answered Jul 30 at 2:53

DanielWainfleet

31.4k31542

answered Jul 30 at 2:53

DanielWainfleet

31.4k31542

31.4k31542

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add a comment |