How would u prove this? (1 Viewer)

eternallyboreduser · Dec 19, 2023

??? Kinda stuck ngl, cant think of any way other than subbing numbers but thats not rlly proving it

Average Boreduser · Dec 19, 2023

eternallyboreduser said:
??? Kinda stuck ngl, cant think of any way other than subbing numbers but thats not rlly proving itView attachment 41871 View attachment 41871

Would graphing be sufficient? Idk why but that looks like smn u could prove using induction too

lettucechamber · Dec 19, 2023

induction

eternallyboreduser · Dec 19, 2023

lettucechamber said:
induction

have not yet learnt that, is there another way?

eternallyboreduser · Dec 19, 2023

oh yeah also tryinf to prove that with the condition that k is between 0 and 1

Average Boreduser · Dec 19, 2023

eternallyboreduser said:
have not yet learnt that, is there another way?

induction is the right method. I'd say any other method wouldn't be strong enough to fully prove the inequal. for all real k.

eternallyboreduser · Dec 19, 2023

Average Boreduser said:
induction is the right method. I'd say any other method wouldn't be strong enough to fully prove the inequal. for all real k.

do you mind working it out for me? i just wanna see how it goes lol, also theres a condition that k lies between zero and one

Average Boreduser · Dec 19, 2023

eternallyboreduser said:
do you mind working it out for me? i just wanna see how it goes lol, also theres a condition that k lies between zero and one

I might be a bit but yeah sure lol. currently running errands.

kendricklamarlover101 · Dec 19, 2023

eternallyboreduser said:
??? Kinda stuck ngl, cant think of any way other than subbing numbers but thats not rlly proving itView attachment 41871 View attachment 41871

is there any information on what n is? i dont think this would be a true statement if theres no restriction on n which can be seen in this desmos graph

kendricklamarlover101 · Dec 19, 2023

for the specific case of n=2 however u could probably just use the binomial theorem writing 3/4 as 1/2 +1/(2^2)
$(\frac{3}{4})^{k}=(\frac{1}{2}+\frac{1}{2^2})^k$
$=\frac{1}{2^k} + \binom{k}{1}\frac{1}{2^{k+1}} + \binom{k}{2}\frac{1}{2^{k+2}} + ... + \frac{1}{2^{2k}}$
$> \frac{1}{2^k} + \frac{1}{2^{2k}$
Hence true for the case that n=2

kendricklamarlover101 · Dec 19, 2023

kendricklamarlover101 said:
for the specific case of n=2 however u could probably just use the binomial theorem writing 3/4 as 1/2 +1/(2^2)
$(\frac{3}{4})^{k}=(\frac{1}{2}+\frac{1}{2^2})^k$
$=\frac{1}{2^k} + \binom{k}{1}\frac{1}{2^{k+1}} + \binom{k}{2}\frac{1}{2^{k+2}} + ... + \frac{1}{2^{2k}}$
$> \frac{1}{2^k} + \frac{1}{2^{2k}$
Hence true for the case that n=2

actually nvm i dont think this proof is valid as well since 0<k<1 oops.

eternallyboreduser · Dec 19, 2023

kendricklamarlover101 said:
is there any information on what n is? i dont think this would be a true statement if theres no restriction on n which can be seen in this desmos graphView attachment 41875

the n is a ‘+ 1’ LOL

Average Boreduser · Dec 19, 2023

eternallyboreduser said:
do you mind working it out for me? i just wanna see how it goes lol, also theres a condition that k lies between zero and one

dyk the basics/fundamentals of induction yet or?

eternallyboreduser · Dec 19, 2023

Average Boreduser said:
dyk the basics/fundamentals of induction yet or?

Nope have not touched that topic yet

Average Boreduser · Dec 19, 2023

eternallyboreduser said:
Nope have not touched that topic yet

without doing the theory, the working will make 0 sense. just leave the qn for now. Or if ur keen, send me an email (via dm), i have some old notes that might help u learn it

eternallyboreduser · Dec 19, 2023

Average Boreduser said:
without doing the theory, the working will make 0 sense. just leave the qn for now. Or if ur keen, send me an email, i have some old notes that might help u learn it

Alright tysm

kendricklamarlover101 · Dec 19, 2023

if you wanted to do the question without induction btw heres the working out:
$\frac{1}{2}+\frac{1}{2^{k+1}}=\frac{1}{2}\left(1+\frac{1}{2^k}\right)$
$=\frac{1}{2}\left(\frac{2^k + 1}{2^k}\right)$
$=\frac{2^k + 1}{2^{k}(2)}$
Notice that $3^k>2^k +1$ by the binomial theorem as:
$3^k =(2+1)^k = 2^k +1+\binom{k}{1}2^{k-1} +\binom{k}{2}2^{k-2}+...$
Now for $0<k<1, 2>2^k$
$\implies \frac{1}{2}<\frac{1}{2^k}$
Hence, $\frac{2^k + 1}{2^{k}(2)} < \frac{3^k}{2^k \times 2^k}$
$=\left(\frac{3}{4}\right)^k$
$\therefore \frac{1}{2}+\frac{1}{2^{k+1}} <\left(\frac{3}{4}\right)^k$

kendricklamarlover101 · Dec 19, 2023

also i dont think u would be able to apply induction here since 0<k<1 and usually u can only apply induction when k is an integer

kendricklamarlover101 · Dec 19, 2023

kendricklamarlover101 said:
if you wanted to do the question without induction btw heres the working out:
$\frac{1}{2}+\frac{1}{2^{k+1}}=\frac{1}{2}\left(1+\frac{1}{2^k}\right)$
$=\frac{1}{2}\left(\frac{2^k + 1}{2^k}\right)$
$=\frac{2^k + 1}{2^{k}(2)}$
Notice that $3^k>2^k +1$ by the binomial theorem as:
$3^k =(2+1)^k = 2^k +1+\binom{k}{1}2^{k-1} +\binom{k}{2}2^{k-2}+...$
Now for $0<k<1, 2>2^k$
$\implies \frac{1}{2}<\frac{1}{2^k}$
Hence, $\frac{2^k + 1}{2^{k}(2)} < \frac{3^k}{2^k \times 2^k}$
$=\left(\frac{3}{4}\right)^k$
$\therefore \frac{1}{2}+\frac{1}{2^{k+1}} <\left(\frac{3}{4}\right)^k$

i just realised this is invalid since u cant apply the binomial theorem once again...
well another approach u could do is RHS-LHS
$RHS-LHS = \frac{3^k}{2^{2k}} - \frac{1}{2^{k+1}} - \frac{1}{2}$
$=\frac{3^k - 2^{k-1} - 2^{2k-1}}{2^{2k}}$
$=\frac{3^k -2^{k-1}\left(1+2^k\right)}{2^{2k}}$
Note that $3^k$ and $2^{k}$ are increasing for all real numbers k
Since 0<k<1, $3^k > 3^0 = 1$ and similarily, $1+2^k>1+2^0=2$
Also, -1<k-1<0, so $2^{k-1} > 2^{-1} = \frac{1}{2}$
Hence $\frac{3^k -2^{k-1}\left(1+2^k\right)}{2^{2k}} > \frac{1 -\frac{1}{2}\left(2\right)}{2^{2k}}=\frac{0}{2^{2k}}$
$\therefore RHS-LHS>0 \implies LHS < RHS$
surely i didnt make another error right

Average Boreduser · Dec 19, 2023

kendricklamarlover101 said:
i just realised this is invalid since u cant apply the binomial theorem once again...
well another approach u could do is RHS-LHS
$RHS-LHS = \frac{3^k}{2^{2k}} - \frac{1}{2^{k+1}} - \frac{1}{2}$
$=\frac{3^k - 2^{k-1} - 2^{2k-1}}{2^{2k}}$
$=\frac{3^k -2^{k-1}\left(1+2^k\right)}{2^{2k}}$
Note that $3^k$ and $2^{k}$ are increasing for all real numbers k
Since 0<k<1, $3^k > 3^0 = 1$ and similarily, $1+2^k>1+2^0=2$
Also, -1<k-1<0, so $2^{k-1} > 2^{-1} = \frac{1}{2}$
Hence $\frac{3^k -2^{k-1}\left(1+2^k\right)}{2^{2k}} > \frac{1 -\frac{1}{2}\left(2\right)}{2^{2k}}=\frac{0}{2^{2k}}$
$\therefore RHS-LHS>0 \implies LHS < RHS$
surely i didnt make another error right

Huge KL fan too. $bitch dont kill my vibe$

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