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February 5

And the winner is...

Which of the following options should be chosen for the best chance of winning, or are they the same:
A) two tries at 10% chance
B) one try at 20% chance
2606:A000:4C0C:E200:211C:FF2A:3329:F571 (talk) 07:28, 5 February 2018 (UTC)[reply]

This is one of those questions that is easier to answer by flipping the question around: Which one has the least chance of losing? You lose with choice A if both your tries fail. The probability of that is 0.9 times 0.9, or 0.81. Anon126 (notify me of responses! / talk / contribs) 07:30, 5 February 2018 (UTC)[reply]

So, inverting that, A has a 19% probability (therefore B is better choice). -Thx, [OP]:2606:A000:4C0C:E200:9895:7442:94B7:AB9C (talk) 19:12, 5 February 2018 (UTC)[reply]

Yes. They may sound similar but they only give the same result if you ask for the expected number of wins, allowing the possibility of 2 wins in A. Then A has 0.1 + 0.1 = 0.2 expected wins, the same as B. "The best chance of winning" asks for the highest probability of at least 1 win and then B is best. PrimeHunter (talk) 20:13, 5 February 2018 (UTC)[reply]

The expeted number of wins is 0.2 for A only if one can win twice (with both tries). Whether the player can win again after winning once is not specified in the question. Therefore B is the safe bet, as it's possibly better and never worse than A. 78.0.213.2 (talk) 05:05, 6 February 2018 (UTC)[reply]

Derivative of a constant raised to a variable exponential?

Like f(x)=26^x. How would you find the derivative a function in that form? Maybe you could use Newton's approximation to determine experimentally. Not quite sure though! Earl of Arundel (talk) 21:37, 5 February 2018 (UTC)[reply]

26^{x}=e^{x\ln(26)}

and so

f^{\prime }(x)=(\ln 26)26^{x}

. 88.109.68.238 (talk) 22:49, 5 February 2018 (UTC)[reply]

You can also derive it from the base rule:

f'(x₀) = lim_x→x₀ [f(x) - f(x₀)]/(x - x₀) =

= lim_x→x₀ (26^x - 26^x₀)/(x - x₀) =

= lim_x→x₀ (e^{x ln 26} - e^{x₀ ln 26})/(x - x₀) =

= lim_x→x₀ (1 + (x ln 26) + (x ln 26)²/2! + (x ln 26)³/3! + ... - 1 - (x₀ ln 26) - (x₀ ln 26)²/2! - (x₀ ln 26)³/3! - ...)/(x - x₀) =

= lim_x→x₀ (1 - 1 + (x ln 26 - x₀ ln 26) + (x ln 26 - x₀ ln 26)(x ln 26 + x₀ ln 26)/2! + (x ln 26 - x₀ ln 26)[(x ln 26)² + (x x₀)(ln 26)² + (x₀ ln 26)²]/3! + ...)/(x - x₀) =

= lim_x→x₀ ln 26 ~~(x - x₀)~~(1 + ln 26(x + x₀)/2! + (ln 26)²(x² + xx₀ + x₀²)/3! + ...)/~~(x - x₀)~~ =

= lim_x→x₀ ln 26 (1 + ln 26(x + x₀)/2! + (ln 26)²(x² + xx₀ + x₀²)/3! + ...) =

= ln 26 (1 + ln 26(2x₀)/2! + (ln 26)²(3x₀²)/3! + ...)

= ln 26 (1 + (ln 26) x₀/1! + (ln 26)²(x₀²)/2! + ... =

= ln 26 e^{(ln 26) x₀}

= 26^x₀ ln 26

78.0.213.2 (talk) 05:28, 6 February 2018 (UTC)[reply]

Interesting, thank you both very much. And is there a general rule that can be used to find the indefinite integral as well? Because the relationship it seems isn't quite as straight-forward as is with plain polynomials. May be a good time to learn some more of those calculus identities come to think of it. Couldn't hurt that's for sure! Earl of Arundel (talk) 10:52, 6 February 2018 (UTC)[reply]

See the third entry at Lists of integrals#Exponential functions. -- ToE 14:18, 6 February 2018 (UTC)[reply]

Note the neither this nor the derivative of a^x are things you need to memorize, but are applications of the chain rule (or its counterpart the substitution rule for integration), as well as change of base via Exponentiation#Powers via logarithms and e^x#Derivatives and differential equations. (The importance of the exponential function in mathematics and the sciences stems mainly from its definition as the unique function which is equal to its derivative and is equal to 1 when x = 0.) Those are what you need to know and become comfortable with their use, then the derivative and integral in question are trivial derivations. -- ToE 14:18, 6 February 2018 (UTC)[reply]

f'(x₀) = lim_x→x₀ [f(x) - f(x₀)]/(x - x₀), is the basic definition of the derivative and you can use it to find the derivative or prove the various rules about derivation (this was a fun exercise BTW, I hadn't done this sort of thing in years!), but as ToE says above for all realistic intents and purposes it's enough to master the basic rules. It is fun to be able to prove that f'(x) is indeed as above, but outside of school/university you won't need that kind of detail. 78.0.197.69 (talk) 01:03, 7 February 2018 (UTC)[reply]

From the differentials of sums and logarithms

d(x+y)=dx+dy

d(\ln x)=x^{-1}dx

follow the differentials of products and powers

d(xy)=xy(x^{-1}dx+y^{-1}dy)

d(x^{y})=x^{y}(yx^{-1}dx+(\ln x)dy)

Bo Jacoby (talk) 02:02, 7 February 2018 (UTC).[reply]

Thanks for the help everyone. The above set of rules should come in handy when it comes to finding the basic inequalities. And ToE, the tip on the chain rule should be helpful too. The derivatives found in the original expression may not be very interesting but then you have something like x^2-x; a function as such as this may have some notable or unusual properties, not to mention its integrals...

Earl of Arundel (talk) 04:17, 8 February 2018 (UTC)[reply]