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experimentX
 3 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.
experimentX
 3 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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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Induction is the best way but I have no idea how the heck I prove it's decreasing.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If an+1 is < an, the the function is decreasing.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0I know the definition ... how to show it? expanding using multinomial theorem is freaking ugly, not sure if induction works so easily.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0are a,b,c,d positive integers?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0yep!! sorry for non specifying that.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0n is a set of natural numbers.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0do 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry, I forgot to put +

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That is the definition but we must prove it.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0nope!! that is always true ... there is ^n to consider of.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0But substituting any real number should yield a decreasing pattern regardless.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok 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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0I 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I still think we have to use induction though. Seems tough though. Even the trivial case.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0sorry ... have to go due to technical reasons. I'll be back in 6 hrs.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0I think induction might work ... but it seems too ugly to work out.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0if anyone found any trick to it .. please update it.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Suppose: \( f(n) = a_n \) Then: \[ f'(n) = (n1)\left( \frac{\sqrt[n]{a}+\sqrt[n]{b}+\sqrt[n]{c}+\sqrt[n]{d}}{4} \right)^{n1} \left( \frac{\sqrt[n]{a}\ln(a) +\sqrt[n]{b}\ln(b)+\sqrt[n]{c}\ln(c)+\sqrt[n]{d}\ln(d)}{4n^2} \right) \]When \(n > 1\) we have \(f'(n)<0\), thus \(f(n)\) will always be decreasing, the same applied for \(a_n\).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0This is because we know \(\sqrt[n]{a}\ln(a) \) is going to be positive if \(a\) and \(n\) are positive.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thus the right term will always be negative, the left term will always be positive, and when multiplied together we get a negative number.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I mean it seems a bit easier to notice that the derivative is always negative than to notice that the function is always decreasing.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@experimentX Tell me your thoughts.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0seems 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0since\(0<1\leq a\), mutliplying by \(a^n\) yields \[\large a^n\leq a^{n+1}\\\large (a^n)^\frac{1}{n(n+1)}<(a^{n+1})^\frac{1}{n(n+1)}\]\[\Large 0<\sqrt[(n+1)]{a}<\sqrt[n]{a}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\large a_{n+1}=\sqrt[(n+1)] a +\sqrt[(n+1)] b +\sqrt[(n+1)] c +\sqrt[(n+1)] d <\sqrt[n] a +\sqrt[n] b +\sqrt[n] c +\sqrt[n] d=a_n\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@experimentX Weird, I used the chain rule to get my derivative and I'm not sure where I went wrong.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Actually, I'm not so sure that I am wrong.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0@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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ah I see, should be doing logarithmic differentiation.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0when 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If a=b=c=d =k then, a_n=k

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0that'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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