experimentX
  • experimentX
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Induction is the best way but I have no idea how the heck I prove it's decreasing.
anonymous
  • anonymous
If an+1 is < an, the the function is decreasing.
experimentX
  • experimentX
I know the definition ... how to show it? expanding using multinomial theorem is freaking ugly, not sure if induction works so easily.

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anonymous
  • anonymous
are a,b,c,d positive integers?
experimentX
  • experimentX
yep!! sorry for non specifying that.
anonymous
  • anonymous
What about the n's ?
experimentX
  • experimentX
n is a set of natural numbers.
anonymous
  • anonymous
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?
anonymous
  • anonymous
sorry, I forgot to put +
anonymous
  • anonymous
That is the definition but we must prove it.
experimentX
  • experimentX
nope!! that is always true ... there is ^n to consider of.
anonymous
  • anonymous
But substituting any real number should yield a decreasing pattern regardless.
anonymous
  • anonymous
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
experimentX
  • experimentX
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.
anonymous
  • anonymous
and a^n
anonymous
  • anonymous
I still think we have to use induction though. Seems tough though. Even the trivial case.
experimentX
  • experimentX
sorry ... have to go due to technical reasons. I'll be back in 6 hrs.
anonymous
  • anonymous
I have to go too
anonymous
  • anonymous
good luck
experimentX
  • experimentX
I think induction might work ... but it seems too ugly to work out.
experimentX
  • experimentX
if anyone found any trick to it .. please update it.
anonymous
  • anonymous
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 \(a_n\).
anonymous
  • anonymous
This is because we know \(\sqrt[n]{a}\ln(a) \) is going to be positive if \(a\) and \(n\) are positive.
anonymous
  • anonymous
Thus the right term will always be negative, the left term will always be positive, and when multiplied together we get a negative number.
anonymous
  • anonymous
I mean it seems a bit easier to notice that the derivative is always negative than to notice that the function is always decreasing.
anonymous
  • anonymous
@experimentX Tell me your thoughts.
experimentX
  • experimentX
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.
sirm3d
  • sirm3d
since\(0<1\leq a\), mutliplying by \(a^n\) yields \[\large a^n\leq a^{n+1}\\\large (a^n)^\frac{1}{n(n+1)}<(a^{n+1})^\frac{1}{n(n+1)}\]\[\Large 0<\sqrt[(n+1)]{a}<\sqrt[n]{a}\]
sirm3d
  • sirm3d
\[\large a_{n+1}=\sqrt[(n+1)] a +\sqrt[(n+1)] b +\sqrt[(n+1)] c +\sqrt[(n+1)] d <\sqrt[n] a +\sqrt[n] b +\sqrt[n] c +\sqrt[n] d=a_n\]
anonymous
  • anonymous
@experimentX Weird, I used the chain rule to get my derivative and I'm not sure where I went wrong.
anonymous
  • anonymous
Actually, I'm not so sure that I am wrong.
experimentX
  • experimentX
@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
  • anonymous
Ah I see, should be doing logarithmic differentiation.
anonymous
  • anonymous
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
anonymous
  • anonymous
If a=b=c=d =k then, a_n=k
experimentX
  • experimentX
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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