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experimentX

  • 2 years ago

How to show that the the sequence \( \{ a_n \} \) is a monotonically decreasing sequence. \[ a_n = { { \left (\frac { \sqrt [ n ]{ a } +\sqrt [ n ]{ b } +\sqrt [ n ]{ c } +\sqrt [ n ]{ d } }{ 4 } \right ) }^{ n } } \] It might not be monotonically decreasing sequence. I might be wrong in my assumption.

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  1. Dido525
    • 2 years ago
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    Induction is the best way but I have no idea how the heck I prove it's decreasing.

  2. incomplte
    • 2 years ago
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    If an+1 is < an, the the function is decreasing.

  3. experimentX
    • 2 years ago
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    I know the definition ... how to show it? expanding using multinomial theorem is freaking ugly, not sure if induction works so easily.

  4. sparik1997
    • 2 years ago
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    are a,b,c,d positive integers?

  5. experimentX
    • 2 years ago
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    yep!! sorry for non specifying that.

  6. Dido525
    • 2 years ago
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    What about the n's ?

  7. experimentX
    • 2 years ago
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    n is a set of natural numbers.

  8. sparik1997
    • 2 years ago
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    do you agree that if wee show that \[ \sqrt[k]{a}+ \sqrt[k]{b} \sqrt[k]{c}+ \sqrt[k]{d}> \sqrt[k+1]{a}+ \sqrt[k+1]{b} \sqrt[k+1]{c}+ \sqrt[k+1]{d}\] then it's solved?

  9. sparik1997
    • 2 years ago
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    sorry, I forgot to put +

  10. Dido525
    • 2 years ago
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    That is the definition but we must prove it.

  11. experimentX
    • 2 years ago
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    nope!! that is always true ... there is ^n to consider of.

  12. Dido525
    • 2 years ago
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    But substituting any real number should yield a decreasing pattern regardless.

  13. sparik1997
    • 2 years ago
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    ok if we put n. we get some number in the brackets. and if we put n+1 , we get a smaller number in the brackets

  14. experimentX
    • 2 years ago
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    I am guessing quite opposite .. (nth root of a's)^(n) < (n+1 th root of s')^(n+1) .... most likely it's due to 4 down below.

  15. sparik1997
    • 2 years ago
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    and a^n<b^n if a<b

  16. Dido525
    • 2 years ago
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    I still think we have to use induction though. Seems tough though. Even the trivial case.

  17. experimentX
    • 2 years ago
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    sorry ... have to go due to technical reasons. I'll be back in 6 hrs.

  18. sparik1997
    • 2 years ago
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    I have to go too

  19. sparik1997
    • 2 years ago
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    good luck

  20. experimentX
    • 2 years ago
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    I think induction might work ... but it seems too ugly to work out.

  21. experimentX
    • 2 years ago
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    if anyone found any trick to it .. please update it.

  22. wio
    • 2 years ago
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    Suppose: \( f(n) = a_n \) Then: \[ f'(n) = (n-1)\left( \frac{\sqrt[n]{a}+\sqrt[n]{b}+\sqrt[n]{c}+\sqrt[n]{d}}{4} \right)^{n-1} \left( -\frac{\sqrt[n]{a}\ln(a) +\sqrt[n]{b}\ln(b)+\sqrt[n]{c}\ln(c)+\sqrt[n]{d}\ln(d)}{4n^2} \right) \]When \(n > 1\) we have \(f'(n)<0\), thus \(f(n)\) will always be decreasing, the same applied for \(a_n\).

  23. wio
    • 2 years ago
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    This is because we know \(\sqrt[n]{a}\ln(a) \) is going to be positive if \(a\) and \(n\) are positive.

  24. wio
    • 2 years ago
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    Thus the right term will always be negative, the left term will always be positive, and when multiplied together we get a negative number.

  25. wio
    • 2 years ago
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    I mean it seems a bit easier to notice that the derivative is always negative than to notice that the function is always decreasing.

  26. wio
    • 2 years ago
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    @experimentX Tell me your thoughts.

  27. experimentX
    • 2 years ago
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    seems legit ... but the derivative is \[ 4^{-n} \left(a^{\frac{1}{n}}+b^{\frac{1}{n}}+c^{\frac{1}{n}}+d^{\frac{1}{n}}\right)^n \\ \left(-\text{Log}[4]-\frac{a^{\frac{1}{n}} \text{Log}[a]+b^{\frac{1}{n}} \text{Log}[b]+c^{\frac{1}{n}} \text{Log}[c]+d^{\frac{1}{n}} \text{Log}[d]}{\left(a^{\frac{1}{n}}+b^{\frac{1}{n}}+c^{\frac{1}{n}}+d^{\frac{1}{n}}\right) n}+\text{Log}\left[a^{\frac{1}{n}}+b^{\frac{1}{n}}+c^{\frac{1}{n}}+d^{\frac{1}{n}}\right]\right) \] now gotta show that this is always negative.

  28. sirm3d
    • 2 years ago
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    since\(0<1\leq a\), mutliplying by \(a^n\) yields \[\large a^n\leq a^{n+1}\\\large (a^n)^\frac{1}{n(n+1)}<(a^{n+1})^\frac{1}{n(n+1)}\]\[\Large 0<\sqrt[(n+1)]{a}<\sqrt[n]{a}\]

  29. sirm3d
    • 2 years ago
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    \[\large a_{n+1}=\sqrt[(n+1)] a +\sqrt[(n+1)] b +\sqrt[(n+1)] c +\sqrt[(n+1)] d <\sqrt[n] a +\sqrt[n] b +\sqrt[n] c +\sqrt[n] d=a_n\]

  30. wio
    • 2 years ago
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    @experimentX Weird, I used the chain rule to get my derivative and I'm not sure where I went wrong.

  31. wio
    • 2 years ago
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    Actually, I'm not so sure that I am wrong.

  32. experimentX
    • 2 years ago
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    @wio there is 'n' inside and outside ... its type x^x ... so you cannot use chain rule. |dw:1358029777544:dw| most likely you should be doing. I got that result from mathematica. Check again. More or less ... i had enough of this.

  33. wio
    • 2 years ago
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    Ah I see, should be doing logarithmic differentiation.

  34. sauravshakya
    • 2 years ago
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    when a=b=c=d=1 a_n = {(1+1+1+1)/4}^n a_n={4/4}^n a_n=1^n a_n=1 So, I dont think it is always decreasing

  35. sauravshakya
    • 2 years ago
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    If a=b=c=d =k then, a_n=k

  36. experimentX
    • 2 years ago
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    that's a special case ... where \( a_{n+1} = a_{n} \) still you can call it monotonically decreasing ... because monotonic means \( a_{n+1} \le a_n \) and strictly decreasing means \( a_{n+1} < a_n \)

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