How to compare the 7 times of 8 with the 8 times of 7

How to compare the 7 times of 8 with the 8 times of 7

8^(7/4)=8×8^(3/4)=8×2^(9/4)=8×2²×2^(1/4)=32×√√2＜32×√1.44=32×1.2=38.4
7^(8/4)=7²=49
Learned: 8 ^ (7 / 4) < 7 ^ (8 / 4)
So: 8 ^ 7 < 7 ^ 8

1.25 () one eighth, compare the size

1.25 (greater than) one eighth

5 / 25 5 / 50 3 / 15 4 / 20 2 / 10 9 / 45

5 / 25 5 / 50 3 / 15 4 / 20 2 / 10 9 / 45
5/25=3/15=2/10=4/20=9/45

Given x-2y = 0, find the value of 5 (x-2y) - 3 (2y-x) - 60

5（x-2y)-3(2y-x)-60
=5x-10y-6y+3x-60
=8x-16y-60
=8(x-2y)-60
When x-2y = 0
Original formula = 8 (x-2y) - 60
=8x0-60
=-60

Power series and function
Sum function of ∑ [(- 1) ^ n / 3 ^ n] x ^ n

N start from 0?
Σ [(- 1) ^ n / 3 ^ n] x ^ n = ∑ [(- X / 3) ^ n, which is an equal ratio series, so when | - X / 3 | < 1, that is | - x | < 3, the power series converges, and its sum function is naturally 1 / [1 - (- X / 3)] = 3 / (3 + x)

SiNx + cosx = 1, then what is the nth power of SiNx + the nth power of cosx? N is a natural number

The answer is 1
Let SiNx be expressed by cosx. If we solve the quadratic equation of one variable, we can see that cosx = 1, SiNx = 0 or cosx = 0, SiNx = 1, so it is the nth power of 1 plus the nth power of 0, so it is 1

Sin θ = √ 5 / 5, then (SiNx) is the fourth power - (cosx) is the fourth power=

The front and back corners are not consistent? Both are treated with X!
(sinx)^4-(cosx)^4
=[(SiNx) ^ 2 + (cosx) ^ 2] [(SiNx) ^ 2 - (cosx) ^ 2] (factorization factor)
=(sinx)^2-(cosx)^2
=2(sinx)^2-1
=2/5-1
= -3/5 .

(SiNx) ^ cosy = (cosx) ^ siny, find dy? (^ for power)

Find ln on both sides, get cosy * ln (SiNx) = siny * ln (cosx), and simplify y = acrtan (lnsinx / lncosx)
Formula: y = arctanx, y '= 1 / 1 + x ^ 2
y'=(1/1+(lnsinx/lncosx)^2))*(cosxlncosx/lnsinx+sinxlnsinx/lncosx)/(lncosx)^2
The result is (cosx * lncosx ^ 2 + SiNx * lnsinx ^ 2) / (lnsinx ^ 3 * lncosx + lnsinx * lncosx ^ 3)

SiNx + cosx = a, find (SiNx) ^ n + (cosx) ^ n

Because a ^ 2 = (SiNx + cosx) ^ 2 = (SiNx) ^ 2 + (cosx) ^ 2 + 2sinx * cosx = 1 + 2sinx * cosx, so SiNx * cosx = (a ^ 2-1) / 2. Combined with SiNx + cosx = a, we get that SiNx and cosx are the roots of the equation T ^ 2-At + (a ^ 2-1) / 2 = 0 about t

Under what conditions must Taylor series converge to f (x)
Since Taylor series does not necessarily converge to f (x) in the convergence domain, under what conditions can it be determined to converge to f (x)

If the Taylor series converges in a neighborhood of x = x0, but it does not necessarily converge to f (x). Theoretically, if the remainder R (x) in the Taylor expansion of F (x) satisfies limr (x) = 0 when n tends to infinity, then