Why binomial expansion approximation works?

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So I have got the expansion of $$ (4-5x)^.5 = 2 + (5/4)x + (25/64)x^2 $$
I am told to use $ x = 1/10 $ to find an approximation of $ sqrt2 $. I can do this, giving $ 181/128 $, however the last part asks: "Explain why substituting $ x = 1/10 $ into this binomial expansion leads to a valid approximation. "

The answer is said to be because $ |x| < 4/5 $. Why?

binomial-theorem

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asked Jul 18 at 10:16

Rudra Mutalik

156

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Do you know about radius of convergence ? This is the key.
  – Claude Leibovici
  Jul 18 at 10:30
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

So I have got the expansion of $$ (4-5x)^.5 = 2 + (5/4)x + (25/64)x^2 $$
I am told to use $ x = 1/10 $ to find an approximation of $ sqrt2 $. I can do this, giving $ 181/128 $, however the last part asks: "Explain why substituting $ x = 1/10 $ into this binomial expansion leads to a valid approximation. "

The answer is said to be because $ |x| < 4/5 $. Why?

binomial-theorem

share|cite|improve this question

asked Jul 18 at 10:16

Rudra Mutalik

156

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Do you know about radius of convergence ? This is the key.
  – Claude Leibovici
  Jul 18 at 10:30
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

So I have got the expansion of $$ (4-5x)^.5 = 2 + (5/4)x + (25/64)x^2 $$
I am told to use $ x = 1/10 $ to find an approximation of $ sqrt2 $. I can do this, giving $ 181/128 $, however the last part asks: "Explain why substituting $ x = 1/10 $ into this binomial expansion leads to a valid approximation. "

The answer is said to be because $ |x| < 4/5 $. Why?

binomial-theorem

share|cite|improve this question

asked Jul 18 at 10:16

Rudra Mutalik

156

So I have got the expansion of $$ (4-5x)^.5 = 2 + (5/4)x + (25/64)x^2 $$
I am told to use $ x = 1/10 $ to find an approximation of $ sqrt2 $. I can do this, giving $ 181/128 $, however the last part asks: "Explain why substituting $ x = 1/10 $ into this binomial expansion leads to a valid approximation. "

The answer is said to be because $ |x| < 4/5 $. Why?

binomial-theorem

share|cite|improve this question

asked Jul 18 at 10:16

Rudra Mutalik

156

share|cite|improve this question

share|cite|improve this question

asked Jul 18 at 10:16

Rudra Mutalik

156

asked Jul 18 at 10:16

Rudra Mutalik

156

asked Jul 18 at 10:16

Rudra Mutalik

156

156

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Do you know about radius of convergence ? This is the key.
  – Claude Leibovici
  Jul 18 at 10:30
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Do you know about radius of convergence ? This is the key.
  – Claude Leibovici
  Jul 18 at 10:30
  

  
  
  
  

Do you know about radius of convergence ? This is the key.
– Claude Leibovici
Jul 18 at 10:30

Do you know about radius of convergence ? This is the key.
– Claude Leibovici
Jul 18 at 10:30

add a comment | 

2 Answers
2

active

oldest

votes

up vote
0
down vote



accepted

The actual explanation has something to do with the convergence of Taylor (or Maclaurin) series but it involves quite a high-level calculus.

A not-so-satisfying answer is that you have to put the LHS into the form $k(1+ax)$ and the approximation is valid for $lvert ax rvert <1$

share|cite|improve this answer

answered Jul 18 at 10:24

Karn Watcharasupat

3,8192426

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am only at A level so i guess the not-so-satisfying answer will have to do for now, thanks
  – Rudra Mutalik
  Jul 18 at 10:57
  

  
  
  
  

add a comment | 

up vote
0
down vote

$$(4-5x)^.5 = 2 + (5/4)x + (25/64)x^2+....$$

You are using a few terms of an infinite series to approximate $sqrt 2$

$$(4-5x)^.5 =2(1-5/4 x)^.5$$

The Newton binomial series $$(1+x)^alpha$$ converges where $|x|<1$ which in this case translates in $|5/4 x|<1$ which is satisfied with $x=1/10.$

share|cite|improve this answer

answered Jul 18 at 11:04

Mohammad Riazi-Kermani

27.5k41852

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

up vote
0
down vote



accepted

The actual explanation has something to do with the convergence of Taylor (or Maclaurin) series but it involves quite a high-level calculus.

A not-so-satisfying answer is that you have to put the LHS into the form $k(1+ax)$ and the approximation is valid for $lvert ax rvert <1$

share|cite|improve this answer

answered Jul 18 at 10:24

Karn Watcharasupat

3,8192426

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am only at A level so i guess the not-so-satisfying answer will have to do for now, thanks
  – Rudra Mutalik
  Jul 18 at 10:57
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

The actual explanation has something to do with the convergence of Taylor (or Maclaurin) series but it involves quite a high-level calculus.

A not-so-satisfying answer is that you have to put the LHS into the form $k(1+ax)$ and the approximation is valid for $lvert ax rvert <1$

share|cite|improve this answer

answered Jul 18 at 10:24

Karn Watcharasupat

3,8192426

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am only at A level so i guess the not-so-satisfying answer will have to do for now, thanks
  – Rudra Mutalik
  Jul 18 at 10:57
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

The actual explanation has something to do with the convergence of Taylor (or Maclaurin) series but it involves quite a high-level calculus.

A not-so-satisfying answer is that you have to put the LHS into the form $k(1+ax)$ and the approximation is valid for $lvert ax rvert <1$

share|cite|improve this answer

answered Jul 18 at 10:24

Karn Watcharasupat

3,8192426

The actual explanation has something to do with the convergence of Taylor (or Maclaurin) series but it involves quite a high-level calculus.

A not-so-satisfying answer is that you have to put the LHS into the form $k(1+ax)$ and the approximation is valid for $lvert ax rvert <1$

share|cite|improve this answer

answered Jul 18 at 10:24

Karn Watcharasupat

3,8192426

share|cite|improve this answer

share|cite|improve this answer

answered Jul 18 at 10:24

Karn Watcharasupat

3,8192426

answered Jul 18 at 10:24

Karn Watcharasupat

3,8192426

answered Jul 18 at 10:24

Karn Watcharasupat

3,8192426

3,8192426

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am only at A level so i guess the not-so-satisfying answer will have to do for now, thanks
  – Rudra Mutalik
  Jul 18 at 10:57
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am only at A level so i guess the not-so-satisfying answer will have to do for now, thanks
  – Rudra Mutalik
  Jul 18 at 10:57
  

  
  
  
  

I am only at A level so i guess the not-so-satisfying answer will have to do for now, thanks
– Rudra Mutalik
Jul 18 at 10:57

I am only at A level so i guess the not-so-satisfying answer will have to do for now, thanks
– Rudra Mutalik
Jul 18 at 10:57

add a comment | 

up vote
0
down vote

$$(4-5x)^.5 = 2 + (5/4)x + (25/64)x^2+....$$

You are using a few terms of an infinite series to approximate $sqrt 2$

$$(4-5x)^.5 =2(1-5/4 x)^.5$$

The Newton binomial series $$(1+x)^alpha$$ converges where $|x|<1$ which in this case translates in $|5/4 x|<1$ which is satisfied with $x=1/10.$

share|cite|improve this answer

answered Jul 18 at 11:04

Mohammad Riazi-Kermani

27.5k41852

add a comment | 

up vote
0
down vote

$$(4-5x)^.5 = 2 + (5/4)x + (25/64)x^2+....$$

You are using a few terms of an infinite series to approximate $sqrt 2$

$$(4-5x)^.5 =2(1-5/4 x)^.5$$

The Newton binomial series $$(1+x)^alpha$$ converges where $|x|<1$ which in this case translates in $|5/4 x|<1$ which is satisfied with $x=1/10.$

share|cite|improve this answer

answered Jul 18 at 11:04

Mohammad Riazi-Kermani

27.5k41852

add a comment | 

up vote
0
down vote

up vote
0
down vote

$$(4-5x)^.5 = 2 + (5/4)x + (25/64)x^2+....$$

You are using a few terms of an infinite series to approximate $sqrt 2$

$$(4-5x)^.5 =2(1-5/4 x)^.5$$

The Newton binomial series $$(1+x)^alpha$$ converges where $|x|<1$ which in this case translates in $|5/4 x|<1$ which is satisfied with $x=1/10.$

share|cite|improve this answer

answered Jul 18 at 11:04

Mohammad Riazi-Kermani

27.5k41852

$$(4-5x)^.5 = 2 + (5/4)x + (25/64)x^2+....$$

You are using a few terms of an infinite series to approximate $sqrt 2$

$$(4-5x)^.5 =2(1-5/4 x)^.5$$

The Newton binomial series $$(1+x)^alpha$$ converges where $|x|<1$ which in this case translates in $|5/4 x|<1$ which is satisfied with $x=1/10.$

share|cite|improve this answer

answered Jul 18 at 11:04

Mohammad Riazi-Kermani

27.5k41852

share|cite|improve this answer

share|cite|improve this answer

answered Jul 18 at 11:04

Mohammad Riazi-Kermani

27.5k41852

answered Jul 18 at 11:04

Mohammad Riazi-Kermani

27.5k41852

answered Jul 18 at 11:04

Mohammad Riazi-Kermani

27.5k41852

27.5k41852

add a comment | 

add a comment | 

 

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