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 one year 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.
 one year 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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Dido525
 one year 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.

incomplte
 one year ago
Best ResponseYou've already chosen the best response.0If an+1 is < an, the the function is decreasing.

experimentX
 one year 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.

sparik1997
 one year ago
Best ResponseYou've already chosen the best response.0are a,b,c,d positive integers?

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

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

sparik1997
 one year 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?

sparik1997
 one year ago
Best ResponseYou've already chosen the best response.0sorry, I forgot to put +

Dido525
 one year ago
Best ResponseYou've already chosen the best response.0That is the definition but we must prove it.

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

Dido525
 one year ago
Best ResponseYou've already chosen the best response.0But substituting any real number should yield a decreasing pattern regardless.

sparik1997
 one year 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
 one year 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.

Dido525
 one year 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
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0I think induction might work ... but it seems too ugly to work out.

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

wio
 one year ago
Best ResponseYou've already chosen the best response.2Suppose: \( f(n) = a_n \) 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 \(a_n\).

wio
 one year ago
Best ResponseYou've already chosen the best response.2This is because we know \(\sqrt[n]{a}\ln(a) \) is going to be positive if \(a\) and \(n\) are positive.

wio
 one year ago
Best ResponseYou've already chosen the best response.2Thus the right term will always be negative, the left term will always be positive, and when multiplied together we get a negative number.

wio
 one year ago
Best ResponseYou've already chosen the best response.2I mean it seems a bit easier to notice that the derivative is always negative than to notice that the function is always decreasing.

wio
 one year ago
Best ResponseYou've already chosen the best response.2@experimentX Tell me your thoughts.

experimentX
 one year 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.

sirm3d
 one year ago
Best ResponseYou've already chosen the best response.0since\(0<1\leq a\), mutliplying by \(a^n\) yields \[\large a^n\leq a^{n+1}\\\large (a^n)^\frac{1}{n(n+1)}<(a^{n+1})^\frac{1}{n(n+1)}\]\[\Large 0<\sqrt[(n+1)]{a}<\sqrt[n]{a}\]

sirm3d
 one year ago
Best ResponseYou've already chosen the best response.0\[\large a_{n+1}=\sqrt[(n+1)] a +\sqrt[(n+1)] b +\sqrt[(n+1)] c +\sqrt[(n+1)] d <\sqrt[n] a +\sqrt[n] b +\sqrt[n] c +\sqrt[n] d=a_n\]

wio
 one year ago
Best ResponseYou've already chosen the best response.2@experimentX Weird, I used the chain rule to get my derivative and I'm not sure where I went wrong.

wio
 one year ago
Best ResponseYou've already chosen the best response.2Actually, I'm not so sure that I am wrong.

experimentX
 one year 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.

wio
 one year ago
Best ResponseYou've already chosen the best response.2Ah I see, should be doing logarithmic differentiation.

sauravshakya
 one year 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

sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.0If a=b=c=d =k then, a_n=k

experimentX
 one year 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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