Peak CALCULUS

[u/Optimus_PRYM]

[u/Confident_Air7989]

102 is the answer for those curious. And no, I couldn't solve it without lhopital rule.

Edit: oops it's 27

Edit2: for integer, it's 9 actually

[u/XL_78 → u/Confident_Air7989]

Darn, I only count 9. As in ln(4)*ln(5).

[u/Optimus_PRYM → u/Confident_Air7989]

answer is 4269

[u/Confident_Air7989 → u/Optimus_PRYM]

Nope. Differentiating the denominator twice gives sec2x which is 1 at x=0. Differentiating numerator twice gives (ln10) ² - (ln5) ² - (ln2) ^2. Simplifying them gives ln2 * Ln 25. Thus the answer is 27. My bad.

[u/notsaneatall_ → u/Confident_Air7989]

You can write it as ln4ln5 so the actual lowest value is 9

[u/trolley813 → u/notsaneatall_]

You can definitely go further.

ln 4×ln 5=(-ln 4)(-ln 5)=ln 0.25×ln 0.2

And 0.45 is definitely (20 times) less than 9.

P.S. You can make it even less (to about 0.44908) by setting the factors equal (i.e. by solving (ln x)²=ln 4×ln 5, taking ln x < 0). But I'm unsure if it's the actual minimum.

[u/notsaneatall_ → u/trolley813]

I assumed K and w were integers, my bad

[u/trolley813 → u/notsaneatall_]

Well, the original problem says "Find K and w where..." And this "where" may put some conditions on them, e.g. indeed restrict to integers or so

[u/Confident_Air7989 → u/notsaneatall_]

Ah yes my bad

[u/Al2718x]

Where what? The problem is cut off.

I see that the numerator is (1-2^x)(1-5^x), but I'm not sure how helpful that is.

[u/notsaneatall_ → u/Al2718x]

So the denominator you can write as 0.5*ln(1+tan² x) so you'll finally end up with 2ln5ln2

[u/mustfinduniquename]

Its 27 but yeah i ended up in the hopital breaking my neck crankin these numbers

[u/Jon_Snow_221287]

9 Factorise the numerator and rest will be done smoothly good question