Prove that $(1-x)^n leq 1 -xn+frac{n(n-1)}{2}x^2$ when $0leq xleq 1$ and $ngeq2$












2












$begingroup$


Reading a book I saw this inequality
$$
(1-x)^n leq 1 -xn+frac{n(n-1)}{2}x^2
$$

when $0leq xleq 1$ and following the author it descended from the inclusion-exclusion principle. I don't understand why. Moreover is there a simple proof of this inequality?

Thanks







I've tried to prove the inequality in the following way:



From the binomial theorem:
$$
(1-x)^n = sum_{k=0}^{n}{{n}choose{k}}(-1)^{k}x^k=1 -xn+frac{n(n-1)}{2}x^2+sum_{k=3}^{n}{{n}choose{k}}(-1)^{k}x^k
$$

and so we need to prove that:
$$
sum_{k=3}^{n}{{n}choose{k}}(-1)^{k}x^kleq0
$$

when $0leq xleq1$.

I've tried to use the inequality
$$
sum_{k=3}^{n}{{n}choose{k}}(-1)^{k}leq0
$$

that I can prove from $0=(1-1)^n$ and then I've tried to group some terms in the sum but without success.










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$endgroup$








  • 1




    $begingroup$
    Perhaps $nleq 2$ should be $n geq 2$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 5 '18 at 10:11










  • $begingroup$
    Yes, you are right! I fixed it, thanks
    $endgroup$
    – Alex
    Dec 5 '18 at 10:13
















2












$begingroup$


Reading a book I saw this inequality
$$
(1-x)^n leq 1 -xn+frac{n(n-1)}{2}x^2
$$

when $0leq xleq 1$ and following the author it descended from the inclusion-exclusion principle. I don't understand why. Moreover is there a simple proof of this inequality?

Thanks







I've tried to prove the inequality in the following way:



From the binomial theorem:
$$
(1-x)^n = sum_{k=0}^{n}{{n}choose{k}}(-1)^{k}x^k=1 -xn+frac{n(n-1)}{2}x^2+sum_{k=3}^{n}{{n}choose{k}}(-1)^{k}x^k
$$

and so we need to prove that:
$$
sum_{k=3}^{n}{{n}choose{k}}(-1)^{k}x^kleq0
$$

when $0leq xleq1$.

I've tried to use the inequality
$$
sum_{k=3}^{n}{{n}choose{k}}(-1)^{k}leq0
$$

that I can prove from $0=(1-1)^n$ and then I've tried to group some terms in the sum but without success.










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Perhaps $nleq 2$ should be $n geq 2$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 5 '18 at 10:11










  • $begingroup$
    Yes, you are right! I fixed it, thanks
    $endgroup$
    – Alex
    Dec 5 '18 at 10:13














2












2








2





$begingroup$


Reading a book I saw this inequality
$$
(1-x)^n leq 1 -xn+frac{n(n-1)}{2}x^2
$$

when $0leq xleq 1$ and following the author it descended from the inclusion-exclusion principle. I don't understand why. Moreover is there a simple proof of this inequality?

Thanks







I've tried to prove the inequality in the following way:



From the binomial theorem:
$$
(1-x)^n = sum_{k=0}^{n}{{n}choose{k}}(-1)^{k}x^k=1 -xn+frac{n(n-1)}{2}x^2+sum_{k=3}^{n}{{n}choose{k}}(-1)^{k}x^k
$$

and so we need to prove that:
$$
sum_{k=3}^{n}{{n}choose{k}}(-1)^{k}x^kleq0
$$

when $0leq xleq1$.

I've tried to use the inequality
$$
sum_{k=3}^{n}{{n}choose{k}}(-1)^{k}leq0
$$

that I can prove from $0=(1-1)^n$ and then I've tried to group some terms in the sum but without success.










share|cite|improve this question











$endgroup$




Reading a book I saw this inequality
$$
(1-x)^n leq 1 -xn+frac{n(n-1)}{2}x^2
$$

when $0leq xleq 1$ and following the author it descended from the inclusion-exclusion principle. I don't understand why. Moreover is there a simple proof of this inequality?

Thanks







I've tried to prove the inequality in the following way:



From the binomial theorem:
$$
(1-x)^n = sum_{k=0}^{n}{{n}choose{k}}(-1)^{k}x^k=1 -xn+frac{n(n-1)}{2}x^2+sum_{k=3}^{n}{{n}choose{k}}(-1)^{k}x^k
$$

and so we need to prove that:
$$
sum_{k=3}^{n}{{n}choose{k}}(-1)^{k}x^kleq0
$$

when $0leq xleq1$.

I've tried to use the inequality
$$
sum_{k=3}^{n}{{n}choose{k}}(-1)^{k}leq0
$$

that I can prove from $0=(1-1)^n$ and then I've tried to group some terms in the sum but without success.







inequality binomial-coefficients inclusion-exclusion






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edited Dec 5 '18 at 10:12







Alex

















asked Dec 5 '18 at 10:10









AlexAlex

32319




32319








  • 1




    $begingroup$
    Perhaps $nleq 2$ should be $n geq 2$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 5 '18 at 10:11










  • $begingroup$
    Yes, you are right! I fixed it, thanks
    $endgroup$
    – Alex
    Dec 5 '18 at 10:13














  • 1




    $begingroup$
    Perhaps $nleq 2$ should be $n geq 2$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 5 '18 at 10:11










  • $begingroup$
    Yes, you are right! I fixed it, thanks
    $endgroup$
    – Alex
    Dec 5 '18 at 10:13








1




1




$begingroup$
Perhaps $nleq 2$ should be $n geq 2$.
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 10:11




$begingroup$
Perhaps $nleq 2$ should be $n geq 2$.
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 10:11












$begingroup$
Yes, you are right! I fixed it, thanks
$endgroup$
– Alex
Dec 5 '18 at 10:13




$begingroup$
Yes, you are right! I fixed it, thanks
$endgroup$
– Alex
Dec 5 '18 at 10:13










4 Answers
4






active

oldest

votes


















2












$begingroup$

If true for some $n$, then
begin{align}
(1-x)^{n+1}&le(1-x)left(1-nx+frac{n(n-1)}2x^2right)\
&=1-(n+1)x+frac{(n+1)n}2x^2-frac{n(n-1)}2x^3\
&le1-(n+1)x+frac{(n+1)n}2x^2.
end{align}

and now use induction.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! Any connection with the inclusion-exclusion principle?
    $endgroup$
    – Alex
    Dec 5 '18 at 10:44






  • 2




    $begingroup$
    @Alex en.wikipedia.org/wiki/…
    $endgroup$
    – Lord Shark the Unknown
    Dec 5 '18 at 11:22



















2












$begingroup$

Interpretation in terms of inclusion-exclusion principle is not too difficult either, though it could become verbose. Consider a coin with probability $x in (0, 1)$ of turning Heads. Suppose you want to find the probability of getting only Tails when tossing $n$ such coins. The probability is obviously $(1-x)^n$ by independence of the events.



OTOH, we can look at the probability of all events, i.e. $1$, and reduce from it the probability of all events where there is a Head in each coin individually. This leads to $1-nx$, as there are $n$ coins. However the second term clearly double counts cases where two Heads simultaneously appears, so we add back these, i.e. $binom{n}{2}x^2$, to get $1-nx+frac12n(n-1)x^2$. We notice that the last term has double counted all cases where three coins had Heads simultaneously, so this is an overestimate, so $(1-x)^n leqslant 1-nx+frac12n(n-1)x^2$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    For $n=2$ the proof is trivial, we consider $n>2$.



    Let $f(x) = (1-x)^n$ and $g(x)=1-nx+frac{n(n-1)}{2}x^2$. You have clearly
    $$f(0) = g(0),quad f'(0) = g'(0),quad f''(0)=g''(0)$$



    However, the third derivative writes
    $$ f'''(x) = -n(n-1)(n-2) (1-x)^{n-3} < 0 = g'''(x).$$
    Given that $f''(0)=g''(0)$ it is easy to see that $forall xin[0,1], f'(x)le g'(x)$. Now use $f(0)=g(0)$ to obtain that $forall xin[0,1], f(x)le g(x)$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you very much
      $endgroup$
      – Alex
      Dec 5 '18 at 10:43



















    1












    $begingroup$

    You can apply Taylor around 0. Let $f = (1-x)^n$, then $f'(0) = -n$, $f''(0) = n (n-1)$, and $f'''(xi) = -n(n-1)(n-2)(1 - xi)^{n-3}$.



    $$(1-x)^n = 1 - nx + frac{n(n-1)}{2} x^2 + frac{f'''(xi)}{6} , x^3,$$



    for some $xi in [0, x]$. Since $x leq 1$, $f'''(xi) leq 0$, and so the remainder term is non-positive.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      up to $f''' <0$ for $x in[0,1]$ =)
      $endgroup$
      – TZakrevskiy
      Dec 5 '18 at 10:52











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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    If true for some $n$, then
    begin{align}
    (1-x)^{n+1}&le(1-x)left(1-nx+frac{n(n-1)}2x^2right)\
    &=1-(n+1)x+frac{(n+1)n}2x^2-frac{n(n-1)}2x^3\
    &le1-(n+1)x+frac{(n+1)n}2x^2.
    end{align}

    and now use induction.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks! Any connection with the inclusion-exclusion principle?
      $endgroup$
      – Alex
      Dec 5 '18 at 10:44






    • 2




      $begingroup$
      @Alex en.wikipedia.org/wiki/…
      $endgroup$
      – Lord Shark the Unknown
      Dec 5 '18 at 11:22
















    2












    $begingroup$

    If true for some $n$, then
    begin{align}
    (1-x)^{n+1}&le(1-x)left(1-nx+frac{n(n-1)}2x^2right)\
    &=1-(n+1)x+frac{(n+1)n}2x^2-frac{n(n-1)}2x^3\
    &le1-(n+1)x+frac{(n+1)n}2x^2.
    end{align}

    and now use induction.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks! Any connection with the inclusion-exclusion principle?
      $endgroup$
      – Alex
      Dec 5 '18 at 10:44






    • 2




      $begingroup$
      @Alex en.wikipedia.org/wiki/…
      $endgroup$
      – Lord Shark the Unknown
      Dec 5 '18 at 11:22














    2












    2








    2





    $begingroup$

    If true for some $n$, then
    begin{align}
    (1-x)^{n+1}&le(1-x)left(1-nx+frac{n(n-1)}2x^2right)\
    &=1-(n+1)x+frac{(n+1)n}2x^2-frac{n(n-1)}2x^3\
    &le1-(n+1)x+frac{(n+1)n}2x^2.
    end{align}

    and now use induction.






    share|cite|improve this answer









    $endgroup$



    If true for some $n$, then
    begin{align}
    (1-x)^{n+1}&le(1-x)left(1-nx+frac{n(n-1)}2x^2right)\
    &=1-(n+1)x+frac{(n+1)n}2x^2-frac{n(n-1)}2x^3\
    &le1-(n+1)x+frac{(n+1)n}2x^2.
    end{align}

    and now use induction.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 5 '18 at 10:41









    Lord Shark the UnknownLord Shark the Unknown

    103k1160132




    103k1160132












    • $begingroup$
      Thanks! Any connection with the inclusion-exclusion principle?
      $endgroup$
      – Alex
      Dec 5 '18 at 10:44






    • 2




      $begingroup$
      @Alex en.wikipedia.org/wiki/…
      $endgroup$
      – Lord Shark the Unknown
      Dec 5 '18 at 11:22


















    • $begingroup$
      Thanks! Any connection with the inclusion-exclusion principle?
      $endgroup$
      – Alex
      Dec 5 '18 at 10:44






    • 2




      $begingroup$
      @Alex en.wikipedia.org/wiki/…
      $endgroup$
      – Lord Shark the Unknown
      Dec 5 '18 at 11:22
















    $begingroup$
    Thanks! Any connection with the inclusion-exclusion principle?
    $endgroup$
    – Alex
    Dec 5 '18 at 10:44




    $begingroup$
    Thanks! Any connection with the inclusion-exclusion principle?
    $endgroup$
    – Alex
    Dec 5 '18 at 10:44




    2




    2




    $begingroup$
    @Alex en.wikipedia.org/wiki/…
    $endgroup$
    – Lord Shark the Unknown
    Dec 5 '18 at 11:22




    $begingroup$
    @Alex en.wikipedia.org/wiki/…
    $endgroup$
    – Lord Shark the Unknown
    Dec 5 '18 at 11:22











    2












    $begingroup$

    Interpretation in terms of inclusion-exclusion principle is not too difficult either, though it could become verbose. Consider a coin with probability $x in (0, 1)$ of turning Heads. Suppose you want to find the probability of getting only Tails when tossing $n$ such coins. The probability is obviously $(1-x)^n$ by independence of the events.



    OTOH, we can look at the probability of all events, i.e. $1$, and reduce from it the probability of all events where there is a Head in each coin individually. This leads to $1-nx$, as there are $n$ coins. However the second term clearly double counts cases where two Heads simultaneously appears, so we add back these, i.e. $binom{n}{2}x^2$, to get $1-nx+frac12n(n-1)x^2$. We notice that the last term has double counted all cases where three coins had Heads simultaneously, so this is an overestimate, so $(1-x)^n leqslant 1-nx+frac12n(n-1)x^2$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Interpretation in terms of inclusion-exclusion principle is not too difficult either, though it could become verbose. Consider a coin with probability $x in (0, 1)$ of turning Heads. Suppose you want to find the probability of getting only Tails when tossing $n$ such coins. The probability is obviously $(1-x)^n$ by independence of the events.



      OTOH, we can look at the probability of all events, i.e. $1$, and reduce from it the probability of all events where there is a Head in each coin individually. This leads to $1-nx$, as there are $n$ coins. However the second term clearly double counts cases where two Heads simultaneously appears, so we add back these, i.e. $binom{n}{2}x^2$, to get $1-nx+frac12n(n-1)x^2$. We notice that the last term has double counted all cases where three coins had Heads simultaneously, so this is an overestimate, so $(1-x)^n leqslant 1-nx+frac12n(n-1)x^2$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Interpretation in terms of inclusion-exclusion principle is not too difficult either, though it could become verbose. Consider a coin with probability $x in (0, 1)$ of turning Heads. Suppose you want to find the probability of getting only Tails when tossing $n$ such coins. The probability is obviously $(1-x)^n$ by independence of the events.



        OTOH, we can look at the probability of all events, i.e. $1$, and reduce from it the probability of all events where there is a Head in each coin individually. This leads to $1-nx$, as there are $n$ coins. However the second term clearly double counts cases where two Heads simultaneously appears, so we add back these, i.e. $binom{n}{2}x^2$, to get $1-nx+frac12n(n-1)x^2$. We notice that the last term has double counted all cases where three coins had Heads simultaneously, so this is an overestimate, so $(1-x)^n leqslant 1-nx+frac12n(n-1)x^2$






        share|cite|improve this answer









        $endgroup$



        Interpretation in terms of inclusion-exclusion principle is not too difficult either, though it could become verbose. Consider a coin with probability $x in (0, 1)$ of turning Heads. Suppose you want to find the probability of getting only Tails when tossing $n$ such coins. The probability is obviously $(1-x)^n$ by independence of the events.



        OTOH, we can look at the probability of all events, i.e. $1$, and reduce from it the probability of all events where there is a Head in each coin individually. This leads to $1-nx$, as there are $n$ coins. However the second term clearly double counts cases where two Heads simultaneously appears, so we add back these, i.e. $binom{n}{2}x^2$, to get $1-nx+frac12n(n-1)x^2$. We notice that the last term has double counted all cases where three coins had Heads simultaneously, so this is an overestimate, so $(1-x)^n leqslant 1-nx+frac12n(n-1)x^2$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 5 '18 at 13:10









        MacavityMacavity

        35.2k52554




        35.2k52554























            1












            $begingroup$

            For $n=2$ the proof is trivial, we consider $n>2$.



            Let $f(x) = (1-x)^n$ and $g(x)=1-nx+frac{n(n-1)}{2}x^2$. You have clearly
            $$f(0) = g(0),quad f'(0) = g'(0),quad f''(0)=g''(0)$$



            However, the third derivative writes
            $$ f'''(x) = -n(n-1)(n-2) (1-x)^{n-3} < 0 = g'''(x).$$
            Given that $f''(0)=g''(0)$ it is easy to see that $forall xin[0,1], f'(x)le g'(x)$. Now use $f(0)=g(0)$ to obtain that $forall xin[0,1], f(x)le g(x)$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you very much
              $endgroup$
              – Alex
              Dec 5 '18 at 10:43
















            1












            $begingroup$

            For $n=2$ the proof is trivial, we consider $n>2$.



            Let $f(x) = (1-x)^n$ and $g(x)=1-nx+frac{n(n-1)}{2}x^2$. You have clearly
            $$f(0) = g(0),quad f'(0) = g'(0),quad f''(0)=g''(0)$$



            However, the third derivative writes
            $$ f'''(x) = -n(n-1)(n-2) (1-x)^{n-3} < 0 = g'''(x).$$
            Given that $f''(0)=g''(0)$ it is easy to see that $forall xin[0,1], f'(x)le g'(x)$. Now use $f(0)=g(0)$ to obtain that $forall xin[0,1], f(x)le g(x)$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you very much
              $endgroup$
              – Alex
              Dec 5 '18 at 10:43














            1












            1








            1





            $begingroup$

            For $n=2$ the proof is trivial, we consider $n>2$.



            Let $f(x) = (1-x)^n$ and $g(x)=1-nx+frac{n(n-1)}{2}x^2$. You have clearly
            $$f(0) = g(0),quad f'(0) = g'(0),quad f''(0)=g''(0)$$



            However, the third derivative writes
            $$ f'''(x) = -n(n-1)(n-2) (1-x)^{n-3} < 0 = g'''(x).$$
            Given that $f''(0)=g''(0)$ it is easy to see that $forall xin[0,1], f'(x)le g'(x)$. Now use $f(0)=g(0)$ to obtain that $forall xin[0,1], f(x)le g(x)$.






            share|cite|improve this answer









            $endgroup$



            For $n=2$ the proof is trivial, we consider $n>2$.



            Let $f(x) = (1-x)^n$ and $g(x)=1-nx+frac{n(n-1)}{2}x^2$. You have clearly
            $$f(0) = g(0),quad f'(0) = g'(0),quad f''(0)=g''(0)$$



            However, the third derivative writes
            $$ f'''(x) = -n(n-1)(n-2) (1-x)^{n-3} < 0 = g'''(x).$$
            Given that $f''(0)=g''(0)$ it is easy to see that $forall xin[0,1], f'(x)le g'(x)$. Now use $f(0)=g(0)$ to obtain that $forall xin[0,1], f(x)le g(x)$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 5 '18 at 10:28









            TZakrevskiyTZakrevskiy

            20.2k12354




            20.2k12354












            • $begingroup$
              Thank you very much
              $endgroup$
              – Alex
              Dec 5 '18 at 10:43


















            • $begingroup$
              Thank you very much
              $endgroup$
              – Alex
              Dec 5 '18 at 10:43
















            $begingroup$
            Thank you very much
            $endgroup$
            – Alex
            Dec 5 '18 at 10:43




            $begingroup$
            Thank you very much
            $endgroup$
            – Alex
            Dec 5 '18 at 10:43











            1












            $begingroup$

            You can apply Taylor around 0. Let $f = (1-x)^n$, then $f'(0) = -n$, $f''(0) = n (n-1)$, and $f'''(xi) = -n(n-1)(n-2)(1 - xi)^{n-3}$.



            $$(1-x)^n = 1 - nx + frac{n(n-1)}{2} x^2 + frac{f'''(xi)}{6} , x^3,$$



            for some $xi in [0, x]$. Since $x leq 1$, $f'''(xi) leq 0$, and so the remainder term is non-positive.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              up to $f''' <0$ for $x in[0,1]$ =)
              $endgroup$
              – TZakrevskiy
              Dec 5 '18 at 10:52
















            1












            $begingroup$

            You can apply Taylor around 0. Let $f = (1-x)^n$, then $f'(0) = -n$, $f''(0) = n (n-1)$, and $f'''(xi) = -n(n-1)(n-2)(1 - xi)^{n-3}$.



            $$(1-x)^n = 1 - nx + frac{n(n-1)}{2} x^2 + frac{f'''(xi)}{6} , x^3,$$



            for some $xi in [0, x]$. Since $x leq 1$, $f'''(xi) leq 0$, and so the remainder term is non-positive.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              up to $f''' <0$ for $x in[0,1]$ =)
              $endgroup$
              – TZakrevskiy
              Dec 5 '18 at 10:52














            1












            1








            1





            $begingroup$

            You can apply Taylor around 0. Let $f = (1-x)^n$, then $f'(0) = -n$, $f''(0) = n (n-1)$, and $f'''(xi) = -n(n-1)(n-2)(1 - xi)^{n-3}$.



            $$(1-x)^n = 1 - nx + frac{n(n-1)}{2} x^2 + frac{f'''(xi)}{6} , x^3,$$



            for some $xi in [0, x]$. Since $x leq 1$, $f'''(xi) leq 0$, and so the remainder term is non-positive.






            share|cite|improve this answer











            $endgroup$



            You can apply Taylor around 0. Let $f = (1-x)^n$, then $f'(0) = -n$, $f''(0) = n (n-1)$, and $f'''(xi) = -n(n-1)(n-2)(1 - xi)^{n-3}$.



            $$(1-x)^n = 1 - nx + frac{n(n-1)}{2} x^2 + frac{f'''(xi)}{6} , x^3,$$



            for some $xi in [0, x]$. Since $x leq 1$, $f'''(xi) leq 0$, and so the remainder term is non-positive.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 5 '18 at 11:31

























            answered Dec 5 '18 at 10:32









            Roberto RastapopoulosRoberto Rastapopoulos

            899424




            899424








            • 1




              $begingroup$
              up to $f''' <0$ for $x in[0,1]$ =)
              $endgroup$
              – TZakrevskiy
              Dec 5 '18 at 10:52














            • 1




              $begingroup$
              up to $f''' <0$ for $x in[0,1]$ =)
              $endgroup$
              – TZakrevskiy
              Dec 5 '18 at 10:52








            1




            1




            $begingroup$
            up to $f''' <0$ for $x in[0,1]$ =)
            $endgroup$
            – TZakrevskiy
            Dec 5 '18 at 10:52




            $begingroup$
            up to $f''' <0$ for $x in[0,1]$ =)
            $endgroup$
            – TZakrevskiy
            Dec 5 '18 at 10:52


















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