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Induction with unknown $n_0$

0

$begingroup$

Proving by induction that $2^ngt n^3 ; ; ; forall ngeq 10$ isn't very difficult, but how would one prove by induction that $2^ngt n^3 ; ; ; forall ngeq n_0$ ? Meaning that one would have to find a suitable $n_0$ for which the base case of the inequality holds, without graphing tools.

The induction step $(nto n+1)$ would be:
$$2^n+1; =; 2cdot2^n$$$$;;;;;;;;;;;;;;;;;;;gt 2 cdot n^3=n^3+n^3 ;;;text(see basis step)$$
$$gt n^3+3n^2+3n+1 $$
$$=(n+1)^3$$

However I am unsure how to formulate the basis step with an unknown $n_0$.

real-analysis

share|cite|improve this question

edited Mar 27 at 18:59

Robert Israel

330k23218473

asked Mar 27 at 18:52

user639631user639631

527

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You don't need to find the smallest possible $n_0$. Just try a few numbers and you'll soon lead yourself to one that's big enough.
  $endgroup$
  – Ethan Bolker
  Mar 27 at 18:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @EthanBolker Sure, but I was hoping for a method that is less trial and error. I already know that the smallest possible $n_0$ is 10
  $endgroup$
  – user639631
  Mar 27 at 18:56
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'm not sure I understand what you're asking. We generally use inductive proofs because there's an easy way to find a base case, and in this case trial and error suffices. If there was an easy general way to find an $n_0$ which serves as a base case this method would serve as a proof and we wouldn't need induction.
  $endgroup$
  – K.Power
  Mar 27 at 19:00
  

  
  
  
  

add a comment | 

0

$begingroup$

Proving by induction that $2^ngt n^3 ; ; ; forall ngeq 10$ isn't very difficult, but how would one prove by induction that $2^ngt n^3 ; ; ; forall ngeq n_0$ ? Meaning that one would have to find a suitable $n_0$ for which the base case of the inequality holds, without graphing tools.

The induction step $(nto n+1)$ would be:
$$2^n+1; =; 2cdot2^n$$$$;;;;;;;;;;;;;;;;;;;gt 2 cdot n^3=n^3+n^3 ;;;text(see basis step)$$
$$gt n^3+3n^2+3n+1 $$
$$=(n+1)^3$$

However I am unsure how to formulate the basis step with an unknown $n_0$.

real-analysis

share|cite|improve this question

edited Mar 27 at 18:59

Robert Israel

330k23218473

asked Mar 27 at 18:52

user639631user639631

527

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You don't need to find the smallest possible $n_0$. Just try a few numbers and you'll soon lead yourself to one that's big enough.
  $endgroup$
  – Ethan Bolker
  Mar 27 at 18:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @EthanBolker Sure, but I was hoping for a method that is less trial and error. I already know that the smallest possible $n_0$ is 10
  $endgroup$
  – user639631
  Mar 27 at 18:56
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'm not sure I understand what you're asking. We generally use inductive proofs because there's an easy way to find a base case, and in this case trial and error suffices. If there was an easy general way to find an $n_0$ which serves as a base case this method would serve as a proof and we wouldn't need induction.
  $endgroup$
  – K.Power
  Mar 27 at 19:00
  

  
  
  
  

add a comment | 

$begingroup$

Proving by induction that $2^ngt n^3 ; ; ; forall ngeq 10$ isn't very difficult, but how would one prove by induction that $2^ngt n^3 ; ; ; forall ngeq n_0$ ? Meaning that one would have to find a suitable $n_0$ for which the base case of the inequality holds, without graphing tools.

The induction step $(nto n+1)$ would be:
$$2^n+1; =; 2cdot2^n$$$$;;;;;;;;;;;;;;;;;;;gt 2 cdot n^3=n^3+n^3 ;;;text(see basis step)$$
$$gt n^3+3n^2+3n+1 $$
$$=(n+1)^3$$

However I am unsure how to formulate the basis step with an unknown $n_0$.

real-analysis

share|cite|improve this question

edited Mar 27 at 18:59

Robert Israel

330k23218473

asked Mar 27 at 18:52

user639631user639631

527

$endgroup$

share|cite|improve this question

edited Mar 27 at 18:59

Robert Israel

330k23218473

asked Mar 27 at 18:52

user639631user639631

527

share|cite|improve this question

edited Mar 27 at 18:59

Robert Israel

330k23218473

asked Mar 27 at 18:52

user639631user639631

527

edited Mar 27 at 18:59

Robert Israel

330k23218473

edited Mar 27 at 18:59

Robert Israel

330k23218473

asked Mar 27 at 18:52

user639631user639631

527

asked Mar 27 at 18:52

user639631user639631

527

2 Answers
2

active

oldest

votes

1

$begingroup$

Finding the exact value of $n_0inBbb R$ such that $2^n_0=n_0^3$ is not analytically possible but since $n_0inBbb N$, we can find its exact value by hand. Note that$$2^9=512<729=9^3\2^10=1024>1000=10^3$$therefore $n_0$ must be greater than or equal to $10$ for our base of indunction to hold. Also note that if $$f(x)=2^x\g(x)=x^3$$then $f'(x)>g'(x)$for all $xge 10$, leading to this result that $f(x)$ grows super faster than $g(x)$ for $xge 10$.

share|cite|improve this answer

answered Mar 27 at 18:59

Mostafa AyazMostafa Ayaz

18.2k31040

$endgroup$

add a comment | 

1

$begingroup$

You could use the binomial theorem: for $n ge 4$

$$ 2^n = (1+1)^n ge n choose 4 = fracn(n-1)(n-2)(n-3)24$$
We want this $> n^3$, but we don't want too much work. So let's see:
if $n ge 6$ we have $n-3 ge n/2$, so
$$ n(n-1)(n-2)(n-3) ge n (n/2)^3 = n^4/8$$
so now we want
$$ fracn^48 cdot 24 > n^3$$
which is true if $n > 8 cdot 24 = 192 $.

share|cite|improve this answer

answered Mar 27 at 19:16

Robert IsraelRobert Israel

330k23218473

$endgroup$

add a comment | 

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1

$begingroup$

Finding the exact value of $n_0inBbb R$ such that $2^n_0=n_0^3$ is not analytically possible but since $n_0inBbb N$, we can find its exact value by hand. Note that$$2^9=512<729=9^3\2^10=1024>1000=10^3$$therefore $n_0$ must be greater than or equal to $10$ for our base of indunction to hold. Also note that if $$f(x)=2^x\g(x)=x^3$$then $f'(x)>g'(x)$for all $xge 10$, leading to this result that $f(x)$ grows super faster than $g(x)$ for $xge 10$.

share|cite|improve this answer

answered Mar 27 at 18:59

Mostafa AyazMostafa Ayaz

18.2k31040

$endgroup$

add a comment | 

1

$begingroup$

Finding the exact value of $n_0inBbb R$ such that $2^n_0=n_0^3$ is not analytically possible but since $n_0inBbb N$, we can find its exact value by hand. Note that$$2^9=512<729=9^3\2^10=1024>1000=10^3$$therefore $n_0$ must be greater than or equal to $10$ for our base of indunction to hold. Also note that if $$f(x)=2^x\g(x)=x^3$$then $f'(x)>g'(x)$for all $xge 10$, leading to this result that $f(x)$ grows super faster than $g(x)$ for $xge 10$.

share|cite|improve this answer

answered Mar 27 at 18:59

Mostafa AyazMostafa Ayaz

18.2k31040

$endgroup$

add a comment | 

$begingroup$

Finding the exact value of $n_0inBbb R$ such that $2^n_0=n_0^3$ is not analytically possible but since $n_0inBbb N$, we can find its exact value by hand. Note that$$2^9=512<729=9^3\2^10=1024>1000=10^3$$therefore $n_0$ must be greater than or equal to $10$ for our base of indunction to hold. Also note that if $$f(x)=2^x\g(x)=x^3$$then $f'(x)>g'(x)$for all $xge 10$, leading to this result that $f(x)$ grows super faster than $g(x)$ for $xge 10$.

share|cite|improve this answer

answered Mar 27 at 18:59

Mostafa AyazMostafa Ayaz

18.2k31040

$endgroup$

share|cite|improve this answer

answered Mar 27 at 18:59

Mostafa AyazMostafa Ayaz

18.2k31040

answered Mar 27 at 18:59

Mostafa AyazMostafa Ayaz

18.2k31040

answered Mar 27 at 18:59

Mostafa AyazMostafa Ayaz

18.2k31040

1

$begingroup$

You could use the binomial theorem: for $n ge 4$

$$ 2^n = (1+1)^n ge n choose 4 = fracn(n-1)(n-2)(n-3)24$$
We want this $> n^3$, but we don't want too much work. So let's see:
if $n ge 6$ we have $n-3 ge n/2$, so
$$ n(n-1)(n-2)(n-3) ge n (n/2)^3 = n^4/8$$
so now we want
$$ fracn^48 cdot 24 > n^3$$
which is true if $n > 8 cdot 24 = 192 $.

share|cite|improve this answer

answered Mar 27 at 19:16

Robert IsraelRobert Israel

330k23218473

$endgroup$

add a comment | 

1

$begingroup$

You could use the binomial theorem: for $n ge 4$

$$ 2^n = (1+1)^n ge n choose 4 = fracn(n-1)(n-2)(n-3)24$$
We want this $> n^3$, but we don't want too much work. So let's see:
if $n ge 6$ we have $n-3 ge n/2$, so
$$ n(n-1)(n-2)(n-3) ge n (n/2)^3 = n^4/8$$
so now we want
$$ fracn^48 cdot 24 > n^3$$
which is true if $n > 8 cdot 24 = 192 $.

share|cite|improve this answer

answered Mar 27 at 19:16

Robert IsraelRobert Israel

330k23218473

$endgroup$

add a comment | 

$begingroup$

You could use the binomial theorem: for $n ge 4$

$$ 2^n = (1+1)^n ge n choose 4 = fracn(n-1)(n-2)(n-3)24$$
We want this $> n^3$, but we don't want too much work. So let's see:
if $n ge 6$ we have $n-3 ge n/2$, so
$$ n(n-1)(n-2)(n-3) ge n (n/2)^3 = n^4/8$$
so now we want
$$ fracn^48 cdot 24 > n^3$$
which is true if $n > 8 cdot 24 = 192 $.

share|cite|improve this answer

answered Mar 27 at 19:16

Robert IsraelRobert Israel

330k23218473

$endgroup$

share|cite|improve this answer

answered Mar 27 at 19:16

Robert IsraelRobert Israel

330k23218473

answered Mar 27 at 19:16

Robert IsraelRobert Israel

330k23218473

answered Mar 27 at 19:16

Robert IsraelRobert Israel

330k23218473