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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
 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
 one year ago

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Dido525Best ResponseYou've already chosen the best response.0
Induction is the best way but I have no idea how the heck I prove it's decreasing.
 one year ago

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

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

sparik1997Best ResponseYou've already chosen the best response.0
are a,b,c,d positive integers?
 one year ago

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

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

sparik1997Best ResponseYou've already chosen the best response.0
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?
 one year ago

sparik1997Best ResponseYou've already chosen the best response.0
sorry, I forgot to put +
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
That is the definition but we must prove it.
 one year ago

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

Dido525Best ResponseYou've already chosen the best response.0
But substituting any real number should yield a decreasing pattern regardless.
 one year ago

sparik1997Best ResponseYou've already chosen the best response.0
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
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
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.
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
I still think we have to use induction though. Seems tough though. Even the trivial case.
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
sorry ... have to go due to technical reasons. I'll be back in 6 hrs.
 one year ago

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

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

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

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

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

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

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

experimentXBest ResponseYou've already chosen the best response.0
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.
 one year ago

sirm3dBest ResponseYou've already chosen the best response.0
since\(0<1\leq a\), mutliplying by \(a^n\) yields \[\large a^n\leq a^{n+1}\\\large (a^n)^\frac{1}{n(n+1)}<(a^{n+1})^\frac{1}{n(n+1)}\]\[\Large 0<\sqrt[(n+1)]{a}<\sqrt[n]{a}\]
 one year ago

sirm3dBest ResponseYou've already chosen the best response.0
\[\large a_{n+1}=\sqrt[(n+1)] a +\sqrt[(n+1)] b +\sqrt[(n+1)] c +\sqrt[(n+1)] d <\sqrt[n] a +\sqrt[n] b +\sqrt[n] c +\sqrt[n] d=a_n\]
 one year ago

wioBest 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.
 one year ago

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

experimentXBest 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.
 one year ago

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

sauravshakyaBest ResponseYou've already chosen the best response.0
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
 one year ago

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

experimentXBest ResponseYou've already chosen the best response.0
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 \( a_{n+1} < a_n \)
 one year ago
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